Analysis of the Navier-Stokes Problem: Solution of a Millennium Problem [2 ed.] 3031307224, 9783031307225

Table of contents :
Preface to the Second Edition
Preface to the First Edition
Contents
About the Author
1 Introduction
2 Brief History of the Navier–Stokes Problem
3 Statement of the Navier–Stokes Problem
4 Theory of Some Hyper-Singular Integral Equations
5 A Priori Estimates of the Solution to the NSP
6 Uniqueness of the Solution to the NSP
7 The Paradox and Its Consequences
8 Logical Analysis of Our Proof
1 Theory of Distributions and Hyper-Singular Integrals
Gamma and Beta Functions
The Laplace Transform
Analysis of the Navier-Stokes Problem. Solution to the Millennium Problem Concerning the Navier-Stokes Equations
Introduction
Derivation of the Integral Inequality
Investigation of Integral Equations and Inequalities with Hyper-Singular Kernel
Uniqueness of the Solution to the NSP
Conclusions
Applications of Analytic Continuation to Tables of Integral Transforms and Some Integral Equations with Hyper-Singular Kernels
Introduction
More Examples
Some Applications
Conclusion
References

Citation preview

Synthesis Lectures on Mathematics & Statistics

Alexander G. Ramm

Analysis of the Navier-Stokes Problem Solution of a Millennium Problem Second Edition

Synthesis Lectures on Mathematics & Statistics Series Editor Steven G. Krantz, Department of Mathematics, Washington University, Saint Louis, MO, USA

This series includes titles in applied mathematics and statistics for cross-disciplinary STEM professionals, educators, researchers, and students. The series focuses on new and traditional techniques to develop mathematical knowledge and skills, an understanding of core mathematical reasoning, and the ability to utilize data in specific applications.

Alexander G. Ramm

Analysis of the Navier-Stokes Problem Solution of a Millennium Problem Second Edition

Alexander G. Ramm Department of Mathematics Kansas State University Oakland, CA, USA

ISSN 1938-1743 ISSN 1938-1751 (electronic) Synthesis Lectures on Mathematics & Statistics ISBN 978-3-031-30722-5 ISBN 978-3-031-30723-2 (eBook) https://doi.org/10.1007/978-3-031-30723-2 1st edition: © Morgan & Claypool Publishers 2021 2nd edition: © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To Luba

Preface to the Second Edition

The second edition of my monograph “The Navier-Stokes problem” has a new title and contains several new features: the Introduction is expanded, Appendices 4 and 5 are added, and the contradictions of the Navier-Stokes equations are emphasized. We solve the millennium problem concerning the Navier-Stokes equations by proving that the Navier-Stokes problem (NSP) in R3 does not have a solution defined for all times t ≥ 0. This follows from the paradox the author found and proved in Chap. 7 of this monograph. Appendix 4 contains a brief but essentially self-contained presentation of our basic ideas and results on the NSP. Mathematical tool we use is the theory of integral equations and inequalities with hyper-singular kernels of special type. This theory was developed by the author for an analysis of the NSP. In Appendix 5, there are some examples of applications of this theory. Oakland, USA

Alexander G. Ramm

vii

Preface to the First Edition

This work is the second edition of the author’s monograph [2]. The aim of this monograph is to prove that the Navier-Stokes problem (NSP) in the space R3 without boundaries is physically and mathematically incorrect and contradictory. Its solution does not exist on all of the time semiaxis t ≥ 0. The contradictory nature of the NSP consists of the following: If one assumes that the initial velocity v0 (x)  ≡ 0 is a smooth and rapidly decaying function as |x| → ∞ and the exterior force f (x, t) = 0, then one proves that v0 (x) = 0. This result, proved by the author, is called the NSP paradox. It shows that there is no solution to the NSP (2) defined for all t > 0. This solves the millennium problem concerning the NSP. The assumption f (x,t) = 0 is made only for simplicity. The ideas of the proof are not changed if this assumption is dropped and one assumes that f (x,t) is a smooth and rapidly decaying function when |x| + t → ∞.   Let W := v(x, t) ∈ H 1 (R3 ) × C(R+ ); ∇ · v = 0 , H 1 (R3 ) is the Sobolev space and C(R+ ) is the space of continuous functions, u C(R3 ) = supt≥0 |u(t)|. + The author proves a priori estimate sup( v + ∇v ) ≤ c,

(1)

t≥0

where v = v(x,t) is the solution to problem (2), the velocity vector of the incompressible viscous fluid, and by c > 0 various constants, independent of the data, are denoted. Here and throughout the book, c > 0 stand for various constants independent of x and t, the norm ||·|| is the L 2 (R3 ) norm, by ∇v the collection of the norms of the first ∂v derivatives ∂xmj is understood, and H 1 (R3 ) is the Sobolev space W21 (R3 ). The NSP consists of solving the equation v  + (v, ∇)v = f − ∇ p + vv, ∇ · v = 0, v(x, 0) = v0 (x),

(2)

ix

x

Preface to the First Edition

where v = v(x, t) is the velocity of the fluid, f = f (x, t) is the (exterior) force, v0 (x) is the initial velocity, p = p(x, t) is the pressure, v = const > 0 is the viscosity coefficient, v = ∇ 2 v is the Laplacean of v. The fluid is assumed non-compressible and viscous, its density is constant ρ = 1, v  := ∂v ∂t . We look for a solution to NSP (2) in the space W. We prove that the NSP problem is equivalent to the integral equation  t  v=F− ds dyG(x − y, t − s)(v, ∇)v, (3) 0

R3

where F = F(x, t) is the data,   t  ds dyG(x − y, t − s) f (y, s) + F= 0

R3

R3

dyg(x − y, t)v0 (y),

(4)

and |x|2

e− 4vt g(x, t) = . (4vtπ)3/2

(5)

The data F depends only on f (x,t) and v0 (x). The function G(x,t) is calculated analytically in Chap. 4. Let us define the Fourier transform  1 F (v) := v˜ = v(ξ, ˜ t) = e−iξ·x v(x, t)d x. (6) (2π)3 R3 Take the Fourier transform of equation (3) and use the known formula F (vw) = (2π)3 F (v)F (w) to get another integral equation equivalent to the NSP:  t ˜ v˜ = F˜ − (2π)3 ds G(ξ, t − s)v(−iξ ˜ v) ˜ (7) 0

where the  denotes the convolution in R3 :  ˜f h˜ = ˜ f˜(λ − μ)h(μ)dμ. R3

(8)

Preface to the First Edition

xi

The Cauchy inequality yields ˜ ≤ f˜ h , ˜ | f˜h|

(9)

where · = · L 2 (R3 ) We prove that ˜ G(ξ, t) =

ξ j ξm  −|ξ|2 t 1  δ e − , jm (2π)3 ξ2

(10)

g˜ (ξ, t) = e−|ξ|

(11)

and 2 νt

.

Therefore, ˜ ≤ ce−|ξ| t , |G| 2

(12)

ξ ξ

j m because |1 − |ξ| 2 | ≤ 2. Equation (7) implies  t  t 2 2 ˜ +c ˜ +c |v| ˜ ≤ | F| dse−ν|ξ| (t−s) v |ξ| ˜ v ˜ ≤ | F| dse−ν|ξ| (t−s) |ξ|v , ˜

0

(13)

0

where the a priori estimate sup v ˜ ≤c

(14)

t≥0

and inequality (9) were used. Estimate (14) is a consequence of the estimate (1) and of the Parseval identities: (2π)3 v ˜ 2 = v ˜ 2 , (2π)3 |ξ|v ˜ 2 = ∇v 2 .

(15)

One can check that e−ν|ξ| t = 2

c c 2 , |ξ|e−ν|ξ| t = 5/4 . t 3/4 t

(16)

Let ˜ b0 (t) := |ξ| F , b(t) = |ξ|v . ˜

(17)

xii

Preface to the First Edition

If 0 ≤ u ≤ w, then u ≤ w Multiply equation (13) by |ξ|, use formula (15), and take the norm · . The result is  t b(s)ds b(t) ≤ b0 (t) + c . (18) 5/4 0 (t − s) The integral in this inequality diverges from the classical analysis point of view (we write sometimes diverges classically). We have to define this integral and derive some estimates for positive solutions to inequality (18). This is done in Chaps. 4 and 5. Together with the inequality (18) we study the corresponding integral equation:  t q(s)ds . (19) q(t) = b0 (t) + c (t − s)5/4 0 We solve equation (19) in closed form under suitable assumptions on the data b0 (t), namely, we assume that b0 (t), is a smooth and rapidly decaying as t → ∞. Moreover, we prove that b(t) ≤ q(t),

(20)

for any non-negative solution to inequality (18). Furthermore, we prove that q(0) = 0

(21)

provided that the data b0 (t) are smooth and rapidly decaying as t → ∞ From Eqs. (20) and (21), it follows that b(0) = 0. This and the definition of b(t) = |ξ|v ˜ imply that v0 (x) = 0, although originally we assumed that v0 (x) = v(x, 0)  ≡ 0. This is the NSP paradox. It shows that the NSP problem is contradictory physically and mathematically and that the NSP with the suitable data does not have a solution defined for all t ≥ 0. Therefore, the millennium problem concerning the Navier-Stokes equations is solved: we prove that this problem does not have a solution. We prove that the NSP with the zero data, that is, f = 0 and v0 (x) = 0 does have a solution v(x, t) = 0 and this solution is unique in the space W . These are our main results concerning the NSP. These results show that physically correct equations should be found for the motion of incompressible viscous fluid. Oakland, USA October 2022

Alexander G. Ramm

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 7

2 Brief History of the Navier–Stokes Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 11

3 Statement of the Navier–Stokes Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 21

4 Theory of Some Hyper-Singular Integral Equations . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23 31

5 A Priori Estimates of the Solution to the NSP . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

6 Uniqueness of the Solution to the NSP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 39

7 The Paradox and Its Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

8 Logical Analysis of Our Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

Appendix 1: Theory of Distributions and Hyper-Singular Integrals . . . . . . . . . . .

45

Appendix 2: Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

Appendix 3: The Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium Problem Concerning the Navier-Stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

Appendix 5: Applications of Analytic Continuation to Tables of Integral Transforms and Some Integral Equations with Hyper-Singular Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87 xiii

About the Author

Alexander G. Ramm was born in Russia, emigrated to USA in 1979, and is a US citizen. He is Professor Emeritus of Mathematics with broad interests in analysis, scattering theory, inverse problems, theoretical physics, engineering, signal estimation, tomography, theoretical numerical analysis, and applied mathematics. He is an author of 716 research papers and 20 research monographs and an editor of 3 books. He has lectured at many universities throughout the world, given more than 150 invited and plenary talks at various Conferences, and supervised 11 Ph.D. students. He was Fulbright Research Professor in Israel and Ukraine, distinguished visiting professor in Mexico and Egypt, Mercator Professor in Germany, Research Professor in France, and invited plenary speaker at the 7-th PACOM; he won Khwarizmi international award in 2004 and received other honors. A. G. Ramm was the first to prove the uniqueness of the solution to inverse scattering problems with fixed-energy scattering data, the first to prove the uniqueness of the solution to inverse scattering problems with non-over-determined scattering data, and the first to study inverse scattering problems with under-determined scattering data. He solved many specific inverse problems and developed new methods and ideas in the area of inverse scattering problems. He introduced the notion of Property C for a pair of differential operators and applied Property C for one-dimensional and multi-dimensional inverse scattering problems. A. G. Ramm solved the many-body wave scattering problem when the bodies are small particles of arbitrary shapes, assuming that a d λ, where a is the characteristic size of small particles, d is the distance between neighboring particles, and λ is the wavelength in the material in which the small particles are embedded. Multiple scattering is essential under these assumptions. He used this theory to give a recipe for creating materials with a desired refraction coefficient and materials with a desired wave-focusing property. These results attracted the attention of the scientists working in nanotechnology. A. G. Ramm gave formulas for the scattering amplitude for scalar and electromagnetic waves by small bodies of arbitrary shapes and formulas for the polarizability tensors for such bodies. A. G. Ramm gave a solution to the Pompeiu problem, proved Schiffer’s conjecture, and gave the first symmetry results in harmonic analysis. xv

xvi

About the Author

A. G. Ramm has developed the Dynamical Systems Method (DSM) for solving linear and non-linear operator equations, especially ill-posed. A. G. Ramm developed a random fields estimation theory. A. G. Ramm has developed a theory of convolution equations with hyper-singular integrals. A. G. Ramm has introduced a wide class of domains with non-compact boundaries. He studied the spectral properties of Schrödinger’s operators in this class of domains and gave sufficient conditions for the absence of eigenvalues on the continuous spectrum of these operators. A. G. Ramm has solved one of the millennium problems concerning the Navier-Stokes problem (NSP) and proved the NSP paradox, which shows the contradictory nature of the NSP and the non-existence of its solution on the interval t ∈ [0, ∞) for the initial data v0 (x)  ≡ 0 and f (x, t) = 0.

1

Introduction

In this work a proof of the author’s basic results concerning the Navier-Stokes problem (NSP) is given. The NSP is: v  + (v, ∇)v = f − ∇ p + νv, ∇ · v = 0, v(x, 0) = v0 (x),

(1.1)

where v = v(x, t) is the velocity of the fluid, f = f (x, t) is the (exterior) force, v0 (x) is the initial velocity, p = p(x, t) is the pressure, ν = const > 0 is the viscosity coefficient, v = ∇ 2 v is the Laplacean of v. The fluid is assumed non-compressible and viscous, its density is constant ρ = 1, v  := ∂v ∂t . We look for a solution to NSP (0.2) in the space W = {v : v ∈ H 1 × C(R+ ), ∇ · v = 0}, where H 1 = W21 (R3 ) is the Sobolev space and C(R+ ) is the space of continuous functions on R+ with the norm bC(R+ ) = supt≥0 |b(t)|. Our results include: (a) The a priori estimate sup[v(x, t) + ∇v(x, t)] ≤ c

(1.2)

t≥0

of the solution to the NSP. Here v is the velocity vector,  ·  is the L 2 (R3 ) norm, by c > 0 various constants independent of t and x are denoted. This a priori estimate is proved in Chap. 5. (b) The theory of integral equations  t 5 (t − s)− 4 b(s)ds (1.3) b(t) = b0 (t) + 0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_1

1

2

1 Introduction

and inequalities



t

b(t) ≤ b0 (t) + 0

5

(t − s)− 4 b(s)ds,

(1.4)

t 5 5 with hyper-singular kernels (t − s)− 4 is developed. Note that the integral 0 (t − s)− 4 b(s)ds diverges classically, that is, from the classical analysis point of view. We define such integrals and solve in closed form the corresponding integral equations with the data b0 (t) smooth and rapidly decaying as t → ∞. We study also integral equations with other hyper-singular kernels. This theory is developed in Appendix 5. (c) This paradox and its consequences are discussed in Chap. 7. This paradox says that if the data (the initial velocity v0 (x) and the force f (x, t) in the Navier-Stokes equations, see Chap. 4), are smooth and rapidly decaying, f = 0 (for simplicity only) but v0 (x) ≡ 0, and the solution to the NPS exists on R+ , then v0 (x) = 0. This means that the solution to the NSP cannot exist. The millennium problem concerning the NSP consists of finding whether the solution to the NSP exists for all t ≥ 0 and is smooth, provided that the data, that is, the initial velocity v(x, 0) and the force f (x, t), are smooth and rapidly decaying as |x| → ∞ and t → ∞. Our proof of the NSP paradox demonstrates that the solution to the NSP with f (x, t) = 0 and u 0 (x) ≡ 0 cannot exist for all t ≥ 0. In paper [1], p. 472, Theorem 2, there is a statement that, for f (x, t) = 0 and u 0 (x) sufficiently small, the solution to the NSP exists for all t ≥ 0 if m ≤ q, where m is the dimension of the space and the solution is in L q . In our case m = 3 and q = 2, so the condition m ≤ q does not hold. Therefore, the claim in [1], p. 472, is not applicable. (d) The uniqueness theorem for the solutions to the NSP is proved in Chap. 6. This theorem says that if the data vanishes, that is, f = 0 and u 0 (x) = 0 then there exists a solution v(x, t) = 0 of the NSP and this solution is unique in the space W := {v : v ∈ H 1 (R3 ) × C(R+ ), ∇ · v = 0}, where H 1 is the usual Sobolev space and C(R+ ) is the space of continuous functions b(t) with the supt≥0 |b(t)| norm. We do not mention here other results. The earlier author’s results [2–5] are used, but our presentation in this monograph is essentially self-contained. The Navier-Stokes equations are discussed in many books and papers. We only mention here [6], [7] and [2]. Our main goal is to present for broad audience the author’s result concerning the NavierStokes problem in R3 without boundaries. This result can be briefly formulated as follows: Assume (for simplicity only) that the exterior force f (x, t) = 0. If the initial velocity v0 (x) := v(x, 0) ≡ 0, ∇ · v0 (x) = 0, v0 (x) is smooth and rapidly decaying as |x| → ∞ and the solution v(x, t) of the NSP exists for all t ≥ 0, then v0 (x) = 0. This result we call the NSP paradox (or just paradox), shows that:

1 Introduction

3

a’) The NSP is not a correct statement of the problem of motion of viscous incompressible fluid; it is neither physically nor mathematically correct statement of the dynamics of incompressible viscous fluid. b’) The NSP does not have a solution unless v0 (x) = 0 and f (x, t) = 0; in this case the solution v(x, t) = 0 for all t ≥ 0. This result solves the millennium problem related to the Navier-Stokes equations. It encourages the search for the correct equations describing the dynamics of viscous incompressible fluid. Let us explain the steps of our proof. The NSP consists of solving the equations (1.1). a) First we reduce the NSP to an equivalent integral equation:  t  ds G(x − y, t − s)(v, ∇)vdy, (1.5) v(x, t) = F − 0

R3

where F = F(x, t) depends only on the data f (x, t) and v0 (x),  t   F(x, t) = ds G(x − y, t − s) f (y, s)dy + g(x − y, t)v0 (y)dy. 0

R3

R3

We assume (for simplicity only and without loss of generality) that f = f (x, t) = 0. Under this assumption one has:  F(x, t) := g(x − y, t)v0 (y)dy, (1.6) R3

where

|x|2

e− 4νt 2 g(x, t) = , t > 0; g(x, t) = 0, t ≤ 0; g˜ = e−|ξ| νt . (4νπt)3/2

(1.7)

The tensor G = G(x, t) = G jm (x, t) is calculated explicitly:    ξ p ξm  ξ p ξm  2 2 −3 eiξ·x δ pm − 2 e−νξ t dξ; G˜ = (2π)−3 δ pm − 2 e−νξ t . G(x, t) = (2π) 3 ξ ξ R (1.8) Let us define the Fourier transform:  v(x, t)e−iξ·x d x. (1.9) v˜ := v(ξ, ˜ t) := (2π)−3 R3

Taking the Fourier transform of formula (1.6), one gets: 2 ˜ t) = (2π)3 g(ξ, ˜ t)v˜0 (ξ) = (2π)3 e−|ξ| νt v˜0 (ξ). F(ξ,

(1.10)

Take the Fourier transform of equation (1.2) and get the integral equation equivalent to this equation:

4

1 Introduction

˜ t) − (2π)3 v(ξ, ˜ t) = F(ξ,



t

˜ ds G(ξ, t − s)v(iξ ˜ v), ˜

(1.11)

0

where  denotes the convolution in R3 . The following inequality, that comes from the Cauchy inequality, is useful: |v(iξ ˜ v)| ˜ ≤ v|ξ| ˜ v. ˜ (1.12) Claim 1 The following a priori estimate holds: sup v ˜ < c.

