Lecture Notes on PDEs

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1 Lecture Notes on PDEs Alberto Bressan February 26, Elliptic equations Let IR n be a bounded open set Given measurable functions a ij, b i, c : IR, consider the linear, second order differential operator Lu = (a ij (xu xi xj + (b i (xu xi + c(xu (11 In this chapter, we study solutions to the boundary value problem Lu = f x, u = 0 x, (12 where f L 2 ( is a given function For future reference, we collect the main assumptions used throughout this chapter (H The domain IR n is open and bounded The coefficients of L in (11 satisfy a ij, b i, c L ( (13 Moreover, the operator L is uniformly elliptic Namely, there exists a constant θ > 0 such that a ij (xξ i ξ j θ ξ 2 for all x, ξ IR n (14 Remark 11 By definition, the uniform ellipticity of the operator L depends only on the coefficielts a ij In the symmetric case where a ij = a ji, the above condition means that for every x the n n symmetric matrix A(x = (a ij (x is strictly positive definite, and its smallest eigenvalue is θ 11 Physical interpretation As an example, consider a fluid moving with velocity b(x = (b 1, b 2, b 3 (x in a domain IR 3 Let u = u(t, x describe the density of a chemical dispersed within the fluid 1

2 Given any subdomain V, we assume that the total amount of chemical contained in V changes only due to the inward or outward flow through the boundary V Namely: d u dx = n (a u ds n (b u ds (15 dt V V Here n(x denotes the unit outer normal to the set V at a boundary point x, while a > 0 is a constant diffusion coefficient The first integral on the right hand side of (15 describes how much chemical enters through the boundary by diffusion Notice that this is positive if n u > 0 Roughly speaking, this is the case if the concentration of chemical outside the domain V is greater than inside The second integral (with the minus sign in front denotes the amount of chemical that moves out across the boundary of V by advection, being transported by the fluid in motion (Fig 1 V n b V Figure 1: As the chemical is transported across the boundary of V, the total amount contained inside V changes in time Using the divergence theorem, from (15 we obtain u t, dx = a u dx V V V div(b u dx (16 Since the above identity holds on every subdomain V, we conclude that u = u(t, x satisfies the parabolic PDE u t a u + div(b u = 0 [diffusion] [advection] A more general model can describe the following situations: The diffusion is not uniform throughout the domain In other words, the coefficient a is not a constant but depends on the location x Moreover, the diffusion is not isotropic: in some directions it is faster than in others All this can be modeled by replacing the constant diffusion matrix A(x ai with a more general symmetric matrix A(x = (a ij (x The total amount of chemical is not conserved Additional terms are present, accounting for linear decay and for an external source In n space dimensions, this leads to a linear evolution equation of the form ( u t (a ij (xu xi + b i (xu = c(x u + f(x x j x i [diffusion] [advection] [decay] [source] (17 2

3 The equation (17 can be used to model a variety of phenomena, including mass transport, heat propagation, etc In many situations, one is interested in steady states, ie in solutions which are independent of time Setting u t = 0 in (17 we obtain the linear elliptic equation (a ij (xu xi + x j ( b i (xu + c(x u = f(x (18 x i 12 Weak solutions By a classical solution of the boundary value problem (12 we mean a function u C 2 ( which satisfies the equation and the boundary conditions at every point In general, due to the lack of regularity in the coefficients of the equation, we do not expect that (12 will have classical solutions A weaker concept of solution is thus needed Definition 11 A weak solution of (12 is a function u H0 1 ( such that ( n a ij u xi v xj b i u v xi + cuv dx = fv dx for all v H0 1 ( (19 Remark 12 The above equality is formally obtained by writing (Lu v dx = f v dx for all v Cc (, (110 and integrating by parts Notice that, if (110 holds for all test functions v Cc (, then by an approximation argument the same integral identity is valid for all v H0 1 ( We observe that a function u H0 1 may not have weak derivatives of second order However, the integral in (19 is always well defined, for all u, v H0 1 A convenient way to reformulate the concept of weak solution is the following On the Hilbert space H0 1 (, consider the bilinear form B[u, v] = a ij u xi v xj b i uv xi + c uv dx (111 A function u H 1 0 is a weak solution of (12 provided that Here and in the sequel we use the notation B[u, v] = (f, v L 2 for all v H 1 0 (112 (f, g L 2 = f g dx (113 for the inner product in L 2 (, to distinguish it from the inner product in H 1 ( (f, g H 1 = f g dx + f xi g xi dx (114 3

