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English Pages [121] Year 2012
Table of contents :
Introduction
Lecture 1
Spaces H(A; )
General notations
Spaces H(A;)
The space H0(A; ).
Extension of functions in Ho(A; ) to Rn.
Lecture 2
Regularization
Problem of local type.
Some generalizations.
Lecture 3
Spaces Hm.
Extension of functions in Ho(A; ) to Rn.
Lecture 4
The mapping .
Lecture 5
General Elliptic Boundary Value Problems
General theory.
Lecture 6
Examples: + , >0, = i=1n {Catalog} ] >> >>2 uxi2.
Lecture 7
Green's kernel.
Relations with unbounded operators.
Lecture 8
Complements on Hm()
Estimates on Hmo().
Regularity of the function in Hm().
Reproducing kernels.
Lecture 9
Complete Continuity.
Applications
Lecture 10
Operators of order 2
V = H1o (), gij constant.
Lecture 11
Formal interpretation:
Complementary results.
Lecture 12
Operators of order 2m
Applications to the Dirichlet's problem.
Lecture 13
Applications
Regularity in the Interior
Statements of theorems
Lecture 14
Some remarks.
Lecture 15
Regularity at the boundary.
Lecture 16
Lecture 17
Completion of the proof of theorem 9.1.
Other results.
An application of theorem 9.1.
Lecture 18
Regularity at the boundary in the case of...
Regularity at the boundary for some more problems.
VisikSoboleff Problems
Lecture 19
Lecture 20
Application.
Aronszajn and Smith Problems[1]
Complements on Hm ()
Lecture 21
AronszajnSmith Problems
Lecture 22
Lecture 23
Regularity of Green's Kernels
Lecture 24
Study at the boundary.
Regularity at the Boundary Problems for...
Lecture 25
Systems
Lectures on Elliptic Partial Differential Equations By
J.L. Lions
Tata Institute of Fundamental Research, Bombay 1957
Lectures on Elliptic Partial Differential Equations By
J. L. Lions
Notes by
B. V. Singbal
Tata Institute of Fundamental Research, Bombay 1957
Introduction In these lectures we study the boundary value problems associated with elliptic equation by using essentially L2 estimates (or abstract analogues of such estimates. We consider only linear problem, and we do not study the Schauder estimates. We give first a general theory of “weak” boundary value problems for elliptic operators. (We do not study the noncontinuous sesquilinear forms; of. Visik [17], Lions [7], VisikLadyzeuskaya [19]). We study then the problems of regularityfirstly regularity in the interior, and secondly the more difficult question of regularity at the boundary. We use the Nirenberg method for Dirichlet and Neumann problems and for more general cases we use an additional idea of AronsazajnSmith. These results are applied to the study of new boundary problems: the problems of VisikSoboleff. These problems are related and generalize the problems of the Italian School (cf. Magenes [11]). We conclude with the study of the Green’s kernels, some indications on unsolved problems and short study of systems. Due to lack of time we have not studied the work of Schechter [15] nor the work of MorreyNirenberg [13] which rots essentially on L2 estimates.
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Contents Introduction 1
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Lecture 1 Spaces H(A; Ω) . . . . . . . . . . . . . . . . . . 1.1 General notations . . . . . . . . . . . . . 1.2 Spaces H(A; Ω) . . . . . . . . . . . . . . 1.3 The space H0 (A; Ω). . . . . . . . . . . . 1.4 . . . . . . . . . . . . . . . . . . . . . . 1.5 Extension of functions in Ho (A; Ω) to Rn .
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Lecture 1.6 1.7 1.8 1.9
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Lecture 2 Spaces H m . . . . . . . . . . . . . . . . . . . . . 2.1 . . . . . . . . . . . . . . . . . . . . . . 2.2 Extension of functions in Ho (A; Ω) to Rn . 2.3 . . . . . . . . . . . . . . . . . . . . . .
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The mapping γ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Lecture 3 General Elliptic Boundary Value Problems . . . . . . . . . . . 3.1 General theory. . . . . . . . . . . . . . . . . . . . . .
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Lecture 2.4 2.5
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CONTENTS 3.2 6
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Lecture 3.4 3.5 3.6
v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . n ∂2 u P .. . . . . . . . . . Examples: ∆ + λ, λ > 0, ∆ = 2 i=1 ∂xi
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19 21 21 25 25 26 27
Lecture 4 Complements on H m (Ω) . . . . . . . . . . 4.1 Estimates on Hom (Ω). . . . . . . . . 4.2 Regularity of the function in H m (Ω). 4.3 Reproducing kernels. . . . . . . . .
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Lecture 5 Complete Continuity. 5.1 . . . . . . . 5.2 . . . . . . . 5.3 Applications
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10 Lecture 6 Operators of order 2 . . . . . . . 6.1 . . . . . . . . . . . . . 6.2 . . . . . . . . . . . . . 6.3 V = Ho1 (Ω), gi j constant. 6.4 . . . . . . . . . . . . .
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11 Lecture 6.5 6.6 6.7
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12 Lecture 6.8 . . . . . . . . . . . . . . . . . . . . . 7 Operators of order 2m . . . . . . . . . . . . . . 7.1 . . . . . . . . . . . . . . . . . . . . . 7.2 Applications to the Dirichlet’s problem. 7.3 . . . . . . . . . . . . . . . . . . . . .
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CONTENTS
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13 Lecture 7.4 Applications . . . . . 8 Regularity in the Interior . . . 8.1 . . . . . . . . . . . . 8.2 Statements of theorems
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15 Lecture 9 Regularity at the boundary. . . . . . . . . . . . . . . . . . . . 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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16 Lecture 9.3 9.4
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17 Lecture 9.5 9.6 9.7
Completion of the proof of theorem 9.1. . . . . . . . . Other results. . . . . . . . . . . . . . . . . . . . . . . An application of theorem 9.1. . . . . . . . . . . . . .
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18 Lecture 9.8 Regularity at the boundary in the case of... . . . . . . 9.9 Regularity at the boundary for some more problems. 10 VisikSoboleff Problems . . . . . . . . . . . . . . . . . . . 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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20 Lecture 10.4 Application. . . . . . . . . . . . . . . . . . . . . . . . 11 Aronszajn and Smith Problems1 . . . . . . . . . . . . . . . . 11.1 Complements on H m (Ω) . . . . . . . . . . . . . . . .
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21 Lecture 11.2
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CONTENTS 22 Lecture 11.3
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23 Lecture 100 12 Regularity of Green’s Kernels . . . . . . . . . . . . . . . . . 100 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 24 Lecture 105 12.3 Study at the boundary. . . . . . . . . . . . . . . . . . 105 13 Regularity at the Boundary Problems for... . . . . . . . . . . . 106 13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 25 Lecture 108 14 Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
Lecture 1 1 Spaces H(A; Ω) 1.1 General notations We shall recall some standard definition and fix some usual notation. Rn (x = (x1 , . . . , xm )) will denote the ndimensional Euclidean space, Ω an open set of Rn , D(Ω) will be the space of all indefinitely differentiable functions (written sometimes C ∞ functions) with compact support in Ω with the usual topology of Schwartz. D ′ (Ω) will be the space of distributions over Ω. L2 (Ω) will be the space of all square integrable functions on Ω. The norm of a function F ∈ L2 (Ω) will be denoted by F o . Derivatives of functions on Ω will always ∂p be taken in sense of distributions; more precisely D p will stand for p ∂x1 , . . . ∂xnpn where p = (p1 , . . . , pn ) with pi nonnegative integers and p = p1 + · · · + pn is the order of D p . If T ∈ D ′ (Ω), hD p T, ϕi = (−1)phT, D p ϕi.
1.2 Spaces H(A; Ω) A defferential operator with constant coefficients A is an expression of the for P A = a p D p where a p are all constants. The highest integer m for which p≤m
there exists an a p , 0 for p = m will be called the order of the operator A.
Definition 1.1. Let A = {A1 , . . . , Aν } be a system of differential operator with constants coefficients. By H(A; Ω) we shall denote the space of u ∈ L2 (Ω) for which Ai u ∈ L2 (Ω).
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1. Spaces H(A; Ω)
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Evidently D(Ω) ⊂ H(A; Ω). On H(A; Ω) we define a sesquilinear form by (u, v)H(A;Ω) = (u, v)o +
γ X
(Ai u, Ai v)o .
(1)
i=1
Theorem 1.1. With the norm defined by (1), H(A; Ω) is a Hilbert space.
2
Proof. It is evident from the expression (1) that (u, v) = (v, u)(u, u) ≥ 0, and that (u, u) = 0 if and only u = 0. So it remains to verify that under the topology defined by the norm, H(A; Ω) is complete. If {un } is any Cauchy sequences in H(A; Ω), from (1) it follows that {un } and {Ai un } are Cauchy sequences in L2 (Ω). Hence {un } and {Ai un } converge to u and vi respectively, say, in L2 (Ω). Since the convergence in L2 (Ω) implies the convergence in D ′ (Ω), {un } and {Ai un } converge to u and vi in D ′ (Ω) respectively. Since Ai are continuous on D ′ (Ω), Ai (uu ) → Ai (u) in D ′ (Ω). Hence Ai (u) = vi which proves that u ∈ H(A; Ω). Proposition 1.1. If W ⊂ Ω and for u ∈ H(A; Ω), uw denotes the restriction of u to W, then (a)uw ∈ H(A; W), and (b) the mapping u → uw is continuous mapping of H(A; Ω) → H(A; W).
1.3 The space H0 (A; Ω). Definition 1.2. H0 (A; Ω) will be the closure of D(Ω) in H(A; Ω).H(A; Ω) will be the “orthogonal complement” of H0 (A; Ω) in H0 (A; Ω). The following question then naturally arises: Problem 1.1. When is H0 (A; Ω) = H(A; Ω)? If A differential operator, let A∗ denote the differential operator defined P by hA∗ T, ϕi = hT, Aϕi. If A = a p D p , then it is easily verified that A∗ = P (−1) p a¯ p D p . Proposition 1.2. H0⊥ (A; Ω) is the space of solution in H(A; Ω) of (1 +
T = 0.
γ P
i=1
A∗i Ai )
Proof. T ∈ H01 (A; Ω) if and only T is orthogonal to every ϕ ∈ D(Ω), i.e., if and only if
(T, ϕ)o +
γ X i=1
(Ai T, Ai ϕ) = 0
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1. Spaces H(A; Ω)
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for all ϕ ∈ D(Ω), which is equivalent to say that X hT + A∗i Ai T, ϕi = 0 for all ϕ ∈ D(Ω),
or that
(1 +
X
A∗i Ai )T = 0.
Some examples. 1) If there is no differential operators, H(A; Ω) = L2 (Ω) = Ho (Λ; Ω). d ; H(A; Ω) = {u/u ∈ L2 and only u′ ∈ L2 . Then dx T ∈ Ho⊥ (A; Ω) if and if T − T ′′ = 0, i.e.,T = λe x + µe−x . Hence Ho⊥ (A; Ω) is space of dimension 2.
2) Let Ω =]0, 1[, x = x1 , A =
3) Let Ω =]0, +∞[, x = x1 , A = Ho⊥ (A; Ω) is of dimension 1. 4) Let Ω =]0, +∞[, x = x1 , A =
d .T = λe x + µe−x is in L2 (Ω) is λ = 0. Hence dx d ⊥ H (A; Ω) = {0}, i.e., dx o
Ho (A; Ω) = H(A; Ω). dm In general it can be proved that if Ω =]0, 1[A = , H ⊥ (A, Ω) is 2mdxm o dimensional.
1.4 We recall some properties of Fourier transformations of distributions. Let S be the space of fastly decreasing functions in Rn , S ′ be the dual of S consisting of tempered distributions. For T ∈ S ′ we shall denote the Fourier Transform of T by F T = Tˆ . We know that L2 (Rn ) ⊂ S ′ and that Tˆ o = T o if T ∈ L2 (Rn ) (Plancherel’s formula). Also F (D p T ) = (2πiξ) p Tˆ , where ξ = (ξ1 , . . . , ξn ) P p and ξ p = ξnp1 . . . ξnpn . Since F is linear, it follows that if A = D is any differential operator with consists coefficients
F (AT ) = A(2πiξ)Tˆ where X p p A j (2πiξ) = a p ξ1 1 · · · · · · ξn n (2πi)p p≤m
=
X
a p (2πiξ) p .
p≤m
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1. Spaces H(A; Ω)
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Proposition 1.3. u ∈ H(A, Rn) if and only if uˆ ∈ L2 (Rn ) and A j (2πiξ)uˆ ∈ L2 (Rn ), j = 1, . . . , N. This is immediate as u ∈ L2 ⇔ uˆ ∈ L2 (Rn ) and A j u ∈ L2 ⇔ A j (2πiξ)uˆ ∈ L (Rn ). 2
Proposition 1.4. Ho (A, Rn ) = H(A, Rn), for any A = {A1 , . . . , Aν } with constant coefficients. From Proposition 1.2 we have T ∈ Ho⊥ (A, Rn ) if and only if T ∈ L2 , A j T ∈ ν P L2 and (1 + A∗j A j )T = 0. By taking Fourier transforms, it follows that j=1 P T ∈ Ho⊥ (A, Rn ) if and only if Tˆ ∈ L2 A j Tˆ ∈ L2 and (1 + A j (2πiξ)2 )Tˆ = 0. P 2 But since (1 + j A j (2πiξ) ) , 0, Tˆ = 0 a.e., and hence T = 0 which proves the proposition.
1.5 Extension of functions in Ho (A; Ω) to Rn . Theorem 1.2. There exists one and only one continuous linear mapping u → u˜ of Ho (A; Ω) into H(A, Rn ) such that if u ∈ D(Ω), u˜ = u. a.e. in Ω. ϕ(x) if x ∈ Ω For ϕ ∈ D(Ω), define ϕ˜ = 0 if x < Ω. n Then ϕ˜ ∈ D(R ) and ϕ ˜ H(A,Rn ) = ϕH(A,Ω) . Hence ϕ → ϕ˜ is a continuous mapping of D(Ω) with the topology induced by H(A; Ω) into H(A, Rn ). This proves the theorem Definition 1.3. If u ∈ Ho (A; Ω), u˜ is called the extension of u to Rn . u(x), x ∈ Ω Remark . If u ∈ H(A; Ω) and we put u˜ (x) = it is not true that 0, x n, X is void. We shall admit, without proof, Theorem 2.2. H m (Ω) = Hom (Ω) if and only if [Ω is m polar.
2.2 Extension of functions in Ho (A; Ω) to Rn . Definition 2.3. An open set Ω ⊂ Rn is said satisfy mextension property if we can find a continuous linear mapping π of H m (Ω) to H m (Rn ) such that πu = u a.e. in Ω. There are examples to show that not all Ω posses this property. For example in the case m = 1, n = 2 take the domain in the figure, which is an open square with the thickened line removed. Let u be a funcLinear tion as indicated in the figure. Let ϕ be ∞ a C functions which vanishes outside the unit circle, is 1 within a smaller circle and 0 ≤ ϕ ≤ 1 elsewhere. Then v = ϕu is zero 0 on the boundary of the given square. We now prove that it is impossible to find to find V such that V = va.e. on Ω. For, if V = v, a.e. on Ω, then ∂v sometimes outside Ω. = ∂u ∂ϕ ∂y ∂y u + ϕ ∂y in Ω. Now
∂V ∂u is a measure supported by th thickened line. Hence is not a ∂y ∂y
function. However, in the following two theorems, it will be proved that some usual domains posses the mextension property. Theorem 2.3. Let Ω = {xn > 0} = R+n . Then Ω has mextension property for any m. ¯ be the restrictions of functions of D(Rn ) to Ω. We require the Let D(Ω) following ¯ is dense in H m (Ω). Lemma. H m (Ω) ∪ D(Ω)
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Assume for the time being this lemma, we shall first complete the proof of the theorem 2.3. It is enough to show that π can be defined continuously on ¯ We do this explicitly as follows: For u ∈ H m (Ω) ∩ D(Ω), ¯ put H m (Ω) ∩ D(Ω). u(x) if xn ≥ 0. π(u(x)) = λ1 u(x′ , −xn ) + · · · + λm u(x′ , − xmn ) = v(x) if xn < 0.
where x′ = (x1 , . . . , xn−1 ). We determine λi in order to ensure that π(u(x)) ∈ H m (Rn ). For that we need verify (v(x′ , 0) = u(x′ , 0),
i.e.λ1 + · · · + λm = 1. .. λm ∂m−1 v ′ ∂m−1 u m−1. (λ1 + · · · m−1 ) = 1. (x , 0) = m−1 , i.e., (−1) ∂m−1 xn ∂ xn m
These equations determine λ′1 s and it is at once seen that D p (π(u)) = D p u for p ≤ m, a.e, on Ω and that the mapping π is continuous. Now we prove the lemma. Let u ∈ H m (Ω); for every ∈> 0, define uǫ (x) = u(x′ , xn + ∈). Let vǫ be the restriction of uǫ to. It is easy to see that u∈ → u in H m (Ω) as ∈→ 0, and so we need prove only that vǫ for every fixed ǫ > 0 can be approximated by functions ¯ i.e., we have to prove that given a function w ∈ H m (Ωα ), of H m (Ω) ∩ D(Ω), where ωα is the domain {xn > −α}, w can be approximated on Ω by functions ¯ Let ∪(xn ) be a C ∞ function defined as follow: θ = 0 for of H m (Ω) ∩ D(Ω). xn < −α, 1 for xn > 0, 0 < θ < 1, elsewhere. Now θw ∈ H m (Rn ) and θw = v a.e. in Ω. However, D(Rn ) is dense in H m (Rn ). Hence there exists a sequences φk ∈ D(Rn ) such that φ → θw in H m (Rn ). Let ϕk be the restriction of φk to Ω. ¯ and ϕk → θw = w in H m (Ω). Then ϕk ∈ H m (Ω) ∩ D(Ω) Remark 1. If Ω has mextension property, then the above lemma holds, i.e., ¯ is dense in H m (Ω). For, since D is dense in H m (Rn ), and since H m (Ω) ∩ D(Ω) there exists a continuous mapping π of H m (Ω) in H m (Rn ), the restrictions of functions of D to Ω are dense in H m (Ω). Remark 2. This lemma holds also, for instance, for starshaped domains.
2.3 Theorem 2.4. Let Ω be an open bounded set such that the boundary of Ω is an (n − 1) dimensional C m manifold ΓΩ lying on one side of Γ. Then Ω has the
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mextension property. Proof. Let Zn be the mdimensional Euclidean space with coordinates ξ, . . . , ξn 0 < ξi < 1 and let W be the open rectangle define by i = 1, . . . , n − 1. Let −1 < ξn < 1 W+ , W− , Wo denote the subsets of W determined by ξ > 0, ξn < 0, ξn = 0, respectively. On account of the hypothesis on Γ, there exists a finite open covering O′ , Oi ¯ and mtimes continuously differentiable functions ψi of W to Oi such that of Ω ¯ and Wo onto Oi ∩Γ, and further, Oi ∩Γ ψi maps W− onto Oi ∩Γ, W+ onto Oi ∩[Ω cover Γ and Let (a′ , ai ) be a C m partition of unity subordinate to this covering. If P u ∈ H m (Ω), then u = a′ u + ai u and ai u have their supports in Oi respectively. Now ψi defines an isomorphism of H m (Oi ) onto H m (W) and of H m (0i ∩ Ω) m onto H m (W− ), which we still denote by ψi . Hence vi = ψ−1 i (ai u) ∈ H (W− ) and vi = 0 near the part of the boundary of W− which is not contained in ξn = 0. Hence vi can be extended to (ξn < 0) by putting it equal to zero outside W− . By theorem 2.3, there exists πvi ∈ H m (Zn ) such that πvi = vi , a.e. on Zn . Let θ(xn ) 2 1 be a C ∞ function which is 0 for ξn > and 1 for ξ < , 0 < θ < 1 elsewhere. 3 3 Let p(ξ) = θπ(vi ). Pi (ξ) has its support in W and is zero near the boundary of W. Let ϕi (x) = ψi (P). Then ϕ(x) ∈ H m (Oi ) and is zero near the boundary of P Oi . Hence π(u) = a′u + ϕ˜ i (x) answers the theorem.
14
Lecture 4 2.4 The mapping γ. ¯ For the function f ∈ H m (Ω), the restriction Let H m (Ω) = H m Ω ∩ D(Ω). of f to the boundary Γ of Ω defines a function γ f on Γ. We wish to know for what for what spaces X(Γ) of function on Γ, this mapping γ of H L (Ω) to X(Γ) is continuous. If γ is continuous, we can extend ν to H 1 (Ω) if, for example, Ω has 1extension property, and γu for u ∈ H ′ (Ω) may be considered as generalized boundary value of u. Theorem 2.5. Let Ω = {xn > 0} so that Γ = {xn = 0}. Let X(Γ) = L2 (Γ) = L2 (Rn−1 ). Then u → γu is a continuous mapping of H 1 (Ω) → L2 (Γ), i.e., there exists a unique mapping γ : H 1 (Ω) → L2 (Γ) which on H 1 (Ω) is restriction. Proof. Let x′ = (x1 , . . . , xn−1 ). Let O(xn ) be a function defined for xn > 0, zero for xn > 1, and 0 < O(xn ) < 1 in (0, 1). We have u(x′ , 0)2 = R∞ ∂ (u(x)u¯ (x)θ(xn ))dx. Hence − 0 ∂xn Z Rn−1
Z
∂ (θ(xn )u(x)u¯ (x))dx ∂xn Ω Z Z ∂u ∂u¯ ′ 2 θ u dx − θ( u¯ + )dx. = ∂xn ∂xn
u(x′ , 0)2dx′ = −
Ω
Ω
So by Schwartz’s inequality, Z Z Z ∂u 2 2 ′ 2 ′ u(x , 0) dx ≤ C( u dx +   dx), ∂x n Ω Γ Ω
12
15
13 ≤ C k u k21 . This means γ is continuous. ∂ ¯ is dense in H(A; Ω), and by the . Then H(A; Ω) ∩ D(Ω) ∂xn same method as above, u → γu is continuous from H(A; Ω) to L2 (Γ).
Remark. Let A =
In the text few propositions, we are going to determine the image and the kernel of γ. We have seen that u ∈ H m (Rn ) if and only if u ∈ L2 and (1 + ξm )uˆ ∈ L2 . Generalizing we define H α (Rn ) for noninteger α > 0. Definition 2.4. u ∈ H α (Rn ) if and only if u ∈ L2 and (1 +  ∈ α )u˜ ∈ L2 . On H α (Rn ), we put the topology defined by the norm (u, u)H α(Rn ) = ((1 + ξα )uˆ )L2 (Rn ) . Theorem 2.6. Let Ω = {xn > 0}. For u ∈ H 1 (Ω), we have 1
1) γu ∈ H 2 (Γ), and 1
2) u → γu is continuous mapping of H 1 (Ω) onto H 2 (Γ). R ′ ′ Proof. Let ξ = (ξ1 , . . . , ξn−1 )n−1 and uˆ (ξ′ , xn ) = e−2πix .ξ u(x′ , xn )dx′ be trunRn−1
cated Fourier transform of u(x). Since ! ∂u = ξi uˆ ξL2 (ξ′ , xn ), i = 1, . . . , n − 1, F ∂xi
we have 1) (1 + ξ′ )u ∈ L2 (ξ′ , xn ). ∂uˆ ∂uˆ Further, since = we have ∂xn ∂xn 2)
∂uˆ ∈ L2 (ξ′ , xn ). ∂xn Now, as in theorem 2.5, uˆ (ξ′ , 0)2 = −
R∞ o
(uˆ u¯ˆ θ)dxn . Hence
R Γ
(1 + ξ)uˆ
R ∂ ¯ (ξ′ , 0)2dξ′ = − (1 + ξ′ ) (uˆ uˆ θ)dx < ∞ by Schwartz’s inequality and (1) ∂xn R+n and (2).
16
14 1
Hence γ u ∈ H 2 (Γ). 1 1 We now prove the second point. f ∈ H 2 (Γ) of and only if (1 + ξ 2 ) fˆ ∈ 2 n 1 L (Z ). We have to look for a function u ∈ H (Ω) such that γu = f . Let
17
U(ξ′ , xn ) = exp(−(+ξ′ )xn ) fˆ(ξ′ ) for xn > 0, and u = Fξ (U(ξ′ , xn )). We prove that uξH 1 and γu = f . The only not com∂u 2 pletely trivial point is to verify that ξL . ∂xn ∂u = −(1 + ξ) exp(−(1 + ξxn ) fˆ(ξ′ ). ∂xn Hence Z
∂u 2 ′  dxn = (1 + ξ2  fˆ(ξ′ )2  ∂xn
Z∞
exp(−(1 + ξ)xn )dxn
o
= (1 + ξ fˆ(ξ′ )2 . 1
Since f ∈ H 2 (Γ), we have
R ∂u 2 dx is finite.  ∂x n Ω
Theorem 2.7. γ u = 0 if and only if u ∈ Ho1 (Ω). Proof. (a) u ∈ Ho1 (Ω) =⇒ u = 0 for we have u = lim ϕk in H 1 (Ω)ϕk ∈ D(Ω).γu = lim γ(ϕk ) in L2 (Γ) = 0. (b) Conversely to prove that γ u = 0 implies u ∈ HO1 (Ω) we require the
¯ Then φu ∈ H 1 (Ω), and Lemma . Let Ω = {xn > 0}, u ∈ HO1 (Ω), φ ∈ D(Ω). γ(φu) = γ(φ)γ(u). Proof. We know H 1 (Ω) is dense in H 1 (Ω). Hence there exists uk ∈ H 1 (Ω), such that uk → u in H 1 (Ω). Now, φuk → φu in H 1 (Ω) and since γ(φuk ) = γ(φ)γ(uk ), we have γ(φu) = γ(φ)γ(u). Coming back to the proof of the theorem, let a(x) be a C ∞ function 1 on the unit ball, 0 outside another ball, and 0 ≤ a(x) ≤ 1 elsewhere. Then if we define a j (x) = a(x/ j), we have a j u → u in H 1 (Ω). Hence if we prove that a j u ∈ H01 (Ω), we shall have proved that u ∈ Ho1 (Ω). Since a j u has compact support, and since γha j ui = a j γu = 0, this means, we may assume, that in
18
15 addition to γu = 0, u has compact defined for xn > 0, 0 Ok (x) = linear 1
¯ Let Ok (xn ) be a function support in Ω. for 0 < xn < 1/k for 1/k ≤ xn ≤ 2/k for xn > 2/k.
