The scientific literature on the HardyLeray inequality, also known as the uncertainty principle, is very extensive and
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English Pages 440 [514] Year 2021
Table of contents :
Preface
Contents
1 Motivation of the Hardy–Leray potential
2 Looking at the Hardy–Leray potential
3 Calderón–Zygmund theory and the Hardy–Leray potential
4 Effect of the Hardy–Leray potential in the solvability of semilinear elliptic equations
5 The Hardy–Leray potential in semilinear heat equations
6 Elliptic equations with a nonlinearity on the gradient and the Hardy–Leray potential
7 The heat equation with nonlinearity on the gradient and the Hardy–Leray potential
8 Fractional Laplacian type operators
9 The fractional Hardy inequality
10 Calderón–Zygmund summability in the fractional setting
11 Fractional semilinear elliptic problems
12 The heat equation with fractional diffusion
13 The influence of the Hardy potential on the linear and semilinear fractional heat equations
Bibliography
Alphabetical Index
Ireneo Peral Alonso, Fernando Soria de Diego Elliptic and Parabolic Equations Involving the HardyLeray Potential
De Gruyter Series in Nonlinear Analysis and Applications

Editorin Chief Jürgen Appell, Würzburg, Germany Editors Catherine Bandle, Basel, Switzerland Alain Bensoussan, Richardson, Texas, USA Manuel del Pino, Santiago de Chile, Chile Avner Friedman, Columbus, Ohio, USA Mikio Kato, Tokyo, Japan Wojciech Kryszewski, Torun, Poland Umberto Mosco, Worcester, Massachusetts, USA Simeon Reich, Haifa, Israel Vicenţiu D. Rădulescu, Krakow, Poland
Volume 38
Ireneo Peral Alonso, Fernando Soria de Diego
Elliptic and Parabolic Equations Involving the HardyLeray Potential 
Mathematics Subject Classification 2010 35A15, 35B25, 35B30, 35B33, 35B40, 35J58, 35B99, 35C15, 35D05, 35D10, 35D30, 35D35, 35J20, 35J25, 35J58, 35J70, 35K05, 35K15, 35K55, 35K57, 35K58, 35R09, 35R11, 45K05, 46E30, 46E35, 47G20 Authors Prof. Dr. Ireneo Peral Alonso Universidad Autonoma de Madrid Departamento de Matemáticas Calle de Tomás y Valiente 7 Ciudad Univ. de Cantoblanco 28049 Madrid Spain [email protected]
Prof. Dr. Fernando Soria de Diego Universidad Autonoma de Madrid Departamento de Matemáticas Calle de Tomás y Valiente 7 Ciudad Univ. de Cantoblanco 28049 Madrid Spain [email protected]
ISBN 9783110603460 eISBN (PDF) 9783110606270 eISBN (EPUB) 9783110605600 ISSN 0941813X Library of Congress Control Number: 2020950485 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2021 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

To our families This book is dedicated to Magdalena and Angelines, for their love, support and perseverance; to our daughters and son, Irene and Magdalena and Irene, Eva and Alberto; to their spouses, David and Carlos and Carlos, Carlos and Irene; and to five important persons, our grandchildren, Iria, Oscar, Laura and Jaime and Emma.
Preface This monograph contains part of the work developed by the first author during the last 25 years in connection with the study of solutions to certain elliptic and parabolic partial differential equations associated to the Hardy–Leray potential. So, the guest star of this story is nothing but 1 , x2
V(x) =
x ∈ ℝN , N ≥ 3,
a potential function that is closely related to the uncertainty principle of Heisenberg in quantum mechanics. The starting point of the theory is Hardy’s inequality in one dimension, the only one properly attributed to Hardy. Our interest in this field came when reading the seminal work of Jean Leray [225] on the Navier–Stokes equations, where one can find the proof of a similar inequality to that by Hardy, but this time in dimension three, the natural dimension for these equations. In the paper [257] this inequality was used to study the stability of the extremal solution of a problem related to a reaction combustion model. The idea behind it was to establish a blowup result. But the first application of what is now called the Hardy– Leray inequality can be seen in the paper by Jean Leray about the Navier–Stokes equations quoted above. The Hardy–Leray inequality also appears as a tool in the study of the Schrödinger equation iut + Δu = V(x, t)u,
V(x, t) = −
(Zq − Q) Q − , 4πx 4πx − v0 t
Z ∈ ℕ, t > 0.
This describes the wave function of an electron under the action of an αparticle with charge Q that, at time t = 0, leaves the nucleus of an atom with velocity v0 ∈ ℝ3 (see [133, Chapter XVIII, page 752–758]). The peculiar behavior of the Hardy–Leray potential is motivated by the fact that it has the same homogeneity as the Dirichlet integral of the Laplacian when rescaling the space variables. In this sense, a similar behavior is observed for V(x) =
1 , xp
x ∈ ℝN \ {0}, N > p > 1,
with respect to the pLaplacian, i. e., the operator Δp (u) = div(∇up−2 ∇u). See the paper [179], where we began to study some nontrivial extensions of the Hardy–Leray inequality. In this monograph we do not follow this direction, but suggest the reader to search for the information about the subsequent results in the bibliography. If we consider V(x) =
1 , x2s
https://doi.org/10.1515/9783110606270201
x ∈ ℝN \ {0}, N > 2s, 0 < s < 1,
VIII  Preface we will find the same behavior of the homogeneity with respect to the socalled fractional Laplacian that was introduced my M. Riesz in [272]. In recent years this operator has attracted the attention of mathematicians and is appearing in a large number of problems. The definition and properties of the fractional Laplacian will be explained in Chapter 8. The main goals of this work are to develop the key ingredients to study how the singular potentials V(x) =
1 x2
and Vs (x) =
1 , 0 < s < 1, N > 2s, x2s
affect some problems modeled by partial differential equations and integrodifferential equations, respectively. It is in this last part in which the cooperation of the authors has been definitively used in a deep way in order to complete the total landscape of both the local and the nonlocal setting. The Hardy–Leray potentials, V(x) and Vs (x), have something fascinating: they complicate the computations and are difficult to handle, but at the same time they provide a very rich set of properties. In this sense V and Vs are both a monster and a friend, as we will try to show in this monograph. We are really in the presence of an old classic theme, especially in the case of V, but with many new results in recent times for both potentials. An important feature about V(x) is its dimensionless character. This is due to the fact that it corresponds, modulo constants, to the Fourier transform of the fundamental solution of the Laplace equation in any dimension greater than or equal to three. That is, in the sense of the Fourier transform defined on 𝒮 (ℝN ), the dual of the Schwartz class, we find ℱ(
b aN )(ξ ) = N2 . ξ  xN−2
The same property clearly holds for the potential Vs . See Chapters 8 and 9 to find the details. There are many reasons, coming particularly from physics, about the importance of studying partial differential equations associated to the Hardy–Leray potential. From a purely point of view, we must say that the potential given by the reciprocal of the square function is a perfect laboratory to study borderline properties or critical behavior in many situations, like the existence of solutions to nonlinear partial differential equations, certain eigenvalue problems, instantaneous and complete blow up and many others. The same can be said for the fractional Hardy potential, Vs , for which the framework is the fractional Laplacian that in the last years has received a lot of attention of researchers due to the analytical interest and the large amount of applications. The reader, again, can find many interesting references about this in the bibliography.
Preface
 IX
The first seven chapters of the monograph are devoted to the classical Hardy– Leray potential V(x) = x1 2 in dimension N ≥ 3. In Chapter 1 we describe several settings in which the Hardy–Leray potential appears in a natural way. In quantum mechanics the Hardy–Leray inequality represents, perhaps, the most elementary way to express the uncertainty principle and provides as well a threshold for many associated results. From the analytical point of view, obN
2 serving that V ∉ Lloc , although V(x) ∈ Lploc for p < N2 , it gives a good example for the spectral theory of the Laplacian in bounded domains. We will see how the Hardy potential appears in the study of the stability of the singular solution of a reaction of combustion by FrankKamenetskii and by Gelfand and collaborators. The Hardy–Leray inequality also appears in [99] to study the stability of the Simmons cones (see [289]). In Section 1.3 we introduce a very useful tool, the Picone inequality. Chapter 2 deals with the elementary properties of the Hardy–Leray potential, the proof of the Hardy–Leray inequality in L2 (ℝN ) and, in general, in Lp (ℝN ), 1 < p < N. In particular the optimal constant is evaluated and its optimality is proved as well as its nonattainability. To this end, we will also present some results on classical symmetrization. The last section is devoted to obtaining the Hardy–Leray inequality with a remainder term, which depends on a weighted L2 norm of the gradient. It is well known, by now, what is the summability of the solution to the problem
−Δu = f ,
in Ω, u = 0, on 𝜕Ω,
in terms of the summability of the datum f . Here we assume that Ω ⊂ ℝN is a bounded domain and 0 ∈ Ω. This can be obtained either as a byproduct of the Calderón– Zygmund theory or as a particular case of the results by Stampacchia about the summability of the solution of the equation in divergence form with measurable bounded coefficients in terms of the datum. Chapter 3 has as an objective to study the effect of a linear term involving the Hardy–Leray potential, that is, to analyze the problem −Δu − λ
u = f, x2
in Ω, u = 0, on 𝜕Ω,
where 0 < λ < ΛN,2 and f ∈ Lm (Ω), m ≥ 1. Summarizing we can say that if we consider the function λ(m) =
N(m − 1)(N − 2m) , m2
then, for 0 < λ < λ(m), the summability is exactly the same as that in the unperturbed problem. Moreover, these results are optimal. For instance, if f ∈ L∞ (Ω) the solution is unbounded and the problem is not well posed in L1 (Ω), the opposite of what happens in the unperturbed problem. A necessary and sufficient condition for existence in weighted L1 spaces, among other results, is obtained in Section 3.4.
X  Preface In the paper by Brezis, Dupaigne and Tesei [89] and by Dupaigne [145], the effect of the Hardy–Leray potential was studied in some semilinear elliptic problems of the type −Δu − λ
u = up + f , x2
u ≥ 0 in Ω, 0 ∈ Ω, u = 0 on 𝜕Ω.
Chapter 4 collects these results and some additional topics of elliptic equations with measure data, potential theory and a proof of the mountain pass theorem by Ambrosetti and Rabinowitz. We start computing the critical exponent for the existence and then we give the existence result for the Dirichlet problem for elliptic equations in divergence form and with measure data. We also present a motivation of the concept of Newtonian capacity and a proof, based on arguments by I. Ekeland [147], of the mountain pass theorem. Chapter 4 includes the proofs of existence under the critical exponent, p+ (λ), and the nonexistence result over the critical exponent. To understand better the nature of the nonexistence result we prove the socalled complete blowup. Following a path to better understanding of the effect of the Hardy–Leray potential, we devote Chapter 5 to studying the behavior of the heat equation with a reaction term of power type. An inspiring result for the linear heat equation perturbed by the Hardy–Leray potential is due to P. Baras and J. Goldstein in [41]. All along the monograph one will find different approaches to this type of result but in different settings. The threshold power for the existence turns out to be the same as the one we have found for the stationary equation in Chapter 4. Some previous results about the existence of weak solution to the heat equation are introduced in order to make the exposition more systematic. The proof of nonexistence if p ≥ p+ (λ) is more involved than in the elliptic case. Moreover the proof of the instantaneous and complete blowup gives precise meaning to the nonexistence phenomenon. The Cauchy problem is also analyzed and one of the new facts that are worth to emphasize is the shift to the right of the Fujita exponent of the heat equation. We explore in Chapter 6 some quasilinear problems, that is, those with a nonlinear term which is a power of the gradient. These kinds of problems appear in several contexts related to Hamilton–Jacobi equations, some growth models, as the stationary part of the Kardan–Parisi–Zhang model, some flame propagation, etc. See [213] for some other physical motivations. These quasilinear problems have an extra degree of difficulty according to the nonlinear gradient term. Because of this, some compactness results have to be used, even more when there is a perturbation by the Hardy–Leray potential. Some analytical tools that will be used along the chapter are discussed as a starting point. Then we will find the critical power of the gradient, q+ (λ), in order to have existence and we prove nonexistence if q ≥ q+ (λ) as well as complete blowup. The arguments in the quasilinear case are more involved than in the sublinear case. Finally the existence result is obtained for 0 < λ < ΛN,2 and 1 < q < q+ (λ) by using the compactness arguments explained in the complementary analytical results at the beginning of Chapter 6.
Preface
 XI
In Chapter 7 we study the parabolic problem associated to the nonlinear elliptic problem analyzed in Chapter 6. As in the sublinear case, the introduction of a linear term with the Hardy–Leray potential gives the same threshold to the existence as in the elliptic case, q+ (λ). Section 7.2 introduces some important analytical tools that will be essential to obtain the results in the chapter. These tools could be useful in other contexts. The nonexistence and instantaneous and complete blowup are obtained if p ≥ q+ (λ). We emphasize here that the Caffarelli–Kohn–Nirenberg inequalities play an important role to prove this nonexistence result. Also, the existence of solution for p ≤ q+ (λ) is studied in detail giving precise results on the regularity of the solutions. The results for the Cauchy problem are really surprising. It is well known that for the quasilinear problem associated to λ = 0, that is, without the additional term of the Hardy potential, there is no Fujita type exponent; see for instance [187]. However the presence of the term depending on the Hardy–Leray potential does produce a Fujita exponent. In the last chapters of this work, Chapters 8 to 13, we will consider the relativistic uncertainty principle; that is, the corresponding fractional Hardy inequality and its consequences in problems with fractional diffusion. Among all the possibilities to define a fractional power, (−Δ)s , 0 < s < 1, of the positive operator given by the Laplacian −Δ, we will use the singular kernel that coincides with that coming from the Fourier transform. This is the point of view of potential theory that was introduced by Marcel Riesz in [272], where he suggested to compute the kernel via an analytical continuation in the complex plane of the exponents. This idea has been well established in the literature; in particular , it can be seen in the book by L. Landkof [223]. Nevertheless, we perform the detailed calculation to obtain the kernel in Chapter 8, where we will also study the Dirichlet problem for a related operator to the fractional Laplacian. The fractional Laplacian is in nature a nonlocal operator and, hence, the estimates are often different from the local case. We have tried to present the techniques and functional framework in this area, many of which come from Fourier analysis and potential theory. We place special emphasis on the regularity of the solution in terms of the regularity of the source term. Related to this, a fractional Picone inequality is proved as a tool that will be used in some subsequent chapters. Chapter 9 represents a central point of the monograph. Here we give a direct proof of the Hardy inequality in this fractional setting. Our proof is inspired by the article of W. Beckner [54]. We give a general result which includes even the cases when s > 1 and with dimension N > 2s. Moreover, we find the value of the optimal constant ΛN,2s and prove that it is not attained. Looking at the Euler–Lagrange equation associated to the Hardy inequality we look at the problems (−Δ)s u − λ
u = 0, x2s
where 0 < λ < ΛN,2s , and we study the local behavior of the solution near 0. This result in the case s = 1, as we have seen, is elementary; in fact, it reduces to the integration of
XII  Preface a linear differential equation of Euler type. The case 0 < s < 1 is more involved and is obtained in detail in the corresponding section of Chapter 9. We complete the results with the ground state representation and the Hardy inequality with remainder term, which is another way to prove that the optimal constant is not attained. Chapters 8 and 9 give a selfcontained approach to the main results with respect to the fractional Laplacian and the proof of the corresponding Hardy inequality. The summability of the solutions of the Dirichlet problem for the fractional Laplacian, studied in Chapter 8, is deeply modified by the presence of a linear term involving the Hardy fractional potential. In Chapter 10 such effect is analyzed. The proofs of the results involve new functional settings such as the weighted Sobolev inequalities that are closely related with the equation obtained by the ground state representation. This new functional setting has an interest in itself. In order to obtain the results, the Harnack inequality for the positive solutions of the equation is obtained as a result of the ground state representation. A necessary and sufficient condition for weighted L1 spaces is also obtained as a consequence of the Hardy inequality. In the last section of Chapter 10 we list some open questions. The effect of the fractional Hardy potential in semilinear problems with fractional diffusion is studied in Chapter 11. Some important tools are presented in order to obtain an almost selfcontained discussion. We concentrate in the first part of the chapter on the socalled concaveconvex problem. This choice is made in order to complete the situation in Chapter 4 for the classical, local case. First, for each λ ∈ (0, ΛN,2s ) we find a critical exponent, p(λ, s), such that if 0 < p < p(λ, s) we are able to find a positive minimal solution. The critical exponent gives, in turn, that the critical Sobolev exponent p∗N,s is less than p(λ, s). Then, if p ≤ p∗N,s we study the existence of a second positive solution. To see that p(λ, s) is the barrier for existence, we prove that if p ≥ p(λ, s) there is no positive solution to the problem. Also in this case, the nonexistence gives a complete blowup behavior. To finish the chapter we give some results with nonlinearities that are singular at the boundary. The last chapters, Chapters 12 and 13, are respectively dedicated to the fractional heat equation and to the effect of the Hardy potential on the existence and regularity of the solution to the evolution equation with fractional diffusion. In Chapter 12 we provide a quite complete exposition of the results of existence and regularity with different kinds of data. We would like to describe here a couple of interesting results related to the Cauchy problem. The first one is a detailed proof of the known asymptotic behavior at infinity of the fundamental solution of the fractional heat equation. See [262] for dimension N = 1 and [63] for N ≥ 1. Our proof is inspired in the probabilistic world (see for instance [84] and the references therein), but translated to pure analytical language. In fact, the proof that we present uses Bernstein’s theorem, which is based on the Helly selection principle. Since these results are in some way hidden in the literature we prove both by completeness. The second point is the uniqueness theorem for positive solutions of the Cauchy problem in the spirit of the classical D. Wid
Preface
 XIII
der theorem for the heat equation. Widder’s uniqueness result shows the coherence of the model with the principles of thermodynamics. We finish the monograph with Chapter 13, where the starting point is to prove the Harnack inequality for the resulting weighted parabolic equation of the ground state representation. Observe that this is an equation with singular coefficients, which makes the proof much more difficult. As a consequence of this Harnack inequality and some extra estimates, we are able to prove the exact behavior of the solution near the pole of the Hardy potential. The first consequence of this provides a complete profile of the existence results in terms of the spectral parameter λ > 0 for the linear fractional heat equation. In this way, we obtain necessary and sufficient conditions for the existence, a result that can be seen as the Baras–Goldstein result in the fractional case. Nevertheless, the arguments given here are slightly different from those in the proof of the Baras–Goldstein result for the heat equation. We also study the effect of the Hardy potential on a semilinear fractional heat equation. The critical power is the same as in the elliptic case, as the reader will find in detail in Section 13.5. One may ask also what happens to the similar problems that contain a power of the norm of the gradient as in the local case. At this point, we can only say, “do not worry,” they are not here. We think that a monograph must be of finite length and should be finished at some point. However, the interested reader can still see some results for the elliptic case along this direction in [26]. As the reader may realize, the problems studied in this monograph require a large variety of techniques, which could be useful in many other different problems. We have explained in detail most of the techniques that we have used. In those cases where we could not do this, we have given precise references for anybody to complete the arguments in their totality if necessary. We hope that the contents of the monograph will be useful to any reader, especially for doctoral students working in some areas of analysis and, more specifically, for those interested in the area of partial differential equations. This monograph is the result of many years of work and academic discussions with colleagues and doctoral students. We are deeply thankful to all our collaborators and colleagues for their continuous support and advice. In particular we point out that the paper with J. L. Vázquez [257] represented a first application of the Hardy– Leray inequality. We would like to thank Juan Luis Vázquez for this collaboration. A bit later, a starting piece of work was introduced in the article written in collaboration with Jesús García Azorero [179]. We deeply thank Jesús for this long scientific journey in which we have enjoyed mathematics and many more things together. We are also thankful to our maestros, A. Ambrosetti, L. Boccardo, L. Caffarelli and Y. Meyer,1 for their friendship and advice. We have learned from them the best flavor of mathematics in many and diverse areas. 1 During the proofreading one of our referents passed away, Antonio Ambrosetti. We will always remember him
XIV  Preface Our special thanks go to Boumediene Abdellaoui and Ana Primo for the work we have shared for more than 15 years, enjoying mathematics in a friendly atmosphere. We thank Luigi Orsina for his friendly collaboration. Special thanks also go to Enrico Valdinoci, Begoña Barrios, Serena Dipierro, Tommaso Leonori, María Medina, Luigi Montoro and Dino Sciunzi for allowing us to join the “fractional” adventure. Young collaborators, as Pablo Ochoa and Leandro del Pezzo, gave us the energy to follow. Thanks go to Magdalena Walias, among many other things, for her collaboration and support. To Juanjo Manfredi and Julio D. Rossi we must thank many helpful discussions. We thank also David Arcoya and David Ruiz, the Granada team, and Andrea Dall’Aglio and Daniela Giachetti for their collaboration and friendly hospitality. Fernando Charro, Eduardo Colorado, Carlos Escudero, Veronica Felli, Fausto Ferrari, Filippo Gazzola, Anna Mercaldo and Susana Merchán, have been close collaborators with whom we have worked on other topics that, still, have influenced this monograph; we thank them deeply. The Department of Mathematics of the UAM has been our scientific home for almost our entire professional life; the Institution and its members deserve our recognition for giving us their support during the shared years. The final work of corrections and adjustments was done during a confinement period in Madrid, due to the Covid19 pandemic. We hope that this critical situation will help us to reflect and improve our human living conditions at all levels throughout the World. With our best wishes for the readers, Madrid May 31, 2020. The Authors
Contents Preface  VII 1 1.1 1.2 1.2.1 1.2.2 1.2.3 1.2.4 1.3
Motivation of the Hardy–Leray potential  1 Introduction  1 Where does the monster appear?  2 The Hardy–Leray potential in quantum mechanics  2 More quantum mechanics: the uncertainty principle  4 FrankKamenetskii model in combustion: a Gelfand problem  5 Stability of the singular solution  12 A Picone type inequality  12
2 2.1 2.2 2.3 2.3.1 2.3.2 2.4 2.5 2.5.1
Looking at the Hardy–Leray potential  15 Properties of the Hardy–Leray potential  15 The Hardy–Leray inequality  15 Optimality and nonattainability of the constant ΛN,2  16 Some preliminaries on symmetrization  17 Constant ΛN,2 is the best constant and is not attained  19 Hardy’s inequality in W 1,p (ℝN ), 1 < p < N  21 Hardy–Leray inequality with remainder term  23 A functional consequence  28
3 3.1 3.1.1 3.1.2 3.2 3.2.1 3.3 3.3.1 3.4
Calderón–Zygmund theory and the Hardy–Leray potential  29 Introduction  29 Unbounded solution for m > N2  30 Nonexistence result for m = 1  30 Summability of finite energy solutions  33 Optimality of the condition on λ  34 Existence of infinite energy solutions  35 Complete blowup for data in L1  38 Necessary and sufficient conditions for solvability in weighted L1 spaces  39 Uniqueness  41 Further results  41 A different way to obtain the critical value for λ  41 A remark in the case λ = ΛN,2  43
3.4.1 3.5 3.5.1 3.5.2 4 4.1 4.2
Effect of the Hardy–Leray potential in the solvability of semilinear elliptic equations  45 Introduction  45 The optimal power  45
XVI  Contents 4.3 4.3.1 4.3.2 4.3.3 4.4 4.5 4.5.1 4.6
Some previous results for linear elliptic equations  47 Elliptic equations with measure data  47 Newtonian capacity  48 The Ambrosetti–Rabinowitz mountain pass theorem  54 Results on existence  62 Results on nonexistence  66 Complete blowup  68 Further results and comments  69
5 5.1 5.2 5.2.1 5.2.2 5.3 5.4 5.4.1 5.4.2 5.5 5.6 5.6.1 5.6.2 5.6.3 5.6.4 5.6.5 5.7 5.7.1 5.7.2
The Hardy–Leray potential in semilinear heat equations  71 Introduction  71 Preliminaries and tools  72 Local behavior of supersolutions of the linear equation  74 A technical remark on existence  77 Nonexistence results: p ≥ p+ (λ)  79 Instantaneous and complete blowup results  83 Blowup for the approximated problems when p ≥ p+ (λ)  83 Blowup when pn → p+ (λ)  84 Existence of solutions: p < p+ (λ)  87 Cauchy problem  89 Subsolution blowup in a finite time for small p  90 Global supersolutions for F (λ) < p < p+ (λ)  92 Local existence results for 1 < p < p+ (λ)  93 Global existence for F (λ) < p < p+ (λ) and small data  94 Blowup result for p < F (λ)  94 Further results and remarks  96 Problems involving the pLaplacian heat equation  96 Problems associated to the Caffarelli–Kohn–Nirenberg inequalities  96 The borderline case p = F (λ)  97
5.7.3 6 6.1 6.2 6.2.1 6.2.2 6.2.3 6.3 6.4 6.5 6.6
Elliptic equations with a nonlinearity on the gradient and the Hardy–Leray potential  103 Introduction  103 Some auxiliary results  104 Comparison results  104 A quantitative version of the maximum principle  112 Caffarelli–Kohn–Nirenberg inequalities  113 Nonexistence results: exponent q ≥ q+ (λ)  114 Blowup result  121 Existence result: 1 < p < q+ (λ) and λ < ΛN,2  126 The critical case, λ ≡ ΛN,2 and p < N+2  132 N
Contents  XVII
6.7 7 7.1 7.2 7.2.1 7.2.2 7.2.3 7.3 7.3.1 7.3.2 7.4 7.5 7.5.1 7.5.2 7.5.3 7.5.4 7.5.5 7.6 8 8.1 8.1.1 8.1.2 8.1.3 8.2 8.2.1 8.3 8.3.1 8.3.2 8.3.3 8.3.4 8.4 8.4.1 8.4.2 8.4.3 8.4.4 8.5
Further remarks  135 The heat equation with nonlinearity on the gradient and the Hardy–Leray potential  137 Introduction  137 Preliminaries and tools  139 Maximum principle and comparison results  140 Local behavior of a very weak supersolution to problem (7.1.1)  145 Passing to the limits in locally truncated problems  147 Nonexistence and blowup results  154 Nonexistence  154 Complete and instantaneous blowup  166 Existence results  168 Cauchy problem  173 A class of subsolutions to (7.5.2) for small p, blowup in a finite time  174 A class of global supersolutions to (7.5.2) for F (λ) < p < q+ (λ)  176 Local existence for 1 < p < q+ (λ)  177 Global existence for small data and F (λ) < p < q+ (λ)  181 Blowup in a finite time for p < F (λ) and any positive initial datum  184 Further remarks  186 Fractional Laplacian type operators  189 Introduction  189 Riesz potentials  192 Analytic continuation and the fractional Laplacian formula  194 Some elementary properties of the fractional Laplacian  197 Analytical preliminaries related to (−Δ)s  199 Some remarks on the regularity of solutions in the whole ℝN  205 The Dirichlet problem: the variational framework  211 Elementary estimates  214 Elliptic Kato inequality  217 Weak maximum principle and comparison results  218 Some interpolation results  220 Elliptic problem: finite energy setting  221 Bounded solutions: Moser and Stampacchia methods  222 N The limit case m = 2s : exponential summability  225 A Calderón–Zygmund type result  226 Further fractional regularity  227 A fractional Picone inequality and applications to sublinear problems  228
XVIII  Contents 8.6 8.6.1 8.7 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7
Nonvariational setting for elliptic problems: weak solutions  232 Calderón–Zygmund type result for weak solutions  236 Further results  237 The fractional Hardy inequality  239 Introduction  239 The fractional Hardy inequality  240 Normalizing the constants  248 Local behavior of solutions of the elliptic equation  250 Ground state representation  252 Hardy’s inequality with remainder terms  255 Further results and comments  261
10 Calderón–Zygmund summability in the fractional setting  263 10.1 Introduction and statement of the problem  263 10.2 Functional setting: inequalities with weights  265 10.2.1 Weighted Sobolev inequalities and applications  269 10.2.2 Some compactness results  277 10.2.3 Some numerical inequalities  278 10.3 Weak Harnack inequality and local behavior of nonnegative supersolutions  280 10.4 Optimal summability in the presence of Hardy potential  291 10.4.1 Regularity of energy solutions  291 10.4.2 About the optimality of the regularity results  295 10.4.3 Nonvariational setting: weak solutions  299 10.4.4 A necessary and sufficient condition for solvability  301 10.5 Further results and comments  303 11 Fractional semilinear elliptic problems  305 11.1 Introduction  305 11.2 Preliminaries  306 11.2.1 A convergence tool by Brezis–Lieb  309 11.2.2 Maximum principle for the fractional Laplacian  310 11.2.3 The Pohozaev identity for the fractional Laplacian  314 11.3 Existence of minimal solutions for 1 < p < p(λ, s)  323 11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1  329 11.4.1 Subcritical case  329 11.4.2 The critical problem: p = 2∗s − 1  342 11.5 Nonexistence for p ≥ p(λ, s): complete blowup  358 11.5.1 Complete blowup  362
Contents  XIX
11.6 11.7 11.7.1
Problems with the Hardy potential and nonlinear terms singular at the boundary  365 Further comments  372 Other operators  372
12 The heat equation with fractional diffusion  373 12.1 Introduction  373 12.2 Finite energy setting  373 12.3 Nonvariational setting: weak solutions  376 12.4 A priori estimates and summability of the solutions  379 12.4.1 Bounded solutions  379 12.4.2 Summability of the solutions outside of the Aronson–Serrin zone  384 12.4.3 Further summability results  386 12.4.4 Parabolic Kato inequality  388 12.5 The Cauchy problem for the fractional heat equation  389 12.5.1 Some remarks on regularity of the solution to the Cauchy problem  398 12.6 Regularity in bounded domains: relation between weak solutions and viscosity solutions  404 12.7 The uniqueness result of Widder type  406 12.7.1 Uniqueness for weak solutions  411 12.7.2 Uniqueness for strong positive solutions  419 12.7.3 About viscosity solutions  424 12.8 Further results  425 12.8.1 Some remarks about the fractional Widder theorem  425 12.8.2 The Fujita exponent for the fractional heat equation  428 13 13.1 13.2 13.3 13.4 13.5 13.5.1 13.5.2 13.6 13.6.1 13.6.2
The influence of the Hardy potential on the linear and semilinear fractional heat equations  433 Introduction  433 Functional framework: some preliminary results  435 Weak Harnack inequality for a weighted problem  441 The linear problem: dependence on the spectral parameter λ  456 Existence and nonexistence results for a semilinear problem  465 Nonexistence results for p > p+ (λ, s), instantaneous and complete blowup  468 Existence results for 1 < p < p+ (λ, s)  471 Further results and comments  473 Fujita exponent depending on λ  473 Other operators  473
XX  Contents Bibliography  475 Alphabetical Index  489
1 Motivation of the Hardy–Leray potential The aim of this chapter is to explain the motivation and basic facts about the Hardy– Leray potential, V(x) =
1 . x2
The space dimension must be N ≥ 3 for integrability reasons. In lower dimensions the setting is different: either we lose homogeneity, or we assume some particular behavior of the elements of the functional space close to the pole of the potential.
1.1 Introduction The one dimensional Hardy inequality is the continuous translation of the following discrete inequality. The discrete Hardy inequality. Let {an }n∈ℕ be a sequence of nonnegative real numbers which is not identically zero; then, for every real number p > 1, one has ∞
∑(
n=1
p
∞ a1 + a2 + ⋅ ⋅ ⋅ + an p ) 0, the condition is satisfied only for a root of the equation, and this fact means that the bounded solution is unique up to a multiplicative constant. More interesting is the case of attractive field, that is, λ < 0. We have two possibilities: 1. If − 41 < λ < 0, then for n = 0 both solutions are unbounded at the origin, but there is one solution with a lower order of singularity at the origin that any other one. By choosing this solution, which in fact is locally of finite energy, it is possible to perform calculations and, as is explained in [247], a formula for the scattered intensity may be obtained. We will find many situations in which this behavior is crucial. Note that 41 is a magic quantity that is related to inequality (1.1.2). 2. If λ < − 41 , then all nontrivial solutions to the equations behave like 1
r − 2 e±iα log r near the origin. That is, in this case infinitely many oscillations happen in any neighborhood of the origin. As is explained in [247] in this case there is no solution to the scattering problem. We will find a lot of anomalies for λ < − 41 . The physicist tells us that in this unbounded interval there is no way do choose a convenient solution. We call the attention of the reader about the existence of a constant that is a borderline in the behavior of the radial equation (1.2.3). We will elucidate this point below. Nevertheless, in [115], a method to select such a solution with ideas from Von Neumann is developed. Such idea is a suitable understanding of the orthogonality of the solutions. To give another idea of the peculiar behavior of the Hardy–Leray potential in quantum mechanics, let us describe an observation by C. Fefferman, in his paper [159].
4  1 Motivation of the Hardy–Leray potential We try to study the negative eigenvalues of the operator L ≡ −Δ + V(x)
in ℝN , N ≥ 3,
where V has a suitable local summability condition. Fefferman refers to a theorem by Cwikel–Lieb–Rosenblum, which gives a precise estimate of the number of negatives eigenvalues of L below the level μ, N(L, μ); see [128], [229], [276] and the paper [166] for some fractional related results. Such theorem establishes that the number of eigenvalues of L below the level μ, N(L, μ), is controlled by a constant depending only on the dimension multiplied by the volume in the phase space, that is, in ℝN × ℝN , of the set A(L, μ) = {(x, ξ ) ∈ ℝN × ℝN  ξ 2 + V(x) < μ}. Namely, N(L, μ) ≤ CN A(L, μ). As a consequence, if for fixed μ, A(L, μ) < CN−1 , then L ≥ μ in the sense of the operators. If we now specialize the result for L ≡ −Δ − xλ 2 , it is clear that A(L, μ) = +∞, so it
could be thought that L is not bounded from below. However, one of the more relevant classical results related to the Hardy potential is inequality (2.2.2), which contradicts the idea, given by the above argument, that L ≡ −Δ −
λ x2
is unbounded from below.
The conclusion is that the Hardy–Leray potential has not the summability required in the theorem of Cwikel–Lieb–Rosenblum; see in particular inequality (1.1) in [166] quantifying Cwikel’s inequality.
1.2.2 More quantum mechanics: the uncertainty principle Hardy’s inequality (1.1.2) is the first mathematical formulation of the Heisenberg uncertainty principle, which, roughly speaking, says that the position and the momentum of a particle cannot be approximated with simultaneous accuracy. Indeed, we can read the Hardy inequality by saying that if u is concentrated close to the origin, that is, we have a very good approximation of the position of the particle, then the momentum has to be big. Note that if we have an inequality of the form C ∫ xσ u2 (x) dx ≤ ∫ ∇u2 dx, ℝN
ℝN
for all u ∈ 𝒞0∞ (ℝN ),
1.2 Where does the monster appear?
 5
then necessarily σ = −2. It is sufficient to check the inequality on the family of rescaled functions ϕμ (x) = ϕ(μx) to find the inverse square potential. That means that the inverse square potential has the same dimensionality of the gradient and this is one of the main characteristics to be important. To see more details and forms of the Heisenberg principle the reader can see the nice lecture notes by R. Frank [165]. 1.2.3 FrankKamenetskii model in combustion: a Gelfand problem The semilinear reaction–diffusion equation ut − Δu = λeu ,
λ > 0,
(1.2.4)
and its stationary counterpart Δu + λeu = 0
(1.2.5)
appear in several contexts. For instance, equation (1.2.5) appears in astrophysics in the description of a ball of isothermal gas in gravitational equilibrium. This was already proposed by Lord Kelvin. Radially symmetric stationary solutions have been studied by Emden (1896), Fowler (1926) and Chandrasekar and Hopf in the 1930s; see Chandrasekar’s book [118]. In combustion the evolution model (1.1) was proposed by FrankKamenetskii [169] and is known as the solid fuel ignition model. A more recent approach could be seen in [53]. In the form Δu + K(x)eu = 0 and space dimension 2, the stationary equation is found in Riemannian geometry in the search for certain Riemannian metrics with given Gaussian curvature; see the paper by Kazdan and Warner [211]. Some pioneering works in the 1950s and 1960s are [182] (in particular, Section 15, by G. I. Barenblatt) and [321]. Important mathematical contributions are due to Fujita [170], who studied the evolution problem with zero boundary data as well. An important contribution is due to Joseph and Lundgren [204], which describes the bifurcation diagram of equation (1.2.5) posed in a ball B with zero boundary data. Now we will try to study the relation of the above equation with exponential reaction term and the Hardy–Leray potential. To answer this question let us recall the main results of the bifurcation analysis in [204]. We will consider the Dirichlet boundary problem, Δu + λeu = 0, u = 0,
x ∈ B(0, 1),
‖x‖ = 1.
(1.2.6)
6  1 Motivation of the Hardy–Leray potential If λ1 is the principal eigenvalue for the Laplacian in B(0, 1) and ϕ1 is a positive eigenfunction, we have λ1 ∫ uϕ1 dx = ∫ u(−Δϕ1 ) dx B(0,1)
B(0,1)
= ∫ (−Δu)ϕ1 dx = λ ∫ ϕ1 eu dx ≥ λ ∫ ϕ1 u dx, B(0,1)
B(0,1)
B(0,1)
which implies that if u is a solution of (1.2.6), then λ < λ1 . By using a monotony argument we are able to prove that there exists an interval [0, λ∗ ] such that for λ ∈ [0, λ∗ ], problem (1.2.6) has a minimal solution. Moreover it is known that: 1. If N ≥ 10, the regular minimal solutions for λ ∈ [0, λ∗ ) converge to a solution u∗ (a priori singular) when λ goes to λ∗ . 2. If N < 10, the limit u∗ is a regular solution. See [176], [240], [241] and [204]. Writing equation (1.2.5) in radial coordinates we find N −1 u (r) = λeu , r
− u (r) −
r > 0.
(1.2.7)
By the classical change of variables given in [204] (see also [182]), s = log r, { { { v = us , { { { 2s u(s) {w = −λe e , we obtain the dynamical system { { { { { { {
dv = w − (N − 2)v, ds dw = (2 + v)w. ds
(1.2.8)
It is easy to check that the autonomous dynamical system (1.2.8) has two isolated critical points, i. e., P1 = (0, 0)
and P2 = −2(1, N − 2).
Since u > 0, we are interested in w < 0. By analyzing carefully the linearized system we find that P1 is an unstable hyperbolic point, independently of the dimension N. The stable manifold is the vaxis and the unstable manifold is tangent to w = Nv. The behavior of P2 depends on the dimension, precisely, P2 is {
a stable nodus if N ≥ 10, a stable focus if 3 ≤ N ≤ 9.
1.2 Where does the monster appear?
 7
Equation (1.2.6) is invariant with respect to the scaling (x, u) → (αx, u(αx) + 2 log α), and then we find a radial singular selfsimilar solution to (1.2.6), that is, u(x) = log(
1 ), x2
λ̃ = 2(N − 2).
The direction field shows that the unstable manifold starting out from P1 remains bounded, and therefore is a stable manifold for P2 . In fact, the direction field implies that either the manifold remains bounded or lims→+∞ w(s) = −∞. But in this case, we have a trajectory corresponding to a radial solution such that if we fix λ > λ∗ , for some s1 , w(s1 ) = −λ. By the invariance of the system under translations in the variable s, we can assume s1 = 0, and then w(0) = −λ = −λeu(0) . Moreover, lims→−∞ w(s) = 0 = lims→−∞ v(s). Then, in the r variable u(1) = 0 and u(r) remains bounded in a neighborhood of the origin. Hence problem (1.2.6) has a solution for λ > λ∗ and this is a contradiction. This argument implies that: 1. There exists an heteroclinic orbit, φ, joining the critical points P1 and P2 . 2. Given λ, problem (1.2.6) has a solution for each intersection of φ with the straight line w = −λ. We need the following results to understand the behavior of problem (1.2.7) in terms of the dynamical system (1.2.8). Lemma 1.2.1. Let u be a radial solution of (1.2.7) and (v, w) the corresponding trajectory of (1.2.8). Then u is a regular solution of (1.2.7) (i. e., lims→−∞ u(s) = A < ∞) if and only if lim (v(s), w(s)) = (0, 0).
s→−∞
8  1 Motivation of the Hardy–Leray potential Proof. It is clear that if u is a regular solution, then lims→−∞ (v(s), w(s)) = (0, 0). Conversely, since the unstable manifold starting from the origin is tangent to the line w = Nv, if lim (v(s), w(s)) = (0, 0),
s→−∞
then lim
s→−∞
w(s) = N, v(s)
and in this way, there exists some negative constant s0 such that if s < s0 , then v < N2 . Hence, w 2w + (
1 2N
(s0 − s).
Finally, we get y(s) >
2 2 1 (( ) + (2y(so ) − ( )e2(s0 −s) )) . 2 N N
But from this inequality it follows easily that u(s) < K. The next step is to identify the selfsimilar solution to (1.2.7) in the dynamical system (1.2.8). Lemma 1.2.2. The unique trajectory of (1.2.8) corresponding to a solution of (1.2.7) such that lim u(s) = +∞
s→−∞
is the critical point P2 .
and
eu ∈ L1loc
1.2 Where does the monster appear?
 9
Proof. Assume that lims→−∞ u(s) = +∞. We organize the proof in several cases. (A) We have lims→−∞ v(s) = v0 . 1. If v0 > −2, then there exists some negative constant s0 < 0 such that if s < s0 , (2 + v) > δ > 0. Taking into account the sign of w, dw < δw, ds
2.
for s < s0 .
Then, by integrating from s to s0 , we get lims→−∞ w(s) = 0. Therefore, = −(N − 2)v0 , and this implies that v0 = 0. lims→−∞ dv ds By Lemma 1.2.1 we reach a contradiction with the hypothesis that u is a singular solution. If v0 < −2, in the same way as before we obtain dw > −δw, ds
= −∞, in contradiction and then lims→−∞ w(s) = −∞. Hence, lims→−∞ dv ds with the hypothesis v(s) → v0 . 3. If v0 = −2, it is possible to prove that lims→−∞ w(s) = −2(N − 2) (it is clear if we assume that the limit exists; we will prove in (C) below that this the case). Then, as P2 is an attractor, necessarily v(s) = −2 and w(s) = −2(N − 2) for all s. We will see that the other possibilities are incompatible with the fact that lim u(s) = ∞.
s→−∞
(B) We have lims→−∞ v(s) = −∞. As in case (A)(2), lims→−∞ w(s) = −∞, which is a contradiction with the nonexistence result for λ large. In fact if λ > λ∗ and we have a trajectory which is a radial solution such that lim (v(s), w(s)) = (−∞, −∞),
s→−∞
we can pick up s1 such that w(s1 ) = −λ. Since the autonomous system is invariant by translation in time we can assume that s1 = 0. Then w(0) = −λeu(0) and in the r variable u(1) = 0. Therefore the elliptic problem has a solution for λ > λ∗ , which is a contradiction. (C) We have lim infs→−∞ v(s) < lim sups→−∞ v(s). Under this hypothesis we have moreover lim inf v(s) ≤ −2 ≤ lim sup v(s) s→−∞
s→−∞
(with strict inequality at least on one side, as a consequence of the arguments in (A)). As an application of the mean value theorem, there exists a decreasing sequence {sn }n∈ℕ such that:
10  1 Motivation of the Hardy–Leray potential (i) limn→∞ sn = −∞; (ii)
dv(sn ) ds
= 0;
(iii) v(s2n ) ≥ −2 ≥ v(s2n+1 ) for all n, with strict inequality at least on one side. Then from (ii), w(sn ) = (N − 2)v(sn ). After a finite number of indexes n, the trajectory must cross the manifold joining P1 and P2 which is the unstable manifold for P1 ; then we must have lim (v(s), w(s)) = (0, 0)
s→−∞
and this is a contradiction with the hypothesis that u is a singular solution. As a conclusion, if u is a radial singular solution of (1.2.7) the only possibility is (A)(3). Then we have the following result. Corollary 1.2.3. The unique unbounded radial singular solution of (1.2.7) is given by λ̃ = 2(N − 2),
u(x) = log(
1 ). x2
By scaling (x, u) → (αx, u(αx) + 2 log α) we can obtain the result in every ball and by the same arguments the unique radial solution for λ fixed of {
−Δu = λeu
1,2 u ∈ Wloc (ℝN )
in ℝN \ {0},
eu ∈ L1loc (ℝN )
is u(x) = log(
λ 1 ). ) − log( 2 2(N − 2) x
The following result describes the structure of the set of solutions in the radial setting. Theorem 1.2.4. 1. If N ≥ 10, then λ∗ = λ;̃ for each λ ∈ (0, λ∗ ) we have a unique radial regular solution, and limλ→λ∗ u(λ) = u∗ is the singular solution. 2. If N ≤ 9, then λ̃ < λ∗ , and for λ = λ,̃ (1.2.7) has infinitely many regular radial solutions, the values at the origin going to infinity. Moreover, in the second case: (a) There exists a positive constant ϵ0 > 0 such that if 0 < λ − λ̃ < ϵ0 , then the corresponding problem (1.2.7) has a finite family of radial solutions. (b) We have limλ→λ∗ u(λ) = u∗ ∈ L∞ .
1.2 Where does the monster appear?
 11
Proof. 1. The trajectory ϕ joining P1 and P2 is a stable manifold for the node P2 . If we show that ϕ is a monotone curve contained in the region −2 < v < 0,
−2(N − 2) < w < 0,
then for each line w = −λ we have a unique point of intersection, and therefore a unique regular radial solution for each λ ∈ (0, 2(N − 2)). First, by the direction field, it is easy to see that ϕ is below the line w = (N − 2)v. Call R the straight line w=
N −2 (v − 2). 2
We will show that if −2 < v < 0, then dw < N−2 along R; therefore, the trajectories (v, w) dv 2 of the phase plane cross R from below, and this implies that the trajectory ϕ cannot cut the line R, since it starts from above. Since in the region −2 < v < 0, we have
dv ds
−2(N − 2) < w < (N − 2)v,
< 0, it is sufficient to check that dw N − 2 dv − > 0, ds 2 ds
when (v, w) ∈ R, −2 < v < 0. But dw N − 2 dv − ds 2 ds = (2 + v)w − (
N −2 )(w − (N − 2)v) 2
= (2 + v)((
N −2 N −2 N −2 )(v − 2)) − (( )(v − 2) − (N − 2)v) 2 2 2
= (2 + v){(
N −2 N −2 )(v − 2) + ( ) }. 2 2
2
Then, since N > 10 and −2 < v < 0, (v − 2) + (
N −2 N −2 ) > −4 + ( ) > 0. 2 2
Therefore, if N ≥ 10, the trajectory ϕ cannot cross the line R. 2. The straight line w = −λ̃ crosses the manifold ϕ infinitely many times. Each point of intersection sj gives us a radial solution of (1.2.7) by scaling s in such a way that for s = 0 we have as initial value sj . The rest is a consequence of the previous analysis.
12  1 Motivation of the Hardy–Leray potential 1.2.4 Stability of the singular solution Finally, this is where the Hardy–Leray potential appears. If we look for the stability of the singular solution corresponding to P2 , −2(N − 2) = w = −λe−2s eu(s) ;
therefore, u(s) = log(2(N − 2)) − log λ − 2s.
Since from the second equation we have u(s) = −2s, we conclude λ = 2(N − 2)
and
u2(N−2) (r) = log(
1 ). r2
The stability depends directly on the sign of the linearized equation. Linearizing equation (1.2.5) in u2(N−2) (r) we find −Δv = 2(N − 2)eu2(N−2) (r) v ≡
2(N − 2) v. x2
This fact is in some sense generic for the unbounded radial solutions to supercritical elliptic problems. The associated energy functional is E(v) = ∫(∇v2 − B1
2(N − 2) 2 v ) dx, x2
and, according to the Hardy inequality (2.2.1), E(v) is positive defined if and only if 2(N − 2) < (
2
N −2 ), 2
that is, if and only if N > 10. In fact, according to the improved Hardy inequalities (see below), in dimension N = 10 the singular solution is also stable. A more general situation in which the Hardy inequality appears in this kind of analysis can be seen in the nice survey [99]. Recently, a general result has been obtained in [102] about the regularity of stable solutions to semilinear elliptic equations up to dimension 9. The dynamical behavior of the unbounded solution can be seen in [257].
1.3 A Picone type inequality We consider for 1 < p < ∞ the space W 1,p (Ω), which is defined as the completion of 𝒞 ∞ (Ω), with respect to the norm 1/p
‖ϕ‖p = (∫(ϕp + ∇ϕp )dx) Ω
,
1.3 A Picone type inequality  13
and we denote by W01,p (Ω) the completion of 𝒞0∞ (Ω) with respect to the norm ‖.‖p . By the Poincaré inequality, this norm is equivalent to the Lp norm of the gradient. We will present in this section an extension of a wellknown Picone identity (see [259]) that will become a handy tool to be used in many situations. See also [14], [34] and [300]. Lemma 1.3.1. Let v > 0, u ≥ 0 be differentiable functions. Then if 1 < p < ∞ ∇up ≥ ∇(
up )∇vp−2 ∇v. vp−1
(1.3.1)
Proof. Let u, v ∈ 𝒞 1 (Ω) be such that u ≥ 0 and v > 0. Then to prove (1.3.1) we just have to prove that R(u, v) = ∇up − ∇(
up )∇vp−2 ∇v ≥ 0. vp−1
By a simple computation we obtain R(u, v) = ∇up + (p − 1)(
up−1 up )∇vp − p( p−1 )∇vp−2 ∇v∇u. p v v
We set a = ∇u and b = ( uv ∇v)p−1 . Then using the Minkowski inequality we get (
up−1 p − 1 up up−1 1 p p−2 p−1 ∇vp . ∇u + )∇v ∇v∇u ≤ ( )∇v ∇u ≤ p p vp vp−1 vp−1
Hence R(u, v) ≥ 0. Note that if R(u, v) = 0, then we obtain ∇u∇v = ⟨∇u, ∇v⟩ and therefore ∇u = uv ∇v. Hence ∇( uv ) = 0 and then u = cv; therefore equality is achieved on straight lines in the function space. We need the following integral form of Lemma 1.3.1. Theorem 1.3.2. Fix p so that 1 < p < ∞. Assume v ∈ W01,p (Ω) verifying − div(∇vp−2 ∇v) = ν, a positive bounded Radon measure, with v𝜕Ω = 0 and v ≩ 0. Then for all u ∈ W01,p (Ω) ∫ ∇up dx ≥ ∫ Ω
Ω
up (− div(∇vp−2 ∇v)) dx. vp−1
Proof. First step. Assume that v ∈ W 1,p (Ω) such that v ≥ δ > 0 in Ω and u ∈ 𝒞0∞ (Ω). Without loss of generality we can assume that u ≥ 0. Then there exists a family of regular functions {vn } such that {
vn → v
in W 1,p (Ω), vn ∈ C 1 (Ω),
vn → v
a. e., and vn >
δ 2
in Ω.
(1.3.2)
14  1 Motivation of the Hardy–Leray potential As a consequence of the continuity of − div(∇(⋅)p−2 ∇(⋅)) as an operator from W 1,p (Ω) p to W −1,p (Ω), p = p−1 , and by (1.3.1) applied to vn we get ∇up ≥ ∇(
up
vnp−1
)∇vn p−2 ∇vn .
Integrating by parts and by the Lebesgue dominated convergence theorem we obtain ∫ ∇up dx ≥ ∫( Ω
Ω
− div(∇vp−2 ∇v) p )u , vp−1
u ∈ 𝒞0∞ (Ω), u ≥ 0.
Second step. We consider the general case, i. e., u ∈ W01,p (Ω), v ∈ W01,p (Ω), v ≥ 0, and − div(∇vp−2 ∇v) ≥ 0
is a bounded Radon measure, v𝜕Ω = 0, v ≥ 0 and not identically zero. By the strong maximum principle (see [310]) v > 0 in Ω. We call vm (x) = v(x) + m1 , m ∈ ℕ. In this way we have − div(∇vm p−2 ∇vm ) = − div(∇vp−2 ∇v), and {vm } converges in W 1,p (Ω) and almost everywhere. Therefore, by the first step, we obtain the result for ϕ ∈ 𝒞0∞ (Ω), ϕ ≥ 0. Now the general case follows by a density argument, namely, if un → u in W01,p (Ω), un ∈ 𝒞0∞ (Ω) and un ≥ 0, then we have ∫ ∇un p dx ≥ ∫( Ω
Ω
− div(∇vn p−2 ∇vn ) vnp−1
)upn dx = ∫(
− div(∇vp−2 ∇v)
Ω
vnp−1
)upn dx.
By the hypothesis and Fatou’s lemma we conclude the proof. Remark 1.3.3. Picone’s inequality has a deep relationship with the spectrum of elliptic operators in divergence form. We think in the simplest case of the principal eigenvalue of the Laplacian; that is, let us assume that for λ > 0 and v ∈ W01,2 (Ω), v > 0, we have a positive Radon measure ν and a positive weight ρ such that −Δv = ν ≥ λρ(x)v,
x ∈ Ω.
By the Picone inequality we find in this case that ∫ ∇u2 dx ≥ λ ∫ ρ(x)u2 dx, Ω
for all u ∈ W 1,2 (Ω),
Ω
which means that λ ≥ λ1 (ρ), the principal eigenvalue of the Laplacian with weight ρ. Note that we can write λ1 (ρ) = sup{λ > 0 : ∃v ∈ W 1,2 (Ω), s. t. − Δv ≥ λρ(x)v}. This observation is used to define the first eigenvalue for elliptic operators in nondivergence form (see [58]) and for fully nonlinear elliptic operators homogeneous of degree one (see [121] and the references therein).
2 Looking at the Hardy–Leray potential This chapter presents the behavior of the Hardy–Leray potential with respect to certain functional inequalities and their consequences.
2.1 Properties of the Hardy–Leray potential Consider the Hardy–Leray potential V(x) =
1 , x2
x ∈ ℝN , N ≥ 3.
(2.1.1)
Lemma 2.1.1. Let V be defined by (2.1.1). Then: N 1. V ∈ ℳ 2 ,∞ (ℝN ), the Marcinkiewicz space; 2. V ∈ Lploc (ℝN ), if p < N2 . N
Proof. We check directly that V ∈ ℳ 2 ,∞ (ℝN ). Indeed, 1 N −2 N −1 −N {x ∈ ℝ : x > μ} = {x ∈ ℝ : x < μ 2 } = ωN−1 μ 2 , N where ωN−1 denotes the measure of the unit sphere. The result in (2) is an elementary consequence. Remark 2.1.2. The lack of integrability of the Hardy potential is the reason for which it does not satisfy the conditions of the above cited Cwikel–Lieb–Rosenblum theorem; see in particular inequality (1.1) in [166].
2.2 The Hardy–Leray inequality The elementary proof of the Hardy–Leray inequality that we present here is due to L. Tartar (see [303]). Theorem 2.2.1 (Hardy–Leray inequality). Assume N ≥ 3. For all ϕ ∈ 𝒞0∞ (ℝN ) the following inequality holds: 2
(
ϕ2 (x) N −2 2 ) ∫ dx ≤ ∫ ∇ϕ(x) dx. 2 x2 ℝN
(2.2.1)
ℝN
Proof. The summability of both terms in the inequality is evident. Consider for λ > 0 0 ≤ ∫ ⟨∇ϕ(x) + λ ℝN
x x ϕ(x), ∇ϕ(x) + λ 2 ϕ(x)⟩ x2 x
ϕ2 (x) x 2 = ∫ ∇ϕ(x) dx + 2λ ∫ ⟨∇ϕ(x), 2 ϕ(x)⟩ + λ2 ∫ dx. x x2 ℝN
https://doi.org/10.1515/9783110606270002
ℝN
ℝN
(2.2.2)
16  2 Looking at the Hardy–Leray potential Now, 2λ ∫ ⟨∇ϕ(x), ℝN
ϕ2 (x) x x 2 ϕ(x)⟩ dx = −2λ ϕ(x) div( ) dx = −λ(N − 2) dx. ∫ ∫ x2 x2 x2 ℝN
ℝN
Therefore, (2.2.2) becomes 2
ϕ (x) 2 dx, 0 ≤ ∫ ∇ϕ(x) dx + (λ2 − λ(N − 2)) x2
∀λ > 0.
ℝN
Since the quadratic function f (λ) = λ2 − λ(N − 2) has a minimum at λ0 = f (λ0 ) = −(
N−2 2
and
2
N −2 ), 2
we conclude with inequality (2.2.1). We will denote ΛN,2 = (
2
N −2 ). 2
(2.2.3)
Definition 2.2.2. We call 𝒟1,2 (ℝN ) the completion of 𝒞0∞ (ℝN ) with respect to the seminorm 1
‖u‖1,2
2 2 = ( ∫ ∇u(x) dx) .
ℝN
Note that by the Sobolev inequality we have 2∗
S( ∫ u dx) ℝN
where 2∗ =
2N , N−2
1 2∗
1
2 2 ≤ ( ∫ ∇u(x) dx) ,
∀u ∈ 𝒟1,2 (ℝN ),
(2.2.4)
ℝN
N ≥ 3.
As a consequence of Theorem 2.2.1 we have the following results.
Corollary 2.2.3. Inequality (2.2.1) holds for all u ∈ 𝒟1,2 (ℝN ). Proof. The density of 𝒞0∞ (ℝN ) in 𝒟1,2 (ℝN ) is sufficient to conclude this.
2.3 Optimality and nonattainability of the constant ΛN,2 We will try to analyze the Hardy–Leray in a closer way. The aim of this section is to prove that ΛN,2 defined in (2.2.3) is the best constant in inequality (2.2.1) and, moreover, that it is not attained in the energy space 𝒟1,2 (ℝN ).
2.3 Optimality and nonattainability of the constant ΛN,2
 17
2.3.1 Some preliminaries on symmetrization We closely follow the classical work by G. Talenti [301] and the references therein (see also [302]). Consider E ⊂ ℝN , a Lebesgue measurable set, and u : E → ℝ, a measurable function. We denote by m(E) the Lebesgue measure of E. We assume that: – either m(E) < ∞ or – u decays at infinity, that is, m({x ∈ E : u(x) > t}) is finite for all t > 0. Definition 2.3.1. Given a measurable function u defined in E, its distribution function is defined as [0, ∞] :
→
[0, ∞],
t:
→
μ(t) = m({x ∈ E : u(x) > t}).
The following properties are easy to check: 1. μ is a nonincreasing function; 2. μ is right continuous; 3. μ(0) is the measure of the support of u and μ(∞) = 0; 4. {t ≥ 0 μ(t) = 0} = [ess sup u, ∞), and as a consequence, support(μ) = [0, ess sup u]; 5. for all t > 0, lim μ(s) = μ(t − ) = m({x ∈ E : u(x) ≥ t}),
s s}). Proposition 2.3.3. The following properties hold: 1. u∗ is nonincreasing; 2. u∗ is right continuous; 3. u∗ (0) = ess sup u and u∗ (+∞) = 0; 4. {s ≥ 0  u∗ (s) = 0} = [m(suppt(u)), +∞);
(2.3.1)
18  2 Looking at the Hardy–Leray potential 5.
{t ≥ 0  μ(t) ≤ s} = [u∗ (s), +∞) for all s ≥ 0, and thus u∗ (s) = min{t ≥ 0 : μ(t) ≤ s};
6. 7.
{s ≥ 0 : u∗ (s) > t} = [0, μ(t)) for all t ≥ 0, and as a consequence we find that u and u∗ are equidistributed; u∗ (μ(t)) ≤ t;
8. μ(u∗ (s)) ≤ s for all s ≥ 0 and μ(u∗ (s)− ) ≥ s for all 0 ≤ s ≤ m(suppt(u)); 9.
[μ(t), μ(t − )) ⫅ {s ≥ 0  u∗ (s) = t} ⫅ [μ(t), μ(t − )].
For the proof we refer the reader to [301]. Note that (8) has the following consequence. Let us take ϕ with ϕ : [0, ∞] → [0, ∞], a continuous, nondecreasing function such that ϕ(0) = 0. Then, ∞
∫ ϕ(u(x)) dx = ∫ ϕ(u∗ (s)) ds.
(2.3.2)
0
E
Next we define the radial rearrangement. Definition 2.3.4. The symmetric rearrangement of u, u , is defined by the expression u (x) = u∗ (ωN xN ),
∀x ∈ ℝN ,
(2.3.3)
where ωN denotes the measure of the unit ball in ℝN . We summarize the relevant properties of u in the following statement. Proposition 2.3.5. The following properties hold: 1. u is nonnegative, radial and radially decreasing with respect to distance to the origin; 2. u and u are equidistributed. It is easy to see that we have the following equivalent representations for u∗ and u , respectively: ∞
u∗ = ∫ χ[0,μ(t)] dt, 0
(2.3.4)
∞
u (x) = ∫ χ{x∈ℝN :ωN xN 0, take the radial function Uϵ (r) = {
AN,ϵ AN,ϵ r
if r ∈ [0, 1], 2−N 2
if r > 1,
−ϵ
(2.3.10)
where AN,ϵ = 2/(N − 2 + 2ϵ), whose derivative is Uϵ (r) = {
0 −r
if r ∈ [0, 1], − N2 −ϵ
if r > 1.
(2.3.11)
20  2 Looking at the Hardy–Leray potential By a direct computation we get ∫ ℝN
Uϵ2 (x) Uϵ2 (x) U 2 (x) dx = ∫ ϵ 2 dx + ∫ dx 2 x x x2 B
ℝN \B
1
A2N,ϵ ωN−1 (∫ r N−3 dr
=
1
0
∞
+ ∫ r −(1+2ϵ) dr) 1
2 = A2N,ϵ ωN−1 ∫ r N−3 dr + A2N,ϵ ∫ ∇Uϵ (x) dx. 0
ℝN
Then, given AN,ϵ < ΛN,2 , ϵ > 0, there exists Uϵ ∈ 𝒟1,2 (ℝN ) such that U 2 (x) 2 A2N,ϵ ∫ ∇Uϵ (x) dx ≤ ∫ ϵ 2 dx. x ℝN
ℝN
Step 2. The same result is true in the unit ball B1 (0) ⊂ ℝN . The argument of optimality in the case of the unit ball proceeds by approximation as follows. First, we remark that by the invariance under dilation the optimal constant will be the same for any ball. Second, let BR be a ball with large radius. We take as test function v(x) = ψ(x)U(x), where U is one of the approximate optimizers explicitly given above and ψ ∈ 𝒞0∞ (BR ) is a cutoff function which is identically 1 on BR−1 with ∇ψ ≤ m. It is easily seen that for R ≫ 1 the influence of ψ in the calculation of step 1 is negligible. It is clear that the same argument works for any bounded domain Ω such that 0 ∈ Ω and for the space W01,2 (Ω). Step 3. Nonattainability of ΛN,2 in ℝN . By a symmetrization argument, for all u ∈ 𝒟1,2 (ℝN ), we have ∫ ℝN
(u (x))2 u(x)2 dx ≤ dx ∫ x2 x2
(2.3.12)
ℝN
and 2 2 ∫ ∇u(x) dx ≥ ∫ ∇u (x) dx.
ℝN
(2.3.13)
ℝN
Then ∫ℝN ∇u(x)2 ∫ℝN
u(x)2 x2
dx
≥
∫ℝN ∇u (x)2 dx ∫ℝN
(u (x))2 x2
dx
≥ ΛN,2 .
(2.3.14)
As a consequence of the arguments above, if the minimizers exist, they must be radial. The corresponding Euler equation in radial coordinates is u (r) + (N − 1)
u u (r) + ΛN,2 2 = 0, r r
(2.3.15)
2.4 Hardy’s inequality in W 1,p (ℝN ), 1 < p < N
 21
which by elementary integration gives u(r) = r −(
N−2 ) 2
(a1 + a2 log(r)),
a1 , a2 ∈ ℝ.
1,2 (ℝN ). In others words, ΛN,2 is not an eigenvalue Note that the minimizers are not in Wloc in bounded domains containing the pole of the potential and with Dirichlet boundary data.
We will call u2 (x) = x−
N−2 2
(2.3.16)
the pseudominimizer. As a consequence of the previous arguments, we find the following result. Proposition 2.3.8. Let Ω ⊂ ℝN be a bounded domain such that 0 ∈ Ω. Consider the linear operator ℒλ u ≡ −Δu −
λ u x2
(2.3.17)
in W 1,2 (Ω). Then: 1. If λ ≤ ΛN,2 , ℒλ is a positive operator. 2. If λ > ΛN,2 , ℒλ is unbounded from below. Proof. (1) This is obvious from Hardy’s inequality. (2) As an easy consequence of the optimality of the constant and a density argument one proves the existence of ϕ ∈ 𝒞0∞ (Ω) such that ⟨ℒλ ϕ, ϕ⟩ < 0. We can assume, without any loss of generality, that ‖ϕ‖2 = 1. Defining uμ (x) = μN/2 ϕ(μx) we have ‖uμ ‖2 = 1 and the homogeneity of the operator allows us to conclude that ⟨ℒλ uμ , uμ ⟩ =
μ2 ⟨ℒλ ϕ, ϕ⟩ < 0.
Remark 2.3.9. It is worth noting that the constant could be attained with other boundary conditions. This result is trivial with Neumann conditions for which the constant of the Rayleigh quotient is zero and the constant becomes the minimizer itself. For mixed boundary conditions, see [7]
2.4 Hardy’s inequality in W 1,p (ℝN ), 1 < p < N The main point of this section is to discuss the following Hardy inequality for 1 < p < N in ℝN , which we include by completeness. For more details and applications, see [179]. Lemma 2.4.1. Assume 1 < p < N. Then if u ∈ W 1,p (ℝN ),
22  2 Looking at the Hardy–Leray potential (1)
u x
∈ Lp (ℝN );
(2) (Hardy inequality) ∫ ℝN
up p dx ≤ Λ−1 N,p ∫ ∇u dx xp ℝN
p p with Λ−1 N,p = ( N−p ) ;
(3) the constant Λ−1 N,p is optimal. Proof. Step 1. A density argument allows us to consider only smooth functions u ∈ C0∞ (ℝN ). Under this hypothesis we have the following identity: ∞
∞
1
1
d p p p−1 u(λx) dλ = −p ∫ u (λx)⟨x, ∇u(λx)⟩ dλ. u(x) = − ∫ dλ By using the Hölder inequality, it follows that ∞
∫ ℝN
u(x)p up−1 (λx) x dx = −p ∫ ∫ ⟨ , ∇u(λx)⟩ dx dλ p x x xp−1 1 ℝN ∞
= −p ∫ 1
≤
dλ
λN−1−p
∫ ℝN
p u(y)p−1 𝜕u(y) u(y)p−1 𝜕u(y) dy = − dy ∫ N −p yp−1 𝜕r yp−1 𝜕r p
(p−1)/p
p u(y) (∫ dy) N −p yp ℝN
ℝN
1/p 𝜕u(y) p ( ∫ dy) , 𝜕r N ℝ
and then we conclude that p
∫ ℝN
p up (x) p dx ≤ ( ) ∫ ∇u(x) dx. xp N −p ℝN
Step 2. Optimality of the constant. We proceed as in the case p = 2. Given ε > 0, consider the radial function { AN,p,ε U(r) = { p−N −ε p { AN,p,ε r
if r ∈ [0, 1], if r > 1,
(2.4.1)
where AN,p,ε = p/(N − p + pε). Observe that its derivative is { 0 U (r) = { − Np −ε { −r
if r ∈ [0, 1], if r > 1.
(2.4.2)
2.5 Hardy–Leray inequality with remainder term 
23
By direct computation we get ∫ ℝN
U p (x) U p (x) U p (x) dx = ∫ dx + ∫ dx p p x x xp B
ℝN −B
1
∞
0
1
= ApN,p,ε ωN−1 (∫ r N−1−p dr + ∫ r −(1+pε) dr) 1
p = ApN,p,ε ωN−1 ∫ r N−1−p dr + ApN,p,ε ∫ ∇U(x) dx. 0
ℝN
We conclude as in the case p = 2. Corollary 2.4.2. The same result is true in the unit ball B ⊂ ℝN . Proof. The proof of the first step is the same. The argument of optimality in the case of the unit ball proceeds by approximation as follows. First, we remark that by the invariance under dilation the optimal constant will be the same for any ball. Second, let BR be a ball with large radius. We take as test function v(x) = ψ(x)U(x), where U is one of the approximate optimizers explicitly given above and ψ ∈ C0∞ (BR ) is a cutoff function which is identically 1 on BR−1 with ∇ψ ≤ m. It is easily seen that for R ≫ 1 the influence of ψ in the calculation of step 2 is negligible. Remark 2.4.3. The nonattainability of the optimal constant can be seen also as a consequence of the Hardy inequality with remainder term obtained in Theorem 2.5.1 below. See [6] for more details. )p . Therefore we have ΛN,p = ( N−p p
We remark that the Euler equation corresponding to the Hardy inequality is −Δp u ≡ − div(∇up−2 ∇u) = ΛN,p
up−1 . xp
See [179] for more details and some applications.
2.5 Hardy–Leray inequality with remainder term The results in the previous section revealed that ΛN,2 is not an eigenvalue, but more precisely, it is the infimum of the essential spectrum of the Laplace operator with respect to the weight given by the Hardy–Leray potential. This fact motivates the presence of a corrector term in inequality (2.2.1). A short history of this question is as follows.
24  2 Looking at the Hardy–Leray potential (a) Brezis–Vázquez in [97] obtained a remainder term for the classical Hardy–Sobolev 2N inequality; precisely, if 1 ≤ q < N−2 , there exists a constant Cq (Ω) > 0 such that 2
2
q u2 N −2 ) ∫ 2 dx ≥ Cq (Ω)(∫ uq dx) , ∫ ∇u dx − ( 2 x
2
Ω
Ω
(2.5.1)
Ω
W01,2 (Ω).
for all u ∈ (b) Vázquez–Zuazua in [311], among other results, improved the previous inequality by showing that if 1 < q < 2, there exists Cq (Ω) > 0 such that, for all u ∈ W01,2 (Ω), 2
2
q u2 N −2 ) ∫ 2 dx ≥ Cq (Ω)(∫ ∇uq dx) . ∫ ∇u dx − ( 2 x
2
Ω
Ω
Ω
(2.5.2)
The improved inequality (2.5.2) proves that H, the completion of 𝒞0∞ (Ω) with respect to the norm ‖u‖2H = ∫ ∇u2 dx − ( Ω
2
N −2 u2 ) ∫ 2 dx, 2 x Ω
is a Hilbert space and, as a consequence, the improved inequality holds for u ∈ H. This kind of inequalities have been extended in several directions, mainly the following. (i) In [29], Adimurthi–Chaudhuri–Ramaswamy obtained by symmetrization arguments an extension of inequality (2.5.1) to the case of the Hardy–Sobolev inequalpN ity in the space W01,p (Ω): if 1 < q < N−p , there exists Cp,q (Ω) > 0 such that p
p
q up N −p ) ∫ p dx ≥ Cp,q (Ω)(∫ uq dx) , ∫ ∇u dx − ( p x
p
Ω
Ω
Ω
for all u ∈ W01,p (Ω).
(ii) By using a change of variables that appeared in [116], Wang–Willem in [314] obtained the optimal remainder term related to (2.5.2) for the Caffarelli–Kohn– Nirenberg inequalities, −1 ∫ ∇u2 x−2γ dx, ∫ u2 x−2(γ+1) dx ≤ λN,2,γ ℝN
ℝN
where λN,2,γ = (
2
N − 2(γ + 1) ), 2
−∞ < γ
0, for all u ∈ W01,2 (Ω). Finally in [6], the following general result is obtained with an elementary proof. Theorem 2.5.1. Let p, γ be such that 1 < p < N and −∞ < γ
0, for some 1,p positive constant ρ(N, p, q, γ), and the weighted Sobolev space 𝒟0,γ (Ω) is defined as the completion of
𝒞0∞ (Ω)
p
1/p
with respect to the norm ‖ϕ‖p,γ = (∫ ∇ϕ x
−pγ
dx)
.
Ω
The details about the functional setting and this general result can be seen in [6]. Here we only point out that for its proof one needs to overcome several difficulties. For instance, in the nonlinear setting, p ≠ 2, Fourier analysis is not allowable and Schwarz symmetrization results are unknown due to the presence of weights. We recall that Schwarz symmetrization is based on an isoperimetric inequality, which in general is unknown in the weighted case. In this section we limit ourselves to prove the result in the case p = 2 and γ = 0. This corresponds to the Hardy–Leray potential and includes an alternative proof of the optimal result in [314]. More precisely, we will prove the following result. Theorem 2.5.2. Assume R > 0 is such that 0 ∈ Ω ⊂ Ω ⊂ BR (0). Then there exists a positive constant C ≡ C(N, Ω) such that for all ϕ ∈ 𝒞0∞ (Ω) we have ∫ ∇ϕ2 dx − ( Ω
2
ϕ2 N −2 R ) ∫ 2 dx ≥ C ∫ ∇ϕ2 (log( )) dx. 2 x x −2
Ω
(2.5.5)
Ω
We prove Theorem 2.5.2 as in [6] but using a different approach, perhaps more elementary than the previous one. This approach uses a Picone type inequality (see Theorem 1.3.2) and convenient test functions that, roughly speaking, approximate the pseudominimizer, the ground state. Proof of Theorem 2.5.2. Consider u2 , the pseudominimizer defined in (2.3.16). Without loss of generality we can assume that R = 1. By a direct computation and using an
26  2 Looking at the Hardy–Leray potential approximation argument we obtain ∫ ∇ϕ2 dx − ( Ω
2 2 ϕ ϕ2 N −2 ) ∫ 2 dx = ∫∇ϕ − ∇u2 dx. 2 u2 x Ω
Ω
Since Ω ⊂ B1 (0), there exists a positive constant C ≡ C(N, Ω) such that −2 2 2 ϕ ϕ 1 ∇ϕ − ≥ C(log( ∇ϕ − ∇u )) ∇u . u2 2 x u2 2
Hence
−2 2 2 ϕ ϕ 1 2 ∇u ∇ϕ − ∇u2 ≥ C(log( )) (∇ϕ + ϕ 2 − 2 ∇u2 ∇ϕ). u2 u2 x u2
Integrating and using Young’s inequality we obtain 2 ϕ ∫∇ϕ − ∇u2 dx u2
Ω
≥ C(1 − ϵ){∫ ∇ϕ2 (log(
ϕ2 1 1 1 )) dx − ΛN,2 ∫ 2 (log( )) dx}. x ϵ x x −2
−2
(2.5.6)
Ω
Ω
Hence, we just need to prove that D1 =
=
inf ∞
ϕ∈𝒞0 (Ω),ϕ=0 ̸
inf
∫Ω ∇ϕ − ∫Ω
ϕ ∇u2 2 dx u2
ϕ2 1 −2 (log( x )) dx x2
)2 ∫Ω ∫Ω ∇ϕ2 dx − ( N−2 2
ϕ∈𝒞0∞ (Ω),ϕ=0 ̸
∫Ω
ϕ2
x2
dx
1 −2 (log( x )) dx
This follows by using the fact that ū 2 (x) = x− to the equation − Δw − ΛN,2
ϕ2 x2
N−2 2
> 0.
(2.5.7)
1
1 (log( x )) 2 is a distributional solution −2
1 w 1 w = (log( )) , 2 2 4 x x x
(2.5.8)
and then using Picone’s inequality with ū 2 . Note that ū 2 ∈ W 1,q (Ω) for all q < 2. By an approximation argument as in [179], we can prove that the optimal constant in (2.5.7) is exactly D1 = 41 and, thus, it is independent of Ω. Therefore by (2.5.6) and (2.5.7) we obtain ∫ ∇ϕ2 dx − ( Ω
2 2 ϕ ϕ2 N −2 ) ∫ 2 dx = ∫∇ϕ − ∇u2 dx 2 u2 x Ω
Ω
≥ C ∫ ∇ϕ2 (log( Ω
1 )) x
−2
dx.
2.5 Hardy–Leray inequality with remainder term 
27
Next we prove that the result in Theorem 2.5.2 is sharp. Theorem 2.5.3. Inequality (2.5.5) in Theorem 2.5.2 is optimal in the sense that 1 −2 1 −2 (log( x )) cannot be replaced by any weight of the form (log( x )) g(x), where g is a positive function such that g(x) → ∞ as x → 0. Proof. First we need the following result in the particular case α = 2. Lemma 2.5.4. Assume that α > 0 and let Ω be a bounded domain such that Ω ⊂ BR (0), where R < 1. We set A=
1 −α )) dx ∫Ω ∇ϕ2 (log( x
inf
ϕ∈𝒞0∞ (Ω),ϕ=0 ̸
∫Ω
ϕ2 1 −α (log( x )) x2
dx
.
Then A = ΛN,2 . Proof. Fix ϕ ∈ 𝒞0∞ (Ω). Then by using Picone’s inequality and a suitable approximation argument we get the following pointwise inequality: ∇ϕ2 (log(
−α
1 )) x
ϕ2 1 )∇u2 (log( )) . u2 x −α
≥ ∇(
Integrating both sides of the previous inequality, we obtain ∫ ∇ϕ2 (log( Ω
−α
1 )) x
dx ≥ ∫ Ω
ϕ2 1 (− div((log( )) ∇u2 )) dx. u2 x −α
Now, by a direct computation we get − div((log(
−α
1 )) ∇u2 ) x −α
= (log(
1 1 )) (−Δu2 ) − α(log( )) x x
−(α+1)
x ∇u . x2 2
Therefore, − div((log(
−α
1 )) ∇u2 ) x
= ΛN,2 (log(
−α
1 )) x
u2 N −2 1 + α( )(log( )) 2 x x2
−α−1
u2 . x2
Hence, using (2.5.9), we conclude that A ≥ ΛN,2 . Note that −α
1 ∫ ∇u2  (log( )) x 2
Ω
R1
1 1 dx ≤ ΛN,2 ∫ (log( )) r r 0
−α
dr < ∞,
(2.5.9)
28  2 Looking at the Hardy–Leray potential where Ω ⊂ BR1 (0) and R1 < 1. Moreover, a direct computation shows that 1 −α ∇u2 2 (log( x )) dx
∫B
R1 (0)
u22 1 −α 2 (log( x )) R1 (0) x
∫B
dx
= ΛN,2 .
Hence, by an approximation argument as in [179], we obtain finally that A = ΛN,2 . We prove the optimality result in Theorem 2.5.3 by the following contradiction argument. Assume by contradiction that there exist a constant C > 0 and a positive function g(x) such that g(x) → ∞ as x → 0, satisfying ∫ ∇ϕ2 dx − ( Ω
2
ϕ2 R N −2 ) ∫ 2 dx ≥ C ∫ ∇ϕ2 (log( )) g(x) dx. 2 x x −2
Ω
Ω
Taking Ω = Bϵ (0) with 0 < ϵ small enough, there exists a positive constant Cϵ such that ΛN,2 Cϵ ≫ D1 = 41 and ∫ ∇ϕ2 dx − ( Bϵ (0)
2
N −2 ) 2
ϕ2 R dx ≥ Cϵ ∫ ∇ϕ2 (log( )) dx. x x2 −2
∫ Bϵ (0)
Bϵ (0)
Using Lemma 2.5.4 with α = 2, we conclude that ∫ ∇ϕ2 dx − ( Bϵ (0)
2
N −2 ) 2
ϕ2 ϕ2 1 dx ≥ C Λ (log( )) dx. ∫ ϵ N,2 x x2 x2 −2
∫ Bϵ (0)
Bϵ (0)
Hence we get a contradiction with the optimality result in (2.5.7). 2.5.1 A functional consequence An important consequence of the Hardy inequality with remainder term is that ‖u‖ℋ(Ω) = (∫(∇u2 − ΛN,2 Ω
1/2
u2 ) dx) x2
is a Euclidean norm on 𝒞0∞ (Ω). Consider the Hilbert space ℋ(Ω) defined as the completion of 𝒞0∞ (Ω) with respect to the norm ‖u‖ℋ(Ω) . Then we have, with strict inclusions,
that
W01,2 (Ω) ⊂ ℋ(Ω) ⊂ ⋂ W01,p (Ω), 1≤p N+2 . This observation will be useful in the next sections.
3 Calderón–Zygmund theory and the Hardy–Leray potential 3.1 Introduction Regularity results for solutions of elliptic equations in terms of the summability of a source term are nowadays classic. In the linear case we have two types of results: those by Stampacchia for variational operators with discontinuous coefficients and those by Calderón and Zygmund for general operators with regular coefficients. The results by Stampacchia (see [291]) can be summarized as follows in the paradigmatic case of the Laplacian. Consider the solution to −Δu = f with homogeneous Dirichlet boundary conditions. Then we know that: –
– –
if f ∈ Lm (Ω), m >
if f ∈ Lm (Ω),
2N N+2
N , 2
then u ∈ W01,2 (Ω) ∩ L∞ (Ω);
≤m≤
if f ∈ Lm (Ω), 1 < m
0. Thus, u ≥ η in Br (0) ⊂ Ω for a suitable η > 0. By weak comparison then u ≥ v in Br (0), where v is the solution to problem η
−Δv = λ x2 { v=0
in Ω, on 𝜕Ω,
which is easily seen to be unbounded. In the following example we can see a precise behavior in terms of λ. Consider the radial equation −u (r) −
λ (N − 1) u (r) − 2 u(r) = 1. r r
We look for a solution in the Sobolev space W01,2 (Ω) such that u(1) = 0. An elementary argument provides that such solution is u(r) = A(r −α− (λ) − r 2 ), where A =
1 . 2N+λ
3.1.2 Nonexistence result for m = 1 The fact that the problem is linear, together with the previous section, shows by duality that, in general, there is no solution for data in L1 (Ω). The meaning of solution here is the following: given f ∈ L1 (Ω) we say that a function u ∈ L1 (Ω) such that xu 2 ∈ L1 (Ω) is a weak solution to equation − Δu − λ
u =f x2
(3.1.5)
3.1 Introduction
 31
if and only if ∫ u(−Δϕ − λ Ω
ϕ − fϕ) dx = 0, x2
for all ϕ ∈ 𝒞0∞ (Ω).
We will say that the problem is well posed in L1 (Ω) if there exists a constant C > 0 such that ‖u‖1 ≤ C‖f ‖1 ,
for all f ∈ L1 (Ω).
Let us prove that problem (3.1.5) is not well posed in L1 (Ω). Assume that for each f ∈ L1 (Ω) we can find u ∈ L1 (Ω), a weak solution to problem (3.1.5). For g ∈ L∞ (Ω) let ψ ∈ W01,2 (Ω) be the solution of −Δψ − λ
ψ = g. x2
Then ∫ fψ dx = ∫ gu dx ≤ ‖g‖∞ ‖u‖1 ≤ C‖g‖∞ ‖f ‖1 . Ω
Ω
But then, taking the infimum on f ∈ L1 (Ω) with ‖f ‖1 ≤ 1, we find that ‖ψ‖∞ ≤ C‖g‖∞ , which is false in general as previously shown. These previous results motivate why we will consider hereafter 1 < m < N2 . We recall the classical Sobolev inequality. Let 1 ≤ p ≤ ∞ and let W 1,p (ℝN ) be the Sobolev space of functions in Lp (ℝN ) with first derivatives in Lp (ℝN ). Proposition 3.1.1. Assume 1 ≤ p < N. Then there exists a constant SN,p such that for every u ∈ W 1,p (ℝN ) p∗
SN,p ( ∫ u
dx)
1 p∗
p
1 p
≤ ( ∫ ∇u dx) ,
where p∗ =
ℝN
ℝN
Np . N −p
(3.1.6)
See for instance [86] and [293] for a couple of different proofs. Considering now 𝒟
1,p
N
(ℝN ) = {f ∈ Lp (ℝN )  ∇f ∈ [Lp (ℝN )] }, ∗
the Sobolev inequality gives the continuous inclusion 𝒟
1,p
(ℝN ) ⊂ Lp (ℝN ), ∗
with SN,p ‖f ‖Lp∗ (ℝN ) ≤ ‖∇f ‖Lp (ℝN ) .
It is not too hard to prove that W 1,p (ℝN ) is dense in 𝒟1,p (ℝN ) with the metric of Lp (ℝN ). By an interpolation argument and the Sobolev inequality we obtain the following result. ∗
32  3 Calderón–Zygmund theory and the Hardy–Leray potential Corollary 3.1.2. Let 1 ≤ p < N. Then W 1,p (ℝN ) ⊂ Lq (ℝN ),
for all q ∈ [p, p∗ ],
with continuous injection. Also, an iteration procedure allows to obtain the following result. Corollary 3.1.3. If p = N, then W 1,N (ℝN ) ⊂ Lq (ℝN ),
for all q ∈ [p, ∞),
with continuous injection. Finally we recall the classical result by Morrey. Proposition 3.1.4. If p > N, then W 1,p (ℝN ) ⊂ L∞ (ℝN ), with continuous injection. Moreover, for all u ∈ W 1,p (ℝN ) we have 1− N u(x) − u(y) ≤ Cx − y p ‖∇u‖p ,
for all x, y ∈ ℝN ,
with C = C(N, p) a universal constant.
The reader can find the main properties of the Sobolev spaces for instance in the books [28], [86] and [239]. The constant of the Sobolev inequality (3.1.6), that is, when 1 ≤ p < N, is the same for all bounded domains Ω. The difference is that in the whole ℝN this constant is attained, while in every bounded domain there is no minimizer. The minimizers in W 1,p (ℝN ) are given, up to dilations and translation, by the function U(x) =
1
p
(1 + x p−1 )
N−p p
.
The existence and uniqueness up to dilations and translation can be seen for instance in [302] and [230]. The exponent p∗ is called the critical Sobolev exponent. Since for p ≥ N the Sobolev exponent is not finite, we can summarize the above by setting Np
p∗ = { N−p +∞
if 1 ≤ p < N, if p > N.
In a domain Ω ⊊ ℝN we define the Sobolev space with null trace, W01,p (Ω), as the completion of 𝒞0∞ (Ω) with respect to the norm of W 1,p (ℝN ). ∗ Finally, when 1 < p < N2 we can consider W 1,p (ℝN ), for which the critical Sobolev exponent is then ∗
(p∗ ) := p∗∗ =
Np . N − 2p
This exponent will appear in the next results. We also recall the classical truncation functions.
3.2 Summability of finite energy solutions  33
Definition 3.1.5. Let Tk : ℝ → ℝ be the truncation defined by Tk (s) = max{min{k, s}, −k}, k > 0. We also consider Gk (s) = s − Tk (s). Note that Tk (s) = s
if s ≤ k,
Tk (s) = k
s s
if s ≥ k.
We will use these truncations in order to obtain some estimates. To that end, we will consider the approximate Dirichlet problems un ∈ W01,2 (Ω) : −Δun = λ
un
x2
+
1 n
+ fn ,
(3.1.7)
where fn (x) = Tn (f (x)). Observe that for every n ∈ ℕ, there is always a unique solution un ∈ W01,2 (Ω) ∩ ∞ L (Ω) of (3.1.7) (see for instance [291]).
3.2 Summability of finite energy solutions 2N < m < N2 . In this case, even though the right In this section we will assume that N+2 hand side of (3.1.1) belongs to the dual space, W −1,2 (Ω), the following surprising phenomenon shows up. A restriction on λ must be assumed in order to have the same results as for the Laplacian. More precisely, for m ≥ 1, let us define
N(m − 1)(N − 2m) . m2
λ(m) =
We will . prove that this is the critical value of λ depending on the summability of f . Theorem 3.2.1. If the right hand side f of equation (3.1.1) belongs to Lm (Ω), m < N2 , and λ < λ(m) =
N(m − 1)(N − 2m) , m2
2N N+2
≤
(3.2.1)
then the corresponding weak solution u belongs to Lm (Ω), where we recall that m∗∗ = Nm . N−2m ∗∗
Proof. Since for any n ∈ ℕ, the solution to (3.1.7), un , belongs to L∞ (Ω), we can use un γ−2 un , with γ = m(N−2) , as a test function in (3.1.7). N−2m Since m ≥
2N N+2
we find that γ ≥ 2. Therefore, γ
1
2
m γ 2 u  2 4(γ − 1) ∫∇un  2 ≤ λ ∫( n ) + ‖fn ‖Lm (∫ un (γ−1)m ) , 2 x γ
Ω
Ω
Ω
and so 1
2
[
m γ 2 4(γ − 1) 2 − λ( ) ] ∫∇un  2 ≤ ‖fn ‖Lm (∫ un (γ−1)m ) . 2 N −2 γ
Ω
Ω
34  3 Calderón–Zygmund theory and the Hardy–Leray potential Note that in order to have an a priori estimate we need 2
4(γ − 1) 2 ), > λ( N −2 γ2
(3.2.2)
that is, (γ − 1)(N − 2)2 > λγ 2 . Given the definition of γ, this is true if λ
N (m−1)(N−2m) m2 coincides with the summability of r α+ and r α+ does not belong to Lm . ∗∗
3.3 Existence of infinite energy solutions As in the previous section we start with a result of existence under the restriction λ < 2N λ(m) and with Lm (Ω) ⊄ W −1,2 (Ω), that is 1 < m < N+2 . More precisely we have the following result. Theorem 3.3.1. If the right hand side f of equation (3.1.1) belongs to Lm , 1 < m < and λ < λ(m) =
2N , N+2
N(m − 1)(N − 2m) , m2
then the weak solution u belongs to W01,m (Ω). ∗
Proof. Define for ϵ > 0 the test function γ−1
vϵ = ((ϵ + un )
− ϵγ−1 )sgn(un ),
where γ =
Use vϵ as test function in (3.1.7) to obtain (γ − 1) ∫ Ω
(N − 2)m . N − 2m
γ γ ∇un 2 4(γ − 1) 2 = ∫∇[(ϵ + un ) 2 − ϵ 2 ] 2−γ 2 (ϵ + un ) γ
Ω
γ
γ
2
(ϵ + un ) 2 − ϵ 2 ) dx ≤ λ ∫( x Ω
+λ∫ Ω
γ γ 2 1 (un vϵ − [(ϵ + un ) 2 − ϵ 2 ] ) dx + ∫ fn vϵ dx. 2 x
Ω
(3.3.1)
36  3 Calderón–Zygmund theory and the Hardy–Leray potential Using the Hardy and Sobolev inequalities we get 2
2∗ 4(γ − 1) λ γ 22 ](∫ (ϵ + u ) 𝒮 [ dx) − n ΛN,2 γ2 ∗
2
Ω
γ γ 2 1 γ−1 ≤ λ ∫ 2 (un (ϵ + un ) − ϵγ−1 − [(ϵ + un ) 2 − ϵ 2 ] ) dx + ∫ fn vϵ dx. x
Ω
Ω
Now for fixed n and ϵ going to zero we have 2
1
∗ 2∗ m 4(γ − 1) λ γ 22 (γ−1)m 𝒮 [ − . ](∫ u  dx) ≤ ‖f ‖ (∫ u  ) n n m n ΛN,2 γ2
2
Ω
Ω
That is, 2
(
2𝒮 N(m − 1)(N − 2m) − λ]‖un ‖Lm∗∗ ≤ ‖fn ‖Lm . )[ N −2 m2
Thus we have proved the following a priori estimate: (
𝒮2
ΛN,2
)[λ(m) − λ]‖un ‖Lm∗∗ ≤ ‖fn ‖Lm .
(3.3.2)
Now we follow [69] in order to show that the sequence {un } is bounded in W01,m (Ω). Fix ϵ = R in (3.3.1) and use (3.3.2). Then we have ∗
∫ Ω
∇un 2 ≤ C(a, m, N, 𝒮 , R) = M, (R + un )(2−γ) 2 m∗
so that (using the Hölder inequality with exponents m∗
∫ ∇un  Ω
∇un m
∗
≤∫ Ω
(R + un )
(2−γ)m∗ 2
(R + un )
(2−γ)m∗ 2
≤M
and
m∗ 2
2 ) 2−m∗
(∫(R + un )
(2−γ)m∗ 2−m∗
2−m∗ 2
)
.
Ω
Thus we can deduce ∫ ∇un m ≤ M ∗
m∗ 2
m∗∗
(∫(R + un )
Ω
2−m∗ 2
)
,
Ω
which implies the boundedness of the sequence {un } in W01,m (Ω). Therefore, since m > 1, there exists u such that, up to a subsequence, ∗
un ⇀ u, and
weakly in W01,m (Ω) and almost everywhere, un u → 2 strongly in L1 (Ω). 2 x x ∗
Hence we can pass to the limit in (3.1.7) to prove that u is a distributional solution to (3.1.1).
3.3 Existence of infinite energy solutions  37
The above result is optimal as the following result shows. Proposition 3.3.2. Let λ≥
N(m − 1)(N − 2m) , m2
1 1 be such that p1 + m1∗∗ = 1, let g be in Lp and let {gn } be a sequence of smooth functions converging to g in Lp . Consider zn ∈ W01,2 (Ω) : −Δzn = λ
zn
x2
+
+ gn .
1 n
Now let fn = zn m −2 zn ; hence, choosing un as a test function in the equation for zn and choosing zn as a test function in (3.1.7), we obtain from (3.3.3)
∫ fn zn = ∫ gn un ≤ ‖gn ‖p ‖un ‖m∗∗ ≤ c0 ‖g‖p ‖fn ‖m . Ω
Ω
Hence, m
∫ zn  Ω
≤ c0 ‖g‖p (∫ zn 
(m −1)m
1 m
) ,
Ω
which implies ‖zn ‖m ≤ c0 ‖g‖p . But then, up to subsequences, zn converges weakly in Lm to z (also in W01,2 (Ω)), the unique weak solution of
z ∈ W01,2 (Ω) : −Δz = λ
z + g, g ∈ Lp . x2
Due to the choice of p, it turns out that z belongs to Lp , and this contradicts Example 3.2.1. Indeed, choosing λ ≥ λ(p), we have λ > λ(p + ϵ) for every ϵ > 0. Hence, the solution in W01,2 (Ω) given by Example 3.2.1 corresponding to the Lp function g(x) = x1 ν , ∗∗
with ν = N α+
N , belongs exactly to the Marcinkiewicz space M N/α+ , and not to LN/α+ . Since p+ϵ
≤ p∗∗ and λ ≥ λ(p), we thus have a contradiction.
38  3 Calderón–Zygmund theory and the Hardy–Leray potential 3.3.1 Complete blowup for data in L1 If f ∈ L1 , the solution of the approximated problems can go to infinity almost everywhere. We show this phenomenon with the following example. Example 3.3.1. Let Ω = B 1 (0), and let un in W01,2 (Ω) be the solution of 2
u
n −Δun = λ 1/n+x 2 + fn
{
in Ω,
un = 0
on 𝜕Ω,
with fn ∈ L∞ being a sequence of nonnegative functions bounded in L1 . Since the solutions un are nonnegative, we have un ≥ zn , where zn is the unique solution in W01,2 (Ω) of −Δzn = fn
in Ω,
{
zn = 0
on 𝜕Ω.
Therefore, by the maximum principle, un ≥ wn , where wn is the unique solution in W01,2 (Ω) of z
−Δwn = λ xn2 + fn { wn = 0
in Ω, on 𝜕Ω.
Supposing that fn is radially symmetric, so will be un , zn and wn . Passing to radial coordinates, we have the following straightforward calculations: 1/2
1 sN−1 2N−2 ‖f ‖ , zn = f (s) ds − ∫ n N −2 (N − 2)ωN−1 n 1 max(ρ, s)N−2 0
and therefore 1/2
1/2
t N−1 fn(t) sN−3 λ ( dt) ds + C ∫ ∫ N −2 max(ρ, s)N−2 max(s, t)N−2
wn (ρ) =
ρ
≥∫ 0
0
0 ρ
s
0
0
1 1 λ ∫ [∫ t N−1 fn (t) dt] ds. N − 2 ρN−2 s
1 , with 1 xN (− ln(x))α
< α < 2, so that f belongs to L1 , and let fn = Tn (f )
Choose now f (x) = as usual. Since fn increases to f , lim ∫ t
n→+∞
0
N−1
t fn(t) s dt) ds + C ∫( N−2 max(ρ, s) max(s, t)N−2
=C+
s
0
1/2
N−3
N−1
s
fn (t) dt = ∫ t N−1 f (t) dt = 0
1 . (α − 1)(− ln(s))α−1
3.4 A necessary and sufficient condition for solvability  39
Hence, for every x = ρ in Ω, ρ
a 1 1 lim w (ρ) ≥ C + = +∞. ∫ N−2 n→+∞ n (N − 2)(α − 1) ρ s(− ln(s))α−1 0
Thus the limit of un is everywhere +∞, and so no estimate can be obtained if the data are assumed to be just in L1 .
3.4 Necessary and sufficient conditions for solvability in weighted L1 spaces Let us present here a result of existence when the second member belongs to a suitable weighted L1 space. Theorem 3.4.1. Let λ ≤ ΛN,2 and suppose that f ∈ L1 (Ω), f ≥ 0. Call γ = α− (λ) defined in (3.1.3). Then problem (3.1.1) has a positive weak solution u if and only if f satisfies ∫ x−γ f dx < +∞, Br (0)
for some Br (0) ⊂⊂ Ω. Moreover, if u is the unique weak solution to (3.1.1), then ∀k ≥ 0m
Tk (u) ∈ W 1,2 (Ω),
u ∈ W01,q (Ω),
with q
0 { { φ { =0
in Ω,
in Ω, on 𝜕Ω.
(3.4.3)
40  3 Calderón–Zygmund theory and the Hardy–Leray potential Taking φn as a test function in (3.1.1), we get ∫ fφn dx ≤ ∫ u dx = C < ∞. Ω
Ω
Since {fφn }n∈ℕ is an increasing sequence uniformly bounded in L1 (Ω), applying the monotone convergence theorem and the local behavior of φ close to the origin, we obtain C̃ ∫ x−γ f dx ≤ ∫ fφ dx ≤ C. Ω
Br (0)
Sufficient condition: Assume that ∫ x−γ f dx < +∞, Br (0)
for some Br (0) ⊂⊂ Ω; let us consider the sequence of energy solutions un ∈ L∞ (Ω) ∩ W 1,2 (Ω) to the following approximating problems: {
−Δun = λ un (x) = 0
un−1 x2 + n1
+ fn
in Ω,
(3.4.4)
on 𝜕Ω,
where {
−Δu0 = f1 u0 (x) = 0
in Ω, on 𝜕Ω,
fn = Tn (f ) and u0 ≤ u1 ≤ un−1 ≤ un in ℝN . Since fn ≥ 0, un (x) ≥ 0 in Ω. Take φ ∈ W01,2 (Ω), the positive energy solution to problem (3.4.3), as a test function in (3.4.4). It follows that ∫ un dx ≤ ∫ fφ dx ≤ C̃ ∫ f x−γ ≤ C. Ω
Ω
Ω
Hence, since the sequence {un }n∈ℕ is increasing, we can define u := limn→∞ un , and conclude that u ∈ L1 (Ω). We claim that xu 2 ∈ L1 (Ω). Indeed, let ψ be the unique bounded positive solution to the problem {
−Δψ = 1 ψ=0
in Ω, in 𝜕Ω.
Then ψ ≥ C in Br (0). By using ψ as a test function in (3.4.4), ∫ Ω
un−1
x2 +
1 n
dx ≤
ψun−1 1 dx + C(r) ∫ C x2 + n1 Br (0)
∫ Ω\Br (0)
un dx ≤ C,
3.5 Further results  41
and thus λ
un−1
x2
+
1 n
+ fn ↗ λ
u +f x2
strongly in L1 (Ω).
Testing with Tk (un ) in (3.4.4) and considering the previous estimates, we easily get u that Tk (un ) ⇀ Tk (u) weakly in W01,2 (Ω). Since the sequence {λ 2n−1 1 + fn }n∈ℕ converges x + n
strongly in L1 (Ω), by the classical results of [291], we reach u ∈ Lσ (Ω) for all σ < N and u ∈ W01,q (Ω) for all q < N−1 .
N N−2
3.4.1 Uniqueness In this section, we discuss the uniqueness of the solutions obtained as limit of solutions of the regular problems (3.1.7) (see [72]). Theorem 3.4.2. Under the assumptions of Theorem 3.2.1 or Theorem 3.3.1, the solution obtained as the limit of solutions of the regular problems (3.1.7) is unique. Proof. Consider the boundary value problems (3.1.7) and zn zn ∈ W01,2 (Ω) : −Δzn = λ + gn , x2 + n1
(3.4.5)
where {gn } is any sequence converging to f in Lm . Therefore, since m > 1, there exists z such that, up to a subsequence, zn converges weakly to z and z is a distributional solution to (3.1.1). Then (un − zn ) ∈ W01,2 (Ω) : −Δ(un − zn ) = λ
(un − zn ) x2 +
1 n
+ fn − gn .
Here, we can repeat the calculation in previous theorems in order to obtain (
𝒮2
ΛN,2
)[λ(m) − λ]‖un − zn ‖Lm∗∗ ≤ ‖fn − gn ‖Lm ,
(3.4.6)
which implies that ‖u − z‖Lm∗∗ = 0, and so the uniqueness follows.
3.5 Further results 3.5.1 A different way to obtain the critical value for λ Consider our problem: −Δu = λ xu 2 + g
{
u=0
in Ω, on 𝜕Ω,
(3.5.1)
42  3 Calderón–Zygmund theory and the Hardy–Leray potential with λ < ΛN,2 . Define v as the ground state transformation of u, that is, v(x) = xγ u(x),
where γ =
N −2 − √ΛN,2 − λ > 0. 2
Then v solves problem − div(x−2γ ∇v) = x−γ g
{
in Ω,
v=0
on 𝜕Ω.
We set f ≡ x−γ g. Note that by definition, 0 0.
1,2 Definition 3.5.3. Consider f ∈ L1 (Ω). We say that u ∈ 𝒯0,γ (Ω) is an entropy solution of
− div(x−2γ ∇v) = f { v=0
in Ω, on 𝜕Ω,
(3.5.3)
1,2 if for every k > 0 and ϕ ∈ 𝒟0,γ (Ω) ∩ L∞ (Ω) we have
∫ x−2γ ⟨∇v, ∇(Tk (v − ϕ))⟩ = ∫ fTk (v − ϕ). Ω
Ω
(3.5.4)
3.5 Further results  43
In the linear case, by uniqueness, the concept of entropy solution is equivalent to that of distributional solution and to the solution obtained by approximations. We refer to [68], [56], [132] and the references therein to justify the use of this concept of solution in a large class of problems. Theorem 3.5.4. Assume f ∈ L1 (Ω); then the problem − div(x−2γ ∇v) = f
{
in Ω,
v=0
(3.5.5)
on 𝜕Ω
has a unique entropy solution. If f ≥ 0, then v ≥ 0. This is a particular case of Theorem 3.4.1, formulated for the ground state transformed problem. It is worth pointing out that in the original problem the result above can be translated as follows. First at all, using the Hölder inequality, we check when f = gx−γ ∈ L1 (Ω): m
1 m
m −γ m−1
∫ f  dx ≤ (∫ g dx) (∫ x Ω
Ω
dx)
m−1 m
.
Ω
m Thus, we need γ m−1 < N. Since γ = N−2 −√ΛN,2 − λ, after some elementary calculations 2 we obtain again the condition λ < λ(m).
3.5.2 A remark in the case λ = ΛN,2 In this subsection we consider the problem in the critical value case λ = ΛN,2 , i. e., −Δu − ΛN,2 xu 2 = g { u=0
in Ω, on 𝜕Ω,
(3.5.6)
where g is a nonnegative function in L1 (Ω). According to Theorem 3.4.1, problem (3.5.6) N−2 is solvable in the weak sense if and only if x− 2 g ∈ L1 (Ω). There is a different approach to the critical problem (3.5.6) under consideration. As a consequence of the improved Hardy inequality proved in Section 2.5 we have defined in Section 2.5.1 a space larger than W01,2 (Ω), denoted ℋ(Ω). In general, if f ∈ ℋ , the dual space, by the Lax–Milgram theorem the problem −Δu −
ΛN,2 = f, x2
u = 0 on 𝜕Ω,
has a unique solution u ∈ ℋ(Ω). Therefore, in particular, for g ∈ Lm (Ω) with m > we can find a solution u ∈ ℋ(Ω).
2N N+2
4 Effect of the Hardy–Leray potential in the solvability of semilinear elliptic equations 4.1 Introduction In this chapter we consider equations of the type −Δu =
λu + up , x2
u ≥ 0,
and we try to understand the conditions relating p and λ for the local solvability of such equations. As in Chapter 3 about the optimal summability, the optimal power for the solvability of the equation depends explicitly on the spectral parameter λ. We will follow the original papers by Brezis, Dupaigne and Tesei, [89] and Dupaigne [145]. The organization of the chapter is as follows. In Section 4.2 we obtain the candidate to be the optimal power. We use a radial argumentation and look for local solutions in the whole ℝN . Section 4.3 will be devoted to collect some known results about elliptic equations and to introduce the Newtonian capacity. In Section 4.4 we discuss the existence problem. Finally in Section 4.5 we prove some nonexistence results that, as a byproduct, imply the optimality of the critical power obtained in Section 4.2.
4.2 The optimal power Consider the equation − Δu = λ
u + up , x2
u ≥ 0.
(4.2.1)
We try to calculate radial solutions, i. e., positive solutions to the ordinary differential equation − u −
(N − 1) u u − λ 2 = up , r r
(4.2.2)
where 0 < λ ≤ ΛN,2 . For homogeneity reasons, we look for a solution of the form ϕ(r) = Ar −β , where A, β > 0. We obtain the algebraic identity Ar −(β+2) (−β(β + 1) + (N − 1)β − λ) = Ap r −pβ , which we solve by taking (β + 2) = pβ https://doi.org/10.1515/9783110606270004
⇔
β=
2 . (p − 1)
(4.2.3)
46  4 The Hardy Potential in Semilinear Elliptic Equations Since A > 0, we need β2 − (N − 2)β + λ < 0. This is equivalent to α− (λ) < β < α+ (λ),
(4.2.4)
where α± (λ) were defined in (3.1.3). Therefore, since (4.2.3) holds, p− (λ) = 1 +
2 2 ≤ p < p+ (λ) = 1 + . α− (λ) α+ (λ)
(4.2.5)
Hence, we find a radial solution to equation (4.2.1) for fixed λ ∈ (0, ΛN,2 ) if p verifies (4.2.5). Some remarks about these critical exponents are in order: p+ (λ) → { p− (λ) → {
2∗ − 1 = ∞ 2 −1= ∗
N N−2
N+2 N−2 N+2 N−2
when λ → λN,2 , when λ → 0, when λ → λN,2 , when λ → 0.
Moreover p+ is decreasing and p− is increasing. In particular, p− < 2 ∗ −1 = 1+
N +2 < p+ N −2
if 0 < λ < ΛN,2 ,
(4.2.6)
2 2 = p− (λ) < p < p+ (λ) = 1 + . α+ (λ) α− (λ)
(4.2.7)
6
p+ (λ)
λ
p− (λ)
ΛN,2