(1.13)

t≥0

By c here and throughout the paper various positive constants, independent of t and x, are denoted. We denote by c1 := |(− 41 )| > 0 the special constant from equation (4.4), and use  ∂v  the following notations: v j,m := ∂xmj , := R3 . Let us write equation (1.1) as v j + vm v j,m = f j − p, j + νv j,mm , v j, j = 0,

(1.14)

where over the repeated indices summation is understood, 1 ≤ j ≤ 3. We assume that v = v(x, t) and other functions in the NSP are real-valued and the following inequality holds:  ∞  f (x, t)dt < c. (1.15) v0  + 0

Here and throughout this paper  ·  is L 2 (R3 ) norm. Proof of Claim 1. Multiply equation (1.14) by v j , integrate over R3 and sum up over j to get 1 (1.16) (v2 ),t ≤ |( f , v)| ≤  f v. 2 where z ,t := ∂z ∂t . In deriving inequality (1.16) we have used integration by parts:     pv j, j d x = 0, νv, j j v j d x = −ν v, j v, j d x ≤ 0, − p, j v j d x = and



1 vm v j,m v j d x = − 2

 vm,m v j v j d x = 0.

From inequality (1.16) it follows that v,t ≤  f . Consequently,  ∞ v ≤ v0  +  f dt < c. 0

1 Introduction

5

This and our assumption (1.15) imply estimate sup v < c. t≥0



By the Parseval equality the desired estimate (1.13) follows. Claim 1 is proved. Inequalities (1.13) and (1.12) imply |v(iξ ˜ v)| ˜ ≤ c|ξ|v. ˜

(1.17)

This inequality is important because it allows one to estimate the nonlinear term in the middle side of equation (0.13) by the linear term on its right side. Equations (1.11) and (1.12) imply inequality ˜ t) + c v(ξ, ˜ t) ≤ F(ξ,

 t 0

˜ + e−ν(t−s)ξ |ξ|vds ˜ ≤ | F| 2

 t 0

e−ν(t−s)ξ b(s)ds, b(s) = |ξ|v ˜ 2

(1.18)

We derive this inequality later. From formula (1.8) it follows that ˜ |G(ξ, t − s)| ≤ ce−ν(t−s)ξ , 2

(1.19)   ξ ξ because | δ pm − pξ 2m | < c. Estimate (1.19) will be used more than once in this book. (b) Secondly, we prove that any solution to Eq. (1.11) satisfies integral inequality (1.20), see below. The integral in this inequality is a convolution with the hyper-singular kernel (t − 5 s)− 4 ; this integral diverges classically, that is, from the classical point of view. Convolution of distributions is defined in [11]. It is proved (see Theorem 7.1.15 in [11], p. 166) that the Fourier transform of the convolution of two distributions equals to the product of their Fourier transforms. The assumptions on the two distributions from Theorem 7.1.15 are satisfied for −5

λ = 0 if t ≤ 0, t λ = t λ if t > 0, the distribution t+ 4 and the function b(t). The distribution t+ + is studied in [12] and [11]. The following inequality is derived in Section 3:  t 5 (t − s)− 4 b(s)ds, (1.20) b(t) ≤ b0 (t) + c 0

where

˜ t), b(t) := |ξ|v(x, ˜ t) ≥ 0. b0 (t) := |ξ| F(x,

(1.21)

6

1 Introduction

Here and below the norm  ·  is the L 2 (R3 ) norm. If the data v0 (x) is smooth and rapidly decaying as |x| → ∞, then |v(ξ)| ˜ ≤ c(1 + |ξ|2 )−m , m > 25 . Therefore, from the definition of b0 (t) it follows that  ∞ 2 e−r νt r 4 (1 + r 2 )−m dr := cI , (1.22) b02 (t) ≤ c 0

where we have used the spherical coordinates. From this relation it follows that the Laplace transform  ∞ e− pt b0 (t)dt

Lb0 :=

0

can be estimated:

|Lb0 | ≤ c(1 + | p|)−1 , Re p > 0. 1 2 Indeed, for t 1 one has: I = 0 e−r νt r 4 (1 + r 2 )−m dr + O(e−νt ) := J + O(e−νt ), J = 5  νt 5 5 (νt)− 2 0 e−ρ ρ2 (1 + νtρ )−m 2ρdρ1/2 = O(t − 2 ). Thus, b0 (t) = O(t − 4 ), t 1 and supt≥0

b0 (t) ≤ c. Consequently, |Lb0 | ≤ c(1 + | p|)−1 , Re p ≥ 0. This estimate is used later. Since the convolution integral in (1.20) diverges classically, we give a new definition of this integral in Section 3 and estimate the solution b(t) to integral inequality (1.20) by the solution q(t) to the integral equation with the same hyper-singular kernel:  t 5 (t − s)− 4 q(s)ds. (1.23) q(t) = b0 (t) + c 0

Namely, we prove the following inequality b(t) ≤ q(t).

(1.24)

We assume that b0 (t) is smooth and rapidly decaying as t → ∞. (c) We prove the following a priori estimate: sup(∇v + v) ≤ c,

(1.25)

t≥0

part of which is inequality v ≤ c, proved earlier. Recall that by the Parseval equality one has: ˜ = v, (2π)3 |ξ|v ˜ 2 = ∇v2 . (2π)3/2 v We prove that Eq. (1.23) has a unique solution in the space C(R+ ), and the following estimate holds: (1.26) sup q(t) ≤ c, t≥0

provided that the datum b0 (t) is smooth and rapidly decaying at infinity.

References

7

Moreover, this solution q(t) is unique and q(0) = 0.

(1.27)

(d) We prove that any solution b(t) ≥ 0 of inequality (1.20), where b0 (t) a smooth rapidly decaying function, satisfies inequality (1.24). Since q(0) = 0 and 0 ≤ b(t) ≤ q(t), it follows that b(0) = 0. This yields the NSP paradox mentioned at the beginning of this section. Indeed, the initial data v0 (x) ≡ 0, so b(0) > 0, but we prove that b(0) = 0. The NSP paradox implies the conclusions we have made: The NSP is physically not a correct description of motion of the incompressible viscous fluid in the whole space R3 without boundaries; the NSP does not have a solution on the whole interval [0, ∞) unless the data are equal to zero; in this case the solution to the NSP does exist on the whole interval [0, ∞) and is identically equal to zero. The uniqueness of the solution to NSP is proved in Chap. 6.

References 1. T. Kato, Strong L p -solutions of the Navier-Stokes equation in Rm , with applications to weak solutions. Math. Z. 187, 471–480 (1984) 2. A.G. Ramm, The Navier-Stokes Problem (Morgan & Claypool publishers, San Rafael, 2021) 3. A.G. Ramm, Comments on the Navier-Stokes problem. Axioms 10, 95 (2021) 4. A.G. Ramm, Navier-Stokes equations paradox. Reports Math. Phys. (ROMP) 88(1), 41–45 (2021) 5. A.G. Ramm, Applications of analytic continuation to tables of integral transforms and some integral equations with hyper-singular kernels. Open J. Optim. 11, 1–6 (2022) 6. O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Fluid (Gordon and Breach, New York, 1969) 7. L. Landau, E. Lifshitz, Fluid Mechanics (Pergamon Press, New York, 1964)

2

Brief History of the Navier–Stokes Problem

Let ρ = ρ(x, t) be the density of the fluid and v(x, t) be its velocity. If D ⊂ R3 is a bounded  domain with a smooth boundary S, then the fluid mass in D is the integral D ρ d x and the  amount of fluid flowing through the boundary is S ρv · N d S, where N is the outer unit normal to S, and S = ∂ D is the boundary of D. The conservation of the mass requires to have   ∂ ρ d x = ρv · N d S. (2.1) − ∂t D S By the divergence theorem one has   ρv · N d S = ∇ · (ρv) d x. S

(2.2)

D

Therefore, (2.1) can be written as  ( D

∂ρ + ∇ · (ρv)) d x = 0. ∂t

(2.3)

∂ρ + ∇ · (ρv) = 0. ∂t

(2.4)

Since D is arbitrary, it follows that

This equation is called equation of continuity. If ρ = const, then (2.4) takes the form ∇ · v = 0.

(2.5)

This is continuity equation for incompressible fluid.  On the fluid in a volume D, the force of the pressure is − S pN d S, where N is the exterior unit normal to S and p = p(x, t) is the pressure. This force can be written by the  divergence theorem as − D ∇ p d x. By Newton’s second law, one has © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_2

9

10

2 Brief History of the Navier-Stokes Problem

ρ

dv = −∇ p + f dt

(2.6)

where f is the exterior force, f = f (x, t). For incompressible fluid, ρ = const. For convenience and without loss of generality, we assume that ρ = 1. (2.7) One has

∂v dv = + (v, ∇)v. dt ∂t

(2.8)

∂v + (v, ∇)v = −∇ p + f . ∂t

(2.9)

From (2.6)–(2.8) it follows that

This is Euler’s equation (1752) for incompressible fluid. D’Alembert’s paradox (1752) proved that Euler’s equation is physically incorrect. Let us write (2.10) v = vjej, where {e j }3j=1 is an orthonormal basis in R3 , (ei , e j ) = δi j , and over the repeated indices in (2.10) summation is understood. Euler’s Equation (2.9) can be written as v j + vi v j,i = − p, j + f j , where v j :=

∂v j , ∂t

v j,i :=

∂v j , ∂ xi

p, j :=

(2.11) ∂p . ∂x j

(2.12)

The Navier–Stokes equation (1822) for viscous incompressible fluid is v j + vi v j,i = p, j + f j + νv j , 1 ≤ j ≤ 3,

(2.13)

where ν = const > 0 is the viscosity coefficient and v j := v j,ii :=

3  ∂ 2v j j=1

∂ xi2

.

Equation (2.7) indicates that the fluid is incompressible. Equation (2.5) should be considered together with the Navier–Stokes equation (2.13). The initial velocity v(x, 0) = v0 (x) (2.14) is given.

Reference

11

The NSP in the whole space R3 consists of solving Eqs. (2.13), (2.5), and (2.14) simultaneously: v  + (v, ∇)v = −∇ p + f + νv, ∇ · v = 0, v(x, 0) = v0 (x).

(2.15)

We assume that v(x, t) decays as |x| → ∞ for any fixed t > 0 and as t → ∞ for any fixed x. We will show in Chap. 7 that the NSP in R3 is contradictory. This means that this problem is physically incorrect. We also prove that although the solution to the NSP is unique, it does not exist on the whole interval t ∈ [0, ∞) unless v0 (x) = f (x, t) = 0; in this case the solution is equal to zero identically, v(x, t) = 0, so it does exist for all t ≥ 0. Claude-Louis Navier (1785–1836) was a French engineer who introduces the Navier– Stokes equations in 1822. George Gabriel Stokes (1819–1903) was Anglo-Irish physicist and mathematician. Stokes graduated in 1841 from Cambridge University. He published the Navier–Stokes equations in 1845 not being familiar with the earlier work of Navier. Among his students are Lord Rayleigh (1842–1919) and H. Lamb (1849–1934). The motion of the fluid presents many unsolved questions. Turbulent motions are an example. In [1] it is mentioned (in the Introduction) that there is no satisfactory theory of the motion of fluids and there are paradoxes related to some concrete problems in this theory. The paradox we construct in Chap. 7 is of a general nature and it shows that the NSP is not physically acceptable.

Reference 1. O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Fluid (Gordon and Breach, New York, 1969)

3

Statement of the Navier–Stokes Problem

The NSP consists of solving the following equations: v  + (v, ∇)v = −∇ p + νv + f ,

in R3 × R+ ,

(3.1)

∇ · v = 0,

in R × R+ ,

(3.2)

v(x, 0) = v0 (x),

in R ,

(3.3)

3 3

where R+ = [0, ∞). The functions v0 (x) and f = f (x, t) are given, ∇ · v0 = 0.

(3.4)

The functions v(x, t) and p = p(x, t), solving Eqs. (3.1)–(3.3), are to be found. These functions decay as |x| → ∞, t → ∞. Our task is to derive an integral equation for v that is equivalent to (3.1)–(3.3). To do this, we construct the Green’s function G(x, t) that solves the linear problem: G  − νG = δ(x)δ(t)δ jm − ∇ pm in R3 × R+ ,

(3.5)

∇ · G = 0,

(3.6)

G(x, 0) = 0,

(3.7)

where δ(x), δ(t) are delta-functions, δ jm = pm = pem ,

 1, j = m 0, j  = m

is the Kronecker’s delta,

p = p(x, t),

(3.8)

the scalar function p(x, t) is to be found together with v,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_3

13

14

3 Statement of the Navier-Stokes Problem

p(x, t) = 0, t < 0, ∂p e j em , ∇ pm = ∂x j

(3.9) (3.10)

over the repeated indices summation is understood, e j em is a tensor and G = G jm (x, t) is a tensor. Let us define the Fourier transform: 1 (2π)3

F(v) := v(ξ, ˜ t) =

 R3

and the inverse Fourier transform

v(x, t)e−iξ·x d x, ξ · x = ξ j x j ,

(3.12)



v(x, t) = Let

(3.11)

R3

v(ξ, ˜ t)eiξ·x dξ.

(3.13)

 H (ξ, t)eiξ·x dξ, G = G jm (x, t),  1 ˜ H (ξ, t) = G(x, t)e−iξ·x d x, H = H jm (ξ, t), H = F(G) = G. (2π)3 R3 G=

R3

(3.14) (3.15)

Taking the Fourier transform of Equations (3.5)–(3.6) one gets: δ(t)δ jm − iξ Pm (ξ, t), (2π)3 ξ · H = 0,

H  + νξ 2 H =

where ξ Pm is a tensor ξ j e j em p(ξ, ˜ t), see formula (3.19) below. From (3.17) one obtains ξ · H  = 0. Vector Pm is

1 Pm (ξ, t) = em (2π)3

 R3

p(x, t)e−iξ·x d x.

(3.16) (3.17)

(3.18)

(3.19)

Multiply (3.16) by ξ and use relations (3.17)–(3.18) to get δ(t)ξm = iξ 2 Pm (ξ, t). (2π)3

(3.20)

δ(t)ξm , (2π)3 ξ 2

(3.21)

Therefore Pm (ξ, t) = −i

ξ2 = ξ j ξ j .

3 Statement of the Navier-Stokes Problem

15

From Eqs. (3.16) and (3.21) one obtains  H jm

 + νξ H jm = 2

ds

δ(t) (2π)3

 δ jm

ξ j ξm − 2 ξ

 ,

(3.22)

From Eq. (3.7) it follows that H (ξ, 0) = 0.

(3.23)

Solving Eqs. (3.22)–(3.23) yields: 1 (2π)3

G˜ = H := H jm (ξ, t) = Indeed, let νξ 2 := a,

1 (2π)3

  ξ j ξm 2 δ jm − 2 e−νξ t . ξ

 δ jm −

ξ j ξm ξ2

(3.24)

 := b.

(3.25)

Then formulas (3.22)–(3.23) can be written as H  + a H = bδ(t),

H |t=0 = 0.

(3.26)

Taking the Laplace transform of (3.26) (see Appendix 3) one gets ( p + a)L(H ) = b.

(3.27)

From (3.27) it follows that L(H ) = Thus,

b . p+a

(3.28)

H = be−at .