4 Remark 13 The minus sign in front of the second order terms in (11 disappears in (111, after a formal integration by parts As it will become apparent later, this sign is chosen so that the corresponding quadratic form B[u, v] can be positive definite Remark 14 (general boundary conditions The homogeneous boundary condition u = 0 on is incorporated in the requirement u H0 1 ( More generally, given a function g H 1 (, one can consider the non-homogeneous boundary value problem Lu = f x, u = g x This can be rewritten as a homogeneous problem for the function ũ = u g, namely Lũ = f Lg x, ũ = 0 x (115 (116 Assuming that Lg L 2, the problem (116 is exactly of the same type as (12 Remark 15 (operators not in divergence form A differential operator of the form Lu = a ij (xu xi x j + b i (xu xi + c(xu can be rewritten as Lu = ( (a ij (xu xi xj + b i (x + a ij x j (x u xi + c(xu j=1 Assuming that a ij, a ij x j, b i, c L (, a weak solution of the corresponding problem (12 can be again obtained by solving (112, where the bilinear form B is now defined by B[u, v] = ( a ij u xi v xj + b i + a ij x j u xi v + c uv dx j=1 As a first example, consider the boundary value problem u + u = f x, u = 0 x (117 Clearly, the operator u = i u x i x i is uniformly elliptic, because in this case the n n matrix A(x = (a ij (x is the identity matrix, for every x The existence of solutions to (117 is provided by Lemma 11 Let IR n be a bounded open set Then for every f L 2 ( the boundary value problem (117 has a unique weak solution u H0 1 ( The corresponding map f u is a compact linear operator from L 2 ( into H0 1( 4

5 Proof By the Rellich-Kondrachov theorem, the canonical embedding ι : H 1 0 ( L2 ( is compact Hence its dual operator ι is also compact Since H 1 0 and L2 are Hilbert spaces, they can be identified with their duals We thus obtain the following diagram: H 1 0 ( H 1 0 ( = [H1 0 (] ι L 2 ( ι [L 2 (] = L 2 ( For each f L 2 (, the definition of dual operator yields (ι f, v H 1 = (f, ιv L 2 = (f, v L 2 for all f L 2 (, v H 1 0 ( By (114 this means that ι f is precisely the weak solution to (117 ( Existence and uniqueness of solutions We begin by studying solutions to the elliptic boundary value problem Lu = f x, u = 0 x, (119 assuming that the differential operator L contains only second order terms: Lu = (a ij (xu xi xj (120 We recall that a weak solution of (119 is a function u H0 1 ( such that B[u, v] = (f, v L 2 = (ι f, v H 1 for all v H 1 0 (, (121 where B : H 1 0 H1 0 IR is the continuous bilinear form B[u, v] = a ij u xi v xj dx (122 Theorem 11 (solution of a linear elliptic boundary value problem - I Let IR n be a bounded open set Let the operator L in (120 be uniformly elliptic, with coefficients a ij L ( Then, for every f L 2 (, the boundary value problem (119 has a unique weak solution u H 1 0 ( The corresponding solution operator, which we denote as L 1 : f u is a compact linear operator from L 2 ( into H 1 0 ( Proof The existence and uniqueness of weak solutions to the elliptic boundary value problem (119 will be achieved by checking that the bilinear form B 0 satisfies all the assumptions of the Lax-Milgram theorem 1 The continuity of B is clear Indeed, B[u, v] a ij u xi v xj dx a ij L u xi L 2 v xj L 2 C u H 1 v H 1 5

