Weak Convergence Methods for Energy Minimization

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1 Weak Convergence Methods for Energy Minimization Bo Li Department of Mathematics University of California, San Diego June 3, 2007 Introduction This compact set of notes present some basic, abstract results with two examples of boundary-value problems on weak convergence methods for the study of existence of minimizers of essentially convex functionals. The method described here is called the direct method in the calculus of variations. The main results are stated in their simplest possible setting. But, they have many variations that can be applied to more complicated problems. The two examples are: () A standard Dirichlet type integral whose minimizers are solutions to the Poisson equation; (2) The elastic energy functional of displacement field whose minimizers are solutions to the fundamental equations in the linear theory of elasticity. The definition and some useful properties of the space H (), H 0(), and H () are given in Appendix (cf. Section A). Some of the properties are proved for special cases. I adopt the following assumptions, notations, and terminologies throughout these notes: () All scalars, functions, and vector spaces are real; (2) The capital Greek letter is used to denote a smooth, bounded, and connected open set in R d with d an integer. The boundary of is denoted by Γ. In several examples, the dimension d = 3; (3) Let K be a nonempty subset of a vector space and I : K R {+ } a given functional. An element u K is a minimizer of I : K R {+ } if I[u] = inf u K I[u].

2 An infimizing sequence of I : K R {+ } is a sequence {u k } k= in K such that 2 Main Theorems lim I[u k] = inf I[u]. k u K Theorem 2. Let X be a reflexive Banach space and K a nonempty, convex, and closed subset of X. Let I : K R be a functional. Assume the following: () Growth condition: there exist p > 0, α > 0, and β R such that I[u] α u p + β u K; (2.) (2) Sequential weak lower semicontinuity: for any u k K (k =,...) and u K, Then there exists u K such that Remarks. u k u (weak convergence) = lim inf k I[u k] I[u]. (2.2) I[u] = min I[v]. (2.3) v K () A Banach space is reflexive if and only if it is locally weakly sequentially compact, i.e., any norm-bounded sequence has a subsequence that converges weakly to some element in the space. This follows from the Banach-Alaoglu Theorem and Eberlein- Šmulian Theorem [3,0,2]; (2) A Hilbert space is a uniformly convex Banach space, and a uniformly convex Banach space is a reflexive Banach space [2]. Proof of Theorem 2.. Let λ = inf u K I[u]. Since K, λ <. By (2.), λ β >. Thus, λ is finite. Let {u k } k= be an infimizing sequence of I : K R. This and the growth condition (2.) imply that { u k } k= is bounded. Since X is reflexive, it is locally weakly sequentially compact. Thus, there exists a subsequence { u ki } i= of { u k } k= such that u k i u as i for some u X. Since K is convex and closed, it is weakly closed. Thus, u K. Now, by (2.2) and the fact that I[u ki ] λ as i, we obtain that This implies (2.3). Q.E.D. λ = lim inf i I[u k i ] I[u] λ. The following result indicates that the strict convexity of a functional defined on a convex set is sufficient for the uniqueness of minimizers, if they exist: 2

3 Proposition 2. Let X be a vector space, K a non-empty, convex subset of X, and I : K R a strictly convex functional. Assume that u K and u 2 K satisfy that Then u = u 2. I[u ] = I[u 2 ] I[u] u K. (2.4) Proof. Assume u u 2. Since K is convex, (u + u 2 )/2 K. The strict convexity of I : K R and (2.4) then imply that [ ] u + u 2 I < 2 2 I[u ] + 2 I[u 2] I[u] u K. This is a contradiction. Q.E.D. Theorem 2.2 Let X be a Banach space and K a nonempty, convex, and closed subset of X. If I : K R is convex and sequentially lower semicontinuous, then it is sequentially weakly lower semicontinuous. Notice that a continuous functional is sequentially lower semicontinuous. Warning! The convexity is far from being necessary for the sequential weak lower semicontinuity for many functionals. In fact, for multiple integrals of gradients of vectorvalued functions, the right notion is the quasiconvexity of the integrand (or energy density), which is weaker than convexity [8]. Proof of Theorem 2.2. Let u k K (k =,...), u K, and u k u. We need only to show that lim inf I[u k] I[u]. (2.5) k If this were not true, then there would exist a subsequence of {u k } k=, not relabeled, and some δ > 0 such that I[u k ] I[u] δ k. (2.6) By Mazur s Theorem [3,0,2], there exist v k = k i= λ (k) i u i co ({u,...,u k }) (2.7) with λ (k) i 0 (i =,...,k) and k i= λ (k) i = 3