(3.29)

Formula (3.29) is identical to (3.24). From (3.24) and (3.14) one derives    ξ j ξm 1 2 δ e−νξ t eiξ·x dξ. G(x, t) = − jm (2π)3 R3 ξ2 Let us define

1 I1 := (2π)3

and I2 := −

1 (2π)3

 R3

 R3

δ jm e−νξ

2 t+iξ·x

dξ

ξ j ξm −νξ 2 t+iξ·x e dξ. ξ2

(3.30)

(3.31)

(3.32)

Integrals (3.31)–(3.32) can be calculated explicitly. Therefore, the function G(x, t) will be calculated explicitly.

16

3 Statement of the Navier-Stokes Problem

Let us calculate I1 . Consider the one-dimensional integral  ∞ 1 2 e−νs t+isq ds. J= 2π −∞

(3.33)

One has (see [1], formula (1.4.11)) 1 J= π



q2

∞

e

−νs 2 t

0

e− 4νt cos(sq) ds = . (4νtπ)1/2

(3.34)

Integral J 3 = I1 , that is I1 = δ jm g(x, t),

(3.35)

e− 4νt , t > 0; lim g(x, t) = g(x, 0) = 0. g(x, t) = (4πνt)3/2 t→0+

(3.36)

where

|x|2

Let us calculate I2 . One has



I2 = ∂ jm  = ∂ jm where ∂ jm =

∂2 ∂x j ∂xm .

1 (2π)3 1 (2π)3

 2 e−νξ t+iξ·x dξ ξ2 R3   ∞  −νtr 2 ir |x| cos θ dr e e sin θ dθ dφ , 

(3.37)

S2

0

Therefore, I2 = ∂ jm

1 (2π)2



∞

dr e−νtr

2



0



∞

1

eir |x|u du

−1 eir |x|

− e−ir |x| ir |x| 0  ∞ 2 −νtr 2 sin(r |x|) ∂ jm dr e . = (2π)2 r |x| 0 = ∂ jm

1 (2π)2

dr e−νtr · 2

(3.38)

We now use a formula from [1], formula (2, 4.21): 

∞ 0

e−ax π sin(x y) d x = Erf x 2 2

where

2 Erf(x) = √ π





x

y √ 2 a



π , 2

, | arg a|
0 various constants are denoted. Lemma 3.5 is proved.



We will use the Parseval’s identity in the Hilbert space L 2 (R3 ): v2 = (2π)3 v ˜ 2.

(3.68)

This result is well known and we omit its proof. Note that      h(ξ − η)g(η) dη  ≤ h g. 

(3.69)

R3

From (3.56), (3.66), and (3.69) one obtains  t 2 v(ξ, ˜ t) ≤ c ds e−σ(t−s)ξ  v(ξ, ˜ s) |ξ|v(ξ, ˜ s) ds 0



+c

t

dse−ν(t−s)ξ | f˜(ξ, s)| ds + v˜0 (ξ, t)e−νtξ . 2

2

(3.70)

0

Lemma 3.6 One has c (t − s)3/4 c 2 . |ξ|e−ν(t−s)ξ  = (t − s)5/4 e−ν(t−s)ξ  = 2

(3.71) (3.72)

Proof Let us calculate the first integral. The second one is calculated similarly. One has for any a > 0   ∞ c 2 −aξ 2 e dξ = 4π e−ar r 2 dr = 3/2 . 3 a 0 R Therefore,

c . a 3/4 Formula (3.72) is derived similarly. The parameter a = ν(t − s). Since ν = const > 0, Lemma 3.6 is proved. e−aξ  = 2

(3.73) 

Assumption 3A. Let us assume in what follows that v˜0 (ξ) and f˜(ξ, t) are smooth and rapidly decaying functions of their arguments.

Reference

21

Then assumptions of Lemma 3.4 are satisfied. Denote |ξ|v ˜ := b(t).

(3.74)

Then inequality (3.70) implies 

t

b(t) ≤ c 0

b(s) ds + b0 (t), (t − s)5/4

(3.75)

where we used estimate (3.59) and denoted b0 (t) := v˜0 (ξ, t)e

−νtξ 2



t

|ξ| + c

2  f˜(ξ, s)e−ν(t−s)ξ ds.

(3.76)

0

From Assumption (1.15) it follows that b0 (t) is a smooth rapidly decaying function as t → ∞. The integral in formula (3.75) diverges classically. We define this integral in Chap. 4.

Reference 1. H. Bateman, A. Erdelyi, Tables of Integral Transforms (McGraw-Hill, New York, 1954)

Theory of Some Hyper-Singular Integral Equations

Consider the integral equation 

t

q(t) = b0 (t) + c

(t − s)λ−1 q(s) ds,

(4.1)

0

where Re λ > 0, c > 0 stands for various constants, b0 (t) is a given function which is assumed to be smooth and rapidly decaying as t → ∞. Taking the Laplace transform of (4.1) one gets (λ) (4.2) L(q) = L(b0 ) + cL(q) λ , p where formula (A2.13) was used: L(t λ−1 ) =

(λ) . pλ

(4.3)

Here (λ) is the gamma function and formula (4.3) is valid classically for Re λ > 0; this formula is valid for all complex λ except λ = 0, −1, −2, . . ., by analytic continuation with respect to λ because (λ) is analytic in C except for the points λ = 0, −1, −2, . . . and p −λ is an entire function of λ; see Appendix 2. In Appendix 3, one finds the results on the theory of the Laplace transform that are used in this book. We are interested in the value λ = − 41 because it appears in inequality (3.75), λ − 1 = 5 − 4 = − 41 − 1, so λ = − 41 . The value (− 41 ) can be calculated: 3 1 (− ) = −4( ) := −c1 , 4 4

3 c1 = 4( ) > 0. 4

(4.4)

For λ = − 41 , formula (4.2) implies L(q) =

L(b0 ) 1

1 + cc1 p 4

, cc1 = const > 0.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_4

(4.5) 23

4

24

4 Theory of Some Hyper-Singular Integral Equations

Theorem 4.1 Assume that |L(b0 )| ≤

c , Re p ≥ 0. 1 + | p|

(4.6)

Equation (4.1) then has a unique solution q, q = q(t) is a bounded function on C([0, ∞)), sup |q(t)| ≤ c,

(4.7)

t≥0

and q(0) = 0.

(4.8)

Proof The statement of Theorem 4.1 concerning continuity of q and estimate (4.7) follow from Theorem 4.1 in Appendix 3. The conclusion (4.8) follows from Theorem A3.2 of Appendix 3. Theorem 4.1 is proved.  Let us establish a relation between the solution to Eq. (4.1) and solutions to inequality (3.75). Theorem 4.2 If b = b(t) solves inequality (3.75) and q(t) solves Eq. (4.1) with λ = − 41 , then b(t) ≤ q(t). (4.9) Proof Proof of Theorem 4.2 requires some preparations. First, let us prove the following lemma. Lemma 4.3 For any T > 0 and for p > −1, the operator A f := Banach space X 0 = C([0, T ]) has spectral radius r (A) = 0.

t

0 (t

− s) p f (s) ds in the

Proof of Lemma 4.3. The spectral radius r (A) of a linear operator A is defined by the formula (see, for example, [1]): 1

r (A) = lim An  n . n→∞

(4.10)

Let us check by induction that An f  X 0 ≤ t n( p+1)

 n ( p + 1)  f X0 . (n( p + 1) + 1)

For n = 0, estimate (4.11) is obvious. For n = 1, one has

(4.11)

4 Theory of Some Hyper-Singular Integral Equations

25

  t p+1    (t − s) p f (s) ds  ≤ t  f X0 .   p+1 0

(4.12)

Suppose estimate (4.11) holds for some n. Let us prove that it holds for n + 1. One has  t An+1 f  X 0 ≤ (t − s) p |An f | ds 0

 ≤

t

(t − s) p s n( p+1) ds

0



t

 (t − s) s

p n( p+1)

ds = t

1

p+n( p+1)+1

0

 n ( p + 1)  f X0 , (n( p + 1) + 1)

(4.13)

(1 − u) p u n( p+1) du

0

= t ( p+1)(n+1)

( p + 1)(n( p + 1) + 1) . ( p + 1 + n( p + 1) + 1)

Thus, An+1 f  X 0 ≤ t (n+1)( p+1)

 n+1 ( p + 1)  f X0 . ((n + 1)( p + 1) + 1)

(4.14)

Applying formula (A2.9) to (4.11), one gets 1/n

lim An f  X 0 = t p+1 ( p + 1)

n→∞

1 1/n lim  f  X 0 lim (n( p + 1) + 1) n→∞

n→∞

=

t p+1 ( p + 1) 1

lim e(n( p+1)+ 2 ) ln(n( p+1)+1)

= 0.

(4.15)

n→∞



By induction, Lemma 4.3 is proved.

Lemma 4.4 Let A be a linear operator in a Banach space Z . If r (A) < 1, then equation f = f0 + A f

(4.16)

is uniquely solvable by iterations and f =

∞ 

A j f0 ,

(4.17)

j=0

where the series converges in Z . Proof Let us show that Eq. (4.15) is an equation with A < 1 in the Banach space Z . If this is done then formula (4.17) follows and the series (4.17) converges in Z .

26

4 Theory of Some Hyper-Singular Integral Equations

If the series (4.17) converges in Z then its sum solves (4.16). Indeed, from (4.17) one derives ∞  A j f0 = f0 + A f . (4.18) f = f0 + A j=0

If r (A) < 1, then for some m Am  ≤ γ < 1.

(4.19)

Let us write the series (4.17) as m−1 

A j f 0 + Am

j=0

m−1 

A j f 0 + A2m

j=0

m−1 

A j f0 + · · · .

(4.20)

j=0

Denote h :=

m−1 

A j f0 .

(4.21)

j=0

Then

∞ 

A j f0 =

∞ 

Amq h.

(4.22)

q=0

j=0

The series (4.22) converges in Z because Amq  ≤ Am q ≤ γ q < 1, 0 < γ < 1.

(4.23)

Therefore the series (4.17) converges and Eq. (4.16) has a solution. Let us prove that this solution is unique. Since A is a linear operator, it is sufficient to prove that the equation f = Af (4.24) has only the trivial solution f = 0. From (4.24) one derives f = Am f .

(4.25)

Am f  ≤ γ f , 0 < γ < 1.

(4.26)

From (4.19) it follows that

This and (4.25) imply f = 0. Lemma 4.4 is proved.



t If A f = 0 (t − s)λ−1 f (s) ds, λ > 0, then by Lemma 4.3, r (A) = 0. Therefore, Eq. (4.16) with this A has a solution, this solution is unique and can be obtained by iterations by formula (4.17).

4 Theory of Some Hyper-Singular Integral Equations

27

In the book [2] it is stated that one can construct a Banach space Z 1 in which A1 < 1 if A is a linear operator and r (A) < 1 in the original Banach space Z . The proof of this statement in [2] is not clear to this author. By this reason we formulated and proved Lemma 4.4. Since this result is used below, let us give an independent proof of it using the specific form of the operator A. Let us prove that if r (A) = 0 then Eq. (4.16) has a solution, this solution is unique in Z , Eq. (4.16) is solvable by iterations f n+1 = f 0 + c A f n ,

f |n=0 = f 0 ,

(4.27)

and f is given by formula (4.17). We took c A because in (3.75) c > 0 is not necessarily equal to 1. Using estimate (4.11) one concludes that the series (4.17) converges in Z . Clearly, its sum solves Eq. (4.16) and is obtained as a limit f = lim f n , n→∞

(4.28)

where f n are iterations defined by formula (4.27). To prove uniqueness of the solution to Eq. (4.16) assume that f 0 = 0, so the Eq. (4.16) is of the form h = c Ah.

(4.29)

One has by formula (4.12) h X 0 ≤ c Choose T so that

T p+1 h X 0 . p+1

cT p+1 < 1. p+1

(4.30)

(4.31)

This is possible since p + 1 > 0. Then estimate (4.30) implies h(t) = 0

∀t ∈ [0, T ].

Continue with this argument and prove (4.32) for any T > 0. Let us now introduce the function λ =

t λ−1 , (λ)

(4.32) 

(4.33)

28

4 Theory of Some Hyper-Singular Integral Equations

where λ t λ = t+ =

Define the convolution:

 λ ∗ f :=

0

t

 0, t λ,

t < 0, t ≥ 0.

(t − s)λ−1 f (s) ds. (λ)

(4.34)

(4.35)

The right side of Equation (4.35) is well defined classically if Re λ > 0 and f (s) is continuous on R+ = [0, ∞). Let us define the right side of (4.35) for Re λ < 0. To do this, consider the Laplace transform (4.36) L(λ ∗ f ) = L(λ )L( f ) = L( f ) p −λ , where formula (A3.21) was used. Formula (4.36) is well defined classically for Re λ > 0 and for Re λ ≤ 0 it is defined by the analytic continuation with respect to λ since L( f ) does not depend on λ and p −λ is an entire function of λ. The convolution λ ∗ f for Re λ < 0 is defined by the formula λ ∗ f = L −1 (L( f ) p λ ).

(4.37)

We are interested in the value λ = − 41 because this value is important for the NSP, see Eq. (3.75). If λ = − 41 then 1 3 (− ) = −4( ) := −c1 , (4.38) 4 4 and formula (4.36) for λ = − 41 takes the form 1

L(− 1 ∗ f ) = p 4 L( f ).

(4.39)

λ ∗ μ = λ+μ

(4.40)

0 (t) = δ(t),

(4.41)

4

Lemma 4.5 One has for any λ, μ ∈ C, and where δ(t) is the delta-function. Proof By formula (4.36) with f = μ one gets L(λ ∗ μ ) =

1 . p λ+μ

(4.42)

By formula (4.37) with f = μ one obtains λ ∗ μ = L −1 (

1 p λ+μ

) = λ+μ .

(4.43)

4 Theory of Some Hyper-Singular Integral Equations

29

Thus, formula (4.40) is proved. If λ + μ = 0, then λ ∗ −λ = 0 = L −1 (1) = δ(t). Here 1=

 1, t ≥ 0,

L(δ(t)) = 1.

0, t < 0,

(4.44)

(4.45) 

Lemma 4.5 is proved. One can give a different proof of formula (4.40). Namely, let u = st . Then 

(t − s)λ−1 s μ−1 ds (λ) (μ) 0  1 1 (1 − u)λ−1 u μ−1 du t λ+μ−1 = (λ)(μ) 0

λ ∗ μ =

t

=

t λ+μ−1 (λ)(μ) (λ)(μ) (λ + μ)

=

t λ+μ−1 (λ + μ)

= λ+μ .

(4.46)

Here we have used the properties of the beta-function; see Appendix 2. While classically the integrals in (4.46) are defined for Re λ > 0, Re μ > 0, they make sense by analytic continuation for all λ, μ ∈ C. For λ + μ → 0 the λ+μ converges to δ(t) in the sense of distribution theory. This was proved in Lemma 4.5. Using formula (4.38) one can rewrite Eq. (3.75) in the form b(t) ≤ b0 (t) − cc1 − 1 ∗ b, c, c1 > 0

(4.47)

q(t) = b0 (t) − cc1 − 1 ∗ q.

(4.48)

4

and Eq. (4.1) as 4

Lemma 4.6 If f ≥ 0 is continuous and λ > 0, then λ ∗ f ≥ 0.

(4.49)

Proof For λ > 0 and f continuous the convolution (4.49) is defined classically and inequality (4.49) is obvious. Lemma 4.6 is proved. 

30

4 Theory of Some Hyper-Singular Integral Equations

Apply the operator  1 ∗ to inequality (4.47) and to Eq. (4.48) and use Eq. (4.41) to get 4

1 ∗ b

≤  1 ∗ b0 − cc1 b,

1 ∗ q

=  1 ∗ b0 − cc1 q,

4

4

1 1  1 ∗ b0 −  1 ∗ b, cc1 4 cc1 4 1 1 q=  1 ∗ b0 −  1 ∗ q, 4 cc1 cc1 4 b≤

4

4

(4.50) (4.51)

Let us now prove Theorem 4.2. Proof of Theorem 4.2. Denote the positive constant cc1 := c2 > 0. By Lemma 4.4 the inequality (4.50) and Eq. (4.51) can be solved by iterations: b≤ q=

∞ 

 (−1)n

n=0 ∞ 



(−1)n

n=0

1 1 ∗ c2 4 1 1 ∗ c2 4

n b0 ,

(4.52)

b0 .

(4.53)

n

From (4.52) and (4.53) the conclusion of Theorem 4.2 follows, in particular inequality (4.9) holds. Theorem 4.2 is proved.  Theorem 4.7 Assume that b0 (t) in Eq. (4.48) is smooth and rapidly decaying as t → ∞. Then Eq. (4.48) has a unique solution in C([0, ∞)) and sup q(t) ≤ c, q(0) = 0.

(4.54)

t≥0

Proof Taking the Laplace transform of Equation (4.48) one gets: 1

L(q) = L(b0 ) − cc1 p 4 L(q). Therefore, L(q) =

L(b0 ) 1

1 + cc1 p 4

(4.55)

.

(4.56)

Under our assumption about b0 (t) one has |L(b0 )| ≤

c , 1 + | p|

Re p > 0,

(4.57) 1

and L(b0 ) is an analytic function of p in the half-plan Re p > 0. Define p 4 in the half-plane Re p > 0 by the formula: 1

1

p 4 = | p| 4 eϕ/4 , ϕ = arg p, −

π π ≤ϕ≤ . 2 2

(4.58)

References

31

1

Then p 4 is analytic in the region Re p > 0; L(q), defined by formula (4.56), is an analytic function in the half-plane Re z > 0 and, using inequality (4.57), one gets |L(q)| ≤ Let us prove that

c   , c2 := cc1 > 0. 1   (1 + | p|) 1 + c2 | p| 4 eiϕ/4 

  1   w := 1 + c2 | p| 4 eiϕ/4  > c, Re p ≥ 0.