6 2 We claim that B is strictly positive definite, ie there exists β > 0 such that B[u, u] β u 2 H 1 ( for all u H 1 0 ( (123 Indeed, since is bounded, Poincaré s inequality yields the existence of a constant κ such that κ u 2 dx for all u H0 1 ( u 2 L 2 ( On the other hand, the uniform ellipticity condition implies B[u, u] = a ij u xi u xj dx Together, the two above inequalities yield θ u 2 x i dx = θ u 2 dx u 2 H 1 = u 2 L 2 + u 2 L 2 (κ + 1 u 2 L 2 κ + 1 θ This proves (123 with β = θ/(κ + 1 B[u, u] 3 By the Lax-Milgram theorem, for every f H 1 0 ( there exists a unique element u H1 0 such that B[u, v] = ( f, v H 1 for all v H 1 0 ( (124 Moreover, the map Λ : f u is continuous, namely u H 1 β 1 f H 1 Choosing f = ι f H0 1 (, defined at (118, we thus achieve B[u, v] = (ι f, v H 1 = (f, v L 2 for all v H 1 0 ( (125 By definition, u is a weak solution of (119 4 To prove that the solution operator L 1 : f u is a compact, consider the the diagram L 2 ( ι H 1 0 ( Λ H 1 0 ( By Lemma 11, the linear operator ι is compact Moreover, Λ is continuous Therefore the composition L 1 = Λ ι is compact 14 Representation of solutions We now study in more detail the structure of the solution operator L 1 for the elliptic boundary value problem (119-(120 Theorem 12 (representation of solutions in terms of eigenfunctions Assume a ij = a ji L ( Then, in the setting of Theorem 11, the linear operator L 1 : L 2 ( L 2 ( is compact, one-to-one, and self-adjoint 6

7 The space L 2 ( admits an orthonormal basis φ k ; L 1, and one has the representation k 1} consisting of eigenfunctions of L 1 f = λ k (f, φ k L 2 φ k (126 The corresponding eigenvalues λ k satisfy lim λ k = 0, λ k > 0 for all k 1 (127 k Proof 1 By Theorem 11, L 1 is a compact linear operator from L 2 ( into itself 2 If u = L 1 f = 0, then 0 = B[u, v] = (f, v L 2 for every v H0 1( In particular, for every test function φ C c ( we have f φ dx = 0 This implies f(x = 0 for ae x one-to-one Hence Ker(L 1 = 0} and the operator L 1 is 3 To prove that L 1 is self-adjoint, assume f, g L 2 (, u = L 1 f, v = L 1 g Recalling the definition of weak solution at (19, this implies u, v H0 1 ( and (L 1 f, g L 2 = u g dx = a ij u xi v xj dx = f v dx = (f, L 1 g L 2 i,j Note that the second equality follows from the fact the v = L 1 g, using u H0 1 ( as a test function The third equality follows from the fact the u = L 1 f, using v H0 1 ( as a test function 4 By the previous steps, we can now apply Theorem?? (Hilbert-Schmidt, providing the existence of an orthonormal basis of eigenfunctions of L 1 This yields (126 By the compactness of the operator L 1, the eigenvalues satisfy λ k 0 Finally, choosing v = φ k as test function in (121, one obtains B[λ k φ k, φ k ] = (φ k, φ k L 2 = 1 Since the quadratic form B is strictly positive definite, we conclude λ k = 1 B[φ k, φ k ] > 0 7