Note the continuity with respect to the semilinear case with λ = 0 and the new feature by the effect of the Hardy–Leray potential, that is, for λ ∈ (0, ΛN,2 ) we have a bounded interval of p for which we find radial solutions. We analyze the local summability of ϕ(r). Since β + 2 < α+ + 2 < N, we have:
4.3 Some previous results for linear elliptic equations  47
–
– –
ϕ(x) ∈ L1loc (ℝN ); ϕ(x) x2
and
∈ L1loc (ℝN )
ϕp (x) ∈ L1loc (ℝN ).
In this chapter we will prove the following important result. The exponent p+ (λ) represents the threshold for weak solvability of the Dirichlet problem for equation (4.2.1) in the case of bounded domains Ω ⊂ ℝN containing the origin.
4.3 Some previous results for linear elliptic equations Before we begin to study the above statement, let us recall some basic result of potential theory and linear partial differential equations. For details we refer to [149, Sections 4.7 and 4.8] and [291], respectively. We will also use the truncations given in Definition 3.1.5. Recall that if 1,2
N
2∗
N
2
N
N
𝒟 (ℝ ) = {f ∈ L (ℝ )  ∇f ∈ [L (ℝ )] },
the Sobolev inequality gives the continuous inclusion 1,2
N
2∗
N
𝒟 (ℝ ) ⊂ L (ℝ ).
More specifically we have the functional inequality S‖f ‖L2∗ (ℝN ) ≤ ‖∇f ‖L2 (ℝN ) . It is not too hard to prove that W 1,2 (ℝN ) is dense in 𝒟1,2 (ℝN ) with the metric of L2 (ℝN ). ∗
4.3.1 Elliptic equations with measure data To complete the landscape of existence results for linear elliptic equations, we also recall the classical result by G. Stampacchia in Section 9 of his fundamental paper [291]. We will denote by ℳ(Ω) the Radon measures in Ω with finite total variation. For μ ∈ ℳ(Ω), μω denotes the total variation of μ. A detailed presentation of the main properties of the Radon measures can be found in [149]. In particular we will use the fact that for all μ ∈ ℳ(Ω) there exists a sequence fn ∈ 𝒞0∞ (Ω) such that fn ⇀ μ weak∗ in ℳ(Ω). We next proceed to estate the classical result by Stampacchia. Theorem 4.3.1. Let A(x) be an N × N symmetric matrix with measurable bounded coefficients in a bounded domain Ω ⊂ ℝN and such that there exists α > 0 verifying ⟨A(x)ξ , ξ ⟩ ≥ αξ 2 ,
for all x ∈ Ω and for all ξ ∈ ℝN .
48  4 The Hardy Potential in Semilinear Elliptic Equations N If μ ∈ ℳ(Ω), then there exists a unique solution u ∈ W 1,q (Ω), q ∈ (1, N−1 ), to the equation
− div(A(x)∇u(x)) = f (x),
x ∈ Ω.
Moreover, there exist C = C(N, q, α) such that ‖u‖W 1,q (Ω) ≤ CμΩ , 0
and for all k > 0, Tk (u) ∈ W01,2 (Ω). The original proof by Stampacchia is a consequence of a duality argument taking advantage of the linearity of the problem. For an alternative proof the reader can see Theorem 11.6 in [64]. The results by Stampacchia are related to the wellknown results by De Giorgi solving the Hilbert problem (see [137]). As a consequence of Theorem 4.3.1 we also obtain in particular the following result. Corollary 4.3.2. If vn ∈ W 1,q (Ω) is a solution to − div(A(x)∇vn (x)) = fn (x), x ∈ Ω, and fn → f in L1 (Ω), then vn → v in L1 (Ω) and v ∈ W01,q (Ω) is the solution to − div(A(x)∇v(x)) = f (x),
x ∈ Ω.
A final remark is that the existence is also true for a larger class of nonlinear elliptic operators; however, the uniqueness for measure data is a different question. See for instance [132] and the references therein. Another ingredient to prove nonexistence results comes from the following Kato inequality. See [209] for the proof in a general setting. Lemma 4.3.3. Let Ω be a bounded domain in ℝN . Assume that u ∈ L1loc (Ω) and Δu ∈ L1loc (Ω). Then Δu+ is a Radon measure and Δu+ ≥ χ{u≥0} Δu
in 𝒟 (Ω).
This inequality has been extended to the case in which Δu is a Radon measure. See Theorem 1.1 in [96] 4.3.2 Newtonian capacity The capacity that we introduce next is related to the Newtonian potential, x2−N , if N ≥ 3. The significance of the concept of capacity from this point of view is contained in the fact that in many circumstances sets of zero capacity and only these can appear as sets of singularities for certain classes of functions or as nullsets that can be ignored. Let us motivate, at least in an informal way, the concept of Newtonian capacity in dimension N = 3. We refer here to the survey by I. Markina [237].
4.3 Some previous results for linear elliptic equations  49
Let D ⊂ ℝ3 be a bounded domain representing a conducting body. Assume that D ⊂ Br0 (0) and consider for R > r0 SR (0) the spherical surface of radius R surrounding D and connected with earth. If the body D is charged with an electrical charge, q, this charge will be distributed over D and will induce a charge on SR (0) just to reach the ⃗ equilibrium. The charged body defines an electric field, F(x), in ℝ3 , that is a potential ⃗ vanishes in ⃗ = −∇ρ(x). The vector field F(x) vector field with potential ρ(x), that is, F(x) D and there exists a distribution of charge in between the surfaces 𝜕D and SR (0) that we suppose have continuous densities, ϕ1 on 𝜕D and ϕ2 on SR (0). Then the potential ρR (x) = ∫ 𝜕D
ϕ1 (y) ϕ2 (y) dσ(y) + ∫ dσ(y), x − y x − y SR (0)
is a continuous function in ℝ3 . ⃗ Since SR (0) is connected to earth we have ρR = 0 on SR (0). Moreover as 0 = F(x) = −∇ρ(x) for x ∈ D, ρR is constant in D and therefore by continuity it is constant in 𝜕D. Call such constant V. The value of ρR (x) is called the difference of potential between 𝜕D and SR (0). It is known that the value of the constant V is proportional to the charge q and the proportionality factor is called the capacity C of the condenser formed by 𝜕D and SR (0), i. e., C=
q . V
Fixing the domain D and R > r0 we can summarize the properties of ρR as follows. Call OR the open set limited by 𝜕D and the spherical surface SR (0). Then ρR verifies ΔρR (x) = 0 { { ρ (x) = V { { R { ρR (x) = 0
x ∈ OR ,
x ∈ 𝜕D,
(4.3.1)
x ∈ SR (0).
By classical elliptic theory, there exists ρ∞ such that limR→∞ ρR (x) = ρ∞ (x) and that Δρ∞ (x) = 0 { { ρ (x) = V { { ∞ { limx→∞ ρ∞ (x) = 0.
x ∈ ℝ3 \ D,
x ∈ 𝜕D,
(4.3.2)
It is natural to expect that for a suitable measure ϕ∞ , ρ∞ (x) = ∫ 𝜕D
dϕ∞ (y) , x − y
with ∫ dϕ∞ (y) = q. 𝜕D
The measure ϕ∞ (y) is called the equilibrium distribution of the charge q and the function ρ∞ the equilibrium potential. Therefore for every charge q distributed on D there is a corresponding equilibrium potential ρ∞ (x) which is ρ∞ (x) = V∞ for all x ∈ D.
50  4 The Hardy Potential in Semilinear Elliptic Equations The constant C∞ = Vq is called the capacity of the body D. Then we can say that the ∞ capacity of a body D is the value of the charge q on D that produces the unit difference of potential between 𝜕D and ∞ under the equilibrium distribution of charge. All the previous considerations can be extended to every N > 2 by considering the Riesz potential x − y2−N , which for N = 3 is x − y−1 . Hereafter we will consider a general dimension N > 2. We will look for a variational definition of the capacity that will give a more useful way to handle the concept. Let ℳ+ (ℝN ) be the set of positive finite Radon measures in ℝN . For μ ∈ ℳ+ (ℝN ) we recall that the Newtonian of μ is given by uμ (x) = ∫
dμ(y) x − yN−2
and
− Δuμ (x) = cN μ(x).
(4.3.3)
The energy of μ is given by ℰ (μ) = ∬
dμ(x) dμ(y) . x − yN−2
The distribution of equilibrium of a charge q distributed on a body D is the measure ν which minimizes the energy, that is, ℰ (ν) ≤ ℰ (μ),
∀μ ∈ ℳ+ (ℝN ) such that ∫ dμ(y) = q.
Note that we can write the energy in terms of the Newtonian potential, ℰ (μ) = ∫ uμ (x) dμ(x),
where uμ (x) = ∫
dμ(y) . x − yN−2
Normalizing by putting C(D) = ν(D) we have uμ (x) = 1
if x ∈ D
and ℰ (ν) = ∫ dν(x) = ν(D) = C(D). D
Therefore, we define the capacity of a compact set D as follows: C(D) = inf{cN ℰ (μ)  μ ∈ ℳ+ (ℝN ), uμ (x) > 1 in D}. Observe that since −Δuμ = cN μ in D and −Δuμ = 0 in ℝN \ D, integrating by parts we get 2 ∫ ∇uμ (x) dx = ∫ uμ (x)(−Δuμ (x)) dx = cN ∫ uμ (x) dμ(x) = cN ℰ (μ) = C(D).
ℝN
ℝN
Let us give a precise definition of capacity.
ℝN
4.3 Some previous results for linear elliptic equations  51
Definition 4.3.4. If A ⊂ ℝN set cap1,2 (A) = inf{ ∫ ∇f 2 dx  f ∈ 𝒟1,2 (ℝN ), A ⊂ Int {f ≥ 1}}.
(4.3.4)
ℝN
We call cap1,2 (A) the Newtonian capacity of A. Using regularization we find that cap1,2 (K) = inf{ ∫ ∇f 2 dx  f ∈ 𝒞0∞ (ℝN ), f ≥ χK } ℝN
for each compact K ⊂ ℝN . It is quite standard to prove the following statement. Lemma 4.3.5. cap1,2 is an outer measure on ℝN . We have, moreover, the following result. Lemma 4.3.6. Call ℋr (A) the rHausdorff measure of A ⊂ ℝN . 1. If ℋN−2 (A) < ∞, with N > 2, then cap1,2 (A) = 0. 2. If cap1,2 (A) = 0, then ℋr (A) = 0 for all r > N − 2. 3. If f ∈ W 1,2 (ℝN ), then f admits a cap1,2 quasicontinuous representative. The proof of this results and other properties of the capacity can be seen in the beautiful handbooks [149] and [236]. Application: eliminable singularities We will also use the following results about eliminable singularities and comparison (see [89] and the references therein). Lemma 4.3.7. Let E ⊂⊂ Ω be a closed set such that cap1,2 (E) = 0 and assume that u, f ∈ L1 (Ω \ E) are two nonnegative functions such that −Δu ≥ f
in 𝒟 (Ω \ E).
Then u, f ∈ L1 (Ω) and in 𝒟 (Ω).
−Δu ≥ f
Furthermore if Ω1 ⊂⊂ Ω is a smooth domain and v ∈ L1 (Ω1 ) is the solution of {
−Δv = f v=0
in Ω1 on 𝜕Ω1 ,
then u≥v
a. e. in Ω1
(4.3.5)
52  4 The Hardy Potential in Semilinear Elliptic Equations Proof. Consider uk = Tk (u). By using Kato’s inequality in Lemma 4.3.3 we obtain − Δuk ≥ fχ{u f (xε )
x∈ℳ
and f (y) ≥ f (xε ) − εd(xε , y). We can read geometrically these conditions by saying that for all ε > 0 the graph of f is above the inverted cone of amplitude ϵ with vertex at (xϵ , f (xϵ )). See the original paper [148]; we follow here the proof in [238]. Theorem 4.3.9. Let ℳ be a complete metric space and let ϕ : ℳ → (−∞, ∞] be a proper function such that: (i) ϕ(y) ≥ β; (ii) ϕ is lower semicontinuous. Given ε > 0 and u ∈ ℳ such that ϕ(u) ≤ inf ϕ + ε, ℳ
there exists v ∈ ℳ such that: 1. ϕ(u) ≥ ϕ(v); 2. d(u, v) ≤ 1; 3. if v ≠ w ∈ ℳ, then ϕ(w) ≥ ϕ(v) − εd(v, w).
4.3 Some previous results for linear elliptic equations  55
Proof. Fixing ε > 0, we define the following order relation on ℳ: we say w ≤ v,
if and only if ϕ(w) + εd(w, v) ≤ ϕ(v).
Consider u0 = u and by recurrence define the sequence {un } as follows: for n ∈ ℕ we take Sn = {w ∈ ℳ : w ≤ un }. Choosing un+1 ∈ Sn such that ϕ(un+1 ) ≤ inf{ϕ} + Sn
1 , n+1
we get un+1 ≤ un and Sn+1 ⊂ Sn . The lower semicontinuity of ϕ implies that Sn is a closed set. Now, if w ∈ Sn+1 , we have w ≤ un+1 ≤ un and then, εd(w, un+1 ) ≤ ϕ(un+1 ) − ϕ(w) ≤ inf{ϕ} + Sn
1 1 − inf{ϕ} = . n + 1 Sn n+1
2 If we set diameter(Sn+1 ) = δn+1 , then δn+1 ≤ ε(n+1) and so, limn→∞ δn+1 = 0. The com∞ pleteness of ℳ implies ⋂n=1 Sn = {v} for some v ∈ ℳ. But, in particular, v ∈ S0 ; hence v ≤ u0 = u, i. e., ϕ(v) ≤ ϕ(u) + εd(u, v) ≤ ϕ(u) and
d(u, v) ≤
ϕ(u) − ϕ(v) ≤ ε−1 (inf{ϕ} + ε − inf{ϕ}) = 1. ℳ ℳ ε
Then, d(u, v) ≤ 1. To get (3), suppose that w ≤ v. Then for all n ∈ ℕ, w ≤ un , i. e., ∞
w ∈ ⋂ Sn , n=1
and then w = v. So we conclude that if w ≠ v, then ϕ(w) ≥ ϕ(v) − εd(v, w). The results that we explain below show how to use the variational principle to find critical points of functionals. This will be used to prove the mountain pass theorem. Corollary 4.3.10. Let 𝒳 be a Banach space, and let φ : 𝒳 → ℝ be a differentiable lower bounded function in 𝒳 . Then for all ε > 0 and for all u ∈ 𝒳 such that φ(u) ≤ inf φ + ε, 𝒳
there exists v ∈ 𝒳 verifying φ(v) ≤ φ(u),
‖u − v‖𝒳 ≤ ε1/2 , 1/2 φ (v)𝒳 ≤ ε .
56  4 The Hardy Potential in Semilinear Elliptic Equations Proof. In Theorem 4.3.9 take ℳ =𝒳 , ϕ = φ, ε > 0, λ = u we obtain v ∈ 𝒳 such that
1 , ε1/2
d(⋅, ⋅) = ‖ ⋅ − ⋅ ‖𝒳 . For each
φ(v) ≤ φ(u),
‖u − v‖𝒳 ≤ ε1/2 ,
and for all w ≠ v φ(w) > φ(v) − ε1/2 (v − w). Taking in particular w = v + th with t > 0 and h ∈ 𝒳 , ‖h‖ = 1. Then φ(v + th) − φ(v) > −ε1/2 t. This implies −ε1/2 ≤ ⟨φ (v), h⟩ for all h ∈ 𝒳 , ‖h‖ = 1, and hence 1/2 φ (v)𝒳 ≤ ε . Corollary 4.3.11. If 𝒳 and φ are as in Corollary 4.3.10, then for all minimizing sequences of φ, {uk } ⊂ 𝒳 , there exists a minimizing sequence {vk } ⊂ 𝒳 such that φ(vk ) ≤ φ(uk ),
‖uk − vk ‖𝒳 → 0, φ (vk )𝒳 → 0,
k → ∞,
k → ∞.
Proof. If φ(uk ) → c = inf𝒳 φ, define εk = φ(uk ) − c if this quantity is positive and εk = k1 if φ(uk ) = c. Then for εk we take the vk that comes from Corollary 4.3.10. In the spirit of the previous corollary, we find a notion of almost minimum for φ. The problem now is to get almost critical points of different types, for instance mountain pass type critical points in the sense of [37]. Here we will describe how the variational principle of Ekeland separates the geometric aspect of the problem of finding critical points of a functional from the analytical one. Clearly, what we will need is some type of compactness property to complete the search of critical points. The mountain pass theorem We start with some hypotheses and notation: H1 Consider 𝒳 , a Banach space, 𝒦 ⊂ ℝN a compact metric space and 𝒦0 ⊂ 𝒦 a closed set. H2 Let χ : 𝒦0 → 𝒳 be a continuous function.
4.3 Some previous results for linear elliptic equations  57
H3 Define ℳ = {g ∈ 𝒞 (𝒦, 𝒳 ) : g(s) = χ(s), s ∈ 𝒦0 }.
We recall the definition of subdifferential in the case of the norm in 𝒳 , 𝜕‖x0 ‖ = {p ∈ 𝒳 : ‖x0 ‖ − ‖x‖ ≤ p(x0 − x), ∀x ∈ 𝒳 }, where 𝒳 denotes the dual space of 𝒳 . We will use the following abstract result. Lemma 4.3.12. Let 𝒳 be a Banach space. Then for the norm ‖⋅‖:𝒳 →ℝ we have 𝜕‖x0 ‖ = {p ∈ 𝒳 : p(x0 ) = ‖x0 ‖, ‖p‖𝒳 = 1}. Proof. Given x0 ∈ 𝒳 by the Hahn–Banach theorem there exists p ∈ 𝒳 such that p(x0 ) = ‖x0 ‖ and ‖p‖𝒳 = 1. For such p we have p(x − x0 ) = p(x) − p(x0 ) ≤ ‖p‖𝒳 ‖x‖ − ‖x0 ‖ ≤ ‖x‖ − ‖x0 ‖, and hence ‖x0 ‖ − ‖x‖ ≤ p(x0 − x) and, therefore, p ∈ 𝜕‖x0 ‖. Reciprocally, if p ∈ 𝜕‖x0 ‖, we get ‖x0 ‖ − ‖x‖ ≤ p(x0 − x) ∀x ∈ 𝒳 . In particular for x = λx0 , λ > 0, we find ‖x0 ‖(1 − λ) ≤ (1 − λ)p(x0 ), so (p(x0 ) − ‖x0 ‖)(1 − λ) ≥ 0 for all λ ∈ ℝ+ . Then p(x0 ) = ‖x0 ‖. Moreover ‖x0 ‖ − ‖x‖ ≤ p(x0 ) − p(x) ≤ ‖x0 ‖ − p(x) implies p(x) < ‖x‖ and then ‖p‖𝒳 ≤ 1 since p(x0 ) = ‖x0 ‖ and ‖p‖𝒳 = 1. Corollary 4.3.13. Let 𝒞 (𝒦, ℝ) with the supremum norm. Denote by ℳ(𝒦) the Radon measures on 𝒦. Then 𝜕‖f ‖∞ = {μ ∈ ℳ(𝒦) : μ ≥ 0, ∫ dμ = 1, supp(μ) ⊂ {t : f (t) = ‖f ‖∞ }}. 𝒦
Proof. It is clear that the Radon measures verifying the conditions above are the subdifferential of the norm. More precisely, by Lemma 4.3.12 we have: (i) If the conditions are satisfied, then ‖μ‖ = 1 and ∫ f (x) dμ = ‖f ‖∞ ∫ dμ = ‖f ‖∞ . 𝒦
supp μ
58  4 The Hardy Potential in Semilinear Elliptic Equations (ii) If ‖μ‖ = 1 and ∫𝒦 f (x) dμ = ‖f ‖∞ , then μ > 0 and supp(μ) ⊂ {s  f (s) = ‖f ‖∞ }. Theorem 4.3.14. Consider 𝒳 , 𝒦, 𝒦0 and ℳ verifying H1H2H3. Let u : 𝒳 → ℝ be a functional such that: (1) u is continuous; (2) u is Gateaux differentiable and, moreover, u : 𝒳 → 𝒳 is continuous from the strong to weak∗ topologies. Consider {
α = infφ∈ℳ maxs∈𝒦 u(φ(s)), α1 = maxχ(𝒦0 ) u,
and assume that α > α1 . Then for all ε > 0 and for all φ ∈ ℳ such that maxs∈𝒦 u(φ(s)) ≤ α + ε there exists vε ∈ 𝒳 such that α − ε ≤ u(vε ) ≤ max u(φ(s)), 1/2
s∈𝒦
d(vε , φ(𝒦)) ≤ ε , 1/2 u (vε ) ≤ ε . Proof. We can assume that 0 < ε < α − α1 . Let φ ∈ ℳ verifying max u(φ(s)) ≤ α + ε. s∈𝒦
Define I : ℳ → ℝ by I(c) = max u(c(s)) s∈𝒦
for c ∈ ℳ. By hypothesis α = infc∈ℳ I(c) > α1 . Moreover I is lower semicontinuous, because if ‖ck − c‖∞ → 0 we have I(c) = u(c(s0 )) = lim u(ck (s0 )) ≤ lim I(ck ). k→∞
k→∞
By the Ekeland variational principle applied to I : ℳ → ℝ, given ε > 0 and φ ∈ ℳ such that I(φ) ≤ α + ε there exists cε ∈ ℳ so that: (i) I(cε ) ≤ I(φ) ≤ α + ε; (ii) I(c) ≥ I(cε ) − ε1/2 ‖c − cε ‖∞ if c ∈ ℳ; (iii) ‖cε − φ‖∞ ≤ ε1/2 .
4.3 Some previous results for linear elliptic equations  59
Consider γ : 𝒦 → 𝒳 continuous such that γ(𝒦0 ) = 0. For h ≠ 0, h ∈ ℝ from (ii) we conclude that I(cε + γh) − I(cε ) ≥ −ε1/2 h‖γ‖∞ . Thus, 1 I(c + γh) − I(cε ) ≥ −ε1/2 ‖γ‖∞ . h ε By the definition of I, I(cε + γh) − I(cε ) = max u(cε + γh) − max u(cε )
= max{u(cε (s)) + h⟨u (cε (s)), γ(s)⟩ + o(h)} − max u(cε (s)). s∈𝒦
s∈𝒦
To simplify the notation we write f (s) = u(cε (s)) and g(s) = ⟨u (cε (s)), γ(s)⟩. We have by hypothesis f ∈ 𝒞 (𝒦, ℝ); also g ∈ 𝒞 (𝒦, ℝ) because: (a) by the Banach–Alaoglu theorem 𝜕‖f ‖∞ is weak∗ compact; (b) by hypothesis if vn → v in 𝒳 , then ⟨u (vn ), y⟩ → ⟨u (v), y⟩ for all y ∈ 𝒳 ; (c) if u (vn ) ⇀ u (v) is weak∗ in 𝒳 and yn → y in 𝒳 , then ⟨u (vn ), yn ⟩ → ⟨u (v), y⟩,
as n → ∞.
Then the inequalities above can be written as 1 {I(cε + hγ) − I(cε )} h 1 = lim {(f + hg)∞ − ‖f ‖∞ } h→0 h
−ε1/2 ‖γ‖∞ ≤ lim
h→0
1
≤ sup{∫ gdμ : μ ∈ 𝜕‖f ‖∞ }. 0
Define F : 𝜕‖f ‖∞ → ℝ by F(μ) = ∫K g dμ and consider a weak∗ convergent sequence μn ⇀ μ in 𝜕‖f ‖∞ . Then F(μn ) = ∫ g dμn → ∫ g dμ = F(μ), 𝒦
𝒦
so the supremum is attained. As a consequence we have −e1/2 ‖γ‖∞ ≤ max{∫{u (cε (s)), γ(s)} dμ : μ ∈ 𝜕‖f ‖∞ }. 𝒦
60  4 The Hardy Potential in Semilinear Elliptic Equations Consider Γ = {γ ∈ 𝒞 (𝒦, 𝒳 ) : γ(s) = 0 if s ∈ 𝒦0 , ‖γ‖∞ ≤ 1}. Then, −ε1/2 ≤ −ε1/2 ‖γ‖∞ ≤ inf{max{∫{u (cε (s)), γ(s)} dμ : μ ∈ 𝜕‖f ‖∞ }}. γ∈Γ
𝒦
Take {μn (γ)} ⊂ 𝜕‖f ‖∞ such that for fixed γ, ∫{u (cε (s)), γ(s)} dμn (γ) → M(γ), n → ∞, 𝒦
where M(γ) = max {∫{u (cε (s)), γ(s)} dμ}. u∈𝜕‖f ‖∞
𝒦
Then by choosing a convenient weak∗ convergent subsequence μn (γ) ⇀ μ(γ) we have ∫{u (cε (s)), γ(s)} dμ(γ) = M(γ). 𝒦
Then by Lemma 4.3.13, −ε1/2 ≤ inf M(γ) = inf ∫{u (cε (s)), γ(s)} dμ(γ) γ∈Γ
γ∈Γ
𝒦
≤ max inf ∫{u (cε (s)), γ(s)} dμ = max (− ∫u (cε (s))𝒳 dμ) μ∈𝜕‖f ‖ μ∈𝜕‖f ‖ γ∈Γ ∞
∞
𝒦
𝒦
= − min{u (cε (s))𝒳 : s ∈ {t ∈ 𝒦 : u(cε (t)) = ‖f ‖∞ }}. Hence, min
1/2 u (cε (s))𝒳 ≤ ε .
s∈{t∈𝒦:u(cε (t))=‖f ‖∞ }
In other words, there exists sε ∈ 𝒦 such that if vε = cε (sε ), then: (i) u(vε ) = maxs∈𝒦 u(cε (s)) = I(cε ) ≤ inf I + ε1/2 ; (ii) ‖u (vε )‖ ≤ ε1/2 ; (iii) u(vε ) ≥ α1 ;
(iv) d(vε , φ(𝒦)) ≤ ε1/2 . The classical condition of compactness is the condition of Palais–Smale that we formalize in the following definition. Definition 4.3.15. Let 𝒳 be a Banach space and U : 𝒳 → ℝ a Gateaux differentiable functional; U verifies the Palais–Smale condition to the level c ∈ ℝ if and only if for all sequences {xk } ⊂ 𝒳 verifying
4.3 Some previous results for linear elliptic equations  61
(i) U(xk ) → c as k → ∞;
(ii) U (xk ) → 0 in 𝒳 for k → ∞, there exists a subsequence xkj → x in 𝒳 .
A sequence verifying (i) and (ii) is called a Palais–Smale sequence for U. With these notions we can formulate the mountain pass theorem.
Theorem 4.3.16. Let 𝒳 be a Banach space and U : 𝒳 → ℝ a continuous functional verifying: (i) U is Gateaux differentiable; (ii) U is continuous from 𝒳 to 𝒳 with the weak∗ topology. Assume that there exist u0 , u1 ∈ 𝒳 and a ball Br , centered in u0 and with radius r such that if u1 ∈ 𝒳 − Br and x − u0  = r, then inf U(x) > max{U(u0 ), U(u1 )}. 𝜕Br
Consider Γ = {g ∈ 𝒞 ([0, 1], 𝒳 ) : g(0) = u0 , g(1) = u1 } and c = inf sup U(f (s)). f ∈Γ s∈[0,1]
If U satisfies the Palais–Smale condition at level c, then c is a critical value for U and c > max{U(u0 ), U(u1 )}. Proof. Take 𝒦 = [0, 1], 𝒦0 = {0, 1}, χ(0) = u0 , χ(1) = u1 and ℳ = Γ; since c0 = inf𝜕Br U(x) > max{U(u0 ), U(u1 )} = c1 and because all paths g ∈ Γ have nonempty intersections with 𝜕Br , Theorem 4.3.14 implies that for a given sequence {εk } with εk → 0 we can find a sequence {xk } ⊂ 𝒳 such that (a) U(xk ) → c; (b) U (xk ) → 0.
By the Palais–Smale condition for level c, there exists a subsequence such that c = lim U(xk ) = U(x), k→∞
0 = lim U (xk ) = U (x). k→∞
This says that x is a critical point at level c.
62  4 The Hardy Potential in Semilinear Elliptic Equations
4.4 Results on existence Consider the equation λu + up + f (x), x2
− Δu =
u ≥ 0 in Ω ⊂ ℝN .
(4.4.1)
We will assume f ∈ L1loc (Ω) and f ≥ 0 in Ω. Let us consider the notion of solution that will be used here. Definition 4.4.1. We say that u ∈ L1loc (Ω), u ≥ 0, is a very weak supersolution (subso
lution) to equation (4.4.1) if we have
u x2
∈ L1loc (Ω), u ∈ Lploc (Ω) and ∀ϕ ∈ C0∞ (Ω) such that ϕ ≥ 0,
∫(−Δϕ)u dx ≥ (≤) ∫(up + λ Ω
Ω
u + f )ϕ dx. x2
If u is a very weak super and subsolution, then we say that u is a very weak solution. The idea in the definition is to give sense to the equation in a distributional sense. According to (4.2.6) the results on existence will be analyzed in separated cases. Case 1. 0 ≤ λ < ΛN,2 , p
α=
inf
u∈H01 (Ω),u=0 ̸
Jλ (u).
By, for instance, Corollary 4.3.11 we find a bounded Palais–Smale sequence {uk }k∈ℕ , i. e., Jλ (uk ) → α
as k → ∞ and Jλ (uk ) → 0 in W −1,2 (Ω).
Therefore there exists a relabeled subsequence {uk }k∈ℕ such that:
4.4 Results on existence
1. 2. 3.
uk ⇀ u weakly in W01,2 (Ω); by the Rellich theorem, uk → u strongly in Lp (Ω) for all 1 ≤ q < uk → u almost everywhere.
 63
2N ; N−2
Next we check the Palais–Smale condition. By hypothesis we have Jλ (uk ) = −Δuk − λ
uk − upk − yk , x2
where yk → 0 in W −1,2 (Ω). Then we test with (u − uk ) and obtain ∫⟨∇uk , ∇(uk − u)⟩ dx = ∫ λ Ω
Ω
uk (u − u) ∫ uk (uk − u) dx + ⟨yk , (uk − u)⟩. x2 k Ω
Adding to both members of the identity the term − ∫⟨∇u, ∇(uk − u)⟩ Ω
and observing that this tends to 0 as k → ∞, we get u 2 ∫∇(uk − u) dx = ∫ λ k2 (uk − u) ∫ uk (uk − u) dx + ⟨yk , (uk − u)⟩ − ∫⟨∇u, ∇(uk − u)⟩. x
Ω
Ω
Ω
Ω
Hence 2 ∫∇(uk − u) dx = o(1), k → ∞,
Ω
and u is a solution to the semilinear problem and we realize a minimum. In this case there are arguments of minimization that arrive at the same result. Note that any minimizing sequence, {un }, is bounded in W01,2 (Ω). Hence, by the Rellich theorem, there exists a subsequence {unk } weakly convergent to some u ∈ W01,2 (Ω). Moreover Jλ (u) is weakly lower semicontinuous (up to a subsequence) and then α = Jλ (u) and Jλ (u) = 0. In particular u is a solution to the semilinear problem. , J has the geometry of mountain pass. Indeed If 1 < p < N+2 N−2 λ p+1 1 λ 1 Jλ (u) ≥ (1 − ) ∫ ∇u2 − Ω1− 2∗ (∫ ∇u2 ) 2 ΛN,2 (p + 1)S
p+1 2
.
Ω
Ω
Hence Jλ (u) ≥ β > 0. Moreover taking v ∈ W01,2 (Ω) with ‖∇v‖2 large enough, Jλ (v) < 0. From Theorem 4.3.16, there exists a Palais–Smale sequence for the minimax level c = inf sup Jλ (f (s)), f ∈Γ s∈[0,1]
64  4 The Hardy Potential in Semilinear Elliptic Equations where Γ = {g ∈ 𝒞 ([0, 1], W01,2 (Ω)) : g(0) = 0, g(1) = v}. N+2 Since 1 < p < N−2 such Palais–Smale sequence admits a convergent subsequence and we conclude with the Ambrosetti–Rabinowitz theorem. As we can see, it is sufficient to reproduce the case 0 < p < 1 to get the Palais–Smale condition.
Remark 4.4.2. In general if N+2 ≤ p by the corresponding Pohozaev identity we have N−2 in general no solution different from zero on starshaped domains. Indeed N (N − 2) 1 𝜕u 2 ⟨x, ν⟩dσ. ∫ up+1 dx − ∫ up+1 dx = ∫ p+1 2 2 𝜕ν Ω
Ω
𝜕Ω
However, we will be able to prove existence for suitable small data f . . By using the improved Hardy inequalities Case 2. 0 ≤ λ = ΛN,2 , p < p+ (ΛN,2 ) = N+2 N−2 (see Theorem 2.5.2) we observed in Section 2.5.1 that ‖u‖2H = ∫(∇u2 − λ
u2 ) dx x2
is a Euclidean norm on 𝒞0∞ (Ω). We have defined in Section 2.5.1 the Hilbert space ℋ(Ω) by completion of 𝒞0∞ (Ω) with respect to such norm. Also by the improved Hardy inequality we have the compact embedding q
ℋ(Ω) → L (Ω),
for all q < 2∗ .
Therefore, as in the first case above, we could find a nontrivial solution by the mountain pass theorem. Case 3. 0 ≤ λ < ΛN,2 , p < p+ (λ). We will study precisely the following problem: −Δu = { { { { u≥0 { { { { { u=0
λu x2
+ up + tf
in Ω, in Ω,
(4.4.2)
on 𝜕Ω,
where Ω ⊂ ℝN , N ≥ 3, is such that 0 ∈ Ω, 0 < λ < ΛN,2 , t > 0 and f ≩ 0. Theorem 4.4.3. Assume the hypotheses above and 0 ≤ p < p+ (λ). Then if t > 0 is small enough, problem (4.4.2) has a distributional solution. Proof. Consider BR (0) ⊃ Ω. Note that since p < p+ (λ), we have α− (λ)p < α− (λ) + 2. Therefore we can choose β ∈ (α( λ)−, α+ (λ)) close to α− (λ) such that βp < β + 2. Fixing β as above, we prove that there exists A > 0 and t > 0 such that w = Ax−β ∈ W 1,2 (Ω) is a supersolution to (4.4.2).
4.4 Results on existence
 65
Indeed, −Δw − λwx2 = −AP(−β)x−(β+2) ,
where P(s) = s(s − 1) + (N − 1)s + λ.
Since β ∈ (α− , α+ ), P(−β) < 0. We need A and t such that 1 − AP(−β)x−(β+2) ≥ Ap x−pβ 2
and
1 − AP(−β)x−(β+2) ≥ tf . 2
Since pβ − β − 2 < 0 the first inequality is verified if 1
p−1 1 A ≤ (− P(−β)Rpβ−β−2 ) . 2
Therefore it is sufficient to pick t such that 1 − AP(−β)R−(β+2) ≥ t‖f ‖∞ . 2 Then we have {
−Δw −
λw x2
≥ wp + tf
w≥0
in Ω, on 𝜕Ω.
Now we use a monotone iteration to construct the solution. Consider u0 , the solution to {
−Δu0 = tf u0 = 0
in Ω, on 𝜕Ω
and define for k ≥ 1 {
−Δuk =
λ u x2 k−1
+ upk−1 + tf
uk = 0
in Ω, on 𝜕Ω.
Since w ≥ u0 > 0, by recurrence we find 0 ≤ uk−1 ≤ uk ≤ w,
for all k ∈ ℕ.
Then there exists u(x) = limk→∞ uk (x) and the convergence is in Lp (Ω) by dominated convergence. Therefore u is a distributional solution. Remark 4.4.4. In fact the minimal solution found above is in the sense of the energy and by the elliptic regularity classical solution in Ω \ {0}. See for details the paper by L. Dupaigne [145]. Theorem 4.4.5. Under the same hypotheses as Theorem 4.4.3 the set I = {t > 0  problem (4.4.2) has a solution} is a bounded interval.
66  4 The Hardy Potential in Semilinear Elliptic Equations Proof. Notice that I is an interval just because if λ1 < λ2 ∈ I and uλ2 is a solution to (4.4.2), then uλ2 is a supersolution to problem (4.4.2) for λ1 , and hence it is sufficient to proceed as in Theorem 4.4.3. Assume that λ ∈ I and uλ is a solution to (4.4.2). Consider ϕ1 > 0 as the eigenfunction corresponding to λ1 , the principal eigenvalue of the Laplacian in Ω. Then testing the equation for uλ with ϕ1 we find, using the Young inequality, ∫ Ω
λ 1 u ϕ + ∫ upλ ϕ1 + λ ∫ fϕ1 = λ1 ∫ uλ ϕ1 ≤ ∫ upλ ϕ1 + C ∫ ϕ1 . 2 x2 λ 1 Ω
Ω
Ω
Ω
Ω
Therefore, ∫ Ω
λ 1 u ϕ + ∫ upλ ϕ1 + λ ∫ fϕ1 ≤ C ∫ ϕ1 , x2 λ 1 2 Ω
Ω
Ω
and then necessarily I is bounded.
4.5 Results on nonexistence We will prove the following nonexistence result. Theorem 4.5.1. Let 0 < λ ≤ ΛN,2 and p ≥ p+ (λ). Assume u ∈ Lp (BR (0) \ {0}), u ≥ 0, satisfies −Δu −
λ u ≥ up , x2
in 𝒟 (BR (0) \ {0}).
Then u ≡ 0. Proof. Assume by contradiction that there exists a nontrivial solution u ≥ 0. Then by Theorem 3.4.1 in particular we have up < ∞. xα− (λ)
∫ Ω
Also by Lemma 4.3.8 in a ball Br (0) and a suitable constant, m, we have, u ≥ mx−α− (λ) . By using the Hölder inequality we obtain ∫ Ω
u
xα− (λ)+2
p
−α− (λ)
≤ (∫ u x Ω
1 p
) (∫ x Ω
p −α− (λ)−2 p−1
p−1 p
)
< ∞,
4.5 Results on nonexistence
 67
p because as p ≥ p+ (λ) and λ < ΛN,2 , −α− (λ) − 2 p−1 > −N. Therefore,
∫ Ω
u < ∞. xα− (λ)−2
If we assume that p > p+ (λ), by using the Picone inequality in Br (0), we find that for all ϕ ∈ 𝒞0∞ (Br (0)) ∫ ∇ϕ2 dx ≥ ∫ ( Br (0)
Br (0)
ϕ2 Δu 2 dx, )ϕ dx ≥ ∫ u xα− (λ)(p−1) Br (0)
and since α− (λ)(p − 1) > α− (λ)(p+ (λ) − 1) = 2, we reach a contradiction with the Hardy inequality. Note that the previous argument does not work in general when p = p+ (λ). To overcome this situation, we consider w(x) = Ax−α− (λ) log(
1 ) x
with A > 0 and we try to find A in such a way that w becomes a subsolution to the equation. By a direct computation we find that −Δw − λ
w = A(2√ΛN,2 − λ)x−α− (λ)−2 . x2
Moreover, −Δu −
λ u ≥ mx−α− (λ)−2 . x2
Taking −1
1 A = m(2√ΛN,2 − λ) + λ log( ) , r
1 C = Ar −α− (λ) log( ) r
and defining z = u + C − w, we have z ∈ L1 (Br (0), x−α− (λ)−2 ). Moreover, by a direct calculation again, −Δz − λ
z ≥ 0 in Br (0), x2
z = C − w ≥ 0 on 𝜕Br ,
68  4 The Hardy Potential in Semilinear Elliptic Equations and, by the strong maximum principle, z > 0 in Br (0). Choosing 0 < A1 < A and r1 < r small enough we prove that u(x) ≥ A1 x−α− (λ) log(
1 ) x
in Br1 (0).
Again by the Picone inequality in Br1 (0), we find that p+ (λ)−1
1 1 Δu log( ) ∫ ∇ϕ dx ≥ ∫ ( )ϕ2 dx ≥ ∫ 2 u x x 2
Br1 (0)
Br1 (0)
Br1 (0)
ϕ2 dx
for all ϕ ∈ 𝒞0∞ (Br1 (0)), which is a contradiction with the Hardy–Leray inequality. 4.5.1 Complete blowup As a byproduct of the previous existence proof and the nonexistence result we are able to prove the complete blowup; that is, the solutions un of any truncated problem verify that un (x) → ∞ as n → ∞ for all x ∈ Ω. Theorem 4.5.2. Assume that 0 < λ ≤ ΛN,2 holds, p > 1 and t > 0. If problem (4.4.2) has no weak solution, then there exists complete blowup. Proof. We argue by contradiction. Consider un as a solution to the truncated problem, −Δun = gn (un ) + λan (x)un + cf { { { un > 0 { { { { un = 0
in Ω, in Ω, on 𝜕Ω,
where an (x) = Tn ( x1 2 ) and gn (s) = Tn (up ). Assume by contradiction that ∫ gn (un )δ(x) dx + ∫ an (x)un δ(x) dx ≤ C < ∞, Ω
for all n ∈ ℕ.
Ω
Consider ψ1 as a solution to the problem {
−Δψ1 = 1
in Ω,
ψ1 = 0
on 𝜕Ω,
and use it as a test function in problem (4.5.1). We get ∫ un (−Δψ1 ) − ∫ an un ψ1 = ∫ gn (un )ψ1 + t ∫ fψ1 . Ω
Ω
Ω
Ω
(4.5.1)
4.6 Further results and comments  69
Hence ∫ un ≤ C
and, since {un }n∈ℕ is nondecreasing, un ↗ u in L1 (Ω) for some u
Ω
by monotone convergence. Since an and gn are nondecreasing, again by monotony, we could pass to the limit in the truncated problems, getting a solution to (4.4.2), which is a contradiction.
4.6 Further results and comments –
The critical case in ℝN , N ≥ 3. It is well known that the minimizers to the variational problem S=
inf
u∈𝒟 1,2 (ℝN )
∫ℝN ∇u2 dx 2N
(∫ℝN u N−2 dx)
N−2 N
(4.6.1)
,
where 𝒟1,2 (ℝN ) is the completion of 𝒞0∞ (ℝN ) with respect to the norm 2
1 2
‖u‖𝒟1,2 (ℝN ) = ( ∫ ∇u dx) , ℝN
are given by the function φ(x) =
N(N − 2) (1 + x2 )
N−2 4 N−2 2
and all its dilations and translations, i. e., φλ,x0 (x) = λ
2−N 2
φ(
(x − x0 ) ). λ
The Euler–Lagrange equation corresponding to the minimization problem (4.6.1) is −Δu = Su2
∗
−1
,
x ∈ ℝN
2∗ =
2N . N −2
Susanna Terracini in [304] studied, among others questions, the problem − Δu − λ
u = Sλ u2∗−1 , x2
0 < λ < ΛN,2 ,
x ∈ ℝN \ {0},
(4.6.2)
for which the positive solutions are ηλ (x) =
(N(N − 2)ρ2 (λ))
1
N−2 4
x(1−ρ(λ)) (1 + x2ρ(λ) )
N−2 2
2 4λ where ρ(λ) = (1 − ) (N − 2)2
70  4 The Hardy Potential in Semilinear Elliptic Equations and 4λ Sλ = S(1 − ) (N − 2)2
N−1 N
,
with S defined in (4.6.1). Note that in a bounded starshaped domain Ω ⊂ ℝN , equation (4.6.2) has no positive solution. To see this, it is sufficient to use a Pohozaev type argument. See [261] and the extensions in [268]. In [10] the reader can find results for the problem λ
p−1
{−Δp u = xp u { {u(x) > 0,
u + up
∗
−1
u, in ℝN , u ∈ 𝒟1,p (ℝN ),
(4.6.3)
where N ≥ 3, λ > 0, 1 < p < N and p∗ = Np/(N −p) is the critical Sobolev exponent. Here 𝒟1,p (ℝN ) denotes the space obtained as the completion of the class of smooth functions with compact support with respect to the norm 1/p
‖u‖𝒟1,p (ℝN ) = ( ∫ ∇up dx)
.
ℝN
–
The potential 1/xp is related to the Hardy–Sobolev inequality in Lemma 2.4.1. Equation (4.6.3) in bounded domains is studied with related nonlinearities, among others, in the articles [14], [87], [179], [184] and in the references quoted in these papers. We may add that there are still many open questions in this framework. Other related problems. One can find similar problems involving the Caffarelli– Kohn–Nirenberg inequalities in [108]; see also Section 6.2.3 of this monograph. More results along these directions can be found in the papers [15], [16], [7] and [5]. The papers [116] and [154] develop the theory in the case of ℝN . The results here are focalized in the attainability of the optimal constant and the symmetry of the minimizers of the Caffarelli–Kohn–Nirenberg inequalities.
5 The Hardy–Leray potential in semilinear heat equations 5.1 Introduction In this chapter we will study the following problem: { { { { { { { { { { { { { { {
ut − Δu = λ xu 2 + up + f
in ΩT ≡ Ω × (0, T),
u(x, t) > 0
in ΩT ,
u(x, t) = 0
on 𝜕Ω × (0, T),
u(x, 0) = u0 (x)
if x ∈ Ω,
(5.1.1)
where either Ω is an open bounded domain in ℝN such that 0 ∈ Ω, or Ω = ℝN , N ≥ 3, p > 1 and λ > 0. We assume that f and u0 are nonnegative measurable functions under some summability hypotheses that will be presented. There exists a great amount of literature dealing with the case λ = 0. See for instance the papers [171], [227], [316], the book [269] and the references therein. In that case, existence and uniqueness of local solutions in time are well known for all p > 1, at least for regular initial data. Moreover, this solution is bounded for small time. On the other hand, Fujita found in [171], in some way, a surprising result, namely, that for 1 < p ≤ 1 + N2 , any positive solution to the semilinear heat equation ((5.1.1) with λ = 0) blows up in a finite time in the L∞ (ℝN )norm. As in the elliptic setting studied in Chapter 4, the case λ > 0 is quite different. 1. In the pioneering work by P. Baras and J. Goldstein, [41], the linear problem is studied. Since in Chapter 13 we study the theory for the fractional heat equation and we introduce a slightly different proof for the linear case that it is applicable to the heat equation, we do not emphasize this point in this chapter. 2. It is not difficult to show that any positive solution of (5.1.1) is unbounded (see Section 5.2.1). 3. Some restrictions on p are needed to insure the existence of solution even in the weakest sense. More precisely, if p+ (λ) = 1 + α 2(λ) , the critical exponent of the − associated elliptic problem, we will prove that for p ≥ p+ (λ) there is no distributional solution. Hereafter, we will say that problem (5.1.1) blows up completely and instantaneously if the solutions to the truncated problems (with the weight 2λ 1 λ ) x2
x + n
instead of converge to infinity for every (x, t) ∈ ΩT as n → ∞. This is the case if p ≥ p+ (λ). 4. There exists a Fujita type exponent, p+ (λ) > F(λ) = 1 + N−α2 (λ) > F(0). That is, − for all 1 < p < F(λ) and for all nonnegative initial data, the solution blows up in a finite time, and F(λ) is optimal with this property. That is, the Hardy potential produces a shift in the Fujita exponent with respect to that corresponding to λ = 0. Note that limλ→0+ F(λ) = F(0). https://doi.org/10.1515/9783110606270005
72  5 The Hardy–Leray potential in semilinear heat equations The objective of this chapter is to explain the influence of the Hardy term on the existence or nonexistence of solutions by proving the results explained above. The results in this chapter were obtained in the paper [21].
5.2 Preliminaries and tools This section contains some preliminaries and certain tools that will be used in the next sections. We start with the different concepts of solution that will be used in the next sections, namely, very weak solution and solution obtained as limit of approximations. Definition 5.2.1. We say that u ∈ 𝒞 ([0, T); L1loc (Ω)) is a very weak supersolution (subsolution) to the semilinear equation in problem (5.1.1) if xu 2 ∈ L1loc (ΩT ), up ∈ L1loc (ΩT ), f ∈ L1loc (ΩT ) and, for all ϕ ∈ 𝒞0∞ (ΩT ) such that ϕ ≥ 0, we have T
T
∫ ∫(−ϕt − Δϕ)u dx dt ≥ (respectively ≤) ∫ ∫(λ 0 Ω
0 Ω
u + up + f )ϕ dx dt. x2
(5.2.1)
If u is a very weak super and subsolution, then we say that u is a very weak solution. In particular if u is a very weak solution to (5.1.1), then u ∈ 𝒞 ([0, T); L1loc (Ω)) ∩ Lp ((0, T); Lploc (Ω)). The previous definition is local in nature, among other things, because there is no reference to the boundary data and because only the regularity is needed to give distributional sense to the equation. We will use the general framework given in Definition 5.2.1 to prove nonexistence with local arguments. To prove existence with L1 data, we will consider solutions obtained as limit of approximations (see for instance [129] and [265]). For the heat operator the solutions obtained as limit of approximations coincide with the socalled renormalized solutions. See [61] for the details. For the reader’s convenience we reformulate the existence and regularity theory with positive L1 data for the heat equation. Recall the usual truncation operator, s
s ≤ n,
n sign(s)
s ≥ n.
Tn (s) = {
Consider f ∈ L1 ((0, T) × Ω), f ≥ 0, and let {fn } be a sequence of bounded functions such that: (i) fn (x, t) ≤ f (x, t), (x, t) ∈ (0, T) × Ω, and (ii) fn → f in L1 ((0, T) × Ω). In a similar way for u0 ∈ L1 (Ω), u0 ≥ 0, let u0,n be a sequence of bounded functions such that:
5.2 Preliminaries and tools  73
(i) u0,n (x) ≤ u0 (x), x ∈ Ω, and (ii) u0,n → u0 in L1 (Ω). Let un be the classical solution to the heat problem { { { { { { { { { { { { {
unt − Δun = fn
in ΩT ≡ Ω × (0, T),
un (x, t) > 0
in ΩT ,
un (x, t) = 0
on 𝜕Ω × (0, T),
un (x, 0) = u0,n (x)
if x ∈ Ω.
(5.2.2)
Then (a) un ∈ L2 ((0, T); W02 (Ω)) ∩ C([0, T]; L2 (Ω)) and unt ∈ L2 ((0, T); W −1,2 (Ω)); (b) un verifies T
T
T
∫⟨unt , ϕ⟩ dt + ∫ ∫⟨∇un , ∇ϕ⟩ dx dt = ∫ ∫ fn ϕ dx dt, 0
(c)
0 Ω
0 Ω
{un } is bounded in L ((0, T); W01,q (Ω)) is bounded in L2 ((0, T); W01,2 (Ω)). q
for all 1 ≤ q
0, {Tk (un )}
As a consequence, up to a subsequence, there exists u ∈ Lq ((0, T); W01,q (Ω)) for all 1 ≤ q < N+2 such that N+1 un ⇀ u weakly in Lq ((0, T); W01,q (Ω)), ∀q ∈ [1,
N +2 ). N +1
The reader can find the details in [42] and in [66] in a more general case. With these previous properties one can prove that ∇un → ∇u,
a. e. in Ω.
Therefore by Fatou’s and Vitali’s theorems, it follows that un → u strongly in Lq ((0, T); W01,q (Ω)), ∀q ∈ [1, and
N +2 ) N +1
Tk (un ) → Tk (u) strongly in L2 ((0, T); W01,2 (Ω)), ∀k > 0. Hence u is a solution to the problem { { { { { { { { { { { { {
ut − Δu = f
in ΩT ≡ Ω × (0, T),
u(x, t) > 0
in ΩT ,
u(x, t) = 0
on 𝜕Ω × (0, T),
u(x, 0) = u0 (x)
if x ∈ Ω.
(5.2.3)
More details can be found, for instance, in [265] and [4], in a more general setting.
74  5 The Hardy–Leray potential in semilinear heat equations By the strong maximum principle for the truncated problems (5.2.2), we find that for f ≥ 0, u0 ≥ 0, then u > 0 in Ω × (0, T). Also, it is easy to check a strong comparison principle for solutions obtained as limit of approximations. In [4] it is proved in particular that the solutions obtained as limit of approximations for the heat equations coincide with the distributional solutions and that problem (5.2.3) has a unique distributional solution. Therefore, in the process of constructing a solution, u, as limit of approximations, we see that all subsequences of {un } converge to u. 5.2.1 Local behavior of supersolutions of the linear equation The behavior of any positive supersolution in a neighborhood of the origin is obtained looking for a selfinvariant solution to the homogenous linear equation v vt − Δv − λ 2 = 0 in ℝN . (5.2.4) x More precisely, we look for a solution to (5.2.4) of the form v(r, t) = t −μ ϕ( trν ). Therefore, by setting s = trν and taking ν = 21 , we find the equation N −1 v )v − λ 2 r r λ (N − 1) s + )ϕ (s) + (μ + 2 )ϕ(s)). ≡ (−t −(μ+1) )(ϕ (s) + ( s 2 s
0 = vt − v − (
(5.2.5)
γ
If we set ϕ(s) = s−α e−βs , then ϕ (s) = (− ϕ (s) = (
α − γβsγ−1 )ϕ(s), s
2
α α − βγ(γ − 1)sγ−2 + ( + βγsγ−1 ) )ϕ(s). s s2
(5.2.6)
From (5.2.5) and (5.2.6), it is sufficient to choose α = α− (λ) given in (3.1.3), β = 41 , γ = 2 and μ =
N−α− (λ) 2
to verify
vt − v − (
N −1 v )v − λ 2 = 0. r r
We find in this way that N
v(x, t) = t − 2 +α− (λ) r −α− (λ) e−
x2 4t
,
and then it is easy to check that ∫ x−α− (λ) v(x, t) dx = C. ℝN
Note that, for λ = 0, we recover the fundamental kernel of the heat equation.
(5.2.7)
5.2 Preliminaries and tools  75
The unboundedness of any very weak supersolution is obtained in the following result, which also gives a quantitative information; see for instance [6]. A perturbation of the linear problem As an example of the borderline behavior of the Hardy potential, we prove the existence of a solution ψ ∈ L2 ((0, T); 𝒟1,2 (ℝN )) to the problem ψ
{ ψt − Δψ − λ x2 = aB(x, t)ψ { { ψ(x, 0) = ψ0 (x),
in ℝN , t ∈ (0, T),
(5.2.8)
N
with supt∈[0,T] ∫ℝN B 2 (x, t) dx ≤ C(T), ψ0 ∈ L1 (ℝN ) ∩ L∞ (ℝN ) and a small. N
The function B is in L∞ ([0, T], L 2 (ℝN )) with respect to the spatial variables, while N
the Hardy potential is only in the Marcinkiewicz space ℳ 2 (ℝN ). The following result shows how B can be interpreted as a small perturbation to the linear problem. Theorem 5.2.2. There exists a0 > 0 such that for all a ≤ a0 , problem (5.2.8) has a unique solution ψ ∈ L2 (0, T, 𝒟1,2 (ℝN )). Proof. We define the following mapping from L2 ((0, T); L2 (ℝN )) into L2 ((0, T); ∗ L2 (ℝN )): ∗
H : L2 ((0, T); L2 (ℝN ))
?
L2 ((0, T); L2 (ℝN )),
v
?
H(v) = u,
∗
∗
where u ∈ L2 ((0, T); L2 (ℝN )) is the solution to ∗
{
ut − Δu − λ xu 2 = aB(x, t)v
in ℝN , t ∈ (0, T),
u(x, 0) = ψ0 (x).
It is clear that u ∈ L2 ((0, T); 𝒟1,2 (ℝN )) and then H is well defined. We claim that for a ≤ a0 small enough the operator H is a contraction. Denote w w = u1 − u2 . It follows that wt − Δw − λ x 2 = aB(x, t)(v1 − v2 ). Hence using w as a test function and by the Hardy inequality it follows that T
ΛN,2 − λ 1 1 ) ∫ ∫ ∇w2 dx dt ∫ w2 (x, T) dx − ∫ w2 (x, 0) dx + ( 2 2 ΛN,2 ℝN
ℝN
T
≤ a ∫ ∫ B(x, t)(v1 − v2 )w dx dt. 0 ℝN
0 ℝN
76  5 The Hardy–Leray potential in semilinear heat equations Applying Hölder’s, Sobolev’s and Young’s inequalities, T
2∗
CS ∫( ∫ w dx) 0
2 2∗
dt
ℝN T
1 2∗
2∗
1 2∗
2∗
2 N
N 2
≤ a ∫( ∫ w dx) ( ∫ (v1 − v2 ) dx) ( ∫ B(x, t) dx) dt 0
ℝN
ℝN T
2∗
≤ aCε ∫( ∫ w dx) 0
ℝN
2 2∗
T
2∗
dt + aC(T)C (ε) ∫( ∫ (v1 − v2 ) dx) 0
ℝN
2 2∗
dt.
ℝN
Hence, choosing a small, we have H(v1 − v2 )L2 ((0,T);L2∗ (ℝN )) ≤ C‖v1 − v2 ‖L2 ((0,T);L2∗ (ℝN )) , with C < 1. Thus the claim follows and then by the classical Banach fixed point theorem we reach the conclusion. The next result will be the key for many arguments in the proofs of nonexistence. Lemma 5.2.3. Assume that u is a nonnegative function defined in Ω such that u ≢ 0, u ∈ L1loc (ΩT ) and xu 2 ∈ L1loc (ΩT ). If u satisfies ut −Δu−λ xu 2 ≥ 0 in 𝒟 (ΩT ) with λ ≤ ΛN,2 and Br1 (0) ⊂⊂ Ω, then there exists a constant C = C(N, r1 , t1 , t2 ) such that for each cylinder Br (0) × (t1 , t2 ) ⊂⊂ ΩT , 0 < r < r1 , u ≥ Cx−α− (λ)
in Br (0) × (t1 , t2 ),
where −α− (λ) is given in (3.1.3). In particular, for r conveniently small we can assume u > 1 in Br (0) × (t1 , t2 ). Proof. Since u ≢ 0, using the strong maximum principle for the heat equation, it follows that for any cylinder Br1 (0) × (T1 , T2 ), there exists η > 0 such that u ≥ η > 0 in Br1 (0) × (T1 , T2 ). Let w ∈ L2 ((T1 , T2 ); W 1,2 (Br1 (0))) be the unique positive solution to the problem w wt − Δw − λ x 2 = 0 { { { w=η { { { { w(x, T1 ) = 0
in Br1 (0) × (T1 , T2 ), on 𝜕Br1 (0) × (T1 , T2 ),
in Br1 (0).
It is clear that w > 0 in Br1 (0) × (T1 , T2 ). We will prove that w(x, t) ≥ Cx−α− (λ)
in Br (0) × (t1 , t2 ) ⊂⊂ Br1 (0) × (T1 , T2 ).
(5.2.9)
5.2 Preliminaries and tools  77
Indeed, define v(x, t) = xα− (λ) w(x, t), that is, the socalled ground state transformation of w. Since w ∈ L2 ((T1 , T2 ); W 1,2 (Br1 (0))), we have v ∈ L2 ((T1 , T2 ); Wα1,2(λ) (Br1 (0))) ∩ 𝒞 ((T1 , T2 ); L2α− (λ) (Br1 (0))), −
where L2α− (λ) (Br1 (0)) and Wα1,2(λ) (Br1 (0)) are the weighted Lebesgue and Sobolev spaces − defined as the complexion of 𝒞0∞ (Br1 (0)) with respect to the norms ‖ϕ‖2L2
α− (λ)
‖ϕ‖2W 1,2
α− (λ)
= ∫ ϕ2 x−2α− (λ) dx, Br1 (0)
= ∫ ϕ2 x−2α− (λ) dx + ∫ ∇ϕ2 x−2α− (λ) dx, Br1 (0)
Br1 (0)
respectively. Moreover v solves the problem x−2α− (λ) vt − div(x−2α− (λ) ∇v) = 0 { { { α (λ) v = ηr1 − { { { { v(x, T1 ) = 0
in Br1 (0) × (T1 , T2 ), on 𝜕Br1 (0) × (T1 , T2 ),
(5.2.10)
in Br1 (0).
Since v is in the corresponding energy space and the weight x−2α− (λ) is in the Muckenhoupt class, we can apply the Harnack inequality obtained in [190] (see also [6] and [123]) and we conclude that v ≥ C in Br (0) × (t1 , t2 ) ⊂⊂ Br1 (0) × (T1 , T2 ). Hence w(x, t) ≥ Cx−α− (λ)
in Br (0) × (t1 , t2 ).
Since u is a supersolution to problem (5.2.9), by using the weak comparison principle we conclude that u ≥ w in Br (0) × (T1 , T2 ), thus u ≥ Cx−α− (λ) in Br (0) × (t1 , t2 ) and the result follows. 5.2.2 A technical remark on existence Note that the direct use of the very weak supersolutions has serious difficulties. For this reason, the following result is an important tool in order to prove the nonexistence result. Roughly speaking, given a very weak supersolution, we will show that there exists a minimal solution to (5.1.1), at least defined in a subcylinder, which is obtained as a limit of solutions to some approximated problems in the sense stated at the beginning of this section. For such a class of solutions, the flexibility of calculus is bigger and allows us to compare and to obtain a priori estimates. ̃ is a very weak supersolution to the equation Lemma 5.2.4. If ū ∈ 𝒞 ([0, T); L1loc (Ω)) 1 in (5.1.1) with λ ≤ ΛN,2 , f ∈ L (ΩT ) and Ω̃ ⊃⊃ Ω, then there exists a minimal solution to problem (5.1.1) obtained as limit of approximations.
78  5 The Hardy–Leray potential in semilinear heat equations Proof. If ū is a supersolution to (5.1.1) with λ ≤ ΛN,2 we construct a sequence {vn } ∈ 𝒞 ([0, T); L1 (Ω)) ∩ Lp ([0, T); Lp (Ω)), starting with v0t − Δv0 = f { { { ̄ 0)) v0 (x, 0) = T1 (u(x, { { { { v0 (x, t) = 0
in Ω × (0, T), if x ∈ Ω,
(5.2.11)
on 𝜕Ω × (0, T).
By the comparison principle for the heat equation, it follows that v0 ≤ ū in Ω × (0, T). By iteration we define v
p vnt − Δvn = λ n−1 2 + 1 + vn−1 + f { x { n { { ̄ 0)) vn (x, 0) = Tn (u(x, { { { { { vn (x, t) = 0
in Ω × (0, T), if x ∈ Ω,
(5.2.12)
on 𝜕Ω × (0, T).
Using again, in a recurrent way, weak comparison for the linear heat equation, it follows that v0 ≤ ⋅ ⋅ ⋅ ≤ vn−1 ≤ vn ≤ ū in Ω × (0, T), so we obtain the pointwise limit v = lim vn that verifies v ≤ ū and vt − Δv = λ xv 2 + vp + f { { { v(x, t) = 0 { { { ̄ 0) { v(x, 0) = u(x,
in Ω × (0, T), on 𝜕Ω × (0, T),
(5.2.13)
if x ∈ Ω,
in the weak sense. Moreover, according to the estimates for the sequence {vn } and using the techniques in [129] and [60], it is easy to check that v has the regularity of a solution obtained as limit of approximations to (5.1.1) in Ω × (0, T). That is, v ∈ ) and Tk (un ) → Tk (u) strongly in L2 ((0, T); W01,2 (Ω)), for Lq ((0, T); W01,q (Ω)), ∀q ∈ [1, N+2 N+1 all k > 0. Remark 5.2.5. Note that if w is a very weak positive supersolution to problem w wt − Δw − λ 2 = g, (5.2.14) x T
then g must satisfy ∫0 ∫B (0) x−α− (λ) g dx < ∞. To see this, it suffices to consider as a r
φ
test function in (5.2.14) a truncation of φ, the solution to the equation −Δφ − λ x2 = 1,
and the result follows. Since we are considering positive solutions to problem (5.1.1), by setting g = up + f , we obtain t2
∫ ∫ x−α− (λ) (up + f ) dx dt < ∞,
for all Br (0) × (t1 , t2 ) ⊂⊂ ΩT .
t1 Br (0)
This necessary condition is inspired in the elliptic Theorem 3.4.1 and will be useful in the forthcoming arguments.
5.3 Nonexistence results: p ≥ p+ (λ)
 79
5.3 Nonexistence results: p ≥ p+ (λ) We will see here that for the heat equation perturbed with the Hardy potential, the critical power to obtain existence is also p+ (λ), the same value as in the elliptic case and obtained in (4.2.5). Recall that the main properties of p− (λ) and p+ (λ) are N N N ∗ p− (λ) → 2 − 1 = N
p+ (λ) → 2∗ − 1 =
+2 as λ → ΛN,2 , −2 +2 as λ → ΛN,2 , −2
p+ (λ) → ∞ as λ → 0, p− (λ) →
N as λ → 0. N −2
6
p+ (λ)
λ
p− (λ)
ΛN,2