(4.59)

(4.60)

One has  1 1 1 1 ϕ ϕ 2 ϕ 2 ϕ  w2 = 1 + c2 | p| 4 cos + ic2 | p| 4 sin  = 1 + c2 | p| 4 cos + c22 | p| 2 sin2 . 4 4 4 4 Thus,

1

1

w2 = 1 + c22 | p| 2 + 2c2 | p| 4 cos

ϕ . 4

(4.61)

Since 0 < cos ϕ4 < 1, it is clear that inf

p≥0,|ϕ|≤ π2

w2 > c > 0.

(4.62)

Therefore, formula (4.59) implies |L(q)| ≤

c , Re p > 0, 1 + | p|

|L(q)| = O( p−5/4 ) as | p| → ∞, Re p > 0.

(4.63) (4.64)

By Theorem 4.1, see formulas (4.7) and (4.8), one gets inequality (4.54) and the relation q(0) = 0. Theorem 4.7 is proved. 

References 1. W. Rudin, Functional Analysis (McGraw-Hill, New York, 1973) 2. M. Krasnoselskii et al., Approximate Solutions of Operator Equations (Walters-Noordhoff, Groningen, 1972). https://doi.org/10.1007/978-94-010-2715-1

5

A Priori Estimates of the Solution to the NSP

One of the a priori estimates was formulated and proved in Lemma 3.4, namely, estimate (3.59): (5.1) sup v(x, t) < c, t≥0

where v(x, t) is a solution to the NSP (3.1)–(3.3). It was proved under the assumption  ∞  f (x, s) ds < c. (5.2) v0 (x, t) + 0

The other basic a priori estimate is formulated in Theorem 4.7. Let us use this result and the Parseval’s equation to derive the following theorem. Theorem 5.1 Denote b(t) := |ξ|v(ξ, ˜ t) and assume that v0 (x) and f (x, t) are smooth functions of their arguments rapidly decaying as |x| → ∞ and t → ∞. Then sup b(t) < c,

(5.3)

t≥0

that is, by the Parseval’s equation sup ∇v(x, t) < c.

(5.3a)

t≥0

Proof Under the assumption of Theorem 5.1 the Assumption 3A (see p. 19) is satisfied and b(t) solves inequality (3.75). Inequality (3.75) is equivalent to inequality (4.47). By Theorem 4.2 of Chap. 4 inequality (4.9) holds, where q(t) solves Eq. (4.48). By Theorem 4.7 of Chap. 4, not only does inequality (5.3) hold since

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_5

33

34

5 A Priori Estimates of the Solution to the NSP

sup b(t) ≤ sup q(t) < c, t≥0

(5.4)

t≥0

but also the relation b(0) = 0,

(5.5)

is valid. Theorem 5.1 is proved.



By the Parseval equality Equation (5.3) is equivalent to sup ∇v < c.

(5.6)

t≥0

The expression ∇v is a tensor: ∇v = v j,m e j em , where v j,m =

∂v j ∂xm

(5.7)

and by the repeated indices one sums up.

Theorem 5.2 Under the assumptions of Theorem 5.1 the following inequalities hold: ξ 2 |v(ξ, ˜ t)| < c, ∀t ≥ 0;

sup t≥0,|ξ|≥0

sup |v(ξ, t)| ≤ c(1 + t).

(5.8)

|ξ|>0

Proof From formula (3.56) one gets ˜ t) − v(ξ, ˜ t) = F(ξ,



t

 ds H (ξ, t − s)(v, ∇)v,

(5.9)

0

where  t −νtξ 2 ˜ + ds H (ξ, t − s) f˜(ξ, s), F(ξ, t) := v˜0 (ξ)e 0   v˜m (ξ − η, s)v˜ j (η, s)iη dη, (v, ∇)v = R3 −νξ 2 (t−s)

|H (ξ, t − s)| ≤ ce

(5.10) (5.11)

,

(5.12)

where inequality (3.66) is identical to inequality (5.12). It follows from (5.9) and (3.69) that ˜ t)| |v(ξ, ˜ t)| ≤ | F(ξ, 

t

+c

ds e 0

−νξ 2 (t−s)

 R3

1/2  |v(ξ ˜ − η, s)| dη ·

1/2 |v(η, ˜ s)| |η| dη 2

R3

2

.

5 A Priori Estimates of the Solution to the NSP

35

Thus, ˜ t)| + c |v(ξ, ˜ t)| ≤ | F(ξ,



t

ds e−νξ

2 (t−s)

|ξ|v(ξ, ˜ s) v(ξ ˜ − η, s)

ds e−νξ

2 (t−s)

,

0

˜ t)| + c ≤ | F(ξ,



t

(5.13)

0

where the a priori estimates (5.1) and (5.1) were used. From inequality (5.13) we derive: −νξ 2 t

˜ t)| + c 1 − e |v(ξ, ˜ t)| ≤ | F(ξ, νξ 2

.

(5.14)

˜ t) depends only on the data f (x, t) and v0 (x). We may assume that the The function F(ξ, data are smooth and rapidly decaying and then ˜ t)| ≤ | F(ξ,

c . 1 + |ξ|2

(5.15)

The last term in (5.14) for |ξ| → 0 is of the order νt, so it is not bounded uniformly in t ∈ [0, ∞). Let us multiply both terms of (5.14) by |ξ|2 and get ˜ t)| + c . |ξ|2 |v(ξ, ˜ t)| ≤ |ξ|2 | F(ξ, ν

(5.16)

By the estimate (5.15) one gets sup

|ξ|2 |v(ξ, ˜ t)| ≤ c.

(5.17)

t≥0,|ξ|≥0

One has

1 − e−νξ sup νξ 2 |ξ|>0

because max u≥0

Thus, Theorem 5.2 is proved.

2t

< ct,

(5.18)

1 − e−u ≤ 1. u 

6

Uniqueness of the Solution to the NSP

Theorem 6.1 There is at most one solution in W21 (R3 ) × C(R+ ) of the NSP (3.1)–(3.3). Proof To prove uniqueness of the solution to the NSP assume that v˜1 and v˜2 solve Eq. (3.56). Let w = v˜1 − v˜2 . We have (with G˜ = H and ∗ the convolution in R3 )  t  t ds H (ξ, t − s)(v˜1 ∗ v˜1 − v˜2 ∗ v˜2 ) = ds H (ξ, t − s)(w ∗ v˜1 + v˜2 ∗ w), w=− 0

0



so |w| ≤ c

t

ds e−νξ

2 (t−s)

(6.1)

(|w ∗ v˜1 | + |v˜2 ∗ w|).

(6.2)

0

One has |w ∗ v˜1 | ≤ wv˜1 ,

(6.3)

|v˜2 ∗ w| ≤ v˜2 w.

(6.4)

By inequalities (5.1), (6.3), and (6.4), one gets from (6.2) the inequality:  t 2 |w(ξ, t)| ≤ c ds e−νξ (t−s) w(ξ, s).

(6.5)

0

Take the norm  ·  of both parts of inequality (6.5) and get  t  t −ν(t−s)ξ 2 w(ξ, t) ≤ c ds e w(ξ, s) ≤ c ds(t − s)−3/4 w(ξ, s). 0

(6.6)

0

Denote w(ξ, t) := z(t).

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_6

(6.7)

37

38

6 Uniqueness of the Solution to the NSP

Then (6.6) takes the form



t

z(t) ≤ c

(t − s)−3/4 z(s) ds.

(6.8)

0

Since v˜1 and v˜2 correspond to the same data, one has z(0) = 0.

(6.9)

Inequality (6.8) and the continuity of z(t) imply condition (6.9) also. Equation (6.8) has only the trivial solution z(t) = 0. (6.10) Indeed, let τ > 0 be so small that  τ c (t − s)−3/4 ds = 4cτ 1/4 := μ < 1

(6.11)

0

Let sup z(t) := θ.

(6.12)

0≤t≤τ

From Eq. (6.8) one obtains θ ≤ μθ.

(6.13)

Since θ ≥ 0 and 0 < μ < 1, it follows that θ = 0.

(6.14)

Therefore, z(t) = w(ξ, t) = 0

∀t ∈ [0, τ ].

(6.15)

Repeating this argument one proves that w(ξ, t) = 0

∀t ≥ 0.

(6.16)

Therefore, v˜1 = v˜2 . Theorem 6.1 is proved.

(6.17) 

Remark 6.2 The uniqueness Theorem 6.1 is established for the first time in W21 (R3 ) × C(R+ ) in this book. In different classes of functions uniqueness theorems were published earlier, see [1, 2]. Remark 6.3 A different proof of Theorem 6.1 is given in Appendix 4, Section 4.4.

References

39

References 1. O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Fluid (Gordon and Breach, New York, 1969) 2. A. Ramm, Concerning the Navier–Stokes problem. Open J. Math. Anal. 4(2), 89–92 (2020). https://doi.org/10.30538/psrp-oma2020.0066

7

The Paradox and Its Consequences

Let the assumption (1.15) p. 4 hold. In this chapter we prove that the NSP (3.1)–(3.3) implies the following. NSP Paradox: If one assumes that the solution v(x, t) to the NSP exists for all t > 0, the initial velocity v0 (x)  ≡ 0, ∇ · v0 = 0, and f (x, t) = 0, then the solution v to the NSP satisfies the condition v0 (x) = v(x, 0) = 0. The consequences of this Paradox: (a) the solution to the NSP (3.1)–(3.3) does not exist on the interval t ≥ 0 except for the case when v0 (x) = f (x, t) = 0; in this case the solution v(x, t) = 0 does exist for all t ≥ 0; and (b) the NSP is physically contradictive, it is physically incorrect. What does it mean that a physical problem is correct? It means that: (i) the equations and the initial and boundary conditions constitute a problem that has a solution and the solution is unique in the functional space in which the solution exists, and (ii) there are no consequences that can be derived from this problem that are contradictory. The NSP fails to satisfy these requirements. This means that the NSP is not correct physically and it is necessary to look for physically correct equations describing the motion of incompressible viscous fluid. There is a discussion of various paradoxes in the existing theories of fluid mechanics. The paradox we described in this chapter follows from Theorem 4.7, namely from the equation q(0) = 0. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_7

41

42

7 The Paradox and Its Consequences

The consequences of the NSP paradox are dramatic: the Navier-Stokes equations were studied from the 19th century and we prove that the NSP is physically (and mathematically) incorrect. The derivations we gave are done for the domain without boundaries, for the whole R3 . If the domain has the boundary, our derivation scheme works in principle: one can construct the Green’s function for the linear problem without the nonlinear term (v, ∇)v and on this basis derive an integral equation. However, technically there will be difficulties since the Green’s function will be more difficult to construct and to estimate; in place of the Fourier transform one has to use expansion in eigenfunctions, etc. It is interesting to find out what relation our Paradox has with the turbulent motions of the fluid.

8

Logical Analysis of Our Proof

The NSP is formulated in Eq. (1.1). Step 1. We derive an integral Equation (3.52) for the fluid velocity v(x, t) which is equivalent to the NSP (Theorem 3.2) and an equivalent Equation (3.56) for v(ξ, ˜ t), for the Fourier transform of v(x, t). The function (tensor) G(x, t) in Eq. (3.52) is constructed for the linear problem (3.5)–(3.8), H (ξ, t) is the Fourier transform of G(x, t). Tensor H (ξ, t) is given explicitly in formulas (3.24), tensor G(x, t) is given explicitly in formulas (3.30) and (3.43). Step 2. Theory of convolution Equation (4.1) with hyper-singular kernel is developed, see Theorems 4.1, 4.2, and 4.7. The solution to the NSP satisfies inequality (3.75). It is proved that a solution to inequality (4.47) is estimated through the solution to Eq. (4.48); see Theorem 4.2. It is proved that Eq. (4.1) with λ = − 41 has a unique solution q(t) and q(0) = 0; see Theorem 4.1. Estimates (4.54) are derived and the basic a priori estimate (5.3a) is proved. Step 3. We derive a priori estimates (5.1) and (5.3)–(5.3a), which prove that v ∈ X := W21 (R3 ) × C(R+ ) a priori, that is, if it exists for all t ≥ 0. Here, W21 (R3 ) is the Sobolev space. In Theorem 6.1 we prove uniqueness in X of the solution to the NSP (3.1)–(3.3). Step 4. We prove the Paradox for the solution v(x, t) to the NSP. This Paradox shows that the solution to the NSP in X does not exist on the region t ∈ [0, ∞) unless the data v0 (x) = f (x, t) = 0. In this case the solution is identically equal to zero for all t ≥ 0. It also proves that the solution to the NSP does not exist in any other functional spaces in R3 × R+ because a priori the solution has to be in X .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2_8

43

1

Theory of Distributions and Hyper-Singular Integrals

Consider the integral



∞

Ia :=

x a f (x) d x,

(A1.1)

0

where f is a smooth function decaying faster than any power of x as x → ∞. If a > −1, then Ia converges classically. How does one define integral Ia for a ≤ −1? The following is the standard method, see [1]. One has   1  ∞ x a+1 1 x a [ f (x) − f (0)] d x + x a f (x) d x + f (0) Ia = a + 1 0 0 1  1  ∞ f (0) = x a [ f (x) − f (0)] d x + x a f (x) d x + , a > −1. (A1.2) a +1 0 1 Assume that f ∈ C 1 ([0, 1]). Then first integral converges classically for a > −2, and is analytic with respect to a in the region Re a > −2. The second integral is analytic with respect to a for any complex a. The third term is analytic for any a except for the point a = −1 at which it has simple pole. Therefore, formula (A1.2) allows one to define Ia in the region Re a > −2 by analytic continuation with respect to a. One can generalize this construction: 

1

Ia = 0

x a [ f (x) −

 ∞ n n   f ( j) (0) j f ( j) (0) 1 x a f (x) d x + x ] dx + , j! j! a + j +1 1 j=0

j=0

(A1.3) where it is assumed that f ∈ C (n+1) (0, 1). The first integral here is analytic with respect to a in the region Re a > −n − 1, the second integral is analytic for all a ∈ C, and the third term is analytic for a ∈ C except for the points a = −1, −2, . . . , −n − 1 at which it has simple poles. Formula (A1.3) allows one to define Ia in the region Re a > −n − 1 by analytic continuation with respect to a. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2

45

46

Appendix 1: Theory of Distributions and Hyper-Singular Integrals

Let  be an open subset of R3 . Define the set C0∞ () := K as the set of test functions ϕ. A distribution (or a generalized function) f is a linear continuous functional ( f , ϕ) on K , ϕ ∈ K is a test function. The linearity means ( f , c1 ϕ1 + c2 ϕ2 ) = c1 ( f , ϕ1 ) + c2 ( f , ϕ2 ),

(A1.4)

where ϕ j ∈ K and c j = const are arbitrary. The continuity means that if ϕn → ϕ, then ( f , ϕn ) → ( f , ϕ). Convergence ϕn → ϕ in K means that all ϕn vanish outside the same ( j) compact domain D and ϕn → ϕ( j) as n → ∞ for any j = 0, 1, 2, . . ., uniformly in D, that is, in C(D). A distribution f = 0 in a neighborhood  of a point x0 if ( f , ϕ) = 0 for all ϕ ∈ K such that ϕ = 0 in  \ . One says that f = f 1 + f 2 if ( f , ϕ) = ( f 1 , ϕ) + ( f 2 , ϕ) ∀ϕ ∈ K . One says that f n → f if limn→∞ ( f n , ϕ) = ( f , ϕ) ∀ϕ ∈ K . Spaces of the test functions other than K can be used. For example, S, the space of C ∞ (R3 ) functions decaying, as |x| → ∞, together will all their derivatives faster than 2 O((1 + |x|2 )−q ) for any q = 0, 1, 2, . . . , see [1]. For instance, e−x ∈ S. One defines the derivative of a distribution f by the formula ( f  , ϕ) = −( f , ϕ ).

(A1.5)

Here ϕ ∈ K ; the functional f  is linear and continuous on K . Let us define the convolution of distributions f ∗ g. For two functions their convolution in R3 is defined by the formula  f ∗g = f (x − y)g(y) dy, (A1.6) R3

and it is assumed that f and g in (A1.6) are continuous and rapidly decaying at infinity function. The assumptions on f and g in (A1.6) can be relaxed. For example, one may assume that f , g ∈ L 2 (R3 ). Let  F(ϕ) := ψ(λ) := ϕ(x)eiλx d x (A1.7) R3

be the Fourier transform of ϕ. If ϕ ∈ K , then ψ(λ) is an entire function of λ. One has the Parseval equality:   R3

|ψ(λ)|2 dλ = (2π)3

R3

|ϕ(x)|2 d x.

(A1.8)

It is easy to check that F(ϕ1 ∗ ϕ2 ) = F(ϕ1 )F(ϕ2 ), ϕ1 , ϕ2 ∈ K .

(A1.9)

Since the product of two distributions is not defined in general, [1] defines the convolution of two distributions using the direct product f × g. The direct product of two distributions is a distribution defined by the formula

Appendix 1: Theory of Distributions and Hyper-Singular Integrals

47

( f × g, ϕ1 (x)ϕ2 (y)) = ( f , ϕ1 )(g, ϕ2 ).