1 n 1 g Then O k u ∈ H (R ). By regularization, we may assume that Ok u ∈ H0 (Ω). 1 We now prove Ok u → u in H (Ω). We have
θk
∂ ∂u ∂u = (θk (u)) → for 1 ≤ i ≤ n − 1. ∂xi ∂xi ∂xi
We have to prove then that θk′ (xn )u+θk 0. Now θk′ (xn )
∂u ∂u → ; that is to say, θk′ (xn )u → ∂xn ∂xn
0 for xn < 1/k and xn > 2/k = k for 1/k < xn < 2/k.
R xn ∂u (x′ , t) dt. Hence 0 ∂t Z xn ∂u  (x′ , t)2 dt u(x′ , xn )2 ≤ xn ∂t 0 Z xn ∂u  (x′ , t)2 dt if xn > 2/k ≤ 2/k ∂t 0 2/k.C if xn > 2/k
Further u(x′ , xn ) = −
Also
(1) 19
Z
θk′ (xn )2 u(x′ , xn )2 dxn
=
Z2/k
θk′ u2 dx
0
≤ 2k
Z2/k 0
Zxn 2 ∂u dxn (x′ , t) dx′ ∂t 0
Z2/k Z2/k ∂u ′ 2 = 2k (x , t) dt dxn ∂t 0
t
(by changing the order of integration)
16
≤4
Z2/k

∂u 2  dxn . ∂xn
0
R
Hence θk′ u2 dx → 0 by (1), which completes the proof. Thus we see that Ho′ (Ω) is the space of functions which are weakly zero on the boundary. The above results can be generalized to spaces H m (Ω). Let Ω = {xn > 0}; u ∈ H m (Ω), if and only if D p (u) ∈ H 1 (Ω), for p ≤ m − 1. Hence γD p u can be defined as above for p ≤ m − 1. we have the Theorem 2.8. u ∈ Hom (Ω) if and only if γ(D p u) = 0 for p ≤ m − 1. In fact, we can say something more, Exercise. Let u ∈ Hom (Ω) and u ∈ D xp , with q arbitrary. Then q
q
γ(D x′ , u) = D x′ , (γu).
2.5 Let Ω be an open set of Rn such that (a)Ω has 1extension property, and (b) the boundary Γ of Ω is an (n − 1) dimensional C 1 manifold. In the case Ω is bounded (b) implies (a). On Γ we have an intrinsic measure. We denote by L2Loc (Γ) the space of square summable functions on every compact of Γ with respect to this measure. Theorem 2.9. Under the above hypothesis on Ω (i.e.) a) Ω possesses 1 extension property b) Γ is an (n − 1) dimensional C 1 manifold, there exists a unique continuous map γ : H 1 (Ω) → L2loc (Γ) which on functions of H 1 (Ω) coincides with the restriction to Γ. Proof. Form (a) it follows that H ′ (Ω) is dense in H ′ (Ω) and hence the uniqueness will follow from the existences. Let Γ2 be any compact of Γ. We observe that by using a C 1 partition of unity the problem is reduced to a local one, that is to say, we may assume that the support of ϕ ∈ H 1 (Ω) is contained in an open set 0 and that there exists a homeomorphism ψ as in theorem 2.4. Further we may assume, without loss of generality, that Γ2 ⊂ 0. Let a be a C ∞ function in Rn with compact support in 0 and which is 1 on γ2 . Then
20
17 auψ−1 ∈ H(Zn ) (in the notation of theorem 2.4), and γauψ−1 ∈ L2 (Wo ). Since ψ defines an isomorphism of L2 (Γ2 ) into L2 (Wo ), ψ(γauψ−1 ) ∈ L2 (Γ2 ). Now on Γ2 we have γu = ψ(γauψ−1 ) which proves that γ is continuous mapping of H 1 (Ω) into L2 (Γ2 ) which completes the proof. Remark. A complete generalization of theorem 2.7 is due to N. Aronszajn [1].
Lecture 5 3 General Elliptic Boundary Value Problems 3.1 General theory. We formulate at the beginning certain problems on topological vector spaces and solve them. Later on we shall show how these answers will help us in solving many of the classical boundary value problems for elliptic differential equations. As a matter of notation, we shall write A ⊂ B, where A and B are two topological vector spaces to mean the injection i : A → B is continuous or that the topology A is finer than the topology induced by B. Let V be a Hilbert space over complex numbers. We shall denote by uV the norm in V. Let Q be a locally convex topological vector space such that 1) V ⊂ Q and V is dense in Q; 2) On Q an involution (i.e., an antilinear isomorphism of order two) f → f¯ is given which leaves V invariant; 3) Let V be given a continuous sesquilinear form a(u, v)(i.e., a(λu, v) = λa ¯ (u, v), and a(u, λv) = λa(u, v). Let Q′ be the dual space of Q. On Q′ an involution is induced by the given one in Q by the following formula < f¯, g >=< f, g¯ >. We raise now the Problem 3.1. Give f ∈ Q′ does there exist a u ∈ V such that 4) a (u, v) =< f, v¯ > for all v ∈ V. 18
21
3. General Elliptic Boundary Value Problems
19
We shall show later that large classes of elliptic problems can be put in this form. Definition 3.1. The space N will consists of all u ∈ V such that the mapping v → a(u, v) is continuous on V with the topology of Q.
22
Since V is dense in Q we can extend this mapping to Q. Hence for every u ∈ N we have an Au ∈ Q′ such that a(u, v) =< Au, v >.
5)
The mapping A : N → Q′ is linear. On N we introduce the upper bound topology to make the mapping N → V and A : N → Q′ continuous. We ask now the Problem 3.2. Is the mapping A onto Q′ ? Lemma 3.1. Problem 1 is equivalent to problem 2. Proof. Let f ∈ Q′ and let u be a solution of problem 1, i.e., a(u, v) =< f, v >. Hence the mapping v → a(u, v) =< f, v > is continuous on V with the topology of Q. Hence u ∈ N. Further < Au, v >= a(u, v) =< f, v > for all v ∈ V, and since V is dense in Q, Au = f . Conversely, let f ∈ Q′ be given and u ∈ N be such that Au = f . Then a(u, v) =< Au, v >=< f, v >, for all v ∈ V, i.e., u is a solution of problem 1.
3.2 We now consider certain sufficient handy condition so that A should be an isomorphism of N onto Q′ Definition 3.2. We shall say that the sesquilinear a(u, v) is elliptic on V, or is Velliptic, if there exists an α > 0 such that Re(a(u, u)) ≥ αu2V for all u ∈ V. Theorem 3.1. Let V, Q, a(u, v) be as given in § 3.1. If a(u, v) is Velliptic, then A is an isomorphism of N onto Q′ . Proof. Let
23
1 a(u, v) + i a(v, u) 2 1 a2 (u, v) = i a(u, v) − a(v, u) . 2 a1 (u, v) =
and
3. General Elliptic Boundary Value Problems
20
Then a1 (u, v) and a2 (u, v) are hermitian and a(u, v) = a1 (u, v) + ia2 (u, v). Put
[u, v] = a1 (u, v).
Since a(u, v) ≤ CuV vV , it follows that [u, u] ≤ Cu2V . On account of the Vellipticity, [u, u] = Re a(u, u) ≥ αu2V . Hence the form [u, v] defines on V an Hilbertian structure equivalent to the one defined by (u, v)V . Now, any f ∈ Q′ defines a continuous semilinear function on V and hence there exists K f such that < f, v >= [K f, v],
K ∈ L (Q′ , V).
For a fixed u ∈ V, the mapping v → a2 (u, v) is a semilinear continuous mapping on V, hence a2 (u, v) = [Hu, v]. Further H is hermitian for the scalar product defined by [u, v]. For [Hu, v] = a2 (u, v) = a2 (v, u) = [Hv, u] = [u, Hv]. Hence and we have to solve i.e.,
a(u, v) = [u, v] + i[Hu, v], a(u, v) =< f, v >= [K f, v], (1 + iH)u = K f.
From Hilbert space theory, we know that if H is hermitian (1 + iH) is nonsingular. Hence u = (1 + iH)−1 K f, which proves that A is an isomorphism.
Lecture 6 3.3 Examples: ∆ + λ, λ > 0, ∆ =
n ∂2 u P . 2 i=1 ∂xi
Let Ω be an open set in Rn and H 1 (Ω), H01 (Ω) be as defined before. Let V be a closed subspace of H 1 such that H01 ⊂ V ⊂ H 1 . The metric on V is one induced by H 1 : (u, v)V = (u, v)1. Let Q be L2 (Ω) with the involution f → f¯. Then V ⊂ Q and is dense in Q. On V consider the sesquilinear form a(u, v) = (u, v)1 + (u, v), λ > 0. Then a(u, v) is continuous on V × V and Re(a(u, u)) = u20 + λu21 ≥ min(1, λ)(u20 + u21 ) = αkuk21 ,
α > 0.
Hence a(u, v) is Velliptic. Hence, for a given f ∈ L2 (Ω) = Q′ we have u ∈ V such that a(u, v) =< f, v¯ > for all v ∈ V. We determine N and A explicitly in this case. Proposition 3.1. 1) A = −∆u + λu for u ∈ Nλ > 0 and u ∈ V, ∆u ∈ L2 2) u ∈ N ⇔ (−∆u, v)0 = (u, v)1 for all v ∈ V.
Proof. We know u ∈ N if and only if u ∈ V and the mapping v → a(u, v) is continuous on V with the topology of Q. Further since a(u, v) is Velliptic,
21
24
22 for f ∈ L2 there exists u ∈ N such that a(u, v) =< Au, v¯ >=< f, v¯ >. Let v = ϕ ∈ D(Ω). Then
Now,
(u, ϕ)1 + λ(u, ϕ)0 = (Au, ϕ)0 . ! ∂2 u ∂u ∂ϕ , =< −Σ 2 , ϕ > (u, ϕ)1 = Σ ∂xi ∂xi 0 ∂xi =< −∆u, ϕ > .
Hence < −∆u, ϕ¯ > +λ < u, ϕ¯ >=< Au, ϕ¯ >=< f, ϕ¯ > for all ϕ ∈ D(Ω). This means A = −∆ + λ and −∆u + λu = f . Since f ∈ L2 and u ∈ L2 we have ∆u ∈ L2 . Further, if u ∈ N, a(u, v) =< Au, v¯ > and hence (u, v)1 + λ(u, v)0 = (−∆u, v)0 + λ(u, v)0 which gives (u, v)1 = (−∆u, v)0. Conversely if u satisfies the above conditions, since −∆u + λu ∈ L2 the mapping v → a(u, v) = (u, v)1 + λ(u, v)0 = (−∆u + λu, v)0, is continuous on V in the topology induced by Q. Hence u ∈ N. Now we give a formal interpretation of u ∈ N. The correct meaning of this interpretation will be brought out later on. Assuming the boundary Γ of Ω to be smooth, we have, by a formal Green’s formula, Z Z ∂u v¯ dσ + (u, v)1 ∆u.¯vdx = (−∆u, v)0 = − Γ ∂n Ω ∂u where is the normal derivative. However if u ∈ N, by proposition 3.1, we ∂n have (−∆u, v)0 = (u, v)1 . R ∂u Hence u ∈ N implies v¯ dσ = 0. ∂n We now take particular cases of V and interpret this formal result. 1) Let V = H 1 . u ∈ H 1 is not a boundary condition, neither is ∆u ∈ L2 . R ∂u v¯ d = 0 for every v ∈ H 1 , is a boundary condition, and However, Γ ∂n ∂u = 0, i.e., u ∈ N implies the normal derivative formally this means ∂n vanishes.
25
23 2) Let V = H01 .u ∈ H01 implies u = 0 on the boundary, and hence is a boundary R ∂u v¯ dσ is always zero condition. ∆u ∈ L2 is not a boundary condition and Γ ∂n for v ∈ H01 .
26
3) Let Γ1 be the subset of Γ and define V to consist of function u ∈ H 1 such that γu = 0 on Γ1 .V is a closed subspace of H 1 . u ∈ N if and only if u ∈ V, that is to say, γu = 0 on Γ1 ; this is a boundary condition, ∆u ∈ L2 which R ∂u v¯ dσ = 0 for v ∈ V, but since γv = 0 is not a boundary condition, and Γ ∂n R ∂u v¯ dσ = 0 for all v ∈ V. This means again formally on Γ1 , this means Γ−Q ∂n ∂u ∂u = 0 on Γ − Γ1 . So formally the condition is u = 0 on Γ1 and = 0 on ∂n ∂n Γ − Γ1 . We call 1), 2) and 3) weak homogeneous, Neumann, Dirichlet and mixed DirichletNeumann problems respectively. We may state the above results in the Theorem 3.2. If λ > 0, Ω an arbitrary open set in Rn , then the equation −∆u + λu = f with f ∈ L2 (Ω) has a unique solution with homogeneous boundary data. Remarks . Nonhomogeneous problems: Corresponding to the homogeneous problems considered above, we may consider nonhomogeneous ones in which not necessarily vanishing boundary values are prescribed. We shall show formally that this can be reduced to a homogeneous case together with a problem of first order partial differential equation. Problem 3.3. Given F ∈ L2 and G ∈ V such that ∆G ∈ L2 determine u such that −∆U + λU = F and U − G ∈ N. Theorem 3.3. Problem 3.3 admits a unique solution for λ > 0. Proof. Put u = U − G. Then we have to seek u such that −∆u + λu = F − (−∆ + λ)G = f, say . 2
Since f ∈ L , there exists unique u ∈ N by theorem 3.2.
∂u on In the case V is as in examples 1), 2) and 3) respectively, this means ∂n ∂U ∂G Γ, G = U on Γ and U = G on Γ1 and = on Γ − Γ1 respectively. The ∂n ∂n
27
24 above solution of problem 3.3 implies then that if we wish to determine U with ∂U , U, given on the boundary, we have only to determine G ∈ L2 satisfying ∂n ∂U ∂G = and U = G on the respective parts of the boundary. ∆G ∈ L2 and ∂n ∂n Remark. If we take V = H01 , Q = H −1 so that Q′ = H −1 we have Theorem 3.4. Given a distribution T ∈ H −1 , there exists a unique solution U ∈ H01 such that −∆u + λu = T . Remark. Roughly speaking we may say that the boundary conditions are introduced by means of the following two conditions :(a)u ∈ V, (b)(−∆u, v)0 = (u, v)1. The two extreme cases are H01 (Dirichlet) and H 1 (Neumann) wherein, in the first case, only u ∈ V is the boundary condition, and in the second one, (−∆u, v)0 = (u, v)1 is the boundary condition. The condition u ∈ V may be considered to be stable and the other one unstable. Heuristically this may be justified as follows: if we consider smooth functions in H 1 (Ω) such that ∂u = 0, on completion this property no longer holds, so we may say this con∂n dition is unstable, while in the second case, the completion of smooth functions vanishing on boundary still possesses this property in a weaker sense. Exercise 1. With the hypothesis as in theorem 3.1, if a(u, v) is Velliptic, the existence of u ∈ V such that a(u, v) =< f, v¯ > for all v ∈ V and any f ∈ Q′ can be carried on the following lines: Since the mappings v →< f, v¯ > and v → a(u, v) are continuous on V with the topology induced by Q, there exists K f and A˜ f in V such that ˜ v)V < f, v¯ >= (K f, v)V . a(u, v) = (Au, ˜ = K f . This is proved if we Hence to solve the problem we require Au prove A˜ is an isomorphism of V onto V. Exercise 2. The same results as in theorem 3.1 is true on a weaker assumption that a(u, v) ≥ αu2V , α > 0.
28
Lecture 7 3.4 Hitherto we considered the particular case where H01 ⊂ V ⊂ H 1 . Now we shall consider a more general case in which D ⊂ V ⊂ Q ⊂ D ′ , D being dense in Q, but not necessarily in V. Involution in Q is as before, viz. f → f¯. Let a(u, v) be a continuous sesquilinear form on V. In this situation the operator A and the space N associated with a(u, v) can be characterized in another way as follows. For a fixed u ∈ V, the mapping ϕ → a(u, ϕ) for ϕ ∈ D is a continuous semilinear form on D(Ω) and hence defines an element A u ∈ D ′ (Ω) so that < A u, ϕ¯ >= a(u, ϕ). This defines a mapping A : V → D ′ (Ω). Let η be the space of u ∈ V such that (a)A u ∈ Q′ and (b) < A u, ϕ¯ >= a(u, v) for all v ∈ V. On η we introduce the topology so as to make both the injection η → V and the mapping A : η → Q′ continuous.
29
Theorem 3.5. η = N and for u ∈′ N, A u = Au. Proof. 1) Let u ∈ N. Then v → a(u, v) is a continuous semilinear form on V with the topology induced by Q and a(u, v) =< Au, v¯ > with Au ∈ Q′ . This holds in particular if v = ϕ ∈ D(Ω). Hence a(u, ϕ) =< Au, ϕ¯ >=< A u, ϕ¯ > for all ϕ ∈ D(Ω). This means A u = Au and that A u ∈ Q′ . Hence < A u, v¯ >=< Au, v¯ >= a(u, v) for all v ∈ V and so u ∈ M. 2) Conversely, let u ∈ M. Then a(u, v) =< A u, v¯ > for all v ∈ V and A u = f ∈ Q′ . Hence a(u, v) =< f, v¯ > so that the mapping v → a(u, v) is continuous on V with the topology induced by Q. Hence u ∈ N and A u = f = Au.
25
30
26 Remark . In practice it is the operator A that is known a priori and A is the restriction of A to N. We agree however to denote A by A itself.
Generalizations. Let ν be an integer. If E is a topological vector space, let E ν be E × . . . × E, the topology on E ν being the product topology. Let V, Q be such that D(Ω)ν ⊂ V ⊂ QD ′ (Ω)ν . Let a(u, v) be a continuous sesquilinear form on V. As in before, we can define the operator A ∈ L (V, D ′ν ). The operator A on D(Ω) may be considered to be a generalisation of differential systems.
General examples: a) An interesting example of the above kind would be where V is the set of functions continuous on a given discrete set in Rn . The solutions of this problem may be considered to be finite difference approximation to boundary value problems. b) Let Ω be an open set in Rn and A1 , . . . , Aν be differential operators with constant coefficients. Let V be such that H 0 (A, Ω) ⊂ V ⊂ H 1 (A, Ω). Let Q = L2 (Ω). Then D(Ω) ⊂ V ⊂ Q ⊂ D ′ (Ω) and D(Ω) is dense in Q. Let Z ν Z X g0 (x)u¯vdx gi j (x)A j (u)Ai (v)dx + a(u, v) = j,i=1
Ω
Ω
with go , gi j ∈ L∞ (Ω). a(u, v) is a continuous sesquilinear form on V. The P corresponding operator A = A∗i (gi j A j ) + g0 .
3.5 Green’s kernel. We have proved that in the case a(u, v) is Velliptic, the operator A is an isomorphism of N onto Q′ . Let G be the inverse operator of A. G is then an isomorphism of Q′ onto N. The restriction of G to D(Ω) is then a continuous mapping of D(Ω) into D ′ (Ω) and conversely the restriction of G to D(Ω) defines G uniquely D is dense in Q′ . Now, L. Schwartz’s kernel Theorem [3] states that any continuous mapping of D into D ′ is defined by an element of D ′ (Ω x ×Ωy ), the space of distributions on Ω x × Ωy . Thus G defines an element G x,y ∈ D ′ (Ω x × Ωy ). Definition 3.3. G x,y defined above is called the Green’s kernel of the form a(u, v) on V.
31
27
3.6 Relations with unbounded operators. Let Ω be an open set in Rn .V, Q be two vector spaces not necessarily of distributions, Q being a Hilbert space and V ⊂ Q. Let a(u, v) be a continuous sesquilinear form on V. As we have seen already in (§ 3.1), this defines a space N and an operator A : N → Q by identifying Q′ and Q. This operator in the topology induced on N by Q is an unbounded operator. Let a∗ (u, v) = a(v, u). On V, a∗ (u, v) is a continuous sesquilinear form. Let the spaces N and operator A associated with a∗ (u, v) be denoted by N ∗ and A∗ , i.e.,
32
u ∈ N ∗ ⇔ v → A∗ (u, v)
is continuous on V with the topology induced by Q and a∗ (u, v) =< A∗ u, v¯ >= (A∗ u, v)Q . We shall give a theorem establishing relationships between usual concepts associated with the unbounded operators and N, A and A∗ . Theorem 3.6. Suppose there exists λ > 0 such that Re a(u, v) + λv2Q ≥ αu2v for all Then (1) N is dense in Q. (2) A is closed. (3) A∗ is the adjoint of A.
u ∈ V.
(definitions will be recalled in the course of proof)
Proof. We first prove that A is closed. We have to prove that if un ∈ DA (the domain of definition of A) and if un → u in Q and Aun → f in Q, then u ∈ DA and Au = f . a(u, v) + λ(u, v) is a continuous sesquilinear form on V and the space and the operator associated with it are N and A + λ respectively. By assumption this form is Velliptic and hence by theorem 3.1, A + λ is an isomorphism of N onto Q′ = Q. Now, (A + λ)un → f + λ u in Q and hence un = (A + λ)−1 (A + λ)un → (A + λ)−1 ( f + λu ) in N. Hence un → (A + λ)−1 ( f + λu) in Q also and so u = (A + λ)−1 ( f + λu), and u ∈ N. Further Aun → Au in Q and so Au = f . Hence A is closed. Now we prove that N is dense in Q. We need prove if f ∈ Q and (u, f )Q = 0 for all u ∈ N. Then f = 0. Since (A + λ) is an isomorphism of N onto Q, there
33
28 exists w ∈ N such that (A + λ)w = f . Hence ((A + λ)w, u)Q = 0 for all u ∈ N. But ((A + λ)w, u)Q = (Aw, u)Q + λ(w, u)Q = a(w, u) + λ(w, u)Q . Taking u = w in particular, we get 0 = Re a(w, w) + λw2Q ≥ αw2 Q. Hence w = 0 and so f = 0. Now we prove that the adjoint of A is A∗ . The domain of the adjoint A˜ of A consists of u ∈ Q such that the mapping v → (Av, u)Q is continuous on N with the topology induced by Q. Since N is dense in Q, this mapping can be ˜ ∈Q extended to a linear form on Q and hence by Riesz’z theorem we have Au such that ˜ Q (Av, u)Q = (v, Au)
for
v ∈ DA
and
u ∈ DA˜ .
This defines A˜ on DA˜ . Since (Av, u)Q = a(v, u) = a∗ (u, v) = (A∗ u, v)Q = (v, A∗ u)Q
for v ∈ N
and u ∈ N ∗ ,
(1)
we have N ∗ ∈ DA˜ , and A˜ = A∗ on N ∗ . We need only prove now DA˜ ⊂ N ∗ . Let u ∈ DA˜ , then there exists u0 ∈ N ∗ such that (A∗ + λ)uo = (A˜ + λ)u, since ∗ A + λ is an isomorphism of N ∗ onto Q on account of Vellipticity of a∗ (u, v). Now, for all v ∈ N ((A + λ)v, u)Q = (v, (A˜ + λ)u)Q = (v, (A ∗ +λ)u0 )Q
= a(v, uo) + λ(v, uo )Q (by(1)) = (Av, uo ) + λ(v, uo )Q since v ∈ N and by definition of A.
Hence for all v ∈ N, ((A+λ)v, u −uo)Q = 0. Since (A+λ) is an isomorphism of N onto Q′ , this means u − u0 = 0, i.e., u ∈ N ∗ , which completes the proof.
