Math 699 Reading Course, Spring 2007 Rouben Rostamian Homogenization of Differential Equations May 11, 2007 by Alen Agheksanterian

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1 . Introduction Math 699 Reading Course, Spring 007 Rouben Rostamian Homogenization of ifferential Equations May, 007 by Alen Agheksanterian In this brief note, we will use several results from functional analysis in a practical context. We will consider the problem of homogenization of a differential equation in -; the differential equation we will consider, taken by itself, is rather trivial; however, the analysis that is provided is helpful in gaining insight and understanding homogenization of PEs in general (where we work in - or 3-). Physically speaking, homogenization entails the replacement of a material with periodic micro-structure (e.g. perforated media) by a homogeneous one. Practically speaking the purpose of the theory is to replace a computationally hard to solve problem by an easier to solve problem whose solution is a good approximation to the solution of the original problem. The mathematical derivations, however, take more time to explain. We will need several tools from functional analysis in the development of the method. Hence, in the Section we provide the functional analysis background needed in the discussion of homogenization. Next, we will discuss homogenization in Section 3 in detail.. Technical Tools Needed This section is devoted to the functional analysis results needed in the development of the idea of homogenization. We will need some basic results on compact operators which will be discussed first. Then, we will discuss Rellich s Theorem and subsequently Riemann-Lebesgue Lemma which are essential in our study... Compact Operators We first start by the following basic definitions. efinition.. (Relatively Compact) Let X be a metric space; A X is relatively compact in X, if Ā is compact in X. efinition.. (Precompact) Let X be a metric space; A X is precompact (also called totally bounded) if for every ɛ > 0, there exist finitely many points x,..., x N in A such that N B(x i, ɛ) covers A. The following Theorem shows that when we are working in a complete metric space, precompactness and relative compactness are equivalent. Theorem.3. Let X be a metric space. If A X is relatively compact then it is precompact. Moreover, if X is complete then the converse holds also. Then, we define a compact operator as below. efinition.4. Let X and Y be two normed linear spaces and T : X Y a linear map between X and Y. T is called a compact operator if for all bounded sets E X, T (E) is relatively compact in Y. By the above definition.4, if E X is a bounded set, then T (E) is compact in Y. The following is a basic Theorem regarding compact operators. Theorem.5. Let X and Y be two normed linear spaces; suppose T : X Y, is a linear operator. Then the following are equivalent.. T is compact.. The image of the open unit ball under T is relatively compact in Y. 3. For any bounded sequence {x n } in X, there exist a subsequence {T x nk } of {T x n } that converges in Y. As it is clear, compact operators have many rich properties; when working in Hilbert spaces, we can get further properties of such operators. We will see shortly that image of a weakly convergent sequence in a Hilbert space under a compact operator is a strongly convergent sequence. To prove that result, however, we need the following lemma which is interesting in its own right. Lemma.6. Let H and H be Hilbert spaces and T : H H a bounded linear mapping. Suppose {x n } is a weakly convergent sequence in H ; that is, x n x. Then, T x n T x.

