3 The variational formulation of elliptic PDEs

Transcription

1 Chapter 3 The variational formulation of elliptic PDEs We now begin the theoretical stuy of elliptic partial ifferential equations an bounary value problems. We will focus on one approach, which is calle the variational approach. There are other ways of solving elliptic problems. The variational approach is quite simple an well suite for a whole class of approximation methos, as we will see later. 3.1 Moel problems Let us start with a few moel problems. The simplest of all is a slight generalization of the Poisson equation with a homogeneous Dirichlet bounary conition. Let us be given a connecte 1 open Lipschitz subset of R, a function c L (), another function f L 2 (). We are looking for a function u: R such that u + cu = f in, (3.1) u = 0 on. We are going to transform the bounary value problem (3.1) into an entirely ifferent kin of problem that is amenable to an existence an uniqueness theory, as well as the efinition of approximation methos. 1 In the context of PDEs, all the open subsets consiere will be connecte, without further mention. It is not that important, but PDEs set on open sets with several connecte components are basically several unrelate PDEs.

2 76 3. The variational formulation of elliptic PDEs Proposition Assume that u H 2 () solves the PDE in problem (3.1). Then, for all v H0 1() u vx+ cuvx = fvx. (3.2) Proof. We take an arbitrary v H0 1 (), multiply the equation by v, ( u)v + cuv = fv, an integrate over. Every term is integrable since u H 2 () hence u L 2 () an v L 2 () imply ( u)v L 1 (), c L (), u L 2 () an v L 2 () imply cuv L 1 (), an f L 2 () implies fv L 1 (). We thus obtain ( u)vx+ cuvx = fvx. We now use Green s formula (2.17), accoring to which ( u)vx= u vx+ γ 1 (u)γ 0 (v)γ, an we conclue since v H 1 0 () is equivalent to γ 0(v)=0. Formulation (3.2) is calle the variational formulation of problem (3.1). Actually, it is not entirely complete, since we have not yet ecie in which space to look for u. In fact, as we have seen in the previous section, the reasonable way to impose the Dirichlet bounary conition is to require that u H0 1 (). The functions v are calle test-functions. Let us rewrite the variational formulation in a stanar, abstract form. We let V = H0 1 (), it is a Hilbert space. Then we have a bilinear form on V V a(u,v)= ( u v + cuv)x an a linear form on V (v)= The variational formulation then reas fvx. v V, a(u,v)=(v), (3.3) an we have shown that a solution of the bounary value problem with the aitional regularity u H 2 () is a solution of the variational problem (3.3). Now what about the reverse implication? Does a solution of the variational problem solve the bounary value problem? The answer is basically yes, the two problems are equivalent.

3 3.1. Moel problems 77 Proposition Assume that u V solves the variational problem (3.3). Then we have u + cu = f in the sense of D () an u L 2 (). Proof. We have D() H0 1 (), therefore we can take v = ϕ D() as testfunction in (3.3). Let us examine each term separately. First of all u ϕ x = i=1 i u i ϕ x = i=1 i u, i ϕ = i=1 by efinition of istributional erivatives. Similarly cuϕ x = cu,ϕ an f ϕ x = f,ϕ. Therefore, we have for all ϕ D() or u + cu f,ϕ = 0 u + cu f = 0 in the sense of D () ii u,ϕ = u,ϕ, an the PDE is satisfie in the sense of istributions The Dirichlet bounary conition is also satisfie by the simple fact that u H0 1 (), hence the bounary value problem is solve. To conclue, we note that u = cu f L 2 (). This also implies that the PDE is satisfie almost everywhere. Remark The two problems are thus equivalent, except for the fact that we have assume u H 2 () in one irection, an only recuperate u L 2 () in the other. 2 Actually, the assumption u H 2 () is somewhat artificial an mae only to make use of Green s formula (2.17). It is possible to ispense with it with a little more work, but that woul take us too far. It shoul be note in any case, that if u H0 1(), u L2 () an is for example of class C 2, then u H 2 (). This is very profoun result in elliptic regularity theory, far beyon the scope of these notes. It is trivial in imension one though. Of course, so far we have no inication that either problem has a solution. The fact is that the variational formulation is significantly easier to treat, once the right point of view is foun. An the right point of view is an abstract point of view, as is often the case. 2 The Laplacian is a specific linear combination of some of the secon orer erivatives. So it being in L 2 is a priori less than all iniviual secon orer erivatives, even those not appearing in the Laplacian, being in L 2.

4 78 3. The variational formulation of elliptic PDEs Before we start elving in the abstract, let us give a couple more moel problems of a ifferent kin. First is a new bounary conition. u + cu = f in, u n = g on. (3.4) This is an example of a Neumann bounary conition. When g = 0, it is naturally calle a homogeneous Neumann bounary conition. In terms of moeling, the Neumann conition is a flux conition. For instance, in the heat equilibrium interpretation, the conition correspons to an impose heat flux through the bounary, as oppose to the Dirichlet conition which imposes a temperature on the bounary. The case g = 0 correspons to perfect thermal insulation. Let us erive the variational formulation informally. Assume first that u H 2, take v H 1 (), multiply, integrate an use Green s formula to obtain v H 1 (), ( u v + cuv)x = fvx+ gγ 0 (v)γ. Note the ifferent test-function space an the aitional bounary term in the right-han sie. The converse is more interesting. Let u H 2 () be a solution of the above variational problem. Taking first v = ϕ D(), we obtain u + cu = f in the sense of D () exactly as in the Dirichlet case. Of course, a test-function with compact support oes not see what happens on the bounary, an no information on the Neumann conition is recovere. Thus, in a secon step, we take v arbitrary in H 1 (). By Green s formula again, we have u vx= ( u)vx+ γ 1 (u)γ 0 (v)γ. Recall that the normal trace γ 1 (u) plays the role of the normal erivative. Since u is a solution of the variational problem, it follows that ( u + cu)vx+ γ 1 (u)γ 0 (v)γ = fvx+ gγ 0 (v)γ. But we alreay know that ( u + cu)vx= fvx, hence we are left with γ 1 (u)γ 0 (v)γ = gγ 0 (v)γ, for all v H 1 (). For simplicity, we assume here that g H 1/2 ( ), the image of the trace γ 0 an that is smooth. Since u H 2 (), it follows that γ 1 (u) =