Lecture 8 4 Complements on H m (Ω) 4.1 Estimates on Hom (Ω). Theorem 4.1. Let Ω be a bounded open set in Rn . Then there exists a c > 0 such that u0 ≤ cu1 for all uǫH01 (Ω). Proof. Since D(Ω) is dense in Ho1 we need prove the inequality for u = ϕǫD ∼ (Ω). Let ϕ = ϕ on Ω and 0 outside Ω in Rn . Since Ω is bounded, we have ∼
ϕ(x) =
Zx1
−∞
∂ ϕ(t, x2 , . . . , xn )dt = ∂x1
Zx1 a
∂ϕ (t, x2 , . . . , xn )dt ∂x1
where a and b are such that Ω is contained in the region determined by ]a, b[xRn−1. By Schwartz’s inequality ∼
2
ϕ(x) ≤ (x1 − a)
Zb a
≤ (b − a)
Zb a
Hence
R
∼
ϕ(x)2 dx ≤ (b − a)2
as it was to be proved.


∂ϕ (t, x2 , . . . , xn )2 dt ∂x1
∂ϕ (t, x2 , . . . , xn )2 dt. ∂x1
R b ∂ϕ ∂ϕ 2 dx, and so ϕ0 ≤ (b − a)  ≤ cϕ1  a ∂x ∂x1 1 29
34
4. Complements on H m (Ω)
30
Remarks. (1) From the above proof it is seen that the theorem remains true even if Ω is bounded only in any one direction. (2) The theorem is not true for H 1 (Ω). Thus, for instance, if we take u = 1, then uǫH 1 (Ω) and u0 = measure of and u1 = 0, so there does not exists c such that u0 ≤ cu1 . (3) The theorem may remain true however for some spaces V such that Ho1 ⊂ V ⊂ H 1 . Thus if Ω is as shown in the figure and V = uǫH 1 such that u(0, x2 ) = 0, then u0 ≤ cu1 .
(4) If u ∈ H0m (Ω), then u0 ≤ cum , uk ≤ cum , for k ≤ m − 1. Applications: We have already proved that for Dirichlet problems (V = H01 (Ω)) the operator −∆ + λ associated with the form a(u, v) = (u, v)1 + λ(u, v)o is an isomorphism of Ho1 onto Ho−1 for λ > 0. We now prove the Theorem 4.2. If Ω is bounded, then −∆ + λ is an isomorphism of Ho1 onto Ho−1 for λ > −α for certain α > 0. Proof. We look for values of λ for which a(u, v) is Velliptic, i.e.,
Re a(u, u) = u21 + λu20 ≥ γu21 .
Since Ω is bounded u21 ≥
1 2 u for some c > 0, and c2 1
u21 + λu20 = u21 + (λ − ǫ)u20 + ǫu20
≥ (1 + (λ − ǫ)c2 )u21 + ǫu20
≥ γkuk21 for positive γi f 1 + (λ − ǫ)c2 > 0,
1 − c2 −1 + ǫc2 . Choosing ǫ sufficiently small, we have α = such i.e., λ > 2 c2 that for λ > −α, a(u, v) is Velliptic and thus the theorem is proved.
35
4. Complements on H m (Ω)
31
Theorem 4.3. For every ǫ > 0, there exists c(ǫ) such that u2k ≤ ǫu2m + c(ǫ)u20 for all uǫH0m (Ω) and 0) ≤ k ≤ m − 1. ∼
Proof. Let u be the function defined in Rn which is equal to u on Ω an 0 else∼ ∼ ∼ ∼ where. We have then um = um. Let u = F (u) be the Fourier transform of u. By Plancherel’s theorem Z X ∼ ∼ ∼ (2π)2k ξ2k u2 dξ = u2k = u2k . u2k = p=k
36
Zn
To verify the stated inequality it is enough to prove that for ǫ > 0 there exists c(ǫ) such that Z Z Z (2π)2k ξ2k uˆ 2 dξ ≤ (2π)2mǫ ξ2m uˆ 2 dξ + c(ǫ) uˆ 2 dξ, Z Z c(ǫ) 2k 2 i.e., ξ uˆ  dξ ≤ (ǫ(2π)2m−2k ξ2m + )uˆ 2 dξ. (2π)2k This will be true if ξ2k ≤ ǫ1 ξ2m + c1 (ǫ)
f or
k ≤ m − 1.
Since k ≤ m − 1, for any ǫ1 > 0, ξ2k ≤ ǫ1 ξ2m for large values of ξ and for remaining necessarily bounded values of ξ, ξ2k − ǫ1 ξ2m is bounded by C1 (ǫ) say. Remark. The status of this theorem is different from that of the theorem 4.1: for it may be sometime true for H m (Ω) also. As we shall see later, this is connected with the problem of mregularity. For example, if Ω =]0, 1[, theorem 4.3 holds for uǫH m (Ω).
4.2 Regularity of the function in H m (Ω). Theorem 4.4. If 2m > n, H m (Ω) ⊂ E o (Ω) algebraically and topologically. Proof. Let uǫH m (Ω). We need prove that for every ϕǫD(Ω), v = uϕǫE o (Ω). Since v vanishes near the boundary of Ω, the function v˜ is in Hom (Rn ). Let v˜ be the Fourier transform of v˜ , then (1 + ξm )ˆvǫL2 . Now, vˆ = (1 + ξm )ˆv.
37
1 . 1 + ξm
4. Complements on H m (Ω)
32
Since 2m > n, Z Z
1 2 dξ = O(  1 + ξm
Z∞
rn−1 dr) < ∞. 1 + rn
o
Hence uˆ ǫL1 . That is to say v is continuous. If now u → 0 in H ′ (Ω) we have vˆ → 0 in L1 . Hence v → 0 in ξo (Ω) and so u → 0 in ξo Remark . Better results valid for more general classes of domains are due to 1 1 1 Soboleff. A typical result is if n ≥ 3, then uǫH ′ (Ω) =⇒ u ∈ Lq (Ω), = − , q 2 n for certain Ω. (viz., DenyLions [7] and also Schwartz [1]).
4.3 Reproducing kernels. Let v be a closed subspace of H m (Ω) such that Hom (Ω) ⊂ V ⊂ H m (Ω) and Q = L2 (Ω). Let a(u, v) be a continuous sesquilinear from on V. Assume now 2m > n. Hence in each class of functions vǫV, there exists a unique continuous function vo (say). Then, for fixed yǫΩ, the mapping v → vo (y) is a continuous semilinear form on V. Hence by Riesz’s theorem, there exists k(y)ǫV such that vo (y) = (k(y), v)V . The mapping y → k(y) is weakly continuous mapping of Ω into V. Definition 4.1. k(y) is called reproducing kernel in V (Aronszajn [2]). If a(u, v) is Velliptic we have by theorem 3.1 Lemma 4.1. For every yǫΩ there exists unique g(y)ǫV such that a(g(y), v) = (k(y), v)V and the mapping y → g(y) of Ω → V is weakly continuous. We now relate the V valued function g(y) with the Green’s operator a(u, v) in V. For every vǫV, we have a(g(y), v) = v(y). Hence, for any ϕǫD(Ω), a(ϕ(y)g(y), v) = ϕ(y)v(y). R Integrating over Ω, a(ϕ(y)g(y), v) = (ϕ, v)o . Hence Ω
Z a g(y)ϕ(y)dy, v = (ϕ, v)o Ω
38
4. Complements on H m (Ω) where
R
33
g(y)ϕ(y) by is a weak integral. Now since a(u, v) is Velliptic, given
Ω
ϕǫD(Ω), there exists uǫV such that Au = ϕ, a(u, v) = (ϕ, v)o for all vǫV, and u = Gϕ. Hence Z g(y)ϕ(y)dy. Gϕ = u = Ω
Theorem 4.5. Let RΩ be an open set in Rn and 2m > n. Let V, Q, a(u, v) be as above. Then Gϕ = g(y)ϕ(y) dy where g(y)V, and is given by a(g(y), v) = v(y). Ω
This is a particular case of Schwartz’s kernel theorem. The kernel G x,y defined by the operator G in 3.5 is g(y)(x). There is yet another way of defining the Vvalued function g(y). Let Q = L2 (Ω) ∩ εo (Ω). On Q we put the upper bound topology of L2 and εo . Since 2m > n any V such that Hom (Ω) ⊂ V ⊂ H m (Ω) is contained in εo (Ω), and hence in Q. Further since D(Ω) is hence in Q. If a(u, v) is a continuous sesquilinear Velliptic from on V, from theorem 3.1 it follows that there exists a space N ⊂ V and an operator A, such that A is an isomorphism of N onto Q′ . Now Q′ = (L2 (Ω))′ + (εo (Ω)) = L2 (Ω) + ε′o (Ω) where ε′o (Ω) is the space of measures with compact support. Let G be the inverse operators of A; G is an isomorphism of Q′ onto N. Then g(y) = G(δy ). Remark. G as defined here, has slightly different meaning from the one defined previously, but the abuse of language is justified since both of these are inverse of the restriction of the same operator A : V → D ′ , see § 3.4.
39
Lecture 9 5 Complete Continuity. 5.1 We recall the definition of a completely continuous operator. Let E and F be two Hilbert spaces, then a continuous linear mapping U of E into F is said to be completely continuous if for any sequence un → 0 weakly in E, U(un ) → 0 strongly in F or equivalently for bounded set B in E, U(B) relatively compact.
40
Theorem 5.1. Let Ω be a bounded open set in Rn . Then the injection Ho1 (Ω) → L2 (Ω) is completely continuous. Proof. We have to prove that if uk → 0 in Ho1 (Ω) weakly, then uk → 0 strongly in L2 (Ω). Let u˜ k be the extension of uk to Rn equal to uk on Ω and 0 elsewhere. Then uk → 0 weakly in Ho1 (Rn ) and hence weakly in L2 (Rn ). Let uˆ k be the Fourier transform of uk , i.e., uˆ k (ξ) = (uk , e2πixξ )o . Since Ω is bounded for every ξ, e2πixξ ∈ L2 (Ω) and hence for fixed ξ, uˆ k (ξ) → 0. Further uk is weakly bounded in Ho1 (Ω) and hence bounded in Ho1 (Ω). So uk o ≤ co , uk 1 ≤ c1 . Hence, by Schwartz’s inequality uˆ k (ξ) ≤ c2 . R To prove uk → 0 strongly in L2 we need prove uˆ k (ξ)2 dξ → 0. Now Z Z Z uˆ k (ξ)2 dξ. uˆ k (ξ)2 dξ + uˆ k (ξ)2 dξ = ξ≥R
ξ 0 we shall prove that we can choose R so large that the second term is less than ∈ /2, and then that we can choose ko such that for k = ko , the first term is less than ∈ /2. This will complete the proof. Now 34
41
5. Complete Continuity. Z
ξ≥R
35
uˆ k (ξ)2 =
Z
(1 + ξ2 )uˆ k (ξ)2 .
ξ≥R
1 ≤ 1 + R2
Z
1 dξ 1 + ξ2
(1 + ξ2 )uˆ k (ξ)2 dξ
ξ≥R
kuk k1 c3 ≤ ≤ . 1 + R2 1 + R2 c3 0.
6.2 Theorem 6.1. Let Ω be a bounded open set in Rn , gi j be constants and V = H 1 (Ω). A necessary and sufficient condition that Re(u, u)g ≥ αu21 for all u ∈ H 1 (Ω) is that
X
(gi j + g¯ i j )pi p¯ i for all complex (pi ).
(1) (2)
46
6. Operators of order 2
40
Proof. (a) Necessity. Let u(x) =
n P
pi xi . Because Ω is bounded u(x) ∈ H 1 (Ω). Hence
i=1
by (1)
Z X Z X 2 Re gi j pi p¯ i dx ≥ α dx, pi  Ω
i.e.,
Re
Ω
P
gi j p j p¯ i ≥ α
P
pi 2
which is (2).
(b) Sufficiency. From (2) we have X ∂u X ∂¯v ∂u (x) (x) ≥ α  (x)2 a.e. (gi j + g¯ ji ) ∂x j ∂xi ∂xi
47
Integrating over XZ
(gi j + g¯ ji )
∂u ∂u¯ dx ≥ αu21 ∂x j ∂xi
Ω
Re(u, u)g ≥ αu21 .
i.e.,
n
Theorem 6.2. Let Ω = R , and gi j be constant. Then a necessary and sufficient condition in order that (1) holds is that X X ξi2 for real ξi and for some α > 0 (3) Re gi j ξi ξ j ≥ α
(We observe (2) =⇒ (3), but converse is not true, e.g., the example quoted above). Proof. By Fourier transform X
Z
gi j 2πiξi uˆ .2πiξ j uˆ dξ Z X = 4π2 gi j ξi ξ j uˆ 2 dξ.
(u, u)g =
Hence (1) is equivalent to Z X 2 Re ξi ξ j uˆ (ξ) dξ ≥ αξ2 uˆ (ξ)2 dξ for all u ∈ H 1 ij
(4)
6. Operators of order 2
41
P Let p(ξ) = Re( gi j ξi ξ j ) − αξ2 . Form (4), (1) is equivalent to Z P(ξ)uˆ (ξ)2 dξ ≥ 0
(5)
We have to prove (5) holds if and only if P(ξ) ≥ 0. Sufficiency is trivial. To see the necessity if P(ξo ) < 0, P(ξ) < 0 in a certain neighbourhood and then to obtain a contradiction we need take u the Fourier transform of which has support in this neighbourhood. The following problem however is not answered: If xi ∈ H 1 (Ω), (Ω) of capacity > 0 is (2) necessary in order that (1) holds for u ∈ H 1 (Ω).
6.3 V = Ho1 (Ω), gi j constant. Theorem 6.3. Let V = Ho1 (Ω) and gi j be constant. A necessary and sufficient condition in order that Re(u, u)g ≥ αu21 for all u ∈ Ho1 (Ω) for some α > 0 is that Re
X
X ξ2 for all ξi ∈ Zn . gi j ξi ξ j ≥ α
(6) (7)
Proof. In order to apply theorem 6.2 we prove that (6) implies that (1) holds for u ∈ Ho1 (Rn ) = H 1 (Rn ). We require a lemma. We may assume without loss of generality that the origin is in Ω. Further, we observe ∪λΩ = Rn . Lemma 6.1. (6) holds if and only if Re(u, u)g ≥ ∂u21 for all u ∈ H01 (Ω) for all λ. Proof. Let u ∈ H01 (Ω). Define uλ (x) = u(λx) for x ∈ Ω.
It is easily seen that uλ ∈ Ho1 . From (6) we get X Z XZ ∂uλ ∂u¯ λ dx ≥ α Re gi j ∂x j ∂xi
(8)
Ω
Since
Ω
2 ∂uλ dx ∂xi
∂u(λx) ∂uλ (x) = λ , from (8) we get ∂xi ∂xi X Z XZ ∂u(λx) ∂u(λx) gi j Re dx ≥ α ∂xi ∂xi Ω
2 ∂u(λx) dx. ∂xi
48
6. Operators of order 2
42
Putting λx = y, we get the required inequality and lemma (6.1) is proved. Returning to the proof of theorem, let ϕ ∈ D(Rn ). There exists λ such that K ⊂ λΩ. Then ϕ ∈ D(Rn ). and hence ϕ ∈ Ho1 (λΩ). This means Re (u, u)g ≥ αu2 , for all ϕ ∈ D(Rn ). Since D(Rn ) is dense in Ho1 (Rn ), we have proved Re (u, u)g ≥ αu21 for all u ∈ Ho1 (Rn ). Theorem 6.2 then gives (7).
6.4 Some problems with variable coefficients : V = H01 (Ω). Theorem 6.4. Let Ω be any open set in Rn and gi j be continuous. If Re(u, u)g ≥ αu21 for all u ∈ H01 (Ω), then Re
X
(9)
n X ξ for all (ξ ) ∈ Rn . gi j (xo )ξi ξ j ≥ α i i i=1
Proof. Given any ∈> 0, let B be a neighbourhood of xo such that (u, u)g(xo) − (u, u)g ≤∈ u21 for all u ∈ Ho1 (B). We need choose Bǫ such that gi j (x) − gi j (xo ) are sufficiently small. (9) gives then Re(u, u)g ≥ (α− ∈ u21 ) for all u ∈ H01 (β∈ ). From theorem 6.3, it follows that X X Re gi j (xo )ξi ξ j ≥ (α − ξ) ξ2i . Since this is true for arbitrarily small ∈, we have X X Re gi j (xo )ξi ξ j ≥ (α − ξ) ξi 2 . Regarding the sufficiency of the above condition, we have
49
6. Operators of order 2
43
Theorem 6.5. Garding’s inequality. If Re Σgi j ξi ξ j ≥ αΣξi 2 for some α > 0 ¯ and Ω is bounded then there exists λ > 0 such that for all x ∈ Ω Re(u, u)g + λu20 ≥ αu21 for all u ∈ Ho1 (Ω). We do not prove this. For a proof, see Yosida [21]. We have a general sufficient condition Theorem 6.6. If Σ(gi j + g¯ ji )p j p¯ i ≥ αΣpi 2 ) for some α > 0 and pi complex a.e. in Ω, then Re(u, u)g ≥ αu21 for all u ∈ H 1 (Ω). Having seen some cases when Re(a(u, u)g) ≥ αu21 we see now some examples when different forms a(u, v) giving rise to the same operator A are Velliptic. 1) Let a(u, v) = (u, v)g + (go u, v)o with go (x) ≥ β > 0. Then
Re(a(u, u)) ≥ αu21 + βu0 2 ≥ min(α, β)u21. Hence a(u, v) is Velliptic for any V such that Ho1 ⊂ V ⊂ H 1 . ! ∂u , v , gi real constants, go (x) ≥ 2) Let a(u, v) = (u, v)g + (go u, v)o + Σ gi ∂xi o 1 1 β > 0. Let V = H0!(Ω). Let V = H0 (Ω). We first observe that for ∂u = 0. For, if ϕ ∈ D(Ω), by integration by parts u ∈ H01 (Ω) Re u, ! !∂xi ∂ϕ ∂ϕ ,ϕ = ϕ and since ∂xi ∂x i o o ! ! ! ∂ϕ ∂ϕ ∂ϕ ,ϕ = , ϕ + ϕ, Re ∂xi ∂xi ∂xi o o o ∂ϕ ) = 0 for all ϕ ∈ D(Ω). Since D(Ω) is dense in Ho1 (Ω)we ∂xi have the result for u ∈ Ho1 (Ω). Hence Re(a(u, u)) = Re(u, u)g + ℜ(go u, u). Hence a(u, v) is Ho1 (Ω) elliptic. we have Re(ϕ,
50
Lecture 11 6.5 We now consider another kind of sesquilinear forms giving rise to the same n P ∂ ∂ ∂ ∂ operator A = (gi j (x) ) + gi (x) + gi + go (x). ∂xi ∂x j ∂xi i, j=1 ∂x j Let Ω be an open set with the boundary Γ having a C 1 (n − 1) dimensional P piece Σ. Let γ u be the extension of functions in H 1 (Ω) to as defined in § 2.4. On H 1 (Ω) consider the sesquilinear form ! X Z X ∂u γ uγ udσ, ,v + gi (go u, v)o + a(u, v) = (u, v)g + ∂xi o Σ
where dσ is the intrinsic measure on Σ. The operator associated with it is the same A as before. To consider the ellipticity of this from we require some definitions. Definition 6.1. Let Ω be a bounded connected open set; we shall say that Ω is of Nykodym type if there exists a constants P(Ω) > 0 such that the following inequality holds for all u ∈ H 1 (Ω). Z Ω
1 u dx − mesΩ 2
Z
udxP(Ω)u21.
(1)
The inequality (1) is called Poincare inequality. We admit without proof the Theorem 6.7. Every Ω with “smooth boundary” is of Nykodym type. (For proof, see Deny [7]). 44
51
45 Another interpretation of the inequality (1 ) is obtained by observing that Z Z 2 1 udx u2 dx − mes Ω is the minimum of + co 0 for all constants c. For Z Z 2 u + c 0 = u2o + c¯ u dx + c u¯ dx + c2 mes Ω ! ! Z Z Z 2 1 1 udx = c + u dx c¯ + u¯ dx + u2o − mes Ω mes Ω Thus (1) means Inf u + c2o ≤ P(Ω)u21.
Theorem 6.8. Let Ω be a domain of Nykodym type with the boundary Γa(n−1) dimensional C 1 manifold. Then the form Z γ uγ v d σ a(u, v) = (u, v)g + Γ
is Velliptic on H 1 (Ω). Proof. Since Re(a(u, u)) = Re(a(u, u))g +
R
γu2 dσ ≥ αu21 +
R Γ
γu2 dσ to
prove the Vellipticity of a(u, v) it is enough to prove that there exists a β > 0 such that Z αu21 + γu2 dσ ≤ βu21 , or that
Z
γu2 dσ + u21 ≥ β1 u21.
R Let [u, v] = (u, v)1 + γuγvdσ. [u, v] is aR continuous sesquilinear form on H 1 (Ω) since [u, u] = 0 implies u21R = 0 and γ u2 dσ = 0, we have u = c, a constant for u21 = 0 and c = 0 for γu2 = 0. That is to say [u, u] = 0 implies u = 0. In fact, we have the Lemma. [u, v] defines a Hilbertian structure on H 1 (Ω). Assuming the lemma for a moment, we see √ √ that on account of the closed graph theorem, the two norms [u, v] and (u, v)V are equivalent. Hence [u, u] ≥ βu21 which was to be proved.
52
46 To complete the proof we have to prove the lemma, i.e., that under the scalar product [ ], H 1 (Ω) is complete. ∂uk ,i = Let uk be a Cauchy sequence for the scalar product [ ]. Then ∂xi 1, . . . , n, and γuk are Cauchy sequences in L2 (Ω), and L2 (Γ) respectively. Hence ∂u → fi , i = 1, . . . , n in L2 (Ω) and γuk → g in L2 (Γ). Since Ω is of Nykodym ∂xi type from (1), we have Z Z 1  u dx2 2 dx ≤ Pu21 uk − mes Ω Ω Z Z 1  uk − ck 2 ≤ Puk 21 where ck = i.e., uk dx. mes Ω
53
Ω
Since uk is a Cauchy sequence in L2 (Ω), uk − ck is a Cauchy sequence in ∂uk ∂v = lim = fi . Hence uk − ck → v L2 (Ω). Hence uk − ck → v in L2 (Ω) and ∂xi ∂xi in H 1 (Ω) and so γ(uk − ck ) → v in L2 (Γ). Since γuk → g in L2 (Γ), ck → c. However uk = (uk − ck ) + ck . Hence uk → v + c in H 1 (Ω) under the norm [ ], which proves the lemma.
