Introduction to Variational Methods in Partial Differential Equations ...

************************************* Introduction to Variational Methods in Partial Differential Equations and Applications A Summer Course at Michigan State University (Math 890, Summer 2008) by Baisheng Yan Department of Mathematics Michigan State University [email protected]

Contents

Course Description A. Motivating Examples

i i

B. Application to Physical Problems

iv

C. Tentative Plans

v

Chapter 1.

Preliminaries

1

§1.1.

Banach Spaces

1

§1.2.

Bounded Linear Operators

5

§1.3.

Weak Convergence and Compact Operators

7

§1.4.

Spectral Theory for Compact Linear Operators

9

§1.5.

Nonlinear Functional Analysis

Chapter 2.

Sobolev Spaces

11 23

§2.1.

Weak Derivatives and Sobolev Spaces

23

§2.2.

Approximations and Extensions

27

§2.3.

Sobolev Imbedding Theorems

39

§2.4.

Equivalent Norms

47

Chapter 3.

Existence Theory for Linear Problems

51

§3.1.

Differential Equations in Divergence Form

51

§3.2.

The Lax-Milgram Theorem

54

§3.3.

G˚ arding’s Estimates and Existence Theory

55

§3.4.

Symmetric Elliptic Operators

60

Chapter 4.

Variational Methods for PDEs

65

§4.1.

Variational Problems

65

§4.2.

Multiple Integrals in the Calculus of Variations

66

§4.3.

Direct Method for Minimization

70

§4.4.

Minimization with Constraints

72

§4.5.

Mountain Pass Theorem

86 iii

iv

Contents

§4.6.

Nonexistence and Radial Symmetry

Chapter 5.

Weak Lower Semicontinuity on Sobolev Spaces

90 95

§5.1.

The Convex Case

95

§5.2.

Morrey’s Quasiconvexity

97

§5.3.

Properties of Quasiconvex Functions

102

§5.4.

Polyconvex Functions and Null-Lagrangians

108

§5.5.

Existence in Nonlinear Elasticity

116

§5.6.

Relaxation Principles

118

Bibliography

123

Index

125

Course Description

This course is intended to give an introduction to some important variational methods for certain problems in partial differential equations (PDE) and applications. The course is suitable for graduate students with some knowledge of partial differential equations. Since this is an introduction course meant to convey important ideas of the methods, the rigorous proof will be kept minimum; some details may be left to students as homework or class presentation. Attendance, class participation and some extra reading of the references may be necessary to truly learn the material.

A. Motivating Examples Variational methods provide a solid basis for the existence theory of PDE and other applied problems. We use some examples to introduce the main content of the course. Example 1 – Dirichlet’s Principle. The starting example of variational method for PDE is the Dirichlet principle for Laplace’s equation: ∆u = 0,

u|∂Ω = f,

where Ω is a given bounded domain in Rn and f is a given function on the boundary of Ω. This principle states that any classical solution u of this problem minimizes the Dirichlet integral: Z I(u) = |∇u(x)|2 dx Ω

among all smooth functions taking f on the boundary ∂Ω. Therefore, in order to solve the problem, one tries to find a minimizer of the functional I among the mentioned class of smooth functions. One of the most important methods for such a minimization problem is the direct method of the calculus of variations, which originates from the Weierstrass theorem. By, such a method, we take a minimizing sequence {uj } in the given class; i.e., lim I(uj ) = inf I(u);

j→∞

the infimum here is taken over all u in the given class. If this infimum is finite, then we know that each component of {∇uj } is a bounded sequence in L2 (Ω). The weak convergence i

ii

Course Description

theorem implies that there exist subsequence {∇ujk } and G = (g1 , · · · , gn ) ∈ L2 (Ω) such that R ∇ujk * G weakly in L2 (Ω) and hence Ω |G|2 dx ≤ inf I(u). Now the question is whether G renders a function in the given class; that is, does there exist a function u in the given class such that G = ∇u? Since, in principle, G is only in L2 (Ω), it is not clear whether such a u should exist or not. For this problem, the smoothness of uj or even the fact that ∇uj converges strongly in L2 (Ω) to G (why?) would not help much. The class in which we seek the minimizers plays an important role in guaranteeing the existence of a minimizer. We need to have a larger class where I(u) is defined and a minimizer can be found by this direct method. A minimizer in this larger class is only a weak solution to the original Laplace equation. Is it smooth and a classical solution of the Laplace equation? This is the regularity problem; we will not focus on such a problem in this course. The extension of the admissible class leads use to the class of functions with gradient in L2 (Ω) and such a class should be also complete (i.e., containing the limits of all its Cauchy sequences). This motivates the study of Sobolev space H 1 (Ω) = W 1,2 (Ω) or the general W m,p (Ω) spaces. Example 2 – Lax-Milgram Method. The second example is on the Hilbert space method (energy method) for second-order linear elliptic equations in divergence form: Lu = f in Ω,

u = 0 on ∂Ω,

where f is a given function in L2 (Ω) and Lu is a second-order linear elliptic operator: Lu = −

n X

ij

(a (x)uxi )xj +

i,j=1

n X

bi (x)uxi + c(x)u,

i=1

with ellipticity condition: for a constant θ > 0 ij

ji

a =a ;

n X

aij (x)ξi ξj ≥ θ|ξ|2 , ∀ a.e. x ∈ Ω, ξ ∈ Rn .

i,j=1

A weak solution u is defined to be a function u ∈ H = H01 (Ω) for which B[u, v] = (f, v)L2 (Ω)

∀ v ∈ H,

where B is the bilinear form associated with L:   Z n n X X  B[u, v] = aij uxi vxj + bi uxi v + cuv  dx, Ω

i,j=1

u, v ∈ H.

i=1

Note that f can also be assumed in the dual space H ∗ = H −1 (Ω); in this case, (f, v)L2 (Ω) above is replaced by the pairing hf, vi between H ∗ and H. Note also that B[u, v] satisfies the G˚ arding’s estimates (energy estimates): |B[u, v]| ≤ αkukkvk,

B[u, u] ≥ βkuk2 − γkuk2L2 (Ω) ,

where α > 0, β > 0 and γ ≥ 0 are constants (assuming aij , bi , c are L∞ (Ω) functions). The Lax-Milgram theorem says that if γ = 0 then for any f ∈ H −1 (Ω) there exists a unique u ∈ H such that B[u, v] = hf, vi for all v ∈ H. If B is also symmetric then the Lax-Milgram theorem is simply the Riesz representation theorem. We will not consider the case when γ > 0; the problem in this case involves the Fredholm alternative theorem.

A. Motivating Examples

iii

Example 3 – Mountain Pass Method. Our second example is the use of critical point theory to find a nontrivial solution to the semilinear elliptic equations of the following type: ∆u + |u|p−1 u = 0 in Ω,

u = 0 on ∂Ω,

where 1 < p < n+2 n−2 (we will see why this special exponent appears here later). If we define a functional on H = H01 (Ω) by Z 1 1 2 p+1 dx, I(u) = |∇u| − |u| p+1 Ω 2 then any critical point of I on H will be a weak solution of the above problem. It can be seen that I does not have finite infimum or superimum on H. Although there is a method solving this problem based on minimizing functional I on a manifold of H (i.e., via minimization with constraints, to be discussed later in the class), to study the critical points of I, we use the mountain pass method, which is a major contribution in the nonlinear functional analysis. The functional I can be proved a C 1 function on H with derivative I 0 : H → H being a locally Lipschitz function. A critical property of I is that it satisfies the so-called PalaisSmale condition: every sequence {uj } in H with {I(uj )} bounded and I 0 (uj ) → 0 in H is precompact in H. Furthermore, there exist positive constants r, a such that I(u) ≥ a

if kuk = r,

and there exists an element v ∈ H with kvk > r and I(v) ≤ 0. Note that I(0) = 0. Therefore, both inside and outside the mountain range kuk = r there are points where I takes a smaller value than it takes on the mountain range. Let Γ be the set of all continuous passes connecting the two lower points 0 and v. Let c = inf max I(u). g∈Γ u∈g

Then the mountain pass theorem says that c is a critical point of I; that is, there exists u ∈ H such that I(u) = c, I 0 (u) = 0. Note that c ≥ a and hence u 6= 0. Example 4 – Weak Lower Semicontinuity. The direct method also works for the minimization problem for many other integral functionals of the type Z I(u) = F (x, u(x), Du(x)) dx, Ω

where u may be even a vector valued function. An important question in this regard is when the functional I is lower semi-continuous with respect to weak convergence of W 1,p (Ω); that is, when we have I(u) ≤ lim inf I(uj ) j→∞

whenever uj * u in W 1,p (Ω).

This will motivate the study of various notion of convexity conditions. In the scalar case, the weak lower semicontinuity of I is equivalent to the convexity of function F (x, u, ξ) on the variable ξ ∈ Rn . In the vectorial case (where, say, u : Ω → RN for some N ≥ 2), the weak lower semicontinuity problem is a difficult problem and involves Morrey’s notion of quasiconvexity, which will also be discussed in the course. But, a sufficient condition for the semicontinuity will be given in terms of null-Lagrangians; this will be the polyconvexity. The vectorial case is closely related to the problems in nonlinear elasticity, harmonic maps, liquid crystals, and other physical problems.

iv

Course Description

B. Application to Physical Problems The problems in nonlinear elasticity, liquid crystals and ferromagnetics will be disucced as some physical problems which can be formulated and solved by variational methods. Problem 1 – Nonlinear Elasticity. In continuum mechanics, a material occupying a domain Ω ⊂ R3 is deformed by a map u to another domain in the same space. The material at position x ∈ Ω is deformed to a point u(x) in the deformed domain u(Ω). The nonlinear elasticity theory postulates that the total stored energy associated with the deformation u is given by Z I(u) =

F (x, u(x), Du(x)) dx, Ω

where Du(x) = (∂ui /∂xj ) is the 3 × 3-matrix of deformation gradient and F is the stored energy density function. here, adj A and det A denote the cofactor matrix and the determinant of matrix A. For elasticity problems, the constraint det Du(x) > 0 for a.e. x ∈ Ω is always assumed; this renders an additional difficulty for the variational problems. Also, material property and frame-indifference often prevent the function F (x, u, A) from being convex in A. Nevertheless, the density function sometime can be written as (or is often assumed to be) F (x, u, A) = W (x, u, A, adj A, det A), where adj A and det A denote the cofactor matrix and the determinant of matrix A, and W (x, u, A, B, t) is a convex function of (A, B, t). Exactly, this means F is polyconvex. For incompressible materials, the constraint det Du(x) = 1 is assumed. The relationship between weak convergence and determinant involves the compensated compactness, and we will discuss this using the null-Lagrangians under a higher regularity assumption. Problem 2 – Liquid Crystals. A liquid crystal is described by the orientation of the line-like molecules. Such an orientation can be thought as a unit vector n(x) ∈ S 2 at each material point x ∈ Ω, the domain occupied by the liquid crystal. The total energy, based on the Oseen-Frank model, is given by Z Z I(n) = WOF (n, Dn) dx = (κ1 (div n)2 + κ2 (n · curl n)2 + κ3 (n × curl n)2 ) dx Ω ZΩ + κ4 (tr((Dn)2 ) − (div n)2 ) dx, Ω

where κi (i = 1, 2, 3) are Frank’s curvature constants; the κ4 -term is a null-Lagrangian, depending only on the boundary data of n. If first three κi > 0, then it can be shown that the first part of WOF (n, A) is convex in A. If all ki are equal to a positive constant κ, then R I(n) reduces to the Dirichlet integral for harmonic maps: I(n) = Ω κ|Dn|2 dx. Note that the constraint |n(x)| = 1 is lower-order in H 1 (Ω; S 2 ) and thus presents no problem when using the direct method, but this nonconvex condition is the main obstacle for uniqueness and regularity. Problem 3 – Micromagnetics. In the theory of micromagnetics, one seeks the magnetization m : Ω ⊂ RN → RN of a body occupying the region Ω by minimizing the energy functional Z Z Z Z 1 α |∇m(x)|2 dx + ϕ(m(x)) dx − H · m(x) dx + |F (z)|2 dz I(m) = 2 Ω 2 N Ω Ω R

C. Tentative Plans

v

among all admissible magnetizations m satisfying m ∈ L∞ (Ω), where F ∈

L2 (RN ; RN )

|m(x)| = 1 a.e. x ∈ Ω,

is the unique field determined by the simplified Maxwell’s equations: curl F = 0,

div(−F + mχΩ ) = 0 in RN .

Here α > 0 is a material constant, ϕ is the density of anisotropy that is minimized along preferred crystallographic directions, H is a given external applied field (typically constant), and F is the induced magnetic field on the whole RN related to m via Maxwell’s equations above. The first term in the energy I(m) is called the exchange energy, the second term called the anisotropy energy, the third the external interaction energy, and the last term called the magnetostatic energy. For large domains, the α-term of the energy can be dropped, which in any case makes the problem similar to the liquid crystal problem. So we assume α = 0. Also, we will restrict attention to the two-dimensional case N = 2, although N = 3 is the most physical case. We will show that the minimization in this simplified case can be reformulated as a minimization problem for nonconvex variational integral of the form Z Z 1 |∇u(x)|2 dx, J(u) = φ(∇u(x)) dx + 2 R2 \Ω Ω where φ is a nonconvex function. Such a functional differs from what we considered above since it is defined also on the complement of Ω. If time permits, I will discuss some new results including the relaxation principle for J(u).

C. Tentative Plans 1. Preliminaries. I plan to start with some preliminary results in functional analysis using some materials of a lecture note by Professor Dunninger; most of the materials can be found in any textbook on real and functional analysis (also in Appendix D of Evans’ book, Partial Differential Equations, AMS Press). 2. Sobolev Spaces. I will do the Sobolev spaces using Evans’ book, Chapter 5, and sometime refer to the new book Sobolev Spaces, Academic Press, by Adams and Fournier. 3. Variational Methods for PDEs. I will follow mainly Chapters 6 and 8 of Evans’ book on the variational methods discussed above, but will add some materials deemed to be useful in understanding and developing the methods. If time permits, the maximum principles and Sections 9.4.2 and 9.5.2 of Evans’ book may also be included. 4. Applications. I will briefly discuss the physical applied problems mentioned above when a method can be applied to the problem. Some of the materials will come from several research articles, including my own. If time permits, I will discuss the nonconvex variational problems and relaxation principle and the Young measure solutions to such problems; explicit example will be the 2-D micromagnetics problem.

Chapter 1

Preliminaries

1.1. Banach Spaces 1.1.1. Vector Spaces. A (real) vector space is a set X, whose elements are called vectors, and in which two operations, addition and scalar multiplication, are defined as follows: (a) To every pair of vectors x and y corresponds a vector x + y in such a way that x+y =y+x

and

x + (y + z) = (x + y) + z.

X contains a unique vector 0 (the zero vector or origin of X) such that x+0 = x for every x ∈ X, and to each x ∈ X corresponds a unique vector −x such that x + (−x) = 0. (b) To every pair (α, x), with α ∈ R and x ∈ X, corresponds a vector αx in such a way that 1x = x,

α(βx) = (αβ)x

and such that the two distributive laws α(x + y) = αx + αy,

(α + β)x = αx + βx

hold. A nonempty subset M of a vector space X is called a subspace of X if αx + βy ∈ M for all x, y ∈ M and all α, β ∈ R. A subset M of a vector space X is said to be convex if tx + (1 − t)y ∈ M whenever t ∈ (0, 1), x, y ∈ M . (Clearly, every subspace of X is convex.) Let x1 , . . . , xn be elements of a vector space X. The set of all α1 x1 + · · · + αn xn , with αi ∈ R, is called the span of x1 , . . . , xn and is denoted by span{x1 , . . . , xn }. The elements x1 , . . . , xn are said to be linearly independent if α1 x1 +· · ·+αn xn = 0 implies that αi = 0 for each i. If, on the other hand, α1 x1 + · · · + αn xn = 0 does not imply αi = 0 for each i, the elements x1 , . . . , xn are said to be linearly dependent . An arbitrary collection of vectors is said to be linearly independent if every finite subset of distinct elements is linearly independent. The dimension of a vector space X, denoted by dim X, is either 0, a positive integer or ∞. If X = {0} then dim X = 0; if there exist linearly independent {u1 , . . . , un } such 1

2

1. Preliminaries

that each x ∈ X has a (unique) representation of the form x = α1 u1 + · · · + αn un

with

αi ∈ R

then dim X = n and {u1 , . . . , un } is a basis for X; in all other cases dim X = ∞. 1.1.2. Normed Spaces. A (real) vector space X is said to be a normed space if to every x ∈ X there is associated a nonnegative real number kxk, called the norm of x, in such a way that (a) kx + yk ≤ kxk + kyk for all x and y in X (Triangle inequality) (b) kαxk = |α|kxk for all x ∈ X and all α ∈ R (c) kxk > 0 if x 6= 0. Note that (b) and (c) imply that kxk = 0 iff x = 0. Moreover, it easily follows from (a) that |kxk − kyk| ≤ kx − yk for all

x, y ∈ X.

1.1.3. Completeness and Banach Spaces. A sequence {xn } in a normed space X is called a Cauchy sequence if, for each > 0, there exists an integer N such that kxm − xn k < for all m, n ≥ N . We say xn → x in X if limn→∞ kxn − xk = 0 and, in this case, x is called the limit of {xn }. X is called complete if every Cauchy sequence in X converges to a limit in X. A complete (real) normed space is called a (real) Banach space. A Banach space is separable if it contains a countable dense set. It can be shown that a subspace of a separable Banach space is itself separable. Example 1.1. Let Ω be an open subset of Rn , n ≥ 1. The set C(Ω) of (real-valued) continuous functions defined on Ω is an infinite dimensional vector space with the usual definitions of addition and scalar multiplication: for f, g ∈ C(Ω), x ∈ Ω

(f + g)(x) = f (x) + g(x)

(αf )(x) = αf (x) for α ∈ R, f ∈ C(Ω), x ∈ Ω. ¯ C(Ω) consists of those functions which are uniformly continuous on Ω. Each such function ¯ C0 (Ω) consists of those functions which are continuous has a continuous extension to Ω. in Ω and have compact support in Ω. (The support of a function f defined on Ω is the closure of the set {x ∈ Ω : f (x) 6= 0} and is denoted by supp(f ).) The latter two spaces are clearly subspaces of C(Ω). For each n-tuple α = (α1 , . . . , αn ) of nonnegative integers, we denote by Dα the partial derivative D1α1 · · · Dnαn , Di = ∂/∂xi of order |α| = α1 + · · · + αn . If |α| = 0, then D0 = I(identity). For integers m ≥ 0, let C m (Ω) be the collection of all f ∈ C(Ω) such that Dα f ∈ C(Ω) for all α with |α| ≤ m. We write f ∈ C ∞ (Ω) iff f ∈ C m (Ω) for all m ≥ 0. For m ≥ 0, define C0m (Ω) = C0 (Ω) ∩ C m (Ω) and let C0∞ (Ω) = C0 (Ω) ∩ C ∞ (Ω). The spaces C m (Ω), C ∞ (Ω), C0m (Ω), C0∞ (Ω) are all subspaces of the vector space C(Ω). Similar defini¯ etc. tions can be given for C m (Ω) For m ≥ 0, define X to be the set of all f ∈ C m (Ω) for which X kf km,∞ ≡ sup |Dα f (x)| < ∞. |α|≤m

Ω

1.1. Banach Spaces

3

Then X is a Banach space with norm k · km,∞ . To prove, for example, the completeness when m = 0, we let {fn } be a Cauchy sequence in X, i.e., assume for any ε > 0 there is a number N (ε) such that for all x ∈ Ω sup |fn (x) − fm (x)| < ε if m, n > N (ε). x∈Ω

But this means that {fn (x)} is a uniformly Cauchy sequence of bounded continuous functions, and thus converges uniformly to a bounded continuous function f (x). Letting m → ∞ in the above inequality shows that kfn − f km,∞ → 0. Note that the same proof is valid for the set of bounded continuous scalar-valued functions defined on a nonempty subset of a normed space X. Example 1.2. Let Ω be a nonempty Lebesgue measurable set in Rn . For p ∈ [1, ∞), we denote by Lp (Ω) the set of equivalence classes of Lebesgue measurable functions on Ω for which Z 1 p p kf kp ≡ |f (x)| dx < ∞. Ω

(Two functions belong to the same equivalence class, i.e., are equivalent, if they differ only on a set of measure 0.) Let L∞ (Ω) denote the set of equivalence classes of Lebesgue measurable functions on Ω for which kf k∞ ≡ ess-supx∈Ω |f (x)| < ∞. Then Lp (Ω), 1 ≤ p ≤ ∞, are Banach spaces with norms k · kp . For p ∈ [1, ∞] we write f ∈ Lploc (Ω) iff f ∈ Lp (K) for each compact set K ⊂ Ω. For the sake of convenience, we will also consider Lp (Ω) as a set of functions. With this convention in mind, we can assert that C0 (Ω) ⊂ Lp (Ω). In fact, if p ∈ [1, ∞), then as we shall show later, C0 (Ω) is dense in Lp (Ω). The space Lp (Ω) is also separable if p ∈ [1, ∞). This follows easily, when Ω is compact, from the last remark and the Weierstrass approximation theorem. Finally we recall that if p, q, r ∈ [1, ∞] with p−1 + q −1 = r−1 , then Hölder’s inequality implies that if f ∈ Lp (Ω) and g ∈ Lq (Ω), then f g ∈ Lr (Ω) and kf gkr ≤ kf kp kgkq . Example 1.3. The Cartesian product X × Y , of two vector spaces X and Y , is itself a vector space under the following operations of addition and scalar multiplication: [x1 , y1 ] + [x2 , y2 ] = [x1 + x2 , y1 + y2 ] α[x, y] = [αx, αy]. If in addition, X and Y are normed spaces with norms k · kX , k · kY respectively, then X × Y becomes a normed space under the norm k[x, y]k = kxkX + kykY . Moreover, under this norm, X × Y becomes a Banach space provided X and Y are Banach spaces.

4

1. Preliminaries

1.1.4. Hilbert Spaces. Let H be a real vector space. H is said to be an inner product space if to every pair of vectors x and y in H there corresponds a real-valued function (x, y), called the inner product of x and y, such that (a) (x, y) = (y, x) for all x, y ∈ H (b) (x + y, z) = (x, z) + (y, z) for all x, y, z ∈ H (c) (λx, y) = λ(x, y) for all x, y ∈ H, λ ∈ R (d) (x, x) ≥ 0 for all x ∈ H, and (x, x) = 0 if and only if x = 0. For x ∈ H we set kxk = (x, x)1/2 .

(1.1)

Theorem 1.4. If H is an inner product space, then for all x and y in H, it follows that (a) |(x, y)| ≤ kxk kyk

(Cauchy-Schwarz inequality);

(b) kx + yk ≤ kxk + kyk (c) kx +

yk2

+ kx −

yk2

(Triangle inequality);

= 2(kxk2 + kyk2 )

(Parallelogram law).

Proof. (a) is obvious if x = 0, and otherwise it follows by taking δ = −(x, y)/kxk2 in 0 ≤ kδx + yk2 = |δ|2 kxk2 + 2δ(x, y) + kyk2 . This identity, with δ = 1, and (a) imply (b). (c) follows easily by using (1.1).

Furthermore, by (d), equation (1.1) defines a norm on an inner product space H. If H is complete under this norm, then H is said to be a Hilbert space. Example 1.5. The space L2 (Ω) is a Hilbert space with inner product Z (f, g) = f (x)g(x) dx for all f, g ∈ L2 (Ω). Ω

Theorem 1.6. Every nonempty closed convex subset S of a Hilbert space H contains a unique element of minimal norm. Proof. Choose xn ∈ S so that kxn k → d ≡ inf{kxk : x ∈ S}. Since (1/2)(xn + xm ) ∈ S, we have kxn + xm k2 ≥ 4d2 . Using the parallelogram law, we see that (1.2)

kxn − xm k2 ≤ 2(kxn k2 − d2 ) + 2(kxm k2 − d2 )

and therefore {xn } is a Cauchy sequence in H. Since S is closed, {xn } converges to some x ∈ S and kxk = d. If y ∈ S and kyk = d, then the parallelogram law implies, as in (1.2), that x = y. If (x, y) = 0, then x and y are said to be orthogonal, written sometimes as x ⊥ y. For M ⊂ H, the orthogonal complement of M , denoted by M ⊥ , is defined to be the set of all x ∈ H such that (x, y) = 0 for all y ∈ M . It is easily seen that M ⊥ is a closed subspace of H. Moreover, if M is a dense subset of H and if x ∈ M ⊥ , then in fact, x ∈ H ⊥ which implies x = 0. Theorem 1.7. (Projection) Suppose M is a closed subspace of a Hilbert space H. Then for each x ∈ H there exist unique y ∈ M , z ∈ M ⊥ such that x = y + z. The element y is called the projection of x onto M .

1.2. Bounded Linear Operators

5

Proof. Let S = {x − y : y ∈ M }. It is easy to see that S is convex and closed. Theorem 1.6 implies that there exists a y ∈ M such that kx − yk ≤ kx − wk for all w ∈ M . Let z = x − y. For an arbitrary w ∈ M , w 6= 0, let α = (z, w)/kwk2 and note that kzk2 ≤ kz − αwk2 = kzk2 − |(z, w)/kwk|2 which implies (z, w) = 0. Therefore z ∈ M ⊥ . If x = y 0 + z 0 for some y 0 ∈ M , z 0 ∈ M ⊥ , then y 0 − y = z − z 0 ∈ M ∩ M ⊥ = {0}, which implies uniqueness. Remark. In particular, if M is a proper closed subspace of H, then there is a nonzero element in M ⊥ . Indeed, for x ∈ H\M , let y be the projection of x on M . Then z = x − y is a nonzero element of M ⊥ .

1.2. Bounded Linear Operators 1.2.1. Operators on a Banach Space. Let X, Y be real vector spaces. A map T : X → Y is said to be a linear operator from X to Y if T (αx + βy) = αT x + βT y for all x, y ∈ D(T ) and all α, β ∈ R. Let X, Y be normed spaces. A linear operator T from X to Y is said to be bounded if there exists a constant m > 0 such that (1.3)

kT xk ≤ mkxk

for all

x ∈ X.

We define the operator norm kT k of T by (1.4)

kT k =

kT xk =

sup x∈X, kxk=1

sup

kT xk.

x∈X, kxk≤1

The collection of all bounded linear operators T : X → Y will be denoted by B(X, Y ). We shall also set B(X) = B(X, X) when X = Y . Observe that kT Sk ≤ kT kkSk

if S ∈ B(X, Y ), T ∈ B(Y, Z).

Theorem 1.8. If X and Y are normed spaces, then B(X, Y ) is a normed space with norm defined by equation (1.4). If Y is a Banach space, then B(X, Y ) is also a Banach space. Proof. It is easy to see that B(X, Y ) is a normed space. To prove completeness, assume that {Tn } is a Cauchy sequence in B(X, Y ). Since (1.5)

kTn x − Tm xk ≤ kTn − Tm kkxk

we see that, for fixed x ∈ X, {Tn x} is a Cauchy sequence in Y and therefore we can define a linear operator T by T x = lim Tn x for all x ∈ X. n→∞

If ε > 0, then the right side of (1.5) is smaller than εkxk provided that m and n are large enough. Thus, (letting n → ∞) kT x − Tm xk ≤ εkxk

for all large enough m.

Hence, kT xk ≤ (kTm k + ε)kxk, which shows that T ∈ B(X, Y ). Moreover, kT − Tm k < ε for all large enough m. Hence, limn→∞ Tn = T . The following theorems are important in linear functional analysis; see, e.g., [3]. Theorem 1.9. (Banach-Steinhaus) Let X be a Banach space and Y a normed space. If A ⊂ B(X, Y ) is such that supT ∈A kT xk < ∞ for each fixed x ∈ X, then supT ∈A kT k < ∞.

6

1. Preliminaries

Theorem 1.10. (Bounded Inverse) If X and Y are Banach spaces and if T ∈ B(X, Y ) is one-to-one and onto, then T −1 ∈ B(Y, X). 1.2.2. Dual Spaces and Reflexivity. When X is a (real) normed space, the Banach space B(X, R) will be called the (normed) dual space of X and will be denoted by X*. Elements of X* are called bounded linear functionals or continuous linear functionals on X. Frequently, we shall use the notation hf, xi to denote the value of f ∈ X* at x ∈ X. Using this notation we note that |hf, xi| ≤ kf k kxk for all f ∈ X*, x ∈ X. Example 1.11. Suppose 1 < p, q < ∞ satisfy 1/p + 1/q = 1 and let Ω be a nonempty Lebesgue measurable set in Rn . Then Lp (Ω)∗ = Lq (Ω). The case of p = ∞ is different. The dual of L∞ is much larger then L1 . The following results can be found in [3]. Theorem 1.12. (Hahn-Banach) Let X be a normed space and Y a subspace of X. Assume f ∈ Y ∗ . Then there exists a bounded linear functional f˜ ∈ X ∗ such that hf˜, yi = hf, yi ∀ y ∈ Y, kf˜kX ∗ = kf kY ∗ . Corollary 1.13. Let X be a normed space and x0 6= 0 in X. Then there exists f ∈ X ∗ such that kf k = 1, hf, x0 i = kx0 k. The dual space X ∗∗ of X ∗ is called the second dual space of X and is again a Banach space. Note that to each x ∈ X we can associate a unique Fx ∈ X ∗∗ by Fx (f ) = hf, xi for all f ∈ X ∗ . From Corollary 1.13, one can also show that kFx k = kxk. Thus, the (canonical) mapping J : X → X**, given by Jx = Fx , is a linear isometry of X onto the subspace J(X) of X ∗∗ . Since J is one-to-one, we can identify X with J(X). A Banach space X is called reflexive if its canonical map J is onto X ∗∗ . For example, all Lp spaces with 1 < p < ∞ are reflexive. We shall need the following properties of reflexive spaces. Theorem 1.14. Let X and Y be Banach spaces. (a) X is reflexive iff X* is reflexive. (b) If X is reflexive, then a closed subspace of X is reflexive. (c) Let T : X → Y be a linear bijective isometry. If Y is reflexive, then X is reflexive. 1.2.3. Bounded Linear Functionals on a Hilbert Space. Theorem 1.15. (Riesz Representation) If H is a Hilbert space and f ∈ H*, then there exists a unique y ∈ H such that f (x) = hf, xi = (x, y)

for all

x ∈ H.

Moreover, kf k = kyk. Proof. If f (x) = 0 for all x, take y = 0. Otherwise, there is an element z ∈ N (f )⊥ such that kzk = 1. (Note that the linearity and continuity of f implies that N (f ) is a closed subspace of H.) Put u = f (x)z −f (z)x. Since f (u) = 0, we have u ∈ N (f ). Thus (u, z) = 0, which implies f (x) = f (x)(z, z) = f (z)(x, z) = (x, f (z)z) = (x, y),

1.3. Weak Convergence and Compact Operators

7

where y = f (z)z. To prove uniqueness, suppose (x, y) = (x, y 0 ) for all x ∈ H. Then in particular, (y − y 0 , y − y 0 ) = 0, which implies y = y 0 . From the Cauchy-Schwarz inequality we get |f (x)| ≤ kxkkyk, which yields kf k ≤ kyk. The reverse inequality follows by choosing x = y in the representation. Corollary 1.16. H is reflexive. Let T : H → H be an operator on the Hilbert space H. We define the Hilbert space adjoint T * : H → H as follows: (T x, y) = (x, T *y)

for all

x, y ∈ H.

The adjoint operator is easily seen to be linear. Theorem 1.17. Let H be a Hilbert space. If T ∈ B(H), then T * ∈ B(H) and kT k = kT *k. Proof. For any y ∈ H and all x ∈ H, set f (x) = (T x, y). Then it is easily seen that f ∈ H ∗ . Hence by the Riesz representation theorem, there exists a unique z ∈ H such that (T x, y) = (x, z) for all x ∈ H, i.e., D(T ∗ ) = H. Moreover, kT ∗ yk = kzk = kf k ≤ kT kkyk, i.e., T ∗ ∈ B(H) and kT ∗ k ≤ kT k. The reverse inequality follows easily from kT xk2 = (T x, T x) = (x, T ∗ T x) ≤ kT xkkT ∗ kkxk.

1.3. Weak Convergence and Compact Operators 1.3.1. Weak Convergence. Let X be a normed space. A sequence xn ∈ X is said to be weakly convergent to an element x ∈ X, written xn * x, if hf, xn i → hf, xi for all f ∈ X ∗. Theorem 1.18. Let {xn } be a sequence in X. (a) Weak limits are unique. (b) If xn → x, then xn * x. (c) If xn * x, then {xn } is bounded and kxk ≤ lim inf kxn k. Proof. To prove (a), suppose that x and y are both weak limits of the sequence {xn } and set z = x − y. Then hf, zi = 0 for every f ∈ X ∗ and by Corollary 1.13, z = 0. To prove (b), let f ∈ X ∗ and note that xn → x implies hf, xn i → hf, xi since f is continuous. To prove (c), assume xn * x and consider the sequence {Jxn } of elements of X ∗∗ , where J : X → X ∗∗ is the bounded operator defined above. For each f ∈ X ∗ , sup |Jxn (f )| = sup |hf, xn i| < ∞ (since hf, xn i converges). By the Banach-Steinhaus Theorem, there exists a constant c such that kxn k = kJxn k ≤ c which implies {xn } is bounded. Finally, for f ∈ X ∗ |hf, xi| = lim |hf, xn i| ≤ lim inf kf kkxn k = kf k lim inf kxn k which implies the desired inequality since kxk = supkf k=1 |hf, xi|.

We note that in a Hilbert space H, the Riesz representation theorem implies that xn * x means (xn , y) → (x, y) for all y ∈ H. Moreover, we have (xn , yn ) → (x, y) if xn * x, yn → y. This follows from the estimate |(x, y) − (xn , yn )| = |(x − xn , y) − (xn , yn − y)| ≤ |(x − xn , y)| + kxn kky − yn k and the fact that kxn k is bounded.

8

1. Preliminaries

The main result of this section is given by: Theorem 1.19. If X is a reflexive Banach space, then the closed unit ball is weakly compact, i.e., the sequence {xn }, with kxn k ≤ 1 has a subsequence which converges weakly to an x with kxk ≤ 1. 1.3.2. Compact Operators. Let X and Y be normed spaces. An operator T : X → Y is said to be compact if it maps bounded sets in X into relatively compact sets in Y , i.e., if for every bounded sequence {xn } in X, {T xn } has a subsequence which converges to some element of Y . Since relatively compact sets are bounded, it follows that a compact operator is bounded. On the other hand, since bounded sets in finite-dimensional spaces are relatively compact, it follows that a bounded operator with finite dimensional range is compact. It can be shown that the identity map I : X → X (kIxk = kxk) is compact iff X is finite-dimensional. Finally we note that the operator ST is compact if (a) T : X → Y is compact and S : Y → Z is continuous or (b) T is bounded and S is compact. One of the main methods of proving the compactness of certain operators is based upon the Ascoli theorem. Let Ω be a subset of the normed space X. A set S ⊂ C(Ω) is said to be equicontinuous if for each ε > 0 there exists a δ > 0 such that |f (x) − f (y)| < ε for all x, y ∈ Ω with kx − yk < δ and for all f ∈ S. Theorem 1.20. (Ascoli) Let Ω be a relatively compact subset of a normed space X and let S ⊂ C(Ω). Then S is relatively compact if it is bounded and equicontinuous. Remark. In other words, every bounded equicontinuous sequence of functions has a uniformly convergent subsequence. Theorem 1.21. Let X and Y be Banach spaces. If Tn : X → Y are linear and compact for n ≥ 1 and if limn→∞ kTn − T k = 0, then T is compact. Thus, compact operators form a closed, but not a dense, subspace of B(X, Y ). Proof. Let {xn } be a sequence in X with M = supn kxn k < ∞. Let A1 denote an infinite set of integers such the sequence {T1 xn }n∈A1 converges. For k ≥ 2 let Ak ⊂ Ak−1 denote an infinite set of integers such that the sequence {Tk xn }n∈Ak converges. Choose n1 ∈ A1 and nk ∈ Ak , nk > nk−1 for k ≥ 2. Choose ε > 0. Let k be such that kT − Tk kM < ε/4 and note that kT xni − T xnj k ≤ k(T − Tk )(xni − xnj )k + kTk xni − Tk xnj k < ε/2 + kTk xni − Tk xnj k. ∞ Since {Tk xni }∞ i=1 converges, {T xni }i=1 is a Cauchy sequence.

Theorem 1.22. Let X and Y be normed spaces. (a) If T ∈ B(X, Y ), then T is weakly continuous, i.e., xn * x

implies

T xn * T x.

(b) If T : X → Y is weakly continuous and X is a reflexive Banach space, then T is bounded. (c) If T ∈ B(X, Y ) is compact, then T is strongly continuous, i.e., xn * x

implies

T xn → T x.

1.4. Spectral Theory for Compact Linear Operators

9

(d) If T : X → Y is strongly continuous and X is a reflexive Banach space, then T is compact. Proof. (a) Let xn * x. Then for every g ∈ Y ∗ hg, T xn i = hT ∗ g, xn i → hT ∗ g, xi = hg, T xi. (b) If not, there is a bounded sequence {xn } such that kT xn k → ∞. Since X is reflexive, {xn } has a weakly convergent subsequence, {xn0 }, and so {T xn0 } also converges weakly. But then {T xn0 } is bounded, which is a contradiction. (c) Let xn * x. Since T is compact and {xn } is bounded, there is a subsequence {xn0 } such that T xn0 → z, and thus T xn0 * z. By (a), T xn * T x, and so T xn0 → T x. Now it is easily seen that every subsequence of {xn } has a subsequence, say {xn0 }, such that T xn0 → T x. But this implies the whole sequence T xn → T x (See the appendix). (d) Let {xn } be a bounded sequence. Since X is reflexive, there is a subsequence {xn0 } such that xn0 * x. Hence T xn0 → T x, which implies T is compact. Theorem 1.23. Let H be a Hilbert space. If T : H → H is linear and compact, then T * is compact. Proof. Let {xn } be a sequence in H satisfying kxn k ≤ m. The sequence {T *xn } is therefore bounded, since T * is bounded. Since T is compact, by passing to a subsequence if necessary, we may assume that the sequence {T T *xn } converges. But then kT *(xn − xm )k2 = (xn − xm , T T *(xn − xm )) ≤ 2mkT T *(xn − xm )k → 0 as m, n → ∞. Since H is complete, the sequence {T *xn } is convergent and hence T * is compact.

1.4. Spectral Theory for Compact Linear Operators 1.4.1. Fredholm Alternative. Theorem 1.24. (Fredholm Alternative) Let T : H → H be a compact linear operator on the Hilbert space H. Then equations (I − T )x = 0, (I − T ∗ )x∗ = 0 have the same finite number of linearly independent solutions. Moreover, (a) For y ∈ H, the equation (I − T )x = y has a solution iff (y, x∗ ) = 0 for every solution x∗ of (I − T ∗ )x∗ = 0. (b) For z ∈ H, the equation (I − T ∗ )x∗ = z has a solution iff (z, x) = 0 for every solution x of (I − T )x = 0. (c) The inverse operator (I − T )−1 ∈ B(H) whenever it exists. 1.4.2. Spectrum of Compact Operators. A subset S of a Hilbert space H is said to be an orthonormal set if each element of S has norm 1 and if every pair of distinct elements in S is orthogonal. It easily follows that an orthonormal set is linearly independent. An P orthonormal set S is said to be complete if x = φ∈S (x, φ)φ for all x ∈ H. It can be shown that (x, φ) 6= 0 for at most countably many φ ∈ S. This series is called the Fourier series for x with respect to the orthonormal set {φ}. Let {φi }∞ i=1 be a countable orthonormal set PN 2 in H. Upon expanding kx − n=1 (x, φn )φn k , we arrive at Bessel’s inequality: ∞ X n=1

|(x, φn )|2 ≤ kxk2 .

10

1. Preliminaries

Let T : D(T ) ⊂ H → H be a linear operator on the real Hilbert space H. The set ρ(T ) of all scalars λ ∈ R for which (T − λI)−1 ∈ B(H) is called the resolvent set of T . The operator R(λ) = (T − λI)−1 is known as the resolvent of T . σ(T ) = R \ ρ(T ) is called the spectrum of T . It can be shown that ρ(T ) is an open set and σ(T ) is a closed set. The set of λ ∈ R for which there exists a nonzero x ∈ N (T − λI) is called the point spectrum of T and is denoted by σp (T ). The elements of σp (T ) are called the eigenvalues of T and the nonzero members of N (T − λI) are called the eigenvectors (or eigenfunctions if X is a function space) of T . If T is compact and λ 6= 0, then by the Fredholm alternative, either λ ∈ σp (T ) or λ ∈ ρ(T ). Moreover, if H is infinite-dimensional, then 0 6∈ ρ(T ); otherwise, T −1 ∈ B(H) and T −1 T = I is compact. As a consequence, σ(T ) consists of the nonzero eigenvalues of T together with the point 0. The next result shows that σp (T ) is either finite or a countably infinite sequence tending to zero. Theorem 1.25. Let T : X → X be a compact linear operator on the normed space X. Then for each r > 0 there exist at most finitely many λ ∈ σp (T ) for which |λ| > r. 1.4.3. Symmetric Compact Operators. The next result implies that a symmetric compact operator on a Hilbert space has at least one eigenvalue. On the other hand, an arbitrary bounded, linear, symmetric operator need not have any eigenvalues. As an example, let T : L2 (0, 1) → L2 (0, 1) be defined by T u(x) = xu(x). Theorem 1.26. Suppose T ∈ B(H) is symmetric, i.e., (T x, y) = (x, T y) for all x, y ∈ H. Then kT k = sup |(T x, x)|. kxk=1

Moreover, if H 6= {0}, then there exists a real number λ ∈ σ(T ) such that |λ| = kT k. If λ ∈ σp (T ), then in absolute value λ is the largest eigenvalue of T . Proof. Clearly m ≡ supkxk=1 |(T x, x)| ≤ kT k. To show kT k ≤ m, observe that for all x, y ∈ H 2(T x, y) + 2(T y, x) = (T (x + y), x + y) − (T (x − y), x − y) ≤ m(kx + yk2 + kx − yk2 ) = 2m(kxk2 + kyk2 ) where the last step follows from the paralleogram law. (kxk/kT xk)T x, then

Hence, if T x 6= 0 and y =

2kxkkT xk = (T x, y) + (y, T x) ≤ m(kxk2 + kyk2 ) = 2mkxk2 which implies kT xk ≤ mkxk. Since this is also valid when T x = 0, we have kT k ≤ m. To prove the ‘moreover’ part, choose xn ∈ H such that kxn k = 1 and kT k = limn→∞ |(T xn , xn )|. By renaming a subsequence of {xn }, we may assume that (T xn , xn ) converge to some real number λ with |λ| = kT k. Observe that k(T − λ)xn k2 = kT xn k2 − 2λ(T xn , xn ) + λ2 kxn k2 ≤ 2λ2 − 2λ(T xn , xn ) → 0. We now claim that λ ∈ σ(T ). Otherwise, we arrive at the contradiction 1 = kxn k = k(T − λ)−1 (T − λ)xn k ≤ k(T − λ)−1 k k(T − λ)xn k → 0.