8 Example 1 Let = ]0, π[ IR and Lu = u xx Given f L 2 (]0, π[, consider the elliptic boundary value problem uxx = f 0 < x < π, (128 u(0 = u(π = 0 As a first step, we compute the eigenfunctions of L 0 Solving the boundary value problem u xx = µu, u(0 = u(π = 0 we find the eigenvalues and the normalized eigenfuntions 2 µ k = k 2, φ k (x = sin kx π Of course, the inverse operator L 1 has the same eigenfunctions φ k, with eigenvalues λ k = 1/k 2 In this special case, the formula (126 yields the well known representation of solutions of (128 in terms of a Fourier sine series: ( u(x = L 1 1 π 2 2 f = λ k (f, φ k L 2 φ k = k 2 f(y sin ky dy sin kx π π = 2 πk 2 ( π 0 f(y sin ky dy sin kx 0 15 General linear elliptic operators The existence and uniqueness result stated in Theorem 11 relied on the fact that, by Poincaré s inequality, the bilinear form B in (121 is strictly positive definite on the space H0 1 ( These properties no longer hold for the more general bilinear form B in (111 For example, if the function c(x is large and negative, one may find some u H0 1 ( such that B[u, u] < 0 Example 2 Consider the one-dimensional open interval = ]0, π[ The operator Lu = u xx 4u is uniformly elliptic on However, the corresponding bilinear form B[u, v] = π 0 u x v x 4uv dx is not positive definite on H0 1 ( For example, taking u(x = sin x we find B[u, u] = π 0 cos 2 x 4 sin 2 x dx = 3π 2 Observe that, if we take f(x = sin 2x, then the boundary value problem uxx 4u = sin 2x x ]0, π[, u(0 = u(π = 0, (129 8

9 has no solutions Indeed, taking v(x = sin 2x, for every u H0 1 ( an integration by parts yields π π ( π B[u, v] = u x v x 4uv dx = 2u x cos 2x 4u sin 2x dx = 0 sin 2 2x dx Therefore, it is not possible to satisfy the identity (112 for all v H0 1 ( In this connection, one should also observe that the corresponding homogeneous problem uxx 4u = 0 x ]0, π[, (130 u(0 = u(π = 0 admits infinitely many solutions: u(x = κ sin 2x, where κ IR is any constant We now study the existence and uniqueness of weak solutions to the more general boundary value problem (12 Before stating a precise result, we outline the two main steps in the approach STEP 1: By choosing a constant γ > 0 sufficiently large, the operator L γ u = Lu + γu (131 is strictly positive definite More precisely, the corresponding bilinear form B γ [u, v] = a ij (x u xi v xj b i (x uv xi + c(x uv + γ uv dx (132 satisfies B γ [u, u] β u 2 H 1 for all u H1 0 (, (133 for some constant β > 0 Using the Lax-Milgram theorem, we conclude that for every f L 2 ( the elliptic equation L γ u = f has a unique solution u H0 1 ( Moreover, the map f u = L 1 γ f is a linear compact operator from L 2 ( into H0 1( L2 ( We regard L 1 γ as a compact operator from L 2 ( into itself STEP 2: The original problem (12 can now be written as Lu = L γ u γu = f Applying the operator L 1 γ to both sides, one obtains u L 1 γ γu = L 1 γ f (134 Introducing the notation K = γl 1 γ, h = L 1 γ f, (135 9

10 we are led to the equation (I Ku = h (136 Since K is a compact operator from L 2 ( into itself, the Fredholm theory can be applied In particular, one has Fredholm s alternative: either (i for every h L 2 ( the equation u Ku = h has a unique solution u L 2 (, or else (ii the equation u Ku = 0 has a nontrivial solution u L 2 ( Calling K : L 2 L 2 the adjoint operator, case (ii occurs if and only if the adjoint equation v K v = 0 has a nontrivial solution v L 2 ( Information about the existence and uniqueness of weak solutions to (12 can thus be obtained by studying the adjoint operator L v = (a ij (x v xj xi b i (xv xi + c(x v (137 In the remainder of this section we shall provide detailed proofs of the above claims Lemma 12 (estimates on elliptic operators Let the operator L in (11 be uniformly elliptic, with coefficients a ij, b i, c L ( Then there exist constants α, β, γ > 0 such that for all u, v H 1 0 ( B[u, v] α u H 1 v H 1, (138 β u 2 H 1 B[u, u] + γ u 2 L 2, (139 Proof 1 The boundedness of the bilinear form B : H0 1 H1 0 ( n B[u, v] = a ij u xi v xj b i uv xi + c uv dx IR follows from a ij L u xi L 2 v xj L 2 + b i L u L 2 v xi L 2 + c L u L 2 v L 2 α u H 1 v H 1 2 Toward the second estimate, using the ellipticity condition (14 and the elementary in- 10