6.6 Formal interpretation: If Ω is of Nykodym type with a smooth (n − 1) dimensional boundary Γ, we have just proved that the form a(u, v) = (u, v)g + (γu, γv)0 !is elliptic on H 1 (Ω). P ∂ ∂ gi j (x) and u ∈ N implies The operator A that it defines is A = − ∂ xi ∂ xj a(u, v) = (Au, v)o for all v ∈ V. Now formally, Z Z ∂u A u v¯ d x = a(u, v) + v¯ d σ ∂ηA Ω
∂u ∂u P = gi j cos(n, xi ), (n, xi ) being the angle between the outer ∂ηA ∂x j ∂u = 0. normal and xi . Thus u ∈ N implies formally ∂ηA
where
6.7 Complementary results. Boundary value problems of oblique type for Ω = {xn > 0}. For general theory,
54
47 see Lions [4]. Let Γ be the boundary of Ω : {xn = 0}. We recall the definition of the spaces H α (Γ) for α real defined in § 2.5. H α (Ω) = { f L2 (Γ)} such that (1 + ξα ) fˆ ∈ L2 (Γ), where fˆ is the Fourier transform of f . We have proved in 1 ¯ 2.4, that there exists a unique mapping γ : H 1 (Ω) → H 2 (Γ) which on D (Ω) is the restriction to Γ and this mapping is onto. Theorem 6.9. The dual of H α (Γ) is H −α (Γ). Proof. Let F (H α (Γ)) be the space of Fourier transforms of H α (Γ). F (H α (Γ)) consists of functions fˆ ∈ L2 (Γ) such that (1 + ξα ) fˆ ∈ L2 (Γ). Hence its 1 dual consists of functions gˆ ∈ L2 (Γ) such that gˆ ∈ L2 (Γ), i.e., 1 + ξα (1 + ξ−α ) gˆ ∈ L2 (Ω). Hence the dual of F (H −α (Γ)) is F (H −α (Γ)) which proves the theorem. Let Λ =
n−1 P i=1
αi
∂ with αi real constants. We call Λ a tangential operator. ∂ xi 1
−1
Lemma 6.2. Λ is a continuous linear mapping of H 2 (Γ) into H 2 (Γ). ∂ is a continuous linear mapping from ∂xi! ∂ 1 1 1 is continuous from F (H 2 (Γ)) into H 2 (Γ) into H − 2 (Γ) or that F ∂xi ! 1 ∂f 1 1 2 −2 ˆ 2 2 = F H (Γ) . Let f ∈ H (Γ). Then (1 + ξ ) f ∈ L (Γ), and so F ∂xi 1 1 2π i ξi fˆ ∈ H − 2 (Γ). Since the mapping g → ξi g is continuous, from F (H 2 (Γ)) 1 into F (H − 2 (Γ)) the proof is complete. Proof. It is enough to prove that
From lemma 6.2 we see that hΛγu, γvi is defined for all u, v ∈ H 1 (Ω). Further we have the Lemma 6.3. Re(Λγ u, γ u) = 0 for all u ∈ H 1 (Ω). For by Fourier transform R ∂ 2 π i ξi  γ uˆ 2 d ξ. Re h γu, γui = Re ∂xi Let a(u, v) = (u, v)1 + λ(u, v)0 + hΛ γ u, γvi for u, v ∈ H 1 (Ω). From lemma 6.2, we see that a(u, v) is a continuous sesquilinear form on H 1 (Ω). Lemma 6.4. If λ > 0, a(u, v) is H 1 (Ω) elliptic. For Re(a(u, u)) = u21 + λu20 ≥ min(λ, 1)u1. From theorem 3.1, we have the
55
48 Theorem 6.10. The operator associated with a(u, v) is −△ + λ and −△ + λ is an isomorphism from N onto L2 (Ω). To get a formal interpretation of the problem, we have to see that u ∈ N means. u ∈ N if and only if ((−△ + λ)u, v)0 = (u, v)1 + (Λ γ u, γu)o + λ(u, v)o. By Green’s formula, (−△ u, v)0 = (u, v)1 + formally
R ∂u v¯ . Hence u ∈ N implies Γ ∂x n
∂u (x1 , . . . , xn , 0) = Λγ u, a condition of oblique derivative. ∂xn
Lecture 12 6.8 Upto now we considered problems in which the space V was a closed subspace of H 1 (Ω). We wish to consider now some cases in which V is not closed in H 1 (Ω). Let Ω = {xn > 0}, p be the boundary of Ω and γ be the mapping of H 1 (Ω) → 1 H 2 (Γ) as defined in § 2.4. Let V = u ∈ H 1 (Ω) such that γ u ∈ H 1 (Γ). On V we introduce the norm  u 2V = u21 + γ uH 1 (Γ)
(1)
Lemma 6.5. (1) defines on V a Hilbert structure. Remark. V is not closed in H 1 (Ω). On V consider the sesquilinear form a(u, v) = (u, v)1 + λ(u, v)0 + (γu, γv)1 , λ > 0. Lemma 6.6. a(u, v) is continuous on V and is elliptic for λ > 0. Let Q = L2 (Ω). Then by theorem 3.1, a(u, v) determines a space N and an operator A which is an isomorphism of N onto L2 . To see what A is we observe a(u, v) = hAu, v¯ i for all ϕ = v ∈ D(Ω). Then a(u, ϕ) = (−△ u + λu, ϕ)o . Hence A = −△ + λ. Further u ∈ N if and only if u ∈ V, −△ u ∈ L2 (Ω) and (Au, v)o = a(u, v) for all v ∈ V. To interpret formally u ∈ N we see that from above we have (−△ u, v)o + λ(u, v)o = (u, v)1 + λ(u, v)o + (γ u, γ v)
49
56
7. Operators of order 2m
50
for all v ∈ V. Applying Green’s formula Z Γ
X ∂2 ∂u ′ (x , 0) γ v dx′ = − γ u v¯ dx′ , where x′ = x1 , . . . , xn−1 ), 2 ∂xn ∂x i i=1 n−1
∂u = −△ x′ u(x′ , 0). ∂xn Before leaving the study of second order equations, we allude to its connections with the theory of semigroups and to mixed problems.
for all v ∈ V. Hence u ∈ N if and only if
a) Let V be such that H01 (Ω) ⊂ V ⊂ H 1 (Ω) and Q = L2 (Ω). Let a(u, v) be a continuous sesquilinear form. Then by theorem 3.1, a space N ⊂ V and an operator A ∈ L (N, L2 ) is defined. If on N we consider the topology induced by L2 (Ω), A is an unbounded operator with domain N. If a(u, v) is elliptic, it is easily proved that there exists ξ so that (A + λ)I has an inverse (A + λ)−1 bounded in norm by 1/λ when λ > ξ, A is an infinitesimal generator of a regular semigroup. b) In mixed boundary value problems we have to consider the following problem: A family of sesquilinear forms Z X ∂u ∂¯v ai j (x, t) (a(u, v, t)) = ∂x j ∂xi are given where ai j (t) are continuous functions from R to L∞ with the weak topology of dual. Let V = H 1 (Ω) and Q = L2 (Ω) and let for every t, a(u, v) be Velliptic. Then for every t, a space N(t) and an operator A(t) is defined such that A(t) is an isomorphism of N(t) onto L2 (Ω). If f ∈ L2 (Ω) and u(t) ∈ N such that At u(t) = f , then u(t) is a continuous function from R into V.
7 Operators of order 2m 7.1 P Definition 7.1. An operator A = (−1)p D p (a pq (x)Dq ), a pq ∈ L∞ (Ω) is called ¯ if there exists an α > 0 such that uniformly elliptic in Ω m X X 2 m p q ¯ and ξ ∈ Rn . Re a pq (x) ξ ξ ≥ α ξi for all x ∈ Ω p, q=m
i=1
57
7. Operators of order 2m
51
We admit without proof (for a proof, see Yosida [21]). Theorem 7.1. Garding’s inequality.
58
If Ω is bounded and A is uniformly elliptic, then there exists a λ > 0 such that
where
Re a(ϕ, ϕ) + λϕ20 ≥ αϕ2m for all ϕ ∈ D(Ω) XZ a pq (x)Dq uD pv dx a(u, v) =
(2) (3)
Ω
7.2 Applications to the Dirichlet’s problem. Theorem 7.2. If Ω is bounded and A is uniformly elliptic, then a) (A + λ) is an isomorphism of Hom (Ω) onto H −m (Ω) for λ large enough; b) (A + λ) is an isomorphism for all λ except for a countable system λ1 , . . . , λn ; such that λn → 0. Proof. (A+λ) is the operator associated with a(u, v)+λ(u, v)o which on account of Garding’s inequality is elliptic on Hom (Ω), for large λ. Hence by theorem 3.1, A + λ is an isomorphism of Hom (Ω) onto H −m (Ω). Further since the injection Hom (Ω) → L2 is completely continuous, we have the second assertion.
7.3 To consider other boundary value problems and specially the Neumann problem it is useful to introduce the motion of mregularity. Let K m (Ω) be the space of all u ∈ L2 (Ω) such that D p u ∈ L2 ( ) for p = m. On K m (Ω) we define the norm u2Km = u2 + u2m .K m (Ω) is a Hilbert space. Trivially H m (Ω) ⊂ K m ( ). However, the inclusion can be strict. Definition 7.2. Ω is said to be mregular if H m (Ω) = K m (Ω) algebraically. For instance, Ω = Rn is mregular for H m (Rn ) = K m (Rn ) as is seen easily by Fourier transformation. Theorem 7.3. If Ω is mregular, then there exists a constant c such that u2k ≤ c (u21 + u2m ) for k = 1, . . . , m − 1.
59
(4)
7. Operators of order 2m
52
Proof. The injection of H m (Ω) into K m (Ω) is onto and continuous. Hence by the closed graph theorem, it is an isomorphism. And so um ≤ c1 (u20 + u2m ), which implies the inequalities (4). Now the problem arises whether if (4) holds Ω is mregular or not. If (4) holds the inclusion mapping is continuous, one to one, and its range is closed. We have to prove then that H m (Ω) is dense in K m (Ω). This is still an unsolved problem. we admit following theorems without proof. Theorem 7.4. Every open set with smooth boundary is mregular. Theorem 7.5. If the injection H 1 (Ω) → L2 (Ω) is completely continuous, then Ω is mregular. Definition 7.3. Ω is strongly mregular, if (a) it is mregular, and (b) for every ∈> 0, there exists a c(∈) such that u2k ≤∈ u2m + c(∈) u20 for k = 1, . . . , m − 1
(5)
for all u ∈ H m (Ω). Proposition 7.1. Ω = Rn is strongly mregular for every m. Proof. By Plancherele’s theorem, we have to prove that given any ∈> 0 there exists c(∈) such that Z Z 2 2 2k uk ∈ uˆ (ξ) ξ d ξ ≤ (∈ ξ2m + c(∈))uˆ 2 dξ for k = 1, . . . , m−1, i.e., ξ2k ≤∈ ξ2m +c(∈) for k = 1, . . . , m−1, which follows from elementary considerations. We do not know however if there exists mregular domain which are not strongly mregular. Theorem 7.6. If the injection H 1 (Ω) → L2 (Ω) is completely continuous, then Ω is strongly mregular. Proof. By theorem 7.5, we see that Ω is mregular. We have now to prove the inequality (5). If it is not true there exists an ∈> 0 and a sequence ui ∈ H m (Ω) and a sequence ci → ∞ such that
Let vi =
ui 2k ≧∈ ui 2m + ci u20 .
ui (ui 2m
1
+ ui o ) 2
. Then vi ∈ K m (Ω) = H m (Ω).
60
7. Operators of order 2m
53
Further vi 2k ≥∈ +(ci − ∈)
ui 20 and c′i = ci − ∈→ ∞. ui 2m + ui 2
Hence vi 2 ≥∈ +c′i vi 20 . 2
Now vi  +
vi 2m
(6) m
= 1, so that vi are bounded in H (Ω) and hence vi k ≤ C. C− ∈ From (6) it follows that vi 2 ≤ , and hence vi → 0 in L2 (Ω). Therec′i fore there exists a sequence vµ converging weakly to 0 in H m−1 (Ω). Since the injection of H 1 (Ω) into L2 (Ω) is completely continuous vµ → 0 strongly in H m−1 (Ω), i.e., vk → 0 which contradicts (6).
Lecture 13 7.4 Applications Let a(u, v) = and
P
p,q≤m
R
a pq Dq (u)D¯ p v dx with a pq ∈ L∞ X A(u, v) = a pq Dq (u) D¯ p v dx
61
p,q=m
be the leading part of a(u, v). Theorem 7.8. Let (a)Ω be strongly mregular and (b) Re A(u, u) ≥ αu2m for some α > 0 and for all u ∈ H m (Ω). Then there exists λ such that Re a(u, u) + λu20 ≥ βu2m for some β > 0, and for all u ∈ H m . Proof. We have Re a(u, v) = Re A(u, u) + Re ρ(u, u) where ρ(u, v) =
X Z
p≤m,
a pq Dq uD p v dx q ≤ m and p + q ≤ 2m − 1.
Every term of ρ(u, v) is majorized by cumum−1 and so Re ρ(u, u) ≤ c1 u m um−1. Hence Re a(u, u) ≥ αu2m − c1 um um−1. We have then to prove that we can find λ such that there exists β satisfying X = αu2m − c1 um um−1 + λu20 ≥ βu2m 54
(1)
8. Regularity in the Interior
55
Since Ω is strongly mregular, using definition for any ∈> 0, there exists c(∈) such that um−1 ≤∈ um +c(∈)uo. Hence c1 umum−1 ≤ c1 ∈ u2m +c(∈ )umuo . Since Ω is mregular also um is equivalent to um + uo . Hence c1 umum−1 ≤ c2 ∈ (um + u20 ) + c′ (∈)(umuo + u1o). So X ≥ αu2m − 1 c2 ∈ (u2m + u2o ) − c′ (∈)(umuo + u2o + u2o . Now 2um uo ≤ ∈1 u2m + u2o ∈1 for any ∈1 . Hence ! ∈1 c′ (∈) 1 X ≥ α − c2 ∈ − u2m + (λ − c′′ (∈) + )u2o. 2 ∈1
62
First we choose ∈ so that α − c2 ∈= α2 . This determines c(∈) and c′ (∈). Then we choose ∈1 so small that ∈1 c′ (∈) < α/4, and then λ so large that 1 > 0. λ − c′′ (∈) + ∈1 Then X ≥ β1 (u2m + u2o ) and by mregularity of Ω, X ≥ βu2m as it was required to be proved.
8 Regularity in the Interior 8.1 Having established the existence and uniqueness of weak solutions of certain elliptic differential equations, we turn now to consider their regularity problem, that is to say, to see whether in the equation Au = f sufficient regularity of f will imply some regularity of u. First, we shall investigate when u is regular in the interior of the given domain Ω and next we shall consider when u is regular ¯ in some sense. in Ω To formulate the problem of interior regularity, we shall require some definitions of new spaces. We recall having defined in § 2.1, that H −r (Ω) = (H r (Ω))′ , for positive r. If 0 is an open set in Ω and if u is a function in Ω, u0 will denote the restriction of u to 0. r Definition 8.1. Let Ω be an open set in Rn .L r = Hloc (Ω) for any integer r, consists of functions u which for any relatively compact 0 ⊂ Ω are such that u0 ∈ H r (0), r integer ≥ 0 or < 0.
Let Kn be an increasing sequence of closures of relatively compact open sets 0n covering Ω. Let pn = u0n r be the norms in H0rn of u0n . p′n s are seminorms in L r . On L r we put the locally compact topology determined by the seminorms pn .
63
8. Regularity in the Interior
56
Definition 8.2. K r r, any integer will denote the space of u ∈ H r with compact support. On K r we put the natural inductive limit topology. A sequence un converges in K r if all un have their supports in a fixed compact A and all un → 0 in H r (A) . We see easily (Z r )′ = K −r . S r Proposition 8.1. E ′ (Ω) = K (Ω) algebraically. r∈Z
Proof. By definition ∪ K (Ω) ⊂ E ′ (Ω). We have to prove only that if T ∈ E ′ then T ∈ K r for some r. Now by a theorem of Schwartz, T ∈ E ′ (Ω) implies P p T = D f p where f p are continuous and have a compact support. Hence r
p≤µ
f p ∈ L2 (Ω) and by theorem 2.1, T ∈ H −µ (Ω). This means that T ∈ K r (Ω), where r = −µ. P p b p (x) D with b p ∈ E . Then B is a continuous Proposition 8.2. Let B = p≤µ
linear mapping of D, E , D ′ , E ′ into itself and also a continuous linear mapping of L r (Ω) into L r−µ (Ω) and K r (Ω) into K r−µ (Ω). Remark. It is not true, however, that B is continuous from H r to H r−µ .
Proof. The first assertion is trivial and the last one follows if we prove the middle one. Let f ∈ L r (Ω). Since b p ∈ E on Ω, b′p s and their derivatives are bounded on 0 so that it is enough to prove that Dµ f ∈ H r−µ (0). We may assume 0 to be an open set with smooth boundary. If µ < r and r > 0 we have the result from the definition. If µ > r, then by integration by parts for g ∈ H µ−r (0). hDµ f, gi = (−1)µ−r hDµ f, Dµ−r gi
exists and hence Dm f is a continuous linear function on H µ−r (0), i.e., Dµ f ∈ H r−µ (0).
8.2 Statements of theorems Let A=
X
(−1) p D p (a pq (x)Dq ), a pq ∈ E (Ω)
(1)
p,q≤m
Definition 8.3. A is uniformly elliptic in Ω if given any compact K ⊂ Ω we have an αK > 0 such that X Re a pq (x)ξ p ξq ≥ α K ξ2m for all x ∈ K and all ξ = (ξ1 , . . . , ξn ) ∈ Rn .
(2)
64
8. Regularity in the Interior
57
Remark. If A is uniformly elliptic, Garding’s inequality (Theorem 7.1) is true on every compact K. Let B=
X
p≤µ
b p (x) D p , b p ∈ E (Ω)
Definition 8.4. B is elliptic in Ω if
P
(3)
b p (x)ξ p = 0 with ξ ∈ Rn , implies ξ = 0.
We see at once that a uniformly elliptic operator is elliptic. The converse, ∂ ∂ +i is elliptic, however, is inexact. For example, in the case n = 2, B = ∂x1 ∂x2 but evidently is not uniformly elliptic being not of even order. In this and the next lecture, we shall prove the following two theorems on the regularity in the interior.
65
Theorem 8.1. Let A be a uniformly elliptic operator of order 2m in Ω. If for some T ∈ D ′ (Ω), A T ∈ L r (Ω) for some fixed r, then T ∈ L r+2m (Ω). Theorem 8.2. Let B be an elliptic differential operator of order µ in Ω. If for some T ∈ D ′ (Ω), B T ∈ L r (Ω) for some fixed r, then T ∈ L r+µ (Ω). From these theorems, the regularity in the classical sense will follows by the Corollary . Let B be an elliptic operator of order µ. If for some T ∈ D ′ , B T ‘ ∈ E , then T ∈ E . For B T ∈ E means B T ∈ L r for all µ. Hence by the theorems T ∈ L r+µ for every r. Hence all the derivatives of T will be functions which means T ∈ E . Before proving these theorems, we shall establish some connections between elliptic and uniformly elliptic operators. Using these, we shall prove that theorem 8.1 implies theorem 8.2, and then we shall occupy ourselves in the proof of theorem 8.1. Proposition 8.3. Let n ≥ 3. If B is elliptic, then B is of even order (See Schechter [15]). P Let x ∈ Ω and b p (x)ξ p = Q(ξ), That B is elliptic at x means that p=µ
only real zero of Q(ξ) is ξ = 0. We prove Q(ξ) must be of even degree. By a nonsingular linear transformation, if necessary, we may assume ξn has degree µ. Let ξ′ = (ξ1 , . . . , ξn−1 ) , 0 be a point in Rn−1 ⊂ C n . Let Q(ξn ) be the
66
8. Regularity in the Interior
58
polynomial in ξn obtained by substituting (ξ1 , . . . , ξn−1 ) by ξ′ . Then Q(ξn ) has n complex roots all of which have imaginary part , 0, for otherwise (ξ′ , ξn ) would be a real nontrivial zero of Q(ξ). Let π+ and π− be number of roots of Q(ξn ), with positive and negative imaginary parts respectively. On account of homogeneity of Q(ξ) if we put ξ′′ = −ξ′ , the number of positive roots of Q(ξ′′ , ξn ) will be π− and negative roots will be π+ . Let ξ′ , ξ′′ be joined by an are not passing through the origin which is possible because n − 1 ≥ 2. From a classical theorem the roots of Q(ξn ) can be continued from ξ′ to ξ′′ continuously. Now at no point on the are ξ′ , ξ′′ can ξn be real on account of ellipticity. So the positive roots at ξ′ are continued into positive roots at ξ′′ . Hence π+ = π− and µ = π+ + π− = 2π+ is even. Proposition 8.4. Let B be an elliptic differential operator with real coefficients. Then B is uniformly elliptic in Ω. P Proof. Let b p (x)ξ p = P(x, ξ). Since B is elliptic, P(x, ξ) = 0 implies p=µ
ξ = 0. Hence on ξ = 1, P(x, ξ) for fixed x keeps same sign which we may P(x, ξ) ≥ α for fixed x. If now K is any compact, P(x, ξ) assume > 0. Hence ξµ P(x, ξ) > α for all x ∈ K is continuous on the compact K × ξ = 1 and hence ξµ and all ξ , 0. If we put −ξ for ξ we get the inequality multiplied by (−1)µ hence µ is even. Proposition 8.5. Theorem 8.1 ⇒ Theorem 8.2 P P b p (x)D p . Let Proof. Let B = b p (x)D p . Put B¯ = p≤µ
¯ = A = BB
X
p=q=µ
67
p≤µ
b p (x) b p (x) D p Dq + · · ·
A is of even order and if P(x, ξ) is its associated form, then P(x, ξ) = Σb p (x)ξ p 2 . Further since B is elliptic, A also is. By proposition 8.4, A is then uniformly elliptic. Let now T ∈ D ′ such that BT ∈ L r . Hence ¯ T ∈ L r−µ . If theorem 8.1 is true, then T ∈ L r+µ proving theorem A T = BB 8.2.
Lecture 14 8.3 We now proceed to prove the theorem 8.1. First we prove a lemma of fundamental character which will help to establish an inductive procedure to prove the theorem.
68
Lemma 8.1. Let A be a uniformly elliptic differential operator of order 2m. Let u ∈ L m (Ω) and let Au ∈ L −m+1 (usually Au ∈ L −m only). Then u ∈ L m+1 . Proof. We prove the lemma in two steps. In the first one it will be shown that it is enough to prove the lemma assuming A and Au to have compact support, for which the assertion will be proved in the second step. Step 1. The lemma is equivalent to “if u ∈ K m and Au ∈ K −m+1 , then u ∈ K m+1 ”. The direct part is evident. To prove the converse, let u ∈ L m be such that Au ∈ L −m+1 . For any ϕ ∈ D(Ω), v = ϕ u ∈ L m . Now Au = A(ϕ u) = P Au+ D p Dq u. p ≤ 2m, q ≤ 2m−1
Since for q ≤ 2m − 1, Dq u ∈ L −m+1 and by assumption, Au ∈ L −m+1 , it follows that Av ∈ L −m+1 . Since ϕ has compact support, v and Av are in K −m+1 . Hence v ∈ K m+1 . Since this is true for every ϕ ∈ D(Ω), v ∈ L −m+1 . Now we prove the Step 2. If u ∈ K m and Au ∈ K −m+1 , then u ∈ K m+1 . ∂u ∈ H m (Ω). A general method to prove this, here ∂xi and in later occasions, will be to estimate the difference quotients of u. Let We have to prove that
59
69
60 1 u(x + h) − u(x) which exists if h is small h = (h, 0, . . . , 0) and uh (x) = h enough. Now we establish the following: P a) A(uh ) − (Au)h = (−1) p D p (ahpq Dq u(x + h)). b) A(uh ) is bounded in K −m . c) uh is bounded in H m . Assuming for a moment that a), b), c) are proved, we complete the proof of the lemma. Since uh is bounded in H m , it is a weakly compact and hence there exists hi → 0 such that uhi → g weakly in H m (Ω). On the other hand, ∂u ∂u uhi → in D ′ . Hence = g ∈ H m , i.e., u ∈ H m+1 . Since u has ∂ xi ∂ xi compact support, u ∈ K m+1 . Now the prove a), b), c). a) We verify easily that (α f )h − α f h = αh f (x + h). Applying this term by term in (Au)h − A(uh ) we obtain (a). b) On account of (a), to prove that A(uh ) is bounded in K −m it is enough to prove that (Au)h and each of D p (ahpq Dq u(x + h)) are bounded in H −m . Since ∂g ∂g Au = g ∈ K −m+1 , ∈ K −m and since (Au)h → , (Au)h is a con∂x1 ∂x1 vergent sequence in K −m and so is bounded, Further, since ahpq ∈ C ∞ as ∂ h → 0, ahpq → a pq (x) ∈ C ∞ uniformly on every compact set. Also ∂x1 Dq u(x + h) → Dq u(x) in L2 . Hence ahpq Dq u(x + h) converge in L2 . Since D p are derivatives of order than or equal to m, D p (ahpq Dq u(x + h)) converge in H −m , and hence in K −m . This proves (b). c) Since by (b), A(uh) is bounded, we have hA uh , u−h i ≤ A(uh)H −m uh m ≤ c1 uh m
On account of Garding’s inequality, we have Re a(uh , uh ) + λuh 20 ≥ αuh 2m on every compact set. As h → 0, we may assume that all uh have their support in a fixed compact set. Hence αuh 2m ≤ λuh 20 + c1 uh m .
70
61 Since u ∈ K m , uh →
∂u in L2 and so uh o is bounded. Further we have ∂xi c1 uh m ≤
c21 α h 2 + u m . 2 2
α Hence uh 2m ≤ c3 which proves uhm is bounded, and this completes the 2 proof of lemma 8.1. Lemma 8.2. Let u ∈ L m , and Au ∈ L −m+1+ j . Then u ∈ L m+ j+1 for every nonnegative integer j. Proof. Lemma 8.1 proves the lemma for the case j = 0; assuming it proved for integers upto j = 1, we prove it for j. Since L −m+ j+1 ⊂ L −m+ j , Au ∈ L −m+ j+1 implies that Au ∈ L −m+ j and hence by induction hypothesis that u ∈ L m+ j . Now D Au − A Du = A′ u where A′ is a differential operator of order 2m. Since Au ∈ L −m+ j+1 , D Au ∈ L −m+ j and since u ∈ L m+ j , Au ∈ L m+ j . Hence A(Du) ∈ L −m+ j . But Du ∈ L m as u ∈ L m+ j , j ≥ 1. Hence by lemma 8.1, Du ∈ L m+ j , i.e., u ∈ L m+ j+1 . Lemma 8.2 can be put in a slightly better form of Lemma 8.2′. Let u ∈ L m and Au ∈ L r , then u ∈ L r+2m . For, if r ≤ −m, the lemma is trivial and if r > −m we have r = −m + j and lemma 8.2′ follows at once from lemma 8.2. Now we complete the proof of theorem 8.1. We have to prove that if T ∈ D ′ and A T ∈ L r , then T ∈ L r+2m . Let O, O1 be two relatively compact open sets such that O ⊂ O1 ⊂ Ω. On account of a theorem of Schwartz, P T 0 = D p f p where f p are continuous in with support contained in O1 . By theorem 2.1, T 0 ∈ H −β (O). Now △m+β where △ is the Laplacian is on account of theorem 1.3 is an isomorphism of H m+β (O) onto Ho−m+β . Hence there exists u ∈ H m+β (0) such that △m+β u = T 0 . Applying lemma 8.2′ to △m+β , we have u ∈ L 2m+β (0) as the order of △ is 2(m + β), T 0 ∈ L −β , and u ∈ L m+β (0). Now (A T 0 ) = (A △m+β u) ∈ L r (0). The order of B = A △m+β is 4m + 2β and B is uniformly elliptic. As u ∈ Z 2m+β and Bu ∈ L r , applying lemma 8.2, we have u ∈ L r+4m+2β . Hence T = △m+β u ∈ L r+2m .