1.5. Nonlinear Functional Analysis

11

Finally, we note that if T φ = µφ, with kφk = 1, then |µ| = |(T φ, φ)| ≤ kT k which implies the last assertion of the theorem. Finally we have the following result. Theorem 1.27. Let H be a separable Hilbert space and suppose T : H → H is linear, symmetric and compact. Then there exists a countable complete orthonormal set in H consisting of eigenvectors of T .

1.5. Nonlinear Functional Analysis In this final preliminary section, we list some useful results in nonlinear functional analysis. Lots of the proof and other results can be found in the volumes of Zeidler’s book [41]. 1.5.1. Contraction Mapping Theorem. Let X be a normed space. A map T : X → X is called a contraction if there exists a number k < 1 such that (1.6)

kT x − T yk ≤ kkx − yk

x, y ∈ X.

for all

Theorem 1.28. (Contraction Mapping) Let T : S ⊂ X → S be a contraction on the closed nonempty subset S of the Banach space X. Then T has a unique fixed point, i.e., there exists a unique solution x ∈ S of the equation T x = x. Moreover, x = limn→∞ T n x0 for any choice of x0 ∈ S. Proof. To prove uniqueness, suppose T x = x, T y = y. Since k < 1, we get x = y from kx − yk = kT x − T yk ≤ kkx − yk. To show that T has a fixed point we set up an iteration procedure. For any x0 ∈ S set xn+1 = T xn , n = 0, 1, ... Note that xn+1 ∈ S and xn+1 = T n+1 x0 . We now claim that {xn } is a Cauchy sequence. Indeed, for any integers n, p kxn+p − xn k = kT

n+p

n

x0 − T x0 k ≤

n+p−1 X

kT j+1 x0 − T j x0 k

j=n

≤

n+p−1 X

k j kT x0 − x0 k ≤

j=n

kn kT x0 − x0 k. 1−k

Hence as n → ∞, kxn+p − xn k → 0 independently of p, so that {xn } is a Cauchy sequence with limit x ∈ S. Since T is continuous, we have T x = lim T xn = lim xn+1 = x n→∞

n→∞

and thus x is the unique fixed point. Note that the fixed point x is independent of x0 since x is a fixed point and fixed points are unique. Our main existence result will be based upon the following so-called method of continuity or continuation method.

12

1. Preliminaries

Theorem 1.29. Let T0 , T1 ∈ B(X, Y ), where X is a Banach space and Y is a normed space. For each t ∈ [0, 1] set Tt = (1 − t)T0 + tT1 and suppose there exists a constant c > 0 such that for all t ∈ [0, 1] and x ∈ X kxkX ≤ ckTt xkY .

(1.7) Then R(T1 ) = Y if R(T0 ) = Y .

Proof. Set S = {t ∈ [0, 1] : R(Tt ) = Y }. By hypothesis, 0 ∈ S. We need to show that 1 ∈ S. In this direction we will show that if τ > 0 and τ c(kT1 k + kT0 k) < 1, then [0, s] ⊂ S

(1.8)

implies

[0, s + τ ] ⊂ S.

(Note that any smaller τ works.) Since τ can be chosen independently of s, (1.8) applied finitely many times gets us from 0 ∈ S to 1 ∈ S. Let s ∈ S. For t = s + τ, Tt x = f is equivalent to the equation Ts x = f + τ T0 x − τ T1 x.

(1.9)

By (1.7), Ts−1 : Y → X exists and kTs−1 k ≤ c. Hence (1.9) is equivalent to x = Ts−1 (f + τ T0 x − τ T1 x) ≡ Ax

(1.10)

and for A : X → X we have for all x, y ∈ X kAx − Ayk ≤ τ c(kT1 k + kT0 k)kx − yk. By the contraction mapping theorem, (1.10) has a solution and this completes the proof. 1.5.2. Nemytskii Operators. Let Ω be a nonempty measurable set in Rn and let f : Ω × R → R be a given function. Assume (i) for every ξ ∈ R, f (x, ξ) (as a function of x) is measurable on Ω (ii) for almost all x ∈ Ω, f (x, ξ) (as a function of ξ) is continuous on R (iii) for all (x, ξ) ∈ Ω × R |f (x, ξ)| ≤ a(x) + b|ξ|p/q where b is a fixed nonnegative number, a ∈ Lq (Ω) is nonnegative and 1 < p, q < ∞, 1/p + 1/q = 1. Note that p/q = p − 1. Then the Nemytskii operator N is defined by N u(x) = f (x, u(x)), x ∈ Ω. We have the following result needed later. Lemma 1.30. N : Lp (Ω) → Lq (Ω) is continuous and bounded with (1.11)

kN ukq ≤ const (kakq + kukp/q p )

for all

u ∈ Lp (Ω)

and Z (1.12)

hN u, vi =

f (x, u(x))v(x)dx

for all

u, v ∈ Lp (Ω).

Ω

Here h·, ·i denotes the duality pairing between Lp (Ω) and Lq (Ω).


13

Proof. Since u ∈ Lp (Ω), the function u(x) is measurable on Ω and thus, and P (ii), the Pn by (i) r function f (x, u(x)) is also measurable on Ω. From the inequality ( i=1 ξi ) ≤ c ni=1 ξir and (iii) we get |f (x, u(x))|q ≤ const(|a(x)|q + |u(x)|p ). Integrating over Ω and applying the above inequality once more yields (1.11), which shows that N is bounded. To show that N is continuous, let un → u in Lp (Ω). Then there is a subsequence {un0 } and a function v ∈ Lp (Ω) such that un0 (x) → u(x) a.e. and |un0 (x)| ≤ v(x) a.e. for all n. Hence Z q kN un0 − N ukq = |f (x, un0 (x)) − f (x, u(x))|q dx Ω Z ≤ const (|f (x, un0 (x))|q + |f (x, u(x))|q )dx ZΩ ≤ const (|a(x)|q + |v(x)|p + |u(x)|p )dx. Ω

By (ii), f (x, u (x)) − f (x, u(x)) → 0 as n → ∞ for almost all x ∈ Ω. The dominating convergence theorem implies that kN un0 − N ukq → 0. By repeating this procedure for any subsequence of un0 , it follows that kN un − N ukq → 0 which implies that N is continuous. Since N u ∈ (Lp (Ω))*, the integral representation (1.12) is clear. n0

Remarks. (a) The following remarkable statement can be proved: If f satisfies (i) and (ii) above and if the corresponding Nemytskii operator is such that N : Lp (Ω) → Lq (Ω), then N is continuous, bounded and (iii) holds. (b) If (iii) is replaced by (iii)0 for all (x, ξ) ∈ Ω × R |f (x, ξ)| ≤ a(x) + b|ξ|p/r where b is a fixed nonnegative number, a ∈ Lr (Ω) is nonnegative and 1 < p, r < ∞, then the above results are valid with q replaced by r. (i.e., N : Lp (Ω) → Lr (Ω).) (c) We say that f satisfies the Caratheodory property, written f ∈ Car, if (i) and (ii) above are met. If in addition (iii) is met, then we write f ∈ Car(p). 1.5.3. Differentiability. Let S be an open subset of the Banach space X. The functional f : S ⊂ X → R is said to be Gateaux differentiable (G-diff) at a point u ∈ S if there exists a functional g ∈ X* (often denoted by f 0 (u)) such that d f (u + tv) − f (u) f (u + tv) = lim = [f 0 (u)]v for all v ∈ X. t→0 dt t t=0

f 0 (u)

The functional is called the Gateaux derivative of f at the point u ∈ S. If f is G-diff at each point of S, the map f 0 : S ⊂ X → X* is called the Gateaux derivative of f on S. In addition, if f 0 is continuous at u (in the operator norm), then we say that f is C 1 at u. Note that in the case of a real-valued function of several real variables, the Gateaux derivative is nothing more than the directional derivative of the function at u in the direction v. Let X, Y be Banach spaces and let A : S ⊂ X → Y be an arbitrary operator. A is said to be Frechet differentiable (F-diff) at the point u ∈ S if there exists an operator

14

1. Preliminaries

B ∈ B(X, Y ) such that lim kA(u + v) − Au − Bvk/kvk = 0.

kvk→0

The operator B, often denoted by A0 (u), is called the Frechet derivative of A at u. Note that if A is Frechet differentiable on S, then A0 : S → B(X, Y ). In addition, if A0 is continuous at u (in the operator norm), we say that A is C 1 at u. Remark. If the functional f is F-diff at u ∈ S, then it is also G-diff at u, and the two derivatives are equal. This follows easily from the definition of the Frechet derivative. The converse is not always true as may be easily verified by simple examples from several variable calculus. However, if the Gateaux derivative exists in a neighborhood of u and if f ∈ C 1 at u, then the Frechet derivative exists at u, and the two derivatives are equal. Example 1.31. (a) Let f (ξ) ∈ C(R). Then for k ≥ 0, the corresponding Nemytskii operator ¯ → C(Ω) ¯ is bounded and continuous. If in addition f (ξ) ∈ C 1 (R), then N ∈ C 1 N : C k (Ω) and the Frechet derivative N 0 (u) is given by [N 0 (u)v](x) = f 0 (u(x))v(x). ¯ C ( Ω)) ¯ with ¯ |N 0 (u)v|0 ≤ |f 0 (u)|0 |v|k and so N 0 (u) ∈ B(C k (Ω), Note that for u, v ∈ C k (Ω), 0 0 0 k ¯ kN (u)k ≤ |f (u)|0 . Clearly N (u) is continuous at each point u ∈ C (Ω). Moreover, Z 1 d 0 |N (u + v) − N u − N (u)v|0 = sup | [ f (u(x) + tv(x)) − f 0 (u(x))v(x)]dt| x 0 dt Z 1 |f 0 (u(x) + tv(x)) − f 0 (u(x))|dt. ≤ |v|0 sup x

The last integral tends to zero since

f0

0

is uniformly continuous on compact subsets of R.

More generally, let f (ξ) ∈ C k (R). Then the corresponding Nemytskii operator N : ¯ is bounded and continuous. If in addition f (ξ) ∈ C k+1 (R), then N ∈ C 1 → C k (Ω) with Frechet derivative given by [N 0 (u)v](x) = f 0 (u(x))v(x). Note that |uv|k ≤ |u|k |v|k for ¯ and since C k (Ω) ¯ ⊂ C(Ω), ¯ the Frechet derivative must be of the stated form. u, v ∈ C k (Ω),

¯ C k (Ω)

(b) Let f (ξ) ∈ C k+1 (R), where k > n/2. Then we claim that the corresponding Nemytskii operator N : H k (Ω) → H k (Ω) is of class C 1 with Frechet derivative given by [N 0 (u)v](x) = f 0 (u(x))v(x). ¯ Then N (u) ∈ C k (Ω) ¯ by the usual chain rule. If u ∈ H k (Ω), First, suppose u ∈ C k (Ω). k ¯ with kum − ukk,2 → 0. Since the imbedding H k (Ω) ⊂ C(Ω) ¯ is continuous, let um ∈ C (Ω) 0 0 um → u uniformly, and thus f (um ) → f (u) and f (um ) → f (u) uniformly and hence in L2 . Furthermore, Di f (um ) = f 0 (um )Di um → f 0 (u)Di u in L1 . Consequently, by Theorem 2.10, we have Di f (u) = f 0 (u)Di u. In a similar fashion we find Dij f (u) = f 00 (u)Di uDj u + f 0 (u)Dij u with corresponding formulas for higher derivatives. 1.5.4. Implicit Function Theorem. The following lemmas are needed in the proof of the implicit function theorem.


15

Lemma 1.32. Let S be a closed nonempty subset of the Banach space X and let M be a metric space. Suppose A(x, λ) : S × M → S is continuous and there is a constant k < 1 such that, uniformly for all λ ∈ M kA(x, λ) − A(y, λ)k ≤ kkx − yk

for all

x, y ∈ S.

Then for each λ ∈ M, A(x, λ) has a unique fixed point x(λ) ∈ S and moreover, x(λ) depends continuously on λ. Proof. The existence and uniqueness of the fixed point x(λ) is of course a consequence of the contraction mapping theorem. To prove continuity, suppose λn → λ. Then kx(λn ) − x(λ)k = kA(x(λn ), λn ) − A(x(λ), λ)k ≤ kA(x(λn ), λn ) − A(x(λ), λn )k + kA(x(λ), λn ) − A(x(λ), λ)k ≤ kkx(λn ) − x(λ)k + kA(x(λ), λn ) − A(x(λ), λ)k. Therefore

1 kA(x(λ), λn ) − A(x(λ), λ)k. 1−k By the assumed continuity of A, the right side tends to zero as n → ∞, and therefore x(λn ) → x(λ). kx(λn ) − x(λ)k ≤

Lemma 1.33. Suppose X, Y are Banach spaces. Let S ⊂ X be convex and assume A : S → Y is Frechet differentiable at every point of S. Then kAu − Avk ≤ ku − vk sup kA0 (w)k. w∈S

In other words, A satisfies a Lipschitz condition with constant q = supw∈S kA0 (w)k. Proof. For fixed u, v ∈ S, set g(t) = A(u + t(v − u)), where t ∈ [0, 1]. Using the definition of Frechet derivative, we have A(u + (t + h)(v − u)) − A(u + t(v − u)) 0 g (t) = lim h→0 h 0 hA (u + t(v − u))(v − u) + kh(v − u)kE = lim h→0 h 0 = A (u + t(v − u))(v − u). Hence kg(0) − g(1)k = kAu − Avk ≤ sup kg 0 (t)k t∈[0,1]

which implies the desired result.

Lemma 1.34. Let X be a Banach space. Suppose A : B(u0 , r) ⊂ X → X is a contraction, with Lipschitz constant q < 1, where r ≥ (1 − q)−1 kAu0 − u0 k. Then A has a unique fixed point u ∈ B(u0 , r). Proof. For u ∈ B(u0 , r) kAu − u0 k ≤ kAu − Au0 k + kAu0 − u0 k ≤ qku − u0 k + (1 − q)r. Since ku − u0 k ≤ r, A maps the ball B(u0 , r) into itself, and the result follows from the contraction mapping theorem.

16

1. Preliminaries

We now consider operator equations of the form A(u, v) = 0, where A maps a subset of X × Y into Z. For a given [u0 , v0 ] ∈ X × Y we denote the Frechet derivative of A (at [u0 , v0 ]) with respect to the first (second) argument by Au (u0 , v0 ) (Av (u0 , v0 )). Theorem 1.35. (Implicit Function) Let X, Y, Z be Banach spaces. For a given [u0 , v0 ] ∈ X × Y and a, b > 0, let S = {[u, v] : ku − u0 k ≤ a, kv − v0 k ≤ b}. Suppose A : S → Z satisfies the following: (i) A is continuous. (ii) Av (·, ·) exists and is continuous in S (in the operator norm) (iii) A(u0 , v0 ) = 0. (iv) [Av (u0 , v0 )]−1 exists and belongs to B(Z, Y ). Then there are neighborhoods U of u0 and V of v0 such that the equation A(u, v) = 0 has exactly one solution v ∈ V for every u ∈ U . The solution v depends continuously on u. Proof. If in S we define B(u, v) = v − [Av (u0 , v0 )]−1 A(u, v) it is clear that the solutions of A(u, v) = 0 and v = B(u, v) are identical. The theorem will be proved by applying the contraction mapping theorem to B. Since Bv (u, v) = I − [Av (u0 , v0 )]−1 Av (u, v) Bv (·, ·) is continuous in the operator norm. Now Bv (u0 , v0 ) = 0, so for some δ > 0 there is a q < 1 such that kBv (u, v)k ≤ q for ku − u0 k ≤ δ, kv − v0 k ≤ δ. By virtue of Lemma 1.33, B(u, ·) is a contraction. Since A is continuous, B is also continuous. Therefore, since B(u0 , v0 ) = v0 , there is an ε with 0 < ε ≤ δ such that kB(u, v0 ) − v0 k ≤ (1 − q)δ for ku − u0 k ≤ ε. The existence of a unique fixed point in the closed ball B(v0 , δ) follows from Lemma 1.34 and the continuity from Lemma 1.32. ¯ and consider the Example 1.36. Let f (ξ) ∈ C 1,α (R), f (0) = f 0 (0) = 0, g(x) ∈ C α (Ω) boundary value problem (1.13)

∆u + f (u) = g(x) in Ω, u|∂Ω = 0. ¯ Y = {u ∈ C 2,α (Ω) ¯ : u|∂Ω = 0} and Set X = Z = C α (Ω), A(g, u) = ∆u + N (u) − g where N is the Nemytskii operator corresponding to f . The operator A maps X × Y into the space Z. Clearly A(0, 0) = 0 (A is C 1 by earlier examples) and Au (0, 0)v = ∆v,

v ∈ Y.

It is easily checked that all the conditions of the implicit function theorem are met. In particular, condition (iv) is a consequence of the bounded inverse theorem. Thus, for a ¯ of sufficiently small norm (in the space C α (Ω)) ¯ there exists a unique function g ∈ C α (Ω) solution of (1.13) which lies near the zero function. There may, of course, be other solutions which are not close to the zero function. (Note that the condition f 0 (0) = 0 rules out linear functions.)


17

¯ : u|∂Ω = 0} would fail ¯ Y = {u ∈ C 2 (Ω) Remark. Note that the choice of X = Z = C(Ω), above since the corresponding linear problem is not onto. An alternate approach would be to use Sobolev spaces. In fact, if we take X = Z = W k−2 (Ω), Y = W k (Ω) ∩ H01 (Ω) with k sufficiently large, and if f (ξ) ∈ C k+1 (R), then as above, we can conclude the existence of a unique solution u ∈ W k (Ω) provided kgkk−2,2 is sufficiently small. Hence, we get existence for more general functions g; however, the solution u ∈ W k (Ω) is not a classical (i.e., C 2 ) solution in general. 1.5.5. Generalized Weierstrass Theorem. In its simplest form, the classical Weierstrass theorem can be stated as follows: Every continuous function defined on a closed ball in Rn is bounded and attains both its maximum and minimum on this ball. The proof makes essential use of the fact that the closed ball is compact. The first difficulty in trying to extend this result to an arbitrary Banach space X is that the closed ball in X is not compact if X is infinite dimensional. However, as we shall show, a generalized Weierstrass theorem is possible if we require a stronger property for the functional. A set S ⊂ X is said to be weakly closed if {un } ⊂ S, un * u implies u ∈ S, i.e., S contains all its weak limits. A weakly closed set is clearly closed, but not conversely. 2 Indeed, the set {sin nx}∞ 1 in L (0, π) has no limit point (because it cannot be Cauchy) so it is closed, but zero is a weak limit that does not belong to the set. It can be shown that every convex, closed set in a Banach space is weakly closed. A functional f : S ⊂ X → R is weakly continuous at u0 ∈ S if for every sequence {un } ⊂ S with un * u0 it follows that f (un ) → f (u0 ). Clearly, every functional f ∈ X* is weakly continuous. A functional f : S ⊂ X → R is weakly lower semicontinuous(w.l.s.c.) at u0 ∈ S if for every sequence {un } ⊂ S for which un * u0 it follows that f (u0 ) ≤ lim inf n→∞ f (un ). According to Theorem 1.18, the norm on a Banach space is w.l.s.c.. A functional f : S ⊂ X → R is weakly coercive on S if f (u) → ∞ as kuk → ∞ on S. Theorem 1.37. Let X be a reflexive Banach space and f : C ⊂ X → R be w.l.s.c. and assume (i) C is a nonempty bounded weakly closed set in X or (ii) C is a nonempty weakly closed set in X and f is weakly coercive on C. Then (a) inf u∈C f (u) > −∞; (b) there is at least one u0 ∈ C such that f (u0 ) = inf u∈C f (u). Moreover, if u0 is an interior point of C and f is G-diff at u0 , then f 0 (u0 ) = 0. Proof. Assume (i) and let {un } ⊂ C be a minimizing sequence, i.e., limn→∞ f (un ) = inf u∈C f (u). The existence of such a sequence follows from the definition of inf. Since X is reflexive and C is bounded and weakly closed, there is a subsequence {un0 } and a u0 ∈ C such that un0 * u0 . But f is w.l.s.c. and so f (u0 ) ≤ lim inf n→∞ f (un0 ) = inf u∈C f (u), which proves (a). Since by definition, f (u0 ) ≥ inf u∈C f (u), we get (b). Assume (ii) and fix u0 ∈ C. Since f is weakly coercive, there is a closed ball B(0, R) ⊂ X such that u0 ∈ B ∩ C and f (u) ≥ f (u0 ) outside B ∩ C. Since B ∩ C satisfies the conditions of (i), there is a u1 ∈ B ∩ C such that f (u) ≥ f (u1 ) for all u ∈ B ∩ C and in particular for u0 . Thus, f (u) ≥ f (u1 ) on all of C.

18

1. Preliminaries

To prove the last statement we set ϕv (t) = f (u0 + tv). For fixed v ∈ X, ϕv (t) has a local minimum at t = 0, and therefore hf 0 (u0 ), vi = 0 for all v ∈ X. The point u0 ∈ X is called a critical point of the functional f defined on X if f 0 (u0 )v = 0 for every v ∈ X. Remark. Even though weakly continuous functionals on closed balls attain both their inf and sup (which follows from the above theorem), the usual functionals that we encounter are not weakly continuous, but are w.l.s.c.. Hence this explains why we seek the inf and not the sup in variational problems. 1.5.5.1. Convex sets. A set C in the real normed space X is called convex if (1−t)u+tv ∈ C for all t ∈ [0, 1], u, v ∈ C. The following result is needed later (see, e.g., [32]). Theorem 1.38. A closed convex set in a Banach space is weakly closed. 1.5.6. Monotone Operators and Convex Functionals. Let A : X → X* be an operator on the real Banach space X. A is monotone if hAu − Av, u − vi ≥ 0

for all

u, v ∈ X.

A is strongly monotone if hAu − Av, u − vi ≥ cku − vkpX

for all

u, v ∈ X

where c > 0 and p > 1. A is coercive if hAu, ui = +∞. kuk→∞ kuk lim

Remark. A strongly monotone operator is coercive. This follows immediately from hAu, ui = hAu − A0, ui + hA0, ui ≥ ckukpX − kA0kkukX . Let C be a convex set in the real normed space X. A functional f : C ⊂ X → R is said to be convex if f ((1 − t)u + tv) ≤ (1 − t)f (u) + tf (v)

for all

t ∈ [0, 1], u, v ∈ C.

In the following we set ϕ(t) = f ((1 − t)u + tv) = f (u + t(v − u)) for fixed u and v. Lemma 1.39. Let C ⊂ X be a convex set in a real normed space X. Then the following statements are equivalent: (a) The real function ϕ : [0, 1] → R is convex for all u, v ∈ C. (b) The functional f : C ⊂ X → R is convex. (c) f 0 : C ⊂ X → X* (assuming f is G-diff on C) is monotone.


19

Proof. Assume ϕ is convex. Then ϕ(t) = ϕ((1 − t) · 0 + t · 1) ≤ (1 − t)ϕ(0) + tϕ(1) for all t ∈ [0, 1], which implies (b). Similarly, if f is convex, then for t = (1 − α)s1 + αs2 , with α, s1 , s2 ∈ [0, 1], we have ϕ(t) = f (u + t(v − u)) ≤ (1 − α)f (u + s1 (v − u)) + αf (u + s2 (v − u)) for all u, v ∈ C, which implies (a). Fix u, v ∈ C. Then ϕ0 (t) = hf 0 (u + t(v − u)), v − ui. If f is convex, then ϕ is convex and therefore ϕ0 is monotone. From ϕ0 (1) ≥ ϕ0 (0) we obtain hf 0 (v) − f 0 (u), v − ui ≥ 0

for all

u, v ∈ C

which implies (c). Finally, assume f 0 is monotone. Then for s < t we have 1 hf 0 (u + t(v − u)) − f 0 (u + s(v − u)), (t − s)(v − u)i ≥ 0. ϕ0 (t) − ϕ0 (s) = t−s Thus ϕ0 is monotone, which implies ϕ, and thus f is convex.

Theorem 1.40. Consider the functional f : C ⊂ X → R, where X is a real Banach space. Then f is w.l.s.c. if any one of the following conditions holds: (a) C is closed and convex; f is convex and continuous. (b) C is convex; f is G-diff on C and f 0 is monotone on C. Proof. Set Cr = {u ∈ C : f (u) ≤ r}. It follows from (a) that Cr is closed and convex for all r, and thus is weakly closed (cf. Theorem 1.38). If f is not w.l.s.c., then there is a sequence {un } ⊂ C with un * u and f (u) > lim inf f (un ). Hence, there is an r and a subsequence {un0 } such that f (u) > r and f (un0 ) ≤ r (i.e., un0 ∈ Cr ) for all n0 large enough. Since Cr is weakly closed, u ∈ Cr , which is a contradiction. Assume (b) holds and set ϕ(t) = f (u + t(v − u)). Then by Lemma 1.39, ϕ : [0, 1] → R is convex and ϕ0 is monotone. By the classical mean value theorem, ϕ(1) − ϕ(0) = ϕ0 (θ) ≥ ϕ0 (0),

0 0 the inequality is false, then there exists a sequence {un } such that (1.17)

kun kY > εkun kX + nkun kZ

for all

n.

Without loss of generality we can assume kun kX = 1. Since the imbedding X ⊂ Y is compact, there is a subsequence, again denoted by {un }, with un → u in Y . This implies un → u in Z. By (1.17), kun kY > ε and so u 6= 0. Again by (1.17), un → 0 in Z, i.e., u = 0, which is a contradiction.

Chapter 2

Sobolev Spaces

This chapter is devoted to a discussion of the necessary Sobolev function spaces which permit a modern approach to the study of differential equations.

2.1. Weak Derivatives and Sobolev Spaces 2.1.1. Weak Derivatives. Let Ω be a nonempty open set in Rn . Suppose u ∈ C m (Ω) and ϕ ∈ C0m (Ω). Then by integration by parts Z Z α |α| (2.1) uD ϕdx = (−1) vϕdx, |α| ≤ m Ω

Ω

where v = Dα u. Motivated by (2.1), we now enlarge the class of functions for which the notion of derivative can introduced. Let u ∈ L1loc (Ω). A function v ∈ L1loc (Ω) is called the αth weak derivative of u if it satisfies Z Z |α| α |α| (2.2) uD ϕdx = (−1) vϕdx for all ϕ ∈ C0 (Ω). Ω

Ω

It can be easily shown that weak derivatives are unique. Thus we write v = Dα u to indicate that v is the αth weak derivative of u. If a function u has an ordinary αth derivative lying in L1loc (Ω), then it is clearly the αth weak derivative. In contrast to the corresponding classical derivative, the weak derivative Dα u is defined globally on all of Ω by (2.2). However, in every subregion Ω0 ⊂ Ω the function Dα u will also be the weak derivative of u. It suffices to note that (2.2) holds for every function |α| ϕ ∈ C0 (Ω0 ), and extended outside Ω0 by assigning to it the value zero. In particular, the weak derivative (if it exists) of a function u having compact support in Ω has itself compact support in Ω and thus belongs to L1 (Ω). We also note that in contrast to the classical derivative, the weak derivative Dα u is defined at once for order |α| without assuming the existence of corresponding derivatives of lower orders. In fact, the derivatives of lower orders may not exist as we will see in a forthcoming exercise. However, in his book, Sobolev shows that if all weak derivatives exist of a certain order, then all lower order weak derivatives exist. 23

24

2. Sobolev Spaces

Example 2.1. (a) The function u(x) = |x1 | has in the ball Ω = B(0, 1) weak derivatives ux1 = sgn x1 , uxi = 0, i = 2, . . . , n. In fact, we apply formula (2.2) as follows: For any ϕ ∈ C01 (Ω) Z Z Z x1 ϕx1 dx x1 ϕx1 dx − |x1 |ϕx1 dx = Ω−

Ω+

Ω

where Ω+ = Ω ∩ (x1 > 0), Ω− = Ω ∩ (x1 < 0). Since x1 ϕ = 0 on ∂Ω and also for x1 = 0, an application of the divergence theorem yields Z Z Z Z ϕdx = − (sgn x1 )ϕdx. ϕdx + |x1 |ϕx1 dx = − Ω−

Ω+

Ω

Ω

Hence |x1 |x1 = sgn x1 . Similarly, since for i ≥ 2 Z Z Z |x1 |ϕxi dx = (|x1 |ϕ)xi dx = − 0ϕdx Ω

Ω

Ω

|x1 |xi = 0 for i = 2, . . . , n. Note that the function |x1 | has no classical derivative with respect to x1 in Ω. (b) By the above computation, the function u(x) = |x| has a weak derivative u0 (x) = sgn x on the interval Ω = (−1, 1). On the other hand, sgn x does not have a weak derivative on Ω due to the discontinuity at x = 0. (c) Let Ω = B(0, 1/2) ⊂ R2 and define u(x) = ln(ln(2/r)), x ∈ Ω, where r = |x| = + x22 )1/2 . Then u 6∈ L∞ (Ω) because of the singularity at the origin. However, we will show that u has weak first partial derivatives.

(x21

First of all u ∈ L2 (Ω), for Z Z 2 |u| dx = Ω

0

2π

Z

1/2

r[ln(ln(2/r))]2 dr dθ

0

and a simple application of L’hopitals rule shows that the integrand is bounded and thus the integral is finite. Similarly, it is easy to check that the classical partial derivative − cos θ , where x1 = r cos θ r ln(2/r)

ux1 =

also belongs to L2 (Ω). Now we show that the defining equation for the weak derivative is met. Let Ωε = {x : ε < r < 1/2} and choose ϕ ∈ C01 (Ω). Then by the divergence theorem and the absolute continuity of integrals Z Z Z Z uϕx1 dx = lim uϕx1 dx = lim − ux1 ϕdx + uϕn1 ds ε→0

ε→0 Ωε

Ω

Ωε

r=ε

where n = (n1 , n2 ) is the unit outward normal to Ωε on r = ε. But (ds = εdθ) Z Z 2π | uϕn1 ds| ≤ |u| |ϕ|εdθ ≤ 2πεc ln(ln(2/ε)) → 0 r=ε

0

as ε → 0. Thus Z

Z uϕx1 dx = −

Ω

ux1 ϕdx. Ω

The same analysis applies to ux2 . Thus u has weak first partial derivatives given by the classical derivatives which are defined on Ω\{0}.

2.1. Weak Derivatives and Sobolev Spaces

25

2.1.2. Sobolev Spaces. For p ≥ 1 and k a nonnegative integer, we let W k,p (Ω) = {u : u ∈ Lp (Ω), Dα u ∈ Lp (Ω), 0 < |α| ≤ k} where Dα u denotes the αth weak derivative. When k = 0, W k,p (Ω) will mean Lp (Ω). It is clear that W k,p (Ω) is a vector space. A norm is introduced by defining kukk,p = kukW k,p (Ω)

(2.3)

Z =(

X

|Dα u|p dx)1/p

Ω |α|≤k

if 1 ≤ p < ∞. The space W k,p (Ω) is known as a Sobolev space of order k. We also introduce the space W0k,p (Ω) which is defined to be the closure of the space C0k (Ω) with respect to the norm k · kk,p . W0k,p (Ω) is also called a Sobolev space of order k. As we shall see shortly, W k,p (Ω) 6= W0k,p (Ω) for k ≥ 1. (Unless Ω = Rn .) Remark. The case p = 2 is special, since the spaces W k,2 (Ω), W0k,2 (Ω) will be Hilbert spaces under the inner product Z X (u, v)k,2 = (u, v)W k,2 (Ω) = Dα uDα vdx. Ω |α|≤k

Since we shall be dealing mostly with these spaces in the sequel, we introduce the special notation: H k (Ω) = W k,2 (Ω), H0k (Ω) = W0k,2 (Ω). Theorem 2.2. W k,p (Ω) is a Banach space under the norm (2.3). If 1 < p < ∞, it is reflexive and if 1 ≤ p < ∞, it is separable. Proof. We first prove that W k,p (Ω) is complete with respect to the norm (2.3). Let {un } be a Cauchy sequence of elements in W k,p (Ω), i.e., X Z p |Dα un − Dα um |p dx → 0 as m, n → ∞. kun − um kk,p = |α|≤k

Ω

Then for any α, |α| ≤ k, when m, n → ∞ Z |Dα un − Dα um |p dx → 0 Ω

and, in particular, when |α| = 0 Z

|un − um |p dx → 0.

Ω

Since Lp (Ω) is complete, it follows that there are functions uα ∈ Lp (Ω), |α| ≤ k such that Dα un → uα (in Lp (Ω)). Since each un (x) has weak derivatives (up to order k) belonging to Lp (Ω), a simple limit argument shows that uα is the αth weak derivative of u0 . In fact, Z Z Z Z α α |α| α |α| uD ϕdx ← un D ϕdx = (−1) ϕD un dx → (−1) uα ϕdx. Ω

Hence

u0

∈

Ω

W k,p (Ω)

and kun −

Consider the map T :

Ω

u0 kk,p

W 1,p (Ω)

→ 0 as n → ∞.

→ (Lp (Ω))n+1 defined by

T u = (u, D1 u, . . . , Dn u).

Ω

26

2. Sobolev Spaces

If we endow the latter space with the norm n+1 X

kvk = (

kvi kpp )1/p

i=1

(Lp (Ω))n+1 ,

for v = (v1 , . . . , vn+1 ) ∈ then T is a (linear) isometry. Now (Lp (Ω))n+1 is reflexive for 1 < p < ∞ and separable for 1 ≤ p < ∞. Since W 1,p (Ω) is complete, its image under the isometry T is a closed subspace of (Lp (Ω))n+1 which inherits the corresponding properties as does W 1,p (Ω) (see Theorem 1.14). Similarly, we can handle the case k ≥ 2. Example 2.3. Let Ω be a bounded open connected set in Rn . Divide Ω into N open disjoint subsets Ω1 , Ω2 , . . . , ΩN . Suppose the function u : Ω → R has the following properties: ¯ (i) u is continuous on Ω. (ii) For some i, Di u is continuous on Ω1 , Ω2 , . . . , ΩN , and can be extended contin¯ 1, Ω ¯ 2, . . . , Ω ¯ N , respectively. uously to Ω (iii) The surfaces of discontinuity are such that the divergence theorem applies. Define wi (x) = Di u(x) if x ∈ ∪N i=1 Ωi . Otherwise, wi can be arbitrary. We now claim that wi ∈ Lp (Ω) is a weak partial derivative of u on Ω. Indeed, for all ϕ ∈ C01 (Ω), the divergence theorem yields Z XZ uDi ϕdx = uDi ϕdx Ω

Ωj

j

X Z

=

uϕni dS −

∂Ωj

j

!

Z

ϕDi udx Ωj

Z = −

ϕDi udx. Ω

Note that the boundary terms either vanish, since ϕ has compact support, or cancel out along the common boundaries, since u is continuous and the outer normals have opposite ¯ and has piecewise continuous derivatives in Ω of order directions. Similarly, if u ∈ C k (Ω) k+1,p k + 1, then u ∈ W (Ω). Remark. More generally, by using a partition of unity argument, we can show the following: If O is a collection of nonempty open sets whose union is Ω and if u ∈ L1loc (Ω) is such that for some multi-index α, the αth weak derivative of u exists on each member of O, then the αth weak derivative of u exists on Ω. Exercise 1. Consider the function u(x) = sgnx1 + sgnx2 in the ball B(0, 1) ⊂ R2 . Show that the weak derivative ux1 does not exist, yet the weak derivative ux1 x2 does exist. Exercise 2. (a) Let Ω be the hemisphere of radius R < 1 in Rn : r2 ≡

n X

x2i ≤ R2 , xn ≥ 0 (n ≥ 3).

i=1

Show that u = (r(n/2−1) `nr)−1 ∈ H 1 (Ω). (b) Let B = B(0, 1) be the open unit ball in Rn , and let u(x) = |x|−α , x ∈ B.

2.2. Approximations and Extensions

27

For what values of α, n, p does u belong to W 1,p (B)?

2.2. Approximations and Extensions 2.2.1. Mollifiers. Let x ∈ Rn and let B(x, h) denote the open ball with center at x and radius h. For each h > 0, let ωh (x) ∈ C ∞ (Rn ) satisfy ωh (x) ≥ 0; ωh (x) = 0 for |x| ≥ h Z Z ωh (x)dx = 1. ωh (x)dx = Rn

B(0,h)

Such functions are called mollifiers. For example, let ( k exp [(|x|2 − 1)−1 ], |x| < 1, ω(x) = 0, |x| ≥ 1, R where k > 0 is chosen so that Rn ω(x) dx = 1. Then, a family of mollifiers can be taken as ωh (x) = h−n ω(x/h) for h > 0. Let Ω be a nonempty open set in Rn and let u ∈ L1 (Ω). We set u = 0 outside Ω. Define for each h > 0 the mollified function Z uh (x) = ωh (x − y)u(y)dy Ω

where ωh is a mollifier. There are two other forms in which uh can be represented, namely Z Z (2.4) uh (x) = ωh (x − y)u(y)dy = ωh (x − y)u(y)dy Rn

B(x,h)

the latter equality being valid since ωh vanishes outside the (open) ball B(x, h). Thus the values of uh (x) depend only on the values of u on the ball B(x, h). In particular, if dist(x, supp(u)) ≥ h, then uh (x) = 0. Theorem 2.4. Let Ω be a nonempty open set in Rn . Then (a) uh ∈ C ∞ (Rn ). (b) If supp(u) is a compact subset of Ω, then uh ∈ C0∞ (Ω) for all h sufficiently small. Proof. Since u is integrable and ωh ∈ C ∞ , the Lebesgue theorem on differentiating integrals implies that for |α| < ∞ Z α D uh (x) = u(y)Dα ωh (x − y)dy Ω

i.e., uh ∈ C ∞ (Rn ). Statement (b) follows from the remark preceding the theorem.

With respect to a bounded set Ω we construct another set Ω(h) as follows: with each point x ∈ Ω as center, draw a ball of radius h; the union of these balls is then Ω(h) . Clearly Ω(h) ⊃ Ω. Moreover, uh can be different from zero only in Ω(h) . Corollary 2.5. Let Ω be a nonempty bounded open set in Rn and let h > 0 be any number. Then there exists a function η ∈ C ∞ (Rn ) such that 0 ≤ η(x) ≤ 1; η(x) = 1, x ∈ Ω(h) ; η(x) = 0, x ∈ (Ω(3h) )c . Such a function is called a cut-off function for Ω.

28

2. Sobolev Spaces

Proof. Let χ(x) be the characteristic function of the set Ω(2h) : χ(x) = 1 for x ∈ Ω(2h) , χ(x) = 0 for x 6∈ Ω(2h) and set Z η(x) ≡ χh (x) = ωh (x − y)χ(y)dy. Rn

Then

Z

ωh (x − y)dy ∈ C ∞ (Rn ), Z ωh (x − y)dy = 1, 0 ≤ η(x) ≤

η(x) =

Ω(2h)

Rn

and (R

Z ωh (x − y)χ(y)dy =

η(x) = B(x,h)

B(x,h) ωh (x

− y)dy = 1,

0,

x ∈ Ω(h) , x ∈ (Ω(3h) )c .

In particular, we note that if Ω0 ⊂⊂ Ω, there is a function η ∈ C0∞ (Ω) such that η(x) = 1 for x ∈ Ω0 , and 0 ≤ η(x) ≤ 1 in Ω. Remark. Henceforth, the notation Ω0 ⊂⊂ Ω means that Ω0 , Ω are open sets and that Ω0 ⊂ Ω. We shall have need of the following well-known result. Theorem 2.6. (Partition of Unity) Assume Ω ⊂ Rn is bounded and Ω ⊂⊂ ∪N i=1 Ωi , where each Ωi is open. Then there exist C ∞ functions ψi (x)(i = 1, . . . , N ) such that (a) 0 ≤ ψi (x) ≤ 1 (b) ψi has its support in Ωi PN (c) i=1 ψi (x) = 1 for every x ∈ Ω. 2.2.2. Approximation Theorems. Lemma 2.7. Let Ω be a nonempty bounded open set in Rn . Then every u ∈ Lp (Ω) is p-mean continuous, i.e., Z |u(x + z) − u(x)|p dx → 0 as z → 0. Ω

Proof. Choose a > 0 large enough so that Ω is strictly contained in the ball B(0, a). Then the function u(x) if x ∈ Ω, U (x) = 0 if x ∈ B(0, 2a) \ Ω ¯ ∈ C(B(0, ¯ 2a)) which satisfies the belongs to Lp (B(0, 2a)). For ε > 0, there is a function U ¯ ¯ inequality kU − U kLp (B(0,2a)) < ε/3. By multiplying U by an appropriate cut-off function, ¯ (x) = 0 for x ∈ B(0, 2a)/B(0, a). it can be assumed that U ¯ (x + z)kLp (B(0,2a)) = kU (x) − U ¯ (x)kLp (B(0,a)) ≤ ε/3. Therefore for |z| ≤ a, kU (x + z) − U ¯ Since the function U is uniformly continuous in B(0, 2a), there is a δ > 0(δ < a) such that ¯ (x + z) − U ¯ (x)kLp (B(0,2a)) ≤ ε/3 whenever |z| < δ. Hence for |z| < δ we easily see that kU ku(x + z) − u(x)kLp (Ω) = kU (x + z) − U (x)kLp (B(0,2a)) ≤ ε. Theorem 2.8. Let Ω be a nonempty open set in Rn . If u ∈ Lp (Ω) (1 ≤ p < ∞), then (a) kuh kp ≤ kukp (b) kuh − ukp → 0

as

h → 0.