11 equality ab 1 2θ a2 + θ 2 b2 +, we obtain θ u xi 2 L 2 = θ u 2 x i dx a ij u xi u xj dx ( n = B[u, u] + b i uu xi cu 2 dx B[u, u] + b i L u L 2 u xi L 2 + c L u 2 L 2 ( 1 B[u, u] + 2θ b i 2 L u 2 L 2 + θ u xi 2 L c L u 2 L 2 Therefore B[u, u] θ u xi 2 L 2 2 C u 2 L 2 for all u H1 0 ( for a suitable constant C Taking β = θ/2 and γ = C + θ/2, the inequality (139 is satisfied Remark 16 By the above lemma, by choosing the constant γ > 0 large enough, the bilinear form B γ in (132 is strictly positive definite Notice that, for γ > 0 large, it would be very easy to show that the bilinear form B γ = B[u, v] + γ(u, vh 1 is strictly positive definite on H0 1 ( However, Lemma 12 shows that we can achieve strict positivity by adding the much weaker term γ(u, v L 2 Let γ be as in (139 and define the bilinear form B γ according to (132 Since B γ is positive definite, we can apply the Lax-Milgram theorem and conclude that, for every f L 2 (, there exists a unique u H0 1 ( such that B γ [u, v] = (f, v L 2 = (i f, v H 1 for all v H 1 0 ( (140 Since the map i is compact, the solution operator f u = L 1 γ f is a linear compact operator from L 2 ( into H0 1( Therefore, it is also a compact operator from L2 ( into itself An entirely similar result holds for the adjoint problem L γ v = g x, v = 0 x, (141 where L is the adjoint operator introduced in (137, and L γu = L u + γu Given g L 2 (, a weak solution of (141 is defined to be a function v H0 1 ( such that Bγ[v, u] = B γ [u, v] = (u, g L 2 for all u H0 1 ( (142 Since B γ is strictly positive definite, for every g L 2 the Lax-Milgram theorem yields a unique weak solution v of (141 The map g v = (L γ 1 g is a linear, compact operator from L 2 ( into itself 11

12 Lemma 13 (adjoint operator In the above setting, the operator (L γ 1 is the adjoint of the operator L 1 γ Proof By definition, for every f, g L 2 ( and u, v H0 1 ( one has [ ] [ ] (f, v L 2 = B γ L 1 f, v, (u, g L 2 = B γ u, (L 1 g γ γ In particular, choosing v = (L γ 1 g and u = L 1 γ f we obtain ( [ f, (L γ 1 g = B γ L 2 L 1 γ ] f, (L 1 g γ = ( L 1 γ f, g L 2 Lemma 14 (representation of weak solutions Given any f L 2 (, a function u L 2 ( is a weak solution of (12 if and ony if (I Ku = h, with K = γl 1 γ, h = L 1 γ f (143 Proof 1 Let u be a weak solution of (12 By definition of weak solution, this implies u H0 1 ( and B γ [u, v] = B[u, v] + γ(u, v L 2 = (f + γu, v L 2 for all v H 1 0 ( Therefore u = L 1 γ (f + γu = h + Ku 2 To prove the converse, let (143 hold Then Moreover, for every v H0 1 ( we have u = γl 1 γ u + L 1 γ f H0 1 ( B[u, v] = B γ [u, v] γ(u, v L 2 = (f + γu, v L 2 γ(u, v L 2 = (f, v L 2 In order to apply Fredholm s theory, together with (12 we consider the homogeneous problem Lu = 0 x, u = 0 x, (144 and the adjoint problem L v = 0 x, v = 0 x, (145 where L is the adjoint linear operator defined at (137 12