2 Proof. Simply note that for all z H T x n, z = x n, T z x, T z = T x, z. Hence, T x n, z T x, z for all z H, and consequently T x n T x. Then, we state and prove the following result which is useful when working with compact operators in Hilbert spaces. Theorem.7. Let H and H be Hilbert spaces and T : H H a bounded linear mapping. Then the following are equivalent:. T is a compact operator. x n x in H implies T x n T x strongly. Proof. () = (): Assume T is a compact operator; let x n x in H but suppose to the contrary that T x n T x. Then there exist ɛ > 0 and {T x nk } such that T x nk T x ɛ for all n k. Without loss of generality suppose, T x n T x ɛ n. (.) Now, x n x implies T x n T x by Lemma.6. Next, we use compactness of T to conclude the existence of {T x nl } such that T x nl y; but this implies (uniqueness of weak limits) that y = T x. Hence, we have T x nl T x 0 as l which clearly contradicts.. Therefore, {T x n } converges strongly. () = (): Now suppose () is true. Let {u n } be a bounded sequence in H. Since, H is a Hilbert space and hence reflexive, we get as a consequence of Banach-Alaoglu Theorem that {u n } has a weakly convergent subsequence, u nk u, but then by () we have that T u nk T u strongly; this in turn implies that T is compact (Theorem.5(3)) While we could say more on compact operators, the machinery developed so far is adequate in our discussion in this note... Rellich s Theorem We will first establish some technical preliminaries. The following Theorem provides a criterion for relative compactness of a set in L p []. Theorem.8. Let A L p. The A is relatively compact in L p if and only if. A is bounded in L p.. lim f(x) p dx = 0 uniformly with respect to f A. R { x >R} 3. lim τ a f = f uniformly with respect to f A. a 0 Note that in the above Theorem, τ a f(x) := f(x a). The proof of above Theorem is rather technical; a clear and concise proof is provided in []. The following two Lemmas are also needed. Lemma.9. Let be a bounded domain and let f H0 (). efine, f := { f on 0 on R d \ Then f H (R d ) and Φ : ( H 0 (),. H 0 ()) ( H (R d ),. H (R d )) is an isometry. Lemma.0. Let f H (R d ) the for every h R d τ h f f L h f L. Next, we state and prove Rellich s Theorem. Theorem.. (Rellich) Let be a bounded domain in R d ; then the inclusion map H 0 () L () is a compact operator. Proof. First define the map, Φ : L (R d ) L () given by Φ(u) = u and note that Φ is clearly continuous; next, we define Ξ : H 0 () L (R d ) by Ξ(f) = f, where f is the extension of f as defined in Lemma.9. We want to show, I : H 0 () L () is compact, where If = f; we can use I = Φ Ξ. Also, we note that the image of relatively compact set under a continuous mapping between Banach spaces

3 is again relatively compact; hence, it is enough to prove that Ξ : f f from H 0 () to L (R d ) is a compact operator. We know by Lemma.9 that Ξ is an isometry such that f H 0 () = Ξ(f) H (R d ). Let B be the closed unit ball in H 0 (). efine, B := Ξ(B) = { f f B}, and note that by Lemma.9 B is contained in the closed unit ball of H (R d ). If we show that B is relatively compact in L (R d ) we are done; to do so we appeal to the criterion for relative compactness in L p provided by Theorem.8. In what follows we will show that items (), (), and (3) of Theorem.8 are satisfied. Boundedness of B in L (R d ) was established above where we noted that B is in the closed units ball of H () L (R d ); from this we get that for any f B, f L f H. Moreover, for any R > 0 such that B(0, R), we have { x >R} f(x) dx = 0, and hence follows the item () of Theorem.8. Thus, it remains to show item (3) of the aforementioned Theorem. Recall that by lemma.0, we have for f B τ h f f L h f L (R d ) h f H (R d ) h, from which we have τ h f f L 0 as h 0. Hence, we also have property (3) of Theorem.8. Therefore, B is relatively compact in L (R d ); this completes the proof. Using the properties of compact operators and Rellich s Theorem we conclude that any bounded set in H 0 () is relatively compact in L (). In particular, the following corollaries are useful in applications. Corollary.. Let be a bounded domain in R d. Any bounded sequence in H0 () has a subsequence that converges strongly in L (). Corollary.3. Let be a bounded domain in R d. Any weakly convergent sequence in H0 () converges strongly in L (). Proof. Use Theorem Riemann-Lebesgue Lemma We first prove the following helping result. Lemma.4. Let R n be a bounded domain. Then, f n f in L () if and only if. f n M for all n.. lim n [ f n (x) f(x) ] dx = 0 for every cube. Proof. Without loss of generality, suppose f = 0. suppose f n 0; then, () follows from the uniform boundedness principle; also, we have () because for any cube, χ L () and using the weak convergence of {f n }, f n(x) dx = χ (x)f n (x) dx 0. Conversely, suppose () and () hold. We need to show, φ(x)f n(x) 0 for all φ L (). Since simple functions are dense in L () and we can write a simple function as a finite sum of characteristic functions, we will prove the result for characteristic functions only. Let φ = χ E for some measurable subset E. Using regularity properties of Lebesgue measure, we can approximate E from inside using cubes; thus, ɛ > 0, there exist, { i } N, such that i E and