4 for all k, such that v k u. Thus, by the convexity of I : K R and (2.6), we have [ k ] k k I[v k ] = I λ (k) i u i λ (k) i I[u i ] λ (k) i (I[u] δ) = I[u] δ k. i= i= It thus follows from the sequential lower semicontinuity of I : K R that i= I[u] lim inf k I[v k] I[u] δ. This is a contradiction. Hence, (2.5) is true. Q.E.D. 3 Examples Example. A Dirichlet type integral and the Poisson equation Let f L 2 () and g H (). Let K = {u H () : u = g on Γ}, where the value on Γ is understood as the trace, cf. Section A. Define I : K R by [ ] I[u] = 2 u(x) 2 f(x)u(x) dx u K. Theorem 3. There exists a unique u K such that I[u] = min I[v]. (3.) v K Moreover, u is characterized by u K and u v dx = fv dx v H0(). (3.2) Remark. If u K satisfies (3.2), then it is called a weak solution of the boundary-value problem of Poisson equation: { u = f in, u = g on Γ. (3.3) It is easy to see that if u is smooth enough, which is indeed true under our assumptions on, f, and g, then u K satisfies (3.2) if and only if u satisfies (3.3). Proof of Theorem 3.. 4

5 Step. We prove the existence of a minimizer of I : K R by verifying that all the assumptions of Theorem 2. are satisfied. Clearly, H () is a Hilbert space. Hence, it is reflexive. Note that g K. So, K. Since the trace operator is linear, K is convex. Let u k K (k =,...) and u k u in H () for some u H (). By the trace inequality (A.3), u g L 2 (Γ) = u u k L 2 (Γ) C u u k H () 0. Hence, u = g on Γ, and K is closed. Let u K. Applying Young s inequality ( ) ε ( εb ab = a ) 2ε a2 + ε 2 b2 for any a,b R and any ε > 0, and the Poincaré inequality (A.4) with the constant C to u g H0(), we have that ( ) I[u] = 2 u 2 fu dx = 4 u 2 dx + 4 (u g) + g 2 dx f(u g)dx fg dx = 4 u 2 dx + 4 (u g) 2 dx + (u g) g dx 2 ( ) f(u g)dx + 4 g 2 fg dx 4 u 2 dx + 4 (u g) 2 dx 8 (u g) 2 dx 8 g 2 dx ( ) 4C 2 f 2 dx 6C u 2 g 2 dx + 4 g 2 fg dx ( ) 3 4 u 2 dx + 6C u 2 g 2 dx 4 g 2 + 4C 2 f 2 + fg dx ( ) ( ) 3 4 u 2 dx + 6C 2 2 u2 g 2 dx 4 g 2 + 4C 2 f 2 + fg dx ( ) ( min 4, 3 u 2H() 32C 2 4 g 2 + ) + 4C 2 f 2 + fg dx. 6C 2g2 This means that I : K R satisfies the growth condition (2.) with ( ) ( p = 2, α = min 4, 3, β = 32C 2 4 g 2 + ) + 4C 2 f 2 + fg dx. 6C 2g2 Let u k K (k =,...), u K, and u k u. Since u k 2 dx 2 ( u k u u 2) dx, 5