13 Theorem 13 (solutions to the elliptic boundary value problem - II Under the basic assumptions (H, the following statements are equivalent: (i For every f L 2 (, the elliptic boundary value problem (12 has a unique weak solution (ii The homogeneous boundary value problem (144 has the only solution u(x 0 (iii The adjoint homogeneous problem (145 has the only solution v(x 0 Proof 1 Since K = γl 1 γ is a compact operator from L 2 ( into itself, Fredholm s theoremcan be applied As a consequence, the linear operator I K is surjective if and only if it is one-to-one, ie if and only if Ker(I K = 0} 2 By Lemma 14, u Ku = 0 if and only if u is a weak solution of (144 An entirely similar argument shows that v K v = 0 if and only if v is a weak solution of (145 By Fredholm s theorem, Ker(I K and Ker(I K have the same dimension We thus obtain a chain of equivalent statements: I K is surjective Ker(I K = 0} Ker(I K = 0} u 0 is the unique solution of (144 v 0 is the unique solution of (145 Theorem 13 covers the situation where I K is one-to-one and Fredholm s first alternative holds In the general case where I K is not necessarily one-to-one, the existence of solutions to u Ku = L 1 γ f can be determined using the identity Range(I K = [Ker(I K ] (146 Theorem 14 (existence of solutions to the elliptic boundary value problem - III Under the assumptions (H, the problem (12 has at least one weak solution if and only if for every weak solution v H0 1 ( of the adjoint problem (145 (f, v L 2 = 0 (147 Proof The boundary value problem (12 has a weak solution provided that L 1 γ f Range(I K By (146, this holds if and only if L 1 γ f is orthogonal to every v Ker(I K, ie to every solution v of the adjoint problem (145 We claim that this holds if and only if f itself is orthogonal to every solution v of (145 Indeed, if v K v = 0, one has Since γ > 0, this proves our claim (f, v L 2 = (f, K v L 2 = (Kf, v L 2 = γ (L 1 γ f, v L 2 13

14 2 Parabolic equations Let IR n be a bounded open set and let L be the operator in (11 In addition to the standard hypotheses (H stated at the beginning of the chapter, we now assume that the coefficients a ij satisfy somewhat stronger regularity condition a ij W 1, ( (21 In this section we study the parabolic initial-boundary value problem u t + Lu = 0 t > 0, x, u(t, x = 0 t > 0, x, u(0, x = g(x x (22 It is convenient to reformulate (22 as a Cauchy problem in the Hilbert space X = L 2 (, namely d u = Au, u(0 = g, (23 dt for a suitable (unbounded linear operator A : L 2 ( L 2 ( More precisely A = L, Dom(A = u H 1 0 ( ; } Lu L 2 ( (24 In other words, u Dom(A if u is the solution to the elliptic boundary value problem (12, for some f L 2 ( In this case, Au = f Our eventual goal is to construct solutions of (23 using semigroup theory We first consider the case where the operator L is strictly positive definite on H0 1 ( More precisely, we assume that there exists β > 0 such that the bilinear form B : H0 1( H1 0 ( IR defined at (111 is strictly positive definite: there exists β > 0 such that B[u, u] β u 2 H 1 for all u H1 0 ( (25 Notice that this is certainly true if b i 0 and c 0 Theorem 21 (semigroup of solutions of a parabolic equation - I Let the standard assumptions (H hold, together with (21 Moreover, assume that the corresponding bilinear form B in (111 is strictly positive definite, so that (25 holds Then the operator A = L generates a contractive semigroup S t ; on L 2 ( t 0} of linear operators Proof To prove that A generates a contraction semigroup on X = L 2 ( we need to check: (i Dom(A is dense in L 2 ( (ii The graph of A is closed (iii Every real number λ > 0 is in the resolvent set of A, and (λi A 1 1 λ 14