It is clear that p+ (λ) and p− (λ) are respectively decreasing and increasing functions on λ and, thus, 1 < p− (λ) ≤ 2∗ − 1 ≤ p+ (λ). Theorem 5.3.1. If p ≥ p+ (λ), then problem (5.1.1) has no positive very weak supersolution. In the case where f ≡ 0, the unique nonnegative very weak supersolution is u ≡ 0. Proof. Without loss of generality, we can assume f ∈ L∞ (ΩT ). We argue by contradiction. Assume that ũ is a very weak supersolution. If λ > ΛN,2 = ( N−2 )2 , then it is sufficient to consider ũ as a very weak supersolution 2 to the problem vt − Δv = λ xv 2 + f1 { { { v>0 { { { { v=0
in Ω × (0, T), in Ω × (0, T),
(5.3.1)
on 𝜕Ω × (0, T),
where f1 (x) = vp + f . Hence the nonexistence result follows using [68] and [6]. For the case λ ≤ ΛN,2 we argue again by contradiction. If ũ is a very weak supersolution to (5.1.1), then ũ t − Δũ − λ xũ 2 ≥ 0 in 𝒟 (ΩT ).
80  5 The Hardy–Leray potential in semilinear heat equations Since ũ is also a very weak supersolution in any BR (0) × (T1 , T2 ) ⊂⊂ ΩT , then by Lemma 5.2.4, the problem { { { { { { { { { { { { { { {
ut − Δu = λ xu 2 + up + f
in BR (0) × (T1 , T2 ),
u(x, t) > 0
in BR (0) × (T1 , T2 ),
u(x, t) = 0
on 𝜕BR (0) × (T1 , T2 ),
̃ T1 ) u(x, t1 ) = u(x,
if x ∈ BR (0),
(5.3.2)
has a minimal solution u obtained by approximation of truncated problems in BR (0) × (T1 , T2 ). In particular u = lim vn , with vn ∈ L∞ (BR (0) × (T1 , T2 )) and vn being a solution to (5.2.12) in BR (0) × (T1 , T2 ). Note that since ut − Δu − λ xu 2 ≥ 0 in 𝒟 (Br1 (0) × (T1 , T2 )) and using Lemma 5.2.3, there exists a cylinder Br (0) × (t1 , t2 ), with 0 < r < r1 < R, 0 < T1 < t1 < t2 < T2 ≤ T and there exists a constant C = C(N, r1 , t1 , t2 ) such that u ≥ Cx−α− (λ) and u > 1 in Br (0)×(T1 , T2 ) for suitable r. In particular, since u ∈ L1loc (ΩT ), we have log(u) ∈ Lp (Br (0) × (t1 , t2 )), for all p ∈ [1, ∞). By a scaling, we can assume that the cylinder is Br (0) × (0, τ). We divide the proof in three cases. ϕ2
Step 1. The case p > p+ (λ). Consider ϕ ∈ 𝒞0∞ (Br (0)) and take v as a test function n in problems (5.2.12). Applying the Picone inequality in Section 1.3, we obtain τ
τ
τ
∫ ∫ vnp−1 ϕ2 dx dt ≤ ∫ ∫ 0 Br (0)
0 Br (0)
ϕ2 ϕ2 vnt dx dt + ∫ ∫ (−Δvn ) dx dt vn vn 0 Br (0) τ
≤ ∫ log vn (x, τ)ϕ2 dx + ∫ ∫ ∇ϕ2 dx dt. 0 Br (0)
Br (0)
Therefore, passing to the limit as n tends to infinity and thanks to Lemma 5.2.3, we have τ
τ
∫ log u(x, τ)ϕ2 dx + ∫ ∫ ∇ϕ2 dx dt ≥ ∫ ∫ up−1 ϕ2 dx dt Br (0)
0 Br (0)
0 Br (0) τ
ϕ2
≥C∫ ∫ 0 Br (0)
x(p−1)α− (λ)
dx dt.
Using the Hölder and Sobolev inequalities, it follows that 2
2 ∗
N 2 ∗ N ∫ log(u(x, τ))ϕ2 dx ≤ ( ∫ ϕ2 dx) ( ∫ log u(x, τ) 2 dx)
Br (0)
Br (0)
Br (0) 2
N N ≤ ( ∫ log u(x, τ) 2 dx) 𝒮 −1 ∫ ∇ϕ2 dx,
Br (0)
Br (0)
5.3 Nonexistence results: p ≥ p+ (λ)
 81
where 𝒮 is the optimal constant in the Sobolev embedding. Thus we have 2
N ϕ2 N [1 + ( ∫ log u(x, τ) 2 dx) 𝒮 −1 ] ∫ ∇ϕ2 dx dt ≥ C ∫ dx dt. (p−1)α − (λ) x
Br (0)
Br (0)
Br (0)
Since p > p+ (λ), we have (p − 1)α− (λ) > 2 and we obtain a contradiction with the Hardy inequality. Step 2. The case p = p+ (λ) and λ < ΛN,2 . 1 β )) + Fix the cylinder Bη (0)×(0, τ) as above, and consider w(x, t) = x−α− (λ) (t 2 (log( x 1) defined for (x, t) ∈ Bη (0) × (0, τ), for some β > 0 that will be chosen below. Since λ < ΛN,2 , we have w ∈ 𝒞 ([0, τ], L2 (Bη (0))) ∩ L2 ((0, τ), W 1,2 (Bη (0))). By a direct computation we get wt − Δw − λ =
t
x2+α−
w x2
{2x2 (log( (λ)
β
β−1
1 1 )) + β(t)(log( )) x x
× [(N − 2 − 2α− (λ)) + (1 − β)((log(
−1
1 )) ]}. x
Note that wp+ (λ) = x−(2+α− (λ)) [t 2 (log(
β
p+ (λ)
1 )) + 1] x
.
If we set h(x, t) = [t 2 (log(
β
1−p+ (λ)
1 )) + 1] x
,
then wp+ (λ) h(x, t) = x−(2+α− (λ)) [t 2 (log(
β
1 )) + 1]. x
Hence, there exists t 2 ∈ (0, τ) such that wt − Δw − λ
w ≤ βh(x, t)wp+ (λ) x2
in Bη (0) × (0, t 2 ).
Denote u1 = c1 u, for a certain constant c1 > 0. Then we have u1t − Δu1 − λ
u1 1−p (λ) p (λ) ≥ c1 + u1 + . x2
Consider c0 > 0 in the same way as C in Lemma 5.2.3 after a scaling. For fixed c1 > 0 such that c1 c0 ≥ 1 and for a suitable small β we have 1−p+ (λ)
c1
≥ β‖h‖L∞ .
82  5 The Hardy–Leray potential in semilinear heat equations Since c1 c0 ≥ 1, we have u1 ≥ w(x, t) on 𝜕Bη (0) × (0, τ), u1 (x, 0) ≥ w(x, 0) for x ∈ Bη (0) and u1t − Δu1 − λ
u1 p (λ) ≥ βh(x, t)u1 + . x2
Claim: We have u1 ≥ w in Bη (0) × (0, t 2 ). Indeed, consider v = w − u1 . Then vt − Δv − λ
v p (λ) ≤ βh(x, t)(wp+ (λ) − u1 + ). x2
Hence using Kato’s inequality (see [255]) we obtain the estimate (v+ )t − Δv+ − λ Since
v+2 x2
v+ ≤ p+ (λ)βh(x, t)wp+ −1 v+ ≤ p+ (λ)βx−2 v+ . x2
(5.3.3)
∈ L1 (Bη (0)), an approximation argument and integration by parts gives v+ ∈ L2 ((0, t 2 ), W01,2 (Bη (0))).
Therefore choosing β small enough such that λ + p+ (λ)β < ΛN,2 and using v+ as a test function in (5.3.3) we deduce that v+ ≡ 0 and the claim is proved. To finish the proof in this step, we use the same argument as in step 1, namely, for all ϕ ∈ 𝒞0∞ (Bη (0)) we arrive at the following inequality: τ
c2 ∫ ∫ u
p−1
2
τ
ϕ dx ≤ ∫ ∫ ∇ϕ2 dx = τ ∫ ∇ϕ2 dx,
0 Bη (0)
0 Bη (0)
(5.3.4)
Bη (0)
where c2 > 0 is independent of ϕ. Using the result of the claim we obtain for r 1, at least for a suitable initial datum.
5.4 Instantaneous and complete blowup results  83
5.4 Instantaneous and complete blowup results The nonexistence result obtained above for p ≥ p+ (λ) is very strong in the sense that a complete and instantaneous blowup phenomenon occurs in two different senses. (a) If un is the solution to problem (5.1.1), where the Hardy potential is replaced by the bounded weight 1/(x2 + n1 ), then un (x, t) → ∞ as n → ∞. (b) If un is the solution to problem (5.1.1), with p = pn < p+ (λ) and pn → p+ (λ) as n → ∞, then un (x, t) → ∞ as n → ∞. In both cases, (x, t) is an arbitrary point in Ω × (0, T). 5.4.1 Blowup for the approximated problems when p ≥ p+ (λ) Let us prove now the blowup behavior for some approximated problems. Theorem 5.4.1. Let un ∈ 𝒞 ((0, T); L1 (Ω)) ∩ Lp ((0, T); Lp (Ω)) be a solution to the problem up
unt − Δun = 1n p + λan (x)un + cf { { 1+ n un { { un (x, t) = 0 { { { { { un (x, 0) = 0 with f ≠ 0, an (x) =
1 x2 + n1
in ΩT ≡ Ω × (0, T), (5.4.1)
on 𝜕Ω × (0, T), if x ∈ Ω,
and p ≥ p+ (λ). Then un (x0 , t0 ) → ∞, ∀(x0 , t0 ) ∈ Ω × (0, T).
Proof. Without loss of generality, we can assume that f ∈ L∞ (ΩT ) and λ ≤ ΛN,2 . The existence of a positive solution to problem (5.4.1) follows using classical subsupersolution arguments. By using the monotonicity of the nonlinear term and the coefficient an , we can assume the existence of a minimal solution un such that un ≤ un+1 , for all n ≥ 1. Therefore to get the blowup result we just have to show the complete blowup for the family of minimal solutions denoted by un . Suppose by contradiction that there exists (x0 , t0 ) ∈ ΩT such that un (x0 , t0 ) → C0 < ∞
as n → ∞.
Using the classical Harnack inequality proved by J. Moser in [244] and J. Nash [250], there exist s > 0 and a positive constant C = C(N, s, t0 , β) such that un , ∫ ∫ un (x, t) dx dt ≤ C ess inf + R
R−
where R− = Bs (x0 )×(t0 − 43 β, t0 − 41 β) and R+ = Bs (x0 )×(t0 − 21 β, t0 + 21 β). We can suppose that 0 ∈ Bs (x0 ), since otherwise, we may consider Bδ (y) ⊂ Bs (x0 ) such that un ≤ C ess inf un , ∫ ∫ un (x, t) dx dt ≤ ∫ ∫ un (x, t) dx dt ≤ C ess inf + + R−y
R−
R
Ry
84  5 The Hardy–Leray potential in semilinear heat equations with R−y = Bδ (y) × (t0 − 43 β, t0 − 41 β) and R+y = Bδ (y) × (t0 − 21 β, t0 + 21 β). In a recurrent way, we get un ≤ Cun (x0 , t0 ) ≤ C , ∫ ∫ un (x, t) dx dt ≤ C ess inf + R
R−
with R− = Br (0) × (t1 , t2 ) and R+ = Br (0) × (t3 , t4 ), t0 ∈ (t3 , t4 ). Therefore, by the monotone convergence theorem, there exists u ≥ 0 such that un ↑ u strongly in L1 (Br (0) × (t1 , t2 )). Let φ be the solution to the problem φt − Δφ = χBr (0)×[T−t2 ,T−t1 ] { { { φ(x, t) = 0 { { { { φ(x, 0) = 0
in ΩT , on 𝜕Ω × (0, T), in Ω,
and consider the translation in time φ(x, t) = φ(T − t, x) in such a way that −φt − Δφ = χBr (0)×[t1 ,t2 ] { { { φ(x, t) = 0 { { { { φ(x, T) = 0
in ΩT , on 𝜕Ω × (0, T), in Ω.
Take φ as a test function in (5.4.1). Then we have
t2
C ≥ ∫ ∫ un (x, t) dx dt t1 Br (0) T
= ∫∫ 0 Ω
T
upn
1 + n1 upn
T
φ dx dt + λ ∫ ∫ an (x)un φ dx dt + c ∫ ∫ fφ dx dt. 0 Ω
0 Ω
By the monotone convergence theorem, as n → ∞, upn
1 + n1 upn
→ up
an (x)un ↗
u x2
in L1loc (Br (0) × (t1 , t2 )), in L1loc (Br (0) × (t1 , t2 )).
Thus it follows that u is a very weak supersolution to (5.4.1) in Br1 (0)×(t 1 , t 2 ) ⊂⊂ Br (0)× (t1 , t2 ), a contradiction with Theorem 5.3.1. 5.4.2 Blowup when pn → p+ (λ) The following blowup result, for a sequence pn ↑ p+ (λ), turns out to be very interesting.
5.4 Instantaneous and complete blowup results  85
Theorem 5.4.2. Assume that pn satisfies pn < p+ (λ), pn → p+ (λ) as n → ∞ and f ≩ 0. Let un ∈ 𝒞 ([0, T); L1loc (Ω)) be a very weak supersolution to the problem p
u
unt − Δun ≥ λ xn2 + unn + f { { { un (x, t) = 0 { { { { un (x, 0) = 0
in ΩT , on 𝜕Ω × (0, T),
(5.4.2)
in Ω.
Then un (x0 , t0 ) → ∞, ∀(x0 , t0 ) ∈ Ω × (0, T). Proof. Without loss of generality we can again assume that f ∈ L∞ (ΩT ). Suppose by contradiction that there exists a subsequence denoted pn and a supersolution un such that for some point (x0 , t0 ) ∈ ΩT we have un (x0 , t0 ) → C0 < ∞, ∀n ∈ ℕ. Without loss 2 of generality we can assume that pn (λ) = 1 + 1 . Thanks to the classical Harnack α− (λ)+ n
inequality, there exists s > 0 and a positive constant C = C(N, s, t0 , β) such that un ≤ CC0 , ∫ ∫ un (x, t) dx dt ≤ C ess inf + R
R−
where R− = Bs (x0 ) × (t0 − 43 β, t0 − 41 β) and R+ = Bs (x0 ) × (t0 − 21 β, t0 + 21 β). As in the discussion of the proof of Theorem 5.4.1, we can suppose that 0 ∈ Bs (x0 ).
If un ∈ 𝒞 ([0, T); L1loc (Ω)) is a very weak supersolution to problem (5.4.2), then there exists a minimal solution to (5.4.2) in Ω1 × (t1 , t2 ) ⊂⊂ ΩT with 0 ∈ Ω1 ⊂⊂ Ω obtained by approximation. Let us denote by vn ≤ un this minimal solution. Then vn solves v
p
vnt − Δvn = λ xn2 + vn n + f { { { vn (x, t) = 0 { { { { vn (x, t1 ) = 0
in Ω1 × (t1 , t2 ), on 𝜕Ω1 × (t1 , t2 ), in Ω1 .
Let ϕ be the solution to the problem −ϕt − Δϕ = 1 { { { ϕ=0 { { { { ϕ(x, t1 ) = 0
in Ω1 × (t1 , t2 ), on 𝜕Ω1 × (t1 , t2 ), in Ω1 .
Using ϕ as a test function in (5.4.3) and since vn ≤ un , we get t2
t2
C ≥ ∫ ∫ vn (x, s) dx ds = ∫ ∫ gn (x, s)ϕ dx ds, t1 Ω1 v
where gn (x, t) = λ xn2 +
p vn n
t1 Ω1
+ f . Thus t2
∫ ∫ gn ϕ dx ds ≤ C t1 Ω1
for all n.
(5.4.3)
86  5 The Hardy–Leray potential in semilinear heat equations Hence, it follows that gn is uniformly bounded in L1loc (Ω1 × (t1 , t2 )) and then gn ⇀ μ in the sense of measures. Let φ ∈ 𝒞0∞ (Ω1 ). Taking Tk (vn )⋅φ as a test function in (5.4.3) and using the previous boundedness, we get t2
2 ∫ Θk (vn (x, t2 ))φ dx + ∫ ∫ ∇Tk (vn ) φ dx ds Ω1
t1 Ω1
t2
t2
+ ∫ ∫ Θk (vn )(−Δφ) dx ds = ∫ ∫ gn (x, s)φTk (vn ) dx ds, t1 Ω1
t1 Ω1
s
where Θk (s) = ∫0 Tk (σ)dσ. Thus, t2
2 ∫ Θk (vn (x, t2 ))φ dx + ∫ ∫ ∇Tk (vn ) φ dx ds Ω1
t1 Ω1
t2
t2
≤ c ∫ ∫ Θk (vn ) dx ds + k ∫ ∫ gn (x, s)φ dx ds ≤ C. t1 Ω1
t1 Ω1
Hence, there exists a nonnegative function v such that Tk (vn ) ⇀ Tk (v),
1,2 weakly in L2loc ((t1 , t2 ), Wloc (Ω1 )).
Moreover, by the results in [66], we obtain vn → v strongly in Lqloc ((t1 , t2 ), Lqloc (Ω1 )) ≡ N Lqloc (Ω1 × (t1 , t2 )) with q < N−2 . that
Let ψ ∈ 𝒞0∞ (Br (0)×(t1 , t2 )) be a nonnegative function. By the Fatou lemma it follows t2
lim ∫ ∫ gn (x, s)ψ dx ds
n→∞
t1 Br (0) t2
t2
t1 Br (0)
t1 Br (0)
≥ ∫ ∫ vp+ (λ) ψ dx ds + λ ∫ ∫
t2
vψ dx ds + ∫ ∫ fψ dx ds. x2 t1 Br (0)
Therefore, using ψ as a test function in (5.4.3) and passing to the limit as n → ∞, it follows that t2
t2
− ∫ ∫ vψt dx ds + ∫ ∫ v(−Δψ) dx ds t1 Br (0) t2
t1 Br (0)
t2
≥ ∫ ∫ vp+ (λ)ψ dx ds + λ ∫ ∫ t1 Br (0)
t1 Br (0)
t2
vψ dx ds + ∫ ∫ fψ dx ds. x2 t1 Br (0)
5.5 Existence of solutions: p < p+ (λ)
 87
Hence v is a very weak supersolution to (5.1.1) obtained by approximation and then we reach a contradiction. Remark 5.4.3. We can also obtain nonexistence results by using the local existence result in Lemma 5.2.4 and some contradiction with the Hardy inequality. Note also that a nonexistence result and Lemma 5.2.4 give us complete and instantaneous blowup.
5.5 Existence of solutions: p < p+ (λ) The goal of this section is to study the complementary interval of powers 1 < p < p+ (λ) and prove that under some suitable hypotheses on f and u0 , problem (5.1.1) has a positive solution. For the existence result we will consider the case f ≡ 0. For the case f ≢ 0, see Remark 5.5.4 below. First of all, recall that if 0 < λ ≤ ΛN,2 and 1 < p < p+ (λ), the stationary problem − Δu = λ
u + up in Ω, x2
u = 0 on 𝜕Ω,
(5.5.1)
has a positive very weak supersolution w in the following cases: (A) For 0 < λ < ΛN,2 , we have the following. N+2 (I) If 1 < p < N−2 and Ω is a bounded domain, there exists a positive solution to problem (5.5.1) using the classical mountain pass theorem in the Sobolev space W01,2 (Ω). See [89]. N+2 (II) If N−2 < p < p+ (λ), there exists a positive solution in ℝN as was found in Chapter 4. N+2 (B) If λ = ΛN,2 , then p+ (ΛN,2 ) = N−2 = p− (ΛN,2 ). Let H(Ω) be the Hilbert space comple∞ tion of 𝒞0 (Ω) with respect to the norm ‖ϕ‖2 = ∫(∇ϕ2 − ΛN,2 Ω
ϕ2 ) dx, x2
N+2 = 2∗ − 1, classical variational introduced in Chapter 2, Section 2.5.1. Since p < N−2 methods in the space H(Ω) allow us to prove the existence of a positive solution w to the stationary problem (5.5.1). See details in [181].
Theorem 5.5.1. Assume that 0 < λ ≤ ΛN,2 and 1 < p < p+ (λ). Suppose that u0 (x) ≤ w, where w is a supersolution to the stationary problem u −Δu = λ 2 + up in Ω, u(x) = 0 on 𝜕Ω. x Then for all T > 0, the problem u − Δu = λ xu 2 + up { { t u(x, t) = 0 { { { u(x, 0) = u0 (x) has a global solution.
in ΩT ≡ Ω × (0, T), on 𝜕Ω × (0, T), if x ∈ Ω
(5.5.2)
88  5 The Hardy–Leray potential in semilinear heat equations Proof. The proof uses the classical subsupersolution argument (see Lemma 5.2.4). Remark 5.5.2. With the results above we find the optimality of the power p+ (λ), which is one of the main objectives. Nevertheless, it will be interesting to know the optimal class of initial data for which there exists a solution and the regularity of such solutions according to the regularity of the data. It is also an interesting question to know the asymptotic behavior as t → ∞ of such solutions. In this direction we have the following necessary condition. Proposition 5.5.3. Assume that λ ≤ ΛN,2 . If problem (5.1.1) has a very weak supersolution, then there exists r > 0 such that the initial value function, u0 , satisfies ∫ x−α− (λ) u0 (x) dx < ∞,
where α− (λ) is defined in (3.1.3).
Br (0)
Proof. We argue by contradiction. Suppose that u is a very weak supersolution to (5.1.1) with u(x, 0) = u0 (x) in Ω such that ∀r > 0, ∫B (0) x−α− (λ) u0 (x) dx = ∞. We consider the r sequence {un } of minimal solutions to the problems u
unt − Δun = λ 2 n 1 + upn { x + n { { { un (x, t) = 0 { { { { { un (x, 0) = Tn (u0 (x))
in Ω × (0, T), (5.5.3)
on 𝜕Ω × (0, T), in Ω,
with 1 < p < p+ (λ) and T depending just on the supersolution. For every n, let us consider the positive eigenfunction φn to the associated problem −Δφn −
λφn
x2 +
1 n
= cn φn ,
with ∫ φ2n dx = 1. Br (0)
Up to a subsequence we find that φn → φ in Ls (Ω), s < −Δφ −
λφ = cφ, x2
2N N−2
and cn → c with
∫ φ2 dx = 1 and φ ≥ Cx−α− (λ) , Br (0)
in a neighborhood of the origin. Define ψn = ψn as a test function in (5.5.3), we get
φn ; then ψn ‖φn ‖1
→ψ=
φ ‖φ‖1
in L1 (Ω). Taking
d ∫ ψn un (x, t) dx + cn ∫ un (x, t)ψn dx ≥ ∫ upn ψn dx. dt Br (0)
Br (0)
Br (0)
Since p > 1, we can apply Jensen’s inequality and conclude that p
d ∫ ψn un (x, t) dx + cn ∫ ψn un (x, t) dx ≥ ( ∫ ψn un (x, t) dx dt) . dt Br (0)
Br (0)
Br (0)
Note that if Yn (t) = ∫B (0) ψn un (x, t) dx, then Yn (t) + cYn (t) ≥ [Yn (t)]p . r
5.6 Cauchy problem  89
Using the hypothesis on u0 and the definition of ψn it follows that limn→∞ Yn (0) = ∞. We set Zn (t) = ect Yn (t). Then Zn (t) ≥ e−c(p−1)t Znp (t). Thus Zn (t) is an increasing function and therefore Zn (t) ≥ Zn (0) = Yn (0) → ∞
as n → ∞ uniformly in t ∈ (0, T).
Integrating the differential equation of Zn , it follows that 1 1 1 1 − )≥ ( (1 − e−c(p−1)t ), p − 1 Znp−1 (0) Znp−1 (t) c(p − 1)
t > 0.
Hence we conclude that Ynp−1 (0) = Znp−1 (0) ≤
1−
c
e−c(p−1)t
< ∞,
∀t > 0,
a contradiction with the hypothesis on the initial datum. Hence the result follows. Remark 5.5.4. 1. An open question that is left here is whether or not the previous necessary condition is also sufficient for local existence. c0 (t) 2. In the presence of a source term f ≩ 0, if f (x, t) ≤ x 2 with c0 (t) sufficiently small, then the above computation allows us to prove the existence of a supersolution. Hence, the existence of a minimal solution to problem (5.1.1) follows for all p < p+ (λ).
5.6 Cauchy problem In [171], H. Fujita considered the initial value problem {
ut = Δu + up ,
x ∈ ℝN , t > 0,
u(x, 0) = u0 (x) ≥ 0,
x ∈ ℝN ,
(5.6.1)
where 1 < p < ∞. Fujita showed that if 1 < p < 1 + N2 , then for all positive initial datum u0 there exists T > 0 such that the solution to problem (5.6.1) satisfies ‖u(⋅, tn )‖∞ → ∞ as tn → T. Moreover, if p > 1 + N2 , then there are both global solutions for small data and nonglobal solutions for large data. The number F(0) = 1 + N2 is often called the critical Fujita blowup exponent for the heat equation. Moreover it is proved that for p = 1 + N2 , a suitable norm of the solution goes to infinity in a finite time. We refer to [316] for a simple proof of this last fact. In this section we study the Fujita type behavior for the Cauchy problem with the Hardy potential. More precisely, we consider ut − Δu = λ
u + up in ℝN , x2
u(x, 0) = u0 (x) ≩ 0 in ℝN ,
(5.6.2)
90  5 The Hardy–Leray potential in semilinear heat equations with 0 < λ ≤ ΛN,2 and 1 < p < p+ (λ). Since, clearly, u ∉ L∞ , in our case we have the blowup in L∞ for free. The problem to study is how to obtain the blowup in a suitable Lebesgue space with a weight. r2
N
We have found in Section 5.2.1 that v(r, t) = t − 2 +α− (λ) r −α− (λ) e− 4t is a selfinvariant solution to the homogenous linear equation (5.2.4), that is, vt − Δv − λ
v =0 x2
in ℝN .
Moreover, ∫ℝN x−α− (λ) v(x, t) dx = C. In particular, for λ = 0 (equivalently α− (λ) = 0), we get the fundamental solution of the heat equation. Incidentally, it is worthy to point out that, in particular for λ = 0, our arguments below give a proof for the Fujita result on the semilinear heat equation.
5.6.1 Subsolution blowup in a finite time for small p According to the behavior of the selfsimilar solution v to (5.2.4) and since any positive solution to (5.6.2) is a supersolution to (5.2.4), it is natural to formulate the following definition. Definition 5.6.1. Considering u(x, t), a positive solution to (5.6.2), we say that u blows up in a finite time if there exists T ∗ < ∞ such that lim ∫ x−α− (λ) u(x, t) dx = ∞,
t→T ∗
Br (0)
for any ball Br (0). In order to find a Fujita type exponent for (5.6.2) where the blowup phenomenon occurs in the sense of Definition 5.6.1, we follow closely the arguments used in [175] which are in some way constructive. More precisely, for r = x, we look for a family of r subsolutions to problem (5.6.2) in the form w(r, t, T) = (T − t)−θ ζ ( (T−t) β ), with θ, β > 0 to be chosen and ζ > 0 a smooth function. Denoting s = wt (r, t) = (T − t)−θ−1 (θζ +
r , (T−t)β
it follows that
βr ζ (s)), (T − t)β
wr (r, t) = (T − t)−θ−β ζ (s), wrr (r, t) = (T − t)−θ−2β ζ (s).
(5.6.3)
We need to have wt − wrr −
(N − 1) w wr − λ 2 ≤ wp . r r
(5.6.4)
5.6 Cauchy problem  91
By setting θ =
1 , p−1
β=
1 2
and replacing the value in (5.6.3), (5.6.4) becomes
−ζ (s) − (
λ N −1 1 − s)ζ (s) − ( 2 − θ)ζ (s) ≤ ζ p (s). s 2 s s2
We will try to find ζ (s) of the form ζ (s) = Aϕ(cs) with ϕ(s) = s−α− (λ) e− 4 , A > 0 and c > 0. Since α− (λ)2 − (N − 2)α− (λ) + λ = 0, we need the inequality −(cs)2 (
1 2 α (λ) c2 (N − 1) c2 1 c2 1 + ) + − c2 α− (λ) − − + + ≤ Ap−1 (cs)−α− (λ)(p−1) e− 4 (cs) (p−1) 4 4 2 2 2 p−1
to be satisfied. Fixing c = 1, there exists s0 such that for s ≥ s0 , γ1 (s) ≡ −s2 (
α (λ) (N − 1) 1 1 1 1 + ) + − c2 α− (λ) − − + + ≤ 0. 4 4 2 2 2 p−1
Set M = max{γ1 (s)  0 ≤ s ≤ s0 }; for A large enough we conclude that 1 2
M < Ap−1 s−α− (λ)(p−1) e− 4 s (p−1)
if 0 ≤ s ≤ s0 .
Then for such an A 1 2
γ1 (s) ≤ Ap−1 s−α− (λ)(p−1) e− 4 s (p−1) . That is, we find a family of subsolutions to (5.6.2) given by 1 − p−1
w(r, t, T) = A(T − t)
(
−α− (λ)
r (T − t)
1 2
)
1 r2
e− 4 T−t .
(5.6.5)
Let Br (0) be a ball in ℝN . Integrating we obtain 1 − p−1 + N2
∫ x−α− (λ) w(x, t, T) dx = C(T − t)
α (λ) − −2
∫ ϕ(s)sN−α− (λ)−1 ds. 0
Br (0) 1 Note that − p−1 +
r (T−t)1/2
N 2
−
α− (λ) 2
Therefore if p < 1 +
< 0 implies p < 1 +
2 , N−α− (λ)
we find
2 . N−α− (λ)
lim ∫ x−α− (λ) w(x, t, T) dx = ∞.
t→T
Br (0)
Hence a candidate to the Fujita exponent for problem (5.6.2) is F(λ) = 1 + clear that if 0 < λ ≤ ΛN,2 , then 1 0 to be
chosen, and where g is a smooth bounded positive function. Denoting s = follows that wt (r, t) = −(T + t)−θ−1 (θg +
r , (T+t)β
it
βr g (s)), (T + t)β
wr (r, t) = (T + t)−θ−β g (s), wrr (r, t) = (T + t)−θ−2β g (s). In order to get homogeneity in equation (5.6.6), it is sufficient to choose θ = β=
1 . 2
Therefore (5.6.6) gives g (s) + (
1 p−1
and
λ N −1 1 + s)g (s) + ( 2 + θ)g(s) + g p (s) ≤ 0. s 2 s
Let us consider α− (λ) < γ < c > 0, then we need
2 . p−1
s2
If we take g(s) = Aϕ(cs), with ϕ(s) = s−γ e− 4 , A > 0,
c2 [γ 2 − (N − 2)γ + λ] c2 1 + (cs)2 [ − ] 2 4 4 (cs) + c2 γ −
c2 c2 (N − 1) γ 1 − − + 2 2 2 p−1
+ Ap−1 (cs)−(p−1)γ e− Let G(c, s) be defined as G(c, s) = c2 (s −
N ) 2
G(1, α− (λ)) = −
−
s 2
(cs)2 4
(p−1)
+
1 . p−1
≤ 0. Since p > F(λ), we have
Nα− (λ) 1 + < 0, 2 p−1
and hence by continuity there exist c < 1 and α− (λ) < γ < 2
c < 1, we obtain (cs)2 [ c4 − 41 ] ≤ 0. Moreover, since γ < choosing A small enough it follows that
2 p−1
2 p−1
such that G(c, γ) < 0. As
and c2 [γ 2 − (N − 2)γ + λ] ≤ 0,
(cs)2 c2 [γ 2 − (N − 2)γ + λ] + Ap−1 (cs)−(p−1)γ e− 4 ≤ 0. 2 (cs)
5.6 Cauchy problem 
93
Therefore, we have found a family of supersolutions w defined by γ
w(r, t, T) = Ac−γ (T + t) 2 Note that w(x, 0, T) = AT
2 x2 4
1 − p−1 −γ
a x−γ e−a
2
1 r − p−1 −γ −c2 4(T+t)
r e
, where a =
c
1
T2
(5.6.7)
. < 1.
Remark 5.6.2. Now we can conjecture that the Fujita exponent is F(λ) = 1 + The next sections will show that this conjecture is true.
2 . N−α− (λ)
5.6.3 Local existence results for 1 < p < p+ (λ) We prove that problem (5.6.2) has a local positive solution for a class of suitable initial data. The main result is the following theorem. Theorem 5.6.3. Assume that 1 < p < p+ (λ) and u0 (x) ≤ T −θ x−β , with T > 0,
α− (λ) < β < α+ (λ) such that β + α− (λ) < N.
1,2 Then problem (5.6.2) has a local positive solution u ∈ L2 (0, T; Wloc (ℝN )). 1 −β Proof. Let w(x, t) = (T−t) with α− (λ) < β < α+ (λ) and β + α− (λ) < N. It follows that θ x β(N − 1) − β(β + 1) − λ > 0. Since p < p+ (λ), we have β < pβ < β + 2. Hence for T > 1, choosing θ large, we get easily that w is a supersolution to (5.6.2) in ℝN × (0, T). It is clear by the hypothesis on u0 that u0 (x) ≤ w(x, 0). To prove the existence of a solution u we follow by approximation. Let Bn be the ball in ℝN with radius n and centered at the origin. We consider
vn ∈ L2 ((0, T), W01,2 (Bn+1 )),
∀T > 0,
the weak solutions to the following approximated problems: p vnt − Δvn = λ 21 1 ṽn−1 + ṽn−1 { x + n { { { vn (x, 0) = u0 (x) { { { { { vn (x, t) = 0
in Bn+1 , t > 0, in Bn+1 , t > 0,
(5.6.8)
on 𝜕Bn+1 , t > 0,
with v0t − Δv0 = 0 { { { v0 (x, 0) = u0 (x) { { { { v0 (x, t) = 0
in B1 , t > 0, in B1 , t > 0, on 𝜕B1 , t > 0,
and ṽn−1 = vn−1 in Bn , ṽn−1 = 0 in ℝN \ Bn . Applying the classical comparison principle, we conclude that 0 < v0 ≤ v1 ≤ ⋅ ⋅ ⋅ ≤ vn−1 ≤ vn ≤ w in Bn+1 × (0, T1 ) with T1 < T. Hence
94  5 The Hardy–Leray potential in semilinear heat equations there exists u ∈ L2 (0, T1 , L2loc (ℝN )) such that vn ↑ u strongly in L2 ((0, T1 ), L2loc (ℝN )) and u ≤ w. Using the monotonicity of vn and the dominated convergence theorem it follows that vn → u strongly Lp (K × (0, T1 )) for all compact sets K ⊂ ℝN . Take ϕ ∈ 𝒞0∞ (ℝN × (0, T1 )). Then using ϕ as a test function in (5.6.8) and by letting n → ∞, we easily get that u solves problem (5.6.2) with u(x, 0) = u0 (x). It is clear that also 1,2 (ℝN )). u ∈ L2 (0, T1 ; Wloc 5.6.4 Global existence for F (λ) < p < p+ (λ) and small data In this subsection we consider the class of initial data 2
α− (λ) −D2 x4
N
ℱD = {u0 : ℝ → ℝ  0 ≤ u0 (x) ≤ Ax
e
}.
Let w(r, t, T) be the family of supersolutions founded in (5.6.7). Choosing D such that 2(γ − α) ) ( 2 D − a2
γα− (λ) 2
e−
γα− (λ) 2
≤ AT
1 − p−1 −γ
a ,
we get u0 (x) ≤ w(x, 0, T). Thus w is a supersolution to (5.6.2). It is clear that w ∈ L2 (0, T; W 1,2 (ℝN )) for all T > 0. Hence using an iteration argument as in Section 5.6.3 we get the existence of a global solution u to problem (5.6.2) with u ≤ w. We also have u ∈ L2 (0, T; W 1,2 (ℝN )) for all T > 0 and u(x, t) → 0 as t → ∞ at least in the pointwise sense. Therefore we obtain the following result. Theorem 5.6.4. Assume that u0 ∈ ℱD . Then for F(λ) < p < p+ (λ), problem (5.6.2) has a minimal positive global energy solution u ∈ L2 (0, T; W 1,2 (ℝN )) for all T > 0 such that u(x, t) → 0 for t → ∞. Remark 5.6.5. It would be very interesting to look for a general class of initial data such that problem (5.6.2) has a global solution. The imposed condition in Theorem 5.6.4 is far from optimal.
5.6.5 Blowup result for p < F (λ) In this subsection we prove that F(λ) behaves like a Fujita type exponent, namely, we will prove that, for p < F(λ), any solution to (5.6.2) blows up in a finite time in an appropriate sense and for any initial datum. Let us begin by analyzing the properties of the subsolutions w(r, t, T) defined in (5.6.5). (i) By construction, w(r, t, T) blows up in a finite time in the sense of the local weighted L1 norm.
5.6 Cauchy problem  95
(ii) Assume that p < F(λ) = 1 + N−α2 (λ) . Let us denote by u(x, t) a time translation of a − solution to (5.6.2), namely, if u(x, t) is a solution to (5.6.2), then u(x, t) = u(x, t + T). Since u(x, t) is a supersolution to the homogenous equation (5.2.4) with the same initial values, it is sufficient to check that v(x, T) ≥ w(r, 0, T), in order to get that u(x, 0) ≥ w(r, 0, T). This immediately follows because p < F(λ) = 1 + N−α2 (λ) . −
(iii) Consider u(x, t) ≥ w(x, t), ∀t < T. Let us write h(x, t) = w(x, t) − u(x, t). Then, ht − Δh ≤ λ xh 2 + wp − up . Applying Kato’s inequality (see [255]), it follows that h+t − Δh+ ≤ λ
h+ + pwp−1 h+ x2
in ℝN , t ∈ (0, T1 ), T1 < T,
h+ (x, 0) = 0.
(5.6.9) h2
2
w p+1 Note that h+ ≤ w. Since λ x ∈ L1 (ℝN × (0, T1 )), we have λ x+2 + pwp−1 h2+ ∈ 2 + pw
L1 (ℝN × (0, T1 )). Therefore using Tk (h+ ) as a test function in (5.6.9), it follows that T1
T1
2
∫ Θk (h+ )(x, T1 ) + ∫ ∫ ∇Tk (h+ ) dx dt ≤ ∫ ∫ (λ 0 ℝN
ℝN
0 ℝN
h2+ + pwp−1 h2+ ) dx dt ≤ C x2
for all k > 0. Hence, letting k → ∞ we obtain h+ ∈ L2 (0, T1 , W 1,2 (ℝN )). Since 2 p < 1 + N−α , we have α1 (p − 1) < 2 and then there exists C(T, T1 ) such that ∀ϵ > 0, 1
wp−1 ≤ ϵ x1 2 + C(T, T1 ). Taking h+ as a test function in (5.6.9), we get 𝜕 ∫ h2+ (x, t) dx ≤ C(T) ∫ h2+ (x, t) dx. 𝜕t ℝN
ℝN
Applying Gronwall’s inequality, we immediately deduce that h+ ≡ 0 and then u(x, t) ≥ w(x, t), ∀t < T. Since we have found a family of subsolutions that has finite blowup, we can deduce that for p < F(λ) = 1 + N−α2 (λ) , the solutions to (5.6.2) blow up in a finite time. − Hence we have shown the following result. Theorem 5.6.6. Assume that p ≤ F(λ) = 1 + the sense of Definition 5.6.1.
2 . N−α− (λ)
Then u blows up in a finite time in
The critical case p = F(λ) is considered in the last section. For small data
Blowup in a finite time
1+
2 N
F =1+
For large data
2 N − α1
p− (λ)
→ →
Global existence
Nonglobal existence
2∗ − 1
Fujita exponent for the heat equation with Hardy potential.
Nonexistence
p+ (λ)
96  5 The Hardy–Leray potential in semilinear heat equations
5.7 Further results and remarks 5.7.1 Problems involving the pLaplacian heat equation In the papers [30] and [179] the following types of problems were considered: ut − Δp u = xλ p up−2 u + μuq−1 u { { { u(x, 0) = f (x) { { { { u(x, t) = 0
x ∈ Ω, t > 0, λ ∈ ℝ, x ∈ Ω,
(5.7.1)
x ∈ 𝜕Ω, t > 0,
where −Δp u ≡ − div(∇up−2 ∇u), f (x) ≥ 0 satisfying convenient regularity assumptions, Ω is a bounded domain in ℝN such that 0 ∈ Ω, q > 0 and 1 < p < N. The elliptic part of each problem is the Euler–Lagrange equation of the Hardy inequality obtained in Section 2.4. The analysis reveals that the behavior depends on p, more precisely on the Harnack inequality for the parabolic operator. This means that there is a different behavior according to whether p ≥ 2 or 1 ≤ p < 2. See for instance [18] and [130] for a more general class of problems involving the Caffarelli–Kohn–Nirenberg inequalities. We recall that for 1 ≤ p < 2 there is the finite time extinction of solutions. For a general study of the pLaplacian heat equation the reader is referred to the classical book by E. Di Benedetto [134]. The existence and regularity results depend in general on the relation between λ and the best constant in Hardy’s inequality defined in Lemma 2.4.1. We omit more details about these kinds of problems and we refer to the quoted articles. 5.7.2 Problems associated to the Caffarelli–Kohn–Nirenberg inequalities The paper by L. Caffarelli, R. Kohn and L. Nirenberg [107] appeared in the year 1982. It contains the closest result up to now to solve the conjecture by J. Leray in [225]. More precisely, what they proved was that the set of singularities, S, of a class of solutions of the Navier–Stokes equations verifies 1
ℋ (S) = 0,
where ℋ1 denotes the Hausdorff 1measure. As a byproduct of the arguments in the proof of this result, the authors also proved in [108] a family of weighted inequalities, often known as Caffarelli–Kohn–Nirenberg inequalities. See also Section 6.2.3. In the article [130] one of the authors analyzed the family of problems p−2
u u − div(x−pγ ∇up−2 ∇u) = λ xu p(γ+1) , { { { t u(x, t) = 0, { { { { u(x, 0) = ψ(x) ≥ 0,
in Ω × (0, ∞), 0 ∈ Ω, on 𝜕Ω × (0, ∞)
(5.7.2)
when 1 < p < N, −∞ < (γ + 1) < Np , under certain hypotheses for the initial data. This family includes linear and quasilinear parabolic evolution problems. Note that
5.7 Further results and remarks  97
if p = 2 and γ = 0 we find again the heat equation perturbed by the Hardy–Leray potential. One can also find results for nonlinear reaction terms in the literature; see for instance [16] and [18]. 5.7.3 The borderline case p = F (λ) We consider in this section the borderline case p = F(λ). The argument will be different and technically more complicated than in the subcritical case. We follow some technical ideas used in [313]. We argue by contradiction. Assume that u is a global solution to problem (5.6.2) in such a way that for some ball Br (0), ∫ x−α− (λ) u(x, t) dx < ∞ for all t > 0. Br (0)
We set v(x, t) = xα− (λ) u(x, t). Therefore v satisfies x−2α− (λ) vt − div(x−2α− (λ) ∇v) = xα− (λ)(p+1) vp ,
(5.7.3)
with ∫ x−2α− (λ) v(x, t) dx < ∞
for all t > 0.
(5.7.4)
Br (0)
1.
2.
Choose θ such that
1 p
< θ < 1. Let ψ ∈ 𝒞 2 (ℝN ) be a cutoff function such that:
ψ = 1 in B1 (0), ψ = 0 in ℝN \B2 (0) and 0 ≤ ψ ≤ 1;
Δψ ≤ C(θ)ψθ .
For each l ∈ ℕ, consider the scaled cutoff function ψl (x) = ψ( xl ). Using ψl as a test function in (5.7.3), it follows that d ∫ x−2α− (λ) vψl dx = − ∫ v div(x−2α− (λ) ∇ψl ) dx + ∫ xα− (λ)(p+1) vp ψl dx. dt ℝN
ℝN
(5.7.5)
ℝN
In the sequel we denote by C any positive constant that is independent of v and l ∈ ℕ. Using the Hölder inequality, we easily find ∫ vdiv(x−2α− (λ) ∇ψl ) dx ℝN
≤ ( ∫ x
α− (λ)(p+1) p
1 p
v ψl dx) ( ∫
≤ Cl
Nα− (λ) −2 p
x
l≤x≤2l
ℝN α− (λ)(p+1) p
( ∫ x ℝN
1 p
v ψl dx) ,
p+1 α (λ) p−1 −
 div(x−2α− (λ) ∇ψl )p
1
ψlp−1
dx)
1 p
98  5 The Hardy–Leray potential in semilinear heat equations where in the last estimate we have used the fact that for 1 ≤ l ≤ x ≤ 2l,  div(x−2α− (λ) ∇ψl )p
1
ψlp−1
p
θp − 1  div(x−2α− (λ) ∇ψl ) ) ψl p−1 =( θ ψl
≤ C(θ)x−2p α− (λ) l−2p ≤ C(θ)l−2p (α− (λ)+1) .
Note that since Then
1 p
< θ < 1, we have θp −
1 p−1
> 0. We call wl (t) = ∫ℝN x−2α− (λ) vψl dx.
p−1 Nα− (λ) 1 d −2 + (xα− (λ)(p+1) vp ψl dx) p ]. wl (t) ≥ (intℝN xα− (λ)(p+1) vp ψl dx) p [−Cl p dt
(5.7.6)
Using again the Hölder inequality, it follows that ∫ xα− (λ)(p+1) vp ψl dx ≥ C0 wlp (t)l−(p−1)(Nα− (λ)) .
(5.7.7)
ℝN
We finally obtain Nα (λ) Nα− (λ) d − − −2 − p−1 (Nα (λ)) wl (t) ≥ Cwl (t)l p [−Cl p + wlp−1 (t)l p − ]. dt
Since p = 1 +
2 , Nα− (λ)
we have
Nα− (λ) p
− 2 = − p−1 (Nα− (λ)), and hence, p
d w (t) ≥ C1 wl (t)l−2 (−C2 + wlp−1 (t)). dt l
(5.7.8)
Note that wl is an increasing function in l. Under the hypothesis that u is a global solution, from (5.7.8) necessarily wlp−1 (t) ≤ C2
for all t > 0 and l > 0.
In fact, by contradiction, if for some ϵ > 0, there exist t0 > 0, l0 > 0 such that wlp−1 (t0 ) ≥ 0
dw
C2 + ϵ, then for any l ≥ l0 we have dt l > 0 in the time in which wl is defined. Consider now the solution to the initial value problem y (t) = C1 y(t)l−2 (−C2 + yp−1 (t)),
1
y(t0 ) = (C2 + ϵ) p−1 .
Since the solution blows up in an explicit finite time, T, obtained by elementary integration and by classical comparison arguments, we conclude that there exists T ∗ ≤ T, such that wl (t) ↑ ∞ as t ↑ T ∗ .
5.7 Further results and remarks  99
But this is a contradiction with (5.7.4), which proves the uniform estimate for wl . Moreover, since C2 is independent of l, by letting l → ∞ we obtain ∫ x−2α− (λ) v(x, t) dx ≤ C2
for all t > 0.
(5.7.9)
ℝN
Also we have ∫ vdiv(x−2α− (λ) ∇ψl ) dx ≤
v
∫ l≤x≤2l
ℝN
≤
 div(x−2α− (λ) ∇ψl ) θ ψl dx ψθl
C(θ) ∫ x−2α− (λ) v(x, t) dx ≤ Cl−2 . l2 ℝN
Hence from (5.7.5), by letting l → ∞ we conclude that 1 d ∫ x−2α− (λ) v(x, t) dx ≥ ∫ xα− (λ)(p+1) vp (x, t) dx, dt 2 ℝN
(5.7.10)
ℝN
in the distributional sense. For τ ∈ (0, 1), we write (5.7.8) as dwl ≥ C1 w1−τ l−2 (−C2 wlτ + wlp−(1−τ) ), dt
(5.7.11)
in the distributional sense. Next, consider ϕl (x) = 1 − ψl (x) and a cutoff function, ξ ∈ 𝒞0∞ (ℝN ), such that 0 ≤ ξ ≤ 1, ξ (x) = 1 for x ≤ 2 and ξ (x) = 0 if x ≥ 3. Define ξk (x) = ξ ( kx ). Using ϕl ξk as a test function in (5.7.3) it follows that d ∫ x−2α− (λ) vϕl ξk dx ≤ ∫ + ∫ xα− (λ)(p+1) vp ϕl ξk dx dt ℝN
ℝN
ℝN
≤ ∫ vdiv(x−2α− (λ) ∇ϕl ) dx + ∫ vdiv(x−2α− (λ) ∇ξk ) dx ℝN
ℝN
+ ∫ xα− (λ)(p+1) vp ϕl ξk dx. ℝN
Since ∇ϕl = −∇ψl , by similar methods as above, we obtain 1
Nα (λ) p −2+ − p ( ∫ xα− (λ)(p+1) vp dx) . ∫ vdiv(x−2α− (λ) ∇ϕl ) dx ≤ Cl
ℝN
ℝN
Now for k ≫ l we have ∫ vdiv(x−2α− (λ) ∇ξk ) dx ≤ Ck −2 ∫ x−2α− (λ) vϕl dx. ℝN
ℝN
100  5 The Hardy–Leray potential in semilinear heat equations Thus, we have d ∫ x−2α− (λ) vϕl ξk dx dt ℝN
≤ Cl
Nα− (λ) p
−2+
(∫ ℝN
1 p
vp
xα− (λ)(p+1)
dx) + Ck −2 ∫ ℝN
vϕl dx + ∫ xα− (λ)(p+1) vp ϕl ξk dx. x2α− (λ) ℝN
Therefore by letting k → ∞ and by Young’s inequality we have d ∫ x−2α− (λ) vϕl dx ≤ 2 ∫ xα− (λ)(p+1) vp dx + C3 l−2p +Nα− (λ) . dt
ℝN
(5.7.12)
ℝN
Consider A = sup wl (t) = sup ∫ x−2α− (λ) v(x, t) dx. t>0
t>0,l>0
ℝN
Then, using (5.7.9) and the fact that u is a global solution, we obtain 0 < A < ∞. Since wl is increasing in l, then for all ϵ > 0 there exist t0 ≥ 0 and l0 ≥ 2 such that w l0 (t0 ) ≥ A − ϵ. 2
From (5.7.10) and by integrating in time, we obtain ∞
1 ∫ ∫ xα− (λ)(p+1) vp dx dt ≤ sup ∫ x−2α− (λ) v(x, t) dx − w l0 (t0 ) ≤ ϵ. 2 2 t>0 t0 ℝN
ℝN
Let s ≥ t0 . By (5.7.12), a direct computation provides ∫ x−2α− (λ) v(x, s)ϕ l0 (x) dx ℝN
2
s
≤ 2 ∫ ∫ x
−2p +Nα− (λ)
l v dx + C4 ( 0 ) 2
α− (λ)(p+1) p
t0 ℝN
−2p +Nα− (λ)
l ≤ 3ϵ + C4 ( 0 ) 2
(s − t0 ) + ∫ x−2α− (λ) v(x, s)ϕ l0 (x) dx 2
ℝN
(s − t0 ).
Now setting l = l0 in (5.7.11) it follows that dwl0 dt
≥ C1 w1−τ l0−2 (−C2 (
τ
∫
l0 ≤x≤2l0
x−2α− (λ) v dx) + wlp−(1−τ) ) 0
−2p +Nα− (λ)
l ≥ C1 w1−τ l0−2 (−C2 (3ϵ + C4 ( 0 ) 2
τ
(s − t0 )) + wlp−(1−τ) ) 0
5.7 Further results and remarks  101
in (t0 , ∞) in the distributional sense. Fix now ϵ0 ∈ (0, A) and M > 0 such that C2 (3ϵ + M)τ ≤ 21 (A − ϵ0 )p−(1−τ) . Then, by letting 2p −(Nα− (λ))
M l0 t1 = t0 + ( ) C4 2
,
it follows that dwl0 (t) dt
≥
C −2 p l w (t) 2 0 l0
for all t ∈ (t0 , t1 ),
in the distributional sense. After integration in time and using the fact that wl0 (t) is increasing for t ∈ (t0 , t1 ), we obtain wl0 (t1 ) ≥ wl0 (t0 ) +
C −2 l (A − ϵ0 )p (t1 − t0 ) 2 0 2p −(Nα− (λ))
M l C ≥ wl0 (t0 ) + l0−2 (A − ϵ0 )p ( 0 ) 2 C3 2 Since p = 1 +
2 , Nα− (λ)
.
we have 2p − (Nα− (λ)) = 2, and therefore wl0 (t1 ) ≥ wl0 (t0 ) +
C M (A − ϵ0 )p 2−2p +(Nα− (λ)) . 2 C3
We set ϱ = C2 (A − ϵ0 )p CM 2−2p +(Nα− (λ)) . Then
3
wl0 (t1 ) ≥ wl0 (t0 ) + ϱ ≥ A − ϵ0 + ϱ. As wl is an increasing function in l, using the same argument as above, we find w2l0 (t2 ) ≥ w2l0 (t1 ) + ϱ ≥ A − ϵ0 + 2ϱ, where t2 = t1 +
M 2p −(Nα− (λ)) l . C4 0
Hence by an iteration argument it follows that w2i−1 l0 (ti ) ≥ w2i−1 l0 (t1 ) + iϱ ≥ A − ϵ0 + iϱ, where ti = ti−1 +
M i−2 2p −(Nα− (λ)) (2 l0 ) . C4
102  5 The Hardy–Leray potential in semilinear heat equations Finally we conclude that sup ∫ x−2α− (λ) v(x, t) dx ≥ w2i−1 l0 (ti ) ≥ iϱ → ∞ t>0
as i → ∞,
ℝN
a contradiction with (5.7.9). Therefore, for all initial data u0 (x) ≥ 0, u0 (x) ≢ 0, there exists T < ∞ such that lim ∫ x−α1 u(x, t) dx = +∞. t↑T
Br (0)
6 Elliptic equations with a nonlinearity on the gradient and the Hardy–Leray potential 6.1 Introduction In this chapter we study the solvability of the quasilinear elliptic equation − Δu = ∇uq + λ
u + cf , x2
x ∈ Ω ⊂ ℝN , N ≥ 3, 1 < q ≤ 2,
(6.1.1)
where Ω is a domain such that 0 ∈ Ω and the existence of solutions in particular for the Dirichlet problem. We assume that λ, c are positive real numbers and f is a nonnegative function under some extra hypotheses that will be presented later. As in the previous chapters, we will see that the behavior of equation (6.1.1) is deeply related to the values α± (λ) defined in (3.1.3), the exponents of the radial weak solutions to −Δu − λ
u = 0, x2
which we know can be written as u(r) = c1 x−α+ (λ) + c2 x−α− (λ) , c1 , c2 ∈ ℝ. As in the semilinear case, the problem under consideration will be to describe, for each λ ∈ (0, ΛN,2 ] fixed, the optimal exponent q+ (λ) for which there is a solution of (6.1.1), at least in a weak sense, when 1 < q < q+ (λ) < 2. It is clear that the variational techniques are not useful in the quasilinear setting and that the difficulties are considerably bigger than in the semilinear case studied in Chapter 4. It is worth pointing out that this type of quasilinear problems appears in several contexts. For instance, for the quadratic case q = 2, in the simplest situation of λ = 0, problem (6.1.1) is the stationary counterpart of the Kardar–Parisi–Zhang model (see [205]) and of some flame propagation models (see [57]). The existence of solutions in this case can be seen for instance in [212], [75] and the references therein. For stronger results on classification and multiplicity of solutions, we refer to [8]. Moreover, equation (6.1.1) can be read as the Hamilton–Jacobi equation ∇uq + λ
u + cf = 0 x2
with the viscosity term given by the Laplacian. As a consequence of the nonexistence results in this paper, the reader could check without difficulty that the vanishing viscosity method by P. Lax does not produce a solution for the first order equation. See for instance [232] for details and applications of this. The physical motivation for q < 2 can be seen in [213]. There is a huge amount of references for the case q ≤ 2; see for instance [74], [75], [76], [78], [156], [157], [214], [249] and [266]. When λ > 0, that is, when the Hardy potential is present, the behavior is different. The study of this behavior is the main goal of this chapter. Section 6.2 starts with some auxiliary results about quasilinear nonlinearities that have some interest https://doi.org/10.1515/9783110606270006
104  6 Quasilinear elliptic equations with the Hardy potential in itself and try to be selfcontained. In Section 6.3 we discuss the candidate to be the critical exponent q+ (λ) and prove that if q > q+ (λ) problem (6.1.1) has not a very weak supersolution in the sense presented below (see Definition 6.3.1). In Section 6.4 we analyze the complete blowup of some approximated problem. This is crucial for a better understanding of the nonexistence results. In order to prove that in fact q+ (λ) is the threshold for existence, we must show existence of solutions for q < q+ (λ). These results are obtained in Sections 6.5 and 6.6. Most of them come from paper [17].
6.2 Some auxiliary results In this section we will obtain some results that will be used later. 6.2.1 Comparison results We describe here two comparison results. The first one is related to finite energy solutions and will use the arguments in [266]. The second one relates to data in L1 (Ω) and was obtained in [31]. We start by proving the comparison for finite energy solutions. More precisely, we prove the following result. Proposition 6.2.1. Let w ∈ W 1,2 (Ω) be such that {
−Δw ≤ ⟨h(x), ∇w⟩
Ω,
w≤0
𝜕Ω,
(6.2.1)
where h ∈ Lq (Ω), q > N. Then w ≤ 0 in Ω. Proof. Taking wk := (w − k)+ as a test function in (6.2.1) and applying the Hölder and Sobolev inequalities, we obtain 2
α
2
1 2
q
1 q
∫ ∇wk  dx ≤ Ω (∫ ∇wk  dx) (∫ h dx) (∫ wk  Ωk
Ωk
Ωk
2N N−2
dx)
N−2 2N
Ωk 1 q
≤ Ωα S(∫ ∇wk 2 dx)(∫ hq dx) , Ωk
(6.2.2)
Ωk
where 𝒮 is the optimal constant in the Sobolev inequality, α = 1 − ( 21 +
1 N
+
N−2 ) 2N
and
Ωk = {x ∈ Ω : w(x) > k, ∇w > 0}. Let us argue by contradiction and assume that sup w > 0. Denote M = sup w. Then either M = +∞ and, since w ∈ W 1,2 (Ω), we have lim Ωk  = 0,
k→M
(6.2.3)
6.2 Some auxiliary results  105
or M < ∞ and again by the fact that w ∈ W 1,2 (Ω), we have ∇w = 0 a. e. in {x ∈ Ω  w = M}. By Stampacchia’s theorem, therefore, we find again (6.2.3). As a consequence, ‖h‖Lq (Ωk ) → 0
as k → M,
and then there exists k0 < M such that
1 Ωα S‖h‖Lq (Ωk ) ≤ . 2
Hence from (6.2.2) we find that
∫ ∇wk 2 dx = 0
and then w ≤ k0 < M,
Ωk0
which gives a contradiction. Lemma 6.2.2. There exists a unique solution u ∈ W01,2 (Ω), with u ≥ 0 in Ω, to the problem {
−Δu = ⟨h(x), ∇u⟩ + f
x ∈ Ω,
u=0
x ∈ 𝜕Ω,
where h : Ω → ℝN is such that h ∈ Lq (Ω), q > N, and f ∈ W −1,2 (Ω). In particular we have ‖v‖W 1,2 (Ω) ≤ C‖f ‖W −1,2 (Ω) .
(6.2.4)
0
Proof. Let us define the operator T : W01,2 (Ω) → W01,2 (Ω), u → T(u) = v,
such that v verifies −Δv = ⟨h(x), ∇u⟩ + f . The continuity of T follows easily. We show that T is compact, i. e., that for any sequence (un ) bounded in W01,2 (Ω) there exists a subsequence of vn = T(un ) which converges strongly to some v ∈ W01,2 (Ω). We note that ⟨h(x), ∇u⟩ ∈ W −1,2 (Ω), since for any φ ∈ W01,2 (Ω) one has ∫⟨h(x), ∇u⟩φ dx Ω
1
1
1 ∗
q 2 2 ∗ q ≤ C(Ω)(∫h(x) dx) (∫ ∇u2 dx) (∫ φ2 dx)
Ω
1 2
≤ C(Ω)(∫ ∇φ2 ) , Ω
Ω
Ω
106  6 Quasilinear elliptic equations with the Hardy potential by the Sobolev inequality. The fact that also f ∈ W −1,2 (Ω) yields that the sequence vn = T(un ) is uniformly bounded in W01,2 (Ω). By Rellich’s theorem we get that for a subsequence, still labeled with n, it holds that strongly in L2 (Ω),
vn → v
∇vn ⇀ ∇v vn → v
N
weakly in (L2 (Ω)) ,
a. e. in Ω,
(6.2.5)
with v ∈ W01,2 (Ω). We take as test function vn − v to arrive at 2 ∫∇(vn − v) dx = ∫⟨h(x), ∇un ⟩(vn − v) dx + ∫ f (vn − v) dx
Ω
Ω
Ω
− ∫ ∇v∇(vn − v) dx Ω
= ∫⟨h(x), ∇v⟩(vn − v) dx + o(1), Ω
due to the weak convergence ∇vn → ∇v in (L2 (Ω))N , the strong convergence vn → v in L2 (Ω) and the fact that f ∈ W −1,2 (Ω). With the remaining term we argue as follows: ∫⟨h(x), ∇v⟩(vn − v) dx ≤ ∫h(x)∇vTm (vn − v) dx Ω
Ω
+ ∫ h(x)∇vvn − v dx := I1 + I2 , Ωm
where Tm (s) = max{−m, min(s, m)} is the usual truncation and Ωm = Ωnm =: Ω ∩ {vn − v > m}. Observe that 1
1
q 2 q I1 ≤ (∫h(x) dx) (∫ ∇v2 dx)
Ω
Ω
1 ∗
2 2∗ × (∫Tm (vn − v) dx) .
Ω
The strong convergence vn → v in L2 (Ω) implies the strong convergence of Tm (vn − v) to zero in Lr (Ω) for any r ≥ 1, in particular for r = 2∗ , thus I1 = o(1) as n → ∞. Concerning the second term, we have 1
1
q 2 q I2 ≤ ( ∫ h(x) dx) ( ∫ ∇v2 dx)
Ωm
Ωm
2∗
× ( ∫ vn − v dx) Ωm
1 2∗
6.2 Some auxiliary results  107 1
1
q 2 q 2 ≤ C( ∫ h(x) dx) ( ∫ ∇(vn − v) dx)
Ωm
Ωm
1 q
q ≤ C( ∫ h(x) dx) . Ωm
Summing up, we have shown that 1
q 2 q ∫∇(vn − v) dx ≤ C sup( ∫ h(x) dx) .
n
Ω
Ωm
We proved that vn − v is uniformly bounded in L2 (Ω) and thus Ωnm  → 0 as n → ∞. Therefore, 2 lim ∫∇(vn − v) dx = 0,
n→∞
Ω
and hence vn is strongly convergent in W01,2 (Ω), which concludes the compactness of T. In order to apply the fixed point theorem, we show that there exists M > 0, such that ‖u‖W 1,2 (Ω) ≤ M, for every u ∈ W01,2 (Ω) and every σ ∈ [0, 1] verifying u = σT(u). 0
Arguing by contradiction, we assume that for every n there exist un ∈ W01,2 (Ω) and
σn ∈ [0, 1] such that un = σn T(un ) and ‖un ‖W 1,2 (Ω) ≥ n. We consider wn = 0
satisfies
−Δwn = σn ⟨h(x), ∇wn ⟩ + σn
un ‖un ‖W 1,2 (Ω) 0
that
f . ‖un ‖W 1,2 (Ω) 0
Since ‖un ‖W 1,2 (Ω) → ∞, applying the Kato inequality (see [209, 96]), this identity im0 plies that for n large −Δwn+ ≤ C h(x)∇wn+ . Then, by Lemma 6.2.1 it follows that w+ ≡ 0, which contradicts the fact that ‖wn ‖W 1,2 (Ω) = 1. 0
We have proved that T has a unique fixed point, which is the desired solution. Finally we show the last estimate of the proposition. Set v1 = ‖f ‖ v−1,2 , then repeating the same computation as in the previous step, we reach
W
(Ω)
‖v1 ‖W 1,2 (Ω) ≤ C(h), 0
and hence (6.2.4) follows. We now formulate and prove the comparison in the weak setting. See also [31], where the result was obtained by using the isoperimetric inequality.
108  6 Quasilinear elliptic equations with the Hardy potential Proposition 6.2.3. Let w ∈ L1 (Ω) be such that Δw ∈ L1 (Ω), satisfying {
−Δw ≤ ⟨h(x), ∇w⟩
Ω,
w≤0
𝜕Ω,
(6.2.6)
where h ∈ (LN+ϵ (Ω))N . Then w ≤ 0 in Ω. To prove Proposition 6.2.3 we will use the following result for the adjoint operator. Lemma 6.2.4. Assume h ∈ Lq (Ω, ℝN ), q > N. Then there exists a unique solution ϕ ∈ W01,2 (Ω) ∩ L∞ (Ω) to the problem − Δϕ = − div(ϕ ⋅ h) + 1, in Ω,
ϕ = 0 on 𝜕Ω.
(6.2.7)
Moreover ϕ > 0 in Ω. Proof. If the norm of h is small enough, this is a direct consequence of the Lax– Milgram result. For the general case we use the Schauder fixed point theorem. Consider the operator T : W01,2 (Ω) → W01,2 (Ω), ϕ → T(ϕ) = v such that −Δv = − div(h(x)ϕ) + 1,
v𝜕Ω ≡ 0,
which is clearly continuous. We need to check that T is compact. Note that div(h(x)ϕ) ∈ W −1,2 (Ω). Indeed, for ψ ∈ W01,2 (Ω), by the Sobolev inequality, we have ∫ div(h(x)ϕ)ψ dx Ω
1 ∗
1
1
q 2 2 ∗ q ≤ (∫h(x) dx) (∫ ϕ2 dx) (∫ ∇ψ2 dx)
Ω
1 2
Ω
Ω
≤ C(∫ ∇ψ2 dx) . Ω
Consider a sequence (ϕn ) bounded in W01,2 (Ω). Set ϕ ∈ W01,2 (Ω) such that for a subsequence ϕn → ϕ strongly in L2 (Ω), ∇ϕn ⇀ ∇ϕ
N
weakly in (L2 (Ω)) ,
ϕn → ϕ a. e. in Ω.
(6.2.8)
6.2 Some auxiliary results  109
Then vn = T(ϕn ) is also a bounded sequence in W01,2 (Ω). Thus, there exists v ∈ W01,2 (Ω) such that, up to a subsequence, vn → v
strongly in L2 (Ω),
∇vn ⇀ ∇v vn → v
N
weakly in (L2 (Ω)) ,
a. e. in Ω.
(6.2.9)
Therefore taking vn − v as a test function we obtain 2 ∫∇(vn − v) dx = ∫ div(h(x)(ϕn − ϕ))(vn − v) dx
Ω
Ω
∫ div(h(x)Tm (ϕn − ϕ))(vn − v) dx + ∫ div(h(x)Gm (ϕn − ϕ))(vn − v) dx Ω
Ωm
≤ ∫ Tm (ϕn − ϕ)⟨h(x), ∇(vn − v)⟩ dx + ∫ Gm (ϕn − ϕ)⟨h(x), ∇(vn − v)⟩ dx Ω
≡ I1 + I2 ,
where Ωm =
Ωnm
Ωm
=: Ω ∩ {ϕn − ϕ > m}.
Note that I1  ≤ ‖h‖q Tm (ϕn − ϕ)2∗ ∇(vn − v)2 Ωα = o(1), Moreover, if α = 1 − ( 21∗ +
1 2
as n → ∞.
+ q1 ) > 0,
I2  = ∫ Gm (ϕn − ϕ)⟨h(x), ∇(vn − v)⟩ dx ≤ Ωm α ‖h‖q ‖ϕn − ϕ‖2∗ ∇(vn − v)2 = o(1), Ωm
because Ωnm  → 0. Finally, let ϕ and σ ∈ [0, 1] satisfy that σT(ϕ) = ϕ, that is − Δϕ = −σ div(h(x)ϕ) + 1.
(6.2.10)
Let us show that there exists a constant M > 0 such that ‖ϕ‖W 1,2 (Ω) ≤ M. We consider 0
v ∈ W01,2 (Ω) a nonnegative function such that
− Δv = σ⟨h(x)∇v⟩ + ϕ2
∗
−2
ϕ,
v𝜕Ω ≡ 0.
(6.2.11)
The existence of such a solution is granted by taking f = ϕ2 −2 ϕ in Lemma 6.2.4. Multiplying by v and ϕ, respectively, equations (6.2.10) and (6.2.11) and subtracting ∗
110  6 Quasilinear elliptic equations with the Hardy potential we get 2∗
2∗
∫ ϕ dx = σ ∫ v dx ≤ C(∫ v dx) Ω
Ω
1 2∗
Ω
1
2 ∗ ≤ C(∫ ∇v2 dx) ≤ C ϕ2 −2 ϕW −1,2 (Ω)
Ω
2∗
≤ C(∫ ϕ dx)
2∗ −1 2∗
,
Ω
where we have used also (6.2.4). Hence, 2∗
(∫ ϕ dx)
1 2∗
≤ C.
(6.2.12)
Ω
To get a bound for the gradient, we multiply (6.2.10) by ϕ. Integrating by parts, we obtain ∫ ∇ϕ2 dx = σ ∫ h(x)ϕ∇ϕ dx + σ ∫ ϕ dx ≤ C(∫ ϕ2 dx) Ω
Ω
Ω
1 2∗
1 2
Ω
1 2
1
q q + (∫ ϕ dx) (∫ ∇ϕ dx) (∫h(x) dx) .
2∗
2
Ω
Ω
Ω
Therefore, 1 ∗
1
1
2 q 2 ∗ q (∫ ∇ϕ2 dx) ≤ (∫ ϕ2 dx) (∫h(x) dx) .
Ω
Ω
Ω
Then ‖ϕ‖W 1,2 (Ω) ≤ C. 0 By the Schauder fixed point theorem (see Theorem 11.3 in [186]), we have shown that there exists a unique solution to (6.2.7), ϕ ∈ W01,2 (Ω). The positivity of ϕ follows easily. To conclude the proof, we show that ϕ ∈ L∞ (Ω). To this end, we argue as in [17] and consider the following test function v = sign(ϕ)(ϕ − k)+ in (6.2.7). We find that ∫ ∇ϕ2 dx − ∫ ⟨h(x), ∇ϕ⟩ϕ dx = ∫ v dx, A(k)
A(k)
(6.2.13)
A(k)
where A(k) = {x ∈ Ω : ϕ(x) > k}. Note that ∇ϕ = ∇v in A(k) and also that v = 0 in Ω \ A(k). Since ϕ ∈ W01,2 (Ω), A(k) → 0 as k → ∞. Our purpose is to show that there
6.2 Some auxiliary results  111
exists k0 such that A(k) = 0 for any k ≥ k0 . We treat first the integral on the right hand side, 1 2∗
1
1− ∗ ∫ v dx ≤ ( ∫ v dx) A(k) 2 . 2∗
A(k)
(6.2.14)
A(k)
Using the Sobolev inequality and the fact that ∇v = ∇ϕ in A(k) we get 1 2
1− 1∗ ∫ v dx ≤ CS( ∫ ∇ϕ dx) A(k) 2 . 2
A(k)
A(k)
By Hölder’s inequality with exponents
1 2
+
1 2∗
+
1 q
+
1 δ
= 1, we find
∫ ⟨h(x), ∇ϕ⟩ϕ dx A(k) 1 ∗
1
q 2 ∗ q ≤ ( ∫ h(x) dx) ( ∫ ϕ2 dx)
A(k)
A(k)
1 2
1 × ( ∫ ∇ϕ2 dx) A(k) δ A(k)
1 2
1
2 1 ≤ C( ∫ ∇ϕ dx) A(k) δ ≤ η( ∫ ∇ϕ2 dx) ,
2
A(k)
A(k)
with η ≪ 1 if k is sufficiently large. Thus we can absorb this term into the first integral in (6.2.13). Then, using (6.2.14) and Sobolev’s inequality we arrive at 2∗
( ∫ v dx) A(k)
1 2∗
1− 1∗ ≤ C A(k) 2 .
(6.2.15)
Since, if 0 < k < m, then A(m) ⊂ A(k), it is easy to deduce that 1 ∗
1 ∗
2 2 1∗ 2∗ 2∗ A(m) 2 (m − k) = ( ∫ (m − k) γ dx) ≤ ( ∫ v dx)
A(m)
A(m)
1 2∗
≤ ( ∫ v2 dx) . ∗
A(k)
Plugging this inequality into (6.2.15) we have shown that μ(A(m)) ≤
C 2∗ −1 , ∗ μ(A(k)) 2 (m − k)
with 2∗ − 1 > 1. Then, the classical Stampacchia theorem (see [291]) implies that there exists k0 such that μ(A(k)) = 0 for any k ≥ k0 , as desired.
112  6 Quasilinear elliptic equations with the Hardy potential To finish this section we proceed to prove the comparison result. Proof of Proposition 6.2.3. Since u, Δu ∈ L1 (Ω), using the representation in terms of the Green function and deriving the integral, we find that ∇u ∈ (L1 (Ω))N (see Lemma 1 in [209]). Using now the Kato inequality [209], we have −Δw+ ≤ ⟨h(x), ∇w+ ⟩ in Ω,
w+ = 0 on 𝜕Ω,
and hence we can assume without loss of generality that w ≥ 0 in Proposition 6.2.3. Taking as a test function ϕ ∈ L∞ (Ω), ϕ ≥ 0, verifying (6.2.7), we find that ∫(−Δw)ϕ dx ≤ ∫⟨h(x), ∇w⟩ϕ dx = − ∫ div(h(x)ϕ)w dx Ω
Ω
Ω
= ∫(−Δϕ)w dx − ∫ w dx. Ω
Ω
Therefore, ∫ w dx ≤ 0,
thus w ≤ 0 in Ω.
Ω
6.2.2 A quantitative version of the maximum principle We present here the version of the maximum principle given by H. Brezis and X. Cabre in [87]. Denote by δ(x) the distance function to the boundary of Ω, that is, δ(x) = d(x, 𝜕Ω). Proposition 6.2.5. Consider h, f (x) ≥ 0 and f ∈ L∞ (Ω). Let v be the solution of {
−Δv(x) = f (x)
in Ω,
v(x) = 0
on 𝜕Ω,
where Ω ⊂ ℝN is a smooth bounded domain. Then, there exists a positive constant c = c(Ω) such that v(x) ≥ c ∫ f (y)δ(y) dy, δ(x)
for all x ∈ Ω.
Ω
. Let {Br (xi )}m Proof. Consider a compact set K ⊂ Ω and 0 < r < d(K,𝜕Ω) 1 be a finite 2 covering of K by balls of radius r, each contained in Ω. Let ϕ1 , ϕ2 ...ϕm be the solutions to the problems {
−Δϕi (x) = χBr (xi ) ϕi (x) = 0
in Ω, on 𝜕Ω,
i = 1, 2, . . . , m,
6.2 Some auxiliary results  113
where χBr (xi ) is the characteristic function of the ball Br (xi ). By the Hopf boundary lemma, there exists a positive constant c such that ϕi (x) ≥ cδ(x),
for all x ∈ Ω and all i = 1, 2, . . . , m,
with δ(x) = d(x, 𝜕Ω). Consider x ∈ K and a ball of the covering, Br (xi ), containing x. Note that Br (xi ) ⊂ B2r (x) ⊂ Ω. Since −Δv > 0 in Ω, that is, v is superharmonic in Ω, we find that v(x) ≥
1 ∫ v(y) dy = c(r) ∫ v(y) dy ≥ c(r) ∫ v(y) dy B2r  B2r (x)
B2r (x)
Br (xi )
= c(r) ∫ v(y)(−Δϕi (y)) dy = c(r) ∫ f (y)ϕi (y) dy ≥ c(r)c ∫ f (y)δ(y) dy. Ω
Ω
Ω
Assume, moreover, that K is such that Ω \ K has a smooth boundary. We have the estimate above and to conclude the global result we have to get the lower estimate of v in Ω \ K. To do that, consider the auxiliary problem −Δψ = 0 { { { ψ(x) = 0 { { { { ψ(x) = 1
in Ω, on 𝜕Ω, on 𝜕K.
By the Hopf lemma we have ψ(x) ≥ cδ(x),
for all x ∈ Ω \ K.
Since v is superharmonic and v(x) ≥ c ∫ f (y)δ(y) dy
on 𝜕K,
Ω
the maximum principle implies that v(x) ≥ c(∫ f (y)δ(y) dy)ϕ(x) ≥ c1 (∫ f (y)δ(y) dy)δ(x) for all x ∈ Ω \ K, Ω
Ω
which proves the result. 6.2.3 Caffarelli–Kohn–Nirenberg inequalities In the proofs of nonexistence that we will perform next, we need some of the inequalities that appear in the paper by Caffarelli, Kohn and Nirenberg [108]. More precisely, ∞ for γ < N−p , we consider the weighted Sobolev space D1,p γ (Ω) as the closure of 𝒞0 (Ω) p
114  6 Quasilinear elliptic equations with the Hardy potential with respect to the norm ‖.‖D1,p (Ω) defined by γ
‖ϕ‖p 1,p = ∫(ϕp + ∇ϕp )x−pγ dx. Dγ
Ω
Using a Poincaré type inequality, we know that the space D1,p 0,γ (Ω) may be also defined as the completion of 𝒞0∞ (Ω) with respect to the Lp (x−pγ dx)norm of the gradient. We recall now the Hardy–Sobolev inequality that we will use systematically in the next chapters. Lemma 6.2.6. Suppose 1 < p < N and −∞ < γ < p −pγ dx, ∫ up x−p(γ+1) dx ≤ Λ−1 N,p,γ ∫ ∇u x
Ω
N−p . p
Then, for all u ∈ 𝒞0∞ (Ω), we have
ΛN,p,γ = (
Ω
p
N − p(γ + 1) ) . p
(6.2.16)
Moreover, Λ−1 N,p,γ is optimal and it is not achieved. As we said, the proof can be seen in [108]. Alternatively, one can try to make a direct proof following the arguments in Chapter 2 for p = 2 and p ≠ 2. The elliptic equations involving coefficients related to Caffarelli–Kohn–Nirenberg inequalities have been studied in the article [15].
6.3 Nonexistence results: exponent q ≥ q+ (λ) The main goal in this section is to show that q+ (λ) represents the threshold for the exponent q, in a such way that problem (6.1.1) has no positive supersolution in the very weak sense if q > q+ (λ). In the whole section, we use the concept of very weak (sub/super)solution, which, roughly speaking, is the more general form for which the equation has a meaning in the distributional sense. We provide a precise definition of this concept that is the translation to the quasilinear case of the same concept for semilinear problems. Definition 6.3.1. We say that u ∈ L1loc (Ω) is a very weak supersolution (subsolution) to equation (6.1.1) when, if xu 2 ∈ L1loc (Ω), ∇uq ∈ L1loc (Ω) and ∀ϕ ∈ C0∞ (Ω) such that ϕ ≥ 0, we have ∫(−Δϕ)u dx ≥ (≤) ∫(∇uq + λ Ω
Ω
u + f )ϕ dx. x2
If u is a very weak super and subsolution, then we say that u is a very weak solution. If λ > ΛN,2 ≡ ( N−2 )2 , the nonexistence result of positive very weak solution to 2 problem (6.1.1) is a consequence of the optimality of ΛN,2 as bound in the Hardy–Leray inequality. See for instance [13]. Then, hereafter we will assume 0 < λ ≤ ΛN,2 .
6.3 Nonexistence results: exponent q ≥ q+ (λ)
 115
We recall the lower estimate of u near the origin obtained in Lemma 4.3.8 and the necessary condition on the source term obtained in Theorem 3.4.1 to have a solution. To find the optimal exponent we search for a solution in the form u(x) = Ax−β of the equation. Hence by a direct computation as in the semilinear case, we obtain β=
2−q q−1
and βq Aq−1 = β(N − β − 2) − λ.
(6.3.1)
Since the left hand side is positive, the right hand side must be positive, but the second member is positive if and only if α− (λ) < β < α+ (λ),
where α± (λ) are defined by (3.1.3).
Since (6.3.1) holds, α− (λ) < β < α+ (λ) is equivalent to q− (λ) ≡
2 + α+ (λ) 2 + α− (λ) q+ (λ). Assume by contradiction that equation (6.1.1) has a very weak supersolution u. Then −Δu − λ
u ⪈ 0. x2
Therefore, by Lemma 4.3.8, there exists a positive constant C and a small ball Br (0) ⊂ q , as a ℝN such that u(x) ≥ Cx−α− (λ) in Br (0). Take ϕ ∈ 𝒞0∞ (Br (0)). Using ϕq , q = q−1 test function in (6.1.1) and by the Hölder and Young inequalities we obtain uϕq dx ≤ ∫ ∇ϕq dx, 2 x
c1 λ ∫ Br (0)
(6.3.3)
Br (0)
where c1 is a positive constant that is independent of u and ϕ. Using the lower estimate for u in Br (0) we obtain ϕq c2 λ ∫ dx ≤ ∫ ∇ϕq dx. 2+α (λ) − x
Br (0)
Br (0)
We recall that q > q+ (λ), and hence 2 + α− (λ) > q and we reach a contradiction with
the classical Hardy inequality for W01,q (Br (0)) proved in Section 2.4. Then, the result follows.
116  6 Quasilinear elliptic equations with the Hardy potential Second step: q = q+ (λ) and λ < ΛN,2 . Again we argue by contradiction. Assume that equation (6.1.1) has a very weak supersolution u. As above, by Lemma 4.3.8 there is a positive constant c0 such that u(x) ≥
c0 xα− (λ)
in some ball Bη (0) ⊂⊂ Ω.
(6.3.4)
Without loss of generality we assume that η = e−1 . Using Lemma 4.3.8 we obtain ∫ ∇uq+ (λ) x−α− (λ) dx < ∞ and
∫ Bη (0)
Bη (0)
u dx < ∞. x2+α− (λ)
(6.3.5)
1 β )) , where β is a positive small constant that we will choose Let w(x) = x−α− (λ) (log( x
below. Since λ < ΛN,2 , we have w ∈ W 1,2 (Bη (0)) and, in particular, w ∈ W 1,q+ (λ) (Bη (0)). By a direct computation we obtain −Δw − λ
β−1
β w 1 = (log( )) x x2 x2+α− (λ)
[(N − 2 − 2α− (λ)) + (1 − β)((log(
1 )) ]. x −1
1 β−1 1 ) + β(log( x )) )), thus Note that ∇w = x−α− (λ)−1 (α− (λ) log( x
∇wq+ (λ) (α− (λ) log(
−1 1−q+ (λ)
1 1 ) + β((log( )) ) x x
= x−α− (λ)−2 (α− (λ) log(
η 1 ) + β((log( )) ). x x −1
Since x ≤ e−1 , choosing β small enough, we conclude that −Δw − λ
1 w ≤ β 2 ∇wq+ (λ) h(x), x2
1 1 −1 1−q+ (λ) where h(x) = (α− (λ) log( x ) + β((log( x )) ) . Observe that this is bounded in the ball Bη (0). Set u1 ≡ c1 u. Then
−Δu1 − λ
u1 1−q (λ) ≥ c1 + ∇u1 q+ (λ) . x2
Let c0 be a fixed constant satisfying (6.3.4) when η = e−1 and take c1 > 0 such that c1 c0 ≥ 1. Then for β suitable small we have 1−q+ (λ)
c1
1
≥ ‖h‖∞ β 2 .
Since c1 c0 ≥ 1 we obtain u1 (x) ≥ w(x) for x = e−1 and, moreover, −Δu1 − λ
1 u1 ≥ β 2 h(x)∇u1 q+ (λ) . 2 x
6.3 Nonexistence results: exponent q ≥ q+ (λ)
 117
Claim. u1 ≥ w. We write v = w − u1 . Using the regularity of w and (6.3.5) we obtain v ∈ W 1,q+ (λ) (Bη (0)), v ≤ 0 on 𝜕Bη (0) and v
∫ Bη (0)
x2+α− (λ)
∫ ∇vq+ (λ) x−α− (λ) dx < ∞.
dx < ∞,
(6.3.6)
Bη (0)
By a direct computation it follows that −Δv − λ
1 v ≤ q+ (λ)h(x)β 2 ∇wq+ (λ)−2 ∇w∇v ≡ a(x)∇v, 2 x 1
where the vector field a(x) = −β 2 q+ (λ) xx 2 belongs to Lq (Bη (0)) for all q < N. Observe that the vector field a(x) does not have enough regularity to apply the comparison argument used in [31]. To overcome this lack of regularity we proceed as follows. Using the Kato inequality (see [209] and the extension in [96]) we get −Δv+ − λ and, since
1 v+ x + q+ (λ)β 2 ⟨ 2 , ∇v+ ⟩ ≤ 0 2 x x
α− (λ) q+ (λ)
2 + α− (λ), depending only on N and λ, such that ∫ Bη (0)
v+ dx < ∞. xσ1
(6.3.9)
Indeed, ∫ Bη (0)
v+ dx = ∫ xσ1
Bη (0)
v+
x
q+ (λ)+α− (λ) q+ (λ)
1
x
q (λ)
≤( ∫ Bη (0)
v++
xq+ (λ)+α− (λ)
σ1 −
q+ (λ)+α− (λ) q+ (λ)
dx)
dx
1 q+ (λ)
1
( ∫ Bη (0)
x
q+ (λ)(σ1 −
q+ (λ)+α− (λ) ) q+ (λ)
dx)
1 (λ) q+
.
118  6 Quasilinear elliptic equations with the Hardy potential Denote θ(σ1 ) = q+ (σ1 − 2 + α− (λ). Hence,
q+ (λ)+α− (λ) ). q+ (λ)
2+α− (λ) , 1+α− (λ)
Since q+ (λ) =
the conjugate of q+ (λ) is q+ =
θ(σ1 ) = (2 + α− (λ))(σ1 − 1) − α− (λ)(1 + α− (λ)). By a direct computation we get θ(2 + α− (λ)) = 2(1 + α− (λ)) = N − 2√ΛN,2 − λ < N. Thus, there exists σ1 > 2 + α− (λ) such that θ(σ1 ) < N and −(q+ (λ)(σ1 −
∫ x
q+ (λ)+α− (λ) )) q+ (λ)
dx < ∞.
Bη (0)
Thus (6.3.9) holds. To finish, the idea should be to use φ, the solution to problem φ
{
1 x2(γ+1)
− div(x−2γ ∇φ) − λ x2(γ+1) =
in Bη (0),
φ=0
on 𝜕Bη (0),
as a test function in (6.3.8). However, a direct calculation shows that φ(x) =
1 1 − , xa ηa
where a =
2
N − 2(γ + 1) √ N − 2(γ + 1) − ( ) − λ, 2 2
which does not have the required regularity to be used directly as a test function in (6.3.8). Therefore, we need to consider the approximating sequence φn (x) =
1 (x +
1 a ) n
1
−
(η + n1 )a
.
We see that φn ∈ 𝒞 1 (Bη (0)), φn = 0 on 𝜕Bη (0), ∇φn (x) = −
a (x +
x x
1 a+1 ) n
and − div(x−2γ ∇φn ) = x−2γ (
a(N − 1 − 2γ)
x(x + n1 )a+1
−
a(a + 1)
(x + n1 )a+2
).
Note that ∫ x−2γ ∇v+ ∇φn  dx < ∞
and
Bη (0)
∫ Bη (0)
v+ φn dx < ∞. x2(γ+1)
Choosing φn as a test function in (6.3.8) we obtain ∫ v(− div(x−2γ ∇φn )) dx − λ ∫ Bη (0)
Bη (0)
v+ φn dx ≤ 0. x2(γ+1)
(6.3.10)
6.3 Nonexistence results: exponent q ≥ q+ (λ)
 119
By the definition of wn we have v+ φn v+ v+ 2 ≤ + . x2(γ+1) xa+2(γ+1) ηa x2(γ+1) Since a + 2(γ + 1) → 2 + α− (λ) as γ → 0, choosing β small we can find γ small such that a + 2(γ + 1) < σ1 . Hence Cv v+ φn ≤ σ+ . 2(γ+1) 1 x x Then by the definition of σ1 and using the dominated convergence theorem, we easily prove that v+ v+ v+ φn 1 dx → ∫ dx − a ∫ dx η x2(γ+1) xa+2(γ+1) x2(γ+1)
∫
as n → ∞.
Bη (0)
Bη (0)
Bη (0)
We deal now with the first term in (6.3.10), a(a + 1)v+ a(N − 1 − 2γ)v+ −2γ − ) v+ div(x ∇φn ) = ( 1+2γ x (x + 1 )a+1 x2γ (x + 1 )a+2 n n ≤
a(N − 1 − 2γ)v+
+
x1+2γ (x + n1 )a+1
a(a + 1)v+
x2γ (x + n1 )a+2
.
As above it is not difficult to see that a(N − 1 − 2γ)v+
x1+2γ (x + n1 )a+1
+
a(a + 1)v+
x2γ (x + n1 )a+2
≤
a(N + a − 2γ)v+ xσ1
and then, by the dominated convergence theorem, we obtain ∫ v+ div(x−2γ ∇φn ) dx → ∫ Bη (0)
Bη (0)
a(N − a − 2(γ + 1))v+ dx x2(1+γ)+a
as n → ∞.
Hence, passing to the limit in (6.3.10) and since a(N − a − 2(γ + 1)) − λ = 0, we have ∫ v(− div(x−2γ ∇φn )) dx − λ ∫ Bη (0)
Bη (0)
v+ φn v+ 1 dx → a ∫ dx, η x2(γ+1) x2(1+γ)
as n → ∞. Thus, according to (6.3.10), ∫B (0) η u1 ≥ w.
Bη (0)
v+ x2(1+γ)
dx ≤ 0, and hence v+ ≡ 0 and then
120  6 Quasilinear elliptic equations with the Hardy potential To finish the proof in this case we use the same argument as in the first step. More precisely, for every ϕ ∈ 𝒞0∞ (Br (0)), 0 < r 0 is independent of ϕ. Using the result of the claim and the fact that q+ (λ) = α− (λ) + 2, we obtain ϕq+ (λ)
β
c2 ∫
xq+ (λ)
Br (0)
(log(
1 )) dx ≤ ∫ ∇ϕq+ (λ) dx, x
Br (0)
1,q (λ)
a contradiction with Hardy’s inequality in W0 + (Br (0)). Hence the result follows. Third step: q = q+ (λ) and λ = ΛN,2 . Assume by contradiction that problem (6.1.1) and q+ (λ) = N+2 . has a positive very weak supersolution u. In this case α− (λ) = N−2 2 N −α− (λ) Hence by Lemma 4.3.8 we obtain u(x) ≥ cx and by Lemma 4.3.8 ∫ ∇uq+ (λ) x−α− (λ) dx < ∞. Bη (0)
We consider ϕ ∈ 𝒞0∞ (Bη (0)) such that ϕ ≥ 0 and ϕ = 1 in Bη1 (0). Then by the regularity of u we have ∫B
η (0)
∇(ϕu)q+ (λ) x−α− (λ) dx < ∞. Since
α− (λ) q+ (λ)
=
the Caffarelli–Kohn–Nirenberg inequalities to obtain
N(N−2) 2(N+2)
< N, we can apply
q (λ) C1 ∫ (ϕu)q+ (λ) x−α− (λ) dx ≤ ∫ ∇(ϕu) + x−α− (λ) dx < ∞, Bη (0)
Bη (0)
∫ uq+ (λ) x−α− (λ) dx < ∞
for η1 < η.
Bη1 (0) 1,q+ (λ) (Bη1 (0)), that we recall is the weighted Sobolev − (λ) ∞ of 𝒞 (Bη (0)) with respect to the norm ‖ ⋅ ‖𝒟1,q+ (λ) , deα (λ)
Therefore we conclude that u ∈ 𝒟α space defined as the completion
−
fined by ‖ϕ‖
q+ (λ)
1,q+ (λ) − (λ)
𝒟α
= ∫ ϕq+ (λ) x−α− (λ) dx + ∫ ∇ϕq+ (λ) x−α− (λ) dx. Bη1 (0)
Bη1 (0) 1,q+ (λ) (Bη (0)) − (λ)
It is not difficult to see that for every ϕ ∈ 𝒟α C2 ∫ Bη1 (0)
we have
ϕq+ (λ) dx ≤ ∫ ϕq+ (λ) x−α− (λ) dx + ∫ ∇ϕq+ (λ) x−α− (λ) dx, xα− (λ)+q+ (λ) Bη1 (0)
Bη1 (0)
6.4 Blowup result  121
where C2 > 0 is independent of ϕ; in particular, ∫ Bη1 (0)
uq+ (λ)
xα− (λ)+q+ (λ)
dx < ∞.
(6.3.12)
Using again the fact that u(x) ≥ cx−α− (λ) and since α− (λ) + q+ (λ) + α− (λ)q+ (λ) = N, we reach a contradiction with (6.3.12). Hence the nonexistence result follows. Remark 6.3.3. 1. Note that q+ (λ) < 2, for all λ ∈ (0, ΛN,2 ]. Hence for q = 2 we easily obtain the nonexistence result from the first step in Theorem 6.3.2. See [19]. Moreover, q+ (λ) → N+2 N if λ → ΛN,2 and q+ (λ) → 2 if λ → 0. As a consequence, we have encountered a discontinuity with the known results for λ = 0. See, for instance, [194]. N 2. If 1 < q ≤ N−1 , then problem (6.1.1) has no very weak positive solution in ℝN . This follows using the results in [31] and [194]. For the reader’s convenience we include a proof. We argue by contradiction. Assume that (6.1.1) has a positive solution u in ℝN with N . It is not difficult to see, using the strong maximum principle, that for 1 < q ≤ N−1 any compact set K ⊂ Ω there exists a positive constant c(K) such that u ≥ c(K). Let ϕ ∈ 𝒞0∞ (Ω). Then using ϕq as a test function in (6.1.1) we obtain q ∫ ∇u∇ϕϕ(q −1) dx ≥ ∫ ∇uq ϕq dx + λ ∫
ℝN
ℝN
ℝN
u ϕq dx. x2
Using Young inequalities we conclude that ∫ ∇ϕq dx ≥ c1 λ ∫
ℝN
3.
ℝN
u ϕq dx. 2 x
(6.3.13)
Since q > N, we have Cap1,q (K) = 0 for any compact set of ℝN . Thus, there exists a sequence {ϕn } ⊂ 𝒞0∞ (ℝN ) such that ϕn ≥ χK and ‖∇ϕn ‖Lq (ℝN ) → 0 as n → ∞. Inserting this in (6.3.13) we reach a contradiction. Despite the previous remark, in bounded domains there are no restrictions on p from below. This follows by the fact that the relative qcapacity, q > N, of a ball with respect to a concentric bigger ball is not zero. See [239, page 106].
6.4 Blowup result As a consequence of the nonexistence result, we obtain the next blowup behavior for approximated problems.
122  6 Quasilinear elliptic equations with the Hardy potential Theorem 6.4.1. Assume that p ≥ q+ (λ). If un ∈ W01,p (Ω) is a solution to problem −Δun = ∇un p + λan (x)un + αf { { { un > 0 { { { { un = 0 with f ≥ 0, f ≠ 0 and an (x) =
1 x2 + n1
in Ω, (6.4.1)
in Ω, on 𝜕Ω,
, then un (x0 ) → ∞, ∀x0 ∈ Ω.
To prove Theorem 6.4.1, we need the following lemma by F. Murat [248], which is a bit more general than Lemma 5.2 in [31]. Lemma 6.4.2. Assume that {un } is a sequence of positive functions such that {un } is uni1,p 1,p formly bounded in Wloc (Ω) for some 1 < p ≤ 2 with un ⇀ u weakly in Wloc (Ω) and that un ≤ u for all n ∈ ℕ. Assume that −Δun ≥ 0 in 𝒟 (Ω) and if p < 2, that the sequence 1,2 {Tk (un )} is uniformly bounded in Wloc (Ω) for k fixed. Then ∇Tk (un ) → ∇Tk (u) strongly in 2 N (Lloc (Ω)) . 1,2 Proof. Note that the hypothesis on the boundedness of {Tk (un )} in Wloc (Ω) is needed just when p < 2. Therefore by hypothesis we conclude that
∇Tk (u)L2 (K) ≤ ∇Tk (un )L2 (K)
for all bounded regular domain K ⊂⊂ Ω.
Let ϕ ∈ 𝒞0∞ (Ω) be a positive function. Since un ≤ u we get 2 ∫ −Δun (Tk (un )ϕ) dx = ∫ ϕ∇Tk (un ) dx + ∫ Tk (un )∇ϕ∇un dx.
Ω
Ω
(6.4.2)
Ω
On the other hand, using again that un ≤ u, ∫ −Δun (Tk (u)ϕ) dx = ∫ ϕ∇un ∇Tk (u) dx + ∫ Tk (u)∇ϕ∇un dx Ω
Ω
Ω
= ∫ ϕ∇Tk (un )∇Tk (u) dx + ∫ Tk (u)∇ϕ∇un dx Ω
≤
Ω
1 1 2 2 ∫ ϕ∇Tk (un ) dx + ∫ ϕ∇Tk (u) dx + ∫ Tk (u)∇ϕ∇un dx. 2 2 Ω
Ω
Ω
Thus by the above computation and (6.4.2), we have 1 1 2 2 ∫ ϕ∇Tk (un ) dx ≤ ∫ ϕ∇Tk (u) dx + ∫(Tk (u) − Tk (un ))∇ϕ∇un dx. 2 2 Ω
Ω
Ω
Hence we conclude that 2 2 lim sup ∫ ϕ(∇Tk (un ) − ∇Tk (u) ) dx ≤ 0 n→∞
Ω
for all positive test functions ϕ.
6.4 Blowup result  123
We set wn = ϕTk (un ) and w = ϕTk (u), where ϕ is a positive test function. Then wn ⇀ w 1,2 (Ω). Note that wn → w strongly in L2loc (Ω). Therefore, using the above weakly in Wloc computation we easily get 0 ≤ lim sup(‖∇wn ‖L2 n→∞
loc
(Ω)
− ‖∇w‖L2
loc
(Ω) )
≤ 0.
Thus, by the definition of the weak limit and using the strong convergence of wn to w in L2loc (Ω) we conclude the proof. Lemma 6.4.3. Let g be a positive function such that g ∈ Lρ (Ω) with ρ > N2 and s > 0. Assume that w1 , w1 are positive functions such that w1 , w1 ∈ W01,2 (Ω) ∩ L∞ (Ω) verifying { −Δw1 ≤ { { w1 = 0
∇w1 p 1+s∇w1 p
+g
∇w2 p 1+s∇w2 p
+g
in Ω,
(6.4.3)
on 𝜕Ω
and { −Δw2 ≥ { { w2 = 0
in Ω,
(6.4.4)
on 𝜕Ω.
Then w2 ≥ w1 in Ω. Proof. Consider w = w1 − w2 . Then w ∈ W01,2 (Ω) ∩ L∞ (Ω). We will prove that w+ = 0. By (6.4.3) and (6.4.4) it follows that −Δw ≤
∇w2 p ∇w1 p − . p 1 + s∇w1  1 + s∇w2 p
Now for x, y ∈ ℝN we define the function γ by setting γ(t) = T(tx + (1 − t)y),
where T(t) =
tp . 1 + stp
For x = ∇w1 and y = ∇w2 we have ∇w1 p ∇w2 p − = γ(1) − γ(0) = γ (θ). p 1 + s∇w1  1 + s∇w2 p Since γ(1) − γ(0) ≤ ∇w1  − ∇w2 T (θ) ≤ ∇w1 − ∇w2 T (θ) p−1
t and T (t) = p (1+st p )2 ≤ C, we conclude that
∇w p ∇w2 p 1 − ≤ C∇w. p 1 + s∇w2 p 1 + s∇w1 
124  6 Quasilinear elliptic equations with the Hardy potential Hence, it follows that w ∈ W01,2 (Ω) ∩ L∞ (Ω).
−Δw ≤ C∇w, Using the Kato inequality we get −Δw+ ≤ C∇w+ ,
0 ≤ w+ ∈ W01,2 (Ω) ∩ L∞ (Ω).
Therefore, from the maximum principle in Proposition 6.2.1 (see also Lemma 4.6 of [31]) it follows that w+ ≡ 0 and then we obtain the comparison result. Now we are able to prove the blowup result. Proof of Theorem 6.4.1. Without loss of generality, we can assume that f ∈ L∞ (Ω) and that λ is small. Assume the existence of x0 ∈ Ω such that un (x0 ) ≤ C for all n. Using the maximum principle in Proposition 6.2.5, there exists a structural positive constant C (independent of un ), such that C ≥ un (x0 ) ≥ C (Ω)δ(x0 ) ∫(λan (x)un + ∇un p + f )δ(x) dx,
(6.4.5)
Ω
where δ(x) = dist(x, 𝜕Ω). Let ϕ ∈ 𝒞0∞ (Ω) be a positive function. Using Tk (un )ϕ as a test 1,2 function in (6.4.1), we can prove that Tk (un ) is uniformly bounded in Wloc (Ω). 1,2 ∞ For n fixed, we consider vj ∈ W0 (Ω) ∩ L (Ω), the minimal positive solution to problem { −Δvj = λan (x)vj + { { vj = 0
∇vj p
1+ 1j ∇vj p
+ αf
in Ω, on 𝜕Ω.
(6.4.6)
Using an iteration argument as in [31], it follows that vj ≤ vj+1 and vj ≤ un for every n. Define wn = lim vj ≤ un . j→∞
Claim. The following statements hold: 1,2 (a) {Tk (wn )} is bounded in Wloc (Ω); 1,p (b) wn ∈ Wloc (Ω); (c) wn is a supersolution to problem (6.4.1); (d) wn ≤ wn+1 . Assume the claim holds. As before, there exists a positive structural constant C such that C ≥ wn (x0 ) ≥ C (Ω)δ(x0 ) ∫(λan (x)wn + ∇wn p + f )δ(x) dx. Ω
6.4 Blowup result  125
Since {wn } is a monotone sequence we conclude that an (x)wn ↗
w x2
in L1loc (Ω) and ∫ ∇wn p δ(x) ≤ C . Ω
1,p Thus {wn } is bounded in Wloc (Ω). Hence using (a) in the claim and Lemma 6.4.2, we conclude that
Tk (wn ) → Tk (w)
1,2 strongly in Wloc (Ω).
Since wn is a supersolution to (6.4.1), letting n → ∞ we obtain that w satisfies also −Δw ≥ ∇wp + λ
w + cf , x2
a contradiction with Theorem 6.3.2. Proof of the claim. Statements (a) and (d) follow directly from equation (6.4.6) by an application of the corresponding inequality of type (6.4.5) and the fact that an is a nondecreasing sequence. To prove (b) we consider separately two cases: (i) p ≤ 2 and (ii) p > 2. For p ≤ 2 we have 1,2 Tk (vj ) → Tk (wn ) as j → ∞ strongly in Wloc (Ω).
(6.4.7)
To obtain the convergence, we use a nonlinear test function as in [70] (see also [74] 1 2 and [71]). Consider ϕ(s) = se 4 s , in such a way that ϕ (s) − ϕ(s) ≥ 21 . For ψ ∈ 𝒞0∞ (Ω), ψ ≥ 0, take ϕ(Tk (vj )−Tk (wn ))ψ(x) as a test function in equation (6.4.6). We obtain from the left hand side ∫ ∇vj ϕ (Tk (vj ) − Tk (wn ))∇(Tk (vj ) − Tk (wn ))ψ dx Ω
2 = ∫∇(Tk (vj ) − Tk (wn )) ϕ (Tk (vj ) − Tk (wn ))ψ dx + o(1). Ω
We set H(∇vj ) =
∇vj p
1+ 1j ∇vj p
. Then, the right hand side could be estimated by
2 ∫ H(∇vj )ϕ(Tk (vj )−Tk (wn ))ψ dx ≤ δ ∫∇Tk (vj )−∇Tk (wn ) ϕ(Tk (vj )−Tk (wn ))ψ dx +o(1),
Ω
Ω
where δ ≤ 1. Since ∫(λan (x)vj + cf )ϕ(Tk (vj ) − Tk (wn ))ψ(x) dx → 0 Ω
as m → ∞,
126  6 Quasilinear elliptic equations with the Hardy potential we conclude the required convergence and in particular the almost everywhere convergence up to a subsequence. In the case p ≥ 2 the result is directly obtained as follows: since 2
1 2
2
2
1 2
∫ ∇vj  dx = ∫(−Δvj )vj dx ≤ ∫(−Δvj )un dx ≤ (∫ ∇vj  dx) (∫ ∇un  dx) , Ω
Ω
Ω
Ω
Ω
we have vj ⇀ wn weakly in W01,2 (Ω) as j → ∞. By using the last inequality and the weak lower semicontinuity of the norm it turns out that ∫ ∇wn 2 dx ≤ lim inf ∫ ∇vj 2 dx ≤ lim sup ∫ ∇vj 2 dx. j→∞
Ω
j→∞
Ω
Ω
Moreover, taking into account that −Δvj ≥ 0, 1 2
1 2
∫ ∇vj 2 dx = ∫(−Δvj )vj dx ≤ ∫(−Δvj )wn dx ≤ (∫ ∇vj 2 dx) (∫ ∇wn 2 dx) . Ω
Ω
Ω
Ω
Ω
Hence lim sup ∫ ∇vj 2 dx ≤ ∫ ∇wn 2 dx. j→∞
Ω
Ω
Then we conclude the strong convergence in W01,2 (Ω). In particular we have the almost everywhere convergence of the gradients. To conclude the proof of (b) it is sufficient to observe that C ≥ un (x0 ) ≥ C (Ω)δ(x0 ) ∫(λan (x)vj + Ω
∇vj p
1 + 1j ∇vj p
≥ C (Ω)δ(x0 ) ∫(λan (x)wn + ∇wn p + f )δ(x) dx,
+ f )δ(x) dx (6.4.8)
Ω
by Fatou’s lemma. To prove (c), we use a nonnegative test function in problem (6.4.6) and we pass to the limit by Fatou’s lemma.
6.5 Existence result: 1 < p < q+ (λ) and λ < ΛN,2 We consider α+ (λ) and α− (λ) defined in (3.1.3). Together with the critical exponent 2+α (λ) 2+α (λ) q+ (λ) ≡ 1+α− (λ) we define q− (λ) ≡ 1+α+ (λ) . Clearly q− (λ) ≤ q+ (λ). −
+
We can now state the following result of existence of solutions in the whole ℝN .
Theorem 6.5.1. Assume that q− (λ) < q < q+ (λ), where q− (λ), q+ (λ) are given above. Then problem (6.1.1) with f ≡ 0 has a very weak solution u > 0 in ℝN .
 127
6.5 Existence result: 1 < p < q+ (λ) and λ < ΛN,2
Proof. We search a solution in the form u(x) = Ax−β . Hence by a direct computation and we obtain β = 2−q q−1 βq Aq−1 = β(N − β − 2) − λ. To have A > 0 we need β ∈ (α− (λ), α+ (λ)), which is equivalent to q− (λ) < q < q+ (λ). Note N that u ∈ L1loc (Ω), xu 2 ∈ L1loc and, since N−1 < q− (λ) < q, we get ∇uq ∈ L1loc (compare with Remark 6.3.3). Hence the result follows. 1,2 Remark 6.5.2. The solution w in Theorem 6.5.1 is in the space Wloc (ℝN ) if and only N+2 N+2 if q > N . Note that for all λ ∈ [0, ΛN,2 ), N ∈ (q− (λ), q+ (λ)) and if λ = ΛN,2 , then N+2 = q− (λ) = q+ (λ). N
We deal now with the existence of solutions to Dirichlet problem in bounded domains. We will use the following result (Lemma 5.3 in [31]). Lemma 6.5.3. If un ⇀ u weakly in W01,2 (Ω) and 1. 0 ≤ un ≤ u; 2. −Δun ≥ 0 in 𝒟 (ω); then ‖∇un − ∇u‖2 → 0
as n → ∞.
Proof. Note that 2
2
1 2
2
1 2
∫ ∇un  = ∫ un (−Δun ) ≤ ∫ u(−Δun ) = ∫⟨∇u∇un ⟩ ≤ (∫ ∇u ) (∫ ∇un  ) . Ω
Ω
Ω
Ω
Therefore, by the weak semicontinuity of the norm, 2
1 2
2
1 2
2
1 2
2
1 2
(∫ ∇u ) ≤ lim inf(∫ ∇un  ) ≤ lim sup(∫ ∇un  ) ≤ (∫ ∇u ) , n→∞
n→∞
which allows us to conclude the proof. Theorem 6.5.4. Assume that 1 < p < q+ (λ), where q+ (λ) = such that if c < c0 and f (x) ≤
{
1 , x2
then problem
2+α− (λ) . 1+α− (λ)
−Δu = ∇up + λ xu 2 + cf
in Ω,
u=0
on 𝜕Ω
has a very weak positive solution u.
Then there exists c0
(6.5.1)
128  6 Quasilinear elliptic equations with the Hardy potential Proof. We start by assuming that for c > 0 and f (x) ≤ positive supersolution w ∈ W ∃s > 0, Consider an (x) =
1 x2 + n1
1,p
1 , x2
we are able to find a radial
(Ω) to problem (6.5.1) such that
for which
w1+s (2−p)s+p , w 2−p ∈ L1 (Ω). x2
(6.5.2)
↑ x−2 , fn = min{f , n} ↑ f . Then problem
{ −Δun = λan (x)un + { { un = 0
∇un p 1+ n1 ∇un p
+ cfn
in Ω,
(6.5.3)
on 𝜕Ω
has a minimal positive solution un ∈ W01,2 (Ω) ∩ L∞ (Ω). By Lemma 6.4.3 and using the comparison principle given in Proposition 6.2.3, we get un ≤ un+1 and un ≤ w for every n. Hence u = limn→∞ un ≤ w. Define ϕn = (1 + un )s − 1, where s is as in (6.5.2). Then using ϕn as a test function in (6.5.3), we have ∫ Ω
∇un 2 dx ≤ C1 , (1 + un )1−s
∫ ∇un p (1 + un )s dx ≤ C2 . Ω
In particular 1 ∫ ∇Tk un 2 ≤ C3 , k Ω
∫ ∇un p ≤ C4 . Ω
We claim now that ∇Tk vn → ∇Tk v
as n → ∞ strongly in W01,p (Ω). 1 2
To prove the claim we consider nonlinear test functions structured by ϕ(s) = se 4 s , which is appropriate because ϕ (s) − ϕ(s) ≥ 21 (see for instance [71]). That is, we consider ϕ(Tk vn − Tk v) as a test function in (6.5.3) and then ∫ ∇vn ∇(Tk vn − Tk v)ϕ (Tk vn − Tk v) dx Ω
=∫ Ω
∇vn p
1 + n1 ∇vn p
ϕ(Tk un − Tk u) dx + ∫(an vn + fn )ϕ(Tk un − Tk u) dx. Ω
It is clear that λ ∫(an vn + fn )ϕ(Tk un − Tk u) dx ≤ λ ∫( Ω
Ω
w + f )ϕ(Tk un − Tk u) dx → 0 x2
as n → ∞.
6.5 Existence result: 1 < p < q+ (λ) and λ < ΛN,2
 129
We deal now with the other terms. We have ∫ ∇vn ∇(Tk vn − Tk v)ϕ (Tk vn − Tk v) dx Ω
= ∫ ∇Tk vn ∇(Tk vn − Tk v)ϕ (Tk vn − Tk v) dx Ω
2 = ∫∇Tk (vn ) − ∇Tk (v) ϕ (Tk vn − Tk v) dx Ω
+ ∫ ∇Tk (v)∇(Tk (vn ) − Tk v)ϕ (Tk vn − Tk v) dx Ω
2 = ∫∇Tk (vn ) − ∇Tk (v) ϕ (Tk vn − Tk v) dx + o(1), Ω
where we have used the monotonicity of {vn }. On the other hand, since p ≤ 2, there exists ε ≤ 1 and a nonnegative constant C(ε) such that ∫ Ω
∇vn p
1 + n1 ∇vn p
ϕ(Tk vn − Tk v)
≤ ∫ ∇Tk vn p ϕ(Tk vn − Tk v) dx Ω
2 ≤ ε ∫∇Tk (vn ) ϕ(Tk vn − Tk v) dx + Cε ∫ϕ(Tk vn − Tk v) dx Ω
Ω
2
≤ ε ∫ ∇Tk vn − ∇Tk v ϕ(Tk vn − Tk v) dx − ε ∫ ∇Tk v2 ϕ(Tk vn − Tk v) dx Ω
Ω
+ 2ε ∫ ∇Tk vn ∇Tk vϕ(Tk vn − Tk v) dx + Cε ∫ϕ(Tk vn − Tk v) dx. Ω
Ω
The fact that Tk vn ⇀ Tk v weakly in ∫ Ω
∇vn p
1 + n1 ∇vn p
W01,2 (Ω)
gives
ϕ(Tk un − Tk u) ≤ ε ∫ ∇Tk vn − ∇Tk v2 ϕ(Tk vn − Tk v) dx + o(1). Ω
Hence it follows that ∫ ∇Tk vn − ∇Tk v2 (ϕ (Tk vn − Tk v) − εϕ(Tk vn − Tk v)) dx ≤ o(1).
Ω
Since ϕ (s) − εψ(s) ≥ 21 , we obtain ∫Ω ∇Tk vn − ∇Tk v2 ≤ o(1) and then ∇Tk vn → ∇Tk v as claimed.
as n → ∞ strongly in W01,p for all k > 0,
130  6 Quasilinear elliptic equations with the Hardy potential Note that, as a consequence, we can find a subsequence such that ∇vnj → ∇v almost everywhere in Ω. Define now ψn = (1 + Gk (vn ))s − 1, where as usual Gk (t) = t − Tk (t), the truncation function. Then using ψn as a test function in (6.5.3), by the Hölder inequality we obtain ∫ Ω
(2−p)s+p ∇Gk (vn )2 w1+s dx + ∫ f (x)(w + 1)s dx}. dx ≤ C0 { ∫ (1 + w) 2−p dx + λ ∫ 1−s (1 + Gk (vn )) x2
vn ≥k
vn ≥k
vn ≥k
It is clear by the hypotheses on w that ∫ (1+w)
(2−p)s+p 2−p
dx +λ ∫
vn ≥k
vn ≥k
w1+s dx + ∫ f (x)(w +1)s dx → 0 x2
as k → ∞ uniformly in n.
vn ≥k
Hence we conclude that s p lim sup ∫ ∇vn p dx ≤ lim sup ∫ ∇Gk (vn ) (1 + Gk (vn )) dx = 0 k→∞
k→∞
vn ≥k
(6.5.4)
Ω
uniformly in n. To prove that ∇vn p → ∇vp
strongly in L1 (Ω),
we will use Vitali’s theorem (see for instance [64] pag. 16). So, we only have to show the equiintegrability of {∇vn p }. Let E ⊂ Ω be a measurable set. Then, ∫ ∇vn p dx ≤ ∫ ∇Tk vn p dx + E
E
∫
∇vn p dx.
{vn ≥k}∩E
By the fact that Tk (vn ) → Tk (v) strongly in W01,p (Ω), the integral ∫E ∇Tk (vn )p dx is uni
formly small if E is small enough. On the other hand, by (6.5.4) we obtain ∫ {vn ≥k}∩E
∇vn p dx ≤ ∫ ∇vn p dx → 0
as k → ∞ uniformly in n.
{vn ≥k}
The equiintegrability of ∇vn q follows immediately, and the strong convergence holds. Hence, since ∇un → ∇u as n → ∞ strongly in Lp (Ω), we conclude that u is a very weak solution to problem (6.5.1). It is worthy to point out that for the values of p for which a supersolution in W01,2 (Ω) exists (in particular if 1 < p ≤ q− (λ)), the proof is easier and, moreover, the solution u ∈ W01,2 (Ω). In this last case it suffices to take un as test function and to use Lemma 6.5.3. To finish the proof we have to find the required supersolution. We will consider two cases:
6.5 Existence result: 1 < p < q+ (λ) and λ < ΛN,2
 131
(i) q− (λ) < p < q+ (λ); (ii) 1 < p ≤ p− (λ).
Case (i). q− (λ) < p < q+ (λ). Consider u, the radial solution obtained in Theorem 6.5.1. 1+s
Then ux2 , u 2−p ∈ L1 (Ω) for all 0 < s < solution to problem (2−p)s+p
{
p(N−1)−N 2−p
−Δv = 0
in Ω,
v=u
on 𝜕Ω.
< 1. Define v(x) to be the unique
Note that v ∈ 𝒞 ∞ (Ω) and 0 < c1 ≤ v ≤ c2 for some positive constants c1 and c2 . We set w = t(u − v), t > 0. Then w ∈ W01,p (Ω), w ≥ 0 in Ω and −Δw − λ
u v v w = t(−Δu − λ 2 ) + tλ 2 = t∇up + tλ 2 x2 x x x ≥ t(
1 + ϵ1 p−1 1 v p −p ∇w t − ( ) ∇vp ) + tλ 2 , 1+ϵ (1 + ϵ)p−1 x
where in the last estimate we have used the following elemental inequality: p−1
1 a + bp ≤ (1 + ϵ)p−1 ap + (1 + ) ϵ Taking t =
1 , 1+ϵ
bp .
we conclude that
−Δw − λ
1 λ w v ≥ ∇wp + ( )∇vp . ) 2 − ( p−1 2 1 + ϵ ϵ (1 + ϵ) x x
Hence choosing ϵ large enough there exists a positive constant c0 such that c 1 λ v − ∇vp ≥ 02 . 1 + ϵ x2 ϵp−1 (1 + ϵ) x Since x2 f (x) < 1, w ∈ W01,p (Ω) is a supersolution to problem (6.5.1) if c < c0 . Case (ii). 1 < p ≤ q− (λ). We start by getting a supersolution in a ball, i. e., Ω = BR (0). Without loss of generality we will assume R = 1. Since p ≤ q− (λ), there exists β ∈ (α− (λ), α+ (λ)), close to α− (λ), such that p(β + 1) < β + 2. Define w(x) ≡ A(x−β − 1). Then w ∈ W01,p (BR (0)) and −Δw − λ
w A = A(β(N − β − 2) − λ)x−β−2 + 2 . x2 x
Since β ∈ (α− (λ), α+ (λ)), we have β(N − β − 2) − λ > 0.
132  6 Quasilinear elliptic equations with the Hardy potential Hence choosing Ap−1 =
β(N−β−2)−λ βp
−Δw − λ
we obtain w A ≥ ∇wp + 2 , x2 x
that is, if c0 = A, w is a supersolution to (6.5.1) in B1 (0) for all c < c0 . In the case of a general domain Ω that contains the origin we consider a ball BR (0) such that Ω ⊂ B R (0). We have the corresponding supersolution in BR (0) found above, 2 for which we perform the same arguments as in the first case.
6.6 The critical case, λ ≡ ΛN,2 and p < Assume that λ = ΛN,2 and p < p+ ≡
N+2 N
x − w(x) = R where A = ( Rr )−
N−2 2
N+2 N
and consider the function
N−2 2
(log(
R )) x
1/2
− A,
R 1/2 (log( r )) . Then w(x) = 0 if x = r. It is easy to check that w ∈
W01,q (Br (0)) for all q < 2 and
−Δw − ΛN,2 N−2
w 1 w R = (log( ) x x2 4 x2
−2
1
+A
ΛN,2 . x2
N
R −1 R 2 Since ∇w(x) = R 2 (log( x )) x− 2 ( N−2 + 21 (log( x )) ) and p < 2 suitable positive constant c
∇wp ≤ c
N+2 , N
we get for a
w R (log( ) . x x2 −2
Hence, up to a positive constant c1 , c1 w is a supersolution to problem {
w −Δw = ∇wp + ΛN,2 x 2 + c0 f
in Br (0),
w=0
on 𝜕Br (0),
(6.6.1)
where x2 f is bounded and c0 is small. To prove the existence of solution, we consider the approximated problems { −Δun = ΛN,2 an (x)un + { { vk = 0
∇vn p 1+ n1 ∇vn p
+ cfn
in Br (0), on 𝜕Br (0),
(6.6.2)
where an (x) = min{n, x1 2 } and fn (x) = Tn (f (x)). It is easy to check that ΛN,2 < λ1 (an ), the principal eigenvalue of the Laplacian with weight an . Then by similar arguments to
6.6 The critical case, λ ≡ ΛN,2 and p
0. Then the existence result follows using the same computation as in the proof of Theorem 6.5.4. Hence the problem {
2.
−Δu = ΛN,2 xu 2 + ∇up + cf
in Ω,
u=0
on 𝜕Ω
has a positive solution u ∈ H(Ω) if x2 f is bounded and c is small. A proper definition of the gradient associated to the operator −Δ − xλ 2 I provides existence of solutions. Indeed, in [15] the following example is studied: −Δu − ΛN,2
2 N−2 u N − 2 u ) = x ∇u + ( x 2 + λf (x), 2 2 2 x x
)2 and with f under some hypotheses of for x in Ω, u = 0 on 𝜕Ω and ΛN,2 = ( N−2 2 summability.
6.7 Further remarks The following interesting questions are still open: 1. Fixing 1 < p < q+ (λ), obtain the optimal class of functions – according to their summability – in order to have existence of a very weak solution to the Dirichlet problem with data in such a class. 2. Assume that, for λ fixed and for f in a given class, we are able to find a very weak solution, u. What is the regularity of u in terms of the regularity of f ? 3. Complete the study of results on uniqueness and nonuniqueness. We recall that for λ = 0 there are some results on multiplicity of unbounded solutions. See for instance [157] (for a ball) and [8], where all the solutions are characterized in a bounded domain.
136  6 Quasilinear elliptic equations with the Hardy potential As in the previous chapters, another line of study is to replace the Laplacian by other elliptic operators, such as the pLaplacian. This problem, with a power of the gradient on the left hand side and with the Hardy–Leray potential, has been studied in [19] and [20].
7 The heat equation with nonlinearity on the gradient and the Hardy–Leray potential 7.1 Introduction This chapter deals with the existence and nonexistence of positive solutions to the following problem: { { { { { { { { { { { { { { {
ut − Δu = ∇up + λ xu 2 + f (x, t)
in ΩT ≡ Ω × (0, T),
u(x, t) > 0
in ΩT ,
u(x, t) = 0
on 𝜕Ω × (0, T),
u(x, 0) = u0 (x)
if x ∈ Ω,
(7.1.1)
where Ω is either an open bounded domain in ℝN such that 0 ∈ Ω, or Ω = ℝN , N ≥ 3, p > 1 and λ > 0. We suppose that f and u0 are positive measurable functions with some hypotheses that will be specified later. For the theory of quasilinear parabolic problem with general data we refer for instance to [61], [62], [67], [70], [77], [265], [267] and the references therein. The general contents of the chapter may be described by saying that we want to understand the effect of the uncertainty principle expressed by the Hardy inequality in this kind of evolution models. Problem (7.1.1) can be seen as a Hamilton–Jacobi equation with a viscosity term; see [127], [232], [55] and the references therein. As we explained in Chapter 6 for the case λ = 0 and p = 2, this model, proposed by Kardar–Parisi–Zhang, appears in the physical theory of growth and roughening of surfaces; see [205]. If p = 2 and λ = 0, a general multiplicity result for positive solutions was proved in [9], where the authors showed a direct connection between problem (7.1.1) and semilinear problems with measure data. For general p > 1 and λ = 0, the model is known as the generalized Kardar–Parisi– Zhang equation. This was introduced by Krug–Spohn in [213]. Existence results are obtained in [187] and [55] for the Cauchy problem (Ω = ℝN ) and for regular initial data u0 and f ≡ 0. Note that the results by B. Gilding, M. Guedda and R. Kersner in [187] show that there is no Fujita type exponent for the Cauchy problem with λ = 0. One of the main results in this chapter is to prove that the optimal power for existence of solutions in the case of the stationary problem studied in Chapter 6, that is, q+ (λ) ≡
2 + α− (λ) , 1 + α− (λ)
0 < λ ≤ ΛN,2 ,
(7.1.2)
with α− (λ) defined in (3.1.3), is also the critical power for the parabolic equation (7.1.1). We recall that the critical power only depends on λ and N. We will also use the related https://doi.org/10.1515/9783110606270007
138  7 Heat equation with a nonlinearity on the gradient parameter q− (λ) =
2 + α+ (λ) , 1 + α+ (λ)
(7.1.3)
with α− (λ) defined again in (3.1.3). Note that q− (λ) ≤ q+ (λ). In order to show that q+ (λ) is the critical power, we will prove that if p ≥ q+ (λ) there is no nontrivial local solution, even in the distributional sense. In the opposite case, if p < q+ (λ), there exist positive solutions under some suitable additional hypotheses on the data. It is due to this behavior depending on the spectral parameter λ that the problem becomes so interesting. Note that the nonlinear perturbations of the heat equation that we are considering do not satisfy the socalled sign condition (see [221]). To prove the existence of solution we use the classical arguments of truncation and compactness that allow to pass to the limit. We need to obtain uniform estimates under the presence of the Hardy potential and the lack of monotonicity of the nonlinear terms. These difficulties make us prove some additional comparison lemmas and new estimates. We will also use the results on almost everywhere convergence of the gradients obtained in [73] (see also [67]). In the second part of this chapter we analyze the behavior of the Cauchy problem (Ω = ℝN ), which brings us some surprises with respect to the case λ = 0. Indeed, if ΛN,2 ≥ λ > 0, there exists a Fujita type exponent. This is a consequence of the singularities introduced by the Hardy potential. The present chapter is organized as follows. In Section 7.2 we present some preliminaries and prove results that will be used in the next sections. Section 7.2.1 is devoted to prove a new maximum principle for the heat equation, which has a precedent in the maximum principles obtained in Chapter 6 for the Laplace equation and has some interest in itself. The techniques for the proof in the parabolic case are very different to those in the stationary case. As a consequence, we obtain a comparison principle and then a uniqueness result for some approximated problems. In Section 7.2.2 we find a lower estimate for the local behavior of a general supersolution and some a priori estimates for very weak solutions in the sense of Definition 7.2.1. Section 7.3 deals with the nonexistence of nontrivial local solutions to problem (7.1.1) for supercritical powers. It is in this place where the Caffarelli–Kohn–Nirenberg inequalities play an important role. In Section 7.4 we prove the existence result in bounded domains for p < q+ (λ) under some additional hypotheses on u0 and f . The idea is to use the fact that the associated elliptic problem has a positive solution. Since the supersolution of the elliptic problem is a supersolution to the parabolic problem under some hypotheses on the data, using an iteration argument and the comparison principle obtained in Section 7.2.1, we get the existence of a positive solution to the parabolic problem in an appropriate sense. Whence, q+ (λ) is the critical exponent for existence of positive solutions to problem (7.1.1).
7.2 Preliminaries and tools  139
In Section 7.5 we analyze the Cauchy problem (7.5.2), with p < p+ (λ). The most relevant fact in this case is the existence of a Fujita type exponent. This result comes in some way as a surprise, because the quasilinear problem with λ = 0 does not present a similar behavior. See [187]. The study is performed as follows. In Section 5.6.1 we find a family of subsolutions to (7.5.2) blowing up in finite time, for 1 < p < F(λ) = 1 + N−α 1(λ)+1 < p+ (λ), in a natural weighted norm. The interval F(λ) < p < p+ (λ) is − studied in Section 7.5.2, where we find a family of supersolutions to problem (7.5.2) defined for all time t > 0. With all these previous results in hand, we conjecture that F(λ) is really the Fujita type exponent for problem (7.5.2). To prove that this is true, we proceed in several steps. In Section 7.5.3 we prove a local in time existence theorem, for a suitable class of initial data and 1 < p < p+ (λ). Moreover in Section 7.5.4, using the global supersolutions found in Section 7.5.2, we obtain that for small initial data and F(λ) < p < p+ (λ), there exist global solutions. Section 7.5.5 is devoted to prove Theorem 7.5.8, the blowup with respect to a natural norm, which in particular shows that F(λ) is the Fujita exponent for (7.5.2). Note that this is a major difference with respect to the case of the heat equation (λ = 0) with a gradient term, for which given any p > 1 and for small initial datum, there exists a global solution; see [187]. Almost all new results on the topic presented in this chapter were obtained in [23].
7.2 Preliminaries and tools As in the elliptic case, despite the fact that weights do not appear in the principal part of the operator in problem (7.1.1), we will need some properties of weighted Sobolev spaces related to the Caffarelli–Kohn–Nirenberg inequalities. See Section 6.2.3. We recall the definition of the ktruncation function Tk (σ), Tk (σ) = {
σ
σ ≤ k,
k sign(σ)
σ > k
(7.2.1)
and Gk (σ) = σ − Tk (σ).
(7.2.2)
As in previous situations, we define the concept of solution in the more general distributional sense. Definition 7.2.1. Consider the equation ut − Δu = ∇up + λ
u + f (x, t) x2
in Ω × (0, T).
(7.2.3)
We say that u ∈ 𝒞 ((0, T); L1loc (Ω)) is a very weak supersolution (subsolution) to probu 1 p 1 1 ∞ lem (7.2.3) if x 2 ∈ Lloc (ΩT ), ∇u ∈ Lloc (ΩT ), f ∈ Lloc (ΩT ) and for all ϕ ∈ 𝒞0 (Ω × (0, T))
140  7 Heat equation with a nonlinearity on the gradient such that ϕ ≥ 0, we have T
T
0 Ω
0 Ω
∫ ∫(−ϕt − Δϕ)u dx dt ≥ (≤) ∫ ∫(∇up + λ
u + cf )ϕ dx dt. x2
(7.2.4)
If u is a very weak super and subsolution, then we say that u is a very weak solution. Note that all these notions of solution are local in nature. The nonexistence result in this framework is the strongest possible. It is clear that if u is a very weak supersolution (subsolution) to (7.1.1), then 1,p u ∈ 𝒞 ((0, T); L1loc (Ω)) ∩ Lp ((0, T); Wloc (Ω)).
Assume u is a solution to problem (7.1.1) such that
u x2
∈ L1 (ΩT ), u ∈ Lp ((0, T);
W01,p (Ω)), f ∈ L1 (ΩT ) and u0 (x) ∈ L1 (Ω). Then u can be recovered as a limit of truncated problems. Indeed, let us define g(x, t) ≡ ∇up + λ xu 2 + f ; then g ∈ L1 (ΩT ). Consider now
{gn }n∈ℕ ⊂ L1 (ΩT ) ∩ L∞ (ΩT ), gn → g in L1 (ΩT ) and {ϕn }n∈ℕ ⊂ L1 (Ω) ∩ L∞ (Ω), ϕn → u0 in L1 (Ω). Solving the problems unt − Δun = gn in ΩT ,
un (x, 0) = ϕn (x) in Ω,
u(x, t) = 0 on (x, t) ∈ 𝜕Ω × (0, T),
the theory of solutions obtained as limit of approximations says that un → u in Lq (0, T; W01,q (Ω)),
q < max{p,
N +2 } N +1
and u ∈ 𝒞 ([0, T], L1 (Ω)). See Section 5.2. Since un ∈ L2 ((0, T); W02 (Ω)) ∩ C([0, T]; L2 (Ω)) and unt ∈ L2 ((0, T); W −1,2 (Ω)) the relevant fact is that we can take test functions ϕ ∈ L2 ((0, T); W01,2 (Ω)) ∩ L∞ (Ω × (0, T)). See for instance [129], [60], [61] and [267] for more general operators. The concept of solutions obtained by approximations is equivalent in this setting to entropy solution or renormalized solution. See, for instance, [258] for a proof of the equivalence. The previous remarks motivate the strategy to follow in order to solve problem (7.1.1). We will consider some truncated problems (see (7.2.10) below) and then we will obtain uniform estimates that permit to pass to the limit and find a solution at least in the renormalized, or entropy, setting. 7.2.1 Maximum principle and comparison results We begin by proving a maximum principle that we will use systematically here. This result is the parabolic counterpart of a result for elliptic equations obtained in Section 6.2; see also [31]. The techniques used in the proof of the parabolic case are completely different from those used in the elliptic case.
7.2 Preliminaries and tools  141
Lemma 7.2.2 (Maximum principle). Let h(x, t) be a measurable function such that h ∈ L2r0 ([0, T]; L2p0 (Ω)), where p0 , r0 > 1 and 2pN + r1 < 1. Assume that w(x, t) ≥ 0 verifies: 1,p
0
0
(i) w ∈ 𝒞 ((0, T); L1 (Ω)) ∩ Lr1 ([0, T]; W0 1 (Ω)), where r1 , p1 ≥ 1 such that 1 N +1 N + > ; 2p1 r1 2
(ii) w is a subsolution to problem wt − Δw ≤ h∇w { { { w(x, t) = 0 { { { { w(x, 0) = 0
in ΩT , (7.2.5)
on 𝜕Ω × (0, T), in Ω.
Then w ≡ 0. 1,p
N Proof. Since w ∈ Lr1 ([0, T]; W0 1 (Ω)) with r1 , p1 ≥ 1 such that 2p + r1 > N+1 , using 2 1 1 the Gagliardo–Nirenberg interpolation inequality (see [173] and [251]) it follows that w ∈ Lr2 ([0, T]; Lp2 (Ω)) with r2 , q2 ≥ 1 and 2pN + r1 > N2 .
As
N 2p0
+
1 r0
< 1, we have
N 2p0
+
1 r0
>
N . 2
2
2
We choose β > 0 such that
N 1 N 1 ( + )> . β + 1 2p0 r0 2 We claim that T
∫ ∫ wβ−1 ∇w2 dx dt < ∞.
(7.2.6)
0 Ω s
s β ) and define Dε (s) = ∫0 mϵ (τ)dτ. Using mε (w) as For ε > 0, consider mϵ (s) ≡ ( 1+εs a test function in (7.2.5), it follows that T
∫ Dε (w(x, T)) dx + β ∫ ∫( 0 Ω
Ω
β−1
w ) 1 + εw
T
β
∇w2 w dx dt ≤ ∫ ∫( ) h∇w dx dt. 1 + εw (1 + εw)2 0 Ω
From the Young inequality we get T
T
β
β−1
w w ) h∇w dx dt ≤ η1 ∫ ∫( ) ∫ ∫( 1 + εw 1 + εw 0 Ω
T
2
∇w dx dt + η2 ∫ ∫ h2 (x, t)wβ+1 dx dt.
0 Ω
0 Ω
On the other hand, by the Hölder inequality, T
T
1 p0
∫ ∫ h2 wβ+1 dx dt ≤ ∫(∫ h2p0 dx) (∫ wp0 (β+1) dx) 0 Ω
0
dt
Ω
Ω T
≤ (∫(∫ h2p0 dx) 0
1 p 0
Ω
r0 p0
1 r0
T
dt) (∫(∫ w 0
Ω
p0 (β+1)
dx)
r0 p 0
1 r 0
dt) .
142  7 Heat equation with a nonlinearity on the gradient Using the hypothesis on h and the choice of β, it follows that T
(∫(∫ h 0
2p0
dx)
r0 p0
1 r0
T
dt) (∫(∫ w 0
Ω
p0 (β+1)
dx)
r0 p 0
dt)
1 r 0
< ∞.
Ω
Letting ε → 0 and using Fatou’s lemma, we find that T
∫ wβ+1 (x, T) dx + β ∫ ∫ wβ−1 ∇w2 dx dt < ∞, 0 Ω
Ω
and then the claim follows. By the assumption on h, there exists q > 2 such that hq ∈ Lr2 ([0, T], Lp2 (Ω)) and 1 N + < 1. 2p2 r2 Let ψ be the solution to the problem ψt − Δψ = h(x, T − t)q ψ { { { ψ(x, t) = 0 { { { { ψ(x, 0) = 1
in ΩT , on 𝜕Ω × (0, T),
(7.2.7)
in Ω.
Using the classical parabolic regularity theory (see for instance [220], [39]), we obtain ψ ∈ L∞ (ΩT ). Considering ψ1 (x, t) = ψ(x, T − t), ψ1 solves dψ
− dt1 − Δψ1 = h(x, t)q ψ1 { { { ψ1 (x, t) = 0 { { { { ψ1 (x, T) = 1
in ΩT , on 𝜕Ω × (0, T),
(7.2.8)
in Ω.
w β ) ψ1 as a test function in (7.2.5), passing to the limit as ε → 0 and using Taking ( 1+εw the estimate (7.2.6), it follows that t
1 ∫ wβ+1 (x, t)ψ1 (x, t) dx + β ∫ ∫ wβ−1 ∇w2 ψ1 dx ds β+1 Ω
t
0 Ω
t
1 + ∫ ∫ wβ+1 ((−ψ1 )t − Δψ1 ) dx ds ≤ ∫ ∫ h∇wwβ ψ1 dx ds β+1 t
0 Ω
0 Ω
= ∫ ∫ h∇wwθ1 +θ2 +θ3 ψ1 dx ds, 0 Ω β−1
β+1
with θ1 = 2 , θ2 = q and θ3 = . Note that θ1 + θ2 + θ3 = β. The Young 2q inequality with θ1 , θ2 , θ3 and the definition of ψ1 gives (q−2)(β+1)
7.2 Preliminaries and tools  143 t
1 ∫ wβ+1 (x, t)ψ1 (x, t) dx + β ∫ ∫ wβ−1 ∇w2 ψ1 dx ds β+1 Ω
0 Ω
t
t
1 + ∫ ∫ wβ+1 hq ψ1 dx ds ≤ η1 ∫ ∫ wβ−1 ∇w2 ψ1 dx ds β+1 t
0 Ω
t
0 Ω
+ η2 ∫ ∫ wβ+1 hq ψ1 dx ds + η3 ∫ ∫ wβ+1 (x, s)ψ1 (x, s) dx ds. 0 Ω
0 Ω
Choosing η1 and η2 small, we get t
1 ∫ wβ+1 (x, t)ψ1 (x, t) dx ≤ η3 ∫ ∫ wβ+1 (x, s)ψ1 (x, s) dx ds. β+1 0 Ω
Ω
Since w ≥ 0 and ψ1 > 0 in Ω × (0, t) for t < T, by Gronwall’s inequality we obtain w ≡ 0. For the reader convenience we recall that a function F : Ω × ℝ × ℝN → ℝ is called a Caratheodory function if is measurable in (x, t) ∈ Ω × ℝ for ξ fixed and continuous in ξ ∈ ℝN for a.e (x, t) ∈ Ω × ℝ. As a consequence of Lemma 7.2.2 we have the following comparison result. Lemma 7.2.3 (Comparison principle). Let u, v ∈ 𝒞 ((0, T); L1 (Ω)) ∩ Lp ((0, T); W01,p (Ω)), for some p > 1, with ut − Δu ∈ L1 (ΩT ) and vt − Δv ∈ L1 (ΩT ). Consider a Caratheodory function H(x, t, s) such that H(x, t, ⋅) ∈ 𝒞 1 (ℝN ) for all (x, t) ∈ ΩT with sup ∇s H(x, t, s) = h(x, t) ∈ L2r ([0, T]; L2p (Ω)),
s∈ℝN
for p, r > 1 and
N 1 + < 1. 2p r
Assume that u and v verify {
ut − Δu ≥ H(x, t, ∇u) + f u(x, 0) = u0 (x)
in Ω,
in Ω,
{
vt − Δv ≤ H(x, t, ∇v) + f
v(x, 0) = v0 (x)
in Ω,
in Ω,
(7.2.9)
where f ∈ L1 (ΩT ), u0 , v0 ∈ L1 (Ω) and v0 (x) ≤ u0 (x) in Ω. Then v ≤ u in ΩT . Proof. Denoting w = v − u, it is sufficient to prove that w+ = 0. For (x, t) ∈ ΩT , consider J(σ) = H(x, t, σ∇v + (1 − σ)∇u), σ ∈ [0, 1]. By hypothesis, there exists θ ∈ (0, 1) such that H(x, t, ∇v) − H(x, t, ∇u) = ⟨∇s H(x, t, θ∇v + (1 − θ)∇u), ∇w⟩. From (7.2.9), we conclude that w satisfies wt − Δw ≤ h(x, t)∇w in ΩT . Therefore, applying Kato inequality (see Lemma 3.2 in [255]) we have w+ − Δw+ ≤ h(x, t)∇w+  { { t+ w (x, t) = 0 { { + { w (x, 0) = 0
in ΩT ,
on 𝜕Ω × (0, T),
in Ω.
By the hypothesis on h and using Lemma 7.2.2, we conclude that w+ ≡ 0.
144  7 Heat equation with a nonlinearity on the gradient The following consequences of this will be very useful. Corollary 7.2.4. Assume u0 ∈ L1 (Ω), f ∈ L1 (ΩT ) and H(s) = consider the problem
unt − Δun − λan (x)un = { { { { un (x, t) = 0 { { { { { un (x, 0) = un0 (x)
∇un p 1+ n1 ∇un p
+ fn
sp , 1+ n1 sp
p > 1. For n ∈ ℕ
in ΩT , (7.2.10)
on 𝜕Ω × (0, T), if x ∈ Ω,
where an (x) = min{ x1 2 , n}, un0 = Tn (u0 ) and fn = Tn (f ). Then (7.2.10) has a unique positive solution un . Moreover, un ≤ un+1 for every n ∈ ℕ.
Corollary 7.2.5. Assume that 1 ≤ p < N+2 and f ∈ L1 (ΩT ). Let u, v be two functions such N+1 p 1,p that u, v ∈ L ((0, T); W (Ω)), ut − Δu, vt − Δv ∈ L1 (ΩT ) and {
ut − Δu ≥ ∇up + f
in ΩT ,
u(x, 0) = u0 (x)
in Ω,
{
vt − Δv ≤ ∇vp + f
in ΩT ,
v(x, 0) = v0 (x)
in Ω,
(7.2.11)
where v ≤ u on 𝜕Ω × (0, T), u0 , v0 ∈ L1 (Ω) and v0 (x) ≤ u0 (x) in Ω. Then v ≤ u in ΩT . Proof. Consider w = v − u. It is clear that w ∈ Lp ((0, T); W 1,p (Ω)), w ≤ 0 on 𝜕Ω × (0, T), w0 (x) ≤ 0 in Ω and wt − Δw ∈ L1 (ΩT ). In order to conclude, it is sufficient to prove that w+ = 0. We have wt − Δw ≤ ∇vp − ∇up in ΩT . Since 1 ≤ p
1 and a(x, t) = 1 if p = 1. Using Kato’s inequality we have (w+ )t − Δw+ ≤ a(x, t)∇w+ , { { { w+ (x, 0) = 0 in Ω, { { { p 1,p { w+ ∈ L (0, T, W (Ω)). Since p < N+2 , we have a(x, t) ∈ L2r ([0, T]; L2q (Ω)) for q, r > 1 and N+1 previous comparison lemma, we conclude that w+ = 0. Remark 7.2.6. As a consequence of Corollary 7.2.5, if 1 ≤ p < ut − Δu = ∇up { { { u=0 { { { { u0 (x) = u0
in ΩT , on 𝜕Ω × (0, T), in Ω
N 2q
N+2 , N+1
+
1 r
< 1. Using the
then the problem
7.2 Preliminaries and tools  145
has a unique positive solution u ∈ 𝒞 ([0, T]; L1 (Ω)) ∩ Lp ((0, T), W01,p (Ω)). It is worthy to point out that since α− (λ) < N, we have N+2 < q+ (λ), where q+ (λ) is the critical power N+1 given in (7.1.2).
7.2.2 Local behavior of a very weak supersolution to problem (7.1.1) We begin by analyzing the behavior of any positive solution in a neighborhood of the origin. According to Lemma 5.2.3, if u is a nonnegative supersolution defined in Ω, that is, u ≩ 0, u ∈ L1loc (ΩT ), xu 2 ∈ L1loc (ΩT ) and u satisfies ut − Δu − λ xu 2 ≥ 0 in 𝒟 (ΩT ) with λ ≤ ΛN,2 and Br1 (0) ⊂⊂ Ω, then there exists a constant C = C(N, r1 , t1 , t2 ) such that for each cylinder Br (0) × (t1 , t2 ) ⊂⊂ ΩT , 0 < r < r1 , u ≥ Cx−α− (λ)
in Br (0) × (t1 , t2 ),
where α− (λ) is given in (3.1.3). By choosing r small enough, we may assume that u > 1 in Br (0) × (t1 , t2 ). We obtain some a priori local integral estimates for a very weak solution to problem (7.2.3). Lemma 7.2.7. Assume that u is a nonnegative very weak supersolution to (7.2.3) with λ ≤ ΛN,2 and α− is as in (3.1.3). Then in each cylinder Br (0) × (t1 , t2 ) with 0 < t1 < t2 ≤ T we have t
1.
∫t 2 ∫B (0) ∇up x−α− (λ) dx dt < ∞;
2.
∫t 2 ∫B (0)
3.
1
t
1
r
r
u x2+α− (λ)
dx dt < ∞;
t ∫t 2 ∫B (0) f x−α− (λ) dx dt 1 r
< ∞.
u 1 Moreover, if ∇up + λ x 2 + f ∈ L (ΩT ) and u is a solution to problem (7.1.1) obtained as a limit of approximations, then
sup ∫ u(x, t)x−α− (λ) dx < ∞.
t∈(t1 ,t2 )
(7.2.12)
Ω
1,p (Ω)), fixing 0 < T1 < t1 < t2 < T2 < T, Proof. Since u ∈ 𝒞 ((0, T); L1loc (Ω)) ∩ Lp ((0, T); Wloc p 1,p in particular, u ∈ L ((T1 , T2 ); W (Br (0))) for all Br (0) ⊂ Ω. Moreover, there exists a positive constant c0 such that u(x, t) ≥ c0 for all (x, t) ∈ Br (0) × (T1 , T2 ). Consider w,̄ the solution to w̄ w̄ t − Δw̄ = λ x 2 + 1 { { { ̄ t) = 0 w(x, { { { ̄ T1 ) = 0 { w(x,
in Br (0) × (T1 , T2 ), on 𝜕Br (0) × (T1 , T2 ), in Br (0),
(7.2.13)
146  7 Heat equation with a nonlinearity on the gradient with λ ≤ ΛN,2 . It is clear that w ∈ L2 (0, T; W01,2 (Ω)) if λ < ΛN,2 and w ∈ L2 (0, T; H(Ω)) if λ = ΛN,2 , where H(Ω) is the Hilbert space defined in Section 2.5.1. According to Lemma 5.2.3, there exists a constant C = C(N, r1 , T1 , T2 ) such that w̄ ≥ Cx−α− in ̄ T2 + T1 − t) satisfies Br (0) × (T1 , T2 ). Then w(x, t) = w(x, w −wt − Δw = λ x 2 + 1 { { { w(x, t) = 0 { { { { w(x, T2 ) = 0
in Br (0) × (T1 , T2 ), on 𝜕Br (0) × (T1 , T2 ), in Br (0).
Observe that w(x, t) ≥ Cx−α− in Br (0) × (t1 , t2 ). Let us consider the sequence {un } of solutions to the approximated problems p
u
+ λ xn2 + Tn (f ) unt − Δun = ∇u { 1+ n1 ∇up { { { un (x, t) = 0 { { { { { un (x, T1 ) = c0
in Br (0) × (T1 , T2 ), on 𝜕Br (0) × (T1 , T2 ),
(7.2.14)
in Br (0).
Since λ ≤ ΛN,2 , using the weak comparison principle for the operator vt − Δv − λ xv 2 , it follows that un ≤ un+1 and un ≤ u, ∀n ∈ ℕ. Taking w as a test function in (7.2.14), we find T2
T2
∇up
∫ ∫ un dx dt = ∫ ∫ T1 Br (0)
T1 Br (0)
1 + n1 ∇up
T2
w dx dt + ∫ ∫ Tn (f )w dx dt. T1 Br (0)
Hence passing to the limit as n → ∞ and using Fatou’s lemma we obtain T2
T2
p
∫ ∫ ∇u w dx dt < ∞ and
∫ ∫ fw dx dt < ∞.
T1 Br (0)
T1 Br (0)
Taking into account the lower estimate of w near the origin given by Lemma 5.2.3, we have t2
p
t2 −α− (λ)
∫ ∫ ∇u x
dx dt < ∞
and
t1 Br (0)
To get the estimate on the term
∫ ∫ f x−α− (λ) dx dt < ∞. t1 Br (0)
u , x2+α− (λ) p
consider vn , the solution to
vnt − Δvn = ∇u +λ { 1+ n1 ∇up { { { vn (x, t) = 0 { { { { { vn (x, T1 ) = c0
Tn (u) x2
+f
in Br (0) × (T1 , T2 ), on 𝜕Br (0) × (T1 , T2 ), if x ∈ Br (0).
(7.2.15)
7.2 Preliminaries and tools  147
It is clear that {vn } is an increasing sequence on n such that vn ≤ u. Hence taking again w as a test function in (7.2.15) and using the same computation as above, it follows that t2
∫ ∫ t1 Br (0)
u
x2+α− (λ)
dx dt < ∞.
We prove now estimate (7.2.12). Fixing t1 , t2 such that (t1 , 2t2 ) ⊂ (0, T), let φ be the solution to the following elliptic problem: φ
{
−Δφ = λ x2 + 1 φ(x) = 0
in Ω,
(7.2.16)
on 𝜕Ω.
Note that φ ≃ Cx−α− (λ) in Br (0). Taking (2t2 − t)φ as a test function in (7.2.15) and integrating in Ω × (t, 2t2 ) with t1 < t < t2 , it follows that 2t2
2t2
− (2t2 − t) ∫ vn (x, t)φ dx + ∫ ∫ vn (x, σ)φ dx dσ + ∫ ∫(2t2 − σ)u(x, σ)( 2t2
t Ω
Ω
t Ω
∇up
λφ + 1) dx dσ x2
2t2
T (u) = ∫ ∫( + λ n 2 )(2t2 − σ)φ dx dσ + c ∫ ∫ f (2t2 − σ)φ dx dσ. x 1 + n1 ∇up t Ω
t Ω
Hence 2t2
2t2 ∫ ∫ un (x, σ)( t1 Ω
λφ +1) dxdσ ≥ (2t2 −t) ∫ u(x, t)φ dx ≥ c(2t2 −t) ∫ un (x, t)x−α− (λ) dx. x2 Ω
Br (0)
Thus by letting n → ∞ we conclude that 2t2
∫ u(x, t)x Br (0)
−α− (λ)
t2 dx ≤ ∫ ∫ u(x, σ)(cx−α− (λ)−2 +1) dx dσ ≤ C (2t2 − t)
for all t ∈ (t1 , t2 ).
t1 Ω
Then (7.2.12) follows. Remark 7.2.8. Since the blowup in the L∞ norm is instantaneous, estimate (7.2.12) suggests the natural norm to study nontrivial blowup results. 7.2.3 Passing to the limits in locally truncated problems Consider for fixed n ∈ ℕ and k ≥ 1 the truncated problems vkt − Δvk = λTn ( x1 2 )vk + { { { { vk (x, t) = 0 { { { { { vk (x, 0) = Tn (ũ 0 )
∇vk p 1+ k1 ∇vk p
+f
in Ω1 × (0, T), on 𝜕Ω1 × (0, T), if x ∈ Ω1 ,
(7.2.17)
148  7 Heat equation with a nonlinearity on the gradient where Ω1 is any bounded regular domain with Ω1 ⊂⊂ Ω, 0 < λ ≤ ΛN,2 , 1 < p < q+ (λ), u0 ∈ L1 (Ω) and f ∈ L∞ (ΩT ). We suppose that ũ ∈ 𝒞 ([0, T); L1loc (Ω)) is a very weak supersolution to (7.2.3) with 1 < p ≤ 2. Then, clearly, ũ is a supersolution to (7.2.17). ̃ for every k. Therefore, Hence, using Lemma 7.2.3 we find that vk ≤ vk+1 and vk ≤ u ̃ a. e. and in L1 (ΩT ). there exists un satisfying un = limk→∞ vk ≤ u For simplicity of notation we set gk (x, t) =
∇vk p
1+
1 ∇vk p k
+ λan (x)vk + f ,
where an = Tn (
1 ). x2
We prove that gk is uniformly bounded in L1 (Ω1 × (0, T), ϕ dx dt), where ϕ is the solution to the problem {
−Δϕ = 1
in Ω1 ,
ϕ=0
on 𝜕Ω1 .
(7.2.18)
Note that ϕ(x) ⋍ c dist(x, 𝜕Ω1 ). Indeed, using ϕ as a test function in (7.2.17) and since ̃ , we have vk ≤ u T
̃ s) dx ds ̃ T)ϕ dx + ∫ ∫ u(x, ∫ u(x, 0 Ω1
Ω1
T
T
≥ ∫ vk (x, T)ϕ dx + ∫ ∫ vk (x, s) dx ds ≥ ∫ ∫ gk (x, s)ϕ dx ds, 0 Ω1
Ω1
0 Ω1
thus T
∫ ∫ gk ϕ dx ds ≤ C
for all k.
0 Ω1
It is not difficult to show that u solves ut − Δu = μ in 𝒟 (Ω1 × (0, T)),
where gk ⇀ μ weakly in the measures sense.
The almost everywhere pointwise convergence of the gradient is well known, but for the selfcontained arguments, we include a proof. To that end, we follow the ideas in [67]. We refer also to the papers [60] and [258], where strong results about the convergence of truncating approximation are obtained. Lemma 7.2.9. Assume that {vk } is defined by (7.2.17). Then ∇vk → ∇un a.e.(x, t) ∈ Ω1 × (0, T).
7.2 Preliminaries and tools  149
Proof. Using Tm (vk ) ⋅ ϕ as a test function in (7.2.17) we get T
2 ∫ Θm (vk (x, T))ϕ dx + ∫ ∫ ∇Tm (vk ) ϕ dx ds 0 Ω1
Ω1 T
T
+ ∫ ∫ Θm (vk )(−Δϕ) dx ds = ∫ ∫ gk (x, s)ϕTm (vk ) dx ds, 0 Ω1
0 Ω1
s
where Θm (s) = ∫0 Tm (σ) dσ. Thus, by the previous boundedness, we find T
2 ∫ Θm (vk (x, T))ϕ dx + ∫ ∫ ∇Tm (vk ) ϕ dx ds 0 Ω1
Ω1 T
T
+ ∫ ∫ Θm (vk ) dx ds ≤ m ∫ ∫ gk (x, s)ϕ dx ds ≤ Cm. 0 Ω1
0 Ω1
Then T
2 ∫ ∫ ∇Tm (vk ) ϕ dx ds ≤ Cm, 0 Ω1
and, therefore, up to a subsequence, ∇Tm (vk ) ⇀ ∇Tm (un )
N
weakly in (L2 (Ω1 × (0, T), ϕ dx dt)) .
Since gk is bounded in L1 (Ω1 × (0, T), ϕ dx dt), we obtain T
∫ ∫ ∇vk q ϕ dx dt < C,
q
0, we have T
θ ∫ ∫ ∇(vk − un ) ϕ dx dt → 0
as k → ∞.
0 Ω1
As in [67], we denote by ω(k, ν, m, ε) any quantity that satisfies lim lim sup lim sup lim sup w(k, ν, m, ε) = 0,
ε→0+
m→∞
ν→∞
k→∞
and by wν,m,ε (k) any quantity that goes to zero as k → ∞ for a fixed ν, m and ε.
150  7 Heat equation with a nonlinearity on the gradient We perform the following Landes–Mustonen time regularization; see [221]. For v ∈ L2 (0, T; W01,2 (Ω1 )), we define t
vν (x, t) = ∫ v(x, s)eν(s−t) ds, −∞
where v(x, t) = {
v(x, t)
if t ∈ [0, T],
0
if t ∈ ̸ [0, T].
It is easy to check that vν → v strongly in L2 ((0, T); W01,2 (Ω1 )) and (vν ) = ν(v − vν ) in the weak sense, i. e., ⟨(vν ) , w⟩ = ν ∫ (v − vν )w dx dt ΩT
for all w ∈ L2 ((0, T); W01,2 (Ω1 )).
See [222] for details. Since ϕ ∈ W01,2 (Ω1 ), we have Tm (vk )ϕ ∈ L2 ((0, T); W01,2 (Ω1 )) and (Tm (vk )ϕ)ν = ϕ(Tm (un ))ν . Consider 0 < 2θ < 1 and the splitting, T
2θ 2θ 2θ ∫ ∫ ∇(vk − un ) ϕ dx dt = ∫ ∇(vk − un ) ϕ dx dt + ∫ ∇(vk − un ) ϕ dx dt un ε} = χ{un −(Tm (un ))ν >ε}
k→∞
for all ν and m ∈ ℕ,
(7.2.19)
7.2 Preliminaries and tools  151
lim χ ν→∞ {un −(Tm (un ))ν >ε}
= χ{un −Tm (un )>ε}
for all m ∈ ℕ.
For simplicity, we set Ω1T ≡ Ω1 × (0, T). Hence we have T
2θ Jm,k ≤ ∫ ∫∇(Tm (vk ) − Tm (un )) ϕχ{vk −(Tm (un ))ν >ε} dx dt 0 Ω
T
2θ + ∫ ∫∇(Tm (vk ) − Tm (un )) ϕχ{vk −(Tm (un ))ν ≤ε} dx dt.
(7.2.20)
0 Ω
q
Since ∇(Tm (vk ) − Tm (un ))2θ ϕ is bounded in L 2θ (Ω1 × (0, T)) and since χ{un −(Tm (un ))ν >ε} goes to zero almost everywhere, we conclude that T
2θ ∫ ∫ ∇(Tm (vk ) − Tm (un )) ϕχ{vk −(Tm (un ))ν >ε} dx dt = ω(k, ν, m, ε).
(7.2.21)
0 Ω1
On the other hand we have T
2θ ∫ ∫ ∇(Tm (vk ) − Tm (un )) ϕχ{vk −(Tm (un ))ν ≤ε} dx dt 0 Ω1
1−θ
≤ C(meas(Ω1 × (0, T)))
T
θ
2 (∫ ∫ ∇(Tm (vk ) − Tm (un )) ϕχ{vk −(Tm (un ))ν ≤ε} dx dt) . 0 Ω1
Note that T
2 ∫ ∫ ∇(Tm (vk ) − Tm (un )) ϕχ{vk −(Tm (un ))ν ≤ε} dx dt 0 Ω1
T
2 ≤ ∫ ∫ (∇vk − Tm (un )) ϕχ{vk −(Tm (un ))ν ≤ε} dx dt 0 Ω1 T
≤ ∫ ∫ ∇vk ∇(vk − Tm (un ))ϕχ{vk −(Tm (un ))ν ≤ε} dx dt 0 Ω1
T
− ∫ ∫ ∇Tm (un )∇(vk − Tm (un ))ϕχ{vk −(Tm (un ))ν ≤ε} dx dt. 0 Ω1
Using the weak convergence of Tm (vk ) and the monotonicity of {vk } we obtain T
∫ ∫ ∇Tm (un )∇(vk − Tm (un ))ϕχ{vk −(Tm (un ))ν ≤ε} dx dt = ων,m,ε (k). 0 Ω
(7.2.22)
152  7 Heat equation with a nonlinearity on the gradient We deal now with the first integral in (7.2.20). Since (Tm (un ))ν ϕ converges strongly to Tm (un )ϕ in L2 ((0, T); W01,2 (Ω1 )) as ν → ∞, it follows that T
2 ∫ ∫ ∇vk − Tm (un )) ϕχ{vk −(Tm (un ))ν ≤ε} dx dt 0 Ω
T
= ∫ ∫ ∇vk ∇(vk − (Tm (un ))ν )ϕχ{vk −(Tm (un ))ν ≤ε} dx dt + ωm,ε (k, ν). 0 Ω
Using Tε (vk − (Tm (un ))ν )ϕ as a test function in (7.2.17), it follows that ⟨(vk )t , Tε (vk − (Tm (un )ν )ϕ⟩ + ∫ ϕ∇vk ∇Tε (vk − (Tm (un ))ν ) dx dt ΩT
T
+ ∫ Tε (vk − (Tm (un ))ν )∇vk ∇ϕ dx dt = ∫ ∫ ϕTε (vk − (Tm (un ))ν )gk dx dt 0 Ω
ΩT T
≤ ε ∫ ∫ gk ϕ dx dt ≤ Cε. 0 Ω
A simple variation of Lemma 3.1 in [67] allows us to prove that ⟨(vk )t , Tε (vk − (Tm (un ))ν )ϕ⟩ ≥ wν,m,ε (k). On the other hand, we can check that T
∫ ∫Tε (vk − (Tm (un ))ν )∇vk ∇ϕ dx dt ≤ Cε. 0 Ω
Thus we conclude that T
∫ ∫ ϕ∇vk ∇Tε (vk − (Tm (un ))ν ) dx dt ≤ w(k, ν, m, ε).
(7.2.23)
0 Ω
Putting together (7.2.21), (7.2.22) and (7.2.23) it follows that Jm,k = w(k, ν, m, ε). Hence, from (7.2.19) and (7.2.24) the result follows. We will apply now Lemma 7.2.9 to obtain the following results.
(7.2.24)
7.2 Preliminaries and tools  153
Lemma 7.2.10. Assume that λ ≤ ΛN,2 and f ∈ L∞ (ΩT ). Let Ω1 be a bounded domain such that 0 ∈ Ω1 ⊂⊂ Ω. Then there exists un , a minimal solution to the problem unt − Δun = λTn ( x1 2 )un + ∇un p + f { { { un (x, t) = 0 { { { { un (x, 0) = Tn (u0 )
in Ω1 × (0, T), (7.2.25)
on 𝜕Ω1 × (0, T), if x ∈ Ω1 .
Moreover if ũ ∈ 𝒞 ([0, T); L1loc (Ω)) is a very weak supersolution to (7.2.3) in Ω × (0, T) with ̃ 0), then un ≤ un+1 ≤ ũ almost everywhere. u0 (x) ≤ u(x, Proof. Consider {vk }, the sequence of solutions to problems (7.2.17) in Ω1 × (0, T), k ≥ 1. ̃ for every k. Since an (x) = Tn ( x1 2 ) ∈ Using Lemma 7.2.3 we find that vk ≤ vk+1 and vk ≤ u L∞ (Ω1 × (0, T)), standard arguments (see for instance [131] and the references therein) show that there exists un such that vk ⇀ un weakly in L2 (0, T; W01,2 (Ω1 )). Therefore, ̃ a. e. and un solves (7.2.25). we have un ∈ L2 (0, T; W01,2 (Ω1 )) ∩ 𝒞 ([0, T], L2 (Ω1 )), un ≤ u Moreover, by construction and by Lemma 7.2.3, we conclude that un is minimal and the same argument shows that un ≤ un+1 ≤ u.̃ Corollary 7.2.11. Let ũ ∈ 𝒞 ([0, T); L1loc (Ω)) be a very weak supersolution to (7.2.3). Then, under the hypotheses of Lemma 7.2.10, there exists a positive function u defined in Ω1 × u 1 p 1 (0, T) such that u ∈ L∞ (0, T; L1 (Ω1 )), x 2 ∈ L (Ω1 × (0, T)), ∇u ∈ Lloc (Ω1 × (0, T)) and T
T
∫ ∫ u(−φt − Δφ) dx dt − ∫ ∫ λ 0 Ω1
0 Ω1
T
T
0 Ω1
0 Ω1
uφ dx dt ≥ ∫ ∫ ∇up φ dx dt + ∫ ∫ fφ x2
holds for all φ ∈ 𝒞0∞ (ΩT ), φ ≥ 0. Moreover, if un is the minimal solution to problem (7.2.25) in Ω1 × (0, T), then: 1. un ↑ u ≤ u,̃ a. e. in Ω1 × (0, T), and in L1 (Ω1 × (0, T), dx dt); 2. an (x)un ↑ xu 2 a. e. and in L1 (Ω1 × (0, T), dx dt); 3. for all t ∈ [0, T]: ̃ t) dx < +∞; (a) limn→∞ ∫Ω un (x, t) dx = ∫Ω u(x, t) dx ≤ ∫Ω u(x, 1
1
(b) limn→∞ ∫Ω an (x)un (x, t) dx = ∫Ω 1
1
u(x,t) x2
1
dx ≤ ∫Ω 1
1
̃ u(x,t) x2
dx < +∞;
4. ∇un → ∇u a. e. and ∇Tm un  → ∇Tm u in L (Ω1 × (0, T), ϕ dx dt), ∀m > 0; 5.
T
T
∫0 ∫Ω ∇up ϕ dx dt ≤ lim infn→∞ ∫0 ∫Ω ∇un p ϕ dx dt ≤ C; 1
1
where ϕ is the solution to the retrograde problem −ϕt − Δϕ = 1 { { { ϕ=0 { { { { ϕ(x, T) = 0.
in Ω1 × (0, T), on 𝜕Ω1 × (0, T),
(7.2.26)
154  7 Heat equation with a nonlinearity on the gradient Proof. Items (1), (2) and (3) are a direct consequence of the monotonicity of the sequence {un } and the fact that un ≤ u.̃ Using Tm (un )ϕ as a test function in (7.2.25) we get T
2 ∫ ∫ ∇Tm (un ) ϕ dx ds ≤ Cm, 0 Ω1
and therefore, up to a subsequence, we have ∇Tm (un ) ⇀ ∇Tm (u) weakly in L2 (Ω1 × (0, T), ϕ dx dt). By Lemma 7.2.9, ∇un → ∇u a. e. Using the Vitali theorem we get the convergence of the gradients in item (4). Finally item (5) is a direct consequence of Fatou’s lemma.
7.3 Nonexistence and blowup results In this section we will assume that λ > 0. To prove nonexistence and blowup results, ̃ 0) ≡ ũ 0 = 0 and f is a nonnegative without loss of generality, we suppose that u(x, function such that f ∈ L∞ (ΩT ). 7.3.1 Nonexistence The main goal in this subsection is to find a necessary and sufficient condition on p in such a way that problem (7.1.1) has no positive solution. We will use sistematically the HardySobolev inequality with weights of CaffarelliKohnNirenberg. See for example [15]. The threshold exponent q+ (λ) coincides with the critical one in the elliptic case, as obtained in (6.3.2). More precisely we will prove the following nonexistence result. Theorem 7.3.1. Let q+ (λ) be defined by (7.1.2). If p ≥ q+ (λ), then problem (7.1.1) has no positive very weak supersolution ũ such that ũ t − Δũ ∈ L1loc (ΩT ). In the case f ≡ 0, the unique nonnegative very weak supersolution is ũ ≡ 0. Proof. We argue by contradiction. Assume that ũ is a very weak supersolution to (7.1.1) such that ũ t − Δũ ∈ L1loc (ΩT ) with ũ 0 ≡ 0 and f ∈ L∞ (ΩT ). If λ > ΛN,2 = ( N−2 )2 , then it is sufficient to consider ũ as a very weak supersolution 2 to the problem v vt − Δv = λ 2 + f1 { { x { { { v>0 { { { { v=0
in Ω × (0, T), in Ω × (0, T),
(7.3.1)
on 𝜕Ω × (0, T),
where f1 (x) = ∇vp + cf . The nonexistence result follows from the result in [68]; see also [6] for an alternative proof.
7.3 Nonexistence and blowup results  155
Let us consider the case λ ≤ ΛN,2 . If ũ is a very weak supersolution to (7.1.1), then by Corollary 7.2.11 there exists a supersolution u which is the limit of the sequence {un } of the minimal solution to the truncated problems (7.2.25). Since ut − Δu − λ xu 2 ≥ 0 in 𝒟 (ΩT ), by Lemma 5.2.3, there exists a cylinder Br (0) × (T1 , T2 ) with 0 < r < r1 0 < T1 < T2 ≤ T and a constant C = C(N, r1 , T1 , T2 ) such that u ≥ Cx−α− and u > 1 in Br (0) × (T1 , T2 ). Fixing 0 < T1 < t1 < t2 < T2 and letting ϕ ∈ 𝒞0∞ (Br (0)), using
in (7.2.25) with p =
p p−1
ϕp um n
as a test function
and 0 < m ≤ 1, passing to the limit as n → ∞, it follows that t2
ϕp −2 ϕ∇ϕ∇u 1 dx dt ∫ u1−m (x, t2 )ϕp dx dt + p ∫ ∫ 1−m um
t1 Br (0)
Br (0)
t2
ϕp ∇up m u
≥∫ ∫ t1 Br (0)
t2
dx dt + λ ∫ ∫ u
1−m ϕ
p
x2
t1 Br (0)
dx dt
if 0 < m < 1
(7.3.2)
and t2
ϕp −2 ϕ∇ϕ∇u dx dt ∫ log(u(x, t2 ))ϕ dx dt + p ∫ ∫ u p
t1 Br (0)
Br (0)
t2
t2
p p ϕ
ϕp dx dt + λ ∫ ∫ dx dt u x2
≥ ∫ ∫ ∇u t1 Br (0)
if m = 1.
(7.3.3)
t1 Br (0)
We start by analyzing inequality (7.3.2). Using the Young inequality, for ε > 0 there exists C(ε) such that t2
p ∫ ∫ t1 Br (0)
t2
t2
ϕp ∇up ∇ϕp ϕp −1 ∇ϕ∇u dx dt ≤ ε ∫ ∫ dx dt + C(ε) ∫ ∫ dx dt. m m u u um
t1 Br (0)
t1 Br (0)
(7.3.4) Since u(x, t) ≥ Cx−α− (λ) in Br (0) × (t1 , t2 ), there exists a positive constant C ≡ C(ε, N, r, t1 , t2 ) such that t2
∇ϕp dx dt ≤ C ∫ ∇ϕp xmα− (λ) dx, m u
∫ ∫ t1 Br (0) t2
λ ∫ ∫ u1−m t1 Br (0)
Br (0)
ϕp ϕp dx dt ≥ C ∫ dx. 2 2+(1−m)α − (λ) x x
Br (0)
(7.3.5)
156  7 Heat equation with a nonlinearity on the gradient Hence, for ε small enough, from (7.3.2), (7.3.4) and (7.3.5), 1 ∫ u1−m (x, t2 )ϕp dx + C ∫ ∇ϕp xmα− (λ) dx 1−m
Br (0)
Br (0)
p
ϕ
≥C ∫ Br (0)
x2+(1−m)α− (λ)
dx.
(7.3.6)
By a similar computation for the case m = 1, using inequality (7.3.3), we obtain p
p
α− (λ)
∫ log(u(x, t2 ))ϕ dx + C ∫ ∇ϕ x Br (0)
Br (0)
ϕp dx. dx ≥ C ∫ x2
(7.3.7)
Br (0)
We divide the proof in four steps. Step 1. p+ (λ) < p ≤ 2. Since p+ (λ) < p, we have p < α− (λ) + 2. There are two possibilities. –
− N α− (λ)
If α− (λ) < p < α− + 2, it follows that x Sobolev inequalities give
p
∈ L1+ε (Ω). Then the Hölder and
∫ log(u(x, t2 ))ϕp dx
Br (0)
≤ ( ∫ x
(p )∗ α− (λ) p
(p )∗
ϕ
dx)
p (p )∗
Br (0)
N p
− N α− (λ)
( ∫ (log u(x, t2 )) x
p
dx)
p N
Br (0) N p
≤ 𝒮 −1 ( ∫ ∇ϕp xα− (λ) dx)( ∫ (log u(x, t2 )) x
− N α− (λ) p
p N
dx) ,
Br (0)
Br (0)
Np where (p )∗ = N−p and 𝒮 is the optimal constant in the Sobolev inequality. Moreover, from (7.3.7) we obtain
p
α− (λ)
( ∫ ∇ϕ x
N p
dx)[𝒮 ( ∫ (log u) x −1
Br (0)
− N α− (λ) p
dx)
p N
+ C(ε, t1 , t2 )]
Br (0)
≥ C(N, r0 , t1 , t2 ) ∫ Br
p
ϕ dx. x2
Hence, ∫ ∇ϕp xα− (λ) dx ≥ C ∫
Br (0)
Br (0)
ϕp dx x2
for all ϕ ∈ 𝒞0∞ (Br (0)),
7.3 Nonexistence and blowup results  157
where the constant C > 0 is independent of ϕ. Since p − α− (λ) < 2, we reach a contradiction with the Hardy–Sobolev inequality, which in this case says p
α− (λ)
∫ ∇ϕ x Br (0)
–
dx ≥ ΛN,p ,α− ∫
Br (0)
ϕp
xp −α− (λ)
dx.
(7.3.8)
The second possibility is the complementary case, p ≤ α− (λ). With this hypothesis we use the family of inequalities (7.3.2). Using the Hölder inequality, since u is the limit of approximations, by the regularity results in Lemma 7.2.7, it follows that 1 ∫ u1−m (x, t2 )ϕp dx 1−m
Br (0)
1−m
≤ ( ∫ u(x, t2 )x−α− (λ) dx)
p
( ∫ ϕ m x
Br (0)
1−m α− (λ) m
m
dx)
Br (0) p
≤ C( ∫ ϕ m x
1−m α− (λ) m
m
dx) .
Br (0)
Consider a = ( m1 − 1)N, b = N −
1 (N m
− p ), θ =
amα− (λ) p
− and γ = α− (λ)[ 1−m m
am ]. p
For these values, it is clear that b < p for all 0 < m < 1. Choosing m such that p < m < 1, we find that a < (p )∗ . Thus a + b = pm and θ + γ = 1−m α (λ). Then m − (p )∗ by the Hölder and Sobolev inequalities we conclude that 1 ∫ u1−m (x, t2 )ϕp dx 1−m
Br (0)
1−m
1 θ( 1−m )
≤ C( ∫ x
dx)
Br (0)
(p )∗
( ∫ ϕ
(p )∗
mα− (λ) p
p
γ bq
x
dx)
ma (p )∗
Br (0) p
γ pb
× ( ∫ ϕ x
dx)
m((p )∗ −a) (p )∗
Br (0) p
≤ CS ( ∫ ∇ϕ x −1
mα− (λ)
Br (0)
≤ C(ε)S−1
dx)
ma p
( ∫ ϕ x
dx)
m((p )∗ −a) (p )∗
Br (0)
p CS−1 ma CS−1 m((p )∗ − a) ∫ ∇ϕp xmα− (λ) dx + ε ∫ ϕp xγ b dx. ∗ q (p )
Br (0)
Br (0)
158  7 Heat equation with a nonlinearity on the gradient Moreover, from the above computation and by (7.3.6), it follows that (C(ε)S−1 + C(ε, t1 , t2 ))( ∫ ∇ϕp xmα− (λ) dx)
Br (0)
≥ C(N, r0 , t1 , t2 ) ∫ Br (0)
ϕp
x2+(1−m)α− (λ)
dx − ε
p CS−1 m((p )∗ − a) ∫ ϕp xγ b dx. ∗ (p )
Br (0)
Let us recall the Hardy–Sobolev inequality with weight that we need in this case, p
mα− (λ)
∫ ∇ϕ x Br (0)
dx ≥ ΛN,p ,m ∫
Br (0)
ϕp
xp −α− (λ)m
dx.
Since 2 + (1 − m)α− (λ) > p − mα− (λ) if and only if q+ (λ) < p, in order to get a contradiction with Hardy’s inequality, it is sufficient to choose the above parameters such that − γpb ≤ 2 + (1 − m)α− (λ). (a) If γ pb ≥ 0, then the result follows directly.
(b) If −γ pb > 0, by the definition of the parameters we have
−γ
p p α− (λ) am 1 − m = [ − ]. b b p m
Hence we need to find the value of m and the decomposition in a, b, θ, γ, such that p α− (λ) am 1 − m [ − ] ≤ 2 + (1 − m)α− (λ). b p m Since (pp )∗ < m < 1, we have b > 0. (i) If γ ≥ 0, then the result follows directly from case (a). (ii) If γ < 0, choosing m close to 1, it follows that
q qα (λ) am 1 − m − ] ≤ 2 + (1 − m)α− (λ). γ = − [ b b q m This finishes case (b). Therefore the nonexistence result follows in this case. Step 2. p = q+ (λ) and λ < ΛN,2 . We will proceed in a similar way as in the elliptic case. Note that, by definition, q+ (λ) ≤ 2. As above from Lemma 5.2.3 and the comments at the beginning of the Section 7.2.2, we know that for a suitable positive constant c0 , u(x, t) ≥
c0 xα− (λ)
in Bη (0) × (t1 , t2 ).
(7.3.9)
7.3 Nonexistence and blowup results  159
Without loss of generality we can assume that η ≤ e−1 . From the construction of u and by using the regularity results in Lemma 7.2.7, we obtain t2
t2
∫ ∫ ∇uq+ (λ) x−α− (λ) dx dt < ∞ and
∫ ∫
t1 Bη (0)
t1 Bη (0)
u dx dt < ∞. x2+α− (λ)
(7.3.10)
1 β Consider the function w(x, t) = x−α− (λ) ((t − t1 )2 (log( x )) + 1) defined in (x, t) ∈ Bη (0) × (t1 , t2 ), with β > 0 that will be chosen below. Since λ < ΛN,2 , we have
w ∈ 𝒞 ([t1 , t2 ], L2 (Bη (0))) ∩ L2 ((t1 , t2 ), W 1,2 (Bη (0))) and by a direct computation we obtain wt − Δw − λ
w x2 β
=
(t − t1 ) 1 1 {2x2 (log( )) + β(t − t1 )(log( )) x x x2+α− × [(N − 2 − 2α− (λ)) + (1 − β)((log(
β−1
1 )) ]}. x −1
Note that ∇w = x−α− (λ)−1 {(t − t1 )2 (α− (λ)(log(
β
β−1
1 1 )) + β(log( )) x x
) + α− (λ)},
thus ∇wq+ (λ) {(t − t1 )2 (α− (λ)(log(
β
1 1 )) + β(log( )) x x
= x−α− (λ)−2 {(t − t1 )2 (α− (λ)(log(
β−1
β
1−q+ (λ)
) + α− (λ)}
1 1 )) + β(log( )) x x
β−1
) + α− (λ)}.
If we set h(x, t) = {(t − t1 )2 (α− (λ)(log(
β
1 1 )) + β(log( )) x x
β−1
1−q+ (λ)
) + α− (λ)}
,
then 0 ≤ h(x, t) ≤ α− (λ)1−q+ (λ) in the same cylinder. Since x ≤ e−1 , choosing β small enough, there exists t1̄ ∈ (t1 , t2 ) such that wt − Δw − λ
w ≤ β∇wq+ (λ) h(x, t) x2
in Bη (0) × (t1 , t1̄ )
and w(x, t1 ) = x−α− (λ) for x ∈ Bη (0). Consider u1 ≡ c1 u. Then (u1 )t − Δu1 − λ
u1 ≥ c11−p ∇u1 q+ (λ) x2
in Bη (0) × (t1 , t1̄ ).
160  7 Heat equation with a nonlinearity on the gradient Let c0 be a constant satisfying (7.3.9) and take c1 > 0 such that c1 c0 ≥ 2. Then for β conveniently small we have 1−q+ (λ)
c1
≥ ‖h‖∞ β.
Since c1 c0 ≥ 2 we obtain u1 (x, t) ≥ w(x, t) for x = η, w(x, t1 ) ≤ u1 (x, t1 ) and (u1 )t − Δu1 − λ
u1 ≥ βh(x, t)∇u1 q+ (λ) . x2
Claim. We have u1 ≥ w in Bη (0) × (t1 , t1̄ ). Set v = w − u1 . Using the regularity of w, by (7.3.10) it follows that v ∈ 𝒞 ([t1 , t1̄ ], L1 (Bη (0))) ∩ Lq+ (λ) ((t1 , t1̄ ), W 1,q+ (λ) (Bη (0))), v(x, t) ≤ 0 on 𝜕Bη (0), v(x, t1 ) ≤ 0 for x ∈ Bη (0) and t1̄
v
∫ ∫ t1 Bη (0)
x2+α− (λ)
t1̄
dx dt < ∞,
∫ ∫ ∇vq+ (λ) x−α− (λ) dx dt < ∞.
(7.3.11)
t1 Bη (0)
By a direct computation we obtain vt − Δv − λ
v ≤ q+ (λ)h(x, t)β∇wq+ (λ)−2 ∇w∇v ≡ a(x, t)∇v, x2
where a(x, t) ≡ βq+ (λ)h(x, t)∇wq+ (λ)−2 ∇w = −βq+ (λ)h(x, t)∇wq+ (λ)−1
x . x
Since h(x, t) = (∇wxα− (λ)+1 )1−q+ (λ) , we have a(x, t) = −βq+ (λ) xx 2 ∈ L∞ ([t1 , t1̄ ]; Lq (Bη (0))) for all q < N. Note that due to the regularity of the vector field a, we cannot apply the comparison principle, Lemma 7.2.3. To overcome this lack of regularity, we proceed as follows. Using the Kato inequality we have v x (v+ )t − Δv+ − λ +2 + q+ (λ)β⟨ 2 , ∇v+ ⟩ ≤ 0 x x Since
α− (λ) q+ (λ)
2 + α− (λ), depending only on N and λ, such that t1̄
∫ ∫ t1 Bη (0)
v+ dx dt < ∞. xσ1
(7.3.14)
Indeed, t1̄
t1̄
v+ dx dt = ∫ ∫ ∫ ∫ xσ1
t1 Bη (0)
t1 Bη (0)
v+
x
q+ (λ)+α− (λ) q+ (λ)
t1̄
1
x
q (λ)
v++
≤ (∫ ∫ t1 Bη (0)
σ1 −
xq+ (λ)+α− (λ)
q+ (λ)+α− (λ) q+ (λ)
dx dt)
1 q+ (λ)
t1̄
1
(∫ ∫ t1 Bη (0)
q+ (λ)+α− (λ) ). q+ (λ)
Denote θ(σ1 ) = q+ (λ)(σ1 −
dx dt
p+ (σ1 −
x
q+ (λ)+α− (λ) ) q+ (λ)
dx dt)
1 (λ) q+
.
2+α (λ)
Observe again that the conjugate of q+ (λ) = 1+α− (λ) is q+ (λ) = 2 + α− (λ). Hence we − have θ(σ1 ) = (2 + α− (λ))(σ1 − 1) − α− (λ)(1 + α− (λ)) and θ(2 + α− (λ)) = 2(1 + α− (λ)) = N − 2√ΛN,2 − λ < N. Thus, there exists σ1 > 2 + α− (λ) such that θ(σ1 ) < N and then −(q+ (λ)(σ1 −
∫ x
q+ (λ)+α− (λ) )) q+ (λ)
dx < ∞.
Bη (0)
Hence (7.3.14) holds. We would like to use φ, the solution to the problem φ
{
− div(x−2γ ∇φ) − λ x2(γ+1) =
φ=0
1 x2(γ+1)
in Bη (0), on 𝜕Bη (0),
as a test function in (7.3.13). However, a direct calculation shows that φ(x) =
ηa 1 1 ( − ), λ xa ηa
where a =
2
N − 2(γ + 1) √ N − 2(γ + 1) − ( ) − λ. 2 2
(7.3.15)
162  7 Heat equation with a nonlinearity on the gradient Thus, φ does not have the required regularity. Instead, we consider the approximated sequence, φn (x) =
ηa 1 1 − ), ( λ (x + 1 )a (η + 1 )a n n
with φn ∈ 𝒞 1 (Bη (0)), φn = 0 on 𝜕Bη (0), ∇φn (x) = − − div(x−2γ ∇φn )) =
ηa a x λ (x+ 1 )a+1 x n
and
ηa −2γ a(N − 1 − 2γ) a(a + 1) x ( − ). λ x(x + n1 )a+1 (x + n1 )a+2
Note that t1̄
t1̄ −2γ
∫ ∫ x t1 Bη (0)
∇v+ ∇φn  dx dt < ∞
and
∫ ∫ t1 Bη (0)
v+ φn dx dt < ∞. x2(γ+1)
Choosing φn as a test function in (7.3.13) and since by construction v+ (x, t1 ) = 0, we get t1̄
t1̄
∫ ∫ v+ (− div(x
−2γ
∇φn )) dx dt − λ ∫ ∫
t1 Bη (0)
t1 Bη (0)
v+ φn dx dt ≤ 0. x2(γ+1)
(7.3.16)
By the definition of φn we have v+ v+ v+ φn ηa 2 ≤ + ). ( λ xa+2(γ+1) ηa x2(γ+1) x2(γ+1) Since a is defined by (7.3.15), we have a + 2(γ + 1) → 2 + α− (λ) as γ → 0. By choosing v φn β small enough, we find γ small in order to have a + 2(γ + 1) < σ1 . Hence x+2(γ+1) ≤ Cv+ xσ1
. Therefore by definition of σ1 and the dominated convergence theorem, we prove that t1̄
t1̄
t1̄
t1 Bη (0)
t1 Bη (0)
t1 Bη (0)
v+ φn v+ v+ ηa 1 dx dt → dx dt − ∫ ∫ dx dt ∫ ∫ ∫ ∫ λ λ x2(γ+1) xa+2(γ+1) x2(γ+1)
We now deal with the first term in (7.3.16), a a(a + 1)v+ η a(N − 1 − 2γ)v+ −2γ − ) v+ div(x ∇φn ) = ( 1+2γ λ x (x + 1 )a+1 x2γ (x + 1 )a+2 n n
≤
ηa a(N − 1 − 2γ)v+ ηa a(a + 1)v+ + . 1 λ x1+2γ (x + )a+1 λ x2γ (x + 1 )a+2 n n
as n → ∞.
7.3 Nonexistence and blowup results  163
As above it is not difficult to see that a(a + 1)v+ ηa a(N − 1 − 2γ)v+ ηa a(N + a − 2γ)v+ + ) ≤ ( , λ x1+2γ (x + 1 )a+1 x2γ (x + 1 )a+2 λ xσ1 n n and then by the dominated convergence theorem again we obtain t1̄
t1̄
∫ ∫ v+ div(x
−2γ
a(N − a − 2(γ + 1))v+ ηa dx dt ∇φn ) dx dt → ∫ ∫ λ x2(1+γ)+a
as n → ∞.
t1 Bη (0)
t1 Bη (0)
Hence passing to the limit in (7.3.16) and since a(N − a − 2(γ + 1)) − λ = 0, we have t1̄
∫ ∫ v+ (− div(x
−2γ
t1 Bη (0)
t1̄
t1̄
t1 Bη (0)
t1 Bη (0)
v+ φn v+ ∇φn )) dx dt − λ ∫ ∫ dx dt → ∫ ∫ dx dt, 2(γ+1) 2(1+γ) x x t̄
v
+ dx dt ≤ 0, and hence v+ ≡ 0. as n → ∞. Thus, according to (7.3.16), ∫t 1 ∫B (0) x2(1+γ) 1 η ̄ Therefore u1 ≥ w in (t1 , t1 ) × Bη (0). To finish the proof we use the same argument as in the first case. More precisely, fixing t1̂ ∈ (t1 , t1̄ ), consider ϕ ∈ 𝒞0∞ (Br (0)) with 0 < r 0 such that p
mα− (λ)
∫ ∇ϕ x
dx ≥ C ∫ Br (0)
Br (0)
βm
ϕp
x2+(1−m)α−
(log (λ)
1 ) x
dx
for all ϕ.
(7.3.18)
Since p = 2 + α− , from the Hardy inequality we know that p
mα− (λ)
∫ ∇ϕ x
dx ≥ ΛN,m,p ∫ Br (0)
Br (0)
ϕp
xp −α− (λ)m
dx = ΛN,m,p ∫ Br (0)
ϕp
x2+(1−m)α− (λ)
dx.
(7.3.19) Putting together (7.3.18) and (7.3.19) we reach a contradiction with the optimality of the Hardy constant. Hence the result also follows in this critical case. Step 3. p = q+ (λ) and λ = ΛN,2 . In this case α− (λ) = N−2 and q+ (λ) = N+2 . We know that u(x, t) ≥ cx−α− (λ) in 2 N Bη (0) × (t1 , t2 ) and t2
∫ ∫ ∇uq+ (λ) x−α− (λ) dx dt < ∞. t1 Bη (0)
We consider ϕ ∈ 𝒞0∞ (Bη1 (0)) such that ϕ ≥ 0 and ϕ = 1 in Bη1 (0), with η1 < η. t
Thanks to the regularity of u, we get ∫t 2 ∫B α− (λ) q+ (λ)
=
N(N−2) 2(N+2)
η (0)
1
∇(ϕu)q+ (λ) x−α− (λ) dx dt < ∞. Since
< N, we can apply the Caffarelli–Kohn–Nirenberg inequalities to obtain
t2
t2
t1 Bη (0)
t1 Bη (0)
q (λ) C1 ∫ ∫ (ϕu)q+ (λ) x−α− (λ) dx dt ≤ ∫ ∫ ∇(ϕu) + x−α− (λ) dx dt < ∞, and hence t2
∫ ∫ uq+ (λ) x−α− (λ) dx dt < ∞
for some η1 < η.
t1 Bη (0) 1 1,p
+ (Bη1 (0))). Therefore we conclude that u ∈ Lq+ (λ) ((t1 , t2 ), 𝒟 α− (λ) q+ (λ)
7.3 Nonexistence and blowup results  165
1,p
+ (Bη1 (0))), we have For all ϕ ∈ Lp+ ((t1 , t2 ), 𝒟 α− (λ) q+ (λ)
t2
t2
t1 Bη (0) 1
t1 Bη (0) 1
ϕq+ (λ) C2 ∫ ∫ dx dt ≤ ∫ ∫ ϕq+ (λ) x−α− (λ) dx dt xα− (λ)+q+ (λ) t2
+ ∫ ∫ ∇ϕq+ (λ) x−α− (λ) dx dt, t1 Bη (0) 1
where C2 > 0 is independent of ϕ. In particular, t2
uq+ (λ)
∫ ∫ t1 Bη (0) 1
xα− (λ)+q+ (λ)
dx dt < ∞.
(7.3.20)
Using the fact that u(x, t) ≥ cx−α− (λ) in Bη1 (0) × (t1 , t2 ) and since α− (ΛN,2 )q+ (ΛN,2 ) + α− (ΛN,2 ) + q+ (ΛN,2 ) = N, we reach a contradiction with (7.3.20). Hence the nonexistence result follows in this case. Step 4. p > 2. This case is now very simple. Indeed, for some constant C(p), we have ∇ũ p ≥ ∇ũ 2 − C(p) and therefore ũ t − Δũ ≥ λ
ũ + ∇ũ 2 − C(p) + f x2
in Ω × (0, T).
By Lemma 5.2.3, there exists a positive constant c0 (N, r, t1 , t2 ) such that u(x, t) ≥
c0 xα− (λ)
in Br (0) × (t1 , t2 ) ⊂⊂ ΩT .
Thus, ũ t − Δũ ≥ (λ − ε)
εc0 ũ + ∇ũ 2 + 2+α − C(p) + f x2 x −
in Br (0) × (t1 , t2 ).
Choosing r small enough, it follows that ũ t − Δũ ≥ (λ − ε)
ũ + ∇ũ 2 + f x2
in Br (0) × (t1 , t2 ).
Since q+ (λ − ε) < 2, we conclude as in the first step. Remark 7.3.2. Note that there is a lack of continuity with respect to the case λ = 0. The same thing happens in the elliptic problem (see Chapter 6). The results in [187], show that under some regularity and boundedness conditions on f and u0 , there is a positive solution to problem ut − Δu = ∇up + f for any p ≥ 1. However, for λ > 0, there is no solution if p ≥ q+ (λ). Note that q+ (λ) → 2 as λ → 0, so the Hardy potential is a singular perturbation of the above problem.
166  7 Heat equation with a nonlinearity on the gradient 7.3.2 Complete and instantaneous blowup As a consequence of the nonexistence result for p ≥ q+ (λ), we obtain instantaneous and complete blowup for the following approximated problems. ̃ n ∈ 𝒞 ((0, T); L1 (Ω)) ∩ Lp ((0, T); W01,p (Ω)) be a solution to the probTheorem 7.3.3. Let u lem ̃ nt − Δu ̃ n = ∇u ̃ n p + λan (x)u ̃n + f u { { { ̃ n (x, t) = 0 u { { { ̃ n (x, 0) = 0 { u
in ΩT , (7.3.21)
on 𝜕Ω × (0, T), if x ∈ Ω,
̃ n (x0 , t0 ) → ∞, ∀(x0 , t0 ) ∈ Ω × (0, T). with f ≠ 0, an (x) = Tn ( x1 2 ) and p ≥ q+ (λ). Then u Proof. Without loss of generality, we can assume that f ∈ L∞ (ΩT ) and λ ≤ ΛN,2 . Let us suppose by contradiction that there exists (x0 , t0 ) ∈ ΩT such that ̃ n (x0 , t0 ) → C0 < ∞, u
n → ∞.
Assume first that q+ (λ) ≤ p ≤ 2. By the same arguments as in Lemma 7.2.10, for all n ∈ ℕ we construct a minimal solution un to (7.3.21) such that un ≤ un+1 ∀n ∈ ℕ and ̃ n (x0 , t0 ) → C0 < ∞. un (x0 , t0 ) ≤ u The classical Harnack inequality for the heat equation implies that there exists s > 0 and a positive constant C = C(N, s, t0 , β) such that un , ∫ ∫ un (x, t) dx dt ≤ C ess inf + R
R−
where R− = Bs (x0 ) × (t0 − 43 β, t0 − 41 β) and R+ = Bs (x0 ) × (t0 + 41 β, t0 + β). We can suppose that 0 ∈ Bs (x0 ), since otherwise, we may consider Bδ (y) ⊂ Bs (x0 ) such that un ≤ C ess inf un , ∫ ∫ un (x, t) dx dt ≤ ∫ ∫ un (x, t) dx dt ≤ C ess inf + + R−y
R
R−
Ry
with R−y = Bδ (y) × (t0 − 43 β, t0 − 41 β) and R+y = Bδ (y) × (t0 + 41 β, t0 + β). In a recurrent way, we get un ≤ Cun (x0 , t0 ) ≤ C , ∫ ∫ un (x, t) dx dt ≤ C ess inf + R
R−
with R− = Br (0) × (t1 , t2 ) and R+ = Br (0) × (t3 , t4 ). Denote gn (x, t) = ∇un p + λan (x)un + f . Then un solves unt − Δun = gn
in Br (0) × (t1 , t2 ).
(7.3.22)
7.3 Nonexistence and blowup results  167
Let ϕ be the solution to the problem −ϕt − Δϕ = 1 { { { ϕ=0 { { { { ϕ(x, t2 ) = 0.
in Br (0) × (t1 , t2 ), on 𝜕Br (0) × (t1 , t2 ),
Using ϕ as a test function in (7.3.22), we have t2
t2
∫ ∫ un (x, s) dx ds = ∫ ∫ gn (x, s)ϕ dx ds, t1 Br (0)
t1 Br (0)
thus T
∫ ∫ gn ϕ dx ds ≤ C
for all n.
0 Ω
Therefore, ∇un ⇀ ∇u weakly in Lp (Br (0)×(t1 , t2 ); ϕ dx dt) and thanks to the monotone convergence theorem, an (x)un ↗ xu 2 in L1 (Br (0) × (t1 , t2 ); ϕ dx dt).
In order to reach a contradiction with Theorem 7.3.1, we prove that u is a supersolution to (7.3.21) in 𝒟 (Br (0) × (t1 , t2 )), i. e., in the sense of Corollary 7.2.11. Since un ↑ u in L1 (Br (0) × (t1 , t2 ); ϕ dx dt), we have −Δun → −Δu in the sense of distributions. Using Lemma 7.2.9, we know that ∇un → ∇u almost everywhere. We can apply Fatou’s lemma and we have T
p
T
∫ ∫ ∇u ϕ dx ≤ ∫ ∫ ∇un p ϕ dx 0 Ω
0 Ω
for all ϕ ∈ 𝒞0∞ (Ω × (0, T)), ϕ ≥ 0. Thus we conclude that u is a supersolution to (7.2.3) in the sense of Corollary 7.2.11 and therefore we reach a contradiction with the nonexistence result of Theorem 7.3.1. ̃ n is a solution to (7.3.21), it follows that u ̃ n is a Suppose now that p > 2. Since u supersolution to the problem wt − Δw = f { { { w(x, t) = 0 { { { { w(x, 0) = 0
in ΩT ≡ Ω × (0, T), on 𝜕Ω × (0, T), if x ∈ Ω.
Moreover using the fact that f ≩ 0, and by the strong maximum principle for the heat equation it follows that ̃ n (x, t) ≥ w(x, t) > η0 u
for all (x, t) ∈ Br (0) × (T1 , T2 ).
168  7 Heat equation with a nonlinearity on the gradient Fixing η0 < 1, since p > 2, there exists η1 > 0 such that ∇un p ≥ η1 ∇un 2 −η0 . Therefore we conclude that ̃ nt − Δu ̃ n ≥ η1 ∇u ̃ n 2 − η0 + λη0 an (x)u ̃n + f u
in Br (0) × (T1 , T2 ).
Since limn→∞ limx→0 an (x) = ∞, for fixed λ > 0, there exists n0 ∈ ℕ and r1 > 0 such ̃ n ≥ λη0 an2(x)̃un in Br1 (0). Thus, from the last that for n ≥ n0 , we have −η0 + λη0 an (x)u computation, it follows that ̃ nt − Δu ̃ n ≥ η1 ∇u ̃ n 2 + λη0 u
̃n an (x)u +f 2
in Br1 (0) × (T1 , T2 ).
̃ n is a supersolution to the problem Therefore for some η2 > 0, η2 u wt − Δw = ∇w2 + λan (x)w + f { { { w(x, t) = 0 { { { { w(x, 0) = 0
in Br1 (0) × (T1 , T2 ),
on 𝜕Br1 (0) × (T1 , T2 ),
(7.3.23)
if x ∈ Br1 (0).
Proceeding as in the case p ≤ 2, we reach the same contradiction. Remark 7.3.4. Note that the nonexistence results of this section are local in nature and then imply also nonexistence for the Cauchy problem, i. e., when Ω = ℝN .
7.4 Existence results In this section we consider 0 < λ ≤ ΛN,2 and 1 < p < q+ (λ), the complementary interval to the nonexistence result studied in the previous section. The goal here is to show that, under the hypothesis of existence of a suitable supersolution w, there exists a positive solution to problem (7.1.1). This result, together with the nonexistence obtained in the previous section, shows that q+ (λ) is the optimal power to have existence. We begin by proving the following general existence result. Theorem 7.4.1. Consider r > 0 such that Ω ⊂⊂ Br (0). Let w be a positive function defined in Br (0) × [0, T1 ) verifying: (i) w ∈ Lp ((0, T1 ); W 1,p (Br (0))) ∩ C([0, T1 ]; L2 (Br (0))); 1+s
(ii) there exists s > 0 such that wx2 , w 2−p ∈ L1 (Br (0) × (0, T)); (iii) 0 ≤ u0 (x) ≤ w(x, 0); (iv) w is a very weak supersolution to equation (7.1.1) with f = 0. (2−p)s+p
Then there exists u, with at least the same regularity of w, which is a solution to the problem ut − Δu = ∇up + λ xu 2 { { { u(x, t) = 0 { { { { u(x, 0) = u0 (x)
in ΩT ≡ Ω × (0, T), on 𝜕Ω × (0, T), if x ∈ Ω.
(7.4.1)
7.4 Existence results  169
Proof. Assume that the initial datum satisfies 0 ≨ u0 (x) ≤ w(x, 0). Since Ω ⊂⊂ Br (0), w is a supersolution to (7.4.1). Moreover, it is clear that v(x, t) = 0 is a strict subsolution to (7.4.1). We construct a solution to (7.4.1) as a limit of approximated problems. Consider vn ∈ L2 ((0, T), W01,2 (Ω)), the weak solution to the following approximated problem: vnt − Δvn = λ 21 1 vn + { { x + n { { vn (x, t) = 0 { { { { { vn (x, 0) = u0 (x)
∇vn p 1+ n1 ∇vn p
in Ω, t > 0, (7.4.2)
on 𝜕Ω, t > 0, in Ω, t > 0.
By Lemma 7.2.3, we conclude that 0 < v1 ≤ ⋅ ⋅ ⋅ ≤ vn−1 ≤ vn ≤ w in Ω×(0, T1 ). Hence there exists u ∈ Lp ((0, T1 ); Lp (Ω)) such that vn ↑ u strongly in L2 (Ω × (0, T1 )) and u ≤ w. Since w1+s ∈ L1 (Ω × (0, T1 )), using the dominated convergence theorem we get 21 1 vn → xu 2 x2 x + n
strongly in L1 (Ω × (0, T1 )). To show that u is a solution to problem (7.4.1), we need to ∇v p prove that 1 n p → ∇up strongly in L1 (Ω×(0, T1 )). Although the arguments are well 1+ n ∇vn 
known we include here the details for completeness. Claim. The sequence {vn } defined in (7.4.2) is uniformly bounded in Lp ((0, T1 ); 1,p W0 (Ω)). 1 To prove the claim we consider h(σ) = (σ + 1)s − 1 and its primitive H(σ) = s+1 (σ + s+1 1) − σ. Using h(vn ) as a test function in (7.4.2), it follows that T1
∫ H(vn (x, T)) dx + s ∫ ∫(vn + 1)s−1 ∇vn 2 dx dt Ω
0 Ω
T1
s
T1
p
≤ ∫ ∫(vn + 1) ∇vn  dx dt + λ ∫ ∫ 0 Ω
0 Ω
T1
(1 + w)s w dx dt + ∫ H(u0 (x)) dx x2 Ω
≤ ε ∫ ∫(vn + 1)s−1 ∇vn 2 dx dt 0 Ω
T1
+ C(ε) ∫ ∫(vn + 1) T1
(2−p)s+p 2−p
T1
dx dt + λ ∫ ∫ 0 Ω
0 Ω
(1 + w)s w dx dt + C(s, u0 ) x2
≤ ε ∫ ∫(vn + 1)s−1 ∇vn 2 dx dt + C(ε, w, u0 ). 0 Ω
Hence we conclude that T1
1 s+1 ∫(vn (x, T) + 1) dx + (s − ε) ∫ ∫(vn + 1)s−1 ∇vn 2 dx dt ≤ C(ε, w, T1 ). s+1 Ω
0 Ω
170  7 Heat equation with a nonlinearity on the gradient Thus T1
s−1
∫ ∫(vn + 1)
T1
2
∇vn  dx dt ≤ C1
∫ ∫(vn + 1)s ∇vn p dx dt ≤ C2 .
and
0 Ω
0 Ω
As a consequence, we have 1
k 1−s
T1
T1
2 ∫ ∫∇Tk (vn ) dx dt ≤ C3
∫ ∫ ∇vn p dx dt ≤ C4 .
and
0 Ω
0 Ω
Hence vn ⇀ u weakly in Lp (0, T1 , W01,p (Ω)) and Tk (vn ) ⇀ Tk (u) weakly in L2 (0, T1 , W01,2 (Ω)). Define θk (σ) = (1 + Gk (σ))s − 1, whose primitive is 0
Θk (σ) = {
1 (1 s+1
if 0 ≤ σ < k,
s+1
+ σ − k)
− (σ − k)
if σ ≥ k.
(7.4.3)
Consider θk (vn ) as a test function in (7.4.2). Then, T1
s−1
∫ Θk (vn (x, T1 )) dx + s ∫ ∫(1 + Gk (vn )) Ω
0 Ω
T1
s
∇Gn (vn )2 dx dt T1
p
≤ ∫ ∫(1 + Gk (vn )) − 1)∇vn  dx dt + λ ∫ ∫ 0 Ω
0 Ω
((1 + Gk (vn ))s − 1))vn dx dt x2
+ ∫ Θk (u0 (x)) dx T1
Ω
T1
s
≤ ∫ ∫((1 + Gk (vn )) − 1)∇vn p dx dt + λ ∫ ∫ 0 Ω
0 Ω
((1 + Gk (w))s − 1))w dx dt x2
+ ∫ Θk (u0 (x)) dx. Ω
Using the regularity of w it follows that T1
∫∫ 0 Ω
(1 + Gk (w))s − 1)w dx dt → 0 x2
as k → ∞.
In the same way, we have ∫Ω Θk (u0 (x)) dx → 0 as k → ∞. Hence T1
s p lim sup ∫ ∫ ∇vn  dx dt ≤ lim sup ∫ ∫∇Gk (vn ) (1 + Gk (vn )) dx dt = 0. p
k→∞
{vn ≥k}
k→∞
0 Ω
(7.4.4)
7.4 Existence results 
171
By Lemma 7.2.9 we obtain ∇vn → ∇u a. e. in Ω × (0, T1 ). Strong convergence of the nonlinear terms. To obtain the L1 strong conver∇v p gence of 1 n p → ∇up , we will use Vitali’s lemma (see for instance [64] page 16). 1+ n ∇vn 
Let E ⊂ Ω × (0, T) be a measurable set such that E < ε. We have ∬ E
∇vn p
1 + n1 ∇vn p
dx dt =
∇vn p
∬ E∩{vn ≥k}
≤
1 + n1 ∇vn p
dx dt +
E∩{vn N+2 . Let w(x) = Ax with β = and βp Ap−1 = β(N − β − 2) − λ. N p−1
1,2 Then β ∈ (α− , α+ ) and A > 0. Now w ∈ Wloc (ℝN ) satisfies the regularity hypotheses of Theorem 7.4.1 and
−Δw = λ
w + ∇wp . x2
Hence the existence result follows. Assume that p− (λ) < p ≤ N+2 . Since λ < ΛN,2 , we fix λ1 such that λ < λ1 < ΛN,2 N N+2 and q+ (λ1 ) > N+2 . Consider p > in such a way that p1 < p+ (λ1 ). As in the previous 1 N N β1 1 and β1p Ap−1 = β1 (N − β − 2) − λ1 . Moreover case, we define w1 (x) ≡ A1 x with β1 = 2−p 1 p −1 1
1,2 w1 ∈ Wloc (ℝN ) satisfies
−Δw1 = λ1
w1 + ∇w1 p1 . x2
Since p < p1 and λ < λ1 , a direct computation shows that −Δw1 ≥ λ
w1 + θ∇w1 p , x2 1
1
where θ > 0. Therefore, setting w2 (x) = θ θ−1 w1 (x) = θ θ−1 A1 xβ1 , we obtain −Δw2 ≥ λ
w2 + ∇w2 p . x2
As above if u0 (x) ≤ w2 (x), then w2 is a supersolution to problem (7.4.1); using Theorem 7.4.1 we conclude the proof.
7.5 Cauchy problem and Fujita exponent  173
Note that if λ = ΛN,2 , then p+ (ΛN,2 ) = N+2 . Therefore, using [14] we get the existence N of a positive constant c1 > 0 such that, if x − w(x) = R
N−2 2
1/2
R (log( )) x
− A,
− N−2 2
r with A = ( ) R
1/2
(log(
R )) , r < R, (7.4.5) r
then c1 w(x) is a supersolution to the problem −Δw = ΛN,2 where 0 ≤ g(x) ≤
1 x2
1 w + ∇wp + cg in Br (0) and wn = 0 on 𝜕Br (0), x2
and c is small enough.
Choosing r > 0 such that Ω ⊂ B r (0), it follows that w ∈ W 1,q (Ω) for all q < 2. In 2 fact, one has w ∈ H(Ω), as defined in Section 2.5.1. It is not difficult to verify that w satisfies the regularity assumptions in Theorem 7.4.1. In this way, we have proved the following result. Theorem 7.4.3. Let λ = ΛN,2 and assume that u0 (x) ≤ w as given in (7.4.5). Then problem (7.4.1) with λ = ΛN,2 has a minimal solution v ∈ 𝒞 ([0, T]; L1 (Ω)) ∩ L2 (0, T; H(Ω)). Remark 7.4.4. We can also prove with similar arguments that the problem ut − Δu = ∇up + λ xu 2 + cf (x, t) { { { u(x, t) = 0 { { { { u(x, 0) = u0 (x)
in ΩT , on 𝜕Ω × (0, T), if x ∈ Ω
has a positive solution if f (x, t) ≤ x1 2 and c ≤ c(T) is a suitable small constant. It would be interesting to find necessary and sufficient conditions on f and u0 to get the existence of a positive solution to the above problem.
7.5 Cauchy problem In [187], the authors consider the initial value problem {
ut = Δu + ∇up , u(x, 0) = u0 (x) ≥ 0,
x ∈ ℝN , t > 0, x ∈ ℝN ,
(7.5.1)
with p > 1. They prove that problem (7.5.1) has a global positive solution under some condition on the initial datum u0 . That means that there is a great difference with the semilinear heat equation because here there is no Fujita type exponent. We will see, however, that the behavior is quite different under the influence of a Hardy term, i. e., if we consider instead the problem ut − Δu = λ
u + ∇up in ℝN , t > 0, ΛN,2 ≥ λ > 0, x2
u(x, 0) = u0 (x) in ℝN .
(7.5.2)
174  7 Heat equation with a nonlinearity on the gradient First of all, from the previous sections, we know that the above equation has no local supersolution for p ≥ q+ (λ). In this section we will prove that a new different feature holds for the Cauchy problem (7.5.2): the existence of a Fujita type exponent F(λ) < q+ (λ) if λ ∈ (0, ΛN,2 ]. In Section 5.2.1 we have found a selfinvariant solution to the homogenous linear equation vt − Δv − λ
v =0 x2
in ℝN × (0, ∞).
(7.5.3)
This gives us a lower bound for the behavior at zero and at infinity of every supersolution to (7.5.2) and, in particular, for λ = 0 we get the fundamental solution of the heat equation. To find the Fujita type exponent we proceed with the same strategy as in the semilinear case.
7.5.1 A class of subsolutions to (7.5.2) for small p, blowup in a finite time To guess the possible Fujita exponent we start by studying the values of p for which there exists a class of subsolutions to equation (7.5.3) blowing up in a finite time. More precisely, we look for a family of subsolutions to problem (7.5.2) in the form w(r, t, T) = (T − t)−θ ζ (
r ), (T − t)β
with θ, β > 0 to be chosen and ζ > 0 a smooth function that will be related to the r selfsimilar solution of the linear problem. Denoting s = (T−t) β , it follows that wt = (T − t)−θ−1 (θζ +
βr ζ (s)), (T − t)β
wr (r, t) = (T − t)−θ−β ζ (s), wrr (r, t) = (T − t)−θ−2β ζ (s).
(7.5.4)
We need to have wt − wrr − By setting θ =
2−p ,β 2(p−1)
=
1 2
(N − 1) w wr − λ 2 ≤ wr p . r r
(7.5.5)
and replacing the value in (7.5.4), inequality (7.5.5) becomes
− ζ (s) − (
N −1 1 λ p − s)ζ (s) − ( 2 − θ)ζ (s) ≤ ζ (s). s 2 s
(7.5.6)
7.5 Cauchy problem and Fujita exponent  175 s2
Assume that ζ (s) = Aϕ(cs) with ϕ(s) = s−α− (λ) e− 4 , A > 0, c > 0. Then (7.5.6) is equivalent to c 1 1 + ) + 2 (cα− (λ)(N − 1) − c2 [α− (λ) + α− (λ)2 ] − λ) 4 4 s c(N − α− (λ)) 1 (2 − p) + − c2 (α− (λ) − ) + 2 2 2(p − 1)
− cs2 (
≤ Ap−1 c(p−1)(1−α− ) (
p
1 2 α− (λ) sc + ) s−α− (λ)(p−1) e− 4 (cs) (p−1) . cs 2
If s is large enough, then the above inequality holds for all A > 0 and c > 0. Now, if s is close to 0, since α− (λ)2 − (N − 2)α− (λ) + λ = 0, choosing c ∈ (0, 1] such that cα− (λ)(N − 1) − c2 [α− (λ) + α− (λ)2 ] − λ ≤ 0, the above inequality holds for all A > 0. It remains to obtain a control in a compact interval, for which it is sufficient to choose A > 0 large enough. Hence we can find a family of subsolutions to (7.5.2) given by 2−p − 2(p−1)
w(r, t, T) = A(T − t)
(
cr
−α− (λ)
(T − t)
1 2
)
1 cr 2
e− 4 T−t .
(7.5.7)
Remark 7.5.1. It is worth to point out that in the case λ = 0, namely, α− (0) = 0, the previous inequality is reduced to −cs2 (
p
1 2 c 1 cN c2 s + )+ + + θ ≤ Ap−1 c2p−1 ( ) e− 4 (cs) (p−1) , 4 4 2 2 2 2
(2−p) which is not true near s = 0 since c2 + cN + 2(p−1) > 0. This gives a quantitative justifi2 cation to the influence of λ > 0, in order to have a Fujita type exponent.
Fix r > 0. Since (2−p) − 2(p−1) + N2
∫ x−α− (λ) w(x, t, T) dx = C(T − t) 1 , N−α− (λ)+1
cr √(T−t)
∫ ϕ(s)sN−α− (λ)−1 ds, 0
Br (0)
for p < 1 +
α (λ) − −2
(2−p) that is, − 2(p−1) +
N 2
−
α− (λ) 2
< 0, it follows that
lim ∫ x−α− (λ) w(x, t, T) dx = ∞.
t→T
Br (0)
We now analyze some properties of the subsolution w(r, t, T) found in (7.5.7). (i) The function w(r, t, T) blows up in a finite time in the sense of the local weighted L1 norm in Definition 7.5.7. (ii) Assume that p < F(λ) = 1 + N−α 1(λ)+1 . − We denote by u(x, t) = u(x, t + T) a time translation of a positive solution u(x, t) to (7.5.2). Since u(x, t) is a supersolution to the homogenous equation (7.5.3) with
176  7 Heat equation with a nonlinearity on the gradient the same initial datum, in order to get u(x, 0) ≥ w(r, 0, T), it is sufficient to check that for v defined in (5.2.7) and some T > 0, one has v(x, T) ≥ w(r, 0, T). This immediately follows because p < F(λ) = 1 + N−α 1(λ)+1 . −
(iii) Call G(x, t) = ∇wp−1 . Since p < F(λ), there exists q > 2 such that qN sup ∫ G(x, t) 2 dx = C(T1 ) < ∞.
t∈[0,T1 ]
ℝN
Remark 7.5.2. It is natural to conjecture that F(λ) = 1 + N−α 1(λ)+1 is optimal with this − blowingup property, i. e., it could be seen as the Fujita exponent for problem (7.5.2). Observe that if 0 < λ ≤ ΛN,2 , then F(0+ ) = lim F(λ) = λ↓0
N +2 N +2 < F(λ) < q− (λ) ≤ ≤ q+ (λ) ≤ 2. N +1 N
The first inequality shows a discontinuity of the Fujita exponent at λ = 0. 7.5.2 A class of global supersolutions to (7.5.2) for F (λ) < p < q+ (λ) Consider F(λ) = 1 + N−α 1(λ)+1 as in the previous section. We look for a family of super− solutions to equation (7.5.2), i. e., we look for w such that wt − wrr −
(N − 1) w p wr − λ 2 ≥ w , r r
(7.5.8)
for F(λ) < p < q+ (λ). r Assume that w(r, t, T) = (T + t)−θ g( (T+t) β ) (see for instance [175] and [174]), with g being a smooth bounded positive function and θ, β > 0 to be conveniently chosen. r Calling s = (T+t) β , it follows that:
–
–
–
wt = −(T + t)−θ−1 (θg +
βr g (s)); (T+t)β
wr (r, t) = (T + t)−θ−β g (s);
wrr (r, t) = (T + t)−θ−2β g (s).
In order to get homogeneity in the equation, it is sufficient to choose θ = β=
1 . 2
Therefore (7.5.8) gives g (s) + (
λ N −1 1 p + s)g (s) + ( 2 + θ)g(s) + g (s) ≤ 0. s 2 s
Consider γ such that α− (λ) < γ
0, c > 0.
(2−p) 2(p−1)
and
(7.5.9)
7.5 Cauchy problem and Fujita exponent  177
Hence substituting in (7.5.9), we have c2 [γ 2 − (N − 2)γ + λ] c2 1 + (cs)2 [ − ] + c2 γ 2 4 4 (cs) −
γ (2 − p) sc c2 c2 (N − 1) γ − − + + Ap−1 cp−1 [ + ] 2 2 2 2(p − 1) cs 2
Since α− (λ) < γ and p > 1 + c2 γ −
1 , N−α− (λ)+1
2 p−1
1
2
(cs)−γ(p−1) e− 4 (cs) (p−1) ≤ 0.
it is sufficient to choose c < 1. Then
c2 c2 (N − 1) γ (2 − p) − − + ≤0 2 2 2 2(p − 1)
Moreover, from the fact that γ < enough we get
p−1
and (cs)2 [
c2 1 − ] ≤ 0. 4 4
and c2 [γ 2 − (N − 2)γ + λ] ≤ 0, choosing A small p−1
γ c2 [γ 2 − (N − 2)γ + λ] sc + Ap−1 cp−1 [ + ] cs 2 (cs)2
1
2
(cs)−γ(p−1) e− 4 (cs) (p−1) ≤ 0.
So, our family of supersolutions is of the form γ
w(r, t, T) = Ac−γ (T + t) 2 Note that w(r, 0, T) = AT
2−p − 2(p−1) −γ
a x−γ e−
−a2 x2 4
2
2−p r − 2(p−1) −γ −c2 4(T+t)
r e
, where a =
c
1
T2
.
(7.5.10)
< 1.
7.5.3 Local existence for 1 < p < q+ (λ) As in Theorem 7.4.1, we begin by proving a local existence result under the hypothesis of existence of a supersolution. Theorem 7.5.3. Let w be a nonnegative function defined in ℝN × [0, T1 ), verifying: 1,p (i) w ∈ Lp ((0, T1 ); Wloc (ℝN )) ∩ C([0, T1 ]; L2loc (ℝN )); (ii) ∃s > 0 such that
w1+s ,w x2
(2−p)s+p 2−p
∈ L1 (K × (0, T)), for all compact sets K ⊂ ℝN ;
(iii) 0 ≤ u0 (x) ≤ w(x, 0) and ∈ L1loc (ℝN ); (iv) w is a very weak supersolution to equation (7.5.2). us+1 0
Then there exists a solution u to (7.5.2) satisfying u(x, 0) = u0 with, at least, the same regularity as w. Proof. The proof follows closely the arguments used in Theorem 7.4.1. The only difference is that in this case the estimates are local in space and we need to use a suitable cutoff function. For the reader’s convenience we give some of the details. Consider an initial datum 0 ≨ u0 (x) ≤ w(x, 0), where w(x, t) satisfies the above hypotheses. We
178  7 Heat equation with a nonlinearity on the gradient construct a global solution as a limit of approximated problems in bounded domains. Let Bn be the ball in ℝN with radius n and centered at the origin. We consider vn ∈ L2 ((0, T); W01,2 (Bn+1 )) ∩ L∞ ((0, T); W01,p (Bn+1 )),
∀T > 0,
the weak solutions to the following approximated problems: vnt − Δvn = λ 21 1 ṽn−1 + { { x + n { { vn (x, 0) = u0 (x) { { { { { vn (x, t) = 0
∇vn p 1+ n1 ∇vn p
in Bn+1 , t > 0, in Bn+1 , t > 0,
(7.5.11)
on 𝜕Bn+1 , t > 0,
with v0t − Δv0 = 0 { { { v0 (x, 0) = u0 (x) { { { { v0 (x, t) = 0
in B1 , t > 0, in B1 , t > 0, on 𝜕B1 , t > 0
and ṽn−1 = vn−1 in Bn , ṽn−1 = 0 in ℝN \ Bn . Applying the comparison Lemma 7.2.3, we conclude that 0 < ṽ0 ≤ ṽ1 ≤ ⋅ ⋅ ⋅ ≤ ṽn−1 ≤ ṽn ≤ w in Bn+1 . Hence there exists u ∈ Lp ((0, T1 ); Lploc (ℝN )) such that ṽn ↑ u strongly in L2 ((0, T1 ); L2loc (ℝN )) and u ≤ w. Since w1+s x2
∈ L1 (K × (0, T1 )) for all compact sets K ⊂ ℝN , using the dominated convergence
theorem we easily get
1 ṽ x2 + n1 n−1
→
u x2
strongly in L1 (K × (0, T1 )).
Note that to conclude we have to prove that ∇vñ p
1 + n1 ∇vñ p
→ ∇up
strongly in L1loc (ℝN × (0, T1 )).
1,p We claim that {ṽn } is uniformly bounded in Lp ((0, T1 ); Wloc (ℝN )). To prove the claim s 2 we use ((vn + 1) − 1)ϕ as a test function in (7.5.11), with ϕ ∈ 𝒞0∞ (ℝN ) being a fixed nonnegative function and n ≥ n0 is chosen such that supp ϕ ⊂⊂ Bn0 (0). 1 Let us consider H(σ) = s+1 (σ + 1)s+1 − σ. It follows that 2
T1
∫ H(vn (x, T))ϕ dx + s ∫ ∫ (vn + 1)s−1 ∇vn 2 ϕ2 dx dt 0 ℝN
ℝN T1
+ 2 ∫ ∫ ((vn + 1)s − 1)ϕ∇vn ∇ϕ dx dt 0 ℝN T1
T1
0 ℝN
0 ℝN
≤ ∫ ∫ (vn + 1)s ∇vn p ϕ2 dx dt + λ ∫ ∫
(1 + w)s w 2 ϕ dx dt + ∫ H(u0 (x))ϕ2 dx x2 ℝN
T1
T1
0 ℝN
0 ℝN
≤ ε ∫ ∫ (vn + 1)s−1 ∇vn 2 ϕ2 dx dt + C(ε) ∫ ∫ (vn + 1)
(2−p)s+p 2−p
ϕ2 dx dt
7.5 Cauchy problem and Fujita exponent  179 T1
T1
0 ℝN
0 ℝN
(1 + w)s w 2 +λ∫ ∫ ϕ dx dt + C(s, u0 ) ≤ ε ∫ ∫ (vn + 1)s−1 ∇vn 2 ϕ2 dx dt x2 + C(ε, w, u0 ). Hence we conclude that T1
1 s+1 ∫ (vn (x, T) + 1) ϕ2 dx + (s − ε) ∫ ∫ (vn + 1)s−1 ∇vn 2 ϕ2 dx dt ≤ C(ε, w, T1 , ϕ). s+1 0 ℝN
ℝN
Thus T1
s−1
∫ ∫ (vn + 1)
2
T1
2
∇vn  ϕ dx dt ≤ C1
and
0 ℝN
∫ ∫ (vn + 1)s ∇vn p ϕ2 dx dt ≤ C2 . 0 ℝN
As a consequence, we have T1
1
k 1−s
2 ∫ ∫ ∇Tk (vn ) ϕ2 dx dt ≤ C3
T1
and
0 ℝN
∫ ∫ ∇vn p ϕ2 dx dt ≤ C4 . 0 ℝN
Hence, using the definition of ṽn we obtain: 1,p (i) ṽn ⇀ u weakly in Lp ((0, T1 ); Wloc (ℝN )) and 1,2 (ii) Tk (ṽn ) ⇀ Tk (u) weakly in L2 ((0, T1 ); Wloc (ℝN )). Let be θk (σ) defined in (7.4.3) and consider θk (vn )ϕ2 as a test function in (7.5.11). Then T1
s−1
∫ θk (vn (x, T1 ))ϕ2 dx + s ∫ ∫ ϕ2 (1 + Gk (vn )) ℝN
0 ℝN
T1
∇Gn (vn )2 dx dt
s
+ 2 ∫ ∫ ((1 + Gk (vn )) − 1)ϕ∇vn ∇ϕ dx dt T1
0 ℝN s
≤ ∫ ∫ ((1 + Gk (vn )) − 1)ϕ2 ∇vn p dx dt + ∫ θk (u0 (x))ϕ2 dx 0 ℝN T1
+λ∫ ∫ T1
0 ℝN
ℝN
((1 + Gk (vn ))s − 1))vn 2 ϕ dx dt x2 s
≤ ∫ ∫ ((1 + Gk (vn )) − 1)ϕ2 ∇vn p dx dt + ∫ θk (u0 (x))ϕ2 dx 0 ℝN T1
+λ∫ ∫ 0 ℝN
ℝN
(1 + Gk (w))s − 1)w 2 ϕ dx dt. x2
180  7 Heat equation with a nonlinearity on the gradient Thus T1
s
∫ ∫ ((1 + Gk (vn )) − 1)∇vn ∇ϕϕ dx dt 0 ℝN T1
s
≤ ε ∫ ∫ ((1 + Gk (vn )) − 1)∇vn ∇ϕϕ dx dt 0 {vn ≥k} T1
T1
s
s 2 ≤ ε ∫ ∫ ((1 + Gk (vn )) ∇Gk (vn ) ϕ2 dx dt + C(ε) ∫ ∫ ((1 + Gk (vn )) − 1)∇ϕ2 dx dt. 0 ℝN
0 {vn ≥k}
Since ϕ ∈ 𝒞0∞ (ℝN ) is fixed, the regularity of w gives T1
T1
∫∫ 0
ℝN
((1 + Gk (w))s − 1))w s ϕ dx dt + ∫ ∫ ((1 + Gk (vn )) − 1)∇ϕ2 dx dt → 0 x2
as k → ∞.
0 {vn ≥k}
In the same way, we have ∫ℝN θk (u0 (x))ϕ2 dx → 0 as k → ∞. Hence we conclude that T1
s p lim sup ∫ ∫ ∇vn  dx dt ≤ lim sup ∫ ∫ ∇Gk (vn ) (1 + Gk (vn )) dx dt = 0. p
k→∞
k→∞
{vn ≥k}
0 ℝN
From Lemma 7.2.9 we obtain that ∇vn → ∇u a. e. To conclude we have to prove the ∇v p strong convergence of 1 n p to ∇up in L1 (K × (0, T1 )). This last statement follows 1+ n ∇vn 
using the same arguments as in the proof of Theorem 7.4.1.
Note that using a separation of variables argument we can find a local supersolution for initial datum u0 such that u0 (x) ≤ Ax−β with β close to α− (λ). More precisely we have the following result. Proposition 7.5.4. Assume that p < q+ (λ) and u(x, 0) ≤ T −θ x−β ,
with T > 0, α− (λ) < β < α+ (λ) such that β + α− (λ) < N.
Then (7.5.2) has a local solution u in ℝN × (0, T) for some T > 0, satisfying that for any ball Br (0), r > 0, ∫ x−α− (λ) u(x, t) dx < ∞,
for each 0 < t < T.
(7.5.12)
Br (0) 1 −β Proof. Consider w(x, t) = (T−t) with α− (λ) < β < α+ (λ) and (β + α− (λ)) < N; it θ x follows that β(N − 1) − β(β + 1) − λ > 0. Moreover, since p < q+ (λ), we have β < (β + 1)p < β + 2. A direct computation shows that for T > 1 and θ large, w is a supersolution to (7.5.2). Since u0 (x) ≤ w(x, 0), using Theorem 7.5.3, there exists a minimal solution u to (7.5.2) such that (7.5.12) holds.
7.5 Cauchy problem and Fujita exponent  181
7.5.4 Global existence for small data and F (λ) < p < q+ (λ) We consider the class of initial data N
ℱD = {u0 : ℝ → ℝ0 ≤ u0 (x) ≤ Ax
2
−α− (λ) −D2 x4
e
}
and the class of supersolutions found in Section 7.5.2 given by γ
w(r, t, T) = Ac−γ (T + t) 2
2
2−p r − 2(p−1) −γ − 4(T+t)
r e
.
In order to have u0 (x) ≤ w(x, 0) for some supersolution, we must choose D verifying 2(γ − α− (λ)) ( ) D2 − a2
γ−α− (λ) 2
e−
γ−α− (λ) 2
≤ AT
p−2 − 2(p−1) −γ
a .
Furthermore, since 2(α− (λ) + 1) < N and p (α− (λ) + 1) < q+ (λ) (α− (λ) + 1) = α− (λ) + 2 < N, for γ > α− (λ), with γ near α− (λ), we have 2(γ + 1) < N and (γ + 1)p < N, so w ∈ L2 ((0, T), W 1,2 (ℝN )) and ∇wp belongs to L1 ((0, T) × ℝN ). Note that γ can be chosen as close to α− (λ) as we want. This fact will be used to reach an extra regularity property on w. Therefore using Theorem 7.5.3 there exists a minimal solution u ≤ w defined in ℝN × (0, T). 2 Since w ∈ L2 ((0, T); W 1,2 (ℝN )) with w 2−p ∈ L∞ ((0, T); L1 (ℝN )), we claim that the solution u, obtained in Theorem 7.5.3, is an energy solution to (7.5.2), i. e., u ∈ L2 ((0, T); W 1,2 (ℝN )) for all T > 0. To prove the claim we consider vn , the solution of (7.5.11). Following the computations of Theorem 7.5.3 and setting s = 1, it follows that T
T
0 Bn
0 Bn
1 1 ∫ vn2 (x, T) dx + 2 ∫ ∫ ∇vn 2 dx dt ≤ ∫ ∫ vn ∇vn p dx dt + ∫ u20 (x) dx 2 2 Bn
T
Bn
T
2
2
≤ ε ∫ ∫ ∇vn  dx dt + C(ε) ∫ ∫ vn2−p dx dt + C(u0 ) 0 Bn T
0 Bn T
2
2
≤ ε ∫ ∫ ∇vn  dx dt + C(ε) ∫ ∫ w 2−p dx dt + C(u0 ). 0 Bn
0 Bn
Hence we conclude that T
T
0 Bn
0 Bn
2 1 ∫ vn2 (x, T) dx + (s − ε) ∫ ∫ ∇vn 2 dx dt ≤ C(ε) ∫ ∫ w 2−p dx dt + C(s, u0 ). 2
Bn
182  7 Heat equation with a nonlinearity on the gradient Since w has an exponential decay at infinity, we have just to consider the behavior of 2γ w near 0. But as 2−p < N, we have T
−2γ 1 ∫ vn2 (x, T) dx + (1 − ε) ∫ ∫ ∇vn 2 dx dt ≤ C1 + C2 ∫ x 2−p dx dt < C. 2
0 Bn
Bn
B1
Therefore {vn } is bounded in L2 ((0, T); W 1,2 (ℝN )) and then vn ⇀ u weakly in L2 ((0, T); W 1,2 (ℝN )). A similar computation allows us to prove that {vnt } is bounded in L2 ((0, T); W −1,2 (ℝN )). We show now that vn → u strongly in L2 ((0, T); W 1,2 (ℝN )). Let ϕ be a positive test function such that 0 ≤ ϕ ≤ 1, ϕ = 1 in BR (0) and ϕ = 0 in ℝN \B2R (0), with R very large. For n >> R, using (vn − u)ϕ as a test function in (7.5.11), it follows that T
T
T
∫ ∫ vnt (vn − u)ϕ dx dt + ∫ ∫ ϕ∇vn ∇(vn − u) dx dt + ∫ ∫ (vn − u)∇ϕ∇vn dx dt 0 ℝN
0 ℝN T
= λ∫ ∫
0 ℝN T
1
(7.5.13)
T T 1 ∇v p n ̃ (vn − u)ϕ dx dt → 0 λ ∫ ∫ v (v − u)ϕ dx dt + ∫ ∫ 1 p x2 + 1 n−1 n ∇v  1 + n n n N N
as n → ∞,
0
x2 +
− u)ϕ dx dt + ∫ ∫
∇vn p
(vn − u)ϕ dx dt.
ℝN
ṽ (v 1 n−1 n n
0
ℝN
1 + n1 ∇vn p
We claim that
0ℝ
0ℝ
independently of the choice of ϕ. To prove the claim we use the fact that vn and u are dominated by w. w2 The first term is dominated by λ x 2 , and hence it is sufficient to apply the dominated convergence theorem. To estimate the second term, we follow the general arguments used above. Indeed, using Hölder’s inequality, we have T T ∇v p n (v − u)ϕ dx dt ≤ ∫ ∫ ∫ ∫ ∇vn p vn − u dx dt 1 + 1 ∇v p n n n N N 0ℝ
0ℝ
p 2
T
T
≤ (∫ ∫ ∇vn 2 dx dt) (∫ ∫ vn − u
2 2−p
dx dt)
2−p 2
.
0 ℝN
0 ℝN
Thanks to the choice of γ and the condition on p, from the dominated convergence T
2
theorem we get ∫0 ∫ℝN vn − u 2−p dx dt → 0 as n → ∞. Since {vn } is bounded in L2 ((0, T); W 1,2 (ℝN )), the claim follows.
7.5 Cauchy problem and Fujita exponent  183
We deal now with the fist term in (7.5.13). It is clear that T
∫ ∫ (vn − u)∇ϕ∇vn dx dt → 0
as n → ∞.
0 ℝN
Hence, T
T
∫ ∫ vnt (vn − u)ϕ dx dt + ∫ ∫ ϕ∇vn ∇(vn − u) dx dt 0 ℝN
0 ℝN
T
T
= ∫ ∫ (vn − u)t (vn − u)ϕ dx dt + ∫ ∫ ut (vn − u)ϕ dx dt 0 ℝN T
T
0 ℝN
2 + ∫ ∫ ϕ∇(vn − u) dx dt + ∫ ∫ ϕ∇u∇(vn − u) dx dt 0 ℝN
=
0 ℝN
T
1 2 ∫ (vn − u)2 ϕ dx + ∫ ∫ ϕ∇(vn − u) dx dt 2 0 ℝN
ℝN T
T
+ ∫ ∫ ut (vn − u)ϕ dx dt + ∫ ∫ ϕ∇u∇(vn − u) dx dt. 0 ℝN
0 ℝN
Using the weak convergence of {vn } and a duality argument, it is not difficult to see that T
T
∫ ∫ ut (vn − u)ϕ dx dt + ∫ ∫ ϕ∇u∇(vn − u) dx dt = o(1), 0 ℝN
0 ℝN
independently of the choice of ϕ. Hence by letting R → ∞ we have T
1 2 ∫ (vn − u)2 dx + ∫ ∫ ∇(vn − u) dx dt → 0 2 ℝN
0
as n → ∞.
ℝN
Thus the strong convergence follows. As a consequence, we get u ∈ L2 ((0, T); W 1,2 (ℝN )) and u ≤ w. Moreover, u(x, t) → 0 as t → ∞ for all x ∈ ℝN \{0}. Hence we have proved the following theorem. Theorem 7.5.5. Assume that F(λ) < p < q+ (λ). Then for all u0 ∈ ℱD , the Cauchy problem (7.5.2) has a positive global energy solution u ∈ L2 (0, T; W 1,2 (ℝN )). Remark 7.5.6. 1. The local existence result proved in Proposition 7.5.4 holds for all p < q+ (λ) at least for a particular class of initial data.
184  7 Heat equation with a nonlinearity on the gradient 2.
The required conditions on the initial data u0 in Proposition 7.5.4 and in Theorem 7.5.5 are far from being optimal.
7.5.5 Blowup in a finite time for p < F (λ) and any positive initial datum The L∞ blowup is instantaneous in problem (7.5.2) because the solutions are unbounded. This means that the L∞ blowup is trivial in this case. Due to the behavior of selfsimilar solutions to (7.5.3) and the fact that any positive solution to (7.5.2) is a supersolution to (7.5.3), it is natural to make the following definition. Definition 7.5.7. Consider u(x, t), a positive solution to (7.5.2). Then we say that u blows up in a finite time if there exists T ∗ < ∞ such that lim ∫ x−α− (λ) u(x, t) dx = ∞,
t→T ∗
Br (0)
for any ball Br (0). Theorem 7.5.8. Assume that p < F(λ) = 1 + N−α 1(λ)+1 . Then any positive solution u to − problem (7.5.2) blows up in a finite time in the sense of Definition 7.5.7. Proof. Consider w(x, t, T), a subsolution of the family defined in (7.5.7), and denote by u(x, t) = u(x, t + T) the time translation of u, which is a supersolution of the linear equation (5.2.4) with the same initial datum. Therefore to see that u(x, 0) ≥ w(x, 0, T) it is sufficient to prove that v defined by (5.2.7), for some T > 0, satisfies v(x, T) ≥ w(x, 0, T). This last inequality follows by the hypothesis p < F(λ) = 1 + N−α 1(λ)+1 . − With these premises, we will prove that u(x, t) ≥ w(x, t, T), ∀t < T. Denote k(x, t) = w(x, t, T) − u(x, t). It follows immediately that kt − Δk ≤ λ
k + ∇wp − ∇up x2
in ℝN , t ∈ (0, T1 ), for every T1 < T.
Since p < F(λ), the right hand side belongs to L1loc (ℝN × (0, T)). Applying the parabolic Kato inequality, we have (k+ )t − Δk+ ≤ λ
k+ + pG∇k+  x2
in ℝN , t ∈ (0, T1 ), T1 < T,
k+ (x, 0) = 0,
(7.5.14)
where G = ∇wp−1 . Using φϵ = 2
1,2
N
k+ 1+ϵk+
as a test function in (7.5.14), one can check that
k+ ∈ L ((0, T1 ); W (ℝ )). Fix γ > 0 such that (γ + 1)λ < ΛN,2 . Then we will prove that T1
∫ ∫ k+γ−1 ∇k+ 2 dx dt < ∞. 0 ℝN
(7.5.15)
7.5 Cauchy problem and Fujita exponent  185
s
s γ To see this, consider ε > 0 and mε (s) ≡ ( 1+εs ) . Define Mε (s) = ∫0 mε (y)dy. Using mε (k+ ) as a test function in (7.5.14), we have T1
∫ Mε (k+ (x, T1 )) dx + γ ∫ ∫ ( 0 ℝN
ℝN T1
γ−1
k+ ) 1 + εk+
∇k+ 2 dx dt (1 + εk+ )2
γ
≤ ∫ ∫( 0 ℝN
k+ ) G∇k+  dx dt. 1 + εk+
Using now Young’s inequality T1
γ
∫ ∫( 0 ℝN
k+ ) G∇k+  dx dt 1 + εk+ T1
γ−1
k+ ≤ ϵ∫ ∫( ) 1 + εk+ 0 ℝN
2
T1
∇k+  dx dt + C(ϵ) ∫ ∫ G2 (x, t)k+γ+1 dx dt. 0 ℝN
Since k+ ≤ w and 2(p − 1)(α− (λ) + 1) + α− (λ)(γ + 1) < N, we get T1
∫ ∫ G2 (x, t)k+γ+1 dx dt < C. 0 ℝN
Hence, by letting ε → 0 and using Fatou’s lemma, we conclude that ∫ ℝN
k+γ+1 dx
T1
+ γ ∫ ∫ k+ γ−1 ∇k+ 2 dx dt < ∞, 0 ℝN
and then (7.5.15) follows. Note that since p < F(λ), there exists q > 2 such that qN sup ∫ G(x, t) 2 dx = C(T1 ) < ∞.
t∈[0,T1 ]
ℝN
For a > 0 sufficiently small and γ as above, i. e., λ(γ + 1) < ΛN,2 , we consider ψ, the solution to the problem ψ
q { −ψt − Δψ − λ(γ + 1) x2 = aG(x, t) ψ { { ψ(x, T1 ) = ϕ(x)
in ℝN , t ∈ (0, T1 ), in ℝN ,
(7.5.16)
where 0 < ϕ ∈ L1 (ℝN ) ∩ L∞ (ℝN ). We refer to Lemma 5.2.2 for the proof of the existence k+ γ of such a ψ. Using ψ( 1+εk ) as a test function in (7.5.14), passing to the limit as ε → 0 +
186  7 Heat equation with a nonlinearity on the gradient and by the summability result in the claim, it follows that T1
1 ∫ k+γ+1 (x, T1 )ψ(x, T1 ) dx + γ ∫ ∫ k+γ−1 ∇k+ 2 ψ dx ds γ+1 0 ℝN
ℝN
+
T1
T1
0
0 ℝN
1 ∫ k+γ+1 (−ψt − Δψ) dx ds ≤ ∫ ∫ G∇k+ k+γ ψ dx ds γ+1
T1
θ +θ2 +θ3
= ∫ ∫ G∇k+ v+1 0
ψ dx ds,
ℝN
, θ2 = γ+1 and θ3 = (q−2)(γ+1) (observe that θ1 + θ2 + θ3 = γ). with θ1 = γ−1 2 q 2q Hence by the definition of ψ and the Young inequality, we have T1
1 ∫ k+γ+1 (x, T1 )ψ(x, T1 ) dx + γ ∫ ∫ k+γ−1 ∇k+ 2 ψ dx ds γ+1 0 ℝN
ℝN
T1
T1
a + ∫ ∫ k+γ+1 Gq ψ dx ds ≤ η1 ∫ ∫ k+γ−1 ∇k+ 2 ψ dx ds γ+1 0 ℝN
0 ℝN
T1
T1
0 ℝN
0 ℝN
+ η2 ∫ ∫ k+γ+1 Gq ψ dx ds + η3 ∫ ∫ k+γ+1 (x, s)ψ(x, s) dx ds. Choosing η1 and η2 small, we get T1
1 ∫ k+γ+1 (x, T1 )ψ(x, T1 ) dx ≤ η3 ∫ ∫ k+γ+1 (x, s)ψ(x, s) dx ds. γ+1 Ω
0 Ω
Since k+ ≥ 0 and ψ > 0 in ℝN × (0, T1 ) for t < T1 , by Gronwall’s inequality we obtain k+ ≡ 0 and therefore w(x, t, T) ≤ u(x, T + t). Since T1 < T is arbitrary and w(x, t, T) blows up as t → T, we conclude that for p < F(λ) = 1 + N−α1 +1 , any positive solution to (7.5.2) blows up in a finite time. 1
7.6 Further remarks –
The arguments used above cannot be applied to deal with the critical exponent p = F(λ) = 1 + N−α 1(λ)+1 . To the best of our knowledge, the analysis of the problem − in the case p = F(λ) remains open.
7.6 Further remarks  187
–
A good example for training could be the problems related to Caffarelli–Kohn– Nirenberg inequalities α
ut − div(x−2γ ∇u) = λ xu2(γ+1) + βx−μ ∇uq in Ω × (0, T), { { { { u ≥ 0 in Ω × (0, T) and u = 0 on 𝜕Ω × (0, T), { { { { { u(x, 0) = u0 (x) in Ω,
(7.6.1)
where Ω ⊂ ℝN (N ≥ 2) is a bounded regular domain such that 0 ∈ Ω, λ > 0, α ≥ 0, −∞ < γ < N−2 , u0 is a positive function such that u0 ∈ Lm (Ω) for some 2 m ≥ 1, μ ∈ ℝ, q > 1 and β = 0 or β = 1. Some particular results for the problem when β = 0 can be seen in [4] and [130]. Note that in [130] problems more general than (7.6.1) are considered. In general, problems with a term in the gradient seem to be interesting and more difficult to tackle than the case p = 2.
8 Fractional Laplacian type operators 8.1 Introduction Let us start this chapter with the observation that if we consider the Laplace operator −Δ and u, v ∈ 𝒞0∞ (ℝN ), then an integration by parts (Green’s formula!) gives ⟨−Δu, v⟩ =
ˆ
In particular, for v = u, ⟨−Δu, u⟩ =
ℝN
(−Δu)v dx =
ˆ ℝN
ˆ ℝN
∇u2 dx > 0,
∇u ⋅ ∇v dx.
if u ≠ 0,
that is, −Δ is a positive operator. Thus, using the spectral theory, we can define fractional powers of such operator. The fractional powers of the Laplacian have been considered from many points of view, including potential theory, Levy processes, semigroup theory and pseudodifferential operators. A reference to find equivalent definitions of the fractional Laplacian is the paper [219]. We will study the potential theory approach and its connection with the Fourier transform. A first appearance of the fractional Laplacian from this point of view is the seminal paper by M. Riesz [272]. To describe correctly this precise framework, let us recall the following basic results of distribution theory. We will denote the nonnegative integers by ℤ+ . Consider 𝒮 (ℝN ), the L. Schwartz class, that is, N
∞
N
N
β α
𝒮 (ℝ ) = {f ∈ 𝒞 (ℝ ) : ∀α, β ∈ ℤ+ , ∃Cα,β > 0, sup x D f (x) ≤ Cα,β }.
x∈ℝN
(8.1.1)
Definition 8.1.1 (Fourier transform). For f ∈ 𝒮 (ℝN ), its Fourier transform is defined as ˆ ℱ (f )(ξ ) = f (x)e−2πi⟨x,ξ ⟩ dx. (8.1.2) ℝN
The inverse Fourier transform is given by ˆ −1 ℱ (f )(ξ ) = f (x)e2πi⟨x,ξ ⟩ dx, ℝN
so that we have the inversion formula −1
ℱ (ℱ (f )) = f ,
and the Parseval–Plancherel identity holds: ˆ ˆ ˆ f (x)h(x) dx = ℱ (f )(ξ )ℱ (h)(ξ ) dξ , ℝN
ℝN
https://doi.org/10.1515/9783110606270008
ℝ
2 f (x) dx = N
ˆ ℝN
2 ℱ (f )(ξ ) dξ . (8.1.3)
190  8 Fractional Laplacian type operators The reader can find the classical theory of Fourier transform in any textbook on real or harmonic analysis; see for instance [189], [144], [231], [270] and [296]. Let 𝒮 (ℝN ) be the dual space of 𝒮 (ℝN ). Given λ > 0, let us denote by δλ the dilation operator that for a function f is defined as δλ f (x) = f (λx). The operator δλ has a natural extension to u ∈ 𝒮 (ℝN ) by means of ⟨δλ u, f ⟩ = ⟨u, λ−N δλ f ⟩, −1
for all f ∈ 𝒮 (ℝN ).
Also, let ρ be an element of the rotation group in ℝN that we identify with the corresponding operator ρf (x) = f (ρx). This defines an operator acting on each u ∈ 𝒮 (ℝN ) by the pairing for all f ∈ 𝒮 (ℝN ).
⟨ρu, f ⟩ = ⟨u, ρ−1 f ⟩,
We say that u ∈ 𝒮 (ℝN ) is a homogeneous distribution of degree α if ⟨u, δλ f ⟩ = λ−N−α ⟨u, f ⟩,
for all f ∈ 𝒮 (ℝN ).
We say that u ∈ 𝒮 (ℝN ) is rotation invariant if ρu = u for all rotation ρ in ℝN . The following result is easy to prove (see [293]). Lemma 8.1.2. Let u ∈ 𝒮 (ℝN ) be a homogeneous distribution of degree α. Then its Fourier transform is a tempered distribution of degree −N − α. If u ∈ 𝒮 (ℝN ) is a rotationinvariant, tempered distribution, its Fourier transform is also rotationinvariant. In the case of radial functions we have explicitly the following formula: let f ∈ L1 (ℝN ) be so that f (x) = f0 (x) for some 1variable function f0 defined on (0, ∞). Then, ℱ (f ) is radial and ˆ ∞ ̂ )r N−1 dr, ℱ (f )(ξ ) = f0 (r)dσ(rξ (8.1.4) 0
̂ denotes the Fourier transform of the surface measure on the unit sphere, where dσ that is, ˆ ˆ ̂ )= dσ(ξ e−2πi⟨ω,ξ ⟩ dσ(ω) = cos(2π⟨ω, ξ ⟩) dσ(ω). (8.1.5) SN−1
SN−1
̂ )= As is well known, we have dσ(ξ
2π ξ 
N−2 2
J N−2 (2πξ ), where Jν denotes the Bessel func2
tion of the first kind and index ν (see [294]). Hence, we have the expression ℱ (f )(ξ ) = 2π
ˆ 0
∞
f0 (r)(
r ) ξ 
N−2 2
J N−2 (2πrξ )r dr. 2
The details can be seen in Chapter 2 of the book [189] or Chapter 4 of [296].
(8.1.6)
8.1 Introduction
 191
A spectral definition of the sfractional Laplacian Note that if u ∈ 𝒮 (ℝN ), then −Δu ∈ 𝒮 (ℝN ), and hence 2
2
ℱ (−Δu)(ξ ) = 4π ξ  ℱ (u)(ξ ).
Therefore, it is natural to define for 0 < s < 1 the sfractional Laplacian by s
2s 2s
2s
ℱ [(−Δ) u](ξ ) := 2 π ξ  ℱ (u)(ξ ).
(8.1.7)
The fractional Laplacian represents one of the simplest examples of pseudodifferential operators, despite its symbol being not regular. From the original contributions by A. P. Calderón and A. Zygmund about singular integral operators we recall the following formula, which gives the relationship between the singular operators and the square root of −Δ. See [110], [111] and [112]. Let Rj , j = 1, 2, . . . , N, be the Riesz transforms, i. e., the singular integral operator
associated to the kernel Kj (x) =
multiplier is then
1 xj , γ xN+1
where γ =
mk (ξ ) = −i
π N+1 . Γ( N+1 ) 2
The corresponding Fourier
ξk . ξ 
By using the Fourier transform we find the following Calderón–Zygmund formula: 1
N
(−Δ) 2 = ∑ Rk 𝜕k . k=1
See [112]. 1 A first consequence of the formula above is that (−Δ) 2 is a nonlocal operator. In the last years an increasing interest to study the fractional Laplacian has appeared. This has been basically motivated by the applications to different areas of mathematics, physics and economy. The fractional Laplacian is a paradigmatic example of Levy processes. Regarding this approach, there is a very extensive literature. We refer for instance to [38], [43], [207] and to the nice survey by E. Valdinoci [309] for more details on this topic. Nonlocal operators related to the fractional Laplacian naturally appear in elasticity problems [285], water waves [125], [126], [297], crystal dislocation [305], thin obstacle problems [103], phase transition [33], [101], [290], flame propagation [104], stratified materials [278], water waves [125], [126], [297] and quasigeostrophic flows [109], [235], among other phenomena. Remark 8.1.3. The general theory of pseudodifferential operators was developed in the 1970s by Calderón, Kohn, Nirenberg, Treves, Hörmander, Fefferman, Beals and KumanoGo, among a large list of researchers. We refer for instance to the books [201], [217], [307] and the articles [252], [253], [254], [52] [50], [51], [113] and [160]. The theory was later extended to include symbols with less regular coefficients by Bony, Meyer, Sjöstrand and others.
192  8 Fractional Laplacian type operators Looking at the identity (8.1.7) and in order to study the behavior of (−Δ)s , 0 < s < 1, let us describe the integral expression of the operator, that is, the kernel corresponding to the multiplier (2πξ )2s . Since ξ 2s is a homogeneous tempered distribution of degree 2s, the sfractional Laplacian should be represented by the convolution (in the principal value sense) with a kernel −(N + 2s)homogeneous of the form KN,s (x) =
CN,s
xN+2s
,
which unfortunately is not a locally integrable function in ℝN . Therefore we must justify the meaning of the operator (−Δ)s acting on functions in 𝒮 (ℝN ). We can find an answer to the above problem in the paper by M. Riesz [272]. We will follow Chapter V in the book by E. M. Stein, [293] and Chapter I of the book by N. Landkof [223], restricting our attention just to the case in which 0 < s < 1. In the next subsection we will explain some classical results about the Riesz potential. They will be the key to formally describe the integral representation of the fractional Laplacian. In the rest of this chapter we will study some integrodifferential operators, with measurable, bounded coefficients, that are nonlocal and have a behavior similar to (−Δ)s . The reader can see the basic results that will be used of Potential Theory, for instance, in [197] and [236]. The study of problems in the framework of integrodifferential equations is quite recent and has raised a great interest particularly in connection with problems involving nonlocal effects. We will analyze the basic theory of these operators related to existence, uniqueness and regularity for the solution to the Dirichlet problem. All this will be formulated in the next section. In this part of the chapter we will follow the exposition in [224].
8.1.1 Riesz potentials Recall the definition of the Gamma function ˆ ∞ Γ(z) = e−s sz−1 ds,
ℜ(z) > 0.
0
Using an iterative argument, the measure of the (N − 1)Hausdorff measure of the unit sphere is given by N
ωN−1 =
2π 2
Γ( N2 )
.
See the details, for instance, on page 74 of [306]. Note that Γ(z +1) = zΓ(z) if z is neither 0 nor a negative integer.
8.1 Introduction  193
Consider the Riesz potentials in ℝN , cN,α kN,α (x) = N−α , x
0 < α < N.
We have kN,α ∈ L1loc (ℝN ) and, therefore, it defines an absolutely continuous measure for which, as we will see next, N
−1
ℱ (kN,α )(ξ ) = cN,α π 2
−α
Γ( α2 )
) Γ( N−α 2
ξ −α .
The constant cN,α will be picked up in order to normalize the previous inverse Fourier transform. To do that, we compute the inverse Fourier transform of the tempered distribution f (x) = x1N−α ; by using polar coordinates, the inverse Fourier transform of f is given by ˆ ˆ ∞ N 2π −1 ℱ (f )(ξ ) = f (y)e2πi⟨ξ ,y⟩ dy = r α− 2 J N −1 (2πξ r) dr N 2 ℝN ξ  2 −1 0 ˆ ∞ Γ( α2 ) 1 N N N 1 = (2π) 2 −α α t α− 2 J N −1 (t)dt = π 2 −α N−α 2 ξ  0 ) ξ α Γ( 2
(see [293]). Hence by taking N
cN,α = π α− 2
Γ( N−α ) 2
we obtain −1
ℱ (kN,α )(ξ ) =
1 . ξ α
Summarizing, calling gα (x) =
1 , xα
(8.1.8)
Γ( α2 )
we have ℱ (gα )(x) = kN,α (x) =
Observe also that cN,α ⋅ cN,N−α = 1.
cN,α
xN−α
. (8.1.9)
Definition 8.1.4. For 0 < α < N we define the Riesz operator of order α, Iα , by means of the Fourier multiplier formula ℱ (Iα (f ))(ξ ) =
1 ℱ (f )(ξ ) ≡ gα (ξ )ℱ (f )(ξ ). ξ α
By the previous calculations we have the convolution integral defining Iα , N−α ˆ f (y) α− N2 Γ( 2 ) Iα f (x) = π dy. Γ( α2 ) ℝN x − yN−α A property of the Riesz potentials that we must have in mind is the semigroup property, that is, Iα (Iβ (f )) = Iα+β (f ),
for all f ∈ 𝒮 (ℝN ), 0 < α, β, α + β < N.
194  8 Fractional Laplacian type operators Remark 8.1.5. Note that the previous computations are done in the distribution sense, that is, the integrability at infinity is provided by the test function and a duality argument. In particular, a solution S ∈ 𝒮 (ℝN ) to the problem (−Δ)s S(x) = δ0 ,
x ∈ ℝN ,
δ0 being the Dirac measure,
is given by 2πξ 2s ℱ (S)(ξ ) = 1. According to the previous calculation S(x) = (2π)−2s cN,2s
1 , xN−2s
where cN,2s is given by (8.1.8).
(8.1.10)
In this context, S represents the fundamental solution of (−Δ)s and coincides with the kernel of the Riesz potential I2s . In particular, if f ∈ 𝒮 (ℝN ) we find that the solution to (−Δ)s u = f , modulo sharmonic functions, is given by u(x) = (−Δ)−s f (x) = f ∗ S(x) = (2π)−2s cN,2s
ˆ ℝN
f (y) dy. x − yN−2s
8.1.2 Analytic continuation and the fractional Laplacian formula We focus now our attention on finding the expression of the fractional Laplacian in terms of a kernel. It is clear that the calculus developed for the Riesz potential does not work mutatis mutandis for the multiplier ξ 2s because the candidate to be the kernel is not locally integrable. The way to solve this problems follows the suggestions in the M. Riesz paper [272], as done in [223]. Since Γ( N−α ) has poles in α = N + 2m, m = 0, 1, 2 . . ., cN,α , as a function in α ∈ ℂ, is 2 regular unless α = N +2m, m = 0, 1, 2 . . . . This remark will be one of the key ingredients to define the sfractional Laplacian. Set c̃N,α = (2π)−α cN,α . Given a function u ∈ 𝒮 (ℝN ), define for x ∈ ℝN fixed ˆ u(x + y) v(x, α) = c̃N,α dy. N yN−α ℝ We see that v(x, α) is an analytic function in ℜα > 0, with singular points at α = N +2m, m = 0, 1, 2 . . . . Following the idea in the paper by M. Riesz, [272], the way to define the kernel expression of (−Δ)s , 0 < s < 1, is to make an analytic extension of v(x, α) to −2 < ℜα < 0 and take such analytic extension as the definition of the fractional
8.1 Introduction
 195
operator. To justify this, note that if ℜα > 0 and α ≠ N + 2m, m = 0, 1, 2 . . ., we can split the integral defining v(x, α) as follows. Consider B the unit ball in ℝN . Then ˆ ˆ ˆ u(x + y) u(x + y) − u(x) 1 ̃ ̃ v(x, α) = c̃N,α dy + c dy + c u(x) dy. N,α N,α N−α N−α yN−α ℝN \B y B B y (8.1.11) Since c̃N,α u(x)
ˆ
1
yN−α
B
dy =
c̃N,α u(x)ωN−1 , α
this term is clearly analytic in ℜα ≤ 0 and, obviously, the first term in the splitting also is analytic in ℜα ≤ 0. It remains to study the second term. To this point we will use P. Pizzetti’s formula in [260]. Proposition 8.1.6. Let u ∈ 𝒮 (ℝN ) and define ˆ 1 Su (r) = u(rσ) dσ. ωN−1 SN−1 Then the following expansion holds: ∞
Su (r) = ∑
k=0
2k k!n(n
r 2k Δk u(0) , + 2) ⋅ ⋅ ⋅ (n + 2k − 2)
where Δ is the classical Laplacian. A detailed proof using the Fourier transform can be found in [228]. Consider the mean value of u(x + ⋅) in y = r, that is, ˆ 1 ̄u(x, r) = u(x + y) dω(y). ωN−1 y=r Then
ˆ B
u(x + y) − u(x) dy = ωN−1 yN−α
ˆ 0
1
̄ r) − u(x))r α−1 dr, (u(x,
(8.1.12)
̄ r) − u(x) = O(r 2 ). Therefore the integral in (8.1.12) is and by the Pizzetti formula, u(x, uniformly bounded. As a consequence, v(x, α) can be analytically continued to −2 < ℜα ≤ 0. Note that if −2 < ℜα ≤ 0, ˆ c̃N,α dy u(x)ωN−1 = −c̃N,α u(x) , N−α N α y ℝ \B then the splitting (8.1.11) can be written as ˆ u(x + y) − u(x) v(x, α) = c̃N,α dy. yN−α ℝN
(8.1.13)
196  8 Fractional Laplacian type operators With these calculations we reach finally some meaning to the operator defined from the kernel corresponding to the multiplier (2πξ )2s , 0 < s < 1: if −α = 2s, then we have the following definition. Definition 8.1.7. For 0 < s < 1 and u ∈ 𝒮 (ℝN ) we define the sfractional Laplacian by s
(−Δ) u(x) =: CN,s P. V.
ˆ ℝN
u(x) − u(y) dy := CN,s lim ϵ→0 x − yN+2s
ˆ x−y>ϵ
u(x) − u(y) dy, x − yN+2s
(8.1.14)
where we have set CN,s
Γ( (N+2s) ) 2
N
= −c̃N,−2s = 4s π − 2
Γ(−s)
.
(8.1.15)
In [223] the case s > 1 is also considered. The first remark is that the operator (−Δ)s , the fractional power of the Laplacian, is nonlocal. For that, remember that a linear operator is local if and only if it is a differential operator. The following result, a Plancherel type identity, justifies that the operator defined in Definition 8.1.7 is really the fractional power of the Laplacian, 0 < s < 1. Lemma 8.1.8. Let N ≥ 1 and 0 < s < 1. Then for all u ∈ 𝒮 (ℝN ) we have ˆ ℝN
1̃ ̂ 2 ξ 2s u(ξ ) dξ = C 2 N,s N
̃ = (2π)−2s C = π 2s− 2 where C N,s N,s
ˆ
ˆ
ℝN
ℝN
u(x) − u(y)2 dx dy, x − yN+2s
(8.1.16)
Γ( (N+2s) ) 2 . Γ(−s)
Proof. Fixing y, we call y = x − y. By the Plancherel identity and taking into account that ? ̂ ), u(⋅ + y)(ξ ) = e2πi⟨ξ ,y⟩ u(ξ we obtain ˆ ℝN
ˆ ℝN
u(x) − u(x + y)2 dx dy = yN+2s
ˆ ℝN
ˆ (
ℝN
2 ̂ 2 y−N−2s e2πi⟨ξ ,y⟩ − 1 dy)u(ξ ) dξ .
Now, if θ = ξ /ξ , using polar coordinates ˆ ℝN
2 y−N−2s e2πi⟨ξ ,y⟩ − 1 dy = ξ 2s
ˆ 0
∞ˆ SN−1
2πir⟨ω,θ⟩ 2 − 1 dθr −2s−1 dr. e
8.1 Introduction  197
Observe that, by rotation invariance, the integral defines an expression independent of θ. To compute this constant we proceed as follows: ˆ ∞ˆ ˆ ∞ˆ 2πir⟨ω,θ⟩ 2 c= − 1 dθr −2s−1 dr = (2 − 2 cos(2πr⟨ω, θ⟩)) dθr −2s−1 dr e 0
=2
ˆ
∞
0
SN−1
0
N 2
− (N−2) 2
(ωN−1 − 2π (2πr)
SN−1
J (N−2) (2πr))r −2s−1 dr, 2
where J (N−2) denotes, as before, the Bessel function of the first kind of degree 2
(see [167] and [27]). Since ωN−1 = 2 tity
ˆ
∞
0
for
N 2
< Re z
1.
ϕ(x) − ϕ(y) − ∇ϕ(x) ⋅ (x − y) dy x − yN+2s
{x−y≥ x } 2
ϕ(x) − ϕ(y) dy = I + II, x − yN+2s
where in I we have used the fact that, by symmetry, ˆ ∇ϕ(x) ⋅ (x − y) dy = 0. x − yN+2s {ϵ 2x, respectively. For the first one we have N+2s ˆ
2 II 1  ≤ ( ) x
y≤2x
(ϕ(x) + ϕ(y)) dy ≤ C
1
N
(x ϕ(x) + xN+2s
ˆ y≤2x
ϕ(y) dy).
1 Thus, II 1  ≤ C xN+2s . For the second piece, we use that ϕ(x) − ϕ(y) ≤ ∇ϕ(η)x − y,
for some η ∈ ℝN with η ∼ y, since in this case y > 2x. Hence, II 2  ≤ C
ˆ y>2x
1 1 x − y dy = 𝒪M ( M+2s−1 ), M N+2s y x − y x
∀M ≫ 1.
This finishes our estimate. According to (8.2.1), it is natural to extend the definition of the operator to a larger class of functions. To that end, consider the space s
N
N
ℒ (ℝ ) := {u : ℝ → ℝ measurable:
endowed with the norm
ˆ ‖u‖ℒs (ℝN ) :=
ℝN
ˆ ℝN
u(x) dx < ∞}, 1 + xN+2s
u(x) dx. 1 + xN+2s
As was pointed out in [287], we have the following result. This will allow us to better understand the pointwise meaning of the fractional Laplacian operator. 2s+γ
Lemma 8.2.1. Let s ∈ (0, 1). If u ∈ ℒs (ℝN ) ∩ 𝒞loc (ℝN ) (or 𝒞 1,2s+γ−1 if 2s > 1) for some γ > 0, then (−Δ)s u is a continuous function and is well defined as the following principal value: ˆ u(x) − u(y) (−Δ)s u(x) := CN,s P. V. dy. N+2s ℝN x − y Proof. Fix a bounded domain Ω ⊂ ℝN . By mollification, there exists a sequence {uk } ⊂ 𝒮 (ℝN ), uniformly bounded in 𝒞 2s+γ (Ω) (or 𝒞 1,2s+γ−1 (Ω) if 2s > 1) and such that uk → u uniformly in Ω and in ℒs (ℝN ). The uniform boundedness in 𝒞 2s+γ (Ω) (or 𝒞 1,2s+γ−1 (Ω) if 2s > 1) and the decay of u implies that ˆ u(x) − u(y) s dy, uniformly in Ω. (−Δ) (uk )(x) → CN,s N+2s N ℝ x − y On the other hand, it is clear that (−Δ)s (uk )(x) → (−Δ)s (u)(x)
in 𝒮 (ℝN ).
8.2 Analytical preliminaries related to (−Δ)s