6 we obtain lim inf k I[u k] lim inf k lim inf k = = I[u]. ( ) 2 u k 2 fu k dx ( u k u 2 ) u 2 fu k dx ) dx ( u u 2 u 2 dx fu Therefore, I : K R is sequentially weakly lower semicontinuous. Now, by Theorem 2., there exists u K that satisfies (3.). Step 2. We prove that I : K R is strictly convex. By Theorem 2.2, this implies that the minimizer of I : K R is unique. Let u,v K and t (0, ). Since K is convex, ( t)u + tv K. Moreover, by direct calculations, I[( t)u + tv] (( t)i[u] + ti[v]) [ = 2 u + t( v u) 2 dx ( t) = t(t ) u v 2 dx 0. ] 2 u 2 dx + t 2 v 2 dx This implies that I : K R is convex. If it is equal to 0 in the above inequality, then (u v) = 0 a.e. in. Since u v = 0 on Γ, the Poincaré inequality (A.4) implies that u = v in H (). Therefore, I : K R is strictly convex. Step 3. We prove that a minimizer u of I : K R is exactly a weak solution defined as u K satisfying (3.2). Let u K be a minimizer of I : K R, i.e., u K satisfies (3.). Let v H 0(). Then, for any t R, u +tv K. The function i : R R defined by i(t) = I[u + tv] I[u] ( = t 2 2 v 2 dx + t u v dx ) fv dx attains its minimum at t = 0. Thus, i (0) = 0. This is (3.2). Conversely, let u K satisfies (3.2). Let w K. Then, v := w u H0(). Thus, by (3.2), u v dx = fv dx. Consequently, I[w] I[u] = I[u + v] I[u] = 6 2 v 2 dx 0.

7 This implies (3.). Q.E.D. Example 2. The elastic energy in the linear theory of elasticity Consider an elastic material subject to applied forces. Its equilibrium can be described by a deformation that minimizes the elastic energy. This energy usually consists of two parts: one is due to the interaction of material particles and the other due to the work done by the applied force. A deformation is mathematically a mapping from a reference state to R 3 with some smoothness. If the deformation gradient is small, then the relative change of material points can be described by infinitesimal strains that are the linearized deformation gradient around the identity mapping, module linearized rigid-body deformations. The resulting theory is the linear theory of elasticity or linear elasticity [2,7,]. Let us denote by R 3 the reference state, and assume that it is a bounded, open, and connected set with a smooth boundary Γ. A displacement of the material is a vectorvalued function u : R 3 with some smoothness. The components of u are u i : R (i =, 2, 3). For i,j =, 2, 3, we denote ε ij (u) = 2 ( ju i + i u j ) and ω ij (u) = 2 ( ju i i u j ), where i = xi is the partial derivative with respect to the ith variable of the position point x = (x,x 2,x 3 ). For each point x, we also denote ε(u) = (ε ij (u)) and ω(u) = (ω ij (u)) the symmetric and skew matrices that consist of entries ε ij (u) and ω ij (u) (i,j =, 2, 3), respectively. The functions ε ij (u) : R (i,j =, 2, 3) are called the infinitesimal strains or linear strains of the displacement u. They measure the relative change of positions of points in the body of elastic material. The skew matrix ω(u) is called the infinitesimal rigid motion. It makes no contributions to the elastic energy. An elastic material is characterized by its stress-strain relation. Stresses are resultant surface forces per unit area at points in the body of material. The stress-strain relation describes how the stress changes as the strain changes, or vice versa. If such responses are linear, then the material is classified as a linear elastic material. In this case, the stress-strain relation is given by σ ij = C ijkl ε kl, i,j =, 2, 3, (3.4) k,l= where, for a given displacement u : R 3, σ ij = σ ij (u) are the Cauchy stresses, ε ij = ε ij (u) are the infinitesimal strains, and C = (C ijkl ) is a symmetric, fourth-order tensor field, called the Cauchy elasticity tenor. Here, symmetry means that at each point in, C ijkl = C klij = C jikl for all i,j,k,l. 7

8 The components C ijkl are called elastic constants or elastic moduli. They have the units same as that of stresses. If the Cauchy elasticity tensor field is constant throughout the reference domain, then the underlying elastic material is called elastically homogeneous. If the Cauchy elasticity tensor is isotropic, i.e., it is such that the stress at each point of the material is invariant under any change of the reference coordinate system by proper rotations, then the material is called elastically isotropic. For a homogeneous and isotropic elastic material which is mostly common in applications, the stress-strain relation (3.4) reduces to ( σ ij = λ ε kk )δ ij + 2µε ij, i,j =, 2, 3, k= where δ ij is if i = j and 0 if i = j, and λ and µ are two material parameters, known as the Lamé constants. We denote by R 3 3 sym the vector space of all 3 3, symmetric matrices with the inner product and norm σ : τ = i,j= σ ij τ ij and σ = σ : σ σ = (σ ij ),τ = (τ ij ) R 3 3 sym. We also denote by L 2 (, R 3 ) and H (, R 3 ) the space of vector-valued functions u : R 3 with each component in L 2 () and H (), respectively. The inner product of u,v L 2 (, R 3 ) is u v dx where the dot denotes the inner product of R 3, and the inner product of u,v H (, R 3 ) is (u v + u : v)dx, where u = ( j u i ) is the matrix-valued, displacement gradient field of u = (u,u 2,u 3 ). The space H 0(, R 3 ) can be defined similarly, cf. Section A. The L 2 -norm of a function u on, scalar or vector-valued or matrix-valued, will all be denoted by u L 2 (). Similar notation will be used for the H -norm. Let now Γ 0 be a smooth subset of Γ and assume that the surface area of Γ 0 is positive. Let Γ = Γ \ Γ 0. Let u 0 H (, R 3 ), f L 2 (, R 3 ), and g L 2 (Γ ). Define the set of admissible displacement fields Define the elastic energy E : A R by E[u] = σ(u) : ε(u)dv 2 A = {u H (, R 3 ) : u = u 0 on Γ 0 }. (3.5) f udv g uds Γ u A, 8