15 1 To prove (i, we observe that, if ϕ C 2 c (, then the regularity assumptions (21 imply f = Lϕ L 2 ( This proves that Dom(A contains the subspace C 2 c ( and is thus dense in L 2 ( 2 If (25 holds, then, by the Lax-Milgram theorem, for every f L 2 ( there exists a unique u H 1 0 ( such that B[u, v] = (f, vl2 for all v H 1 0 ( The map f u = L 1 f is a bounded linear operator from L 2 ( into L 2 ( We now observe that the pair of functions (u, f is in the graph of A if and only if ( f, u lies in the graph L 1 Since L 1 is a continuous operator, its graph is closed Hence the graph of A is closed as well 3 According to the definition of A, to prove (iii we need to show that, for every λ > 0, the operator λi A has a bounded inverse with operator norm (λi A 1 1/λ Equivalently, for every f L 2 (, we need to show that the problem λu + Lu = f x, (26 u = 0 x, has a weak solution satisfying u L 2 1 λ f L 2 (27 By the Lax-Milgram theorem, there exists a unique u H0 1 ( such that Taking v = u in (28 we obtain (λu, v L 2 + B[u, v] = (f, v L 2 for all v H 1 0 ( (28 λ u 2 L 2 + B[u, u] = (f, u L 2 f L 2 u L 2 Since we are assuming B[u, u] 0, this yields proving (27 λ u L 2 f L 2, By the generation theorem, the linear operator A generates a contractive semigroup 21 Representation of solutions in terms of eigenfunctions Consider the special case where a ij = a ji and L is the operator in (120, containing only second order terms In this case, by Poincare s inequality, the bilinear form B in (122 is strictly positive definite and Theorem 21 can be applied Relying on Theorem 12, we can provide a representation of the semigroup trajectories, in terms of the orthonormal basis φ k ; k 1} consisting of eigenfunctions of the compact self-adjoint operator L 1 By construction, for every k 1 one has L 1 φ k = λ k φ k, 15

16 where λ k > 0 is the corresponding eigenvalue Therefore φ k Dom(L, Lφ k = µ k φ k µ k = 1 λ k (29 Notice that λ k 0 and µ k +, as k u(t provides a C 1 solution to the Cauchy problem For every k 1, the function = e µ kt φ k d dt u(t = Lu(t, u(0 = φ k Hence, by the uniqueness of semigroup trajectories, one must have S t φ k = e µ kt φ k By linearity, for any coefficients c 1,, c N IR one has ( N S t c k φ k = N c k e µ kt φ k Since S t is a bounded linear operator, decomposing an arbitrary function g L 2 ( along the orthonormal basis φ k ; k 1}, we thus obtain S t g = e µ kt (g, φ k L 2 φ k (210 The above representation of semigroup trajectories is valid for every g L 2 ( and every t 0 Lemma 21 Let L = L 0 be the operator at (120 Then for every g L 2 ( the formula (210 defines a map t u t = S t g from [0, [ into L 2 ( This map is continuous for t [0, [ and continuously differentiable for t > 0 Moreover, u(t Dom(L H0 1 ( for every t > 0 and there holds d u(t = Lu(t for all t > 0 (211 dt Proof 1 Let g L 2 ( Since µ k > 0 for every k, it is clear that Therefore, e µ kt (g, φ k L 2 2 (g, φ k 2 L 2 e µ kt (g, φ k L 2 2 φ k k 1 k 1(g, 2 L 2 = g 2 L 2 < Hence the series in (210 is convergent, uniformly for t 0 In particular, since the partial sums are continuous functions of time, the map t S t g is continuous as well 16

17 2 We claim that, even if g / H0 1 (, one always has S t g Dom(L H 1 0 ( for all t > 0 (212 Indeed, a function u = k c kφ k lies in Dom(L if and only if the coefficients c k satisfy (c k µ k 2 < In the case where c k (t k k = e µ kt (g, φ k L 2, we have the estimate (µ k c k (t 2 sup k (µ k e µ 2 kt (g, φ k 2 L 2 (213 An elementary calculation now shows that, for ξ 0 and t fixed, the function ξ ξ e tξ attains its global maximum at ξ = 1/t In particular, k µ k e µ kt Using this bound in (213 we obtain max ξ 0 ξe tξ = 1 et ( 2 1 µ k c k (t e 2 t 2 g 2 L 2 Hence the series defining Lu(t is convergent This implies u(t Dom(L, for each t > 0 3 Differentiating the series (210 term by term, and observing that the series of derivatives is also convergent, we achieve (211 Example 83 As in Example 81, let = ]0, π[ IR and L 0 u = u xx Given g L 2 (]0, π[, consider the parabolic initial-boundary value problem u t = u xx t > 0, 0 < x < π, u(0, x = g(x 0 < x < π, u(t, 0 = u(t, π = 0 In this special case, the formula (210 yields the solution as the sum of a Fourier sine series: u(t, x = ( 2 π t π e k2 g(y sin ky dy sin kx 0 22 More general operators To motivate the following construction, we begin with a finite dimensional example Let L be an n n matrix and consider the linear ODE on IR n d x(t = Lx(t (214 dt 17