(A1.10)

One has f ∗ g = g ∗ f . If f and g are smooth and decaying at infinity functions, then     ( f ∗ g, ϕ(x)) = f (x − y)g(y) dy · ϕ(x) d x = f (z)g(y)ϕ(z + y) dz dy, R3

R3

R3

R3

so ( f ∗ g, ϕ) = ( f × g, ϕ(z + y)) := ( f , (g, ϕ(z + y)).

(A1.11)

The function ϕ(z + y) may not belong to K even if ϕ(x) ∈ K . The definition (A1.11) makes sense only if ϕ(z + y) ∈ K . This happens, according to [1], if the supports of f and g are bounded from the same side. It also happens if one of the distributions f or g is compactly supported. The support of a distribution f is the set of points in a neighborhood of which f = 0. Assume that the support of f is the set supp f = [0, ∞), the support of g is also the set [0, ∞) and that x ∈ R1 . If ϕ(x) ∈ K then ϕ(x) has compact support, but the function (g, ϕ(x + y)) := ϕ1 (x) does not have compact support if the support of g is [0, ∞). Indeed, one can choose x negative so that x + y belongs to the support of ϕ(z), z = x + y, so that the ∞ integral 0 g(y)ϕ(x + y)dy is not equal to zero. Thus, the function ϕ1 (x) is not compactly supported, it does not belong to K . In [1] it is claimed that since supp f ∩ supp ϕ1 is a compact set (due to our assumption supp f = [0, ∞)), the formula (A1.11) makes sense and defines the convolution f ∗ g of two distributions (with the properties that supp f and supp g are bounded on the same side). This is not clear because ϕ1 (x) on the support of f may be not vanishing in a neighborhood of the boundary of the support of f .

Reference 1. I. Gel’fand, G. Shilov, Generalized Functions, vol. 1, (GIFML, Moscow, 1959). (in Russian)

2

Gamma and Beta Functions

The Gamma function is the special function defined by the integral  ∞ t z−1 e−t dt. (z) =

(A2.1)

0

This integral is defined classically for Re z > 0. One has ∞   ∞  tz 1 ∞ z −t 1 (z) = t z−1 e−t dt = e−t  + t e dt = (z + 1), z z z 0 0 0 so (z + 1) = z(z), and (z) =

(z + n) . z(z + 1) · · · (z + n − 1)

(A2.2) (A2.3)

The right side of formula (A2.3) is analytic for z = 0, −1, . . . , −n + 1, Re (z + n) > 0. At the points z = − j the function (z) has simple poles. In this argument n > 0 can be as large as one wishes. Therefore (z), defined originally by formula (A2.1) in the region Re z > 0, can be analytically continued to the whole complex plane of z with the exception of the points n z = − j, j = 0, 1, 2, . . .. At these points (z) has simple poles with the residues (−1) n! . The formula for the residue can be easily derived from formula (A2.3): multiply formula (A2.3) by z + , let z → − and pass to the limit. The result is Res (z) =

z=−

Thus, the function

1 (z)

(−1) . !

is an entire function of z.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2

49

50

Appendix 2: Gamma and Beta Functions

One has (z)(1 − z) =

π , sin(πz)

(A2.4)

1 1 22z−1 (z)(z + ) = π 2 (2z). 2

(A2.5)

These formulas are proved, e.g., in [1] . It follows from (A2.2) that

Let z =

1 2

(n + 1) = 1 · 2 · . . . · n = n!

(A2.6)

√ 1 ( ) = π. 2

(A2.7)

in (A2.2). Then

Thus, √ 1 1 1 1 3 1 1 · 3 · . . . · (2n − 1) π . (n + ) = (n − )(n − ) = (n − )(n − ) · · · = 2 2 2 2 2 2 2n (A2.8) One can prove (see [1]) the Stirling formula 1

(z) = e(z− 2 ) ln z−z+ In particular, n! =

ln(2π) 2

[1 + o(1)], |z| → ∞, | arg z| ≤

π . 2

√ 1 2πn n+ 2 e−n [1 + o(1)], n → +∞.

(A2.9)

(A2.10)

Let us define beta function: 

1

B(x, y) =

t x−1 (1 − t) y−1 dt.

(A2.11)

0

This integral is defined classically for Re x > 0, Re y > 0. If u =  B(x, y) = 0

Note the formula



∞ 0

∞

u x−1 du. (1 + u)x+y

e− pt t z−1 dt =

(z) , pz

t 1−t ,

then (A2.12)

(A2.13)

which is derived easily from (A2.1) if p > 0 by the substitution pt = s. For complex p, Re p > 0, the result holds by analytic continuation with respect to p. For p > 0 and Re z > 0 formula (A2.13) holds classically; for complex z, z = 0, −1, −2, . . ., formula (A2.13) holds by analytic continuation with respect to z since (z) is analytic for z = 0, −1, −2, . . . and p z is an entire function of z for p > 0.

Appendix 2: Gamma and Beta Functions

51

Using formula (A2.13) one gets 1 1 = (1 + u)x+y (x + y)



∞

e−(1+u)t t x+y−1 dt.

(A2.14)

0

If one substitutes formula (A2.14) into (A2.12), one gets  ∞  ∞ 1 du u x−1 e−(1+u)t t x+y−1 dt B(x, y) = (x + y) 0 0  ∞  ∞ 1 = e−t t x+y−1 dt du u x−1 e−ut (x + y) 0 0  ∞ (x) = e−t t y−1 dt (x + y) 0 =

(x)(y) . (x + y)

(A2.15)

1 Thus, using the fact that (x) is analytic for complex x, x = 0, −1, −2, . . . and (x) is an entire function of x ∈ C, where C is the complex plane, one concludes that B(x, y) is analytic for x, y ∈ C except at the points x, y = 0, −1, −2, . . .. Therefore, integral (A2.11), defined classically for Re x > 0, Re y > 0, admits analytic continuation on the complex plane x, y ∈ C with the exception of the points x, y = 0, −1, −2, . . .. One has

B(x, y) =

(x)(y) . (x + y)

(A2.16)

Reference 1. N. Lebedev, Special Functions and Their Applications (Dover, New York, 1972)

3

The Laplace Transform

Let f ∈ L 1 [0, ∞), f (t) = 0 for t < 0, and  ∞ e− pt f (t) dt, Re p := σ ≥ 0, p = σ + is. L( f ) := F( p) :=

(A3.1)

0

Formula (A3.1) defines the Laplace transform L( f ) := F( p) of f . The function F( p) is analytic in the half-plane Re p > 0 and continuous up to the imaginary axis p = is. If f  ∈ L 1 (R+ ), then integration by parts yields:  f (0) 1 ∞ − pt  e f (t) dt. (A3.2) F( p) = + p p 0 Therefore, under these assumptions one has lim p F( p) = f (0).

(A3.3)

p→∞

Therefore, if f (0) = 0, then lim p→∞ p F( p) = 0. The converse is also true; see Theorem A3.2. If F( p) = F(is) is known on the imaginary axis, then  ∞ F(is) 1 ds (A3.4) F( p) = 2π −∞ is − p by the Cauchy formula provided that |F(is)| ≤ 

∞

F(is) =

c 1+|s|a , a

> 0. From (A3.1) it follows that

e−ist f (t) dt.

(A3.5)

0

Thus, by the inverse Fourier transform one gets © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2

53

54

Appendix 3: The Laplace Transform

1 2π   Therefore, if F( p)  f (t) =



∞ −∞

σ=0

1 2πi

F(is)eist ds =



i∞

e pt F( p) d p,

−i∞

p = is.

(A3.6)

is known, then f (t) is uniquely determined by formula (A3.6) and

F( p) is uniquely determined by formula (A3.5), provided that f is sufficiently smooth and decaying at infinity sufficiently fast. ∞ If f (t) ∈ L 2 ([0, ∞)), then −∞ |F(is)|2 ds < ∞, F( p) is analytic in the half-plane Re p > 0, the limit limσ→0 F(σ + is) = F(is) exists for almost every s and F(is) ∈ L 2 (−∞, ∞). Theorem A3.1 If F( p) is analytic in the half-plane Re p > 0, continuous up to the imaginary axis, 1 c , < a, Re p > 0, (A3.7) |F( p)| ≤ 1 + | p|a 2 then F( p) = L( f ), where f (t) = L 2 ([0, ∞)), 1 f (t) = 2π



∞

1 e F(is) ds = 2πi −∞ ist



i∞

−i∞

e pt F( p) d p,

p = is.

(A3.8)

If a > 1, then f (t) is a continuous bounded function, supt≥0 | f (t)| ≤ c. Proof Let Rn → ∞. Consider the contour Cn = (−i Rn , i Rn ) ∪ Cn , Cn := Rn eiϕ , − π2 ≤ ϕ ≤ π2 . By the analyticity of F( p) and by the Cauchy formula, one has 1 2πi If t < 0, then



i Rn −i Rn

e pt F( p) d p +

1 n→∞ 2πi lim

1 2πi

 Cn

e pt F( p) d p = 0.

(A3.9)

 Cn

e pt F( p) d p = 0, t < 0.

(A3.10)

Therefore, (A3.9) implies f (t) =

1 2πi



i∞

−i∞

e pt F( p) d p = 0 t < 0.

(A3.11)

If t > 0 then, by formula (A3.6), one gets: f (t) =

1 2πi



i∞ −i∞

e pt F( p) d p, t > 0.

(A3.12)

The integral in this formula converges in L 2 sense as the Fourier transform of L 2 (−i∞, i∞) function F( p).

Appendix 3: The Laplace Transform

55

Assumption (A3.7) guarantees that F( p) ∈ L 2 (−i∞, i∞). If a > 1, then | f (t)| ≤  ∞ ds c −∞ 1+|s|a ≤ c and f (t) is continuous. Theorem A3.1 is proved.  Theorem A3.2 Assume that F( p) is analytic in the half-plane Re p > 0, condition (A3.7) holds and (A3.13) lim | p|b |F( p)| = 0, b > 1, Re p ≥ 0. | p|→∞

Then F( p) = L( f ) and f (0) = 0.

(A3.14)

Proof From the assumption (A3.7) and the analyticity of F( p) in the half-plane Re p > 0 is follows by Theorem A3.1 that F( p) = L( f ). From condition (A3.13) and Theorem A3.1 it follows that f ∈ C(R+ ) and sup | f (t)| < c. t≥0

Putting t = 0 in formula (A3.8), one gets  ∞ 1 F(is) ds. f (0) = 2π −∞

(A3.15)

Since F( p) is analytic in the half-plane Re p > 0 and assumption (A3.13) holds, one can use the Cauchy formula (A3.9) with t = 0 

i Rn

−i Rn

Let n → ∞. Then  lim | n→∞

Cn

 F(is) d p +

 F( p) d p| ≤ lim

n→∞ C  n

Consequently,

 lim

i Rn

n→∞ −i R n

Cn

F( p) d p = 0.

c |d p| = 0, b > 1. 1 + | p|b

1 F(is) ds = 2π



∞ −∞

F(is) ds = 0.

From (A3.15) and (A3.18) the conclusion (A3.14) follows. Theorem A3.2 is proved.

(A3.16)

(A3.17)

(A3.18)



Remark A3.1 Assume that F( p) is analytic in the half-plane Re p > 0 and |F( p)| ≤

c , Re p > 0. 1 + | p|

(A3.19)

56

Appendix 3: The Laplace Transform

Consider the function

F( p) . 1 + p 1/4

F1 ( p) =

(A3.20)

Then F1 ( p) = L( f 1 ), supt≥0 | f 1 (t)| < c and f 1 (0) = 0. Proof From the analyticity of F( p) in the region Re p > 0 and from (A3.19) it follows that F1 ( p) is analytic in the half-plane Re p > 0, − π2 < arg p < π2 . By Theorem A3.1, F1 ( p) = L( f 1 ) and by Theorem A3.2 f 1 (0) = 0 since condition (A3.13) with b = 45 holds. Remark A3.1 is proved.  Theory of the Laplace transform of distributions and a table of the Laplace transforms of distributions are given in [3]. Let us calculate the Laplace transform L(t λ−1 ): L(t

λ−1



∞

)=

e

− pt λ−1

t

 dt =

0

∞

s λ−1 ds (λ) = λ . p λ−1 p p

e−s

0

(A3.21)

In this formula the integral converges classically if Re λ > 0, but the right side in formula (A3.21) admits analytic continuation on the whole complex plane λ ∈ C except points λ = 0, −1, −2, . . ., at which (λ) has simple poles (see Appendix 2). Therefore, by analytic continuation with respect to λ, one obtains the formula we use often: (λ) (A3.22) L(t λ−1 ) = λ , λ ∈ C, λ = 0, −1, −2, . . . p Define λ =  where

t λ−1

:=

λ−1 t+

:=

t λ−1 , (λ)

(A3.23)

t λ−1 , t ≥ 0 0,

t < 0.

We have: L(λ ) =

1 , ∀λ ∈ C. pλ

(A3.24)

Let us formulate some known results from the theory of the Laplace transform: 



L( f ) = pL( f ) − f (0); L

(a)

f ds 0

Proof



∞ 0



t

=

1 L( f ). p

(A3.25)

∞ e− pt f  (t) dt = e− pt f (t) 0 + pL( f ) = pL( f ) − f (0).

The function h(t) := follows.

t 0

f ds has the property h(0) = 0. So the second formula in (A3.25) 

Appendix 3: The Laplace Transform

57

F (n) ( p) = (−1)n L(t n f (t)).

(b) Proof Let n = 1. Then





∞

F ( p) =

(A3.26)

e− pt (−t f (t)) dt.

0

Similarly, formula (A3.26) is proved for n > 1.  (c)

L

f (t) t





 =

∞

F(q) dq.

(A3.27)

p

Proof 

∞



p



∞

F(q) dq =

∞

dq p

e−qt f (t) dt =



0

∞

 dt f (t)

0

∞

dq e−qt =

p



∞

dt 0

f (t) − pt e . t

 (d) Proof



∞

L( f (t − a)) = e

e− pt f (t − a) dt =

0



∞

− pa

L( f ).

(A3.28)

e− p(y+a) f (y) dy = e− pa L( f ).

0

Here we used the assumption f (t) = 0 for t < 0. (e)



L( f ∗ g) = L( f )L(g),

(A3.29)

where the convolution is defined by the formula:  t f ∗ g := f (s)g(t − s) ds. 0

Proof 

∞ 0

e− pt



t

 f (s)g(t − s) ds =

0

0

 =

0

 =

∞ ∞ ∞



∞

ds f (s) 

s

e− pt g(t − s) dt

∞

e− p(u+s) g(u) du  ∞ f (s)e− ps ds e− pu g(u) du

ds f (s)

0

= L( f )L(g).

0

0



We especially will be interested in the case when g(t) is hyper-singular, g(t) = t λ−1 , λ < 0. For Re λ > 0 one has classically (see formula (A3.22)):

58

Appendix 3: The Laplace Transform

L( f ∗ t λ−1 ) = L( f )

(λ) . pλ

By analytic continuation with respect to λ this formula remains valid for λ = 0, −1, −2, . . . because (λ) is analytic in this region and p −λ is an entire function of λ. If one uses the convolution f ∗ λ , where λ is defined in formula (A3.23), then formula L( f ∗ λ ) = L( f )

1 pλ

(A3.30)

holds for all λ ∈ C by analytic continuation with respect to λ from the region Re λ > 0. Note that f ∗g = g∗ f. (f) Efros’s theorem (see [1], p. 477). Let L( f ) = F( p), and G( p) and q( p) are analytic functions in the region Re p > 0 such that (A3.31) G( p)e−sq( p) = L(g(t, s)). 

Then

∞

F(q( p))G( p) = L

 f (s)g(t, s) ds .

(A3.32)

0

Proof One has 

∞

L

 f (s)g(t, s) ds

 =

0

0

 =

0

 =

∞ ∞ ∞

dt e− pt



∞ 0



∞

ds f (s)

f (s)g(t, s) ds dt e− pt g(t, s)

0

ds f (s)G( p)e−sq( p)

0

= G( p)F(q( p)).  Example A3.1 (see [1], p. 478). Let G( p) =

1 p 1/2

Then

g(t, s) = L

where formula (A3.33) is taken from [2].

, q( p) = p 1/2 .

e−sp p 1/2

1/2



s2

e− 4t = √ , πt

(A3.33)

Appendix 3: The Laplace Transform

59

By formula (A3.31) one gets ⎞ ⎛ 2   √  ∞ − s4t F( p) s2 1 e f (s)e− 4t ds . = L⎝√ ∗ f⎠ = L √ √ p πt πt 0 Let F( p) =

e−ap p .

Then

√ F( p) √ p

=

√

e−a p

p

. By formula (A3.28) one gets:

e−ap = L(1(t − a)), p where 1(t) =

 1, t ≥ 0 0, t < 0 √

e−a p

p

 =L

(A3.34)

(A3.35)

. Therefore, by formulas (A3.33)–(A3.34) one gets

1 √ πt



∞

e





2

− s4t

=L

ds

a

2 √ π





∞ a √ 2 t

e

−x 2

dx ,

s where x = 2√ . t The special functions Erf(x) and Erfc x := 1 − Erf x are defined in [2]:  x 2 2 e−s ds, Erf(x) = √ π 0  ∞ 2 2 e−s ds. Erfc(x) = √ π x

Thus, 2 √ π



e−x d x = Erfc 2

a √ 2 t

(g) If F( p) =

∞ 



a √ 2 t

(A3.36)

(A3.37)

 .

c j p− j ,

(A3.38)

(A3.39)

j=1

where the series (A3.39) converges for all sufficiently large p, | p| > R, then F( p) = L( f ), f (t) =

∞  j=1

and f (t) is an entire function of t.

cj t j−1 , ( j − 1)!