5 3.2. Abstract variational problems 79 i=1 γ 0( i u)n i H 1/2 ( ). Therefore, there exists v H 1 () such that γ 0 (v)= γ 1 (u) g. With this choice of v, we obtain (γ 1 (u) g) 2 Γ = 0, hence γ 1 (u)=g which is the Neumann conition. The last hypotheses mae are for brevity only. They are not at all necessary to conclue. Another problem of interest is the non homogeneous Dirichlet problem. u + cu = f in, (3.5) u = g on, with g H 1/2 ( ). This problem is reuce to the homogeneous problem by taking a function G H 1 () such that γ 0 (G)=g an setting U = u G. Then clearly U H0 1 () an U + cu = u + cu + G cg = f + G cg. Then we just write the variational formulation of the homogeneous problem for U with right-han sie F = f + G cg. The Dirichlet an Neumann conitions can be mixe together, but not at the same place on the bounary, yieling the so-calle mixe problem. More precisely, let Γ 1 an Γ 2 be two subsets of such that Γ 1 Γ 2 = /0 an Γ 1 Γ 2 =. Then the problem u + cu = f in, u = g 1 on Γ 1, (3.6) u n = g 2 on Γ 2. The variational formulation for the mixe problem (in the case g 1 = 0 for brevity, if not follow the above route) is to let V = {v H 1 ();γ 0 (v)=0 on Γ 1 } an v V, ( u v + cuv)x = fvx+ g 2 γ 0 (v)γ, Γ 2 with u V. Remark An important rule of thumb to be remembere from the above examples is that (homogeneous) Dirichlet conitions are taken into account in the test-function space, whereas Neumann bounary conitions are taken into account in the linear form via bounary integrals. 3.2 Abstract variational problems We escribe the general abstract framework for all variational problems. Let us start with a quick review of Hilbert space theory. Let H be a real Hilbert space with scalar prouct ( ) H an norm H. We first recall the Cauchy-Schwarz inequality, which is really a hilbertian property.

6 80 3. The variational formulation of elliptic PDEs Theorem For all u,v H, we have (u,v) H u H v H. One of the most basic results in Hilbert space theory is the orthogonal projection theorem, which we recall below. Theorem Let C be non empty, convex, close subset of H. For all x H, there exists a unique p C (x) C such that x p C (x) H = inf y C x y H. The vector p C (x) is calle the orthogonal projection of x on C. It is also characterize by the inequality y C, (x p C (x) y p C (x)) H 0. In aition, if C is a close vector subspace of H, then p C is a continuous linear mapping from H to C which is also characterize by the equality y C, (x p C (x) y) H = 0. C p C (x) y x Figure 1. The orthogonal projection on a close convex C. The orthogonal projection of x on C is thus the element of C closest to x an the angle between x p C (x) an y p C (x) is larger than π 2. In particular, if x C, then p C (x)=x. An important consequence of the last characterization in the case of a close vector subspace is that we can write H = C C with continuous orthogonal projections on each factor. Inee, we have x = p C (x)+(x p C (x) with p C (x) C by construction an x p C (x) C by the secon characterization. Hence H = C +C. To show that the sum is irect, it suffices to note that C C = {0} which is obvious since x C C implies 0 =(x x) H = x 2 H.

7 3.2. Abstract variational problems 81 E x p E (x) E p E (x) 0 Figure 2. The orthogonal projection on a close vector subspace E. Another important consequence is a characterization of ense subspaces. Lemma A vector subspace E of H is ense in H if an only of E = {0}. Proof. For any vector subspace, it is always true that E =(Ē). Let E be a ense subspace, i.e., Ē = H. Then, of course E = H = {0}. Conversely, if E = {0}, this implies that (Ē) = {0} an since H =(Ē) Ē, it follows that Ē = H, an E is ense in H. The Riesz theorem provies a canonical way of ientifying a Hilbert space an its ual. Theorem (Riesz) Let H be a Hilbert space an an element of its ual H. There exists a unique u H such that Moreover v H, (v)=(u v) H. H = u H an the linear mapping δ : H H, u, is an isometry.

8 82 3. The variational formulation of elliptic PDEs Proof. If = 0, then we set u = 0 to be the unique u in question. Let = 0. It is thus a nonzero continuous linear form, hence its kernel ker is a close hyperplane of H. Let us choose u 0 (ker) with u 0 H = 1 (this is possible since ker is not ense). Since u 0 / ker, we have (u 0 ) = 0 an for all v H, we can set w = v (v) (u 0 ) u 0 so that (w)= v (v) (u 0 ) u 0 = (v) (v) (u 0 ) (u 0)=0, an w ker. Now writing v = (v) (u 0 ) u 0 + w an setting u = (u 0 )u 0 (ker), we obtain (v) (v u) H = (u 0 ) u 0u +(w u) H = (v)(u 0 u 0 ) H = (v), H hence the existence of u. For the uniqueness, assume u 1 an u 2 are two solutions, then for all v H, (v u 1 u 2 ) H = 0. This is true in particular for v = u 1 u 2, hence u 1 = u 2. The mapping δ is thus well efine an obviously linear. Finally, for the isometry, we have on the one han H = sup (v) = sup (v u) H v H u H u H, v H 1 v H 1 by the Cauchy-Schwarz inequality. On the other han, equality trivially hols for = 0, an for = 0, choosing v = u u H yiels (v u) H = u H with v H = 1, hence the equality in this case too. Remark The Riesz theorem shows that the ual of a Hilbert space is also a Hilbert space for the scalar prouct ( 1 2 ) H =(δ 1 δ 2 ) H, since it is not obvious a priori that the ual norm is hilbertian. It is often use to ientify H an H via the isometry δ or δ 1. This ientification is not systematic however. For example, when we have two Hilbert spaces H an V such that V H an V is ense in H, the usual ientification is to let V H = H V using the Riesz theorem for H, which is calle the pivot space, but not for V. Such is the case for H = L 2 (), V = H0 1 () in which case we have an ientification of V as the space H 1 (). Inee, the scalar prouct use in the ientification of the pivot space an its ual is the uality bracket of an L 2 function seen as a istribution an a D test-function. This is not the case for any of the two scalar proucts that H0 1 comes equippe with, an an ientification using these scalar proucts, which is also legitimate, oes not yiel a space of istributions.