29

¯ where Ω ¯ is compact, then If u ∈ C k (Ω), (c) kuh − ukC k (Ω¯ 0 ) → 0

h → 0,

as

where Ω0 ⊂⊂ Ω. 1/p 1/q

Proof. If 1 < p < ∞, let q = p/(p − 1). Then ωh = ωh ωh and Hölder’s inequality implies Z p/q Z p p |uh (x)| ≤ ωh (x − y)|u(y)| dy ωh (x − y)dy Ω Ω Z ≤ ωh (x − y)|u(y)|p dy Ω

which obviously holds also for p = 1. An application of Fubini’s Theorem gives Z Z Z Z p p |uh (x)| dx ≤ ωh (x − y)dx |u(y)| dy ≤ |u(y)|p dy Ω

Ω

Ω

Ω

which implies (a). To prove (b), let ω(x) = hn ωh (hx). Then ω(x) ∈ C ∞ (Rn ) and satisfies ω(x) ≥ 0; ω(x) = 0 for |x| ≥ 1 Z Z ω(x)dx = ω(x)dx = 1. Rn

B(0,1)

Using the change of variable z = (x − y)/h we have Z uh (x) − u(x) = [u(y) − u(x)]ωh (x − y)dy B(x,h) Z = [u(x − hz) − u(x)]ω(z)dz. B(0,1)

Hence by H¨ older’s inequality p

Z

|u(x − hz) − u(x)|p dz

|uh (x) − u(x)| ≤ d B(0,1)

and so by Fubini’s Theorem Z Z p |uh (x) − u(x)| dx ≤ d Ω

Z ( |u(x − hz) − u(x)|p dx)dz.

B(0,1)

Ω

The right-hand side goes to zero as h → 0 since every u ∈ Lp (Ω) is p-mean continuous. We now prove (c) for k = 0. Let Ω0 , Ω00 be such that Ω0 ⊂⊂ Ω00 ⊂⊂ Ω. Let h0 be the shortest distance between ∂Ω0 and ∂Ω00 . Take h < h0 . Then Z uh (x) − u(x) = [u(y) − u(x)]ωh (x − y)dy. B(x,h)

¯ 0 , then in the above integral y ∈ Ω ¯ 00 . Now u is uniformly continuous in Ω ¯ 00 and If x ∈ Ω ωh ≥ 0, and therefore for an arbitrary ε > 0 we have Z |uh (x) − u(x)| ≤ ε ωh (x − y)dy = ε B(x,h)

provided h is sufficiently small. The case k ≥ 1 is handled similarly and is left as an exercise.

30

2. Sobolev Spaces

Remark. The following example shows that in (c) we cannot replace Ω0 by Ω. Let u ≡ 1 for R1 Rh x ∈ [0, 1] and consider uh (x) = 0 ωh (x − y)dy, where ωh (y) = ωh (−y). Now −h ωh (y)dy = 1 and so uh (0) = 1/2 for all h < 1. Thus uh (0) → 1/2 6= 1 = u(0). Moreover, for x ∈ (0, 1) R x+h and h sufficiently small, (x − h, x + h) ⊂ (0, 1) and so uh (x) = x−h ωh (x − y)dy = 1 which implies uh (x) → 1 for all x ∈ (0, 1). Corollary 2.9. Let Ω be a nonempty open set in Rn . Then C0∞ (Ω) is dense in Lp (Ω) for all 1 ≤ p < ∞. Proof. Suppose first that Ω is bounded and let Ω0 ⊂⊂ Ω. For a given u ∈ Lp (Ω) set u(x), x ∈ Ω0 v(x) = 0, x ∈ Ω\Ω0 . Then

Z

p

Z

|u|p dx.

|u − v| dx = Ω

Ω\Ω0

By the absolute continuity of integrals, we can choose Ω0 so that the integral on the right is arbitrarily small, i.e., ku − vkp < ε/2. Since supp(v) is a compact subset of Ω, Theorems 2.4(b) and 2.8(b) imply that for h sufficiently small, vh (x) ∈ C0∞ (Ω) with kv − vh kp < ε/2, and therefore ku − vh kp < ε. If Ω is unbounded, choose a ball B large enough so that Z |u|p dx < ε/2 Ω\Ω0 where Ω0 = Ω ∩ B, and repeat the proof just given.

We now consider the following local approximation theorem. Theorem 2.10. Let Ω be a nonempty open set in Rn and suppose u, v ∈ L1loc (Ω). Then v = Dα u iff there exists a sequence of C ∞ (Ω) functions {uh } with kuh − ukL1 (S) → 0, kDα uh − vkL1 (S) → 0 as h → 0, for all compact sets S ⊂ Ω. Proof. (Necessity) Suppose v = Dα u. Let S ⊂ Ω, and choose d > 0 small enough so that the sets Ω0 ≡ S (d/2) , Ω00 ≡ S (d) satisfy Ω0 ⊂⊂ Ω00 ⊂⊂ Ω. For x ∈ Rn define Z Z uh (x) = ωh (x − y)u(y)dy, vh (x) = ωh (x − y)v(y)dy. Ω00 ∞ n C (R ) for

Clearly, uh , vh ∈ kuh − ukL1 (Ω00 ) → 0.

Ω00

h > 0. Moreover, from Theorem 2.8 we have kuh − ukL1 (S) ≤

Now we note that if x ∈ Ω0 and 0 < h < d/2, then ωh (x − y) ∈ C0∞ (Ω00 ). Thus by Theorem 2.4 and the definition of weak derivative, Z Z α α |α| D uh (x) = u(y)Dx ωh (x − y)dy = (−1) u(y)Dyα ωh (x − y)dy 00 Ω00 ZΩ = ωh (x − y) · v(y)dy = vh (x). Ω00

Thus,

kDα uh

− vkL1 (S) → 0. |α|

(Sufficiency) Choose ϕ ∈ C0 (Ω) and consider a compact set S ⊃ supp(ϕ). Then as h→∞ Z Z Z Z uDα ϕdx ← uh Dα ϕdx = (−1)|α| ϕDα uh dx → (−1)|α| vϕdx S

S

S

S


31

which is the claim.

If u is equal to a constant(a.e.) in Ω, then u has the weak derivative Dα u = 0, |α| > 0. An application of Theorem 2.10 yields the converse: Theorem 2.11. Let Ω be a bounded open connected set in Rn . If u ∈ L1loc (Ω) has a weak derivative Dα u = 0 whenever |α| = 1, then u =const. a.e. in Ω. Proof. Let Ω0 ⊂⊂ Ω. Then for x ∈ Ω0 and with uh as in Theorem 2.10, Dα uh (x) = (Dα u)h (x) = 0 for all h sufficiently small. Thus uh = const = c(h) in Ω0 for such h. Since kuh − ukL1 (Ω0 ) = kc(h) − ukL1 (Ω0 ) → 0 as h → 0, it follows that kc(h1 ) − c(h2 )kL1 (Ω0 ) = |c(h1 ) − c(h2 )|mes(Ω0 ) → 0 as h1 , h2 → 0. Consequently, c(h) = uh converges uniformly and thus in L1 (Ω0 ) to some constant. Hence u = const(a.e.) in Ω0 and therefore also in Ω, by virtue of it being connected. We now note some properties of W k,p (Ω) which follow easily from the results of this and the previous section. (a) If Ω0 ⊂ Ω and if u ∈ W k,p (Ω), then u ∈ W k,p (Ω0 ). (b) If u ∈ W k,p (Ω) and |a(x)|k,∞ < ∞, then au ∈ W k,p (Ω). In this case any weak derivative Dα (au) is computed according to the usual rule of differentiating the product of functions. (c) If u ∈ W k,p (Ω) and uh is its mollified function, then for any compact set S ⊂ Ω, kuh − ukW k,p (S) → 0 as h → 0. If in addition, u has compact support in Ω, then kuh − ukk,p → 0 as h → 0. More generally, we have the following global approximation theorems. (See Meyers and Serrin H = W . The proofs make use of a partition of unity argument.) Theorem 2.12. Assume Ω is bounded and let u ∈ W k,p (Ω), 1 ≤ p < ∞. Then there exist functions um ∈ C ∞ (Ω) ∩ W k,p (Ω) such that um → u in W k,p (Ω). In other words, C ∞ (Ω) ∩ W k,p (Ω) is dense in W k,p (Ω). Theorem 2.13. Assume Ω is bounded and ∂Ω ∈ C 1 . Let u ∈ W k,p (Ω), 1 ≤ p < ∞. Then ¯ such that there exist functions um ∈ C ∞ (Ω) um → u in W k,p (Ω). ¯ is dense in W k,p (Ω). In other words, C ∞ (Ω) Exercise 1. Prove the product rule for weak derivatives: Di (uv) = (Di u)v + u(Di v) where u, Di u are locally Lp (Ω), v, Di v are locally Lq (Ω) (p > 1, 1/p + 1/q = 1). ¯ prove that uv ∈ W k,p (Ω). Exercise 2. (a) If u ∈ W k,p (Ω) and v ∈ C k (Ω), 0

(b) If u ∈

W k,p (Ω)

and v ∈

0

C0k (Ω),

prove that uv ∈

W0k,p (Ω).

32

2. Sobolev Spaces

2.2.3. Chain Rules. Theorem 2.14. Let Ω be a bounded open set in Rn . Let f ∈ C 1 (R), |f 0 (s)| ≤ M for all s ∈ R and suppose u has a weak derivative Dα u for |α| = 1. Then the composite function f ◦u has a weak derivative Dα (f ◦u) = f 0 (u)Dα u. Moreover, if f (0) = 0 and if u ∈ W 1,p (Ω), then f ◦ u ∈ W 1,p (Ω). Proof. According to Theorem 2.10, there exists a sequence {uh } ⊂ C 1 (Ω) such that kuh − ukL1 (Ω0 ) → 0, kDα uh − Dα ukL1 (Ω0 ) → 0 as h → 0, where Ω0 ⊂⊂ Ω. Thus Z Z 0 |f (uh ) − f (u)|dx ≤ sup |f | |uh − u|dx → 0 as h → 0 Ω0

Ω0

Z Ω0

0

0

0

Z

|Dα uh − Dα u|dx |f (uh )D uh − f (u)D u|dx ≤ sup |f | Ω0 Z |f 0 (uh ) − f 0 (u)||Dα u|dx. + α

α

Ω0

Since kuh − ukL1 (Ω0 ) → 0, there exists a subsequence of {uh }, which we call {uh } again, which converges a.e. in Ω0 to u. Moreover, since f 0 is continuous, {f 0 (uh )} converges to f 0 (u) a.e. in Ω0 . Hence the last integral tends to zero by the dominated convergence theorem. Consequently, the sequences {f (uh )}, {f 0 (uh )Dα uh } tend to f (u), f 0 (u)Dα u respectively, and the first conclusion follows by an application of Theorem 2.10 again. Since f (0) = 0, the mean value theorem implies |f (s)| ≤ M |s| for all s ∈ R. Thus, |f (u(x))| ≤ M |u(x)| for all x ∈ Ω and so f ◦ u ∈ Lp (Ω) if u ∈ Lp (Ω). Similarly, f 0 (u(x))Dα u ∈ Lp (Ω) if u ∈ W 1,p (Ω), which shows that f ◦ u ∈ W 1,p (Ω). Corollary 2.15. Let Ω be a bounded open set Dα u, |α| = 1, then so does |u| and   Dα u α 0 D |u| =  −Dα u

in Rn . If u has an αth weak derivative if u > 0 if u = 0 if u < 0

i.e., Dα |u| = (sgn u)Dα u for u 6= 0. In particular, if u ∈ W 1,p (Ω), then |u| ∈ W 1,p (Ω). Proof. The positive and negative parts of u are defined by u+ = max{u, 0}, u− = min{u, 0}. If we can show that Dα u+ exists and that α D u if u > 0 α + D u = 0 if u ≤ 0 then the result for |u| follows easily from the relations |u| = u+ − u− and u− = −(−u)+ . Thus, for h > 0 define 1 (u2 + h2 ) 2 − h if u > 0 fh (u) = 0 if u ≤ 0. Clearly fh ∈ C 1 (R) and fh0 is bounded on R. By Theorem 2.14, fh (u) has a weak derivative, and for any ϕ ∈ C01 (Ω) Z Z Z uDα u α α fh (u)D ϕdx = − D (fh (u))ϕdx = − ϕ 1 dx. Ω Ω u>0 (u2 + h2 ) 2


33

Upon letting h → 0, it follows that fh (u) → u+ , and so by the dominating convergence theorem Z Z Z α + α ϕD udx = − vϕdx u D ϕdx = − u>0

Ω

where

Ω

Dα u if u > 0 0 if u ≤ 0 + which establishes the desired result for u .

v=

Remark. Since u = u+ + u− , we have ∂u/∂xi = ∂u+ /∂xi + ∂u− /∂xi . Consequently, ∂u/∂xi = 0 a.e. on {u = c} = {x ∈ Ω : u(x) = c}. The following result is of independent importance. Theorem 2.16. un * u in W k,p (Ω), if and only if Dα un * Dα u in Lp (Ω) for all |α| ≤ k. Theorem 2.17. Let f : R → R be Lipschitz continuous with f (0) = 0. Then if Ω is a bounded open set in Rn , 1 < p < ∞ and u ∈ W01,p (Ω), we have f ◦ u ∈ W01,p (Ω). Proof. Given u ∈ W01,p (Ω), let un ∈ C01 (Ω) with kun − uk1,p → 0 and define vn = f ◦ un . Since un has compact support and f (0) = 0, vn has compact support. Also vn is Lipschitz continuous, for |vn (x) − vn (y)| = |f (un (x)) − f (un (y))| ≤ c|un (x) − un (y)| ≤ cn |x − y|. Hence vn ∈

Lp (Ω).

Since vn is absolutely continuous on any line segment in Ω, its partial derivatives (which exist almost everywhere) coincide almost everywhere with the weak derivatives. Moreover, we see from above that |∂vn /∂xi | ≤ cn for 1 ≤ i ≤ n, and as Ω is bounded, ∂vn /∂xi ∈ Lp (Ω). Thus vn ∈ W 1,p (Ω) and has compact support, which implies vn ∈ W01,p (Ω). From the relation |vn (x) − f (u(x))| ≤ c|un (x) − u(x)| it follows that kvn − f ◦ ukp → 0. Furthermore, if ei is the standard ith basis vector in Rn , we have |vn (x + hei ) − vn (x)| |un (x + hei ) − un (x)| ≤c |h| |h| and so ∂vn ∂un lim sup k kp ≤ c lim sup k kp . ∂xi ∂xi n→∞ n→∞ But, {∂un /∂xi } is a convergent sequence in Lp (Ω) and therefore {∂vn /∂xi } is bounded in Lp (Ω) for each 1 ≤ i ≤ n. Since kvn k1,p is bounded and W01,p (Ω) is reflexive, a subsequence of {vn } converges weakly in W 1,p (Ω), and thus weakly in Lp (Ω) to some element of W01,p (Ω). Thus, f ◦ u ∈ W01,p (Ω). Remark. In terms of the trace operator (defined later) we have γ0 (f ◦ u) = f ◦ γ0 (u). Corollary 2.18. Let u ∈ W01,p (Ω). Then |u|, u+ , u− ∈ W01,p (Ω). Proof. We apply the preceding theorem with f (t) = |t|. Thus |u| ∈ W01,p (Ω). Now u+ = (|u| + u)/2 and u− = (u − |u|)/2. Thus u+ , u− ∈ W01,p (Ω).

34

2. Sobolev Spaces

2.2.4. Extensions. If Ω ⊂ Ω0 , then any function u(x) ∈ C0k (Ω) has an obvious extension U (x) ∈ C0k (Ω0 ). From the definition of W0k,p (Ω) it follows that the function u(x) ∈ W0k,p (Ω) and extended as being equal to zero in Ω0 \Ω belongs to W0k,p (Ω0 ). In general, a function u ∈ W k,p (Ω) and extended by zero to Ω0 will not belong to W k,p (Ω0 ). (Consider the function u(x) ≡ 1 in Ω.) However, if u ∈ W k,p (Ω) has compact support in Ω, then u ∈ W0k,p (Ω) and thus the obvious extension belongs to W0k,p (Ω0 ). We now consider a more general extension result. Theorem 2.19. Let Ω be a bounded open set in Rn with Ω ⊂⊂ Ω0 and assume k ≥ 1. (a) If ∂Ω ∈ C k , then any function u(x) ∈ W k,p (Ω) has an extension U (x) ∈ W k,p (Ω0 ) into Ω0 with compact support. Moreover, kU kW k,p (Ω0 ) ≤ ckukW k,p (Ω) where the constant c > 0 does not depend on u. ¯ has an extension U (x) ∈ C k (Ω0 ) (b) If ∂Ω ∈ C k , then any function u(x) ∈ C k (Ω) 0 into Ω0 with compact support. Moreover, kU kC k (Ω¯ 0 ) ≤ ckukC k (Ω) ¯ , kU kW k,p (Ω0 ) ≤ ckukW k,p (Ω) where the constant c > 0 does not depend on u. (c) If ∂Ω ∈ C k , then any function u(x) ∈ C k (∂Ω) has an extension U (x) into Ω ¯ Moreover which belongs to C k (Ω). kU kC k (Ω) ¯ ≤ ckukC k (∂Ω) where the constant c > 0 does not depend on u. ¯ Proof. Suppose first that u ∈ C k (Ω). Let y = ψ(x) define a C k diffeomorphism that 0 0 straightens the boundary near x = (x1 , . . . , x0n ) ∈ ∂Ω. In particular, we assume there is a ball B = B(x0 ) such that ψ(B ∩ Ω) ⊂ Rn+ (i.e., yn > 0), ψ(B ∩ ∂Ω) ⊂ ∂Rn+ . (e.g., we could choose yi = xi − x0i for i = 1, . . . , n − 1 and yn = xn − ϕ(x1 , . . . , xn−1 ), where ϕ is of class C k . Moreover, without loss of generality, we can assume yn > 0 if x ∈ B ∩ Ω.) Let G and G+ = G ∩ Rn+ be respectively, a ball and half-ball in the image of ψ such 0 that ψ(x0 ) ∈ G. Setting u ¯(y) = u ◦ ψ −1 (y) and y = (y1 , . . . , yn−1 , yn ) = (y , yn ), we define ¯ (y) of u an extension U ¯(y) into yn < 0 by ¯ (y 0 , yn ) = U

k+1 X

0

ci u ¯(y , −yn /i), yn < 0

i=1

where the ci are constants determined by the system of equations (2.5)

k+1 X

ci (−1/i)m = 1, m = 0, 1, . . . , k.

i=1

Note that the determinant of the system (2.5) is nonzero since it is the Vandemonde deter¯ is continuous with all derivatives minant. One verifies readily that the extended function U up to order k in G. For example, ¯ (y) = lim U

y→(y 0 ,0)

k+1 X i=1

ci u ¯(y 0 , 0) = u ¯(y 0 , 0)


35

by virtue of (2.5) with m = 0. A similar computation shows that ¯y (y) = u lim U ¯yi (y 0 , 0), i = 1, . . . , n − 1. i

y→(y 0 ,0)

Finally ¯yn (y) = lim U

y→(y 0 ,0)

k+1 X

¯yn (y 0 , 0) ci (−1/i)¯ uyn (y 0 , 0) = u

i=1

by virtue of (2.5) with m = 1. Similarly we can handle the higher derivatives. ¯ ◦ ψ ∈ C k (B 0 ) for some ball B 0 = B 0 (x0 ) and w = u in B 0 ∩ Ω, (If x ∈ B 0 ∩ Ω, Thus w = U + ¯ (ψ(x)) = u then ψ(x) ∈ G and w(x) = U ¯(ψ(x)) = u(ψ −1 ψ(x)) = u(x)) so that w provides a C k extension of u into Ω ∪ B 0 . Moreover, sup |¯ u(y)| = sup |u(ψ −1 (y))| ≤ sup |u(x)| G+

G+

Ω

and since x ∈ B 0 implies ψ(x) ∈ G ¯ (ψ(x))| ≤ c sup |¯ sup |U u(y)| ≤ c sup |u(x)|. B0

G+

Ω

Since a similar computation for the derivatives holds, it follows that there is a constant c > 0, independent of u, such that kwkC k (Ω∪B 0 ) ≤ ckukC k (Ω) ¯ ¯ . Now consider a finite covering of ∂Ω by balls Bi , i = 1, . . . , N , such as B in the preceding, and let {wi } be the corresponding C k extensions. We may assume the balls Bi 0 are so small that their union with Ω is contained in Ω . Let Ω0 ⊂⊂ Ω be such that Ω0 and the balls Bi provide a finite open covering of Ω. Let {ηi }, i = 1, . . . , N , be a partition of unity subordinate to this covering and set X w = uη0 + wi η i 0

with the understanding that wi ηi = 0 if ηi = 0. Then w is an extension of u into Ω and has the required properties. Thus (b) is established. ¯ such that (a) If u ∈ W k,p (Ω), then by Theorem 2.13, there exist functions um ∈ C ∞ (Ω) k,p 00 0 um → u in W (Ω). Let Ω ⊂ Ω ⊂ Ω , and let Um be the extension of um to Ω00 as given in (b). Then kUm − Ul kW k,p (Ω00 ) ≤ ckum − ul kW k,p (Ω) which implies that {Um } is a Cauchy sequence and so converges to a U ∈ W0k,p (Ω00 ), since Um ∈ C0k (Ω00 ). Now extend Um , U by 0 to Ω0 . It is easy to see that U is the desired extension. (c) At any point x0 ∈ ∂Ω let the mapping ψ and the ball G be defined as in (b). By 0 ¯ 0 , yn ) = u definition, u ∈ C k (∂Ω) implies that u ¯ = u◦ψ −1 ∈ C k (G∩∂Rn+ ). We define Φ(y ¯(y ) ¯ ◦ ψ(x) for x ∈ ψ −1 (G). Clearly, Φ ∈ C k (B) ¯ for some ball B = B(x0 ) in G and set Φ(x) = Φ and Φ = u on B ∩ ∂Ω. Now let {Bi } be a finite covering of ∂Ω by balls such as B and let Φi be the corresponding C k functions defined on Bi . For each i, we define the function Ui (x) as follows: in the ball Bi take it equal to Φi , outside Bi take it equal to zero if x 6∈ ∂Ω and equal to u(x) if x ∈ ∂Ω. The proof can now be completed as in (b) by use of an appropriate partition of unity.

36

2. Sobolev Spaces

2.2.5. Trace Theorem. Unless otherwise stated, Ω will denote a bounded open connected set in Rn , i.e., a bounded domain. ¯ and has the representation Let Γ be a surface which lies in Ω xn = ϕ(x0 ), x0 = (x1 , . . . , xn−1 ) ¯ . Here U is the projection of Γ onto the coordinate where ϕ(x0 ) is Lipschitz continuous in U plane xn = 0. Let p ≥ 1. A function u defined on Γ is said to belong to Lp (Γ) if Z 1 kukLp (Γ) ≡ ( |u(x)|p dS) p < ∞ Γ

where Z Γ

|u(x)|p dS =

Z

|u(x0 , ϕ(x0 ))|p [1 +

U

n−1 X

(

i=1

∂ϕ 0 2 1 0 (x )) ] 2 dx . ∂xi

reduces to a space of the type Lp (U ) where U is a domain in Rn−1 . ¯ its values γ0 u ≡ u|Γ on Γ are uniquely given. The function For every function u ∈ C(Ω), γ0 u will be called the trace of the function u on Γ. Note that u ∈ Lp (Γ) since γ0 u ∈ C(Γ).

Thus

Lp (Γ)

On the other hand, if we consider a function u defined a.e. in Ω (i.e., functions are considered equal if they coincide a.e.), then the values of u on Γ are not uniquely determined since meas(Γ) = 0. In particular, since ∂Ω has measure 0, there exist infinitely many ¯ that are equal a.e. We shall therefore introduce the concept of trace for extensions of u to Ω 1,p ¯ the new definition of trace reduces functions in W (Ω) so that if in addition, u ∈ C(Ω), to the definition given above. ¯ Lemma 2.20. Let ∂Ω ∈ C 0,1 . Then for u ∈ C 1 (Ω), kγ0 ukLp (∂Ω) ≤ ckuk1,p

(2.6)

where the constant c > 0 does not depend on u. Proof. For simplicity, let n = 2. The more general case is handled similarly. In a neighborhood of a boundary point x ∈ ∂Ω, we choose a local (ξ, η)-coordinate system, where the boundary has the local representation −α ≤ ξ ≤ α

η = ϕ(ξ), with the

C 0,1

function ϕ. Then there exists a β > 0 such that all the points (ξ, η) with −α ≤ ξ ≤ α,

ϕ(ξ) − β ≤ η ≤ ϕ(ξ)

¯ Let u ∈ C 1 (Ω). ¯ Then belong to Ω. ϕ(ξ)

Z u(ξ, ϕ(ξ)) =

uη (ξ, η)dη + u(ξ, t) t

where ϕ(ξ) − β ≤ t ≤ ϕ(ξ). Applying the inequality (a + b)p ≤ 2p−1 (ap + bp ) together with Hölder’s inequality we have Z ϕ(ξ) p p−1 p−1 |u(ξ, ϕ(ξ))| ≤ 2 β |uη (ξ, η)|p dη + 2p−1 |u(ξ, t)|p . ϕ(ξ)−β

An integration with respect to t yields Z p p−1 β|u(ξ, ϕ(ξ))| ≤ 2

ϕ(ξ)

ϕ(ξ)−β

[β p |uη (ξ, η)|p + |u(ξ, η)|p ]dη.


37

Finally, integration over the interval [−α, α] yields Z Z α p p−1 (β p |uη |p + |u|p )dξdη (2.7) β|u(ξ, ϕ(ξ))| dξ ≤ 2 −α

S

where S denotes a local boundary strip. Suppose ϕ(·) is C 1 . Then the differential of arc length is given by ds = (1 + ϕ02 )1/2 dξ. Addition of the local inequalities (2.7) yields the assertion (2.6). Now if ϕ(·) is merely Lipschitz continuous, then the derivative ϕ0 exists a.e. and is bounded. Thus we also obtain (2.6). ¯ = W 1,p (Ω), the bounded linear operator γ0 : C 1 (Ω) ¯ ⊂ W 1,p (Ω) → Lp (∂Ω) Since C 1 (Ω) 1,p can be uniquely extended to a bounded linear operator γ0 : W (Ω) → Lp (∂Ω) such that (2.6) remains true for all u ∈ W 1,p (Ω). More precisely, we obtain γ0 u in the following ¯ with kun − uk1,p → 0. Then way: Let u ∈ W 1,p (Ω). We choose a sequence {un } ⊂ C 1 (Ω) kγ0 un − γ0 ukLp (∂Ω) → 0. The function γ0 u (as an element of Lp (∂Ω)) will be called the trace of the function u ∈ W 1,p (Ω) on the boundary ∂Ω. (kγ0 ukLp (∂Ω) will be denoted by kukLp (∂Ω) .) Thus the trace of a function is defined for any element u ∈ W 1,p (Ω). The above discussion partly proves the following: Theorem 2.21. (Trace) Suppose ∂Ω ∈ C 1 . Then there is a unique bounded linear operator 1,p (Ω), and γ (au) = γ a·γ u ¯ γ0 : W 1,p (Ω) → Lp (∂Ω) such that γ0 u = u|∂Ω for u ∈ C(Ω)∩W 0 0 0 1,p 1 1,p ¯ u ∈ W (Ω). Moreover, N (γ0 ) = W (Ω) and R(γ0 ) = Lp (∂Ω). for a(x) ∈ C (Ω), 0

¯ ∩ W 1,p (Ω). Then by Theorem 2.19, u can be extended into Proof. Suppose u ∈ C(Ω) ¯ 0 ) ∩ W 1,p (Ω0 ). Let Uh (x) be the mollified Ω0 (Ω ⊂⊂ Ω0 ) such that its extension U ∈ C(Ω function for U . Since Uh → U as h → 0 in both the norms k · kC(Ω) ¯ , k · kW 1,p (Ω) , we find that as h → 0, Uh |∂Ω → u|∂Ω uniformly and Uh |∂Ω → γ0 u in Lp (∂Ω). Consequently, γ0 u = u|∂Ω . ¯ u ∈ W 1,p (Ω) and consequently, γ0 (au) is defined. Let Now au ∈ W 1,p (Ω) if a ∈ C 1 (Ω), ¯ with kun − uk1,p → 0. Then {un } ⊂ C 1 (Ω) γ0 (aun ) = γ0 a · γ0 un and the desired product formula follows by virtue of the continuity of γ0 . If u ∈ W01,p (Ω), then there is a sequence {un } ⊂ C01 (Ω) with kun − uk1,p → 0. But un |∂Ω = 0 and as n → ∞, un |∂Ω → γ0 u in Lp (∂Ω) which implies γ0 u = 0. Hence W01,p (Ω) ⊂ N (γ0 ). Now suppose u ∈ N (γ0 ). If u ∈ W 1,p (Ω) has compact support in Ω, then by an earlier remark, u ∈ W01,p (Ω). If u does not have compact support in Ω, then it can be shown that there exists a sequence of cut-off functions ηk such that ηk u ∈ W 1,p (Ω) has compact support in Ω, and moreover, kηk u − uk1,p → 0. By using the corresponding mollified functions, it follows that u ∈ W01,p (Ω) and N (γ0 ) ⊂ W01,p (Ω). Details can be found in Mikhailov [26]. To see that R(γ0 ) = Lp (∂Ω), let f ∈ Lp (∂Ω) and let ε > 0 be given. Then there is a ¯ be the extension of u into Ω, ¯ u ∈ C 1 (∂Ω) such that ku − f kLp (∂Ω) < ε. If we let U ∈ C 1 (Ω) 1,p then clearly kγ0 U − f kLp (∂Ω) < ε, which is the desired result since U ∈ W (Ω).

38

2. Sobolev Spaces

¯ and its trace on ∂Ω is Remark. We note that the function u ≡ 1 belongs to W 1,p (Ω) ∩ C(Ω) 1,p 1. Hence this function does not belong to W0 (Ω), which establishes the earlier assertion that W01,p (Ω) 6= W 1,p (Ω). Let u ∈ W k,p (Ω), k > 1. Since any weak derivative Dα u of order |α| < k belongs to this derivative has a trace γ0 Dα u belonging to Lp (∂Ω). Moreover

W 1,p (Ω),

kDα ukLp (∂Ω) ≤ ckDα uk1,p ≤ ckukk,p for constant c > 0 independent of u. Assuming the boundary ∂Ω ∈ C 1 , the unit outward normal vector n to ∂Ω exists and is bounded. Thus, the concept of traces makes it possible to introduce, for k ≥ 2, ∂u/∂n for u ∈ W k,p (Ω). More precisely, for k ≥ 2, there exist traces of the functions u, Di u so that, if ni are the direction cosines of the normal, we may define γ1 u =

n X

(γ0 (Di u))ni , u ∈ W k,p (Ω), k ≥ 2.

i=1

The trace operator γ1 : W k,p (Ω) → Lp (∂Ω) is continuous and γ1 u = (∂u/∂n)|∂Ω for u ∈ ¯ ∩ W k,p (Ω). C 1 (Ω) ¯ we define the various traces of normal derivatives given by For a function u ∈ C k (Ω) γj u =

∂j u |∂Ω , 0 ≤ j ≤ k − 1. ∂nj

Each γj can be extended by continuity to all of W k,p (Ω) and we obtain the following: Theorem 2.22. (Trace) Suppose ∂Ω ∈ C k . Then there is a unique continuous linear Q k−1−j,p (∂Ω) such that for u ∈ C k (Ω) ¯ operator γ = (γ0 , γ1 , . . . , γk−1 ) : W k,p (Ω) → k−1 j=0 W ∂j u |∂Ω , j = 1, . . . , k − 1. ∂nj Q k−1−j,p (∂Ω). Moreover, N (γ) = W0k,p (Ω) and R(γ) = k−1 j=0 W γ0 u = u|∂Ω , γj u =

The Sobolev spaces W k−1−j,p (∂Ω), which are defined over ∂Ω, can be defined locally. 2.2.6. Green’s Identities. In this section we assume that p = 2. Theorem 2.23. (Integration by Parts) Let u, v ∈ H 1 (Ω) and let ∂Ω ∈ C 1 . Then for any i = 1, . . . , n Z Z Z (2.8) vDi udx = (γ0 u · γ0 v)ni dS − uDi vdx. Ω

∂Ω

Ω

(Di u, Di v are weak derivatives.) ¯ with kun − ukH 1 (Ω) → Proof. Let {un } and {vn } be sequences of functions in C 1 (Ω) 0, kvn − vkH 1 (Ω) → 0 as n → ∞. Formula (2.8) holds for un , vn Z Z Z vn Di un dx = un vn ni dS − un Di vn dx Ω

∂Ω

and upon letting n → ∞ relation (2.8) follows. Corollary 2.24. Let ∂Ω ∈ C 1 .

Ω

2.3. Sobolev Imbedding Theorems

39

(a) If v ∈ H 1 (Ω) and u ∈ H 2 (Ω) then Z Z Z v∆udx = γ0 v · γ1 udS − (∇u · ∇v)dx (Green’s 1st identity). Ω

∂Ω

Ω

(b) If u, v ∈ H 2 (Ω) then Z Z (v∆u − u∆v)dx = (γ0 v · γ1 u − γ0 u · γ1 v)dS (Green’s 2nd identity). Ω

∂Ω

In these formulas ∇u ≡ (D1 u, . . . , Dn u) is the gradient vector and ∆u ≡ Laplace operator.

Pn

i=1 Dii u

is the

Proof. If in (2.8) we replace u by Di u and sum from 1 to n, then Green’s 1st identity is obtained. Interchanging the roles of u, v in Green’s 1st identity and subtracting the two identities yields Green’s 2nd identity. Exercise 1. Establish the following one-dimensional version of the trace theorem: If u ∈ W 1,p (Ω), where Ω = (a, b), then kukLp (∂Ω) ≡ (|u(a)|p + |u(b)|p )1/p ≤ const kukW 1,p (Ω) where the constant is independent of u.

2.3. Sobolev Imbedding Theorems We consider the following question: If a function u belongs to W k,p (Ω), does u automatically belong to certain other spaces? The answer will be yes, but which other spaces depend upon whether 1 ≤ kp < n, kp = n, n < kp < ∞. The result we want to prove is the following: Theorem 2.25. (Sobolev-Rellich-Kondrachov) Let Ω ⊂ Rn be bounded and open with ∂Ω ∈ C 1 . Assume 1 ≤ p < ∞ and k is a positive integer. (a) If kp < n and 1 ≤ q ≤ np/(n − kp), then W k,p (Ω) ⊂ Lq (Ω) is a continuous imbedding; the imbedding is compact if 1 ≤ q < np/(n − kp). Moreover, (2.9)

kukLq (Ω) ≤ CkukW k,p (Ω) where the constant C depends only on k,p,n and Ω. (b) If kp = n and 1 ≤ r < ∞, then W k,p (Ω) ⊂ Lr (Ω) is a compact imbedding and

(2.10)

kukLr (Ω) ≤ CkukW k,p (Ω) where the constant depends only on k,p,n and Ω.

40

2. Sobolev Spaces

(c) If kp > n and 0 ≤ α ≤ k − m − n/p, then ¯ W k,p (Ω) ⊂ C m,α (Ω) is a continuous imbedding; the imbedding is compact if 0 ≤ α < k − m − n/p. Moreover, kukC m,α (Ω) ¯ ≤ CkukW k,p (Ω)

(2.11)

where the constant C depends only on k,p,n,α and Ω. (d) Let 0 ≤ j < k, 1 ≤ p, q < ∞. Set d = 1/p − (k − j)/n. Then W k,p ⊂ W j,q is a continuous imbedding for d ≤ 1/q; the imbedding is compact for d < 1/q. The above results are valid for W0k,p (Ω) spaces on arbitrary bounded domains Ω. Remark. It is easy to check that the imbedding W 1,p (Ω) ⊂ Lp (Ω) is compact for all p ≥ 1 and all n. A series of special results will be needed to prove the above theorem. Only selected proofs will be given to illustrate some of the important techniques. 2.3.1. Gagliardo-Nirenberg-Sobolev Inequality. Suppose 1 ≤ p < n. Do there exist constants C > 0 and 1 ≤ q < ∞ such that kukLq (Rn ) ≤ Ck∇ukLp (Rn )

(2.12) for all u ∈

C0∞ (Rn )?

The point is that the constants C and q should not depend on u.

We shall show that if such an inequality holds, then q must have a specific form. For this choose any u ∈ C0∞ (Rn ), u 6≡ 0, and define for λ > 0 (x ∈ Rn ).

uλ (x) ≡ u(λx) Now

Z

Z

Z 1 |uλ | dx = |u(λx)| dx = n |u(y)|q dy n n n λ IR IR IR and Z Z Z λp p p p |∇uλ | dx = λ |∇u(λx)| dx = n |∇u(y)|p dy. λ IRn IRn IRn Inserting these inequalities into (2.12) we find q

1 λn/q

kukLq (Rn ) ≤ C

q

λ λn/p

k∇ukLp (Rn )

and so (2.13)

kukLq (Rn ) ≤ Cλ1−n/p+n/q k∇ukLp (Rn ) .

But then if 1 − n/p + n/q > 0(< 0), we can upon sending λ to 0 (∞) in (2.13) obtain a contradiction (u = 0). Thus we must have q = p* where np (2.14) p* = n−p is called the Sobolev conjugate of p. Note that then 1 1 1 (2.15) = − , p∗ > p. ∗ p p n Next we prove that the inequality (2.12) is in fact correct.


41

Lemma 2.26. (Gagliardo-Nirenberg-Sobolev Inequality) Assume 1 ≤ p < n. Then there is a constant C, depending only on p and n, such that kukLp∗ (Rn ) ≤ Ck∇ukLp (Rn )

(2.16) for all u ∈ C01 (Rn ).

Proof. First assume p = 1. Since u has compact support, for each i = 1, . . . , n we have Z xi uxi (x1 , . . . , xi−1 , yi , xi+1 , . . . , xn )dyi u(x) = −∞

and so Z

∞

|∇u(x1 , . . . , yi , . . . , xn )|dyi

|u(x)| ≤

(i = 1, . . . , n).

−∞

Consequently |u(x)|

(2.17)

n n−1

≤

n Z Y

∞

|∇u(x1 , . . . , yi , . . . , xn )|dyi

1 n−1

.

−∞

i=1

Integrate this inequality with respect to x1 : Z

∞

|u(x)|

n n−1

Z

∞

dx1 ≤

−∞

n Z Y

−∞ i=1

Z

∞

1 n−1

|∇u|dyi

dx1

−∞

∞

1 n−1

Z

∞

|∇u|dy1

= −∞

Z

−∞ i=2

∞

≤

n Z Y

1 n−1

n Z Y

|∇u|dy1 −∞

i=2

∞

−∞

∞

1 n−1

|∇u|dyi

dx1

−∞

Z

!

∞

1 n−1

|∇u|dx1 dyi −∞

the last inequality resulting from the extended Hölder inequality in the appendix. We continue by integrating with respect to x2 , . . . , xn and applying the extended Hölder inequality to eventually find (pull out an integral at each step) Z |u(x)|

n n−1

dx ≤

IRn

n Z Y

∞

Z ···

−∞

i=1

∞

Z

1 n−1

|∇u|dx1 . . . dyi . . . dxn −∞

n n−1

|∇u|dx

= IRn

which is estimate (2.16) for p = 1. Consider now the case that 1 < p < n. We shall apply the last estimate to v = |u|γ , where γ > 1 is to be selected. First note that (γuγ−1 Di u)2 if u ≥ 0 γ 2 (Di |u| ) = = (γ|u|γ−1 Di u)2 . γ−1 2 (−γ(−u) Di u) if u ≤ 0 Thus v ∈ C01 (Rn ), and

42

2. Sobolev Spaces

Z |u(x)|

γn n−1

n−1

Z

n

|∇|u|γ |dx

≤

dx

IRn

IRn

Z

|u|γ−1 |∇u|dx

= γ IRn

Z ≤ γ

|u|

p(γ−1) p−1

p−1 Z p

|∇u| dx

dx

IRn

p

1

p

.

IRn

We set γ=

p(n − 1) >1 n−p

in which case γn p(γ − 1) np = = = p∗ . n−1 p−1 n−p Thus, the above estimate becomes Z 1∗ Z 1 p p p∗ p |u| dx ≤C |∇u| dx . IRn

IRn

Theorem 2.27. Let Ω ⊂ Rn be bounded and open, with ∂Ω ∈ C 1 . Assume 1 ≤ p < n, and ∗ u ∈ W 1,p (Ω). Then u ∈ Lp (Ω) and (2.18)

kukLp∗ (Ω) ≤ CkukW 1,p (Ω)

where the constant C depends only on p,n and Ω. Proof. Since ∂Ω ∈ C 1 , there exists an extension U ∈ W 1,p (Rn ) such that U = u in Ω, U has compact support and (2.19)

kU kW 1,p (Rn ) ≤ CkukW 1,p (Ω) .

Moreover, since U has compact support, there exist mollified functions um ∈ C0∞ (Rn ) such that um → U in W 1,p (Rn ). Now according to Lemma 2.26, kum − ul kLp∗ (Rn ) ≤ Ck∇um − ∇ul kLp (Rn ) ∗

for all l, m ≥ 1; whence um → U in Lp (Rn ) as well. Since Lemma 2.26 also implies kum kLp∗ (Rn ) ≤ Ck∇um kLp (Rn ) we get in the limit that kU kLp∗ (Rn ) ≤ Ck∇U kLp (Rn ) . This inequality and (2.19) complete the proof.

Theorem 2.28. Let Ω ⊂ Rn be bounded and open. Assume 1 ≤ p < n, and u ∈ W01,p (Ω). Then u ∈ Lq (Ω) and kukLq (Ω) ≤ Ck∇ukLp (Ω) for each q ∈ [1, p∗ ], the constant C depending only on p,q,n and Ω.


43

Proof. Since u ∈ W01,p (Ω), there are functions um ∈ C0∞ (Ω) such that um → u in W 1,p (Ω). ¯ and apply Lemma 2.26 to discover (as above) We extend each function um to be 0 in Rn \Ω kukLp∗ (Ω) ≤ Ck∇ukLp (Ω) . Since |Ω| < ∞, we furthermore have kukLq (Ω) ≤ CkukLp∗ (Ω) for every q ∈ [1, p∗ ].