201
Therefore, by the uniqueness of the limit, (−Δ)s (u)(x) = CN,s
ˆ
u(x) − u(y) dy x − yN+2s
ℝN
if x ∈ Ω.
Since Ω is arbitrary we conclude that (−Δ)s u(x) ∈ 𝒞 (ℝN ). If u ∈ ℒs (ℝN ), then (−Δ)s u can be defined in the weak sense as a tempered distribution, that is, we can compute the duality product ⟨(−Δ)s u, φ⟩, for every φ in the Schwartz class. With more generality we have the following definition. Definition 8.2.2. Given f ∈ L1 (ℝN ) we say that a measurable function u : ℝN → ℝ is a weak solution to (−Δ)s u = f if one has ˆ ℝN
(−Δ)s uϕ dx =
ˆ ℝN
for all ϕ ∈ 𝒮 (ℝN ).
fϕ dx,
We observe the following properties, which can be obtained by an easy calculation. Assume that u, v ∈ 𝒮 (ℝN ). 1. It holds that ˆ ˆ ˆ (u(x) − u(y))(v(x) − v(y)) 1 ⟨(Δ)s u, v⟩ := (−Δ)s uv dx = CN,s P. V dy dx. 2 x − yN+2s ℝN ℝN ℝN 2.
As a consequence, (−Δ)s is a symmetric operator, ⟨(−Δ)s u, v⟩ = ⟨u, (−Δ)s v⟩.
3.
Also, ⟨(−Δ)s u, v⟩ is a definite positive bilinear form in 𝒮 (ℝN ). The associated Gagliardo seminorm is defined by 1 ‖u‖2s = CN,s P. V 2
ˆ
ˆ
ℝN
(u(x) − u(y))2 dy dx. x − yN+2s
ℝN
We define Ḣ s (ℝN ) as the homogeneous Hilbert space that results by completion with respect to the seminorm ‖ ⋅ ‖s . We also consider H s (ℝN ) as the completion of 𝒮 (ℝN ) with respect to the norm ‖u‖H s = ‖u‖s = (‖u‖22 +
CN,s P. V 2
ˆ ℝN
ˆ ℝN
1/2
(u(x) − u(y))2 dy dx) . x − yN+2s
The space H s (ℝN ) is a Hilbert space. Therefore, we can extend Lemma 8.1.8 to H s (ℝN ). In this way, we have the following result.
202  8 Fractional Laplacian type operators Proposition 8.2.3. Let N ≥ 1 and 0 < s < 1. Then for all u ∈ H s (ℝN ) ˆ ℝN
2 ξ 2s ℱ (u)(ξ ) dξ =
̃ ˆ ˆ u(x) − u(y)2 C N,s dx dy, 2 ℝN ℝN x − yN+2s
̃ is the constant defined in Lemma 8.1.8. where C N,s The class 𝒞0∞ (ℝN ) is dense in H s (ℝN ) (see for instance Theorem 7.38 in [28]). Moreover, the associated scalar product in H s (ℝN ) can be reformulated, modulo constants, as follows: ⟨u, v⟩H s (ℝN ) := ⟨(−Δ)s u, v⟩ + (u, v) ˆ ˆ ˆ CN,s (u(x) − u(y))(v(x) − v(y)) P. V. dx dy + uv dx. := 2 x − yN+2s ℝN ℝN ℝN Summarizing, we obtain the following useful formulation, which includes the corresponding integration by parts (see also [138]). Proposition 8.2.4. Let s ∈ (0, 1) and u ∈ H s (ℝN ). Then, s s 2 2 ⟨(−Δ)s u, u⟩ = (−Δ) 2 uL2 (ℝN ) = (2πξ ) ℱ uL2 (ℝN ) .
The dual space of H s (ℝN ) is defined by H −s (ℝN ) = {f ∈ 𝒮 (ℝN )/ξ −s ℱ (f ) ∈ L2 (ℝN )}. The following properties are now immediate: 1. (−Δ)s : H s (ℝN ) → H −s (ℝN ) is a continuous operator; 2. (−Δ)s is a symmetric operator in H s (ℝN ), that is, ⟨(−Δ)s u, v⟩ = ⟨u, (−Δ)s v⟩, 3.
u, v ∈ H s (ℝN );
denoting also by ⟨⋅, ⋅⟩ the natural duality product between H s (ℝN ) and H −s (ℝN ), we have s ⟨(−Δ) u, v⟩ ≤ ‖u‖H s (ℝN ) ‖v‖H s (ℝN ) .
The regional Sobolev space in a domain Ω ⊂ ℝN is defined by H s (Ω) = {u ∈ L2 (Ω) 
u(x) − u(y) x − y
N 2
+s
∈ L2 (Ω × Ω)},
with the norm ‖u‖H s (Ω) =
(‖u‖2L2 (Ω)
CN,s + 2
ˆ ˆ Ω
Ω
1
2 u(x) − u(y)2 dy dx) . N+2s x − y
One natural problem is to know when the restriction to Ω of functions of H s (ℝN ) coincides with H s (Ω). According to this, let us make the following definition.
8.2 Analytical preliminaries related to (−Δ)s

203
Definition 8.2.5. Given Ω ⊂ ℝN , an open set, we say that Ω has the extension property for H s (Ω) if for any u ∈ H s (Ω) there exists ũ ∈ H s (ℝN ) such that ũ Ω = u and ‖u‖̃ s ≤ C(n, s, Ω)‖u‖H s (Ω) . The following result gives a class of domains with the extension property with respect to H s (Ω). Theorem 8.2.6. Assume Ω ⊂ ℝN , an open set with Lipschitz bounded boundary. Then H s (Ω) is continuously embedded in H s (ℝN ), that is, Ω has the extension property with respect to H s (Ω). A proof can be seen, for example, in [138] and more details can be found in the references therein. The next step is to study the following fractional Sobolev inequality, which will be used in a systematic way hereafter. The elementary proof that we present here appears in Proposition 15.5 of the book [264], where a more general case is studied. This proof is due to H. Brezis. Theorem 8.2.7. Let s ∈ (0, 1) and N > 2s. There exists a constant S(N, s) such that for any u ∈ H s (ℝN ) we have ‖u‖2L2∗s (ℝN ) ≤ S(N, s)
ˆ ℝN
ˆ ℝN
u(x) − u(y)2 dx dy, x − yN+2s
where 2∗s =
2N N − 2s
is the Sobolev critical exponent. Proof. By density it is sufficient to do the proof for u ∈ 𝒞0∞ (ℝN ). Given x, y ∈ ℝN we trivially have u(x) ≤ u(x) − u(y) + u(y). Consider a ball Br (x) and take the average in Br (x) on both sides of the inequality. Then u(x) ≤
Br (x)
u(x) − u(y) dy +
Br (x)
u(y) dy.
By the Hölder inequality we have 1
1
Br (x)
u(x) − u(y) dy ≤ (
Br (x)
2 2 N+2s u(x) − u(y) dy) ≤ (r
Br (x)
2 u(x) − u(y)2 dy) N+2s x − y
204  8 Fractional Laplacian type operators and for any p > 1, 1
Br (x)
u(y) dy ≤ (
Br (x)
p p u(y) dy) .
Hence, ˆ
1 s u(x) ≤ r ( ω
N
1
ℝN
2 u(x) − u(y)2 1 −N dy) + r p ( N+2s ωN x − y
ˆ
1
p p u(y) dy) . N
(8.2.2)
ℝ
Since, in particular, u is bounded and with compact support, u ∈ Lp (ℝN ) for all p ≥ 1. Therefore the right hand side of the previous inequality is finite for all x ∈ ℝN . Call f (r) = Ar s + Br
− Np
,
where 1 A=( ωN
ˆ ℝN
1
2 u(x) − u(y)2 dy) x − yN+2s
1 and B = ( ωN
ˆ
1
ℝN
p p u(y) dy) .
Minimizing in r > 0 we obtain that the minimum is attained at p
B ps+N = C(N, p, s)( ) A
rmin and, substituting in (8.2.2), we get ˆ u(x) ≤ C1 ( If we choose p such that ˆ ℝN
1 p
2
ℝN
u(x) − u(y) dy) x − yN+2s
=
1 2
−
ˆ
p p u(x) dx ≤ C1 (
ℝN
s , N
1 2p s +1 N p
ˆ (
ℝN
p u(y) dy)
s pN s +1 N p
.
then
ˆ
u(x) − u(y)2 dy dx)( x − yN+2s
ℝN
ˆ ℝN
2s
N p u(y) dy) ,
which implies ‖u‖
2N N−2s
ˆ ≤ S(N, s)(
ℝN
ˆ ℝN
1
2 u(x) − u(y)2 dy dx) . N+2s x − y
As an immediate consequence of Theorem 8.2.7 we can formulate the following results. Corollary 8.2.8. If 0 < s < 1 and N > 2s, then H s (ℝN ) ⊂ Lq (ℝN ) if q ∈ [2, 2∗s ], with continuous embedding.
8.2 Analytical preliminaries related to (−Δ)s

205
Proof. This result follows just from applying Sobolev embedding and interpolation ∗ between L2 (ℝN ) and L2s (ℝN ). Corollary 8.2.9. Consider s ∈ (0, 1) and N > 2s. Let Ω ⊂ ℝN be a domain with the extension property respect to H s (Ω). Then there exists C = C(N, s, q, Ω) such that for any u ∈ H s (Ω) ‖u‖q ≤ C‖u‖H s (Ω) ,
for any q ∈ [2, 2∗s ].
Moreover, if Ω is bounded, then H s (Ω) is continuously embedded in Lq (Ω) for q ∈ [1, 2∗s ]. Proof. To verify the result we apply the extension theorem, the Sobolev inequality and again interpolation. In Theorem 7.1 and Corollary 7.2 of [138] we find the compactness result extending to the fractional case Rellich results. We will use precisely the following result. Theorem 8.2.10. Let 0 < s < 1 and N > 2s. Let q ∈ [1, 2∗s ) and let Ω ⊂ ℝN be a bounded domain verifying the extension property for H s (Ω). Assume that 𝒜 ⊂ L2 (Ω) satisfies: 1. There exists a constant M > 0 such that ‖u‖2 ≤ M for any u ∈ 𝒜. 2. It holds that sup
ˆ ˆ
u∈𝒜
Ω
Ω
u(x) − u(y)2 dy dx < +∞. x − yN+2s
Then 𝒜 is precompact in Lq (Ω). We refer to [138] for the proof. More functional properties related to fractional Sobolev spaces can be seen for instance in [28], [138], [239] and the references therein. 8.2.1 Some remarks on the regularity of solutions in the whole ℝN We will try to summarize the behavior of the sLaplacian operating in spaces of Hölder functions. We follow closely Section 2 in the paper by L. Silvestre [287]. The first result shows the behavior of (−Δ)s with respect to Hölder continuous function spaces. We call 𝒞 0,α (ℝN ), 0 < α < 1, the class of bounded continuous functions such that the seminorm [u]0,α = sup x =y̸
u(x) − u(y) < ∞. x − yα
The norm in 𝒞 0,α (ℝN ) is defined by ‖u‖0,α = ‖u‖∞ + [u]0,α .
206  8 Fractional Laplacian type operators For instance, the class of continuous functions with compact support and with the αmodulus of continuity are a subspace of 𝒞 0,α (ℝN ). In a similar way we denote by 𝒞 k,α (ℝN ), k ∈ ℕ, 0 < α < 1, the class of functions which have continuous bounded derivatives until the order k and the derivative of order k belongs to 𝒞 0,α (ℝN ). The norm in 𝒞 k,α (ℝN ) is defined, for instance, by k
‖u‖k,α = ∑ Dj u0 + [Dk ]0,α , j=0
where j β D u0 = max supD u and β=k N ℝ
[Dk ]0,α = max[Dβ u]0,α . β=k
More properties of the spaces of Hölder continuous functions can be seen in the literature, for instance on page 51 and onwards of [186]. Proposition 8.2.11. Consider 1 ≥ α > 2s > 0. Then there exists a positive constant c = c(N, s, α) such that, for all u ∈ 𝒞 0,α , one has (−Δ)s u ∈ 𝒞 0,α−2s and s (−Δ) u𝒞 0,α−2s ≤ cu𝒞 0,α .
Proof. The difference (−Δ)s u(x1 ) − (−Δ)s u(x2 ) is given by ˆ u(x ) − u(x + y) − u(x ) + u(x + y) s s 1 1 2 2 dy. (−Δ) u(x1 ) − (−Δ) u(x2 ) = CN,s ℝN yN+2s
For a ball Br with r > 0 we split the integral in two pieces, the integral over Br and the integral over its complementary set, that is, we consider ˆ u(x ) − u(x + y) − u(x ) + u(x + y) 1 1 2 2 dy I1 = CN,s Br yN+2s and
ˆ u(x1 ) − u(x1 + y) − u(x2 ) + u(x2 + y) I2 = CN,s dy. ℝN \Br yN+2s
To estimate I1 the regularity of u gives u(xi ) − u(xi + y) ≤ u𝒞 0,α yα . Then ˆ yα dy = cu𝒞 0,α r α−2s . I1 ≤ CN,s u𝒞 0,α N+2s Br y For I2 we use that u(x1 + y) − u(x2 + y) ≤ u𝒞 0,α x1 − x2 α ; it follows that ˆ yα I2 ≤ CN,s 2u𝒞 0,α x1 − x2 α N+2s dy ≤ CN,s 2u𝒞 0,α x1 − x2 α r −2s . y ℝN \Br In particular, choosing r = x1 − x2  we conclude that I1 + I2 ≤ c(N, s, α)x1 − x2 α−2s .
8.2 Analytical preliminaries related to (−Δ)s

207
As a direct consequence of Proposition 8.2.11 and the commutation of the fractional Laplacian with derivatives we have the following result. Corollary 8.2.12. Assume 1 ≥ α > 2s > 0. There exists a positive constant c1 = c1 (α, s, N) such that for every u ∈ 𝒞 1,α , s (−Δ) u𝒞 1,α−2s ≤ c1 u𝒞 1,α . Since we are considering fractional powers, it is possible to consider Hölder regularity 0 < α < 2s < 2. In this case we have the following result. Proposition 8.2.13. If 0 < α < 2s and 0 < s < 1, then there exists a constant c1 = c1 (α, s, N) such that for every u ∈ 𝒞 1,α , s (−Δ) u𝒞 0,α−2s+1 ≤ c1 u𝒞 1,α . Proof. – If s
k, we have Ah ⊂ Ak and Gk (u)χAh ≥ (h − k) and thus 1
(h − k)Ah  2∗s ≤
𝒮2
λ
‖f ‖Lm (Ω) Ak 
1− 21∗ − m1 s
.
Manipulating the above inequality we deduce that ∗
Ah  ≤ Since m >
N 2s
2∗
𝒮 22s ‖f ‖Lsm (Ω) Ak  ∗ λ2s (h
−
2∗s (1− 21∗ − m1 ) s
.
∗ k)2s
we have 2∗s (1 −
1 1 − ) > 1. 2∗s m
Now we apply Lemma 8.4.2 with the choice ψ(σ) = Aσ . Consequently there exists k0 such that ψ(k) ≡ 0 for any k ≥ k0 and thus ess supΩ u ≤ k0 . 8.4.2 The limit case m =
N 2s :
exponential summability
As in the local case, we stress that for the limit case f ∈ LN/2s (Ω) the boundedness of the solutions cannot be expected, but solutions satisfy an exponential summability. Here we state the result for such a case. N
Theorem 8.4.3. Assume that f ∈ L 2s (Ω). Then the unique energy solution u to (8.4.1) satisfies the following estimate: ˆ ∃α > 0 s. t. eαu dx < ∞. (8.4.7) Ω
q
In particular, u ∈ L (Ω) for every q < ∞. Proof. In order to prove the regularity of the energy solution (the existence follows by Theorem 8.3.5), let us consider for any T > 0 the following convex function: Φ(σ) = ΦT (σ) = {
eασ − 1 αT
αe (σ − T) + e
αT
if 0 ≤ σ < T, −1
if σ ≥ T,
where α > 0 will be fixed later. Let u be the unique H0s (Ω) solution to (8.4.1) and since ΦT (σ) is a convex and globally Lipschitz function with ΦT (0) = 0, ΦT (u) is an admissible test function. Thus, according to the Sobolev inequality and (8.3.15) we deduce that ˆ ˆ 2 λ𝒮 Φ(u)L2∗s (Ω) ≤ Φ (u)Φ(u)ℒu = Φ (u)Φ(u)f . (8.4.8) Ω
Ω
226  8 Fractional Laplacian type operators Observe that for u < T we have Φ (u) = αeαu = αΦ(u) + α, and so by (8.4.8) we split the last integral in two parts: ˆ 2 [Φ(u)2 f + Φ(u)f ] dx λ𝒮 Φ(u)L2∗s (Ω) ≤ α {x∈Ω:u 0 such that the unique energy solution to
{
ℒu = f
u=0
in Ω, in ℝN \ Ω
satisfies ‖u‖Lm∗∗ ≤ c‖f ‖Lm (Ω) , s (Ω)
where m∗∗ s =
mN . N − 2ms
(8.4.10)
Proof. The existence of a unique energy solution was obtained in Theorem 8.3.5. Hence we just have to deal with the regularity of the solution. Define the following convex, differentiable function: σβ
Φ(σ) = { with β =
m∗∗ s 2∗s
> 1.
βT
β−1
(σ − T) + T
β
if 0 ≤ σ ≤ T, if σ > T,
(8.4.11)
8.4 Elliptic problem: finite energy setting
 227
Note that since Φ is convex and Φ(0) = 0, according to Proposition 8.3.6, Φ(u) is an admissible test function. Hence by (8.3.15) we have
ℒΦ(u) ≤ f (x)Φ (u)
weakly in Ω.
Multiplying the above inequality one more time by Φ(u), we deduce that ˆ ℰ (Φ(u), Φ(u)) ≤ f Φ(u)Φ (u) dx. Ω
The Hölder and Sobolev inequalities yield 1 ˆ m m 2 ∗ λ𝒮 Φ(u)L2s (Ω) ≤ ‖f ‖Lm (Ω) ( Φ (u)Φ(u) ) .
Ω
(8.4.12)
An elementary computation shows that there exists a constant c, independent of T, such that Φ (σ)Φ(σ)m ≤ cΦ(σ)2s ,
∀σ ≥ 0,
∗
so that ˆ
λ𝒮 (
2 ∗
Ω
2∗ 2s Φ(u) s )
Finally, since β2∗s = m (2β − 1) = m∗∗ s and as T diverges, that
2 2∗s
−
1 m
−
⩽ c‖f ‖Lm (Ω) .
1 m
=
1 , we conclude, taking the limit m∗∗ s
‖u‖Lm∗∗ ⩽ c‖f ‖Lm (Ω) , s (Ω) where c depends on λ, 𝒮 , s, N, m and Ω. 8.4.4 Further fractional regularity This section is devoted to prove an interesting regularity property of the energy solutions of (8.4.1). Assume that Ω has a bounded Lipschitz boundary, so that we ensure that Ω has the extension property. By Theorem 8.2.6 any function in H0s (Ω) extended as 0 outside Ω belongs to W s,2 (ℝN ). Thus, we can apply the interpolation result that we stated in Theorem 8.3.15 and we get the following regularity result. Theorem 8.4.5. Let u be an energy solution of (8.4.1) with f in Lm (Ω), m ≥ (2∗s ) . Then u belongs to W0σ,p (Ω), where σ = θs
and
1 1 1 1 2s = + θ( − ) − (1 − θ), p m 2 m N
θ ∈ (0, 1).
(8.4.13)
228  8 Fractional Laplacian type operators The Sobolev space W0σ,p (Ω), σ ∈ (0, 1), p > 1, is defined as the completion of 𝒞0∞ (Ω) with respect to the seminorm upW σ,p := 0
ˆ ˆ Ω
Ω
u(x) − u(y)p dxdy. x − yN+pσ
The reader can see the functional properties of this space for example in [28] and in [138]. Below, in Chapter 10, we will return to these Sobolev spaces. Proof. The proof easily follows applying Theorem 8.3.15 and interpolating between ∗∗ the Sobolev space H0s (Ω) ≡ W0s,2 (Ω) (i. e., θ = 1) and Lms (Ω) (i. e., θ = 0). Observe that the regularity given by the above theorem does not appear in the literature for the local case (s = 1), where one looks for estimates either on u or on Δu.
8.5 A fractional Picone inequality and applications to sublinear problems We formulate an extension of an inequality obtained by Picone in [259]. To be more precise, Picone considers that if u, v ∈ 𝒞 2 (a, b) and u > δ > 0 in (a, b), then 2
(v ) ≥ (
v2 ) u . u
The corresponding extension of this pointwise inequality to higher dimensions is that if Ω ⊂ ℝN , u, v ∈ 𝒞 2 (Ω) and u > δ > 0 on Ω, then ∇v2 ≥ ⟨∇(
v2 ), ∇u⟩. u
Moreover, for any 1 < p < ∞, ∇vp ≥ ⟨∇(
vp ), ∇up−2 ∇u⟩; up−1
see Section 1.3. The useful integral form was obtained in a quite general framework in [14], where also some interesting applications were obtained. It is precisely this integral form that we prove for the fractional operator ℒ. More precisely, we obtain the following result. Theorem 8.5.1 (Picone inequality). Consider u, v ∈ H0s (Ω) and assume that ℒu is a positive bounded Radon measure in Ω, with u ⪈ 0. Then, ˆ ℒu 2 v dx ≤ ℰ (v, v). (8.5.1) Ω u
8.5 Fractional Picone inequality  229
Proof. Let us recall first that for any ϕ, ψ ∈ H0s (Ω), we have, thanks to the symmetry of ℒ, ˆ ˆ −1 ϕℒψ dx = ℒϕψ dx = ℰ (ϕ, ψ) ≤ λ ‖ϕ‖H s (Ω) ‖ψ‖H s (Ω) . (8.5.2) 0 0 Ω
Ω
2
v We set, for any k, η > 0, vk = Tk (v), ũ = u + η and we define w = uk̃ ; actually it is not s s hard to see that w ∈ H0 (Ω). We want to prove that, ∀u, v ∈ H0 (Ω) and ∀k, η > 0 and thanks to (8.5.2),
ℰ (u, w) =
ˆ
ℒu
vk2 2 ≤ ℰ (vk , vk ) ≤ λ−1 Tk (v)H s (Ω) . 0 ũ
(8.5.3)
Once we have proved such an inequality, (8.5.1) follows by letting k diverge and η vanish, using the monotone convergence theorem and Fatou’s lemma (this is why we ask ℒu to be a positive Radon measure). Observe that vk2 (x) vk2 (y) − ) ̃ ̃ u(x) u(y) ̃ ̃ u(y) u(x) = vk2 (x) + vk2 (y) − vk2 (x) − vk2 (y) . ̃ ̃ u(x) u(y)
̃ − u(y))( ̃ (u(x) − u(y))(w(x) − w(y)) = (u(x)
(8.5.4)
Thus, by the Young inequality we deduce vk2 (x)
̃ ̃ u(y) u(x) + vk2 (y) ≥ 2vk (y)vk (x) ̃ ̃ u(x) u(y)
and so, gathering together the above inequality with (8.5.4), we get 2
(u(x) − u(y))(w(x) − w(y)) ≤ (vk (x) − vk (y)) . Hence, (8.5.3) follows by changing the above inequality into the corresponding integral against the kernel 𝒦(x, y). Some extensions of the Picone inequality The general Picone inequality in Section 1.3 can be extended to the fractional framework by observing the next result. Lemma 8.5.2. Let 1 < p < ∞, 1 < q ≤ p and let u1 , u2 be two measurable functions with u2 ⩾ 0 and u1 > 0. Then the following inequality holds: u1 (x) − u1 (y)p−2 (u1 (x) − u1 (y))[
u2 (x)q u2 (y)q − ] u1 (x)q−1 u1 (y)q−1
⩽ u2 (x) − u2 (y)q u1 (x) − u1 (y)p−q .
(8.5.5)
230  8 Fractional Laplacian type operators
by
Now we can state Picone’s inequality related to the fractional pLaplacian defined (−Δ)sp u(x)
:= P. V.
ˆ ℝN
u(x) − u(y)p−2 (u(x) − u(y)) dy. x − yN+ps
(8.5.6)
Proposition 8.5.3. Consider u, v ∈ W0s,p (Ω) with u ≥ 0. Assume that (−Δ)sp u is a positive bounded Radon measure in Ω. Then ˆ vp 2 (−Δ)sp u p−1 ⩽ ‖v‖pW s,p (Ω) . (8.5.7) u 0 Ω The detailed proofs can be found in [49]. An application Next we present an application of the Picone inequality to a semilinear problem first studied by Brezis–Oswald in [95], and then by Brezis–Kamin in [90]. To be more precise, consider the semilinear problem ℒu = f (x, u) in Ω, { { { u>0 in Ω, { { { in ℝN \ Ω, {u ≡ 0
(8.5.8)
where f : Ω×[0, ∞) → ℝ is a Caratheodory function. The aim is to give some condition on f to have uniqueness of solutions to (8.5.8). Observe that if ℒ is a linear differential operator of second order, this problem is classical and was solved by H. Brezis and L. Oswald in 1986 (see [95] and [90]). Indeed, requiring that the nonlinearity satisfies σ→
f (x, σ) σ
is decreasing for a. e. x ∈ Ω,
(8.5.9)
they proved uniqueness for solutions of −Δu = f (x, u) in Ω, { { { u>0 in Ω, { { { on 𝜕Ω. {u = 0 Here we prove the analog for problem (8.5.8), associated to the operator ℒ. First, we need to recall the following definition. Definition 8.5.4. A subsolution (supersolution) to (8.5.8) is a positive function u ∈ H0s (Ω) ∩ L∞ (Ω) such that f (x, u) ∈ L1 (Ω) and ˆ ˆ uℒφ ≤ (respectively ≥) f (x, u)φ, ∀φ ∈ H0s (Ω), φ ≥ 0. (8.5.10) Ω
Ω
A solution is both a subsolution and a supersolution.
8.5 Fractional Picone inequality  231
Now we can state our Brezis–Kamin–Oswald type result. Theorem 8.5.5. Let u and v be a sub and a supersolution, respectively, to (8.5.8). Assume that (8.5.9) holds true. Then u ≤ v a. e. in Ω. Remark 8.5.6. Before proving the comparison result that in particular proves the uniqueness, observe that solutions of problem (8.5.8) exist under different conditions. For instance, according to [95], the following hypotheses are sufficient: (f1) the function x → f (x, σ) belongs to L∞ (Ω) for each σ ⩾ 0; (f2) f (x,σ) → a0 (x) as σ → 0, where −∞ < −c ⩽ a0 (x) ⩽ ∞ and f (x,σ) → 0, as σ → ∞, σ σ uniformly in Ω; (f3) there exists a function ϕ ∈ H0s (Ω) ∩ L∞ (Ω) such that ‖ϕ‖2H s (Ω) −
ˆ
0
{ϕ=0} ̸
2 a0 (x)ϕ(x) dx < 0.
Proof. Define the subset Ω0 = {x ∈ Ω  u(x) > v(x)}. We have to prove that Ω0  = 0. ϕ ϕ Set ϕ = (u2 − v2 )+ . Then u and v are admissible test functions, thanks to the Picone inequality (8.5.1). Hence ˆ
ϕ uℒ( ) dx − u Ω
ˆ
ϕ vℒ( ) dx ≤ v Ω
ˆ Ω
(
f (x, u(x)) f (x, v(x)) − )ϕ(x) dx. u(x) v(x)
(8.5.11)
Note that the right hand side is negative due to (8.5.9), so if we show that the left hand side is positive, then ϕ ≡ 0 and we conclude the proof. To prove that the left hand side is nonnegative, observe first that if we call u0 = uχΩ0 , v0 = vχΩ0 and ϕ = u20 − v02 , it suffices to prove the following pointwise inequality: (v(x) − v(y))(
u20 (x) − v02 (x) u20 (y) − v02 (y) − ) v(x) v(y)
≤ (u(x) − u(y))(
u20 (x) − v02 (x) u20 (y) − v02 (y) − ). u(x) u(y)
(8.5.12)
By rearranging the terms, (8.5.12) is equivalent to v2 (x) v02 (y) u2 (x) u20 (y) (v(x) − v(y))( 0 − ) + (u(x) − u(y))( 0 − ) v(x) v(y) u(x) u(y) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (A)
(B)
≤ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (u(x) − u(y))(u0 (x) − u0 (y)) + (v(x) − v(y))(v0 (x) − v0 (y)) . ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (C)
D
By symmetry it is sufficient to consider the following three cases: (i) x, y ∉ Ω0 ; the inequality is trivial because the terms (A), (B), (C) and (D) are identically zero; (ii) x, y ∈ Ω0 ; by the pointwise Picone inequality we have (A) ≤ (C) and (B) ≤ (D);
232  8 Fractional Laplacian type operators (iii) x ∈ Ω0 and y ∉ Ω0 ; in this case, {
(A) + (B) = u20 (x) − u20 (x) uv(y) − v02 (x) uu(y) + v02 (x), (x) (x) 0
0
(C) + (D) = u20 (x) − u0 (x)u(y) + v02 (x) − v0 (x)v(y).
Therefore (A) + (B) ≤ (C) + (D) is equivalent to u(y)(u0 (x) −
v02 (x) u2 (x) ) ≤ v(y)( 0 − v0 (x)), u0 (x) v0 (x)
which itself is equivalent to v(y) u(y) ≤ , u0 (x) v0 (x) that is, u(y)v0 (x) ≤ v(y)u0 (x). This is true because x ∈ Ω0 and y ∉ Ω0 , and then u(y) ≤ v(y) and v0 (x) ≤ u0 (x). So, we conclude the proof. Remark 8.5.7. A natural extension to the case of the fractional pLaplacian can be found using Proposition 8.5.3.
8.6 Nonvariational setting for elliptic problems: weak solutions 2N we cannot expect a solution of finite energy. When f ∈ Lm (Ω) with 1 ≤ m < N+2s However, as in the local case, by relaxing the meaning of solution we can show the existence of weaker solutions for nonvariational data. Consider the problem
{
ℒu = f
u=0
in Ω,
(8.6.1)
in ℝN \ Ω.
In the case of Ω = ℝN the existence and uniqueness of renormalized solutions is obtained in [2] (see also [206]). However we consider a weak solution to (8.6.1) in a very elementary context, in bounded domains and making explicit the summability of the solution. Definition 8.6.1. We define the class of test functions N
∞
𝒯 (Ω) = {ϕ  ℒ(ϕ) = ψ in Ω, ϕ = 0 in ℝ \ Ω, ψ ∈ 𝒞0 (Ω)}.
(8.6.2)
Note that if v ∈ 𝒯 (Ω), then, using the results in the previous section, v ∈ H0s (Ω) ∩ L∞ (Ω). Moreover, according to the regularity theory developed in [287], if Ω is smooth enough, there exists a constant β > 0 (that depends only on the structural constants) such that v ∈ 𝒞 β (Ω) (see also [207]).
8.6 Dirichlet problem: Nonvariational setting