9 3.2. Abstract variational problems 83 We now come to abstract variational problems an how they are solve. This is the Lax-Milgram theorem. This theorem is important, not because it is in any way ifficult, which it is not, but because it has a very wie range of applicability as we will see later. Theorem (Lax-Milgram) Let V be a Hilbert space, a be a bilinear form an be a linear form. Assume that i) The bilinear form a is continuous, i.e., there exists a constant M such that a(u,v) Mu V v V for all u,v V, ii) The bilinear form a is V-elliptic 3, i.e., there exists a constant α > 0 such that a(v,v) αvv 2 for all v V, iii) The linear form is continuous, i.e., there exists a constant C such that (v) Cv V for all v V. There exists a unique u V that solves the abstract variational problem: Fin u V such that v V, a(u,v)=(v). (3.7) Proof. Let us start with the uniqueness. Let u 1 an u 2 be two solutions of problem (3.7). Since a is linear with respect to its first argument, it follows that a(u 1 u 2,v)=0 for all v V. In particular, for v = u 1 u 2, we obtain 0 = a(u 1 u 2,u 1 u 2 ) αu 1 u 2 2 V, so that u 1 u 2 V = 0 since α > 0. We next prove the existence of a solution. We first note that for all u V, the mapping v a(u,v) is linear (by bilinearity of a) an continuous (by i) continuity of a). Therefore, there exists a unique element Au of V such that a(u,v) = Au,v V,V. Moreover, the bilinearity of a shows that the mapping A: V V thus efine is linear. It is also continuous since for all v V, v V 1, so that Au,v V,V = a(u,v) Mu V v V Mu V Au V = sup Au,v V,V Mu V. v V 1 We rewrite the variational problem as: Fin u V such that v V, Au,v V,V = 0 or Au =, 3 This conition is also sometimes calle coerciveness.

10 84 3. The variational formulation of elliptic PDEs an this is where the continuity of is use Thus, proving the existence is equivalent to showing that the mapping A is onto. We o this in two inepenent steps: we show that ima is close on the one han an that it is ense on the other han. 4 For the closeness of the image, we use assuption ii) of V-ellipticity. Let n be a sequence in ima such that n in V. We want to show that ima, which will imply that ima is close. The sequence n is a Cauchy sequence in V, an for all n, there exists u n V such that Au n = n. By V -ellipticity, u n u m 2 V 1 α a(u n u m,u n u m )= 1 α Au n Au m,u n u m V,V = 1 α n m,u n u m V,V 1 α n m V u n u m V, by the efinition of the ual norm. Therefore, if u n u m V = 0 we are happy, otherwise we ivie by u n u m V an in both cases u n u m V 1 α n m V, so that u n is a Cauchy sequence in V. Since V is complete, there exists u V such that u n u in V. Since A is continuous, it follows that n = Au n Au in V. Hence = Au ima. To show the ensity, we show that (ima) = {0}. We note that (Au ) V = (δau δ) V = Au,δ V,V = a(u,δ). Let (ima). For all u V, we thus have a(u,δ)=0. In particular, for u = δ, we obtain 0 = a(δ,δ) αδv 2 by V -ellipticity. Since α > 0, it follows that = 0. Remark It shoul be note that the Lax-Milgram theorem is not a particular case of the Riesz theorem. It is actually more general, since it applies to bilinear forms that are not necessarily symmetric, an it implies the Riesz theorem when the bilinear form is just the scalar prouct. Sometimes when the bilinear form a is symmetric, people think it avantageous to apply Riesz s theorem in place of the Lax-Milgram theorem. This is usually an illusion: inee, if a new scalar prouct efine by the bilinear form is introuce, in orer to apply Riesz s theorem, it is necessary to show that the space equippe with the new scalar prouct is still a Hilbert space, i.e., is complete. This is one by V -ellipticity, hence nothing is gaine (although this is the part that people who think they are seeing a goo eal usually forget). The continuity of the linear form for the new norm must also be checke, which amounts to having the bilinear form an linear form continuous for the original norm, again, no gain. 4 This is a pretty common strategy, to be kept in min.

11 3.2. Abstract variational problems 85 The only case when Riesz s theorem can be eeme avantageous over the Lax-Milgram theorem, is when both above facts to be checke are alreay known. An example is the bilinear form a(u,v)= u vx on V = H1 0 (). Remark In the case of complex Hilbert spaces an complex-value variational problems, the Lax-Milgram theorem still hols true for a bilinear or sesquilinear form. The V -ellipticity assumption can even be relaxe to only involve the real part of a : R(a(u,u)) αu 2 (or the imaginary part), which is rather useful as the imaginary part can then be pretty arbitrary (exercise). The linear form in the right-han sie of a variational problem shoul be thought of as ata. In this respect, the solution epens continuously on the ata. Proposition The mapping V V, u efine by the Lax-Milgram theorem is linear an continuous. Proof. Let 1, 2 be two linear forms an u 1,u 2 the corresponing solutions to the variational problem. For all λ R, we have for all v V, a(u 1 + λu 2,v)=a(u 1,v)+λa(u 2,v)= 1 (v)+λ 2 (v)=( 1 + λ 2 )(v) hence the linearity by uniqueness of the solution. For the continuity, we have αu 2 V a(u,u)=(u) V u V, hence u V 1 α V, an we have the continuity, with continuity constant 1 α. Proposition Let the hypotheses of the Lax-Milgram theorem be satisfie. Assume in aition that the bilinear form a is symmetric. Then the solution u of the variational problem (3.7) is also the unique solution of the minimization problem: J(u)=inf J(v) with J(v)=1 a(v,v) (v). (3.8) v V 2 Proof. Let u be the Lax-Milgram solution. For all v V, we let w = v u an J(v)=J(u + w)= 1 2 a(u,u)+1 2 a(u,w)+1 2 a(w,u)+1 a(w,w) (u) (w) 2 = J(u)+a(u,w) (w)+ 1 2 a(w,w) J(u),