2.3.2. Morrey’s Inequality. We now turn to the case n < p < ∞. The next result shows that if u ∈ W 1,p (Ω), then u is in fact Hölder continuous, after possibly being redefined on a set of measure zero. Theorem 2.29. (Morrey’s Inequality) Assume n < p < ∞. Then there exists a constant C, depending only on p and n, such that kuk

(2.20)

C

0,1− n p

(Rn )

∀ u ∈ C 1 (Rn ).

≤ CkukW 1,p (Rn ) ,

Proof. We first prove the following inequality: for all x ∈ Rn , r > 0 and all u ∈ C 1 (Rn ), Z Z |Du(y)| rn dy. (2.21) |u(y) − u(x)| dy ≤ n B(x,r) |x − y|n−1 B(x,r) To prove this, note that, for any w with |w| = 1 and 0 < s < r, Z s d |u(x + sw) − u(x)| = u(x + tw)dt dt Z0 s ∇u(x + tw) · wdt = Z 0s |∇u(x + sw)| dt. ≤ 0

Now we integrate w over ∂B(0, 1) to obtain Z Z |u(x + sw) − u(x)| dS ≤ ∂B(0,1)

0

sZ

|∇u(x + sw)| dSdt ∂B(0,1)

Z = B(x,s)

Z ≤ B(x,r)

|∇u(y)| dy |x − y|n−1 |∇u(y)| dy. |x − y|n−1

Multiply both sides by sn−1 and integrate over s ∈ (0, r) and we obtain (2.21). To establish the bound on kukC 0 (Rn ) , we observe that, by (2.21), for x ∈ Rn , Z Z 1 1 |u(x)| ≤ |u(y) − u(x)| dy + |u(y)| dy |B(x, 1)| B(x,1) |B(x, 1)| B(x,1) ! p−1 Z Z 1/p

|∇u(y)|p dy

≤ C Rn

≤ CkukW 1,p (Rn ) .

|y − x| B(x,1)

(1−n)p p−1

p

dy

+ CkukLp (Rn )

44

2. Sobolev Spaces

To establish the bound on the semi-norm [u]γ , γ = 1 − np , take any two points x, y ∈ Rn . Let r = |x − y| and W = B(x, r) ∩ B(y, r). Then Z Z 1 1 |u(x) − u(z)| dz + |u(y) − u(z)| dz. (2.22) |u(x) − u(y)| ≤ |W | W |W | W R R R Note that |W | = βrn , r = |x − y| and W ≤ min{ B(x,r) , B(y,r) }. Hence, using (2.21), by Hölder’s inequality, we obtain Z Z Z rn |u(x) − u(z)| dz ≤ |u(x) − u(z)| dz ≤ |Du(z)||x − z|1−n dz n B(x,r) W B(x,r) !1/p Z ! p−1 Z p (1−n)p rn |∇u(z)|p dz |z − x| p−1 dz ≤ n B(x,r) B(x,r) Z r p−1 p (1−n)p n n−1 p−1 ≤ C r k∇ukLp (Rn ) s ds s 0

≤ Cr where γ = 1 − np ; similarly, Z

n+γ

k∇ukLp (Rn ) ,

|u(y) − u(z)| dz ≤ C rn+γ k∇ukLp (Rn ) .

W

Hence, by (2.22), |u(x) − u(y)| ≤ C |x − y|γ k∇ukLp (Rn ) . This inequality and the bound on kukC 0 above complete the proof.

Theorem 2.30. (Estimates for W 1,p , n < p ≤ ∞) Let Ω ⊂ Rn be bounded and open, 0,1− n ¯ a.e. and p (Ω) with ∂Ω ∈ C 1 . Assume n < p < ∞, and u ∈ W 1,p (Ω). Then u ∈ C kuk

C

0,1− n p

¯ (Ω)

≤ CkukW 1,p (Ω)

where the constant C depends only on p, n and Ω. Proof. Since ∂Ω ∈ C 1 , there exists an extension U ∈ W 1,p (Rn ) such that U = u in Ω, U has compact support and (2.23)

kU kW 1,p (Rn ) ≤ CkukW 1,p (Ω) .

Moreover, since U has compact support, there exist mollified functions um ∈ C0∞ (Rn ) such that um → U in W 1,p (Rn ) (and hence on compact subsets). Now according to Morrey’s inequality, kum − ul kC 0,1−n/p (Rn ) ≤ Ckum − ul kW 1,p (Rn ) for all l, m ≥ 1; whence there is a function u∗ ∈ C 0,1−n/p (Rn ) such that um → u∗ in C 0,1−n/p (Rn ). Thus u∗ = u a.e. in Ω. Since we also have kum kC 0,1−n/p (Rn ) ≤ Ckum kW 1,p (Rn ) we get in the limit that ku∗ kC 0,1−n/p (Rn ) ≤ CkU kW 1,p (Rn ) . This inequality and (2.23) complete the proof.


45

2.3.3. General Cases. We can now concatenate the above estimates to obtain more complicated inequalities. Assume kp < n and u ∈ W k,p (Ω). Since Dα u ∈ Lp (Ω) for all |α| ≤ k, the SobolevNirenberg-Gagliardo inequality implies kDβ ukLp∗ (Ω) ≤ CkukW k,p (Ω) ∗

if |β| ≤ k − 1, and so u ∈ W k−1,p (Ω). Moreover, kukk−1,p∗ ≤ ckukk,p . Similarly, we find ∗∗ u ∈ W k−2,p (Ω), where 1 1 1 2 1 = ∗− = − . ∗∗ p p n p n Moreover, kukk−2,p∗∗ ≤ ckukk−1,p∗ . Continuing, we find after k steps that u ∈ W 0,q (Ω) = Lq (Ω) for 1 1 k = − . q p n The stated estimate (2.9) follows from combining the relevant estimates at each stage of the above argument. In a similar manner the other estimates can be established. 2.3.4. Compactness. We now consider the compactness of the imbeddings. Before we present the next result we recall some facts that will be needed. A subset S of a normed space is said to be totally bounded if for each ε > 0 there is a finite set of open balls of radius ε which cover S. Clearly, a totally bounded set is bounded, i.e., it is contained in a sufficiently large ball. It is not difficult to see that a relatively compact subset of a normed space is totally bounded, with the converse being true if the normed space is complete. Moreover, a totally bounded subset of a normed space is separable. Theorem 2.31. (Rellich-Kondrachov) Let Ω ⊂ Rn be bounded and open. Then for 1 ≤ p < n: (a) The imbedding W01,p (Ω) ⊂ Lq (Ω) is compact for each 1 ≤ q < np/(n − p). (b) Assuming ∂Ω ∈ C 1 , the imbedding W 1,p (Ω) ⊂ Lq (Ω) is compact for each 1 ≤ q < np/(n − p). (c) Assuming ∂Ω ∈ C 1 , γ0 : W 1,p (Ω) → Lp (∂Ω) is compact. If p > n, then ¯ is compact for each 0 ≤ (d) Assuming ∂Ω ∈ C 1 , the imbedding W 1,p (Ω) ⊂ C α (Ω) α < 1 − (n/p). Proof. We shall just give the proof for p = q = 2. The other cases are proved similarly. (a) Since C01 (Ω) is dense in H01 (Ω), it suffices to show that the imbedding C01 (Ω) ⊂ L2 (Ω) is compact. Thus, let S = {u ∈ C01 (Ω) : kuk1,2 ≤ 1}. We now show that S is totally bounded in L2 (Ω). For h > 0, let Sh = {uh : u ∈ S}, where uh is the mollified function for u. We claim that Sh is totally bounded in L2 (Ω). Indeed, for u ∈ S, we have Z |uh (x)| ≤ ωh (z)|u(x − z)|dz ≤ (sup ωh )kuk1 ≤ c1 (sup ωh )kuk1,2 B(0,h)

and |Di uh (x)| ≤ c2 sup |Di ωh |kuk1,2 , i = 1, . . . , n ¯ Thus by the Ascoli Theorem, so that Sh is a bounded and equicontinuous subset of C(Ω). ¯ Sh is relatively compact (and thus totally bounded) in C(Ω) and consequently also in L2 (Ω).

46

2. Sobolev Spaces

Now, by earlier estimates, we easily obtain Z Z 2 2 kuh − uk2 ≤ ωh (z) |u(x − z) − u(x)| dx dz B(0,h)

Ω

and 2 Z Z 1 du(x − tz) |u(x − z) − u(x)| dx = dt dx dt Ω Ω 0 2 Z Z 1 (−∇u(x − tz) · z)dt dx = 0 Ω Z Z 1 2 2 |z| |∇u(x − tz)| dt dx ≤ |z|2 kuk21,2 . ≤

Z

2

Ω

0

Consequently, kuh − uk2 ≤ h. Since we have shown above that Sh is totally bounded in L2 (Ω) for all h > 0, it follows that S is also totally bounded in L2 (Ω) and hence relatively compact. (b) Suppose now that S is a bounded set in H 1 (Ω). Each u ∈ S has an extension U ∈ H01 (Ω0 ) where Ω ⊂⊂ Ω0 . Denote by S 0 the set of all such extensions of the functions u ∈ S. Since kU kH 1 (Ω0 ) ≤ ckuk1,2 , the set S 0 is bounded in H01 (Ω0 ). By (a) S 0 is relatively compact in L2 (Ω0 ) and therefore S is relatively compact in L2 (Ω). ¯ the inequality (2.7) with (c) Let S be a bounded set in H 1 (Ω). For any u(x) ∈ C 1 (Ω), p = 2 yields c1 (2.24) kuk2L2 (∂Ω) ≤ kuk22 + c2 βkuk21,2 β where the constants c1 , c2 do not depend on u or β. By completion, this inequality is valid for any u ∈ H 1 (Ω). By (b), any infinite sequence of elements of the set S has a subsequence {un } which is Cauchy in L2 (Ω): given ε > 0, an N can be found such that for all m, n ≥ N, kum − un k2 < ε. Now we choose β = ε. Applying the inequality (2.24) to um − un , it follows that the sequence of traces {γ0 un } converges in L2 (∂Ω). (d) By Morrey’s inequality, the imbedding is continuous if α = 1 − (n/p). Now use the fact that C β is compact in C α if α < β. Remarks. (a) When p = n, we can easily show that the imbedding in (a) is compact for all 1 ≤ q < ∞. Hence, it follows that the imbedding W01,p (Ω) ⊂ Lp (Ω) is compact for all p ≥ 1. (b) The boundedness of Ω is essential in the above theorem. For example, let I = (0, 1) and Ij = (j, j + 1). Let f ∈ C01 (I) and define fj to be the same function defined on Ij by translation. We can normalize f so that kf kW 1,p (I) = 1. The same is then true for each fj and thus {fj } is a bounded sequence in W 1,p (R). Clearly f ∈ Lq (R) for every 1 ≤ q ≤ ∞. Further, if kf kLq (R) = kf kLq (I) = a > 0 then for any j 6= k we have kfj − fk kqLq (R) =

Z j

j+1

|fj |q +

Z

k+1

|fk |q = 2aq

k

and so fi cannot have a convergent subsequence in Lq (R). Thus none of the imbeddings W 1,p (R) ⊂ Lq (R) can be compact. This example generalizes to n dimensional space and to open sets like a half-space.

2.4. Equivalent Norms

47

2.4. Equivalent Norms Two norms k · k and | · | on a vector space X are equivalent if there exist constants c1 , c2 ∈ (0, ∞) such that kxk ≤ c1 |x| ≤ c2 kxk

for all x ∈ X.

Note that the property of a set to be open, closed, compact, or complete in a normed space is not affected if the norm is replaced by an equivalent norm. A seminorm q on a vector space has all the properties of a norm except that q(u) = 0 need not imply u = 0. 2.4.1. Equivalent Norms of W 1,p (Ω). The next result plays a crucial part in the next chapter. Theorem 2.32. Let ∂Ω ∈ C 1 and let 1 ≤ p < ∞. Set !1/p Z X n kuk = |Di u|p dx + (q(u))p Ω i=1

where q : W 1,p (Ω) → R is a seminorm with the following two properties: (i) There is a positive constant d such that for all u ∈ W 1,p (Ω) q(u) ≤ dkuk1,p . (ii) If u = constant, then q(u) = 0 implies u = 0. Then k · k is an equivalent norm on W 1,p (Ω). Proof. First of all, it is easy to check that k · k defines a norm. Now by (i), it suffices to prove that there is a positive constant c such that kuk1,p ≤ ckuk

(2.25)

for all

u ∈ W 1,p (Ω).

Suppose (2.25) is false. Then there exist un (x) ∈ W 1,p (Ω) such that (kun k1,p > nkun k. Set vn = un /kun k1,p .) kun k1,p = 1

(2.26)

and

1 > nkun k.

According to Theorem 2.31, there is a subsequence, call it again {un }, which converges to u in Lp (Ω). From (2.26) we have kun k → 0 and therefore un → u in W 1,p (Ω) and q(un ) → 0. Hence, kuk1,p = 1 and by continuity, q(u) = 0. However, since Di u = 0 a.e. in Ω, u = const., which must be zero by (ii) and this contradicts kuk1,p = 1. Example 2.33. Let ∂Ω ∈ C 1 . Assume a(x) ∈ C(Ω), σ(x) ∈ C(∂Ω) with a ≥ 0 (6≡ 0), σ ≥ 0 (6≡ 0). Then the following norms are equivalent to k · k1,p on W 1,p (Ω): Z p !1/p Z X n R (2.27) kuk = |Di u|p dx + udx with q(u) = Ω udx . Ω i=1

(2.28)

kuk =

Z X n Ω i=1

(2.29)

kuk =

Z X n Ω i=1

Ω

Z p |Di u| dx +

∂Ω

p

Z

|Di u| dx +

p !1/p γ0 udS !1/p p

σ|γ0 u| dS ∂Ω

R with q(u) = ∂Ω γ0 udS .

with q(u) =

R

σ|γ0 u|p dS

1/p

.

48

2. Sobolev Spaces

(2.30)

Z X n

kuk =

p

|Di u| dx +

Ω i=1

!1/p

Z

p

a|u| dx

with q(u) =

1/p p . Ω a|u| dx

R

Ω

In order to verify condition (i) of Theorem 2.32, one uses the trace theorem in (2.28) and (2.29). R 2.4.2. Poincar´ e’s Inequalities. If u has zero mean value, i.e., Ω udx = 0, then from (2.27) it follows that Z Z X n (2.31) |u(x)|p dx ≤ c |Di u|p dx, u ∈ W 1,p (Ω) Ω

Ω i=1

where the constant c > 0 is independent of u. This inequality is often referred to as Poincar´ e’s inequality. We also note that (2.28) implies that (2.31) holds for u ∈ W01,p (Ω), and therefore !1/p Z X n p kuk1,p,0 = |Di u| dx Ω i=1

defines an equivalent norm on

W01,p (Ω).

2.4.3. Fourier Transform Methods. For a function u ∈ L1 (Rn ), we define the Fourier transform of u by Z 1 e−ix·y u(x) dx, ∀ y ∈ Rn , u ˆ(y) = (2π)n/2 Rn and the inverse Fourier tranform by Z 1 u ˇ(y) = eix·y u(x) dx, (2π)n/2 Rn

∀ y ∈ Rn .

Theorem 2.34. (Plancherel’s Theorem) Assume u ∈ L1 (Rn ) ∩ L2 (Rn ). Then u ˆ, u ˇ∈ L2 (Rn ) and kˆ ukL2 (Rn ) = kˇ ukL2 (Rn ) = kukL2 (Rn ) . Since L1 (Rn ) ∩ L2 (Rn ) is dense in L2 (Rn ), we can use this result to extend the Fourier transforms on to L2 (Rn ). We still use the same notations for them. Then we have Theorem 2.35. (Property of Fourier Tranforms) Assume u, v ∈ L2 (Rn ). Then R R v dx = Rn u ˆv¯ ˆ dy, (i) Rn u¯ α u(y) = (iy)α u d (ii) D ˆ(y) for each multiindex α such that Dα u ∈ L2 (Rn ), ˇ (iii) u = u ˆ. Next we use the Fourier transform to characterize the spaces H k (Rn ). Theorem 2.36. Let k be a nonnegative integer. Then, a function u ∈ L2 (Rn ) belongs to H k (Rn ) if and only if (1 + |y|k )ˆ u(y) ∈ L2 (Rn ). In addition, there exists a constant C such that C −1 kukH k (Rn ) ≤ k(1 + |y|k ) u ˆkL2 (Rn ) ≤ C kukH k (Rn ) for all u ∈ H k (Rn ).

2.4. Equivalent Norms

49

Using the Fourier transform, we can also define fractional Sobolev spaces H s (Rn ) for any 0 < s < ∞ as follows H s (Rn ) = {u ∈ L2 (Rn ) | (1 + |y|s ) u ˆ ∈ L2 (Rn )}, and define the norm by kukH s (Rn ) = k(1 + |y|s ) u ˆkL2 (Rn ) . From this we easily get the estimate kukL∞ (Rn ) ≤ kˆ ukL1 (Rn ) = k(1 + |y|s )ˆ u (1 + |y|s )−1 kL1 (Rn ) ≤ k(1 + |y|s )ˆ ukL2 (Rn ) k(1 + |y|s )−2 k2L1 (Rn ) ≤ C kukH s (Rn ) , where C = k(1 + |y|s )−2 k2L1 (Rn ) < ∞ if and only if s > n2 . Therefore we have an easy embedding, which is known valid for integers s by the previous Sobolev embedding theorem, H s (Rn ) ⊂ L∞ (Rn )

if s > n2 .

Chapter 3

Existence Theory for Linear Problems

3.1. Differential Equations in Divergence Form Henceforth, Ω ⊂ Rn denotes a bounded domain with boundary ∂Ω ∈ C 1 . 3.1.1. Linear Elliptic Equations. Consider the (Dirichlet) boundary value problem (BVP) (3.1)

Lu = f in Ω,

u = 0 on ∂Ω.

Here f is a given function in L2 (Ω) (or more generally, an element in the dual space of H01 (Ω)) and L is a formal second-order differential operator in divergence form given by

Lu ≡ −

n X

Di (aij (x)Dj u) +

i,j=1

n X

bi (x)Di u + c(x)u

i=1

with real coefficients aij (x), bi (x) and c(x) only in L∞ (Ω) and aij (x) = aji (x) (i, j = 1, . . . , n). Moreover, L is assumed to be uniformly elliptic in Ω, i.e., there exists a number θ > 0 such that for every x ∈ Ω and every real vector ξ = (ξ1 , . . . , ξn ) ∈ Rn n X

(3.2)

aij (x)ξi ξj ≥ θ

i,j=1

n X

|ξi |2 .

i=1

A function u ∈ H01 (Ω) is called a weak solution of (3.1) if the following holds   Z Z n n X X  (3.3) B1 (u, v) ≡ aij Dj uDi v + ( bi Di u + cu)v  dx = f vdx, ∀ v ∈ H01 (Ω). Ω

i,j=1

i=1

Ω

Much of the material in this chapter is devoted to proving the existence of weak solutions; the regularity problem will not be studied. Other problems can also be formulated in the weak sense as above. 51

52

3. Existence Theory for Linear Problems

Exercise 1. Consider the following weak formulation: Given f ∈ L2 (Ω). Find u ∈ H 1 (Ω) satisfying Z Z f vdx for all v ∈ H 1 (Ω). ∇u · ∇vdx = Ω

Ω

Find the boundary value problem solved by u. What is the necessary condition for the existence of such a u? 3.1.2. General Systems in Divergence Form. For N unknown functions, u1 , · · · , uN , we can write u = (u1 , · · · , uN ) and say u ∈ X(Ω; RN ) if each uk ∈ X(Ω), where X is a symbol of any function spaces we learned. For example, if u ∈ W 1,p (Ω; RN ) then we use Du to denote the N × n Jacobi matrix (∂uk /∂xi )1≤k≤N,1≤i≤n . The (Dirichlet) BVP for a most general system of partial differential equations in divergence form can be written as follows: (3.4)

− div A(x, u, Du) + b(x, u, Du) = F in Ω,

u = 0 on ∂Ω,

where A(x, s, ξ) = (Aki (x, u, ξ)), 1 ≤ i ≤ n, 1 ≤ k ≤ N, and b(x, s, ξ) = (bk (x, u, ξ)), 1 ≤ k ≤ N, are given functions of (x, u, ξ) ∈ Ω × RN × MN ×n satisfying some structural conditions, and F = (f k ), 1 ≤ k ≤ N , with each f k being a given functional in the dual space of W01,p (Ω). The structural conditions will generally assure that both |A(x, u, Du)| 0 p and |b(x, u, Du)| belong to Lp (Ω) for all u ∈ W 1,p (Ω; RN ), where p0 = p−1 . A function u ∈ W01,p (Ω; RN ) is then called a weak solution of (3.4) if the following holds # Z "X n (3.5) Aki (x, u, Du)Di ϕ + bk (x, u, Du)ϕ dx = hf k , ϕi Ω

i=1

for all ϕ ∈ W01,p (Ω) and each k = 1, 2, · · · , N. The system (3.4) is said to be linear if both A and b are linear in the variables (u, ξ); that is, (3.6)

Aki (x, u, Du)

X

=

l akl ij (x) Dj u

+

1≤l≤N, 1≤j≤n

X

k

b (x, u, Du) =

N X

l dkl i (x) u ,

l=1 l bkl j (x) Dj u

1≤j≤n, 1≤l≤N

+

N X

ckl (x) ul .

l=1

In this case, as above for single equations, we work in the Hilbert space H01 (Ω; RN ), which has the inner product defined by Z X (u, v) ≡ Di uk Di v k dx. 1≤i≤n, 1≤k≤N

Ω

The pairing between H01 (Ω; RN ) and its dual is given by hF, ui =

N X

hf k , uk i

if F = (f k ) and u = (uk ).

k=1

Then a weak solution of linear system (3.4) is a function u ∈ H01 (Ω; RN ) such that Z l k kl l k kl l k kl l k (3.7) B2 (u, v) ≡ akl D u D v + d u D v + b D u v + c u v dx = hF, vi i i j ij j i j Ω

3.1. Differential Equations in Divergence Form

53

holds for all v ∈ H01 (Ω; RN ). Here the conventional summation notation is used. There are some ellipticity conditions for the system (3.5) in terms of the leading coefficients A(x, u, ξ). Assume A is smooth on ξ and define Akl ij (x, u, ξ) =

∂Aki (x, u, ξ) , ∂ξjl

ξ = (ξjl ).

The system (3.5) is said to satisfy the (uniform, strict) Legendre ellipticity condition if there exists a ν > 0 such that, for all (x, s, ξ), it holds (3.8)

n X N X

k l 2 Akl ij (x, s, ξ) ηi ηj ≥ ν |η|

for all N × n matrix η = (ηik ).

i,j=1 k,l=1

A weaker condition, obtained by setting η = q ⊗ p = (q k pi ) with p ∈ Rn , q ∈ RN , is the following (strong) Legendre-Hadamard condition: (3.9)

n X N X

k l 2 2 Akl ij (x, s, ξ) q q pi pj ≥ ν |p| |q|

∀ p ∈ Rn , q ∈ RN .

i,j=1 k,l=1

Note that for systems with linear leading terms A given by (3.6), the Legendre condition and Legendre-Hadamard condition become, respectively, n X N X

(3.10)

k l 2 akl ij (x) ηi ηj ≥ ν |η|

∀ η;

i,j=1 k,l=1

n X N X

(3.11)

k l 2 2 akl ij (x) q q pi pj ≥ ν |p| |q|

∀ p, q.

i,j=1 k,l=1

Remark. The Legendre-Hadamard condition does not imply the Legendre ellipticity condition. For example, let n = N = 2 and define constants akl ij by 2 X

k l 2 akl ij ξi ξj ≡ det ξ + |ξ| .

i,j,k,l=1

Since 2 X

k l 2 2 2 akl ij pi pj q q = det(q ⊗ p) + |q ⊗ p| = |p| |q| ,

i,j,k,=1

the Legendre-Hadamard condition holds for all > 0. But, if 0 < < 1/2, then the Legendre ellipticity condition fails. Exercise 1. Prove that the Legendre condition holds for this system if and only if > 1/2.

54


Remark. Let u = (v, w) and (x1 , x2 ) = (x, y). Then the system of differential equations defined by akl ij given above is ∆v + wxy = 0,

∆w − vxy = 0.

This system reduces to two fourth-order equations for v, w (where ∆f = fxx + fyy ): 2 ∆2 v − vxxyy = 0,

2 ∆2 w + wxxyy = 0.

We can easily see that both equations are elliptic if and only if > 1/2. Exercise 2. Formulate the biharmonic equation ∆2 u = f as a linear system and find the appropriate B2 (u, v) in the definition of weak solutions.

3.2. The Lax-Milgram Theorem Let H denote a real Hilbert space with inner product (·, ·) and norm k · k. A map B : H × H → R is called a bilinear form if B(αu + βv, w) = αB(u, w) + βB(v, w) B(w, αu + βv) = αB(w, u) + βB(w, v) for all u, v, w ∈ H and all α, β ∈ R. Our first existence result is frequently referred to as the Lax-Milgram Theorem. Theorem 3.1. (Lax-Milgram Theorem) Let B : H → H be a bilinear form. Assume (i) B is bounded; i.e., |B(u, v)| ≤ αkukkvk, and (ii) B is strongly positive; i.e., B(u, u) ≥ βkuk2 , where α, β are positive constants. Let f ∈ H ∗ . Then there exists a unique element u ∈ H such that (3.12)

B(u, v) = hf, vi,

Moreover, the solution u satisfies kuk ≤

1 β

∀ v ∈ H.

kf k.

Proof. For each fixed u ∈ H, the functional v 7→ B(u, v) is in H ∗ , and hence by the Riesz Representation Theorem, there exists a unique element w = Au ∈ H such that B(u, v) = (w, v) ∀ v ∈ H. It can be easily shown that A : H → H is linear. From (i), kAuk2 = B(u, Au) ≤ αkukkAuk, and hence kAuk ≤ αkuk for all u ∈ H; that is, A is bounded. Furthermore, by (ii), βkuk2 ≤ B(u, u) = (Au, u) ≤ kAukkuk and hence kAuk ≥ βkuk for all u ∈ H. By the Riesz Representation Theorem again, we have a unique w0 ∈ H such that hf, vi = (w0 , v) for all v ∈ H and kf k = kw0 k. We will show that the equation Au = w0 has a (unique) solution. There are many different proofs for this, and here we use the Contraction Mapping Theorem. Note that the solution u to equation Au = w0 is equivalent to the fixed-point of the map T : H → H defined by T (v) = v − tAv + tw0 (v ∈ H) for any fixed t > 0. We will show for t > 0 small enough T is a contraction. Note that for all v, w ∈ H we have kT (v) − T (w)k = k(I − tA)(v − w)k. We compute that for all u ∈ H k(I − tA)uk2 = kuk2 + t2 kAuk2 − 2t(Au, u) ≤ kuk2 (1 + t2 α2 − 2βt).

3.3. G˚ arding’s Estimates and Existence Theory

55

We now choose t such that 0 < t < α2β2 . Then the expression in parentheses is positive and less than 1. Thus the map T : H → H is a contraction on H and therefore has a fixed point. This fixed point u solves Au = w0 and thus is the unique solution of (3.12); moreover, we have kf k = kw0 k = kAuk ≥ βkuk and hence kuk ≤ β1 kf k. The proof is complete. For the bilinear forms B1 , B2 defined above, one can easily show the boundedness: |Bj (u, v)| ≤ αkukkvk for all u, v in the respective Hilbert spaces H = H01 (Ω) or H = H01 (Ω; RN ) for j = 1, 2. The strong positivity (also called coercivity) for both B1 and B2 is not always guaranteed and involves estimating on the quadratic form Bj (u, u); this is usually called G˚ arding’s estimates. We will derive these estimates for both of them and state the corresponding existence theorems below.

3.3. G˚ arding’s Estimates and Existence Theory In the following, we assume all coefficients involved in the problems are in L∞ (Ω). 3.3.1. Theory for B1 . Theorem 3.2. Assume the ellipticity condition (3.2) holds. Then, there are constants β > 0 and γ ≥ 0 such that B1 (u, u) ≥ βkuk2 − γkuk2L2 (Ω)

(3.13) for all u ∈ H = H01 (Ω).

Proof. Note that, by the ellipticity, Z X Z X n n B1 (u, u) − ( bi Di u + cu)u dx ≥ θ |Di u|2 dx. Ω i=1

Ω i=1

Let m = max{kbi kL∞ (Ω) | 1 ≤ i ≤ n}. Then |(bi Di u, u)2 | ≤ mkDi uk2 kuk2 ≤ (m/2)(εkDi uk22 + (1/ε)kuk22 ) where in the last step we used the arithmetic-geometric inequality |αβ| ≤ (ε/2)α2 +(1/2ε)β 2 . Combining the estimates we find B1 (u, u) ≥ (θ − mε/2)kDuk2L2 (Ω) − (k0 + mn/2ε)kuk2L2 (Ω) . By choosing ε > 0 so that c − mε > 0 we arrive at the desired inequality, using the Poincare inequality. Theorem 3.3. There is a number γ ≥ 0 such that for each λ ≥ γ and for each function f ∈ L2 (Ω), the boundary value problem Lu + λu = f (x) in Ω,

u = 0 on ∂Ω

has a unique weak solution u ∈ H = H01 (Ω) which satisfies kukH ≤ ckf k2 where the positive constant c is independent of f .

56


Proof. Take γ from (3.13), let λ ≥ γ and define the bilinear form B λ (u, v) ≡ B1 (u, v) + λ(u, v)2

for all

u, v ∈ H

which corresponds to the operator Lu + λu. Then B λ (u, v) satisfies the hypotheses of the Lax-Milgram Theorem. Example 3.4. Consider the Neumann boundary value problem ∂u (3.14) −∆u(x) = f (x) in Ω, = 0 on ∂Ω. ∂ν A function u ∈ H 1 (Ω) is said to be a weak solution to (3.14 if Z Z f v dx, ∀ v ∈ H 1 (Ω). ∇u · ∇v dx = (3.15) Ω

Ω

Obviously, taking v ≡ 1 ∈ H 1 (Ω), a necessary condition to have a weak solution is R Ω f (x) dx = 0. We show this is also a sufficient condition for existence of the weak solutions. Note that, if u is a weak solution, then u + c, for all constants c, is also a weak solution. Therefore, to fix the constants, we consider the vector space Z H = u ∈ H 1 (Ω) u(x) dx = 0 Ω

equipped with inner product Z ∇u · ∇v dx.

(u, v)H = Ω

By the theorem on equivalent norms, it follows that H with this inner product, is indeed a Hilbert space, and (f, u)L2 (Ω) is a bounded linear functional on H: |(f, u)L2 (Ω) | ≤ kf kL2 (Ω) kvkL2 (Ω) ≤ kf kL2 (Ω) kvkH . Hence the Riesz Representation Theorem implies that there exists a unique u ∈ H such that (3.16)

(u, w)H = (f, w)L2 (Ω) ,

∀ w ∈ H.

It follows that u is a weak solutionR to the Neumann problem since for any Rv ∈ H 1 (Ω) we 1 take w = v − c ∈ H, where c = |Ω| Ω vdx, in (3.16) and obtain (3.15) using Ω f dx = 0. Example 3.5. Let us consider the nonhomogeneous Dirichlet boundary value problem −∆u = f in Ω,

(3.17)

u|∂Ω = ϕ

where f ∈ L2 (Ω) and ϕ is the trace of a function w ∈ H 1 (Ω). Note that it is not sufficient to just require that ϕ ∈ L2 (∂Ω) since the trace operator is not onto. If, for example, ¯ which is the desired w. ϕ ∈ C 1 (∂Ω), then ϕ has a C 1 extension to Ω, The function u ∈ H 1 (Ω) is called a weak solution of (3.17) if u − w ∈ H01 (Ω) and if Z Z ∇u · ∇vdx = f vdx for all v ∈ H01 (Ω). Ω

Ω

Let u be a weak solution of (3.17) and set u = z + w. Then z ∈ H01 (Ω) satisfies Z Z (3.18) ∇z · ∇vdx = (f v − ∇v · ∇w)dx for all v ∈ H01 (Ω). Ω

Ω

Since the right hand side belongs to the dual space H01 (Ω)∗ , the Lax-Milgram theorem yields the existence of a unique z ∈ H01 (Ω) which satisfies (3.18). Hence (3.17) has a unique weak solution u.


57

Example 3.6. Now let us consider the boundary value (also called Dirichlet) problem for the fourth order biharmonic operator: ∆2 u = f in Ω,

u|∂Ω =

∂u |∂Ω = 0. ∂n

We take H = H02 (Ω). By the general trace theorem, H = H02 (Ω) = {v ∈ H 2 (Ω) : γ0 v = γ1 v = 0}. Therefore, this space H is the right space for the boundary conditions. Accordingly, for f ∈ L2 (Ω), a function u ∈ H = H02 (Ω) is a weak solution of the Dirichlet problem for the biharmonic operator provided Z Z f vdx ∀ v ∈ H. ∆u∆vdx = Ω

Ω

Consider the bilinear form Z ∆u∆vdx.

B(u, v) = Ω

Its boundedness follows from the Cauchy-Schwarz inequality |B(u, v)| ≤ k∆uk2 k∆vk2 ≤ dkuk2,2 kvk2,2 . Furthermore, it can be shown that k∆uk2 defines a norm on H02 (Ω) which is equivalent to the usual norm on H 2 (Ω). (Exercise!) Hence B(u, u) = k∆uk22 ≥ ckuk22,2 and so, by the Lax-Milgram theorem (in fact, just the Riesz Representation Theorem), there exists a unique weak solution u ∈ H. Exercise 1. Denote by Hc1 the space Hc1 = {u ∈ H 1 (Ω) : γ0 u = const}. Note that the constant may be different for different u’s. (a) Prove that Hc1 is complete. (b) Prove the existence and uniqueness of a function u ∈ Hc1 satisfying Z Z (∇u · ∇v + uv)dx = f vdx for all v ∈ Hc1 Ω

Ω

¯ where f ∈ C(Ω). ¯ satisfies the equation in (b), find the underlying BVP. (c) If u ∈ C 2 (Ω) Exercise 2. Let Ω ⊂ Rn . Show that if u, v ∈ H02 (Ω), then Z n Z X ∆u∆vdx = Dij uDij vdx. Ω

i,j=1 Ω

Hence, deduce that k∆uk2 defines a norm on H02 (Ω) which is equivalent to the usual norm on H02 (Ω). Exercise 3. Let Ω = (1, +∞). Show that the BVP −u00 = f ∈ L2 (Ω), u ∈ H01 (Ω) does not have a weak solution.

58


3.3.2. Theory for B2 . We will derive the G˚ arding estimates for B2 (u, u). For simplicity, 1 N let H = H0 (Ω; R ) and let (u, v)H and kukH be the inner product and norm defined above on H. Define the bilinear form of the leading terms by n X N Z X l k A(u, v) = akl ij (x) Dj u Di v dx. i,j=1 k,l=1 Ω kl Theorem 3.7. Assume that either coefficients akl ij satisfy the Legendre condition or aij are all constants and satisfy the Legendre-Hadamard condition. Then

A(u, u) ≥ ν kuk2H ,

∀ u ∈ H.

Proof. In the first case, the conclusion follows easily from the Legendre condition. We prove the second case when Akl ij are constants satisfying the Legendre-Hadamard condition n X N X

k l 2 2 akl ij q q pi pj ≥ ν |p| |q| ,

∀p ∈ Rn , q ∈ RN .

i,j=1 k,l=1

We prove n X N Z X

A(u, u) =

l k akl ij Dj u Di u dx ≥ ν

Z

i,j=1 k,l=1 Ω

|Du|2 dx

Ω

for all u ∈ C0∞ (Ω; RN ). For these test functions u, we extend them onto Rn by zero outside Ω and thus consider them as functions in C0∞ (Rn ; RN ). Define the Fourier transforms for such functions u by Z −n/2 u ˆ(y) = (2π) e−i y·x u(x) dx; y ∈ Rn . Rn

Then, for any u, v ∈

C0∞ (Rn ; RN ), Z

Z u(x) · v(x) dx =

Rn

u ˆ(y) · vˆ(y) dy, Rn

ck [ k D j u (y) = i yj u (y); c the last identity can also be written as Du(y) = iu ˆ(y) ⊗ y. Now, using these identities, we have Z Z k l [k [l akl D u (x) D u (x) dx = akl i j ij ij Di u (y) Dj u (y) dy n n RZ Z R kl kl b b c c k l k l = aij yi yj u (y) u (y) dy = Re aij yi yj u (y) u (y) dy . Rn

Rn

Write u ˆ(y) = η + iξ with η, ξ ∈ RN . Then c b k l Re u (y) u (y) = η k η l + ξ k ξ l . Therefore, by the Legendre-Hadamard condition, n X N X kl c b k l Re aij yi yj u (y) u (y) ≥ ν |y|2 (|η|2 + |ξ|2 ) = ν |y|2 |ˆ u(y)|2 . i,j=1 k,l=1

Hence, A(u, u) =

n X N Z X n i,j=1 k,l=1 R

k l akl ij Di u (x) Dj u (x) dx


n X N Z X

= Re

Rn

i,j=1 k,l=1

Z

2

59

akl ij yi yj

ck (y) ubl (y) dy u

Z

2

|iˆ u(y) ⊗ y|2 dy

|y| |ˆ u(y)| dy = ν

≥ν Rn

Z =ν Rn

2 c |Du(y)| dy = ν

ZR

n

|Du(x)|2 dx.

Rn

The proof is complete.

Theorem 3.8. Let B2 (u, v) be defined by (3.7). Assume ¯ 1) akl ∈ C(Ω), ij

2) the Legendre-Hadamard condition holds for all x ∈ Ω; that is, k l 2 2 akl ij (x) q q pi pj ≥ ν |p| |q| ,

∀p ∈ Rn , q ∈ RN .

kl kl ∞ 3) bkl i , c , di ∈ L (Ω). Then, there exist constants λ0 > 0 and λ1 ≥ 0 such that

B2 (u, u) ≥ λ0 kuk2H − λ1 kuk2L2 ,

∀ u ∈ H01 (Ω; RN ).

Proof. By uniform continuity, we can choose a small > 0 such that ν kl ¯ |x − y| ≤ . |akl ∀ x, y ∈ Ω, ij (x) − aij (y)| ≤ , 2 We claim Z Z ν k l (3.19) akl (x) D u D u dx ≥ |Du(x)|2 dx i j ij 2 Ω Ω for all test functions u ∈ C0∞ (Ω; RN ) with the diameter of the support diam(supp u) ≤ . To see this, we choose any point x0 ∈ supp u. Then Z Z kl k l k l aij (x) Di u Dj u dx = akl ij (x0 ) Di u Dj u dx Ω Ω Z kl k l + akl ij (x) − aij (x0 ) Di u Dj u dx supp u Z Z ν 2 ≥ν |Du(x)| dx − |Du(x)|2 dx, 2 Ω Ω ¯ with finitely many open balls {B/4 (xm )} with xm ∈ Ω which proves (3.19). We now cover Ω and m = 1, 2, ..., M. For each m, let ζm ∈ C0∞ (B/2 (xm )) with ζm (x) = 1 for x ∈ B/4 (xm ). ¯ we have at least one m such that x ∈ B/4 (xm ) and thus ζm (x) = 1, we Since for any x ∈ Ω may therefore define ϕm (x) = Then thus (3.20)

PM

2 m=1 ϕm (x)

ζm (x) PM 2 1/2 , j=1 ζj (x)

m = 1, 2, ..., M.

= 1 for all x ∈ Ω. (This is a special case of partition of unity.) We have

k l akl ij (x) Di u Dj u =

M X

2 k l akl ij (x) ϕm Di u Dj u

m=1

and each term (no summation on m) 2 k l kl k l akl ij (x) ϕm Di u Dj u = aij (x) Di (ϕm u ) Dj (ϕm u )

60


l k k l k l −akl (x) ϕ D ϕ u D u + ϕ D ϕ u D u + D ϕ D ϕ u u . m i m i m i m j i m j m ij Since ϕm u ∈ C0∞ (Ω ∩ B/2 (xm ); RN ) and diam(Ω ∩ B/2 (xm )) ≤ , we have by (3.19) Z Z ν k l akl (x) D (ϕ u ) D (ϕ u ) dx ≥ |D(ϕm u)|2 dx. i m j m ij 2 Ω Ω Note also that |D(ϕm u)|2 = ϕ2m |Du|2 + |Dϕm |2 |u|2 + 2 ϕm Di ϕm uk Di uk . P 2 Therefore, we have by (3.20) and the fact that M m=1 ϕm = 1 on Ω, Z k l akl ij (x) Di u Dj u dx Ω Z ν |Du|2 dx − C1 kukL2 kDukL2 − C2 kuk2L2 . ≥ 2 Ω The terms in B2 (u, u) involving b, c and d can be estimated by kukL2 kDukL2 and kuk2L2 . Finally, by all of these estimates and the inequality 1 2 ab ≤ a2 + b 4 we have B2 (u, u) ≥ λ0 kuk2H −λ1 kuk2L2 for all u ∈ H01 (Ω; RN ). This completes the proof. Note that the bilinear form B λ (u, v) = B2 (u, v) + λ (u, v)L2 satisfies the condition of the Lax-Milgram theorem on H = H01 (Ω; RN ) for all λ ≥ λ1 ; thus, by the Lax-Milgram theorem, we easily obtain the following existence result. Theorem 3.9. Under the hypotheses of the previous theorem, for λ ≥ λ1 , the Dirichlet problem for the linear system (3.21)

− div(A(x, u, Du)) + b(x, u, Du) + λu = F,

u|∂Ω = 0

H01 (Ω; RN )

has a unique weak solution u in for any bounded linear functional F on H. Moreover, the solution u satisfies kukH ≤ C kF k with a constant C depending on λ and the L∞ -norms of the coefficients of linear terms A(x, s, ξ) and b(x, s, ξ) given above. Corollary 3.10. Given λ ≥ λ1 as in the theorem, the operator K : L2 (Ω; RN ) → L2 (Ω; RN ), where, for each F ∈ L2 (Ω; RN ), u = KF is the unique weak solution to the BVP (3.21) above, is a compact linear operator. Proof. By the theorem, kukH01 (Ω;RN ) ≤ CkF kL2 (Ω;RN ) . Hence K is a bounded linear operator from L2 (Ω; RN ) to H01 (Ω; RN ), which, by the compact embedding theorem, is compactly embedded in L2 (Ω; RN ). Hence, as a linear operator from L2 (Ω; RN ) to L2 (Ω; RN ), K is compact.