8.4 Some remarks. We remark that theorem 8.2 implies theorem 8.1 trivially though in the course of the proof, we proved theorem 8.1, before proving theorem 8.2. This raises
71
62 a vague question what properties which are true for uniformly elliptic differential equations can be upheld for the elliptic ones. For instance, we know for Dirichlet’s problem for bounded domains with smooth boundary Fredholm’s alternative holds if the operator is uniformly elliptic. In the case n = 2, we have the following counter example of Bicadze [4]. Consider the Dirichlet problem in the unit circle for the operator A = !2 1 ∂ ∂ +i . A is elliptic but is not uniformly elliptic, for the associ4 ∂x ∂y ated form has ξ2 − η2 , as its real part. We prove that the space of u such that Au = 0, u = 0 on the boundary is not finite dimensional and hence that Fred∂ ∂ ∂ ∂2 u = +i holm alternative does not hold. Au = 0 means 2 = 0, where ∂¯z ∂ x ∂ y ∂¯z ∂u and hence is holomorphic in the unit circle. Hence u = f + z¯g where f and ∂¯z g are holomorphic in the unit circle. But u = 0 on the boundary z¯z = 1 . Hence 0 = z u = z f + g on the boundary, and hence g = −z f everywhere as f and g are holomorphic in the unit circle. Thus u = (1 − z¯z) f (z) is a solution of the above problem for any holomorphic f (z) which shows that the space of u such that Au = 0, u = 0, on the boundary, is not finite dimensional. For complementary results, see Schechter [15] and a forthcoming paper by Agmon, Douglis, Nirenberg.
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Lecture 15 9 Regularity at the boundary. In the last lecture we dealt with the regularity in the interior or local regularity of the solutions of the elliptic differential equations. Now we wish to consider ¯ In a sense such solutions can be extended to the regularity of the solutions in Ω. the boundary. These should not be confused with problems in which boundary values to be attained are given. These will be considered in a general set up under the name of VisikSobolev problems.
9.1 Theorem 9.1. Let Ω be a bounded open set in Rn with a boundary which is an n − 1 dimensional C ∞ manifold. Let X Z a(u, v) = a pq Dq u D p v dx p,q≤m
¯ be given such that Re (a(u, u)) ≥ α u2m for some α > 0 with a pq ∈ E (Ω) and for all u ∈ H m (Ω). Let V = H m (Ω) and Q = L2 (Ω) and let A and N be as determined in theorem 3.1. If f ∈ L2 (Ω) and u ∈ N is such that Au = f , then u ∈ H 2m (Ω). Remark . If we do not take any condition on the boundary (eg., u ∈ N) then we can assert only that u ∈ L 2m (Ω) and cannot assert in general that u ∈ H 2m (Ω). The proof of this theorem is fairly complicated and will be broken in several steps. 63
73
9. Regularity at the boundary.
64
Step 1. First we reduce the problem to one in a cube in the following way: Let Oi be a finite covering by relatively compact open sets ofthe boundary Γ 0 < ǫi < 1, ¯ = such that there exists C ∞ homeomorphisms ψi of Oi to W −1 < ǫn < n 0 < ǫi < 1, i = 1, . . . n − 1 such that ψi maps 0i ∩ Ω onto W+ = i = 0 < ǫn < 1 1, . . . , n − 1 and Γ ∩ Oi onto W0 = W ∩ {ξn = O} . Since the regularity is the interior of u ∈ H m (Ω) has been already proved to prove that u ∈ H 2m (Ω), it remains only to prove the restrictions of u to Oi , i.e., uOi ∈ H 2m (Oi ). The homeomorphisms ψi define isomorphisms of H m (Oi ∩ Ω) onto H m (W+ ). Let u1 , v1 ∈ H m (W+ ). Define a0 (u, v) = a(ψ−1 (u1 ), ψ−1 (v1 )). (We drop i from the suffix). This definition is possible as A is an operator of local type, more precisely Z a(ψ−1 (u1 )), (ψ−1 (v1 )) = a pq (x) Dq (ψ−1 (u1 )D p (ψ−1 (v1 )dx.
74
O
a0 (u, v) is a continuous sesquilinear form on H m (W+ ). Now by theorem 3.1, a(u, v) = ( f, v)o for all v ∈ H m (Ω). Let, in particular, v vanish near the boundary of O − Γ ∩ O, and have its support in O. Then a(u, v) − ao (u, v) = ( f, v)0 . Hence if v1 is in H m (W+ ), and vanishes, near the boundary of W+ − Γ, then a(ψ(u), v) = (ψ( f ), v)0 , when ψ( f ) ∈ L2 (W+ ). 2m ǫ ǫ If we prove now that ψ(u) ∈ H (W ) for every ∈> 0, where W = 1− ∈< ǫ 0, where ∈ < xi < 1− ∈, i = 1, . . . , n − 1 Ωǫ = 0 < x < 1 . n
We shall prove this in two steps. First we consider the derivatives of u in the direction parallel to xn axis, which we call tangential derivatives and p denote them by Dτ (u) with p = (p1 , . . . , pn−1 , 0). By an induction argument and considering difference quotients as in the previous lecture, we shall prove that D p u ∈ H m (Ω). In the next section we shall consider Dmxn u. p=m
Proposition 9.1. Under the hypothesis of the reduced problem Dτp u ∈ p=m
H m (Ωǫ ).
P Proof. If u ∈ H m (Ω) is such that v = 0 near ∂Ω − , then we denote by 1 vh (x) = [v(x + h) − v(x)] which is defined for sufficiently small h, where h h = (h, 0, . . . , 0). We note two simple identities relating to vh . R R P 1. Ω uh v dx + Ω uvh dx = 0 where u and v both vanish near ∂Ω − . 2. (au)−h = a u−h + a−h u(x − h).
¯ vanishing near ∂Ω − P. u is in H m (Ω) and Let φ be a function in D(Ω) vanishes near the boundary. Using Leibnitz’s formula, it is seen at once that to prove u ∈ H m (Ω), it is enough to show that Dτ (φu) ∈ H m (Ω). We shall prove first that (φ u)−h is bounded. Let
76
b(u, v) = a(φ u, v) − a(v, φu) Z X = b pq (x) dq u D p v dx
(2)
p≤m,q≤m,p+q≤2m−1
where b pq (x) are products of derivatives of φ with a′pq s, and so vanish near P ¯ Using (1) and (2), we have ∂Ω − and are in D(Ω). a(φ u)−h , v) = [a((φ u)−h , v) + a(φu, vh)] − b(u, vh) − ( f, φvh )o .
Now we prove three lemmas. Lemma 9.1.  a((φ u)−h , v) + a(φ u, vh ) ≤ c1 um.
(3)
9. Regularity at the boundary.
66
Lemma 9.2.  b(u, vh )  ≤ c2 vm. Lemma 9.3.  ( f, φ vh )0  ≤ c3 vm. Using these in (3) with v = (φ u)−h and using ellipticity condition, we have α  (φ u)−h 2m ≤ c4  (φ u)−h m . Hence (φ u)−h is bounded in H m . Since bounded sets in H m are weakly compact, there exists a sequence hi such that (φ u)−hi converges weakly to a ∂u ∂u in D, we have ∈ Hm. function g ∈ H m . However since (φ u)−hi → ∂ xi ∂ x1 This proves the proposition 9.1, that DT u ∈ H m . It remains to prove the above lemma 9.1, 9.2 and 9.3. Proof of Lemma 9.1. a((φ u)−h , v)) + a(φu, vh) consists of sums of terms like Z Z a(x) Dq (φ u)−h D p v dx + X= a(x)Dq (φ u) D p vh dx p=m,q=m
= =
Z
Z
=−
a(x) Dq (φ u)−h D p v dx − a(x)Dq (φ u)−h D p v dx − Z
Z Z
p=m,q=m
((a(x)Dq(φ u)) D p v dx [a(x)Dq(φ u)−h
+ a−h (x)Dq (φ u)(x − h)] D p v dx a−h Dq (φ u)(x − h)D p v dx.
Since a−h are bounded and translations are continuous in H m and q ≤ m, we have, by using Schwartz’s lemma. X ≤ cD p vo ≤ c1 vm. Proof of Lemma 9.2. By definition, b(u, vh) =
PR
b pq (x)Dq uD p vn dx. If p ≤
p≤m,q≤m,p+q≤2m−1
m − 1, then as vR ∈ H m , D p vh is boundedR in L2 . If p = m we have q ≤ m − 1, and b pq (x)Dq uD p vh dx = − (b pq Dq u)−h D p vdx, and since ¯ and u ∈ H m (Ω), we have (b pq Dq u)−h bounded in L2 ; so that b pq ∈ D(Ω) at any rate b(u, vh) ≤ cvm. Proof of Lemma 9.3. This follows easily, for as h → 0, vh → Dτ v in L2 and hence ( f, φvh )o ≤ cDτ vL2 ≤ cvm.
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Lecture 16 9.3 We continue with the proof of theorem 9.1. Having proved that Dτ u ∈ H m (Ωǫ ) we proceed to prove now the Proposition 9.2. Dτp u ∈ H m (Ωǫ ) for p ≤ m. We prove this by induction. Assume it to have been proved for p = 1, . . ., ¯ vanishing near ∂Ω − P. r − 1; we prove it for p = r. Let φ be as before in D(Ω) To prove that Dµτ u ∈ H m (Ω∈ ), it is enough to prove that φDǫτ u ∈ H m (Ω), where φ is as in proposition 9.1. This will follow from the weak compactness argument used previously if we prove that (φDµτ u)−h is bounded in H m (Ω) and this itself will follow on account of ellipticity if we have proved that a((φDµτu)−h , v) ≤ cvm.
(1)
To prove (1), we write as before µ
µ
µ
a((φDu)−h, v) = (a(φDτ u)−h , v) + a((φDτ u), vh ) − a(Dµ u, vh ) − b(Dτ uvh ) where b(u, v) = a(φu, v) − a(uφv). We prove now in the following three lemmas saying that each of the terms above is bounded in H m (Ω). Lemma 9.4. a((φDµτ u)−h , v) + a(φDµτ u−h , v) ≤ c1 vm . Lemma 9.5. a(φDµτ uφvh ) ≤ c1 vm. Lemma 9.6. b(φDµτ uvh ) ≤ c3 vm. 67
78
68 We begin with 9.4. The expression to be estimated consists of sums of terms like Z X = (α(x)D p (φDrτ u)−h Dq v + (x)d p (φDµτ u)Dq vh dx Z = ((α[D p (φDrτ u)−h − (αD p (φDrτ u))−h ]Dq vdx by using (1) Z = α−h D p φDrτ u(x − h)Dq v by using (2) Since r = k−1, by induction hypothesis, Drτ u(x−h) is bounded in H 2m (Ω∈ ) −h
and α D
p
(φDrτ u(x
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2
− h) in L as p ≤ m, which proves 9.4.
Now we prove 9.5. We have a(Drτuφvh ) = a(Drτ uφvh ) − (−1)k−1 a(u (−1)k−1a(u, Drτ(φv)h ). From 9.2, we have
Drτ (φu)h)+
a(u, Drτ(φv)h ) = ( f, Drτ (φv)h )o , for r = k − 1 ≤ m − 1, and hence a(u, Drτ(φv)h ) ≤ cvm. It remains to consider the first difference, which consists of finite sum of terms Z Z q r k−1 p h αD Dτ uD φv dx − (−1) αDq Drτ uD p φvh dx Z= q≤m,p≤m,r=k−1
q≤m,p≤m,r=k−1
By induction hypothesis, if q ≤ m − 1, and p ≤ m − 1, D p φvh and D (Drτ φv−h ) are bounded in L2 . So we consider the terms where p = q = m. Now Z Z k−1 q p r h Drτ (αDq u)Dq φvh dx αD uD (Dτ φv )dx = (−1) p
and terms in Z with p = q = m become XZ j q q h βDr− τ D uD φv dx j≥1
which proves that Z ≤ cvm. R Finally we prove 9.6 b(Dr u, vh ) is a sum of terms like αDq Drτ uD p φvh dx. If p ≤ m−1, since vh are bounded in H m , D p vh are bounded in L2 . If p = m−1, Z Z βDq Drτ uDq φvh dx = − (βDq Drτ u)−h D p vdx
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69 and vh are bounded in L2 . Hence in any case b(Drτ u, vh ) ≤ cvm. Upto now we followed the proof given by Nirenberg [2]. In the following, the proof will be slightly more complicated than his, but will prove slightly more. Another proof is briefly indicated in Browder [5].
9.4 We still require a few preparatory lemmas before taking up the proof proper of the theorem. 2 Lemma 9.7. Let Ω =]0, 1[n be ncube and u ∈ L2 (Ω) be such that Dm i u ∈ L m ∂ (exactly mth derivatives in each variable). Then u ∈ H m (Ω). where Dm i = ∂xm i
Remark. This lemma is related to the theory of coercive forms of Aronszajn [1]. This lemma will be proved in two steps. (a) We prove first Dki ∈ L2 (Ω)for k ≤ m − 1.
Let K m (Ω) be the space of u ∈ L2 such that Dki u ∈ L2 (Ω). This is a space of type H(Ω, A) and hence is a Hilbert space with its usual norm. By using Fourier transforms, we see that on D(Ω) the K m metric and H m metric are equivalent. Hence the closure of D(Ω) in K m (Ω) is Hom (Ω). From prop. 1.3, we have K m (Ω) = Hom (Ω) ⊕ H , P where f ∈ H if and only if (−1)m D2m i f + f = 0. Now as the operator P (−1)m D2m i is uniformly elliptic, we have f ∈ E (Ω).
To prove (a) we have to prove, say Dk1 f (x1 , . . . , xn ) ∈ L2 (Ω), f ∈ E (Ω) ∩ K m . From a classical inequality, we have Z1−ǫ ǫ
Dk1 f (x1 , . . . , xn )dx1
Z1−ǫ 2 ≤ c ( f 2 + Dm 1 f  )dx1 , ǫ
where c is independent of ∈. Integrating over the remaining variables, we have Z Z 2 Dk1 f (x1 , . . . , xn )2 dx ≤ c ( f 2 + Dm 1 f  )dx Ωǫ
Ωǫ
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70 for every ∈> 0. Hence Z Dk1 f (x1 , . . . , x)2 dx ≤ cu2K m for f ∈ H . Ωǫ
We have proved then if u ∈ K m (Ω), then Dki u ∈ L2 for all k ≤ m − 1. Corollary. If O ∈ D(Ω) and u ∈ Km , then Ou ∈ Km .
This follows at once from Leibnitz’s formula, and (a). Now the second step is to prove (b) Let
−1 < x1 < 1 Ω = 0 < xi < 1 ′
, i = 2, . . . , n.
Then for every u ∈ K m (Ω), there exists U ∈ K m (Ω′ ) such that U = ua. e. on Ω. Assuming this for a moment, we finish the proof of the lemma, Applying ¯ ∩ Q and (b) to each of variables xi , we get an open cube Q such that Ω m m such that for every u ∈ K (Ω), there exists U ∈ K (Ω) with U = u a.e. on Ω. Let O be a function in D(Q) which is 1 on Ω. Then by the corollary to (a), θU ∈ K m (Q) and having compact support is in Hom (Ω). Hence its restriction to Ω which is u is in H m (Ω). Now to prove (b), we require ¯ is dense in K m (Ω). (c) D(Ω) We may obviously assume Ω =] − 1, 1[n ·. Let u ∈ K m (Ω) and define 1 vt (x) = v(tx), for t < 1 for all x ∈ Ω such that x ∈ Ω. t Let ut be the restriction of vt to Ω. Then it is easily seen that as t → 1, ut (x) → u(s) in K m (Ω). Hence to prove (c) it is enough to prove that each ¯ ¯ ⊂ Ω′ , let θ be a ut (x) can be approached by functions of D(Ω). Since Ω m ¯ ¯ function in D(Ω) which is 1 on Ω. Then ω = θvτ (x)K (Ω′ ) and has compact support. Hence ω ∈ H0m (Ω′ ) and so ω = lim ϕk in Hom (Ω′ ) where ϕk ∈ D(Ω′ ). Hence restrictions of ϕk to Ω converge to uτ = restriction of ω to Ω in K m (Ω). ¯ to K m (Ω′ ) Now we prove (b). It is enough to define a map from D(Ω) m ¯ ¯ which is continuous in D(Ω) with the topology of K (Ω). Let u(x) ∈ D(Ω) ′ and let Ω be as in (b). Define u(x) in Ω. U(x) = λ1 u(x′ , −xn ) + · · · + λn u(x′ − xn ) n
82
71 where x′ = (x1 , . . . , xn−1 ) and (x′ , xn ) ∈ Ω′ − Ω. We find λi suitably so that all the Derivatives of U on Σ are well defined. (See § 2.5). This mapping u → U of E (Ω) with the topology of K m (Ω) to K m (Ω′ ) is seen at once to be continuous. This finishes the poof of lemma 9.7.
Lecture 17 9.5 Completion of the proof of theorem 9.1. We are now in a position to complete the proof of theorem 9.1. Our problem is to prove that if u is such that Au = f ∈ H m (Ωǫ ), then u ∈ H 2m (Ωǫ ) for every ¯ We have already proved ∈> 0, where A = Σ(−1)p D p (a pq Dq ) with a pq ∈ D(Ω). p that Dτ u ∈ L2 (Ω) for p ≤ m. We now have to consider D p u. We denote the derivatives with respect to xn by Dy . In this part of proof only ellipticity of A is required, and boundary conditions will not be necessary. We write also Ω for Ωǫ . Now X X p (1) Au = (−1)m Dm (x)Dm yg+ yD u where
g=
X
p≤m
r≤m−1,p≤2m−r
p
(x)D u = β(x)Dm y u+···
(2)
We prove now Lemma 9.8. Re β(x) ≥ α > 0. Lemma 9.9. g ∈ H m (Ω).
Lemma 9.10. Dym+1 u ∈ L2 (Ω).
Using these lemmas and lemmas 9.7, since already it is proved that Dτp u ∈ H (Ω), for p ≤ m, we have the m
Corollary 1. u ∈ H m+1 (Ω).
Proof of the lemma 9.8 It is easily checked that β(x) = aρρ (x), ρ = (0, . . . , 0, m). Since Re Σa pq (x)ξ p ξq ≥ αξ2m , taking ξ = (0, . . . , 0, 1), we have Re β(x) = a pq (x) ≥ α > 0. 72
83
73 Proof of the lemma 9.9 On account of lemma 9.7, it is enough to prove (a) 2 λ 2 Dm y g ∈ L (Ω), (b) Dτ g ∈ L (Ω), λ = m. q
q′
a) follows from (1) for Au ∈ L2 and Dry Dτp u = Dry Dτ (Dτ u) ∈ L2 (Ω), r ≤ m − 1, p = 2m − rq + r ≤ m, q′  ≤ m for by proposition 9.2, q′ Dτ u ∈ H m (Ω) and q + r ≤ m. P αD p Dλτ u and Dλτ u ∈ b) follows from (2) since we have Dλτ g = p
m,
84
m
H m (Ω) by proposition 9.2.
Proof of lemma9.10 From (2), we have Dy g = βDym+1 u + (Dy β)Dm yu+
X
α(x)D p u.
p≤m+1,pn ≤m 2 m From lemma 9.9, Dy g ∈ L2 ; Dm y u ∈ L as u ∈ H (Ω) and the last sum is 2 in L as seen in lemma 9.9. Now, by lemma 9.8, we have Dym+1 u ∈ L2 .
Thus, having proved that u ∈ H m+1 (Ω). There are two ways in which we could possibly carry the induction. However, the easier one of proving that u ∈ H m+k (Ω) ⇒ u ∈ H m+k+1 (Ω) does not work for if we take Dyk+1 g we get P p terms like αDy Du about which we cannot say anything at once ρ≤k+1,p≤m,:pn ≤m−1
unless k = 0. We proceed in a slightly different way. We prove first a) Dλτ Dym+1 u ∈ L2 (Ω) with λ = k, and
b) assuming Dλτ Dym+1 ρu ∈ L2 (Ω) for λ ≤ k − ρ + 1, we prove that Dµτ Kym+ρ+1 u ∈ L2 for µ ≤ k − ρ. (a) From (2) we have Dλτ Dy g = βDλτ Dym+1 u +
X
αD p u.
p≤m+k+1,pn ≤m
By lemma 9.9, Dλτ Dy g ∈ L2 . Since k + 1 ≤ m, and pn ≤ m, D p u = Dρτ Dq u with q ≤ m. Hence the last sum is in L2 , and Dλτ Dym+1 u ∈ L2 (Ω) as Re β(x) ≥ α > 0. b) Again from (2), X µ m+ρ+1 Dµτ Dρ+1 u+ αDq u. y g = βDτ Dy R q≤m+µ+ +1,qn ≤m+p
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74 We have q ≤ m + 1k + 1, qn ≤ m + ρ; hence by induction hypothesis, ρ+1 2 the sum is in L2 (Ω). Since µ + ρ + 1 ≤ k + 1 ≤ m, Dm τ Dy g ∈ L . Hence µ m+ρ+1 2 Dy Dy u ∈ L2 . This proves D2m y u ∈ L . p Since we have already proved Dτ u ∈ L2 , p ≤ 2m, by lemma 9.7, we have u ∈ H 2m (Ω).
9.6 Other results. Theorem 9.1 is but a first step in considering the regularity at the boundary. We prove now the Theorem 9.2. Hypothesis being same as in theorem 9.1, if f ∈ H k (Ω) and Au = f , then u ∈ H 2m+k (Ω). Theorem 9.1 corresponds to the case k = 0. The proof of this theorem is similar in its development to the proof of theorem 9.1. First by making use of local mappings we prove that it is enough to make the proof in the case of a cube ]0, 1[n . Next to prove u ∈ H 2m+k we have to prove Dτp u and Dypn u are in H m (Ω) for p ≤ m + k and pn ≤ m + k respectively. The third step not involving the boundary conditions is essentially the same as in the previous considerations. We consider briefly the second step by proving the p
Lemma 9.11. Dτ u ∈ H m (Ωǫ ) for p ≤ m + k.
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We have proved this lemma for k = 0; we assume it to be true for 1, . . . , k − 1, and prove it for k. As before we consider the difference quotients (Drτ u)−n with r = m + k + 1, and prove that they are bounded in H m (Ωǫ ). It is actually ¯ vanishing near enough to show that (φDr u)−h is bounded where φ ∈ D(Ω) P ∂Ω − . As before, we consider the identity a((φDrτu)−h v) = a((φDrτu)−h , v) + a(φDrτ u, vh ) − b(Drτ u, v) − a(Drτ u, φvh )
where b(u, v) = a(φu, v) − a(u, φv). p By induction hypothesis we may assume that Dτ u ∈ H m (Ωǫ ) for p ≤ m + k − 1. Using this and almost the same manipulation as in proposition 9.1, we prove that a((φDrτu)−h , v) + a(φDrτ u, vh ) and b(Drτ u, v) are bounded in H m (Ωǫ ) by cvm. To prove a(Drτ u, φvh ) is bounded we write h i a(Drτ u, φvh ) = a(Drτ u, φvh ) + (−1)r−1a(u, Drτ(φv)h )
+ (−1)r−1a(u, Dr (φv)h ).
75 The first sum is proved to be bounded again by the same methods as those used in proposition 9.1. However, to prove e(u, Drτφvh ) is bounded, we cannot q use at once a(u, Drτφvh ) = ( f, Drτ φvh ), for Dτ v is not necessarily in H m (Ω) for q ≤ r. However, by regularization, it is seen that such v that Dqτ v ∈ H m (Ω) for q ≤ r are dense in H m (Ω). It is enough then to prove a(Drτ u, φvh) is bounded for such v′ s and then we have a(u, Drτ(φvh ))o , r = m + k − 1. Now, since f ∈ H k we can integrate the last expression ktimes by parts and obtain
87
a(u, Drτ(φv)h ) = (−1)k (Dkτ f, Dτm−1 (φv)h ) ≤ cvo ≤ cvm. Having proved then that a((φDrτ u)−h , v) is bounded by cvm in H m (Ω) by putting v = φ(Drτ u)−h , we obtain, as usual, by ellipticity, that φ(Drτu)−h m ≤ c and by now standard arguments that Dτm+1 u ∈ H m (Ωǫ ). From theorem 4.4 we have H ρ (Ω) ⊂ E ◦ (Ω) if aρ > n. Further if Ω has ¯ Hence, by using theρextension property from theorem 4.4 H ρ (Ω) ⊂ E ◦ (Ω). orem 9.2, we have the Theorem 9.3. Under the hypothesis of theorem 9.1, if 2k > n, then u is in ¯ E 2m (Ω). In this case u is a usual solution of Neumann problem. Remarks. Analogous proof applies for Dirichlet’s problem. Now the question arises for what spaces V such that Hon ⊂ V ⊂ H m can we apply the above methods for proving regularity at the boundary. One of the crucial steps in above proof was the manipulation of difference quotients vh and hence the subspace of V consisting of functions which vanish near the boundary ∂Ω − Σ must be invariant for translations. For spaces V given by conditions like ∂u ∂k u = 0, k ≤ n − 1 , this condition is satisfied. However, for u, , . . . , ∂η ∂η k ∂u = 0 on spaces V ⊂ H m , m ≥ 2, determined by conditions like α(x)u + β(x) ∂xn Σ, this condition is obviously not satisfied. Nevertheless by changing a little the method of proof the regularity theorems have been proved by Aronsza jn∂u Smith for the spaces V given by conditions like α(x)u + β(x) = 0. We shall ∂xn consider these methods in later lectures.