12 86 3. The variational formulation of elliptic PDEs since a(w,w) 0. Make note of where the symmetry is use. Hence, u minimizes J on V. Conversely, assume that u minimizes J on V. Then, for all λ > 0 an all v V, we have J(u + λv) J(u). Expaning the left-han sie, we get so that iviing by λ We then let λ 0, hence a(u,u)+λa(u,v)+λ a(v,v) (u) λ(v) J(u) 2 an finally change v in v to obtain a(u,v) l(v)+ λ a(v,v) 0. 2 a(u,v) l(v) 0, a(u,v) l(v)=0, for all v V. Remark Taking λ > 0, iviing by λ an then letting λ 0 is quite clever, an known as Minty s trick. Remark When the bilinear form a is not symmetric, we can still efine the functional J in the same fashion as before an try to minimize it. It is clear from the above proof that the minimizing element u oes not solve the variational problem associate with a but the variational problem associate with the symmetric part of a. Of course, when both variational problems are translate into PDEs, we get entirely ifferent equations. 3.3 Application to the moel problems, an more We now apply the abstract results to concrete examples. We start with the first moel problem (3.1). Proposition Let be a boune open subset of R n, f L 2 (), c L (). Assume that c 0. Then the problem: Fin u V = H0 1 () such that v V, ( u v + cuv)x = fvx, has one an only one solution.

13 3.3. Application to the moel problems, an more 87 Proof. We alreay that V is a Hilbert space, for both scalar proucts that we efine earlier. Of course a(u,v)= ( u v + cuv)x clearly efines a bilinear form on V V an (v)= fvx a linear form on V. Hence, we have an abstract variational problem. Let us try an apply the Lax-Milgram theorem. We nee to check the hypotheses. For efiniteness, we choose to work with the full H 1 norm. First of all, for all (u,v) V V, a(u,v) = ( u v + cuv)x u v + cuv x u v x+ cuv x u L 2 () v L 2 () + c L ()u L 2 () v L 2 () max 1,c L () uh 1 () v H 1 (), by Cauchy-Schwarz to go from the thir line to the fourth line, hence the continuity of the bilinear form a. Next is the V -ellipticity. For all v V, we have a(v,v)= ( v 2 + cv 2 )x v 2 x αv 2 H 1 () with α =(C 2 + 1) 1/2 > 0 by Corollary where C is the Poincaré inequality constant, an since c 0. Finally, we check the continuity of the linear form. For all v V, (v) = fvx f L 2 () v L 2 () f L 2 () v H 1 () by Cauchy-Schwarz again. All the hypotheses of the Lax-Milgram theorem are satisfie, therefore there is one an only one solution u V. Remark Now is a time to celebrate since we have successfully solve our first bounary value problem in arbitrary imension. Inee, we have alreay

14 88 3. The variational formulation of elliptic PDEs seen that any solution of the variational problem is a solution of the PDE in the istributional sense an in the L 2 sense. The solution u epens continuously in H 1 on f in L 2. Note that we have also solve the non homogeneous Dirichlet problem at the same time. It is an instructive exercise to reo the proof using the H 1 semi-norm in place of the full norm. The same ingreients are use, but not at the same spots. This is a case of a symmetric bilinear form, therefore the solution u also minimizes the so-calle energy functional J(v)= 1 2 ( v 2 + cv 2 )x fvx over V. Remark It shoul be note that the positivity conition c 0 is by no means a necessary conition for existence an uniqueness via the Lax-Milgram theorem. With a little more work, it is not too har to allow the function c to take some negative values. However, we have seen an example at the very beginning of these notes with a negative function c for which existence an uniqueness fails. One shoul also be aware that there is an existence an uniqueness theory that goes beyon the Lax-Milgram theorem, which is only a sufficient conition for existence an uniqueness. Let us now consier the non homogeneous Neumann problem (3.4). The hypotheses are slightly ifferent. Proposition Let be a boune open subset of R n, f L 2 (), c L (), g L 2 ( ). Assume that there exists a constant η > 0 such that c η almost everywhere. Then the problem: Fin u V = H 1 () such that v V, has one an only one solution. ( u v + cuv)x = fvx+ gγ 0 (v)x, Proof. We have a ifferent Hilbert space (but known to be Hilbert, nothing to check here), the same bilinear form an a ifferent linear form (v)= fvx+ gγ 0 (v)x.