4 µ(e \ N i i ) < ( ɛ M ). For convenience, denote := N i i and := E \ N i i. Then, we have χ E f n dx = = < E f n dx f n dx + f n dx ( f n dx + dx f n dx + Mµ( ) f n dx + ɛ. ) ( ) fn dx From which the result follows; note that we have used Cauchy-Schwarz along with boundedness of {f n }. Now, we are ready to state and prove the main result. The proof that follows is based on the more general argument in [3]. n Theorem.5. (Riemann-Lebesgue) Let = (a i, b i ) and let f L (). Extend f by periodicity from to R n. Let f m (x) = f(mx). Then, as m, f m f in L (), where f := µ() f(x) dx. Proof. By the previous lemma, it is sufficient to show the following to obtain the result.. f m M for all m (for some M R). [. lim f m (x) f(x) ] dx = 0 for every cube. m To show the first item, we proceed (having periodicity of f in mind) as follows: f m (x) dx = f(mx) dx = f(x) dx m = f(x) dx = f, from which () follows. To show (), we first need to have a suitable representation for a typical cube in ; for any such, we can find α and β in R such that = α + β, where α + β = n (α i + β i a i, α i + β i b i ). We will also develop the notation [x] for the integer part of x which will become useful in a minute. We proceed as follows. Let be any arbitrary cube in. Then, we have (f m (x) f) dx = (f(mx) f) dx α+β = = [mβ]n = mα+[mβ] (f(x) f) dx + (f(x) f) dx + mα+(mβ [mβ]) (f(x) f) dx. mα+(mβ [mβ]) mα+(mβ [mβ]) (f(x) f) dx (f(x) f) dx

5 In the above calculation we used the fact that f is periodic and that (f(x) f) dx = 0. Next, we conclude, (f m (x) f) dx = (f(x) f) dx mα+(mβ [mβ]) f(x) f dx 0, as m and thus, () is proved also. Therefore, by Lemma.4, we have that f m f as m. Note that the above Theorem can be generalized in the following way; suppose the hypotheses of the above Theorem hold. Then, let be any bounded domain in R n, in the first place, since is bounded K for some K > 0; then it is straight-forward matter to check that f m L () is bounded (similar argument to the one in the proof of above Theorem). Moreover, any cube can be obtained by proper scaling and translation of that is = α+β for some α and β in R n as we saw in the proof of the previous Theorem. Then, we can use a similar argument to show that for all cubes, [ f m (x) f ] dx 0 as m. This leads to the following corollary. n Corollary.6. Let = (a i, b i ) and let f L (). Extend f by periodicity from to R n. Let f m (x) = f(mx). Also, let be a bounded domain in R n. Then, as m, f m f in L (), where f := µ() f(x) dx.

6 3. Homogenization of a ifferential Equation in - The problem we seek to solve is the following (a(x)u (x)) = f(x), x (0, ) (3.) u(0) = u() = 0. (3.) In the above equation f L [0, ] and a L (R) is a -periodic function such that there exist α, β R so that 0 < α a(x) β. In figure 3. we have depicted an example of an acceptable function for a(x) Fig. 3.. An example for a(x). As a first step in our analysis we look at the following problem: where a ɛ is defined as a ɛ (x) = a( x ɛ ). Our goal is to get to the following limiting problem. (a ɛ (x)u ɛ(x)) = f(x), x (0, ) (3.3) u ɛ (0) = u ɛ () = 0. (3.4) (āū (x)) = f(x), x (0, ) (3.5) ū(0) = ū() = 0. (3.6) There are some questions that one needs to address here. In the first place, what does ā look like? Moreover, how do we know a ū exists and satisfies (3.5)? Furthermore, we need to show that u ɛ somehow converges to ū as ɛ 0. We will address all this questions in what follows. Froow on, we denote = (0, ) as our domain. Proposition 3.. The sequence {u ɛ } has a weakly convergent subsequence in H0 () norm. Proof. efine, the functional J ɛ (u) by the following. J ɛ (u) = a ɛ (x)u (x) dx f(x)u(x) dx. (3.7) We know that u ɛ (x) is the unique minimizer of J ɛ (u) in H 0 (). The first thing we want to show is the the functional J ɛ (u) is bounded from below: J ɛ (u) = a ɛ (x)u (x) dx f(x)u(x) dx α α α ( a ɛ (x)u (x) dx ( u (x) dx f(x) dx f(x) dx ) ( ) ( u(x) ) u(x) ) u (x) dx f(x) dx c u(x) dx c u (x) dx f(x) dx c c c 0 u (x) dx (P oincare).