9 where σ(u) = Cε(u) : R 3 3 sym has the components given as in the right-hand side of (3.4), f is the applied body force per unit volume, and g is the applied traction per unit surface area. Here, We consider a general Cauchy elasticity tensor field C = (C ijkl ). We assume that each C ijkl C () and that there exists a constant γ > 0 such that Finally, we denote Cε : ε γ ε 2 ε R 3 3 sym. (3.6) H Γ 0 (, R 3 ) = {u H (, R 3 ) : u = 0 on Γ 0 }. Theorem 3.2 There exists a unique u A such that E[u] = min E[v]. (3.7) v A Moreover, this unique energy minimizer u is characterized by u A and σ(u) : ε(v)dv = f v dv + g v ds v HΓ 0 (, R 3 ). (3.8) Γ If u A satisfies (3.8), then it is called a weak solution of the boundary-value problem divσ(u) = f in, u = u 0 on Γ 0, (3.9) σ(u)n = g on Γ, where n is the unit normal along the boundary Γ and σ(u)n represents the normal component of the Cauchy stress. Here, divσ(u) : R 3 has components 3 j= jσ ij. The first equation in (3.9) is then the same as j σ ij (u) = f i, i =, 2, 3, in, j= where f i is the ith component of f. These are the fundamental equations in the linear theory of elasticity that describe the force balance of an underlying elastic body in equilibrium. For elastically homogeneous and isotropic material, these equations are known as Navier equations. Let u A satisfy (3.8). Assume u is smooth enough, say, all the second order partial derivatives of all the components of u are in L 2 (). This is indeed true under usual assumptions on the smoothness of the data, u 0, f, and g, together with the assumption that Γ is a subset of Γ \ Γ 0 with the closures of Γ 0 and Γ do not intersect. Then u satisfies (3.9). We shall prove the regularity result for the weak solution u A. But, to see the idea how a weak solution defined by (3.8) is equivalent to the classical solution of the boundary-value problem (3.9), we shall assume such regularity. 9

10 Proposition 3. If u A is smooth and satisfies (3.8), then it satisfies (3.9). Conversely, if u : R 3 is smooth and satisfies (3.9), then u A and u satisfies (3.8). Proof. Let u A be a smooth solution of (3.8). Let v H Γ 0 (, R 3 ). Since σ(u) is symmetric, σ(u) : ε(v) = σ(u) : [ε(v) + ω(v)] = σ(u) : v. Thus, by (3.8) and integration by parts, we obtain divσ(u) v dv + σ(u)n v ds = Γ f v dv + g v ds. Γ (3.0) This is true for all v H0(, R 3 ) HΓ 0 (, R 3 ), and thus implies the first equation in (3.9). Using this first equation in (3.9) and (3.0) for general v HΓ 0 (, R 3 ), we obtain then σ(u)n v ds = Γ g v ds Γ v HΓ 0 (, R 3 ). This leads to the last equation in (3.9). Since u A, the second equation in (3.9) is satisfied. The reverse part is trivial. Q.E.D. To prove Theorem 3.2, we need a Poincaré type inequality that uses the strains ε(u) to control the full H norm of a displacement field. This turns out to be quite non-trivial in general. The following lemma gives the inequality or ellipticity that we need; its proof is postponed: Lemma 3. Let A be defined as in (3.5). Then, there exists a constant C > 0 such that Proof of Theorem 3.2. v H (,R 3 ) C ε(v) L 2 (,R 3 ) v H Γ 0 (, R 3 ). (3.) Step. Clearly, H (, R 3 ) is a Hilbert space and A is a nonempty, convex, closed subset of H (, R 3 ), cf. the proof of Theorem 3.. By Lemma 3. and the same argument as in the proof of Theorem 3., we obtain the growth condition E[v] α v 2 H (,R 3 ) + β v A for some constants α > 0 and β R. Note that for any v,w H (, R 3 ), σ(v) : ε(w) = σ(w) : ε(v) = Cε(v) : ε(w) = ε(v) : Cε(w). 0