15 3.3. Application to the moel problems, an more 89 We have alreay shown that the bilinear form is continuous in the H 1 norm 5. The V -ellipticity is clear since, for all v V, a(v,v)= ( v 2 + cv 2 )x v 2 x+ η v 2 x min(1,η)v 2 H 1 (), with min(1,η) > 0. The continuity of the linear form is also clear (v) f L 2 () v L 2 () + g L 2 ( ) γ 0(v) L 2 ( ) ( f L 2 () +Cg L 2 ( ) )v H 1 () by Cauchy-Schwarz, where C is continuity constant of the trace mapping. The mixe problem (3.6) is practically entirely ientical to the Neumann problem. Proposition Same hypotheses as in Proposition an let Γ 1 an Γ 2 be two subsets of such that Γ 1 Γ 2 = /0 an Γ 1 Γ 2 =. Then the problem: Fin u V = {v H 1 ();γ 0 (v)=0 on Γ 1 } such that v V, ( u v + cuv)x = fvx+ gγ 0 (v)γ, Γ 2 has one an only one solution. Proof. The only real ifference with Proposition lies with the space V, which we o not know yet to be a Hilbert space. It suffices to show that V is a close subspace of H 1 (). Let v n be a sequence in V such that v n v in H 1 (). By continuity of the trace mapping, we have γ 0 (v n ) γ 0 (v) in L 2 ( ). Therefore, there exists a subsequence γ 0 (v np ) that converges to γ 0 (v) almost everywhere on. Since γ 0 (v n )=0 almost everywhere on Γ 1, it follows that γ 0 (v)=0 almost everywhere on Γ 1, hence v V, which is thus close. A natural question arises about the Neumann problem without a strictly positive boun from below for the function c, in particular for c = 0. Now, this is an entirely ifferent problem from the previous ones. First we have to fin the variational formulation of the bounary value problem an show that it is equivalent to the bounary value problem, then we have to apply the Lax-Milgram theorem. Let us thus consier the Neumann problem u = f in, u n = g on, (3.9) 5 If we ha worke with the semi-norm for the Dirichlet problem, we woul have ha to o the continuity all over again here...

16 90 3. The variational formulation of elliptic PDEs in a Lipschitz open set in R. We see right away that things are going to be ifferent since we o not have uniqueness here. Inee, if u is a solution, then u+s is also a solution for any constant s. Furthermore, by Green s formula (2.17) with v = 1, it follows that if there is a solution, then, necessarily fx+ gγ = 0. (3.10) If the ata f,g oes not satisfy the compatibility conition (3.10), there is thus no solution. The two remarks, non uniqueness an non existence, are actually ual to each other. There are several ways of going aroun both problems, thus several variational formulations 6. We choose to set, V = v H 1 (); vx= 0. This is well efine, since is boune an we thus have H 1 () L 2 () L 1 (). Note that V is the L 2 -orthogonal to the one-imensional space of constant functions, which are the cause of non uniqueness. Lemma The space V is a Hilbert space for the scalar prouct of H 1 (). Proof. It suffices to show that V is close. Let v n be a sequence in V such that v n v in H 1 (). Of course, v n v in L 2 () an by Cauchy-Schwarz, v n v in L 1 (). Therefore 0 = v n x vx, an v V. Proposition Assume that f L 2 (), g L 2 ( ) satisfy the compatibility conition (3.10). Then the triple V, a(u,v) = u vx, (v) = fvx+ gγ 0(v)Γ efines a variational formulation for the Neumann problem (3.9). Proof. Multiplying the PDE by v V an using Green s formula, we easily see that if u solves problem (3.9), then we have for all v V, a(u,v)=(v). Conversely, let us be given a function u V such that for all v V, a(u,v)= (v). We woul like to procee as before an take v D() to euce the PDE. This oes not work here because D() V. For all ϕ D(), we set ψ = ϕ 1 meas ϕ(x)x, 6 We have always sai the variational formulation, but there is no evience that it is unique in general.

17 3.3. Application to the moel problems, an more 91 so that ψ V an we can use ψ as a test-function. Now ϕ an ψ iffer by a constant k = meas 1 ϕ(x)x, therefore ψ = ϕ. We thus obtain, u ϕ x = f ψ x+ gψ Γ = f (ϕ + k)x+ g(ϕ + k)γ = f ϕ x+ k fx+ gγ = f ϕ x, since ϕ vanishes on an f,g satisfy conition (3.10). So we can euce right away that u = f in the sense of istributions, an since f L 2 () in the sense of L 2 () as well. We next pick an arbitrary v V an apply Green s formula again. This yiels fvx+ gγ 0 (v)γ = ( u)vx+ γ 1 (u)γ 0 (v)γ. Hence, taking into account that u = f, we obtain (g γ 1 (u))γ 0 (v)γ = 0 for all v V. Now it is clear that γ 0 (V )=H 1/2 ( ). Inee, let is pick a θ D() such that θ x = 1. Then, for all w H1 (), v = w wxθ V an γ 0 (v)=γ 0 (w). Therefore, there are enough test-functions in V to conclue that γ 1 (u)=g, since H 1/2 ( ) is ense in L 2 ( ). Remark We i not insist on the regularity neee to apply Green s formula or to efine γ 1 (u) as an element of H 1/2, because it is possible to write own slightly more complicate arguments that completely o away with such artificial hypotheses. To apply the Lax-Milgram theorem, we nee a new inequality. Theorem (Poincaré-Wirtinger inequality) Let be a Lipschitz open subset of R. There exists a constant C epening on such that, for all v H 1 (), v 1 vx meas C v L 2 L () 2 (). (3.11) Proof. We amit the Poincaré-Wirtinger inequality.