9.7 An application of theorem 9.1. Let A and N be as in theorem 9.1. On N we consider the metric uN = um + Auo and on H 2m ∩ N, the metric u2m + um + Auo which defines on H 2m ∩ N
88
76 the upper bound topology. The inclusion mapping H 2m ∩ N → N is continuous and since from theorem 9.1 it is onto, it is an isomorphism. Hence
i.e. ,
u2m + um + Auo ≥ c(uo + Auo ). Auo + um ≥ γ′ u2m.
This is equivalent to Auo + uo ≥ γ′ u2m, in the case of strongly mregular open sets (which is the case in theorem 9.1). This is proved directly by Ladyzenskya for the case m = 2, and by Guseva for the general case. [ ]. To obtain the regularity at the boundary from these inequalities, one has to prove moreover a nontrivial density theorem.
Lecture 18 9.8 Regularity at the boundary in the case of problem of oblique type 1
Let Ω = {xn > 0}. In § 2.4, we have defined map γ of H 1 (Ω) onto H 2 (Γ). Let ∂u Λu = Σαi where αi are real constants, and let a(u, v) = (u, v)1 + λ(u, v)o + < ∂xi Λγu, γ¯v > with λ > 0 be a sesquilinear form on H 1 (Ω). In § 6.7, we have proved that the form a(u, v) is H 1 (Ω) elliptic and that the operator A associated with it is −△ +λ. We gave there a formal interpretation of the space N. Now we prove some regularity theorems justifying the formal interpretation in regular cases.
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Theorem 9.4. If f ∈ L2 and u ∈ N is such that Au = f , then u ∈ H 1 (Ω).
As it is by now usual we consider the difference quotients uh (x) = u(x + h) − u(x) and prove that they are bounded in H 1 (Ω). This will imh ∂u ∂u ∈ H 1 for i = 1, . . . , n − 1. Next we consider . We know ply that ∂xi ∂xn a(u, v) = ( f, v)o for all v ∈ H 1 (Ω). Hence a(u, vh ) = ( f, vh )o for all v ∈ H 1 (Ω). h −h Since a(u, v) has constant coefficients we have a(u, v ) = −a(u , v). Hence −h h −h h a(u , v) = ( f, v )o and so a(u , v) ≤ cv o ≤ cv1. Taking v = u−h we have αu−h 2 ≤ a(u−h , u−h ) ≤ cu−h 1 . Hence u−h 1 ≤ c. Next −∆u + λu = f ∂2 u + tangential derivatives. Since ∆u ∈ L2 , u ∈ L2 , f ∈ L2 and and ∆u = ∂x2n ∂2 u as has been proved the tangential derivative are in L2 we have 2 ∈ L2 , this ∂xn complete the proof that u ∈ H 1 (Ω). The same proof can be adopted to prove the 77
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Corollary. If f ∈ H k , then u ∈ H k+2 .
If k is large enough, say 2k > n, then we have proved in that H k (Ω) ⊂ ¯ Hence the formal interpretation given E o (Ω). Hence for 2k > n + 2, u ∈ E 2 (Ω). § 6.7 for u ∈ N is a genuine one and we have if 2k > n + 2 and if f ∈ H k and ∂u . u ∈ N is such that Au = f , then u satisfies Λγu = γ ∂xn
9.9 Regularity at the boundary for some more problems. In the § 6.86.8 we have considered the case where V consists of u ∈ H 1 (Ω) such that γu ∈ H 1 (Γ), the topology on V being given by the norm u1 + γu1 . If a(u, v) = (u, v)1 + λ(u, v)o + (γu, γv) with λ > 0, then we have proved that a(u, v) is Velliptic, that the operator defined by a(u, v) is −∆ + λ and that the ∂u ′ boundary value problem solved formally was (x , 0) = ∆Γ u. We prove now ∂xn the Theorem 9.5. If f ∈ L2 and u ∈ N is such that Au = f , then u ∈ H 1 (Ω) and γu ∈ H 1 (Γ). Proof. First of all we observe that V is closed for translations, i.e., u ∈ V ⇒ vh ∈ V for sufficiently small h. Now we know a(u, v) = ( f, v)o for all v ∈ V and hence a(u, vh) = ( f, vh )o . Since a(u, v) is with constant coefficients −a(u−h, v) = +a(u, vh) = ( f, vh )o . Hence a(u−h , v) ≤ cvV . Putting v = u−h we obtain u−h 21 + γu−h 21 ≤ cu−h 1 . Hence uh and γuh are bounded so that as usual, Dτ u ∈ H 1 (Ω) and Dτ γu ∈ H 1 (Γ). Further since −∆u ∈ L2 and Dτ u ∈ H 1 (Ω), ∂2 u ∈ L2 . Hence u ∈ H 2 (Ω). we have ∂xn Corollary. If f ∈ H k , then u ∈ H k+ .
n For sufficiently large k, e.g., 2k > n, we have H k ⊂ E o . Hence for k > +1, 2 the formal boundary condition becomes a genuine one, and we have ∂u ′ (x , 0) − ∆Γu = 0. ∂xn
10 VisikSoboleff Problems 10.1 In a sense these problems generalize nonhomogeneous boundary value problems, e.g., such ones in which solutions of Au = f are sought which would
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79
attain in some sense boundary values given a priori. However, since not until late this aspect of the problem will be evident from the way we shall formulate the problem, and since the hypothesis we shall have to assume in order to ensure the existence and the uniqueness of solutions will not be obvious, in this lecture we prefer to discuss the development of the problem and deduce theorems as consequences thereof. Let Ω be an open set in Rn andR V be such that Hom (Ω) ⊂ V ⊂ H m ( ). P Let Q = L2 (Ω) and a(u, v) = Ωa pq Dqu D p vdx+ some surface integrals p,q≤m
for u, v ∈ V. (However in the sequel we shall drop surface integrals as their inclusion only complicates the technical details. ). As in theorem 3.1, we P define the spaces N and the operator Λ = p,q≤m(−1) p D p (a pq Dq ). We shall assume a(u, v) to be Velliptic, i. e. , a(u, u) ≥ αu2m . for some α > 0 and all u ⊂ V. In this case it is known that A is an isomorphism of N onto L2 . Let a∗ (u, v) = a(v,¯ u). Then a∗ (u, u) ≥ αu2m for all u ∈ V and the P operator A∗ = (−1)p(D p aqp (x)Dq ) it defines is an isomorphism of N ∗ onto 2 Q=L . Suppose now there exists Aqp ∈ DL∞ (Rn ) such that A pq = a pq on Ω and let P A = (−1)p D p (A pq (x)Dq ). We remark that though A is elliptic, A need not be elliptic. Let for f ∈ L2 (Ω), u ∈ N be such that Au = f . Let f˜ and u˜ be the extensions of f and u respectively obtained by defining them to be zero outside Ω. Of course, we do not have A u˜ = f˜. The difference A u − f is given by the Proposition 10.1. If u ∈ N be such that Au = f , then for every v ∈ H 2m (Rn ) such that vΩ ∈ N ∗ we have < A u − f, v¯ >= 0, where vΩ is the restriction of v to Ω. Proof. < A u − f, v¯ >=< u˜ , A ∗ v > − < f˜, v¯ > for v ∈ H 2m (Rn ).
Now since u vanishes outside Ω, we have < u˜ , A ∗ v >= (u, A∗ vΩ )o = (A∗ vΩ , uo ). Since vΩ ∈ N ∗ we have (A∗ vΩ , u)o = φa∗ (vΩ , u) = a(u, vΩ) = (Au, vΩ )o . Further < f˜, v¯ >= ( f, vΩ ) as f vanishes outside Ω. Hence < A u˜ − f˜, v¯ >=< Au − f, v¯ >= 0, for v ∈ H 2m (Rn ) such that vΩ ∈ N ∗ . Now arises the converse problem. Let w ∈ L2 (Rn ) be such that the support ¯ and let there exist f ∈ L2 (Ω) such that < A w − f, v¯ >= 0 of w is contained in Ω m n for all v ∈ H (R ) such that vΩ ∈ N ∗ . Does there exist u ∈ N such that w = u˜ and Au = f . Let uo ∈ N be the solution of Auo = f . By proposition 10.1,
= 0 for v ∈ H m (Rn ) such that vΩ ∈ N ∗ . Hence < A −w−u˜ o, u¯ >= 0, i.e., < −(w − u˜ o ), A ∗ v >= 0. Since and u˜ o have their support in Ω, the above means (w − u˜ o , A ∗ vΩ ) = 0. In order to have w − u˜ o = 0, we have to secure that Λ∗ vΩ be dense in L2 (Ω). ∗ A being an isomorphism of N ∗ onto L2 (Ω) we must consider when the solution x ∈ N ∗ of Ax = g is restriction of a v ∈ H 2m (Rn ). This would follow (a) if we should apply the theory of § 9. Then it would follow that x ∈ H 2m (Ω), and (b) if Ω had 2mextension property, then there would exist x ∈ H 2m (Rn ), such that (πx)Ω = x. In other words, for every g ∈ L2 (Ω) there exists vΩ such that A∗ vΩ = g if the above two conditions are satisfied. We have proved then the
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Proposition 10.2. Besides the hypothesis of Proposition 10.1. , assume 1) A∗ u = g with u ∈ N ∗ and g ∈ H o implies u ∈ H 2m (Ω), 2) Ω has 2mextension property, and ¯ contained in Ω ¯ and 3) there is given ∈ H o (Rn ) such that the support of Ω 2m n ∗ < A w − f, v¯ >= 0 for all v ∈ H (R ) such that vΩ ∈ N . Then wuo , uo ∈ N being the solution of Auo = f . Remarks. Since D is dense in L2 (Ω) instead of assuming the theory of § 9, it would be enough to assume that A∗ x = g, g ∈ D(Ω) implies x ∈ H 2m (Ω). (2) It is not known whether (1) and (2) in proposition 10.2 are independent or not, or whether (2) is a consequence of (1) . The condition (3) can be put more succinctly by making the following Definition 10.1. M o is the subspace of H −2m (Rn ) consisting of distribution T such that < T, v¯ >= 0 for all v ∈ H 2m (Rn ) such that vΩ ∈ N ∗ . It is easily seen that M o is a closed subspace of H −2m (Rn ) and that the support of T ∈ M o is contained in ⌈. We may summarize the proposition 10.1 and 10.2 in the following. Theorem 10.1. Under the hypothesis of Proposition 10.1 and 10.2 the boundary value problem “Given f ∈ L2 (Ω), find u ∈ N such that Au = f ′′ is equivalent to “Given f ∈ L2 (Ω), find w ∈ H o (Rn ) such that A w − f˜ ∈ M o ”.
10.2 Now the second formulation has an advantage over the first one that it can be generalized. In the first instance we notice that instead of f we could take any
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1 T ∈ HΩ−2m and raise the problem ¯
Problem 10.1. Given T ∈ HΩ−2m does there exist w ∈ L2 (Rn ) with the support ¯ o in Ω such that A w − T ∈ M . Similarly a much general problem could be formulated by defining new spaces M k . Definition 10.2. M k is the subspace of H −(k+2m) (Rn ), k being a nonnegative integer, such that < T, v¯ >= 0 for all v ∈ H k+2m (Ω) with vΩ ∈ N ∗ . Lemma 10.1. M k is a closed subspace of H −k+2m . Proof. We prove only that the support of T ∈ M k is contained in the other ¯ take v = ϕ. assertion being then obvious. If ϕ ∈ D(σΩ), ˜ Then vΩ = 0, and ∗ hence is in N . Then < T, ϕ¯ >=< T, vΩ >= 0. Hence the support of T is ¯ If now ϕ ∈ D(Ω), then again let v = ϕ. contained in Ω. ˜ Now v ∈ H k+2m (Rn ) and vΩ = ϕ ∈ N ∗ . Hence < T, ϕ >= 0. This proves that the support of T is contained in Γ. We have now the Problem 10.2. Given T ∈ H −(k+2m) does there exist U ∈ H −k (Rn ) with support ¯ such that A U − T ∈ M k . For K = 0 we get the problem 10.1 in Ω
1 H −m (Ω)
¯ consists of u ∈ H −m (Rn ) such that the support of u ⊂ Ω
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Lecture 19 The problem of formulated in the last lecture would loose its interest if A ∪ were not independent of the extension A of A that we have chosen. We prove that in fact this is the case for some kinds of domains. Let A and A ′ be two extensions of A. Let ∈ HΩ−k ¯ . We have then hA ∪
−A ∪, v¯ i = h∪., (A ∗ − A ∗ )v > for v ∈ H k+2m (Rn ). Since A ∗ = A ∗ on Ω, w = (A ∗ −A ∗ )v is such that wΩ = 0. Now in order that A ∪ = A ′ ∪ it is sufficient ′ to assume some sort of density of (A ∗ − A ∗ )v, for v ∈ H k+2m (Rn ) in H k (Rn ), i. e. , of w ∈ H k (Rn ) such that wΩ = 0. This can be done by having the following definition. Definition 10.3. Ω is ksufficiently regular if w ∈ H k (Rn ) is such that wΩ = 0. ¯ such that = g˜ . Then there exists g ∈ Hok (σΩ) ′
Assuming then Ω to be sufficiently regular, we have A ∗ − A ∗ v = w = ¯ with ϕ j ∈ D([Ω). ¯ But since ∪ = 0 on [Ω, ¯ h∪, ϕ¯ j i = 0. Hence lim ϕi in Hok ([Ω) h(A − A ∗ )∪, v¯ i = 0 for all v ∈ H k+2m (Rn ), and so A u = A ′ ∪. Definition 10.4. If Ω is ksufficiently regular, the problem will be called VisikSoboleff problems.
10.3 We prove now the uniqueness and existence theorem for the VisikSoboleff problems. Theorem 10.2. (1) Let Ω be a domain in Rn such that (a) Ω has k and k + 2m extension property; 82
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83 (b) Ω is k and k + 2m sufficiently regular. (2) Let V be such that Hom (Ω) ⊂ V ⊂ H m (Ω) and a(u, v) be a sesquilinear form on V satisfying a(u, v) ≥ αu2m and such that there exists A pq ∈ DL∞ (Ω) such that A pq = O pq on Ω.
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(3) Let the operator A∗ defined by a∗ (u, v) = a(v, u) be such that A∗ u ∈ H k (Ω) imply u ∈ H k+2m . Then VisikSoboleff problem admits a unique solution ¯ there exists a unique u ∈ H −k (Ω) ¯ such that i. e. , given T ∈ H −k+2m (Ω, k Au − T ∈ M . Proof. To prove this theorem we shall require some lemmas. Let (H k (Ω))′ be the dual of H k (Ω). We do not identify (H k (Ω))′ with any space of distributions for D(Ω) is not dense in general in H k (Ω). We know the restriction map v → vΩ of H k (Rn ) into H k (Ω) is continuous. The transpose of this mapping is a mapping of H k (Ω))′ into (H k (Rn )) = (Hok (Rn )) = (Hok (Rn )) = H −k (Rn ), given explicitly by hπk , T, v¯ i = (T v¯ Ω ) for v ∈ H k (Rn ). Further if vΩ = 0, hπk T, v¯ i = 0, ¯ so that the support of πk is contained in Ω. ¯ Hence πk is a i. e. , πk T = 0 on [Ω continuous mapping of (H k (Ω))′ into HΩ−k . ¯ Using (1)  (a), and (b) of the theorem we prove the fundamental Lemma. The mapping T → πk T of (H k (Ω))′ into HΩ−k ¯ is an isomorphism. Proof. We build explicitly the inverse. On account of the mextension property of Ω, there exists a continuous mapping u → P(u) of H k (Ω) into H k (Rn ), with ¯ Pµ = u a. e. on Ω. Let S ∈ HΩ−k ¯ . Then the semilinear form u → hS , Pui k k ′ is continuous on H (Ω) and hence defines an element ωS ¯ ∈ (H (Ω)) so that ¯ k S ) (u¯ ) hS , P(u)i = (ω The mapping S → ω ¯ k S is obviously continuous. The lemma will be proved if we prove a) ω ¯ k −−k T = T , and b) πk ω ¯ kS = S . a) We have ω ¯ k πk T (u¯ ) = hπk T, Pui = T ((Pu)Ω ) = T (u¯ ). b) For v ∈ H k (Rn ) we have ¯ Ω )i. hπk ω ¯ k S , v¯ i = (ω ¯ k S )(¯vΩ ) = hS , P(v Let w = P(v ) we have to prove hS , wi ¯ = hS , v¯ i.
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84 Let g = w − v; we have g ∈ H k (Rn ) and gΩ = WΩ = 0. By 1b) we have P ¯ g = h˜ with h ∈ Hoq ( (Ω). Hence ˜ = limhS , ϕ j i, ϕ j ∈ D(∈ Ω) ¯ hS , g¯ i = hS , hi 0
since ζ ∈ HΩ−k ¯ . Hence hS , wi = hS , v¯ i which completes the proof of lemma. To complete ¯ we have to determine ∪ ∈ the proof of the theorem, given T ∈ H −k+2m (Ω) H −k )Ω¯ such that A V − T ∈ M k ; hA u − T, v¯ i = 0 for v ∈ H k+2m (Rn ) such that vΩ ∈ N ∗ ; or such aU that (1) hU, A ∗ vi = hT, v¯ i. By hypothesis 1) (a), (b), on account of the above lemma, there exist isomorphisms ω ¯ k, ω ¯ k+2m of HΩ−k and HΩ−(k+2m) into (H k (Ω))′ and (H k+2m (Ω))′ re¯ spectively. Let ω ¯ k U = u, ω ¯ k+2m T = t. Then hU, A ∗ vi = hπk u, A ∗ vi = u((A ∗ v)Ω ) = u(A∗ vΩ ) and hT, v¯ i = hπk+2m t, v¯ i = t(¯vΩ ), so that from (1) our problem will be solved if given t = ω ¯ k+2m T we can determine u ∈ (H k (Ω))′ such that u(A∗ vΩ ) = t(¯vΩ ), for all v ∈ H k+2m (Rn ) such that vΩ ∈ N ∗ .
(2)
We prove that the problem can be still simplified in as much as we need prove (2) only for w ∈ H k+m (Ω) ∈ N∗. Indeed on account of (k + 2m) extension property of Ω, w ∈ H k+2m (Ω) ∩ N ∗ is a vΩ when v = P(w). Hence our problem is reduced to: Given t = ω ¯ k+2m T determine u ∈ (H k (Ω))′ such that u(A ∗ v) = t(¯v) for all v ∈ H k+2m (Ω) ∩ N ∗ . Now we use (3). Let f ∈ H k . Then A∗ v = f has a unique solution G∗ f ∈ N ∗ which on account of the hypothesis (3) of the theorem is in H k+2m (Ω). If f → 0 in H k (Ω), G∗ f → 0 in H k+2m (Ω). Hence by the closed graph theorem, G∗ is a continuous mapping of H k into H k+2m (Ω) ∩ N ∗ in the topology of H k+2m (Ω). Since t (H k+2m (Ω))′ f → t(G∗ f ) is a continuous semilinear mapping on H k (Ω) and hence there exists a unique u ∈ (H k (Ω))′ such that u(A∗ v) = u( f¯) = t(G∗ f ) = t(¯v). Remark. U depends continuously on T .
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Lecture 20 10.4 Application. We now consider some applications of the above theory bringing out how the usual nonhomogeneous boundary value problems are particular case of VisikSoboleff problems. Let V be such that Ho1 (Ω) ⊂ V ⊂ H 1 (Ω). and a(u, v) = (u, v)1 + λ(u, v)o for λ > 0. The operator A associated with a(u, v) is then by § 3.5 F − △ + λ. Let A = −△ + λ. Since a(u, v) is hermitian A = A∗ and n = N ∗ . VisikSoboleff 2 ¯ problem reads now for M o as: Given T ∈ HΩ−2 ¯ determine U ∈ L (Ω) such that ◦ −△u + λv − T ∈ . From theorem 10.2, it follows that this problem admits a solution, say for example, if Ω has smooth boundary. Now we take a particular T = f˜ + S where f ∈ L2 (Ω) and S ∈ HΓ−2 . Since the support of A u − T is in Γ restricting to Ω we see Au = f where u = UΩ . Further hA U − T, v¯ i = 0 for all v ∈ H 2 (Rn ) such that vΩ ∈ N ∗ . Hence for such v, h−△U + λU, v¯ i = hT, v¯ i = h f˜, v¯ i + hS , v¯ i g+ λu), v¯ i + hS , v¯ i. = (−(△U)
Formally, by Green’s formula, Z Z U(−△ + λ)¯vdx −(△ + λ)U v¯ dx = h−△U + λU, v¯ i = Ω Ω Z Z Z Z ∂¯vΩ ∂u u.(−△ + λ)¯vdx = − u = dσ + v¯ Ω dσ + (−△ + λ)u¯vΩ dx ∂η ∂η Ω Ω R g and h(−△u + λu), v¯ i = Ω (−△ + λ)u¯vdx. 85
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86 Hence the original problem is formally equivalent to: given S ∈ H −2 , find ! R ∂u ∂¯v dσ = hS , v¯ i where v ∈ H 2 (Rn ) such that v¯ − u u ∈ L2 (Ω) such that Γ ∂η ∂η¯ vΩ ∈ N ∗ . We now take some particular cases of S . R ∂ϕ¯ R ¯ − h d 1) Let g and h ∈ L2 (Γ). If ϕ ∈ D(Rn ) the mapping ϕ → Γ gϕdσ ∂η is a continuous linear mapping on D(Rn ) with the topology of H 2 (Rn ). For ∂ϕ if ϕ → 0 in H 2 (Rn ), → 0 in H 1 (Rn ) and since the mapping γ from ∂η R R ∂ϕ¯ H 1 (Rn ) to L2 (Γ) is continuous, Γ gϕdσ ¯ and Γ h( )dσ tend to zero in ∂η L2 (Ω). Hence this mapping defines S ∈ H −2 (Γ). Then (1) reads ! Z Z Z ∂¯v ∂u ∂¯v g¯vd − h dσ dσ = v¯ − u (2) ∂n Γ ∂n Γ ∂n Γ Let now R ∂u R ∂v a) V = H 1 (Ω). Then = 0 and (1) means Γ v¯ dσ = Γ g¯vdσ for all v. ∂n ∂n ∂u Hence = g, on Γ. Hence the problem solved is ∂n △
λ
− u + u = f,
∂u = g on Γ. ∂n
(3)
b) V = Ho1 (Ω). Then γv = 0, and (2) becomes −△u + λu = f, u = h on Γ.
(4) 1
−3/2 2) Another example would be to take (Ω) and h ∈ H − 2 (Γ). Then the + g∈H ∂ϕ¯ is continuous on D(Rn ) with the topology mapping ϕ → hg, ϕi ¯ − hh, γ ∂n ! ∂ϕ →0 of H 2 (Rn ) for as we shall prove later on γϕ → 0 in H 3/2 (Ω) and γ ∂n 1 in H 2 (Ω), as ϕ → 0 in H 2 (Rn ). This defines a S ∈ H −2 (Γ). With this S and
a) V = H 1 , the formal problem solved is −△u + λu = f, 1
∂u 3 = g ∈ H − 2 (ϕ). ∂n
1
b) V = H 2 , . . . − ∆u + λu = f, u = h ∈ H − 2 (Ω). These are the problems studied by the Italian School. The principal problem is to give a precise meaning to (3), (4) and so on. (See Magenes [11].
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11 Aronszajn and Smith Problems1 11.1 Complements on H m (Ω) 1 ¯ = {xn > In § 2.4, we have defined a mapping γ of H 1 (Ω) onto H 2 (Γ) where Ω o} and Γ = {xn = 0}. Now we prove the
102
Proposition 11.1. Let Ω = {xn > 0} and Γ = {xn = 0}. Then the mapping γ 1 maps H m (Ω) onto H m− 2 (Γ) for all m. Proof. We denote x′ = (x1 , . . . , xn−1 ), xn = y, ξ′ = (ξ1 , . . . , ξn−1 ).
Here we shall prove that the mapping is into. That it is onto will follow from a more general theorem to be proved later on. Since Ω is mextendible ¯ the restrictions of functions u(x′ , y) of D(Rn ) is u(x′ , 0), and since γ on D(Ω), it is enough to prove that the mapping u → u(x′ , 0) is a continuous mapping 1 of D(Rn ) with the topology of Hom (Rn ), into H m− 2 (Γ). This we do by using Fourier transform. R LetRF (u(x)) = u¯ (ξ) = e−2πix. u(x)dx R be′ the Fourier transform of u. Then u(x) = e2πix. v(ξ)dξ and so u(x′ , 0) = e2πξ . v(ξ, ξn )dξdξn . Hence Z F x′′ u(x′ , 0) = v(ξ′ ξn )dξn . (1) We have now to prove that the mapping F (u) = vF x′ (u(x′ , 0)) is contin1 ˆ uous from F (D) with the topology of H(m) into F (H m− 2 (Γ)) or that v → m− 12 ′ (1 + ξ )F x′ (u(x , 0)) is continuous from F (D) with the topology of Hˆ m into L2 . Hence we have to prove, using (1), that Z Z Z 2 2 2m−1 ′ ′ (1 + ξ dξ v(ξ , ξn )dξn ≤ c (1 + ξ2m )v(ξ) dξ. Now
Z 2 Z v(ξ′ , ξn )dξn = v(ξ′ , ξn )(1 + ξm )(1 + ξm ) − 1dξn 2 Z dξn ≤ v(ξ′ , ξn )2 (1 + ξm )2 dξn (1 + ξm )2 Z Z dξn ≤c v2 (1 + ξm )2 dξn . (1 + ξ2 )m
(2)
1 The author’s thanks are due to Professors Aronszajn and Smith for lending him an unpublished manuscript concerning these problems.