11 Moreover, since all C, σ, and ε are symmetric and linear, we have by (3.6) that 2 σ(v) : ε(v) + σ(w) : ε(w) σ((v + w)/2) : ε(v + w)/2) 2 = σ(v w) : ε(v w) 4 = Cε(v w) : ε(v w) 4 γ 4 ε(v w) 2. This, together with Lemma 3., implies that the energy functional E : A R is strict convex. Clearly, this quadratic energy functional is continuous. Therefore, it is sequentially lower semicontinuous. The existence and uniqueness of u A that satisfies (3.7) follows from Theorem 2., Theorem 2.2, and Proposition 2.. The rest can be proved by arguments similar to that used in the proof of Theorem 3.. Q.E.D. We now come back to the discussion of Lemma 3.. The inequality (3.) is closely related to Korn s inequalities. In the special case that Γ 0 = Γ and u 0 = 0, the result, stated below, is not difficult to prove. Theorem 3.3 (First Korn s inequality) We have u L 2 () 2 ε(u) L 2 () u H 0(, R 3 ). (3.2) Proof. By the definition of the space H0(), we need only to consider u Cc (, R 3 ). By integration by parts, we have for i j that j u i i u j dx = u i j i u j dx = u i i j u j dx = i u i j u j dx. Consequently, we obtain by straight forward calculations that 4 ε(u) 2 dx = = = i,j= + [ ( j u i ) 2 + ( i u j ) j u i i u j ] dx u 2 dx + u 2 dx + 3 i,j=,i j i,j= i= [ ( i u j ) j u i i u j ] dx ( i u i ) 2 dx + i u i j u j dx + i,j=,i j i,j=,i j ( i u j ) 2 dx j u i i u j dx

12 = u 2 dx + 3 ( i u i ) 2 dx + i= [ ( i u i ) 2 + ( j u j ) 2] dx 2 2 = i,j=,i j u 2 dx + u 2 dx, i= leading to the inequality in (3.2). ( i u i ) 2 dx Q.E.D. i,j=,i j ( i u j ) 2 dx i,j=,i j [ ( j u i ) 2 + ( i u j ) 2] dx The general case known as Korn s inequality or Korn s second inequality is more delicate. See [4,9] for a proof. Theorem 3.4 (Second Korn s inequality) There exists a constant C > 0 such that u H () C ( ) u L 2 () + ε(u) L 2 () u H (, R 3 ). Proof of Lemma 3.. Suppose (3.) were not true for any C > 0. Then there exist v k H Γ 0 (, R 3 ) (k =,...) such that v (k) L 2 () = and ε ( v (k)) L 2 () < k k. (3.3) By the second Korn s inequality (cf. Theorem 3.4), {v (k) } k= is bounded in H (, R 3 ). Therefore, the Rellich-Kondrachov compact embedding theorem (cf. Theorem A.5) implies that this sequence has a subsequence, not relabeled, converging in L 2 norm. Thus, again by the second Korn s inequality, v (k) v in H (, R 3 ) for some v = (v,v 2,v 3 ) H (, R 3 ). The trace theorem (Theorem A.6) implies that v H Γ 0 (, R 3 ), and (3.3) implies v L 2 () = and ε(v) = 0 a.e.. By the definition of ε(v), we have i v i = 0 a.e. for i =, 2, 3. Hence, v i is independent of x i (i =, 2, 3). Consider i,j {, 2, 3} with i j. We have 2ε ij (v) = j v i + i v j = 0 a.e., (3.4) Since v j is independent of x j, j i v j exists, and is 0 a.e.. Thus, by (3.4), j v i is differentiable with respect to x j. Consequently, since i v j is independent of x j, j j v i = 0 a.e., and v i is linear in x j. Thus, there exists a 3 3 matrix A and a vector a R 3 such that v(x) = Ax + a a.e.. By modifying the value of v on a set of measure 0, we have v(x) = Ax + a for any x. Again by ε(v) = 0 in, the matrix A is skew. Fix a point z Γ 0. Since Γ 0 has a positive area, {x z : x Γ 0 } expands a two-dimensional space. Since v = 0 on Γ 0, A(x z) = 0 for all x Γ 0. Thus, the rank of A can be at most. Since A is skew, A must be in fact the zero matrix. Now, v(x) = a on. But, v = 0 on Γ 0. Thus a = 0, and hence v = 0 on. This contradicts the fact that v L 2 () =. Q.E.D. 2