(A3.40) (A3.41)

60

Appendix 3: The Laplace Transform

Proof Formula (A3.39) follows from (A3.41) and (A3.22) because the series (A3.41) con verges for any t ∈ R+ . (h) Example A3.2 (see [1], p. 488). Let F( p) =

1 √ . p+ p

(A3.42)

We want to find f (t) such that L( f ) = F( p) where F( p) is given by formula (A3.42). One has √ F1 ( p) 1 = , (A3.43) F( p) = √ √ √ p(1 + p) p where F1 ( p) =

1 = L(e−t ). 1+ p

(A3.44)

By formula (A3.32) one can derive from formula (A3.43) the representation    ∞ 2 1 1 − s4t −s e e ds . = L √ √ p+ p πt 0

(A3.45)

Let us simplify the integral in (A3.45). One has s2 +s = 4t Introducing the new variable x = 1 =L √ p+ p



s √ 2 t

2et √ π



 √ 2 s √ + t − t. 2 t √ + t one rewrites (A3.45) as



∞ √ t

e

−x 2

 dx

√ = L(et Erfc( t)),

where Erfc(x) is defined by formula (A3.37). Since 1 1 1 =√ − √ √ , p+ p p 1+ p one concludes that

1 √ = L( 1 ) = L 2 p

1 √ =L 1+ p





t −1/2



( 21 )

 √ 1 t √ − e Erfc( t) . πt

 =L

(A3.46)

1 √ πt

 (A3.47)

(A3.48)

The above formulas are known, see [1, 2]. (i) One checks that  ∞    t   f (s) f (s) 1 ∞ 1 p F(q) dq, L F(q) dq. (A3.49) ds = ds = L s p p s p 0 0 t

Appendix 3: The Laplace Transform

61

Proof One has by formula (A3.27):   ∞  f (t) = F(q) dq. L t p 

This and formula (A3.25) yield formula (A3.49). (j) One has



∞

L 0

f (s) ds s



Proof Clearly: L(c) = If c =

∞ 0

f (s) s

ds, then c =

∞ 0



1 = p

∞

F(q) dq.

(A3.50)

0

c . p

F(q) dq as follows from formula (A3.27).



Theorem A3.3 Assume that F( p) = L( f ), | f (t)| ≤ ceat and lim p b |F( p)| = 0,

p→+∞

b ≥ 1.

(A3.51)

Then f (0) = 0.

(A3.52)

Proof One has 

ε

p b F( p) = p b

e− pt f (t) dt + p b

0

 ε

∞

e− pt f (t) dt := I1 + I2 ,

where ε > 0 is a small number and  ∞ e−( p−a)ε b e−( p−a)t dt = cp b −→ 0. |I2 | ≤ cp p − a p→+∞ ε On the other hand,  ε e− pt f (t) dt = lim lim p b p→+∞

0

p→+∞

f (0)(1 + a|ε|) p b

1 − e− pε , b ≥ 1, p

(A3.53)

(A3.54)

(A3.55)

where lim aε = 0.

(A3.56)

1 ε= √ . p

(A3.57)

ε→0

Choose

Then the limit in (A3.55) can be equal to zero only if (A3.52) holds. Theorem A3.3 is proved.



62

Appendix 3: The Laplace Transform

Lemma A3.1 Let F( p) be analytic in the region D := {|ϕ| ≤ α, p = | p|eiϕ , α < π2 , | p| > 0}. Assume that (a)

|F( p) ≤ Ae B| p| , | arg p| ≤ α,

(b)

|F( p) ≤ 1,

p = | p|e

±iα

(A3.58)

.

(A3.59)

Then |F( p)| ≤ 1 ∀ p ∈ D. Proof Let 1 < λ < One has

π 2α ,

(A3.60)

p λ := | p|eiλϕ , |ϕ| < α. λ

λ cos(λα)

|e p | = e| p|

≥ 1.

(A3.61)

If | p| = r , p ∈ D, then λ

|e p | = er

λ cos(λϕ)

≥ er

λ cos(λα)

.

(A3.62)

λ

Consider F( p)e−ε p , ε = const > 0. Let p0 ∈ D be arbitrary. Choose r = | p| such that r > | p0 |, Then |F( p0 )e

−ε pλ

One has

 | ≤ max

r>

ln A , B

 r>

  λ  max F( p)e−ε p  ,

p∈∂ D

2B ε cos(λα)

max

| p|=r , p∈D



1 λ−1

.

   −ε pλ  F( p)e  .

  λ  max F( p)e−ε p  ≤ 1

p∈∂ D

due to (A3.59), and, due to (A3.63), e Br > A, r λ ε cos(λα) > 2Br , so   λ  max F( p)e−ε p  ≤ max Ae Br −εr cos(λϕ) ≤ Ae−Br ≤ 1. | p|=r , p∈D

| p|=r , p∈D

(A3.63)

(A3.64)

(A3.65)

(A3.66)

Thus, by the maximum principle   λ  F( p0 )e−ε p0  ≤ 1.

(A3.67)

Since ε > 0 is arbitrarily small and p0 ∈ D is arbitrary, it follows from (A3.67) that (A3.60) holds. Lemma A3.1 is proved. 

Appendix 3: The Laplace Transform

63

Lemma A3.2 Assume that F( p) is analytic in the half-plane  := Re p > 0, |F( p)|∂ ≤ 1,

(A3.68)

, p ∈ , |F( p)| ≤ Ae ln |F(r )| ≤ 0, r := | p|. lim sup r r →∞

(A3.69)

B| p|

(A3.70)

Then |F( p)| ≤ 1 ∀ p ∈ .

(A3.71)

Proof Consider the function F1 ( p) := F( p)e−η p e−εz

3π 3/2 e−i 8

, η > 0, ε > 0,

(A3.72)

in the region 0 := {0 ≤ ϕ ≤ π2 , r := | p| ≥ 0}. One has π

|F1 ( p)|ϕ=0 = |F(r )|e−ηr e−εr cos 8 ≤ |F(r )|e−ηr ≤ cη , π 2

iπ 2

|F1 ( p)|ϕ= π2 = |F(r ei )|e−ηr e e



3π −εr 3/2 cos 3π 4 − 8



≤ 1,

(A3.73) (A3.74)

where cη := maxr ≥0 |F(r )|. The number cη must be not more than 1 since max |F1 ( p)| cannot be attained at an interior point of . On the boundary Re p = 0 one has |F1 ( p)| ≤ 1 by the assumption (A3.68). On the line i p = 0 max |F1 ( p)| cannot be attained at infinity due to the assumption (A3.70). Thus, (A3.75) max |F( p)e−η p | ≤ 1. p∈

Since η > 0 can be arbitrarily small, it follows that (A3.71) holds. Lemma A3.2 is proved.



Lemmas A3.1–A3.2 are taken essentially from [3], Vol. 1., Part III, Sect. 6.5, problem 325. Theorem A3.4 Assume that F( p) is analytic in  := Re p > 0, (A3.69) and (A3.70) hold. In place of assumption (A3.68) we assume |F( p)| ≤

3 c , 1 0 denotes various constants. The left inequality (A3.80) is trivial. The right one can be checked:  

c|1 + | p|a eiaϕ | = c 1 + | p|2a + 2| p|a cos(aϕ) ≥ c 1 + | p|2a + 2| p|a cos

One can check that  1 + | p|2a

+ 2| p|a

 3π cos ≥ c3 1 + | p|2a , c3 = 4

3π . 4 (A3.81)

 1 1− √ , 2

where c3 is the largest number for which inequality (A3.82) holds. Thus, 3 c , ∀ p ∈ , 1 < a < . |F( p)| ≤ 1 + | p|a 2

(A3.82)

(A3.83)

If F( p) is analytic in Re p > 0 and (A3.83) holds, then F( p) = L( f ), where f = f (t) ∈ C(R+ ), as follows from Theorem A3.1. From Theorem A3.2 it follows that f (0) = 0. Theorem A3.4 is proved. 

References 1. M. Lavrentiev, B. Shabat, Methods of the Theory of Functions of a Complex Variable (Nauka, Moscow, 1958) 2. H. Bateman, A. Erdelyi, Tables of Integral Transforms (McGraw-Hill, New York, 1954) 3. G. Polya, G. Szego, Problems and Theorems from Analysis (Nauka, Moscow, 1956)

4

Analysis of the Navier-Stokes Problem. Solution to the Millennium Problem Concerning the Navier-Stokes Equations

Introduction In this paper a proof of the author’s basic results concerning the Navier-Stokes problem (NSP) is given. These results include: (a) the a priori estimate of the solution to the NSP, (b) the theory of integral equations and inequalities with hyper-singular kernels, (c) the NSP paradox, Theorems 1 and 2, (d) the uniqueness result, Theorem 3, and other results. The earlier author’s results are used, but our presentation in this paper is essentially self-contained. Suppose that the data (the initial velocity v0 (x) and the force f (x, t) in equation (A4.1), see below) are smooth and rapidly decaying. The millennium problem concerning the NSP consists of finding whether the solution to equation (A4.1) exists for all t ≥ 0 and is smooth, provided that the data, that is, the initial velocity v(x, 0) and the force f (x, t), are smooth and rapidly decaying as |x| → ∞ and t → ∞. Our proof of the NSP paradox demonstrates that the solution to (A4.1) with f (x, t) = 0 and v0 (x) ≡ 0 (v0 (x) is smooth and rapidly decaying at infinity), cannot exist for all t ≥ 0. In paper [4], p. 472, Theorem 2, there is a statement that, for f (x, t) = 0 and u 0 (x) sufficiently small, the solution to the NSP exists for all t ≥ 0 if m ≤ q, where m is the dimension of the space and the solution is in L q . In our case m = 3 and q = 2, so the condition m ≤ q does not hold. Therefore, the claim in [4], p. 472, is not applicable. The Navier-Stokes equations are discussed in many books and papers. We only mention here [5, 6, 8, 9]. Our goal is to present for broad audience the author’s result concerning the Navier-Stokes problem in R3 without boundaries. This result can be briefly formulated as follows: © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2

65

66

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

Assume (for simplicity only) that the exterior force f (x, t) = 0. If the initial velocity v0 (x) := v(x, 0) ≡ 0, ∇ · v0 (x) = 0, v0 (x) is smooth and rapidly decaying as |x| → ∞ and the solution v(x, t) of the NSP exists for all t ≥ 0, then v0 (x) = 0. This result, that we call the NSP paradox, shows that: a) The NSP is not a correct statement of the problem of motion of viscous incompressible fluid; it is neither physically nor mathematically correct statement of the dynamics of incompressible viscous fluid. b) The NSP does not have a solution unless v0 (x) = 0 and f (x, t) = 0; in this case the solution v(x, t) = 0 for all t ≥ 0. This result solves the millennium problem related to the Navier-Stokes equations. It encourages the search for correct equations describing the dynamics of viscous incompressible fluid. Let us explain the steps of our proof. The NSP consists of solving the equations: v  + (v, ∇)v = −∇ p + νv + f , x ∈ R3 , t ≥ 0, ∇ · v = 0, v(x, 0) = v0 (x), (A4.1) see, for example, books [6, 9]. Here v = v(x, t) is the velocity of incompressible viscous fluid, a vector function, v  := vt , p = p(x, t) is the pressure, a scalar function, f = f (x, t) is the exterior force, ν = const > 0 is the viscoucity coefficient, v0 = v0 (x) is the initial velocity, ∇ · v0 = 0. The data v0 and f are given, the v and p are to be found. The fluid’s density ρ = 1. a) First we reduce the NSP to an equivalent integral equation:  t  ds G(x − y, t − s)(v, ∇)vdy, (A4.2) v(x, t) = F − 0

R3

where F = F(x, t) depends only on the data f (x, t) and v0 (x),  t   F(x, t) := ds G(x − y, t − s) f (y, s)dy + g(x − y, t)v0 (y)dy. R3

0

R3

We assume (for simplicity only and without loss of generality) that f = f (x, t) = 0. Under this assumption one has:  g(x − y, t)v0 (y)dy, (A4.3) F(x, t) := R3

where |x|2

e− 4νt 2 g(x, t) = , t > 0; g(x, t) = 0, t ≤ 0; g˜ = e−|ξ| νt . 3/2 (4νπt)

(A4.4)

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

67

The tensor G = G(x, t) = G jm (x, t) is calculated explicitly in [9]: G(x, t) = (2π)−3



  ξ p ξm  ξ p ξm  2 2 eiξ·x δ pm − 2 e−ν|ξ| t dξ; G˜ = (2π)−3 δ pm − 2 e−ν|ξ| t . 3 ξ ξ R

(A4.5)

˜ Let us define the Fourier transform F v = v: v˜ := v(ξ, ˜ t) := (2π)−3

 R3

v(x, t)e−iξ·x d x.

(A4.6)

One has:

F (vw) = (2π)3 F (v)F (w), (2π)3 F (v)2 = v2 ,

(A4.7)

where  denotes the convolution in R3 . Taking the Fourier transform of formula (A4.3), one gets: ˜ t) = (2π)3 g˜ (ξ, t)v˜0 (ξ) = (2π)3 e−|ξ|2 νt v˜0 (ξ). (A4.8) F(ξ, Take the Fourier transform of equation (A4.2) and get the integral equation equivalent to equation (A4.2):  t ˜ ˜ t) − (2π)3 ds G(ξ, t − s)v(iξ ˜ v). ˜ (A4.9) v(ξ, ˜ t) = F(ξ, 0

The following inequality, that comes from the Cauchy inequality, is useful: |v(iξ ˜ v)| ˜ ≤ v|ξ| ˜ v. ˜

(A4.10)

Claim 1 The following a priori estimate holds: ˜ < c. sup v

(A4.11)

t≥0

By c here and throughout the paper various positive constants, independent of t and x, are denoted. We denote by c1 := |(− 41 )| > 0 the special constant and use the following  ∂v  notations: v j,m := ∂xmj , := R3 . Let us write equation (A4.1) as v j + vm v j,m = f j − p, j + νv, j j , v j, j = 0,

(A4.12)

where over the repeated indices summation is understood, 1 ≤ j ≤ 3. We assume that v = v(x, t) and other functions in the NSP are real-valued and the following inequality holds:  ∞  f (x, t)dt < c. (A4.13) v0  + 0

Here and throughout this paper  ·  is L 2 (R3 ) norm.

68

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

Proof of Claim 1 Multiply equation (A4.12) by v j , integrate over R3 and sum up over j to get 1 (A4.14) (v2 ),t ≤ |( f , v)| ≤  f v. 2 where z ,t := ∂z ∂t . In deriving inequality (A4.14) we have used integration by parts:     pv j, j d x = 0, νv, j j v j d x = −ν v, j v, j d x ≤ 0, − p, j v j d x = and

 vm v j,m v j d x = −

1 2

 vm,m v j v j d x = 0.

From inequality (A4.14) it follows that v,t ≤  f . Consequently,  ∞ v ≤ v0  +  f dt < c. 0

This and our assumption (A4.13) imply estimate supt≥0 v < c. By the Parseval equality the desired estimate (A4.11) follows. Claim 1 is proved.  Inequalities (A4.10) and (A4.11) imply |v(iξ ˜ v)| ˜ ≤ c|ξ|v. ˜

(A4.15)

This inequality is important because it allows one to estimate the nonlinear term on the left side of equation (A4.15) by the linear term on its right side. Equations (A4.9) and (A4.15) imply the first inequality used in Sect. 2. From formula (A4.4) it follows that ˜ |G(ξ, t − s)| ≤ ce−ν(t−s)ξ , 2

(A4.16)

  ξ ξ because | δ pm − pξ 2m | < c. Estimate (A4.16) will be used more than once in this paper. b) Secondly, we prove that any solution to equation (A4.9) satisfies integral inequality (A4.17), see below. The integral in this inequality is a convolution with the hyper-singular 5 kernel (t − s)− 4 ; this integral diverges classically, that is, from the classical point of view. Convolution of distributions is defined in [3]. It is proved (see Theorem 7.1.15 in [3], p.166) that the Fourier transform of the convolution of two distributions equals to the product of their Fourier transforms. The assumptions on the two distributions from Theorem 7.1.15 −5

λ = 0 if t ≤ 0, are satisfied for the distribution t+ 4 and the function b(t). The distribution t+ λ = t λ if t > 0, is studied in [2, 3, 8]. t+ The following inequality is derived in Sect. 2:  t 5 b(t) ≤ b0 (t) + c (t − s)− 4 b(s)ds, (A4.17) 0

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

where

˜ t), b(t) := |ξ|v(ξ, ˜ t) ≥ 0. b0 (t) := |ξ| F(ξ,

69

(A4.18)

Here and below the norm  ·  is the L 2 (R3 ) norm. If the data v0 (x) is smooth and rapidly decaying as |x| → ∞, then |v(ξ)| ˜ ≤ c(1 + |ξ|2 )−m , m >

5 . 2

Therefore, from equation (A4.8) and the definition of b0 (t) it follows that  ∞ 2 2 e−r νt r 4 (1 + r 2 )−m dr := cI (t), b0 (t) ≤ c

(A4.19)

0

where we have used the spherical coordinates. From this relation it follows that the Laplace transform Lb0 (see formula (A4.33) below) can be estimated: |Lb0 | ≤ c(1 + | p|)−1 , Re p > 0. ∞ ∞ Claim 2 Assume that 0 [|h(t)| + |h  (t)|]dt < ∞. Then L(h) := 0 e− pt h(t)dt is analytic function of p in the region Re p > 0, continuous up to the imaginary axis Re p = 0, and |L(h)| ≤ c(1 + | p|)−1 . Proof of Claim 2 Let p = a + is. One has 

∞

sup |L(h)| ≤ a≥0

|h(t)|dt ≤ c.