233
Definition 8.6.2. We say that u ∈ L1 (Ω) is a weak solution to (8.6.1) if for f ∈ L1 (Ω) we have ˆ ˆ uψ dx = fϕ dx Ω
Ω
for any ϕ ∈ 𝒯 (Ω) with ψ ∈ 𝒞0∞ (Ω). Remark 8.6.3. We recall that if in the definition of the test function space, we consider ψ ∈ Lm (Ω), m > N/2, we recover the classical definition of duality solution introduced by Stampacchia in [291]. Indeed, using Theorem 8.4.1, ϕ is bounded and consequently all the terms in the above identity make sense. We also stress that such a definition can be extended to equations with measure data if we have a regularity result that guarantees the continuity, up to the boundary, of the solution of the problem ℒ(ϕ) = ψ in Ω, ϕ = 0 in ℝN \ Ω. Results in this direction can be found in [206]. Now we can state the existence result for L1 data. Theorem 8.6.4. For any f belonging to L1 (Ω) there exists a unique weak solution to (8.6.1). Moreover, ∀k ≥ 0,
Tk (u) ∈ H0s (Ω),
u ∈ Lq (Ω),
∀q ∈ (1,
and s r (−Δ) 2 u ∈ L (Ω),
(8.6.3)
N ), N − 2s
∀r ∈ (1,
N ). N −s
(8.6.4)
(8.6.5)
Remark 8.6.5. The philosophy behind the above result follows the ideas that come from the analog in the local case (see, for instance [291] and [69] for the nonlinear case). Indeed, as already observed, we cannot expect solutions to belong to the Sobolev space H0s (Ω) since L1 (Ω) is not embedded into H −s (Ω). Anyway, we prove that any truncation at level k does belong to the energy space. On the other hand, observe that if s = 1 we recover the existence and regularity results that are, nowadays, well known for second order elliptic operators in divergence form with L1 data. In the local case, according to the results mentioned above, weak N N ), and u ∈ Lq (Ω), ∀q ∈ (1, N−2 ). solutions u are such that ∇u ∈ Lr (Ω), ∀r ∈ (1, N−1 Before we begin with the proof, it is interesting to observe that the two estimates (8.6.4) and (8.6.5) are quite different from one another. This is because since u has been defined as zero outside Ω, (8.6.4) turns out to be a global estimate on u. On the other hand, the a priori estimate on the s/2derivative of u cannot be extended to s a global one. Also, the nonlocal nature of the equation makes (−Δ) 2 u to be different from zero in ℝN \ Ω, but since the integrodifferential equation is set only in Ω we cannot get summability information outside Ω.
234  8 Fractional Laplacian type operators Proof of Theorem 8.6.4. The proof of the above theorem is split into two parts. We first prove uniqueness of a weak solution. We stress that the proof follows the idea of the one by Stampacchia. Subsequently we construct the solution by approximation with solutions of problems with smooth data. We prove suitable a priori estimates (that, in particular, imply (8.6.3)–(8.6.5)) and compactness of the approximating sequence in L1 (Ω). Uniqueness. Assume that u is a weak solution to (8.6.1) with f = 0. Then ˆ uψ dx = 0 for any ψ ∈ 𝒞0∞ (Ω). Ω
Therefore u ≡ 0. Existence. We obtain the solution to (8.6.1) as a limit of solutions to approximated problems. Consider fn ∈ L∞ (Ω) such that fn → f in L1 (Ω) and let un be the solution to the problem ℒun = fn { un = 0
in Ω,
(8.6.6)
in ℝN \ Ω.
Then we prove the existence of solution in several steps. Step 1. There exists a positive constant c, only depending on N, Ω and s, such that ‖un ‖Lq (Ω) ≤ c‖fn ‖L1 (Ω) ,
∀q ∈ (1,
N ). N − 2s
(8.6.7)
Let us multiply the equation in (8.6.6) by Tk (un ), for k ≥ 0, and let us integrate over Ω. Thanks to Proposition 8.3.6 we obtain 2 λTk (un )L2∗s (Ω) ≤ 𝒮 2 k‖fn ‖L1 (Ω) .
(8.6.8)
Moreover by using the Sobolev inequality and an elementary estimate we have N−2s 2 λk 2 An,k (un ) N ≤ λTk (un )L2∗s (Ω) ≤ 𝒮 2 k‖fn ‖L1 (Ω) , where An,k (un ) = {x ∈ Ω : un (x) ≥ k}. It follows that N
‖fn ‖L1 (Ω) N−2s ) . An,k (un ) ≤ c( k
(8.6.9) N
It means that un is bounded in the Marcinkiewicz space ℳ N−2s (Ω) and consequently (8.6.7) holds true. Step 2. There exists a positive constant c, just depending on q, N, Ω and s, such that s (−Δ) 2 un Lr (Ω) ≤ c‖fn ‖L1 (Ω) ,
∀r ∈ (1,
N ). N −s
(8.6.10)
8.6 Dirichlet problem: Nonvariational setting

235
We fix λ > 0 and, for any positive k, we want to estimate the measure of the following set: s {x ∈ Ω : (−Δ) 2 un ≥ λ} s s = {x ∈ Ω : (−Δ) 2 un ≥ λ, un < k} ∪ {x ∈ Ω : (−Δ) 2 un ≥ λ, un ≥ k}.
Consequently, s s {x ∈ Ω : (−Δ) 2 un ≥ λ} ⊂ {x ∈ Ω : (−Δ) 2 un ≥ λ, un < k} ∪ An,k (un ).
Since s 1 {x ∈ Ω : (−Δ) 2 un ≥ λ, un < k} ≤ 2 λ
ˆ {x∈Ω,un 0, N
s ‖fn ‖L1 (Ω) N−2s k ) . ({x ∈ Ω : (−Δ) 2 un ≥ λ}) ≤ 2 ‖fn ‖L1 (Ω) + c( k λ s
N−2s
Minimizing in k we find that the minimum is achieved at k = λ N−s ‖fn ‖LN−s 1 (Ω) , thus we have N
s ‖fn ‖L1 (Ω) N−s ) . ({x ∈ Ω : (−Δ) 2 un ≥ λ}) ≤ c( λ
(8.6.11) N
This means that (−Δ)s/2 un  is bounded in the Marcinkiewicz space ℳ N−s (Ω) and consequently (8.6.10) holds true. Step 3. Passing to the limit. Before passing to the limit in the equation, we need to determine the a. e. limit of un . Using the linearity of the equation, for any m and n ∈ ℕ, un − um solves ℒ(un − um ) = fn − fm
{
un = 0, um = 0
in Ω,
in ℝN \ Ω.
Hence, choosing, for any k > 0, Tk (un − um ) as a test function in the weak formulation of the above problem, we deduce, by repeating the computations of step 1, that N
‖fn − fm ‖L1 (Ω) N−2s ) . ({x ∈ Ω : un − um  ≥ k}) ≤ c( k
236  8 Fractional Laplacian type operators Since the right hand side of the above inequality is small for n and m large enough, it follows that {un } is a Cauchy sequence in measure. Consequently, up to subsequences (not relabeled), it converges in Ω almost everywhere, toward a function u. By step 1 we also deduce (using the embedding of the ℳp (Ω) spaces into Lp (Ω), for N . Note that this is all p finite) that un also converges to u in Lq (Ω), for any 1 ≤ q < N−2s sufficient to pass to the limit in the equation and obtain a weak solution of (8.6.1). Observe that, by the uniqueness, the whole sequence converges to u in Lq (Ω) and that (8.6.4) holds. Since (8.6.11) holds, we have N
s ‖fn − fm ‖L1 (Ω) N−s ) , ({x ∈ Ω : (−Δ) 2 (un − um ) ≥ λ}) ≤ c( λ s
thus (−Δ) 2 (un ) is a Cauchy sequence in measure in Ω. Therefore, up to a subsequence, s (−Δ) 2 (un ) converges a. e. in Ω. 8.6.1 Calderón–Zygmund type result for weak solutions In the local case it is well known that the Calderón–Zygmund results are true if the right hand side is Lm (Ω) for 1 < m < ∞ (see for instance [186, Chapter 9] and [69]). The existence of a unique weak solution of (8.4.1) is a consequence of Theorem 8.6.4. Thus we can define the operator T as in (8.4.3) and we can get the Calderón–Zygmund type results in such a range of m by using the interpolation Theorem 8.3.13. The statement of our last result concerning solutions of elliptic problems is the following. 2N Theorem 8.6.6. Let f belong to Lm (Ω), with 1 < m < N+2s . Then there exists a unique weak solution of (8.4.1). Moreover there exists a constant c, only depending on N, Ω, m and s, such that
‖u‖Lm∗∗ ≤ c‖f ‖Lm (Ω) s (Ω)
(8.6.12)
s/2 ≤ c‖f ‖Lm (Ω) , (−Δ) u m∗∗ L s/2 (Ω)
(8.6.13)
and
where 1 1 2t = − , m∗∗ m N t
∀t ∈ (0, 1).
Proof. In order to get the estimate we argue by interpolation, namely, we apply the above interpolation theorem. We define the map T as the inverse of ℒ (see (8.4.3)), and N due to Theorem 8.6.4 we know that it acts between L1 (Ω) and Lr (Ω), for any r < N−2s .
8.7 Further results  237
On the other hand, estimate (8.6.12) in the case (2∗s ) = m tells us that it acts between
L(2s ) (Ω) and L2s (Ω), too. Consequently ∗
∗
‖un ‖Lp (Ω) = ‖Tfn ‖Lp (Ω) ≤ c‖fn ‖Lq (Ω) , where 1 1−θ θ = ∗ + p 2s r
and
1 1−θ = + θ, q (2∗s )
i. e., 1 1 2s − = , q p N
which implies
1 1 = . p m∗∗ s
On the other hand, arguing in the same way with the norms of (−Δ)s/2 un  we deduce (8.6.13).
8.7 Further results Consider the minimization problem (8.4.2), where f is a measurable function defined in Ω and the hypotheses about the kernel K hold. In order to have that for all u ∈ H0s (Ω), 2N
fu ∈ L1 (Ω), we must assume that f ∈ L N+2s (Ω) and, passing to the Euler–Lagrange equation, we want to find the natural conditions to have bounded solutions of the problem {
ℒu = f
u=0
in Ω, in ℝN \ Ω.
(8.7.1)
N in From the results in Section 8.4.1 we know that f must be in Lm (Ω) with m > 2s order to have bounded solutions. This condition is natural in the De Giorgi program. The solution to the 19th Hilbert problem proposed in [200] was achieved by Ennio De Giorgi in his famous paper [137]. The De Giorgi program has two main steps. Given a variational solution to problem (8.7.1), Step 1. L2 estimates implies L∞ estimates; Step 2. bounded solutions of finite energy become 𝒞 α continuous.
The result by De Giorgi was also obtained by J. F. Nash in [250] by different methods and, later on, was reproved by J. Moser in [245] using a different approach. The De Giorgi proof has been applied in many different settings, for instance in [185] and [109]. The book [188] could be a good reference to see the modern development of the De Giorgi ideas in the local case. For the nonlocal case there are many references and the following is a (not exhaustive) list of them: [109], [124], [136], [207], [208], [273], [288].
238  8 Fractional Laplacian type operators In our case, the first step is solved by the methods of Stampacchia and Moser in N Section 8.4.1 when f ∈ Lm (Ω), m > 2s . That is N s the solution u ∈ H0 (Ω) to problem (8.7.1) with f ∈ Lm (Ω) for m > 2s is bounded. Regarding the second step, the following result holds. Theorem 8.7.1. Let u ∈ H0s (Ω) ∩ L∞ (Ω) be the solution to problem (8.7.1). Then u is Hölder continuous in Ω. More precisely, there exist α and C, both depending only on N, s and the ellipticity constants, such that if B2r (x0 ) ⊂ Ω, then α
ρ oscBρ (x0 ) (u) ≤ C( ) [‖u‖L2 (B2r (x0 )) + ‖f ‖Lm (B2r (x0 )) ]. r The main difficulty in the proof is to find local estimates for solutions of a nonlocal operator and then find similar results as in the local case. In the references cited above the reader can find detailed proofs of the 𝒞 α regularity. In references [109], [124] and [136] the arguments are extensions of those by De Giorgi. In references [207] and [208] the arguments are close to the Moser techniques, while in [273] the method of Krylov–Safonov is used. In [288] general results are presented and some interesting applications are obtained.
9 The fractional Hardy inequality 9.1 Introduction In this chapter we will study Hardy’s inequality for the fractional 2s Laplacian and the influence of the Hardy potential in semilinear problems. The motivation comes from several points of view, but we take here potential theory and theoretical physics as our main references. – Assume 0 < s < 1. Let us try to find a potential x−α , with α > 0, for which there exists C > 0 such that C∫ ℝN
u2 2 dx ≤ ∫ ξ 2s ℱ (u)(ξ ) dξ , xα
u ∈ 𝒞0∞ (ℝN ).
ℝN
Then, by homogeneity as in the local case s = 1, we must have α = 2s. By the analysis done in Chapter 8, 2 s/2 2 2s 2s (−Δ) uL2 (ℝN ) = (2π) ∫ ξ  ℱ (u)(ξ ) dξ ,
u ∈ 𝒞0∞ (ℝN ).
ℝN
Thus, we have the equivalent inequality in terms of fractional derivatives C ∫ ℝN
–
u2 2 dx ≤ (−Δ)s/2 uL2 (ℝN ) , x2s
u ∈ 𝒞0∞ (ℝN ).
Therefore the fractional Hardy inequality is just the extension of the classical Hardy inequality studied in Chapter 2. From theoretical physics, the previous inequality can be seen as the uncertainty principle with respect to the 2sfractional momentum. The inequality plays an important role in the study of the stability of matter from a relativistic point of view, as is explained in the fundamental paper by R. Frank, E. Lieb and R. Seiringer [167]. We recommend the reader to see also the references in this paper.
The goal in this chapter is to prove the following inequality, in which we include the explicit calculation of the optimal constant. Fractional Hardy inequality. Given u ∈ 𝒞0∞ (ℝN ), ΛN,2s ∫ ℝN
u2 ̂ 2 dx ≤ ∫ ξ 2s u (ξ ) dξ , x2s ℝN
where 0 < s < 1, N > 2s and ΛN,2s = π −2s https://doi.org/10.1515/9783110606270009
Γ2 ( N+2s ) 4
Γ2 ( N−2s ) 4
.
240  9 The fractional Hardy inequality As we said above, this inequality is a cornerstone in the proof of the relativistic stability of matter as is described in [167], and it has become a classical tool in quantum mechanics for years. Despite this, the main results from an analysis point of view are more recent. Perhaps the first work in which the optimal constant ΛN,2s is obtained is the paper by Herbst in [199] and then by Beckner in [54] and Yafaev in [319]. We will present in this chapter a proof of Hardy’s inequality that is inspired in the articles [54] and [295] and, as these, uses classical tools from potential theory. Our approach, however, is more direct and elementary. We have made it also selfcontained. Extensions of the Hardy inequality to the fractional quasilinear framework can be found in the literature; see for instance [168] and the references therein.
9.2 The fractional Hardy inequality The functional inequality known as the fractional Hardy inequality is the following. Theorem 9.2.1. Assume 2s < N and 0 < s < 1. Then the following inequality holds: ΛN,2s ∫ ℝN
u2 2 dx ≤ ∫ ξ 2s ℱ (u)(ξ ) dξ , x2s
u ∈ 𝒞0∞ (ℝN ),
(9.2.1)
ℝN
where ΛN,2s = π −2s
Γ2 ( N+2s ) 4
) Γ2 ( N−2s 4
(9.2.2)
.
Remark 9.2.2. In the complementary case, N ≤ 2s, a similar inequality to (9.2.1) holds on the space 𝒞0∞ (ℝN \ {0}). See [167]. It is worth pointing out that 𝒞0∞ (ℝN \ {0}) is dense in H s (ℝN ) if N > 2s. If N ≤ 2s the completion of 𝒞0∞ (ℝN \ {0}) is H s (ℝN \ {0}). See [140] and [139]. Remark 9.2.3. Since 2 s/2 2 2s 2s (−Δ) uL2 (ℝN ) = (2π) ∫ ξ  ℱ (u)(ξ ) dξ ,
u ∈ 𝒞0∞ (ℝN ),
(9.2.3)
ℝN
we can write the Hardy inequality as 2 s/2 2 ̃ N,2s ∫ u (x) dx, (−Δ) uL2 (ℝN ) ≥ Λ x2s ℝN
̃ N,2s = 22s where Λ
Γ2 ( N+2s ) 4
) Γ2 ( N−2s 4
.
u ∈ 𝒞0∞ (ℝN ),
(9.2.4)
9.2 The fractional Hardy inequality  241
The result in Theorem 9.2.1 is a direct consequence of a family of Hardy inequalities that have many applications to other areas of analysis, including certain estimates for higher order fractional and local operators. The notation for the Riesz potentials and their corresponding kernels is the same as used in Section 8.1. We will use repeatedly that the Fourier transform is an isomorphism on the Schwartz class. We will also use the following elementary result of measure theory. Lemma 9.2.4. Let dμ be a probability measure and u ∈ L2 ( dμ), u ≥ 0. If 2
(∫ u(x) dμ) = ∫ u2 (x) dx,
(9.2.5)
then u(x) = ‖u‖L2 ( dμ) for dμalmost x. Proof. Assume that ‖u‖L2 ( dμ) ≠ 0, otherwise there is nothing to prove. Hence by normalization we can assume that ‖u‖L2 ( dμ) = 1. Therefore by hypothesis, 1 2
2
∫ u(x) dμ = (∫ u (x) dμ) = 1. Since 0 ≤ u(x) ≤ 21 u2 (x) +
1 2
for all x, we see that
1 1 1 = ∫ u(x) dμ ≤ ∫( u2 (x) + ) dμ = 1. 2 2 Hence, we must have u(x) = dμalmost x.
1 2 u (x) 2
1 2
+
for dμalmost x, and therefore u(x) = 1 for
A more general form of Hardy’s inequality is given by the following. Theorem 9.2.5. Assume 0 < α < N. Then 2 2 ds α ∫ ℱ (f )(s) α ≤ Λ−1 N,α ∫ x f (x) dx, s
for all f ∈ 𝒮 (ℝN ),
(9.2.6)
ℝN
ℝN
where Λ−1 N,α = Iα (g N+α )(ω).
(9.2.7)
2
Here ω is an arbitrary unit vector and g N+α , Iα were defined in (8.1.9). 2
Moreover, with cN, N+α defined in (8.1.8), 2
2 α Λ−1 N,α = [cN, N+α ] = π [ 2
Γ( N−α ) 4
Γ( N+α ) 4
2
]
is the best constant. The identity in (9.2.6) is never attained, except for f ≡ 0 a. e.
(9.2.8)
242  9 The fractional Hardy inequality Proof. We present the proof in three steps. First step. Proof of (9.2.7) and (9.2.8). Inequality (9.2.6) by the Plancherel theorem is equivalent to 2 2 α ∫ I α (f )(x) dx ≤ Λ−1 N,α ∫ x f (x) dx. 2
ℝN
(9.2.9)
ℝN
Since I α is selfadjoint, by the semigroup property the left hand side of (9.2.9) can be 2 written as follows: ∫ I α (f )(x)I α (f )(x) dx = ∫ Iα (f )(x)f (x) dx ℝN
2
2
ℝN
= cN,α ∫ ∫ f (x) ℝN
Call h(x) = f (x)x to
N+α 2
(9.2.10)
, that is, f (x) = h(x)g N+α (x). Then, inequality (9.2.9) is equivalent 2
N−α
cN,α
ℝN
1 f (y) dy dx. x − yN−α
N−α
x 2 y 2 dy dx 2 dx h(y) N N ≤ Λ−1 . ∫ ∫ h(x) N,α ∫ h(x) x − yN−α y x xN
ℝN ℝN
(9.2.11)
ℝN
We set N−α
K(x, y) = cN,α
N−α
x 2 y 2 x − yN−α
and dμ(x) =
dx . xN
Then, the left hand side of (9.2.11) can be estimated as follows: ∫ ∫ h(x)h(y)K(x, y) dμ(x) dμ(y) ℝN ℝN 1
1
2 2 2 2 ≤ ( ∫ h(x) dμ(x)) ( ∫ ∫ h(y)K(x, y) dμ(x) dμ(y)) N N N ℝ
ℝ ℝ
1
2 2 ≤ ‖h‖L2 ( dμ) ( ∫ ( ∫ h(y) K(x, y) dμ(y))( ∫ K(x, z) dμ(z)) dμ(x))
ℝN ℝN
ℝN 1
2 2 = ‖h‖L2 ( dμ) ( ∫ h(y) ( ∫ K(x, y)( ∫ K(x, z)dμ(z)) dμ(x)) dμ(y)) .
ℝN
ℝN
(9.2.12)
ℝN
For the inequalities we have used simply Cauchy–Schwarz and the last identity is just the application of Fubini’s theorem. The proof will conclude if we are able to prove that ∫ K(x, z) dμ(z) = Λ−1 N,α , ℝN
independent of x.
(9.2.13)
9.2 The fractional Hardy inequality  243
Note that, by symmetry, (9.2.13) is the same as ∫ K(x, y) dμ(x) = Λ−1 N,α ,
independent of y.
ℝN
To prove (9.2.13) we write x = xx with x ∈ SN−1 and make the change of variable z ; therefore w = x N−α
ℝN
N−α
x 2 z 2 dz z N−α N−α x − x zN  x
∫ K(x, z) dμ(z) = cN,α ∫ ℝN
N−α
dw 1 dw w 2 = cN,α ∫ = Iα (g N+α )(x ). 2 x − wN−α wN x − wN−α w N+α 2 N
= cN,α ∫ ℝN
ℝ
We need to evaluate Iα (g N+α )(x ), which, clearly, is independent of x because g N+α is a 2 2 radial function. Now, ℱ (Iα (g N+α ))(ξ ) = 2
1 1 1 ℱ (g N+α )(ξ ) = c N+α = cN, N+α g N+α (ξ ). 2 2 2 ξ α ξ α N, 2 ξ  N−α 2
That means that g N+α is invariant for the operator h → ℱ (Iα )(h). Therefore, 2
Iα (g N+α )(x) = cN, N+α ℱ −1 (g N+α )(x) = [cN, N+α ]2 2
2
2
2
1 x
N−α 2
.
This completes the proof of (9.2.8). According with the previous estimates and (9.2.12) the inequality (9.2.6) follows. Note that, by density, inequality (9.2.6) holds for all f ∈ L2 (ℝN , xα dx). Second step. Optimality of Λ−1 N,α . Proving the fact that ΛN,α is sharp is equivalent to proving the following statement. If for 0 < α < N and for every f ∈ L2 (ℝN , xα dx) we have 2 2 ∫ I α (f )(x) dx ≤ C0 ∫ f (x) xα dx, 2
ℝN
(9.2.14)
ℝN
then C0 ≥ Λ−1 N,α . Before proving the statement, we perform a formal calculation to motivate the rigorous proof and, as a byproduct, to find a pseudominimizer as in the local case. Indeed, consider f (x) = g N+α (x) := g(x). Then 2
ℱ (I α )(f )(ξ ) = 2
1 ξ 
α 2
cN, N+α 2
1 ξ 
N−α 2
= cN, N+α 2
1
N
ξ  2
.
244  9 The fractional Hardy inequality Hence α
I α (f )(x) = cN, N+α ℱ −1 (g N )(x) = cN, N+α g N (x) = cN, N+α f (x)x 2 , 2
2
2
2
2
2
that is, we obtain the following pointwise identity: α 2 α 2 −1 I (f )(x) = ΛN,α f (x) x . 2 Since f (x) = g N+α (x) ∈ ̸ L2 (ℝN , xα dx), we cannot obtain what we want, but we have 2 two consequences. The first one is that g N+α (x) behaves really like a candidate to be 2 a minimizer. The problem is that it does not belong to the space that we need. For this reason we will call g N+α (x) a pseudominimizer. The second conclusion is that a 2 complete proof will be obtained doing the same estimates on truncates of our pseudominimizer. This is essentially the same kind of argument that we used in Chapter 2, Section 2.3.2 to study the classical Hardy–Leray inequality. We proceed now to prove the previous statement and then that ΛN,α is the best constant in inequality (9.2.6). For R large enough, take ψR ∈ 𝒮 (ℝN ) such that χ[ 1 ,R] (x) ≤ ψR (x) ≤ χ[ 1 ,2R] (x). R
2R
We will multiply the pseudominimizer by the cutoff ψR , that is, we consider f (x) =
ψR (x) x
N+α 2
.
Clearly, we have f ∈ 𝒮 (ℝN ). We can compute directly both sides of inequality (9.2.14) for f , 2 2 1 dx ≤ ∫ f (x) xα dx = ∫ ψ(x) xN
ℝN
ℝN
∫ 1 0 such that q
2 q
2( q1 − 2∗1 )
(∫ u dx) ≤ Cq,s,N Ω Ω
s
2N . N−2s
Then there exists a constant
̂ 2 dξ − ΛN,2s ∫ ( ∫ ξ 2s u ℝN
ℝN
u2 dx), x2s
∀u ∈ 𝒞0∞ (Ω). (9.6.1)
Remark 9.6.2. Note that ̂ 2 dξ − ΛN,2s ∫ hs (u) := ( ∫ ξ 2s u ℝN
ℝN
u2 dx), x2s
∀u ∈ 𝒞0∞ (Ω),
256  9 The fractional Hardy inequality is the square of a preHilbert seminorm, associated to the following scalar product uψ ̂ −Λ ̂ ψdξ ⟨u, ψ⟩ := ∫ ξ 2s u dx = ⟨u, Hs ψ⟩, N,2s ∫ x2s
∀u, ψ ∈ 𝒞0∞ (Ω),
ℝN
ℝN
ψ
where Hs (ψ) = (−Δ)s ψ − ΛN,2s x2s . Notice that the brackets in the last term means the
duality product between H s (ℝN ) and its dual H −s (ℝN ). The result in Theorem 9.6.1, in particular, shows that hs (u) is in fact a Hilbertian norm in a bigger space than H s (ℝN ) and, has a byproduct, that the optimal constant is not attained. To prove Theorem 9.6.1 we need some previous results. Lemma 9.6.3. Let 0 ≤ χ ≤ 1 be a smooth function in ℝ+ of compact support with χ(r) = 1 for r ≤ 1. Define ψλ (x) := χ(
x )x−α λ
(9.6.2)
for 0 < α < (N−2s) and λ > 0. Then, ψλ ∈ H s (ℝN ) for 0 < s < 1 and, moreover, for all 2 ϵ > 0 there exists a λϵ = λϵ (α, N, χ) such that for any λ ≥ λϵ , (((−Δ)s − ΛN,s x−2s )ψλ )(x) ≤ −(ΦN,s (α) − ϵ)x−α−2s ,
∀x ∈ B,
(9.6.3)
in the sense of distributions, and with B representing the unit ball in ℝN . Proof. The proof that ψλ ∈ H s (ℝN ) is a direct computation that follows from [167, Proposition 4.1]. Consider ϕ ∈ 𝒞0∞ (Ω). Using (9.5.1) we find ⟨ψλ , ((−Δ)s − ΛN,s x−2s )ϕ⟩ = ΦN,s (α)⟨x−α−2s , ϕ⟩ − ⟨(1 − χ(
x )x−α ), (−Δ)s ϕ⟩. λ
̃ (x) = (1 − χ( x ))x−α . By (9.3.13) we obtain Call ψ λ λ ⟨(1 − χ(
̃ (y)ϕ(x) ψ x )x−α ), (−Δ)s ϕ⟩ = − 2aN,s ∫ ∫ λ N+2s dy dx ≥ λ x − y ℝN ℝN
− sup 2aN,s xα+2s ∫ x≤1
ℝN
̃ (y) ϕ ϕ(x) λ dy ∫ dx. x − yN+2s x − yN+2s ℝN
Set ρ(λ) = sup 2aN,s xα+2s ∫ x≤1
Now
ℝN
̃ (y) ϕ (1 − ϕ(y)) λ dy = sup 2aN,s xα+2s ∫ α dy. 1 x − yN+2s y x − yN+2s x≤ λ
ℝN
9.6 Hardy’s inequality with remainder terms  257
(i) ρ(λ) < ∞ if λ > 1; (ii) ρ(λ) ↓ 0 as λ → ∞. Hence, given ϵ > 0 we can choose λε such that ρ(λϵ ) = ϵ. Since ΦN,s (α) < 0 by Lemma 9.5.1, we obtain (9.6.3). Lemma 9.6.4. Let 0 ≤ q < 1 and u ∈ Lq (ℝN ) be a positive, radially symmetric, decreasing function. Then, denoting by B the unit ball, ‖u‖q ≤
1 − q1 −N B ∫ u(x)x q dx. q
(9.6.4)
ℝN
Proof. If u is the characteristic function of a ball, a simple computation shows that the above is in fact an identity. For general positive, radially symmetric, decreasing u we have for every t > 0 that the level set {x : u(x) > t} is a ball. We denote by χt (x) its characteristic function. Then, ∞
u(x) = ∫ χt (x) dt. 0
Therefore by Minkowski’s inequality we conclude that ∞
∞
0
0 ℝN
1 −1 1 −1 −N −N ‖u‖q ≤ ∫ ‖χt ‖q dt = B q ∫ ∫ χt (x)x q dx dt = B q ∫ u(x)x q dx, q q ℝN
where the last identity is just Fubini’s theorem. Proof of Theorem 9.6.1. By decreasing rearrangement and scaling, we can assume that Ω is the unit ball centered at the origin, B, and then perform the proof in the radial setting. By Lemmas 9.6.3 and 9.6.4, if u is a symmetric function in B, ‖u‖q ≤
1 − q1 1 −1 −N −1 B ∫ u(x)x q dx ≤ 2 B q ΦN,s (α) ⟨u, Hs ψ⟩. q q ℝN
Here ψ ≡ ψλϵ is as in Lemma 9.6.3 with ϵ = inequality,
ΦN,s (α) . 2
Finally by Cauchy–Schwarz’s
2 ⟨u, Hs ψ⟩ ≤ hs (u)hs (ψ) and we conclude the proof. We also have the following result obtained in [24]. See also [164] for a different proof.
258  9 The fractional Hardy inequality Theorem 9.6.5. Let 0 < s < 1 and N > max{2s, 1}. Assume that Ω ⊂ ℝN is a bounded domain. Then for all 1 ≤ q < 2, there exists a positive constant C = C(Ω, q, N, s) such that for all u ∈ 𝒞0∞ (Ω), aN,s ∫ ∫ ℝN ℝN
(u(x) − u(y))2 u2 (x) dx dx dy − Λ ∫ N,2s x2s x − yN+2s
≥ C(Ω, q, N, s) ∫ ∫ Ω Ω
ℝN
2
(u(x) − u(y)) dx dy. x − yN+qs
(9.6.5)
Proof. We follow the same strategy as in [167], that is, the ground state representation. u(x) . Then w(x) = x−α and v(x) = w(x) . Thus, For simplicity, we set α = N−2s 2 u(y) ((u(x) − u(y)) − w(y) (w(x) − w(y)))2 w(y) (v(x) − v(y))2 f1 (x, y) := w(x)w(y) = . w(x) x − yN+2s x − yN+2s
In the same way, thanks to the symmetry of f1 (x, y), it immediately follows that u(x) ((u(y) − u(x)) − w(x) (w(y) − w(x)))2 w(x) (v(x) − v(y))2 w(x)w(y) = ≡ f2 (x, y). w(y) x − yN+2s x − yN+2s
Then, hs (u) =
cN,2s ( ∫ ∫ f1 (x, y) dx dy + ∫ ∫ f2 (x, y) dx dy). 2 ℝN ℝN
ℝN ℝN
Since f1 and f2 are positive functions, it follows that hs (u) ≥
cN,2s (∫ ∫ f1 (x, y) dx dy + ∫ ∫ f2 (x, y) dx dy). 2 Ω Ω
Ω Ω
Note that for all (x, y) ∈ Ω × Ω and q < 2, we have K(x, y) ≡ It is clear that K(x, y)( w(x) + w(y)
w(y) ) w(x)
−2
≥
C x−yN+qs
and
w(x)w(y) 1 ≤ . w2 (x) + w2 (y) 2
= 1. Hence,
f1 (x, y) ≥ C1 K(x, y) ×[
1 x−yN+2s
w(y) w(x)
2
u(y) (w(x) − w(y))2 (u(x) − u(y))2 +( ) N+qs w(y) x − y x − yN+qs
u(y) (u(x) − u(y))(w(x) − w(y)) ] w(y) x − yN+qs
9.6 Hardy’s inequality with remainder terms  259
and f2 (x, y) ≥ C1 K(x, y) ×[ −2
w(x) w(y)
2
(u(y) − u(x))2 u(x) (w(y) − w(x))2 ) + ( w(x) x − yN+qs x − yN+qs
u(x) (u(y) − u(x))(w(y) − w(x)) ]. w(x) x − yN+qs
Therefore, hs (u) ≥
C1 w(y) w(x) (u(x) − u(y))2 dx dy + ) ∫ ∫ K(x, y)( 2 w(x) w(y) x − yN+qs Ω Ω
− C1 ∫ ∫ K(x, y)[ Ω Ω
u(y) (u(x) − u(y))(w(x) − w(y)) w(x) x − yN+qs
u(x) (u(y) − u(x))(w(y) − w(x)) + ] dx dy w(y) x − yN+qs
≥
C1 (u(x) − u(y))2 dx dy − C1 ∫ ∫ g1 (x, y) dx dy − C1 ∫ ∫ g2 (x, y) dx dy, ∫∫ 2 x − yN+qs Ω Ω
Ω Ω
Ω Ω
with g1 (x, y) = K(x, y)
u(y) (u(x) − u(y))(w(x) − w(y)) ) ( w(x) x − yN+qs
g2 (x, y) = K(x, y)
u(x) (u(y) − u(x))(w(y) − w(x)) ). ( w(y) x − yN+qs
and
Using the fact that u ∈ 𝒞0∞ (Ω) it follows that ∫Ω ∫Ω gi (x, y) dx dy < ∞ for i = 1, 2. Since g2 (x, y) = g1 (y, x), we just have to estimate ∫Ω ∫Ω g2 (x, y) dx dy. Note that g2 (x, y) =
w(x)u(x) (u(y) − u(x))(w(y) − w(x)) . + w2 (y) x − yN+qs
w2 (x)
Hence, using the Young inequality, ∫ ∫g2 (x, y) dx dy Ω Ω
≤ ϵ∫∫ Ω Ω
w2 (x)u2 (x)(w(x) − w(y))2 (u(x) − u(y))2 dx dy + C(ϵ) ∫ ∫ 2 dx dy. N+qs x − y (w (x) + w2 (y))2 x − yN+qs Ω Ω
We claim that I ≡ ∫∫ Ω Ω
w2 (x)u2 (x)(w(x) − w(y))2 dx dy ≤ Chs (u). (w2 (x) + w2 (y))2 x − yN+qs
260  9 The fractional Hardy inequality It is clear that I = ∫ u2 (x)[∫ Ω
Ω
(xα − yα )2 y2α dy] dx. (x2α + y2α )2 x − yN+qs
To compute the above integral, we use a similar argument as in the proof of Theorem 1.1 in [158]. We set y = ρy and x = rx . Then, since Ω ⊂ BR (0) for a suitable R, we get R
(r α − ρα )2 ρN+2α−1 dH N−1 (y ) ( ∫ )dρ dx, 2α 2α 2 (r + ρ ) ρy − rx N+qs
I ≤ ∫ u2 (x) ∫ Ω
0
y =1
where dH N−1 is the (N − 1) dimensional Hausdorff measure on the sphere. Let ρ = rσ. Then R r
dH N−1 (y ) u2 (x) (1 − σ α )2 σ N+2α−1 ( )dσ dx I ≤ ∫ qs ∫ ∫ x (1 + σ 2α )2 σy − x N+qs Ω
0
y =1
∞
≤∫ Ω
u2 (x) (1 − σ α )2 σ N+2α−1 K(σ)dσ, dx ∫ qs x (1 + σ 2α )2 0
where K(σ) = 2
N−1 2
π
) Γ( N−1 2
π
∫ 0
∞ (1−σ α )2 σ N+2α−1 K(σ)dσ. (1+σ 2α )2
Let us analyze ∫0
sinN−2 (θ) (1 − 2σ cos(θ) + σ 2 )
N+qs 2
dθ.
We have
∞
1
∞
0
0
1
(1 − σ α )2 σ N+2α−1 (1 − σ α )2 σ N+2α−1 (1 − σ α )2 σ N+2α−1 K(σ)dσ = K(σ)dσ + K(σ)dσ. ∫ ∫ ∫ (1 + σ 2α )2 (1 + σ 2α )2 (1 + σ 2α )2 By setting ξ = we get
1 σ
in the first integral and taking into account that K( ξ1 ) = ξ N+qs K(ξ ),
∞
∞
∞
0
1
1
(ξ α − 1)2 ξ qs−1 (1 − σ α )2 σ N+2α−1 (σ α − 1)2 σ N+2α−1 K(σ)dσ = K(ξ )dξ + K(σ)dσ. ∫ ∫ ∫ (1 + σ 2α )2 (1 + ξ 2α )2 (1 + σ 2α )2 Note that K(σ) ⋍ σ −N−qs as σ → ∞, thus both integrals converge near ∞. On the other hand and following the computation of [158] (estimates (2.1) to (2.10), see also the definition of the function H in formula (2.6)), we reach K(σ) ≤ C(σ 2 − 1)−1−2s as σ → 1. Hence, as σ, ξ → 1+ , 2
(σ α − 1) K(σ) ≤ C(σ − 1)1−2s ∈ L1 (1, 2). Thus, combining the above estimates, we get I ≤ C ∫Ω
u2 (x) xqs
dx.
9.7 Further results and comments 
261
Since q < 2, using the Hölder inequality and (9.6.1), it follows that I ≤ Chs (u). Then the claim follows. As a conclusion, we obtain hs (u) ≥ C(Ω, q, N, s) ∫ ∫ Ω Ω
(u(x) − u(y))2 dx dy, x − yN+qs
which finishes the proof.
9.7 Further results and comments The Hardy inequality has its counterpart with homogeneity sp > N; see for instance [168]. More precisely in this paper the following result is obtained. Theorem 9.7.1 (Hardy inequality). Let N ≥ 1 and 0 < s < 1. Then for all u ∈ Ẇ s,p (ℝN ) in case 1 ≤ p < Ns and for all u ∈ Ẇ s,p (ℝN \ {0}) in case p > Ns , ∫ ∫ ℝN ℝN
u(x)p u(x) − u(y)p dx dy ≥ Λ dx, ∫ N,s,p xps x − yN+ps
(9.7.1)
ℝN
with 1
N−ps p ΛN,s,p := 2 ∫ r ps−1 1 − r p ΦN,s,p (r) dr,
(9.7.2)
0
and 1
2 N−3 2
(1−t ) { N+ps dt {ωN−2 ∫−1 (1−2rt+r 2 ) 2 ΦN,s,p (r) = { { 1 1 {( (1−r)1+ps + (1+r)1+ps )
if N ≥ 2, if N = 1.
(9.7.3)
The constant ΛN,s,p is optimal. If p > 1 the inequality is strict for any function 0 ≠ u ∈ Ẇ s,p (ℝN ) in case 1 ≤ p < Ns , and for all 0 ≠ u ∈ Ẇ s,p (ℝN \ {0}) in case p > Ns . If p = 1, equality holds if and only if u is proportional to a symmetric decreasing function. In the same paper [168] Frank and Seiringer find a general Hardy inequality and prove for p ≥ 2 a Hardy inequality with remainder term. The previous inequality in the above theorem is related to the following integrodifferential problem. Defining f (x, u) by means of the difference ∫ ℝN
u(x) − u(y)p−2 (u(x) − u(y)) u(x)p−2 u dy − λ = f (x, u), N+ps xps x − y
find lower estimates for f depending on the value of λ with respect to ΛN,s,p . The function f could also be dependent on ∇u if s > 21 .
10 Calderón–Zygmund summability in the fractional setting 10.1 Introduction and statement of the problem In Chapter 3 we have established how the Hardy potential affects the Calderón– Zygmund summability of the solutions to the linear problem (3.1.1). In this chapter we will study the following problem: (−Δ)s u − λ xu2s = f (x) { { { u>0 { { { { u=0
in Ω, (10.1.1)
in Ω, in ℝN \ Ω,
where s ∈ (0, 1) is such that 2s < N, Ω ⊂ ℝN is a bounded regular domain containing the origin and f is a measurable function satisfying suitable hypotheses of integrability. The fractional sLaplacian has been defined in Chapter 8, Sections 8.1 and 8.2 in different settings. Recall that we define the fractional Laplacian as the following integral operator: (−Δ)s u(x) := CN,s P.V. ∫ ℝN
u(x) − u(y) dy, x − yN+2s
s ∈ (0, 1), u ∈ 𝒮 (ℝN ),
(10.1.2)
where CN,s is given by (9.3.12). The asymptotics of the constant CN,s can be found in [138, Section 4]. Due to its second term, problem (10.1.1) is related to the following fractional Hardy inequality, proved in Chapter 9: ∫ ξ 2s û 2 dξ ≥ ΛN,2s ∫ x−2s u2 dx ℝN
∀u ∈ 𝒞0∞ (ℝN ),
(10.1.3)
ℝN
where ΛN,2s :=
2 N+2s 2s Γ ( 4 ) 2 ) Γ2 ( N−2s 4
(10.1.4)
is optimal and not attained. See Theorem 9.2.1. Moreover, lim ΛN,2s = ( s→1
2
N −2 ), 2
the classical Hardy constant. We can also rewrite (10.1.3) in the form aN,s ∫ ∫ ℝN ℝN
u2 u(x) − u(y)2 dx dy ≥ ΛN,2s ∫ 2s dx, N+2s x x − y
https://doi.org/10.1515/9783110606270010
ℝN
u ∈ 𝒞0∞ (ℝN ),
aN,s =
CN,s , 2
264  10 Calderón–Zygmund summability in the fractional setting which we will use along this chapter. See also [167]. This critical value ΛN,2s will also play a fundamental role concerning the regularity of the solution in terms of the summability of the datum. In particular, we already know that for λ > ΛN,2s problem (10.1.1) has no positive supersolution (see for example [46, 171]). Hence, from now on we will assume 0 < λ ≤ ΛN,2s . Remark 10.1.1. The behavior of changingsign solutions in problems involving the Hardy potential is quite different. In fact some obstructions for the existence disappear (see [311, Section 8]). Roughly speaking, the oscillation can produce cancellations near the origin that kill the critical behavior of the potential. Since in this book we are interested in the critical behavior we will focus on nonnegative solutions. The goal in this chapter will be to obtain the optimal summability of u according to the summability of the datum f and the parameter λ. The case λ = 0 was studied in Chapter 8, Sections 8.4 and 8.6 (see also [224]), where some Calderón–Zygmund type results were obtained. Since this problem is linear, we will assume, without loss of generality, that the datum f is positive and we will deal with positive solutions. The influence of the Hardy potential in the local case (s = 1) was studied in Chapter 3 (see also [80]). The results in the nonlocal case, 0 < s < 1, are more involved and need a change in the functional setting. In order to find an analogous optimal condition for problem (10.1.1), we will need a weak Harnack inequality for a singular weighted nonlocal operator associated with the ground state transform that we have defined in Section 9.5 and we will recall in (10.2.3) below. This study, performed in Section 3, requires the combination of techniques on elliptic operators and the behavior of nonlocal radial integrals studied in Chapter 9, and it provides the precise behavior of the solutions around the origin. This result will be the key in the proofs of existence and regularity in the next sections. This chapter is organized as follows. In Section 10.2 we precise the meaning of solutions that will be used along this part of the work, with the corresponding functional setting. In Section 10.2.1 we prove some useful tools as a weighted Sobolev inequality, compactness results and certain algebraic inequalities. In Section 10.3 we prove a weighted singular version of the Harnack inequality. Note that using the ground state transformation stated in Lemma 10.2.4, the weak Harnack inequality gives the exact blowup rate for the positive supersolutions to (10.1.1) near the origin. This result will be the key for the optimality in the results of the following sections. Finally, in Section 10.4 we prove the main results. According to λ and the summability of f , we obtain the optimal summability of the solution u for certain values of the spectral parameter λ. In particular, we see that the local techniques applied in Chapter 3 do not give complete information in this framework, leaving the optimality for certain ranges of λ as an open problem. Some results in this chapter were obtained in [12].
10.2 Functional setting: inequalities with weights  265
10.2 Functional setting: inequalities with weights The Sobolev spaces H s (ℝN ), H s (Ω), their duals, the definition of the fractional Laplacian and its fundamental properties have been introduced in Chapter 8. In this section we will concentrate on the specific functional tools to study the summability of the solutions to (10.1.1). This new framework is suggested by the meaning of solutions for problem (10.1.1): (i) energy solutions when the variational framework can be used, that is, when the datum belongs to the dual space; (ii) weak solutions for data that are integrable but not in the dual space. Recall that we also assume 0 ∈ Ω to avoid the classical situation of a positive bounded potential. Definition 10.2.1. Assume 0 < λ < ΛN,2s . For f ∈ H −s (Ω) we say that u ∈ H0s (Ω) is a finite energy solution to (10.1.1) if ⟨(−Δ)s u, w⟩ − λ ∫ Ω
uw dx = ⟨f , w⟩, x2s
∀w ∈ H0s (Ω).
If λ < ΛN,2s , then existence and uniqueness of a solution u ∈ H0s (Ω) for all f ∈ H −s (Ω) easily follows from the Lax–Milgram theorem. Remark 10.2.2. If λ = ΛN,2s , the same result holds but in a space H(Ω) defined as the completion of 𝒞0∞ (Ω) with respect to the norm ‖ϕ‖2H(Ω) := aN,s ∬ DΩ
ϕ2 ϕ(x) − ϕ(y)2 dx. dx dy − Λ ∫ N,2s x2s x − yN+2s Ω
Using the Hardy inequality with remainder term obtained in Section 9.6 (see also [19] and the references therein), we get that H(Ω) is a Hilbert space and H0s (Ω) ⊊ H(Ω) ⊊ W0s,q (Ω), for all q < 2. This mechanism is similar to the one used in the local case, s = 1. To deal with the case of a general f ∈ L1 (Ω), we need to define the notion of weak solution, where we only request the regularity needed to give weak sense to the equation. Since the operator is nonlocal, we need to detail the class of test functions to be considered, which is precisely N
s
∞
α
N
𝒯 := {ϕ : ℝ → ℝ  (−Δ) ϕ = φ, φ ∈ L (Ω) ∩ 𝒞 (Ω), 0 < α < 1, ϕ = 0 in ℝ \ Ω}.
(10.2.1)
Note that every ϕ ∈ 𝒯 belongs in particular to L∞ (Ω) (see Chapter 8) and, moreover, is a strong solution to the equation (−Δ)s ϕ = φ. See [282] and [287].
266  10 Calderón–Zygmund summability in the fractional setting Definition 10.2.3. Assume f ∈ L1 (Ω). We say that u ∈ L1 (Ω) is a weak supersolution (subsolution) of problem (10.1.1) if xu2s ∈ L1 (Ω), u = 0 in ℝN \Ω, and for all nonnegative ϕ ∈ 𝒯 , the following inequality holds: ∫ u(−Δ)s ϕ dx − λ ∫ Ω
Ω
uϕ dx ≥ (≤) ∫ fϕ dx. x2s Ω
If u is a super and subsolution, then we say that u is a weak solution. Recall that if u ∈ 𝒞0∞ (ℝN ), in Section 9.5 we have proved, following [167], the ground state representation. N−2s . 2
Lemma 10.2.4 (Ground state representation). Let 0 < γ < v(x) := xγ u(x), then
If u ∈ 𝒞0∞ (ℝN ) and
v(x) − v(y)2 dx dy ̂ 2 2 , ) dξ − (ΛN,2s + ΦN,s (γ)) ∫ x−2s u(x) dx = aN,s ∬ ∫ ξ 2s u(ξ x − yN+2s xγ yγ
ℝN
ℝ2N
ℝN
where ΦN,s (γ) = 22s (
Γ( γ+2s )Γ( N−γ ) 2 2
Γ( N−γ−2s )Γ( γ2 ) 2
−
) Γ2 ( N+2s 4 Γ2 ( N−2s ) 4
)
and aN,s is defined in (9.3.14). In particular this representation proves that the constant ΛN,2s is optimal and not attained. Using this representation with λ = ΛN,2s + ΦN,s (γ), we obtain that if u solves {
(−Δ)s u − λ xu2s = f (x) u=0
in Ω, in ℝN \ Ω,
then v satisfies {
Lγ,Ω v = x−γ f (x)x−γ =: g(x) v=0
in Ω, in ℝN \ Ω,
(10.2.2)
with Lγ,Ω v := CN,s P.V. ∫ ℝN
v(x) − v(y) dy . x − yN+2s xγ yγ
(10.2.3)
Observe that if γ → 0, then ΦN,s (γ) → −ΛN,2s and λ → 0. On the other hand, if γ → N−2s , then ΦN,s (γ) → 0 and λ → ΛN,2s . 2
10.2 Functional setting: inequalities with weights  267
To analyze the behavior and the regularity of u, we deal with the same questions for v. Thus, we need to work in fractional Sobolev spaces with admissible weights. For simplicity, we denote dμ :=
dx x2γ
and dν :=
dx dy . x − yN+2s xγ yγ
(10.2.4)
dx Remark 10.2.5. Note that dμ := x 2γ is a positive Radon measure with the doubling property, that is, there exists a positive constant C such that
B2r (x)dμ ≤ C Br (x)dμ ,
for each ball Br (x) ⊂ ℝN .
This property will be relevant to obtain the Harnack inequality for operators with weights. For Ω ⊆ ℝN , we define the regional, weighted, fractional Sobolev space W s,γ (Ω) as follows: 2
W s,γ (Ω) := {ϕ ∈ L2 (Ω, dμ) s. t. ∫ ∫(ϕ(x) − ϕ(y)) dν < +∞}. Ω Ω
It is clear that W s,γ (Ω) is a Hilbert space endowed with the norm 1
‖ϕ‖
W s,γ (Ω)
2 2 2 := (∫ϕ(x) dμ + ∫ ∫(ϕ(x) − ϕ(y)) dν) .
Ω
Ω Ω
The following extension lemma can be proved by using the same arguments of [28] (see also [138]). Lemma 10.2.6. Let Ω ⊂ ℝN be a smooth domain. Then for all w ∈ W s,γ (Ω), there exists w̃ ∈ W s,γ (ℝN ) such that w̃ Ω = w and ̃ W s,γ (ℝN ) ≤ C‖w‖W sγ (Ω) , ‖w‖ where C := C(N, s, Ω, γ) > 0. If Ω = ℝN , we have the next result. Lemma 10.2.7. If w(x) := x−θ , with 0 < θ < (N − 2s − 2γ), then there exists a positive constant C := C(N, s, γ, θ) such that Lγ,ℝN (w)(x) = C
w(x) x2s+2γ
a. e in ℝN .
Proof. In ℝN , the operator has the form Lγ,ℝN (w)(x) := CN,s P.V. ∫ ℝN
(w(x) − w(y)) dy. xγ yγ x − yN+2s
268  10 Calderón–Zygmund summability in the fractional setting We closely follow the arguments used in [158]. By setting r := x and ρ := y, we have x = rx and y = ρy , where x  = y  = 1. Thus, +∞
Lγ,ℝN (w)(x) = Set now σ :=
ρ . r
CN,s (r −θ − ρ−θ )ρN−1 dH N−1 (y ) ) dρ. ( ∫ ∫ ρ γ γ N+2s x ρ r x − r y N+2s 0
y =1
Then +∞
CN,s w(x) dH N−1 (y ) −θ N−γ−1 (1 − σ )σ ( ) dσ. Lγ,ℝN (w)(x) = ∫ ∫ x2s+2γ x − σy N+2s 0
y =1
Define D(σ) := ∫ y =1
Then D(σ) = 2
π
N−1 2
Γ( N−1 ) 2
π
∫ 0
dH N−1 (y ) . x − σy N+2s
sinN−2 (η) (1 − 2σ cos(η) + σ 2 )
Thus Lγ,ℝN (w) = ΛN,2s,γ
N+2s 2
dη.
(10.2.5)
w(x) , x2s+2γ
where we define +∞
ΛN,2s,γ = CN,s ∫ (σ θ − 1)σ N−γ−θ−1 D(σ) dσ. 0
In [158] (see also [223]) they proved the behavior of D near σ = 1 and at +∞, namely, D(σ) ≤ C(σ 2 − 1)−1−2s close to σ = 1 and D(σ) ≈ σ −N−2s as σ → ∞. Then we obtain ΛN,2s,γ  < ∞. To conclude we just have to show that ΛN,2s,γ > 0. Since D( s1 ) = sN+2s D(s) for all s > 0, we get 1
∞
ΛN,2s,γ = ∫(σ θ − 1)σ N−γ−θ−1 D(σ) dσ + ∫ (σ θ − 1)σ N−γ−θ−1 D(σ) dσ 0
∞
1 ∞
1
1
= − ∫ (ξ θ − 1)ξ 2s+γ−1 D(ξ ) dξ + ∫ (σ θ − 1)σ N−γ−θ−1 D(σ) dσ ∞
= ∫ D(σ)(σ θ − 1)(σ N−γ−θ−1 − σ 2s+γ−1 ) dσ. 1
Since 0 < θ < N − 2s − 2γ, by formulas (2.5) and (2.6) in [158], we have D(σ) > 0 in (1, ∞), thus ΛN,2s,γ > 0.
10.2 Functional setting: inequalities with weights  269
10.2.1 Weighted Sobolev inequalities and applications In [3], the authors proved a weighted Sobolev inequality for general 1 < p. Here we will prove the result in the Hilbertian case, which is the case that we will use in this book. To be more precise, we will prove the following weighted Sobolev inequality. Theorem 10.2.8. Given 0 < s < 1, 0 < β < constant C0 = C(N, s, β) so that p
(∫( ℝN
1/p
u(x) ) dx) xβ
≤ C0 ( ∬ ℝN ×ℝN
N−2s 2
and p = 2∗s =
2N , N−2s
there exists a
1/2
u(x) − u(y)2 dy dx ) , x − yN+2s yβ xβ
(10.2.6)
for all u, say, in the Schwartz class. Proof. From the usual, unweighted, Sobolev inequality we only need to prove ∬ ℝN ×ℝN
u(y) 2  yβ yN+2s
 u(x) − xβ x −
dy dx ≤ C ∬ ℝN ×ℝN
u(x) − u(y)2 dy dx , x − yN+2s yβ xβ
(10.2.7)
for suitable constant C independent of u. In order to do this, let us consider the following algebraic identities with A = u(x), B = u(y), a = xβ and b = yβ : (A − B) a1 + B( a1 − b1 ), A B − ={ a b (A − B) b1 + A( a1 − b1 ). Hence, for every t ∈ [0, 1] we have 1 1 A B t 1−t ) + (tB + (1 − t)A) − . − ≤ A − B( + a b a b a b Choosing t =
a a+b
and squaring both sides, we get
2 2 2 2 2 A B 2 B b − a 1 A b − a 2 ) + ( ) − ≤ C(A − B ( + ( ) ). a b a+b b a + b a a + b
Integrating in L2 (ℝN × ℝN ) against the kernel K(x, y) = we have
1 x−yN+2s
and using symmetry
u(x) u(y) 2 1 u(x) − u(y)2 dy dx ∬ β − K(x, y) dy dx ≤ C ∬ β x x − yN+2s (xβ + yβ )2 y N N N N
ℝ ×ℝ
ℝ ×ℝ
+ 2C ∬ ℝN ×ℝN
2
u(x)2 xβ − yβ ( ) K(x, y) dy dx. x2β xβ + yβ
270  10 Calderón–Zygmund summability in the fractional setting Now, using that
1 (xβ +yβ )2
≤
1 4xβ yβ
we observe that the first term in the previous sum
is controlled by the expression on the right hand side of (10.2.7). In order to estimate the second term in the sum we make use of the following identity: 2
∫ ℝN
xβ − yβ 1 1 1 xβ − yβ dy = C dy, ( ) ∫ 3 xβ xβ + yβ x − yN+2s xβ yβ x − yN+2s
(10.2.8)
ℝN
for some C3 > 0. This can be easily seen by homogeneity and the change of variables z = y/x in both integrals. This gives, aside from the constants, the identities for the second term 2
∬ ℝN ×ℝN
u(x)2 xβ − yβ 1 dy dx ( ) x2β xβ + yβ x − yN+2s
= C3 ∬ ℝN ×ℝN
= C3
u(x)2 xβ − yβ 1 dy dx xβ xβ yβ x − yN+2s
1 1 u(x)2 u(y)2 dy dx − )(xβ − yβ ) ∬ ( N+2s β β 2 x − y x y yβ xβ ℝN ×ℝN
≤ C3
u(x) − u(y)2 dy dx 1 . ∬ 2 x − yN+2s yβ xβ ℝN ×ℝN
The last inequality follows from the trivial pointwise estimate (
b a A2 B2 − )(a − b) = A2 + B2 − A2 − B2 ≤ A2 + B2 − 2AB = (A − B)2 . a b a b
This finishes the proof of the theorem. Remark 10.2.9. Note that inequality (10.2.6) can be read as a weighted Caffarelli– Kohn–Nirenberg inequality. Observe that the argument used here is reminiscent of the one used to prove Picone’s lemma replacing K(x, y) = x−y1N+2s by the symmetric kernel Kβ (x, y) =
1 1 1 . N+2s x − y yβ xβ
One can check that symmetry is actually the only requirement needed for such a lemma to work. s,γ We define the space W0 (Ω) as the completion of 𝒞0∞ (Ω) with respect to the norm s,γ s,γ of W (Ω). We obtain the following Sobolev inequality in W0 (Ω).
10.2 Functional setting: inequalities with weights  271
Proposition 10.2.10. Let Ω be a bounded Lipschitz domain and suppose that the hys,γ potheses of Theorem 10.2.8 hold. Then, for all v ∈ W0 (Ω), there exists a positive constant v ∈, there exists a positive constant S = S(N, s, γ, Ω) such that 2
v(x) − v(y)2 dx dy v(x)2s 2∗s ≥ S(∫ ) . ∗ γ γ N+2s x y x − y x2s γ ∗
aN,s ∫ ∫ Ω Ω
(10.2.9)
Ω
s,γ Proof. Let v ∈ W0 (Ω) and define ṽ to be the extension of v to ℝN given in Lemma 10.2.6. Then using Theorem 10.2.8 and an argument of density, we get ∗
C‖v‖W s,γ (Ω) ≥ aN,s 0
2
2∗ s ̃ − v(y) ̃ 2 dx dy ̃ 2s v(x) v(x) ≥ S(N, s, γ)( dx) . ∫ ∫ ∫ ∗γ γ γ N+2s 2 x y x − y x s
ℝN ℝN
ℝN
Since ṽΩ = v we reach 2
2∗ s v(x)2s v(x) − v(y)2 dx dy ≥ S(N, s, γ, Ω)(∫ aN,s ∫ ∫ dx) ∗ γ γ N+2s 2 γ s x y x − y x ∗
Ω
Ω Ω
and the result follows. If Ω is a bounded regular domain, then as in the case γ = 0, we have the next Poincaré inequality. Theorem 10.2.11. There exists a positive constant C := C(Ω, N, s, γ) such that for all s,γ ϕ ∈ W0 (Ω), we have 2
C ∫ ϕ2 (x) dμ ≤ ∫ ∫(ϕ(x) − ϕ(y)) dν. Ω
Ω Ω
Proof. By density it is sufficient to consider ϕ ∈ 𝒞0∞ (Ω). If ϕ ≡ 0, the inequality follows trivially. Thus, let us define λ1 (Ω) :=
inf ∞
{ϕ∈𝒞0 (Ω),ϕ≡0} ̸
∫Ω ∫Ω (ϕ(x) − ϕ(y))2 dν ∫Ω ϕ2 (x) dμ
.
Hence, to prove the lemma we need to check that λ1 (Ω) > 0. We argue by contradiction, that is, let us suppose λ1 (Ω) = 0. Then there exists a sequence {ϕn }n∈ℕ ⊂ 𝒞0∞ (Ω) such that ∫ ϕ2n (x) dμ = 1
Ω
and
2
∫ ∫(ϕn (x) − ϕn (y)) dν → 0 as n → ∞. Ω Ω s,γ
It is clear that ‖ϕn ‖W s,γ (Ω) ≤ C and, hence, there exists a function ϕ̄ ∈ W0 (Ω) such that 0 s,γ ϕ ⇀ ϕ̄ weakly in W (Ω). n
0
272  10 Calderón–Zygmund summability in the fractional setting From the Sobolev inequality (10.2.9) it follows that ϕn 2s dx ≤ C(N, s, Ω, γ)‖ϕn ‖W s,γ (Ω) ≤ C,̄ ∗ 0 x2s γ ∗
∫ Ω
with C̄ independent of n. Then, from [28] (see also [138]) we have ϕn → ϕ̄ strongly in L2 (Ω). Hence, combining the estimates above and using Vitali’s lemma we obtain, up to a subsequence, ϕn → ϕ̄ strongly in L2 (Ω, dμ), and thus, ∫ ϕ̄ 2 (x) dμ = 1.
(10.2.10)
Ω
̄ s,γ ≤ ‖ϕ ‖ s,γ and Since ‖ϕ‖ n W (Ω) W (Ω) 0
0
2
∫ ∫(ϕn (x) − ϕn (y)) dν → 0
as n → ∞,
Ω Ω
we get s,γ ϕn → ϕ̄ strongly in W0 (Ω),
2 ̄ ̄ thus ∫ ∫(ϕ(x) − ϕ(y)) dν = 0. Ω Ω
s,γ Hence ϕ̄ ≡ C. Now, since ϕ̄ ∈ W0 (Ω), necessarily ϕ̄ ≡ 0, a contradiction with (10.2.10).
As a direct application of Theorem 10.2.11 we obtain that if Ω is a bounded regular s,γ domain, then every w ∈ W0 (Ω) satisfies 1 2
̃ W s,γ (Ω) ≤ C( ∫ ∫(w(x) − w(y)2 dν) , ‖w‖ 0
Ω Ω
where C := C(N, s, Ω, γ) > 0 and w̃ is the extension of w given in Lemma 10.2.6. As a s,γ consequence we reach that if Ω is a bounded domain, we can consider W0 (Ω) with the equivalent norm ‖ϕ‖
s,γ W0 (Ω)
2
1 2
:= ( ∫ ∫(ϕ(x) − ϕ(y) dν) . Ω Ω
We state now a weighted version of the Poincaré–Wirtinger inequality.
10.2 Functional setting: inequalities with weights  273
s,γ
Theorem 10.2.12. Let w ∈ W0 (Ω) and assume that ψ is a radial decreasing function such that supp ψ ⊂ B1 and 0 ≨ ψ ≤ 1. Define Wψ :=
∫B w(x)ψ(x) dμ 1
∫B ψ(x) dμ
.
1
Then, there exists C := C(N, s, ψ) > 0 such that 2
2
∫(w(x) − Wψ ) ψ(x) dμ ≤ C ∫ ∫(w(x) − w(y)) min{ψ(x), ψ(y)} dν. B1
B1 B1
Proof. Consider Ψ(x) := Corollary 6] we get
ψ(x) , x2γ
which is a radial decreasing function. Then using [146,
(w(x) − w(y))2 2 min{Ψ(x), Ψ(y)} dx dy, ∫(w(x) − W̄ Ψ ) Ψ(x) dx ≤ C ∫ ∫ x − yN+2s
B1
B1 B1
where W̄ Ψ =
∫B w(x)Ψ(x) dx 1
∫B Ψ(x) dx
.
1
Then, we get 2
2
∫(w(x) − W̄ Ψ ) Ψ(x) dx = ∫(w(x) − Wψ ) ψ(x) dμ, B1
B1
and ∫∫ B1 B1
(w(x) − w(y))2 min{Ψ(x), Ψ(y)} dx dy x − yN+2s
= ∫∫ B1 B1
ψ(x) ψ(y) (w(x) − w(y))2 min{ 2γ , 2γ } dx dy. x y x − yN+2s
Hence, to finish we just have to show that min{
min{ψ(x), ψ(y)} ψ(x) ψ(y) , }≤ xγ yγ x2γ y2γ
in B1 × B1 .
Without loss of generality we can assume that x ≥ y. ψ(s) Define H(s) := s2γ , which is again a decreasing function in (0, 1). Let s1 := x and s2 := y. Then min{
ψ(x) ψ(y) , } = H(s1 ). x2γ y2γ
274  10 Calderón–Zygmund summability in the fractional setting Using that ψ is decreasing, we obtain ψ(s1 ) ≤ ψ(s2 ). Thus min{ψ(x), ψ(y)} ψ(s1 ) = γ γ . xγ yγ s1 s2 Since s2 ≤ s1 ≤ 1, we conclude that H(s1 ) ≤
ψ(s1 ) γ γ s1 s2
and the result follows.
Finally, we define s,γ
Wloc (Ω) := {u ∈ L2loc (Ω, dμ) s. t. ∀Ω1 ⊂⊂ Ω, ∫ ∫ Ω1 Ω1
u(x) − u(y)2 dx dy < +∞}. x − yN+2s xγ yγ
Remark 10.2.13. The definition for γ = 0 is similar. In this case we will denote the s associated space simply as Hloc (Ω). We consider the following natural definition. s,γ
Definition 10.2.14. Let v ∈ Wloc (Ω). We say that v is a supersolution to problem (10.2.2) if 1 ∬ (v(x) − v(y))(φ(x) − φ(y)) dν ≥ ∫ gφ dx 2 DΩ1
(10.2.11)
Ω1
s,γ
for every nonnegative φ ∈ W0 (Ω1 )) and every Ω1 ⊂⊂ Ω. Consider the operator Lγ,Ω (w)(x) := CN,s P.V. ∫(w(x) − w(y))K(x, y) dy,
(10.2.12)
Ω
where K(x, y) :=
1 . xγ yγ x − yN+2s
Note that Picone’s inequality in Section 8.5 is based on a pointwise inequality and is valid for any positive symmetric kernel. Therefore we can formulate the following result. s,γ
Theorem 10.2.15 (Picone type inequality). Let w ∈ W0 (Ω) be such that w > 0 in Ω, s,γ and assume that Lγ,Ω (w) = ν with ν ∈ L1loc (ℝN ) and ν ≩ 0. Then for all v ∈ W0 (Ω) we have aN,s ∫ ∫ Ω Ω
v(x) − v(y)2 dx dy v2 ≥ ⟨L (w), ⟩, γ,Ω w x − yN+2s xγ yγ
where Lγ,Ω w is defined by (10.2.12).
(10.2.13)
10.2 Functional setting: inequalities with weights  275
The proof is the same as the one in Section 8.5 (see also [224]). As a consequence we get the next weighted Hardy inequality. Theorem 10.2.16. There exists a positive constant C(N, s, γ) such that for all ϕ ∈
𝒞0∞ (ℝN ), we have
C∫ ℝN
ϕ2 (x) 2 dx ≤ ∫ ∫ (ϕ(x) − ϕ(y)) dν. x2s+2γ ℝN ℝN
Proof. Let ϕ ∈ 𝒞0∞ (ℝN ) and define w(x) := x−θ , with 0 < θ < Lemma 10.2.7, Lγ,ℝN (w)(x) = C
w(x) x2s+2γ
N−2s−2γ . 2
Then, by
a. e. in ℝN .
Using (10.2.13) it follows that aN,s ∫ ∫ ℝN ℝN
ϕ2 ϕ2 (x) ϕ(x) − ϕ(y)2 dx dy ≥ ⟨L ⟩ = C(N, s, γ) dx. ∫ N (w), γ,ℝ γ γ w x2s+2γ x − yN+2s x y ℝN
Hence we conclude the proof. Remark 10.2.17. It should be interesting from the analytical point of view to find the value of the optimal constant. If Ω is a bounded domain, we have a similar result. Theorem 10.2.18. There exists a positive constant C(Ω, N, s, γ) such that for all ϕ ∈ 𝒞0∞ (Ω), we have C∫ Ω
ϕ2 (x) 2 dx ≤ ∫ ∫(ϕ(x) − ϕ(y)) dν. x2s+2γ Ω Ω
Proof. Let ϕ ∈ 𝒞0∞ (Ω) and define ϕ̃ to be the extension of ϕ to ℝN given in Lemma 10.2.6. Then from Theorem 10.2.16, we get ∫ ∫ ℝN ℝN
2 ̃ ̃ ϕ(x) − ϕ(y) ϕ̃ 2 (x) dx dy dx. ≥ C(N, s, γ) ∫ x2s+2γ x − yN+2s xγ yγ ℝN
Now, using the fact that ϕ̃ Ω = ϕ and combining the results of Lemma 10.2.6 and Theorem 10.2.11, we reach the desired result. In the case of nonzero boundary conditions, we obtain the following version of the Hardy inequality.
276  10 Calderón–Zygmund summability in the fractional setting Theorem 10.2.19. There exists a positive constant C(Ω, N, s, γ) such that for all ϕ ∈ s,γ W0 (Ω), we have C∫ Ω
ϕ2 (x) 2 dx ≤ ∫ ∫(ϕ(x) − ϕ(y)) dν + ∫ ϕ2 (x) dμ. x2s+2γ Ω Ω
Ω
s,γ
Proof. Fix ϕ ∈ W0 (Ω) and define ϕ̃ as the extension of ϕ to ℝN given in Lemma 10.2.6. Then from Theorem 10.2.16, we get ∫ ∫ ℝN ℝN
2 ̃ ̃ ϕ̃ 2 (x) ϕ(x) − ϕ(y) ϕ2 (x) dx dy ≥ C(N, s, γ) dx ≥ C(N, s, γ) dx. ∫ ∫ x2s+2γ x2s+2γ x − yN+2s xγ yγ Ω
ℝN
̃ s,γ N ≤ C(Ω)‖ϕ‖ s,γ , the result follows. Since ‖ϕ‖ W (Ω) W (ℝ ) Moreover, in the particular case ϕ ∈ W s,γ (BR ), as an application of Theorem 10.2.19 we can prove the following inequality. Theorem 10.2.20. Let R > 0 and ϕ ∈ W s,γ (BR ). Then, there exists C = C(N, s, R, γ) > 0 such that 2
2∗ s ϕ2s 2 C(∫ 2∗ γ dx) ≤ ∫ ∫ (ϕ(x) − ϕ(y)) dν + R−2s ∫ ϕ2 dμ. x s ∗
BR BR
BR
(10.2.14)
BR
Proof. We prove the result for R = 1, and then (10.2.14) follows by a scaling argument. ϕ(x) We set ϕ1 (x) := xγ . Then we have 2∗s
C(∫ ϕ1  dx)
2 2∗ s
(ϕ1 (x) − ϕ1 (y))2 dx dy + ∫ ϕ21 dx. x − yN+2s
≤ ∫∫ B1 B1
B1
(10.2.15)
B1
See [28] for more details. To finish the proof we just have to estimate the term ∫∫ B1 B1
(ϕ1 (x) − ϕ1 (y))2 dxdy. x − yN+2s
Since 2
(ϕ1 (x) − ϕ1 (y)) =
ϕ2 (x) ϕ2 (y) (ϕ(x) − ϕ(y))2 1 1 +( − )( γ − γ ), γ γ γ γ x y x y x y
it follows that ∫∫ B1 B1
(ϕ1 (x) − ϕ1 (y))2 (ϕ(x) − ϕ(y))2 ϕ2 (x) dxdy ≤ ∫ ∫ γ γ dxdy + ∫ L (x−γ ) dx. N+2s N+2s xγ 0,B1 x − y x y x − y B1 B1
B1
10.2 Functional setting: inequalities with weights  277
Proceeding as in the proof of Lemma 10.2.7, since 0 < γ < L0,B1 (x−γ ) ≤ xCγ+2s , and hence ∫∫ B1 B1
N−2s , 2
we realize that
(ϕ1 (x) − ϕ1 (y))2 (ϕ(x) − ϕ(y))2 ϕ2 dx. dxdy ≤ dxdy + C ∫ ∫ ∫ x2s+2γ x − yN+2s xγ yγ x − yN+2s B1 B1
B1
Finally, using Theorem 10.2.19 and substituting ϕ(x) = xγ ϕ1 (x), we reach (10.2.14). Remark 10.2.21. By interpolation of the Sobolev and Hardy weighted inequalities, we can arrive at a family of fractional Caffarelli–Kohn–Nirenberg type inequalities. See [3]. We will not go into more detail in this direction. Remark 10.2.22. As a consequence of the weighted Picone inequality, we have the next comparison principle, which extends to the weighted fractional framework the classical one obtained by Brezis and Kamin in [90] for the local case and the extension to the nonlocal case studied in Section 8.5. Lemma 10.2.23. Let Ω be a bounded domain and let f be a nonnegative continuous s,γ function such that f (x, σ) > 0 if σ > 0 and f (x,σ) is decreasing. Let u, v ∈ W0 (Ω) be such σ that u, v > 0 in Ω and {
Lγ,Ω (u) ≥ f (x, u)
in Ω,
Lγ,Ω (v) ≤ f (x, v)
in Ω.
Then, u ≥ v in Ω. Proof. Since the proof relies on pointwise inequalities, the result follows in a similar way to the constant coefficients case, obtained in Section 8.5. This result will be applied in the next chapter. 10.2.2 Some compactness results Here, we will need the next compactness result. s,γ
Lemma 10.2.24. Let {un }n∈ℕ ⊂ W0 (Ω) be an increasing sequence of nonnegative funcs,γ tions such that Lγ,Ω (un ) ≥ 0. Assume that un ⇀ u weakly in W0 (Ω). Then, un → u s,γ strongly in W0 (Ω). Proof. Using the hypothesis Lγ,Ω (un ) ≥ 0, we have ⟨Lγ,Ω (un ), wn ⟩ ≤ 0, where wn := un − u. Thus ∫∫ Ω Ω
(un (x) − un (y))(wn (x) − wn (y)) dx dy ≤ 0. xγ yγ x − yN+2s
278  10 Calderón–Zygmund summability in the fractional setting Since 2
(un (x) − un (y))(wn (x) − wn (y)) = (un (x) − un (y)) − (un (x) − un (y))(u(x) − u(y)) by the Young inequality we conclude that ∫∫ Ω Ω
(un (x) − un (y))2 dx dy (u(x) − u(y))2 dx dy ≤ . ∫ ∫ x − yN+2s xγ yγ x − yN+2s xγ yγ Ω Ω
s,γ
Therefore ‖un ‖W s,γ (Ω) ≤ ‖u‖W s,γ (Ω) and hence un → u strongly in W0 (Ω). 0
0
Likewise, we have the following local version of Lemma 10.2.24. s,γ
Lemma 10.2.25. Let {un }n∈ℕ ⊂ W0 (Ω) be an increasing sequence of nonnegative funcs,γ tions such that Lγ,Ω (un ) ≥ 0. Assume that {un }n∈ℕ is uniformly bounded in Wloc (Ω). Then s,γ s,γ there exists u ∈ Wloc (Ω) such that un → u strongly in Wloc (Ω). Proof. By using a straightforward modification of Lemma 5.3 in [138] and multiplying the sequence {un }n∈ℕ by a Lipschitz cutoff function ψ such that ψ ≡ 1 on Ω ⊂ Ω, we can apply Lemma 10.2.24 to conclude. Remark 10.2.26. Note that Lemma 10.2.24 and Lemma 10.2.25 also hold for γ = 0, i. e., s for the spaces H0s (Ω) and Hloc (Ω). 10.2.3 Some numerical inequalities The next elementary algebraic inequality in [208] will be used in some arguments. See [2]. For the reader’s convenience we give a complete proof here. Lemma 10.2.27. Let s1 , s2 ≥ 0 and a > 0. Then (s1 − s2 )(sa1 − sa2 ) ≥ Proof. Since
4a (a+1)2
a+1 a+1 4a 2 (s1 2 − s2 2 ) . 2 (a + 1)
(10.2.16)
≤ 1 for 0 ≤ a, if s1 = 0 or s2 = 0 the inequality trivially follows.
Hence, we can assume s1 > s2 > 0. Thus, setting x := (1 − x)(1 − xa ) ≥
a+1 2 4a (1 − x 2 ) 2 (a + 1)
s2 , s1
(10.2.16) is equivalent to
for all x ∈ (0, 1).
(10.2.17)
We set h(x) := (1 − x)(1 − xa )(a + 1)2 − 4a(1 − x
a+1 2
2
),
and then we just have to show that h(x) ≥ 0 for all x ∈ (0, 1). Moreover, h can be written as h(x) = (a − 1)2 (1 − x
a+1 2
2
1
a
2
) − (a + 1)2 (x 2 − x 2 ) .
10.2 Functional setting: inequalities with weights  279
First, we assume a > 1. We claim that (a − 1)(1 − x
a+1 2
1
a
) ≥ (a + 1)(x 2 − x 2 ).
In fact, let us define h1 (x) := (a − 1)(1 − x a−1
1
a+1 2
(−(a − 1)x 2 − x− 2 + ax so that h1 (x) = (a+1) 2 Using the Young inequality, we obtain a
x 2 −1 ≤
1
a
) − (a + 1)(x 2 − x 2 ),
a−2 2
).
a − 1 a−1 1 1 x 2 + x− 2 . a a
Thus h1 (x) ≤ 0 and hence h1 (x) ≥ h1 (1) = 0. Therefore h(x) ≥ 0 and the result follows in this case. Consider now the case a < 1. We prove the result if we show (1 − a)(1 − x
a+1 2
a
1
) ≥ (a + 1)(x 2 − x 2 ).
Defining h2 (x) := (1 − a)(1 − x
a+1 2
a
1
) − (a + 1)(x 2 − x 2 )
and using again the Young inequality we obtain h2 (x) ≤ 0 for all x ∈ (0, 1). Thus h2 (x) ≥ h2 (1) = 0. Hence h(x) ≥ 0 and we conclude the proof. Lemma 10.2.28. For a > 0, b > 0 the following identities hold: ∞ 1 1 (log a − log b)2k (a − b)2 = (a − b)( − ) = (log a − log b)2 + 2 ∑ . ab b a (2k)! k=2
(10.2.18)
As a consequence (a − b)2 1 1 = (a − b)( − ) ≥ (log a − log b)2 . ab b a
(10.2.19)
Proof. The first identity is evident. With respect to the second, without any loss of generality we can assume that a > b > 0. For x = ba > 1 the second identity is equivalent to (log x)2k . 2k! k=1 ∞
(x − 1)2 = 2x ∑
Calling t = log x ∈ [0, ∞) the identity above is equivalent to 2
t 2k . 2k! k=1 ∞
(et − 1) = 2et ∑
280  10 Calderón–Zygmund summability in the fractional setting The identity follows by noting that the functions f (t) = (et − 1)2 and g(t) = t 2k 2e ∑∞ k=1 2k! verify f (0) = g(0) = 0 and t
∞ t 2k−1 t 2k + 2et ∑ ≡ 2et (et − 1) = f (t). 2k! (2k − 1)! k=1 k=1 ∞
g (t) = 2et ∑
Inequality (10.2.19) is an immediate consequence.
10.3 Weak Harnack inequality and local behavior of nonnegative supersolutions 2N and u, the energy solution to problem (10.1.1) with Consider f ∈ Lm (Ω), m ≥ N+2s 0 < λ < ΛN,2s . By setting v(x) := xγ u(x), where γ is defined in (9.4.9), it follows that v solves
{
Lγ,Ω v(x) = x−γ f (x) v=0
in Ω, in ℝN \ Ω,
(10.3.1)
with 0 < γ < N−2s and Lγ,Ω defined in (10.2.3). Hence, to study the behavior of u near the 2 origin, we may deal with the same question for v. More precisely, we want to prove that the weighted operator Lγ v satisfies a suitable weak Harnack inequality. Note that the natural functional framework for the new equation of v is the space Wγs (ℝN ) defined in Section 10.2. The statement of the result that we want to prove is the following. Theorem 10.3.1 (Weak Harnack inequality). Let r > 0 such that B2r ⊂ Ω. Assume that f ≥ 0 and let v ∈ W s,γ (ℝN ) be a supersolution to (10.3.1) with v ≩ 0 in ℝN . Then, for N there exists a positive constant C = C(N, s, γ) such that every q < N−2s q
1 q
(∫ v dμ) ≤ C inf v. Br
B3r 2
Note that in the nonlocal framework we must have v ≥ 0 in the whole space. Otherwise it is possible to find counterexamples (see the work by Kassmann [207] and the references therein). The proof follows classical arguments by Moser and Krylov–Safonov (see [150] for the local case with weights). For the nonlocal case we have the precedents of [135, 215], where the kernel is comparable to a fractional Laplacian and the operator considered is of fractional pLaplacian type. Remark 10.3.2. The ground state representation has been considered for the case of the fractional pLaplacian, p > 2 and the corresponding Hardy’s inequality in [168].
10.3 Weak Harnack inequality  281
When p > 2 the identity of Lemma 10.2.4 becomes an inequality. However, the corresponding operator, that is, (−Δ)sp,γ u(x) := P.V. ∫ ℝN
u(x) − u(y)p−2 (u(x) − u(y)) dy , xγ yγ x − yN+ps
with 0 ≤ γ < N−ps , 1 < p < N, s ∈ (0, 1) and ps < N, is studied in [1, 2, 3]. In these 2 references, the authors prove the existence of solution for general data using suitable approximation arguments. They extend the results of [218] using a different approach and they also get a weak Harnack inequality partially extending the results of [135]. The behavior of the radial solution in the local case for p ≠ 2 is quite involved (see [10, Section 3.2]). For the fractional pLaplacian this seems to be an open problem. Since the kernel defined in (10.2.3) is singular we have to check the arguments step by step. That is, our result can be seen as the fractional counterpart of [150]. Note that it is enough to consider the case Br (x0 ) = Br (0). For simplicity of typing, we will write Br instead of Br (0). We start by proving the following estimate. Lemma 10.3.3. Let R > 0 such that BR ⊂ Ω, and assume that v ∈ W s,γ (ℝN ) with v ≩ 0 is a supersolution to (10.3.1). Let k > 0 and suppose that for some σ ∈ (0, 1] we have Br ∩ {v ≥ k}dμ ≥ σBr dμ with 0 < r
0 in BR (otherwise we can s,γ deal with v + ε and let ε → 0 at the end). Let ψ ∈ W0 (BR ) be such that 0 ≤ ψ ≤ 1, C supp ψ ⊂ B7r , ψ = 1 in B6r and ∇ψ ≤ r . Using ψ2 v−1 as a test function in (10.3.1), it follows that ∫ ∫ (v(x) − v(y))(ψ2 (x)v−1 (x) − ψ2 (y)v−1 (y)) dν ≥ 0. ℝN ℝN
Thus 0 ≤ ∫ ∫ (v(x) − v(y))( B8r B8r
+2 ∫ ℝN \B
ψ2 (x) ψ2 (y) − ) dν v(x) v(y)
∫ (v(x) − v(y)) 8r
B8r
ψ2 (x) dν. v(x)
(10.3.3)
282  10 Calderón–Zygmund summability in the fractional setting Denote x = xx and y = ρy , where x  = y  = 1 and ρ := y. We have ∫
∫ (v(x) − v(y))
ℝN \B8r B8r
ψ2 (x) dν ≤ v(x)
∫
∫ ψ2 (x) dν
ℝN \B8r B8r ∞
≤ ∫ B7r
Setting here τ :=
ψ2 (x) ρN−γ−1 dy ) dρ dx. ∫ N+2s ( ∫ ρ γ x x  x y − x N+2s 8r
𝕊N−1
ρ , x ∞
∫
∫ (v(x) − v(y))
ℝN \B8r B8r
ψ2 (x) ψ2 (x) dy dν ≤ C ∫ 2γ+2s ∫ τN−γ−1 ( ∫ ) dτ dx v(x) x τy − x N+2s B7r
8 7
𝕊N−1
∞
ψ2 (x) ≤ C ∫ 2γ+2s ∫ τN−γ−1 D(τ) dτ dx, x B7r
8 7
where D(τ) := 2
π
N−1 2
) Γ( N−1 2
π
∫ 0
sinN−2 (θ) (1 − 2σ cos(θ) + τ2 )
N+2s 2
dθ.
Considering the behavior of D near 0 and 1 and at ∞ as in [158] (see also the proof of Lemma 10.2.7), we obtain ∞
∫ τN−γ−1 D(τ) dτ ≤ C, 8 7
and therefore we conclude that ∫ (v(x) − v(y))
∫ ℝN \B
8r
B8r
ψ2 (x) dν ≤ Cr N−2s−2γ . v(x)
(10.3.4)
Moreover ∫ ∫ (v(x) − v(y))( B8r B8r
ψ2 (x) ψ2 (y) − ) dν v(x) v(y)
= ∫ ∫ (v(x) − v(y))( B6r B6r
+
∬
(v(x) − v(y))(
B8r ×B8r \B6r ×B6r
≤ ∫ ∫ (v(x) − v(y))( B6r B6r
ψ2 (x) ψ2 (y) − ) dν v(x) v(y) ψ2 (x) ψ2 (y) − ) dν v(x) v(y)
1 1 − ) dν + Cr N−2s−2γ , v(x) v(y)
(10.3.5)
10.3 Weak Harnack inequality  283
where the last inequality follows as a consequence of ψ ≡ 1 in B6r and the fact that the integral in (B8r × B8r ) \ (B6r × B6r ) can be estimated in the same way as (10.3.4). If (x, y) ∈ (B6r × B6r ) (v(x) − v(y))(
ψ2 (x) ψ2 (y) 1 1 − ) = (v(x) − v(y))( − ) v(x) v(y) v(x) v(y) 2
≤ −(log(v(x)) − log(v(y))) ,
(10.3.6)
where the last inequality is a consequence of (10.2.19). Hence from (10.3.6) we deduce ∫ ∫ (v(x) − v(y))( B6r B6r
1 1 2 − ) dν ≤ − ∫ ∫ (log(v(x)) − log(v(y))) dν, v(x) v(y)
(10.3.7)
B6r B6r
and thus, putting together (10.3.3), (10.3.4), (10.3.5) and (10.3.7), it follows that 2
∫ ∫ (log(v(x)) − log(v(y))) dν ≤ Cr N−2s−2γ .
(10.3.8)
B6r B6r
Let δ ∈ (0, 1/4). We set w(x) := min{log( 2δ1 ), log( kv )}+ . Now, since w is a truncation of log( kv ), from (10.3.8) we obtain 2
∫ ∫ (w(x) − w(y)) dν ≤ Cr N−2s−2γ . B6r B6r
Call ⟨w⟩B6r :=
1 ∫ w(x) dμ. B6r dμ B6r
Thus, using the Hölder and Poincaré–Wirtinger inequalities (Theorem 10.2.12), ∫ w(x) − ⟨w⟩B6r dμ ≤ CB6r dμ .
(10.3.9)
B6r
Note that {x ∈ Ω : w(x) = 0} = {x ∈ Ω : v(x) ≥ k}, and then from (10.3.2) we have σ B6r ∩ {w = 0}dμ ≥ N−2γ B6r dμ . 6
As a consequence of this, we can see that 1 6N−2γ ∫ w(x) − ⟨w⟩B6r dμ, B6r ∩ {w = log( )} ≤ 2δ dμ σ log( 1 ) 2δ B6r
and, hence, we conclude the result by applying (10.3.9) and the fact that {B6r ∩ {v ≤ 2δk}} = {B6r ∩ {w = log(
1 )}}. 2δ
284  10 Calderón–Zygmund summability in the fractional setting As a consequence we have the next estimate on infB4r v. Lemma 10.3.4. Assume that the hypotheses of Lemma 10.3.3 are satisfied. Then, there exists δ ∈ (0, 21 ), depending only on N, s, σ and γ, such that inf v ≥ δk.
(10.3.10)
B4r
s,γ W0 (Bρ ) with r
Proof. We set w := (l − v)+ , where l ∈ (δk, 2δk), and we let ψ ∈ ≤ ρ < 6r. Using wψ2 as a test function in (10.3.1) and using similar arguments to those in [135, Lemma 3.2], we reach 2
∫ ∫ (w(x)ψ(x) − w(y)ψ(y)) dν Bρ Bρ 2
2
≤ C1 ∫ ∫ max{w(x), w(y)} (ψ(x) − ψ(y)) dν Bρ Bρ
+ l2 Bρ ∩ {v < l}dμ ×
sup
{x∈supp(ψ)}
∫ ℝN \Bρ
dy . x − yN+2s
(10.3.11)
We define now the sequences {lj }j∈ℕ , {ρj }j∈ℕ and {ρj̄ }j∈ℕ by setting ρj := 4r + 21−j r,
lj := δk + 2−j−1 δk,
ρ̄ j :=
ρj + ρj+1 2
.
Likewise, let us denote wj := (lj − v)+ ,
Bj := Bρj ,
and let ψj ∈ 𝒞0∞ (Bρj̄ ) be such that 0 ≤ ψ ≤ 1, ψ ≡ 1 in Bj+1 and ∇ψj  ≤ 2j+3 /r. Using the Sobolev inequality stated in Proposition 10.2.10 we obtain wj ψj 2s
∗
C(N, s, γ)(∫
dx)
xγ2s
∗
Bj
2 2∗ s
2
≤ ∫ ∫(wj (x)ψj (x) − wj (y)ψj (y)) dν. Bj Bj
Hence, using the facts that wj ψj ≥ (lj − lj+1 )
in Bj+1 ∩ {v < lj+1 }
and x−2s γ ≥ Cr̄ −(2s −2)γ x−2γ ∗
∗
in Bj ,
with C̄ independent of j, it follows that wj ψj 2s
∗
(∫ Bj
γ2∗s
x
dx)
2 2∗ s
≥
r
C
(2∗s −2)γ
2
2∗ (lj − lj+1 )2 Bj+1 ∩ {v < lj+1 }dμs .
10.3 Weak Harnack inequality  285
Since Bj+1 dμ = Cr N−2γ , we have ∗
r (2s −2)γ (Bj+1 dμ )
1
1
2 2∗
= Cr −(N−2s−2γ) r 4sγ( N−2s − N ) ≤ C(N, s, γ, Ω)r −(N−2s−2γ) .
Hence we get 2
(lj − lj+1 ) (
Bj+1 ∩ {v < j + 1}dμ Bj+1 dμ
)
2 2∗ s
2
≤ C(N, s, γ, Ω)r −(N−2s−2γ) ∫ ∫(wj (x)ψj (x) − wj (y)ψj (y)) dν. Bj Bj
Applying (10.3.11) to wj , we conclude that 2
(lj − lj+1 ) (
Bj+1 ∩ {v < j + 1}dμ Bj+1 dμ
)
2 2∗ s
C(N, s, γ) 2 2 (C1 ∫ ∫ max{wj (x), wj (y)} (ψj (x) − ψj (y)) dν r (N−2s−2γ)
≤
Bj Bj
+ lj2 Bj ∩ {v < lj }dμ
sup
{x∈supp(ψj )}
∫ ℝN \B
j
dy ). x − yN+2s
(10.3.12)
We have 2
2
∫ ∫ max{wj (x), wj (y)} (ψj (x) − ψj (y)) dν Bj Bj
≤ lj2 ‖∇ψj ‖2L∞ (Bj ) ≤ C22j lj2 r −2s
∫ Bj ∩{v 0. Set ṽ := (v + d), and assume that ψ ∈ Wγ,0 (Ω) is such
that supp(ψ) ⊂ Bτr , ψ = 1 in Bτ r and ∇ψ ≤
C , (τ−τ )r
ṽ1−q ψ2 as a test function in (10.2.11), we obtain ̃ − v(y))( ̃ 0 ≤ ∫ ∫ (v(x) Bτr Bτr
where
ψ2 (x) ψ2 (y) − ) dν + 2 ∫ ṽq−1 (x) ṽq−1 (y)
1 2
≤ τ < τ < 32 . Then using
̃ − v(y)) ̃ ∫ (v(x)
ℝN \Bτr Bτr
ψ2 (x) dν. ṽq−1 (x)
Since x < y in Bτr × (ℝN \ Bτr ), using the positivity of ṽ it follows that ∫
̃ − v(y)) ̃ ∫ (v(x)
ψ2 (x) dν ≤ ( ∫ ṽ2−q ψ2 dμ)( sup ∫ ṽq−1 (x) {x∈supp(ψ)} Bτr
ℝN \Bτr Bτr
ℝN \Bτr
dy ). x − yN+2s
Furthermore, by the pointwise inequality of [155, Lemma 3.3(i)], there exist positive constants C1 and C2 , depending on q, such that ̃ − v(y))( ̃ ∫ ∫ (v(x) Bτr Bτr
≤ −C1 ∫ ∫ (ṽ
2−q 2
ψ2 (y) ψ2 (x) − q−1 ) dν q−1 ṽ (x) ṽ (y)
(x)ψ(x) − ṽ
2−q 2
2
(y)ψ(y)) dν
Bτr Bτr 2
+ C2 ∫ ∫ ((ṽ2−q (x) + ṽ2−q (y))(ψ(x) − ψ(y)) dν. Bτr Bτr
By symmetry we have 2
2
∫ ∫ ((ṽ2−q (x) + ṽ2−q (y))(ψ(x) − ψ(y)) dν = 2 ∫ ∫ (ṽ2−q (x)(ψ(x) − ψ(y)) dν, Bτr Bτr
Bτr Bτr
and we obtain 2
∫ ∫ ((ṽ2−q (x) + ṽ2−q (y))(ψ(x) − ψ(y)) dν ≤ Bτr Bτr
See [11, Lemma 4.6].
Cr −2s ∫ ṽ2−q dμ. (τ − τ )2 Bτr
10.3 Weak Harnack inequality  287
Since sup
{x∈Supp(ψ)}
∫ ℝN \Bτr
dy ≤ Cr −2s , x − yN+2s
combining the estimates above we reach ∫ ∫ (ṽ
2−q 2
(x)ψ(x) − ṽ
2−q 2
2
(y)ψ(y)) dν ≤
Cr −2s ∫ ṽ2−q dμ. (τ − τ )2 Bτr
Bτr Bτr
Hence, from the previous inequality and the Sobolev inequality in Proposition 10.2.10, we get (
1 Bτ r dμ
∫ ṽ
(2−q)N N−2s
dμ)
N−2s N
≤(
Bτ r
≤
1 Bτ r dμ
∫ (ṽ
2−q 2
2∗s
ψ) dμ)
N−2s N
Bτr
C ∫ ṽ2−q dμ. Bτr dμ (τ − τ )2 Bτr
N Since q ∈ (1, 2) is arbitrary and N−2s > 1 by using the Hölder inequality we obtain the estimate (10.3.14) for ṽ = v + d with α1 and α2 in the hypotheses. Finally letting d → 0 and by the monotone convergence theorem we conclude the proof.
In order to obtain the weak Harnack inequality, we need to prove the following estimate as in the De Giorgi seminal work [137]. Lemma 10.3.6. Let r > 0 such that Br ⊂ Ω. Assume that v is a supersolution to (10.3.1). Then, there exists a constant η ∈ (0, 1) depending only on N, s and γ such that 1
η 1 ( ∫ vη dμ) ≤ C inf v. Br Br dμ
Br
To prove Lemma 10.3.6 (see [213], [135, Lemma 4.1] and the pioneering work [137]) we use the next covering result in the spirit of Krylov–Safonov theory; see [216]. Recall that in general, we say that a measure ω in a metric space has the doubling property if there exists a constant C > 0 such that ω(B2r (x)) ≤ Cω(Br (x)) for all x and all r > 0. Note that we are working here with μ, a doubling measure on bounded domains of ℝN . Lemma 10.3.7. Assume that E ⊂ Br (x0 ) is a measurable set. For δ ∈ (0, 1), we define [E]δ := ⋃ {B3ρ (x) ∩ Br (x0 ), x ∈ Br (x0 ) : E ∩ B3ρ (x)dμ > δBρ (x)dμ }. ρ>0
Then, either
288  10 Calderón–Zygmund summability in the fractional setting 1. 2.
̃
[E]δ dμ ≥ Cδ Edμ , or [E]δ = Br (x0 );
where C̃ is the doubling constant of dμ. Proof. Define the maximal operator M : Br (x0 ) { { { { x {
→
ℝ,
→
M(x) =
E ∩ B3ρ (x)dμ
sup
Bρ (x)dμ
ρ>0, so that x−x0 ≤min{r,3ρ}
.
(10.3.15)
It is clear that [E]δ = {x ∈ Br (x0 ), M(x) > δ},
∀δ ∈ (0, 1).
Assume that alternative 2 does not hold, that is, Bρ (x) \ [E]δ ≠ 0. The set [E]δ is open by definition. We consider a covering of [E]δ by balls Brx (x) such that x ∈ [E]δ and rx = 1 dist(x, Bρ (x) \ [E]δ ). By the Besicovitch–Vitali theorem (see [192, Chapter 1] and [117, 2 Chapter 3]) there are countably many pairwise disjoint balls {Bri (xi )}i∈ℕ ,
ri ≡ rxi , such that [E]δ ⊂ ⋃ B5ri (xi ). i∈ℕ
Then B5ri (xi ) ∩ (Bρ (x) \ [E]δ ) ≠ 0,
i = 1, 2, 3 . . . ,
and there exists a point yi ∈ B5ri (xi ) ∩ (Bρ (x)\ [E]δ ), that is, M(yi ) ≤ δ for all i = 1, 2, 3 . . .. Since xi ∈ B5ri (yi ) we get E ∩ B5ri (xi )dμ ≤ δB 5 ri (xi )dμ ≤ CδBri (xi )dμ , 3 where C is the constant of the doubling property. For y, a density point of E, we have lim inf ρ→0
E ∩ B3ρ (y)dμ Bρ (y)dμ
≥ lim sup ρ→0
E ∩ Bρ (y)dμ Bρ (y)dμ
= 1 > δ.
By the Lebesgue theorem dμalmost every point of E is a density point, and hence dμalmost every point of E is in [E]δ , for every 0 < δ < 1. Therefore ∞
Edμ = E ∩ [E]δ dμ ≤ ∑E ∩ B5ri (xi )dμ i=1
∞
∞
i=1
i=1
≤ Cδ ∑E ∩ Bri (xi )dμ ≤ Cδ ∑Bri (xi )dμ = Cδ[E]δ dμ , which implies alternative 1, as we wanted to prove.
10.3 Weak Harnack inequality  289
We next prove Lemma 10.3.6. Proof of Lemma 10.3.6. Note that, for any η > 0, ∞
Br ∩ {v > t}dμ 1 dt. ∫ vη dμ = η ∫ t η−1 Br dμ Br dμ
(10.3.16)
0
Br
Then, for t > 0 and i ∈ ℕ, we set Ait := {x ∈ Br : v(x) > tδi }, where δ is given by i Lemma 10.3.4. Note that Ai−1 t ⊂ At . Let ρ > 0 and x ∈ Br such that B3ρ (x) ∩ Br ⊂ [Ai−1 t ]δ̄ . Thus, ̄ i−1 ̄  = δ B  . At ∩ B3ρ (x)dμ > δB 3ρ dμ ρ dμ 3N−2γ Hence, using Lemma 10.3.4, we reach v(x) > δ(tδi−1 ) = tδi
for all x ∈ Br ,
i i−1 and therefore [Ai−1 t ]δ̄ ⊂ At , being [At ]δ̄ as in Lemma 10.3.7. This fact, together with Lemma 10.3.7, allows us to deduce that
Ait = Br
C̃ or Ait dμ ≥ Ai−1 . ̄δ t dμ
(10.3.17)
Thus, if for some m ∈ ℕ we have m δ̄ 0 At dμ > ( ) Br dμ , C̃
(10.3.18)
then Am t = Br . If not, it follows from (10.3.17) that C̃ m−1 m At dμ ≥ At dμ . δ̄ Since Ai−1 ⊂ Am t t ⊊ Br for all i ≤ m, the second point of the alternative (10.3.17) holds i−1 for At and then m−1 −1 C̃ m−2 C̃ C̃ m−1 0 At dμ ≥ At dμ ⋅ ⋅ ⋅ ≥ ( ) At dμ > ( ) Br dμ . δ̄ δ̄ δ̄ m m Thus Am t dμ > Br dμ , a contradiction with the fact that At ⊊ Br . Hence At = Br . It is clear that (10.3.18) holds if
m>
1
̄
log( Cδ̃ )
log(
A0t dμ Br dμ
),
(10.3.19)
290  10 Calderón–Zygmund summability in the fractional setting and consequently, fixing m to be the smallest integer such that (10.3.19) holds, we have m ≥ 1 and 1
0≤m−1≤
̄
log( Cδ̃ )
log(
A0t dμ Br dμ
).
Thus, using the fact that δ ∈ (0, 21 ), it can be checked that m
inf v > tδ =≥ tδ( Br
A0t dμ Br dμ
1 β
) ,
̄
log( δ )
C̃ . with β := log(δ) Set now ξ := infBr v. Then,
Br ∩ {v > t}dμ Br dμ
=
A0t dμ Br dμ
̃ −β t −β ξ β . ≤ Cδ
Going back to (10.3.16), we have a
∞
1 ∫ vη dμ ≤ η ∫ t η−1 dt + ηC̃ ∫ t η−1 δ−β t −β ξ β dt Br dμ a
0
Br
η−β
̃ −β ξ β a = aη − ηCδ . η−β β
Choosing a := ξ and η := 2 , we reach the result. After this result, we can prove the weighted weak Harnack inequality in Theorem 10.3.1. Proof. Using Lemma 10.3.6 we obtain 1
η 1 ( ∫ vη dμ) ≤ C inf v Br Br dμ
Br
for some η ∈ (0, 1). Fixing 1 ≤ q < follows that
N , N−2s
by Lemma 10.3.5 with α1 = η and α2 = q, it
1
1
q η 1 1 ( ∫ vq dμ) ≤ C( ∫ vη dμ) . Br dμ B 3 r dμ
Br
2
B3r 2
Hence 1
(
q 1 ∫ vq dμ) ≤ C inf v B3r Br dμ
Br
and we conclude the proof.
2
10.4 Optimal summability in the presence of Hardy potential  291
As a consequence of the previous Harnack inequality, we get much information about the behavior of the supersolutions to (10.1.1) around the origin. In particular, we see that any of them must be unbounded, even if f ∈ L∞ (Ω). Lemma 10.3.8. Let λ ≤ ΛN,2s . Assume that u is a nonnegative function defined in Ω such that u ≢ 0, u ∈ L1 (Ω), xu2s ∈ L1 (Ω) and u ≥ 0 in ℝN \ Ω. If u satisfies (−Δ)s u − λ xu2s ≥ 0 in the weak sense in Ω, then there exist δ > 0 and a constant C = C(N, δ, γ) such that for each ball Bδ (0) ⊂⊂ Ω, u ≥ Cx−γ
in Bδ (0),
where γ = N−2s − α is defined in (9.4.9). In particular, for δ conveniently small, we can 2 assume that u > 1 in Bδ (0). Proof. Considering v := xγ u, v ≩ 0 and satisfies Lγ v ≥ 0, with Lγ defined in (10.2.3). Hence using the weak Harnack inequality in Theorem 10.3.1, we conclude that infBr (0) v ≥ C. Thus u(x) ≥ Cx−γ in Br (0) and the result follows.
10.4 Optimal summability in the presence of Hardy potential In this section we analyze the question of the optimal summability of the solution to the problem {
(−Δ)s u − λ xu2s = f u=0
in Ω, in ℝN \ Ω,
(10.4.1)
with 0 < λ < ΛN,2s in terms of the summability of f . We try to obtain the corresponding results to those in Chapter 3 for fractional elliptic operators. In this fractional framework the results are more involved and up to now the optimality is not well understood.
10.4.1 Regularity of energy solutions 2N Along this subsection we will assume that f ∈ Lm (Ω) with m ≥ N+2s , and thus the s corresponding solution, u, belongs to H0 (Ω). We have seen in Chapter 8 that in the unperturbed problem, λ = 0, if f ∈ Lm (Ω) N with m > 2s , then, as in the classical case, u ∈ L∞ (Ω). However, as a consequence of Lemma 10.3.8, this feature is no longer true for λ > 0, and actually u(x) ≥ Cx−γ in a neighborhood of the origin. Hence, the first natural question here is whether this rate is exactly the rate of growth of u, and the answer is yes for regular data, as the following theorem shows.
292  10 Calderón–Zygmund summability in the fractional setting N Theorem 10.4.1. Let f ∈ Lm (Ω), m > 2s . Let us consider u ∈ H0s (Ω), the unique energy solution to problem (10.4.1), with λ < ΛN,2s . Then u(x) ≤ Cx−γ in ℝN .
Proof. Since the problem is linear, without loss of generality we can assume f ≥ 0. Defining v(x) := xγ u(x), it can be checked that it solves {
Lγ,Ω (v) = x−γ f v=0
in Ω,
(10.4.2)
in ℝN \ Ω,
where the operator Lγ,Ω (v) was defined by (10.2.3). Consider now Gk (v(x)), specified in (8.3.12), with k > 0 as test function in (10.4.2). Hence, aN,s ∬ DΩ
(v(x) − v(y))(Gk (v(x)) − Gk (v(y))) dx dy = ∫ x−γ fGk (v) dx. xγ yγ x − yN+2s Ω
Since for any σ ∈ ℝ, σ = Tk (σ) + Gk (σ), we have (v(x) − v(y))(Gk (v(x)) − Gk (v(y))) 2
= (Gk (v(x)) − Gk (v(y))) + (Tk (v(x)) − Tk (v(y)))(Gk (v(x)) − Gk (v(y))). Moreover, by Proposition 8.3.6, we know that (Tk (v(x)) − Tk (v(y)))(Gk (v(x)) − Gk (v(y))) ≥ 0, and therefore aN,s ∬ DΩ
Gk (v(x)) − Gk (v(y))2 dx dy dx ≤ ∫ fGk (v(x)) γ . γ γ N+2s x y x x − y Ω
Let us denote Ak := {x ∈ Ω : v(x) ≥ k}. Applying the weighted Sobolev inequality on the left hand side we obtain
2
𝒮 Gk (v)L2∗s (Ω,x−γ dx) ≤ ∫ fGk (v(x)) Ak
dx , xγ
and using Hölder’s inequality on the right hand side, dx 1− 1∗ − 1 ∫ fGk (v(x)) γ ≤ ‖f ‖Lm (Ω) Gk (v)L2∗s (Ω,x−γ dx) Ak  2s m . x Ak
Thus we have 1− 1∗ − 1 −1 Gk (v)L2∗s (Ω,x−γ dx) ≤ 𝒮 ‖f ‖Lm (Ω) Ak  2s m .
10.4 Optimal summability in the presence of Hardy potential  293
On the other hand, since Ω is bounded, there exists a constant c > 0 such that Gk (v)L2∗s (Ω,x−γ dx) ≥ cGk (v)L2∗s (Ω) . Moreover, for any z > k, we have Az ⊂ Ak and Gk (s)χAz ≥ (z − k) for every s ∈ ℝ and thus 1
(z − k)Az  2∗s ≤
1 1− 1 − 1 ‖f ‖Lm (Ω) Ak  2∗s m . c𝒮
Manipulating the inequality above, we deduce that 2∗
Az  ≤
‖f ‖Lsm (Ω) Ak  2∗s
2∗s (1− 21∗ − m1 )
(c𝒮 ) (z −
s
∗ k)2s
.
Hence we apply Lemma 8.4.2 with the choice ψ(s) := As . Since m > 2∗s (1 −
N 2s
we have
1 1 − ) > 1. ∗ 2s m
Consequently, there exists k0 such that ψ(k) ≡ 0 for any k ≥ k0 and thus ess sup v ≤ k0 . Ω
We next try to obtain conditions on λ, under which the Calderón–Zygmund summability holds for the rest of the Lebesgue spaces contained in the dual of H0s (Ω), N 2N ≤ m < 2s . Then we have the following result. that is, when f ∈ Lm (Ω), where N+2s Theorem 10.4.2. Let f be a positive function f ∈ Lm (Ω), with λ < ΛN,2s
2N N+2s
≤m
0 such that the unique energy solution of problem (10.4.1) verifies ‖u‖Lm∗∗ ≤ c‖f ‖Lm (Ω) , s (Ω)
where m∗∗ s =
mN . N − 2ms
(10.4.4)
Proof. Since f ∈ H −s (Ω), the existence and uniqueness of an energy solution is a direct consequence of the Lax–Milgram theorem applied in H0s (Ω). To study the regularity of the solution, for every k ∈ ℕ, we consider uk ∈ L∞ (ℝN ) ∩ H0s (Ω), the solution to the following approximated problem: u
s k−1 { (−Δ) uk − λ x2s + 1 = fk (x) k { u = 0 k {
where fk (x) := min{f (x), k} and u0 = 0. In this way we obtain the following properties:
in Ω, in ℝN \ Ω,
(10.4.5)
294  10 Calderón–Zygmund summability in the fractional setting (i) {uk } is an increasing sequence; (ii) each uk is bounded; (iii) uk → u, the unique solution to problem (10.4.1), in Lp (Ω), for every 1 ≤ p ≤ 2∗s . Define β =
β uk
2∗s 2m −2∗s
=
N(m−1) N−2ms
≥ 1. Then βm =
as a test function in (10.4.1), obtaining aN,s ∬
β
β
(uk (x) − uk (y))(uk (x) − uk (y)) x −
DΩ
yN+2s
(β+1) ∗ 2s . 2
Since uk is bounded we can take
dx dy ≤ λ ∫ Ω
β+1
uk
x2s
β
dx + ∫ fk uk dx. Ω
By Hölder’s inequality, β ∫ fk uk Ω
dx ≤
m β ‖fk ‖Lm (Ω) (∫ uk Ω
dx)
1 m
(β+1)2∗ s 2
≤ ‖f ‖Lm (Ω) (∫ uk
dx)
β 2 β+1 2∗ s
.
Ω
Now, by the algebraic inequality in (10.2.16), we get β
β
(uk (x) − uk (y))(uk (x) − uk (y)) ≥
β+1 β+1 4β 2 (uk 2 (x) − uk 2 (y)) , 2 (β + 1)
and, hence, using Hardy’s inequality again, we conclude that β+1
β+1
β
2
(β+1)2∗ β+1 2∗ s (uk 2 (x) − uk 2 (y))2 s 4β λ 2 m (Ω) (∫ u ) . dx) dx dy ≤ ‖f ‖ aN,s ( − ∬ L k (β + 1)2 ΛN,2s x − yN+2s
DΩ
Ω
On the other hand, hypothesis (10.4.3) is equivalent to (
4β λ − ) > 0. (β + 1)2 ΛN,2s
Thus, by the Sobolev inequality, we reach (β+1)2∗ s 2
(∫ uk
dx)
2 2∗ s (β+1)
≤ C‖f ‖Lm (Ω) .
Ω
Furthermore, the proof.
(β+1)2∗s 2
=
mN N−2sm
= m∗∗ s , and therefore passing to the limit we conclude
Remark 10.4.3. Note that, making s → 1, the condition over λ becomes λ
J1 (m) and Ω = B1 (0), there exists a suitable radial ∗∗ function f ∈ Lm (Ω) such that the solution u does not belong to Lm (Ω). This proves the optimality of the curve λ = J1 (m). In the nonlocal case the situation is more delicate. Indeed, we will see that in this case, the previous example does not provide the optimality of the curve Js (m). It proves that the m∗∗ s summability does not hold above a curve, which we will call Ps (m), which is in general above Js (m). Thus, to the best of our knowledge, the optimality of Js (m) for 2N N every N+2s ≤ m < 2s remains open. In order to define such curve Ps (m), let us first recall that for all λ ∈ (0, ΛN,2s ] there exists a unique nonnegative constant α ∈ [0, N−2s ) such that 2 λ = λ(α) =
)Γ( N+2s−2α ) 22s Γ( N+2s+2α 4 4 )Γ( N−2s−2α ) Γ( N−2s+2α 4 4
,
(10.4.7)
2N N and λ(α) is a decreasing function of α (see Section 9.4 and [11]). Hence, for m ∈ [ N+2s , 2s ) fixed, we consider
α0 (m) :=
N + 2s N − 2 m
and Ps (m) := λ(α0 (m))
(10.4.8)
given by (10.4.7). Then we have the following result. Lemma 10.4.4. Assume that f (x) = x1 ν with ν = N−ε , for some ε > 0. Let λ1 ∈ (0, ΛN,2s ] m be such that λ1 ≥ Ps (m). If u is the unique solution of (10.4.1) with λ = λ1 , then u ∉ ∗∗ Lms (B1 (0)). Proof. Note that f ∈ Lm (B1 (0)). From (10.4.7) we know that λ1 = λ(α1 ) for some α1 ∈ ). Since λ1 ≥ Ps (m), using the fact that λ(α) is a decreasing function we reach [0, N−2s 2 α1 ≤ α0 (m) =
N + 2s N − . 2 m
Define v(x) := C(
1 1 − ), xγ xν−2s
(10.4.9)
N N where γ = N−2s − α1 . Since γ ≥ m − 2s, using the fact that ν < m it follows that γ > ν − 2s. 2 Thus, v ≥ 0 in B1 (0). Hence, choosing a suitable positive constant C, we reach
(−Δ)s v − λ1
v =f x2s
in B1 (0).
296  10 Calderón–Zygmund summability in the fractional setting Since v ≤ 0 in ℝN \B1 (0), by comparison it follows that v ≤ u, where u is the unique ∗∗ N solution of (10.4.1) for λ = λ1 . Since γ ≥ m − 2s, we have v ∉ Lms (B1 (0)) and we conclude. As a direct application of this lemma we can prove the optimality of the curve Js (m) in a particular case. Lemma 10.4.5. Assume that the hypotheses of Lemma 10.4.4 hold and let u be the solu∗∗ 2N tion of problem (10.4.1). If m = N+2s and λ ≥ Js (m), then u ∉ Lms (B1 (0)). Proof. Note that Ps (m), defined in (10.4.8), can be rewritten as Ps (m) = and in the particular case of m =
N N )Γ( 2m ) 2m , N N )Γ( 2m − s) 2m
− 22s Γ( N+2s 2 Γ( N2 −
2N , N+2s
(10.4.10)
it satisfies
Js (m) = Ps (m). Hence λ ≥ Ps (m) by hypothesis, and we conclude applying Lemma 10.4.4. Remark 10.4.6. Note that in the local case, s = 1, α = √ΛN,2 − λ and P1 (m) = J1 (m) for
2N N , 2 ), and thus Lemma 10.4.4 holds in the whole range. all m ∈ [ N+2
Next we show that for radial functions the result in Lemma 10.4.4 cannot be improved. In other words the optimality of the curve Js (m) cannot be proved with radial functions. We start by proving the following result (for the properties of the Gamma function we refer to [27]). 2N N Lemma 10.4.7. Assume that s ∈ (0, 1) and m ∈ [ N+2s , 2s ). Then
Js (m) ≤ Ps (m) and equality holds in (10.4.11) if and only if m =
(10.4.11)
2N . N+2s
Proof. Using (10.4.10) and the definition of ΛN,2s (see Chapter 9 and (10.1.4)), it easily follows that (10.4.11) is equivalent to D(m) := ≥
N N Γ( N+2s − 2m )Γ( 2m ) m2 2 (m − 1)(N − 2sm) Γ( N − N )Γ( N − s)
Γ2 ( N+2s ) 4
2
2m
4N =: Θ(N, s). (N − 2s)2 Γ2 ( N−2s ) 4
Furthermore, as we saw in the proof of Lemma 10.4.5, D(
2N ) = Θ(N, s), N + 2s
2m
(10.4.12)
10.4 Optimal summability in the presence of Hardy potential  297
and thus (10.4.12) holds if we prove that D is an increasing function. Let us denote D(m) = D1 (m)D2 (m), where D1 (m) :=
Γ( N+2s − 2
Γ( N2 −
N N )Γ( 2m ) 2m N N )Γ( 2m − s) 2m
and D2 (m) :=
m2 . (m − 1)(N − 2sm)
On the other hand, it is known that for t > 0 one has Γ (t) = ψ(t)Γ(t), where ψ(t), the socalled digamma function, is given by ∞ 1 1 ψ(t) := − − C0 + t ∑ , t n(n + t) n=1
with C0 being the Euler constant. Hence, it follows that D1 (m) =
N D (m)K(m), 2m2 1
where K(m) = [ψ(a) − ψ(b) + ψ(c) − ψ(d)] and a :=
N + 2s N − , 2 2m
b :=
N N − , 2 2m
c :=
N − s, 2m
d :=
N . 2m
Thus D (m) = D1 (m)
(m − 2)N + 2sm N m ( + K(m)). (m − 1)(N − 2sm) (m − 1)(N − 2sm) 2m
The first two terms here are positive, and then to analyze the sign of D we have to study the function H(m) := (
N (m − 2)N + 2sm + K(m)). (m − 1)(N − 2sm) 2m
By definition, there holds K(m) = −
4sm3 (m − 2)N + 2sm N (m − 1)(N − 2sm)((m − 1)N + 2sm) ∞
+ s ∑( n=1
1 1 − ), (n + a)(n + b) (n + c)(n + d)
and noting that a = b + s and d = c + s, we have 1 1 1 1 = ( − ) (n + a)(n + b) s n + b n + b + s
298  10 Calderón–Zygmund summability in the fractional setting and 1 1 1 1 = ( − ). (n + c)(n + d) s n + c n + c + s Thus, 1 1 1 1 1 1 1 − = (( − )+( − )) (n + a)(n + b) (n + c)(n + d) s n + b n + c n+c+s n+b+s 1 1 b−c ( − ). =− s (n + c)(n + b) (n + c + s)(n + b + s) Hence H(m) = ((m − 2)N + 2sm)H1 (m), where H1 (m) :=
1 1 N ∞ 1 − − ]. ∑[ 2 (m − 1)N + 2sm 4m n=1 (n + c)(n + b) (n + c + s)(n + b + s)
2N N Using the fact that m ∈ ( N+2s , 2s ), we obtain (m − 2)N + 2sm > 0 and therefore the sign of H is the sign of H1 . On the other hand, since s ∈ (0, 1), we have ∞
∑[
n=1
1 1 1 − ]≤ , (n + c)(n + b) (n + c + s)(n + b + s) (1 + c)(1 + b)
whence we conclude that
1 1 N − 2 (m − 1)N + 2sm 4m (1 + c)(1 + b) N 1 − . = (m − 1)N + 2sm (N + 2m − 2sm)((m − 1)N + 2m)
H1 (m) ≥
Now, using the fact that s ∈ (0, 1), we get H1 (m) > 0 and the result follows. As a consequence we get the next regularity result. N 2N , 2s ), s < 1 and f (x) = x1 ν with Proposition 10.4.8. Let s ∈ (0, 1). Assume that m ∈ ( N+2s N−ε ν= for some small ε > 0. Then there exists λ̃ ∈ (J (m), P (m)) such that the solution s
m
∗∗ u of problem (10.4.1) with λ = λ̃ satisfies u ∈ Lms (B1 (0)).
s
−α Proof. Fix δ > 0 small enough so that if α ∈ (α0 (m), α0 (m) + δ), then γ := N−2s 2 satisfies γ > ν − 2s and γm∗∗ < N. s Define λ̃ = λ(α). Since α > α0 , we have λ̃ < Ps (m). Thus, due to the continuity of λ as a function of α, choosing δ small enough and applying Lemma 10.4.7 we deduce Js (m) < λ̃ < Ps (m). Let u be the unique solution of problem (10.4.1) with λ = λ̃ and consider the function v defined in (10.4.9). Since γ > ν − 2s, by similar arguments as in the proof of Lemma 10.4.4 we conclude that v ≤ u. By setting w := u − v, it follows that {
(−Δ)s w − λ xw2s = 0 w = −v ≥ 0
in Ω, in ℝN \ Ω.
10.4 Optimal summability in the presence of Hardy potential  299
Thus, by Lemma 10.3.8 and Theorem 10.4.1, w ≃ x−γ close to the origin. Since v ≃ x−γ , m∗∗ s (B (0)) and also u ≃ x−γ . By the definition of γ, we obtain γm∗∗ 1 s < N, thus u ∈ L then we conclude the proof. 10.4.3 Nonvariational setting: weak solutions In this subsection we consider f ∈ Lm (Ω), with 1 < m
1, the Lebesgue theorem
implies that un ↑ u a. e. and strongly in L (Ω) for all 1 ≤ σ ≤ u x2s
(10.4.17)
1
(β+1)2∗s 2
and
un x2s
↑
strongly in L (Ω). Therefore, u is a weak solution of (10.4.1). Moreover, using Fatou’s lemma, (10.4.16) and (10.4.17) in (10.4.15) we obtain ∬ DΩ
(u(x) − u(y))(uβ (x) − uβ (y)) dx dy ≤ C. x − yN+2s
(10.4.18)
In addition, we know that u is the unique weak solution. Indeed if u2 is another solution of (10.4.1) with the regularities above, then setting w := u2 − u, we conclude that {
(−Δ)s w − λ xw2s = 0 w=0
in Ω, in ℝN \ Ω.
By testing with ϕ ∈ 𝒯 defined in (10.2.1), with (−Δ)s ϕ = φ > 0, we obtain that w = 0, and hence u2 = u. To finish, we prove the regularity of the fractional gradient. Fix s1 < s and let q = m∗s < 2. Call uβ (x) − uβ (y) { {( ) dx dy dσ := { u(x) − u(y) { {0
if u(x) ≠ u(y), if u(x) = u(y),
and note that dσ is positive. Therefore, by Hölder’s inequality ∫∫ Ω Ω
u(x) − u(y)q dx dy x − yN+qs1
= ∫∫ Ω Ω
u(x) − u(y) u(x) − u(y)q uβ (x) − uβ (y) )×( β ( ) dx dy u(x) − u(y) x − yN+qs1 u (x) − uβ (y) q
2
2 u(x) − u(y)2 u(x) − u(y) 2−q dσ ) ) ≤ (∫ ∫ dσ) × (∫ ∫( β N+2s x − y u (x) − uβ (y) x − yN−θ
Ω Ω
Ω Ω
β
q
β
2 (u(x) − u(y))(u (x) − u (y)) = (∫ ∫ dx dy) N+2s x − y
Ω Ω
q
u(x) − u(y) 2−q dx dy ) ) × (∫ ∫( β u (x) − uβ (y) x − yN−θ Ω Ω
2−q 2
,
2−q 2
10.4 Optimal summability in the presence of Hardy potential  301
1) where θ := 2(s−s . The first term is bounded by (10.4.18), and the second one can be 2−q estimated as follows. Since β < 1, we have
0≤
u(x) − u(y) 1 ≤ (u1−β (x) + u1−β (y)), β β u (x) − u (y) β
and hence q
(1−β)q (1−β)q C dx dy u(x) − u(y) 2−q dx dy ) ≤ ∫ ∫(u 2−q (x) + u 2−q (y)) ∫ ∫( β β N−θ β u (x) − u (y) x − y x − yN−θ
Ω Ω
Ω Ω
(1−β)q dy 2C ) dx. ≤ ∫ u 2−q (x)(∫ β x − yN−θ
Ω
Ω
Note that sup (∫
{x∈Ω}
Ω
dy ) ≤ C, x − yN−θ
hence q
(1−β)q u(x) − u(y) 2−q dx dy 2−q (x) dx. ≤ C u ) ∫ ∫ ∫( β 1 u (x) − uβ (y) x − yN−θ
Ω
Ω Ω
Now, since
(1−β)q 2−q
= m∗∗ s , the result follows using (10.4.13).
10.4.4 A necessary and sufficient condition for solvability As in the local problem studied in Chapter 3, in the case where no condition is imposed on λ, an additional condition on f is needed. The next theorem gives a necessary and sufficient condition to ensure the existence of a weak solution. Theorem 10.4.10. Let λ ≤ ΛN,2s and suppose that f ∈ L1 (Ω), f ≥ 0. Then u is a positive weak solution to problem (10.4.1) if and only if f satisfies ∫ x−γ f dx < +∞, Br (0)
for some Br (0) ⊂⊂ Ω with γ defined in Lemma 10.2.4. Moreover, if u is the unique weak solution to (10.4.1), then ∀k ≥ 0,
Tk (u) ∈ H0s (Ω),
u ∈ Lq (Ω),
∀q ∈ (1,
N ), N − 2s
(10.4.19)
302  10 Calderón–Zygmund summability in the fractional setting s r (−Δ) 2 (u) ∈ L (Ω),
∀r ∈ (1,
s ,q
N ), N −s
(10.4.20)
N . N−s
and u ∈ W01 1 (Ω) for all s1 < s and for all q1
0 { { { { φ=0
in Ω, in Ω,
(10.4.21)
in (ℝN \ Ω).
Taking φn as a test function in (10.4.1), we get ∫ fφn dx ≤ ∫ u dx = C < ∞. Ω
Ω
Hence, {fφn }n∈ℕ is an increasing sequence uniformly bounded in L1 (Ω). Applying the monotone convergence theorem and Lemma 10.3.8 we obtain C̃ ∫ x−γ f dx ≤ ∫ fφ dx ≤ C. Ω
Br (0)
Sufficient condition: Assume that ∫ x−γ f dx < +∞, Br (0)
for some Br (0) ⊂⊂ Ω. Let us consider the sequence of energy solutions un ∈ L∞ (Ω) ∩ H0s (Ω) to the following approximated problems: u
s n−1 { (−Δ) un = λ x2s + 1 + fn n { u (x) = 0 { n
in Ω, in ℝN \ Ω,
(10.4.22)
10.5 Further results and comments  303
where {
(−Δ)s u0 = f1 u0 (x) = 0
in Ω, in (ℝN \ Ω),
with fn = Tn (f ) and u0 ≤ u1 ≤ un−1 ≤ un in ℝN . Since fn ≥ 0, un (x) ≥ 0 in Ω. Take φ ∈ H0s (Ω), the positive energy solution to (10.4.21), as a test function in (10.4.22). As a consequence of Lemma 10.4.1, it follows that ∫ un dx ≤ ∫ fφ dx ≤ C̃ ∫ f x−γ ≤ C. Ω
Ω
Ω
Hence, since the sequence {un }n∈ℕ is increasing, we can define u := limn→∞ un , and conclude that u ∈ L1 (Ω). We claim that xu2s ∈ L1 (Ω). Indeed, let ψ be the unique bounded positive solution to the problem {
(−Δ)s ψ = 1
in Ω,
ψ=0
in ℝN \ Ω.
Then ψ ≥ C in Br (0). By using ψ as a test function in (10.4.22), ∫ Ω
un−1
x2s +
1 n
dx ≤
ψun−1 1 dx + C(r) ∫ C x2s + n1 Br (0)
un dx ≤ C,
∫ Ω\Br (0)
and thus λ
un−1
x2s +
1 n
+ fn ↗ λ
u +f x2s
strongly in L1 (Ω).
Testing with Tk (un ) in (10.4.22) and by the previous estimates, we easily get u Tk (un ) ⇀ Tk (u) weakly in H0s (Ω). Since the sequence {λ 2sn−1 1 + fn }n∈ℕ converges x + n
strongly in L1 (Ω), by the results of [224], we reach that u ∈ Lσ (Ω) for all σ < s 2
N N−2s
N and (−Δ) (u) ∈ Lr (Ω) for all r ∈ (1, N−s ). Moreover, according to Theorem 5 (C) of s ,q N Chapter 5 in [293], we conclude that u ∈ W01 1 (Ω) for all s1 < s and for all q1 < N−s . See also [2, 218] for a simple proof.
Remark 10.4.11. As a consequence of this result, together with the weak Harnack inequality, one can easily prove nonexistence for λ > ΛN,2s .
10.5 Further results and comments 1.
An intellectual challenge here is to investigate what happens in the context of variational solutions in the interval Js (m) < λ < Ps (m),
304  10 Calderón–Zygmund summability in the fractional setting
2. 3.
defined in Section 10.4.2. Independently of the result, we believe that the techniques needed to attack this problem would be very interesting. It would also be interesting to find negative results to the summability for λ > Js (m) 2N if m ∈ (1, N+2s ). Another direction of research comes from studying the analysis of other kinds of nonlocal elliptic operators. However, there is no substantial difference for those operators with the kernels of Definition 8.3.2.
11 Fractional semilinear elliptic problems 11.1 Introduction During the last 20 years much attention has been devoted to understand the role of the Hardy–Leray potential in the solvability of sublinear and quasilinear elliptic problems. References to study the evolution of the research on elliptic problems involving the inverse square potential can be found in the papers [13, 14, 15, 17, 87, 88, 89, 99, 145, 184, 179, 181, 257, 311], among others. In this chapter we will study the effect of the extension of the Hardy inequality studied in Chapter 9 in the semilinear problem involving the fractional Hardy potential. There exist many recent results related to the Dirichlet problem for the fractional Laplacian with semilinear perturbations; see [44, 279, 280, 281, 283]. M. M. Fall, in [151], has extended for the nonlocal case some results given by Brezis–Dupaigne– Tesei in [89]. The problem he analyzes is u (−Δ)s u − λ 2s = up { { x { { (P0 ) = { u > 0 { { { { u=0
in Ω, in Ω, in ℝN \ Ω,
for which he finds the critical power p to have solvability. In his proof, he uses the extension given by L. Caffarelli and L. Silvestre in [105]. We will also discuss the problem (P0 ) but without such extension. In the spirit of the classical paper by H. Brezis and L. Nirenberg [94], along this chapter we will analyze the solvability of the Dirichlet problem for the nonlocal operator (−Δ)s and a zero order convexconcave term. In particular, we will analyze the existence of nontrivial solutions for the following problem with the presence of the Hardy potential: (−Δ)s u − λ xu2s = up + μuq { { { (Pμ ) = { u > 0 { { { u=0
in Ω, in Ω,
(11.1.1)
N
in ℝ \ Ω,
where Ω is a bounded Lipschitz domain of ℝN satisfying the exterior ball condition and 0 ∈ Ω, s ∈ (0, 1), N > 2s, μ > 0, 0 < q < 1, λ < ΛN,2s . Here p > 1 but smaller than an upper bound that we will explain right after. The case q = 0 can be extended in an easy way to the case with a source term, f (x), with suitable integrability. We recall that a domain Ω satisfies the exterior ball condition if there exists a positive radius ρ∗ such that all the points on 𝜕Ω can be touched by some exterior ball of radius ρ∗ . The hypotheses that we have presented above are supposed to hold in https://doi.org/10.1515/9783110606270011
306  11 Fractional semilinear elliptic problems what follows. We remark again that we will study directly the nonlocal operator, without the harmonic extension in an extra space dimension as in the Caffarelli–Silvestre argument. The organization of the chapter is as follows. Section 11.2 is devoted to the development of some important tools for the fractional Laplacian: the maximum principle (L. Silvestre [287]) and the Pohozaev type inequality (X. RosOton and J. Serra [274]). In Section 11.3 we deal with the existence of at least one solution for the whole range 1 < p < p(λ, s), where p(λ, s) is the threshold for the existence of a positive radial weak solution of the equation in the whole ℝN (see (11.2.1)). Proceeding by a monotonicity argument, if 1 < p ≤ 2∗s − 1 we will reach a finite energy solution, while if 2∗s − 1 < p < p(λ, s) we find a weak solution. In Section 11.4.1, we prove the existence of a second solution in the subcritical case, that is, for 1 < p < 2∗s − 1, taking advantage of the corresponding variational framework. For μ small, we find two solutions using first a minimization procedure and later applying the mountain pass theorem in [37]. For μ large, we will need to prove that the solution obtained in Section 2 is a local minimum, using an argument due to Alama (see [32]). This generalization is not immediate because, due to the nonlocal behavior of our operator, the bounded support of the test functions is not preserved. In Section 11.5 we will study the nonexistence of solution for the case p ≥ p(λ, s). In fact, using the nonlocal version of Lemma 3.2 in [87] (see Lemma 11.5.3 below), we will prove that the solutions to the truncated problems blow up for every x in Ω. Finally in Section 11.6 the problem with nonlinearities singular on the boundary is analyzed.
11.2 Preliminaries In the first place, let us guess the candidate to be an upper bound for p. To that end, we look for a radial solution to the problem in the whole ℝN : (−Δ)s w − λ If we choose w = Ax
2s−N 2
+β
w = wp . x2s
, with A a positive constant, we have
Aγβ x−2s+
2s−N 2
+β
− λAx−2s+
2s−N 2
+β
= Ap x(
2s−N 2
+β)p
,
where γβ :=
N+2s−2β N+2s+2β )Γ( 4 ) 4 N−2s−2β N−2s+2β Γ( 4 )Γ( 4 )
22s Γ(
was calculated in Chapter 9. Hence, in order to have homogeneity, we need 2s − N −2s +β= 2 p−1
and, therefore, the equation becomes γβ − λ = Ap−1 .
11.2 Preliminaries  307
Since A > 0, we need γβ − λ > 0. Note that the map γ : [0,
N − 2s ) 2 β
→
(0, ΛN,2s ],
→
γβ
by Lemma 9.5.1 is decreasing (see also [167]). Therefore there exists a unique αλ such that γαλ = λ. −2s + N−2s and, consequently, Thus γβ − λ > 0 is equivalent to αλ > β, that is, αλ > p−1 2 p
2∗s −1, we are concerned with a supercritical problem and there is no possible variational formulation. In this supercritical context we give the following definition. Definition 11.2.2. We say that u ∈ L1 (Ω) is a weak solution of (Pμ ) if u ≥ 0 a. e., u = 0 in ℝN \ Ω, and it satisfies ∫ (λ Ω
u + up + μuq )δs dx < ∞ x2s
and ∫ u(−Δ)s φ dx = ∫ (λ ℝN
Ω
u + up + μuq )φ dx, x2s
for all φ ∈ C 2s+β (Ω) ∩ C s (Ω), β > 0, with φ = 0 in ℝN \ Ω and δ(x) := dist(x, 𝜕Ω).
(11.2.5)
11.2 Preliminaries  309
Note that the right hand side makes sense because, since φ ∈ C s (Ω), it follows that φ(x) ≤ Cδs (x). See also [287, Proposition 2.4]. It is interesting to remark that for the existence results, the size of the biggest exponent, the smallest power q (that introduces an appropriate behavior of the nonlinearity that allows the use of monotonicity arguments) and the size of λ are relevant. Also, as an application of the weak setting, we will study a semilinear problem that is singular at the boundary. More precisely, we will consider the problem (−Δ)s u − λ xu2s = { { { { u>0 { { { { { u=0
h(x) uσ
in Ω, in Ω,
(11.2.6)
N
in ℝ \ Ω.
The local case (s = 1) with λ = 0 was studied in [79]. Here the authors proved that for all h ∈ L1 (Ω), there exists at least one distributional solution. Regularity is obtained according to the regularity of h and the value of σ. The fractional case when λ = 0 was studied in [45] (see also the recent paper [163] for some extensions of [46]).
11.2.1 A convergence tool by Brezis–Lieb For completeness we include the following result by H. Brezis and E. Lieb [91]. Theorem 11.2.3. Assume 1 ≤ p < ∞. Let {fn }n∈ℕ ⊂ Lp (Ω) be a sequence such that: (a) fn (x) → f (x) almost everywhere as n → ∞; (b) ‖fn ‖ ≤ M < ∞. Then f ∈ Lp (Ω) and lim ∫(fn p − fn − f p ) dx = ∫ f p dx.
n→∞
Ω
Ω
Moreover, if ‖fn ‖p → ‖f ‖p as n → ∞, then ‖fn − f ‖p → 0
as n → ∞.
Proof. Note that Fatou’s lemma gives ∫ f p dx ≤ lim inf ∫ fn p dx ≤ M. n→∞
Ω
Ω
Hence, f ∈ Lp (Ω). Fixing p ∈ [1, ∞) we have t + 1p − tp − 1 = 0, t→∞ tp lim
310  11 Fractional semilinear elliptic problems and hence for all ϵ > 0 there exists Cϵ > 0 such that t + 1p − tp − 1 ≤ ϵtp + Cϵ ,
∀t ∈ ℝ.
Therefore by an elementary argument, for all ϵ > 0 there exists Cϵ > 0 such that p p p p p a + b − a − b ≤ ϵa + Cϵ b ,
for all a, b ∈ ℝ.
We use the previous inequality with a = fn − f and b = f . Then, for ϵ > 0 and the corresponding Cϵ , ϕn (x) =: (fn p − f p − fn − f p − ϵfn − f p ) ≤ Cϵ f p . By hypothesis (a), ϕn (x) → 0 almost everywhere as n → ∞ and 0 ≤ ϕn ≤ Cϵ f p . Hence, by the dominated convergence theorem, limn→∞ ∫Ω ϕn = 0. Since p p p p fn  − f  − fn − f  ≤ ϕn + ϵfn − f  , lim sup ∫fn p − f p − fn − f p dx ≤ ϵ ∫ fn − f p dx ≤ ϵ(2M)p , n→∞ Ω
for all ϵ > 0,
Ω
which implies the result.
11.2.2 Maximum principle for the fractional Laplacian We follow in this section the ideas in [287]. Consider the fundamental solution of the sfractional Laplacian defined in (9.3.10), SN,s (x) =
cN,2s
xN−2s
,
where cN,2s is defined in (9.3.9). Consider a paraboloid PN,s (x) tangent from below to SN,s on x = 1 and define PN,s (x) if x ≤ 1
Ψ(x) = {
SN,s (x)
if x ≥ 1.
Hence Ψ ∈ 𝒞 1,1 (ℝN ). For λ > 0 define Ψλ (x) =
1
λN−2s
x Ψ( ) ∈ 𝒞 1,1 (ℝN ). λ
By Proposition 8.2.11, (−Δ)s Ψ is a continuous function. Moreover we have the following properties of Ψ:
11.2 Preliminaries  311
1.
We have (−Δ)s Ψ(x) ≥ 0. By the regularity of Ψ we can write (−Δ)s Ψ(x0 ) = CN,s p.v. ∫ ℝN
–
If x0  > 1, Ψ(x0 ) = SN,s (x0 ) and for y ≠ x0 , Ψ(y) ≤ SN,s (y). Then (−Δ)s Ψ(x0 ) = CN,s ∫ ℝN
–
Ψ(x0 ) − Ψ(y) dy. x0 − yN+2s
SN,s (x0 ) − SN,s (y) Ψ(x0 ) − Ψ(y) dy ≥ CN,s ∫ dy = 0. N+2s x0 − y x0 − yN+2s ℝN
If 0 < x ≤ 1 we choose a point x1 in such a way that SN,s (x0 − x1 ) < Ψ(x0 ) and ∇SN,s (x0 − x1 ) = ∇Ψ(x0 ). Then we choose δ > 0 such that SN,s (x − x1 ) + δ touches Ψ(x) from above at x0 . Therefore (−Δ)s Ψ(x0 ) = CN,s ∫ ℝN
≥ CN,s ∫
Ψ(x0 ) − Ψ(y) dy x0 − yN+2s SN,s (x0 − x1 ) − SN,s (y − x1 )
ℝN
x0 − yN+2s
dy = 0.
–
2.
If x0 = 0 then Ψ attains its maximum at x0 , that is, Ψ(x0 ) − Ψ(y) ≥ 0 that implies (−Δ)s Ψ(x0 ) ≥ 0. We have (−Δ)s Ψ(x) ∈ L1 (ℝN ). Moreover s (−Δ) Ψ1 = 1. We compute the L1 norm of (−Δ)s Ψ(x). Consider a cutoff function η ∈ 𝒞0∞ (ℝN ) such that η(x) ≤ 1
for all x ∈ ℝN , supp(η) ⊂ B2 and η(x) = 1, x ∈ B1 .
Define ηR (x) = η( Rx ). Since Ψ(x) − SN,s (x) ∈ L1 (ℝN ) is compact supported and (−Δ)s ηR (x) → 0 as R → ∞ on compact sets, we find ∫ (−Δ)s Ψ(x) dx − 1 = lim ∫ ((−Δ)s Ψ(x) − (−Δ)s SN,s (x))ηR (x) dx ℝN
R→∞
ℝN
= lim ∫ (Ψ(x) − SN,s (x))(−Δ)s ηR (x) dx = 0. R→∞
ℝN
Define γλ (x) = (−Δ)s Ψλ (x).
312  11 Fractional semilinear elliptic problems By the previous remarks γλ is a compact supported function and ∫ γ1 (x) dx = 1. ℝN
On the other hand by a simple scaling we find that x γλ (x) = λ−N γ( ). λ Thus, {γλ (x)}λ>0
is an approximation to the identity.
In particular if u ∈ ℒs , where we recall s
N
ℒ := {u : ℝ → ℝ measurable: ∫ ℝN
u(x) dx < ∞} (1 + xN+2s )
is a Banach space endowed with the norm ‖u‖ℒs := ∫ ℝN
u(x) dx, (1 + xN+2s )
we have γλ ∗ u(x) → u(x),
λ → 0, a. e.
The next result plays the same role as the mean inequality in the classical case. Proposition 11.2.4. Let u ∈ ℒs be such that (−Δ)s u(x) ≥ 0 in an open set Ω. Then u is lower semicontinuous in Ω and u(x0 ) ≥ ∫ u(x)γλ (x − x0 ) dx,
(11.2.7)
ℝN
for any x0 ∈ Ω and λ ≤ dist(x0 , 𝜕Ω). Proof. Call r = dist(x0 , 𝜕Ω) and take r > λ1 > λ2 > 0. Then Ψλ2 −Ψλ1 ≥ 0 is a 𝒞 1,1 function supported in Br . Since (−Δ)s u(x) ≥ 0 in Br (x0 ) we get 0 ≤ ⟨(−Δ)s u(x), Ψλ2 (x − x0 ) − Ψλ1 (x − x0 )⟩ = CN,s ∫ u(x)(γλ2 (x − x − 0) − γλ1 (x − x0 )) dx, ℝN
and hence γλ2 ∗ u(x0 ) ≥ γλ1 ∗ u(x0 ).
11.2 Preliminaries 
313
For each λ > 0, γλ ∗ u is a continuous function and is an increasing family in λ. Given that the family γλ is an approximation of the identity u is the increasing limit of continuous functions and hence is lower semicontinuous up to a set of zero measure. Taking limits for λ2 → 0, u(x0 ) ≥ ∫ u(x)γλ (x − x0 ) dx. ℝN
Proposition 11.2.4 is the main tool to prove the following maximum principle. Note that the setting in which we find the result is very general. Proposition 11.2.5. Assume that Ω is a bounded open set in ℝN . Let u be lower semicontinuous in Ω and such that (i) (−Δ)s u ≥ 0 in Ω; (ii) u ≥ 0 in ℝN \ Ω. Then u ≥ 0 in ℝN . Moreover if u(x) = 0 in a point x ∈ Ω, then u = 0 in ℝN . Proof. Since, by hypothesis, u ≥ 0 in ℝN , if by contradiction we suppose that there exists a point y ∈ ℝN such that u(y) < 0, then y ∈ Ω. By the hypothesis of lower semicontinuity u attains its minimum in x0 ∈ Ω. If we assume that x0 ∈ Ω and u(x0 ) < 0, by Proposition 11.2.4 there is a λ > 0 such that u(x0 ) ≥ ∫ u(x)γλ (x − x0 ) dx. ℝN
Since γλ is positive and has integral equal to 1, we have 0 ≥ ∫ (u(x) − u(x0 ))γλ (x − x0 ) dx, ℝN
which is a contradiction with the fact that the second member is positive. If there exists x0 ∈ Ω such that u(x0 ) = 0, we find 0 ≥ ∫ u(x)γλ (x − x0 ) dx, ℝN
which, jointly with the nonnegativity of u, implies 0 = ∫ u(x)γλ (x − x0 ) dx. ℝN
Since γλ (y) > 0 we conclude that u = 0.
314  11 Fractional semilinear elliptic problems 11.2.3 The Pohozaev identity for the fractional Laplacian The classical paper by S. Pohozaev [261] studies solutions u to the problem −Δu = f (u) in Ω, { u=0 on 𝜕Ω, where f is a continuous function and Ω is a smooth starshaped bounded domain in ℝN . Assuming a suitable regularity of the solution u and by using the multiplier ⟨x, ∇u(x)⟩, i. e., ∫⟨x, ∇u(x)⟩(−Δu(x)) dx = ∫⟨x, ∇u(x)⟩f (u(x)) dx, Ω
Ω
an integration by parts gives ∫⟨x, ∇u(x)⟩(−Δu(x)) dx = Ω
2 − N 1 2 2 ∫∇u(x) dx − ∫ ∇u(x) ⟨x, ν⟩ dσ 2 2 Ω
𝜕Ω
and u
∫⟨x, ∇u(x)⟩f (u(x)) dx = N ∫ F(u(x)) dx, Ω
where F(u) = ∫ f (s) ds. 0
Ω
On the other hand, multiplying the equation by u and integrating by parts we find that ∫ ∇u2 dx = ∫ uf (u) dx, Ω
Ω
which brings the final form of the Pohozaev identity 1 N −2 2 ∫ uf (u) dx − N ∫ F(u) dx = ∫ ∇u(x) ⟨x, ν⟩ dσ. 2 2 Ω
Ω
𝜕Ω
If we try to find similar identities for solutions to the problem (−Δ)s u = f (u) in Ω,
(Ps ) {
u=0
in ℝN \ Ω,
(11.2.8)
by using the multiplier of Pohozaev, the first difficulty is to calculate the integral ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx Ω
and find the local behavior of the solution on the boundary 𝜕Ω. The solution to this problem is a deep result found by X. RosOton and J. Serra in [274]. We will present an approach to this in a sketchy way, with the recommendation for the reader to read the original paper for more details.
11.2 Preliminaries  315
Regularity up to the boundary There are several key points in the research of the fractional Pohozaev identity. The first one is the paper [273], in which RosOton and Serra give the precise behavior of a solution on the boundary. As usual, we denote δ(x) = d(x, 𝜕Ω). Let β > 0 be so that β ∉ ℕ. Denote by [v]𝒞 β (O) the classical seminorm of Hölder spaces in an open set 𝒪. That is, if k is the integral part of β and β = β − k, then [v]𝒞 β (𝒪) = [v]𝒞 k,β (𝒪) =
sup
x,y∈𝒪,x =y̸
Dk u(x) − Dk u(y) . x − yβ
We are in a position to formulate the regularity result. Assume in problem (11.2.8) that Ω is a bounded and 𝒞 1,1 domain and f is a locally Lipschitz function. If u is a bounded solution to (11.2.8) and g = f (u), then u ∈ 𝒞 s (ℝN ) and ‖u‖𝒞 s (ℝN ) ≤ C‖g‖L∞ (Ω) , where C = C(Ω, s). Moreover, there exists α = α(Ω, s) > 0, α < min{s, (1 − s)}, such that u ≤ C‖g‖L∞ (Ω) , s δ 𝒞 s (ℝN )
u δs
∈ 𝒞 α (Ω) and
C = C(Ω, s).
The last inequality tells the precise behavior of u at the boundary and gives some kind of Hopf lemma for the fractional Laplacian. The following extra interior regularity of u, a bounded solution to (11.2.8), is also obtained: (i) If β ∈ [s, 1 + 2s), u ∈ 𝒞 β (Ω) and [u]𝒞 β ({x∈Ωδ(x)>ρ}) ≤ Cρs−β . (ii) If β ∈ [α, s + α], [
u ] ≤ Cρα−β . δs 𝒞 β ({x∈Ωδ(x)>ρ})
The constant C depends on β, f , ‖u‖∞ , Ω and s. The precise statement of the previous regularity results and the proofs can be seen in [273]. Key estimates to prove the fractional Pohozaev identity Lemma 11.2.6. Let Ω be a bounded and 𝒞 1,1 domain, and let u be a function such that u = 0 in ℝN \ Ω that satisfies the same regularity as the bounded solution of prob
316  11 Fractional semilinear elliptic problems lem (11.2.8). Then, there exists a 𝒞 α (ℝN )extension v of s
(−Δ) 2 u(x) =
) Γ(1 + s) sin( sπ 2 π
[log− δ(x) +
u  δs Ω
such that
π χΩ (x)]v(x) + h(x), ) tan( sπ 2
where h ∈ 𝒞 α (ℝN ). Moreover, if β ∈ (0, 1 + s), s
[(−Δ) 2 u]𝒞 β ({x∈ℝN δ(x)>ρ}) ≤ Cρ−β
for all ρ ∈ (0.1),
with C independent of ρ. We also need the following technical result. Lemma 11.2.7. Let a, b ∈ ℝ and ϕ(t) = a log− t − 1 + bχ[0,1] (t) + h(t), where log− τ = min{0, log τ} and h is a real function satisfying that there exist α, γ ∈ (0, 1) and C0 > 0 such that: (i) ‖h‖𝒞 α [0,∞) ≤ C0 ; (ii) for all β ∈ [γ, γ + 1], ‖h‖𝒞 β (0,1−ρ∪(1+ρ,2)) ≤ C0 ρ−β , ρ ∈ (0, 1); (iii) h (t) ≤ C0 t −2−γ and h (t) ≤ C0 t −3−γ , for all t > 2.
Then ∞
t d ∫ ϕ(λt)ϕ( ) dt = a2 π 2 + b2 . dt t=1+ λ 0
See the proofs of the two previous lemmas in [274]. The fractional Pohozaev identity by RosOton and Serra in strictly starshaped domains Next we formulate the fractional Pohozaev in starshaped domains with respect to the origin. The reader can also find in the paper [274] the proof of the identity in general 𝒞 1,1 domains. Theorem 11.2.8. Let Ω ⊂ ℝN be a bounded starshaped and 𝒞 1,1 domain, f : ℝ → ℝ a locally Lipschitz and u a bounded solution to problem (11.2.8). Then δus has a continuous extension to Ω, that is, 𝒞 α (Ω), for some α ∈ (0, 1). Moreover, the following identity holds: (2s − N) ∫ uf (u) dx + 2N ∫ F(u) = Γ(1 + s)2 ∫ ( Ω
Ω
𝜕Ω
2
u ) ⟨x, ν⟩ dσ, δs
where ν is the unit outward normal to 𝜕Ω at x, δ(x) is the distance to the boundary, Γ is u the Gamma function and F(u) = ∫0 f (τ) dτ.
11.2 Preliminaries  317
Proof. The basic idea is to use the same Pohozaev multiplier as in the local case, ⟨x, ∇u(x)⟩. In the fractional case the integration by parts must be carefully done. This integration by parts will be done step by step. Step 1. Define uλ (x) = u(λx). Then ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx = Ω
d ∫ uλ (x)(−Δ)s u(x) dx. dλ λ=1+
(11.2.9)
Ω
By the regularity results, in particular we know that (−Δ)s u(x) is pointwise defined and bounded. Call w(x) = (−Δ)s u(x) and note that if λ > 1, then λ1 Ω ⊂ Ω, and hence supp uλ = λ1 Ω. Then d u(λx) − u(x) w(x) dx. ∫ uλ (x)(−Δ)s u(x) dx = lim ∫ λ↓1 dλ λ=1+ λ−1 Ω
Ω
Making the change of variables y = λx, we obtain lim ∫ λ↓1
Ω
u(y) − u( yλ ) y u(λx) − u(x) w(x) dx = lim λ−N ∫ w( ) dy λ↓1 λ−1 λ−1 λ λΩ
= lim ∫ λ↓1
λΩ
u(y) − u( yλ ) λ−1
−u( λy ) y y w( ) dy + lim ∫ w( ) dy. λ↓1 λ λ−1 λ λΩ\Ω
Consider β = 1 in the regularity result. Then ∇u(x) ≤ δ(x)s−1 ≤ λ1−s δ(x)s−1 for all x in the segment joining y and yλ . Observe that δ(x)s−1 is integrable. Therefore we can apply the dominated convergence theorem to find lim ∫ λ↓1
λΩ
u(y) − u( yλ ) λ−1
y w( ) dy = ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx. λ Ω
To finish the first step we have to prove that lim ∫ λ↓1
λΩ\Ω
−u( λy ) λ−1
y w( ) dy = 0. λ
But this follows immediately from the facts that λΩ\Ω ≤ C(λ −1) and since u ∈ 𝒞 s (ℝN ) and u = 0 in ℝN \ Ω, we have ‖u‖L∞ (λΩ\Ω) → 0 as λ → 1+ . Step 2. Under the general hypotheses the following identity holds: ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx = Ω
2
(2s − N) Γ(1 + s)2 u ∫ u(−Δ)s u dx − ∫ ( s ) ⟨x, ν⟩ dσ. 2 2 δ Ω
𝜕Ω
318  11 Fractional semilinear elliptic problems Note that with the change of variables y = √λx, we have s
s
∫ uλ (x)(−Δ)s u(x) dx = ∫ (−Δ) 2 uλ (x)(−Δ) 2 u(x) dx = λ Ω
2s−N 2
ℝN
∫ w√λ w ℝN
1 √λ
dx.
Therefore, according to the first step ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx = Ω
2s−N d {λ 2 ∫ w√λ w 1 dx} √λ dλ λ=1+ ℝN
=
d (2s − N) I ∫ w2 dx + 2 dλ λ=1+ √λ N ℝ
=
(2s − N) 1 d I , ∫ u(−Δ)s u dx + 2 2 dλ λ=1+ λ ℝN
where Iλ = ∫ w√λ w ℝN
1 √λ
dx.
(11.2.10)
So, to finish step 2 we need to evaluate the derivative of Iλ at t = 1 on the right. Since Ω is a starshaped and 𝒞 1,1 domain we can define the radial projection of x and since Ω is starshaped with 𝕊N−1 into 𝜕Ω. More precisely, if x ∈ 𝜕Ω, then θ = x respect to the origin there exists a unique ρ(θ) > 0 such that ρ(θ)θ ∈ 𝜕Ω. Then we define the radial projection by ϝ : 𝕊N−1
→
𝜕Ω,
x x
→
x = ϝ(θ) = r(θ)θ.
θ=
Writing the integral (11.2.10) in spherical coordinates and performing the change of r variables t = r(θ we get ∞
d r d Iλ = ∫ dθ ∫ r N−1 w(λrθ)w( θ)dr + + dλ λ=1 dλ λ=1 λ 0
𝕊N−1
∞
=
d r(θ)t N−1 θ) dt ∫ ∫ (r(θ)t) w(λr(θ)tθ)w( dλ λ=1+ λ 𝕊N−1 0
∞ d tx x = ∫ ⟨ , ν⟩ dσ(x) ∫ t N−1 w(λtx)w( ) dt, dλ λ=1+ x λ N−1 𝕊
0
where the identity r(θ)N−1 dθ =
1 x ⟨ , ν⟩ dσ(x) r(θ) x
11.2 Preliminaries 
319
follows from the change of variables associated to the radial projection on 𝜕Ω. For x0 ∈ 𝜕Ω fixed and recalling the meaning of the function w, we define ϕ(t) = t
N−1 2
s
(−Δ) 2 u(tx0 ).
By Lemma 11.2.6 we find ϕ(t) = c1 [log− δ(tx0 ) + c2 χ[0,1] ]v(tx0 ) + h0 (t), where v is a 𝒞 α (ℝN )extension of δus Ω and h0 is a 𝒞 α ([0, ∞))function. By the regularity and the geometry of the domain Ω the authors in [274] are able to prove that the function ϕ(t) satisfies the hypotheses of Lemma 11.2.7. Therefore ∞
d t 2 ∫ ϕ(λt)ϕ( ) dt = (v(x0 )) c12 (π 2 + c22 ) dλ λ=1+ λ 0
and then ∞
d t 2 ∫ t N−1 w(λtx0 )w( x0 ) dt = (v(x0 )) c12 (π 2 + c22 ). dλ λ=1+ λ 0
Now, taking into account that v is a 𝒞 α (ℝN )extension of
u  , we find that for x0 δs Ω
u(x) u (x ) = lim . x→x0 ,x∈Ω δs (x) δs 0 We conclude that ∞ tx x0 d d , ν⟩ dσ(x) ∫ t N−1 w(λtx0 )w( 0 ) dt I = ∫⟨ dλ λ=1+ λ dλ λ=1+ x0  λ N−1 0
𝕊
= c12 (π 2 + c22 ) ∫ ⟨ 𝜕Ω
2
x0 u , ν⟩( s (x0 )) dσ(x0 ). x0  δ
By Lemma 11.2.6 we have c1 =
Γ(1 + s) sin( sπ ) 2 π
,
and
c2 =
and then 2
c12 (π 2 + c22 ) = (Γ(1 + s)) , which finishes step 2.
π , ) tan( sπ 2
∈ 𝜕Ω
320  11 Fractional semilinear elliptic problems Step 3. Consider u, a bounded solution to problem (11.2.8). Multiplying the equation by ⟨x, ∇u(x)⟩ and integrating in Ω we get ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx = ∫⟨x, ∇u(x)⟩f (u(x)) dx = ∫⟨x, ∇F(u(x))⟩ dx. Ω
Ω
Ω N
Integrating by parts, since u = 0 in ℝ \ Ω, ∫⟨x, ∇F(u(x))⟩ dx = −N ∫ F(u(x)) dx. Ω
Ω
Moreover multiplying the equation by u and integrating we have ∫ u(x)(−Δ)s u(x) dx = ∫ u(x)f (u(x)) dx. Ω
Ω
Hence summarizing the results in the previous steps and the last calculation we have −N ∫ F(u(x)) dx = ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx Ω
Ω
=
2
(2s − N) Γ(1 + s)2 u ∫ u(−Δ)s u dx − ∫ ( s ) ⟨x, ν⟩ dσ 2 2 δ Ω
=
𝜕Ω
2
Γ(1 + s)2 u (2s − N) ∫ u(x)f (u(x)) dx − ∫ ( s ) ⟨x, ν⟩ dσ. 2 2 δ Ω
𝜕Ω
Pohozaev with the Hardy potential For the application to solutions of problems involving power nonlinearities of the type (−Δu)s − λ
u = f (u), x2s
u ≥ 0 x ∈ Ω,
u=0
x ∈ ℝN \ Ω, 0 < λ < ΛN,2s , (11.2.11)
we need to adapt the arguments above. We will begin with a solution u ∈ H0s (Ω) that satisfies the same interior regularity as the solutions to (11.2.8) referring to Ω \ {0} and (ii) assume that the function δus has a continuous extension to Ω \ Br (0), that is, 𝒞 α (Ω \ 2N Br (0)), for some α ∈ (0, 1). Since u ∈ H0s (Ω), f (u) ∈ Lm (Ω) with m ≥ N+2s , and then 1 uf (u) ∈ L (Ω). The first difference is the term involving the Hardy potential. The second difference is that the positive solutions to problem (11.2.11) are in general unbounded. However, we have extra information about the precise behavior of the solution in Br (0) given in Theorem 10.4.1. To handle the term involving the Hardy potential we will proceed as follows. First note that ∫⟨x, ∇u⟩λ Ω
u dx = lim ρ→0 x2s
∫ Ω\Bρ (0)
λ x ⟨ 2s ∇u2 ⟩ dx. 2 x
11.2 Preliminaries 
321
This last term is calculated as follows. First we compute the integral in the perforated domain, ∫ Ω\Bρ (0)
u 1 ⟨x, ∇u⟩ = − 2 x2s =−
=−
u2 div(
∫ Ω\Bρ (0)
N − 2s 2 N − 2s 2
x 1 )+ 2 x2s
𝜕(Ω\Bρ (0))
2
∫ Ω\Bρ (0)
1 u + x2s 2
Ω\Bρ (0)
u 1 + x2s 2
u2 ⟨ν, x⟩ dσ x2s
∫ 𝜕(Ω\Bρ (0))
2
∫
u2 ⟨ν, x⟩ dσ x2s
∫
∫
u2 ⟨ν, x⟩ dσ, x2s
𝜕Bρ (0)
using the boundary condition. Taking into account the local behavior of u, the last integral can be estimated as 1 2
ρ u2 ⟨ν, x⟩ dσ ≤ C 2γ+2s ρN−1 , 2s x ρ
∫ 𝜕Bρ (0)
and we have N − 2γ − 2s > 0. Therefore ∫⟨x, ∇u⟩λ Ω
u dx = lim ρ→0 x2s
∫ Ω\Bρ (0)
N − 2s λ x u2 ⟨ 2s ∇u2 ⟩ dx = − ∫ 2s dx, 2 x 2 x Ω
by the monotone convergence theorem. Since the zero order term can be evaluated as in the case λ = 0, it remains to calculate ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx. Ω
We will proceed by approximation of the solution u by bounded functions. Call h(x) = f (u(x)) > 0. We consider w0 the solution to (−Δ)s w0 = T 1 (h(x)) with w0 = 0 in ℝN \ Ω 2 and by recurrence we define wk , k = 1, 2, . . ., the solution to w
{
(−Δ)s wk = λ xk−1 2s + hk (x)
wk = 0
in Ω, in ℝN \ Ω,
where hk = Tk (h). According to Corollary 8.3.12, we easily find w0 ≤ w1 ≤ ⋅ ⋅ ⋅ ≤ wk−1 ≤ wk ≤ ⋅ ⋅ ⋅ ≤ u.
(11.2.12)
322  11 Fractional semilinear elliptic problems Moreover, testing in problem (11.2.12) with wk we have ∫(−Δ)s wk wk dx = λ ∫ Ω
Ω
wk−1 w dx + ∫ hk wk dx x2s k Ω
w2 ≤ λ ∫ k2s dx + ∫ hk u dx x Ω
≤ λ∫ Ω
Ω
2
u dx + ∫ f (u)u dx < ∞. x2s Ω
Then wk ⇀ v weakly in H0s (Ω) and it is standard to see that v solves the problem (−Δ)s v − λ
v = h(x) in Ω, x2s
v = 0 in ℝN \ Ω.
By the uniqueness of the linear problem we conclude that u = v. We will prove that wk → u strongly in H0s (Ω). Indeed, u − wk−1 2 s (u−wk ) dx+∫(h−hk )(u−wk ) dx. (u−wk )H s (Ω) = ∫(−Δ) (u−wk )(u−wk ) dx = λ ∫ 0 x2s Ω
Ω
Ω
The second member converges pointwise to zero and it is uniformly bounded by the function F(x) = 4(λ
u2 + f (u)u) ∈ L1 (Ω), x2s
and hence by the dominated convergence theorem we conclude that ‖(u − wk )‖H0s (Ω) → 0
as k → ∞.
Applying the estimate by RosOton and Serra to each wk we find ∫⟨x, ∇wk (x)⟩(−Δ)s wk (x) dx = Ω
2
w Γ(1 + s)2 2s − N ‖wk ‖2H s (Ω) − ∫ ( sk ) ⟨x, ν⟩ dσ 0 2 2 δ 𝜕Ω
and passing to the limit when k → ∞ ∫⟨x, ∇u(x)⟩(−Δ)s u(x) dx = Ω
2
Γ(1 + s)2 u 2s − N ‖u‖2H s (Ω) − ∫ ( s ) ⟨x, ν⟩ dσ. 0 2 2 δ 𝜕Ω
Multiplying equation (11.2.11) by u, integrating and doing the same calculations as in the case λ = 0, we find the Pohozaev identity with Hardy potential (2s − N) ∫ uf (u) dx + 2N ∫ F(u) = Γ(1 + s)2 ∫ ( Ω
Ω
𝜕Ω
2
u ) ⟨, x, ν⟩ dσ, δs
(11.2.13)
11.3 Existence of minimal solutions for 1 < p < p(λ, s)
 323
u
where, as above, F(u) = ∫0 f (t) dt. As an immediate consequence of (11.2.13) we find that problem (11.2.11) in a starshaped domain has no positive solution if f (u) = up with p ≥ 2∗s − 1. Assume 0 < λ ≤ ΛN,2s . If f (u) = up with 0 < p < 2∗s −1 we can find a positive solution to (11.2.13) by application of classical results of critical point theory, minimization if q < 1 and mountain pass arguments if 2∗s − 1 > p > 1. The case p = 1 corresponds to finding the eigenvalues of the coercive operator, with compact inverse, (−Δ)s (⋅) − λ x12s (⋅).
11.3 Existence of minimal solutions for 1 < p < p(λ, s) We will study the existence of solution to problem (11.1.1). We use the comparison results obtained in Section 8.3.3 and for the supercritical case with respect to the Sobolev exponent we will need the following version of the results for weak solutions in the sense of Definition 8.6.2. This result is the extension of [88, Lemma 1] to the nonlocal framework. Lemma 11.3.1. Let f ∈ L1 (Ω, δs (x) dx), where δ(x) = dist(x, 𝜕Ω). Then there exists a unique v ∈ L1 (Ω), which is a weak solution of the problem (−Δ)s v = f
{
v=0
in Ω,
in ℝN \ Ω
and there exists a constant C > 0 such that ‖v‖L1 (Ω) ≤ C‖f ‖L1 (Ω,δs (x) dx) . Moreover if f ≥ 0 a. e. in Ω, then v ≥ 0 a. e. in Ω. Proof. The uniqueness and the existence are obtained like in Theorem 8.6.4. By the linearity of the problem we can assume without loss of generality that f ≥ 0. We prove that v≥0
a. e. in Ω.
(11.3.1)
Indeed, given ψ ∈ C0∞ (Ω), ψ ≥ 0, define φψ as the solution of (−Δ)s φψ = ψ ≥ 0 in Ω, { φψ = 0 in ℝN \ Ω. Since φψ ≥ 0 in ℝN , considering the definition of weak solution (Definition 11.2.2) associated to this problem we obtain ∫ vϕ(x) dx ≥ 0. Ω
324  11 Fractional semilinear elliptic problems Hence v ≥ 0 a. e. in Ω. With this observation the estimate follows immediately. Consider (−Δ)s ϕ = 1
in Ω,
{
in ℝN \ Ω.
ϕ=0
Then ‖v‖L1 (Ω) = ∫ v dx = ∫ v(−Δ)s (ϕ) dx = ∫(−Δ)s (v)ϕ dx = ∫ fϕ dx ≤ C ∫ fδs dx Ω
Ω
Ω
Ω
Ω
= ‖f ‖L1 (Ω,δs dx) , according to the regularity until the boundary obtained in [273]. The first important result in this section is the following. Theorem 11.3.2. Consider 1 < p < p(λ, s). Let M be defined by M = sup{μ > 0 : problem (11.1.1) has a solution}.
(11.3.2)
Then 0 < M < ∞. Proof. We use the same kind of ideas as in [65], that is, we have to construct a subsolution and a supersolution to problem (11.2.2) for μ small enough, in such a way that pointwise the subsolution is less than or equal to the supersolution. Step 1. Finding a subsolution. For the subsolution, consider the eigenvalue problem (−Δ)s φ1 = λ1,(−Δ)s φ1 (P1) = { φ1 = 0
in Ω,
(11.3.3)
in ℝN \ Ω.
By the regularity theory in Chapter 8 (see also [283, Proposition 4]), the nonnegative function φ1 belongs to H0s (Ω) ∩ L∞ (Ω). Hence, taking t small enough, we have, in Ω, (−Δ)s (tφ1 ) = λ1,(−Δ)s tφ1 ≤ μ(tφ1 )q ≤ μ(tφ1 )q + (tφ1 )p + λ
tφ1 . x2s
Thus, u := tφ1 is a subsolution of (Pμ ). Step 2. Finding a supersolution. To build the supersolution, we need to deal with the subcritical, critical and supercritical cases separately. (i) Subcritical and critical case: 1 < p ≤ 2∗s − 1. We look for a supersolution of the form w(x) := Ax−β , where A > 0 and β is a positive real parameter that verifies β
0. For μ small enough, taking u := C1 w with 0 < C1 < 1 a suitable constant such that 1 , (11.3.7) ηp−q ≥ μ 1−q C (1 − C p ) it follows that (−Δ)s u − λ
u ≥ up + μuq . x2s
Thus, we have obtained a supersolution of (Pμ+ ) for 1 < p ≤ 2∗s − 1. Moreover, by (11.3.5) we also have u2 ∈ L1 (Ω). x2s
u ∈ Lp+1 (Ω) and
(11.3.8)
Choosing the parameter t of u := tφ1 small enough, it yields u ≥ u. Now, since p ≤ 2∗s − 1, we can build a nonnegative sequence {uk } in H0s (Ω) of solutions to the iterated problems u
(Pk ) = {
(−Δ)s uk = λ xk−12s + upk−1 + μuqk−1
uk = 0
in Ω, in ℝN \ Ω,
for k ≥ 1 and u0 := u. By Corollary 8.3.12 it can be easily checked that u ≤ u1 ≤ ⋅ ⋅ ⋅ ≤ u. Hence, we can define uμ := limk→∞ uk in L1 (Ω). Moreover, by (11.3.8), u u s/2 2 dx + ∫ uk upk−1 dx + μ ∫ uk uqk−1 dx (−Δ) uk L2 (ℝN ) = λ ∫ k k−1 x2s Ω
Ω
Ω
u2 ≤ λ ∫ 2s dx + ∫ up+1 dx + μ ∫ uq+1 dx ≤ C. x Ω
Ω
(11.3.9)
Ω
Therefore, up to a subsequence, we know that uk ⇀ uμ in H0s (Ω). In fact, by monotony, the whole sequence converges weakly. Hence, since p ≤ 2∗s − 1, we can pass to the limit in the iterated problems to conclude that uμ ≥ 0 is a minimal energy solution of (11.2.2) and, obviously, of (11.1.1).
326  11 Fractional semilinear elliptic problems (ii) Supercritical case: 2∗s − 1 < p < p(λ, s). Firstly, as we explained in the introduction, if p < p(s, λ), p(s, λ) as in (11.2.1), there −2s
exists a radial function u(x) := Ax p−1 , with A a positive constant, satisfying (−Δ)s u − λ
Since p > 2∗s − 1 >
u = up x2s
in ℝN .
N , N−2s
u ∈ Lploc (ℝN )
and
u ∈ L1loc (ℝN ). x2s
(11.3.10)
Again, taking u = C1 u, with C1 > 0 a suitable constant (see (11.3.7)), we get (−Δ)s u − λ xu2s ≥ up + μuq
{
u>0
in Ω,
in ℝN \ Ω.
Moreover by (11.3.10), in particular u ≥ 0 satisfies (11.2.5). Hence, by Lemma 11.3.1 we can define {uk } to be the weak solutions to (Pk ), and we will prove by induction that, in fact, uk ∈ L1 (ℝN )
0 ≤ u ≤ uk−1 ≤ uk ≤ u a. e. in Ω, for every k ∈ ℕ.
and
For u there is nothing to prove. Suppose the result is true up to order k − 1, that is, u ≤ uj−1 ≤ uj ≤ u for j ≤ k − 1 a. e. in Ω. Then (−Δ)s uk = λ
uk−1 u + upk−1 + μuqk−1 ≤ λ 2s + up + μuq . 2s x x
Hence, by (11.3.10), since the right hand side is in L1 (Ω), by Lemma 11.3.1 uk , in particular, is integrable too. Moreover, by the induction hypothesis, (−Δ)s (uk − uk−1 ) = λ
(uk−1 − uk−2 ) + (upk−1 − upk−2 ) + μ(uqk−1 − uqk−2 ) ≥ 0 x2s
and (−Δ)s (u − uk ) = λ
(u − uk−1 ) + (up − upk−1 ) + μ(uq − uqk−1 ) ≥ 0. x2s
Therefore, by (11.3.1), we have 0 ≤ u ≤ uk−1 ≤ uk ≤ u a. e. in Ω. By the monotone convergence argument we conclude that {uk } converges in L1 (ℝN ) to a weak nonnegative solution uμ of (Pμ ) for 2∗s − 1 < p < p(λ, s).
11.3 Existence of minimal solutions for 1 < p < p(λ, s)
 327
Therefore, for μ small enough, we have built a minimal solution in the subcritical, critical and supercritical cases. That is, we have proved that M > 0. To finish the proof we need to check that M < ∞. Consider 1 < p ≤ 2∗s − 1 and the eigenvalue problem with the Hardy potential given by ϕ
(−Δ)s ϕ1 − λ x12s = λ1 ϕ1
(P2) = {
ϕ1 = 0
in Ω,
in ℝN \ Ω.
Note that since λ < ΛN,2s , this problem is well defined and, following the same ideas as in the proof of assertion (b) of [281, Lemma 9], ϕ1 ∈ H0s (Ω). Suppose that u is a solution to problem (11.2.2). By the maximum principle we know that u > 0 in Ω. Then taking ϕ1 as a test function in this problem we get aN,s ∫ Q
(ϕ1 (x) − ϕ1 (y))(u(x) − u(y)) ϕu dx dy − λ ∫ 12s dx x x − yN+2s Ω
p
q
= ∫ u ϕ1 dx + μ ∫ u ϕ1 dx. Ω
Ω
Using now that ϕ1 is a solution of (P2) it follows that ∫ (up + μuq )ϕ1 dx = λ1 ∫ uϕ1 dx.
(11.3.11)
Ω
Ω
If 2∗s − 1 < p < p(λ, s), we consider φ1 ≥ 0, a solution to (P1 ), as a test function in (11.1.1). Then ∫ u(−Δ)s φ1 dx = ∫(λ Ω
Ω
u + up + μuq )φ1 dx x2s
≥ ∫(up + μuq )φ1 dx.
(11.3.12)
Ω
Moreover, since φ1 is also an energy bounded solution of the eigenvalue problem (11.3.3), φ1 is also a classical solution (see [273, Remark 2.1]). Hence from (11.3.12) we also get λ1 ∫ uφ1 dx ≥ ∫(up + μuq )φ1 dx. Ω
(11.3.13)
Ω
Since there exist structural positive constants c0 , c1 such that t p + μt q > c0 μc1 t,
for every t > 0,
we obtain from (11.3.11) and (11.3.13) that c0 μc1 < λ1 . This implies that M < ∞ for p < p(λ, s).
328  11 Fractional semilinear elliptic problems Notice that the previous result is true in the local problem and is independent of the Hardy potential term. This is motivated by the structure of the nonlinearity, with a concave part and a convex part, and holds for a larger class of second order operators for which the concaveconvex structure must be well understood. For the plaplacian case see [180]. The next result shows that the set of parameters μ for which there exists a solution is in fact the whole interval (0, M]. Proposition 11.3.3. Problem (11.1.1) has at least one positive solution for every 0 < μ < M. In fact, the sequence {uμ } of minimal solutions is increasing with respect to μ. If μ = M problem (11.1.1) admits at least one weak solution. Proof. Since M > 0, we can find a solution for a value of μ as close as we want to M. Denote this value by μ and the associated minimal solution by uμ . Then, for all μ < μ, uμ is a supersolution for the problems (11.1.1) with μ < μ. Furthermore, for every μ, the solution to problem (11.3.3) can be modified if necessary to a subsolution to (11.1.1) in such a way that it is pointwise less than or equal to uμ . Following the same procedure as in the previous proof, we conclude that there exists a solution uμ for all μ ∈ (0, μ), and therefore for all μ ∈ (0, M). For the case μ = M, the idea consists of passing to the limit when μn ↗ M on the sequence {un } = {uμn }, where uμn is the minimal solution of (11.2.2) with μ = μn . Since we are considering minimal solutions un ≤ un+1 for all n ∈ ℕ. Consider the solution φ1 to the eigenvalue problem (11.3.3) as a test function in (11.2.2). Since supp un ⊆ Ω we get λ1 ∫ un φ1 dx = λ ∫ Ω
Ω
un φ1 dx + ∫ upn φ1 dx + μ ∫ uqn φ1 dx. x2s Ω
(11.3.14)
Ω
By Young’s inequality again and by Hopf’s lemma (see [273, Lemma 3.2]) we can conclude that λ∫ Ω
un δs dx + ∫ upn δs dx + μ ∫ uqn δs dx ≤ C, x2s Ω
(11.3.15)
Ω
where C is a constant independent of n and δ(x) := dist(x, 𝜕Ω), x ∈ Ω. Let now ξ1 be the solution to the linear problem (−Δ)s ξ1 = 1 in Ω, { ξ1 = 0 in ℝN \ Ω. Using ξ1 as a test function of (11.2.2), since supp(un ) ⊆ Ω, by [273, Proposition 1.1] and (11.3.15), we obtain ∫ un dx = λ ∫ Ω
Ω
un ξ1 dx + ∫ upn ξ1 dx + μ ∫ uqn ξ1 dx x2s Ω
Ω
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
≤ C(λ ∫ ≤ C.
Ω
 329
un δs dx + ∫ upn δs dx + μ ∫ uqn δs dx) x2s Ω
Ω
Hence, {un } converges in L1 (Ω) to a limit uM ≥ 0. Then, since {un } is an increasing sequence that is uniformly bounded in L1 (Ω), by the monotone convergence theorem, taking the limit when n → ∞, we conclude that uM is actually a weak solution of (PM ). Remark 11.3.4. The results obtained in this section can be easily translated for the case of q = 0, that is, considering a function f with appropriate growth conditions instead of the concave term uq . In particular, one could obtain the same existence results given in Section 4.4 in the nonlocal framework.
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1 Problem (11.1.1) is the Euler–Lagrange equation of the energy functional 𝒥μ (u) := aN,s ∫ Q
−
u(x) − u(y)2 dx dy x − yN+2s
μ λ (u+ )2 1 dx − ∫ ∫ up+1 ∫ uq+1 + dx − + dx, 2 x2s p+1 q+1 Ω
H0s (Ω)
Ω
(11.4.1)
Ω
2∗s
which is well defined in if 1 < p ≤ − 1. There are two different cases for the verification of the Palais–Smale condition: (i) the subcritical case, 1 < p < 2∗s − 1, in which the compactness follows from the Rellich theorem; (ii) the critical case, p = 2∗s − 1. In this case the verification of the Palais–Smale condition is more involved and it follows from the concentration compactness arguments by P. L. Lions in [233, 234]. Both cases will be studied in the following paragraphs. 11.4.1 Subcritical case Consider 1 < p < 2∗s − 1 and 0 < λ < ΛN,2s hereafter in this section. Taking advantage of the variational structure of (11.1.1) we will prove the existence of, at least, two positive solutions as critical points of the functional 𝒥μ (u) defined in (11.4.1). We begin with the parameter μ sufficiently small. In this situation a local minimization argument allows to find the first solution, and the mountain pass theorem
330  11 Fractional semilinear elliptic problems guarantees the existence of the second one. This behavior in the local case and for the pLaplacian operator and λ = 0 was first proved in [177] even for the critical Sobolev exponent. The proof of the existence of two positive solutions in the whole interval (0, M) was proved by Ambrosetti, Brezis and Cerami in the case of the Laplacian in [35]. For radial solutions and the pLaplacian operator the reader is referred to [36] and for the general case to [180]. In this section we follow the reference [46]. In order to use the mountain pass theorem, we have to check: (i) the mountain pass geometry and (ii) the compactness of the functional, that is, the Palais–Smale condition. Proposition 11.4.1. There exists a μ0 > 0 such that for any 0 < μ < μ0 , there exist α > 0 and β > 0 such that: (a) 𝒥μ (u) ≥ β, for any u ∈ H0s (Ω) with ‖u‖H0s (Ω) = α and μ small enough; (b) there exists u1 ∈ H0s (Ω) positive such that ‖u1 ‖H0s (Ω) > α and 𝒥μ (u1 ) < β. Proof. (a) Since q + 1 < p + 1 < 2∗s , by (9.2.4) and Theorem 8.2.7, it can be checked that 𝒥μ (u) ≥ g(‖u‖H s (Ω) ), 0
c1 t 2 − c2 t+p+1 − μc3 t+q+1 , for some positive constants c1 , c2
where g(t) = and c3 . Then, choosing μ small enough, there exists α > 0 such that if β := g(α) > 0, therefore 𝒥μ (u) ≥ β for u ∈ H0s (Ω) with ‖u‖H s (Ω) = α. 0 (b) Fix u0 ∈ H0s (Ω) positive such that ‖u0 ‖H0s (Ω) = 1 and take t > 0. Since p > 1, it is clear that lim 𝒥μ (tu0 ) = −∞.
t→∞
Therefore, there exists t0 large enough, such that, defining u1 := t0 u0 , it follows that ‖u1 ‖H0s (Ω) > α and 𝒥μ (u1 ) < β. By a similar argument, we obtain lim 𝒥μ (tu0 ) = 0− .
t→0+
(11.4.2)
Finally, we need to check that the functional satisfies the Palais–Smale condition. First we prove the following. Proposition 11.4.2. Let {un } be a bounded sequence in H0s (Ω) such that 𝒥μ (un ) → 0 in H −s (Ω) as n → ∞. Then there exists u ∈ H0s (Ω) such that, up to a subsequence, ‖un −u‖H0s (Ω) → 0 as n → ∞. Proof. The condition 𝒥μ (un ) → 0 in H −s (Ω) is equivalent to saying that (−Δ)s un − λ
(un )+ − μ(un )q+ − (un )p+ = χn , x2 s
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
 331
with χn → 0 as n → ∞ in H −s (Ω). Since {un } is uniformly bounded in the Hilbert space H0s (Ω) with the norm ‖⋅‖H0s (Ω) , there exists C > 0 such that ‖un ‖2∗ := (aN,s ‖un ‖2H s (Ω) − λ ∫ 0
Ω
(un )2+ dx) ≤ aN,s ‖un ‖2H s (Ω) ≤ C. 0 x2s
Since, by (9.2.4), ‖ ⋅ ‖2∗ is a norm equivalent to ‖ ⋅ ‖2H s (Ω) , we consider H0s (Ω) endowed 0 with this norm and hence up to a subsequence, there exists u ∈ H0s (Ω) such that un ⇀ u un → u
weakly in H0s (Ω) with the norm ‖ ⋅ ‖∗ , r
N
in L (ℝ ), 1 ≤ r
2 > q + 1, by (9.2.4) and the Sobolev inequality, Theorem 8.2.7, we obtain 1 1 c + o(1) = aN,s ( − )‖un ‖2H s (Ω) 0 2 p+1
(u )2 1 1 1 1 ) ∫ n2s+ dx − μ( − ) ∫(un )q+1 − λ( − + dx 2 p+1 q+1 p+1 x
≥
C1 ‖un ‖2H s (Ω) 0
Ω
−
Ω
C2 ‖un ‖q+1 , H0s (Ω)
with C1 and C2 positive constants. Therefore ‖un ‖H0s (Ω) ≤ C. Applying Proposition 11.4.2 we conclude the strong convergence in the space H0s (Ω). Now we can already state the following multiplicity theorem for μ sufficiently small. Theorem 11.4.4. For μ small enough, the problem (Pμ ) has at least two solutions. Proof. We construct the first one by minimization following the construction done in [177]. Indeed, as we saw in Proposition 11.4.1, there exists α > 0 such that 𝒥μ (u) ≥ β > 0 for all u ∈ H0s (Ω) with ‖u‖H0s (Ω) = α. Thus we can choose α1 = {inf α : 𝒥μ (u) > 0 for all u ∈ H0s (Ω) with ‖u‖H0s (Ω) = α}. α∈ℝ
We know that α1 > 0, because near the origin the functional is negative. We choose now α2 > α1 so close that g(‖u‖H0s (Ω) ), the lower radial bound of 𝒥μ (u), is nondecreasing for u with α1 ≤ ‖u‖H0s (Ω) ≤ α2 . We define now a smooth function τ as 1
τ(t) := {
t ≤ α1 ,
0 t ≥ α2 ,
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
 333
and we consider the truncated functional 𝒥2 (u) :=
1 a ‖u‖2 s 2 N,s H0 (Ω) τ(‖u‖H0s (Ω) ) u2 μ λ − ∫ +2s dx − ∫ up+1 ∫ uq+1 + dx − + dx. 2 x p+1 q+1 Ω
Ω
Ω
By definition, 𝒥2 (u) = 𝒥μ (u)
whenever ‖u‖H0s (Ω) ≤ α1
and 𝒥2 (u) =
u2 μ λ 1 aN,s ‖u‖2H s (Ω) − ∫ +2s dx − ∫ uq+1 + dx 0 2 2 x q+1 Ω
Ω
whenever ‖u‖H0s (Ω) ≥ α2 . Note that, by the Sobolev embedding theorem, Theorem 8.2.7, since q + 1 < 2 the functional 𝒥2 is coercive. The lower semicontinuity is given because H0s (Ω) is a Hilbert space. Therefore there exists u1 , a minimum of 𝒥2 with negative energy, that is also a minimum of 𝒥μ . Then, we have already found a first solution to (11.1.1). For the second one, we have proved that if μ is small enough the functional 𝒥μ has the suitable geometry, Proposition 11.4.1, and satisfies the Palais–Smale condition, Proposition 11.4.3. Then if we consider Γ := {γ ∈ C 0 ([0, 1], X0s (Ω)) : γ(0) = 0, γ(1) = u1 }, where u1 is the local minimum found above, and C := inf sup 𝒥 (γ(t)), γ∈Γ t∈[0,1]
the Ambrosetti–Rabinowitz mountain pass theorem (see Section 4.3.3 and [37]) gives us a solution u ∈ H0s (Ω) satisfying 𝒥 (u) = C ≥ β > 0.
Here β is specified in Proposition 11.4.1. Note that this solution and the one obtained before are different because the previous one had negative energy. Therefore, for μ small enough, problems (Pμ+ ) and, consequently, (Pμ ) have at least two solutions. Now we will prove that in fact problem (Pμ ) has two solutions for every μ ∈ (0, M). With this purpose we will generalize a result by Alama in [32] to the fractional Laplacian to check that the minimal solution obtained is a local minimum, which will allow us to apply the mountain pass lemma. Due to the involved computations we provide a detailed proof.
334  11 Fractional semilinear elliptic problems Proposition 11.4.5. Let p ∈ (1, 2∗s − 1). Then for 0 < μ < M, where M is defined in (11.3.2), problem (Pμ ) has at least two solutions. Proof. Let μ0 ∈ (0, M) and take μ1 such that μ0 < μ1 < M. Then, by Proposition 11.3.3, we can consider wμ0 and wμ1 , the minimal solutions to problems (Pμ0 ) and (Pμ1 ), respectively, which satisfy wμ0 < wμ1 . Now we define W = {w ∈ H0s (Ω) : 0 ≤ w ≤ wμ1 }. Since W is a closed convex set of H0s (Ω), and 𝒥μ0 is bounded from below and semicontinuous in W, therefore there exists w ∈ W such that 𝒥μ0 (w) = infw∈W 𝒥μ0 (w). Let w0 ∈ H0s (Ω) be the positive solution to w
(−Δ)s w0 − λ x02s = μ0 w0q
{
w0 = 0
in Ω,
in ℝN \ Ω.
The uniqueness of this solution is a consequence of the extension of the Brezis–Kamin result to the nonlocal setting, obtained in Theorem 8.5.5. Note that, since λ < ΛN,2s , the existence is given by minimization. Then, for 0 < ε 2 we have 0≤
1 [(w + wn )p+1 − wp+1 ] − wp wn ≤ C(p)(wp−1 wn2 + wnp+1 ). p+1
(11.4.31)
Therefore, from (11.4.27), by (11.4.30) and (11.4.31) it follows that 2
𝒥μ0 (vn ) ≥ (1 − q)(aN,s ‖wn ‖H s (Ω) − 0
+ 𝒥μ0 (zn ) + ∫ C(p)(−w
2 λ wn (x) dx) ∫ 2 x2s Ω
p−1
wn2 − wnp+1 )(x) dx,
S̃n
≥ C2 ‖wn ‖2X s (Ω) + 𝒥μ0 (zn ) + ∫ C(p)(−wp−1 wn2 − wnp+1 )(x) dx, 0
(11.4.32)
S̃n
where C2 = (1 − q)C1 > 0. What remains to prove now is lim S̃n  = 0.
n→∞
Let ϵ, δ > 0, and define An = {x ∈ Ω : vn (x) ≥ wμ1 (x) and wμ1 > w + δ},
Bn = {x ∈ Ω : vn (x) ≥ wμ1 (x) and wμ1 ≤ w + δ}.
(11.4.33)
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1