3.4. Symmetric Elliptic Operators 3.4.1. Symmetric Elliptic Operators. In what follows, we assume Ω is a bounded domain. We consider the operator Lu = − div A(x, Du) + c(x)u,

u ∈ H01 (Ω; RN ),

where A(x, Du) is a linear system defined with A(x, ξ), ξ ∈ MN ×n , given by X l Aki (x, ξ) = akl ij (x)ξj . 1≤l≤N, 1≤j≤n

3.4. Symmetric Elliptic Operators

61

∞ Here akl ij (x) and c(x) are given functions in L (Ω).

The bilinear form associated to L is   Z N X n X l k   dx, (3.22) B(u, v) = akl ij (x)Dj u Di v + c(x)u · v Ω

u, v ∈ H01 (Ω; RN ).

k,l=1 i,j=1

In order for B to be symmetric on H01 (Ω; RN ), that is, B(u, v) = B(v, u) for all u, v ∈ H01 (Ω; RN ), we need the following symmetry condition: (3.23)

lk akl ij (x) = aji (x),

∀ i, j = 1, 2, · · · , n; k, l = 1, 2, · · · , N.

We also assume the G˚ arding inequality holds (see Theorem 3.8 for sufficient conditions): (3.24)

B(u, u) ≥ σkuk2H 1 − µkuk2L2 , 0

∀ u ∈ H01 (Ω; RN ),

where σ > 0 and µ ∈ R are constants. 3.4.2. The Compact Inverse. For each F ∈ L2 (Ω; RN ), define u = KF to be the unique weak solution in H01 (Ω; RN ) of the BVP Lu + µu = F

in Ω,

u|∂Ω = 0.

By Theorem 3.9 and Corollary 3.10, this K is well defined and is a compact linear operator on L2 (Ω; RN ). Sometime, we write K = (L+µI)−1 . Here I denotes the identity on L2 (Ω; RN ) and also the identity embedding of H01 (Ω; RN ) into L2 (Ω; RN ). Theorem 3.11. K : L2 (Ω; RN ) → L2 (Ω; RN ) is symmetric and positive; that is, (KF, G)L2 = (KG, F )L2 ,

(KF, F )L2 ≥ 0,

∀ F, G ∈ L2 (Ω; RN ).

Furthermore, given λ ∈ R and F ∈ L2 (Ω; RN ), u ∈ H01 (Ω; RN ) is a weak solution of Lu − λu = F if and only if [I − (λ + µ)K]u = KF. Proof. Let u = KF and v = KG. Then (u, G)L2 = B(v, u) + µ(v, u)L2 = B(u, v) + µ(v, u)L2 = (v, F )L2 , proving the symmetry. Also, by (3.24), (3.25)

(KF, F )L2 = (u, F )L2 = B(u, u) + µkuk2L2 ≥ σkuk2H 1 = σkKF k2H 1 ≥ 0. 0

0

H01 (Ω; RN )

Finally, u ∈ is a weak solution of Lu − λu = F if and only if Lu + µu = F + (λ + µ)u, which is equivalent to the equation u = K[F + (λ + µ)u] = KF + (λ + µ)Ku; that is, [I − (λ + µ)K]u = KF. 3.4.3. Orthogonality Conditions. From Theorem 3.11 above, the BVP (3.26)

Lu = F,

u|∂Ω = 0

has a solution if and only if KF ∈ R(I − µK) = [N (I − µK)]⊥ . If µ = 0, the Lax-Milgram theorem already implies that the problem (3.26) has a unique weak solution u = KF. If µ 6= 0, then it is easy to see that N = N (I − µK) = {v ∈ H01 (Ω; RN ) | Lv = 0}. Note that, by the Fredholm Alternative Theorem, this null space N is of finite dimension k, and let {v1 , v2 , · · · , vk } be a basis of this null space. Then (3.26) is solvable if and only if the following orthogonality condition holds: Z F · Gi dx = 0, i = 1, 2, · · · , k, Ω

62


where Gi = Kvi ; that is, Gi is the unique weak solution to the BVP: LGi + µGi = vi , in H01 (Ω; RN ). 3.4.4. Eigenvalue Problems. A number λ ∈ R is called a (Dirichlet) eigenvalue of operator L if the BVP problem Lu − λu = 0,

u|∂Ω = 0

has nontrivial weak solutions in H01 (Ω; RN ); these nontrivial solutions are called the eigenfunctions corresponding to eigenvalue λ. From Theorem 3.11, we see that λ is an eigenvalue of L if and only if equation (I − (λ + 1 µ)K)u = 0 has nontrivial solutions u ∈ L2 (Ω; RN ); this exactly says that λ 6= −µ and λ+µ is an eigenvalue of operator K. Since, by (3.25), K is strictly positive, all eigenvalues of K consist of a countable set of positive numbers tending to zero and hence the eigenvalues of L consist of a set of numbers {λj }∞ j=1 with −µ < λ1 ≤ λ2 ≤ · · · ≤ λj → ∞. Theorem 3.12. (Eigenvalue Theorem) Assume (3.23) and (3.24) with µ = 0. Then the eigenvalues of L consist of a countable set Σ = {λk }∞ k=1 , where 0 < λ1 ≤ λ2 ≤ λ3 ≤ · · · and lim λk = ∞.

k→∞

Note that λ1 is called the (Dirichlet) principal eigenvalue of L. Let wk be an eigenfunction corresponding to λk satisfying kwk kL2 (Ω;RN ) = 1. Then {wk }∞ k=1 forms an orthonormal 2 N basis of L (Ω; R ). Note that (3.27)

λ1 =

min

u∈H01 (Ω;RN ), kukL2 =1

B(u, u).

Theorem 3.13. Let N = 1 and let w1 be an eigenfunction corresponding to the principal eigenvalue λ1 . Then, either w1 (x) > 0 for all x ∈ Ω or w1 (x) < 0 for all x ∈ Ω. Moreover, the eigenspace corresponding to λ1 is one-dimensional. Proof. The first part of the theorem relies on the characterization (3.27) and a maximum principle which we do not study here. We prove only the second part. Let w be another eigenfunction. Then, either w(x) > 0 for all x ∈ Ω or w(x) < 0 for all x ∈ Ω. Let t ∈ R be such that Z Z w(x) dx = t w1 (x)dx. Ω

Ω

Note that u = w − tw1 is also a solution to Lu = λ1 u. We claim u ≡ 0 and hence w = tw1 , proving the eigenspace is one-dimensional. Suppose u 6= 0. Then u is another eigenfunction corresponding to λ1 . Then, by theR theorem, we would have either u(x) > 0 for all x ∈ Ω or u(x) < 0 for all x ∈ Ω and hence Ω u(x)dx 6= 0, which is a contradiction.

3.4. Symmetric Elliptic Operators

63

Remark. Why is the eigenvalue problem important? In one specific case, one can use eigenvalues and eigenfunctions to study some evolution (i.e. time-dependent) problems. For example, to solve the initial boundary value problem (IBVP) for the time-dependent parabolic equation: (3.28)

ut + Lu = 0,

u|∂Ω = 0,

u(x, 0) = ϕ(x),

one can try to find the special solutions of the form: u(x, t) = e−λt w(x); this reduces to the eigen-problem: Lw = λw. Therefore, for each pair (λi , wi ) of eigenvalue and eigenfunction, one obtains a special solution to the evolution equation given by ui (x, t) = e−λi t wi (x). Then one proceeds to solve the IBVP (3.28) by finding the solution of the form X u(x, t) = ai e−λi t wi (x), i

where ai is determined by the eigen-expansion of the initial data ϕ: X ϕ(x) = ai wi (x). i

However, this course does not study the parabolic or other time-dependent problems.

Chapter 4

Variational Methods for PDEs

4.1. Variational Problems This chapter and the next will discuss some new methods for solving the boundary value problem for some partial differential equations. All these problems can be written in the abstract form:

(4.1)

A[u] = 0

in Ω, B[u] = 0

on ∂Ω,

where A[u] denotes a given PDE for unknown u and B[u] is a given boundary value condition. There is, of course, no general theory for solving such problems. The Calculus of Variations identifies an important class of problems which can be solved using relatively simple techniques from nonlinear functional analysis. This is the class of variational problems, where the operator A[u] can be formulated as the first variation (“derivative”) of an appropriate “energy” functional I(u) on a Banach space X; that is, A[u] = I 0 (u). In this way, A : X → X ∗ and the equation A[u] = 0 can be formulated weakly as hI 0 (u), vi = 0,

∀ v ∈ X.

The advantage of this new formulation is that solving problem (4.1) (at least weakly) is equivalent to finding the critical points of I on X. The minimization method for a variational problem is to solve the problem by finding the minimizers of the related energy functional. In this chapter and the next, we shall only study the variational problems on the Sobolev space X = W 1,p (Ω; RN ). We should also mention that many of the physical laws in applications arise directly as variational principles. However, although powerful, not all PDE problems can be formulated as variational problems; there are lots of other (nonvariational) important methods for studying PDEs. 65

66

4. Variational Methods for PDEs

4.2. Multiple Integrals in the Calculus of Variations Consider the multiple integral functional Z F (x, u(x), Du(x)) dx, (4.2) I(u) = Ω

where F (x, s, ξ) is a given function on Ω × RN × MN ×n . 4.2.1. First Variation and Euler-Lagrange Equations. Suppose F (x, s, ξ) is contin¯ RN )) minimizer of uous and is also smooth in s and ξ. Assume u is a nice (say, u ∈ C 1 (Ω; I(u) with its own boundary data; that is, u is a map such that I(u) ≤ I(u + tϕ) C0∞ (Ω; RN ).

R1

for all t ∈ and ϕ ∈ Then by taking derivative of I(u + t ϕ) at t = 0 we see that u satisfies Z (4.3) Fξk (x, u, Du) Di ϕk (x) + Fsk (x, u, Du) ϕk (x) dx = 0 Ω

i

for all ϕ ∈ C0∞ (Ω; RN ). (Summation notation is used here.) The left-hand side is called the first variation of I at u. Since this holds for all test functions, we conclude after integration by parts: − div A(x, u, Du) + b(x, u, Du) = 0,

(4.4) where A, b are defined by

Aki (x, s, ξ) = Fξk (x, s, ξ),

(4.5)

i

bk (x, s, ξ) = Fsk (x, s, ξ).

The coupled differential system in divergence form for u is called the system of EulerLagrange equations for the functional I(u). 4.2.2. Second Variation and Legendre-Hadamard Conditions. If F, u are sufficiently smooth (e.g. of class C 2 ) then, at the minimizer u, we have d2 I(u + tϕ) ≥ 0. dt2 t=0 This implies (4.6) Z h i Fξk ξl (x, u, Du) Di ϕk Dj ϕl + 2Fξk sl (x, u, Du) ϕl Di ϕk + Fsk sl (x, u, Du) ϕk ϕl dx ≥ 0 Ω

i

j

i

for all ϕ ∈ C0∞ (Ω; RN ). The left-hand side of this inequality is called the second variation of I at u. We can extract useful information from (4.6). Note that routine approximation argument shows that (4.6) is also valid for all functions ϕ ∈ W01,∞ (Ω; RN ) (that is, all Lipschitz functions vanishing on ∂Ω). Let ρ : R → R be the periodic zig-zag function of period 1 defined by (4.7) (4.8)

ρ(t + 1) = ρ(t) ρ(t) = t

(t ∈ R),

if 0 ≤ t ≤ 12 ; ρ(t) = 1 − t

1 2 ≤ t ≤ 1. W01,∞ (Ω; RN )

if

Given any vectors p ∈ Rn , q ∈ RN and > 0, define ϕ ∈ x·p ϕ(x) = ρ( )ζ(x)q, ∀ x ∈ Ω,

by

4.2. Multiple Integrals in the Calculus of Variations

67

k where ζ ∈ C0∞ (Ω) is a given scalar test function. Note that Di ϕk (x) = ρ0 ( x·p )pi q ζ + O() as → 0+ . Substitute this ϕ into (4.6) and let → 0+ and we obtain   Z n X N X  Fξk ξl (x, u, Du) pi pj q k q l  ζ 2 dx ≥ 0. i

Ω

j

i,j=1 k,l=1

Since this holds for all ζ ∈ C0∞ (Ω), we deduce (4.9)

n X N X

Fξk ξl (x, u, Du) pi pj q k q l ≥ 0, i

∀ x ∈ Ω, p ∈ Rn , q ∈ RN .

j

i,j=1 k,l=1

This is the weak Legendre-Hadamard condition for F at the minimum point u. 4.2.3. Ellipticity Conditions and Convexities. We now consider the ellipticity of the Euler-Lagrange equation (4.4), where A(x, s, ξ), b(x, s, ξ) are given by (4.5) and F (x, s, ξ) is C 2 in ξ. In this case, the Legendre ellipticity condition (3.8) and the Legendre-Hadamard condition (3.9) defined in the previous chapter reduce to, respectively: (4.10)

Fξk ξl (x, s, ξ) ηik ηjl ≥ ν |η|2

(4.11)

Fξk ξl (x, s, ξ) q k q l pi pj ≥ ν|p|2 |q|2

i

i

j

j

∀η ∈ MN ×n ; ∀q ∈ RN , p ∈ Rn .

Obviously, (4.10) implies (4.11). We also have the following equivalent conditions. Lemma 4.1. Let F be C 2 in ξ. Then, the conditions (4.10) and (4.11) are equivalent to the following conditions, respectively: ν (4.12) F (x, s, η) ≥ F (x, s, ξ) + Fξk (x, s, ξ) (ηik − ξik ) + |η − ξ|2 ; i 2 (4.13)

F (x, s, ξ + q ⊗ p) ≥ F (x, s, ξ) + Fξk (x, s, ξ) pi q k + i

ν 2 2 |p| |q| 2

for all x ∈ Ω, s, q ∈ RN , ξ, η ∈ MN ×n and p ∈ Rn . Proof. Let ζ = η − ξ and f (t) = F (x, s, ξ + tζ). Then, by Taylor’s formula, Z 1 0 f (1) = f (0) + f (0) + (1 − t) f 00 (t) dt. 0

Note that f 0 (t) = Fξk (x, s, ξ + tζ) ζik , i

f 00 (t) = Fξk ξl (x, s, ξ + tζ) ζik ζjl . i j

From this and the Taylor formula, inequalities (4.12) and (4.13) are equivalent to (4.10) and (4.11), respectively. A function F (x, s, ξ) is said to be convex in ξ ∈ MN ×n if F (x, s, tξ + (1 − t)η) ≤ tF (x, s, ξ) + (1 − t)F (x, s, η) for all x, s, ξ, η and 0 ≤ t ≤ 1. While F (x, s, ξ) is said to be rank-one convex in ξ if the function f (t) = F (x, s, ξ + t q ⊗ p) is convex in t ∈ R1 for all x, s, ξ and q ∈ RN , p ∈ Rn . Obviously, a convex function is always rank-one convex. We easily have the following result.

68


Lemma 4.2. Let F (x, s, ξ) be C 2 in ξ. Then the convexity of F (x, s, ξ) in ξ is equivalent to (4.10) with ν = 0, while the rank-one convexity of F (x, s, ξ) in ξ is equivalent to (4.11) with ν = 0. Remark. Rank-one convexity does not imply convexity. For example, take n = N ≥ 2, and F (ξ) = det ξ. Then F (ξ) is rank-one convex but not convex in ξ. (Exercise!) Later on, we will study other convexity conditions related to the energy functionals given by (4.2). 4.2.4. Structural Conditions. Distributional solutions to the Euler-Lagrange equations (4.4) can be defined as long as A(x, u, Du) and b(x, u, Du) are in L1loc (Ω; RN ), but we need some structural conditions on F (x, s, ξ) so that the weak solutions to the BVP (4.14)

− div A(x, u, Du) + b(x, u, Du) = 0

in Ω, u = ϕ on ∂Ω,

can be defined and studied in W 1,p (Ω; RN ). These conditions are also sufficient for the functional I to be Gateaux-differentiable on W 1,p (Ω; RN ). Standard Growth Conditions. We assume F (x, s, ξ) is C 1 in (s, ξ) and c2 ∈ L1 (Ω);

|F (x, s, ξ)| ≤ c1 (|ξ|p + |s|p ) + c2 (x),

(4.15)

p

(4.16)

|Ds F (x, s, ξ)| ≤ c3 (|ξ|p−1 + |s|p−1 ) + c4 (x), c4 ∈ L p−1 (Ω);

(4.17)

|Dξ F (x, s, ξ)| ≤ c5 (|ξ|p−1 + |s|p−1 ) + c6 (x), c6 ∈ L p−1 (Ω),

p

where c1 , c3 , c5 are constants. Theorem 4.3. Under the standard conditions above, the functional I : W 1,p (Ω; RN ) → R is Gateaux-differentiable and, for u, v ∈ W 1,p (Ω; RN ), the directional derivative hI 0 (u), vi is exactly given by Z 0 (4.18) hI (u), vi = Fξk (x, u, Du) Di v k (x) + Fsk (x, u, Du) v k (x) dx i

Ω

(as usual, summation notation is used here). Proof. Given u, v ∈ X = W 1,p (Ω; RN ), let h(t) = I(u + tv). Then, by (4.15), h is finite valued, and for t 6= 0 we have Z Z h(t) − h(0) F (x, u + tv, Du + tDv) − F (x, u, Du) = dx ≡ F t (x) dx, t t Ω Ω where for almost every x ∈ Ω, 1 1 F (x) = [F (x, u + tv, Du + tDv) − F (x, u, Du)] = t t t

Z 0

t

d F (x, u + sv, Du + sDv) ds. ds

Clearly, lim F t (x) = Fξk (x, u, Du) Di v k (x) + Fsk (x, u, Du) v k (x) a.e.

t→0

i

We also write Z i 1 th Fξk (x, u + sv, Du + sDv) Di v k (x) + Fsk (x, u + sv, Du + sDv) v k (x) ds. F t (x) = i t 0 Using conditions (4.16), (4.17), and Young’s inequality: ab ≤ obtain that, for all 0 < |t| ≤ 1,

ap p

|F t (x)| ≤ C1 (|Du|p + |Dv|p + |u|p + |v|p ) + C2 (x),

+

bq q ,

where

1 p

C2 ∈ L1 (Ω).

+

1 q

= 1, we

4.2. Multiple Integrals in the Calculus of Variations

69

Hence, by the Lebesgue dominated convergence theorem, Z Z h i h0 (0) = lim F t (x) dx = Fξk (x, u, Du) Di v k (x) + Fsk (x, u, Du) v k (x) dx. t→0 Ω

i

Ω

This proves the theorem.

Dirichlet Classes. Given ϕ ∈ W 1,p (Ω; RN ), we define the Dirichlet class of ϕ to be the set Dϕ = {u ∈ W 1,p (Ω; RN ) | u − ϕ ∈ W01,p (Ω; RN )}. 4.2.5. Weak Solutions of Euler-Lagrange Equations. Under the standard growth conditions, we say u ∈ W 1,p (Ω; RN ) is a weak solution to the BVP of Euler-Lagrange equations (4.14) if u ∈ Dϕ and Z Fξk (x, u, Du) Di v k (x) + Fsk (x, u, Du) v k (x) dx = 0 (4.19) i

Ω

for all v ∈ W01,p (Ω; RN ). Theorem 4.4. Under the standard growth conditions above, any minimizer u ∈ Dϕ of I(u) = min I(v) v∈Dϕ

is a weak solution of the BVP for the Euler-Lagrange equation (4.14). Proof. This follows from Theorem 4.3.

4.2.6. Minimality and Uniqueness of Weak Solutions. We study the weak solutions of Euler-Lagrange equations under the hypotheses of convexity and certain growth conditions. For this purpose, we consider a simple case where F (x, s, ξ) = F (x, ξ) satisfies, for some 1 < p < ∞, (4.20)

|Fξk (x, ξ)| ≤ µ (χ(x) + |ξ|p−1 ) i

∀ x ∈ Ω, ξ ∈ MN ×n ,

p

where µ > 0 is a constant and χ ∈ L p−1 (Ω) is some function. Let Z I(u) = F (x, Du(x)) dx. Ω

C2

Theorem 4.5. Let F (x, ξ) be and convex in ξ. Let u ∈ W 1,p (Ω; RN ) be a weak solution of the Euler-Lagrange equation of I and I(u) < ∞. Then u must be a minimizer of I in the Dirichlet class Du of W 1,p (Ω; RN ). Furthermore, if F satisfies the Legendre condition (4.10) (with ν > 0), then u is the unique minimizer of I in Du . Proof. Since u is a weak solution of the Euler-Lagrange equation of I, it follows that Z (4.21) Fξk (x, Du(x)) Di v k dx = 0 Ω

i

p

for all v ∈ W01,p (Ω; RN ). The growth condition (4.20) implies Fξk (x, Du) ∈ L p−1 (Ω). Now i

let v ∈ W 1,p (Ω; RN ) with v − u ∈ W01,p (Ω; RN ). By the convexity condition, we have ν (4.22) F (x, η) ≥ F (x, ξ) + Fξk (x, ξ) (ηik − ξik ) + |η − ξ|2 , ∀ ξ, η, i 2

70


where ν = 0 if F is only convex in ξ and ν > 0 if F satisfies the Legendre condition. This implies Z Z Z F (x, Dv) dx ≥ F (x, Du) dx + Fξk (x, Du) Di (v k − uk ) dx i Ω Ω Ω Z ν + |Du − Dv|2 dx. 2 Ω Since u is a weak solution, we have Z Fξk (x, Du) Di (v k − uk ) dx = 0. Ω

i

Therefore, it follows that I(v) ≥ I(u) +

(4.23) W 1,p (Ω; RN )

ν 2

Z

|Du − Dv|2 dx ≥ I(u)

Ω

W01,p (Ω; RN ).

for all v ∈ with v − u ∈ This shows that u is a minimizer of I in the Dirichlet class Du . Now assume ν > 0 and v ∈ Du is another such minimizer of I. Then from (4.23) we easily obtain Du = Dv in Ω and hence v ≡ u since u − v ∈ W01,p (Ω; RN ). The proof is now completed.

4.3. Direct Method for Minimization 4.3.1. Weak Lower Semicontinuity. Let X = W 1,p (Ω; RN ). A set C ⊂ X is called weakly closed if {uν } ⊂ C, uν * u implies u ∈ C. For instance, by Theorem 1.38, all Dirichlet classes Dϕ are weakly closed. ¯ is called weakly lower semicontinuous(w.l.s.c.) on X if for A functional I : X → R every u0 ∈ X and every sequence {uν } weakly convergent to u0 in X it follows that I(u0 ) ≤ lim inf I(uν ). ν→∞

I is called weakly coercive on a (unbounded) set C in X if I(u) → ∞ as kuk → ∞ on C. 4.3.2. Direct Method in the Calculus of Variations. The following theorem is a special case of the generalized Weierstrass theorem. Theorem 4.6. (Existence of Minimizers) Let I : C ⊆ X = W 1,p (Ω; RN ) → R be w.l.s.c. and weakly coercive on a nonempty weakly closed set C in X. Assume 1 < p < ∞. Then there is at least one u0 ∈ C such that I(u0 ) = inf u∈C I(u). In this case we say u0 ∈ C is a minimizer of I on C. Proof. This theorem is a special case of Theorem 1.37 under assumption (ii) there; the proof involves the basic ideas of what is known as the direct method of calculus of variations. We explain this method by giving a proof of this theorem. First of all, by the weak coercivity, one can easily show that inf u∈C I(u) is finite; so take a sequence {uν }, called a minimizing sequence, such that lim I(uν ) = inf I(u). ν→∞

u∈C

Then the weak coercivity condition implies that {uν } must be bounded in X. Since 1 < p < ∞ and thus X = W 1,p (Ω; RN ) is reflexive, there exists a subsequence of {uν }, denoted

4.3. Direct Method for Minimization

71

by {uνj }, and u0 ∈ X such that uνj * u0 weakly in X. The weak closedness of C implies u0 ∈ C. Now the w.l.s.c. of I implies I(u0 ) ≤ lim inf I(uνj ) = inf I(u). j→∞

u∈C

This implies u0 is a minimizer of I over C.

4.3.3. An Example: p-Laplace Equations. As an example, we consider the BVP for nonlinear p-Laplace equations (p ≥ 2): (4.24)

−

n X

Di (|∇u|p−2 Di u) + f (x, u) = 0

in Ω, u = 0

on ∂Ω,

i=1

where f : Ω × R → R be a given function satisfying the Carath´ eodory property, i.e., for every s ∈ R, f (x, s) (as a function of x) is measurable on Ω, and for almost all x ∈ Ω, f (x, s) (as a function of s) is continuous on R. Note that when p = 2, (4.24) becomes the semilinear elliptic problem. Define

Z F (x, s) =

s

f (x, t) dt 0

and the functional I on X = W01,p (Ω) by Z 1 p I(u) = |∇u| + F (x, u) dx. Ω p Assume the following growth conditions are satisfied: (4.25)

F (x, s) ≥ −c1 |s| − c2 (x),

(4.26)

|F (x, s)| ≤ c3 (x) + c4 |s|p ,

(4.27)

|f (x, s)| ≤ c5 (x) + c6 |s|p−1

for all x ∈ Ω and s ∈ R, where c1 , c4 , c6 are nonnegative constants, and c2 , c3 ∈ L1 (Ω), and p c5 ∈ L p−1 (Ω) are given functions. Note that (4.27) implies (4.26). Theorem 4.7. Under these conditions, the functional I has a minimizer on X = W01,p (Ω) and hence (4.24) has a weak solution. Proof. We only need to check I is w.l.s.c and weakly coercive on X. We write Z Z 1 I(u) = I1 (u) + I2 (u) = |∇u|p dx + F (x, u) dx. p Ω Ω Note that pI1 (u) = kukp1,p,0 on X which is w.l.s.c. And since the embedding X ⊂ Lp (Ω) is always compact, by (4.26), we can show that I2 is in fact continuous under the weak convergence. Hence I is w.l.s.c. on X. By (4.25), we have 1 1 I(u) ≥ kukp1,p,0 − c1 kukL1 (Ω) − C ≥ kukp1,p,0 − ckuk1,p,0 − C, p p and hence I(u) → ∞ if kukX = kuk1,p,0 → ∞. This proves the weak coercivity of I. The result follows then from Theorem 4.6. Exercise 1. Let n ≥ 3 and 2 ≤ p < n. Show the theorem is valid if (4.27) above is replaced by |f (x, s)| ≤ a(x) + b |s|q ,

72


where b ≥ 0 is a constant, a ∈ L

q+1 q

(Ω) and 1 ≤ q < p∗ − 1.

4.4. Minimization with Constraints In some cases, we need to minimize functional I under certain constraints. If I is a multiple integral functional on X = W 1,p (Ω; RN ) defined before, there may be constraints given in terms of one of the following: R (4.28) J(u) = 0, where J(u) = Ω G(x, u) dx. ∀ a.e. x ∈ Ω.

(4.29)

h(u(x)) = 0,

(4.30)

M (Du(x)) = 0,

∀ a.e. x ∈ Ω.

All these lead to some PDEs involving the Lagrange multipliers. For different types of constraints, the Lagrange multiplier comes in significantly different ways. 4.4.1. Nonlinear Eigenvalue Problems. The following theorem was proved in the Preliminaries. Theorem 4.8. Let X be a Banach space. Let f, g : X → R be of class C 1 and g(u0 ) = c. Assume u0 is a local extremum of f with respect to the constraint g(u) = c. Then either g 0 (u0 )v = 0 for all v ∈ X, or there exists λ ∈ R such that f 0 (u0 )v = λg 0 (u0 )v for all v ∈ X; that is, u0 is a critical point of f − λg. If u0 6= 0 then the corresponding λ is called an eigenvalue for the nonlinear eigenvalue problem: f 0 (u) = λg 0 (u) and u0 is the corresponding eigenfunction. We have following applications. ¯ with l(x) > 0 on Ω. ¯ Then, for each R ∈ (0, ∞), the Theorem 4.9. Let k(x), l(x) ∈ C(Ω) problem (4.31)

∆u + k(x)u + λl(x)|u|τ −1 u = 0 in Ω, u|∂Ω = 0

has a (weak) solution pair (uR , λR ) with

1 τ +1

τ +1 dx Ω l(x)|uR |

R

(1 < τ
0. First of all, by Hölder’s inequality we have Z τ +1 Z τ |u| dx ≤ |Ω| |u|τ +1 dx Ω Ω Z |Ω|τ ≤ l(x)|u|τ +1 dx ≤ Cg(u) α Ω

4.4. Minimization with Constraints

73

where α = inf l(x). Hence kuk1 is uniformly bounded if g(u) ≤ const. Furthermore, by Ehrling’s inequality, for each ε > 0, there is an absolute constant c(ε) > 0 such that kuk22 ≤ εkuk21,2,0 + c(ε)kuk21

for all

u ∈ H01 (Ω).

Thus 1 1 f (u) ≥ (1 − ε max |k|)kuk21,2,0 − c(ε) max |k|kuk21 . 2 2 If we choose ε small enough so that 1 − ε max |k| > 0, then f (u) → ∞ as kuk1,2,0 → ∞ with g(u) ≤ const. It easily follows that g 0 (u) = 0 iff u = 0. Moreover, since the range of the functional g(u) is [0, ∞), we can apply Theorem 4.8 to conclude that for each R ∈ (0, ∞) there exists a nontrivial weak solution (uR , λR ) of (4.31) with g(uR ) = R. Moreover, if βR = f (uR ) = R min f (u) subject to g(u) = R, then it is easily seen that λR = (τ2β +1)R . Corollary 4.10. For 1 < τ < (n + 2)/(n − 2) there exists a nontrivial weak solution of ∆u + |u|τ −1 u = 0 in Ω, u|∂Ω = 0.

(4.32)

Proof. By Theorem 4.9, there exist uR 6= 0, λR > 0 such that ∆uR + λR |uR |τ −1 uR = 0 in Ω, uR |∂Ω = 0. Set uR = kv (k > 0 is to be determined). Then k∆v + λR k τ |v|τ −1 v = 0, and so if we choose k to satisfy λR k τ −1 = 1, it follows that ∆v + |v|τ −1 v = 0 in Ω, v|∂Ω = 0 has a nontrivial weak solution in H01 (Ω).

Remarks. (i) Another method to show the existence of nontrivial weak solution to (4.32) is given later by the Mountain Pass Theorem. (ii) The nonexistence of nontrivial classical solutions to problem (4.32) when τ ≥ will be studied later for certain domains with simple topological property.

n+2 n−2

(iii) For domains like annulus, problem (4.32) always has nontrivial solutions for all τ > 1. See the following exercise. Exercise 1. Let Ω be the annulus 0 < a < r < b, r = |x| and suppose τ > 1. Prove that the BVP ∆u + |u|τ −1 u = 0 in Ω,

u|∂Ω = 0

has a nontrivial solution. Hint: Minimize the functional Z f (u) =

b

(u0 )2 rn−1 dr, r = |x|

a

for all u in the set C=

u∈

Z

H01 (a, b)

a

b

|u|

τ +1 n−1

r

dr = 1 .

74


4.4.2. Harmonic Maps and Liquid Crystals. We now consider the Dirichlet energy Z 1 I(u) = |Du|2 dx 2 Ω for u ∈ H 1 (Ω; RN ) with point-wise constraint |u(x)| = 1 for almost every x ∈ Ω. Let C = {u ∈ Dϕ | |u(x)| = 1 a.e.}, where Dϕ is a Dirichlet class. We assume C is non-empty. Then C is weakly closed in H 1 (Ω; RN ). We have the following result. Theorem 4.11. There exists u ∈ C satisfying I(u) = min I(v). v∈C

Moreover, u satisfies Z for each v ∈

Z Du · Dv dx =

(4.33) H01 (Ω; RN )

∩

Ω ∞ L (Ω; RN ).

|Du|2 u · v dx

Ω

Remark. In this case, we see u is a weak solution to the harmonic map equation: −∆u = |Du|2 u in Ω. The Lagrange multiplier in this case is the function λ = |Du|2 , instead of a constant. Proof. The existence of minimizers follows by the direct method as above. Given any v ∈ H01 (Ω; RN ) ∩ L∞ (Ω; RN ), let be such that ||kvkL∞ (Ω) ≤ 12 . Define w (x) =

u(x) + v(x) , |u(x) + v(x)|

h() = I(w ).

Note that w ∈ C and h(0) = I(u) = minC I ≤ h() for sufficiently small ; hence, h0 (0) = 0. Computing this h0 (0) and using the fact (Du)T u = 0 (which follows from |u| = 1) yield the weak form of harmonic equation (4.33). Liquid Crystals. Let n = N = 3. Then the harmonic map Dirichlet energy can be considered as a special case of the Oseen-Frank energy for liquid crystals defined earlier: Z Z I(u) = F (u, Du) dx = (κ1 (div u)2 + κ2 (u · curl u)2 + κ3 (u × curl u)2 ) dx Ω ZΩ + κ4 (tr((Du)2 ) − (div u)2 ) dx, Ω R3 ,

where Ω is a bounded domain in u : Ω → R3 , div u = trDu is the divergence of u and curl u = ∇ × u denotes the curl vector of u in R3 . In the first part of the total energy I(u), the κ1 -term represents the splay energy, κ2 -term represents the twist energy and κ3 -term represents the bending energy, corresponding to the various deformations of the nematic director u with |u(x)| = 1; the κ4 -term is a null-Lagrangian, depending only on the boundary data of u (see Chapter 5). From the algebraic relation |Du − (Du)T |2 = 2| curl u|2 one easily sees that |Du|2 = | curl u|2 + tr((Du)2 ). Furthermore, since |u(x)| = 1, it easily follows that | curl u|2 = (u · curl u)2 + |u × curl u|2 .


75

If κ = min{κ1 , κ2 , κ3 } then it easily follows that Z Z (4.34) I(u) ≥ κ |Du|2 dx + (κ4 − κ) [tr((Du)2 ) − (div u)2 ] dx. Ω

Ω

If κ > 0, then it can be shown that the first part of F (u, ξ) is convex and quadratic in ξ, and hence the first part of the energy I is w.l.s.c. on H 1 (Ω; R3 ) by Tonelli’s theorem (Theorem 5.1 of Chapter 5). By (4.34), one can easily obtain that Z |Du(x)|2 dx ≤ c0 I(u) + Cϕ , ∀ u ∈ Dϕ ⊂ H 1 (Ω; R3 ). Ω

Therefore, by the direct method, we have established the following existence result for minimizers. Theorem 4.12. Let κi > 0 for i = 1, 2, 3. If C = {u ∈ Dϕ | |u(x)| = 1 a.e.} 6= ∅, then there exists at least one u ∈ C such that I(u) = min I(v). v∈C

1 Remark. If all four κi are R equal2 to 2 , then I(u) reduces to the Dirichlet integral for har1 monic maps: I(u) = 2 Ω |Du| dx.

4.4.3. Stokes’ Problem. Let Ω ⊂ R3 be open, bounded and simply connected. Given f ∈ L2 (Ω; R3 ), the Stokes’ problem −∆u = f − ∇p,

div u = 0

in Ω,

u=0

on ∂Ω,

can be solved by minimizing the functional Z 1 I(u) = ( |Du|2 − f · u) dx 2 Ω over the set C = {u ∈ H01 (Ω; R3 ) | div u = 0 in Ω}. The function p in the equation is the corresponding Lagrange multiplier, known as the pressure. Pressure does not appear in the variational problem, but arises due to the constraint div u = 0. 4.4.4. Incompressible Elasticity. Suppose u : Ω → R3 represents the displacement of an elastic body occupying the domain Ω ⊂ R3 before deformation. Assume the body is incompressible, which means det Du(x) = 1 ∀ x ∈ Ω. The stored energy is given by an integral functional Z I(u) = F (x, u, Du) dx. Ω

The suitable space to work in this case is W 1,p (Ω; R3 ) with p ≥ 3. For further details, see next chapter.

76


4.4.5. A Nonlocal Problem: Ferromagnetism. (See James & Kinderlehrer [23], and Pedregal & Yan [34] for more results.) We study a minimization problem in ferromagnetism which involves a constraint and a nonlocal term. After some simplifications, the total energy I(m) in ferromagnetism is given by Z Z 1 ϕ(m(x)) dx + (4.35) I(m) = |∇u(x)|2 dx, 2 n Ω R where Ω is a bounded domain with C 1 boundary ∂Ω, m ∈ L∞ (Ω; Rn ) with |m(x)| = 1 a.e. x ∈ Ω (that is, m ∈ L∞ (Ω; S n−1 )), representing the magnetization of a ferromagnetic material occupying the domain Ω, ϕ is a given function representing the anisotropy of the material, and u is the nonlocal stray energy potential determined by m over the whole space Rn by the simplified Maxwell equation: (4.36)

in Rn ,

div(−∇u + mχΩ ) = 0

1 (Rn ) with ∇u ∈ L2 (Rn ; Rn ). The where χΩ is the characteristic function of Ω, and u ∈ Hloc equation (4.36) is understood in the weak sense: Z Z ∇u · ∇ζ dx = m · ∇ζ dx ∀ ζ ∈ H01 (Rn ). Rn

Ω

Note that any two weak solutions of (4.36) can only differ by a constant. Let B be a fixed ¯ in Rn . We have the following result. open ball containing Ω Theorem 4.13. For each m R∈ L2 (Ω; Rn ), there exists a unique weak solution u = T m ∈ 2 n 1 1 (Rn ) to (4.36) such that Hloc B u(x)dx = 0. Moreover, T : L (Ω; R ) → H (B) is linear and bounded with kT mkH 1 (B) ≤ C kmkL2 (Ω;Rn ) ,

∀ m ∈ L2 (Ω; Rn ).

Proof. Let X = {u ∈

1 (Rn ) | Hloc

2

n

n

Z

∇u ∈ L (R ; R ),

u(x) dx = 0}. B

Given m ∈ L2 (Ω; Rn ), define a functional J : X → R by Z Z 1 J(u) = |∇u|2 dx − m · ∇u dx, 2 Rn Ω

u ∈ X.

We solve the minimization problem: inf u∈X J(u). Note that X is simply a nonempty set of functions and has no topology defined. But we can always take a minimizing sequence uk ∈ X such that lim J(uk ) = inf J(u) < ∞. k→∞

u∈X

By Cauchy’s inequality with , it follows that Z Z Z 1 2 2 J(u) ≥ |∇u| dx − |∇u| dx − C |m|2 dx. 2 Rn Rn Ω Taking =

1 4

yields

Z 1 |∇u|2 dx − C, ∀ u ∈ X. 4 Rn Therefore {∇uk } is bounded in L2 (Rn ; Rn ) and hence, via a subsequence, converges weakly to some F ∈ L2 (Rn ; Rn ); this F can be written as F = ∇¯ u for a unique u ∈ X. (The last fact needs a little more analysis; see the Hodge decompositions below!) Using this J(u) ≥


77

weak convergence, we see that u ¯ ∈ X is a minimizer of J over X. It also satisfies the Euler-Lagrange equation Z Z m · ∇ζ ∀ζ ∈ X. ∇¯ u · ∇ζ = (4.37) Rn

Ω 1 n H0 (R ).

This equation also holds for all ζ ∈ From this equation, the uniqueness of minimizers and the linear dependence of u ¯ on m follow. Define u ¯ = T m. Furthermore, using ζ=u ¯ ∈ X in (4.37), we also have Z Z 2 |∇¯ u| dx = m · ∇¯ u dx ≤ kmkL2 (Ω;Rn ) k∇¯ ukL2 (B;Rn ) . Rn

Ω

Hence k∇¯ ukL2 (B;Rn ) ≤ kmkL2 (Ω;Rn ) . Since

R B

u ¯ dx = 0, by Poincaré’s inequality, k¯ ukH 1 (B) ≤ C kmkL2 (Ω;Rn ) .

The theorem is proved.

From (4.37), we also see that Z

1 I(m) = ϕ(m) dx + 2 Ω

(4.38)

Z m · ∇¯ u dx, Ω

where u ¯ = T m is defined in the theorem above. The following result is a special case of some general theorem. We leave the easy proof to the student. Lemma 4.14. Let T : L2 (Ω; Rn ) → H 1 (B) be defined above. Then T mk * T m weakly in H 1 (B) whenever mk * m weakly in L2 (Ω; Rn ). Theorem 4.15. Assume ϕ(m) ≥ 0 on S n−1 and the minimal set ϕ−1 (0) contains at least {±m0 }. Then inf I(m) = 0. m∈L∞ (Ω;S n−1 )

Proof. It suffices to find a sequence {mk } in L∞ (Ω; S n−1 ) such that Z Z 1 mk (x) · ∇uk (x) dx → 0 I(mk ) = ϕ(mk (x))dx + 2 Ω Ω as k → ∞, where uk = T mk is the solution in X of the simplified Maxwell equation (4.36). To this end, let η ∈ S n−1 be such that η ⊥ m0 and define mk (x) = ρ(kx · η)m0 + (1 − ρ(kx · η))(−m0 ),

x ∈ Ω,

where ρ(t) is a periodic function of period 1 and 1 1 0 ≤ t ≤ ; ρ(t) = 0 < t < 1. 2 2 It is easy to check wk * 0 weakly in L2 (Ω; Rn ). Let uk = T wk . Then, by Lemma 4.14 above, uk * 0 in H 1 (B). Note that, by compact embedding theorems, H 1 (B) ⊂⊂ L2 (Ω) and H 1 (B) ⊂⊂ L2 (∂Ω) are compact; hence uk → 0 strongly in both L2 (Ω) and L2 (∂Ω). We now compute by the divergence theorem that Z Z Z mk · ∇uk dx = uk mk · ν dS − uk div mk dx, ρ(t) = 1

Ωj

∂Ωj

Ωj

78


where ν is the unit outward normal on the boundary and the formula is valid on each piece ¯ where mk is constant m0 or −m0 ; hence div mk = 0 on each Ωj . Moreover, mk ·ν = 0 Ωj of Ω on ∂Ωj \ ∂Ω. Hence we have Z Z mk · ∇uk dx ≤ |uk (x)| dS ≤ |∂Ω|1/2 kuk kL2 (∂Ω) → 0 as k → ∞. Ω

∂Ω

Finally, noting that ϕ(mk (x)) = 0, we arrive at Z Z 1 ϕ(mk (x))dx + I(mk ) = mk (x) · ∇uk (x) dx → 0 2 Ω Ω as k → ∞.