13 A The Spaces H (), H 0(), and H () We consider R d with any d. We define H () = {u L 2 () : j u L 2 (), j d}, where j u denotes the weak partial derivative of u with respect to x j, the jth component of the position vector x = (x,...,x d ). We recall that the weak derivative j v for any v L loc () is defined by jv L loc () and j uφdx = u j φdx φ Cc (). Two functions in H () are identical if and only if they are equal to each other almost everywhere in. Clearly, H () is a linear subspace of L 2 (). Define u,v = (uv + u v)dx u,v H (). This is an inner-product of H (). The corresponding norm is u H () = ( u 2 + u 2 )dx u H (). We now give some of the useful properties of the space H (). Their proofs can be found in standard references, e.g., [,5,6]. Theorem A. The space H () is a separable Hilbert space. Theorem A.2 (Approximation by smooth functions) The space C () is dense in H () with respect to the H ()-norm. Theorem A.3 (Extension) Let be any open set of R d such that. Then there exists a bounded linear operator E : H () H (R d ) such that for any u H (), Eu = u a.e. in, supp u, and u H (R d ) C u H () for some constant C > 0 depending only on and. Theorem A.4 (Sobolev embedding) Let d 3 and q = 2d/(d 2). Let u H (). Then u L q () and u L q () C u H () for some constant C > 0 independent of u. 3

14 Theorem A.5 (Rellich-Kondrachov compact embedding) Let d 3. Then any bounded sequence in H () has a subsequence that is convergent in L q () for any q [, 2d/(d 2)). Remark. Theorem A.4 and Theorem A.5 hold true for d = 2 with q [, ) and d = with L q () replaced by C(). Let u H (). What is the value of u on the boundary Γ? Since two functions in H () are regarded as the same whenever they are equal to each other almost everywhere in, and since the boundary Γ has the R d Lebesgue measure 0, the value of u H () on Γ must be carefully defined. This can be done through approximations by functions in C () which have well-defined values on the boundary Γ. Lemma A. There exists a constant C > 0 such that u L 2 (Γ) C u H () u C (). (A.) Proof of Lemma A. for a special case: = B is the unit ball of R d. Let φ : [0, ] R be a smooth function such that φ(t) = t for t [/2, ]. Define ψ : B R by ψ(x) = φ( x ). Fix u C (B), where B is the unit ball of R d. By Green s formula, we have ψ ( B n u2 ds = u 2 ψ + ψ (u 2 ) ) dv, (A.2) B where n is the unit exterior normal along B, the boundary of B. By the definition of ψ, ψ/ n = on B. Thus, we obtain B u 2 ds sup ψ(x) x B sup ψ(x) x B leading to (A.) with = B and B B u 2 dv + sup x B ψ(x) u 2 dv + sup ψ(x) x B B B 2 u u dv ( u 2 + u 2) dv, C 2 = sup x B ψ(x) + sup ψ(x). x B Q.E.D. Fix now u H (). Let u k C () (k =,...) be such that u k u in H (). By Lemma A., {u k } k= is a Cauchy sequence in L2 (Γ). Therefore, it converges to some Tu L 2 (Γ). It can be proved that this function Tu does not depend on the choice of the sequence {u k } k= in C () that converges to u in H (). We can now define the boundary value of u H () on Γ to be Tu. We summarize our discussions in the following theorem: 4