341
Since ∞ 1 0 = {x ∈ Ω : wμ1 (x) < w} = ⋂{x ∈ Ω : wμ1 (x) < w + } j j=1 1 = lim {x ∈ Ω : wμ1 (x) < w + }, j→∞ j
(11.4.34)
for j0 large enough and δ < j1 we have {x ∈ Ω : wμ1 (x) < w + δ} ≤ ϵ2 . 0 Therefore Bn  ≤ ϵ2 . Moreover, by (11.4.9), lim ‖vn − w‖H0s (Ω) = 0.
n→∞
Then, by Theorem 8.2.7 and the Hölder inequality, it follows that lim ‖vn − w‖L2 (Ω) = 0.
n→∞
That is, for n ≥ n0 large enough we get δ2 ε ≥ ∫ vn − w2 dx ≥ ∫ vn − w2 dx ≥ δ2 An . 2 An
Ω
Therefore An  ≤ ε2 , for n ≥ n0 . Since S̃n ⊂ Bn ∪ An we conclude that S̃n  ≤ ε for n ≤ n0 . Hence (11.4.33) follows. Then, by (11.4.33), the Sobolev inequality and Theorem 8.2.7, we obtain ). ∫ wnp+1 (x) + wp−1 wn2 (x) dx ≤ o(1)(‖wn ‖2H s (Ω) + ‖wn ‖p+1 H s (Ω) 0
S̃n
0
Therefore from (11.4.32) we conclude that 2
p+1
2
𝒥μ0 (vn ) ≥ C2 ‖wn ‖H s (Ω) + 𝒥μ0 (zn ) − o(1)(‖wn ‖H s (Ω) + ‖wn ‖H s (Ω) ). 0
0
0
Hence, for n large enough, since zn ∈ W and w was the infimum of 𝒥μ0 over W, this implies 𝒥μ0 (vn ) ≥ 𝒥μ0 (zn ) ≥ 𝒥μ0 (w),
which is a contradiction with (11.4.9). Hence w is a minimum. Remark 11.4.6. If λ = ΛN,2s and 1 < p < 2∗s − 1, using the improved Hardy–Sobolev inequality given in section 9.6, (see too [24] and [168]), we could try to prove the existence of at least two positive solutions in a suitable bounded interval μ ∈ (0, M). For the local case these kinds of results were obtained in [181] even in the framework of the Caffarelli–Kohn–Nirenberg inequalities in [108].
342  11 Fractional semilinear elliptic problems 11.4.2 The critical problem: p = 2∗s − 1 In Section 4.6 we have described the results about the critical problem with the Hardy potential for the Laplacian in the whole ℝN obtained by S. Terracini in [304]. In the article [140] the corresponding problem for the fractional Laplacian is studied. More precisely, consider the problem (−Δ)s u = λ with N > 2s, 0 < s < 1, 2∗s =
2N N−2s
∗ u + u2s −1 , 2s x
u ∈ Ḣ s (ℝN ),
(11.4.35)
and 0 < λ < ΛN,2s . The main results are the following.
Theorem 11.4.7. Let 0 ≤ λ < ΛN,2s . Then problem (11.4.35) has a positive solution u ∈ Ḣ s (ℝN ) which is a minimizer of the problem SN,s (λ) :=
inf
u∈Ḣ s (ℝN )\{0}
Qλ (u),
(11.4.36)
where Qλ (u) :=
(aN,s ∫ℝN ∫ℝN
u(x)−u(y)2 x−yN+2s
u2 x2s
dx dy − λ ∫ℝN
∫ℝN u2s dx
dx)
∗
2∗ s 2
.
(11.4.37)
Moreover: (1) If u ∈ Ḣ s (ℝN ) is a positive solution to (11.4.35), then u is radial and radially decreasing with respect to the origin. Namely, there exists some strictly decreasing function v : (0, +∞) → (0, +∞) such that u(x) = v(r),
r = x.
(2) If u ∈ Ḣ s (ℝN ) is a positive solution to (11.4.35), then there exist two positive constants c and C such that c (x1−ηλ (1
+
x2ηλ ))
N−2s 2
≤ u(x) ≤
C (x1−ηλ (1
+ x2ηλ ))
N−2s 2
,
x ∈ ℝN \ {0}, (11.4.38)
where ηλ = 1 −
2αλ , N − 2s
(11.4.39)
αλ ∈ (0, (N−2s) ) verifies the equation λ(α) = ΨN ( N−2s − α) and Ψs,N (t) is defined 2 2 in (9.4.5). (3) There exists a function ρ(λ, N, s), (1 − Λλ ) ≤ ρ(λ, N, s) < 1, such that N,s
SN,s (λ) = ρ(λ, N, s)SN,s . We refer the reader to [140] for the proof and some related details.
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1

343
Remark 11.4.8. Note that in Theorem 11.4.7 the uniqueness of solutions up to dilations remains open. Now we come back to the problem under study, (−Δ)s u = λ xu2s + u2s −1 + θuq , ∗
u(x) ≥ 0,
x ∈ Ω, x ∈ Ω,
(11.4.40)
N
u(x) = 0,
x ∈ ℝ \ Ω,
2N with N > 2s, 0 < s < 1, 2∗s = N−2s , 0 < q < 1, θ > 0 and 0 < λ < ΛN,2s . See [284] for a related result in the whole ℝN . The solutions in H0s (Ω) (11.4.40) are critical points of the energy functional,
𝒥θ (u) := aN,s ∫ Q
−
u(x) − u(y)2 dx dy x − yN+2s
1 θ λ (u+ )2 2∗ dx − ∗ ∫ u+s dx − ∫ ∫ uq+1 + dx, 2s 2 x 2s q+1 Ω
Ω
(11.4.41)
Ω
which is well defined in H0s (Ω). The existence of a minimal nontrivial solution has been proved in Section 11.3. Remark 11.4.9. If u is a weak solution to (11.4.41), then by Lemma 10.3.8, for each ball Bδ (0) ⊂⊂ Ω, u ≥ C(N, δ, γ)x−γ
in Bδ (0),
where γ = N−2s − α(λ) is defined in (9.4.9). By using the ground state transformation 2 in problem (11.4.40), that is, by putting u = xγ v with γ defined in (9.4.9), we arrive at equation (10.3.1) with a second member, that is, vq v2s −1 fθ (v) = ( 2∗ γ + θ (q+1)γ ). x s x ∗
One can prove that any variational positive solution to the transformed problem is bounded. The idea is to use the fractional Caffarelli–Kohn–Nirenberg inequalities presented in Section 10.2 and then use the Moser type argument introduced by N. Trudinger to study the Yamabe problem in [308]. The details are left to the interested reader. These arguments were used in [256] for the pLaplacian case and in [44] for problem (11.4.40) with λ = 0. Note that positive solutions to problem (11.4.40) are unbounded in any neighborhood of the origin, but with the previous observations they are bounded in the complement set to any ball centered at 0.
344  11 Fractional semilinear elliptic problems We will use the following concentration compactness, which can be found in the P. L. Lions article [234]. As is customary, δx means the Dirac delta concentrated at the point x. Lemma 11.4.10. Assume {un }n∈ℕ ⊂ H0s (Ω) verifies un → u a. e., and un → u in Lα (Ω), α ∈ [1, 2∗s ). If s (i) (−Δ) 2 un 2 ⇀ dμ;
(ii) un 2s ⇀ dν; ∗
(iii)
u2n x2
⇀ dγ;
where dμ, δν and dγ are Radon measures, then there exists an index set 𝒥 at most countable such that s 1. dμ ≥ (−Δ) 2 u2 + ∑j∈𝒥 μj δxj + μ0 δ0 ; 2. 3.
2 2∗
dν = u2 + ∑j∈𝒥 νj δxj + ν0 δ0 with Sνj s ≤ μj for all j ∈ 𝒥 ∪ {0}, where S is the optimal Sobolev constant; u2 ⇀ dγ = x 2 + γ0 δ0 , with γ0 ΛN,s ≤ μ0 . ∗
The three main steps of the proof are the following. Step 1. The local Palais–Smale condition. Here the main difference with respect to the subcritical case is the lack of compacteness of the critical term. That is, passing to the limit in the critical term cannot be done directly since the Sobolev embedding is not compact. We need to have a level of energy strictly less than the one given by the concentration compactness by P. L. Lions in [233, 234]. The following lemmas are needed to find the first level of concentration. See Lemmas 2.8 and 2.9 in [44]. These results are the extension to the fractional framework of wellknown results for the local case. See for instance [177]. Lemma 11.4.11. Let ϕ ∈ 𝒞0∞ (ℝN ) be a radial nonincreasing test function and ϕϵ (x) = ϕ( xϵ ). Assume {un }n∈ℕ ⊂ H0s (Ω) verifying the hypotheses of Lemma 11.4.10. Then:
1.
2.
s
s
limϵ→0 limn→∞  ∫ℝN un (−Δ) 2 ϕϵ (−Δ) 2 un dx = 0; s 2
limϵ→0 limn→∞  ∫ℝN (−Δ) un ∫ℝN
(ϕϵ (x)−ϕϵ (y))(un (x)−un (y)) x−yN+2s
dy dx = 0.
Lemma 11.4.12. Let {un }n∈ℕ ⊂ H0s (Ω) be a Palais–Smale sequence for 𝒥θ , namely, 𝒥θ (un ) → c < ∞, 𝒥θ (un ) → 0.
Let M = M(Ω, s, N, q) be the constant defined by the identity 2∗ s
min f (t) = f (t0 ) = −Mθ 2∗s −q , t>0
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
with f (t) =
 345
2s −q s 2∗s 1 1 t − ( − )θΩ 2∗s t q . If N q 2 ∗
c < c∗ =
2s 2s N N s s SN,s (λ) 2s − Mθ 2∗s −q = (SN,s ρ(λ, N, s)) 2s − Mθ 2∗s −q , N N ∗
∗
then {un }n∈ℕ has a strongly convergent subsequence in H0s (Ω). Proof. Assume that {un }n∈ℕ verifies the hypotheses. Then the sequence is bounded in H0s (Ω). To see this fact, it is sufficient to note that applying Hardy’s, Hölder’s and Sobolev’s inequalities, it follows that c = lim (𝒥θ (un ) − n→∞
1 ⟨𝒥 (u ), u ⟩) 2∗s θ n n
q+1 1 1 λ 1 1 ≥ ( − ∗ )(1 − )aN,s ‖u‖2H s (Ω) − ( − ∗ )K‖u‖H2s (Ω) , 0 0 2 2s ΛN,2s q + 1 2s
where K = K(s, N, Ω, q). This implies the boundedness. As a consequence we find a subsequence, relabeled to {un }n∈ℕ , that verifies un ⇀ u as n → ∞ weakly in H0s (Ω), [ [ ‖un − u‖Lr (Ω) → 0 as n → ∞ if 1 ≤ r < 2∗ − 1, s [ [ [ u (x) → u(x) as n → ∞ almost everywhere in Ω and [ n the hypotheses of Lemma 11.4.10. [
(11.4.42)
Hence we have, in particular, the representation of the measures of Lemma 11.4.10. N
2s . Claim. The set J is finite and if j ≠ 0, either νj = 0 or νj ≥ SN,s
Consider a positive function ϕ(x) ∈ 𝒞0∞ (ℝN ) such that ϕ(x) = 1 in B1 (0) and ϕ(x) = 0 in ℝN \ B2 (0) and define ϕϵ,j (x) = ϕ(
x − xj ϵ
ϵ > 0, j ∈ J.
),
Note that we have s
s
(−Δ) 2 ϕϵ,j (x) = ϵ−s (−Δ) 2 ϕ(
x − xj ϵ
(11.4.43)
).
Since 𝒥θ (un ) → 0 in H −s we have, using the Kato inequality, 0 = lim ⟨𝒥θ (un ), (ϕϵ,j un )⟩ ≥ lim ⟨𝒥θ ((un )+ ), (ϕϵ,j (un )+ )⟩ n→∞
n→∞
= ∫ (ϕϵ,j (un )+ )(−Δ)s ((un )+ ) dx − ∫ λϕϵ,j ℝN
B2ϵ (xj )
(un )2+ 2∗ dx − θ ∫ (un )q+ − ∫ ϕϵ,j (un )+s . 2s x B2ϵ (xj )
B2ϵ (xj )
346  11 Fractional semilinear elliptic problems Hence s
s
lim ( ∫ (un )+ (−Δ) 2 (un )+ (−Δ) 2 ϕϵ,j dx
n→∞
ℝN s
− 2aN,s ∫ (−Δ) 2 (un )+ ∫
x − yN+2s
ℝN
ℝN
≤ lim (λ ∫ ϕϵ,j n→∞
(ϕϵ,j (x) − ϕϵ,j (y))((un )+ (x) − (un )+ (y))
s (un )2+ 2∗ dx + θ ∫ (un )q+ dx + ∫ ϕϵ,j (un )+s − ∫ (−Δ) 2 ϕϵ,j ). 2s x
B2ϵ (xj )
B2ϵ (xj )
dy dx)
B2ϵ (xj )
B2ϵ (xj )
(11.4.44)
Suppose first that j ≠ 0. Therefore s
s
lim lim ( ∫ (un )+ (−Δ) 2 (un )+ (−Δ) 2 ϕϵ,j dx
ϵ→0 n→∞
ℝN s
− 2 ∫ (−Δ) 2 (un )+ aN,s ∫ ℝN
x − yN+2s
ℝN
≤ lim (λ ∫ ϕϵ,j ϵ→0
(ϕϵ,j (x) − ϕϵ,j (y))((un )+ (x) − (un )+ (y))
(u)2+ dx + θ ∫ (u)q+ dx + ∫ ϕϵ,j dν − ∫ ϕϵ,j dμ). x2s B2ϵ (xj )
B2ϵ (xj )
dy dx)
B2ϵ (xj )
B2ϵ (xj )
Since by Lemma 11.4.11 the left hand is zero we obtain 0 ≤ lim (λ ∫ ϕϵ,j ϵ→0
B2ϵ (xj )
(u)2+ dx + θ ∫ (u)q+ dx + ∫ ϕϵ,j dν − ∫ ϕϵ,j dμ) = νj − μj . x2s B2ϵ (xj )
B2ϵ (xj )
B2ϵ (xj )
2 2∗
Now, by the P. L. Lions lemma, Lemma 11.4.10, Sνj s ≤ μj and the measure dν is finite, thus the claim follows. We study next the concentration at the pole of the Hardy potential. We consider the test functions ϕϵ,0 as above. We have Q(un ϕϵ,0 ) ≥ SN,s (λ), where Q is defined in (11.4.37) and SN,s (λ) by (11.4.36). That is, aN,s ∫ ∫
un (x)ϕϵ,0 (x) − un (y)ϕϵ,0 (y)2 x − yN+2s
ℝN ℝN
ℝN
(un ϕϵ,0 )2 λ dx ∫ 2 x2s ℝN
2 2∗ s
≥ SN,s (λ)( ∫ un ϕϵ,0 2s dx) . ∗
dx dy −
(11.4.45)
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1

347
From the nonlocal Leibniz formula in (8.1.17), with (un (x) − un (y))(ϕϵ,0 (x) − ϕϵ,0 (y))
Bs (un , ϕϵ,0 )(x) = CN, s ∫ 2
x − yN+s
ℝN
dy,
we find s 2 (−Δ) 2 (un ϕϵ,0 )2 s s 2 = ϕϵ,0 (−Δ) 2 (un ) + un (−Δ) 2 (ϕϵ,0 ) − Bs (un , ϕϵ,0 )2 s s 2 2 2 = (−Δ) 2 (un )2 + un (−Δ) 2 (ϕϵ,0 )2 + Bs (un , ϕϵ,0 )2 s
s
+ 2 ∫ ϕϵ,0 (−Δ) 2 (un )un (−Δ) 2 (ϕϵ,0 ) dx ℝN s
+ 2 ∫ ϕϵ,0 (x)(−Δ) 2 (un )(x)Bs (un , ϕϵ,0 ) dx ℝN s
+ 2 ∫ un (x)(−Δ) 2 (ϕϵ,0 )(x)Bs (un , ϕϵ,0 ) dx ℝN
≡ I1 + I2 + I3 + I4 + I5 + I6 .
(11.4.46)
We have to estimate limϵ→0 limn→∞ Ij , j = 1, 2, 3, 4, 5, 6. We have lim lim I1 = lim ∫ ϕ2ϵ,0 (x) dμ = μ0 .
ϵ→0 n→∞
ϵ→0
ℝN
Next we deal with I2 : lim lim I2
ϵ→0 n→∞
s 2 = lim lim ∫ u2n (x)(−Δ) 2 (ϕϵ,0 (x)) dx
ϵ→0 n→∞
ℝN
s 2 2 ≤ 2 lim lim ∫un (x) − u(x) (−Δ) 2 (ϕϵ,0 (x)) dx ϵ→0 n→∞
Ω
s 2 2 + 2 lim lim ∫u(x) (−Δ) 2 (ϕϵ,0 (x)) dx
ϵ→0 n→∞
Ω
s 2 ≤ A1 lim lim ϵ−s ‖un − u‖22 + A2 lim lim ∫ u2 (x)(−Δ) 2 (ϕϵ,0 (x)) dx
ϵ→0 n→∞
ϵ→0 n→∞
ℝN
2 = A2 lim ∫ u2 (x)(−Δ) (ϕϵ,0 (x)) dx ≡ J1 . ϵ→0
s 2
ℝN
348  11 Fractional semilinear elliptic problems N
Since u2 ∈ L N−2s (Ω), J1 ≤ A3 lim (‖u‖ = A4 ‖u‖
s (−Δ) 2 (ϕϵ,0 )
N
L N−2s (Ω)
ϵ→0
N
s 2
lim (−Δ) (ϕϵ,0 )
L N−2s ϵ→0
N
L 2s (Ω)
N
L 2s (Ω)
)
= A4 ‖u‖
lim ϵs = 0,
N
L N−2s ϵ→0
and therefore lim lim I2 = 0.
ϵ→0 n→∞
To prove that lim lim I3 = 0,
ϵ→0 n→∞
we follow the arguments as in the proof of Lemma 2.9 of [44]. The triangular inequality gives us Bs (un , ϕϵ,0 )2 ≤ Bs (un − u, ϕϵ,0 )2 + Bs (u, ϕϵ,0 )2 and we evaluate separately both terms. By the Cauchy–Schwarz inequality we have 2 Bs (un − u, ϕϵ,0 )2 ≤ ∫ Bs (un − u, un − u)(x)Bs (ϕϵ,0 , ϕϵ,0 )(x) dx, ℝN
and it is easy to check that ϵ−s x ≡ Cϵ−s ψϵ (x) ≤ Cϵ−s , Bs (ϕϵ,0 , ϕϵ,0 )(x) = ϵ−s Bs (ϕ0 , ϕ0 )( ) ≤ C ϵ 1 +  xϵ N+s where ψϵ (x) :=
1 . 1+ xϵ N+s
(11.4.47)
Hence
2 Bs (un − u, ϕϵ,0 )2 s
≤ Cϵ−s ∫ Bs (un − u, un − u)(x) dx = Cϵ−s ∫ (un − u)(−Δ) 2 (un − u) dx ℝN
ℝN
s ≤ Cϵ ‖un − u‖2 (−Δ) 2 (un − u)2 ≤ C1 ϵ−s ‖un − u‖2 .
(11.4.48)
−s
In a similar way, using (11.4.47), we find 2 −s Bs (u, ϕϵ,0 )2 ≤ ∫ Bs (u, u)(x)Bs (ϕϵ,0 , ϕϵ,0 )(x) dx ≤ Cϵ ∫ Bs (u, u)(x)ψϵ (x) dx. ℝN
ℝN
We have the following elementary identity: s
s
∫ u2 (x)(−Δ) 2 ψϵ (x) dx = ∫ ψϵ (x)(−Δ) 2 (u2 )(x) dx ℝN
ℝN
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1