103
11. Aronszajn and Smith Problems1
88
Putting ξn =
Z q 1 + ξ′2 t,
dξn (1 +
2 1 ξ )m− 2
Hence from (2) and (3), Z Z 2 (1 + ξ′ 2m−1 ) v(ξ′ , ξn )dξn Z ≤ c (1 + ξ′ 2m−1 ) ≤c
Z
=
Z
dt (1 + t2 )m
dξ′ 1
(1 + ξ′ 2 )m− 2
Z
(3)
v2 (1 + ξm )2 dξn
v2 (1 + ξm )2 dξ
as was to be proved. ∂u Now, if u ∈ H m (Ω), we have ∈ H m−1 (Ω) and by the above proposition, ∂x n ! ∂u 1 ∈ H m− 2 (Γ). Hence we have the γ ∂xn j
1
Corollary. γ∂ (u) = γ(Dy u) ∈ H m− j− 2 (Γ) for j = 1, . . . , m − 1. →
1
1
Now, let γ(u) = (γo u, . . . , γm−1 u) and F = H m− 2 (Γ) × . . . × H m− j− 2 (Γ) with the product Hilbertia an structure. →
Theorem 11.1. The mapping u → γ(u) of H m (Ω) onto F with kernel Hom (Ω). →
From the above proposition, it follows that γ maps H m (Ω) into F and that →
→
its kernel is Hom (Ω). To prove that γ is onto it is enough to show that if f = 1 (0, . . . , f j , . . . , 0) ∈ F with f j ∈ H m− j− 2 (Γ), then there exists u ∈ H m (Ω) such that γ j u = f j and γk u = 0 for k , j and k ≤ m − 1. Taking Fourier transforms in x′ , with ξ′ = (ξ1 , . . . , ξn−1 ) we have to find ′ v(ξ , y) such that 1) (1 + ξm )v(ξ′ , y) ∈ L2 (, y). ′ 2 ′ 2) Dm y v(ξ , y) ∈ L (ξ , y), and
3) (a) Dyj v(ξ′ , 0) = fˆj (ξ′ ) (b) Dky v(ξ′ , 0) = 0 for k , j, k ≤ m − 1.
104
11. Aronszajn and Smith Problems1
89
1 j −1+ξ′ y ˆ ′ ye f j (ξ ). j! ′ Let t = (1 + ξ )y. Then put Put
φ(ξ′ , y) =
v(ξ′ , y) = φ(ξ′ , y)(1 + α1 t + · · · + αm− j−1 tm− j−1 ). By direct computation, we have Dky φ(ξ′ , 0) = 0 for k ≤ j − 1, Dyj φ(ξ′ , 0) = fˆj (ξ′ ), and 1 j j+1 Dy (ξ′ , 0) = C j+1 D j (y j )(−1)l (1 + ξ′ )l fˆj (ξ′ ). j! Hence Dky v(ξ′ , 0) = 0, for k ≤ j − 1, Dyj (ξ′ , 0) = fˆj (ξ), and j j Dyj+1 v(ξ′ , 0) = C j+1 (−1)l (1 + ξ′ )l F j ( ) + ( j + 1)C j+l−1 (1 + ξ′ )l−1 (1 1 + ξ′ ) fˆj (ξ′ ).
α1 , . . . , αm− j−1 are determined by m − j − 1 conditions that Dyj+1 v(ξ′ , 0) = 0 for l = 1, . . . , m − j − 1, .α1 , . . . , αm− j−1 are then welldetermined independent of j ξ′ , e. g. , ( j + 1)α1 = C j+1 and so on. It remains to verify (a) (1 + ξ′ m )tk φ(ξ′ , y) ∈ L2 for k ≤ m − j − 1, and k ′ 2 (b) Dm y (t φ(ξ , y) ∈ L .
(a) We have to consider (1 + ξ′ m tk φ(ξ′ , y) . o
Z
Z ∞ 2 ′ ′ ′ 2k ′ ˆ 1 + ξ  f j (ξ ) (1 + ξ  )dξ y2 j+2k e−(1+ξ )y dy o Z Z ∞ t j+2ke −tdt ′ 2m ˆ ′ 2 ′ 2k ′ = (1 + ξ   f j (ξ ) (1 + ξ  )dξ ′ 2 j+2k+1 o (1 + ξ  by putting (1 + ξ′ )y = t. Z 1 ≤ c (1 + ξ′ 2m−2 j−1  fˆj (ξ′ )2 dξ′ < ∞, since f j ∈ H m− j− 2 (Γ). ′ 2m
k j −(1+ξ′ )y ˆ ′ b) We have to consider Dm f j (ξ )). This is a sum of terms y (t y e ′
tk−r (1 + ξr )y j−1 (1 + ξm−r−1 e(1+ξ )y fˆj (ξ′ ) for r = l, . . . , k. 1=1,..., j.
105
11. Aronszajn and Smith Problems1
90
Hence we have to consider Z Z ′ (1 + ξ′ )2k−2r (1 + ξ′ )2m−2l  fˆj 2 dξ′ y2 j+2l+2k−2r e−2(1+ξ y) dy Z 1 = (1 + ξ)2k−2r (1 + ξ′ )2m−2l  fˆj 2 dξ′ ′ 2 (1 + ξ ) j−2l+2k−2r+1 2 j−2l+2k−2r −2t t e d by putting t = (1 + ξ′ )y Z ≤ c (1 + ξ′ )2m−2 j−1  fˆj 2 dξ′ < ∞, which proves the theorem.
Lecture 21 11.2 AronszajnSmith Problems We prove a lemma which will be required often.
106 m− 12
n
Lemma 11.1. Let Ω =]0, 1L and Γ = {Ω ∩ {xn = 0}}. Let F = H (Γ) × . . . × 1 H 2 (Γ). Let fα be a bounded set in F such that all fα have their support in a P fixed compact set in Γ. Then there exists a bounded set vα in H m (Ω) such P → → that γvα = fα , γ being as defined in 11.1 above, and vα ≡ 0 near ∂Ω − . →
Proof. By Theorem 11.1, γ : H m (Ω) → F is onto with kernel H◦m . Hence → → γ induces an isomorphism γ 1 of H m /Hom onto F. Since fα is a bounded set →−1
in F, γ ( fα ) is bounded in H m /H◦m . Therefore we can choose a bounded ¯ be zero near ∂Ω − P set ωα ∈ H m such that γωα = f . Next let ϕ ∈ D(Ω) P P and ϕ = 1 on . Then vα = ϕvα are bounded, vanish near ∂Ω − and →
→
→
→
γvα = γ(ϕ) γ(ωα ) = f α .
In § 9, we considered regularity at the boundary of some problems related to the operator A defined by a sesquilinear form a(u, v) on V such that Hom ⊂ V ⊂ H m . Now we take up a particular example of a different space V. In this case the technique used in § 9 is not at once applicable. Since the preliminary step of using local maps is at any rate permissible we assume Ω =]0, 1[n. Further to avoid technical details, we assume that m = 2. Let ϑ be the subspace of H 2 (Ω) consisting of functions P a) vanish near ∂Ω − ,
b) Bu = 0 on where
X ∂u ∂u ′ (x , 0) + αi (x′ ) (x′ , 0) + αo (x′ )u(x′ , 0) ∂xn ∂x i i=1 n
Bu =
91
92 with αo , . . . , αn−1 ∈ D(ξ). P R ¯ Let a(u, v) = a (x)Dq uD p vdx with a pq ∈ E (Ω). Ω pq
107
p,q≤
Let Re a(u, u) ≥ αu22 for all u ∈ ϑ. In this case according to the theory of § 3, as transformed by local maps as in §, for a given f ∈ L2 (Ω), there exists u ∈ N such that a(u, v) = ( f, v)o for all v ∈ ϑ. We prove now Theorem 11.2. Let u ∈ H 2 (Ω) with Bu = 0 and a(u, v) = ( f, v)o for all v ∈ ϑ, f ∈ L2 . Then u ∈ H 4 (Ωǫ ) for every ∈> 0. Proof. Though a shorter proof by induction is possible in order to bring out the significance of the method we give a direct proof. Since after having proved 2 that DTp u ∈ H 2 for p ≤ 2, to prove Dm y u ∈ H no use of boundary conditions need be made as in §9, to prove the theorem, we have to prove DτP u ∈ H 2 for ¯ ϕ ≡ 0, near ∂− − P, to prove Dτp u ∈ H 2 for p ≤ 2, p ≤ 2. Further if φ ∈ D Ω), p it is enough to prove Dτ (φu) ∈ H 2 for p ≤ 2. We break this in two steps. Step 3. D1τ (φu) ∈ H 2 . As usual we need prove (φu−h ) is bounded in H 2 (Ω) by cu2, and for this we consider a((φu)−h, v). Lemma 11.2. a(φu)−h, v) ≤ cv . 2
We write
a((φu)−h, v) = [a((φu)−h, v) + a(φu, vh)] − b(u, vh) − a(u, φvh) where b(u, v) = a(φu, v) − a(u, φv). As in § 9, we can estimate a((φu)−h, v) +a(φu, v−h) and b(u, v) by almost the same methods. It remains to be proved that a(u, φvh ≤ cv2. We cannot put a(u, φvh) = ( f, φvh )o as ϑ is not necessarily closed for translations. However by “correcting” φvh with a “compensating” function ωh we prove that a(u, φvh) < cv2. More precisely we prove the Lemma 11.3. There exists wh in H 2 (Ω) such that (a) φvh − wh ∈ ϑ (b) wh 2 ≤ cv2 . Assuming for a moment the lemma 11.3, we prove lemma 11.2. We have a(u, φvh) = a(u, φvh − wh ) + a(u, wh ).
108
93 h h h Since φv φv − wh ) = ( f, φv − wh )o . − wh ∈ ϑ, a(u, Hence a(u, φvh − wh ) ≤  f o φvh − wh o ≤ c(v1 + wh o ) ≤ cv2 . Further a(u, wh ) ≤ cwh 2 ≤ cv2, whence the lemma 11.2.
Now we prove lemma 11.3. We have to find wh such that φv j − wh ∈ ϑ, i. e. , B(φvh − wh ) = 0,
∂wh ′ P ∂wh ′ (x , 0) + αi (x′ ) (x , 0) + αo (x′ )wh (x′ , 0) = B(φvh). ∂xn ∂xi ∂wh ′ This holds if (x , 0) = B(φvh), and wh (x′ , 0) = 0. ∂xn 1 If we prove that B(φvh ) is bounded in H 2 by c′ v2 , by using lemma 11.1, we can find wh bounded by cv2 , such that γwh = 0 and γ1 wh = B(φvh) which will prove the lemma. Now i.e.,
B(φvh) = gh + kh gh (x′ , 0) = φ(x′ , 0)Bvh, P ∂ϕ ′ ∂ϕ ′ and kh (x′ , 0) = (x , 0)vh (x′ , 0) + (x , 0)αi vh (x′ , 0). ∂xn ∂xi Since ϕ has compact support all kh have support in a fixed compact. ∂ϕ Further since v ∈ H 2 , we have vh (x′ , 0) ∈ H 3/2 (ΓΓ) and since are ∂xh 3/2 h ′ ′ smooth, we have kh (x) ∈ H (Γ). Now as h → 0, v (x , 0) → Dτ v(x , 0), hence 1 kh is bounded by cv2 in H 2 (Γ). 1 It remains to see that gh is bounded in H 2 (Γ) by cv2 . Since Bv = 0, (Bv)h = 0, and since where
(Bv)h = Bvh +
X
αhi
∂v ′ (x + h, 0) + αho v(x′ + h, 0) ∂xi
P ∂v ′ we have gh = ϕBvh = −φ(x′ , 0)( αhi (x + h, 0) + αho v(x′ + h, 0)). ∂xi ∂v 1 As h → 0, αhi are uniformly bounded and are bounded in H 2 (Γ) as ∂xi translations are continuous. 1 This proves then that B(φvh ) is bounded in H 2 (Γ) and the proof of lemma 11.3 and hence that of lemma 11.2 is complete. Now we are in a position to prove the Lemma 11.4. (φu)−h2 ≤ c.
109
94 We cannot prove this as in § 9 by taking v = (φu)−h in lemma 11.2, and using ellipticity for (φu)−h does not necessarily belong to ϑ. We again correct this by Lemma 11.5. There exists wh ∈ H 2 (Ω) such that a)(φu)−h − wh ∈ ϑ and b)wh 2 ≤ cu2 = c′ (since u is fixed). To prove this we note (φu)−h = φu−h + φu−h (x − h) and from lemma 11.3, there exists w′h such that φu−h − wh ∈ ϑ, and wh 2 ≤ cu2. We have only to look then for w(2) h such that φ−h u(x − h) − w(2) h ∈ ϑ, and w(2) h 2 ≤ cu2 .
2 We have to find w(2) h bounded in H (Ω) by cu2 and such that ′ −h ′ w(2) h (x , 0) = φ u(x − h, 0) ∂wh ′ ∂ −h ′ (x , 0) = (φ u(x − h, xn ) xn = 0. ∂xn ∂xn
Hence, by lemma 11.1 such w(2) h as required above exist and the lemma 11.5 is proved. To prove lemma 11.4 consider now a((φu)−h − w′h , (φu)−h − w′h ) = a((ϕu)−h, (φu)−h − w′h ).a(w′h , (φu)−h − w′h ), = Xh − Yh . By lemma 11.2, we have Xh  ≤ c(φu)−h − w′h 2
Yh  ≤ cw′h 2 (φu)−h − w′h 2 ≤ c′ (φu−h − w′h 22 . On account of ellipticity, a((φu)−h − w′h , (φu)−h − w′h ) ≥ α(φu)−h − w′h 22 . Hence (φu)−h − w′h 2 ≤ c and since w′h 2 ≤ c we get the lemma. This completes the first step of the proof, viz. Dτp u ∈ H 2 (Ω), p = 1.
110
Lecture 22 11.3 Now we come to the Second step. We wish to prove (φDτ u)−h is bounded in H 2 (Ω). To do this we consider a((φDτu)−h , v) and prove the Lemma 11.6. a((φDτu)−h , v) ≤ cv2 for v ∈ ϑ such that Dτ u¯ ∈ H 2 . Proof. We write
a((φDτ u)−h , v) = a((φDτu)−h , v + a(φDt auu, vh) − b(Dτ u, vh ) − a(Dτ u, φvh).
As in the previous cases, we have straight forward estimates except for a(Dτ u, φvh ). Now a(Dτ u, φvh ) = a(Dτ u, φvh ) + a(u, Dτ(φvh )) − a(u, Dτ(φvh )) which exists since D τ v ∈ H 2 . Again as in the previous cases, only nontrivial part is to prove that a(u, Dτ(φvh )) ≤ cv2. To do this we have to correct Dτ (φvh ) by the
Lemma 11.7. There exists wh ∈ H 2 (Ω), wh = 0 near ∂Ω − i) Dτ (φvh ) − wh ∈ ϑ. ii) wh 1 ≤ cv2. iii) a(u, wh ) ≤ cv2. 95
P
such that
111
96 Admitting this for a moment, we have a(u, Dτ(φvh )) = a(u, wh ) + a(u, Dτ(φvh ) − wh ) = a(u, wh ) + ( f, Dτ (φvh ) − wh )
and we have the lemma 11.6 as usual. So we have to prove now lemma 11.7. If wh have to verify i) Then we can write : B(wh ) = BDτ (φvh ) = fh +
n−1 X i=1
Di gih , Di =
∂ , say . ∂xi
We shall prove that one can choose fh and gih such that
112 1
a) fh have their support in a fixed compact and are bounded in H 2 (Γ) by cv2 . 1
b) gih ∈ H 3/2 (Ω), and are bounded in H 2 (Ω) by cv2 with support in a fixed compact. Assuming (a) and (b) we prove that wh satisfying (i), (ii), and (iii) can be found. For by (a) and lemma. 11.1, there exists woh ∈ H 2 (Ω) such that ∂ o ′ woh (x′ .0) = 0, w (x , 0) = fh , woh are bounded in H 2 (Ω) by cv2 and woh ∂xnP h vanish near ∂Ω − . Similarly on account of (b) and lemma 11.1, there exists ∂ i ′ w′h ∈ H 2 (Ω) with wih (x′ , 0) = 0, wh (x , 0) = gih , wih bounded in H 2 (Ω) by ∂x n P P i cv2 and wih ≡ 0 near ∂Ω − . Setting wh = woh + n−1 i=1 Di w h we see that (i) and (ii) are at once satisfied. Further to verify (iii) we have a(u, woh ≤ cv2 and it remains to estimate a(u, Di wih ). But since Dτ u ∈ H 2 and since wih = 0 P near ∂Ω − by integration by parts, we get a(u, Di wih ) ≤ cv2. We have still to verify (a) and (b), we indicate which parts of B(Dτ(φvh )) are to be taken as fh and which as gih and prove each time that they are bounded by cv2 in the appropriate spaces. We have B(Dτ(φvh ) = B((Dτφ)vh ) + B(φDτvh ) ! ! X ∂ ∂ B((Dτφ)vh ) = (Dτ φ)Bvh + Dτ φ v h + αi (Dτ φ) vh , ∂xn ∂xi and Bvh = −
X
αhi
∂v ′ (x + h, 0), since (Bv)h = 0. ∂xi
97 1
Then we take B((Dτφ)vh ) as part of fh ; it is H 2 and is bounded by cv2 . ∂φ ′ P ∂φ Now B(φ(Dτvh )) = φB(Dτvh ) + (x , 0)Dτvh + α Dτ v h . ∂xn i∂xi We consider each of the summands separately : ! ! ∂φ ∂φ h ∂φ h (Dτ vh ) = Dτ v − Dτ v. ∂xn ∂xn ∂xn ∂φ h )v as part of fn and it is seen that it satisfies (a). For We take Dτ ∂xn ! ∂φ h Dτ v we take it as part of Dτ gih . It is also seen that gih satisfies (b). ∂xn X ∂φ Similarly we consider Dτ vh . It remains only to consider φB(Dτvh ). ∂x ∂ i x i i Now since (Bv)h = 0, we have ! X ∂vh + Dτ αo vh ), φB(Dτvh ) = −φ Dτ αi ∂xi and they are to be taken as parts of fh . This completes the proof of lemma 11.6. To complete the proof of the theorem, we require the Lemma 11.8. Dτ (φu )−h 2 ≤ c. Again we require corrections w′h as follows: Lemma 11.9. There exists w′h ∈ H 2 (Ω) vanishing near ∂Ω − (1) Dτ (φu)−h − w′h ∈ ϑ
P
and such that
(2) w′h 2 ≤ c. Assuming the existence of such w′h and considering a(Dτ (φu)−h − w′h , Dτ (φu−h − w′h ) and using the ellipticity of a(u, u) we obtain lemma 11.8, as in the lemma 11.4, after observing that lemma 11.6 can be applied though v = Dτ (φu)−h − w′h is such that Dτ v < H 2 (Ω). To see this last point we prove that in ϑ, v′ s such that Dτ v ∈ H 2 (Ω) are dense. Let v ∈ ϑ and v(x′ , 0) = f . Let ψ ∈ D(Ω) be such that ϕα f in H 3/2 (Γ), and P ∂ϕ let ψα = − α − αo ϕα . ∂xi
113
98 ¯ such that vα (x′ , 0) = ϕα , ∂v = Now it is possible to find vα ∈ H 2 (Ω)∩ϑ(Ω) ∂xn P ϕα , v ≡ 0 near ∂Ω − and vα → v in H 2 . By the choice of ψα , v belongs to ϑ and the result follows. To prove lemma 11.9, we observe
114
(φu)−h = φu−h + φ−h u(x′ − h, xn ), and
Dτ (φu)−h = (Dτ φ)u−h + φDτ u−h + (Dτ φ−h u(x′ − h, xn ) + φ−h Dτ u(x − h). We first define w1h such that a) (Dτ φ−h )u(x − h) + φ−h Dτ u(x − h) − w1h ∈ ϑ, and b) w1h 2 ≤ c. To verify (a) we choose w1h so that w1h (x′ , 0) = ((Dτ φ)−h u(x − h) + φ−h Dτ u(x − h)) xn = 0. ! ∂ω′h (x′ , 0) ∂ −h ∂ −h = u(x − λ) xn = ∂ u(x − λ) + φ 0xn ∂xn ∂xn
and
Since Dτ u ∈ H 2 (Ω) the right hand side in the first expression ie in H 3/2 (Γ), 1 and is bounded. Similarly the one in the second expression is in H 2 (Γ), and is bounded, and by lemma 11.1 the existence of w1h is proved. Next we find w2h so that a) (Dτ φ)u−h − w2h ∈ ϑ and b) w2h 2 ≤ c. The existence of such w2h is assured by lemma 11.4. Finally we define w3h so that a) σDτ u−h − w3h ∈ ϑ and b) w3h 2 ≤ c. To verify (a), we should have B(φ(DT u−h )) = Bw3h . But B(φ(Dτu−h )) = φB(DT u−h ) +
X ∂φ ∂φ Dτ u−h + αi Dτ u−h . ∂xn ∂xi
Since Bu = 0 and Dτ Bu = 0, we have
115
99
B(Dτu) +
n−1 X
(Dτ αi )Di u + (Dτ αo )u = 0.
i=1
Hence on the support of φ, X (B(Dτu))−h + ((Dτ αi )Di u)−h + ((Dτ αo )u)−h = 0, i.e., X X (Dτ αi Di u)−h + ((Dτ αo )u)−h = 0; (B(Dτu−h + α−h i Dτ u(x − h) + so that we have to find w3h satisfying X X Bw3h = −φ α−h ((Dτ αi )Di u)−h − φ((Dτ αo )u)−h i Dτ u(x − h) − φ X ∂φ ∂φ Dτ u−h + αi + Dτ u−h = gh . + ∂xn ∂xi Hence it is enough to find w3h such that 3 ′ wh (x , 0) = 0 3 ∂wh (x′ , 0) = gh defined above. ∂xn
1
Since Dτ u ∈ H 2 (Ω), gh ∈ H 3/2 (Γ) and is bounded in H 2 (Γ). Hence by lemma 11.1 we have the existence of such w3h . This completes the proof of the theorem 11.2. Final Remarks. (1) By using Stampanhia [17] and Lions [6], Campanato [6] has proved the regularity at the boundary for Picone problems. (2) For another method for Dirichlet conditions with constant coefficients in two dimensions, and very general conditions on the boundary, see Agmon [1].
Lecture 23 12 Regularity of Green’s Kernels 12.1 In § 3.5 we have defined Green’s kernel of the operator A associated with an elliptic sesquilinear from a(u, v) on V such that Hom (Ω) ⊂ V ⊂ H m (Ω), Q being L2 (Ω) say. We recall that a(u, v) begin V elliptic, A is an isomorphism of N onto Q′ . Hence A−1 = G is an isomorphism of Q′ onto N. Since D(Ω) is dense in Q′ , by Schwartz’s kernel theorem A−1 = G is given by G x,y ∈ D ′ (Ω x × Ωy ).G x,y is called the kernel of the operator A. Let a∗ (u, v) = a(v, u); a∗ (u, v) is also V elliptic and defines a space N ∗ and an operator A∗ . Let its kernel be G∗x,y . If T x,y ∈ D ′ (Ω x , Ωy ), then T y,x will be defined by setting on the everywhere dense set D(Ω x ) × D(Ωy ) in D(Ω x × Ωy ).