0

Thus, L(h) is analytic in the region a > 0 and continuous up to the imaginary axis a = 0. Let us prove the estimate |L(h)| ≤ c(1 + | p|)−1 . Integrate by parts and get   e− pt 1 ∞ − pt  h(0) 1 ∞ − pt  L(h) = − + e h (t)dt = e h (t)dt := J . h(t)|∞ + 0 p p 0 p p 0 For a > 0 and | p|  1 one has J = O( | 1p| ). This and the estimate |L(h)| ≤ c for a ≥ 0 imply the desired estimate: |L(h)| ≤ c(1 + | p|)−1 . Claim 2 is proved.  Since the convolution integral in (A4.17) diverges classically, we give a new definition of this integral in Sect. 3 and estimate the solution b(t) to integral inequality (A4.17) by the solution q(t) to the integral equation with the same hyper-singular kernel:  t 5 (t − s)− 4 q(s)ds. (A4.20) q(t) = b0 (t) + c 0

70

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

Namely, we prove the following inequality 0 ≤ b(t) ≤ q(t).

(A4.21)

We assume that b0 (t) ≥ 0 is smooth and rapidly decaying as t → ∞. c) We prove the following a priori estimate: sup(∇v + v) ≤ c,

(A4.22)

t≥0

part of which is inequality (A4.11). Recall that by the Parseval equality one has: (2π)3/2 v ˜ = v, (2π)3 |ξ|v ˜ 2 = ∇v2 .

(A4.23)

We prove that equation (A4.20) has a unique solution in the space C(R+ ), and the following estimate holds: (A4.24) sup q(t) ≤ c, t≥0

provided that the datum v0 (x) is smooth and rapidly decaying at infinity. Moreover, this solution q(t) is unique and q(0) = 0.

(A4.25)

d) We prove that any solution b(t) ≥ 0 of inequality (A4.17), where b0 (t) ≥ 0 a smooth rapidly decaying function, satisfies inequality (A4.21). Since q(0) = 0 and 0 ≤ b(t) ≤ q(t), it follows that b(0) = 0. This yields the NSP paradox mentioned at the beginning of this section. Indeed, the initial data v0 (x) ≡ 0, so b(0) > 0, but we prove that b(0) = 0. The NSP paradox implies the conclusions we have made: The NSP is physically not a correct description of motion of incompressible viscous fluid in R3 without boundaries; the NSP does not have a solution on the whole interval [0, ∞) unless the data are equal to zero ( in this case the solution to the NSP does exist on the whole interval [0, ∞) and is identically equal to zero). The uniqueness of the solution to NSP is proved in Sect. 4, see Theorem 3.

Derivation of the Integral Inequality Take absolute value of both sides of equation (A4.9), then use inequalities (A4.11) and (A4.16) to get  t  t 2 2 u(ξ, t) ≤ μ(ξ, t) + c e−ν(t−s)ξ u|ξ|uds ≤ μ + c e−ν(t−s)ξ b(s)ds, 0

0

b(s) := |ξ|u(ξ, s),

(A4.26)

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

71

where the Parseval formula and the estimates (2π)3/2 v ˜ = v < c, supt≥0 v(x, t) < c were used. We denoted: ˜ t)| := μ(ξ, t) := μ. |v(ξ, ˜ t)| := u(ξ, t) := u, | F(ξ,

(A4.27)

If 0 ≤ u ≤ w, then u ≤ w. Multiply inequality (A4.26) by |ξ|, take the norm  ·  of both sides of the resulting inequality and get inequality (A4.17). In this calculation one uses the second of the following formulas e−ν(t−s)ξ  = 2

c c 2 , |ξ|e−ν(t−s)ξ  = , 0 ≤ s < t, 3/4 (t − s) (t − s)5/4

(A4.28)

which are easy to derive. The constants c > 0 are different in our formulas. To study integral equation (A4.20) and integral inequality (A4.17) we need to define the hyper-singular integrals in these equations. To do this, we continue analytically the Laplace transform of some convolution operators with hyper-singular kernels, see Sect. 3. Let us define the function t λ−1 , (A4.29) λ := (λ) where (λ) is the gamma function. Here and throughout t = t+ , that is, t = 0 for t < 0, t := t for t ≥ 0. It is known that (λ) is an analytic function of λ ∈ C except for the points 1 is entire function of λ, λ = 0, −1, −2, ...., at which it has simple poles; the function (λ) see [7]. Consider the convolution operator  t λ (t − s)b(s)ds. (A4.30) λ  b := 0

For Reλ > 0 the integral in (A4.30) is understood classically. For Reλ < 0 this integral is understood as analytic continuation with respect to λ from the region Reλ > 0. One has  t 5 1 1 3 (t − s)− 4 b(s)ds = (− )− 1  b = −c1 − 1  b, c1 := |(− )| = 4( ) > 0, 4 4 4 4 4 0 where  denotes the convolution on R+ (that is, the convolution of λ and b(t) both of which have supports on R+ ). Inequality (A4.17) can be written as b(t) ≤ b0 (t) − cc1 − 1  b, b(t) ≥ 0, b0 (t) ≥ 0. 4

(A4.31)

The integral equation (A4.20) for q takes the form: q(t) = b0 (t) − cc1 − 1  q. 4

(A4.32)

72

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

In the next section we define the hyper-singular convolution integrals λ  q for λ = − 41 . This will be done by taking the Laplace transform of λ  q for Reλ > 0. This Laplace transform is classically well defined. We continue analytically the result of the Laplace transform in the region Reλ ≤ 0, so that λ = − 41 belongs to the region where the analytic continuation is defined. Then we prove, under natural assumptions, that this analytic continuation is a Laplace transform of a uniformly bounded function of t, t ∈ [0, ∞), and take the inverse Laplace transform.

Investigation of Integral Equations and Inequalities with Hyper-Singular Kernel In this section we solve equation (A4.32) analytically and prove estimate (A4.21). First, let us define the hyper-singular integral ψ := λ  q. We are especially interested in the value λ = − 41 because it appears in equation (A4.32). For λ > 0 the convolution λ  q is defined classically and one has L(ψ) = L(q) p −λ , where L is the Laplace transform operator defined as  ∞

L(q) :=

e− pt q(t)dt,

(A4.33)

0

which is analytic in the region Re p > 0 if q ∈ L 2 (R+ ). If q(t)e−at ∈ L 2 (R+ ) and a = const > 0, then L(q) is an analytic function of p in the region Re p > a. The Laplace transform is injective on the domain of its definition. Therefore the inverse operator L −1 is well defined on the range of L. In this paper we prove that the function q(t) is bounded on R+ = [0, ∞) and q(0) = 0, provided that the data are smooth and rapidly decaying, which implies that b0 (t) is bounded and rapidly decaying as t → ∞. The inversion formula for the Laplace transform is known:  1 −1 e pt L( p)d p, (A4.34) q(t) := L L( p) := 2πi Cσ where Cσ is the straight line σ = const > a, p = σ + iω, ω runs from −∞ to ∞ and L( p) is the Laplace transform of q(t). Let us give a sufficient condition for a function Q( p) to be the Laplace transform of a a function q ∈ L 2 (R+ ). Lemma 1 If Q( p) is analytic in the region Re p > 0 and |Q( p)| < c(1 + | p|)−β , β > 1/2, | p|  1, Re p > 0, then

1 q(t) = 2π



∞

−∞

eiωt Q(iω)dω,

p = σ + iω,

(A4.35)

(A4.36)

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

73

where q(t) ∈ L 2 (R+ ) and L(q) = Q( p). If β > 1, then q ∈ C(R+ ), supt≥0 |q(t)| < c, and q(0) = 0. In Lemma 1 sufficient conditions are given for a function, analytic in the region Re p > 0, to be the Laplace transform of an L 2 (R+ ) function or an C(R+ ) function. For convenience of the reader we prove the part of Lemma 1 that is used in this paper in Theorem 1. Proof Assume that F( p), p = σ + iω, is analytic in σ > 0 and continuous up to the line C0 := { p : σ = 0, −∞ < ω < ∞}. Denote by C0n the subset of C0 such that {−n ≤ ω ≤ n}, by γn the semicircle closing the segment C0n from the right. If β > 1 then  limn→∞ γn F( p)d p = 0. Define f (t) =

1 2πi

 e pt F( p)d p = C0

1 2π

 eiωt F(iω)dω.

(A4.37)

C0

This integral converges absolutely since β > 1, so f (t) ∈ C(R+ ), supt≥0 | f (t)| < c and   1 1 F(iω)dω = lim F(iω)dω := I . (A4.38) f (0) = n→∞ 2π C 2π C0 0n We now prove that f (0) = 0.

(A4.39)

To do this, denote by L n a closed contour consisting of the union of C0n and γn , oriented counterclock-wise. Use the analyticity of F, the assumption β > 1 and the Cauchy formula to get   F( p)d p = 0,

0= Ln

lim

n→∞ γ n

F( p)d p = 0.

(A4.40)



Therefore, lim

n→∞ C 0n

F(iω)dω = I = 0.

(A4.41)

From equations (A4.38) and (A4.41) the desired conclusion (A4.39) follows. The proof is finished.  Lemma 2 One has

L(λ  q) = L(q) p −λ .

(A4.42)

Here we have used the formula (which is proved below): L(λ ) = p −λ , ∀λ ∈ C,

(A4.43)

L(λ  q) = L(λ )L(q).

(A4.44)

and the formula

74

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

For Reλ > 0 and q ∈ L 2 (R+ ) this formula can be understood classically. For Reλ < 0 formula (A4.44) is defined by the analytic continuation with respect to λ ∈ C, where L(λ ) is given in formula (A4.43). Formula (A4.44) is valid for all λ ∈ C by the analytic continuation with respect to λ from the region Reλ > 0, where it is valid classically. Note that the function L(q) does not depend on λ and the function (A4.43) is an entire function of λ for p = 0 since p −λ = e−λ ln p . Let us define the convolution ψ := λ  q by the formula:   (A4.45) ψ(t) := L −1 L(q) p −λ . The expression under the sign L −1 in formula (A4.26) is an entire function of λ. For Reλ > 0 the ψ(t) is well defined classically if q ∈ C(R+ ) ∩ L 2 (R+ ). The function L(ψ) admits analytic continuation with respect to λ to the whole complex plane C. Therefore, the convolution ψ is defined for all λ ∈ C. We are interested in the value λ = − 41 because it appears in equations (A4.31) and (A4.32). To illustrate the argument with the analytic continuation, consider a simple example:  ∞  ∞ z−1 − pt t e dt = s z−1 e−s dsp −z = (z) p −z , (A4.46) 0

0

where we set s = pt, p > 0. Formula (A4.46) is valid classically for Rez > 0, but remains valid for all z ∈ C, z = 0, −1, −2......, by the analytic continuation with respect to z. Indeed, (z) is analytic for all z ∈ C except for the points z = 0, −1, −2, ...., and p −z = e−z ln p is an entire function of z if p = 0. Formula (A4.43) follows from (A4.46) immediately: just t z−1 . divide both sides of (A4.46) by (z) and remember that z (t) = (z) The integral (A4.46) diverges classically for Rez ≤ 0, but formula (A4.46) is valid by analytic continuation for all z ∈ C, z = 0, −1, −2, ..... In [2] a regularization method is described for defining divergent integrals. This method ∞ is much less convenient for our purposes in this paper because (z) = 0 t z−1 e−t dt is known to be analytic for all z ∈ C, except for the points z = −n, n = 0, 1, 2, ..., and it is not convenient and not advisable to use the regularization method, described in [2], for ∞ defining the integral 0 t z−1 e− pt dt for z = − 41 . Lemma 3 One has λ  μ = λ+μ .

(A4.47)

0 (t) = δ(t),

(A4.48)

for any λ, μ ∈ C. If λ + μ = 0 then

where δ(t) is the Dirac distribution.

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

75

Proof By formulas (A4.42) and (A4.43) one gets 1 . p λ+μ

(A4.49)

 1  = λ+μ . p λ+μ

(A4.50)

L(λ  μ ) = By formula (A4.43) one has L −1

This proves formula (A4.47). If λ + μ = 0, then p −(λ+μ) = 1 and L −1 1 = δ(t).

(A4.51)

This proves formula (A4.48). Lemma 3 is proved.  Our plan is to prove that equation (A4.32) has a solution q(t) ∈ C(R+ ) provided that b0 (t) is smooth and rapidly decaying as t → ∞. Moreover, this solution is unique in C(R+ ) and q(0) = 0. We also prove that any solution b(t) ≥ 0 to inequality (A4.31) satisfies the relation b(t) ≤ q(t). Since q(0) = 0, it follows that b(0) = 0. If b(0) = 0, then we have the NSP paradox, because a priori it was assumed that v0 (x) ≡ 0 and, therefore, b(0) = 0. To realize this plan, let us investigate equation (A4.32). First, let us apply to equation (A4.32) the operator 1/4  and use Lemma 3 to get

This implies

1/4  q = 1/4  b0 − cc1 q.

(A4.52)

  q = c2 1/4  b0 − 1/4  q , c2 := (cc1 )−1 > 0.

(A4.53)

Take the Laplace transform of (A4.53) to get 1

1

L(q) = c2 L(b0 ) p − 4 − c2 L(q) p − 4 . Therefore L(q) =

c2 L(b0 ) , |Lb0 | ≤ c(1 + | p|)−1 , Re p ≥ 0. p 1/4 + c2

(A4.54)

(A4.55)

The same result we would obtain if we applied the Laplace transform to equation (A4.32). The main reason for using equation (A4.53) is the possibility to apply Lemma 4 to this equation, see formulas (A4.59)–(A4.60) below. Let us now prove that the right side of formula (A4.55) is a Laplace transform of a bounded on [0, ∞) function which vanishes at t = 0. First, we prove that the function p1/41+c is 2

analytic in the region Re p > 0. Let φ be the argument of p. The function p 1/4 := r eiφ is an analytic function of p in the region −π/2 ≤ φ ≤ π/2. One can check that the function

76

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

> 0, is an analytic function of p in the region Re p > 0 and it is bounded in this region. To check this, write 1 ,c p1/4 +c2 2

   2 |r eiφ/4 + c2 |2 = r 2 + 2r c2 cos(φ/4) + c22 = r 2 1 − cos2 (φ/4) + r cos(φ/4) + c2 > c > 0.

This inequality is valid for all −π/2 ≤ φ ≤ π/2, r ≥ 0. The function L(b0 ) is analytic in the region Re p > 0 by our assumption. Therefore, the function L(q) in formula (A4.55) is analytic in this region. We assumed that b0 (t) is smooth and rapidly decaying as t → ∞. Therefore, |L(b0 )| ≤ c(1 + | p|)−1 , Re p > 0, as we have proved earlier. Consequently, see formula (A4.55), the function L(q) is analytic in the region Re p > 0 and 5 |L(q)| ≤ c(1 + | p|)− 4 , Re p > 0, | p|  1. (A4.56) By Lemma 1, the function L(q) is the Laplace transform of the function q(t) ∈ C(R+ ) and q(0) = 0. We have proved the following result: Theorem 1 Assume that v0 (x) is smooth and rapidly decaying as |x| → ∞, f (x, t) = 0 and x ∈ R3 . Then equation (A4.32) is solvable in C(R+ ), its solution q(t) is unique in this space and q(0) = 0. Let us now prove Theorem 2. Theorem 2 Any solution b(t) ≥ 0 of inequality (A4.31) satisfies the estimate b(t) ≤ q(t). Proof of Theorem 2 requires the following lemma: t Lemma 4 The operator A f := 0 (t − s)a f (s)ds in the space X := C(0, T ) for any fixed T ∈ [0, ∞) and a > −1 has spectral radius r (A) equal to zero. The equation f = A f + h is uniquely solvable in X . Its solution can be obtained by iterations f n+1 = A f n + h,

f 0 = h;

lim f n = f =

n→∞

∞ 

A j h,

j=0

for any h ∈ X and the convergence holds in X . Proof The spectral radius of a linear operator A is defined by the formula r (A) = lim An 1/n . n→∞

(A4.57)

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

77

By the mathematical induction one proves that An f  ≤ T n(a+1)

 n (a + 1)  f  X , n ≥ 1. (n(a + 1) + 1)

(A4.58)

From this formula and the known asymptotic of the Gamma function (z) for z → ∞, namely: (z) = e(z−0.5) ln z−z+0.5 ln(2π) [1 + O(|z|−1 ], |z|  1, |ar gz| ≤ π − δ, δ > 0, the conclusion r (A) = 0 follows. If r (A) = 0, then the solution to equation f = A f + h is unique and can be calculated by formula (A4.39). Lemma 4 is proved.  Note that the operator A preserves inequality sign if a > −1. By Lemma 4 the solution to equation (A4.53) can be obtained as q=

∞ 

(−c2 1/4  ) j c2 1/4  b0 ,

(A4.59)

j=0

and any solution to inequality (A4.31) satisfies the inequality b≤

∞ 

(−c2 1/4  ) j c2 1/4  b0 ,

(A4.60)

j=0

which is checked by iterations. It is now easy to prove Theorem 2. Proof of Theorem 2. From formulas (A4.59) and (A4.60) the inequality b(t) ≤ q(t) follows. Theorem 2 is proved.  It follows from Theorems 1 and 2 that sup q(t) < c, b(t) ≤ q(t).