349
s
= ∫ ψϵ (x){2u(x)(−Δ) 2 u(x) − Bs (u, u)} dx. ℝN
Therefore s s 2 −s 2 Bs (u, ϕϵ,0 )2 ≤ Cϵ ∫ {−u (x)(−Δ) 2 ψϵ (x) + 2ψϵ (x)u(x)(−Δ) 2 u(x)} dx.
ℝN
As in the estimate of I2 , s s Cϵ−s ∫ −u2 (x)(−Δ) 2 ψϵ (x) dx ≤ C1 ϵ−s u2 N (−Δ) 2 ψϵ (x) N ≤ C1 u2 N ϵs . L 2s L N−2s L N−2s N ℝ
(11.4.49) N
Since u2 ∈ L N−2s (Ω), for all η > 0 there exists g ∈ 𝒞0∞ (Ω) such that 2 < η. N u − g N−2s (Ω) L Then the remaining term can be estimated as follows: s
2Cϵ−s ∫ ψϵ (x)u(x)(−Δ) 2 u(x) dx ℝN
s ≤ 2Cϵ−s (−Δ) 2 u2 ψϵ (x)u2
1
2 ≤ C1 ϵ ψϵ (x)u2 = C1 ϵ−s ( ∫ [(u2 − g)ψ2ϵ + gψ2ϵ ] dx)
−s
ℝN
≤ C2 ϵ−s [u2 − g
N
L N−2s 1
1 1 2 −s 2s N 2 ψϵ 2sN + ‖g‖∞ ‖ψϵ ‖2 ] 2 ≤ ϵ [C3 ϵ η + C4 ϵ ] 2 L
≤ [C3 η + C4 ϵN−2s ] 2 .
(11.4.50)
Putting together (11.4.48), (11.4.49) and (11.4.50) we find that 1
0 ≤ lim lim I3 ≤ C5 η 2 , ϵ→0 n→∞
and since η is an arbitrary positive number, we conclude. The limits of I4 , I5 and I6 are zero as an easy consequence of the previous analysis on the behavior of the three first terms and the Cauchy–Schwarz inequality. As a consequence we find that taking limits in (11.4.45), 2 2∗
μ0 − λγ0 ≥ SN,s (λ)ν0s
2 2∗
with γ0 ΛN,2s ≤ μ0 and SN,s ν0s ≤ μ0 .
On the other hand, by the hypotheses, lim lim ⟨𝒥θ (un ), un ϕϵ ⟩ = 0,
ϵ→0 n→∞
350  11 Fractional semilinear elliptic problems and therefore μ0 − λγ0 ≤ ν0 . N
As a consequence, either ν0 = 0 or ν0 ≥ SN,s (λ) 2s . Assuming that u ∈ H0s (Ω) and the weak limit of the sequence and that there exists a concentration point, by Lemma 11.4.10 we find N 1 s 1 c = lim 𝒥θ (un ) ≥ 𝒥θ (u) + ( − ∗ ) ∑ νj ≥ 𝒥θ (u) + SN,s (λ) 2s , n→∞ 2 2s N
that is, since c < c∗ = consequence,
N s S (λ) 2s N N,s
2∗ s
− Mθ 2∗s −q , 𝒥θ (u) < 0, and in particular u ≠ 0. As a
∗ 1 1 1 s c = lim (𝒥θ (un ) − ⟨𝒥θ (un ), un ⟩) = lim ( ∫ un 2s dx − ( − )θ ∫ uqn dx) n→∞ n→∞ N 2 q 2
Ω
Ω
N ∗ 1 1 s s ≥ ∫ u2s dx + SN,s (λ) 2s − ( − )θ ∫ uq dx. N N q 2
Ω
Ω
By Hölder’s inequality 2s −q 1 1 1 1 − )θ ∫ uq dx ≤ ( − )θΩ 2∗s ‖u‖q2∗ . s q 2 q 2 ∗
(
Ω
Note that the function in the statement, f (t) =
2s −q s 2∗s 1 1 t − ( − )θΩ 2∗s t q , N q 2 ∗
1
1
attains its negative minimum in ℝ+ at t0 = [( N−2s )θ( 2−q )] 2∗s −q Ω 2∗s , that is, 2s 2 2∗ s
f (t) ≥ f (t0 ) = −Mθ 2∗s −q
for all t > 0, where M = M(Ω, s, N, q).
Hence c≥
2s N s SN,s (λ) 2s − Mθ 2∗s −q N ∗
and we arrive at a contradiction with the hypothesis on c. Therefore we have ‖un ‖2∗s → ‖u‖2∗s
as n → ∞
and since un → u almost everywhere, by the result of Brezis–Lieb in Theorem 11.2.3 we conclude that ‖un − u‖2∗s → 0
as n → ∞.
Therefore, using the continuity of the fractional Laplacian from H −s (Ω) to H0s (ω) we find the Palais–Smale condition.
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
 351
Corollary 11.4.13. For all λ ∈ [0, ΛN,2s ), there exists a θ0 (λ) > 0 such that for every 0 < θ < θ0 (λ) there exists u0 ∈ H0s (Ω) which is a local minimum of 𝒥θ (u). Proof. We estimate from above the functional 𝒥θ (u) in the following elementary way. Call qλ (u) = (aN,s ‖u‖2H s − λ ∫ Ω
u2 dx). x2s
1 2
Recall that qλ (u) := ‖u‖2H s (Ω) is an equivalent norm in H0s (Ω). Since Sn,s (λ) is defined 0 by (11.4.36), we have 2 2∗
s ‖u‖22∗s ≤ qλ (u). SN,s
Therefore, 𝒥θ (u) ≥
2∗ q+1 s 1 1 q+1 1 θ q 2 . qλ (u) − ∗ qλ (u) 2 − Ω(1− 2∗ −s ) 2 2s SN,s (λ) q+1 SN,s λ
1
Putting r = qλ (u) 2 we have a radial minorant of 𝒥θ (u) in the Sobolev space, given by 𝒥θ (u) ≥
1 2 r − 2
1 2∗s S
r 2s − C(N, s, q, Ω) ∗
2∗ s 2
θ q+1 r := hλ,θ (r). q+1
For all 0 < λ < ΛN,2s and θ small enough it is easy to check that there exist 0 < rmin < r0 < rmax < r1 such that: – hλ,θ (rmin ) ≤ hλ,θ (r) < 0 for all 0 < r < r0 ; – hλ,θ (rmax ) ≥≤ hλ,θ (r) for all 0 < r < r1 ; – hλ,θ (r0 ) = hλ,θ (r1 ) = 0, hλ,θ (r) < 0 if r > r1 and limr→∞ hλ,θ (r) = −∞. Consider τ : ℝ+ → [0, 1], a nonincreasing function, τ ∈ 𝒞 ∞ (ℝ+ ) such that τ(r) = 1 if 0 < r ≤ r0 and τ(r) = 0 if r > r1 . Let ψ(u) := τ(qλ (u)) and consider the perturbed functional ∗ 1 1 θ F(u) = qλ (u) − ∗ ∫ ψ(u)u2s dx − ∫ uq+1 dx. 2 2s q+1
Ω
It is clear that F has the same regularity as 𝒥θ , that is, F ∈ C 1 (H0s (Ω), ℝ). Moreover, 1
(i) if F(u) ≤ 0, then qλ2 ≤ r0 and F(u) = 𝒥θ (u) in a neighborhood of u; (ii) there exists N
2∗ s
a θ1 > 0 such that for all 0 < θ < θ1 , Ns SN,s (λ) 2s − Mθ 2∗s −q > 0 and (iii) F(u) is bounded from below. Let c0 = infu∈H0s (Ω) F(u). Then c0 < 0 and any Palais–Smale sequence to this level has a strongly convergent subsequence. So there exists a u0 ∈ H0s (Ω) such that c0 = F(u0 ) is a minimum. Since F(u) = 𝒥θ (u) in a neighborhood of u0 , u0 is a local minimum for 𝒥θ and 𝒥θ (u0 ) = c0 < 0.
352  11 Fractional semilinear elliptic problems It is possible to read the result in Lemma 11.4.12 in a more accurate way to find the local Palais–Smale condition in order to have a second positive solution (see [178]). Lemma 11.4.14. Let {un }n∈ℕ ⊂ H0s (Ω) be a Palais–Smale sequence for 𝒥θ , namely, 𝒥θ (un ) → c < ∞, 𝒥θ (un ) → 0.
Suppose θ > 0 is small enough and take u0 ∈ H0s (Ω) such that c0 = 𝒥θ (u0 ) < 0 is a local minimum. If c satisfies c < c ∗ = c0 +
N N s s S (λ) 2s = 𝒥θ (u0 ) + SN,s (λ) 2s , N N,s N
then {un }n∈ℕ has a strongly convergent subsequence in H0s (Ω). Proof. We follow the arguments in [178]. First, note that as in the subcritical case {uj } is bounded in H0s (Ω). Then we obtain a subsequence, again relabeled by {uj }, such that it verifies the properties in Lemma 11.4.10 and Lemma 11.4.11. For 0 < λ < ΛN,2s , recall that ‖u‖2H s (Ω) := aN,s ‖u‖2H s (Ω) − λ ∫ 0
0
Ω
u2 x2s
by the Hardy inequality (9.2.1) is an equivalent norm in H0s (Ω). With this notation we can handle simultaneously the principal term and the term associated to the Hardy potential. By using again Lemma 11.4.10, it is easy to prove that the regular part verifies ⟨𝒥θ (u), u⟩ ≤ 0. Assuming the existence of a nontrivial singular part, we get a contradiction with N the hypothesis c < c0 + Ns SN,s (λ) 2s . Recall that u is the weak limit of the sequence uj . Then c0 +
N s 1 SN,s (λ) 2s > lim 𝒥θ (un ) = lim 𝒥θ (un ) − ∗ ⟨𝒥θ (un ), un ⟩ n n N 2s
= lim n
≥
1 1 s ‖un ‖2H s (Ω) − θ( − ∗ ) ∫ un q dx 0 N q 2s Ω
s 1 1 ‖u‖2H s (Ω) − θCq ( − ∗ )‖u‖q/2 . H0s (Ω) 0 N q 2s
In particular, as c0 < 0 we obtain N 1 1 s s ‖u‖2H s (Ω) − θCq ( − ∗ )‖u‖q/2 < SN,s (λ) 2s . s H (Ω) 0 0 N q 2s N
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
 353
N
Therefore, ‖u‖H0s (Ω) < α(θ)SN,s (λ) 4s , where α(θ) → 1 as θ → 0. Taking g(t) = (t ‖u‖u s
(i) g (‖u‖H0s (Ω) ) =<
H (Ω) 0
) we have:
𝒥θ (u), ‖u‖u s H (Ω) 0
>≤ 0;
(ii) g(‖u‖H0s (Ω) ) ≥ f (‖u‖H0s (Ω) ), where f (x) = 21 x2 −
2∗
s 1 S (λ)− 2s 2∗s N,s
x 2s − qθ Cq x q . ∗
The function f is a radial minorant of the functional 𝒥θ in the Sobolev space H0s (Ω) (see [177] for some applications of this argument in a different setting). N Moreover f (α(θ)SN,s (λ) 4s ) → Ns SN,s (λ)N/2s as θ → 0; therefore, for θ small enough N
we have f (α(θ)SN,s (λ) 4s ) > 0. Consider the set A defined by A = {t ≥ 0 
g (t) ≤ 0}.
Consider I1 , the connected component of A containing t = 0, and I2 , the intersection of A with the set where g(t) ≥ 0. Then ‖u‖H0s (Ω) ∈ I1 ∪ I2 . In any case, 𝒥θ (u) = g(‖u‖H s (Ω) ) ≥ co . 0
Then, if there are K Dirac masses we have c0 +
N N N s s s SN,s (λ) 2s > lim 𝒥θ (un ) ≥ 𝒥θ (u) + K SN,s (λ) 2s ≥ c0 + K SN,s (λ) 2s . n N N N
Therefore K = 0 and we get strong convergence. Step 2. The mountain pass lemma. The geometry is the same as in the subcritical case, so that for θ > 0 small enough we find a local minimum. Therefore the application of the mountain pass lemma in order to obtain a second positive solution depends of the existence of a Palais–Smale level verifying the smallness stated in Lemma 11.4.12. Step 3. The energy estimate. This step is just to construct a Palais–Smale sequence verifying the hypotheses of Lemma 11.4.14. At this point the sublinear term and the truncation of the minimizers are crucial in order to break down the N energy under the critical level c∗ = c0 + Ns SN,s (λ) 2s . Following [35], the idea is to take vε = u0 + tuε , where uε = ρ(x)Uε (x),
ρ(x) being a cutoff function centered at 0,
(11.4.51)
and Uε is a minimizer of (11.4.37), for which we have found the behavior given in (11.4.38). To complete the proof we need to estimate carefully the energy of this function vε . In the fractional case we have an extra difficulty given that we do not have the explicit expression of the minimizer but only the precise behavior, so the determination of his sLaplacian cannot be done directly. At this point we will
354  11 Fractional semilinear elliptic problems use the strategy developed by Luis Caffarelli and Luis Silvestre in [105], which we recall briefly. Consider the Poisson kernel ps (x, y) =
1 x k( ), N y y
where k(x) =
αN,s (1 + x2 )
N+2s 2
(11.4.52)
and αN,s > 0 is chosen in order to have ∫ k(x) dx = 1. ℝN
Given u ∈ H s (ℝN ), define U(x, y) =: ∫ ps (x − z)u(z)dz.
(11.4.53)
ℝN
Then one can check (see [105]) that U solves the following elliptic problem: {
− div(y1−2s ∇U) = 0
in ℝN+1 + ,
(11.4.54)
on ℝN ,
U(x, 0) = u(x)
and moreover the Dirichlet–Neumann transformation gives lim y1−2s
y→0
𝜕U (x, y) = −ds (−Δ)s u(x), 𝜕y
ds = 22s−1
Γ(s) . Γ(1 − s)
The last identity establishes the connection between the problem for the sLaplacian in ℝN and the weighted local problem (11.4.54) in ℝN+1 + . If we consider the weighted Sobolev space, X s (ℝN+1 ), as the completion of + 𝒞0∞ (ℝN+1 + ) with respect to the seminorm
1
‖U‖X s := (ds ∫ y
2 2 ∇U(x, y) dx dy) ,
1−2s
ℝN+1 +
then we have ‖U‖X s = ‖u‖H s .
(11.4.55)
Since we know the behavior given by (11.4.38) of the minimizers uϵ of (11.4.37), the isometric identity (11.4.55), it is sufficient to evaluate the behavior of ‖Uϵ ‖X s and then ‖uϵ ‖H s , where, as above, uε = ρ(x)Uε , with ρ(x) being a cutoff function centered at the point 0. Lemma 11.4.15. Let μλ be defined by (11.4.39) and N > 2s. The following estimates hold:
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
 355
(i) ‖uϵ ‖2H s (Ω) = ‖U1 ‖2H s (ℝN ) + 𝒪(ϵμλ (N−2s) ); 0
(ii)
u2 ∫Ω xϵ2 s
U12 dx x2 s N+2s , N−2s
dx = ∫ℝN
(iii) letting 0 < q ≤
+ 𝒪(ϵμλ (N−2s) );
N−2s
2N 1 (N−2s) (1+μλ ) 2N 1 − 1, (N−2s) (1+μλ ) 2N 1 − 1; (N−2s) (1+μλ )
0 < q < qλ =
Cϵ 2 (q+1)μλ , { { { { N−2s ∫ uϵ q+1 dx ≥ { CϵN− 2 (q+1) log( ϵ1 ), { { { Ω N− N−2s (q+1) 2 , { Cϵ
q = qλ = q > qλ =
− 1,
(iv) moreover, ∫Ω uϵ 2s dx = ∫ℝN U1 2s dx + 𝒪(ϵμλ N ). ∗
∗
Proof. See [284] to find the details. We are looking for an estimate of the type sup 𝒥θ (u0 + tuε ) < 𝒥θ (u0 ) + t>0
s S (λ)N/2s , N N,s
for ϵ > 0 small. Then for this value of ϵ we can choose vϵ = u0 + tuε , with t large enough, such that 𝒥θ (vϵ ) < c0 . Given v ∈ H0s (Ω) such that 𝒥θ (v) < c0 there exists a sequence {uj } ∈ H0s (Ω) such that (i)
𝒥θ (uj ) → 0,
(ii) 𝒥θ (uj ) → c = inf sup 𝒥θ (γ(t)), γ∈𝒞 t∈[0,1]
where 𝒞 = {γ ∈ C([0, 1], H0s (Ω))  γ(0) = u0 , γ(1) = v}. Therefore, to obtain a critical point we must find v ∈ H0s (Ω) such that c < c0 + Ns SN,s (λ)N/2s . Lemma 11.4.16. Suppose θ > 0 is small enough and take u0 ∈ H0s (Ω) such that c0 = 𝒥θ (u0 ) < 0 is a local minimum. Let uϵ be defined in (11.4.51). Then sup 𝒥θ (u0 + tuε ) < 𝒥θ (u0 ) + t>0
s S (λ)N/2s . N N,s
Proof. Note that by the scalar product in H0s (Ω), we have ‖u0 + tuϵ ‖2H s = ‖u0 ‖2H s + t 2 ‖uϵ ‖2H s + 2t ∫ ∫ ∫ Ω
u0 + tuϵ 2 u 2 dx = ∫ 02s dx + t 2 ∫ 2s x x Ω
Ω
(u0 (x) − u0 (y))(uϵ (x) − uϵ (y)) dx dy, x − yN+2s
ℝN ℝN uϵ 2 dx x2s
+ 2t ∫ Ω
u0 uϵ dx. x2s (11.4.56)
356  11 Fractional semilinear elliptic problems By using the elementary inequality for the critical exponent 2∗s = (1 + τ)2s ≥ 1 + τ2s + 2∗s τ + 2∗s τ2s −1 + Cτγ , ∗
∗
∗
2N , N−2s
τ ≥ 0,
valid for γ ∈ (1, 2∗s − 1], we can estimate from below the critical term. Indeed, ∫ u0 + tuϵ 2s dx ≥ ∫ u0 2s dx + t 2s ∫ uϵ 2s dx ∗
∗
Ω
∗
∗
Ω
Ω
+
∗ 2∗s t ∫ u0 2s −1 uϵ
Ω
2∗ −1
dx + 2∗s t ∫ u0 uϵs Ω
2∗ −γ γ uϵ .
+ C ∫ u0s Ω
(11.4.57)
For 0 < q < 1, the mean value theorem provides the inequality q ∫(u0 + tuε )q+1 dx ≥ ∫ uq+1 0 dx + (q + 1)t ∫ u0 uε dx.
Ω
Ω
(11.4.58)
Ω
Since u0 is a local minimum that is, in particular, a solution of problem (11.4.40), by inequalities (11.4.56), (11.4.57) and (11.4.58), we find ∗ u 2 1 ∗ t2 (aN,s ‖uϵ ‖2H s − λ ∫ ϵ2s dx) − ∗ t 2s ∫ uϵ 2s dx 2 2s x
𝒥θ (u0 + tuε ) ≤ 𝒥θ (u0 ) +
Ω
+
2∗ −1 2∗s t ∫ u0 uϵs
+
Ω
Ω
2∗ −γ C ∫ u0s uγϵ . Ω
We have to estimate the last two terms. Since 2∗s − 1 > qλ and u0 is bounded in the complementary of any ball centered at the origin, we find 2∗ −1
∫ u0 uϵs
dx ≥ Cϵ
N−2s μλ 2
,
Ω N N+2s according to Lemma 11.4.15. Taking a γ ∈ ( N−2s , N−2s ) we can have γ < qλ and then, again by the boundedness of u0 outside of the origin, 2∗ −γ γ uϵ
∫ u0s
Ω
≥ CϵN−
(N−2s) γ 2
,
by Lemma 11.4.15. Therefore we obtain 𝒥θ (u0 + tuε ) ≤ 𝒥θ (u0 ) + t
− Ct
2∗s −1
ϵ
21
2
(aN,s ‖U1 ‖2H s − λ ∫
N−2s μλ 2
+ o(ϵ
N−2s μλ 2
Ω
).
∗ U1 2 1 ∗ dx) − ∗ t 2s ∫ U1 2s dx 2s x2s
Ω
11.4 Existence of at least two nontrivial variational solutions if 1 < p ≤ 2∗s − 1
 357
To arrive to the strict inequality let us consider the function N−2s ∗ ∗ U 2 1 1 ∗ hϵ (t) := t 2 (aN,s ‖U1 ‖2H s − λ ∫ 12s dx) − ∗ t 2s ∫ U1 2s dx − Ct 2s −1 ϵ 2 μλ . 2 2s x
Ω
Ω
Putting, as above, qλ (u) = 21 (aN,s ‖U1 ‖2H s − λ ∫Ω
2
U1  x2s
dx), we can write
N−2s ∗ ∗ t 2s t2 hϵ (t) = qλ (U1 ) − ∗ ∫ U1 2s dx − Ct 2s −1 ϵ 2 μλ . 2 2s ∗
Ω
For ϵ = 0, h0 (t) achieves a maximum at t0 = [ minimizer of (11.4.37), h0 (t0 ) =
qλ (U1 )
2∗ ‖U1 ‖2s∗ s
1
] 2∗s −2 and moreover, since U1 is a
N s SN,s (λ) 2s . N N s S (λ) 2s , N N,s
For all t > 0, ϵ > 0 we have hϵ (t) < h0 (t) ≤
and hence
max hϵ (t) = hϵ (tϵ ) < h0 (tϵ ) ≤ h0 (t0 ) = t>0
N s SN,s (λ) 2s . N
Note that the point tϵ where hϵ achieves its maximum satisfies 0 < tϵ < t0 and tϵ → t0 as ϵ → 0. Therefore we write tϵ = xϵ t0 with xϵ → 1 as ϵ → 0. To finish the proof we have to prove that the error term is negligible. Since we have hϵ (tϵ ) = 0, we get 2∗ −1 2∗ −1
2∗
2∗ −2 2∗ −2
t0 xϵ qλ (U1 ) − t0s xϵs ‖U1 ‖2s∗ = C(2∗s − 1)t0s xϵs ϵμλ s
N−2s 2
.
Including the precise value of t0 and after some computations we arrive at 2∗ −2
1 − xϵs
2∗ −3 μλ N−2s 2
= Axϵs
ε
,
where A = A(N, s, λ, ‖U1 ‖2∗s ). By Taylor’s expansion, 2∗ −3
(1 − xε )(2∗s − 2)xε s N−2s
N−2s
Therefore, 1 − xε = Mεμλ 2 + o(εμλ 2 ), for M = Finally, this identity allows us to prove hε (tε ) =
2∗ −3 μλ N−2s 2
+ o(1 − xε ) = Axεs
ε
.
A . 2∗s −2
N−2s N−2s s N/2s 2∗ −1 S − Ct0s εμλ 2 + o(εμλ 2 ), N
which shows the strict inequality. An immediate consequence of the previous results is the following result.
358  11 Fractional semilinear elliptic problems Theorem 11.4.17. For every λ ∈ [0, ΛN,2s ), there exists a θ0 (λ) such that for all 0 < θ < θ0 (λ), problem (11.4.40) has at least two positive variational solutions of finite energy. Proof. From Corollary 11.4.13, for θ < θλ the functional 𝒥θ attains a local minimum in u0 in such a way that 𝒥θ (u0 ) < 0. Then by Lemma 11.4.16 we find a Palais–Smale sequence under the concentration level and then by construction we find a mountain pass critical point that allows us to conclude the proof. Remark 11.4.18. The result of existence of at least two positive solutions in the whole range of θ for which we find a minimal solution will be considered in a forthcoming work, with slightly different techniques.
11.5 Nonexistence for p ≥ p(λ, s): complete blowup The main objective in this section is to prove that for p ≥ p(λ, s) problem (11.1.1) has no weak solutions. To prove the nonexistence result we will use in a suitable way the Picone inequality of Lemma 8.5.2. Theorem 11.5.1. Let Ω be a bounded domain of ℝN , such that 0 ∈ Ω. Assume that 0 < λ ≤ ΛN,2s and p ≥ p(λ, s). Then, there does not exist any positive weak solution in the sense of Definition 11.2.2 to problem (11.1.1). Proof. We argue by contradiction. Assume that there exists u > 0, a weak solution to problem (11.1.1). We can approximate u by a sequence of approximate solutions, for instance, the solutions uk ∈ H0s (Ω) to the sequence of problems u
(Pk ) = {
(−Δ)s uk − λ xk2s = Tk (u)p + μTk (u)q uk = 0
in Ω, in ℝN \ Ω.
(i) We consider first the case λ < ΛN,2s . By the Hardy inequality and a slight modification of the weak maximum principle proved in Theorem 8.3.11 we obtain u1 < u2 < ⋅ ⋅ ⋅ < uk < ⋅ ⋅ ⋅ < u, and then we can conclude that u(x) = limk→∞ uk (x) pointwise and in Lp (Ω). (a) Assume p > p(λ, s). By Lemma 10.3.8 there exists C > 0 such that u(x) ≥
C , xγ
for all x ∈ Br (0),
N−2s−2α
λ with γ = . By the Picone inequality applied to uk and to any ϕ ∈ 2 ∞ 𝒞0 (Br (0)) we find
∫ Br (0)
Tk (u)p 2 (−Δ)s uk 2 ϕ dx ≤ ∫ ϕ ≤ ‖ϕ‖2H s (Ω) , 0 uk uk Br (0)
11.5 Nonexistence for p ≥ p(λ, s): complete blowup 
359
and therefore passing to the limit as k → ∞ and by the local estimate of u, it follows that C ∫ Br (0)
ϕ2 dx ≤ ∫ up−1 ϕ2 dx ≤ ‖ϕ‖2H s (Br (0)) . 0 x(p−1)γ Br (0)
Since p > p(λ, s), (p − 1)γ > (p(λ, s) − 1)γ = ( = 2s.
N + 2s − 2αλ N − 2s − 2αλ 4s − 1)γ = ( ) N − 2s − 2αλ N − 2s − 2αλ 2
But then, the inequality (p − 1)γ > 2s is a contradiction with the Hardy inequality. (b) If p = p(λ, s) the argument above does not give a contradiction in an easy way. The idea is to obtain a stronger singularity to reach the contradiction. To be systematic, the calculations are divided into several steps. 1 )) = Step 1. For 0 < s < 1, (−Δ)s (log( x
aN,s C0 x2s
1 ) yN+2s N+2s y
log y(1 −
C0 = 2 ∫ y>1
x −
with dy > 0.
Note that ∫ (log ℝN
where x =
1 1 y 1 dy dy = − log ) ∫ log y N+2s , x y x − yN+2s xN+2s x x − x  ℝN
x x
∈ 𝕊N−1 . By the change of variable z = ∫ (log ℝN
y x
we obtain
C 1 1 dy − log ) = 0 , x y x − yN+2s x2s
where C0 = ∫ log y ℝN
x
dy , − yN+2s
which is constant on x =∈ 𝕊N−1 . We check next the sign of C0 . We have C0 = ∫ log z ℝN
dz dy dz = ∫ log y + ∫ log y . x − yN+2s x − yN+2s x − yN+2s y>1
y≤1
360  11 Fractional semilinear elliptic problems The first term is positive. With respect to the second term, changing to polar coordinates we obtain 1
∫ log y
t N−1 dy = ∫ ∫ log t dt dy N+2s x − y x − ty N+2s 𝕊N−1 0
y≤1
1
= ∫ ∫ log t 𝕊N−1 0 ∞
= ∫ ∫ log 𝕊N−1 1
t 2s  xt
y≤1
Therefore C0 = 2 ∫y>1
x −yN+2s
)
−
y N+2s
dt dy y
1 r 2s−N dr dy r x − ry N+2s
= − ∫ log y 1 log y(1− N+2s y
1
x
dy . − yN+2s
dy > 0.
1 1 Step 2. If 0 < s < 1 and log+ ( x ) = max{log x , 0}, then
(−Δ)s (log+ (
0 1 )) ≤ { C 0 x 2s x
if x > 1 if x < 1.
It is sufficient that the function ψ(u) = u+ is convex and Lipschitz, in fact, ψ (u) = χ(0,∞) . Hence (−Δ)s (log+ (
C 1 1 1 )) ≤ ψ (log )(−Δ)s (log ) = 02s χ(0,1) . x x x x
Step 3. According to (8.1.17), if two functions verify that u(x) ≤ u(y) if and only if v(x) ≤ v(y), then (−Δ)s (uv) ≤ u(−Δ)s v + v(−Δ)s u. Note that the condition holds if both functions are radial monotone. Then for w(x) =
1 1 ), (log+ γ x x
γ=
N − 2s − 2αλ , 2
we find 1 1 1 1 ) log+ + (−Δ)s (log+ ) xγ x x xγ C0 λ ≤ 2s w(x) + γ+2s χ{y 2s + γ. Consider w(x) = Aw with A > 0 such that AC0 < C1p . Then (−Δ)s w(x) −
C1p AC0 λ λ w(x) ≤ ≤ ≤ up (x) ≤ (−Δ)s u(x) − 2s (x), x2s xγ+2s x2s+γ x
in x ∈ B1 (0). Since w(x) = 0 ≤ u(x) in ℝN ⊂ B1 (0) we conclude that u(x) ≥ w(x) = A
1 1 (log+ ). xγ x
As in the previous case for ϕ ∈ 𝒞′∞ (B1 (0)), by the application of the Picone inequality we find Ap(λ,s)−1 ∫ B1 (0)
p(λ,s)−1
ϕ2 1 (log+ ) x x2s
which contradicts the Hardy inequality.
dx ≤ ∫ up−1 ϕ2 dx ≤ ‖ϕ‖2H s (B1 (0)) , B1 (0)
0
362  11 Fractional semilinear elliptic problems (ii) Now we consider the case λ = ΛN,2s . This case is easier. Assume that there exist a positive solution u to problem (11.1.1) with λ = ΛN,2s and p ≥ p(ΛN,2s , s). By Theorem 10.4.10, in particular, we must have ∫ Br (0)
up dx < ∞, xγ
r > 0.
But by Lemma 10.3.8 there exists C > 0 such that u(x) ≥ and therefore ∫ Br (0)
C , xγ
for all x ∈ Br (0),
1 up dx ≥ C p ∫ dx, xγ x(p+1)γ Br (0)
and since for ΛN,2s , p(ΛN,2s , s) =
N+2s N−2s
(p + 1)γ ≥
and γ =
N−2s , 2
we have
2N N − 2s = N, N − 2s 2
which contradicts the necessary integrability condition. Remark 11.5.2. Nonexistence results, using the Caffarelli–Silvestre extension in [105], were proved in [152, Theorem 0.2]. Our proof is more direct, using the natural homogeneities associated to the problem. 11.5.1 Complete blowup As a byproduct of the nonexistence result we are able to prove complete blowup, which is a common fact in all the nonexistence results associated to the Hardy potential. That is, we will see that the solutions un of the truncated problems verify that un (x) → ∞ when n → ∞ and x ∈ Ω. Before proving the complete blowup result, we need the following auxiliary lemma, which is an extension to the fractional semilinear framework of Proposition 6.2.5 (see the original in [87, Lemma 3.2]). Lemma 11.5.3. Let F(x, u) ≥ 0 in L∞ (Ω), and let u be the solution of (−Δ)s u = F(x, u) in Ω, (PF ) = { u=0 in ℝN \ Ω. Then, u(x) ≥ C ∫ F(x, u)δs (x) dx, δs (x)
x ∈ Ω,
Ω
where δ(x) = dist(x, 𝜕Ω) and C is a constant depending only on Ω.
(11.5.2)
11.5 Nonexistence for p ≥ p(λ, s): complete blowup
 363
Proof. First of all we will prove (11.5.2) for points that belong to a compact set K ⊂ Ω. Let x0 ∈ K. By the mean value inequality in Proposition 11.2.4 (see [287]), we have u(x0 ) ≥ ∫ u(x)γr (x − x0 ) dx = ∫ u(x)γr (x − x0 ) dx > 0, Ω
ℝN
for any x0 ∈ Ω, r ≤ dist(x0 , 𝜕Ω) and γr = (−Δ)s Γr , where Γr is a C 1,1 paraboloid that matches outside the ball B(0, r) with the fundamental solution. Then there exist a positive constant c > 0 and a compact set K ⊆ Ω such that u(x0 ) > c for every x0 ∈ K. That is, u(x0 ) > M ∫ u(x) dx,
x0 ∈ K,
(11.5.3)
Ω
where −1
M = c(∫ u(x) dx) . Ω
Consider now the function ξ1 such that (−Δ)s ξ1 = 1
(PF=1 ) = {
ξ1 = 0
in Ω,
(11.5.4)
in ℝN \ Ω.
Using ξ1 as a test function in (PF ), by (11.5.3) we obtain u(x0 ) ≥ M ∫ u(x) dx = M ∫ F(x, u)ξ1 (x), Ω
x0 ∈ K.
(11.5.5)
Ω
Then, by Hopf’s lemma (see [273, Lemma 3.2]), u(x0 ) ≥ C ∫ F(x, u)δs (x) dx,
x0 ∈ K.
Ω
Moreover, since c1 ≤ δs (x0 ) ≤ c2 for x0 ∈ K, we have u(x0 ) ≥ C̃ ∫ F(x, u)δs (x) dx, δs (x0 )
x0 ∈ K.
Ω
Let now w satisfy (−Δ)s w = 0 { { { w=0 { { { { w=1
in Ω \ K, in ℝN \ Ω, in K.
(11.5.6)
364  11 Fractional semilinear elliptic problems We define v(x) =
u(x) , ̃ C ∫Ω F(x, u)δs (x) dx
x ∈ ℝN .
Therefore (−Δ)s v ≥ 0 { { { v=0 { { { { v≥1
in Ω, in ℝN \ Ω, in K.
Then, by the strong maximum principle of Section 11.2.2, we have v(x) ≥ w(x) for x ∈ Ω \ K. Since by Hopf’s lemma w(x) ≥ Cδs (x), it follows that u(x0 ) ≥ C ∫ F(x, u)δs (x) dx, δs (x0 )
x0 ∈ Ω \ K.
(11.5.7)
Ω
Hence, by (11.5.6) and (11.5.7), we obtain the desired estimate given in (11.5.2). We state now the blowup result. Theorem 11.5.4. Assume that 0 < λ ≤ ΛN,2s . Let p ≥ p(λ, s) and μ > 0. Then, there exists complete blowup of the problem (Pμ ). Proof. We argue by contradiction. Consider the minimal solution un ≥ 0 to the truncated problem (Pn ) = {
(−Δ)s un = λan (x)Tn (un ) + gn (un ) + μfn (un ) un = 0
in Ω, in ℝN \ Ω,
where an (x) := Tn (
1 ), x2s
gn (u) := Tn (up+ ) and
fn (u) := Tn (uq+ ),
with Tn (x) := {
x
if x ≤ n,
x n x
if x > n.
Note that we can affirm that this minimal solution exists because, since λan (x)Tn (un ) + gn (un ) + μfn (un ) ≤ Cn , we can consider, for a suitable c > 0, u := Cn ξ1 and u := cφ1 as wellordered superand subsolutions of (Pn ), respectively. Here φ1 is the nonnegative first eigenfunction of the fractional Laplacian defined in (11.3.3) and ξ1 is given in (11.5.4).
11.6 Hardy potential and nonlinear term singular at the boundary  365
We suppose that ∫ (λan (x)Tn (un ) + gn (un ) + μfn (un ))δs (x) dx ≤ C < ∞,
n ∈ ℕ,
Ω
with C independent of n. Using ξ1 given in (PF=1 ) as a test function in problem (Pn ), by [273, Proposition 1.1], we obtain ∫ un = ∫ un (−Δ)s ξ1 ℝN
ℝN
= λ ∫ an Tn (un )ξ1 + ∫ gn (un )ξ1 + μ ∫ fn (un )ξ1 Ω
≤ C.
Ω
Ω
Hence, up to a subsequence, {un } converges in L1 (Ω) to a nonnegative limit u. Then, since an (x)Tn (un ) + gn (un ) + μfn (un ) increases to xu2s + up + μuq in Ω, by the monotone convergence theorem we can pass to the limit in (Pn ) obtaining a nonnegative weak solution of the problem {
(−Δ)s u − λ xu2s = up + μuq
u=0
in Ω, in ℝN \ Ω.
But this is a contradiction with the nonexistence results above (see also [151, Theorem 0.2]) and, therefore, we conclude applying Lemma 11.5.3. Indeed, since un (x) ≥ C ∫ (λan Tn (un ) + gn (un ) + μfn (un )) δs (x)dx, δs (x) Ω
hence un (x) → ∞, as n → ∞, for all x ∈ Ω.
11.6 Problems with the Hardy potential and nonlinear terms singular at the boundary The results in this section have some partial precedents in [1] and are also applicable to the local case. The aim will be to study the problem u h(x) { (−Δ)s u = λ 2s + σ { { { u x { { u>0 { { { { { u=0
in Ω, in Ω,
(11.6.1)
in ℝN \ Ω,
where h is a nonnegative function, σ > 0 and λ ≥ 0. As we pointed out in Remark 10.4.11 it can be easily checked that problem (11.6.1) has no positive solution for λ > ΛN,2s . Hence we will assume λ ≤ ΛN,2s .
366  11 Fractional semilinear elliptic problems Remark 11.6.1. Call μ := −σ. We know by the previous sections that problem (11.6.1) has no positive solution for μ > p+ (λ), where γ is defined in (9.4.9). Therefore, finding a solution of (11.6.1) can be seen as proving that there is no lower threshold for the power to solve the semilinear problem. The main result in this section is the following. Theorem 11.6.2. Assume that σ ≥ 1 and λ ≤ ΛN,2s . Then, for all h ∈ L1 (Ω), problem (11.6.1) has a positive weak solution. More precisely, 1. if σ = 1, then u ∈ H0s (Ω) when λ < ΛN,2s , and u ∈ W0s,q (Ω) for all q < 2 if λ = ΛN,2s ; 2.
σ+1
s if σ > 1, then u ∈ Hloc (Ω) with Gk (u) ∈ H0s (Ω) and Tk 2 (u) ∈ H0s (Ω). Moreover if 4σ [ (σ+1) 2 −
λ ] ΛN,2s
> 0, then u
σ+1 2
∈ H0s (Ω).
Proof. Let hn := Tn (h), the usual truncation of h, and define un to be the unique positive solution to the approximated problem u
(−Δ)s un = λ xn2s + { { { { un > 0 { { { { { un = 0
hn (x) (un + n1 )σ
in Ω, (11.6.2)
in Ω, in (ℝN \ Ω).
The existence follows by minimization and the uniqueness by using the result in Lemma 10.2.23. Since Tn (h) is an increasing function in n, again by Lemma 10.2.23 we conclude that {un }n∈ℕ is an increasing function in n. We divide the proof in two cases. First case: σ = 1 and λ < ΛN,2s . Taking un as a test function in (11.6.2) and using the Hardy inequality, we obtain aN,s h u λ (1 − )‖un ‖2H s (Ω) ≤ ∫ n n1 dx ≤ ∫ h dx = C. 0 2 ΛN,2s un + n
Ω
Ω
Thus {un }n∈ℕ is bounded in H0s (Ω) and then there exists u ∈ H0s (Ω) such that, up to a subsequence, un ⇀ u weakly in H0s (Ω) and un ↑ u strongly in Lη (Ω) for all η < 2∗s . Since (−Δ)s un ≥ 0, using the monotonicity of {un }n∈ℕ and the compactness lemma, Lemma 10.2.24, we easily obtain that un → u strongly in H0s (Ω). Hence we conclude that u solves problem (11.6.1). Second case: σ > 1 and λ < ΛN,2s . Using now Gk (un ) as a test function in (11.6.2) we have aN,s ∬ DΩ
Gk (un (x)) − Gk (un (y))2 u G (u ) h G (u ) dx dy − λ ∫ n k 2s n dx ≤ ∫ n k 1 n dx x x − yN+2s (un + n )σ Ω
and ∫ Ω
hn Gk (un ) (un +
1 σ ) n
dx ≤
Ω
1 ∫ h dx. k σ−1 Ω
11.6 Hardy potential and nonlinear term singular at the boundary  367
Moreover, un Gk (un ) = Gk2 (un ) + kGk (un ), and thus aN,s ∬ DΩ
Gk2 (un ) Gk (un (x)) − Gk (un (y))2 dx dx dy − λ ∫ x2s x − yN+2s Ω
G (u ) 1 ≤ λk ∫ k 2sn dx + σ−1 ∫ h dx. k x Ω
Ω
Since λ < ΛN,2s , by the Hardy–Sobolev inequality we obtain C∬ DΩ
G (u ) Gk (un (x)) − Gk (un (y))2 dx dy ≤ λk ∫ k 2sn dx + C(k, h), x x − yN+2s Ω
and applying the Young and Hardy–Sobolev inequalities in the integral on the right hand side we reach ∬ DΩ
Gk (un (x)) − Gk (un (y))2 dx dy ≤ C(k, λ, ΛN,2s , h). x − yN+2s
Therefore {Gk (un )}n∈ℕ is uniformly bounded in H0s (Ω), and again by the Hardy– Sobolev inequality, ∫ Ω
Gk2 (un (x)) dx ≤ C(k, λ, ΛN,2s , h). x2s
Then we get ∫ Ω
Tk2 (un (x)) Gk2 (un (x)) u2n (x) T (u )G (u ) dx = dx + dx + 2 ∫ k n 2sk n dx ≤ C(k, λ, ΛN,2s , h). ∫ ∫ 2s 2s 2s x x x x Ω
Ω
Ω
Likewise, testing with Tkσ (un ) in (11.6.2), it follows that aN,s ∬ DΩ
(Tkσ (un (x)) − Tkσ (un (y)))(un (x) − un (y)) dx dy x − yN+2s
≤ λ∫ Ω
un Tkσ (un ) hn Tkσ (un ) dx dx + ∫ x2s (un + 1 )σ
≤ k σ−1 λ ∫ Ω
u2n x2s
Ω
n
dx + ∫ hn dx ≤ C(k, λ, ΛN,2s , h), Ω
and applying Lemma 10.2.27 we conclude σ+1
∬ DΩ
σ+1
(Tk 2 (un (x)) − Tk 2 (un (y)))2 x − yN+2s
dx dy ≤ C(k, λ, ΛN,2s , h, σ).
368  11 Fractional semilinear elliptic problems σ+1
Thus {Tk 2 (un )}n∈ℕ is bounded in H0s (Ω). Furthermore, the strong maximum principle provides that un ≥ u1 ≥ c(K) > 0,
for any compact set K ⊂ Ω.
s Claim. The sequence {Tk (un )}n∈ℕ is bounded in Hloc (Ω). Since {un }n∈ℕ is an increasing sequence, we have Tk (un ) ≥ Tk (u1 ), and for all Ω ⊂⊂ Ω, u1 ≥ C(Ω ). Thus,
Tk (un ) ≥ min{k, C(Ω )} =: C0 . T (u (x))
Tk (un (y)) . It is clear that αn , βn C0
For (x, y) ∈ Ω × Ω , we define αn := k Cn and βn := 0 Therefore the following inequality holds: σ+1
σ+1
2
(αn − βn )2 ≤ (αn 2 − βn 2 ) .
≥ 1.
(11.6.3)
Indeed, if αn = βn the estimate is trivial. Otherwise, without loss of generality we can β assume αn > βn ≥ 1. Let 0 < x := αn < 1. Since σ > 1, we easily obtain n
0≤1−x ≤1−x
σ+1 2
,
and hence (1 − x)2 ≤ (1 − x
σ+1 2
2
).
Clearly αn2 < αnσ+1 , and thus αn2 (1 − x)2 ≤ αnσ+1 (1 − x
σ+1 2
2
).
Recalling the definition of x, (11.6.3) follows. Finally, by the definition of αn and βn , we conclude that for (x, y) ∈ Ω × Ω , we have 2
σ+1
σ+1
2
(Tk (un (x)) − Tk (un (y))) ≤ C0 1−σ (Tk 2 (un (x)) − Tk 2 (un (y))) . σ+1
Thus the claim follows using the boundedness of {Tk 2 (un )}n∈ℕ in H0s (Ω). Hence combining the estimates above, we obtain that {un }n∈ℕ is bounded in s s Hloc (Ω) and then, up to a subsequence, there exists u ∈ Hloc (Ω) such that s un ⇀ u weakly in Hloc (Ω),
Gk (un ) ⇀ Gk (u) weakly in H0s (Ω) σ+1 2
σ+1 2
Tk (u) ⇀ Tk (u) weakly in H0s (Ω).
and
11.6 Hardy potential and nonlinear term singular at the boundary  369
Applying the compactness result in Lemma 10.2.25 we obtain that un → u strongly in s Hloc (Ω). Thus un ↑ u strongly in Lr (Ω) for all 1 ≤ r < 2∗s . Let ϕ ∈ 𝒯 , where 𝒯 was defined in (8.6.2). Testing with ϕ in (11.6.2), it follows that ∫(−Δ)s un ϕ dx = λ ∫ Ω
Ω
un ϕ ϕhn (x) dx. dx + ∫ 2s x (un + 1 )σ
(11.6.4)
n
Ω
By the estimates above we reach that when n goes to +∞, λ∫ Ω
un ϕ ϕhn (x) uϕ ϕh(x) dx → λ ∫ 2s + ∫ +∫ dx < +∞. 1 2s σ uσ x x (un + ) Ω
n
Ω
Ω
Moreover, we have ∫(−Δ)s un ϕ dx = ∫ un (−Δ)s ϕ dx → ∫ u(−Δ)s ϕ dx, Ω
Ω
Ω
as n → +∞. Therefore, passing to the limit in (11.6.4), ∫(−Δ)s uϕ dx = λ ∫ Ω
Ω
h(x)ϕ uϕ dx + ∫ dx, 2s uσ x Ω
i. e., u is a weak solution. 4σ Finally, for every λ < ΛN,2s take σ such that  (σ+1) 2 −
function in (11.6.2), it follows that aN,s ∬ DΩ
λ  ΛN,2s
> 0. By using uσn as a test
uσ+1 (un (x) − un (y))(uσn (x) − uσn (y)) n dx dy ≤ λ dx + ∫ hn dx. ∫ x2s x − yN+2s Ω
Ω
By Lemma 10.2.27, we get (un (x) − un (y))(uσn (x) − uσn (y)) ≥
σ+1 σ+1 4σ 2 2 2 (u (x) − u (y)) , n n (σ + 1)2
and hence, by Hardy’s inequality, σ+1
σ+1
(u 2 (x) − un 2 (y))2 4σ λ aN,s ( − )∬ n dx dy ≤ ‖h‖L1 (Ω) . 2 ΛN,2s (σ + 1) x − yN+2s DΩ
Therefore u
σ+1 2
∈ H0s (Ω) and this is the sense how the boundary value is reached. 2∗ s
We now deal with the case σ < 1. If h ∈ L( 1−σs ) (Ω), the existence of a positive energy solution can be proved proceeding as in the case σ = 1. However, our goal from now on will be to study the solvability when h has less regularity. Indeed, we have the following result.
370  11 Fractional semilinear elliptic problems Theorem 11.6.3. Assume σ < 1, λ < ΛN,2s and h ∈ L1 (Ω, x−(1−σ)γ dx), h ⪈ 0. Then problem (11.6.1) has at least a weak solution. Proof. Let {hn }n∈ℕ be such that hn ≥ 0 and hn ↑ h strongly in L1 (Ω). Define un as the unique positive solution to the approximated problem (11.6.2). Then by setting vn := xγ un , it follows that vn satisfies Lγ (vn ) = x−γ
hn (x) −γ (x vn + n1 )σ
≤ x−(1−σ)γ
hn vnσ
in Ω,
(11.6.5)
where Lγ was defined in (10.2.3). Using vnσ as a test function in (11.6.5) we obtain σ+1
σ+1
hn (v 2 (x) − vn 2 (y))2 dx dy 4σ aN,s dx ≤ C, ≤ ∫ (1−σ)γ ∬ n 2 xγ yγ (σ + 1) x − yN+2s x DΩ
Ω
with C independent of n.
σ+1
Thus we conclude that the sequence {vn 2 }n∈ℕ is bounded in the weighted Sobolev s space Wγ,0 (Ω) and hence there exists v0 such that, up to a subsequence, σ+1
σ+1
vn 2 ⇀ v0 2
s weakly in Wγ,0 (Ω) σ+1
< 1, also Lγ (vn 2 ) ≥ 0. Hence by the monoSince Lγ (vn ) ≥ 0, using the fact that σ+1 2 tonicity of vn and Lemma 10.2.24, we obtain σ+1
σ+1
vn 2 → v0 2
s strongly in ϕ ∈ Wγ,0 (Ω).
Passing to the limit in (11.6.5), it follows that v0 solves −(1−σ)γ h { Lγ (v0 ) = x v0σ { { v0 = 0
in Ω, in ℝN \ Ω, u
in the weak sense. Defining u0 := x−γ v0 , then λ x02s ∈ L1 (Ω) and u0 is a weak solution of problem (11.6.1). To end this section, we consider the problem in the whole space, that is, Ω = ℝN . Then we will work in the space Ḣ s (ℝN ) defined by the completion of 𝒞0∞ (ℝN ) with respect to the Gagliardo seminorm 1
[ϕ]Ḣ s (ℝN ) = ( ∫ ∫ ℝN ℝN
2 (ϕ(x) − ϕ(y))2 dx dy) . N+2s x − y
We obtain the following existence result.
11.6 Hardy potential and nonlinear term singular at the boundary  371
Theorem 11.6.4. Consider the problem {
(−Δ)s u = λ xu2s +
h(x) uσ
u>0
in ℝN ,
(11.6.6)
in ℝN .
Then: (i) If σ = 1, then for all h ∈ L1 (ℝN ), problem (11.6.6) has a solution u ∈ Ḣ s (ℝN ). (ii) If σ > 1, then for all h ∈ L1 (ℝN ), problem (11.6.6) has a weak solution u such that σ+1 Gk (u) ∈ Ḣ s (ℝN ) and T 2 (u) ∈ Ḣ s (ℝN ), for all k > 0. Moreover, if k
λ 4σ − > 0, (σ + 1)2 ΛN,2s
(11.6.7)
σ+1
then u 2 ∈ Ḣ s (ℝN ). 2∗s ) , then problem (11.6.6) has a solution u such (iii) If σ < 1 and h ∈ Lm (ℝN ) with m = ( 1−σ that u ∈ Ḣ s (ℝN ). (iv) If σ < 1 and (11.6.7) holds, then for all h ∈ L1 (ℝN ) problem (11.6.6) has a weak σ+1 solution u such that u 2 ∈ Ḣ s (ℝN ). Proof. Consider un to be the unique positive solution to the approximated problem u
(−Δ)s un = λ 2sn 1 + { x + n { { { un > 0 { { { { { un = 0
hn (x) (un + n1 )σ
in Bn (0), in Bn (0),
(11.6.8)
in ℝN \ Bn (0).
It is clear that un is increasing with n. If σ = 1, taking un as a test function in (11.6.8) and using the Hardy–Sobolev inequality it follows that aN,s λ (1 − )[un ]2Ḣ s (ℝN ) ≤ C. 2 ΛN,2s Hence {un }n∈ℕ is uniformly bounded in Ḣ s (ℝN ) and then, up to a subsequence, un ⇀ u weakly in Ḣ s (ℝN ), where u solves (11.6.6). Using the monotonicity of un and a straightforward adaptation of Lemma 10.2.24 we can prove that un → u strongly in Ḣ s (ℝN ), which proves (i), and (iii) similarly follows. To prove (ii) we take Gk (un ) as a test function in (11.6.8) and performing the same computations as in the proof of Theorem 11.6.2 we conclude. Finally, (iv) follows closely using the arguments in the proof of Theorem 11.6.2. In particular, the existence of u ∈ L1 (Ω) is a consequence of the uniform bounds of σ+1 {Gk (un )}n∈ℕ and {T 2 (un )}n∈ℕ in Ḣ s (ℝN ) and the fact that σ+1 < 1. k
2
372  11 Fractional semilinear elliptic problems Remark 11.6.5. In a similar way to the results in [46] the problem (−Δ)s u = λ xu2s + μ h(x) + up { uσ { { { u>0 { { { { { u=0
in Ω, in Ω,
(11.6.9)
in ℝN \ Ω
can be analyzed. In fact the existence of a minimal solution for p < p+ (λ) and λ ≤ ΛN,2s and μ small can be done with minor analytical changes. Theorem 11.6.6. Assume that in problem (11.6.9) one of the following conditions holds: 2∗s (i) σ < 1, p < 2∗s − 1 and h ∈ Lλ (Ω) for some λ ≥ 2∗ −(1−σ) ; 2∗s
2∗ s 2∗ s −(1−σ1 )
s
(ii) σ ≥ 1, p < − 1 and h ∈ L (Ω) for some σ1 < 1; (iii) σ > 0, 2∗s − 1 ≤ p < p+ (λ) and h ∈ L∞ (Ω). Then there exists μ∗ > 0 such that for all μ < μ∗ , problem (11.6.9) has a minimal weak solution u and, moreover, for all μ > μ∗ , problem (11.6.9) has no positive solution. The existence of a second positive solution in cases (i) and (ii) for μ small enough is easy to obtain. The result for all μ < μ∗ by a direct method seems to be an open problem.
11.7 Further comments 11.7.1 Other operators The results above could be studied also for a larger class of linear operators like those studied in Chapter 8. Another case that seems very interesting comes from considering the fractional pLaplacian operator and the corresponding Hardy inequality. See [168]. The techniques to attack this problem may have some mathematical interest for applications.
12 The heat equation with fractional diffusion 12.1 Introduction This chapter is dedicated to the evolution problem associated to the nonlocal elliptic operators studied in Chapter 8, that is, we will study the following parabolic problem: ut + ℒu = f in QT , { { { { { u(x, t) = 0 in (ℝN \ Ω) × [0, T), { { { { { {u(x, 0) = u0 (x) in Ω × {0},
(12.1.1)
where ℒ is given by (8.3.9) with a kernel as in Definition 8.3.2, u0 and f are functions defined in suitable Lebesgue spaces, T > 0 and Ω is a bounded subset of ℝN , N ≥ 2, with Lipschitz boundary. We will study in particular the behavior of the problem when ℒ = (−Δ)s , giving a complete proof of the behavior of the kernel and some related results. This kind of problems appears in different situations. They arise, for instance, in probabilistic processes in which a particle moves randomly in the space subject to a probability that allows long jumps with a polynomial tail. See for instance [98]. This type of problems also appears in fluid mechanics, mathematical economy and, in general, in problems with a fractionallike diffusion. We refer to the introduction in Chapter 8 for motivations and references.
12.2 Finite energy setting As in the elliptic case, we are interested in the existence, uniqueness and summability of the solution u with respect to the summability of the datum f . First, we define the meaning of a finite energy solution. Definition 12.2.1. We say that u ∈ L2 (0, T; H0s (Ω)) ∩ 𝒞 ([0, T], L2 (Ω)) with ut ∈ L2 (0, T; H −s (Ω)) is a finite energy solution for the parabolic problem ut + ℒu = f in QT , { { { { { u(x, t) = 0 in (ℝN \ Ω) × [0, T), { { { { { {u(x, 0) = u0 (x) in Ω × {0}, with f ∈ L2 (0, T; H −s (Ω)) and u0 ∈ L2 (Ω) if it satisfies T
T
T
∫ ∫ ut w dx dt + ∫ ℰ (u(x, t), w(x, t)) dt = ∫⟨f , w⟩ dt, 0 Ω 2
for any w ∈ L
(0, T; H0s (Ω)),
0
0
where ℰ (⋅, ⋅) is the bilinear form defined in (8.3.10).
https://doi.org/10.1515/9783110606270012
374  12 The heat equation with fractional diffusion We recall the following classical result. Theorem 12.2.2. Let H be a Hilbert space such that dense
V → H → V . Let u ∈ Lp (a, b; V) be such that ut , defined in the distributional sense, belongs to Lp (a, b; V ). Then u belongs to C([a, b]; H). For the proof see, for instance, [133, Chapter XVIII, Section 2, Theorem 1]. The particular case p = 2 can be seen in [148, Theorem 3 in Section 5.9]. Remark 12.2.3. Observe that we ask u to belong to 𝒞 ([0, T], L2 (Ω)) in order to have the identity u(x, 0) = u0 (x) in the L2 (Ω)sense. Anyway, Theorem 12.2.2 guarantees that if u ∈ L2 (0, T; H0s (Ω)) and ut ∈ L2 (0, T; H −s (Ω)), then u ∈ 𝒞 ([0, T]; L2 (Ω)). The existence and uniqueness of an energy solution to problem (12.1.1) for f in the dual space L2 (0, T; H −s (Ω)) and u0 ∈ L2 (Ω) can be proved by means of a direct abstract Hilbert space approach. For the reader’s convenience, we include here the proof inspired by the one of A. N. Milgram (see [242]) for the local case, which is based on a method by Vishik (see [312] and [226]). Theorem 12.2.4. Assume that f ∈ L2 (0, T; H −s (Ω)). Then for any u0 ∈ L2 (Ω) problem (12.1.1) has a unique finite energy solution. Moreover if f is also a nonnegative function and u0 ≥ 0, such a solution is nonnegative too. Proof. Let 𝒞∗∞ (Ω × [0, T]) denote the 𝒞 ∞ (Ω × [0, T]) functions that vanish in (ℝN \ Ω) × [0, T] and in Ω × {T}. Consider ϕ ∈ 𝒞∗∞ (Ω × [0, T]), u ∈ L2 (0, T; H0s (Ω)), and define the operator T
T
Lϕ (u) := ∫ ∫ −uϕt dx dt + ∫ ℰ (u(x, t), ϕ(x, t)) dt. 0 Ω
0
Note that u is an energy solution to (P) with f ∈ L2 (0, T; H −s (Ω)) if and only if T
Lϕ (u) = ∫⟨f , ϕ⟩ dt + ∫ u(x, 0)ϕ(x, 0) dx. 0
Ω
We also define the following inner product: T
1 ⟨φ, ϕ⟩∗ = ⟨φ(x, 0), ϕ(x, 0)⟩L2 (Ω) + ∫ ℰ (φ(x, t), ϕ(x, t)) dt, 2
(12.2.1)
0
and denote by H ∗ (Ω×[0, T]) the Hilbert space built as the completion of 𝒞∗∞ (Ω×[0, T]) with the norm ‖ϕ‖∗ induced by the inner product (12.2.1).
12.2 Finite energy setting
 375
Observe that for any φ ∈ L2 (0, T; H0s (Ω)), by the Hölder and Sobolev inequalities, we have −1 Lϕ (φ) ≤ cϕ (‖φ‖L2 (0,T,L2 (Ω)) + λ ‖φ‖L2 (0,T,H0s (Ω)) ) ≤ c̃ϕ ‖φ‖∗ . Therefore, Lϕ is a linear continuous functional in H ∗ (Ω × [0, T]), and by the Fréchet– Riesz theorem, there exists 𝒯 ϕ ∈ H ∗ (Ω × [0, T]) such that Lϕ (φ) = ⟨φ, 𝒯 ϕ⟩∗
for all φ ∈ H ∗ (Ω × [0, T]).
Since 𝒯 is a linear operator in H ∗ (Ω × [0, T]), it is injective too. Moreover, T
1 Lϕ (ϕ) = ∫ ϕ2 (x, 0) dx + ∫ ℰ (ϕ(x, t), ϕ(x, t)) dt = ‖ϕ‖2∗ , 2 0
Ω
and consequently, ⟨ϕ, 𝒯 ϕ⟩∗ = ‖ϕ‖2∗ . Thus, by the Cauchy–Schwarz inequality, ‖ϕ‖2∗ ≤ ‖ϕ‖∗ ‖𝒯 ϕ‖∗ ,
i. e.,
‖ϕ‖∗ ≤ ‖𝒯 ϕ‖∗ .
Therefore, this implies that 𝒯 is bijective and its inverse 𝒯 −1 has a norm less than or equal to 1, and can be extended to the closure M of Rank(𝒯 ). On the other hand, let us define T
Bu0 ,f (ϕ) := ∫ u0 ϕ(x, 0) dx + ∫ ∫ ϕf dx dt. 0 Ω
Ω
Denoting ϕ0 := ϕ(x, 0), by the Hölder inequality we have Bu0 ,f (ϕ) ≤ ‖u0 ‖L2 (Ω) ‖ϕ0 ‖L2 (Ω) + ‖f ‖L2 (0,T;L2 (Ω)) ‖ϕ‖L2 (0,T;L2 (Ω)) ≤ cu0 ,f ,λ ‖ϕ‖∗ , and thus, −1 −1 Bu0 ,f (𝒯 ψ) ≤ c𝒯 ψ∗ ≤ c‖ψ‖∗ . Therefore, by applying the Fréchet–Riesz theorem again, there exists a unique u ∈ M such that Bu0 ,f (𝒯 −1 ψ) = ⟨ψ, u⟩∗ for every ψ ∈ M. Calling ϕ = 𝒯 −1 ψ this means Bu0 ,f (ϕ) = ⟨𝒯 ϕ, u⟩∗ = Lϕ (u), that is, T
T
T
∫ ∫ −uϕt dx dt + ∫ ℰ (u(x, t), ϕ(x, t)) dt = ∫ ∫ fϕ dx dt + ∫ u(x, 0)ϕ(x, 0) dx, 0 Ω
0
0 Ω
Ω
376  12 The heat equation with fractional diffusion where ϕ ∈ L2 (0, T; H0s (Ω)) and ϕt ∈ L2 (0, T; H −s (Ω)). Finally, by a density argument, one can conclude, integrating by parts, that in fact u ∈ L2 (0, T; H0s (Ω)), ut ∈ L2 (0, T; H −s (Ω)) and T
T
T
∫ ∫ ut ϕ dx dt + ∫ ℰ (u(x, t), ϕ(x, t)) dt = ∫ ∫ fϕ dx dt. 0 Ω
0
0 Ω
Thus u(x, t) is an energy solution of (P).
12.3 Nonvariational setting: weak solutions We define the notion of weak solution just asking the regularity needed to give distributional sense to the equation. Let us consider the following class of test functions: 1
β
𝒫 (QT ) = {ϕ(⋅, t) ∈ C ([0, T), 𝒞0 (Ω)), a solution to (P)},
(12.3.1)
−ϕt + ℒϕ = ψ in QT , { { { { { where (P) = {ϕ(x, t) = 0 in (ℝN \ Ω) × [0, T), { { { { in Ω, {ϕ(x, T) = 0 with ψ ∈ 𝒞0∞ (QT ). Note that the existence of a regular solution is guaranteed by the result by Felsinger and Kassmann in [169] (for a kernel 𝒦 more general than ours). Definition 12.3.1. We say that u ∈ C([0, T]; L1 (Ω)) is a weak solution to (12.1.1) for f ∈ L1 (QT ) if for all ϕ ∈ 𝒫 (QT ), ϕ = 0 on (ℝn \ Ω) × [0, T) and ϕ(x, T) = 0 in Ω, we have ∫ uψ dx dt = ∫ fϕ dx dt + ∫ u0 (x)ϕ(x, 0) dx. QT
fine
QT
Ω
Henceforth, let us introduce the following notations: for any measurable v we deAk (v) = {(x, t) ∈ QT : v(x, t) > k},
and for a. e. t ∈ (0, T), Atk (v) = {x ∈ Ω : v(x, t) > k}.
(12.3.2)
Theorem 12.3.2. Let f ∈ L1 (QT ) and u0 ∈ L1 (Ω). Then there exists a unique weak solution to (12.1.1). Moreover, ∀k ≥ 0,
Tk (u) ∈ L2 (0, T; H0s (Ω)),
u ∈ Lq (QT ), and
∀q ∈ (1,
(12.3.3)
N + 2s ) N
s N + 2s r ). (−Δ) 2 u ∈ L (QT ), ∀r ∈ (1, N +s
(12.3.4)
12.3 Nonvariational setting: weak solutions  377
Remark 12.3.3. Observe that if s = 1 the above results correspond to the classical ones for second order parabolic problems (see for instance [67]). Proof. Uniqueness. Let w be a very weak solution of (12.1.1) with f = 0 and u0 = 0, i. e., wt + ℒw = 0 in QT , { { { { { w=0 in (ℝN \ Ω) × [0, T), { { { { { in Ω; {w(x, 0) = 0 we want to prove that w ≡ 0. Consider F ∈ 𝒞0∞ (QT ), and let ϕF be the solution of the backward problem −(ϕF )t + ℒϕF = F { { { { { ϕF (x, t) = 0 { { { { { {ϕF (x, T) = 0
in QT , in (ℝN \ Ω) × [0, T),
(12.3.5)
in Ω.
Choosing ϕF as a test function we deduce that for any F ∈ 𝒞0∞ (QT ), T
∫ ∫ wF dx dt = 0, 0 Ω
which means w = 0 in 𝒟 (QT ). Existence. We obtain the solution as a limit of solutions to approximated problems. Let us consider un ∈ L2 (0, T; H0s (Ω)) ∩ L∞ (QT ), the solution to the approximated problems (un )t + ℒun = fn in Ω × (0, T), { { { { { un (x, t) = 0 in (ℝN \ Ω) × (0, T), { { { { { {un (x, 0) = un0 (x) in Ω,
(12.3.6)
where fn = Tn (f (x, t)) and un0 = Tn (u0 (x)) are smooth functions. Step 1. There exists a positive constant c, depending only on N, Ω and s, such that ‖un ‖Lq (QT ) ≤ c‖fn ‖L1 (QT ) ,
∀q ∈ (1,
N + 2s ). N
(12.3.7)
We take Tk (un ) as a test function in (12.3.6) and, by a standard argument, we deduce that there exists a positive constant c, just depending on Ω and ‖fn ‖L1 (QT ) , such that ‖un ‖L∞ (0,T,L1 (Ω)) ≤ c
and Tk (un )L2 (0,T,H s (Ω)) ≤ ck. 0
378  12 The heat equation with fractional diffusion Observe moreover that by interpolation we have (1−θ)σ
σ ∫Tk (un ) dx ≤ (∫Tk (un ) dx)
Ω
Ω
Ω
with θ ∈ (0, 1) such that
1 σ
=1−θ+
θ . 2∗s
This and the Sobolev inequality lead to (1−θ)σ
σ ∫Tk (un ) dx ≤ c(∫Tk (un ) dx) Ω
σθ ∗
2s 2∗ (∫Tk (un ) s dx) ,
Ω
σθ
(∫Tk (un ) dx + Tk (un )H s (Ω) ) , 0 Ω
for a c > 0. Choosing θσ = 2, σ =
2(N+s) , N
after integration over (0, T), we get
T
σ ∫ ∫Tk (un ) dx dt ≤ ck. 0 Ω
Therefore, k σ m{(x, t) ∈ Ω × (0, T), un (x, t) ≥ k} ≤ CM k, N+2s
and then un is bounded in the Marcinkiewicz space ℳ N (QT ). Here m(A) denotes the Lebesgue measure of the set A in the corresponding Euclidean space. Step 2. There exists a positive constant c, depending on q, N, Ω and s, such that s q (−Δ) 2 un Lq (QT ) ≤ c‖fn ‖L1 (Q ) , T
∀q ∈ (1,
N + 2s ). N +s
(12.3.8)
For any λ > 0, we have s {(x, t) ∈ QT : (−Δ) 2 un ≥ λ} s s = {(x, t) ∈ QT : (−Δ) 2 un ≥ λ, un  < k} ∪ {(x, t) ∈ QT : (−Δ) 2 un ≥ λ, un  ≥ k},
so that s s {(x, t) ∈ QT : (−Δ) 2 un ≥ λ} ⊂ {(x, t) ∈ QT : (−Δ) 2 un ≥ λ, un  < k} ∪ Atk (un ),
where Atk (un ) has been defined in (12.3.2). We have s m({(x, t) ∈ QT : (−Δ) 2 un ≥ λ, un  < k})
T
1 ≤ 2∫ λ
∫
0 {(x,t)∈ΩT ,un  0 ‖fn ‖L1 (QT ) s k ) m({(x, t) ∈ QT : (−Δ) 2 un ≥ λ}) ≤ 2 ‖fn ‖L1 (QT ) + c( k λ N
N+2s N
. s
Minimizing in k we find that the minimum is achieved in k = λ N+s ‖f ‖LN+s 1 (Q ) . Then the T above inequality becomes N+2s
‖f ‖L1 (QT ) N+s s m({(x, t) ∈ QT : (−Δ) 2 un ≥ λ}) ≤ c( ) . λ
(12.3.9)
Step 3. Passing to the limit. We have obtained in step 1 that the sequence {un } is bounded in the Marcinkiewicz N+2s space ℳ N (QT ). Using the linearity of the operator, we conclude that the sequence N+2s un a. e. converges to a function u in QT . By the embedding of ℳ N (QT ) in Lq (QT ) for any q < N+2s and the uniqueness of the solution, the whole sequence un converges to N u in Lq (QT ). Then, it follows that u is a weak solution to (12.1.1). As in the elliptic case, from (12.3.9) and the linearity of the operator we conclude s s that the sequence (−Δ) 2 un converges a. e. in QT to (−Δ) 2 u. Therefore, from step 2, using Fatou’s lemma, u verifies (12.3.4)
12.4 A priori estimates and summability of the solutions In this section we deal with the summability of the solutions that we have found in the previous section when the right hand side has extra integrability. More precisely, we will study the optimal summability of the solution in terms of the summability of the data. Let us represent the summability of the datum f ∈ Lr (0, T; Lq (Ω)) in a diagram with axes q1 and 1r . Since r, q ∈ [1, +∞], all possible cases of summability are inside of the square [0, 1] × [0, 1] (we use the notation
1 ∞
= 0).
12.4.1 Bounded solutions N The curve 1r + 2qs = 1 will be called the Aronson–Serrin (AS) curve to be consistent with the local case. The zone under the AS curve, called AS zone, corresponds to the set
380  12 The heat equation with fractional diffusion 1
2 6 1 L L L L L L 1 L L r L L L L L 2 L 1 L L 2 L L L AS L L 1 L L 2s N
1 2
1 2
1
N+2s 2N