116
T y,x (ϕ(x) · ψ(y)) = T x,y (ψ(x)ϕ(y)). We denote D(Ω x ) by D x , D(Ω x × Ωy ) by D x,y and so on. We have the Proposition 12.1. G xy = G∗y,x . ¯ = hϕ, G∗ ψi. Let Gϕ = Let ϕ, ψ ∈ D(Ω x ). We have to verify that hGϕ, ψi ∗ ∗ ∗ u ∈ N and G ψ = w ∈ N . Then ϕ = Au and ψ = A w. Hence we have to verify that hu, A∗ wi = hAψ, ωi. ¯ This follows since hu, A∗ wi = a∗ (w, u) = a(u, w) = hAu, wi. ¯ Definition 12.1. An element G x,y in Q ′ (Ω x × Ωy ) will be called a kernel. Definition 12.2. A kernel is semiregular, with respect to x, if G x,y is given by a C ∞ function of Ω x into Dy′ . We write it then as G(x)y or Gy (x). 100
117
12. Regularity of Green’s Kernels
101
Definition 12.3. A kernel is regular if it is semiregular with respect to x and y. Definition 12.4. A kernel is very regular if it is regular and a C ∞ function outside the diagonal. ′
′
ˆ If G x,y is semiregular it is an element R of E x ⊗Dy = E (x, Dy ). Hence it defines a mapping G : Dy → Ez given by G(x)y ϕ(y)dy ∈ E (x) for ϕ(y) ∈ Dy . Conversely by Schwartz’s kernel theorem, any linear mapping G : Dy → E x is given by a semiregular kernel. We now with to consider conditions on a(u, v) so that the kernel G x,y be very regular. Definition 12.5. A partial differential operator A defined in Ω with C ∞ coef′ ficients is said to by hypoelliptic if for s ∈ D (Ω)(AS )θ ∈ E (θ) implies that S ∈ O, for every 0 ⊂ Ω. For example, if a(u, v) is V elliptic, Ω is bounded with smooth boundary, then by the results of § 8, it follows that A is hypoelliptic. Theorem 12.1. Let a(u, u) ≥ αu2V and the operators A and A∗ be hypoelliptic. Then G x,y is very regular. Proof. First we prove G x,y is regular. We know G is an isomorphism of Q′ onto N. Let ϕ ∈ D(Ω). Then Gϕ = u ∈ N and Au = ϕ. By hypoellipticity of A it follows that u ∈ E . Hence G defines a mapping of D(Ω) onto N ∩ E . By closed graph theorem this mapping is continuous and hence Schwartz’s kernel theorem is given by a semiregular kernel G(x)y . Hence G x,y = G∗ (x)y . Similarly, since A∗ is hypoelliptic G∗x,y = G∗ (x)y . But by proposition 12.1, G x,y = G∗y,x and hence G x,y = G∗x (y). This shows that G x,y is regular. To complete the proof we have to show that outside the diagonal it is a C ∞ function. ′ Let R O1 and O2 ⊂ Ω be two open sets such that O1 ∩O2 = φ. Let T ∈ E (O2 ). Then G x (y)T y dµ is defined and is an element of D x say S x such that AS = T . Restricting A, S , T to O1 since T = R0 on O1 by hypoellipticity of A on O1 , we have S O1 ∈ E (O2 ). Hence T → G x (y)T y dµ is a mapping of E ′ (O2 ) onto E (O1 ), which on account of the closed graph theorem, is continuous. Hence ˆ (O1 = E (θ1 × θ2 ). G ∈ L(E ′ (O2 ); E (O1 ) ≃ E (O1 )⊗E That is to say G is C ∞ in O1 × O2 . Since O1 and O2 are any two open set such that O1 ∩ O2 = φ, G is a C ∞ function outside the diagonal.
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12. Regularity of Green’s Kernels
102
Remark. For a more detailed study, see Malgrange [12]. 2. The extension of the mapping G : Q′ → N. Under the hypothesis of theorem 12.1, G defines an algebraic isomorphism of Q′ ∩ E onto N ∩ E . For if f ∈ Q′ ∩ E and G f = uE N . Then Au = f . Hence by hypoellipticity of A it follows that u ∈ E . Conversely if u ∈ N ∩ E , then Au = f ∈ Q′ ∩ E and G f = u. If we could apply closed graph theorem, then it would follow that G is a topological isomorphism of Q′ ∩ E onto N ∩ E , the intersections being given as usual the upper bound topology. This is so, for example, if Q is a Banach space. We have then the Theorem 12.2. If the closed graph theorem is applicable G ∈ L (Q′ ∩E , N∩E ) and is an isomorphism. Similarly G∗ ∈ L (Q′ ∩ E , N ∩ E ) and is an isomorphism.
119
12.2 Now we wish to consider the transpose G. Let us first consider the Dirichlet problem so that V = Hom = Q · D(Ω) is dense in V and N = V. Hence by transposing G we have an isomorphism t G : V ′ +E ′ → V ′ +E ′ . However since D(Ω) is not always dense in N, the dual of N is not a space of distributions and hence we do not consider directly the transport of G. Here the sums of locally convex topological vector space A and B subspaces of an algebraic vector space F is topologized as follows : we consider the mapping (a, b) → a + b of A × B onto A + B and put on A + B the finest locally convex topology such that this mapping is continuous. If Z is the kernel, then A + B ≈ A × B/Z. Theorem 12.3. Under the hypothesis of theorem 12.1, if further, for every S ∈ Q′ ∩ E ′ there exists a sequence ϕn ∈ D(Ω) such that ϕn → S in Q′ ∩ E ′ , then G : Q′ → N can be extended by continuity to G : Q′ + E ′ → N + E ′ . Proof due to L. Schwartz (unpublished). We define first G on Q + E ′ . G ′ Ris already defined on Q . By theorem 12.1 is very regular and is given by G(x)y ϕ(ϕ)dyϕ(ϕ) ∈ Q′ We cannot use this at once to define it on E ′ , for then the integral itself is not in E ′ . We proceed then as follows : Let α(x, y) ∈ E (Ω x × Ωy ) be a function with support in a neighbourhood of diagonal and equal to 1 in another neighbourhood of diagonal. Let H x,y = α(x, y)G x,y = H(x)y = H x (y). Hence H x,y is regular. It is easily seen, since G x,y is a C ∞ function outside the diagonal that A x H x,y − δ(x)y = L(x, y) ∈ E (Ω x × Ωy ). Now let T ∈ E ′ (Ω)
120
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103
R with compact support K say. Since ρ(x)y T = T , we have T = LT + AHT where Z LT x = L(x, y)T y dy ∈ E . Z (H, T ) x = H x (y)T y dy ∈ D ′ . But the supports of the mapping y → H x (y) and y → L(x, y) are contained in the support of α(x, y). By choosing the support of near enough the diagonal, we may have the support of HT and LT in any arbitrary neighbourhood of the ˜ = HT + GLT ∈ support of T . Hence HT ∈ E ′ and LT ∈ D. We define GT ˜ E ′ + N. We have to verify that if ϕ ∈ D(Ω), then G(ϕ) = G(ϕ). This follows as in general ˜ = AHT + AGLT = T − LT + LT = T. AGT
˜ = Hϕ + GLϕ ∈ D+N ⊂N and AGϕ ˜ = ϕ. Since A is an If ϕ ∈ D(Ω), Gϕ ˜ isomorphism, Gϕ = Gϕ. ′ Now G is continuous from ξk (Ω) into E ′ + N. This proves that G˜ does not depend on α and G˜ can be extended to E ′ (Ω) so that G˜ : E ′ (Ω) → E ′ (Ω) + N is continuous. We denote now G˜ be G itself. G defines then a continuous mapping θ from Q′ × ǫ ′ → N + ǫ ′ by θ( f, s) = G f + GS . If we prove θ is zero on the kernel of Q′ × E ′ → Q′ + E ′ , we shall have proved that θ defines a mapping G of Q′ + ǫ ′ → N + E ′ as required. Let f ∈ Q′ and S ∈ E ′ such that f + S = 0. Hence f ∈ Q′ ∩ E ′ . By assumption (2), there exists ϕn ∈ D(Ω) which converges in Q′ and E ′ to f . Hence  ϕn converges to S in E ′ and we have G f + GS = lim(Gϕn + G(ϕn )) = 0. Corollary. We have A.G = IQ′ +E ′ and G.A = IN+E ′ where IQ′ +E ′ and IN+E ′ are the identity maps Q′ + E ′ and N + E ′ respectively. For A(G f + GS ) = f + AGS = f + S and G.A(u + S ) = GAu + G A S = u + G A S . Since GAS = S on D(Ω) it is so on E ′ . This proves the Theorem 12.4. Under the hypothesis of theorem 12.3, A is a topological isomorphism from N + E ′ to Q′ + E ′ . The uniqueness of G x (y) is given by the following
121
12. Regularity of Green’s Kernels
104
Theorem 12.5. Under the hypothesis of theorem 12.3, for y ∈ Ω, G x (y) is defined as the solution of A x (G x (y) = δ x (y) G x (y) ∈ N + E ′ . Consider y → δ x (y) ∈ E (Ωy , E ′ (Ω x )). Let G(δ x (y)) = G x (y). Then y ∈ G x (y) is a C ∞ function from Ω → N + ǫ ′ we have A x (G x (y)) = δ x (y), and G x (y) is the only distribution to verify the equation in N + E ′ .
Lecture 24 12.3 Study at the boundary. Definition 12.6. We say that a(u, v) is regular at the boundary if u ∈ N is such that a(u, v) = 0 for every v ∈ V vanishing outside a neighbourhood of some compact K ⊂ Ω then u is C ∞ in a neighbourhood of Γ. P If V = Hom (Ω) or H m (Ω) and a(u, v) = a pq Dq D p v then the results on regularity at the boundary of § 9 state that under the conditions specified in theorem 9.1, a(u, v) is regular at the boundary. Theorem 12.6. Under the hypothesis of theorem 12.3, if further a(u, v) is regular at the boundary, then for fixed y, G x (y) is C ∞ in a neighbourhood of Γ. This means in this case G(x, y) for fixed y is a usual function in a neighbourhood of Γ satisfying usual boundary conditions. By theorem 12.5, G x (y) = G(δ x (y)) = S + u with S ∈ E ′ and u ∈ N. Hence Au + AS = δ x (y), i.e., Au = δ x (y) − AS = T for T ∈ E ′ . Let K be the support of T , which on account of the splitting proved in theorem 12.1 can be taken in any arbitrary neighbourhood of y. Let ∈ V such that v = 0 in neighbourhood of K. Now by regularization we can find ϕn vanishing on the support of v such that T = lim ϕn in E ′ . Then hAu, v¯ i = limhϕn , v¯ i. Hence a(u, v) = 0 for all v ∈ V vanishing in a neighbourhood of K. By regularity at the boundary of K, u is C ∞ in a neighbourhood of Γ. This completes the proof of the theorem.
105
122
13. Regularity at the Boundary Problems for...
106
13 Regularity at the Boundary Problems for General Decompositions. 13.1 Hitherto we considered boundary value problems for differential operators in the space H m (Ω). For this we obtained A as the operator associated with a form a(u, v) on H m (Ω) is the space of type H({A}, Ω) where {A} stands for the system D(p) . More generally we consider now what problems are solved by considering A as the operator associated with sesquilinear forms a(u, v) on spaces H({Ai }, Ω). That this solves now problems can be seen from the following example. Let A = ∆2 + 1. Consider on H(∆, Ω) the sesquilinear form a(u, v) = (∆u, ∆v)o + (u, v)o. The operator A associated with a(u, v) is ∆2 + 1.a(u, v) is H(∆, Ω) elliptic and hence for f ∈ Q′ where Q is such that H(∆, Ω) is dense in Q, we have u ∈ N such that Au = f . Firstly we observe that H 2 (Ω) may be contained in H(∆, Ω) strictly. For o example, if Ω is a domain such that for a given T ∈ HΩ−2 ¯ , there exists ∈ H such o that −∆U − T ∈ M , i.e., for which VisikSoboleff problem is soluble, then there exists u ∈ H(∆, Ω) such that u < H 1 (Ω). For, let T ∈ HΩ−2 ¯ be defined by R 1 2 hT, ϕi ¯ = r f (γϕ)dσ ¯ for f ∈ L (Γ) and such that f < H 2 (Γ) = γ(H 1 (Ω)). Now if ∪ is the corresponding solution, let u be its restriction to Ω. We have u ∈ L2 (Ω) and −∆u = u by § 10. Hence u ∈ H(∆, Ω) : If u were in H 1 (Ω), 1 then γu = f would be in H 2 (Γ) contrary to the assumption. Another more elementary example can be given for a circle. It is easy to construct examples such that u ∈ L2 and ∆u = 0, but γu < H 1 . Thus Hadamard’s classical example P with u = an rn einθ with suitable an is of this type. However it is true that Ho2 (Ω) = H(∆, Ω). For, by Plancherel’s formula, the two norms are equivalent on D(Ω). This raises in fact the question : To determine the conditions on Ai and Ω so that H(A; Ω) = H m (Ω) where m = highest of orders of the operators Ai . Now we interpret formally the boundary value problems that are solved on V ∈ H(∆, Ω). We write first of all Green’s formula Z Z Z Z ∂¯v ∂∆u .¯vdσ − ∆u. dσ + ∆u.∆vdx ∆2 u · v¯ dx = (1) ∂n Γ Γ ∆n a) Let V = Ho (∆, Ω). Since H2 (∆, Ω) = Ho2 (Ω) no new problem is solved. b) V = H(∆, Ω). Given f ∈ Q′ there exists u ∈ N such that a(u, v) = ( f, v)o for all v ∈ V. Further (∆2 + 1)u = f (2)
123
124
13. Regularity at the Boundary Problems for... Hence
R
Ω
¯ (∆2 u − ∆vdx +
Using (1) and (2), Z Γ
R
Ω
107
R u, v¯ dx = ( f − v¯ )dx
∂v ∂∆u .¯vdσ + ρ ∈ ∆u. d = 0 for all v ∈ V. ∂n ∂n
Formally this means ∆u = 0, and
∂∆u = 0. ∂n
c) V = Closure in H(∆, Ω) of continuous function with u = 0. Then u ∈ N implies uΓ = 0 and ∆uΓ = 0. d) V = Closure in H(∆, Ω) of continuous function with implies
∂u ∂∆u = 0 and = 0. ∂n Γ ∂n Γ
∂u = 0 Then u ∈ N ∂n Γ
∂∆u ∂u = 0 and ∆u = 0 or u = 0 and = 0 are However, problems in which ∂n ∂n not solved by this method.
Lecture 25 Now we consider the regularity at the boundary of solutions so determined. This means we want to determine whether if f ∈ H k (Ω) implies u ∈ H k (Ω). The solution of this problem in full generality is not known though it would be desirable to know it, for in that case, for large k weak solutions would be usual ones. We shall show that this is the case with certain kind of operators in Ω = {xn > 0} with constant coefficients. Let
125
m−1 ∧1 u + Dym−2 ∧2 u + · · · + ∧m u, Ω = {xn > 0} and Bu = Dm y u + Dy
where ∧m are partial differential in x1 , . . . , xn−1 operators with constant coefficients of order ≤ k. Let V = H(B, Ω) and a(u, v) = (Bu, Bv)o + (u, v)o be a sesquilinear form on V, a(u, v) is Velliptic. Let Q = L2 (Ω). If f ∈ L2 (Ω), by § 3, there exists u ∈ N such that a(u, v) = ( f, v)o for all v ∈ V. Further (B∗ B + 1)u = f . To consider the regularity of u we consider first its tangential derivatives and next the normal ones. Proposition 13.1. Let DTp f ∈ L2 for all p ≤ µ for any positive integer µ. Then 1 Dτp u and BDτp u are in L2 for p ≤ µ. Let vh (x) = (v(x + h) − v(x)) where h h = (0, . . . , h, . . . , 0), hn = 0. Since B is with constant coefficients, vh ∈ V if v ∈ V. Hence a(u, vh) = ( f, vh )o , i.e., (Bu, Bvh)o + (u, vh )o = ( f, vh )o for all v ∈ V. Since B is with constant coefficients (Bhh, Bv) + (u−h , v)o = ( f −h , v)o . Putting v = u−h ,
(1)
Bu−h2o + u−h 2o ≤ C f −h o u−h o .
If Dτ f in L2 , Bu−h and u−h are bounded in L2 which means BDτ u and Dτ u ∈ 108
126
109 L2 . Letting h → 0 in (1), (B(Dτu), Bv)o + (Dτ u, v)o = (Dτ f, v)o . µ
(µ)
If now Dt f ∈ L2 we can repeat the process proving that if Dτ f ∈ L2 , then and BDµτ u ∈ L2 . Now we consider normal derivatives.
Dµτ u
Theorem 13.1. Let µ = m. Under the hypothesis of proposition 13.1 u ∈ H m (Ω). We use the Lemma 13.1. Let Ω = {y > 0}. Consider µ such that p Dτ u ∈ L2 for p ≤ k p 2 D D u ∈ L for p ≤ k − 1 y τ (1) and . .. .. . Dk−1 Dτp u ∈ L2 for p ≤ 1 y
(2)
−m+k Dm . yu∈ H
Then Dky ∈ L2 . If we denote by E(Ω) the space defined by all the conditions above, the lemma means E(Ω) = H k (Ω). In general, i.e., for arbitrary Ω, H k (Ω) ⊂ E(Ω). This lemma should hold for Ω with smooth boundary, though as yet it is not proved. Assuming the lemma for a moment, we complete the proof of the theorem. m−1 We have Bu = Dm + Dym−1 ∧1 u + · · · + ∧m u. y u + Dy p From proposition 13., Dτ u ∈ L2 and Dτp Bu ∈ L2 for p ≤ m. Hence ∧k u ∈ H o (Ω) and Dm−k ∧k u ∈ H −m+1 . Hence by lemma 13.1 Dy u ∈ L2 . Next we consider Dτ Bu which is in L2 . m−1 Dτ Bu = Dm ∧1 Dτ u + · · · y Dτ u + D −m+1 This gives Dm . But D p (Dτ u)L2 , p ≤ 1. By lemma 13.1 again y Dτ u ∈ H 2 Dy Dτ u ∈ L . Proceeding similarly we obtain Dky u ∈ L2 . Hence u ∈ H k (Ω). Now we prove if Ω = {xn > 0}, then E(Ω) = H k (Ω).
Lemma 13.2. If Ω = Rn , H k = E(Rn ).
127
110 Let by Fourier transformation x1 , . . . , xn−1 go into ξ1 , . . . , ξn−1 and xn into −m+k ξn . Actually we need use only Dτp u ∈ L2 p ≤ k and Dm (Ω); yu∈ H i.e., ηm uˆ ∈ L2 (1) (1 + ξk )uˆ ∈ L2 and m−k 1 + ξ + ηm−k
We may also assume m > k. We have to conclude that ηk uˆ ∈ L2 . Now we use the following inequality ηk ≤ c1 (1 + ξk ) + c2
ηm . 1 + ξm−k + ηm−k
For then ηk uˆ ∈ L2 by (1). To prove the inequality we have to prove that ηk + ξm−k ξk + ηm ≤ c1 (1 + ξk )(1 + ξm−k + ηm−k ) + c2 ηm . Since m > k, ηk ≤ c3 ηm 1 + c. Hence we need prove ηm + ξm−k ηk ≤ c1 (1 + ξk )(1 + ξm−k + ηm−k ) + c2 ηm . ! a p bq 1 1 ηkp η(m−k)q ab ≤ + + , + =1 . But ξm−k ηk ≤ p q p q p q Hence ηm + ξm−k ηk ≤ ηm + ξm . This is trivially less than right hand side of the inequality. Lemma 13.3. If ρ ∈ DL ∞ (Ω) and u ∈ E(Ω), then ρu ∈ E(Ω). This follows from the definition of E(Ω) itself. Lemma 13.4. E(Rn )Ω, i.e., restrictions of E(Rn ) to Ω is dense in E(Ω). Let ut (x) = u(x′ , y + t) for t > 0. Let vt = ut (x)Ω. vτ → u in E(Ω). Let 0 for y < −t ρ(y) = 1 for y > 1 ρ(y) ∈ DL∞ (Ω) 0 < y < 1 elsewhere.
ρuτ ∈ V by lemma 13.3, and (ρuτ )Ω = vτ . The extension f ρuc of δut are in E(Rn ) and their restrictions vτ are dense in E(Ω). ¯ is dense in E(Ω). Lemma 13.5. E(Ω) ∩ D(Ω)
From lemma 13.4, we need prove that if u = vΩ with u ∈ E(Rn ) then u can ¯ For ∪ = lim ∪×ρn with ρn → δ. be approached by function from E(Ω)∩D(Ω). The restrictions of ∪ ∗ ρn → u. To complete the proof of the lemma 13.1, then, we prove
128
14. Systems
111
¯ Then Lemma 13.6. Let u ∈ E(Ω) ∩ D(Ω).
u(x), U(x) = λ1 u(x′ , . . . , y) + λ2 u(x′ , . . . , − y ) + · · · + λn (x′ , . . . , − y )xn < 0 2 n
xn ≥ 0
is in E(Rn ) for suitable λ′ s.
¯ is dense in E(Ω), we have a continuous If we prove this since E(Ω) ∩ D(Ω) mapping π : E(Ω) → E(Rn ) = H k (Rn ). Hence E(Ω) ⊂ H k (Ω), which proves ∂R ∪ should be equal that H k (Ω) = E k (Ω). λ′ s are determined so that on y = 0, ∂y from above and below. A simple argument shows that ∪ ∈ E (Rn ).
14 Systems 14.1 We shall consider briefly systems. We shall denote in this article H m (Ω) by H m . Let H (m) = H m1 × · · · × H mν with the usual product Hilbert structures. In H (m) the closure of (D(Ω))ν is Hom1 × · · · × Hom . An element of H (m) (Ω) we → denote by u = (u1 , . . . , uν ) with ui ∈ H mi . Let V be such that Ho(m) ⊂ V ⊂ H (m) . Let XZ a(u, v) = a pq,λu (x)Dq uµ D p vλ dx λ = 1, . . . , ν µ=1,...,ν
be a sesquilinear form with a pq,λµ ∈ L∞ (Ω) and a pq,λu = 0, if p ≥ mλ or q > mµ . This last condition assures that a(u, v) is continuous on V × V. → → If a( u, u) ≥ αu2 for all u ∈ V and for α > 0 and if Q = L2 then from the general theory of § 3, there exists a space N and an operator A which establishes an isomorphism of N onto Q′ so that → →
→ →
→
hA u, ϕi = a( u, ϕ) for ϕ ∈ (D(Ω))ν, i.e.,
PR →→ →→ hA1 u ϕ 1 i + · · · + hAν u ϕ 1 i = a pqλµ Dq uµ Dλp dx. Hence X → Aλ ( u) = (−1) p D p (a pqλµ Dq uµ ) p,q,µ
129
14. Systems
112
Hence the theory solves the differential systems →
→
Aλ u = f . The variety of boundary value problems solved is much larger; e.g., if ν = 2, m1 = m2 and V may be defined as consisting of (u1 , u2 ) such that γu1 = γu2 .
14.2 We now give, following Nirenberg [13] an example which presents a little strange behaviour. We take n = 2, ν = 2, m1 = 1, m2 = 3. We write x1 = x and x2 = y, so that V = H 1 × H 3 . Let L1 , M2 , L3 , M3 , Nz be the differential operators the order of which is equal to the index. Let a(u, v) = (D x u1 , D x v1 ) + (Dy u1 , Dy v1 ) + (u1 , L∗1 v1 )+ + (−D3y u2 , Dy v1 ) + (L3 u2 , v1 ) + (Dy u1 , D3y v2 )+ + (u1 , M3∗ v2 ) + (D3x u2 , D3x v2 ) + (D3y u2 , D3y v2 )+ + 3(D2x Dy u2 , D2x Dy v2 ) + 3(D x D2y u2 , D x D2y v2 ) + (M2 u2 , N3 v2 ). Lemma 14.1. If Ω is three strongly regular, → →
→ →
a( u, v ) + λ( u, v ) elliptic for λ large enough. The system A associated with a(u, v) is →
A1 ( u) = −(D2x + D2y )u1 + L1 u + D4y u3 + L3 u2 →
A2 ( u) = −D4y u1 + M3 u1 − (D6x + D6y + 3D2x D4y )u2 + N3 M2 u1 . From the underlined term in the operator it would look like as if we have to assume v1 ∈ H 2 and u2 ∈ H 2 . While existence and uniqueness in ensured in H 1 × H 3 itself, i.e., we require four conditions on boundary while from the → → differential equation it looks as if we require five conditions. Further a( u, v ) is not elliptic on H 2 × H 3 . This happens because in computation of the real part → → of a( u, v ) the terms involving D2y u1 , e.g., (−D3y u2 , Dy v1 ) + (Dy u1 , D3y u2 ) = e, give zero real part as they are of the form z − z¯. To see that the form is H 1 × H 3 is straight forward by using the definition of strong regularity and the above remarks.
130
Bibliography [1] S. Agmon: Comm. Pure Applied Maths., X (1957), p.179239. [2] N. Aronszajn 1. Theory of Coerciveness, Laurence Tech. Report, (1954). 2. Theory of Reproducing Kernels, Trans. Amer. Math. Soc., (1950). [3] N. Aronszajn and K.T.Smith : Regularity at the boundary (forthcoming). [4] Bicadze : On the uniqueness of the solutions of the Dirichlet problem for elliptic partial differential equations, Us tekhi Math. Nauk, 3 (1948), p. 211212. [5] F.E.Browder : On the regularity properties of solutions of elliptic differential equations, Comm. Pure Appl. Math., IX(1956), p. 351361. [6] Campanato : Forthcoming paper on the regularity at the boundary for Picone problems. [7] J. Deny and J.L. Lions : Surless espaces de BeppoLevi, Amales Inst. Fourier, 1955. [8] L.Garding : Dirichlet’s problem for linear elliptic partial differential equations, Math. Scand. Vol. 1 (1953), p. 5572. [9] L.Hormander and J.L.Lions : Completion par la xxxxx Dirichlet, Math, Scand. (1956). [10] J.L.Lions: 1. Problems aux limits en theoie des distributions, Acta Math. 94 (1955), p. 13153. 2. Sur certain problemes aux limits, S.M.F. (1955). 113
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