18 92 3. The variational formulation of elliptic PDEs Remark Even though there is a certain similarity with the Poincaré inequality, there are major ifferences. In particular, the Poincaré-Wirtinger inequality fails for open sets that are not regular enough (Lipschitz is sufficient for it to hol), whereas no regularity is neee for Poincaré. Instea of proving it, let us just note that the Poincaré-Wirtinger is at least reasonable, since both sies vanish for constant functions v. Proposition Let be a Lipschitz open subset of R, f L 2 () an g L 2 ( ). Then the problem: Fin u V such that v V, u vx= fvx+ gγ 0 (v)γ, has one an only one solution. Proof. We have alreay shown that V is a Hilbert space for the H 1 scalar prouct. The continuity of both bilinear an linear forms have also alreay been prove. Only the V -ellipticity remains. For all v V, we have vx= 0, hence by the Poincaré-Wirtinger inequality (3.11), v 2 H 1 () = v2 L 2 () + v2 L 2 () (C2 + 1) v 2 L 2 (). Therefore, a(v,v)= v 2 L 2 () αv2 H 1 () with α = 1 (C 2 +1) > 0. Remark The compatibility conition (3.10) plays no role in the application of the Lax-Milgram theorem. So exercise: What happens when it is not satisfie? What exactly are we solving then? Since the space V is a hyperplane of H 1 that is L 2 orthogonal to the constants, it follows that the general solution of the Neumann problem is of the form v + s, where v V is the unique solution of the variational problem above an s R is arbitrary. We now introuce a new kin of bounary conition, the Fourier conition (also calle the Robin conition or the thir bounary conition). The bounary value problem reas u + cu = f in, bu + u n = g on, (3.12) where b an c are given functions. When b = 0, we recognize the Neumann problem (an, in a sense, when b =+ the Dirichlet problem). This conition

19 3.3. Application to the moel problems, an more 93 is calle after Fourier who introuce it in the context of the heat equation. In the heat interpretation, u n represents the heat flux through the bounary. Let us assume that we are moeling a situation in which the bounary is actually a very thin wall that insulates from the outsie where the temperature is 0. If g = 0, the Fourier conition states that u n = bu, that is to say that the heat flux passing through the wall is proportional to the temperature ifference between the insie an the outsie. For this interpretation to be physically reasonable, it is clearly necessary that b 0, i.e., the heat flows inwars when the outsie is warmer than the insie an conversely. It thus to be expecte that the sign of b will play a role. We follow the same pattern as before: First fin a variational formulation for the bounary value problem (3.12), secon apply Lax-Milgram to prove existence an uniqueness. Proposition Assume that f L 2 (), g L 2 ( ), c L () an b L ( ). Then the triple V = H 1 (), a(u,v)= ( u v + cuv)x+ (v)= fvx+ gγ 0 (v)γ, bγ 0 (u)γ 0 (v)γ, efines a variational formulation for the Fourier problem (3.12). Proof. As always, we multiply the PDE by v V an use Green s formula, ( u v + cuv)x = fvx+ γ 1 (u)γ 0 (v)γ = fvx+ (g bγ 0 (u))γ 0 (v)γ, hence ( u v + cuv)x+ bγ 0 (u)γ 0 (v)γ = fvx+ gγ 0 (v)γ, (3.13) for all v V. Conversely, let us be given a solution u of the variational problem (3.13). Taking first v = ϕ D(), all the bounary integrals vanish an we obtain u + cu = f exactly as before. Taking then v arbitrary, using Green s formula an the PDE just obtaine, we get γ 1 (u)γ 0 (v)γ + bγ 0 (u)γ 0 (v)γ = gγ 0 (v)γ, so that (γ 1 (u)+bγ 0 (u) g)γ 0 (v)γ = 0, for all v V = H 1 (), hence the Fourier bounary conition.

20 94 3. The variational formulation of elliptic PDEs Remark A natural question to ask is why not keep the term γ 1 (u) in the bilinear form? The answer is that, while it is true that γ 1 (u) exists when u is a solution of either the bounary value problem or the variational problem, it oes not exist for a general v H 1 (), hence cannot appear in a bilinear form that is efine on H 1 () H 1 (). Besies, how woul b appear otherwise? Let us give a first existence an uniqueness result. Proposition Let be a Lipschitz open subset of R, f L 2 (), g L 2 ( ), c L () an b L ( ). Assume that c η > 0 for some constant η an that b L ( ) < min(1,η), where C Cγ 2 γ0 is the continuity constant of the trace mapping. 0 Then the problem: Fin u V such that v V, ( u v + cuv)x+ bγ 0 (u)γ 0 (v)γ = fvx+ gγ 0 (v)γ, has one an only one solution. Here b = min(0,b) enotes the negative part of b. Proof. We check the hypotheses of the Lax-Milgram theorem. We alreay know that V is a Hilbert space. The continuity of the bilinear form a has also alreay been checke, except for the bounary integral terms bγ 0 (u)γ 0 (v)γ b L ( )γ 0 (u) L 2 ( ) γ 0(v) L 2 ( ) C 2 γ 0 b L ( )u H 1 () v H 1 () for all u an v. The linear form is also known to be continuous. Let us check the V -ellipticity. Obviously b b, thus ( v 2 + cv 2 )x+ bγ 0 (v) 2 Γ min(1,η)v 2 H 1 () b L ( )γ 0 (v) 2 L 2 ( ) min(1,η) Cγ 2 0 b L ( ) v 2 H 1 (), hence the V -ellipticity. Remark Uner the previous hypotheses, we have existence an uniqueness via Lax-Milgram provie b is not too negative in some sense.

21 3.3. Application to the moel problems, an more 95 All these hypotheses only give sufficient conitions. Let us give another set of such hypotheses. Proposition Same hypotheses except that we assume that c 0 an that b µ > 0 for some constant µ. Then the Fourier problem has one an only one solution. Proof. The only point to be establishe is V-ellipticity. We use a compactness argument by contraiction. For this we amit that Rellich s theorem is also true in imension in a Lipschitz open set. We have ( v 2 + cv 2 )x+ bγ 0 (v) 2 Γ v 2 x+ µ γ 0 (v) 2 Γ. Let us assume for contraiction that there is no constant α > 0 such that v 2 x+ µ γ 0 (v) 2 Γ αv 2 H 1 (). This implies that for all n N, there exists v n H 1 () such that v n 2 x+ µ γ 0 (v n ) 2 Γ < 1 n v n 2 H 1 (). We can assume without loss of generality that v n 2 H 1 () = 1, (3.14) an we have v n 2 x+ µ γ 0 (v n ) 2 Γ 0. (3.15) Now v n is boune in H 1 () by (3.14), thus relatively compact in L 2 () by Rellich s theorem. We may extract a subsequence, still enote v n, an v L 2 () such that v n v in L 2 (). By (3.15), v n L 2 () 0, therefore, since v n v in D (), we have v = 0 an v is constant on each connecte component of. Moreover, v n v 2 H 1 () = v n 2 L 2 () + v n v 2 L 2 () 0 (3.16) so that, by continuity of the trace mapping γ 0 (v n ) γ 0 (v) in L 2 ( ). By (3.15) again, we also have γ 0 (v n ) L 2 ( ) 0 since µ > 0 an therefore γ 0(v) =0. It follows that v being a constant with zero trace vanishes in each connecte component, i.e., v = 0. We now realize that (3.14) an (3.16) contraict each other, therefore our premise that there exists no V -ellipticity constant α is false. Remark This is a typical compactness argument: we can prove that the constant exists but we have no iea of its value.