7 Hence, choosing c appropriately and renaming constants, we have k and k positive such that. J ɛ (u) k u (x) dx k f(x) dx k f(x) dx. This in particular gives that the sequence {J ɛ (u ɛ )} ɛ>0 is bounded from below; that is there exist K = inf {J ɛ(u ɛ )}. ɛ Take a minimizing sequence {J ɛj (u ɛj )}. We know from the calculations above that k u ɛ j (x) dx J ɛj (u ɛj ) + k f(x) dx (3.8) Next, recall that by Poincare Inequality, u ɛ j dx c 0 (u ɛ j ) dx (3.9) By (3.8) and (3.9) we have that {u ɛj } is bounded in H 0 () norm. Then, since H 0 () is a Hilbert space and hence reflexive we know (as a consequence of Banach Alaoglu Theorem) that {u ɛj } has a weakly convergent subsequence. We write (Wlog) that u ɛj ū in H 0 () norm. This completes the proof. Next, we note immediately that as a result of Rellich s compact imbedding Theorem (Theorem.) and Theorem.7 the following: u ɛj ū strongly in L () norm. (3.0) Now our goal is to show that ū satisfies the differential equation (3.5). To do so, we need one more technical result. For convenience, denote σ ɛj = u ɛ j. (3.) Proposition 3.. The sequence {σ ɛj } has a weakly convergent subsequence in H 0 (). Proof. First note that { } is bounded in L () and {u ɛ j } is bounded in L () (using (3.8)). So, {σ ɛj } is bounded in L (). Then, we note that σ ɛ j = f; (3.) therefore, {σ ɛj } is bounded in H 0 (). Consequently, {σ ɛj } has a weakly convergent subsequence in H 0 (). Using the previous Proposition, we can write, without loss of generality, that σ ɛj σ in H 0 (). Therefore, again using Rellich s Theorem and properties of compact operators we have σ ɛj σ strongly in L (). Consequently, we have that u ɛ j ū in L (). Now we use the following trick; u ɛ j = aɛ j σ ɛj ū. On the other hand, using Riemann-Lebesgue Lemma, we know that Also as we saw before, σ ɛj σ in L (). Therefore, σ ɛj ū, and σ ɛj σ ā, u ɛ j ū. Therefore, a(x) dx in L ().

8 where, we have denoted ā := dx. Hence, we have σ ā = ū from which it follows that σ = (āū ). Also, a(x) using (3.), σ = f and hence (āū ) = f. That is ū also satisfies the problem (3.5), which brings our analysis to an end. 4. Final Remarks We saw that using homogenization, when we have a problem of form (a ɛ (x)u ɛ(x)) = f(x), x (0, ) (4.) u ɛ (0) = u ɛ () = 0. (4.) with a ɛ (x) as we saw before (see problem 3.3), in the case of small ɛ, one can solve a nearby limiting problem of the following form: Where in the limiting problem ā has the unexpected form below: (āū (x)) = f(x), x (0, ) (4.3) ū(0) = ū() = 0. (4.4) ā = a(x) dx. The advantage in doing so becomes clear when we think about the numerical cost of solving the original problem. Note that if we have a small ɛ, we will need a very fine mesh to capture the periodic microstructure appropriately to get convergence to the true solution. On the other hand, the limiting problem is a second order differential equation with constant coefficient which can easily be solved (in the case of our - problem, we can easily solve the problem analytically). We also note that such computational gains become more pronounced when we work in higher dimensions (where we have PEs). In terms of the mathematics involved, the analysis is elegant in the sense that we link several ideas from functional analysis to reach a result that even in the case of a simple - problem is not obvious. 5. References. Hirsch, F. and Lacombe, G. Elements of Functional Analysis. Springer (999). Adams, R. A. Sobolev Spaces. Academic Press (975) 3. acorogna, B. irect Methods in Calculus of Variations. Springer (989).