1 q
Figure 12.4.1: The Aronson–Serrin (AS) zone is delimitated by the line line, in zone 2, we have the straight line line
1 r
=
N 1 N−2 q
−
2 N−2
N 2qs
+
1 r
= 1+
N 4s
1 r
+
N 2qs
= 1. Parallel to this AS
for r < 2. Zones 1 and 2 are separated by the
with r > 2. Finally, zone 1 is delimited on the right by the line q =
2N . N+2s
of data for which we find bounded solutions; in the case of second order differential operators (as, for instance, the Laplacian) such a result was obtained by D. G. Aronson and J. Serrin in [39]. The solutions for zones 1 and 2 are not expected to be bounded, but we can determine their summability in some Lebesgue space (for s = 1, the local case, we recover the regularity results in [81]). Let us consider the following approximated problems: unt + ℒun = fn { { { { { un (x, t) = 0 { { { { { {un (x, 0) = un0
in QT , in (ℝN \ Ω) × [0, T),
(12.4.1)
in Ω × {0},
where fn (x, t) = Tn (f (x, t)) and f ≥ 0 in a suitable Lebesgue space. Observe that since fn is bounded, the existence of a sequence {un } of energy solutions to (12.4.1) is guaranteed by Theorem 12.2.4. The main result in this paragraph is the L∞ estimate for un . Theorem 12.4.1. Assume f ∈ Lr (0, T; Lq (Ω)) with r, q satisfying 1 N + k > 0. Then Ah ⊂ Ak , where Ah is as in (12.3.2), and μ(h) ≤
c μ(k)1+χ . (h − k)r ̂
Applying Lemma 8.4.2, we conclude that there exists a constant d, depending only on q, r, ‖f ‖Lr (0,T;Lq (Ω)) and s, such that μ(d) = 0, that is, ‖un ‖L∞ (Ω×[0,τ]) ≤ d. Iterating this procedure in the sets Ω × [τ, 2τ], . . . , Ω × [jτ, T], where T − jτ ≤ τ, we can conclude that ‖un ‖L∞ (QT ) ≤ C
uniformly in n ∈ ℕ.
There are several interesting consequences of the above estimate; here we stress that it directly implies existence of a solution, with the definition of solution changing according to the summability of f . Corollary 12.4.2. Let f be a function f ∈ Lr (0, T; Lq (Ω)) with r, q satisfying N 1 + < 1. r 2qs Then, the solution obtained as limit of the solutions to (12.4.1) satisfies:
(12.4.5)
12.4 A priori estimates and summability of the solutions  383
1. 2. 3.
‖u‖L∞ (QT ) ≤ c; ‖u‖L2 (0,T;H s (Ω)) ≤ c; and 0 moreover, (a) if r ≥ 2, then ut ∈ L2 (0, T; H −s (ℝN )), and u is an energy solution; (b) if 1 < r < 2, then the equation is satisfied in the following weak sense: T
− ∫ uϕt dx dt + ∫ ℰ (u, ϕ) dt = ∫ fϕ dx dt + ∫ u(x, 0)ϕ(x, 0) dx, 0
QT
QT
Ω
for any ϕ ∈ L2 (0, T; H0s (Ω)) ∩ L∞ (QT ) ∩ 𝒞 1 ([0, T); L1 (Ω)), with ϕ(x, T) = 0. Proof. We first observe that since the fn are increasing, by the weak comparison principle (using that the equation is linear) we deduce that the sequence un is increasing, too. Thus, the uniform estimate gives (1). Moreover, testing equations (12.4.1) with un , part (2) easily follows. ∗ Noting that if r ≥ 2, then f ∈ L2 (0, T; L(2s ) (Ω)), we obtain (3a). Finally (3b) is obtained by a simple integration by parts argument in (12.4.1) and passing to the limit. Going further into the summability of the solution, we consider the borderline N case in which 1r + 2qs = 1. Having in mind what happens in the elliptic case, we prove that solutions are not anymore bounded, but have an exponential summability. Namely, we prove the following result. Theorem 12.4.3. Assume that f belongs to Lr (0, T; Lq (Ω)), with r ≥ 1, q >
N 2s
such that
N 1 + = 1, r 2qs
(12.4.6)
and let un be the solution of (12.4.1) with un0 (x) ≡ 0. Then there exists an α > 0 such that αun e L∞ (0,T;L2 (Ω)) ≤ c‖f ‖Lr (0,T;Lq (Ω)) .
(12.4.7)
Consequently, un is bounded in Lp (QT ), for every p ≥ 1. Proof. We only sketch the proof, since it is very similar to the elliptic one that we have proved in Section 8.4 (see Theorem 8.4.3). Consider the following auxiliary convex function: Φk (σ) = {
eασ − 1 αk
αe (σ − k) + e
if 0 ≤ σ < k, αk
−1
if σ ≥ k
for k > 0 and for some α to be fixed later. Thus we have T
1 2 + λ‖un ‖2L2 (0,T;H s (Ω)) ≤ ∫ ∫ Φk (un )Φk (un )f dx dt. Φ (u ) ∞ 2 0 2 k n L (0,T;L (Ω)) 0 Ω
384  12 The heat equation with fractional diffusion Now, observe that for 0 ≤ un ≤ k we have Φk (un ) = α(eαun − 1 + 1) = αΦk (un ) + α. Hence, T
∫ ∫ Φk (un )Φk (un )f dx dt 0 Ω
T
T
2
≤ α ∫ ∫ Φk (un ) f + α ∫ ∫ Φk (un )f + αe 0 {un ≤k}
0 {un ≤k}
αk
T
∫ ∫ Φk (un )f . 0 {un >k}
From now on, we follow the same ideas as in the elliptic case, and we get (12.4.7).
12.4.2 Summability of the solutions outside of the Aronson–Serrin zone Outside the AS zone, the solutions are not expected to be bounded. We obtain the following summability results for zones 1 and 2 of Figure 12.4.1. Theorem 12.4.4. Assume un0 (x) ≡ 0 and f ∈ Lr (0, T; Lq (Ω)) with r > 1, q > 1, satisfying 1
1 such that 1 1 C ≤ Pα (x, 1) ≤ , C (1 + xN+α ) (1 + xN+α )
for all x ∈ ℝN .
(12.5.6)
To prove Theorem 12.5.2 we will use a celebrated result by S. N. Bernstein (see [59]) about the characterization of completely monotone functions via Laplace transform. To try to make the presentation selfcontained we will show the proof of the Bernstein result and then we will recall the Helly selection principle [198], which is a classical tool used in harmonic analysis, probability theory and real analysis. With our current perspective, this can be seen as a Rellich compactness result in the space of bounded variation functions. We will formulate the result as we will need it, and later we will give the general version. Lemma 12.5.3 (The Helly selection principle). Let {ρn }n∈ℕ be a uniformly bounded sequence of non decreasing functions in a given interval I ⊂ ℝ. Then there exists a subsequence converging pointwise to an increasing function. Proof. We apply the Cantor diagonal principle to obtain a subsequence {ρnk }nk ∈ℕ ⊂ {ρn }n∈ℕ that converges at any rational r ∈ I ∩ ℚ. As every ρnk is non decreasing, the limit function ρ : I ∩ℚ → ℝ is also non decreasing. To extend ρ to the whole interval I we proceed as follows: given a point x ∈ I \ ℚ we consider an increasing sequence of rational numbers rk → x as
12.5 The Cauchy problem for the fractional heat equation
 391
k → ∞ and define ρ(x) = lim ρ(rk ); k→∞
therefore ρ is well defined in I and again is non decreasing. Hence the discontinuities of ρ are finite jumps. The points of discontinuity of ρ are contained in the set ∞
D = ⋃ Dn , 1
where Dn = {x ∈ I, ρ(x+ ) − ρ(x− ) ≥
1 }. n
Note that since ρ is bounded, each Dn must be finite, and then D is at most countable. To finish we have to prove that there exists a subsequence that converges pointwise in I. If x is a continuity point of ρ, then for rl < x < rm rational numbers we find ρnk (rl ) − ρ(rm ) ≤ ρnk (x) − ρ(x) ≤ ρnk (rm ) − ρ(rl ). Choosing rl , rm rational numbers such that for ϵ > 0 ρ(rm ) − ρ(rl ) < ϵ, we obtain −ϵ ≤ lim inf(ρnk (x) − ρ(x)) ≤ lim sup(ρnk (x) − ρ(x)), nk →∞
nk →∞
and since ϵ is arbitrary we conclude the proof. Assume, to finish, that for x1 ∈ D, by the Bolzano–Weierstrass theorem there exists a subsequence of the previous subsequence that converges in x1 and since D is numerable, iterating the Cantor process we obtain a subsequence that converges in each x ∈ D. Recall that if I = [a, b] is a bounded interval in the real line, a function f :I→ℝ
is of bounded variation if and only if f (x) = p(x) − n(x),
where p and n are monotone functions (see for instance [275]). As customary we denote by BV(I) the class of bounded variation functions in I = [a, b]. For a partition 𝒫 = {a = x0 < x1 < ⋅ ⋅ ⋅ < xn = b}
k
we define v𝒫 (f ) = ∑ f (xi ) − f (xi−1 ) 1
and the total variation of f ∈ BV(I) by VI (f ) = sup{v𝒫 (f )  𝒫
partition of I}.
We can formulate now the general version of the Helly principle. Theorem 12.5.4. Let {fm }n∈ℕ be a sequence in BV(I) such that: (i) There exists C1 such that supI fm (x) < C1 for all m ∈ ℕ. (ii) There exists C2 such that VI (fm ) ≤ C2 for all m ∈ ℕ.
392  12 The heat equation with fractional diffusion Then there exist a function f ∈ BV(I) and a subsequence {fnk }nk ∈ℕ ⊂ {fn }n∈ℕ such that lim f (x) nk →∞ nk
= f (x),
for all x ∈ I.
Proof. We find a couple of sequences {pm }m∈ℕ and {nm }m∈ℕ such that fn = pm − nm . From Lemma 12.5.3 we find a subsequence {pmk }mk ∈ℕ ⊂ {pm }m∈ℕ and a monotone function p such that pmk (x) → p(x)
as mk → ∞.
Starting now with the sequence {nmk }mk ∈ℕ , again from Lemma 12.5.3 we find a subsequence {nmk }mk ∈ℕ ⊂ {nmk }mk ∈ℕ and a monotone function n(x) such that j
j
nmk (x) → n(x) j
as mkj → ∞ for all x ∈ I.
Therefore, for all x ∈ I, fmk (x) = pmk (x) − nmk (x) → f (x) = p(x) − n(x), j
j
j
as mkj → ∞,
and hence f ∈ BV(I). Some applications of the Helly selection principle to the study of trigonometic series can be found in the classical book by A. Zigmund [322]. More general details about the above result understood as a compactness result can be found in [149]. We return to the study of the result by Bernstein by introducing the precise meaning of completely monotone function. Definition 12.5.5. Let g : [0, ∞) → [0, ∞) be a continuous function which is additionally in 𝒞 ∞ ((0, ∞)). We say that g is completely monotone when ∀n ≥ 0, ∀t > 0,
(−1)n g (n) (t) ≥ 0.
In particular if g is completely monotone g (0) := g ≥ 0 on [0, ∞), g (1) := g ≤ 0 on (0, ∞), g (2) := g ≥ 0 on (0, ∞), etc. And moreover, g(+∞) := limt→+∞ g(t) = 0 exists and g is bounded. Theorem 12.5.6 (Bernstein). The following statements are equivalent: 1. g is completely monotone, that is, ∀n ≥ 0, ∀t > 0, 2.
(−1)n g (n) (t) ≥ 0;
g is the Laplace transform of a nonnegative finite Borel measure μ on ℝ+ : ∞
∀t ∈ ℝ+ ,
g(t) = ∫ e−xt dμ(x). 0
12.5 The Cauchy problem for the fractional heat equation
 393
Proof. That (2) implies (1) follows from the fact that g
(n)
n
∞
(t) = (−1) ∫ xn e−tx dμ(x). 0
Now we will prove that (1) implies (2). Since (−1)n g (n) is nonnegative and nonincreasing, we have for any t > 0 and n ≥ 1, t
2 (n) n (n) n (n) g (t) = (−1) g (t) ≤ ∫ (−1) g (u) du = t t/2
2 (n−1) (t) − g (n−1) (t/2). g t
By induction, it follows then that ∀n ≥ 1,
g (n) (t) = o(t −n ),
as t → +∞.
By integration by parts (the boundary terms vanish thanks to the former result), ∞
g(x) − g(+∞) = − ∫ g (t) dt x
.. .
∞
(−1)n+1 = ∫ (t − x)n g (n+1) (t) dt n! x
n+1
∞
(−1) n n = ∫ (1 − x/(tn)) (nt)n g (n+1) (nt) dt n! x/n
∞
t
0 ∞
0
(−1)n+1 n = ∫ φn (x/t) d( ∫(nu)n g (n+1) (nu) du) n! = ∫ φn (xt) dσn (t), 0
where φn (x) := (1 − x/n)n+ and where ∞
σn (t) :=
1 ∫ (−1)n+1 n(nu)n g (n+1) (nu) du. n! 1/t
Integration by parts again gives ∞
∞
0
0
1 (−1)n ∫ xn g (n+1) (x) dx = ∫ x n−1 g (n) (x) dx n! (n − 1)!
394  12 The heat equation with fractional diffusion .. .
∞
= − ∫ g (x) dx = g(0) − g(+∞). 0
Therefore, the total variation of σn on [0, +∞) is ∞
1 ∫ n(nt)n g (n+1) (nt) dt = g(0) − g(+∞). n! 0
By Lemma 12.5.3, there exists a subsequence σnk (t) that converges almost everywhere to a bounded nonnegative nondecreasing function σ(t) on [0, +∞). Since φn (x) → e−x uniformly on [0, ∞) as n → ∞, it follows that ∞
g(x) − g(+∞) = ∫ e−tx dσ(t). 0
Finally, if we set μ := σ + g(+∞)δ0 , δ0 being the Dirac measure concentrated at 0, then we get for all x ∈ [0, +∞), ∞
g(x) = ∫ e−tx dμ(t), 0
as we want to prove. We will also need the following. Corollary 12.5.7. Assume that g(t) is completely monotone and define, for ξ ∈ ℝN , the function F(ξ ) = g(ξ 2 ). Then, the inverse Fourier transform of F is given by a positive, radially decreasing function. Proof. The proof is a simple consequence of Bernstein’s theorem. By hypothesis, there exists a finite Borel measure μ so that ∞
2
F(ξ ) = g(ξ 2 ) = ∫ e−ξ  s dμ(s). 0
Using the Fubini theorem, we have P(x) = (2π)
−N/2
∫ F(ξ )e
ix⋅ξ
∞
dξ = (2π)
ℝN ∞
=∫ 0
2 1 e−x /4s dμ(s). (2s)N/2
This clearly gives the result.
−N/2
2
∫ ∫ e−ξ  s eix⋅ξ dξ dμ(s) 0 ℝN
(12.5.7)
12.5 The Cauchy problem for the fractional heat equation
In particular, for the η = ηα , the inverse Laplace Transform of e−A Pα (x, 1) = (2π)
∫e
−N
i⟨x,ξ ⟩−ξ α
∞
dξ = ∫ 0
ℝN
 395
α/2
2 1 − x 4s η(s) ds. e (4πs)N/2
So, Pα (x, 1) > 0 for all x ∈ ℝN . Moreover, the corollary tells us that Pα (x, 1) is radially decreasing and that 0 < Pα (x, 1) ≤ Pα (0, 1),
∀x.
Therefore we have Pα (x, t) ≤ Pα (y, t),
if x > y.
By the scaling N
Pα (t, x) = t − α Pα (
x
1
tα
, 1),
for t > s we find N
s α Pα (x, t) ≥ ( ) Pα (x, s). t Before we go to the proof of Theorem 12.5.2 let us present another result that will be used later. Corollary 12.5.8. Assume that Pα (0, t) ≤ 1 and δ ≥ 2. Then Pα (
(x − y) , t) ≥ Pα (x, t)Pα (y, t). δ
Proof. It is clear that the numerical inequality 1 2 2 x − y ≤ max{ x, y} ≤ max{x, y} δ δ δ holds. Hence 1 Pα ( x − y, t) ≥ Pα (max{x, y}, t) ≥ Pα (x, t)Pα (y, t), δ where the last inequality is a consequence of the radial decrease of the kernel Pα and the hypothesis Pα (0, t) ≤ 1. The conclusion of Theorem 12.5.2 is thus a direct consequence of the positivity of the kernel, which the Bernstein theorem provides, and Theorem 12.5.1 by Polya (N = 1) and Blumenthal–Getoor (N ≥ 1). Let us prove the latter in a straightforward manner.
396  12 The heat equation with fractional diffusion γ
Proposition 12.5.9. Given as before 0 < γ < 1, g(t) = e−t , if we denote by Pγ (x) the in2γ
verse Fourier transform of F(ξ ) = g(ξ 2 ) = e−ξ  , then Pγ satisfies the following asymptotic behavior: lim xN+2γ Pγ (x) =
x→∞
γ N + 2γ Γ( ). Γ(1 − γ) 2
Proof. Using that g (n) → 0 as t → ∞, ∀n = 0, 1, . . . , integration by parts and an induction argument we can write ∞
∞
g(t) = − ∫ g (y) dy = ∫ (y − t)g (y) dy
t ∞
t ∞
(−1)n 1 = ∫ (y − t)2 g (y) dy ⋅ ⋅ ⋅ = ∫ (y − t)n g (n+1) (y) dy 2 n! t
t
n+1
=
(−1) n!
y=n/u
=
n
∞
t ∫ (1 − ) yn g (n+1) (y) dy y t
n/t
n
∞
n (−1)n+1 tu nn+1 ∫ (1 − ) n+2 g (n+1) ( ) du = ∫ En (tu)ηn (u) du, n! n u u 0
(12.5.8)
0
where En (s) = (1 −
n
s ) n +
and ηn (s) =
kγ
(−1)n+1 nn+1 (n+1) n 1 n+1 n+1 n 1 n g ( ) = ∑ c ( ) g( ). n! sn+2 s n! k=1 k s s s
(12.5.9)
Let us also set Gn (s) = s1+γ ηn (s).
(12.5.10)
Observe that Gn (s) = (An + with An =
c1n+1 n!
1 n+1 n+1 n ∑c ( ) n! k=2 k s
(k−1)γ
n )g( ), s
and, therefore, Gn is bounded and lim Gn (s) = An .
s→∞
(12.5.11)
12.5 The Cauchy problem for the fractional heat equation
 397
We recall, from (12.5.8), that we have for all n = 1, 2, . . . ∞
g(t) = ∫ En (tu)ηn (u) du.
(12.5.12)
0
Differentiating both sides we get ∞
∞
0
0
u u −γt γ−1 g(t) = − ∫ Ẽn (tu)uηn (u) du = − ∫ Ẽn (u) 2 ηn ( ) du, t t where Ẽn (u) = (1 − un )n−1 + . Hence, ∞
1 u g(t) = ∫ Ẽn (u)u−γ Gn ( ) du. γ t 0
Letting t → 0+ and using the dominated convergence theorem, we get ∞
1=
1 A ∫ Ẽn (u)u−γ du. γ n 0
Therefore, −1
∞
lim An = γ( ∫ e u
−u −γ
n→∞
du)
0
=
γ . Γ(1 − γ)
Finally, using identity (12.5.12) and the argument in Corollary 12.5.7 we have ∞
Pγ (x) = ∫ 0
1
Fn (x/u uN/2
1/2
)ηn (u) du,
(12.5.13)
where Fn denotes the radial function described by the inverse Fourier transform of En (ξ 2 ), that is, Fn (x) = (2π)−N/2 ∫ℝN En (ξ 2 )eix⋅ξ dξ . From the general theory of Bochner–Riesz multiplier operators, each Fn (t) is in L1 (ℝ+ , t N−1 dt) for n greater than the critical index (N − 1)/2 (see [294]). Making the change of variables u = x2 /s in (12.5.13) we get ∞
Pγ (x) = ∫ 0 ∞
x2 x2 sN/2 Fn (s1/2 )ηn ( ) 2 ds N s s x N+2γ
s 2 x2 ds ) . = ∫ N+2γ Fn (s1/2 )Gn ( s s x 0
(12.5.14)
398  12 The heat equation with fractional diffusion By the dominated convergence theorem again we have ∞
lim xN+2γ Pγ (x) = ∫ s
x→∞
N+2γ 2
ds , s
Fn (s1/2 )An
0
∀n ∈ ℕ.
Letting n → ∞ and using the above estimates we conclude lim x
N+2γ
x→∞
∞
N+2γ γ N + 2γ γ ds = Γ( ). Pγ (x) = ∫ s 2 e−s Γ(1 − γ) s Γ(1 − γ) 2
0
End of the proof of Theorem 12.5.2. Taking 0 < ϵ < x > R,
1 γ , 2 Γ(1−γ)
by Proposition 12.5.9 there exists an R > 0 such that if
γΓ(1 − γ) − ϵ ≤ (1 + xN+α )Pα (x) ≤
γ + ϵ. Γ(1 − γ)
According to the Bernstein theorem, the kernel Pα (x, 1) is positive in ℝN , and then we easily conclude the proof.
12.5.1 Some remarks on regularity of the solution to the Cauchy problem In this paragraph we summarize some regularity results with respect to the solution of the Cauchy problem in spaces of Hölder continuous functions. We follow the appendix in the paper by L. Caffarelli and A. Figalli [106]. More precisely, we will study the behavior of the fractional heat equation in the following spaces: α,β – v ∈ 𝒞x,t ([t0 , t1 ] × ℝN ) if ‖v‖𝒞 α,β ([t x,t
–
N 0 ,t1 ]×ℝ )
:= ‖v‖L∞ ([t0 ,t1 ]×ℝN ) +
sup
[t0 ,t1 ]×ℝN
v ∈ Lipt 𝒞xα ([t0 , t1 ] × ℝN ) if ‖v‖Lipt 𝒞xα ([t0 ,t1 ]×ℝN ) := ‖v‖L∞ ([t0 ,t1 ]×ℝN ) +
–
v(x, t) − v(x , t ) < +∞; x − x α + t − t β
sup
[t0 ,t1 ]×ℝN
v(x, t) − v(x , t ) < +∞; x − x α + t − t 
v ∈ log Lipt 𝒞xα ([t0 , t1 ] × ℝN ) if ‖v‖log Lipt 𝒞xα ([t0 ,t1 ]×ℝN ) := ‖v‖L∞ ([t0 ,t1 ]×ℝN ) + < +∞.
sup
[t0 ,t1 ]×ℝN
v(x, t) − v(x , t ) x − x α + t − t (1 + log(t − t ))
12.5 The Cauchy problem for the fractional heat equation
α,β−0+
 399
α,β−ϵ
By v ∈ 𝒞x,t ([t0 , t1 ] × ℝN ) we will understand that v ∈ 𝒞x,t ([t0 , t1 ] × ℝN ) for all ϵ > 0, and similarly in the other spaces. Note that from the result on the regularity and decay of the kernel P2s (x, t) we find by scaling, for instance, the following estimates: P2s (x, t) ≈
N+2s 2s
t
,
+ xN+2s 𝜕 2 1 ( ) P2s (x, t) ≤ ( ) N+2s 𝜕t t t 2s t
𝜕 1 , (12.5.15) P2s (x, t) ≤ N+2s 𝜕t t 2s + xN+2s 𝜕 1 1 1 , ∇x P2s (x, t) ≤ ( ) N+2s . x t 2s + xN+2s 𝜕t + xN+2s (12.5.16)
We assume that the initial data v(x, 0) = v0 (x) ∈ 𝒞 2 (ℝN ∩ W 1,∞ (ℝN ) and that ‖D2 v0 ‖L∞ (ℝN ) + ‖(−Δ)2 v0 ‖𝒞x1−s (ℝN ) < ∞ in order to avoid technicalities at t = 0. Then the solution to the linear problem vt + (−Δ)s v = f ,
if (x, t) ∈ ℝN × [0, T], v(x, 0) = v0 (x)
is given by the convolution with the fractional heat kernel, i. e., t
v(x, t) = ∫ P2s (x − y, t)v0 (y) dy + ∫ ∫ P2s (x − y, t − τ)f (y, τ) dy dτ.
(12.5.17)
0 ℝN
ℝN
As a consequence we obtain s
s
t
−(−Δ) v(x, t) = − ∫ P2s (x−y, t)(Δ) v0 (y) dy+ ∫ ∫ ℝN
0 ℝN
𝜕 P (x−y, t −τ)[f (y, τ)−f (x, τ)] dy dτ. 𝜕t 2s
Note that the first term on the right hand side is regular according to the regularity of v0 . Then studying the regularity of (−Δ)s v is reduced to the following result. Theorem 12.5.10. The following inequalities hold: 1. If 0 < β < 2s, then
2.
t ∫ ∫ 𝜕 P2s (x − y, t − τ)[f (y, τ) − f (x, τ)] dy dτ ≲ ‖f ‖L∞ (0,T],𝒞 β (ℝN ) . β−0+ , β x N 𝜕t 𝒞 2s N ×[0,T]) (ℝ x,t 0ℝ (12.5.18) If 1 > β ≥ 2s, then t ∫ ∫ 𝜕 P2s (x − y, t − τ)[f (y, τ) − f (x, τ)] dy dτ ≲ ‖f ‖L∞ (0,T],𝒞 β (ℝN ) . x N 𝜕t logLipt 𝒞xβ−0+ (ℝN ×[0,T]) 0ℝ (12.5.19)
400  12 The heat equation with fractional diffusion Before proving this result we state the main consequences. Corollary 12.5.11. The following estimates hold: 1. If 0 < β < 2s, then (−Δv)
β β−0+ , 2s
2.
≲ (1 + ‖f ‖L∞ (0,T],𝒞 β (ℝN ) ). x
(ℝN ×[0,T])
𝒞x,t
(12.5.20)
If 1 > β ≥ 2s, then (−Δv)
β−0+
(ℝN ×[0,T])
logLipt 𝒞x
≲ ‖f ‖L∞ (0,T],𝒞 β (ℝN ) . x
(12.5.21)
And finally, taking into account the equation, we obtain ‖vt ‖
β−0+
L∞ (0,T],𝒞x
+ (−Δ)s v
(ℝN )
β−0+
L∞ (0,T],𝒞x
(ℝN )
≲ (1 + ‖f ‖L∞ (0,T],𝒞 β (ℝN ) ), x
(12.5.22)
for all β ∈ (0, 1). In order to prove Theorem 12.5.10 we will use the following technical lemma. Lemma 12.5.12. Let 0 < β < 1. The following estimates hold: 1. There exists a constant C > 0 such that for all h ∈ (0, 1] ∫ ℝN
2.
min{yβ , 1} h
N+2s 2s
+
β
dy ≤ C(1 + h 2s −1 ).
yN+2s
(12.5.23)
Assume that t − τ < 1. Then there exists a constant C > 0 such that for all h > 0, t
∫ 0
dτ
(t − τ)
N+2s 2s
hN+2s
+
≤ C min{
1 1 , }. hN hN+2s
Proof. We start by proving (12.5.23). We have ∫ ℝN
min{yβ , 1} h
N+2s 2s
+ yN+2s
= ∫ ℝN \B
h
1
1 + yN+2s
dy
yβ
+ ∫ Bh1/2s
+
N+2s 2s
dy
h
N+2s 2s
∫ B1 \Bh1/2s
+ yN+2s
dy
yβ h
Then
+ yN+2s 1
I1 ≤ ∫ ℝN \B
N+2s 2s
yN+2s 1
dyI1 + I2 + I3 .
dy ≤ 1,
(12.5.24)
12.5 The Cauchy problem for the fractional heat equation
I2 ≤
1
h
 401
β
N+2s 2s
∫ yβ dy ≲ h 2s −1 Bh1/2s
and I3 ≤
β
yβ−N−2s dy ≲ 1 + h 2s −1 .
∫ B1 \Bh1/2s
To prove (12.5.24), note that the estimate is trivial if h ≥ 1. For h ∈ (0, 1] the result follows by writing t
∫ 0
t−h2s
dτ
(t − τ)
N+2s 2s
hN+2s
+
= ∫
(t − τ)
0
t
dτ
N+2s 2s
+
hN+2s
+ ∫ t−h2s
dτ
(t − τ)
N+2s 2s
+ hN+2s
.
The first integral can be estimated by a constant times h−(N+2s) and the second, as above, by t
t−h2s
–
t
dτ
∫
(t − τ)
N+2s 2s
+
hN+2s
≤ ∫ t−h2s
1 dτ ≤ . hN+2s hN
Proof of the regularity with respect to time in (12.5.20). Assume without loss of generality that σ < t. We check directly the difference s s (−Δ) v(x, t) − (−Δ) v(x, σ) t 𝜕 = ∫ ∫ P2s (x − y, t − τ)[f (y, τ) − f (x, τ)] dy dτ 𝜕t 0 ℝN σ
−∫ ∫ 0 ℝN
𝜕 P2s (x − y, σ − τ)[f (y, τ) − f (x, τ)] dy dτ 𝜕t
t
𝜕 ≤ ∫ ∫ P2s (x − y, t − τ)f (y, τ) − f (x, τ) dy dτ 𝜕t σ N ℝ σ
𝜕 𝜕 + ∫ ∫ P2s (x − y, t − τ) − P2s (x − y, σ − τ) min{1, x − yβ } dy dτ 𝜕t 𝜕t N 0ℝ
= I1 + I2 . We estimate I1 by using (12.5.15) and (12.5.23): t
I1 ≤ ∫ ∫ σ ℝN
min{1, x − yβ } (t − τ)
N+2s 2s
+ x − yN+2s β 2s
≤ (t − σ) + (t − σ) .
t
β
dy dτ ≤ ∫(1 + (τ − τ) 2s −1 ) dτ σ
402  12 The heat equation with fractional diffusion To estimate I2 we use (12.5.15), (12.5.16) and (12.5.23): σ
I2 ≤ ∫ ∫ min{ 0 ℝN σ−(t−σ)
≲
∫ 0
t−σ 1 , 1} min{x − yβ , 1} dy dt N+2s σ−τ N+2s 2s + x − y (σ − τ)
β t−σ (1 + (σ − τ) 2s −1 ) dτ + σ−τ
σ
β
∫ (1 + (σ − τ) 2s −1 ) dτ σ−(t−σ) β
≲ (t − σ) + (t − σ) log t − σ + (t − σ) 2s . –
Therefore we conclude the regularity with respect to t. Proof of spatial regularity in (12.5.20) and (12.5.21). We evaluate the variation with respect to the spatial variable with fixed t, namely, s s (−Δ) v(x, t) − (−Δ) v(z, t) t 𝜕 = ∫ ∫ P2s (x − y, t − τ)[f (y, τ) − f (x, τ)] dy dτ 0 ℝN 𝜕t σ
𝜕 − ∫ ∫ P2s (z − y, t − τ)[f (y, τ) − f (z, τ)] dy dτ. 𝜕t 0 ℝN
We decompose the spatial integral in two complementary sets, A1 := {x − z ≤
x − y } and 2
A2 := {x − z >
x − y }. 2
In A2 , using again (12.5.15), we find 𝜕 𝜕 P2s (x − y, t − τ)[f (y, τ) − f (x, τ)] − P2s (z − y, t − τ)[f (y, τ) − f (z, τ)] 𝜕t 𝜕t 𝜕 𝜕 ≤ P2s (x − y, t − τ) − P2s (x − y, σ − τ) min{1, x − yβ } 𝜕t 𝜕t ≤
x − yβ
(t − τ)
N+2s 2s
+ x − yN+2s
+
z − yβ
(t − τ)
N+2s 2s
+ z − yN+2s
.
Therefore, since A2 ⊂ B3x−z (x) ∩ B3x−z (z), using again (12.5.24), we obtain t ∫ ∫ 𝜕 P2s (x − y, t − τ)[f (y, τ) − f (x, τ)] dy dτ 𝜕t 0 A2 σ
−∫∫ 0 A2
𝜕 P2s (z − y, t − τ)[f (y, τ) − f (z, τ)] dy dτ 𝜕t
12.5 The Cauchy problem for the fractional heat equation t
≲∫
x − yβ
∫
0 B3x−z (x)
=
∫
(t − τ) β
∫
+ x − yN+2s
t
x − y ∫ 0
B3x−z (x)
≲
N+2s 2s
(t − τ)
x − yβ min{
B3x−z (x)
≲
∫
N+2s 2s
 403
dy dτ
1 + x − yN+2s
dτ dy
1 1 , } dy x − yN x − yN+2s
x − yβ−N dy ≲ x − zβ .
B3x−z (x)
To finish the proof we have to evaluate the integral in A1 . On A1 the following quantities are comparable: (t − τ)
N+2s 2s
1 + x − yN+2s
,
(t − τ)
N+2s 2s
1 + z − yN+2s
.
Hence, the integrand on A can be estimated by 𝜕 𝜕 P2s (x − y, t − τ)  [f (y, τ) − f (x, τ)] − P2s (z − y, t − τ)[f (y, τ) − f (z, τ)] 𝜕t 𝜕t 𝜕 𝜕 ≤ f (y, τ) − f (x, τ) P2s (x − y, t − τ) − P2s (z − y, t − τ) 𝜕t 𝜕t 𝜕 + f (x, τ) − f (z, τ) P2s (x − y, t − τ) := F(x, y, z, t, τ). 𝜕t Using (12.5.24), we obtain t
∫ ∫ F(x, y, z, t, τ) dy dτ 0 A2 t
1 x − z + x − zβ ) ∫ dτ dy × ∫( N+2s 1−β x − y (t − τ) 2s + x − yN+2s 0 A 2
≲ ∫ x − zβ min{ A2
≤ x − zβ (
1 1 , } dy x − yN x − yN+2s
∫
x − yN dy + ∫ x − yN − 2s dy)
x−z< x−y ≤1 2
× x − zβ log x − z ≲ x − zβ−ϵ
x−y >1 2
for all ϵ > 0.
Putting together the previous estimates we conclude the proof of Theorem 12.5.10.
404  12 The heat equation with fractional diffusion
12.6 Regularity in bounded domains: relation between weak solutions and viscosity solutions The regularity of the solutions to the problem {
ut + (−Δ)s u = f (x, t)
in ℝN × (0, T), if x ∈ ℝN
u(x, 0) = u0 (x)
(12.6.1)
according to the regularity of the data can be seen in [106], where the properties of the fundamental solution play a crucial role. Our aim here is to prove a similar result for the case of a bounded domain. In particular, we want to see that provided that our right hand side is good enough, the solution is regular and can be understood in a pointwise sense. More precisely, we have the following theorem. Theorem 12.6.1. Assume that Ω is a 𝒞 1,1 bounded domain in ℝN . Let f ∈ L∞ (Ω × (0, T)) be Hölder continuous in space–time and let u be the energy solution to the problem ut + (−Δ)s u = f (x, t) { { { u(x, t) = 0 { { { { u(x, 0) = 0
in Ω × (0, T), in (ℝN \ Ω) × (0, T),
(12.6.2)
in Ω.
Then, u is a strong solution in the following sense: (i) ut ∈ 𝒞 (Ω × (0, T)); (ii) u ∈ 𝒞 (Ω × [0, T)); (iii) the equation holds pointwise, i. e., ut (x, t) + CN,s P. V. ∫ ℝN
u(x, t) − u(y, t) dy = f (x, t) x − yN+2s
for every (x, t) ∈ Ω × (0, T).
The proof of this result is based on the relation between energy and viscosity solutions, and will require the following regularity result [161, Theorem 1.1 and Corollary 4.1]. Theorem 12.6.2. Assume that Ω is a 𝒞 1,1 bounded domain in ℝN and let v0 ∈ L2 (Ω). Consider v, the unique energy solution to problem vt + (−Δ)s v = 0 { { { v=0 { { { { v(x, 0) = v0 (x)
in Ω × (0, +∞), in ℝN \ Ω × (0, +∞), in Ω.
Then, for each ϵ > 0, (i) supv(⋅, t)𝒞 s (ℝN ) ≤ C1 (ϵ)‖v0 ‖L2 (Ω) ; t>ϵ
12.6 Relation between weak and viscosity solutions  405
(ii) denoting δ(x) := dist(x, 𝜕Ω), v(⋅, t) sup s ≤ C2 (ϵ)‖v0 ‖L2 (Ω) t>ϵ δ (⋅) 𝒞 s−η (Ω)
for any η > 0.
Both constants C1 and C2 blow up when ϵ → 0. In addition, for every j ∈ ℕ, 𝜕j v(⋅, t) ≤ Cj (ϵ)‖v0 ‖L2 (Ω) . sup t>ϵ 𝜕t j 𝒞 s (ℝN ) This result in particular provides continuity up to the boundary of the domain, and thus we can formulate an extension result. Corollary 12.6.3. Let v be as in Theorem 12.6.2. Then v can be extended as a continuous function to ℝN × (0, +∞). We can now prove Theorem 12.6.1. Proof of Theorem 12.6.1. Let τ > 0. By Theorem 12.6.2 and Corollary 12.6.3, the solution v to the problem vt (x, τ, t) + (−Δ)s v(x, τ, t) = 0 { { { v(x, τ, t) = 0 { { { { v(x, τ, 0) = f (x, τ)
in Ω × (τ, +∞), in ℝN \ Ω × (τ, +∞), in Ω
is continuous up to the boundary and can be extended to the whole space ℝN ×(τ, +∞) (we still name it v). Moreover, applying Duhamel’s principle, it is easy to see that t
u(x, t) := ∫ v(x, t, τ) dτ 0
is the extension to ℝN × (0, T) of the unique solution to (12.6.2). The goal now is to prove that u is also a viscosity solution. For the definition of this concept and the description of the class of test functions (that we will call 𝒮 ) we refer to [120, Definition 2.3]. Let (x0 , t0 ) ∈ Ω×(0, T). Due to the stability of viscosity solutions by uniform limits in compact sets (see [120, Theorem 5.3]), by a mollification argument we can assume that u is regular in a neighborhood of (x0 , t0 ) (see for example [282], where this argument is detailed in the elliptic case). Hence, consider a test function ϕ ∈ S verifying (i) u(x0 , t0 ) = ϕ(x0 , t0 ); (ii) u(x, t) < ϕ(x, t) in some Br (x0 ) × (t0 − ε, t0 ]; (iii) u(x, t) = ϕ(x, t) in (ℝN \ Br (x0 )) × (t0 − ε, t0 ].
406  12 The heat equation with fractional diffusion Then, ϕt − (x0 , t0 )+(−Δ)s ϕ(x0 , t0 ) = lim+ h→0
ϕ(x0 , t0 ) − ϕ(x0 , t0 − h) ϕ(x0 , t0 ) − ϕ(y, t0 ) dy + CN,s P. V. ∫ h x0 − yN+2s ℝN
u(x0 , t0 ) − u(x0 , t0 − h) u(x0 , t0 ) − u(y, t0 ) dy ≤ lim+ + CN,s P. V. ∫ h h→0 x0 − yN+2s ℝN
= f (x0 , t0 ), and we conclude that u is a viscosity subsolution (note that the singular integrals are well defined since we assume u regular). Analogously, it can be seen that u is a viscosity supersolution and therefore we conclude that u is a viscosity solution (12.6.2) and we can use the regularity results of [119, 120, 203]. In particular, by using [203, Corollaries 2.6 and 2.7], for a second member Hölder continuous in space–time, we find that the equation is verified in the pointwise sense. Remark 12.6.4. The interior regularity for a general class of operators has been studied in [162]. The study also considers regularity up to the boundary. We recommend the reader to check reference [162] for the details.
12.7 The uniqueness result of Widder type The heat equation has the hyperplane t = 0 as a characteristic surface and therefore the Cauchy problem with initial data on t = 0 is not well posed in general. However, there is a clear agreement between the principles of thermodynamics and the model of transfer of heat given by such an equation. This is reflected in the fact that the initial temperature evolves in time as the convolution with a kernel, giving rise to an average, and smoothing out in this way the potential effect of sharp thermal differences (this is in agreement with the entropy effect of the Second Principle of Thermodynamics); also, uniqueness holds true under a positivity assumption on the function (this is in agreement with the socalled Third Principle of Thermodynamics, according to which temperatures are always positive if measured in the Kelvin scale). In this sense D. V. Widder in [317] proved the following classical result: Assume that u : ℝN × [0, T) ⊂ ℝN+ → ℝ is so that u(x, t) ≥ 0,
u ∈ 𝒞 (ℝN × [0, T)),
ut , uxi xi ∈ 𝒞 (ℝN × (0, T)),
and satisfies ut (x, t) − Δu(x, t) = 0, in the classical sense.
(x, t) ∈ ℝN × (0, T),
12.7 The uniqueness result of Widder type  407
Then u(x, t) =
1 (4πt)
N 2
∫ u(y, 0)e(
−x−y2 4t
)
dy.
ℝN
In this chapter we obtain a similar result for the nonlocal heat equation, that is, the equation ut + (−Δ)α/2 u = 0
for (x, t) ∈ ℝN × (0, T), 0 < α < 2,
(12.7.1)
in which the diffusion is given by a power of the Laplacian (we refer to [138] and [294] for basic definitions and properties of the fractional Laplace operator). It is worth noting that for every α ∈ (0, 2] the operator in (12.7.1) does not satisfy the socalled Hadamard condition, that is, the Cauchy problem is ill posed (see [193] or [243, page 290, Theorem 5.3] for more details). As a consequence we face here a problem similar to the one of the heat equation. In this section we will present the results of the article [48]. As is well known, the operator in (12.7.1) and its inverse are nonlocal. In a sense, equation (12.7.1) is perhaps one of the simplest classical examples of pseudodifferential operators. In Section 12.5 we have studied the problem ut + (−Δ)α/2 u = 0
for (x, t) ∈ ℝN × (0, T), 0 < α < 2,
with initial datum u(x, 0) = u0 (x), assuming, say, u0 ∈ 𝒞 (ℝN ) ∩ L∞ (ℝN ) has a solution obtained by the convolution with the kernel Pt (x) =
1 x P( 1/α ), t N/α t
(12.7.2)
where α
P(x) := (2π)−N ∫ eix⋅ξ −ξ  dξ .
(12.7.3)
ℝN
Note that in this section, for simplicity we will denote by P the kernel Pα of Section 12.5. Therefore, a solution is given by u(x, t) = ∫ Pt (x − y)u0 (y) dy.
(12.7.4)
ℝN
We look here for a class of solutions of the fractional parabolic equation, such that the sign condition u(x, t) ≥ 0 implies that u, necessarily, is of the form (12.7.4) with u0 (x) replaced by the trace u(x, 0). This is the type of extension that we propose for the classical result of Widder to the nonlocal equation in (12.7.1). In this section we will describe several interpretations of what solution to equation (12.7.1) means, and consider weak, viscosity and strong solutions, in a sense that we introduce next.
408  12 The heat equation with fractional diffusion Weak solutions Consider the space α/2
ℒ
(ℝN ) := {u : ℝN → ℝ measurable: ∫ ℝN
u(x) dx < ∞}, 1 + xN+α
endowed with the norm ‖u‖ℒα/2 (ℝN ) := ∫ ℝN
u(x) dx. 1 + xN+α
If u ∈ ℒα/2 (ℝN ), then (−Δ)α/2 u can be defined in the weak sense as a distribution, that is, we can compute the duality product ⟨(−Δ)α/2 u, φ⟩, for every φ in the Schwartz class. Definition 12.7.1. We say that u(x, t) is a weak solution of the fractional heat problem {
ut + (−Δ)α/2 u = 0 u(x, 0) = u0 (x)
for (x, t) ∈ ℝN × (0, T), in ℝN
if the following conditions hold: (i) u ∈ L1 ([0, T ], ℒα/2 (ℝN )) for every T < T; (ii) u ∈ 𝒞 ((0, T), L1loc (ℝN )); (iii) for every test function φ ∈ C0∞ (ℝN × [0, T)) and 0 < T < T one has T
∫ ∫ [−u(x, t)φt (x, t) + u(x, t)(−Δ)α/2 φ(x, t)] dx dt
(12.7.5)
0 ℝN
= ∫ u(x, T )φ(x, T ) dx − ∫ u0 (x)φ(x, 0) dx. ℝN
ℝN
Condition (ii) is imposed so that the right hand side of equality (12.7.5) makes sense for every T . Note that the continuity would follow in any case from the left hand side of (12.7.5) due to the integrability condition in (i). Viscosity solutions Consider Q = ℝN × (0, T) and 1,2
𝒞p (Q) = {f : Q → ℝ  ft ∈ 𝒞 (Q), fxi ,xj ∈ 𝒞 (Q) p and sup f (x, t) ≤ C(1 + x) }. t∈(0,T)
12.7 The uniqueness result of Widder type  409
Definition 12.7.2. A function u ∈ 𝒞 (Q) is a viscosity subsolution (respectively supersolution) of ut + (−Δ)α/2 u = 0
(12.7.6)
in Q if for all (x,̂ t)̂ ∈ Q and φ ∈ 𝒞p1,2 (Q) such that u − φ attains a local maximum (minimum) at (x,̂ t)̂ one has φt (x,̂ t)̂ + (−Δ)α/2 φ(x,̂ t)̂ ≤ 0
(respectively ≥ 0).
We say that u ∈ 𝒞 (Q) is a viscosity solution of (12.7.6) in Q if it is both a viscosity subsolution and a viscosity supersolution. See [202], [119], [120] and [203] for more details about these types of solutions. Strong solutions From the results in Section 8.2.1 (see also [286, Proposition 2.1.4]) we know that if u ∈ α+γ ℒα/2 (ℝN ) ∩ 𝒞loc (ℝN ) (or 𝒞 1,α+γ−1 if α > 1), for some γ > 0, then (−Δ)α/2 u is a continuous function and it is well defined by an integral representation as the principal value of (−Δ)α/2 u(x) := CN,α/2 P. V. ∫ ℝN
u(x) − u(y) dy, α ∈ (0, 2). x − yN+α
See Chapters 8 and 9. In order to allow a larger class of functions, and following standard procedures in the theory of singular integrals, we will define hereafter the principal value as the twosided limit P. V. ∫ ℝN
u(x) − u(y) dy = lim ϵ→0 x − yN+α
∫ {y:ϵ 0, α ∈ (0, 2) and let u be a weak solution of the fractional heat equation {
ut + (−Δ)α/2 u = 0 u(x, 0) = 0
for (x, t) ∈ ℝN × (0, T), in ℝN .
(12.7.7)
Then u(x, t) = 0 for every t ∈ (0, T) and a. e. x ∈ ℝN . Proof. We must show that u(x, t0 ) = 0 for an arbitrary t0 ∈ (0, T) and x ∈ ℝn . For this, we fix R0 > 0 and θ ∈ 𝒞0∞ (BR0 ) and we will prove that ∫ u(x, t0 )θ(x) dx = 0. ℝN
For any t ∈ [0, t0 ), we define φ(x, t) := (θ(⋅) ∗ Pt0 −t (⋅))(x), where Pt is the kernel defined in (12.7.2) and (12.7.3). By the results in Section 12.5 (see also [210], [134] and [100]) we know that 1 1 C ≤ P(x) ≤ . C 1 + xN+α 1 + xN+α Therefore α
α
̂ )e−(t0 −t)ξ  = C(ξ )etξ  . ̂ , t) = θ(ξ φ(ξ Since ̂ , t), φ̂ t (ξ , t) = ξ α φ(ξ
(12.7.8)
412  12 The heat equation with fractional diffusion we have {
φt − (−Δ)α/2 φ = 0
for (x, t) ∈ ℝN × [0, t0 ), in ℝN .
φ(x, t0 ) = θ(x)
(12.7.9)
Now we claim that C1 , φ(x, t) = θ(x) ∗ Pt0 −t (x) ≤ 1 + xN+α
(12.7.10)
where C1 depends of n, α, R0 , M := ‖θ‖L∞ (BR ) and t0 . 0 Indeed, considering without loss of generality that 2R0 > 1, we will distinguish two cases. When x ≤ 2R0 , using that Pt is a summability kernel in L1 we have ∫ Pt0 −t (y)θ(x − y) dy ≤ M ∫ Pt0 −t (y) dy = M N N ℝ
ℝ
1 + (2R0 )N+α 1 + xN+α c1 ≤ , 1 + xN+α ≤M
(12.7.11)
where c1 = c1 (n, α, R0 , M). Consider now x > 2R0 . Note that from (12.7.2) and (12.7.8), we obtain Pt (y) ≤
C
N α
t (1 +
yN+α t
N+α α
Then, since for x − y ≤ R0 it follows that y ≥ Pt0 −t (y) ≤
≤ ) x , 2
Ct . yN+α
(12.7.12)
we have
2N+α C(t0 − t) . xN+α
As a consequence, ∫ Pt0 −t (y)θ(x − y) dy = N ℝ
∫
x−y≤R0
Pt0 −t (y)θ(x − y) dy
t0 − t xN+α t0 ≤ 2N+α 2CMBR0  1 + xN+α c2 ≤ , 1 + xN+α ≤ 2N+α CMBR0 
(12.7.13)
where c2 = c2 (N, α, R0 , M, t0 ) and BR0  denotes, as usual, the Lebesgue measure of the ball. Hence, (12.7.10) follows from (12.7.11) and (12.7.13).
12.7 The uniqueness result of Widder type
 413
Also, applying (12.7.13) to the derivatives of θ ∈ 𝒞0∞ (BR0 ), we have C2 , ∇φ(x, t) = ∇θ(x) ∗ Pt0 −t (x) ≤ 1 + xN+α
(12.7.14)
where C2 depends also on n, α, R0 , M := ‖∇θ‖L∞ (BR ) and t0 . 0
Then, from (12.7.10) and the fact that u ∈ L1 ([0, t0 ], ℒα/2 (ℝN )) we deduce that t0
M2 := ∫ ∫ u(x, t)φ(x, t) dx dt < ∞. 0
(12.7.15)
ℝN
Let now ϕ ∈ 𝒞 ∞ (ℝ) be such that χB1/2 ≤ ϕ ≤ χB1 .
(12.7.16)
For R > 2R0 we define ϕR (x) = ϕ( Rx ) and ψ(x, t) := φ(x, t)ϕR (x) as a test function of problem (12.7.7). Also, by the nonlocal Leibniz formula in (8.1.17), one sees that, for any f , g ∈ ℒα/2 (ℝN ), (−Δ)α/2 (fg)(x) = f (x)(−Δ)α/2 g(x) + g(x)(−Δ)α/2 f (x) − B(f , g)(x), with B(f , g) the bilinear form given by B(f , g)(x) := CN,α/2 ∫ ℝN
(f (x) − f (y))(g(x) − g(y)) dy. x − yN+α
Applying this formula (for a fixed t) to the functions φ and ϕR and recalling (12.7.9) we obtain (−Δ)α/2 ψ = φ(−Δ)α/2 ϕR + ϕR φt − B(φ, ϕR ).
(12.7.17)
Moreover, the function uψ evaluated at t0 is u(x, t0 )θ(x)ϕR (x), thanks to the terminal time condition in (12.7.9). Thus, considering ψ as a test function in (12.7.7) and using that φ is a solution of problem (12.7.9), we obtain ∫ u(x, t0 )θ(x)ϕR (x) dx N ℝ
t0 = ∫ ∫ [uφt (x, t)ϕR (x) − u(−Δ)α/2 (ψ(x, t))] dx dt 0 ℝN
(12.7.18)
414  12 The heat equation with fractional diffusion t0 = ∫ ∫ [uB(φ, ϕR )(x, t) − uφ(x, t)(−Δ)α/2 ϕR (x)] dx dt 0 ℝN t0
≤ ∫ ∫ uφ(x, t)(−Δ)α/2 ϕR (x) dx dt 0 ℝN
t0
+ ∫ ∫ u(x, t)B(φ, ϕR )(x, t) dx dt. 0 ℝN
Then, since θ is supported in BR0 and R0 < R/2, recalling (12.7.16), we conclude t
0 α/2 ∫ u(x, t0 )θ(x) dx ≤ ∫ ∫ uφ(x, t)(−Δ) ϕR (x) dx dt N N
0ℝ
ℝ
t0
+ ∫ ∫ u(x, t)B(φ, ϕR )(x, t) dx dt 0 ℝN
=: I1 (R) + C(n, α)I2 (R).
(12.7.19)
It remains to show that lim I1 (R) + I2 (R) = 0.
R→∞
Indeed, since x α/2 −α α/2 −α (−Δ) ϕR (x) = R ((−Δ) ϕ)( ) ≤ C0 R , R by (12.7.15), it follows that t0
I1 (R) ≤ C0 R
−α
∫ ∫ uφ(x, t) dx dt ≤ C0 M2 R−α 0 ℝN
and so lim I1 (R) = 0.
R→∞
(12.7.20)
Now we are going to estimate I2 (R). For this we cover ℝN ×ℝN with six domains suitably described by the radii R/4, R/2, R and 2R and represented (for N = 1) in the following picture:
12.7 The uniqueness result of Widder type
A4
 415
C
2R A2
C A5
R R/2 R/4
A5
C
A3 A1
C R 4
R
R 2
2R
Therefore 5
ℝ2N = ( ⋃ Ak ) ∪ C, k=1
where A1 := {(x, y) : x > R/2, y ≤ R/4},
A2 := {(x, y) : x ≤ R/4, y > R/2},
A3 := {(x, y) : x ≥ 2R, R/4 < y < R},
A4 := {(x, y) : R/4 < x < R, y ≥ 2R},
A5 := {(x, y) : R/4 < x < 2R, R/4 < y < 2R}
and C := {(x, y) : x ≤ R/2, y ≤ R/2} ∪ {(x, y) : x ≥ R, y ≥ R}. From (12.7.16), we know that ϕR (x) − ϕR (y) = 0 if (x, y) ∈ C, and so t0
φ(x, t) − φ(y, t)ϕR (x) − ϕR (y) I2 (R) = ∫ ∫ u(x, t) ∫ dy dx dt x − yN+α 0 ℝN 5
ℝN
A
≤ ∑ I2 k (R), k=1
where A I2 k (R)
for k = 1, . . . , 5.
t0
φ(x, t) − φ(y, t)ϕR (x) − ϕR (y) dy dx dt, = ∫ ∫ u(x, t) x − yN+α 0 Ak
(12.7.21)
416  12 The heat equation with fractional diffusion We are going to estimate each of these five integrals separately. For (x, y) ∈ A1 we get x − y ≥ Cx. Moreover by (12.7.10) it follows that C . φ(x, t) + φ(y, t) ≤ 1 + yN+α Therefore A I2 1 (R)
t0
≤∫ ∫ 0 x>R/2
u(x, t) xN+α
t0
≤C∫ ∫ 0 x>R/2
∫ y≤R/4
C dy dx dt 1 + yN+α
u(x, t) dx dt. xN+α
(12.7.22)
Following the same ideas, since for (x, y) ∈ A2 we obtain C φ(x, t) + φ(y, t) ≤ N+α x
and x − y ≥ Cy,
(12.7.23)
we have t0
A
I2 2 (R) ≤ ∫
∫
0 x≤R/4
u(x, t) xN+α
t0
≤ CR−α ∫
∫
0 x≤R/4
∫ y>R/2
C dy dx dt yN+α
u(x, t) dx dt. xN+α
(12.7.24)
Also, since for (x, y) ∈ A3 (12.7.23) is satisfied, we have t0
A
I2 3 (R) ≤ CR−α ∫ ∫
0 x≥2R
u(x, t) dx dt. xN+α
(12.7.25)
Similarly, using again the good decay of φ and the fact that x − y ≥ Cy for every (x, y) ∈ A4 , we obtain A I2 4 (R)
t0
≤ CR
−α
∫
∫
0 R/4 0.
(12.7.41)
Then, (x0 , t0 ) cannot lie in (𝜕Ω×[0, T))∪(Ω×{0}), since v ≤ 0 there, thanks to the boundary conditions in (12.7.40). As a consequence, (x0 , t0 ) lies in Ω × (0, T ] and therefore vt (x0 , t0 ) = 0. Therefore the equation in (12.7.40) implies that 0 ≥ (−Δ)α/2 v(x0 , t0 ) = CN,α/2 P. V. ∫ ℝN
= CN,α/2 (P. V. ∫ Ω
v(x0 , t0 ) − v(y, t0 ) dy y − x0 N+α
v(x0 , t0 ) − v(y, t0 ) v(x0 , t0 ) − v(y, t0 ) dy + ∫ dy) N+α y − x0  y − x0 N+α
≥ CN,α/2 P. V. ∫ ℝN \Ω
ℝN \Ω
v(x0 , t0 ) − v(y, t0 ) dy. y − x0 N+α
Since v(y, t0 ) ≤ 0 for y ∈ ℝN \ Ω, thanks to (12.7.40), the latter integrand is strictly positive, due to (12.7.41), and this is a contradiction. Now we are able to prove that strong solutions are upper bounds for the kernel convolutions. Lemma 12.7.9. Let (x, t) ∈ ℝn × (0, T). If u(x, t) ≥ 0 is a strong solution of the fractional heat equation, then I := ∫ Pt (x − y)u(y, 0) dy ≤ u(x, t),
(12.7.42)
ℝN
where Pt (x) is the function defined in (12.7.2). Proof. First of all, observe that the integral I = I(x, t) exists, for it is given by the integration of two (measurable) positive functions, although we do not know a priori that
12.7 The uniqueness result of Widder type
 421
I is finite. However, this will be a consequence of our result that gives the inequality I ≤ u(x, t) for every (x, t) ∈ ℝN × [0, T). To this aim, we let 1 { { { ϕR (x) := { R − x { { { 0
x ≤ R − 1, R − 1 ≤ x ≤ R,
(12.7.43)
x > R.
We define vR (x, t) := ∫ Pt (x − y)ϕR (y)u(y, 0) dy = (Pt (⋅) ∗ ϕR u(⋅, 0))(x). ℝN
Then 𝜕vR + (−Δ)α/2 vR = 0 { { { { 𝜕t { vR (x, t) ≥ 0 { { { { vR (x, 0) = ϕR (x)u(x, 0)
for (x, t) ∈ ℝN × (0, T), for (x, t) ∈ ℝN × (0, T), in ℝN .
Let x > R. As u(x, t) ∈ 𝒞 (ℝN × [0, T)) we can define the real number MR := sup u(y, 0) < ∞. y 0 it follows that 0 ≤ vR (x, t) ≤ ε,
for any x ≥ ρ, t ∈ (0, T),
where 1
C(T, MR , N)RN N+α ) > 0. ρ=R+( ε
(12.7.44)
422  12 The heat equation with fractional diffusion Moreover, as u(x, t) ≥ 0 in ℝN × [0, T) we obtain 0 ≤ vR (x, t) ≤ ε ≤ ε + u(x, t),
for any x ≥ ρ, t ∈ (0, T),
(12.7.45)
and vR (x, 0) = ϕR (x)u(x, 0) ≤ u(x, 0) ≤ ε + u(x, 0),
for any x ≤ ρ.
(12.7.46)
Consider the cylinder Dρ,T = Bρ × (0, T). We define the function w(x, t) := vR (x, t) − u(x, t) − ε. Then, by (12.7.45) and (12.7.46), we get w(x, t) ≤ 0 in ℝN × [0, T) \ Dρ,T . Therefore, since u(x, t) is a strong solution of the fractional heat equation in ℝN × [0, T), applying Lemma 12.7.8 in Dρ,T to the function w(x, t) we have vR (x, t) ≤ ε + u(x, t),
for x ≤ ρ and t ∈ [0, T).
Therefore, from (12.7.45), it follows that vR (x, t) ≤ ε + u(x, t),
for x ∈ ℝN and t ∈ [0, T).
Since ε is fixed but arbitrary, the previous inequality implies that vR (x, t) ≤ u(x, t),
for every x ∈ ℝN and t ∈ [0, T).
Finally, by the monotone convergence theorem, as limR→∞ ϕR = 1, we conclude that 0 ≤ v(x, t) = lim vR (x, t) = ∫ Pt (x − y)u(y, 0) dy ≤ u(x, t). R→∞
ℝN
By a simple time translation, we obtain from Lemma 12.7.9 the following. Corollary 12.7.10. Let 0 < τ < T and (x, t) ∈ ℝn × (0, T − τ). If u(x, t) ≥ 0 is a strong solution of the fractional heat equation, then ∫ Pt (x − y)u(y, τ) dy ≤ u(x, t + τ),
(12.7.47)
ℝN
where Pt (x) is the function defined in (12.7.2). As a consequence, for every x ∈ ℝN and t ∈ [0, T − τ) we have T−t
T−t
∫ ∫ Pt (x − y)u(y, τ) dy dτ ≤ ∫ u(x, t + τ) dτ. 0 ℝN
0
(12.7.48)
12.7 The uniqueness result of Widder type
 423
Moreover we have the following. Corollary 12.7.11. Let (x, t) ∈ ℝn ×(0, T). If u(x, t) ≥ 0 is a strong solution of the fractional heat equation, then u(⋅, t) ∈ ℒα/2 (ℝN ). Proof. Let 0 < T < T. Taking t = T − T in (12.7.47), from (12.7.8), we get u(y, τ) T − T dy ≤ ∫ PT−T (x − y)u(y, τ) dy ∫ N+α C (T − T ) α + x − yN+α N N ℝ
ℝ
≤ u(x, T − T + τ) < ∞.
(12.7.49)
Let C(T) := min{1,
1 (T − T )
N+α α
}.
Then, since 1
(T −
T )
N+α α
+
yN+α
≥ C(T)
1 , 1 + yN+α
taking x = 0 in (12.7.49), we conclude that ∫ ℝN
u(y, t) dy < ∞. (1 + yN+α )
We also have the following corollary. Corollary 12.7.12. Let (x, t) ∈ ℝN × (0, T). If u(x, t) ≥ 0 is a strong solution of the fractional heat equation, then u(⋅, t) ∈ L1 ([0, T ], ℒα/2 (ℝN )) for every 0 < T < T. Proof. Take an arbitrary 0 < T < T. By (12.7.48) with t = T − T we have T
T
∫ ∫ PT−T (x − y)u(y, τ) dy dτ ≤ ∫ u(x, τ) dτ. 0 ℝN
0
Then, as u ∈ 𝒞 (ℝN × [0, T ]) we get T
∫ ∫ PT−T (x − y)u(y, τ) dy dτ < ∞. 0 ℝN
Therefore proceeding as in the proof of Corollary 12.7.11 we get T
∫∫ 0 ℝN
u(y, t) dy dt < ∞. (1 + yN+α )
That is, u ∈ L1 ([0, T ], ℒα/2 (ℝN )) for every 0 < T < T.
424  12 The heat equation with fractional diffusion Note that Corollary 12.7.12 affirms that if u(x, t) ≥ 0 is a strong solution of the fractional heat equation, then u is also a weak solution of the same equation. Now we are able to prove our main result. Proof of Theorem 12.7.5. By Corollary 12.7.12 we get u ∈ L1 ([0, T ], ℒα/2 (ℝN )) for every 0 < T < T. Moreover if we define p(x, t) := ∫ Pt (x − y)u(y, 0) dy, ℝN
by (12.7.48) with t = T − T , we also have p ∈ L1 ([0, T ], ℒα/2 (ℝN )). Let now the function w(x, t) := u(x, t) − p(x, t) ≥ 0. It is clear that w is a strong solution of the fractional heat equation. Moreover, as w ∈ L1 ([0, T ], ℒα/2 (ℝN )), w(x, t) is also a solution in the weak sense with zero initial datum. Therefore applying Theorem 12.7.6 we conclude that w(x, t) = 0 for every (x, t) ∈ ℝN × [0, T). 12.7.3 About viscosity solutions As said at the beginning of this work, it is natural to consider viscosity solutions of the fractional heat equation. Our purpose here is to describe some cases in which a positive viscosity solution has the unique representation in terms of the kernel Pt . Proposition 12.7.13. Let {un } be a sequence of nonnegative, strong solutions of the fractional heat equation converging uniformly over compact sets to a given function u. Then u ≥ 0 is a viscosity solution of (12.7.1) satisfying ∫ Pt (x − y)u(y, 0) dy ≤ u(x, t),
(x, t) ∈ ℝN × (0, T)
(12.7.50)
ℝN
and u ∈ L1 ([0, T ], ℒα/2 (ℝN ))
for every 0 < T < T.
(12.7.51)
That is, the conclusions of Lemma 12.7.9 and Corollary 12.7.12 are satisfied. Proof. By Lemma 12.7.9 it follows that ∫ Pt (x − y)un (y, 0) dy ≤ un (x, t),
(x, t) ∈ ℝN × (0, T).
ℝN
Applying the Fatou lemma we obtain (12.7.50). Therefore, proceeding as in the proof of Corollary 12.7.12 we conclude (12.7.51). Note also that, by the comparison principle [286,
12.8 Further results  425
Corollary 2.1.6], un ≥ 0 is a viscosity solution of (12.7.1) for every n ∈ ℕ. Therefore, since u is the uniform limit over compact sets of viscosity solutions, u is also a viscosity solution of (12.7.1). Note that to conclude the equality in (12.7.50) we would need to know that u is a weak solution. If we add a monotonicity condition over the sequence {un }, then we obtain the following. Proposition 12.7.14. Let {un } be a monotone sequence of nonnegative, strong solutions of the fractional heat equation converging uniformly over compact sets to a given function u. Then u ≥ 0 is a strong solution of (12.7.1) satisfying ∫ Pt (x − y)u(y, 0) dy = u(x, t),
(x, t) ∈ ℝN × (0, T).
(12.7.52)
ℝN
Proof. Since, for every n ∈ ℕ, un ≥ 0 satisfies Theorem 12.7.5, by the monotone convergence theorem we obtain (12.7.52). Clearly this implies that u ≥ 0 is a strong solution of (12.7.1). Remark 12.7.15. It would be interesting to find the biggest class of positive viscosity solutions for which the representation property (12.7.52) holds. To the best of our knowledge, this is an open problem.
12.8 Further results 12.8.1 Some remarks about the fractional Widder theorem As in the local case (see [318]), given a solution u, one can define the enthalpy t
v(x, t) = ∫ u(x, s) ds, 0
which is also a solution of the fractional heat equation. Indeed we have the following result. Lemma 12.8.1. Let (x, t) ∈ ℝn × [0, T). If u(x, t) is a strong solution of the fractional heat equation with vanishing initial condition, then the enthalpy term t
v(x, t) = ∫ u(x, s) ds 0
is also a strong solution of the fractional heat equation. Moreover if u is positive the function v is increasing in t for x fixed and α/2subharmonic as a function of x in the weak sense.
426  12 The heat equation with fractional diffusion Proof. Let (x, t) ∈ ℝN × [0, T). First of all note that v(x, t) satisfies conditions (i)–(iii) of Definition 12.7.3. Therefore, since u(x, t) is a strong solution of the fractional heat equation, by the Fundamental Theorem of Calculus and the Fubini theorem it follows that CN,α/2 P. V. ∫ ℝN
v(x, t) − v(y, t) dy x − yN+α t
t
= CN,α/2 P. V. ∫
∫0 u(x, s) ds − ∫0 u(y, s) ds x − yN+α
ℝN t
= CN,α/2 P. V. ∫ ∫ ℝN 0
dy
u(x, s) − u(y, s) ds dy x − yN+α
t
= ∫ (−Δ)α/2 u(x, s) ds 0
t
= − ∫ us (s, t) ds 0
= −u(x, t) = −vt (x, t), for any (x, t) ∈ ℝN × (0, T]. Then, v(x, t) satisfies the fractional heat equation in the strong sense. Also if u ≥ 0 from the calculations done before we deduce that −(−Δ)α/2 v(x, t) = vt (x, t) = u(x, t) ≥ 0. So v is α/2subharmonic as a function of x and increasing in t for x fixed. Lemma 12.8.1 shows that the enthalpy belongs to the special class of positive strong solutions that are also α/2subharmonic. This class naturally satisfies a polynomial estimate, as shown by the following result. Proposition 12.8.2. Let (x, t) ∈ ℝn × [0, T). Let u(x, t) ≥ 0 such that: (i) u(x, t) is an α/2subharmonic function with respect to the variable x; (ii) u(x, t) is a strong solution of the fractional heat equation. Then u(x, t) ≤ C(t)(1 + xN+α ) if (x, t) ∈ ℝN × [0, T). Proof. By hypothesis it is clear that u is an α/2subharmonic function with respect to the variable x for t fixed and therefore u is increasing in time. Let now 0 < t1 < T and 0 < t0 < T − t1 . By Corollary 12.7.10 we have ∫ Pt (x − y)u(y, t1 ) dy ≤ u(x, t + t1 ), ℝN
for any 0 < t < T − t1 .
12.8 Further results  427
Therefore Mt0 := ∫ Pt0 (y)u(y, t1 ) dy ≤ u(0, t0 + t1 ) < ∞. ℝN
Our objective is to show that N+α u(x, t1 ) ≤ C(1 + x ).
(12.8.1)
Once this is done, using that u is increasing in time, we would get 0 ≤ u(x, t) ≤ u(x, t1 ) ≤ C(1 + xN+α ) for every (x, t) ∈ ℝN × (0, t1 ). But, since t0 and t1 are fixed but arbitrary, we would conclude that N+α u(x, t) ≤ C(1 + x ). So, we are left to showing that (12.8.1) is true. Note that from Corollary 12.7.11 we have u ∈ ℒα/2 (ℝN ). Then as u is α/2subharmonic, by Proposition 11.2.4 (see also [286, Proposition 2.2.6]) u also satisfies the following mean value property: u(x0 , t) ≤ ∫ γλ (y − x0 )u(y, t) dy
for every x0 ∈ Ω ⊆ ℝN and λ ≤ dist(x0 , 𝜕Ω), (12.8.2)
ℝN
where, as in Proposition 11.2.4, γλ (x) := (−Δ)α/2 Γλ (x) and Γλ (x) :=
1
λN−α
x Γ( ). λ
Here Γ is a C 1,1 function that coincides with Φ(x) := cxα−N outside the ball Bλ and Γ is a paraboloid inside this ball. Recall that Φ is the fundamental solution of (−Δ)α/2 . From (12.8.2) we have u(x, t1 )(1 + xN+α )
−1
≤ (1 + xN+α )
∫ γλ (y)u(x − y, t1 ) dy
−1
ℝN
= (1 + x
N+α −1
)
γλ (y)u(x − y, t1 ) dy
∫ {y:y≥λ}
+ (1 + xN+α )
−1
∫
γλ (y)u(x − y, t1 ) dy
{y:y≤λ}
:= I1 (x) + I2 (x).
(12.8.3)
428  12 The heat equation with fractional diffusion Choosing λ = x/4,
(12.8.4)
we have x − y ≤ 5y. Therefore, we obtain I1 (x) ≤ C(1 + xN+α )
−1
∫ {y:y≥λ}
≤ C(1 + xN+α )
−1
∫ ℝN
u(x − y, t1 ) dy yN+α
u(x − y, t1 ) dy x − yN+α
≤ C‖u‖ℒα/2 (ℝN ) := C1 .
(12.8.5)
Moreover, using that the fractional Laplacian of a paraboloid is bounded, by (12.7.8) and (12.8.4), it follows that I2 (x) ≤ C(1 + xN+α )
−1
∫
u(x − y, t1 ) dy
{y:y≤λ} N+α −1
≤ C(1 + x
)
u(z, t1 )dz
∫ {z:x−z≤λ}
N+α −1
≤ C(1 + x
)
∫
u(z, t1 )P1 (z)
{z:z≤2x} N+α
N+α −1
≤ C(1 + x
) (1 + (2x)
)
1 dz P1 (z)
∫
u(z, t1 )P1 (z)dz
{z:z≤2x} N+α −1
≤ CM1 (1 + x
) (1 + (2x)
N+α
)
≤ C(N, α)u(0, 1 + t1 ) := C2 .
(12.8.6)
By (12.8.3), (12.8.5) and (12.8.6), we obtain (12.8.1) and we conclude the proof. Remark 12.8.3. It would be interesting to prove that the pointwise behavior in Proposition 12.8.2 and some comparison arguments provide an alternative proof to our main result, namely, the representation of every solution as a convolution with the associated Poisson kernel.
12.8.2 The Fujita exponent for the fractional heat equation In Chapter 5 we have studied the H. Fujita result in [170] for the heat equation. A similar result for the fractional heat equation was obtained by S. Sugitani in [299]. The estimates on the kernel of the fractional heat equation and the uniqueness of the positive solution allow us to study the behavior of the positive solution to the following
12.8 Further results  429
semilinear problem: {
ut + (−Δ)s u = u1+β ,
u(x, t) > 0, x ∈ ℝN , t > 0 x ∈ ℝN ,
u(x, 0) = u0 (x),
(12.8.7)
where 0 < s < 1, β > 0 and u0 ∈ L∞ (ℝN ) ∩ 𝒞 (ℝN ), u0 (x) ⪈ 0. Given the positive solution to problem (12.8.7), u(x, t) ≥ 0, we can write by the representation formula t
u(x, t) = ∫ P2s (x − y, t)u0 (y) dy + ∫ ∫ P2s (x − y, t − τ)u1+β (y, τ) dy dτ, 0
ℝN
(12.8.8)
ℝN
in the times t in which the function u is finite. That is, we consider the normalization ∫ P2s (y, 1) dy = 1. ℝN
Note that by using Corollary 12.5.8 we can find t0 > 0, c > 0 and γ > 0 such that u(x, t0 ) ≥ cP2s (x, γ).
(12.8.9)
Indeed, choosing t0 > 0 such that P2s (0, t0 ) ≤ 1, we have P2s (x − y, t0 ) = P2s (
t0 2x − 2y )P2s (2y, t0 ), , t0 ) ≥ P2s (2x, t0 )P2s (2y, t0 ) ≥ 2−N P2s (x, 2s 2 2
and therefore, u(x, t0 ) ≥ (2−N ∫ P2s (2y, t0 )u0 (y) dy)P2s (x, ℝN
t0 ). 22s
That is, (12.8.9) is satisfied for c = (2−N ∫ P2s (2y, t0 )u0 (y) dy), ℝN
γ=
t0 . 22s
We remark that the lower estimate (12.8.9) can be seen as a proof of the non finite speed of propagation. Now by using the semigroup property, t
u(x, t + t0 ) = ∫ P2s (x − y, t)u(y, t0 ) dy + ∫ ∫ P2s (x − y, t − τ)u1+β (y, τ + t0 ) dy dτ. (12.8.10) ℝN
0 ℝN
430  12 The heat equation with fractional diffusion Plugging inequality (12.8.9) in the identity (12.8.10), we obtain u(x, t + t0 ) ≥ c ∫ P2s (x − y, t)P2s (x − y, t)P2s (y, γ) dy ℝN t
+ ∫ dτ ∫ P2s (x − y, t − τ)u1+β (y, τ + t0 ) dy 0
ℝN t
= cP2s (x, t + γ) + ∫ dτ ∫ P2s (x − y, t − τ)u1+β (y, τ + t0 ) dy. 0
(12.8.11)
ℝN
Call f (t) = ∫ P2s (x, t)u(x, t + t0 ) dx. ℝN
Multiplying inequality (12.8.11) by P2s (x, t), integrating in the whole space and using Fubini’s theorem, we obtain f (t) ≡ ∫ P2s (x, t)u(x, t + t0 ) dx ℝN t
≥ c ∫ P2s (x, γ)p(x, t + τ) + ∫ dτ ∫ P2s (x, t) ∫ P2s (x − y, t − τ)u1+β (y, τ + t0 ) dy dx 0
ℝN
ℝN
ℝN
t
= cP2s (0, 2t + γ) + ∫ dτ ∫ P2s (y, 2t − τ)u1+β (y, τ + t0 ) dy 0
≥ cP2s (0, 1)
1
ℝN t
N
(2t + γ) 2s
N
2s τ ) ∫ P2s (y, τ)u1+β (y, τ + t0 ) dy. + ∫ dτ( 2t − τ
0
(12.8.12)
ℝN
Since dμ := P2s (x, t) dx is a probability measure for all t, we can apply the Jensen inequality and conclude the following inequality: f (t) ≥ cP2s (0, 1)
t
1
N
(2t + γ) 2s
N
τ 2s + ∫( ) f 1+β (τ) dτ. 2t
(12.8.13)
0
N
Take δ > 0 and consider t > δ; then multiplying inequality (12.8.13) by t 2s we find N 2s
t f (t) ≥ cP2s (0, 1)
δ
t
N
(2δ + γ) 2s
N
τ 2s + ∫( ) f 1+β (τ) dτ. 2 0
(12.8.14)
12.8 Further results  431
N
Calling h(t) = t 2s f (t), inequality (12.8.14) can be written as h(t) ≥ cP2s (0, 1)
t
δ
N
(2δ + γ) 2s
N
N 1+β τ 2s + ∫( ) (h(τ)τ− 2s ) dτ. 2
0
Considering the function g(t) satisfying g(t) = cP2s (0, 1)
δ
t
N
(2δ + γ) 2s
N
N 1+β τ 2s + ∫( ) (g(τ)τ− 2s ) dτ, 2
0
h(t) ≥ g(t), and then if g blows up in a finite time, also h and f blow up in finite time. Now g(t) is the solution to the problem N
g(t) 1+β t { { g (t) = ( 2 ) 2s ( t 2sN ) , { N { δ ) 2s := g0 . g(0) = cP2s (0, 1)( 2δ+γ {
(12.8.15)
Since g0 > 0, integrating the ordinary differential equation, g(t) = (
1 β
1
g0 − ( −β
β βN/2s N )t 1− 2s β
) ,
which blows up in a finite time if Nβ < 2s. As a conclusion we have obtained the following result. Lemma 12.8.4. Assume that β < 2s . Then if u(x, t) is a positive solution to probN lem (12.8.7), there exists a time T such that ∫ P2s (x, t)u(x, t) dx → ∞
as t → T.
ℝN
The result can be made even stronger. We have proved the blowup of the L1 norm with respect to the probability measure dμ := P2s (x, t) dx. But next we will prove that there is a complete blowup, in the sense that for any positive solution to problem (12.8.7), there exists a time T1 such that u(x, t) = ∞ for every t > T1 and every x ∈ ℝN . The complete blowup is a consequence of the following result. Lemma 12.8.5. Assume that u(x, t) ≥ 0 is a solution to (12.8.7) and f (t) = ∫ P2s (x, t)u(x, t) dx → ∞
as t → T.
ℝN
Then there exists a T1 > 0 such that u(x, t) → ∞
as t → T1 , for all x ∈ ℝN .
432  12 The heat equation with fractional diffusion Proof. We can assume that P2s (0, T1 ) ≤ 1 and therefore P2s (0, t) ≤ 1 if t > T1 . Consider T1 ≤ t ≤ τ ≤
23 t. +1
2β
By using again the homogeneity of the kernel and Corollary 12.5.8, we find P2s (x − y, 23 t − τ) = P2s (x − y, τ( N
23 t − τ )) τ
2s τ τ ) P2s (( 3 )(x − y), τ) =( 3 2 t−τ 2 t−τ N
2s τ ≥( 3 ) P2s (x, τ)P2s (y, τ). 2 t−τ
Then N
2s τ ) P2s (x, τ)f (τ) = ∞. ∫ P2s (x − y, 2 t − τ)u(y, τ) dy ≥ ( 3 2 t−τ
3
ℝN
Finally, using the Jensen inequality we conclude that 23 t 2β +1
u(x, 23 t) ≥ ∫ dτ ∫ P2s (x − y, 23 t − τ)u1+β (y, τ) dy t 23
2β +1
ℝN t 3
1+β
≥ ∫ dτ( ∫ P2s (x − y, 2 t − τ)u(y, τ) dy) t
= ∞.
ℝN
Summarizing, u(x, t) = ∞,
if t > 23 T1 for all x ∈ ℝN .
Remark 12.8.6. In [299] the blowup result is obtained for positive increasing convex functions, F(u), such that for some β ∈ (0, 2s ], N lim
u→∞
Note that the extreme β = functions.
2s N
F(u) = γ ∈ (0, ∞). u1+β
lives in the exponent where blowup occurs for all positive
13 The influence of the Hardy potential on the linear and semilinear fractional heat equations 13.1 Introduction We will study in this chapter the solvability of the linear problem u ut + (−Δ)s u = λ 2s + f in Ω × (0, T), { { x { { { { { u(x, t) > 0 in Ω × (0, T), { { { { u(x, t) = 0 in (ℝN \ Ω) × [0, T), { { { if x ∈ Ω { u(x, 0) = u0 (x) and of the semilinear problem { { { { { { { { { { { { { { {
ut + (−Δ)s u = λ
u + up + f x2s
(13.1.1)
in Ω × (0, T),
u(x, t) > 0
in Ω × (0, T),
u(x, t) = 0
in (ℝN \ Ω) × [0, T),
u(x, 0) = u0 (x)
if x ∈ Ω,
(13.1.2)
where Ω is a C 1,1 bounded domain in ℝN , N > 2s, 0 < s < 1, p > 1 and c and λ are positive constants. To avoid the trivial case, we assume 0 ∈ Ω. We suppose that f and u0 are nonnegative functions satisfying some hypotheses that we will precise later. We recall the integral expression of the fractional Laplacian of order 2s obtained in Chapter 8, that is, (−Δ)s u(x) := CN,s P. V. ∫ ℝN
u(x) − u(y) dy, s ∈ (0, 1), x − yN+2s
(13.1.3)
where CN,s is defined in (9.3.12). The local case, s = 1, has been studied in Chapter 5, and there problems (13.1.1) and (13.1.2) correspond to the classical heat equation. The linear problem (13.1.1) was first studied by P. Baras and J. Goldstein in [41] and (13.1.2), the semilinear problem, has been studied in [19]. There are many other related papers in the literature. For s ∈ (0, 1), the fractional setting, there exists also a large literature dealing with the case λ = 0. We refer to Chapter 12 and to the articles [106, 169] and the references therein for the basic theory of existence. A result on the uniqueness of a positive regular solution to the linear problem in the whole space was studied in Section 12.7 (see also the original article [48]). The case under consideration in this chapter, λ > 0 and s ∈ (0, 1), presents many new difficulties; for instance as in the elliptic case, any positive supersolution to problem (13.1.1) is unbounded close to the origin, even for nice data. In the nonlocal framework, similarly to the elliptic problem, the precise rate of growth of the solutions near https://doi.org/10.1515/9783110606270013
434  13 Fractional Heat equation and Hardy potential the origin will be the key to obtain the optimal results. It is worth pointing out here the difference with the local case, where this rate is obtained just by solving an elementary linear differential equation. The main results in this chapter can be summarized as follows. – First, to study the local behavior of the solutions we need some sharp local estimates that are based on a Harnack inequality for a related problem resulting from the ground state transformation by Frank, Lieb and Seiringer [167], studied in Section 9.5. This weak Harnack inequality will give the exact blowup rate for the positive supersolutions near the taxis. For the proof of this result, we closely follow the work of Felsinger and Kassmann [155], where the authors develop a weak parabolic Harnack inequality for a general type of nonlocal operators with bounded coefficients. Since the resulting problem after the ground state transformation has singular coefficients, this result does not apply straightforward to our singular operator, so, based on their scheme, we need to check every step in the Moser protocol (see [244, 246]). The Harnack inequality for singular weights can be useful in related problems. The results in this chapter for the linear problem (13.1.1) can be seen as the extension to the fractional setting of those for the heat equation developed by P. Baras and J. A. Goldstein in [41] and studied in Chapter 5. Nevertheless, the proofs that we present are significantly different. More precisely, we obtain the optimal summability required for the data in order to solve the problem and to prove the instantaneous and complete blowup for λ > ΛN,2s by using the results of Section 13.3. As a byproduct, we also prove the optimality of the power in the p Hardy potential term, i. e., that p > 1 is a supercritical power for xu 2s . This result in the local framework was obtained by Brezis and Cabré in [87]. – Second, concerning the semilinear problem (13.1.2), the main result that we obtain in this chapter is that for all 0 < λ < ΛN,2s there exists a threshold exponent, p+ (λ, s), for the existence of positive solutions. As before, by threshold we mean that when we consider an exponent p > p+ (λ, s), there are no positive supersolutions even in the weak sense, while if p < p+ (s, λ), it is possible to establish a suitable class of nonnegative data for which we can find a positive solution. We will see in particular that the threshold exponent p+ (s, λ) is the same as in the elliptic case (see [46, 171]). In fact, this critical power is related to the possibility of finding a supersolution to the elliptic problem in the whole ℝN . As in the linear problem the main ingredient is the local estimates of the solutions close to the origin. The chapter is organized as follows. In Section 7.2.21 we describe the natural functional framework associated to problems (13.1.1) and (13.1.2). We define the two notions of solution we will use along the paper: weak solutions and energy solutions. Moreover, we prove some interesting comparison principles. In Section 3 we describe the radial solutions of the corresponding homogeneous elliptic problem in ℝN . These solutions will
13.2 Functional framework: some preliminary results  435
allow us to detail the singularity of the supersolutions to problem (13.1.1) near the origin. This is a key point to perform the socalled ground state transformation (see [167]), and to obtain the nonexistence results afterwards. Section 13.3 is devoted to obtaining the weak Harnack inequality for the positive supersolutions of the problem resulting from the ground state transformation by Frank, Lieb and Seiringer, that introduces the difficulty of dealing with a kernel with singular coefficients. The goal of Section 13.4 is to study the linear problem (13.1.1) and some consequences. Finally, in Section 13.5 we consider the semilinear problem (13.1.2). Furthermore, we prove an instantaneous and complete blowup phenomenon.
13.2 Functional framework: some preliminary results Along this chapter we will always assume that Ω is a C 1,1 bounded domain of ℝN such that 0 ∈ Ω. We consider the fractional Sobolev space H0s (Ω) defined in Chapter 8 as H0s (Ω) := {u ∈ H s (ℝN ) with u = 0 a. e. in ℝN \ Ω}, endowed with the norm 1/2
‖u‖H0s (Ω) := (aN,s ∬ DΩ
u(x) − u(y)2 dx dy) , x − yN+2s
where DΩ = ℝ2N \(𝒞 Ω× 𝒞 Ω). The pair (H0s (Ω), ‖⋅‖H0s (Ω) ) yields a Hilbert space. Moreover, as we point out in Chapter 8, (−Δ)s : H0s (Ω) → H −s (Ω) is a continuous operator, where (−Δ)s is defined in (13.1.3). In what follows we will use the relation between the norm in the space H0s (Ω) and the L2 norm of the fractional Laplacian (see [138, Proposition 3.6]) 2 ‖u‖2H s (Ω) = (−Δ)s/2 uL2 (ℝN ) . 0
(13.2.1)
It is easy to check that for u and φ smooth enough, with vanishing conditions outside Ω, we have the following duality product: ∫ u(−Δ)s φ dx = aN,s ∫ ℝN
DΩ
(u(x) − u(y))(φ(x) − φ(y)) dx dy, x − yN+2s
which in particular implies the selfadjointness of (−Δ)s in H0s (Ω). See Section 8.2. We will use the Sobolev inequality in Theorem 8.2.7. The parabolic problems studied in this article are related to the fractional Hardy inequality studied in Section 9.2. Let us recall some relevant facts about the Hardy inequality.
436  13 Fractional Heat equation and Hardy potential 1. 2.
3.
The criticality of inequality (9.2.1) is motivated by the coincidence of homogeneity between the fractional Laplacian and the inverse 2spotential. As we see in Chapter 9, ΛN,2s → ΛN,2 := ( N−2 )2 , the classical Hardy constant, when 2 s tends to 1. And by scaling we know that the optimal constant is the same for every domain containing the pole of the Hardy potential. The optimal constant defined in (9.2.2) coincides for every bounded domain Ω containing the pole of the Hardy potential. That is, if 0 ∈ Ω, using (13.2.1) we can rewrite the Hardy inequality (10.1.3) as aN,s ∫ DΩ
u2 u(x) − u(y)2 dx, u ∈ H0s (Ω). dx dy ≥ Λ ∫ N,2s x2s x − yN+2s
(13.2.2)
Ω
The optimality of ΛN,2s here follows again by a scaling argument. 4. In [167] the fractional Hardy inequality plays an important role in the proof of the stability of relativistic matter in a very general setting. Consider the parabolic problem ut + (−Δ)s u = f (x, t, u) { { { (P) { u(x, t) = 0 { { { u(x, 0) = u0 (x)
in Ω × (0, T), in (ℝN \ Ω) × [0, T), if x ∈ Ω.
Let α, β ∈ (0, 1). Denote N
s
𝒯 := {ϕ : ℝ × [0, T] → ℝ, s. t. − ϕt + (−Δ) ϕ = φ, φ ∈ L (Ω × (0, T)) ∩ 𝒞 ∞
α,β
(Ω × (0, T)),
N
ϕ = 0 in (ℝ \ Ω) × (0, T], ϕ(x, T) = 0 in Ω}. Note that every ϕ ∈ 𝒯 belongs in particular to L∞ (Ω × (0, T)); see Section 12.4.1 and [224]. Moreover, according to the results in Section 12.6, ϕ ∈ 𝒯 satisfies the equation in a pointwise sense (see [48, Definition 1.3]). We define the meaning of weak solution. Definition 13.2.1. Assume u0 ∈ L1 (Ω). We say that u ∈ 𝒞 ([0, T); L1 (Ω)) is a weak supersolution (subsolution) of problem (P) if f (x, t, u) ∈ L1 (Ω × [0, T)), u ≥ (≤)0 in (ℝN \ Ω) × [0, T), u(x, 0) ≥ (≤)u0 (x) in Ω, and for all nonnegative ϕ ∈ 𝒯 we have T
s
T
∫ ∫ u(−ϕt + (−Δ) ϕ) dx dt ≥ (≤) ∫ ∫ f (x, t, u)ϕ dx dt + ∫ u0 (x)ϕ(x, 0) dx. 0 Ω
0 Ω
(13.2.3)
Ω
If u is a super and subsolution, then we say that u is a weak solution to (P). Weak solutions will be considered to formulate the optimal nonexistence results. For existence results, we will consider the classical notion of finite energy solutions.
13.2 Functional framework: some preliminary results  437
Definition 13.2.2. Assume u0 (x) ∈ L2 (Ω). We say that u ∈ L2 (0, T; H s (ℝN )) with ut ∈ L2 (0, T; H −s (Ω)) is a finite energy supersolution (respectively subsolution) of (P) if f (x, t, u) ∈ L2 (0, T; H −s (Ω)) and it satisfies T
T
∫ ∫ ut φ dx dt + aN,s ∫ ∫ 0 DΩ
0 Ω
(u(x, t) − u(y, t))(φ(x, t) − φ(y, t)) dx dy dt x − yN+2s
T
≥ (≤) ∫ ∫ f (x, t, u)φ dx dt, 0 Ω
for any nonnegative φ ∈ L2 (0, T; H0s (Ω)), φ = 0 in (ℝN \ Ω) × (0, T). If u is a super and subsolution, then u is a finite energy solution. Remark 13.2.3. If u ∈ L2 (0, T; H0s (Ω)) and ut ∈ L2 (0, T; H −s (Ω)), then by approximating with smooth functions and taking advantage of the Hilbertian structure of the space, it can be checked that u ∈ 𝒞 ([0, T]; L2 (Ω)). Note that both definitions can be considered, by scaling, in ℝN × [T1 , T2 ) with [T1 , T2 ) ⊂ [0, T). The existence and uniqueness of an energy solution to problem (P) when f does not depend on u and is in the dual space L2 (0, T; H −s (Ω)) was obtained in section 12.2 by means of a direct Hilbert space approach due to A. N. Milgram in [242] which are based on a method of Vishik in [312], see too [226]. Such arguments are essentially an extension of the Lax–Milgram theorem to parabolic problems. We recall such result. Theorem 13.2.4. Let f ∈ L2 (0, T; H −s (Ω)). Then problem (P) has a unique finite energy solution. See Chapter 12 and [224, Theorem 26] for a detailed proof in this fractional framework. Remark 13.2.5. Note that by defining T
T
Lϕ (u) := ∫ ∫ −uϕt dx dt + aN,s ∫ ∫ 0 Ω
0 DΩ T
− λ∫∫ 0 Ω
(u(x, t) − u(y, t))(ϕ(x, t) − ϕ(y, t)) dx dy dt x − yN+2s
uϕ dx dt x2s
and λ 1 ⟨φ, ϕ⟩∗ := ⟨φ(x, 0), ϕ(x, 0)⟩L2 (Ω) + aN,s (1 − )⟨φ, ϕ⟩L2 (0,T;H s (Ω)) , 0 2 ΛN,2s
(13.2.4)
438  13 Fractional Heat equation and Hardy potential thanks to Hardy’s inequality one can reproduce the proof of section 12.2 to ensure the existence and uniqueness of an energy solution to the problem ut + (−Δ)s u − λ xu2s = F(x, t) { { { (Pλ ) { u(x, t) = 0 { { { u(x, 0) = u0 (x)
in Ω × (0, T), in (ℝN \ Ω) × [0, T), if x ∈ Ω,
for F ∈ L2 (0, T; H −s (Ω)), u0 ∈ L2 (Ω) and λ < ΛN,2s . For the case λ = ΛN,2s , consider the Hilbert space H(Ω) defined as the completion of 𝒞0∞ (Ω) with respect to the norm ‖u‖2H(Ω) := aN,s ‖u‖2H s (Ω) − ΛN,2s ∫ 0
Ω
u2 dx. x2s
(13.2.5)
In [164], the author proves the following improved Hardy inequality: aN,s ‖u‖2H s (Ω) − ΛN,2s ∫ 0
Ω
u2 dx ≥ C(Ω, q, N, s)‖u‖2W τ,2 (Ω) , 0 x2s
(13.2.6)
for all s/2 < τ < s (see also [171, 22] for alternative proofs without using the Fourier transform). In Remark 10.2.2 we have defined H(Ω) as the completion of 𝒞0∞ (Ω) with respect to the norm defined in (13.2.5). Thus we see that H(Ω) ⊂ W0τ,2 (Ω) and therefore, H(Ω) is compactly embedded in Lp (Ω) for all 1 ≤ p < 2∗s (see [138, Corollary 7.2]). Therefore, the proof remains the same considering Lϕ (u) as in (13.2.4) (setting λ = ΛN,2s ), and defining the scalar product ⟨⋅, ⋅⟩∗ as 1 ⟨φ, ϕ⟩∗ := ⟨φ(x, 0), ϕ(x, 0)⟩L2 (Ω) + ⟨φ, ϕ⟩L2 (0,T;H(Ω)) , 2 where the last term follows from (13.2.5). In order to study monotonicity approaches, we will need to prove comparison results for both kind of solutions. Lemma 13.2.6 (Weak comparison principle). Let 0 ≤ λ ≤ ΛN,2s and let u, v ∈ 𝒞 ([T1 , T2 ); L1 (Ω)) be weak solutions to the problems ut + (−Δ)s u − λ xu2s = f1 { { { u = g1 { { { {u(x, T1 ) = h1 (x) v s {vt + (−Δ) v − λ x2s = f2 { { v = g2 { { { {v(x, T1 ) = h2 (x)
in Ω × (T1 , T2 ),
in (ℝN \ Ω) × [T1 , T2 ), in Ω,
in Ω × (T1 , T2 ),
in (ℝN \ Ω) × [T1 , T2 ), in Ω,
13.2 Functional framework: some preliminary results  439
respectively, where f1 , f2 ∈ L1 (Ω × (T1 , T2 )), g1 , g2 ∈ L1 ((ℝN \ Ω) × (T1 , T2 )) and h1 , h2 ∈ L1 (Ω). If f1 ≤ f2 in Ω × (T1 , T2 ), g1 ≤ g2 in (ℝN \ Ω) × [T1 , T2 ) and h1 ≤ h2 in Ω, then u ≤ v in ℝN × (T1 , T2 ). Proof. Define w = v − u. Hence, w is a weak solution of wt + (−Δ)s w − λ xw2s = f2 − f1 ≥ 0 in Ω × (T1 , T2 ), { { { w = g2 − g1 ≥ 0 in (ℝN \ Ω) × [T1 , T2 ), { { { in Ω. {w(x, T1 ) = h2 − h1 ≥ 0 Consider now Φ ∈ 𝒞0∞ (Ω × (T1 , T2 )), Φ ≥ 0 and the solution φn to the problem φ
−(φn )t + (−Δ)s φn = λ 2sn−1 1 + Φ { x + n { { { φn = 0 { { { { { φn (x, T2 ) = 0
in Ω × (T1 , T2 ), (13.2.7)
in (ℝN \ Ω) × [T1 , T2 ), in Ω,
with −(φ0 )t + (−Δ)s φ0 = Φ { { { φ0 = 0 { { { { φ0 (x, T2 ) = 0
in Ω × (T1 , T2 ),
in (ℝN \ Ω) × [T1 , T2 ),
(13.2.8)
in Ω.
Note that the existence of {φn }n∈ℕ follows from Theorem 13.2.4 after an appropriate change of variable in time. Since φn is regular in Ω×[T1 , T2 ) and bounded in ℝN ×(T1 , T2 ) (see Section 12.6), this equation can be understood in a pointwise sense. Moreover, by the weak comparison principle, we know that φn ≥ 0 and φn−1 ≤ φn in ℝN × [T1 , T2 ) for all n ∈ ℕ. Hence, by the definition of weak solutions and using that w ≥ 0 and (−Δ)s φn ≤ 0 in (ℝN \ Ω) × [T1 , T2 ), it follows that T2
T2
T2
s
T2
∫ ∫ wΦ dx dt = ∫ ∫ w(−φn )t dx dt + ∫ ∫ w(−Δ) φn dx dt − λ ∫ ∫ T1 Ω
T1 Ω
T1 Ω
T1 Ω
T2
T2
T2
T1 Ω
T1 ℝN
T1 Ω
wφn−1
x2s +
≥ ∫ ∫ w(−φn )t dx dt + ∫ ∫ w(−Δ)s φn dx dt − λ ∫ ∫
1 n
dx dt
wφn dx dt x2s
T2
= ∫ ∫(f2 − f1 )φn dx dt + ∫ w(x, T1 )φn (x, T1 ) dx ≥ 0, T1 Ω
Ω
for all Φ ∈ 𝒞0∞ (Ω × (T1 , T2 )), Φ ≥ 0. Thus, w ≥ 0 in ℝN × (T1 , T2 ), and therefore u ≤ v in ℝN × (T1 , T2 ).
440  13 Fractional Heat equation and Hardy potential Corollary 13.2.7 (Uniqueness of weak solutions for the linear problem). Let us suppose F ∈ L1 (Ω × (0, T)). Then problem (Pλ ) has at most one weak solution. The comparison result for energy solutions can be proved in a standard way, so we skip the proof (see for example [46] for a proof in the elliptic case). Lemma 13.2.8 (Energy comparison principle). Let 0 ≤ λ < ΛN,2s and let u, v ∈ L2 (T1 , T2 ; H s (ℝN )) with ut , vt ∈ L2 (T1 , T2 ; H −s (Ω)) be finite energy solutions to the problems ut + (−Δ)s u − λ xu2s = f1 { { { u = g1 { { { {u(x, T1 ) = h1 (x) λ xv2s
vt + (−Δ)s v − { { { v = g2 { { { {v(x, T1 ) = h2 (x)
= f2
in Ω × (T1 , T2 ),
in (ℝN \ Ω) × [T1 , T2 ), in Ω,
in Ω × (T1 , T2 ),
in (ℝN \ Ω) × [T1 , T2 ), in Ω,
respectively, where f1 , f2 ∈ L2 (T1 , T2 ; H −s (Ω)), g1 , g2 ∈ L2 (T1 , T2 ; L2 (ℝN \ Ω)) and h1 , h2 ∈ L2 (Ω). If f1 ≤ f2 in Ω × (T1 , T2 ), g1 ≤ g2 in (ℝN \ Ω) × [T1 , T2 ) and h1 ≤ h2 in Ω, then u ≤ v in ℝN × (T1 , T2 ). Remark 13.2.9. Note that if λ = ΛN,2s , we can obtain the same result for u, v ∈ L2 (T1 , T2 ; H(Ω)), where H(Ω) was defined in (13.2.5), only by exactly repeating this proof. Finally, consider the problem ut + (−Δ)s u = 0 { { { u(x, t) = 0 { { { { u(x, 0) = u0 (x) ⪈ 0
in Ω × (0, T), in (ℝN \ Ω) × [0, T),
(13.2.9)
if x ∈ Ω.
We enunciate a weak Harnack inequality that we will use along the section (see [155, Theorem 1.1] for a more general setting). Lemma 13.2.10 (Weak Harnack inequality). Let t0 ∈ (0, T) and r, β > 0 such that Br (0)× (t0 − 43 β, t0 + 43 β) ⊂ Ω × (0, T). If u is a nonnegative supersolution of (13.2.9) in Ω × (0, T), then there exists a positive constant C = C(N, s, r, t0 , β) such that u), ∬ u(x, t) dx dt ≤ C(ess inf + R−
R
where R− := Br (0) × (t0 − 43 β, t0 − 41 β), R+ := Br (0) × (t0 + 41 β, t0 + 43 β). As a consequence of this lemma, we can formulate the strong maximum principle. Theorem 13.2.11 (Strong maximum principle). If u is a nonnegative supersolution of (13.2.9), then u(x, t) > 0 in Ω × (0, T).
13.3 Weak Harnack inequality for a weighted problem
 441
13.3 Weak Harnack inequality for a weighted problem According to Lemma 9.4.2 we know that if λ ≤ ΛN,2s (−Δ)s u = λ has radial solutions v±α = x−
N−2s ±α 2
u x2s
in (ℝN \ {0})
(13.3.1)
if and only if α is obtained by the identity
λ = λ(α) = λ(−α) =
)Γ( N+2s−2α ) 22s Γ( N+2s+2α 4 4 )Γ( N−2s−2α ) Γ( N−2s+2α 4 4
(13.3.2)
.
In Section 9.5 we have studied the ground state representation obtained by Frank, Lieb and Seiringer in [167, Proposition 4.1]. We recall the representation result. Lemma 13.3.1 (Ground state representation). Let 0 < γ < ̄ ϕ(x) := xγ ϕ(x), then
N−2s . 2
If ϕ ∈ C0∞ (ℝN ) and
̂ 2 2 ) dξ − (ΛN,2s + ΦN,s (γ)) ∫ x−2s ϕ(x) dx ∫ ξ 2s ϕ(ξ ℝN
ℝN
2 ̄ ̄ ϕ(x) − ϕ(y) dx dy , = aN,s ∬ N+2s xγ yγ x − y
(13.3.3)
ℝ2N
where ΦN,s (γ) := 22s (
Γ( γ+2s )Γ( N−γ ) 2 2 )Γ( γ2 ) Γ( N−γ−2s 2
−
Γ2 ( N+2s ) 4 ) Γ2 ( N−2s 4
).
(13.3.4)
We will use the monotonicity property of ΨN,s obtained in Lemma 9.4.2. The result −α) and γ is defined follows from the proof of Lemma 9.4.2 noting that λ(α) = ΨN,s ( N−2s 2 in (9.4.9). Thus, for any α ∈ (0, N−2s ) thanks to Lemma 13.3.1 we know that 2 v(x, t) − v(y, t)2 dx dy ̂ 2 2 , t) dξ − λ(α) ∫ x−2s u(x, t) dx = aN,s ∬ , ∫ ξ 2s u(ξ xγ yγ x − yN+2s
ℝN
ℝN
ℝ2N
where v(x, t) := xγ u(x, t). Consider now the operator defined as Lγ (v(x)) := CN,s P. V. ∫ (v(x) − v(y))K(x, y) dy, ℝN
and K(x, y) =
1 1 1 , xγ yγ x − yN+2s
0 0 : U(1) ∩ {log w > s}dμ×dt ≤ s Then there exists C(θ, η, c0 , m, p0 ) such that (∬ w
p0
dμ dt)
U(θ)
1 p0
1
p0 . ≤ C U(1)dμ×dt
Hereafter, we will make use of the following notation. Given r > 0, we define I− (r) := (−r 2s , 0),
I+ (r) := (0, r 2s ),
Q− (r) := Br (0) × I− (r),
(13.3.9)
Q+ (r) := Br (0) × I+ (r).
(13.3.10)
The first step to prove Theorem 13.3.3 is to establish the next estimate (see [155, Proposition 3.4]). Note that we just have to consider the case where Br (x0 ) = Br (0). For simplicity, we will write Br instead of Br (0). Lemma 13.3.5. Assume that to (13.3.6). Then
1 2
≤ r < R ≤ 1 and let p > 0. Consider v ≩ 0, a supersolution
(∬ v Q− (r)
where τ := 1 +
2s N
−τp
1 τ
dμ dt) ≤ A ∬ v−p dμ dt,
(13.3.11)
Q− (R)
and
A := A(N, s, p, r, R, γ) = C(N, s, γ)(p + 1)2 (
τ
1 1 + ) . (R − r)2s R2s − r 2s
13.3 Weak Harnack inequality for a weighted problem
 445
Proof. Without loss of generality we can assume that v ≥ ε > 0 in Q− (R) (otherwise s,γ we can deal with v + ε and let ε → 0 at the end). Let q > 1 and ψ ∈ W0 (BR ) ∩ L∞ (BR ) be a nonnegative radial cutoff function such that supp(ψ) ⊆ BR with r < R, 0 ≤ ψ ≤ 1, ψ = 1 in Br and (ψ(x) − ψ(y))2 C ≤ . x − y2 (R − r)2
(13.3.12)
Using ψq+1 v−q as a test function in (13.3.6), it follows that ∫ ψq+1 v−q vt dμ + aN,s ∫ ∫ (v(x, t) − v(y, t))(ψq+1 (x)v−q (x, t) − ψq+1 (y)v−q (y, t)) dν ≥ 0. BR
ℝN ℝN
Hence, using a pointwise inequality proved in [155, Lemma 3.3], it can be deduced that 1−q 1−q 1 1 2 aN,s 2 ∫ ∫(v 2 (x, t) − v 2 (y, t)) dν ∫ ψq+1 (v1−q )t dμ + q−1 q−1
BR
≤ CqaN,s ∫ ∫ (( ℝN ℝN
Br Br
1−q
v(x, t) ) ψ(x)
+(
1−q
v(y, t) ) ψ(y)
2
)(ψ(x) − ψ(y)) dν.
Furthermore, 1−q
∫ ∫ (( ℝN ℝN
v(x, t) ) ψ(x)
≤2∫ ∫ BR BR
1−q
+(
v(y, t) ) ψ(y)
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy xγ yγ x − yN+2s
+4 ∫ ∫ ℝN \BR BR
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy . xγ yγ x − yN+2s
We set I=∫∫ BR BR
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy xγ yγ x − yN+2s
and J=
∫ ∫ ℝN \BR BR
2
)(ψ(x) − ψ(y)) dν
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy . xγ yγ x − yN+2s
446  13 Fractional Heat equation and Hardy potential Let us begin by estimating the term J. Since using Fubini’s theorem, we reach J≤ ∫ BR
v1−q (x, t) x2γ
+∫
v
1−q
BR
1 yγ
∫ {ℝN \BR }∩{x−y>R−r}
(x, t) x2γ
≤
1 xγ
for x ∈ BR and y ∈ ℝN \BR and
(ψ(x) − ψ(y))2 dy dx x − yN+2s
∫ {ℝN \BR }∩{x−y≤R−r}
(ψ(x) − ψ(y))2 dy dx x − yN+2s
≤ J1 + J2 . Setting ρ = x − y, we get J1 ≤ 4 ∫ BR
≤4∫ BR
≤
v1−q (x, t) x2γ
∫ {ℝN \BR }∩{x−y>R−r}
dy dx x − yN+2s
v1−q (x, t) dx ∫ ρ−1−2s dρ x2γ ∞
R−r
1−q
C v (x, t) dx. ∫ (R − r)2s x2γ BR
We estimate now the term J2 . Using (13.3.12), it follows that J2 ≤
v1−q (x, t) C ∫ (R − r)2 x2γ BR
∫ {ℝN \BR }∩{x−y≤R−r}
x − y2 dy dx x − yN+2s
R−r
C v1−q (x, t) ≤ dx ∫ ρ1−2s dρ ∫ (R − r)2 x2γ BR
≤
0
1−q
v (x, t) C dx, ∫ (R − r)2s x2γ BR
for some constant C possibly changing from line to line. Hence J≤
C ∫ v1−q (x, t) dμ. (R − r)2s BR
We deal now with I. Using the definition of ψ, we easily get I=
∬ BR ×BR \Br ×Br
= ∫ ∫ BR \Br Br
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy xγ yγ x − yN+2s
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy + ∫ xγ yγ x − yN+2s
∫
BR \Br BR \Br
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy xγ yγ x − yN+2s
13.3 Weak Harnack inequality for a weighted problem
+∫ ∫ Br BR \Br
 447
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy xγ yγ x − yN+2s
= I1 + I2 + I3 . Let us begin estimating I1 . Since (x, y) ∈ Br × BR \Br , we have x ≤ y, and hence I1 ≤ ∫ ∫ BR \Br Br
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy x2γ x − yN+2s
v1−q (x, t) x2γ
≤ ∫ Br
+∫
v
∫ {BR \Br }∩{x−y>R−r}
1−q
(x, t) x2γ
Br
(ψ(x) − ψ(y))2 dy dx x − yN+2s (ψ(x) − ψ(y))2 dy dx x − yN+2s
∫ {BR \Br }∩{x−y≤R−r}
≤ I11 + I12 . As in the previous computations, setting ρ = x − y and using the fact that ψ is bounded, we conclude that C v1−q (x, t) v1−q (x, t) −1−2s ρ dρ ≤ dx. I11 ≤ 4 ∫ ∫ ∫ x2γ (R − r)2s x2γ ∞
R−r
Br
Br
In the same way and using the fact that
2
(ψ(x)−ψ(y)) x−y2
≤
C , (R−r)2
we get
R−r
I12 ≤
C C v1−q (x, t) v1−q (x, t) dx. ∫ ρ1−2s dρ ≤ ∫ ∫ 2 2γ 2s (R − r) x (R − r) x2γ 0
Br
Br
Therefore, I1 ≤
C v1−q (x, t) dx. ∫ (R − r)2s x2γ BR
We deal now with I2 . Since I2 ≤ 2γ ∫
1 2
≤ r ≤ x, y ≤ R < 1, we have
∫
BR \Br BR \Br
≤
∫ BR \Br
v1−q (x, t) x2γ
+ ∫ BR \Br
≤
v1−q (x, t) (ψ(x) − ψ(y))2 dx dy x2γ x − yN+2s
v
1−q
∫ {BR \Br }∩{x−y>R−r}
(x, t) x2γ 1−q
∫ {BR \Br }∩{x−y≤R−r}
C v (x, t) dx, ∫ (R − r)2s x2γ BR
(ψ(x) − ψ(y))2 dy dx x − yN+2s (ψ(x) − ψ(y))2 dy dx x − yN+2s
448  13 Fractional Heat equation and Hardy potential by repeating the computations performed for I1 . Let us consider now the term I3 , which is the most complicated. We have I3 =
∫ r≤x≤R
+
(ψ(x) − ψ(y))2 v1−q (x, t) ( ∫ dy) dx γ x x − yN+2s yγ v
∫ r≤x≤R
y≤ x 2
1−q
(x, t) ( xγ
= I31 + I32 . If y ≤
x , 2
then x − y ≥
x 2
≥
I31 ≤ C
r 2
∫ x ≤y≤r 2
(ψ(x) − ψ(y))2 dy) dx x − yN+2s yγ
≥ 41 , and thus, v1−q (x, t) 1 ( ∫ dy) dx γ x yγ
∫ r≤x≤R
y≤ x 2 x 2
≤C
v1−q (x, t) ( ∫ ρN−1−γ dρ) dx xγ
∫ r≤x≤R
≤
0
1−q
C v (x, t) dx. ∫ (R − r)2s x2γ BR
1 yγ
To estimate I32 , we use the fact that I32 ≤ C
∫ r≤x≤R
≤C
∫ r≤x≤R
+C
v1−q (x, t) ( x2γ v1−q (x, t) ( x2γ
∫ r≤x≤R
∫ x ≤y≤r 2
≤
2γ , xγ
and hence
(ψ(x) − ψ(y))2 dy) dx x − yN+2s ∫
≤y≤r}∩{x−y>R−r} { x 2
v1−q (x, t) ( x2γ
(ψ(x) − ψ(y))2 dy) dx x − yN+2s
∫
{ x ≤y≤r}∩{x−y≤R−r} 2
(ψ(x) − ψ(y))2 dy) dx. x − yN+2s
Using the same computations as in the estimates of I11 and I12 it follows that I32 ≤
C v1−q (x, t) dx. ∫ (R − r)2s x2γ BR
Combining the estimates above, we get ∫ ψq+1 (v1−q )t dμ + aN,s ∫ ∫(v
BR
Br Br
1−q 2
(x, t) − v
Cq2 ≤ a ∫ v1−q (x, t) dμ. (R − r)2s N,s BR
1−q 2
2
(y, t)) dν
13.3 Weak Harnack inequality for a weighted problem
 449
2s
Set now θ(t) := [min{ Rt+R 2s −r 2s , 1}]+ . Then, multiplying the last inequality by θ, integrating in time in (−R2s , t) with t ∈ (−r 2s , 0) and noting that θ(t) = 1 for t ≥ −r 2s and θ (t) ≤ R2s 1−r2s , it follows that sup ∫(v1−q ) dμ + aN,s ∬ ∫(v
t∈I− (r)
Br
1−q 2
(x, t) − v
1−q 2
2
(y, t)) dν dt
Q− (r) Br
≤ Cq2 aN,s (
1 1 + ) ∬ v1−q (x, t) dμ dt. (R − r)2s R2s − r 2s
(13.3.13)
Q− (R)
, let us define w := v Recalling that τ := 1 + 2s N we have
1−q 2
. Then, applying the Hölder inequality,
4s
∬ w2τ dμ dt = ∬ w2 w N dμ dt Q− (r)
Q− (r) 2s N
2
2∗s
≤ ∫ (∫ w dμ) (∫ w dμ) I− (r) Br
2 2∗ s
dt.
Br
Since γ > 0 and R ≤ 1, we conclude that 2s N
2
2∗ s w2s ∬ w dμ dt ≤ ∫ (∫ w (x, t) dμ) (∫ 2∗ γ dx) dt. x s
2τ
2
I− (r) Br
Q− (r)
∗
Br
Now, using the Sobolev inequality obtained in Theorem 10.2.20, 2τ
2
∬ w dμ dt ≤ C sup (∫ w (x, t) dμ) t∈I− (r)
Q− (r)
2s N
Br 2
× ( ∬ ∫(w(x, t) − w(y, t)) dν dt + r −2s ∬ w2 dμ dt). Q− (r) Br
Q− (r)
Applying (13.3.13) twice in this inequality and recalling that 21 ≤ r ≤ 1, it can be checked that τ
∬ w2τ dμ dt ≤ A(q, r, R, N, s)( ∬ w2 dμ dt) , Q− (r)
Q− (R)
where A(q, r, R, N, s) = C(N, s, γ)q2 ( Setting p := q − 1, we conclude the proof.
τ
1 1 + ) . (R − r)2s R2s − r 2s
450  13 Fractional Heat equation and Hardy potential As an application of the previous estimate, we reach a control of supQ− (r) v−1 . More precisely, we have the following result. Lemma 13.3.6. Assume that 21 ≤ r < R ≤ 1 and p ∈ (0, 1]. Then, there exists a constant C = C(N, s, γ) > 0 such that every v ⪈ 0, supersolution to problem (13.3.6), satisfies 1
sup v
−1
Q− (r)
1
p p C ) ( ∬ v−p dμ dt) , ≤( α(r, R)
(13.3.14)
Q− (R)
where α(r, R) = {
(R − r)N+2s (R2s − r 2s )
N+2s 2s
if s ≥ 21 , if s < 21 .
Proof. We consider ℳ(r, p) := ( ∬ v
−p
1 p
dμ dt) .
Q− (r)
By Lemma 13.3.5, we have 1
ℳ(r, τp) ≤ A p ℳ(R, p),
. Construct now the sequences {ri }i∈ℕ and {pi }i∈ℕ by setting r0 := R > where τ := 1 + 2s N r1 > r2 > ⋅ ⋅ ⋅ > r and pi := pτi . Using again Lemma 13.3.5, we obtain 1 m
ℳ(r, pm+1 ) ≤ ℳ(rm+1 , pm+1 ) ≤ Amτ p ℳ(rm , pm ), 2s 2s −1 with Am := C(pm + 1)2 ((rm − rm+1 )−2s + (rm − rm+1 ) ). Iterating this and following the arguments in [155, Theorem 3.5], we conclude the proof.
We prove now a control of a small positive exponent of u. Lemma 13.3.7. Suppose that 21 ≤ r < R ≤ 1 and fix q ∈ (0, τ−1 ], with τ := 1 + v ⪈ 0 is a supersolution to (13.3.6), we have 1 τ
( ∬ vqτ dμ dt) ≤ α ∬ vq dμ dt, Q+ (r)
Q+ (R)
where α = α(N, s, r, R, γ) = C(N, s, γ)(
1 1 + ). (R − r)2s R2s − r 2s
2s . N
Then, if
(13.3.15)
13.3 Weak Harnack inequality for a weighted problem
 451
Proof. The proof follows similarly to the one of Lemma 13.3.5. We set a := (1 − q) ∈ 1−a [1 − τ−1 , 1) and w(x, t) := v 2 . Then, using v−a ψ2 (with ψ defined in the proof of Lemma 13.3.5) as a test function in (13.3.6), we reach 2
sup ∫ w2 (x, t) dμ + C(q)aN,s ∫ ∫(w2 (x, t) − w2 (y, t)) dν dt
t∈I+ (r)
Br
Q+ (r) Br
≤ C(q)aN,s (
1 1 + ) ∫ w2 (x, t) dμ dt. (R − r)2s R2s − r 2s Q+ (R)
Using Theorem 10.2.20 and proceeding as in Lemma 13.3.5, we get (13.3.15). Define 1 q
q
ℋ(r, q) = ( ∫ v dμ dt) . Q+ (r) 1
From (13.3.15), we get ℋ(r, τq) ≤ α q ℋ(R, q). Let us define qj := τ−j and R−r
{r + 2j rj := { R2s −r 2s 1/2s 2s {(r + 2j )
if s ≥ 21 , if s < 21 .
By Lemma 13.3.7 for rn and rn−1 , it follows that τ
ℋ(rn , q1 τ) ≤ αn ℋ(rn−1 , q1 ), 1
where αn = C(N, s, γ)( (r that r ≥
1 , 2
we get
2s
n−1 −rn )
+
1 ). 2s rn−1 −rn2s
(13.3.16)
By using the definition of rn and considering
αn ≤ C(N, s, γ)
22ns . (R − r)2s
Hence ℋ(rn , q1 τ) ≤ (C(N, s, γ)
τ
22ns ) ℋ(rn−1 , q1 ). (R − r)2s
(13.3.17)
Iterating this inequality (see [155, Theorem 3.7]), we reach the next result. Lemma 13.3.8. Assume that 21 ≤ r < R ≤ 1 and q ∈ (0, τ−1 ), with τ := 1 + supersolution v ⪈ 0 of problem (13.3.6) satisfies ∬ v dμ dt ≤ ( Q+ (r)
C
Q+ (1)dμ×dt α(r, R)
1−q q
)
q
1 q
( ∬ v dμ dt) , Q+ (R)
2s . N
Then, every
(13.3.18)
452  13 Fractional Heat equation and Hardy potential where C = C(N, s, γ) > 0 and if s ≥ 21 ,
(R − r)ω1
α(r, R) = {
if s < 21 ,
(R2s − r 2s )ω2
with ω1 , ω2 > 0 depending only on s and N. In order to apply Lemma 13.3.4 we need to estimate Q+ (1) ∩ {log v < −m − a}dμ×dt
and
Q− (1) ∩ {log v > m − a}dμ×dt ,
(13.3.19)
where m, a > 0 and v is a supersolution to (13.3.6). To do this, we need the following auxiliary result, whose proof immediately follows from Lemma 4.1 in [155]. See Section 10.3 for the elliptic case. Lemma 13.3.9. Let I ⊂ ℝ and let ψ : ℝN → [0, +∞) be a continuous function satisfying supp(ψ) = BR for some R > 0 and ‖ψ‖Y s,γ (ℝN ) ≤ C. Then, for v : ℝN × I → [0, +∞), the 0 following inequality holds: ∫ ∫ (v(x, t) − v(y, t))(−ψ2 (x)v−1 (x, t) + ψ2 (y)v−1 (y, t)) dν ℝN ℝN
≥ ∫ ∫ ψ(x)ψ(y)(log BR BR
2
v(y, t) v(x, t) 2 − log ) dν − 3 ∫ ∫ (ψ(x) − ψ(y)) dν. ψ(y) ψ(x) ℝN ℝN
With this result, following the proof of Moser in [244] we can establish the estimates in (13.3.19) (see [155, Proposition 4.2] for more details). Lemma 13.3.10. Assume that v is a supersolution to (13.3.6) in the cylinder Q := B2 × (−1, 1). Then there exists a positive constant C = C(N, s, γ) such that for some constant a = a(v), we have CB1 dμ ∀m > 0 : Q+ (1) ∩ {log v < −m − a}dμ×dt ≤ m
(13.3.20)
CB1 dμ ∀m > 0 : Q− (1) ∩ {log v > m − a}dμ×dt ≤ . m
(13.3.21)
and
Proof. Suppose that v ≥ ε > 0 in Q. Let ψ be such that ψ2 = [min{ 32 −
us denote w(x, t) := − log 2
v(x,t) . ψ(x)
Using
2
ψ v
x , 1}]+ , 2
and let
as a test function in (13.3.6) and noting that
supp(ψ ) ⊆ B3/2 , by Lemma 13.3.9 and the fact that ‖ψ‖W s,γ (ℝN ) ≤ C, it follows that 0
2
∫ ψ2 wt dμ + aN,s ∫ ∫ ψ(x)ψ(y)(w(x, t) − w(y, t)) dν B3/2
B3/2 B3/2 2
≤ 5aN,s ∫ ∫ (ψ(x) − ψ(y)) dν ≤ C. ℝN ℝN
13.3 Weak Harnack inequality for a weighted problem
We set W(t) :=
ψ2 w(x,t) dμ
∫B
3/2
∫B
3/2
rem 10.2.12, we reach
ψ2 dμ
 453
. Then by the Poincaré type inequality obtained in Theo
2
∫ ψ2 wt dμ + C ∫ (w(x, t) − W(t)) ψ(x)2 dμ ≤ C. B3/2
B3/2
Let (t1 , t2 ) ⊂ (−1, 1). Integrating in time the previous inequality, dividing by ∫B ψ2 dμ 3/2
and noting that
∫ ψ2 dμ ≤ 2N−2γ B1 dμ , B3/2
one gets t2
W(t2 ) − W(t1 ) C1 2 + ∫ ∫(w(x, t) − W(t)) dμ ≤ C2 . t2 − t1 B1 dμ (t2 − t1 ) t1 B1
We can suppose that W is differentiable. In the contrary case, it is possible to follow a discretization argument as in [155, Proposition 4.2]. By letting t2 → t1 , we get W (t) +
C1 2 ∫(w(x, t) − W(t)) dμ ≤ C2 B1 dμ
a. e. in (−1, 1).
(13.3.22)
B1
̃ ̃ t) = v(x, t) − C2 t. From (13.3.22), it follows that Define W(t) = V(t) − C2 t and w(x, W̃ (t) +
C1 2 ̃ ̃ t) − W(t)) dμ ≤ 0 ∫(w(x, B1 dμ
a. e. in (−1, 1).
(13.3.23)
B1
Note that from (13.3.23) we deduce W̃ (t) ≤ 0, and therefore calling a(v) := W(0) we get ̃ W(t) ≤ W(0) =: a(v)
for all t ∈ (0, 1).
Let t ∈ (0, 1). Then if we define + ̃ t) > m + a}, Gm (t) := {x ∈ B1 (0) : w(x, + for x ∈ Gm (t), we have
̃ ̃ ̃ t) − W(t) w(x, ≥ m + a − W(t) > 0. Thus W̃ (t) +
C1 + 2 ̃ ≤ 0. G (t) (m + a − W(t)) B1 dμ m dμ
454  13 Fractional Heat equation and Hardy potential Hence C1 + −W̃ (t) ≥ G (t) . 2 ̃ B1 dμ m dμ (m + a − W(t)) Integrating the previous differential inequality for t ∈ (0, 1) and substituting w̃ by its value yields C1 B1 dμ . Q+ (1) ∩ {log v + C2 t < −m − a}dμ×dt ≤ m Now m Q+ (1) ∩ {log v < −m − a}dμ×dt ≤ Q+ (1) ∩ {log v + C2 t < − − a} dμ×dt 2 m + Q+ (1) ∩ {C2 t > } 2 dμ×dt CB1 dμ , ≤ m which finishes the proof of (13.3.20). Estimate (13.3.21) follows using the same approach. We are now able to prove the weighted weak Harnack inequality. Proof of Theorem 13.3.3. Roughly speaking, the key to prove this result will be to define appropriate functions and parameters so that we can deduce the result from Lemma 13.3.4. We divide the proof in two cases. Let 0 < r < 1 such that Br ⊂ Ω. 1. Assume first that s ≥ 21 . We set θ1 = θ2 = 21 and define U1 (r) = Br × (1 − r 2s , 1), U2 (r) = Br × (−1, −1 + r 2s ). In the same way we consider U1 (1) = Q+ (1) and U2 (1) = Q− (1). Let w1 := e−a v−1 , w2 := ea v, where a = a(v) is defined in Lemma 13.3.10. From this result we obtain CB1 dμ Q+ (1) ∩ {log w1 > m}dμ×dt ≤ m and CB1 dμ . Q− (1) ∩ {log w2 > m}dμ×dt ≤ m Using Lemma 13.3.6, it follows that (w1 , U1 (r)) satisfies the conditions of Lemma 13.3.4 with p0 = ∞ and η any positive constant. Moreover, by Lemma 13.3.8, N (w2 , U2 (r)) satisfies the same conditions with p0 = 1 and η = N+2s < 1. Hence we conclude that sup w1 ≤ C
U1 ( 21 )
and ‖w2 ‖L1 (U ( 1 ), dμ) ≤ C.̃ 2 2
13.3 Weak Harnack inequality for a weighted problem
 455
Using these estimates and the definitions of w1 and w2 , we get ‖v‖L1 (U ( 1 ),dμ) ≤ C inf v, 2 2
2.
U1 ( 21 )
and the result follows in this case. If 0 < s < 21 , we have to change the domains by setting θ1 = θ2 = ( 21 )2s and U1 (r) = B 2s1 × (1 − r, 1), U2 (r) = B 2s1 × (−1, −1 + r). Then the same arguments as in r r the previous case allow us to conclude the proof.
From the weighted weak Harnack inequality, we immediately deduce the next corollary. Corollary 13.3.11. Let λ ≤ ΛN,2s . Assume that u is a nonnegative function such that u ≢ 0, u ∈ L1 (Ω × (0, T)) and xu2s ∈ L1 (Ω × (0, T)). If u satisfies ut + (−Δ)s u − λ xu2s ≥ 0 in the weak sense in Ω × (0, T), then there exist r1 > 0, t2 > t1 > 0 and a constant C = C(N, r1 , t1 , t2 ) such that for each cylinder Br (0) × (t1 , t2 ) ⊂⊂ Ω × (0, T), 0 < r < r1 , u ≥ Cx−
N−2s +α 2
in Br (0) × (t1 , t2 ),
where α is the singularity of the homogeneous problem given in Lemma 9.4.3. In particular, for r conveniently small we can assume that u > 1 in Br (0) × (t1 , t2 ). Finally, to end this section, we can establish a boundedness condition on the solutions of (13.3.6). s,γ
Proposition 13.3.12. Let v ∈ 𝒞 ([0, T); L2 (Ω, dμ)) ∩ L2 (0, T; W0 (Ω)) be a solution N to (13.3.6) with u0 ∈ L∞ (Ω). If f ∈ Lr (0, T; Lq (Ω)) with r, q > 1 and 1r + 2qs < 1, then ∞ v ∈ L (Ω × (0, T)). Proof. The proof follows the same idea of the classical result by D. G. Aronson and s,γ J. Serrin in [39] as in Theorem 12.4.1. We test with Gk (v) ∈ W0 (Ω) in (13.3.6), and defining 2 2 ‖Gk (v)‖2 := Gk (v)L∞ (0,T;L2 (Ω,dμ)) + Gk (v)L2 (0,T;W s,γ (Ω)) , 0
the result is obtained in a similar way as Theorem 12.4.1. See too [224, Theorem 29]. The presence of the singular term is handled in a straightforward way, so we skip the details. Remark 13.3.13. Apart from the integral version we proved, if the solution is bounded we can prove the strong Harnack inequality by classical arguments as in the elliptic case. We skip the details because we are not using this inequality in the applications. Corollary 13.3.14. If u is a solution to (13.1.1) with a sufficient regular datum f , then u ≤ N−2s Cx− 2 +α in Ω × (0, T).
456  13 Fractional Heat equation and Hardy potential
13.4 The linear problem: dependence on the spectral parameter λ Along this section we will study the problem ut + (−Δ)s u = λ xu2s + g(x, t) { { { u(x, t) = 0 { { { { u(x, 0) = u0 (x) ⪈ 0
in Ω × (0, T), in (ℝN \ Ω) × [0, T), if x ∈ Ω,
(13.4.1)
0 ∈ Ω,
where g(x, t) is a nonnegative function. The goal will be to establish some necessary and sufficient conditions on g and u0 in order to find solutions of this problem. These results correspond to the ones obtained by P. Baras and J. A. Goldstein for the heat equation in the presence of the inverse square potential (see [41]). First, we deal with the necessary summability conditions on g and u0 . Theorem 13.4.1. Let 0 < λ ≤ ΛN,2s . Assume that ũ is a positive weak supersolution to problem (13.4.1). Then g and u0 must satisfy t2
∫ ∫ x−γ g dx dt < +∞, t1 Br (0)
∫ x−γ u0 dx < +∞ Br (0)
for any cylinder Br (0) × (t1 , t2 ) ⊂⊂ Ω × (0, T), where γ was defined in (9.4.9). Proof. Fix ε > 0. Let us consider φn , the positive solution to φ
−(φn )t + (−Δ)s φn = λ 2sn−1 1 + 1 { x + n { { { φn = 0 { { { { { φn (x, T) = 0
in Ω × (−ε, T), in (ℝN \ Ω) × ( − ε, T],
(13.4.2)
in Ω,
with −(φ0 )t + (−Δ)s φ0 = 1 { { { φ0 = 0 { { { { φ0 (x, T) = 0
in Ω × (−ε, T), in (ℝN \ Ω) × ( − ε, T],
(13.4.3)
in Ω.
Note that φ0 is a strong solution in Ω × ( − ε, T], and therefore every φn is a strong solution too. Furthermore, φn−1 ≤ φn ≤ φ, where φ is the positive solution to φ
−φt + (−Δ)s φ − λ x2s = 1 { { { φ=0 { { { { φ(x, T) = 0
in Ω × (−ε, T), in (ℝN \ Ω) × (−ε, T], in Ω.
As a consequence of Corollary 13.3.11, for any cylinder Cr1 ,t1 ,t2 := Br (0) × (t1 , t2 ) ⊂⊂ Ω × (−ε, T), 0 < r < r1 , there exists a constant A = A(N, s, Cr1 ,t1 ,t2 ) such that φ(x, t) ≥
A , xγ
(x, t) ∈ Cr1 ,t1 ,t2 ,
γ=
N − 2s − α. 2
(13.4.4)
13.4 The Linear Problem, dependence on λ
 457
Since φn is regular and bounded we can use it as a test function in (13.4.1); hence we get T
∫ ∫ gφn dx dt + ∫ u0 φn (x, 0) dx 0 Ω
Ω
T
T
T
s ̃ n )t dx dt + ∫ ∫ u(−Δ) ̃ ≤ − ∫ ∫ u(φ φn dx dt − λ ∫ ∫ 0 Ω
0 ℝN
0 Ω
T
T
T
s ̃ n )t dx dt + ∫ ∫ u(−Δ) ̃ ≤ − ∫ ∫ u(φ φn dx dt − λ ∫ ∫ T
0 Ω
0 Ω
0 Ω
̃ n uφ dx dt x2s ̃ n−1 uφ
x2s +
1 n
dx dt
= ∫ ∫ ũ dx dt = C < +∞, 0 Ω
where we have used the fact that ũ is a weak supersolution (see Definition 13.2.1). Since both integrals on the left hand side are positive, in particular each one is uniformly bounded. Thus, {gφn } is an increasing sequence uniformly bounded in L1 (Ω × (0, T)), and applying the monotone convergence theorem and (13.4.4) we get t2
+α − N−2s 2
C ∫ ∫ x
t2
g dx dt ≤ ∫ ∫ gφ dx dt t1 Br (0)
t1 Br (0)
T
T
≤ ∫ ∫ gφ dx dt = lim ∫ ∫ gφn dx dt < +∞. n→∞
0 Ω
0 Ω
Likewise, {u0 φn (x, 0)} is also an increasing sequence, uniformly bounded in L1 (Ω), and thus, choosing t1 and t2 so that 0 ∈ (t1 , t2 ) ⊊ (−ε, T) as above, we conclude C̃ ∫ x−
N−2s +α 2
u0 (x) dx ≤ ∫ u0 φ(x, 0) dx < +∞. Ω
Br (0)
Conversely, we would like to find the optimal summability conditions on g and u0 to prove existence of a weak solution. In this direction, note that if g ∈ L2 (0, T; H −s (Ω)) and u0 ∈ L2 (Ω), by Remark 13.2.5 we can ensure the existence of an energy solution of problem (13.4.1) when λ < ΛN,2s , and in H(Ω) (see Remark 10.2.2 and (13.2.5)) for λ = ΛN,2s . A sharper result, for a more general class of data, is the following. Theorem 13.4.2. Assume that 0 < λ ≤ ΛN,2s and that g and u0 satisfy u ∫ 0γ dx < +∞, x
Ω
T
∫∫ 0 Ω
g dx dt < +∞, xγ
where γ is defined in (9.4.9). Then problem (13.4.1) has a positive weak solution.
458  13 Fractional Heat equation and Hardy potential Proof. Consider the approximated problems { { { { { { { { { { { { { { {
unt + (−Δ)s un = λ un (x, t) > 0
un−1 x2s + n1
+ gn
in Ω × (0, T), in Ω × (0, T),
un (x, t) = 0
in (ℝN \ Ω) × [0, T),
un (x, 0) = Tn (u0 (x))
if x ∈ Ω,
(13.4.5)
where { { { { { { { { { { { { {
u0t + (−Δ)s u0 = g1
in Ω × (0, T),
u0 (x, t) > 0
in Ω × (0, T),
u0 (x, t) = 0
in (ℝN \ Ω) × [0, T),
u0 (x, 0) = T1 (u0 (x))
if x ∈ Ω,
(13.4.6)
with gn = Tn (g) and as customary, Tn (g) = {
g
if g ≤ n,
g n g
if g > n.
By Lemma 13.2.6, it follows that u0 ≤ u1 ≤ ⋅ ⋅ ⋅ ≤ un−1 ≤ un in ℝN × (0, T). Note that since the right hand sides of these problems are bounded, every un is actually an energy solution. Consider φ, the solution of the problem { { { { { { { { { { { { { { {
φ
−φt + (−Δ)s φ − λ x2s = 1
in Ω × (−ε, T),
φ>0
in Ω × (−ε, T),
φ=0
on (ℝN \ Ω) × (−ε, T],
φ(x, T) = C
in Ω,
(13.4.7)
where C > 0. As a consequence of the weak Harnack inequality, Theorem 13.3.3 and Proposition 13.3.12, for any cylinder Br (0) × [t1 , t2 ] ⊂ Ω × (−ε, T) we find c1 , c2 > 0 such that c c1 ≤ φ(x, t) ≤ 2γ . xγ x
(13.4.8)
Since φ also belongs to L2 (0, T; H0s (Ω)), we can use it as a test function in (13.4.5). Thus, T
T
T
∫ ∫ (un )t φ dx dt + ∫ ∫ un (−Δ)s φ dx dt = λ ∫ ∫ 0 Ω
0 ℝN
0 Ω
un−1 φ
x2s +
1 n
T
dx dt + ∫ ∫ gn φ dx dt 0 Ω
13.4 The Linear Problem, dependence on λ T
T
0 Ω
0 Ω
 459
u φ ≤ λ ∫ ∫ n 2s dx dt + ∫ ∫ gn φ dx dt. x Integrating in time and applying (13.4.7) and (13.4.8), we conclude that T
T
C ∫ un (x, T) dx + ∫ ∫ un dx dt ≤ ∫ ∫ gn φ dx dt + ∫ Tn (u0 (x))φ(x, 0) dx 0 Ω
Ω
0 Ω
Ω
T
≤ ∫ ∫ gφ dx dt + ∫ u0 (x)φ(x, 0) dx 0 Ω
T
≤ C∫∫ 0 Ω
Ω
u (x) g dx dt + C ∫ 0 γ dx < +∞, xγ x Ω
by hypothesis. Hence, since the sequence {un }n∈ℕ is increasing, we can define u := limn→∞ un and conclude that u ∈ L1 (Ω×(0, T)) by applying the monotone convergence theorem. Note that using the same computations as above and integrating in Ω × [0, t] with t ≤ T, by considering the estimates on {un }n∈ℕ , we reach T
sup ∫ un (x, t) dx + ∫ ∫ un dx dt ≤ C
t∈[0,T]
for all n.
(13.4.9)
0 Ω
Ω
Fix T1 > T and define φ̃ as the unique solution to the problem { { { { { { { { { { { { {
−φ̃ t + (−Δ)s φ̃ = 1
in Ω × (−ε, T1 ),
φ̃ > 0
in Ω × (−ε, T1 ),
φ̃ = 0
on (ℝN \ Ω) × (−ε, T1 ],
φ(x, T1 ) = 0
in Ω.
(13.4.10)
̃ t) ≥ C̄ > 0 for all (x, t) ∈ Br (0)×[0, T], where It is clear that φ̃ ∈ L∞ (Ω×(−ε, T1 )) and φ(x, Br (0) ⊂⊂ Ω. Now, using φ̃ as a test function in (13.4.5) and integrating in Ω × (0, T), it follows that T
T
̃ T) dx dt + ∫ ∫ un dx dt ≥ λ ∫ ∫ ∫ un (x, T)φ(x, 0 Ω
Ω
0 Ω
un−1 φ̃
x2s +
1 n
dx dt.
Thus T
λ∫∫ 0 Ω
un−1 φ̃
x2s +
1 n
T
dx dt ≤ C sup ∫ un (x, t) dx + ∫ ∫ un dx dt ≤ C {t∈[0,T]}
Ω
0 Ω
for all n.
460  13 Fractional Heat equation and Hardy potential Hence T
un−1
∫∫ 0 Ω
x2s +
T 1 n
dx dt = ∫ ∫ 0 Br (0) T
≤ C∫∫ 0 Ω
T
un−1
1 n
x2s + un−1 φ̃
x2s +
1 n
dx dt + ∫
un−1
∫
0 Ω\Br (0)
x2s +
1 n
dx dt
T
dx dt + C ∫ ∫ un−1 dx dt ≤ C. 0 Ω
Therefore, by the monotone convergence theorem we find un−1
x2s
+
1 n
+ gn ↑
u +g x2s
strongly in L1 (Ω × (0, T)).
To conclude that u is a weak solution to problem (13.4.1), it remains to check that u ∈ 𝒞 ([0, T); L1 (Ω)). We claim that {un }n∈ℕ is a Cauchy sequence in 𝒞 ([0, T]; L1 (Ω)), and hence the result follows. In order to prove this claim, we closely follow the arguments in [267]. For n, m ∈ ℕ such that n ≥ m, denote un,m := un − um and gn,m := gn − gm . Clearly, un,m , gn,m ≥ 0. We set t
Cn,m := λ ∫ ∫ 0 Ω
t
un,m dx dτ + ∫ ∫ gn,m T1 (un,m ) dx dτ, x2s 0 Ω
and then Cn,m → 0 as n, m → ∞. By the definition of the approximated problems in (13.4.5) and the linearity of the operator, for t ≤ T, t
t
∫ ∫(un,m )t T1 (un,m ) dx dτ + ∫ ∫(−Δ)s (un,m )T1 (un,m ) dx dτ ≤ Cn,m . 0 Ω 2
Since un,m ∈ L
0 Ω
(0, T; H0s (Ω)), t
it follows that (see Chapter 12 for details) t
2 ∫ ∫(−Δ) (un,m )T1 (un,m ) dx dτ ≥ ∫T1 (un,m )H s (Ω) dτ ≥ 0, s
0 Ω
0
and therefore, t
∫ ∫(un,m )t T1 (un,m ) dx dτ ≤ Cn,m . 0 Ω
Let us define s
Ψ(s) := ∫ T1 (σ) dσ. 0
0
13.4 The Linear Problem, dependence on λ
 461
Since un ∈ 𝒞 ([0, T]; L2 (Ω)), we have t
∫ ∫(un,m )t T1 (un,m ) dx dτ = ∫(Ψ(un,m )(t) − Ψ(un,m )(0)) dx. 0 Ω
Ω
Thus ∫ Ψ(un,m )(t) dx ≤ Cn,m + ∫ Ψ(un,m )(0) dx. Ω
Ω
Taking into account that Ψ(un,m )(0) = Ψ(Tn (u0 ) − Tm (u0 )) and by noting that Ψ(s) ≤ s and Tn (u0 ) − Tm (u0 ) → 0 strongly in L1 (Ω) as n, m → ∞, we obtain ∫ Ψ(un,m )(t) dx → 0 as n, m → ∞,
uniformly in t.
Ω
Now, since ∫ un,m 1
Ω
we conclude that un,m → 0 uniformly in t. Thus {un }n∈ℕ is a Cauchy sequence in 𝒞 ([0, T]; L1 (Ω)) and passing to the limit in the weak formulation of the approximated problems, one obtains that u is a positive weak solution of problem (13.4.1) in Ω × (0, T). Next, we see that ΛN,2s provides a real obstruction on λ. Proposition 13.4.3. If λ > ΛN,2s , problem (13.4.1) has no positive weak supersolution. Proof. Consider ũ as a weak supersolution to the problem ut + (−Δ)s u − ΛN,2s xu2s = (λ − ΛN,2s ) xu2s + g { { { u(x, t) > 0 { { { { u(x, t) = 0
in Ω × (0, T), in Ω × (0, T),
(13.4.11)
N
in (ℝ \ Ω) × [0, T).
Since in the left hand side the constant is ΛN,2s , we are in the case α = 0, and by Theorem 13.4.1, necessarily ((λ − ΛN,2s )
N−2s ũ + g)x− 2 ∈ L1 (Br (0) × (t1 , t2 )), 2s x
for any Br (0) × (t1 , t2 ) ⊂⊂ Ω × (0, T) small enough. In particular, this implies (λ − ΛN,2s )
N−2s ũ x− 2 ∈ L1 (Br (0) × (t1 , t2 )), 2s x
462  13 Fractional Heat equation and Hardy potential and hence, applying Corollary 13.3.11 again, (λ − ΛN,2s )x−N ∈ L1 (Br (0) × (t1 , t2 )), which is a contradiction. Therefore, there does not exist a positive supersolution if λ > ΛN,2s . Remark 13.4.4. The previous nonexistence result implies that for λ > ΛN,2s an instantaneous and complete blowup phenomenon occurs. The proof is a simple adaptation of Theorem 13.5.4, where this result is proved for a more involved semilinear problem. Furthermore, we can state a nonexistence result that shows the optimality of the p power p = 1 in the singular term xu 2s . The proof for this nonlocal problem closely follows the classical case, due to H. Brezis and X. Cabré (see [87]). Theorem 13.4.5. Let p > 1, and let u ≥ 0 satisfy ut + (−Δ)s u ≥
up x2s
in Ω × (0, T),
in the weak sense. Then u ≡ 0. Proof. Consider a cylinder Bτ (0) × (t1 , t2 ). If u ⪈ 0, by the maximum principle (Theorem 13.2.11), we know that there exists ε > 0 so that u≥ε
in Bτ (0) × (t1 , t2 ).
Let us define 1 { (p−1)εp−1 − ϕ(s) := { 1 { εp (s − ε)
1 (p−1)sp−1
if s ≥ ε, if s < ε.
Note that 0 ≤ ϕ < +∞ in [ε, +∞), ϕ(ε) = 0, ϕ (ε) = 0, and ϕ is a 𝒞 1 function satisfying ϕ (s) = s1p for s ≥ ε. Moreover, since ϕ is concave, it follows that (−Δ)s (ϕ(u)) ≥ ϕ (u)(−Δ)s u and thus, (ϕ(u))t + (−Δ)s (ϕ(u)) ≥ ϕ (u)(ut + (−Δ)s u) ≥
1 x2s
in Bτ (0) × (t1 , t2 ),
with u ≥ ε. Without loss of generality we can assume that τ = 1. Define w(x, t) := (t − t1 )ϑ(x), where ϑ(x) := {
1 ) log( x
0
if x < 1, if x ≥ 1.
Then w(x, t1 ) = 0 in B1 (0) and w(x, t) = 0 in ℝN \B1 (0).
13.4 The Linear Problem, dependence on λ
 463
We claim that wt + (−Δ)s w ≤
C x2s
in B1 (0) × (t1 , t2 ),
where C ≡ C(t1 , t2 ) > 0. Indeed, we have wt + (−Δ)s w = ϑ(x) + (t − t1 )(−Δ)s ϑ. C1 x2s
It is clear that ϑ(x) ≤ show that
in B1 (0) × (t1 , t2 ), and hence to prove the claim we have to C2 x2s
(−Δ)s ϑ(x) ≤
for all x ∈ B1 (0).
In fact, (−Δ)s ϑ(x) = aN,s ∫ ℝN
(ϑ(x) − ϑ(y)) dy x − yN+2s
= aN,s ∫ {y1}
= I1 (x) + I2 (x).
We follow the arguments in [158] to estimate the integrals above. By setting r := x and ρ := y, we have x = rx , y = ρy , where x  = y  = 1. We avoid the constant aN,s in the next computations. Thus, 1
ρ dH N−1 (y ) I1 (x) = ∫ log( )ρN−1 ( ∫ ) dρ. r rx − ρy N+2s 0
y =1
ρ
Calling σ := r , I1 (x) =
D1 (x) , x2s
where 1 r
D1 (r) = ∫ log(σ)σ N−1 K(σ) dσ, 0
and N−1
K(σ) := ∫ y =1
π
sinN−2 (η) dH N−1 (y ) π 2 = 2 dη. ∫ N+2s x − σy N+2s Γ( N−1 ) (1 − 2σ cos(η) + σ 2 ) 2 2 0
464  13 Fractional Heat equation and Hardy potential In the same way we have +∞
D (x) I2 (x) = 2 2s , x
where D2 (r) = − log(r) ∫ σ N−1 K(σ) dσ. 1 r
Combining the estimates above, we get D(x) x2s
(−Δ)s ϑ(x) = with 1 r
D(r) := D1 (r) + D2 (r) = ∫ log(σ)σ
N−1
+∞
K(σ) dσ − log(r) ∫ σ N−1 K(σ) dσ.
0
1 r
Note that K(σ) ≤ C1 − σ−1−2s as σ → 1 and K( σ1 ) = σ N+2s K(σ) for all σ > 0. If s ≤ 21 , then using the behavior of K at +∞ we can easily prove that D1 (r) + D2 (r) ≤ C for all r ≤ 1. To study the general case we need to do some sharp computations. Since +∞
D(r) ≤ ∫ log(σ)σ N−1 K(σ) dσ = D,̄ 0
to finish we have to show that D < ∞. Note that 1
+∞
0
1
D̄ = ∫ log(σ)σ N−1 K(σ) dσ + ∫ log(σ)σ N−1 K(σ) dσ. Putting θ := that
1 σ
in the first integral and using the fact that K( θ1 ) = θN+2s K(θ), it follows +∞
D̄ = ∫ K(σ) log(σ)(σ N−1 − σ 2s−1 ) dσ. 1
Now, due to the behavior of K near 1 and ∞ we obtain 0 < D̄ < ∞. Thus (−Δ)s ϑ(x) ≤
D̄ . x2s
Hence wt + (−Δ)s w ≤
C x2s
in B1 (0) × (t1 , t2 ),
where C = C1 + (t2 − t1 )D̄ and the claim follows.
13.5 Semilinear problem, existence and nonexistence
 465
Fixing ε > 0 such that εC ≪ 1, we obtain (ϕ(u) − εw)t + (−Δ)s (ϕ(u) − εw) ≥ 0
in B1 (0) × (t1 , t2 )
in the weak sense. Since (ϕ(u) − εw)(x, t1 ) ≥ 0 in B1 (0) and (ϕ(u) − εw)(x, t1 ) ≥ 0 in ℝN \B1 (0)×(t1 , t2 ), the comparison principle implies that ϕ(u)−εw ≥ 0 in Bτ (0)×(t1 , t2 ). Now, w is unbounded and this reaches a contradiction with the fact that ϕ is, however, a bounded function. The proof is finished. Thus, as a straightforward consequence we obtain the following result. Corollary 13.4.6. Let g ∈ L1 (Ω × (0, T)), g ≥ 0, λ > 0 and p > 1. Therefore, the problem p
ut + (−Δ)s u = λ xu 2s + g { { { { u(x, t) = 0 { { { { { u(x, 0) = u0 (x) ⪈ 0
in Ω × (0, T), in (ℝN \ Ω) × [0, T), if x ∈ Ω
has no positive weak solution.
13.5 Existence and nonexistence results for a semilinear problem The goal of this section is to study how the addition of a semilinear term of the form up , with p > 1, interferes with the solvability of the previous problems. As in the classical heat equation (see [19]), the relevant feature is that, as in the elliptic case, for every 0 < λ < ΛN,2s there exists a threshold for the existence, p+ (λ, s), that depends on the spectral parameter. Indeed, we will consider the problem { { { { { { { { { { { { { { {
ut + (−Δ)s u = λ xu2s + up + f
in Ω × (0, T),
u(x, t) > 0
in Ω × (0, T),
u(x, t) = 0
in (ℝN \ Ω) × [0, T),
u(x, 0) = u0 (x)
if x ∈ Ω,
(13.5.1)
with p > 1, 0 < λ < ΛN,2s , f (x, t) ≥ 0 and u0 (x) ≥ 0. By weak or energy solutions of this problem, we mean solutions in the sense of Definition 13.2.1 and Definition 13.2.2 by fixing F = λ xu2s + up + cf . We will prove that there exists such critical exponent p+ (λ, s) so that one can prove existence of solution for problem (13.5.1) when 1 < p < p+ (λ, s), and nonexistence for p > p+ (λ, s). Following the same ideas as in Chapter 11, one can expect p+ (λ, s) to depend on s and λ, and in particular to satisfy p+ (λ, s) = 1 +
2s
N−2s 2
−α
=1+
2s . γ
466  13 Fractional Heat equation and Hardy potential Note that if λ = ΛN,2s , namely, α = 0, then p+ (λ, s) = 2∗s − 1, and if λ = 0, i. e., α = N−2s , 2 then p+ (λ, s) = ∞. We will need some auxiliary results that allow us to build a solution whenever we have a supersolution. To prove existence of a weak solution to (13.5.1) with L1 data from a weak supersolution, we will consider the solution obtained as limit of solutions of approximated problems (see for instance [129] in the local parabolic operators case). Lemma 13.5.1. If ū ∈ 𝒞 ([0, T); L1 (Ω)) is a weak positive supersolution to the equation in (13.5.1) with λ ≤ ΛN,2s and f ∈ L1 (Ω × (0, T)), then there exists a positive minimal weak solution to problem (13.5.1) obtained as limit of solutions of approximated problems. Proof. If ū is a positive supersolution to (13.5.1) with λ ≤ ΛN,2s , we construct a sequence {un }n∈ℕ starting with { { { { { { { { { { { { {
u0t + (−Δ)s u0 = T1 (f )
in Ω × (0, T),
u0 (x, t) > 0
in Ω × (0, T),
u0 (x, t) = 0
in (ℝN \ Ω) × [0, T),
u0 (x, 0) = T1 (u0 (x))
if x ∈ Ω.
(13.5.2)
By the weak comparison principle (Lemma 13.2.6), it follows that u0 ≤ ū in ℝN × (0, T). Define in a recurrence way { { { { { { { { { { { { { { {
unt + (−Δ)s un = λ un (x, t) > 0
un−1 x2s + n1
+ upn−1 + Tn (f )
in Ω × (0, T), in Ω × (0, T),
un (x, t) = 0
in (ℝN \ Ω) × [0, T),
un (x, 0) = Tn (u0 (x))
if x ∈ Ω.
(13.5.3)
We have {un }n∈ℕ ⊂ 𝒞 ([0, T); L1 (Ω)) ∩ Lp ([0, T); Lp (Ω)) (see [224]). As before, it follows that u0 ≤ ⋅ ⋅ ⋅ ≤ un−1 ≤ un ≤ ū in ℝN × (0, T), so we obtain the pointwise limit u := lim un that verifies u ≤ ū and { { { { { { { { { { { { { { {
ut + (−Δ)s u = λ xu2s + up + f
in Ω × (0, T),
u(x, t) > 0
in Ω × (0, T),
u(x, t) = 0
in (ℝN \ Ω) × [0, T),
u(x, 0) = u0 (x)
if x ∈ Ω
(13.5.4)
in the weak sense. The fact that u ∈ 𝒞 ([0, T); L1 (Ω)) follows as in the proof of Theorem 13.4.2. Finally, we prove the minimality of the solution found above. Assume that v is another positive weak solution of problem (13.5.4). Then, in particular it is a positive weak supersolution. By the weak comparison principle we can prove by recurrence that the approximated solution verifies that un ≤ v in any (x, t) ∈ Ω × (0, T) and for all n ∈ ℕ. Therefore u ≤ v.
13.5 Semilinear problem, existence and nonexistence
 467
Likewise, if the supersolution belongs to the energy space, the minimal solution obtained by approximations as above is also an energy solution. Lemma 13.5.2. If ū ∈ L2 (0, T; H0s (Ω)) with ū t ∈ L2 (0, T; H −s (Ω)) is a positive finite energy supersolution to (13.5.1) with λ ≤ ΛN,2s and f ∈ L2 (0, T; H −s (Ω)), then there exists a positive minimal energy solution to problem (13.5.1) obtained as limit of solutions of the approximated problems. Proof. Proceeding as in the proof of Lemma 13.5.1 and using the comparison principle for energy solutions (Lemma 13.2.8), we can build a sequence {un }n∈ℕ of energy solutions of the approximated problems (13.5.3), so that u0 ≤ u1 ≤ ⋅ ⋅ ⋅ ≤ un ≤ ⋅ ⋅ ⋅ ≤ ū
in ℝN × (0, T).
Hence, by the monotone convergence theorem we can define u := limn→∞ un ≤ u.̄ Moreover, applying the energy formulation of un , ‖un ‖2L2 (0,T;H s (Ω)) 0
T
T
(u (x, t) − un (y, t))2 2 = ∫ un (⋅, t)H s (Ω) dt = ∫ ∫ n dx dy dt 0 x − yN+2s 0
0 Q
T
T
0 Ω
0 Ω
u2 {∫ ∫(λ n2s + up+1 = n + Tn (f )un ) dx dt − ∫ ∫(un )t un dx dt} aN,s x 1
=
T
1
aN,s
{∫ ∫(λ 0 Ω
Ω
1 ∫ un (x, 0)2 dx} 2
+
Ω
≤
u2n 1 2 + up+1 n + Tn (f )un ) dx dt − ∫ un (x, T) dx 2 x2s
1
aN,s
≤ C.
T
{∫ ∫(λ 0 Ω
1 ū 2 ̄ dx dt + ∫ u(x, ̄ 0)2 dx} + ū p+1 + f u) 2 x2s Ω
Thus, up to a subsequence, we know that un ⇀ u in L2 (0, T; H0s (Ω)). Since for every 0 ≤ t ≤ T, (un )t H −s (Ω) =
sup ∫ (un )t φ dx ‖φ‖H s (Ω) ≤1 0 Ω
≤ aN,s
(u (x) − u (y))(φ(x) − φ(y)) n dx dy sup ∫ n N+2s x − y ‖φ‖H s (Ω) ≤1 0 Q
+
sup {∫ upn φ dx + λ ∫
‖φ‖H s (Ω) ≤1 0
Ω
Ω
un φ dx + ∫ Tn (f )φ} x2s Ω
468  13 Fractional Heat equation and Hardy potential
≤
sup {‖un ‖H0s (Ω) ‖φ‖H0s (Ω) + ∫ ū p φ dx + λ ∫
‖φ‖H s (Ω) ≤1
Ω
0
≤ C(‖un ‖H0s (Ω) +1 + ‖f ‖L2 (Ω) ).
Ω
̄ uφ dx + ∫ fφ dx} x2s Ω
Hence, T
2 ∫ (un )t H −s (Ω) dt ≤ C(‖un ‖2L2 (0,T;H s (Ω)) +1 + ‖f ‖2L2 (0,T;L2 (Ω)) ) ≤ C, 0
0
and therefore, up to a subsequence, (un )t ⇀ ut in L2 (0, T; H −s (Ω)), and we can pass to the limit to conclude that u is a finite energy solution to (13.5.1). The minimality follows as in the case of weak solutions.
13.5.1 Nonexistence results for p > p+ (λ, s), instantaneous and complete blowup Assume first that p is greater than the threshold exponent p+ (λ, s). Then we can formulate the nonexistence result as follows. Theorem 13.5.3. Let λ ≤ ΛN,2s and suppose that (f , u0 ) ∈ L1 (Ω × (0, T)) × L1 (Ω). If p > p+ (λ, s), then problem (13.5.1) has no positive weak supersolution. In the case where f ≡ 0, the unique nonnegative supersolution is u ≡ 0. Proof. Without loss of generality, we can assume f ∈ L∞ (Ω × (0, T)). We argue by contradiction. Assume that ũ is a positive weak supersolution of (13.5.1). Then ũ t + (−Δ)s ũ − λ xũ 2 ≥ 0 in Ω × (0, T) in the weak sense. Since ũ is also a weak supersolution in any BR (0) × (T1 , T2 ) ⊂⊂ Ω × (0, T), by Lemma 13.5.1, the problem { { { { { { { { { { { { { { {
ut + (−Δ)s u = λ xu2s + up + f u(x, t) > 0
in BR (0) × (T1 , T2 ), in BR (0) × (T1 , T2 ),
u(x, t) = 0
in (ℝN \ BR (0)) × [T1 , T2 ),
̃ T1 ) u(x, T1 ) = u(x,
if x ∈ BR (0)
(13.5.5)
has a solution u obtained by approximation of truncated problems in BR (0)×(T1 , T2 ). In particular u = lim un , with un ∈ L∞ (BR (0)×(T1 , T2 )) being the energy solution to (13.5.3) in BR (0) × (T1 , T2 ). Note that ut +(−Δ)s u−λ xu2s ≥ 0 in Br1 (0)×(T1 , T2 ) in the weak sense, and therefore, by Corollary 13.3.11, for any cylinder Br (0) × (t1 , t2 ), with 0 < r < r1 < R, 0 < T1 < t1 < N−2s t2 < T2 ≤ T, there exists a constant C = C(N, r1 , t1 , t2 ) such that u ≥ Cx− 2 +α and then for r small enough u > 1 in Br (0) × (T1 , T2 ).
 469
13.5 Semilinear problem, existence and nonexistence
In particular, since u ∈ L1 (Ω × (0, T)), using the fact that u ≥ 1 in Br (0) × (T1 , T2 ), we reach log(u) ∈ Lp (Br (0) × (t1 , t2 )), for all p ≥ 1. By a suitable scaling, we can assume that the cylinder is Br (0) × (0, τ). ϕ2
Let ϕ ∈ 𝒞0∞ (Br (0)). Then using u as a test function in the approximated probn lems (13.5.3) and applying the Picone (Theorem 8.5.1) and Sobolev (Theorem 8.2.7) inequalities, τ
τ
τ
2 ∫ ∫ up−1 n ϕ dx dt ≤ ∫ ∫ 0 Br (0)
0 Br (0)
ϕ2 ϕ2 unt dx dt + ∫ ∫ (−Δ)s un dx dt un un 0 Br (0)
≤ ∫ log un (x, τ)ϕ2 dx + C (N, s, τ)‖ϕ‖2H s (Ω) . 0 Br (0)
Therefore, passing to the limit as n tends to infinity and considering that u ≥ Cx− in Br (0) × (0, τ), we obtain
N−2s +α 2
τ
∫ log u(x, τ)ϕ2 dx + C (N, s, τ)‖ϕ‖2H s (Ω) ≥ ∫ ∫ up−1 ϕ2 dx dt 0 0 Br (0)
Br (0)
τ
ϕ2
≥ C∫ ∫
x(p−1)(
0 Br (0)
N−2s −α) 2
dx dt.
Using the Hölder and Sobolev inequalities, it follows that 2s
2 ∗
N 2s ∗ N ∫ log(u(x, τ))ϕ2 dx ≤ ( ∫ ϕ2s dx) ( ∫ log u(x, τ) 2s dx)
Br (0)
Br (0)
Br (0)
2s
N N ≤ ( ∫ log u(x, τ) 2s dx) 𝒮 ‖ϕ‖2H s (Ω) , 0
Br (0)
where 𝒮 is the optimal constant in the Sobolev embedding. Thus we have 2s
N N [C (N, s, τ) + ( ∫ log u(x, τ) 2s dx) 𝒮 ]‖ϕ‖2H s (Ω) ≥ C ∫ 0
Br (0)
Br (0)
ϕ2 x(p−1)(
N−2s −α) 2
dx.
Since p > p+ (λ, s), for a cylinder small enough, (p − 1)( N−2s − α) > 2s and we obtain a 2 contradiction with the Hardy inequality. The previous nonexistence result is very strong in the sense that a complete and instantaneous blowup phenomenon occurs. That is, if un is the solution to the approximated problems (13.5.3), then un (x, t) → ∞ as n → ∞, where (x, t) is an arbitrary point in Ω × (0, T).
470  13 Fractional Heat equation and Hardy potential Theorem 13.5.4. Let un be a solution to problem (13.5.3) with p > p+ (λ, s). Then un (x0 , t0 ) → ∞, for all (x0 , t0 ) ∈ Ω × (0, T). Proof. Without loss of generality, we can assume that λ ≤ ΛN,2s . The existence of a positive solution to problem (13.5.3) is clear and, due to comparison principle, we know that un ≤ un+1 for all n ∈ ℕ. Suppose by contradiction that there exists (x0 , t0 ) ∈ Ω × (0, T) such that un (x0 , t0 ) → C0 < ∞
as n → ∞.
By using the Harnack inequality (see Lemma 13.2.10), there exists s0 > 0 and a positive constant C = C(N, s0 , t0 , β) such that un ≤ C, ∬ un (x, t) dx dt ≤ C ess inf + R0
R−0
where R−0 = Bs0 (x0 ) × (t0 − 43 β, t0 − 41 β) and R+0 = Bs0 (x0 ) × (t0 + 41 β, t0 + 43 β). Without loss of generality, we can suppose x0 = 0. Otherwise, we can find a finite M sequence of points {xi }M i=0 , ending with xM = 0, and of radii {si }i=0 such that Bsi (xi ) ⊂ Ω, Bsi (xi ) ∩ Bsi+1 (xi+1 ) ≠ 0, for all i = 0, . . . , M, and, by the Hanarck inequality, un , ∬ un (x, t) dx dt ≤ C ess inf + Ri
R−i
where R−i = Bsi (xi ) × (ti − 43 β, ti − 41 β) and R+i = Bsi (xi ) × (ti + 41 β, ti + 43 β), ti ∈ (0, T), and β is small enough so that ti − 43 β > 0 and ti + 43 β < T for all i = 0, . . . , M. Let us choose now ti = ti−1 − β for i = 1, . . . , M. Note that in this case (ti +
1 3 3 1 β, t + β) = (ti−1 − β, ti−1 − β), 4 i 4 4 4
and in particular, R+i ∩ R−i−1 ≠ 0. Thus, un (x, t) ≤ ess ∬ un (x, t) dx dt ≤ ess inf + RM
R−M
≤
≤
1 R+M ∩ R−M−1 
∬
inf
R+M ∩R−M−1
un (x, t)
un (x, t) dx dt
R+M ∩R−M−1
1 ∬ un (x, t) dx dt R+M ∩ R−M−1  R−M−1
≤ . . . ≤ C ∬ un (x, t) dx dt ≤ C.̃ R−0
13.5 Semilinear problem, existence and nonexistence
 471
Therefore, supposing x0 = 0, by the monotone convergence theorem there exists u ≥ 0 such that un ↑ u strongly in L1 (Br (0) × (t1 , t2 )). Let now φ be the solution to the problem { { { { { { { { { { { { {
−φt + (−Δ)s φ = χBr (0)×[t1 ,t2 ]
in Ω × (0, T),
φ(x, t) > 0
in Ω × (0, T),
φ(x, t) = 0
in (ℝN \ Ω) × [0, T),
φ(x, T) = 0
in Ω.
(13.5.6)
Note that due to the regularity of the right hand sides of problems (13.5.3) and (13.5.6), both un and φ are in the energy space, and thus both can be used as test functions in the energy formulation of the problems. Indeed, considering first un as test function in (13.5.6) and then, after integrating by parts, φ in (13.5.3), and defining η := infBr (0)×(t1 ,t2 ) φ(x, t), we have t2
C ≥ ∫ ∫ un (x, t) dx dt t1 Br (0) T
≥ λ∫∫ 0 Ω
un−1
x2s +
t2
≥ λη ∫ ∫ t1 Br (0)
T
T
φ dx dt + ∫ ∫ upn−1 φ dx dt + ∫ ∫ Tn (f )φ dx dt 1
n
un−1
x2s +
0 Ω
1 n
t2
0 Ω
t2
dx dt + η ∫ ∫ upn−1 dx dt + η ∫ ∫ Tn (f ) dx dt. t1 Br (0)
t1 Br (0)
By the monotone convergence theorem, upn−1
un−1 x2s + n1
Tn (f )
→
up in L1 (Br (0) × (t1 , t2 )),
↗
u x2s
→
f in L1 (Br (0) × (t1 , t2 )).
in L1 (Br (0) × (t1 , t2 )),
Thus it follows that u is a weak supersolution to (13.1.1) in Br (0)×(t1 , t2 ), a contradiction with Theorem 13.5.3.
13.5.2 Existence results for 1 < p < p+ (λ, s) The goal now is to consider the complementary interval of powers, 1 < p < p+ (λ, s), and to prove that under some suitable hypotheses on f and u0 , problem (13.5.1) has a positive solution. We will consider here the case f ≡ 0. For the case f ≢ 0, see Remark 13.5.6.
472  13 Fractional Heat equation and Hardy potential First of all, note that if 0 < λ ≤ ΛN,2s and 1 < p < p+ (λ, s), the stationary problem (−Δ)s u = λ
u + up in Ω, x2s
u > 0 in Ω,
u = 0 in ℝN \ Ω
(13.5.7)
has a positive supersolution w, depending on the following cases. (A) If 0 < λ < ΛN,2s , in Section 11.3 (see also [46, Proposition 2.3]), we find a positive solution to the problem (−Δ)s u = λ
u + up + μuq in Ω, x2s
u > 0 in Ω,
u = 0 in ℝN \ Ω,
(13.5.8)
for μ small enough, 0 < q < 1 and 1 < p < p+ (λ, s). In particular, for every μ ≥ 0 this solution is a supersolution of (13.5.7). Note that for 1 < p ≤ 2∗s − 1 this supersolution is in the energy space, and for 2∗s − 1 < p < p+ (λ, s), it is a weak positive supersolution. (B) If λ = ΛN,2s , then p+ (λ, s) = 2∗s − 1. Thus, instead of H0s (Ω), we consider the Hilbert space H(Ω) defined in (13.2.5). Since H(Ω) is compactly embedded in Lp (Ω) for all 1 ≤ p < 2∗s , classical variational methods in the space H(Ω) allow us to prove the existence of a positive solution w to the stationary problem (13.5.7). Theorem 13.5.5. Assume that 0 < λ ≤ ΛN,2s and 1 < p < p+ (λ, s). Suppose that u0 (x) ≤ w, where w is a supersolution to the stationary problem (−Δ)s w = λ
w + wp in Ω, x2s
w(x) > 0 in Ω,
w(x) = 0 on ℝN \ Ω.
Then for all T > 0, the problem ut + (−Δ)s u = λ xu2s + up { { { u(x, t) = 0 { { { { u(x, 0) = u0 (x)
in Ω × (0, T), in (ℝN \ Ω) × [0, T),
(13.5.9)
if x ∈ Ω
has a global positive solution. If w is a weak supersolution, the solution will also be weak, and similarly, if w is an energy supersolution, problem (13.5.9) will have an energy solution. Proof. Since w(x) ≥ u0 (x) for all x ∈ Ω, w is a positive supersolution to problem (13.5.9). Hence, we conclude just by applying Lemma 13.5.1, when w is a weak supersolution, or Lemma 13.5.2, if w is an energy supersolution. Remark 13.5.6. (I) With the results above we find the optimality of the power p+ (λ, s), which was our main aim. Nevertheless, it could be interesting to know the optimal class of data for which there exists a solution and the regularity of such solutions
13.6 Further results and comments  473
according to the regularity of the data. In this direction, considering g = up + cf in problem (13.4.1), Theorem 13.4.1 establishes that, necessarily, t2
∫ ∫ x−
N−2s +α 2
g dx dt < +∞
and
t1 Br (0)
∫ x−
N−2s +α 2
u0 dx < +∞,
Br (0)
if we expect to find a solution of problem (13.5.1). c0 (t) In the presence of a source term f ≩ 0, if f (x, t) ≤ x 2s with c0 (t) bounded and sufficiently small, then the computation above allows us to prove the existence of a supersolution. Then the existence of solution to problem (13.1.1) follows for all p < p+ (λ, s). (III) The asymptotic behavior of the solution remains essentially open.
(II)
13.6 Further results and comments 13.6.1 Fujita exponent depending on λ We have analyzed the Fujita phenomenon for the fractional heat equation in Sec. Similarly to the local tion 12.8.2; the Fujita exponent happens to be F(0, s) = 1 + 2s N case (see Section 5.6), the Hardy potential produces a deviation to the right of the Fujita exponent obtained in Section 12.8.2. Precisely, with the parameters defined in Section 9.4, the following result holds. 2s Theorem 13.6.1. Suppose that 1 < p < F(λ, s) := 1 + N−γ , where γ = N−2s − α, and let u 2 be a positive solution to problem (7.5.2). Then there exists T ∗ := T ∗ (u0 ) such that
lim ∫ x−γ u(x, t) dx = ∞.
t→T ∗
ℝN
See [25] for the proof. Note also that F(0, s) < F(λ, s) < p+ (λ, s). 13.6.2 Other operators As in the elliptic case, the results above could be extended for a larger class of linear operators, like those studied in Chapter 8. Another case that could have some interest would be the elliptic part of the fractional pLaplacian operator taking into consideration the corresponding Hardy inequality. See [168]. The techniques to be used, such as Harnack’s inequality, may have a genuine mathematical interest. These aspects have not been considered in this book.
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Alphabetical Index 2soperator, composition with a convex Lipschitz function 217 2soperators, weak maximum principle 218 Ambrosetti–Rabinowitz, the mountain pass theorem 56, 61 an application of the fractional weighted Harnack inequality 291 Barenblatt, G. I. 5 Bernstein theorem 392 Brezis–Kamin–Oswald comparison, extension to the fractional setting 230 Brezis–Lieb, a convergence result 309 Caffarelli–Kohn–Nirenberg inequalities 42, 113 Calderón–Zygmund 29 Calderón–Zygmund formula 191 completely monotone function 392 Cwikel–Lieb–Rosenblum, theorem 4 Dirichlet problem for 2soperator, bounded solution via Moser method 222 Dirichlet problem for 2soperator, bounded solution via Stampacchia method 224 Dirichlet problem for 2soperator, Calderón–Zygmund type result 226 Dirichlet problem for 2soperator, Calderón–Zygmund type result for weak solutions 236 Dirichlet problem for 2soperator, existence and regularity for datum in L1 (Ω) 233 Dirichlet problem for 2soperator, exponential summability 225 Dirichlet problem for 2soperator, finite energy setting 221 Dirichlet problem for 2soperator, further fractional regularity 227 Dirichlet problem for 2soperator, nonvariational setting 232 Dirichlet problem for 2soperator, weak solution 233 domain with extension property for the Sobolev space 202 doubling property of dμ 267 Ekeland, variational principle 54 Ekeland, variational principle finding critical points of functionals 55
elliptic equation, measure data 47 elliptic linear problem, alternative method to find the critical λ(m) 43 elliptic linear problem, critical λ(m) 33 elliptic linear problem, entropy solutions 42 elliptic linear problem, existence for λ < λ(m) 35 elliptic linear problem, necessary and sufficient condition for solvability in weighted L1 spaces 39 elliptic linear problem, nonexistence in L1 30 elliptic linear problem, optimality of λ(m) 34, 37 elliptic linear problem, solvability of the problem for ΛN,2 43 elliptic linear problem, uniqueness 41 elliptic linear problem, summability of finite energy solutions, condition on λ 33 elliptic problem, behavior at the origin of the positive supersolutions 53 elliptic problem, eliminable singularities 51 elliptic problem, quantitative maximum principle 112 energy solution to the fractional problem, upper estimate 291 extension of Sobolev spaces 267 extension Theorem 203 first eigenvalue, elliptic operators in non divergence form 14 Fourier transform 189, 248 Fourier transform, of a homogeneous distribution 190 Fourier transform, of a rotation invariant distribution 190 fractional elliptic equation, local behavior of solutions 250 fractional elliptic problem, about the optimality of sufficient condition for Calderón–Zygmund summability 295 fractional elliptic problem, finite energy solution 265 fractional elliptic problem, ground state representation 252, 254, 266 fractional elliptic problem, necessary and sufficient condition for existence of weak solution 301 fractional elliptic problem, regularity up to the boundary 315
490  Alphabetical Index
fractional elliptic problem, some examples for the nonoptimality 295 fractional elliptic problem, sufficient condition for Calderón–Zygmund summability of energy solutions 293 fractional elliptic problem, sufficient condition for Calderón–Zygmund summability of weak solutions 299 fractional elliptic problem, test function class 265 fractional elliptic problem, weak Harnack inequality 264 fractional elliptic problem, weak super(sub)solution 265 fractional elliptic problem, weights of the ground state representation 267 fractional elliptic problem with weights in ℝN 267 fractional Hardy inequality, for the fractional pLaplacian 261 fractional Hardy inequality, relationship with the classical inequalities 247 fractional Hardy inequality, remainder terms 255, 257 fractional heat Cauchy kernel, direct analytical proof of the asymptotic behavior 390 fractional heat Cauchy problem, Blumenthal–Getoor theorem 389 fractional heat Cauchy problem, Caffarelli–Figalli regularity of the solution 398 fractional heat Cauchy problem, Fujita type exponent 428 fractional heat Cauchy problem, kernel 389 fractional heat Cauchy problem, proof of the Widder type result 424 fractional heat Cauchy problem, representation formula 419 fractional heat Cauchy problem, representation theorem 410 fractional heat Cauchy problem, strong solution 409 fractional heat Cauchy problem, uniqueness of some viscosity solutions 424 fractional heat Cauchy problem, uniqueness of weak solution 411 fractional heat Cauchy problem, upper bounds for kernel convolutions 420 fractional heat Cauchy problem, viscosity solution 408
fractional heat Cauchy problem, weak maximum principle 419 fractional heat Cauchy problem, weak solution 408 fractional heat Cauchy problem, Widder type uniqueness result 406 fractional heat equation, a priori estimates 379 fractional heat equation, Aronson–Serrin curve 379 fractional heat equation, Aronson–Serrin zone 379 fractional heat equation, bounded solutions 379 fractional heat equation, Cauchy problem 389 fractional heat equation, existence and uniqueness of finite energy solution 374 fractional heat equation, existence and uniqueness of weak solution 376 fractional heat equation, exponential summability on the Aronson–Serrin curve 383 fractional heat equation, finite energy setting 373 fractional heat equation, finite energy solution 373 fractional heat equation, further summability results 387 fractional heat equation, summability of solution outside of Aronson–Serrin zone 384 fractional heat equation, test functions class 376 fractional heat equation, weak solution 376 fractional Hilbert space 201 fractional Laplacian 189 fractional Laplacian, analytic continuation 194 fractional Laplacian, fundamental solution 194 fractional linear heat equation, influence of the Hardy potential 433 fractional linear heat equation, strong maximum principle 440 fractional linear heat equation, weak Harnack inequality 440 fractional linear heat equation with Hardy potential, comparison principle in the finite energy setting 440 fractional linear heat equation with Hardy potential, existence of finite energy solutions 437
Alphabetical Index  491
fractional linear heat equation with Hardy potential, finite energy solutions 436 fractional linear heat equation with Hardy potential, round state representation 441 fractional linear heat equation with Hardy potential, test functions class 436 fractional linear heat equation with Hardy potential, weak comparison principle 438 fractional linear heat equation with Hardy potential, weak solutions 436 fractional Pohozaev identity by RosOton and Serra 316 fractional Pohozaev identity with Hardy potential 320 fractional Poincaréinequality 211 fractional Poisson kernel 209 fractional semilinear concaveconvex elliptic problem, existence of at least two nontrivial variational solutions if 1 < p < 2∗s − 1 329 fractional semilinear concaveconvex elliptic problem, existence of at least two nontrivial variational solutions if p = 2∗s − 1 342 fractional semilinear elliptic problem, complete blow up for p ≥ p(λ, s) 362 fractional semilinear elliptic problem, critical power for existence 306 fractional semilinear elliptic problem, criticalsubcritical problem respect to Sobolev embedding 307 fractional semilinear elliptic problem, energy functional 308 fractional semilinear elliptic problem, energy solution 307 fractional semilinear elliptic problem, existence and non existence for problems with nonlinear term singular at the boundary 372 fractional semilinear elliptic problem, existence and regularity for problems with nonlinear term singular at the boundary 366 fractional semilinear elliptic problem, existence of a minimal weak solution for 1 < p < p(λ, s) 324, 328 fractional semilinear elliptic problem, nonexistence of positive weak solution for p ≥ p(λ, s) 358 fractional semilinear elliptic problem, problems with nonlinear term singular at the boundary 365, 369
fractional semilinear elliptic problem, singular at the boundary 309 fractional semilinear elliptic problem, weak solution 308
fractional semilinear heat equation, influence of the Hardy potential 433 fractional Sobolev, Rellich compactness 205 fractional Sobolev embedding 204 fractional Sobolev inequality 203
fractional Sobolev spaces, interpolation results 220 fractional weighted elliptic problem, weak Harnack inequality 280
fractional weighted heat equation with Hardy potential, nonexistence for p > 1 462 fractional weighted linear heat equation 456
fractional weighted linear heat equation, proof of the weak Harnack inequality 454 fractional weighted linear heat equation, weak Harnack inequality 443 fractional weighted linear heat equation, weak super (sub)solutions 443 fractional weighted linear heat equation with Hardy potential, instantaneous and complete blow up if λ > ΛN,s 462
fractional weighted linear heat equation with hardy potential, necessary conditions on the data for solvability 456
fractional weighted linear heat equation with Hardy potential, nonexistence for λ > ΛN,2s 461
fractional weighted linear heat equation with Hardy potential, sufficient condition on the data for solvability 457 fractional weighted problem, De Giorgi type inequality 287
fractional weighted semilinear heat equation with Hardy potential 465 fractional weighted semilinear heat equation with Hardy potential, critical power for existence p+ (λ, s) 465 fractional weighted semilinear heat equation with Hardy potential, existence if p < p+ (λ, s) 472
fractional weighted semilinear heat equation with Hardy potential, instantaneous and complete blow up if p > p+ (λ, s) 469
492  Alphabetical Index
fractional weighted semilinear heat equation with Hardy potential, nonexistence if p > p+ (λ, s) 468 fractional weighted semilinear heat equation with Hardy potential, solutions obtained as limits of approximations 466 FrankKamenetski, model of combustion 5 Gagliardo–Nirenberg inequality 220 Gelfand problem 5 Gelfand problem, dynamical system 6 Gelfand problem, stability of the singular solution 12 Hardy inequality, a general inequality 241 Hardy inequality, discrete 1 Hardy inequality, fractional 239, 240 Hardy inequality, nonattainability of Λ−1 N,α 246 Hardy inequality, one dimensional 1 Hardy inequality, optimality of Λ−1 N,α 243 Hardy inequality in W 1,p (ℝN ) 21 Hardy–Leray, inequality 1 Hardy–Leray inequality 15 Hardy–Leray inequality, non attainability of ΛN,2 in ℝN 20 Hardy–Leray inequality, optimality of ΛN,2 19 Hardy–Leray inequality, pseudominimizer 21 Hardy–Leray inequality with remainder term 23 Hardy–Leray potential 1 Hardy–Leray potential, basic properties 15 Hardy–Littlewood, theorem 19 heat equation, fractional diffusion 373 heat linear problem, local behavior of nonnegative supersolutions 74 heat linear problem, perturbation of the linear problem 75 heat linear problem, technical remark on existence 77 Helly selection principle 390 homogeneous fractional Hilbert space 201 interpolation result for parabolic Sobolev spaces 387 Joseph–Lundgren 5 Kardar–Parisi–Zhang 137 Kardar–Parisi–Zhang, generalized equation 137 Kato, inequality in the local case 48
Kato inequality, elliptic 217 Kato inequality, parabolic 388 Krylov–Safonov type covering result 287 limit space ℋ(Ω) associated to ΛN 28 linear elliptic problem, complete blowup for datum in L1 38 linear elliptic problem, unbounded solution for m > N2 30 linear heat problem, existence results 72 maximum principle for fractional Laplacian 310, 313 mean property for supersolution of the fractional Laplacian 312 mountain pass theorem, constructing Palais–Smale sequences 58 Murat, F., compactness Lemma 122 Newtonian capacity, basic properties 51 Newtonian capacity, definition 50 Newtonian capacity, motivation 48 nonlocal 2skernel 213 nonlocal 2soperators 213 nonlocal operator 191 normalized constants 248 Palais–Smale, condition 60 Palais–Smale, sequence 60 Picone inequality 12 Picone inequality, fractional setting 228 Picone inequality, some fractional extensions 229 Pizzetti formula 195 Polya–Szego, theorem 19 quasilinear Cauchy problem, blowup in a finite time for p < F (λ) 184 quasilinear Cauchy problem, global existence for small data and F (λ) < p < q+ (λ) 181 quasilinear Cauchy problem, global supersolutions for F (λ) < p < q+ (λ) 176 quasilinear Cauchy problem, local existence for 1 < p < q+ (λ) 177 quasilinear Cauchy problem, meaning of the blowup 184 quasilinear Cauchy problem, subsolutions blowingup in a finite time 174 quasilinear elliptic equation, heuristic determination of q+ (λ) 115
Alphabetical Index  493
quasilinear elliptic equation, nonexistence for q ≥ q+ (λ) 115 quasilinear elliptic problem 103
quasilinear elliptic problem, comparison results 104
quasilinear elliptic problem, existence if λ ≡ ΛN,2 132 and p < N+2 N
quasilinear elliptic problem, existence results for 1 < p < q+ (λ), λ < ΛN,2 126
quasilinear elliptic problem, the optimal exponent q+ (λ) 103 quasilinear elliptic problem, very weak supersolution (subsolution) 114
quasilinear elliptic problem: nonexistence results and critical power q+ (λ) 114
quasilinear heat equation 137
quasilinear heat equation, a comparison principle 143
quasilinear heat equation, a general existence result 168 quasilinear heat equation, a pointwise convergence result of the gradients 148 quasilinear heat equation, a priori integral estimates 145
quasilinear heat equation, Cauchy problem 173 quasilinear heat equation, complete and instantaneous blowup 166
quasilinear heat equation, critical exponent for existence, q+ (λ) 137 quasilinear heat equation, existence result if λ < ΛN,2 and 1 < p < q+ (λ) 171
quasilinear heat equation, Fujita type exponent F (λ) < q+ (λ) 174
quasilinear heat equation, local existence result for a the truncated Hardy potential 152
quasilinear heat equation, nonexistence of very weak supersolution if p ≥ q+ (λ) 154 quasilinear heat equation, passing to the limits in truncated problems 147 quasilinear heat equation, uniqueness for truncated problems 144 quasilinear heat equation, very weak subsolution (supersolution) 139
quasilinear heat equation with sublinear growth, maximum principle 140
quasilinear problem, blowup of solutions to approximated problems 121
relation between energy and viscosity solutions 404 Riesz, theorem 19 Riesz potentials 192 Riesz potentials, multiplier 193 Riesz transforms 191 sfractional Green operator 212 sfractional Laplacian, analytical properties 199 sfractional Laplacian, elementary properties 197 sfractional Laplacian, integration by parts 197 sfractional Laplacian, nonlocal operator 196 sfractional Laplacian, pointwise meaning 200 sfractional Laplacian, regularity in ℝN 206 sfractional Laplacian, singular kernel definition 196 sfractional Laplacian, spectral definition 191 sfractional Laplacian, variational Dirichlet problem 211 sfractional Laplacian, weak solution 201 scattering of atomic particles 2 semilinear elliptic problem, complete blowup 68 semilinear elliptic problem: existence of distributional solution p < p+ (λ) 64 semilinear elliptic problem: existence of variational solutions for λ < ΛN,2 62 semilinear elliptic problem: existence of variational solutions for λ = ΛN,2 64 semilinear elliptic problem: nonexistence for 0 < λ≤ΛN,2 and p ≥ p+ (λ) 66 semilinear heat Cauchy problem, blowup for p < F (λ) 94 semilinear heat Cauchy problem, existence for small time if F (λ) < p < p+ (λ) 93 semilinear heat Cauchy problem, Fujita exponent F (0) 89 semilinear heat Cauchy problem, global existence for F (λ) < p < p+ (λ) and small data 94 semilinear heat Cauchy problem, global supersolutions for F (λ) < p < p+ (λ) 92 semilinear heat Cauchy problem, shifted Fujita exponent F (λ) 91 semilinear heat Cauchy problem, subsolution blowingup for small p 90 semilinear heat Cauchy problem, the case p = F (λ) 97
494  Alphabetical Index
semilinear heat equation 71 semilinear heat equation, Cauchy problem 89 semilinear heat equation, critical power for existence p+ (λ) 71 semilinear heat equation, existence of solutions for p < p+ (λ) 87 semilinear heat equation, instantaneous and complete blow up for p ≥ p+ (λ) 83 semilinear heat equation, nonexistence results for p ≥ p+ (λ) 79 semilinear heat problem, critical power p+ (λ) 45, 46 Sobolev, inequality 31 Sobolev, space 31 Sobolev, weighted inequality 42 Sobolev critical exponent 32 Sobolev minimizers 32 Sobolev space, continuous injection 31, 32 s (Ω) 270 Sobolev space Wγ,0 Sobolev spaces with weights 267 Stampacchia 29 stationary Schrödinger equation 2
supersolution of weighted fractional problem, a compactness result 277, 278 symmetrization 17 Terracini, the semilinear critical case in ℝN , N ≥ 3 69 test functions for 2soperator 232 truncation functions 32, 214 truncation functions, elementary estimates 214 uncertainty principle 4 weighted fractional elliptic problem, Brezis–Kamin–Oswald comparison type 277 weighted fractional Hardy inequality 275 weighted fractional Picone inequality 274 weighted fractional problem, supersolution 274 s (Ω) 271 weighted Poincaré inequality in Wγ,0 weighted Poincaré–Wirtinger inequality 272 weighted Sobolev inequality 269 s,γ weighted Sobolev inequality in W0 (Ω) 270 weighted Sobolev space 442