22 96 3. The variational formulation of elliptic PDEs 3.4 General secon orer elliptic problems Up to now, the partial ifferential operator always was the Laplacian. Let us rapily consier more general secon orer elliptic operators in a Lipschitz open subset of R. We are given a matrix-value function A(x)=(a ij (x)) with a ij C 1 ( ). Let u C 2 () (we can lower this regularity consierably), then A u is a vector fiel with components whose ivergence is given by iv(a u)= = (A u) i = i=1 i (A u) i a ij ij u + i, j=1 a ij j u j=1 i a ij j u. j=1 i=1 The principal part of this operator i, j=1 a ij ij is of the secon orer. We will consier the bounary value problem iv(a u)+cu = f in, u = h on Γ 0, (3.17) bu + n A u = g on Γ 1, where c, b, f, g an h are given functions an Γ 0, Γ 1 a partition of as in the mixe problem. When A = I, we recognize iv(a u)= u an n A u = u n, so that we are generalizing all the moel problems seen up to now. First of all, we reuce the stuy to the case h = 0 by subtracting a function with the appropriate trace, as before. Proposition Assume that f L 2 (), g L 2 (Γ 1 ), c L () an b L (Γ 1 ). Then the triple V = {v H 1 ();γ 0 (v)=0 on Γ 1 }, a(u,v)= (A u v + cuv)x+ bγ 0 (u)γ 0 (v)γ, Γ 1 (v)= fvx+ gγ 0 (v)γ, Γ 1 efines a variational formulation for problem (3.17).

23 3.4. General secon orer elliptic problems 97 Proof. The proof is routine, but we write it partially own for completeness. Multiply the PDE by v V an integrate by parts. This yiels first then i=1 i (A u) i vx+ cuvx = fvx, i=1 (A u) i i vx Γ 1 i=1 (A u) i n i γ 0 (v)γ + an finally (A u v + cuv)x+ bγ 0 (u)γ 0 (v)γ = Γ 1 We leave the converse to the reaer. cuvx = fvx, fvx+ gγ 0 (v)γ. Γ 1 Proposition Let be a Lipschitz open subset of R, f L 2 (), g L 2 (Γ 1 ), c L () an b L (Γ 1 ). We assume that the matrix A is uniformly elliptic, that is to say that there exists a constant α > 0 such that a ij (x)ξ i ξ j αξ 2 i, j=1 for all x an all ξ R. We assume in aition that c η > 0 for some constant η an that b 0. Then the problem: Fin u V = {v H 1 ();γ 0 (v)= 0 on Γ 1 } such that v V, (A u v + cuv)x+ bγ 0 (u)γ 0 (v)γ = fvx+ gγ 0 (v)γ, Γ 1 Γ 1 has one an only one solution. Proof. That V is a Hilbert space an that is continuous are alreay known facts. We leave the proof of the continuity of the bilinear form, which is implie by the bouneness of the matrix coefficients a ij (x). The V-ellipticity is also quite obvious, since a(v,v)= (A v v + cv 2 )x+ bγ 0 (v) 2 Γ Γ 1 α v 2 x+ η v 2 x min(α,η)v 2 H 1 (), hence the existence, uniqueness an continuous epenence of the solution on the ata by Lax-Milgram.

24 98 3. The variational formulation of elliptic PDEs Remark When the matrix A is not symmetric, neither is the bilinear form a, even though the principal part of the operator is symmetric since i, j=1 a ij ij = a ij +a ji i, j=1 2 ij ue to the fact that ij = ji. When A is symmetric, then so is the bilinear form an we have an equivalent minimization problem with J(v)= 1 (A v v + cv 2 )x+ bγ 0 (v) 2 Γ fvx gγ 0 (v)γ, 2 Γ 1 Γ 1 minimize over V. It is quite clear that we can reuce the regularity of A own to L without loosing the existence an uniqueness of the variational problem. The interpretation in terms of PDEs stops at the ivergence form iv(a u)+cu = f since we cannot evelop the ivergence using Leibniz formula in this case. Such lack of regularity is useful to moel heterogeneous meia. We now give an example of a non symmetric problem, the convection-iffusion problem. Let us be given a vector fiel σ. The convection-iffusion problem reas u + σ u + cu = f in, (3.18) u = 0 on. We have a iffusion term u an a transport term σ u in the same equation that compete with each other. Proposition Assume that f L 2 (), σ C 1 ( ;R ) an c L (). Then the triple V = H0 1 (), a(u,v)= u v +(σ u + cu)v x, (v)= fvx, efines a variational formulation for problem (3.18). Proof. This is really routine now... Note that the bilinear form is not symmetric. Proposition Let be a boune open subset of R, f L 2 (), σ C 1 ( ;R ) an c L (). We assume that c 1 2 ivσ 0. Then the problem: Fin u V such that v V, u v +(σ u + cu)v x = fvx, has one an only one solution.