Theorem 4.16. Assume ϕ−1 (0) = {±m0 }. Then there exists no minimizer of I(m) on L∞ (Ω; S n−1 ). Proof. Suppose m ¯ is a minimizer. Then I(m) ¯ = inf I = 0 and hence ϕ(m(x)) ¯ = 0 and Tm ¯ = 0. Therefore, m(x) ¯ = χE (x)m0 + (1 − χE (x))(−m0 ),

div(mχ ¯ Ω ) = 0,

where E = {x ∈ Ω | m(x) ¯ = m0 } is a measurable subset of Ω. Therefore Z (4.39) f (x)∇ζ(x) · m0 dx = 0 ∀ ζ ∈ C0∞ (Rn ), Rn

where f (x) = χΩ (x)[2χE (x) − 1] = 2χE (x) − χΩ (x). Let x = x0 + tm0 where x0 ⊥ m0 and write f (x) = g(x0 , t). In (4.39), by change of variables, we have Z g(x0 , t)ζt (x0 , t) dx0 dt = 0 ∀ ζ ∈ C0∞ (Rn ). Rn

This implies the weak derivative gt (x0 , t) ≡ 0 on Rn , and hence g(x0 , t) = h(x0 ) is independent of t. But h(x0 ) = f (x0 + tm0 ) vanishes for large t and hence h ≡ 0. So f ≡ 0 on Rn . However f (x) ∈ {±1} for x ∈ Ω. This is a contradiction. 4.4.6. The Hodge Decompositions on Rn . We now present a short theory for Hodge decompositions on Rn with elementary proofs. See [38] for some related results. Let X = L2 (Rn ; Rn ) denote the Hilbert space with the inner product and norm defined by Z

1 1

n n

Z

(u v + · · · + u v ) dx =

(u, v) = Rn

u · v dx;

kuk = (u, u)1/2 .

Rn

For u ∈ L2 (Rn ; Rn ), we define div u, curl u = (curl u)ij as distributions as follows: Z Z 1 n hdiv u, ϕi = − (u ϕx1 + · · · + u ϕxn ) dx = − u · ∇ϕ dx; Rn Rn Z h(curl u)ij , ϕi = − (ui ϕxj − uj ϕxi ) dx, ∀ ϕ ∈ C0∞ (Rn ). Rn

1,1 Note that, if u ∈ Wloc (Rn ; Rn ), then div u = tr Du and curl u = Du − (Du)T . In the case n = 2 or n = 3, the operator curl u can be identified as follows:

curl u ≈ ∇⊥ · u = ∇ · u⊥ = div(u⊥ ) = u1x2 − u2x1 curl u ≈ ∇ × u =

(u3x2

−

u2x3 , u1x3

−

u3x1 , u2x1

−

u1x2 )

(n = 2); (n = 3).


79

Note that in the sense of distribution, the Laplacian of m ∈ X (for each component of m) can be written as ∆m = ∇(div m) + div(curl m),

(4.40) which means i

∆m = ∂xi (div m) +

n X

∂xj (curl m)ij

∀ i = 1, 2, · · · , n.

j=1

Define the subspaces of divergence-free and curl-free fields as follows: Xdiv = {u ∈ L2 (Rn ; Rn ) | div u = 0 in the sense of distribution}; Xcurl = {u ∈ L2 (Rn ; Rn ) | curl u = 0 in the sense of distribution}. It is easy to see they are closed linear subspaces of X = L2 (Rn ; Rn ). First we have the following result. Lemma 4.17. Xdiv ∩ Xcurl = {0}. Proof. Let m ∈ Xdiv ∩ Xcurl . Then curl m = div m = 0 in the sense of distributions. Hence, by (4.40) above, ∆m = 0 holds also in the sense of distributions. Hence m ∈ C ∞ (Rn ; Rn ) and each of its components mi is a harmonic function in Rn . Sine each mi also belongs to L2 (Rn ), the mean value property and Hölder’s inequality imply that Z 1 |mi | dy ≤ c R−n/2 kmk |mi (x)| ≤ |BR (x)| BR (x) for any x ∈ Rn and R > 0, where BR (x) = {y ∈ Rn | |y − x| < R} denotes the ball of radius R and center x. Letting R → ∞ shows mi = 0 and hence m = 0. To characterize the spaces Xdiv and Xcurl and prove the Hodge decomposition theorem, we introduce the linear function spaces: Y = {f ∈ L2loc (Rn ) | ∇f ∈ X} and, for n = 3, M = {m ∈ L2loc (R3 ; R3 ) | ∇ × m ∈ L2 (R3 ; R3 )}. It is easy to see that ∇f ∈ Xcurl for all f ∈ Y and, for n = 3, ∇ × m ∈ Xdiv for all m ∈ M. The converse is also true; we have the following results. Theorem 4.18. There exists a uniform constant Cn > 0 such that, for each v ∈ Xcurl , there exists a f ∈ Y satisfying that Z Z 1 2 |f | dx ≤ Cn |v|2 dx, (4.41) (a) v = ∇f ; (b) sup n+2 R≥1 R BR (0) Rn where BR (y) = {x ∈ Rn | |x − y| < R} denotes the ball centered y of radius R. Theorem 4.19. Let n = 3. There exists a uniform constant C > 0 such that, for every w ∈ Xdiv , there exists a m ∈ M satisfying that Z Z 1 (4.42) (a) w = ∇ × m; (b) sup 3 |m|2 dx ≤ C |w|2 dx, R n R≥1 QR (0) R where QR (y) = {x ∈ R3 | |xi −yi | < R, i = 1, 2, 3} denotes the cubes centered y of side-length 2R.

80


The estimate (4.41b) in Theorem 4.18 above is not sharp as we will obtain some better estimates later. However, the estimate (b) in both (4.41) and (4.42) does provide a way to represent a curl-free v or divergence-free field w by some local function f or m, with f being viewed as the potential function of v and m the velocity of w; initially, these local functions have only been defined as the Schwartz distributions. The estimate (4.41b) also suggests that we equip the space Y with the norm kf k∗ defined by Z Z 1 2 2 (4.43) kf k∗ = |∇f (x)| dx + sup n+2 |f (x)|2 dx. R≥1 R Rn BR Let Y1 be the subspace of Y defined by 1 Y1 = {f ∈ Hloc (Rn ) | kf k∗ < ∞}.

It can be easily shown that Y1 is indeed a Banach space with the norm defined. We shall try to find the minimal subspace Z of Y1 for which the gradient operator ∇ : Z → Xcurl is bijective. In fact such a Z can be completely determined when n ≥ 3. Theorem 4.20. Let Y2 be the closure of C0∞ (Rn ) in Y1 . Then, for n ≥ 3, 2n

Y2 = {f ∈ L n−2 (Rn ) | ∇f ∈ L2 (Rn ; Rn )}. Furthermore, Y2 has the equivalent norms kf k∗ ≈ k∇f kL2 (Rn )

∀ f ∈ Y2 ,

and the gradient operator ∇ : Y2 → Xcurl is bijective. The proof of this theorem relies on the Sobolev-Gagliardo-Nirenberg inequality for H 1 (Rn ) functions when n ≥ 3 (see [2, 16, 21]); note that in this case the finite num2n ber 2∗ = n−2 is the Sobolev conjugate of 2. In the case n = 2, there is no such a SobolevGagliardo-Nirenberg inequality; instead, there is a John-Nirenberg-Trudinger type of BMOestimates for functions with gradient in L2 (R2 ) (see [21, 35]. However, we shall try to avoid the BMO-estimates. One of the minimal subspaces of Y1 on which the gradient operator is bijective can be characterized as follows. Theorem 4.21. Let n = 2 and Z be the closure in Y1 of the subspace S = {ϕ − hϕiρ | ϕ ∈ H 1 (R2 )}, where hϕiρ = (4.44)

R

R2

ϕ(x)ρ(x) dx and ρ(x) is the weight function defined by 1 1 χ{|x|≤1} + 4 χ{|x|>1} . ρ(x) = 2π |x|

Then Z has the equivalent norms kf k∗ ≈ k∇f kL2 (R2 ) ∇ : Z → Xcurl is bijective.

∀ f ∈ Z, and the gradient operator

Given any u ∈ X = L2 (Rn ; Rn ), for each R > 0, let BR = BR (0) and consider the following minimization problem: Z (4.45) inf |∇ϕ − u|2 dx. ϕ∈H01 (BR ) BR

By the direct method, this problem has a unique solution, which we denote by ϕR , also extended by zero to all Rn . This sequence {ϕR } is of course uniquely determined by u ∈ X. It also satisfies the following properties: R 1 (4.46) Rn (∇ϕR − u) · ∇ζ dx = 0 ∀ ζ ∈ H0 (Ω), Ω ⊆ BR , (4.47)

k∇ϕR kL2 (Rn ) ≤ kuk.


81

Lemma 4.22. Given u ∈ X, it follows that ∇ϕR → v in X as R → ∞ for some v ∈ X and that v ∈ Xcurl is uniquely determined by u. Moreover, this v satisfies kv − uk = 0 min kv 0 − uk; v ∈Xcurl

therefore, v = u if u ∈ Xcurl . Proof. First of all, we claim ∇ϕR * v weakly in X as R → ∞. Let v 0 , v 00 be the weak limits of any two subsequences {∇ϕR0 } and {∇ϕR00 }, where R0 , R00 are two sequences going to ∞. We would like to show v 0 = v 00 , which shows that ∇ϕR * v as R → ∞. Note that it is easily seen that v 0 , v 00 ∈ Xcurl and, by (4.46) above, Z Z (v 00 − u) · ∇ζ = 0 ∀ Ω ⊂⊂ Rn , ∀ ζ ∈ H01 (Ω). (v 0 − u) · ∇ζ = (4.48) Rn

Rn

div(v 0

div(v 00

This implies − u) = − u) = 0 and hence div(v 0 − v 00 ) = 0; therefore, v 0 − v 00 ∈ Xdiv ∩ Xcurl = {0} by Lemma 4.17 above. We denote this weak limit by v ∈ Xcurl . Note that, by (4.48), div(v − u) = 0. Hence if curl u = 0 then v − u ∈ Xdiv ∩ Xcurl = {0}; hence v = u. We now prove ∇ϕR → v in X as R → ∞. Taking ζ = ϕR ∈ H01 (BR ) in (4.48) and letting R → ∞ we have Z (4.49) (v − u) · v dx = 0. Rn

Using ζ = ϕR in (4.46), taking R → ∞ and by weak limit, we have Z Z Z 2 lim |∇ϕR | dx = v·u= |v|2 dx. R→∞ Rn

Rn

This implies ∇ϕR → v strongly in

L2 (Rn ; Rn ).

Rn

We now show that

kv − uk = 0 min kv 0 − uk.

(4.50)

v ∈Xcurl

v0

Given any ∈ Xcurl , choose the sequence ϕ˜R corresponding to v 0 − v. Since div(v − u) = 0, it easily follows that (v 0 − v, v − u) = lim (∇ϕ˜R , v − u) = 0. R→∞

Hence kv 0 − uk2 = kv − v 0 k2 + 2(v 0 − v, v − u) + kv − uk2 ≥ kv − uk2 ; this proves (4.50). We can now prove the following Hodge decomposition theorem. Theorem 4.23. (Hodge decomposition) For each u ∈ X, there exist unique elements ⊥ , X ⊥ v ∈ Xcurl , w ∈ Xdiv such that u = v + w. Moreover, Xcurl = Xdiv div = Xcurl . Consequently, X = L2 (Rn ; Rn ) = Xdiv ⊕ Xcurl . Proof. Given u ∈ X, let v ∈ Xcurl be defined as above, and let w = u − v. Then u = v + w and, by (4.48) above, w ∈ Xdiv . We now show that v, w are unique. Suppose u = v 0 + w0 for another pair v 0 ∈ Xcurl and w0 ∈ Xdiv . Then m = v − v 0 = w0 − w ∈ Xcurl ∩ Xdiv . Hence v 0 = v and w0 = w. ⊥ , given any v ∈ X 1 To show Xcurl = Xdiv curl , w ∈ Xdiv , let ϕR ∈ H0 (BR ) be the sequence determined by u = v as above. Since ∇ϕR → v and Z (w, ∇ϕR ) = w(x) · ∇ϕR (x) = 0, Rn ⊥ . This shows X ⊥ ⊥ it follows easily that (w, v) = 0; hence v ∈ Xdiv curl ⊆ Xdiv . Assume u ∈ Xdiv . We will show u ∈ Xcurl . By the first part of this result, let u = v + w, where v ∈ Xcurl ,

82


w ∈ Xdiv . Then 0 = (u, w) = (v, w) + |w|2 = |w|2 ; hence w = 0 and u = v ∈ Xcurl . This ⊥ . Similarly, it follows X ⊥ proves Xcurl = Xdiv div = Xcurl . Theorem 4.24. Let Y1 be the space with the norm k·k∗ defined above. Then Y1 is a Banach space. Moreover, the gradient operator ∇ : Y1 → Xcurl is surjective; more precisely, for any v ∈ Xcurl , there exists a f ∈ Y1 such that v = ∇f,

kf k∗ ≤ Cn kvk.

Proof. The proof that Y1 is a Banach space follows directly by the definition and will not be given here. We prove the rest of the theorem. Given v ∈ Xcurl , let v = v ∗ ρ be the smooth approximation of v. Then v ∈ Xcurl ∩ C ∞ (Rn ; Rn ). Define Z 1 v (tx) · x dt. f (x) = 0

Then one can easily compute that Z 1 Z 1 j ∂xj f (x) = v (tx)dt + (∂xj v )(tx) · tx dt 0 0 Z 1 Z 1 j = v (tx)dt + ∇vj (tx) · tx dt 0 0 Z 1 Z 1 d j j = v (tx)dt + [v (tx)]t dt dt 0 0 Z 1 Z 1 1 j j j = v (tx)dt + v (tx)t 0 − v (tx)dt 0

0

= vj (x), where we have used the condition ∂xj v = ∇vj since curl v = 0. This proves ∇f (x) = v (x) for all x ∈ Rn . Therefore, for all x, y ∈ Rn , Z 1 f (x + y) − f (y) = v (y + tx) · x dt. 0

Hence 2

|f (x + y) − f (y)|

1

Z =

0 2

2 v (y + tx) · x dt Z

≤ |x|

1

|v (y + tx)|2 dt.

0

Integrating this inequality over x ∈ BR (0) = BR , we obtain Z Z 1 Z 2 2 2 |f (z) − f (y)| dz ≤ R |v (y + tx)| dx dt BR (y)

0

= R2

Z

BR 1

0

= Rn+2

!

Z

|v (z)|2 dz

t−n dt

BtR (y)

Z 0

1

1 |BtR (y)|

≤ Rn+2 M(|v |2 )(y),

Z BtR (y)

! |v (z)|2 dz

dt


83

where M(h) is the maximal function of h (see Stein [35]). Since |v |2 ∈ L1 (Rn ), it follows that Z Z 5n 5n n 2 2 m{y ∈ R | M(|v | )(y) > α} ≤ |v | dx ≤ |v|2 dx. α Rn α Rn Let E = {y ∈ B1 | M(|v |2 )(y) ≤ α0 }, where we choose

2 · 5n α0 = |B1 |

Z

|v|2 dx.

Rn

Then it follows that |E | ≥ 12 |B1 | for all . Therefore, it is a simple exercise to show that there exists a sequence k → 0 and a point y0 ∈ B1 such that y0 ∈ ∩∞ k=1 Ek ; that is, 2 · 5n kvk2 , |B1 |

M(|vk |2 )(y0 ) ≤ α0 =

∀ k = 1, 2, · · · .

Using this y0 we define a new sequence z ∈ Rn .

gk (z) = fk (z) − fk (y0 ), Then, for all R ≥ 1, we have Z Z 2 |gk (z)| dz ≤

|gk (z)|2 dz ≤ (2R)n+2

B2R (y0 )

BR

2 · 5n kvk2 . |B1 |

By using diagonal subsequences, there exists a subsequence gkj and a function f ∈ L2loc (Rn ) such that gkj * f weakly as kj → ∞ on all balls BR (0), R > 0. This function f must satisfy ∇f = v ∈ L2 (Rn ; Rn ) and Z Z 1 2 sup n+2 |f (x)| dx ≤ Cn |v(x)|2 dx; R≥1 R BR Rn hence kf k∗ ≤ Ckvk. This completes the proof.

Proof of Theorem 4.19. Given w ∈ Xdiv , let w = w ∗ ρ be the smooth approximation of w. Then w ∈ Xdiv ∩ C ∞ (R3 ; R3 ). Define x ∈ R3 , c ∈ [−1, 1],

m (x, c) = (p (x, c), q (x, c), 0), where Z (4.51)

x3

p (x, c) =

w2 (x1 , x2 , s) ds −

c

w3 (x1 , t, c) dt;

0

Z (4.52)

x2

Z

q (x, c) = −

x3

w1 (x1 , x2 , s) ds.

c

Since div w = 0, one can easily verify that w (x) = ∇ × m (x, c),

∀ c ∈ [−1, 1].

We now estimate 2

|q (x, c)| ≤ (|x3 | + 1)

2

Z

|x3 |+1

−|x3 |−1

|w1 (x1 , x2 , s)|2 ds.

Integrating this inequality over the cube QR (0) = {x ∈ R3 | |xi | < R, i = 1, 2, 3}, we obtain Z Z 2 2 (4.53) |q (x, c)| dx ≤ 2R(R + 1) |w1 (x)|2 dx. QR (0)

R3

84


Next we write p (x, c) = g (x, c) − f (x0 , c) with x0 = (x1 , x2 ), where Z x2 Z x3 2 0 w3 (x1 , t, c) dt. w (x1 , x2 , s) ds; f (x , c) = (4.54) g (x, c) = 0

c

For g (x, c), we have the same estimate as q (x, c): Z Z 2 2 |g (x, c)| dx ≤ 2R(R + 1) (4.55) R3

QR (0)

For f (x0 , c), we easily estimate that Z Z 0 2 0 2 (4.56) |f (x , c)| dx ≤ 2R |x0i | α} ≤ H (c) dc = |w3 (x)|2 dx. α R α R3 R Let E = {c ∈ [−1, 1] | H (c) ≤ α0 }, where α0 = R3 |w3 (x)|2 dx. Then it follows that |E | ≥ 1 for all . Therefore, as above, there exists a sequence k → 0 and a point c0 ∈ [−1, 1] such that c0 ∈ ∩∞ k=1 Ek ; that is, Z Hk (c0 ) ≤ |w3 (x)|2 dx ∀ k = 1, 2, · · · . Note that H (c) =

R

R2

R3

Hence by (4.56) Z

2

|f (x, c0 )| dx ≤ 4R

(4.57)

3

Z R3

QR (0)

|w3 (x)|2 dx.

Using this c0 we define a new sequence uk (x) = mk (x, c0 ). Then we have wk = curl uk and, for all R ≥ 1, by (4.53)-(4.56), Z (4.58) |uk (x)|2 dx ≤ CR3 kwk2 . QR (0)

By using diagonal subsequences, there exists a subsequence ukj and a function m ∈ L2loc (R3 ) such that ukj * m weakly as kj → ∞ on all cubes |xi | < R, R > 0. This field m must satisfy w = curl m ∈ L2 (R3 ; R3 ); hence m ∈ M. Moreover, by (4.58), Z Z 1 |m(x)|2 dx ≤ C |w(x)|2 dx. sup 3 R 3 R≥1 QR (0) R This completes the proof.

Proof of Theorem 4.20. Let n ≥ 3 and Y1 , Y2 be the spaces defined above, and define 2n

W = {f ∈ L n−2 (Rn ) | ∇f ∈ L2 (Rn ; Rn )}. We proceed with several lemmas to prove Theorem 4.20. Lemma 4.25. Y2 ⊆ W. Moreover (4.59)

kf k∗ ≤ Ck∇f kL2 (Rn )

∀ f ∈ Y2 .


85

Proof. Let f ∈ Y2 . Then there exists a sequence fj ∈ C0∞ (Rn ) such that kfj − f k∗ → 0 as j → ∞. Therefore k∇fj kL2 → k∇f kL2 . By Sobolev-Galiardo-Nirenberg inequality, kfj kL2∗ (Rn ) ≤ C k∇fj kL2 (Rn )

∀ j.

∗

Hence fj * g ∈ L2 (Rn ). Since fj → f in L2 (BR ) for all R > 0. We have f = g. Hence f ∈ W. Furthermore, by H¨ older’s inequality, kfj kL2 (BR ) ≤ cn R2 kfj kL2∗ (BR ) ≤ C R2 k∇fj kL2 (Rn ) . Hence, by taking limits as j → ∞, it follows that Z Z 1 |∇f (x)|2 dx, sup n+2 |f |2 dx ≤ C R n R≥1 BR R which proves (4.59).

Lemma 4.26. W ⊆ Y2 . Proof. Let f ∈ W. Define fj = f ρj , where ρj ∈ W 1,∞ (Rn ) defined by ρj (x) = 1 on |x| ≤ j, ρj (x) = 0 on |x| ≥ 2j and ρj (x) is linear in |x| for j ≤ |x| ≤ 2j. Then fj ∈ Y2 . It can be easily shown that lim kfj − f k∗ = 0, j→∞

which proves f ∈ Y2 and hence W ⊆ Y2 .

Lemma 4.27. ∇ : Y2 → Xcurl is surjective. Proof. Given any v ∈ Xcurl , let ϕR ∈ H01 (BR ) be the function determined as in the minimization problem (4.45) above with u = v. Then ∇ϕR → v in X = L2 (R2 ; R2 ) as R → ∞. Lemma 4.25 implies that {ϕR } is a Cauchy sequence in Y2 and hence its limit f belongs to Y2 and satisfies ∇f = v. This completes the proof. Proof of Theorem 4.21. We first prove the following result. Lemma 4.28. Let ρ(x) be defined as above. Then for all ϕ ∈ H 1 (R2 ), Z Z 2 (4.60) |ϕ(x) − hϕiρ | ρ(x) dx ≤ β |∇ϕ(x)|2 dx. R2

R2

Proof. First of all, by Poincaré’s inequality, Z Z 2 (4.61) |ψ(x) − (ψ)0 | dx ≤ C B1

|∇ψ(x)|2 dx

B1

for all ψ ∈ H 1 (B1 ), where (ψ)0 is the average value of ψ on B1 ; that is, Z Z 1 (ψ)0 = ψ(x) dx = 2 ψ(x)ρ(x) dx. π B1 {|x|1}

{|y|>1}

where (ψ)0 =

1 π

Z ψ(x) dx = B1

1 π

Z {|y|>1}

ϕ(y)|y|−4 dy = 2

Z ϕ(x)ρ(x) dx. {|x|>1}

Combining (4.61) for ψ(x) = ϕ(x) with (4.62) we obtain (4.60).

86


Note that, for all f ∈ Y1 , 1 4 R≥1 R

(4.63)

Z

sup

|f (x)|2 dx ≤ 2π

Z

|f (x)|2 ρ(x) dx.

R2

BR

Let S = {ϕ − hϕiρ | ϕ ∈ H 1 (R2 )}. Then the previous lemma and (4.63) imply Z Z Z 1 2 2 sup 4 |f | dx ≤ 2π |f | ρ dx ≤ C |∇f |2 dx R R≥1 BR R2 R2

∀ f ∈ S.

We have thus proved the following result. Corollary 4.29. Let Z be the closure of S in Y1 . Then kf k∗ ≈ k∇f kL2 for all f ∈ Z. Furthermore, for all f ∈ Z, Z Z Z 2 f (x)ρ(x) dx = 0, |f (x)| ρ(x) dx ≤ C |∇f |2 dx, R2

R2

R2

where C is a constant independent of f. Finally, we prove the following result to complete the proof of Theorem 4.21. Corollary 4.30. ∇ : Z → Xcurl is surjective. Proof. The proof is similar to that of Lemma 4.27 above. Given any v ∈ Xcurl , let ϕR ∈ H01 (BR ) be the function determined as in the minimization problem (4.45) above with u = v. Let fR = ϕR − hϕR iρ . Then fR ∈ S and ∇fR = ∇ϕR → v in X = L2 (R2 ; R2 ) as R → ∞. Corollary 4.29 implies that {fR } is a Cauchy sequence in Y1 and hence its limit f belongs to the closure Z of S and satisfies ∇f = v. This completes the proof.

4.5. Mountain Pass Theorem In this section we consider another approach to noncoercive problems of the type dealt with in the previous section. 4.5.1. The Palais-Smale Condition. We first introduce the following Palais-Smale compactness condition (PS), which plays an essential role in the results to follow. The Palais-Smale Condition (PS). Let E : X → R be G-diff on the Banach space X. We say E satisfies the Palais-Smale condition (PS) if the following holds: whenever {un } is a sequence in X such that E(un ) is bounded and kE 0 (un )k → 0, then {un } has a convergent subsequence. Remark. The (PS) condition is not satisfied by very smooth functions very often. For example, the function E : R → R with E(u) = cos u does not satisfy (PS), which can be easily seen by considering the sequence un = nπ. Similarly, the function E(u) = c does not satisfy (PS). It can be shown that if E is F-diff on the Banach space X (not necessarily reflexive) and is bounded below and satisfies (PS), then E attains its minimum value. But we study the case where functionals are neither bounded from above or below.

4.5. Mountain Pass Theorem

87

4.5.2. Deformation Lemma. Lemma 4.31. (Deformation) Let E : X → R be a C 1 functional satisfying (PS). Let c, s ∈ R and define Kc = {u ∈ X : E(u) = c, E 0 (u) = 0} As = {u ∈ X : E(u) ≤ s}. Assume Kc = ∅. Then there exists an ε¯ > 0 and a continuous function η : [0, 1] × X → X such that for all 0 < ε ≤ ε¯, (i) η(0, u) = u

for all

u ∈ X,

(ii) η(t, u) = u

for all

t ∈ [0, 1], u 6∈ E −1 ([c − ε, c + ε]),

(iii) E(η(t, u)) ≤ E(u)

t ∈ [0, 1], u ∈ X,

for all

(iv) η(1, Ac+ε ) ⊂ Ac−ε . This lemma shows that if c is not a critical level, then we can nicely deform the set Ac+ε into Ac−ε for some ε > 0. 4.5.3. The Mountain Pass Theorem. We now prove the main theorem of this section. Theorem 4.32. (Ambrosetti-Rabinowitz) Let E : X → R be a C 1 functional satisfying (PS) on the real Banach space X. Let u0 , u1 ∈ X, c0 ∈ R and R > 0 be such that (i) ku1 − u0 k > R (ii) E(u0 ), E(u1 ) < c0 ≤ E(v) for all v such that kv − u0 k = R. Then E has a critical point u with E(u) = c, c ≥ c0 ; the critical value c is defined by (4.64)

c = inf sup E(p(t)) p∈K t∈[0,1]

where K denotes the set of all continuous maps p : [0, 1] → X with p(0) = u0 and p(1) = u1 . Remark. Think of the graph of E as a landscape with a low spot at u0 , surrounded by a ring of mountains. Beyond these mountains lies another low point at u1 . Note that every path p connecting u0 to u1 has to cross the sphere {v : kv − u0 k = R} since u1 lies outside the sphere. Moreover, on this sphere the value of E is at least c0 . Hence the maximum value of E(p(t)) for any such p is at least c0 . Hence c ≥ c0 . An important aspect of the MPT is that the critical point u is distinct from u0 and u1 . Hence, if u0 already satisfies E 0 (u0 ) = 0 by some other method, then u will be a second solution of E 0 (u) = 0. Proof. Let c be as defined in (4.64). If it is not a critical value, then Kc = ∅. Let η and ε¯ be as in the deformation lemma. Now by condition (ii) of the theorem, we can choose ε small enough so that 0 < ε < ε¯ and E(u0 ), E(u1 ) 6∈ [c − ε, c + ε] (since c ≥ c0 ). Let p ∈ K and define the path ξ : [0, 1] → X by ξ(t) = η(1, p(t)). Since p(0) = u0 and p(1) = u1 , it follows, by the choice of ε, that ξ(0) = η(1, u0 ) = u0 , ξ(1) = η(1, u1 ) = u1 using condition (ii) of the lemma. Thus ξ ∈ K. Now, we can choose p ∈ K such that max E(p(t)) < c + ε. t∈[0,1]

88


Since p(t) ∈ Ac+ε , by (iv) of the lemma, ξ(t) ∈ Ac−ε . Thus max E(ξ(t)) ≤ c − ε t∈[0,1]

which contradicts the definition of c. Hence Kc 6= ∅.

4.5.4. Saddle Point Solutions. In order to illustrate these ideas we apply the MPT to the problem (4.65)

∆u + f (x, u) = 0 in Ω,

u|∂Ω = 0

where f ∈ Car satisfies the growth condition |f (x, z)| ≤ c(1 + |z|τ ), |fz (x, z)| ≤ c(1 + |z|τ −1 ) with 1 < τ < (n + 2)/(n − 2). Moreover, we assume f (x, 0) = 0 and 0 ≤ F (x, z) ≤ γf (x, z)z Rz for some constant γ < 1/2, where F (x, z) = 0 f (x, s)ds. Finally we assume a|z|τ +1 ≤ |F (x, z)| ≤ A|z|τ +1 for constants 0 < a ≤ A. Associated with (4.65) is the functional I defined by Z 1 I(u) = |∇u|2 dx − J(u), 2 Ω where Z J(u) =

F (x, u)dx. Ω

The following lemma simplifies some of the technical details that follow. Lemma 4.33. Let I and J be defined on H = H01 (Ω) as above. Then both I and J are of C 1 on H and (a) J 0 : H → H ∗ is compact. (b) If {un } is a bounded sequence in H such that I 0 (un ) → 0 as n → ∞, then {un } has a convergent subsequence. R Proof. The norm Ω |∇u|2 dx is obviously of C ∞ on H. J is a Nemytskii operator with F being C 1 in u, and hence is of C 1 on H with J 0 : H → H ∗ given by Z 0 hJ (u), vi = f (x, u(x)) v(x) dx, ∀ u v ∈ H. Ω

We have the estimate kJ 0 (u) − J 0 (w)kH ∗

=

sup

|hJ 0 (u) − J 0 (w), vi|

v∈H, kvk≤1

≤ kf (·, u) − f (·, w)k(τ +1)/τ kvkτ +1 ≤ C kf (·, u) − f (·, w)k(τ +1)/τ ,

∀ u w ∈ H.

Let {un } be a bounded sequence in H. Then by choosing a subsequence if necessary, un * u in H and un → u in Lτ +1 (Ω). So J 0 (un ) converges to J 0 (u) and hence J 0 : H → H ∗ is compact. Let A : H → H* denote the duality map defined by Z hAu, vi = ∇u · ∇v dx for all Ω

u, v ∈ H.

4.5. Mountain Pass Theorem

89

Since hI 0 (u), vi = hAu, vi − hJ 0 (u), vi, and A−1 exists and is bounded on H ∗ (by the LaxMilgram theorem), it follows that A−1 I 0 (u) = u − A−1 J 0 (u). Let {un } be a bounded sequence in H with I 0 (un ) → 0 as n → ∞. Since A−1 is continuous and J 0 is compact, by passing to a subsequence if necessary, we have un = A−1 I 0 (un ) + A−1 J 0 (un ) converges. Remark. From the above proof we see that the compactness of the embedding H = H01 (Ω) ⊂ Lτ +1 (Ω) is necessary for proving (PS). Theorem 4.34. The boundary value problem (4.65) has at least one nontrivial solution u ∈ H01 (Ω). Proof. Consider the C 1 functional I(u) above, which we rewrite as R 1 I(u) = kuk21,2,0 − J(u), where kuk21,2,0 = Ω |∇u|2 dx, 2 1 for u ∈ H0 (Ω). In order to apply MPT, we first choose u0 = 0. Clearly, I(u0 ) = 0. Now we show that I|∂B(0,R) ≥ c0 for some R, c0 > 0. In view of the embedding H01 (Ω) ⊂ Lτ +1 (Ω), we have Z +1 . J(u) ≤ A |u|τ +1 dx ≤ ckukτ1,2,0 Ω

Thus for all u satisfying kuk1,2,0 = R R2 − cRτ +1 ≥ c0 > 0 2 provided we take R sufficiently small. Next, let v = tu, where u 6= 0 is a fixed function in H01 (Ω), and t > 0 is to be selected. Then I(u) ≥

t2 kuk21,2,0 − J(tu) 2 Z t2 2 τ +1 ≤ |u|τ +1 dx kuk1,2,0 − at 2 < 0 ¯ R). Finally, we must verify for t > 0 and sufficiently large. Moreover, v ∈ H01 (Ω)verb|kB(0, 1 the Palais-Smale condition. Let {un } ⊂ H0 (Ω) be a sequence such that |I(un )| ≤ c for all n and I 0 (un ) → 0. If we can show that {kun k1,2,0 } is bounded, then according to Lemma 4.33, I satisfies (PS). Thus, for the given sequence Z 1 2 I(un ) = kun k1,2,0 − F (x, un )dx 2 Ω and Z 0 2 hI (un ), un i = kun k1,2,0 − un f (x, un )dx I(v) =

Ω

and so

Z

1 kun k21,2,0 ≤ I(un ) + γ un f (x, un )dx 2 Ω ≤ c + γ(kun k21,2,0 + kI 0 (un )k kun k1,2,0 ). Now suppose kun k1,2,0 → ∞. Then from the above inequality we arrive at the contradiction 1/2 ≤ γ. Thus, by MPT, there exists a critical point u 6= 0 of I which is a weak solution of (4.65).

90


4.6. Nonexistence and Radial Symmetry We now establish a nonexistence result which shows that the boundary value problem (4.31) (with k = 0, l = 1) does not have a positive solution for τ > (n + 2)/(n − 2). 4.6.1. Pohozaev’s Identity. We shall need the following fundamental identity: ¯ u|∂Ω = 0. Then Theorem 4.35. (Pohozaev) Let u ∈ C 2 (Ω), Z Z (4.66) u2n (x · ν)dS = [2(∇u · x) + (n − 2)u]∆udx ∂Ω

Ω

where ν = (ν1 , · · · , νn ) denotes the outward unit normal to ∂Ω and un = denotes the normal derivative of u on ∂Ω.

∂u ∂ν

= ∇u · ν

Proof. By an easy calculation we have the following identity 2∆u(∇u · x) = div[2(∇u · x)∇u − |∇u|2 x] + (n − 2)|∇u|2 . An application of the divergence theorem gives Z Z [2(∇u · x)un − |∇u|2 x · ν]dS = [2∆u(∇u · x) − (n − 2)|∇u|2 ]dx. ∂Ω

Ω

R

R

|∇u|2 dx

Since u = 0 on ∂Ω, Ω u∆udx + Ω = 0 and uxi = νi un , i.e., ∇u = un ν, which combined with the above integral identity yields (4.66). ¯ be a solution of the boundary value problem Corollary 4.36. Let u ∈ C 2 (Ω) ∆u + f (u) = 0 in Ω, u|∂Ω = 0. Let F (u) = (4.67)

Ru 0

f (t)dt. Then u satisfies the equation Z Z Z 2−n 1 u2n (x · ν)dS = n F (u)dx + uf (u)dx. 2 ∂Ω 2 Ω Ω

Proof. From (4.66) we have Z Z Z 1 2−n 2 u (x · ν)dS = ∆u(∇u · x)dx + uf (u)dx 2 ∂Ω n 2 Ω Ω so it suffices to show that Z

Z ∆u(∇u · x)dx = n

Ω

F (u)dx. Ω

In this direction, we first note that ∇F = f ∇u; hence ∆u(∇u · x) = −f (u)(∇u · x) = −(∇F · x). But ∇ · (xF ) = nF + (x · ∇F ) and therefore Z Z Z ∆u(∇u · x)dx = nF dx − ∇ · (xF )dx. Ω

Ω

Ω

However, the second integral on the right is zero, as may be seen by applying the divergence theorem and noting that F = 0 on ∂Ω since u is. 4.6.2. Star-shaped Domains. Let Ω be an open set containing 0. We say Ω is star¯ the line segment {λx : 0 ≤ λ ≤ 1} lies in shaped (with respect to 0) if, for each x ∈ Ω, ¯ Ω.

4.6. Nonexistence and Radial Symmetry

91

Remark. Clearly if Ω is convex and 0 ∈ Ω, then Ω is star-shaped. An annulus is not star-shaped since x · ν < 0 on the boundary of the inner circle. Lemma 4.37. Assume ∂Ω is C 1 and Ω is star-shaped with respect to 0 ∈ Ω. Then x·ν(x) ≥ 0 for all x ∈ ∂Ω, where ν(x) is the outward unit normal at x ∈ ∂Ω. Proof. Given x0 ∈ ∂Ω, since ∂Ω is C 1 , there exists a ball B = Bε (x0 ) and a C 1 function φ on B such that Ω ∩ B = {x ∈ B | φ(x) < 0},

∂Ω ∩ B = {x ∈ B | φ(x) = 0}.

∇φ(x0 ) Note that the outward unit normal at x0 is given by ν(x0 ) = |∇φ(x . Let δ > 0 be 0 )| ¯ ∩ B for all sufficiently small so that λx0 ∈ B and hence, from Ω being star-shaped, λx0 ∈ Ω λ ∈ [1 − δ, 1]. Consider h(λ) = φ(λx0 ) defined on λ ∈ [1 − δ, 1]. Then h has the maximum 0 at right-end point 1 and hence

h0 (1− ) = ∇φ(x0 ) · x0 ≥ 0. This proves x0 · ν(x0 ) ≥ 0 for all x0 ∈ ∂Ω.

4.6.3. Nonexistence of Classical Solutions. Theorem 4.38. Let Ω be star-shaped with respect to x = 0. Then the problem ∆u + |u|τ −1 u = 0 in Ω, u|∂Ω = 0

(4.68)

(τ >

n+2 ) n−2

¯ solution. Furthermore, the problem has no nontrivial C 2 (Ω) ∆u + uτ = 0 in Ω, u|∂Ω = 0

(4.69)

(τ ≥

n+2 ) n−2

¯ solution. has no positive C 2 (Ω) ¯ solution of (4.68). Applying formula (4.67) with Proof. Suppose u is a nontrivial C 2 (Ω) 1 τ −1 f (u) = |u| u (and thus F (u) = τ +1 |u|τ +1 ), and using Lemma 4.37, we obtain Z Z Z 1 |u|τ +1 2−n 2 (4.70) u (x · ν)dS = n dx + |u|τ +1 dx 2 ∂Ω n τ + 1 2 Ω Ω Z n n−2 τ +1 (4.71) = − |u| dx ≥ 0, τ +1 2 Ω which yields

n τ +1

n−2 n+2 2 and hence a contradiction: τ ≤ n−2 . ¯ solution of (4.69) with τ = n+2 . Since positive C 2 (Ω) n−2

≥

If u is a u = 0 on ∂Ω, by a sharp maximum principle (see Serrin’s Maximum Principle below), we have un ≤ −σ on ∂Ω for a positive number σ. By (4.71), we obtain Z Z σ2 x · ν dS ≤ u2n (x · ν) dS = 0, ∂Ω

∂Ω

from which we have a desired contradiction: Z Z |x|2 0= x · ν dS = ∆( ) dx = n|Ω|. 2 ∂Ω Ω This completes the proof.

92


4.6.4. Radial Symmetry of Solutions on a Ball. Let Ω = B be the open unit ball in Rn . We shall study the nonlinear problem (4.72)

∆u + f (u) = 0 in Ω,

u|∂Ω = 0.

We are interested in positive solutions: u > 0 in Ω. We assume that f : R → R is Lipschitz, but is otherwise arbitrary. Our intention is to prove that u is necessarily radial, i.e., u(x) depends only on r = |x|. This is a remarkable conclusion, since we are making essentially no assumptions on the nonlinearity. The technique we use is due to Gidas, Ni & Nirenberg [20], and is based upon an extension of the maximum principle and the method of moving planes. 4.6.5. Serrin’s Maximum Principle. Lemma 4.39. (Serrin’s Maximum Principle) Let Ω be a bounded domain in Rn . As¯ satisfies sume u ∈ C 2 (Ω) ∆u + a(x)u ≤ 0, u ≥ 0 (6≡ 0) in Ω where a(x) is bounded. If x0 ∈ ∂Ω, u(x0 ) = 0, and Ω satisfies the interior ball condition at x0 , then the normal derivative un (x0 ) < 0. Moreover, u > 0 in Ω. Proof. Set a = a+ + a− , where a+ = max(a, 0), a− = min(a, 0). Thus ∆u + a− (x)u ≤ −a+ u ≤ 0 in Ω and the conclusions follows from the strong maximum principle and the boundary point principle. 4.6.6. Moving Plane Method. Let Ω+ denote the open upper half ball Ω ∩ {xn > 0}, and Ω− the open lower half ball Ω ∩ {xn < 0}. ¯ be a positive solution of (4.72). Then Lemma 4.40. Let u ∈ C 2 (Ω) uxn < 0 in Ω near ∂Ω+ . Proof. Fix any point x0 ∈ ∂Ω+ and let ν = (ν1 , ..., νn ) denote the outer unit normal to ∂Ω+ at x0 . Note that νn > 0. We claim that uxn < 0 in Ω near x0 . We shall give the proof under the assumption f (0) ≥ 0. Then 0 = −∆u − f (u) + f (0) − f (0) Z 1 ∂ ≤ −∆u − f (su(x))ds 0 ∂s ≤ −∆u + cu R1 for c(x) = − 0 f 0 (su(x))ds. According to Lemma 4.39, uν (x0 ) = ∇u(x0 ) · ν < 0. Since ∇u is parallel to ν on ∂Ω and νn > 0, we conclude that uxn (x0 ) < 0, and thus uxn < 0 in Ω near x0 . We now apply the method of moving planes. (i) If 0 ≤ λ ≤ 1, set Pλ ≡ {x ∈ Rn : xn = λ}. (ii) Let xλ ≡ (x1 , . . . , xn−1 , 2λ − xn ) denote the reflection of x in Pλ . (iii) Let Eλ ≡ {x ∈ Ω : λ < xn < 1}.

4.6. Nonexistence and Radial Symmetry

93

¯ be a positive solution of (4.72). Then u is radial, i.e., Theorem 4.41. Let u ∈ C 2 (B) u(x) = v(r)

(r = |x|)

for some strictly decreasing function v : [0, 1] → [0, ∞). Proof. Consider for each 0 ≤ λ < 1 the statement (4.73)

u(x) < u(xλ )

for all

x ∈ Eλ .