15 Theorem A.6 ( [,5,6] ) There exists a unique linear operator T : H () L 2 (Γ) that satisfies the following: () Tu = u Γ for any u C (); (2) There exists a constant C > 0 such that u L 2 () C u H () u H (). (A.3) The operator T is called the trace operator, the image Tu for u H () is called the trace of u, and the inequality (A.3) is often called the trace inequality. This version of trace inequality is not optimal in that the norm L 2 (Γ) can be replaced by one with a stronger topology. When no confusion arises, we often write u instead of Tu as the trace of u. We define the space H 0() to be the closure of the space C c () in H (). We also define the space H () to be the dual space of H 0(). Theorem A.7 ( [,5,6] ) We have H 0() = {u H () : u = 0 on Γ}. Theorem A.8 (Poincaré inequality) There exists a constant C > 0 such that u L 2 () C u L 2 () u H 0(). (A.4) Proof of the Poincaré inequality (A.4) for a special case: = B is the unit ball of R d. By the fact that Cc() is dense in H0(), we need only to prove the Poincaré inequality for Cc() functions. So, let u Cc(). Define ψ : B R by ψ(x) = x 2 /(2d) for any x B. Clearly, ψ = and ψ /d in B. Since u = 0 on Γ, we thus obtain by Green s formula (cf. (A.2)) that u 2 dv ψ (u 2 ) dv 2 u u dv ( ) d B d B 2 u u 2 dv. B This leads to B B ( 2d u 2 dv max 2d, 2 ) u 2 dv. d B Q.E.D. The Poincaré inequality can be generalized to functions in H () that vanish on a fixed subset of Γ. 5

16 Theorem A.9 (Poincaré inequality again) Let Γ 0 be smooth subset of Γ and assume that the surface area of Γ 0 is positive. Let Then there exists a constant C > 0 such that H Γ 0 () = {u H () : u = 0 on Γ 0 }. u L 2 () C u L 2 () u H Γ 0 (). (A.5) Proof. Suppose this were not true, then for each k there would exist u k H Γ 0 () such that u k L 2 () > k u k L 2 () k =,... Let v k = u k / u k L 2 () H Γ 0 (). Then v k L 2 () = and v k L 2 () < k k =,... (A.6) Hence, {v k } k= is bounded in H (). It then follows from Theorem A.5 that there exists a subsequence of {v k } k=, not relabeled, that converges in H () to some v H (). By (A.6), we thus have v L 2 () = and v L 2 () = 0. (A.7) From the second equation in (A.7), we see that v is a constant function on. But, by Theorem A.6 and the fact that each v k HΓ 0 (), we infer that v = 0 on Γ 0. Since the surface are of Γ 0 is positive, v = 0 on Γ. Hence, v = 0 on. This contradicts the first equation in (A.7). Q.E.D. References [] R. Adams. Sobolev Spaces. Academic Press, New York, 975. [2] P. G. Ciarlet. Mathematical Elasticity. Volume : Three Dimensional Elasticity. North-Holland, 988. [3] J. B. Conway. A Course in Functional Analysis. Springer, 2nd edition, 990. [4] G. Duvaut and J. L. Lions. Inequalities in Mechanics and Physics. Springer-Verlag, 976. [5] L. C. Evans. Partial Differential Equations, volume 9 of Graduate Studies in Mathematics. AMS, Providence, 998. This follows from the fact that a locally integrable function with all its first-order weak derivatives vanishing a.e. on an open and connected set is a constant. 6

17 [6] D. Gilbarg and N. S. Trudinger. Elliptic Partial Differential Equations of Second Order, volume 224 of Grundlehren Der Mathematischen Wissenschaften. Springer- Verlag, 998. [7] M. E. Gurtin. The linear theory of elasticity. In S. Flügge and C. Truesdell, editors, Handbuch der Physik, Vol. 2, pages 295, 972. [8] Bo Li. Weak Convergence Methods for Variational Models of Crystalline Solids. Springer, [9] J. A. Nitsche. On Korn s second inequality. R.A.I.R.O. Numer. Anal., 5(3): , 98. [0] W. Rudin. Functional Analysis. McGraw-Hill, 2nd edition, 99. [] I. S. Sokolnikoff. Mathematical Theory of Elasticity. McGraw-Hill, 956. [2] K. Yosida. Functional Analysis. Springer, 6th edition,