25 3.4. General secon orer elliptic problems 99 Proof. We just prove the V -ellipticity. We have a(v,v)= v 2 + cv 2 +(σ v)v x. Let us integrate the last integral by parts (σ v)vx= σ i i v vx i=1 = i (σ i v) vx= i=1 i σ i v 2 x i=1 σ i i v vx i=1 = ivσv 2 x (σ v)vx, since all bounary terms vanish, so that Therefore a(v,v)= (σ v)vx= 1 2 ivσv 2 x. v 2 + c 1 2 ivσ v 2 x v 2 H 1 (), hence the result by the equivalence of the H 1 semi-norm an the H 1 norm on H0 1(). Remark We thus have existence an uniqueness if c = 0 an ivσ = 0. The case ivσ = 0 is interesting because if σ represents the velocity fiel of such a flui as air or water, the ivergence free conition is the expression of the incompressibility of the flui. Uner usual experimental conitions, both fluis are in fact consiere to be incompressible. Let us now give a fourth orer example, even though only secon orer problems were avertise in the section title. We consier a slight variant of the plate problem with homogeneous Dirichlet bounary conitions 2 u + cu = f in, u = 0 on, (3.19) u n = 0 on, with boune in R ( = 2 in the actual case of a plate).

26 The variational formulation of elliptic PDEs The erivation of a variational formulation is again fairly routine, but since this is our first (an only) fourth orer problem, we give some etail. The variational space for this Dirichlet problem is V = H0 2 () which incorporates the two bounary conitions. Assume that u H 4 () H0 2(). Then u H2 () an we can use Green s formula ( 2 u)vx= = ( ( u))vx u vx+ (γ 0 (v)γ 1 ( u) γ 1 (v)γ 0 ( u))γ = u vx since γ 0 (v)=γ 1 (v)=0 for all v H0 2 (). So we have our variational formulation v V, ( u v + cuv)x = fvx, (3.20) which is easily checke to give rise to a solution of the bounary value problem. Let 2 v enote the collection of all 2 secon orer partial erivatives of v. We have Lemma The semi-norm 2 v L 2 () is a norm on H2 0 () that is equivalent to the H 2 norm. Proof. It is enough to establish a boun from below. Let v H0 2 (). Then we have i v H0 1() for all i. Therefore ( iv) 2 L 2 () C2 i v 2, by Poincaré. Now H 1 () of course i v 2 H 1 () = ( iv) 2 L 2 () + iv 2 L 2 (), so summing in i, we get 2 v 2 L 2 () = i=1 ( i v) 2 L 2 () C2 2 v 2 L 2 () + v 2 H 1 () C 2 2 v 2 L 2 () +C4 v 2 H 1 () C4 v 2 H 2 () since v H0 1 () an C 1. Proposition Let be a boune open subset of R, f L 2 () an c L (). We assume that c 0. Then problem (3.20) has one an only one solution.

27 3.4. General secon orer elliptic problems 101 Proof. We just prove the V -ellipticity. We have a(v,v) ( v) 2 x. We argue by ensity. Let ϕ D(), since ϕ = i=1 iiϕ, we can write ( ϕ) 2 x = i=1 ii ϕ jj ϕ x = j=1 = i, j=1 ii ϕ jj ϕ x i, j=1 i ϕ ijj ϕ x = ij ϕ ij ϕ x i, j=1 with two successive integrations by parts, the first with respect to x i an the secon with respect to x j. Hence, for all ϕ D(), we obtain ( ϕ) 2 x = ( ij ϕ) 2 x = 2 ϕ 2 L i, j=1 2 (). Now, by efinition, H0 2() is the closure of D() in H2 (), thus for all v H0 2(), there exists a sequence ϕ n D() such that ϕ n v in H 2 (). Passing to the limit in the above equality, we thus get ( v) 2 x = 2 v 2 L 2 (), since ij ϕ n ij v in L 2 (), hence the result by Lemma Remark Notice the trick use in the above proof. To establish an equality for H 2 functions, we nee to use thir erivatives, which make no sense as functions in this context. However, the formula is vali for smooth functions, for which thir erivatives are not a problem, an since in the en the formula in question oes not involve any erivatives of orer higher than two, it extens to H 2 by ensity. The formula is actually surprising, since v oes not contain any erivative ij v with i = j, an only the sum of all ii v erivatives. Its L 2 norm square is nonetheless equal to the sum of the L 2 norms square of all iniviual secon erivatives. This is relate to elliptic regularity, which was mentione in passing before. Remark This is another symmetric problem, hence we have an equivalent energy minimization formulation with J(v)= 1 ( v) 2 x fvx, 2 to be minimize on H 2 0 ().

28 The variational formulation of elliptic PDEs To conclue this section, we iscuss the general three point strategy for solving elliptic problems that was repeately applie here. First we establish a variational formulation: (homogeneous) Dirichlet bounary conitions are enforce by the testfunction space, which is inclue in H 1 for secon orer problems; we multiply the PDE by a test-function possibly assuming aitional regularity on the solution an use integration by parts or Green s formula to obtain the variational problem. The bilinear form must be well-efine on the test-function space. The secon point is to check that the variational formulation actually gives rise to a solution of the bounary value problem. This point is usually itself in two steps: first obtain the PDE in the sense of istributions by using test-functions in D, secon retrieve Neumann or Fourier bounary conitions by using the full test-function space. The first two points can appear somewhat formal because of the assume regularity on the solution that is not always easily obtaine in the en. This is not a real problem, since it is possible to write rigorous arguments, at the expense of more theory than we nee here. The final thir point is to try an apply the Lax-Milgram theorem, by making precise regularity an possibly sign assumptions on the various functions that act as ata. Here we prove existence an uniqueness of the solution to the variational problem. A question that can be aske is what is the relevance of such solutions to a bounary value problem, in which the partial erivatives are taken in a rather weak sense. This is where elliptic regularity theory comes into play. Using elliptic regularity, it is possible to show that the variational solution is inee the classical solution, provie the ata (coefficients, right-han sie, bounary of ) is smooth enough.