(A4.61)

t≥0

This and the Parseval equality implies sup ∇v < c.

(A4.62)

t≥0

Together with estimate (A4.11) this proves a priori estimate (A4.22). So, solutions to equation (A4.9) belong to H 1 (R3 ) × C(R+ ), where H 1 (R3 ) is the Sobolev space. In Sect. 4 we prove that the NSP does not have more than one solution in the space W := {v(x, t) ∈ H 1 (R3 ) × C(R+ ); ∇ · v = 0}, C(R+ ) is the space of continuous functions, uC(R3 ) = supt≥0 |u(t)|. +

78

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

Uniqueness of the Solution to the NSP Theorem 3 There is no more than one solution to the NSP in the space W . Proof Let there be two solutions v˜ and w˜ to (A4.9) and z := v˜ − w, ˜ z = z(ξ, t). Then, subtracting from the first equation the second, one gets:  t   ˜ z=− ds G(ξ, t − s) z(iξ v) ˜ + w(iξz) ˜ .

(A4.63)

0

Using estimates (A4.15) and (A4.22), one obtains from (A4.63) the following inequality:  t 2 e−ν(t−s)ξ η(s)ds, η := z + |ξ|z. (A4.64) |z| ≤ c 0

From (A4.64), taking the norm  · , one obtains:  t 3 (t − s)− 4 η(s)ds. z ≤ c

(A4.65)

0

Multiply (A4.64) by |ξ|, take the norm  ·  and get:  t 5 (t − s)− 4 η(s)ds. |ξ|z ≤ c

(A4.66)

0

Taking the Laplace transform of (A4.65) and of (A4.66) and summing the results yields:     1 1 1 1 1 L(η) ≤ c (− ) p 4 + p − 4 (1/4) L(η) = c −c1 p 4 + p − 4 (1/4) L(η), c1 > 0. 4 (A4.67) Since L(η) ≥ 0, one concludes that   1 1 1 ≤ c − c1 p 4 + p − 4 (1/4) . (A4.68) This is a contradiction: if one takes p → +∞ then the above inequality yields 1 ≤ −∞. This contradiction proves that L(η) = 0, so η = 0 and z = 0. Therefore, z = 0 and v˜ = w. ˜ Theorem 3 is proved.  Theorem 3 is new but it is not used in the derivation of our NSP paradox. Earlier uniqueness theorems were proved under different assumptions on the spaces to which the solution to the NSP belongs, see [5, 9].

Appendix 4: Analysis of the Navier-Stokes Problem. Solution to the Millennium …

79

Conclusions It is proved that if the initial velocity is a smooth rapidly decaying function v0 (x) ≡ 0 and the force is zero, and v(x, t) exists for all t ≥ 0, then v0 (x) ≡ 0. This paradox (the NSP paradox) shows that the NSP is not a correct description of the dynamics of viscous incompressible fluid. From the NSP paradox we conclude that the NSP is physically and mathematically contradictive and is not a correct description of the dynamics of incompressible viscous fluid. The solution to the NSP does not exist unless the initial velocity and the exterior force are zeroes, in which case the solution to the NSP is equal to zero identically by Theorem 3. This solves the millennium problem concerning the Navier-Stokes equations.

References 1. Yu. Brychkov, A. Prudnikov, Integral Transforms of Generalized Functions (Nauka, Moscow, 1977) (in Russian) 2. I. Gel’fand, G. Shilov, Generalized Functions, vol. 1 (GIFML, Moscow, 1959). (in Russian) 3. L. Hormander, ¨ The Analysis of Linear Partial Differential Operators, vol. 1 (Springer, New York, 1983) 4. T. Kato, Strong L p -solutions of the Navier-Stokes equation in Rm , with applications to weak solutions. Math. Z. 187, 471–480 (1984) 5. O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Fluid (Gordon and Breach, New York, 1969) 6. L. Landau, E. Lifshitz, Fluid Mechanics (Pergamon Press, New York, 1964) 7. N. Lebedev, Special Functions and Their Applications (Dover, New York, 1972) 8. A.G. Ramm, Theory of hyper-singular integrals and its application to the Navier-Stokes problem. Contrib. Math. 2, 47–54 (2020) 9. A.G. Ramm, The Navier-Stokes Problem (Morgan & Claypool publishers, 2021) 10. A.G. Ramm, Navier-Stokes equations paradox. Reports Math. Phys. (ROMP) 88(1), 41–45 (2021) 11. A.G. Ramm, Comments on the Navier-Stokes problem, axioms. Open Access J 10, 95 (2021) 12. A.G. Ramm, Applications of analytic continuation to tables of integral transforms and some integral equations with hyper-singular kernels. Open J. Optim. 11, 1–6 (2022)

5

Applications of Analytic Continuation to Tables of Integral Transforms and Some Integral Equations with Hyper-Singular Kernels

Introduction This Appendix is based on paper [2]. In [1] one finds several formulas of integral transforms the validity of which can be greatly expanded by analytic continuation with respect to a parameter. This is of interest per se, but also is important in applications. Analytic continuation with respect to parameter can be used in a study of integral equations with hyper-singular kernels. This is done in Sect. 3. The examples of the integral equations are chosen to demonstrate that some integral equations, which do not make sense classically (that is, from the classical point of view), can be understood using the analytic continuation. Moreover, they can be solved analytically and the properties of their solutions can be studied. In Sects. 1 and 2 examples of the formulas from tables of integral transforms are discussed. The number of such examples can be increased greatly. The author wants to emphasize the principle based on the analytic continuation. The choice of the parameter λ = − 41 is motivated by the role playing by the corresponding integral equations in the Navier-Stokes problem. The choice of the parameter λ = − 21 is motivated by the novel feature in the investigation, the pole in the Laplace transform of the solution. Example 1 In [1] formula (1) in Sect. 2.3. is given in the form:  0

∞

x −ν sin(x y)d x = y ν−1 (1 − ν) cos(

νπ ), 2

y > 0, 0 < Re ν < 2.

(A5.1)

Here and below (ν) is the Gamma function. Classically the integral on the left in (A5.1) diverges if Re ν < 0 or Re ν > 2. On the other hand, in many applications, one has to consider ν outside the region specified in (A5.1). The right side of formula (A5.1) admits analytic continuation with respect to ν. Indeed, if y > 0, which we assume throughout, then y ν−1 = e(ν−1) ln y is an entire function of ν. The (z) is analytic function of z on the © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2

81

82

Appendix 5: Applications of Analytic Continuation to Tables of Integral Transforms …

complex plane z except for a discreet set of points z = 0, −1, −2, ...., at which it has simple poles with known residues, see [3]. Therefore (1 − ν) is an analytic function of ν except for the points ν = 1, 2, 3, ..... The function cos( νπ 2 ) is an entire function of ν. Therefore, the right side of formula (A5.1) admits analytic continuation on the complex plane ν except for 1 the points ν = 2, 3, 4, ..... The function cos( νπ 2 ) = 0 if ν = n + 2 , where n is an integer. νπ Therefore, the zeros of cos( 2 ) do not eliminate the poles of the (1 − ν). We have proved the following theorem. Theorem 1 Formula (A5.1) remains valid by analytic continuation with respect to ν for all complex ν = 2, 3, 4, ..... 

More Examples Consider two more examples of a similar nature. The number of such examples can be increased. In [1], formula (7) in Section 2.4. is:  ∞ 0

ν

x −ν e−ax sin(x y)d x = (ν)(a 2 + y 2 )− 2 sin[νar ctg(y/a)],

Re a > 0,

Re ν > −1.

(A5.2)

In [1], formula (4) in Sect. 2.5. is: 

∞

x −ν ln x sin(x y)d x =

0

π y −ν [ψ(ν) + 0.5πctg(νπ/2) − ln y], |Re ν| < 1, 2(1 − ν) cos(νπ/2)

(A5.3)  (z) . where ψ(z) := (z) We leave for the reader to discuss the analytic continuation of formulas (A5.2) and (A5.3) with respect to ν.

Some Applications Consider an integral equation 

t

q(t) = h(t) +

5

(t − s)− 4 q(s)ds.

(A5.4)

(t − s)λ−1 q(s)ds.

(A5.5)

0

More generally, consider the equation  q(t) = h(t) +

t

0

This equation has a hyper-singular kernel: the integral in this equation diverges if Reλ ≤ 0 classically (that is, from the point of view of classical analysis).

Appendix 5: Applications of Analytic Continuation to Tables of Integral Transforms …

83

Our goal is to give sense to this equation and solve it analytically. One knows that L(t λ−1 ) = (λ) p −λ ,

(A5.6)

∞

where L(h) := 0 e− pt h(t)dt is the Laplace transform. For Reλ > 0 formula (A5.6) is known classically. For Re p > 0 the function p −λ = e−λ ln p is an entire function of λ on the complex plane C of λ. The function (λ) is analytic on C except for the points 0, −1, −2, ..... Therefore, formula (A5.6) is valid by analytical continuation with respect to λ from the region Reλ > 0 to the C except for the points 0, −1, −2, ..... In the region Reλ > 0 one can take the Laplace transform of equation (A5.5) classically and get (A5.7) L(q) = L(h) + (λ) p −λ L(q). From (A5.7) one gets L(q) =

L(h) . 1 − (λ) p −λ

(A5.8)

The question is: under what conditions the right side of formula (A5.8) is the Laplace transform of a function q from some functional class? The answer to this question depends on λ and h. Equation (A5.7) makes sense by an analytic continuation with respect to λ for λ ∈ C except for the points λ = 0, −1, −2, ..... Therefore, equation (A5.4) can be considered as a particular case of equation (A5.5) with λ = −1/4. At this value of λ this equation is well defined by analytic continuation, although classically its kernel is hyper-singular and the integral in (A5.4) diverges classically. To solve equation (A5.4), we apply the Laplace transform and continue analytically the result with respect to λ, as was explained above. For λ = −1/4 equation (A5.7) yields: L(q) =

L(h) 1 − (− 41 ) p 1/4

.

(A5.9)

One has (z + 1) = z(z). For z = − 41 , one has ( 21 ) 1 (− ) = = −4π 1/2 := −b, b > 0. 4 − 41

(A5.10)

Let R+ = [0, ∞). We assume for simplicity that h(t) is a smooth rapidly decaying function. Then c , (A5.11) |L(h)| ≤ (1 + | p|)−1 where c > 0 does not depend on p. From formulas (A5.9) and (A5.10) the following theorem follows.

84

Appendix 5: Applications of Analytic Continuation to Tables of Integral Transforms …

Theorem 2 Assume that (A5.11) holds. Then equation (A5.4) has a solution q(t) in C(R+ ), q(0) = 0, this solution is unique in C(R+ ) and can be calculated by the formula q(t) = L −1



L(h)  , 1 + bp 1/4

(A5.12)

where b > 0 is defined in (A5.10). Proof To prove Theorem 2 it is sufficient to check that the expression transform of a function q(t) ∈ C(R+ ). We also prove that q(0) = 0. Consider the function  ∞ 1 L(h) q(t) = eist ds, p = is. 2π −∞ 1 + bp 1/4 This is the inverse of the Laplace transform of

L(h) 1+bp1/4

L(h) 1+bp1/4

is the Laplace

(A5.13)

since d p = ids. The integral (A5.13)

converges absolutely under our assumptions since the integrand is O( | p|15/4 ) for | p|  1. Therefore, q ∈ C(R+ ). To prove that q(0) = 0, let us check that  ∞ L(h) eist ds|t=0 = 0. (A5.14) 1 + bp 1/4 −∞ The function

L(h) 1+bp1/4

is analytic in Re p > 0 and is O( | p|15/4 ) for | p|  1. One checks that

with b > 0 is a uniformly bounded analytic function of p in the half-plane Re p ≥ 0. Let L n be a closed contour, oriented counterclockwise, consisting of the segment [in, −in] and half a circle γn = {neiφ }, − π2 ≤ φ ≤ π2 . By the Cauchy theorem,  L(h) d p = 0, (A5.15) 1/4 L n 1 + bp 1 1+bp1/4



and lim

n→∞ γ n

L(h) d p = 0. 1 + bp 1/4

(A5.16)

Consequently, from (A5.15) and (A5.16) it follows that  lim

in

n→∞ −in

L(h) d p = 0. 1 + bp 1/4

(A5.17)

This relation is equivalent to (A5.14). Theorem 2 is proved.  Consider equation (A5.5) with λ = − 21 and with the minus sign in front of the integral. By formula (A5.7) with the minus sign in front of the (λ) the Laplace transform of the solution is: 1 L(h) , a = |(− )| > 0. L(q) = (A5.18) 1 − ap 1/2 2

Appendix 5: Applications of Analytic Continuation to Tables of Integral Transforms …

85

The new feature, compared with Theorem 2, is the existence of the singularity at p = a12 . We assume for simplicity that h(t) ∈ C(R+ ) has compact support. In this case L(h) is an entire function of p and the behavior for large t of the solution q(t), found in Theorem 3 (see below), is easy to estimate. Let us investigate the function  −1 + p 1/2  1/2 1 1 −1 a −2 −1 p = −a = −a − a . 1 − ap 1/2 p − a −2 p − a −2 p − a −2 One has:

(A5.19)

 L −1 − a −2

 1 −2 = −a −2 ea t . −2 p−a   t So, using the known formula L( f )L(g) = L 0 f (τ )g(t − τ )dτ , we derive:  L −1 − L(h)a −2

 1 = −a −2 p − a −2



t

h(τ )ea

−2 (t−τ )

(A5.20)

dτ .

(A5.21)

0

Consider the last term in (A5.19). In [1] formula (22) in Sect. 5.3 is:   1 ( p − β)−1 p 1/2 = L (πt)− 2 + β 1/2 eβt Er f (β 1/2 t 1/2 ) ,

1

Er f (x) := 2π − 2



x 0

Therefore, taking β = L where

−1



a −2 ,

− L(h)a

e−t dt. 2

(A5.22)

one derives:

−1

p 1/2  = −a −1 p − a −2



t

h(τ )g(t − τ )dτ ,

(A5.23)

0

1

g(t) := (πt)− 2 + β 1/2 eβt Er f (β 1/2 t 1/2 ) β = a −2 .

(A5.24)

From formulas (A5.18), (A5.19), (A5.21), (A5.23) it follows that  t  t −2 h(τ )ea (t−τ ) dτ − a −1 h(τ )g(t − τ )dτ . q(t) = −a −2

(A5.25)

0

0

We have proved the following theorem Theorem 3 Assume that h is compactly supported. Then equation (A5.5) with λ = − 21 is uniquely solvable and its solution is given by formula (A5.25), where g(t) is given by formula (A5.24). The behavior of the solution for t → ∞ depends on λ, on the sign in front of the integral equation (A5.5) and on h. Theorems 2 and 3 are examples of a study of the solution to equation (A5.5).

86

Appendix 5: Applications of Analytic Continuation to Tables of Integral Transforms …

Conclusion It is proved in this paper that the validity of some formulas in the tables of integral transforms can be greatly expanded by the analytic continuation with respect to parameters. This idea is used for investigation of some integral equations with hyper-singular kernels. Such an equation, see (A5.4), plays a crucial role in the author’s investigation of the Navier-Stokes problem.

References 1. H. Bateman, A. Erdelyi, Tables of Integral Transforms (McGraw Hill, New York, 1954) 2. A.G. Ramm, Applications of analytic continuation to tables of integral transforms and some integral equations with hyper-singular kernels. Open J. Optim. 11, 1–6 (2022). Open access journal 3. W. Rudin, Real and Complex Analysis (McGraw Hill, New York, 1974)

References

1. T. Kato, Strong L p -solutions of the Navier-Stokes equation in Rm , with applications to weak solutions. Math. Z. 187, 471–480 (1984) 2. A.G. Ramm, The Navier-Stokes Problem (Morgan & Claypool publishers, 2021) 3. Yu. Brychkov, A. Prudnikov, Integral Transforms of Generalized Functions (Nauka, Moscow, 1977). https://doi.org/10.1007/bf01262407 4. M. Lavrentiev, B. Shabat, Methods of the Theory of Functions of a Complex Variable (Nauka, Moscow, 1958) 5. A.G. Ramm, Navier-Stokes equations paradox. Reports Math. Phys. (ROMP) 88(1), 41–45 (2021) 6. H. Bateman, A. Erdelyi, Tables of Integral Transforms (McGraw-Hill, New York, 1954) 7. G. Polya, G. Szego, Problems and Theorems from Analysis (Nauka, Moscow, 1956) 8. A.G. Ramm, Applications of analytic continuation to tables of integral transforms and some integral equations with hyper-singular kernels. Open J. Optim. 11, 1–6 (2022) 9. O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Fluid (Gordon and Breach, New York, 1969) 10. L. Landau, E. Lifshitz, Fluid Mechanics (Pergamon Press, New York, 1964) 11. L. Hormander, ¨ The Analysis of Linear Partial Differential Operators, vol. 1 (Springer, New York, 1983) 12. I. Gel’fand, G. Shilov, Generalized Functions, vol. 1 (GIFML, Moscow, 1959). ((in Russian)) 13. N. Lebedev, Special Functions and Their Applications (Dover, New York, 1972) 14. A.G. Ramm, Comments on the Navier-Stokes problem, axioms. Open Access J 10, 95 (2021) 15. W. Rudin, Real and Complex Analysis (McGraw Hill, New York, 1974)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. G. Ramm, Analysis of the Navier-Stokes Problem, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-30723-2

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