According to Lemma 4.40, we see that this statement is valid for each λ < 1, λ sufficiently close to 1. Set λ0 ≡ inf{0 ≤ λ ≤ 1 : (4.73) holds for each λ ≤ µ < 1}. We will show that λ0 = 0. If to the contrary λ0 > 0, set w(x) = u(xλ0 ) − u(x)

(x ∈ Eλ0 ).

Then −∆w = f (u(xλ0 )) − f (u(x)) = −cw in Eλ0 for

1

Z c(x) = −

f 0 (su(xλ0 ) + (1 − s)u(x))dx.

0

As w ≥ 0 in Eλ0 , we deduce from Lemma 4.39 (applied to Eλ0 ) that w > 0 in Eλ0 , wxn > 0 on Pλ0 ∩ Ω. Thus (4.74)

u(x) < u(xλ0 ) in Eλ0

and uxn < 0 on Pλ0 ∩ Ω.

(4.75)

Using (4.75) and Lemma 4.40 we conclude (4.76)

u(xn ) < 0 on Pλ0 −ε ∩ Ω

for all

0 ≤ ε ≤ ε0

if ε0 is sufficiently small. Then (4.74) and the continuity of u imply (4.77)

u(x) < u(xλ0 −ε ) in Eλ0 −ε

for all

0 ≤ ε ≤ ε0

if ε0 is small enough. Assertion (4.77) contradicts our choice of λ0 . Since λ0 = 0, we see that u(x1 , . . . , xn−1 , −xn ) ≥ u(x1 , . . . , xn )

for all

x ∈ Ω+ .

for all

x ∈ Ω+ .

A similar argument in Ω− proves u(x1 , . . . , xn−1 , −xn ) ≤ u(x1 , . . . , xn ) Thus u is symmetric in the plane P0 and uxn = 0 on P0 . This argument applies as well after any rotation of coordinate axes, and so the Theorem follows.

Chapter 5

Weak Lower Semicontinuity on Sobolev Spaces

As discussed before, in many variational problems, weak lower semicontinuity is an essential condition for existence of minimizers using the minimization method. In this chapter, we study the conditions for weak lower semicontinuity of a multiple integral functional I(u) in the Sobolev space W 1,p (Ω; RN ). Assume Z I(u) = F (x, u, Du) dx. Ω

Recall that I is called weakly lower semicontinuous on W 1,p (Ω; RN ) if (5.1)

I(¯ u) ≤ lim inf I(uν ) ν→∞

whenever uν * u ¯ weakly in W 1,p (Ω; RN ).

5.1. The Convex Case 5.1.1. Tonelli’s Theorem. We first prove a semicontinuity result of Tonelli. Theorem 5.1. Let F (x, s, ξ) ≥ 0 be smooth and convex in ξ. Assume F, Fξ are both continuous in (x, s, ξ). Then the functional I(u) defined above is sequentially weakly (weakly star if p = ∞) lower semicontinuous on W 1,p (Ω; RN ) for all 1 ≤ p ≤ ∞. Proof. We need only to prove I(u) is w.l.s.c. on W 1,1 (Ω; RN ). To this end, assume {uν } is a sequence weakly convergent to u in W 1,1 (Ω; RN ). We need to show I(u) ≤ lim inf I(uν ). ν→∞

By the Sobolev embedding theorem it follows that (via a subsequence) uν → u in L1 (Ω; RN ). We can also assume uν (x) → u(x) for almost every x ∈ Ω. Now, for any given δ > 0 we choose a compact set K ⊂ Ω such that (i) uν → u uniformly on K and |Ω \ K| < δ (by Egorov’s theorem); (ii) u, Du are continuous on K (by Lusin’s theorem). 95

96

5. Weak Lower Semicontinuity on Sobolev Spaces

Since F (x, s, ξ) is smooth and convex in ξ, it follows that F (x, s, η) ≥ F (x, s, ξ) + Fξk (x, s, ξ) (ηik − ξik ) i

∀ξ, η ∈ MN ×n .

Therefore, since F ≥ 0, Z I(uν ) ≥

F (x, uν , Duν ) dx K

Z h

i F (x, uν , Du) + Fξk (x, uν , Du) (Di ukν − Di uk ) i ZK Z = F (x, uν , Du) + Fξk (x, u, Du) (Di ukν − Di uk ) i K K Z [Fξk (x, uν , Du) − Fξk (x, ukν , Du)] (Di ukν − Di uk ). + ≥

i

K

i

Since F (x, s, ξ) is uniformly continuous on bounded sets and uν (x) → u(x) uniformly on K we have Z Z lim F (x, uν , Du) dx = F (x, u, Du) dx, ν→∞ K

K

lim kFξk (x, uν , Du) − Fξk (x, ukν , Du)kL∞ (K) = 0.

ν→∞

i

i

Now since Fξk (x, u, Du) is bounded on K and Di ukν converges to Di uk weakly in L1 (Ω) as i ν → ∞, we thus have Z Fξk (x, u, Du) (Di ukν − Di uk ) dx = 0. lim ν→∞ K

i

From these estimates, noting that, for any two sequences {aν }, {bν }, lim inf (aν + bν ) ≥ lim inf aν + lim inf bν ,

(5.2)

ν→∞

ν→∞

ν→∞

we have Z lim inf I(uν ) ≥ ν→∞

F (x, u, Du). K

If F (x, u, Du) ∈ L1 (Ω), i.e., I(u) < ∞, then for any given > 0, we use Lebesgue’s absolute continuity theorem to determine δ > 0 so that Z Z F (x, u, Du) ≥ F (x, u, Du) − , ∀E ⊂ Ω, |Ω \ E| < δ. E

Ω

On the other hand, if I(u) = ∞ then for any given large number M > 0 we choose δ > 0 so that Z F (x, u, Du) dx > M, ∀E ⊂ Ω, |Ω \ E| < δ. E

In any of these two cases, by letting either → 0 or M → ∞, we obtain lim inf I(uν ) ≥ I(u). ν→∞

The theorem is proved.

5.2. Morrey’s Quasiconvexity

97

5.1.2. Existence in the Convex Case. Using the theorem, we obtain the following existence result for convex functionals. Theorem 5.2. In addition to the hypotheses of the previous theorem, assume there exists 1 < p < ∞ such that F (x, s, ξ) ≥ c |ξ|p − C(x), where c > 0, C ∈ L1 (Ω) are given. If for some ϕ ∈ W 1,p (Ω; RN ), I(ϕ) < ∞, then minimization problem inf u∈Dϕ I(u) has a minimizer in the Dirichlet class Dϕ . Proof. This follows from the abstract existence Theorem 4.6 in the previous chapter.

Remark. Both theorems in this section hold for more general functions F (x, s, ξ). For example, we can replace the continuity condition by the Carathéodory condition.

5.2. Morrey’s Quasiconvexity In this section, we will derive a condition which, under the mild general assumption, will be the “right” (necessary and sufficient) condition for the weak lower semicontinuity for integral functionals on Sobolev spaces. This will be Morrey’s quasiconvexity condition; see Morrey [27, 28]. Please be warned that there is at least one other quasiconvexity in the analysis that has a totally different meaning. 5.2.1. Lipschitz Convergence. Note that W 1,∞ (Ω; RN ) can be identified with the space of all Lipschitz maps from Ω to RN . A sequence {uν } converges to u in the weak star topology of W 1,∞ (Ω; RN ) if and only if {uν } converges to u in the sense of Lipschitz convergence; that is, ¯ RN ); 1) uν → u uniformly in C(Ω; 2) the Lipschitz norms of uν and u are bounded. 5.2.2. Quasiconvexity as Necessary Condition. The following result, mainly due to Morrey [27], gives the necessary condition for the lower semicontinuity under the Lipschitz convergence of the multiple integral Z I(u) = F (x, u(x), Du(x)) dx. Ω

¯ × RN × MN ×n . Assume the functional Theorem 5.3. Assume F (x, s, ξ) is continuous on Ω I(u) is s.l.s.c. with respect to Lipschitz convergence on W 1,∞ (Ω; RN ). Then the following condition holds for all x0 ∈ Ω, s0 ∈ RN , ξ0 ∈ MN ×n and all φ ∈ W01,∞ (Ω; RN ) : Z (5.3) F (x0 , s0 , ξ0 ) ≤ − F (x0 , s0 , ξ0 + Dφ(x)) dx. Ω

¯ with center x Proof. Let Q be a fixed open cube containing Ω ¯ and side-length 2L. We prove this theorem by several lemmas. Lemma 5.4. Suppose (5.4)

Z F (x0 , s0 , ξ0 ) ≤ − F (x0 , s0 , ξ0 + Dφ(x)) dx Q

holds for all φ ∈ W01,∞ (Ω; RN ). Then (5.3) holds.

98


Proof. For any φ ∈ W01,∞ (Ω; RN ), we extend φ by zero onto Q; then φ ∈ W01,∞ (Q; RN ). Inserting it into (5.4) yields (5.3). In the following, let x0 ∈ Ω, s0 ∈ RN , ξ0 ∈ MN ×n be given. Define u ˜(x) = s0 + ξ0 · (x − x0 ). Let also φ ∈ W01,∞ (Q; RN ) be given. Assume Q0 ⊂⊂ Ω is an arbitrarily given cube containing x0 with side-length 2l. For any positive integer ν we divide each side of Q0 into 2ν intervals of equal length, each being equal to 2−ν+1 l. This divides Q0 into 2nν small cubes {Qνj } with j = 1, 2, ..., 2nν . Denote the center of each cube Qνj by x ¯νj and define a function uν : Ω → RN as follows: ( nν u ˜(x) if x ∈ Ω \ ∪2j=1 Qνj ; uν (x) = ν −ν ¯ + 2 l L (x − x ¯νj ) if x ∈ Qνj , 1 ≤ j ≤ 2nν . u ˜(x) + 2 L l φ x We easily see that uν → u ˜ uniformly on Ω. Moreover, nν ξ0 if x ∈ Ω \ ∪2j=1 Qνj ; ν Duν (x) = ¯νj ) if x ∈ Qνj , 1 ≤ j ≤ 2nν . ξ0 + Dφ x ¯ + 2 l L (x − x Therefore {Duν } is uniformly bounded and, by definition, {uν } converges to u ˜ as ν → ∞ in the sense of Lipschitz convergence. Note that Z I(˜ u) = F (x, u ˜(x), ξ0 ) dx Ω

and that

Z I(uν ) =

F (x, uν (x), Duν ) dx Z F (x, u ˜, ξ0 ) dx + F (x, uν , Duν ) dx. Ω

Z = Ω\Q0

Q0

Therefore, by the lower semicontinuity of I, we thus have Z Z (5.5) F (x, u ˜, ξ0 ) dx ≤ lim inf F (x, uν , Duν ) dx. ν→∞

Q0

Q0

From the uniform continuity of F (x, s, ξ) on bounded sets and the fact that uν → u ˜ uniformly on Ω we have Z Z (5.6) lim inf F (x, uν , Duν ) dx = lim inf F (x, u ˜, Duν ) dx. ν→∞

ν→∞

Q0

Q0

We now compute Z 2nν Z X F (x, u ˜, Duν ) dx = Q0

j=1 nν

=

2ν L F x, u ˜, ξ0 + Dφ x ¯+ (x − x ¯νj ) dx l Qνj

Z 2 X j=1

2ν L F x ¯νj , u ˜(¯ xνj ), ξ0 + Dφ x ¯+ (x − x ¯νj ) dx + o(1) l Qνj nν

n Z 2 X l = F x ¯νj , u ˜(¯ xνj ), ξ0 + Dφ(y) dy + o(1) ν 2 L Q j=1

nν

(5.7)

=

2 X j=1

F˜ (¯ xνj ) |Qνj | + o(1),


99

where o(1) → 0 as ν → ∞, and Z F˜ (x) = − F (x, u ˜(x), ξ0 + Dφ(y)) dy. Q

Q0

This function is continuous on and the sum in (5.7) is simply the Riemann sum of the integral of F˜ over Q0 . Therefore, we arrive at Z Z F (x, u ˜, Duν ) dx = F˜ (x) dx, lim ν→∞ Q0

Q0

which by (5.6) implies Z

Z Q0

F (x, u ˜(x), ξ0 ) dx ≤

F˜ (x) dx.

Q0

This inequality holds for any cube Q0 ⊂⊂ Ω containing x0 ; therefore, F (x0 , u ˜(x0 ), ξ0 ) ≤ F˜ (x0 ). This is nothing but Z F (x0 , s0 , ξ0 ) ≤ − F (x0 , s0 , ξ0 + Dφ(y)) dy. Q

Finally, the proof of Theorem 5.3 is complete.

Motivated by this theorem, we have the following definition of quasiconvex functions in the sense of Morrey. ¯ is called quasiconvex (in the sense Morrey’s Quasiconvexity. A function F : MN ×n → R of Morrey) if Z (5.8) F (ξ) ≤ − F (ξ + Dφ(x)) dx Ω

holds for all φ ∈

W01,∞ (Ω; RN ).

5.2.3. Quasiconvexity as Sufficient Condition. The following result, due to Meyers, will be useful to relax the zero boundary condition on φ. Theorem 5.5. Let F : MN ×n → R be continuous and quasiconvex. For every bounded set Q ⊂ Rn and every sequence {zν } in W 1,∞ (Q; RN ) converging to zero in the sense of Lipschitz convergence, we have Z F (ξ) ≤ lim inf − F (ξ + Dzν (x)) dx ν→∞

for every ξ ∈

Q

MN ×n .

Proof. Let Qk = {x ∈ Q | dist(x, ∂Q) > 1/k}. Then Qk ⊂⊂ Q and |Q\Qk | → 0 as k → ∞. Choose a cut-off function ζk ∈ C0∞ (Q) such that 0 ≤ ζk ≤ 1, ζk = 1, Mk = kDζk kL∞ < ∞. Qk

Since zν → 0 uniformly on Q we can choose a subsequence {νk } such that kzνk kL∞ ≤ (Mk + 1)−1

∀ k = 1, 2, ...

and we may also assume Z Z lim F (ξ + Dzνk (x)) dx = lim inf F (ξ + Dzν (x)) dx. k→∞ Q

ν→∞

Q

100


Define φk = ζk zνk . Then φk ∈ W01,∞ (Q; RN ) and we can use them as test functions in the definition of quasiconvexity to obtain Z F (ξ + Dφk (x)) dx |Q| F (ξ) ≤ Q

Z

Z

F (ξ + ζk Dzνk + zνk ⊗ Dζk )

F (ξ + Dzνk ) +

=

Q\Qk

Qk

Z F (ξ + Dzνk (x)) dx + k ,

= Q

where

Z [F (ξ + ζk Dzνk + zνk ⊗ Dζk u) − F (ξ + Dzνk (x))] dx.

k = Q\Qk

Since F (ξ) is bounded on bounded sets and |Q \ Qk | → 0 as k → ∞, we easily have k → 0 as k → ∞. Therefore, Z |Q| F (ξ) ≤ lim inf F (ξ + Dzν (x)) dx. ν→∞

Q

This completes the proof.

We now prove the sufficiency of quasiconvexity for the lower semicontinuity of the functional Z I(u) =

F (x, u(x), Du(x)) dx Ω

under the Lipschitz convergence on Ω. ¯ × RN × MN ×n and is quasiconvex in Theorem 5.6. Assume F (x, s, ξ) is continuous on Ω ξ. Then the functional I defined above is s.l.s.c. with respect to Lipschitz convergence on Ω. Proof. Let {zk } be any sequence converging to 0 in the sense of Lipschitz convergence on Ω, and let u ∈ W 1,∞ (Ω; RN ) be any given function. We need to show Z Z (5.9) F (x, u, Du) ≤ lim inf F (x, u + zk , Du + Dzk ). k→∞

Ω

Ω

For any given > 0, we choose finitely many disjoint cubes Qj contained in Ω such that Z I(u) ≤ F (x, u, Du) dx + ∪Qj

and

Z I(u + zk ) ≥

F (x, u + zk , Du + Dzk ) dx − , ∪Qj

for all k = 1, 2, · · · . In what follows, we prove for each cube Q = Qj Z IQ (u) ≡ F (x, u, Du) dx ≤ lim inf IQ (u + zk ). k→∞

Q

This, by (5.2), will certainly imply the conclusion of the theorem. To this end, for each positive integer ν, we divide Q into small cubes {Qνj } with center x ¯νj as in the proof of Theorem 5.3: nν 2[ Q= Qνj ∪ E, |E| = 0. j=1


Define

101

Z (u)νj = − u(x) dx, Qνj

and

Z (Du)νj = − Du(x) dx,

nν

ν

U (x) =

2 X

Qνj

nν

(u)νj

· χQνj (x),

2 X M = (Du)νj · χQνj (x). ν

j=1

j=1

Note that kU ν kL∞ + kM ν kL∞ ≤ kukW 1,∞ and that the sequences {U ν } and {M ν } converge almost everywhere to u and Du on Q as ν → ∞, respectively. We now estimate IQ (u + zk ). Z F (x, u + zk , Du + Dzk ) = ak + bνk + cνk + dν + IQ (u), IQ (u + zk ) = Q

where Z [F (x, u + zk , Du + Dzk ) − F (x, u, Du + Dzk )] dx,

ak = Q

nν

bνk

=

Z 2 X j=1

Qνj

[F (x, u, Du + Dzk ) − F (¯ xνj , (u)νj , (Du)νj + Dzk )] dx,

nν

cνk

=

Z 2 X j=1

Qνj

[F (¯ xνj , (u)νj , (Du)νj + Dzk ) − F (¯ xνj , (u)νj , (Du)νj )] dx,

nν

ν

d

=

Z 2 X j=1

Qνj

[F (¯ xνj , (u)νj , (Du)νj ) − F (x, u, Du)] dx.

By the uniform continuity of F (x, s, ξ) on bounded sets and the pointwise convergence of {U ν } and {M ν } we have lim ak = 0, lim dν = 0 ν→∞

k→∞

and limν→∞ bνk = 0 uniformly with respect to k. We apply Theorem 5.5 to each Qνj to obtain, by (5.2), lim inf cνk ≥ 0 k→∞

for all ν = 1, 2, · · · . Therefore, again by (5.2), lim inf IQ (u + zk ) ≥ IQ (u), k→∞

as desired. The proof is complete.

5.2.4. Weak Lower Semicontinuity on Sobolev Spaces. Quasiconvexity is also the “right” condition for weak lower semicontinuity of integral functionals on W 1,p (Ω; RN ). The most general theorem in this direction is the following theorem due to Acerbi & Fusco [1]. Theorem 5.7. (Acerbi-Fusco’s Theorem) Let F (x, s, ξ) be a Carathéodory function. Assume for some 1 ≤ p < ∞ 0 ≤ F (x, s, ξ) ≤ a(x) + C (|s|p + |ξ|p ), where C > 0 isR a constant and a(x) ≥ 0 is a locally integrable function in Ω. Then functional I(u) = Ω F (x, u, Du) dx is w.s.l.s.c. on W 1,p (Ω; RN ) if and only if F (x, s, ξ) is quasiconvex in ξ.

102


5.2.5. Existence in the General Case. Theorem 5.8 (Existence of minimizers). Let F (x, s, ξ) be Carathéodory and quasiconvex in ξ and satisfy max{0, c |ξ|p − C(x)} ≤ F (x, s, ξ) ≤ a(x) + C (|s|p + |ξ|p ) for some 1 < p < ∞, where c > 0 is a constant and C(x), a(x) are given integrable functions in Ω. Then, for any ϕ ∈ W 1,p (Ω; RN ), the minimization problem Z F (x, u(x), Du(x)) dx min u∈Dϕ

Ω

has a minimizer in the Dirichlet class Dϕ .

5.3. Properties of Quasiconvex Functions 5.3.1. Domain Independence. We prove that quasiconvexity is independent of the domain Ω. ¯ be such that (5.8) holds for all φ ∈ W 1,∞ (Ω; RN ). Then Theorem 5.9. Let F : MN ×n → R 0 for any bounded open set G ⊂ Rn with |∂G| = 0 one has Z (5.10) F (ξ) ≤ − F (ξ + Dψ(y)) dy, ∀ξ ∈ MN ×n G

holds for all ψ ∈

W01,∞ (G; RN ).

Proof. Let G ⊂ Rn be any bounded open set with |∂G| = 0, and ψ ∈ W01,∞ (G; RN ). Assume y¯ ∈ G. For any x ∈ Ω and > 0 let G(x, ) = {z ∈ Rn | z = x + (y − y¯) for some y ∈ G}. Then there exists an x > 0 such that x ∈ G(x, ) ⊂ Ω for all x ∈ Ω and 0 < < x . This means the family {G(x, ) x ∈ Ω, 0 < < x } covers Ω in the sense of Vitali covering. Therefore, there exists a countable disjoint subfamily {G(xj , j )} and a set E of measure zero such that (5.11)

Ω=

∞ [

G(xj , j ) ∪ E.

j=1

We now define a function φ : Ω → RN as follows. ( S 0 if x ∈ ∞ j=1 ∂[G(xj , j )] ∪ E, φ(x) = x−xj j ψ y¯ + j if x ∈ G(xj , j ) for some j. One can verify that φ ∈ W01,∞ (Ω; RN ) and x − xj Dφ(x) = Dψ y¯ + j

∀x ∈ G(xj , j ).

Therefore, from (5.8), it follows that Z F (ξ) |Ω| ≤

F (ξ + Dφ(x)) dx Ω

=

∞ Z X j=1

x − xj F ξ + Dψ y¯ + dx j G(xj ,j )

5.3. Properties of Quasiconvex Functions

=

∞ X

103

Z

nj

F (ξ + Dψ(y)) dy G

j=1

Z |Ω| F (ξ + Dψ(y)) dy, |G| G P n where the last equality follows since, by (5.11), ∞ j=1 j = |Ω|/|G|. We have thus proved (5.10). =

In the following, let Σ be the unit cube in Rn ; that is Σ = {x ∈ Rn | 0 < xα < 1, α = 1, 2, · · · , n}. Note that |Σ| = 1. Let f : Rn → RN be any given function. We say f is Σ-periodic if f (· · · , xα , · · · ) is 1-periodic in xα for all α = 1, 2, · · · , n. Quasiconvexity can be also characterized by the following condition. Theorem 5.10. Let F : MN ×n → R be continuous. Then F is quasiconvex if and only if Z (5.12) F (ξ) ≤ F (ξ + Dφ(x)) dx ∀ξ ∈ MN ×n Σ

for all Σ-periodic Lipschitz functions φ ∈ W 1,∞ (Rn ; RN ). Proof. Since any function ψ ∈ W01,∞ (Σ; RN ) can be extended as a Σ-periodic function on Rn , we easily see that (5.12) implies (5.10) for G = Σ thus the quasiconvexity of F . We have only to prove (5.12) holds if F is quasiconvex. Let φ ∈ W 1,∞ (Rn ; RN ) be a Σ-periodic function. Define 1 φj (x) = φ(jx) j for all j = 1, 2, · · · . It is easily seen that φj → 0 in the sense of Lipschitz convergence on W 1,∞ (Σ; RN ). Therefore the theorem of Meyers, Theorem 5.5, and the quasiconvexity of F implies Z F (ξ) ≤ lim inf j→∞

Note that

F (ξ + Dφj (x)) dx. Σ

Z Σ

Z F (ξ + Dφj (x)) dx = F (ξ + Dφ(jx)) dx Σ Z = j −n F (ξ + Dφ(y)) dy jΣ

and that, besides a set of measure zero, n

jΣ =

j [

(¯ xν + Σ),

ν=1

where x ¯ν are the left-lower corner points of the subcubes obtained by dividing the sides of jΣ into j-equal subintervals. Since Dφ(x) is Σ-periodic, we thus have Z Z jn Z X n F (ξ + Dφ(y))dy = F (ξ + Dφ(y))dy = j F (ξ + Dφ(x))dx, jΣ

ν=1

x ¯ν +Σ

and therefore

Σ

Z F (ξ) ≤

F (ξ + Dφ(x)) dx Σ

as needed; the proof is complete.

104


5.3.2. Convexity vs Quasiconvexity. Lemma 5.11 (Jensen’s inequality). Let (E, µ) be a measure space with total mass µ(E) = 1 and let h : E → RL be an integrable function on E. If G : RL → R is a convex function, then Z Z G h(x) dµ ≤ G(h(x)) dµ. E

E

Proof. Let F = E h(x) dµ. Since G : RL → R is convex, there exists lF ∈ RL (note that lF = DG(F ) for almost every F ) such that R

G(A) ≥ G(F ) + lF · (A − F )

∀A ∈ RL .

Then G(h(x)) ≥ G(F ) + lF · (h(x) − F ) for all x ∈ E, and integrating over x ∈ E yields Z Z G(h(x)) dµ ≥ G(F ) + lF · (h(x) − F ) dµ = G(F ), E

E

which proves Jensen’s inequality.

Theorem 5.12. If F : MN ×n → R is convex, then F is quasiconvex. Proof. This follows easily from Jensen’s inequality and the divergence theorem given above. ¯ is rank5.3.3. Quasiconvexity vs Rank-one Convexity. Recall that F : MN ×n → R N ×n N n one convex if for any ξ ∈ M , q ∈ R , p ∈ R the function f (t) = F (ξ + tq ⊗ p) is a convex function of t ∈ R. We have the following result. Theorem 5.13. Every finite-valued quasiconvex function is rank-one convex. Proof. Let F be quasiconvex. We need to show that for any ξ ∈ MN ×n , q ∈ RN , p ∈ Rn the function f (t) = F (ξ + tq ⊗ p) is a convex function of t ∈ R; that is, for all 0 < θ < 1 and t, s ∈ R, f (θt + (1 − θ)s) ≤ θf (t) + (1 − θ)f (s).

(5.13) which is equivalent to

˜ ≤ θF (ξ˜ + aq ⊗ p) + (1 − θ)F (ξ˜ + bq ⊗ p), F (ξ)

(5.14) where

ξ˜ = ξ + [θt + (1 − θ)s]q ⊗ p, a = (1 − θ)(t − s), b = θ(s − t). Assume t > s; so a > 0 and b < 0. Let ζ(τ ) be the periodic Lipschitz function of peroid 1 on R satisfying ζ(τ ) = aτ for 0 ≤ τ ≤ θ and ζ(τ ) = b(τ − 1) for θ ≤ τ ≤ 1. Let G be a cube of unit volume which is bounded between two planes {x · p = 0} and {x · p = 1}. For x ∈ Rn , define |x|∞ = max{|xi | | 1 ≤ i ≤ n}, Then δ

∈ W01,∞ (G) and ∇δ(x) uk (x) = k −1 ζ(kx · p) q,

δ(x) = dist∞ (x, ∂G) = inf{|x − y|∞ | y ∈ ∂G}.

∈ {±ei | i = 1, 2, · · · , n}. Define functions by φk (x) = min{k −1 ζ(kx · p), δ(x)} q,

Note that Duk (x) ∈ {aq ⊗ p, bq ⊗ p}, φk ∈ W01,∞ (G; RN ) and lim |{x ∈ G | Duk (x) 6= Dφk (x)}| = 0.

k→∞

k = 1, 2, · · · .


105

Hence, by (5.8) and noting that Duk (x) and Dφk (x) take only on finitely many values, Z ˜ F (ξ˜ + Dφk (x)) dx F (ξ) ≤ lim k→∞ G Z F (ξ˜ + Duk (x)) dx = lim k→∞ G

= θF (ξ˜ + aq ⊗ p) + (1 − θ)F (ξ˜ + bq ⊗ p). This completes the proof.

Remark. Another proof in the case of continuous F is as follows. Let F be quasiconvex. If F is of class C 2 , then Z F (ξ + tDφ(x)) dx f (t) = Ω

takes its minimum at t = 0. Therefore f 00 (0) ≥ 0; that is, Z Fξk ξl (ξ) Di φk (x)Dj φl (x) dx ≥ 0 Ω

i j

for all φ ∈ C0∞ (Ω; RN ). This implies the weak Legendre-Hadamard condition: Fξk ξl (ξ) q k q l pi pj ≥ 0 ∀ q ∈ RN , p ∈ Rn , i j

which is equivalent to that F is rank-one convex. For a continuous F , let F = F ∗ ρ be the regularization of F . Then F is of class C ∞ and can be shown to be quasiconvex, and hence F satisfies (5.14) for all > 0; letting → 0 yields that F satisfies (5.14) and thus is rank-one convex. Lemma 5.14. Let n ≥ 2 and let A, B ∈ MN ×n be such that rank(A − B) = 1. Define ( 0 ξ ∈ {A, B}, F (ξ) = ∞ ξ∈ / {A, B}. Then F is quasiconvex convex, but not rank-one convex. Proof. The rank-one convexity of F would imply 0 ≤ F (λA + (1 − λ)B) ≤ λF (A) + (1 − λ)F (B) = 0 for all 0 < λ < 1. Hence F is not rank-one convex. To see F is quasiconvex, given ξ ∈ MN ×n and φ ∈ W01,∞ (Σ; RN ), we need to show Z (5.15) F (ξ) ≤ F (ξ + Dφ(x)) dx. Σ

Since the integral on the right-hand side takes only two values of {0, ∞}, we only need to prove the inequality when Z F (ξ + Dφ(x)) dx = 0. Σ

In this case, we must have ξ + Dφ(x) ∈ {A, B} for almost every x ∈ Σ. We claim, in this case, one must have ξ ∈ {A, B} and hence (5.15) is valid. To prove this, let ξ + Dφ(x) = χE (x)A + (1 − χE (x))B, where E = {x ∈ Σ | ξ + Dφ(x) = A}. Integrating this over Σ yields ξ = λA + (1 − λ)B, where λ = |E| ∈ [0, 1]. Hence Dφ(x) ∈ {A − ξ, B − ξ} = {(1 − λ)(A − B), λ(A − B)} S n−1 .

S n−1

a.e.

x ∈ Σ.

Let A − B = q ⊗ p with p ∈ Let e ∈ be a vector perpendicular to p. Then Dφ(x) e = 0 for almost every x ∈ Σ. This implies φ(x + te) is independent of t as long as

106


¯ hence, choosing t such that x + te ∈ ∂Σ, we have φ(x) = 0 for all x ∈ Σ. Hence, x + te ∈ Σ; ξ ∈ {A, B}, as claimed. Theorem 5.15. Every finite-valued quasiconvex function is locally Lipschitz continuous. Proof. Let F : MN ×n → R be quasiconvex. Then f (t) = F (ξ + tq ⊗ p) is a finite convex function of t ∈ R; hence f is locally Lipschitz continuous. From this the theorem follows. Lemma 5.16. If F is a polynomial of degree two (quadratic polynomial) then F is quasiconvex if and only if F is rank-one convex. Proof. Assume F is a rank-one convex quadratic polynomial. We show F is quasiconvex. Since subtraction of an affine function from a function does not change the quasiconvexity or rank-one convexity, we thus assume F is a homogeneous quadratic polynomial given by k l F (ξ) = Akl ij ξi ξj

(summation notation is used here and below)

with Akl ij are constants. Note that the rank-one convexity is equivalent to the weak LegendreHadamard condition: n X N X k l Akl ij q q pi pj ≥ 0. i,j=1 k,l=1

Also, k l kl l k kl k l F (ξ + η) = F (ξ) + Akl ij ξi ηj + Aij ξj ηi + Aij ηi ηj .

(5.16)

αβ αβ satisfies the LegendreHence the linear system defined by constants A˜αβ ij = Aij + δij δ Hadamard condition, where > 0 and δij , δ αβ are the usual delta notation. Using the Fourier transform as before, we can show that Z k l Akl ij Di φ (x)Dj φ (x) dx ≥ 0 Rn

for all φ ∈ C0∞ (Rn ; RN ). Hence by (5.16) Z Z F (ξ + Dφ(x)) dx = F (ξ) + Rn

Σ

k l Akl ij Di φ (x)Dj φ (x) dx ≥ F (ξ),

holds for all φ ∈ C0∞ (Σ; RN ). This proves the quasiconvexity of F.

Using the definition of null-Lagrangians given later, we have the following. Lemma 5.17. A rank-one convex third degree polynomial must be a null-Lagrangian and thus quasiconvex. Proof. Let F be a rank-one convex third degree polynomial. Then the polynomial f (t) = F (ξ + tq ⊗ p) is convex and of degree ≤ 3 in t, and hence the degree of f (t) cannot be 3. Note that the coefficient of t2 term in f is half of (5.17)

Fξk ξl (ξ) pi pj q k q l ≥ 0, i j

which holds for all ξ, p, q. Since Fξk ξl (ξ) is linear in ξ, condition (5.17) implies i j

Fξk ξl (ξ) pi pj q k q l ≡ 0. i j

Therefore f (t) = F (ξ +tq⊗p) is affine in t and hence F is rank-one affine. Consequently, the result follows from the fact that a rank-one affine function must be a null-Lagrangian.


107

ˇ ak settles a long-standing open ˇ 5.3.4. Sver´ ak’s Example. The following example of Sver´ problem raised by Morrey. ˇ ak [36]). If n ≥ 2, N ≥ 3 then there exists a rank-one convex function Theorem 5.18 (Sver´ N ×n F: M → R which is not quasiconvex. Proof. We only prove the theorem for n = 2, N = 3. Consider the periodic function u : R2 → R3 by 1 u(x) = (sin 2πx1 , sin 2πx2 , sin 2π(x1 + x2 )). 2π Then   cos 2πx1 0 . 0 cos 2πx2 Du(x) =  cos 2π(x1 + x2 ) cos 2π(x1 + x2 ) Let L be the linear subspace of M = M3×2 ; that is,     r 0   L = [r, s, t] ≡  0 s  r, s, t ∈ R .   t t Note that a matrix ξ = [r, s, t] ∈ L is of rank ≤ 1 if and only if at most one of {r, s, t} is nonzero. Define g : L → R by g([r, s, t]) = −rst. Using formula 2 cos α · cos β = cos(α + β) + cos(α − β), we easily obtain 1 1 g(Du(x)) = − − sin 4π(x1 + x2 ) + cos 4πx1 + cos 4πx2 4 4 and hence by a direct computation that Z Z 1Z 1 1 (5.18) g(Du(x)) dx = g(Du(x)) dx1 dx2 = − . 4 Σ 0 0 We now extend g to the whole M. Let P : M → L be the orthogonal projection onto L and consider the fourth degree polynomials: F,k (ξ) = g(P ξ) + (|ξ|2 + |ξ|4 ) + k |ξ − P ξ|2 .

(5.19)

Lemma 5.19. For each > 0 there exists a k = k > 0 such that F,k is rank-one convex. Proof. We use contradiction method. Suppose otherwise; then there exists an 0 > 0 such that F0 ,k is not rank-one for each k > 0. Hence there exist sequences ξk ∈ M, pk ∈ R2 , qk ∈ R3 with |pk | = |qk | = 1 such that (5.20)

∂ 2 F0 ,k (ξk ) ∂ξiα ∂ξjβ

qkα qkβ pki pkj ≡ D2 F0 ,k (ξk ) [qk ⊗ pk , qk ⊗ pk ] ≤ 0,

∀ k = 1, 2, · · · .

Computing f 00 (0) for f (t) = F,k (ξ + tη) yields D2 F,k (ξ) [η, η] = f 00 (0) = D2 g(P ξ) [P η, P η] + 2 |η|2 + (4|ξ|2 |η|2 + 8(ξ · η)2 ) + 2k |η − P η|2 . The term D2 g(P ξ) is linear in ξ; the third term is quadratic and positive definite in ξ if η 6= 0 (this is the reason the |ξ|4 -term is needed for F,k ). Using this formula with ξ = ξk and η = qk ⊗ pk and from (5.20), we deduce {ξk } is bounded as k → ∞. Assume, via subsequence, ¯ qk → q¯, pk → p¯ as k → ∞. ξk → ξ,

108


Since D2 F,j (ξ) [η, η] ≤ D2 F,k (ξ) [η, η] for all k ≥ j, we deduce (5.21)

¯ [P (¯ D2 g(P ξ) q ⊗ p¯), P (¯ q ⊗ p¯)] + 20 + 2j |P (¯ q ⊗ p¯) − q¯ ⊗ p¯ | ≤ 0

for all j = 1, 2, · · · . Hence P (¯ q ⊗ p¯) = q¯ ⊗ p¯; that is, q¯ ⊗ p¯ ∈ L. This implies q¯ ⊗ p¯ = [a, b, c], where at most one of a, b, c is nonzero. Therefore, function t 7→ g(P (ξ¯ + t¯ q ⊗ p¯)) = g(P ξ¯ + t¯ q ⊗ p¯) is affine in t, and hence the first term in (5.21) vanishes. This yields the desired contradiction 0 ≤ 0. The lemma is proved. ˇ ak’s theorem. Let u be the periodic function above. We now complete the proof of Sver´ We choose > 0 small enough such that Z 1 (|Du(x)|2 + |Du(x)|4 ) dx < . 4 Σ Let F (ξ) = F,k (ξ) be a rank-one function determined by the previous lemma. Since Du(x) ∈ L, it follows by (5.18) that Z Z Z F (Du(x)) dx = g(Du) + (|Du|2 + |Du|4 ) < 0 = F (0). Σ

Σ

Σ

This shows that F is not quasiconvex by Theorem 5.10 above. The theorem is now proved.

5.4. Polyconvex Functions and Null-Lagrangians Unlike the convexity and rank-one convexity, quasiconvexity is a global property since the inequality (5.8) is required to hold for all test functions. It is thus generally impossible to verify whether a given function F (ξ) is quasiconvex. We have already seen that every convex function is quasiconvex. However, there is a class of functions which are quasiconvex but not necessarily convex. This class, mainly due to Morrey, has been called the polyconvex functions in J. M. Ball [4]. In order to introduce the polyconvex functions, we need some notation. 5.4.1. Determinant and Adjugate Matrix. Let ξ be a n×n square matrix. We denote by det ξ and adj ξ the determinant and adjugate matrix of ξ, respectively, which satisfy the following relation: ξ (adj ξ) = (adj ξ) ξ = (det ξ) I, where I is the n × n identity matrix. From this relation, we have ∂ det ξ = (adj ξ)T , ∂ξ

(5.22)

where η T is the transpose matrix of η. ¯ Rn ) it follows that div(adj Du(x))T = 0; that is, Lemma 5.20. For all u ∈ C 2 (Ω; n X i=1

Di [(adj Du(x))ik ] = 0

(k = 1, 2, · · · , n).

5.4. Polyconvex Functions and Null-Lagrangians

109

Proof. This result follows from a general result on null-Lagrangians later, but we present a direct proof here. Note that by the identity above, we have (det ξ)I = ξ T (adj ξ)T . For ξ = Du(x) this implies (det Du(x)) δij =

n X

Di (uk (x)) (adj Du(x))jk ,

i, j = 1, 2, · · · , n.

k=1

Differentiating this identity with respect to xj and summing over j = 1, 2, · · · , n, we have n X

k δij (adj Du)m k Dj (Dm u )

j,k,m=1

=

n X

(Dj Di uk ) (adj Du)jk + (Di uk ) Dj [(adj Du)jk ]

k,j=1

for i = 1, 2, · · · , n. This identity simplifies to read   n n X X (Di uk )  Dj [(adj Du)jk ] = 0 k=1

(i = 1, 2, · · · , n).

j=1

In short, this can be written as Du(x)T [div(adj Du(x))T ] = 0,

(5.23)

x ∈ Ω.

Now, if det Du(x0 ) 6= 0 then by (5.23), div(adj Du(x0 ))T = 0. If instead det Du(x0 ) = 0, we choose a sequence ν → 0 such that det(Du(x0 ) + ν I) 6= 0 for all ν. Use (5.23) with u ˜ = u + ν x we have det D˜ u(x0 ) 6= 0 and hence 0 = div(adj D˜ u(x0 ))T = div(adj(Du(x0 ) + ν I))T = 0 for all ν → 0. Hence div(adj(Du(x0 )))T = 0.

5.4.2. Subdeterminants. Let σ = min{n, N }. Given an integer k ∈ [1, σ], for any two ordered sequences of integers 1 ≤ i1 < i2 < · · · < ik ≤ N,

1 ≤ α1 < α2 < · · · < αk ≤ n, i

k Jαi11iα2 2···i ···αk (ξ)

be the determinant of the k × k matrix whose (q, p) position element is ξαqp let for all 1 ≤ p, q ≤ k. Note that, by the usual notation, iq ∂(ui1 , ui2 , · · · , uik ) ∂u i1 i2 ···ik Jα1 α2 ···αk (Du(x)) = = det . ∂(xα1 , xα2 , · · · , xαk ) ∂xαp k Let J (ξ) be the collection of all Jαi11iα2 2···i ···αk (ξ) for all k ∈ [1, σ] with the same ordered integral sequences {iq }, {αp } for all ξ. In this way, J defines a function from MN ×n to RL , where σ X N n L = L(n, N ) = . k k

k=1

Theorem 5.21. Let J (ξ) be defined as above, and let Σ be the unit cube in ⊂ Rn . Then it follows that Z J ξ + Dφ(x) dx = J (ξ) Σ

for all φ ∈

C0∞ (Σ; RN )

and ξ ∈ MN ×n .

Proof. Since each J (ξ) is given by a k × k-determinant, without loss of generality, we 12···k (ξ), where 1 ≤ k ≤ σ = min{n, N }. For only prove this identity for J (ξ) = Jk (ξ) = J12···k simplicity, set u(x) = ξx + φ(x). Let x0 = (x1 , · · · , xk ),

x00 = (xk+1 , · · · , xn )

if k + 1 ≤ n.

110


Let Σ0 , Σ00 be the unit cubes in x0 , x00 variables, respectively. Fix x00 ∈ Σ00 , for t ≥ 0, consider maps Vt , Ut : Σ0 → Rk such that Vti (x0 ) = txi + (ξx)i ,

Uti (x0 ) = txi + ui (x0 , x00 ).

We can choose t > 0 sufficiently large so that Vt , Ut are both diffeomorphisms on Σ0 , and therefore Z Z Z Z 0 0 0 0 det(DUt (x )) dx = dy = dy = det(DVt (x0 )) dx0 . Σ0

Ut (Σ0 )

Vt (Σ0 )

Σ0

Since both sides are polynomials of t of degree k, it follows that this equality holds for all t. When t = 0 this implies Z Z 0 00 0 (5.24) Jk (ξ + Dφ(x , x )) dx = Jk (ξ) dx0 . Σ0

Σ0

Integrating (5.24) over x00 ∈ Σ00 we deduce Z (5.25) Jk (ξ + Dφ(x)) dx = Jk (ξ), Σ

completing the proof.

5.4.3. Polyconvex Functions. A function F (ξ) on MN ×n is called a polyconvex func¯ such that F (ξ) = G(J (ξ)) for all tion if there exists a convex function G : RL → R N ×n N ×n ξ∈M ; that is, F = G ◦ J on M . Remark. For a polyconvex function we may have different convex functions in its representation. For example, let n = N = 2 and F (ξ) = |ξ|2 − det ξ. In this case, consider J (ξ) = (ξ, det ξ) ∈ R5 . Then we have F (ξ) = G1 (J (ξ)),

F (ξ) = G2 (J (ξ)),

where G1 (ξ, t) = |ξ|2 − t,

G2 (ξ, t) = (ξ11 − ξ22 )2 + (ξ12 + ξ21 )2 + t

are both convex functions of (ξ, t). Theorem 5.22. A polyconvex function is quasiconvex. Proof. Let F : MN ×n → R be a polyconvex function. Then there exists a convex function G : RL → R such that F (ξ) = G(J (ξ)) for all ξ. Given ξ ∈ MN ×n and φ ∈ C0∞ (Σ; RN ), let h(x) = J (ξ + Dφ(x)). Then Jensen’s inequality implies Z Z G h(x) dx ≤ G(h(x)) dx. Σ

Σ

By the theorem above, the left-hand side is G(J (ξ)) = F (ξ) and therefore Z Z Z F (ξ) ≤ G(h(x)) dx = G(J (ξ + Dφ(x))) dx = F (ξ + Dφ(x)) dx, Σ

proving that F is quasiconvex.

Σ

Σ


111

5.4.4. Null-Lagrangians. A smooth function F : Ω × RN × MN ×n → R is called a nullLagrangian if the system of Euler-Lagrange equations (4.4) for the energy Z I(u) = F (x, u, Du) dx Ω

¯ RN ), the equation is satisfied by all smooth functions u : Ω → RN ; that is, for all u ∈ C 2 (Ω; n X ∂ ¯ F k (x, u(x), Du(x)) = Fsk (x, u(x), Du(x)), ∀ x ∈ Ω, (5.26) ∂xi ξi i=1

holds for all k = 1, 2, · · · , N. Example 5.23. Let f : Rn → R be C 1 , and let F (s, ξ) = f (s) det ξ for s ∈ Rn , ξ ∈ Mn×n . Then, by (5.22), Fξk (s, ξ) = f (s) (adj ξ)ik .

Fsk (s, ξ) = fsk (s) det ξ,

i

¯ Rn ), by Lemma 5.20 above, we have Hence for any u ∈ C 2 (Ω; n n X X ∂ ∂ Fξk (u(x), Du(x)) = [f (u(x)) (adj Du(x))ik ] ∂xi i ∂xi i=1

i=1 n X

=

=

fsj (u(x)) ujxi (adj Du(x))ik

i,j=1 n X

fsj (u(x)) (det Du(x))δjk

j=1

= fsk (u(x)) (det Du(x)) for all k = 1, 2, · · · , n. Hence F (s, ξ) = f (s) det ξ is a null-Lagrangian. We now prove the following boundary dependence property for null-Lagrangians. R Theorem 5.24. Let I(u) = Ω F (x, u, Du) dx and F be a null-Lagrangian. Assume u, v ∈ ¯ RN ) satisfy u(x) = v(x) for all x ∈ ∂Ω. Then I(u) = I(v). Moreover, if F satisfies C 2 (Ω; |F (x, s, ξ)| ≤ c(|s|p + |ξ|p ) + C(x) with c > 0 and C ∈ L1 (Ω), then I(u) = I(v) for all u, v ∈ W 1,p (Ω; RN ) with u − v ∈ W01,p (Ω; RN ). Proof. Define h(t) = I(U (t, ·)) for t ∈ [0, 1], where U (t, x) = tu(x) + (1 − t)v(x). Then Z h0 (t) = Fξk (x, U (t, x), DU (t, x))Di (uk − v k ) + Fsk (x, U (t, x), DU (t, x))(uk − v k ) dx. Ω

i

¯ RN ) and u − v = 0 on ∂Ω, using the divergence theorem and Since u − v, U (t, ·) ∈ C 2 (Ω; (5.26) with u = U (t, ·) we have h0 (t) = 0 for all t ∈ (0, 1) and hence h(0) = h(1) and the first part of the result follows. The second part follows by the continuity of I on W 1,p (Ω; RN ) under the given assumption. We now study the null-Lagrangians depending only on variable in MN ×n . Theorem 5.25. Let F : MN ×n → R be continuous and let Z IΩ (u) = F (Du(x)) dx, Ω

where Ω is any smooth bounded domain in Rn . Then the following conditions are equivalent: (1) F is of C 2 and is a null-Lagrangian;

112


+ Dϕ(x)) dx for all ξ ∈ MN ×n and ϕ ∈ C0∞ (Σ; RN ); ¯ RN ) and φ ∈ C ∞ (Ω; RN ); (3) IΩ (u) = IΩ (u + φ) for all u ∈ C 1 (Ω; 0 (2) F (ξ) =

R

Σ F (ξ

(4) IΩ is continuous with respect to the Lipschitz convergence on W 1,∞ (Ω; RN ); (5) there exists a linear function L : RL → R such that F (ξ) = L(J (ξ)) for all ξ ∈ MN ×n . Proof. We will not show the equivalence of (5) to other conditions, but it is important to know that null-Lagrangians can only be the linear combination of subdeterminants. We prove other equivalent conditions. It is easy to see that (3) implies (2). Note that (2) is equivalent to that both F and −F are quasiconvex; the latter is equivalent to that both IΩ and −IΩ are lower semicontinuous with respect to the Lipschitz convergence on W 1,∞ (Ω; RN ). Therefore, (4) is equivalent to (2). It remains to show (2) implies (1) implies (3). Let us first show that (1) implies (3). ¯ RN ) and φ ∈ C ∞ (Σ; RN ), let To this end, given u ∈ C 2 (Σ; 0 Z f (t) = F (Du(x) + tDφ(x)) dx. Σ

Then, by (1), f 0 (t) =

Z Σ

Fξαi (Du(x) + tDφ(x)) Dα φi (x) dx = 0.

Hence f is a constant function; hence f (0) = f (1), which proves (3). The proof of that (2) implies (1) will follow from several lemmas proved below. Lemma 5.26. If F is of C 2 , then (2) implies (1). Proof. Note that (2) implies Z F (ξ + Dφ(x)) dx = F (ξ) |Ω|

∀ φ ∈ C0∞ (Ω; RN ).

Ω

Since F is of C 2 , this implies that, for any φ, ψ ∈ C0∞ (Ω; RN ), the function Z f (t) = F (t Dφ(x) + Dψ(x)) dx Ω

is constant and of

C 2.

f 0 (0)

Therefore, = 0, which gives Z Fξk (Dψ(x)) Di φk (x) dx = 0. Ω

i

¯ RN ), we can select a sequence {ψν } in C ∞ (Ω; RN ) such that ψν → u Now given u ∈ C 2 (Ω; 0 2 N in C (supp φ; R ). Using the identity above with ψ = ψν and letting ν → ∞ yield Z Fξk (Du(x)) Di φk (x) dx = 0, Ω

i

which, by the divergence theorem, shows the Euler-Lagrange equation for I is satisfied by ¯ RN ), and hence F is a null-lagrangian. u ∈ C 2 (Ω; We say F is rank-one affine if F (ξ + tq ⊗ p) is affine in t for all ξ ∈ MN ×n , q ∈ RN , p ∈ Rn . Lemma 5.27. If F satisfies (2) then F is rank-one affine.


113

Proof. Since (2) is equivalent to that F and −F are rank-one convex, which is equivalent to that F is rank-one affine. We need some notation. Let µiα = ei ⊗ eα , where {ei } and {eα } are the standard bases of RN and Rn , respectively. For each 1 ≤ k ≤ σ = min{n, N } and 1 ≤ i1 , · · · , ik ≤ N, 1 ≤ α1 , · · · , αk ≤ n, we define, inductively, Fαi11 (ξ) = F (ξ + µiα11 ) − F (ξ), i ···i

i ···i

1 1 k−1 k−1 ik k Fαi11···i ···αk (ξ) = Fα1 ···αk−1 (ξ + µαk ) − Fα1 ···αk−1 (ξ).

Note that if F is a polynomial it follows k i1 ik k Fαi11···i ···αk (ξ) = ∂ F (ξ)/∂ξα1 · · · ∂ξαk .

Indeed, we have the same permutation invariance property. Lemma 5.28. Let {10 , · · · , k 0 } be any permutation of {1, · · · , k}. Then i 0 ···i

1 k k Fαi11···i ···αk (ξ) = Fα10 ···αk0 (ξ). 0

Proof. We use induction on k. Assume this is true for all k ≤ s − 1. Let {10 , 20 , · · · , s0 } be a permutation of {1, 2, · · · , s}. We need to show i 0 ···i

s 1 s Fαi11···i ···αs (ξ) = Fα10 ···αs0 (ξ).

(5.27)

0

By definition i ···i

i ···i

1 s−1 1 s−1 is s Fαi11···i ···αs (ξ) = Fα1 ···αs−1 (ξ + µαs ) − Fα1 ···αs−1 (ξ).

By induction assumption, (5.27) holds if s0 = s. We thus assume s0 < s. In this case, by induction assumption, i ···î 0 ···is0 (ξ) s0 ···αs0

i ···î 0 ···is0 (ξ s0 ···αs0

1 s s Fαi11···i ···αs (ξ) = Fα1 ···α ˆ

+ µiαss ) − Fα11 ···αˆs

(where the m ˆ means omitting m) i ···î 0 ··· (ξ s0 ···

= Fα11 ···αˆs

i

0

i

0

i

0

i ···î 0 ··· (ξ s0 ···

+ µiαss + µαss0 ) − Fα11 ···αˆs

i ···î 0 ··· (ξ s0 ···

+ µαss0 ) + Fα11 ···αˆs

i ···î 0 ···is (ξ s0 ···αs

+ µαss0 ) − Fα11 ···αˆs

−Fα11 ···αˆs = Fα11 ···αˆs

+ µiαss )

i ···î 0 ··· (ξ) s0 ··· i ···î 0 ···is (ξ), s0 ···αs

which, by induction assumption, equals i 0 ···i

0

i

0

i 0 ···i

0

i 0 ···i

0

(ξ + µαss0 ) − Fα110 ···α(s−1) (ξ) = Fα110 ···αss0 (ξ). Fα110 ···α(s−1) (s−1)0 (s−1)0 This proves the induction procedure and hence the lemma.

k Lemma 5.29. If F is rank-one affine, then all Fαi11···i ···αk are also rank-one affine. Moreover,

(5.28)

j i1 ···ik j i1 ···ik k Fαi11···i ···αk (ξ + t µβ ) = Fα1 ···αk (ξ) + t Fα1 ···αk β (ξ).

Proof. Use induction again on k. Lemma 5.30. If F satisfies (2) then F (ξ) is a polynomial in ξ.

114


Proof. If F satisfies (2) then F is rank-one affine. Write ξ=

N X n X

ξαi µiα .

i=1 α=1

Then a successive use of the previous lemma shows that F (ξ) is a polynomial of degree at k most nN in ξ with coefficients determined by Fαi11···i ···αk (0). Remark. The function F (ξ) = (tr ξ)2 − tr ξ 2 is a null-Lagrangian on Mn×n . To see this, note n n n n P P P P ξik ξki = Pik (ξ), where Pik (ξ) = ξii ξkk − ξki ξik . For each that F (ξ) = ( ξii )( ξkk ) − i=1

i,k=1

k=1

i,k=1

pair (i, k), i 6= k, Pik (ξ) is a 2 × 2 subdeterminant of ξ. Hence F (ξ) is a null-Lagrangian. 5.4.5. Compensated Compactness of Null-Lagrangians. We prove a compensated compactness property of the null-Lagrangians. For nonlinear operators with similar properties we refer to Coifman et al [12]. Theorem 5.31. Let Jk (Du) be any k × k subdeterminant. Let {uj } be any sequence weakly convergent to u ¯ in W 1,k (Ω; RN ) as j → ∞. Then Jk (Duj ) → Jk (D¯ u) in the sense of distribution in Ω; that is, Z Z lim Jk (Duj (x)) φ(x) dx = Jk (D¯ u(x)) φ(x) dx j→∞ Ω

for all φ ∈

Ω

C0∞ (Ω).

Proof. We prove this by induction on k. Obviously, the theorem is true when k = 1. Assume it holds for Js with s ≤ k − 1. We need to show it also holds for s = k. Without loss of generality, we may assume Jk (Du(x)) =

∂(u1 , u2 , · · · , uk ) . ∂(x1 , x2 , · · · , xk )

For any smooth function u, we observe that Jk (Du) is actually a divergence: k ν+1 ∂(u2 , · · · , uk ) X ∂ 1 (−1) (5.29) Jk (Du(x)) = u , ∂xν ∂(x1 , · · · , xˆν , · · · , xk ) ν=1

where xˆν , again, means deleting xν . Let (ν)

Jk−1 (Du(x)) =

(−1)ν+1 ∂(u2 , · · · , uk ) . ∂(x1 , · · · , xˆν , · · · , xk )

Then (5.29) implies Z k Z X (ν) (5.30) Jk (Du(x)) φ(x) dx = u1 (x) Jk−1 (Du(x)) Dν φ(x) dx. Ω

ν=1

Ω

By density argument, this identity still holds if u ∈ W 1,k (Ω; RN ). Suppose uj * u ¯ in W 1,k (Ω; RN ). By the Sobolev embedding theorem, uj → u ¯ in Lk (Ω; RN ). Moreover, by the (ν) (ν) induction assumption, Jk−1 (Duj ) → Jk−1 (D¯ u) in the sense of distribution. Note that since k

(ν)

k

sequence {Jk−1 (Duj )} is also bounded in L k−1 (Ω) it also weakly converges in L k−1 (Ω); by (ν)

density the weak limit must be Jk−1 (D¯ u). We can then apply (5.30) to conclude Z Z lim Jk (Duj (x)) φ(x) dx = Jk (D¯ u(x)) φ(x) dx, j→∞ Ω

Ω


115

as desired. The proof is complete.

Corollary 5.32. Let Jk (Du) be any k × k subdeterminant and p > k be a number. Let {uj } be any sequence weakly convergent to u ¯ in W 1,p (Ω; RN ) as j → ∞. Then Jk (Duj ) * Jk (D¯ u) p k weakly in L (Ω). p Proof. Note that fj = Jk (Duj ) is bounded in L k (Ω) and hence, via a subsequence, fj * f¯ p for some f¯ weakly in L k (Ω), which also implies fj → f¯ in distribution. Hence f¯ ≡ Jk (Du) p and the whole sequence fj * Jk (Du) in L k (Ω).

Remark. The following example shows that the weak convergence uj * u ¯ in W 1,k (Ω; RN ) 1 does not imply the weak convergence Jk (Duj ) * Jk (D¯ u) in L (Ω). Example 5.33. (Ball and Murat [7]). Let B be the j = 1, 2, · · · , the radial mappings   jr Uj (|x|) 2 − jr uj (x) = x, Uj (r) =  |x| 0

unit open ball in Rn . Consider, for if 0 ≤ r ≤ 1/j, if 1/j ≤ r ≤ 2/j, if 2/j ≤ r < 1.

Computation shows that uj * 0 in W 1,n (B; Rn ) as j → ∞. But det Duj (x) = (Uj (r)/r)n−1 Uj0 (r) for a.e. x ∈ B, where r = |x|, and hence Z | det Duj (x)| dx = C |x| ¯ (2) F (x, s, ξ) is continuous in (x, s) ∈ Ω×R 0; that is, there exists a continuous function W (x, s, ξ, η, τ ) which is convex in (ξ, η, τ ) ∈ M3×3 × M3×3 × R+ such that, for all x, s, F (x, s, ξ) = W (x, s, ξ, adj ξ, det ξ)

for all ξ ∈ M3×3 with det ξ > 0;

(3) F (x, s, ξ) ≥ c0 |ξ|3 − C(x) for all ξ ∈ M3×3 , where c0 > 0 and C ∈ L1 (Ω) are given. Note that these assumptions are compatible with the assumption (5.35) above. 5.5.3. Existence Theorem. Let I(u) be the energy functional defined by Z I(u) = F (x, u(x), Du(x)) dx, Ω

where F satisfies the previous assumptions. Let ϕ ∈ W 1,3 (Ω; R3 ) be a given function such that I(ϕ) < ∞. Then we have the following existence result. Theorem 5.35. (Existence in Nonlinear Elasticity) There exists at least one minimizer u ¯ ∈ Dϕ with det D¯ u(x) > 0 for almost every x ∈ Ω such that I(¯ u) = min I(u). u∈Dϕ

Proof. Let {uν } be a minimizing sequence in Dϕ . Since lim I(uν ) = inf Dϕ I ≤ I(ϕ) < ∞, from Assumption (1), it follows that det Duν (x) > 0

a.e.

Now Assumption (3) implies that {|Duν |} is bounded in L3 (Ω) and hence from the fixed Dirichlet boundary condition, {uν } is bounded in W 1,3 (Ω; R3 ). Therefore, via a subsequence, we assume uν * u ¯ in W 1,3 (Ω; R3 ) as ν → ∞. Hence, by Corollary 5.32 and Theorem 5.34, (5.36)

adj Duν * adj D¯ u in L3/2 (Ω),

det Duν * det D¯ u in L1 (Ω).

It is easy to see that det D¯ u(x) ≥ 0 for almost every x ∈ Ω. We now show det D¯ u(x) > 0 a.e. in Ω. (We follow Pedregal.) Let ¯ × R3 × M3×3 , det ξ = τ }. h(τ ) = inf{F (x, s, ξ) | (x, s, ξ) ∈ Ω Then h(τ ) is continuous and h(τ ) = ∞ for τ ≤ 0. Note that Z Z sup h(det Duν (x)) dx ≤ F (x, uν (x), Duν (x)) dx < ∞. ν

Ω

Ω

Let E = {x ∈ Ω | det D¯ u(x) = 0}. Define w(x) = lim inf det Duν (x), ν→∞

By Fatou’s lemma, Z Z 0≤ w(x)dx ≤ lim inf E\{w=0}

ν→∞

E\{w=0}

x ∈ E.

Z det Duν (x)dx =

det D¯ u(x)dx = 0. E\{w=0}

118


Hence w(x) = 0 a.e. in E. Using Fatou’s lemma again, Z Z lim inf h(det Duν (x)) dx h(0) = E ν→∞ E Z ≤ lim inf h(det Duν (x))dx < ∞. ν→∞

E

Since h(0) = ∞, we must have |E| = 0. This proves that det D¯ u(x) > 0 a.e. in Ω. Note that at every point x where det Duν (x) > and det D¯ u(x) > 0, by the polyconvex convexity assumption (2), there exists a function W (x, s, J) which is, for all given (x, s), convex on J = (ξ, η, τ ), τ > 0 such that F (x, s, ξ) = W (x, s, J(ξ)) for all x, s, ξ with det ξ > 0, where J(ξ) = (ξ, adj ξ, det ξ). Hence F (x, uν , Duν ) − F (x, u ¯, D¯ u) = [W (x, uν , J(D¯ u)) − W (x, u ¯, J(D¯ u))] + [W (x, uν , J(Duν )) − W (x, uν , J(D¯ u))] ≥ [W (x, uν , J(D¯ u)) − W (x, u ¯, J(D¯ u))] + WJ (x, uν , J(D¯ u)) · (J(Duν ) − J(D¯ u)). We can then follow the similar steps as in the proof of the Tonelli theorem (Theorem 5.1) to show that I(¯ u) ≤ lim inf I(uν ). ν→∞

This proves that u ¯ is a minimizer.

Remark. For more general cases, see M¨ uller, Tang & Yan [31].

5.6. Relaxation Principles There are many application problems which do not satisfy the convexity conditions. In such cases, minimizers may not be found as we did before using the direct method, but the variational methods may still help to study the problem. 5.6.1. Non-quasiconvex Problems. In general, we consider the multiple integral functional Z I(u) = F (x, u(x), Du(x)) dx, Ω

where Ω is a bounded domain in Rn , u : Ω → RN , and F (x, s, ξ) is a given function not quasiconvex in ξ. We give two examples of one-dimensional problem. Example 5.36. Consider the classical example due to Bolza: to minimize the functional Z 1 I(u) = [((u0 )2 − 1)2 + u2 ]dx 0

among all Lipschitz functions u satisfying boundary conditions u(0) = u(1) = 0. Let 1 [kx] 1 uk (x) = − x − − , 2k k 2k where [a] stands for the integer part of a ∈ R. Then uk → 0 in the Lipschitz convergence, and I(uk ) → 0 as k → ∞. Hence the infimum of I over all such u is 0. Note that I(0) = 1. Hence I(0) > lim inf I(uk ), k→∞

and so I is not lower semicontinuous in the Lipschitz convergence. Also I does not have minimizers since I(u) = 0 would imply |u0 | = 1 a.e and u ≡ 0.

5.6. Relaxation Principles

119

Example 5.37. Let f (ξ) = (ξ 2 − 1)2 for ξ ∈ R and consider the functional Z 1 Z 1 ((u0 )2 − 1)2 dx f (u0 (x)) dx = I(u) = 0

0

and minimize it over all Lipschitz functions u satisfying boundary conditions u(0) = 0, u(1) = a. We show that this nonconvex problem has minimizers for all values of a; moreover, the minimizer is unique if |a| ≥ 1 and there exist infinitely many minimizers if |a| < 1. Proof. Let first of all |a| < 1. Then it can be easily seen that there are infinitely many Lipschitz functions u such that u0 (x) ∈ {−1, 1} with u(0) = 0 and u(1) = a; any of such functions is a minimizer of I with minimum value zero. Now let |a| ≥ 1. Note that the strict inequality f (ξ) − [f (a) + f 0 (a)(ξ − a)] > 0 holds for all ξ 6= a if |a| > 1, or holds for all ξ ∈ / {−1, 1} if |a| = 1. Therefore, for all Lipschitz functions u with u(0) = 0, u(1) = a it follows that Z 1 Z 1 I(u) = f (u0 (x)) dx ≥ f (a) + f 0 (a)(u0 (x) − a) dx = f (a); 0

0

hence the minimum value of I for such u’s is f (a) and, certainly, u ¯(x) = ax is a minimizer. We show this is the unique minimizer. Suppose v is any minimizer of I over all Lipschitz functions u with u(0) = 0, u(1) = a; that is, Z 1 I(v) = f (v 0 (x)) dx = f (a) 0

with v(0) = 0 and v(1) = a. If a = 1 then since f (1) = 0 it must follow that v 0 (x) ∈ {−1, 1} R1 a.e. x ∈ (0, 1) and 0 (1−v 0 (x)) dx = 0 and hence v 0 (x) = 1 on (0, 1). This implies v(x) = x. Similarly if a = −1 then v(x) = −x. Now assume |a| > 1. Since f (ξ)−f (a)−f 0 (a)(ξ −a) > 0 for all ξ 6= a, from Z 1 [f (v 0 (x)) − f (a) − f 0 (a)(v 0 (x) − a)] dx = 0, 0

it follows that

v 0 (x)

= a for a.e. x ∈ (0, 1). Hence v(x) = ax.

For this example, we see that the function ( Z 1 f (ξ) if |ξ| ≥ 1 0 c g(ξ) = inf f (ξ + w (x)) dx = f (ξ) = 0 if |ξ| < 1 w∈W01,∞ (0,1) 0 is the convexification of function f ; that is, the largest convex function less than or equal to f . 5.6.2. Quasiconvexification. Given a function F : MN ×n → R, we call the largest quasiconvex function less than or equal to F the quasiconvexication or quasiconvex envelope of F. We denote the quasiconvexification of F by F qc . Theorem 5.38. Let F ≥ 0 be continuous. Then Z qc (5.37) F (ξ) = inf F (ξ + Dφ(x)) dx, φ∈W01,∞ (Σ;RN )

Σ

ξ ∈ MN ×n .

120


Proof. For any quasiconvex function G ≤ F, Z Z G(ξ) ≤ G(ξ + Dφ(x)) dx ≤ F (ξ + Dφ(x)) dx. Σ

Σ

Hence G(ξ) ≤ F qc (ξ) by the definition of F qc . It remains to prove that F qc is itself quasiconvex. We first observe that Z qc − F (ξ + Dφ(x)) dx (5.38) F (ξ) = inf φ∈W01,∞ (Ω;RN )

Ω

for any open set Ω ⊂ Rn with |∂Ω| = 0. This can be proved by using the Vitali covering argument as before. We now claim that, for all piecewise affine Lipschitz continuous functions φ ∈ W01,∞ (Σ; RN ), it follows that Z qc F qc (ξ + Dφ(x)) dx. (5.39) F (ξ) ≤ Σ

To see this, let Σ = ∪i Ωi ∪ E be such that |E| = 0 and each Ωi is an open set with |∂Ωi | = 0 and such that φ ∈ W01,∞ (Σ; RN ) takes constant gradients Dφ = Mi on each Ωi . Let > 0 be given. By the above remark on the definition of F qc , there exists ψi ∈ W01,∞ (Ωi ; RN ) such that Z F qc (ξ + Mi ) ≥ − F (ξ + Mi + Dψi (x)) dx − . Ωi

P With each ψi being extended to Σ, we set ψ = φ + i ψi . Then ψ ∈ W01,∞ (Σ; RN ) and we have Z X F qc (ξ + Dφ(x)) dx = |Ωi |F qc (ξ + Mi ) Σ

i

Z ≥

F (ξ + Dψ(x)) dx − ≥ F qc (ξ) − ,

Σ

and the (5.39) follows. From this we can show F qc is rank-one convex (using “sawtooth” like piecewise affine functions) and therefore is locally Lipschitz continuous (since it is convex in each coordinate direction). Since piece-wise affine Lipschitz functions are dense in W01,p (Σ; RN ), (5.39) holds for all φ ∈ W01,p (Σ; RN ), and hence F qc is quasiconvex. The proof of the theorem is complete. 5.6.3. Weak Lower Semicontinuous Envelopes. Suppose F (x, s, ξ) is a Caratheodory function and satisfies ∀x ∈ Ω, s ∈ RN , ξ ∈ MN ×n ˜ on W 1,p (Ω; RN ) for some 1 ≤ p < ∞. We are interested in the largest w.l.s.c. functional I(u) which is less than or equal to the functional Z I(u) = F (x, u(x), Du(x)) dx. 0 ≤ F (x, s, ξ) ≤ C (|ξ|p + |s|p + 1)

Ω

˜ This functional I(u) is called the (weak lower semicontinuous) envelope or relaxation of I in the weak topology of W 1,p (Ω; RN ). It turns out under the condition given above, ˜ I(u) is given by the integral functional of the quasiconvexification of F with respect to ξ. ˜ Theorem 5.39. Under the given condition, the envelope I(u) of I in the weak topology of 1,p N W (Ω; R ) is given by Z ˜ I(u) = F qc (x, u(x), Du(x)) dx, Ω

5.6. Relaxation Principles

121

where F qc (x, s, ·) is the quasiconvexification of F (x, s, ·) for given (x, s). ˆ Proof. Let I(u) be the integral of F qc (x, u(x), Du(x)) over Ω. Since F qc is quasiconvex and ˆ satisfies the same growth condition as F , the functional I(u) is thus (sequential) w.l.s.c. on 1,p N ˜ To prove the other W (Ω; R ) by the theorem of Acerbi and Fusco. Therefore, Iˆ ≤ I. direction, we first assume there exists a Caratheodory function g(x, s, ξ) such that Z ˜ g(x, u(x), Du(x)) dx ∀ u ∈ W 1,p (Ω; RN ). (5.40) I(u) = Ω

ˆ Then g(x, s, ·) must be quasiconvex and g ≤ F , and thus g ≤ F qc ; this proves I˜ ≤ I. However, the proof of integral representation (5.40) is beyond the scope of this lecture and is omitted; see e.g., Acerbi & Fusco [1], Buttazzo [9], or Dacorogna [13]. 5.6.4. Relaxation Principle. We have the following theorem; the proof of the theorem is also omitted; see, e.g., Acerbi & Fusco [1], Buttazzo [9], or Dacorogna [13]. Theorem 5.40 (Relaxation Principle). Assume c |ξ|p ≤ F (x, s, ξ) ≤ C (|ξ|p + |s|p + 1) holds for some constants c > 0, C > 0, p > 1. Then Z Z inf F (x, u, Du) dx = min F qc (x, u, Du) dx u∈Dϕ

for any ϕ ∈

W 1,p (Ω; RN ),

u∈Dϕ

Ω

Ω

where Dϕ is the Dirichlet class of ϕ.

Remark. The passage from F to F qc is called relaxation. The relaxation principle above seems too nice for the nonconvex variational problems since it replaces a nonconvex problem which may not have any solution by a quasiconvex problem that has solutions. But, there are costs for this: we may lose some information about the minimizing sequences; a minimizer so obtained for the relaxed problem may not characterize what seems more interesting in applications the finer and finer patterns of minimizing sequences. In the phase transition problems for certain materials, it accounts for the loss of information about the microstructures by a macroscopic effective processing (relaxation). For more information, we refer to Ball & James [6], M¨ uller [30] and the references therein. 5.6.5. Existence of Minimizers for a Nonconvex Problem. Consider Z I(u) = f (∇u(x)) dx Ω

where Ω is a bounded domain in given continuous function.

Rn ,

u : Ω → R is a scalar function, and f : Rn → R is a

We define the subgradient of f at a point ξ by ∂f (ξ) = {l ∈ Rn | f (η) ≥ f (ξ) + l · (η − ξ),

∀ η ∈ Rn }.

Each l ∈ ∂f (ξ) is called a subdifferential of f at ξ. Note that ∂f (ξ) may be an empty set for some ξ. For a set of points ξ1 , ξ2 , · · · , ξq in Rn , we denote by co{ξ1 , · · · , ξq } the convex hull defined by ( q ) q X X conv{ξ1 , · · · , ξq } = λi ξi ∀ λi ≥ 0, λi = 1 . i=1

i=1

122


The following theorem is due to A. Cellina [10, 11]. We do not give a proof of this theorem, but simply remark that even without convexity assumption the minimization may still have solutions. However, the existence can not follow from the direct method, but has to rely on different methods. Theorem 5.41. Given ξ ∈ Rn , the minimization problem Z f (∇u)dx inf u∈W 1,1 (Ω), u|∂Ω=ξx

Ω

has a minimizer if and only if either ∂f (ξ) 6= ∅, or there exist ξ1 , ξ2 , · · · , ξq such that ξ ∈ int conv{ξ1 , · · · , ξq }

and

∩qi=1 ∂f (ξi ) 6= ∅.

Remark. By the relaxation principle, any minimizer u of the above problem must be a minimizer for the relaxed problem and satisfy ∇u(x) ∈ K = {ξ ∈ Rn | f (ξ) = f # (ξ)},

(5.41) f#

a.e. x ∈ Ω,

where is the convexification of f . The proof of Cellina’s theorem relies on constructing minimizers u of the relaxed problem satisfying the first-order partial differential relation (5.41).

Bibliography

[1] E. Acerbi and N. Fusco, Semicontinuity problems in the calculus of variations, Arch. Rational Mech. Anal., 86 (1984), 125–145. [2] R. A. Adams and J. Fournier, “Sobolev Spaces,” Second Edition. Academic Press, New York, 2003. [3] A. Bruckner, J. Bruckner and B. Thomson, “Real Analysis,” Prentice-Hall, 1997. [4] J. M. Ball, Convexity conditions and existence theorems in nonlinear elasticity, Arch. Rational Mech. Anal., 63 (1977), 337–403. [5] J. M. Ball, J. C. Currie and P. J. Olver, Null Lagrangians, weak continuity, and variational problems of arbitrary order, J. Funct. Anal., 41 (1981), 135–174. [6] J. M. Ball and R. D. James, Proposed experimental tests of a theory of fine microstructures and the two well problem, Phil. Trans. Roy. Soc. London, A 338 (1992), 389–450. [7] J. M. Ball and F. Murat, W 1,p -Quasiconvexity and variational problems for multiple integrals, J. Funct. Anal., 58, (1984), 225–253. [8] V. Barbu and Th. Precupanu, “Convexity and Optimization in Banach Spaces,” Reidel, Boston, 1986. [9] G. Buttazzo, “Semicontinuity, Relaxation and Integral Representation in the Calculus of Variations,” Pitman Res. Notes Math. Ser. 207, Harlow, 1989. [10] A. Cellina, On minima of a functional of the gradient: necessary conditions, Nonlinear Anal., 20 (1993), 337–341. [11] A. Cellina, On minima of a functional of the gradient: sufficient conditions, Nonlinear Anal., 20 (1993), 343–347. [12] R. Coifman, P.-L. Lions, Y. Meyer and S. Semmes, Compensated compactness and Hardy spaces, J. Math. Pures Appl., 72(9) (1993), 247–286. [13] B. Dacorogna, “Direct Methods in the Calculus of Variations,” Springer, New York, 1989. [14] D. Dunninger, Lecture Notes on Applied Analysis, (unpublished) Michigan State University [15] L. C. Evans, Quasiconvexity and partial regularity in the calculus of variations, Arch. Rational Mech. Anal., 95 (1986), 227–252. [16] L. C. Evans, “Partial Differential Equations,” AMS-GSM, Vol. 19, 1998. [17] M. Giaquinta, “Multiple Integrals in the Calculus of Variations and Nonlinear Elliptic Systems,” Princeton University Press, 1983. [18] M. Giaquinta, “Introduction to Regularity Theory for Nonlinear Elliptic Systems,” Birkh¨ auser, 1993. [19] M. Giaquinta and E. Giusti, On the regularity of the minima of certain variational integrals, Acta Math., 148 (1982), 31–46. [20] B. Gidas, W.-M. Ni and L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Physics 68 (1979), 209–243.

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[21] D. Gilbarg and N. S. Trudinger, “Elliptic Partial Differential Equations of Second Order,” 2nd ed., Springer-Verlag, Berlin, 1984. [22] R. Hart, D. Kinderlehrer and F.H. Lin, Existence and partial regularity of static liquid crystal configurations, Commun. Math. Phys., 105 (1986), 547–570. [23] R. D. James and D. Kinderlehrer, Frustration in ferromagnetic materials, Cont. Mech. Thermodyn., 2 (1990), 215–239. [24] F.-H. Lin, On nematic liquid crystals with variable degree of orientation, Comm. Pure Appl. Math., 44(3) (1991), 453–486. [25] N. Meyers, Quasi-convexity and lower semicontinuity of multiple variational integrals of any order, Trans. Amer. Math. Soc., 119 (1965), 125–149. [26] V. P. Mikhailov, “Partial Differential Equations,” Mir, 1978. [27] C. B. Morrey, Jr., Quasiconvexity and the lower semicontinuity of multiple integrals, Pacific J. Math., 2 (1952), 25–53. [28] C. B. Morrey, Jr., “Multiple Integrals in the Calculus of Variations,” Springer, New York, 1966. [29] S. M¨ uller, Higher integrability of determinants and weak convergence in L1 , J. Reine Angew. Math., 412 (1990), 20–34. [30] S. M¨ uller, Variational models for microstructure and phase transitions, in “Calculus of Variations and Geometric Evolution Problems,” (Cetraro, 1996), pp. 85–210, Lecture Notes in Math. 1713, Springer, 1999. [31] S. M¨ uller, Q. Tang and B. Yan, On a new class of elastic deformations not allowing for cavitation, Ann. Inst. Henri Poincaré, Analyse non linéaire, 11(2) (1994), 217–243. [32] L. Nirenberg, “Functional Analysis.” Courant Institute Lecture Notes, New York University, 1960. [33] P. Pedregal, “Parametrized Measure and Variational Principles.” Birkhauser, Basel, 1997. [34] P. Pedregal and B. Yan, On two-dimensional ferromagnetism, Preprint. [35] E. Stein, “Harmonic Analysis.” Princeton University Press, Princeton, 1993. ˇ ak, Rank-one convexity does not imply quasiconvexity, Proc. Roy. Soc. Edinburgh Sect. A, [36] V. Sver´ 120 (1992), 185–189. [37] L. Tartar, The compensated compactness method applied to systems of conservation laws, in “Systems of Nonlinear Partial Differential Equations”, NATO ASI Series, Vol. 103, D. Reidel, Dordrecht, 1983. [38] R. Temam, “Navier-Stokes Equations: Theory and Numerical Analysis.” Revised Edition. Elsevier Science Publishers, Amsterdam, 1984. [39] K. Yosida, “Functional Analysis,” 6th edition, Springer, New York, 1980. [40] L. C. Young, “Lectures on the Calculus of Variations and Optimal Control Theory,” Saunders, 1969 (reprint by Chelsea, 1980). [41] E. Zeidler, “Nonlinear Functional Analysis and its Applications,” Vol.1-4, Springer.

Index

p-Laplace equations, 71 Acerbi-Fusco’s Theorem Theorem, 101 adjugate matrix, 108 Ambrosetti-Rabinowitz Theorem, 87 Ascoli Theorem, 8 Banach space reflexive, 6 separable, 2 Banach-Steinhaus Theorem, 5 Bessel’s inequality, 9 biharmonic, 57 bilinear form, 54 bounded, 54 strongly positive, 54 Bounded Inverse Theorem, 6 Caratheodory property, 13 Cartesian product, 3 Cauchy sequence, 2 Cauchy-Schwarz inequality, 4 characteristic function, 116 coercivity, 55 continuation method, 11 Contraction Mapping Theorem, 11 convex, 18, 67 convex hull, 121 convexification, 119, 122 critical point, 18, 20 critical points, 65 Deformation Lemma, 87 determinant, 108 differential operator uniformly elliptic, 51 direct method, 70 Dirichlet class, 69 dual space, 6 second dual space, 6 Ehrling’s inequality, 21 eigenfunctions, 10, 62 eigenspace, 62

eigenvalue, 62 Eigenvalue Theorem Theorem, 62 eigenvalues, 10 eigenvectors, 10 equicontinuous, 8 Estimates for W 1,p , n < p ≤ ∞ Theorem, 44 Euler-Lagrange equations, 66 Existence in Nonlinear Elasticity Theorem, 117 Existence of Minimizers Theorem, 70 extended-valued, 116 extreme, 19 maximal, 20 minimal, 20 ferromagnetism, 76 first variation, 65, 66 fixed point, 11 Fourier series, 9 Fourier transform, 48 Fourier transforms, 58 Frechet differentiable, 13 Fredholm Alternative Theorem, 9 function cut-off function, 27 mollified function, 27 p-mean continuous, 28 support, 2 trace, 36, 37 functional convex, 18 weakly coercive, 17, 70 weakly continuous, 17 weakly lower semicontinuous, 17 G˚ arding’s estimates, 55 Gagliardo-Nirenberg-Sobolev Inequality Lemma, 41 Gateaux differentiable, 13 Gateaux-differentiable, 68 Green’s 1st identity, 39 Green’s 2nd identity, 39 Hahn-Banach Theorem, 6 harmonic map equation, 74 harmonic maps, 75

125

126

Hilbert space, 4 Hodge decomposition Theorem, 81 Hodge decomposition theorem, 81 Hodge decompositions, 76 hyperelastic material, 116 Implicit Function Theorem, 16 inner product, 4 inverse Fourier tranform, 48 Lagrange multiplier, 20 Lagrange multipliers, 72 Lagrange Theorem, 20 Lax-Milgram Theorem, 54 Lax-Milgram Theorem Theorem, 54 Legendre ellipticity condition, 53 Legendre-Hadamard, 106 Legendre-Hadamard condition, 53, 67 linear functionals bounded linear functionals, 6 continuous linear functionals, 6 Lipschitz convergence, 97 liquid crystals, 74 Ljusternik Theorem, 21 method of continuity, 11 minimization method, 65 minimizer, 70 minimizing sequence, 70 Morrey’s Inequality Theorem, 43 Neumann boundary, 56 nonexistence, 73 nonlinear eigenvalue problem, 72 nonlinear functional analysis, 11 norm, 2 equivalent, 47 null-Lagrangian, 74, 111 null-Lagrangians, 106 operator bounded, 5 compact, 8 contraction, 11 Hilbert space adjoint, 7 Laplace operator, 39 linear operator, 5 Nemytskii operator, 12 operator norm, 5 resolvent, 10 strongly continuous, 8 weakly continuous, 8 ordinary point, 20 orientation-preserving, 116 orthogonal, 4 orthogonal complement, 4 orthogonality condition, 61 orthonormal, 9 complete, 9 orthonormal basis, 62 Oseen-Frank energy, 74 Palais-Smale compactness condition, 86 Parallelogram law, 4 Partition of Unity Theorem, 28

Index

Plancherel’s Theorem Theorem, 48 Pohozaev Theorem, 90 Poincar´ e’s inequality, 48 point spectrum, 10 polyconvex, 108, 110 principal eigenvalue, 62 projection, 4 Projection Theorem, 4 Property of Fourier Tranforms Theorem, 48 quasiconvex, 99 quasiconvex envelope, 119 quasiconvex functions, 99 quasiconvexication, 119 quasiconvexity, 97 rank-one convex, 67, 104 regularity, 51 relaxation, 121 Rellich-Kondrachov Theorem, 45 resolvent set, 10 Riesz Representation Theorem, 6 second variation, 66 second-order differential operator in divergence form, 51 semilinear, 71 seminorm, 47 sequence weakly convergent, 7 Serrin’s Maximum Principle Lemma, 92 set convex, 1 totally bounded, 45 weakly closed, 17, 70 simplified Maxwell equation, 76 Sobolev conjugate, 40 Sobolev space of order k, 25 Sobolev-Rellich-Kondrachov Theorem, 39 spectrum, 10 star-shaped, 90 stored energy density function, 116 structural conditions, 68 subdifferential, 121 subgradient, 121 total stored energy, 116 Trace Theorem, 37, 38 Triangle inequality, 2, 4 variational problems, 65 vector space, 1 Banach space, 2 basis, 2 complete, 2 dimension, 1 inner product space, 4 normed space, 2 subspace, 1 Vitali covering, 102 weak derivative, 23 weak solution, 51, 69 weakly compact, 8 weakly lower semicontinuous, 70, 95

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