18 If L is positive definite, ie if Lx, x 0 for all x IR n, then L generates a contractive semigroup Indeed d d dt x(t 2 = 2 dt x(t, x(t = 2 Lx(t, x(t 0, showing that the Euclidean norm of a solution does not increase in time Next, let L be an arbitrary matrix We can then find a number γ 0 large enough so that the matrix L + γi is positive definite (and hence it generates a contractive semigroup In this case, if x(t = e tl x(0 is a solution to (214, writing L = γi (L + γi one obtains x(t = e Lt x(0 = e γi (L+γI}t x(0 = e γt e (L+γIt x(0 e γt x(0 (215 According to (215, the operator L generates a semigroup of type γ We shall work out a similar construction in the case where L is a general elliptic operator, as in (11, and the corresponding bilinear form B[u, v] in (111 is not necessarily positive definite According to Lemma 12, there exists a constant γ > 0 large enough so that the bilinear form B γ [u, v] = B[u, v] + γ(u, v L 2 (216 is strictly positive definite on H0 1 ( We can thus define L γ u = Lu + γu, B γ [u, v] The parabolic equation in (22 can now be written as u t = L γ u + γu = B[u, v] + γ(u, v L 2 By the previous analysis, the operator A γ = (L + γi generates a contractive semigroup of linear operators, say S (γ t ; t 0} Therefore, the operator A = L = γi L γ defined as in (24 generates a semigroup of type γ Namely S t ; t 0}, with S t = e γt S (γ t t 0 Summarizing the above analysis, we have Theorem 84 (semigroup of solutions of a parabolic equation - II Let IR n be a bounded open set Assume that the operator L in (11 satisfies the regularity conditions (21 and the uniform ellipticity condition (14 Then the operator A defined at (24 generates a semigroup S t ; on L 2 ( t 0} of linear operators Having constructed a semigroup S t ; t 0} generated by the operator A, one needs to understand in which sense a trajectory of the semigroup t u(t = S t f provides a solution to the parabolic equation (22 In the case where L = L 0 is the symmetric operator defined at (120, the representation (210 provides all the needed information Indeed, according to 18

19 Lemma 83, for every initial data g L 2 ( the solution t u(t = S t g is a C 1 map, which takes values in Dom(L 0 and satisfies (211 for every t > 0 A similar result can be proved for general elliptic operators of the form (11 However, this analysis is beyond the scope of the present notes Here we shall only make a few remarks: (1 - Initial condition The map t u(t = S t g is continuous from [0, [ into L 2 ( and satisfies u(0 = g The initial condition in (22 is thus satisfied as an identity between functions in L 2 ( (2 - Regular solutions If g Dom(A, then u(t = S t g Dom(A for all t 0 Moreover, the map t u(t is continuously differentiable and satisfies the ODE (23 at every time t > 0 Since Dom(A H0 1 (, this also implies that u(t satisfies the correct boundary conditions, for all t 0 (3 - Distributional solutions Given a general initial condition f L 2 (, one can construct a sequence of initial data f m Dom(A such that f m f L 2 0 as m In this case, if the semigroup is of type γ, we have S t f m S t f L 2 e γt f m f L 2 Therefore the trajectory t u(t = S t f is the limit of a sequence of C 1 solutions t u m (t = S t f m The convergence is uniform for t in bounded sets Relying on these approximations, we now show that the function u = u(t, x provides a solution to the parabolic equation u t = (a ij (xu xi xj b i (xu xi c(xu (217 in distributional sense Namely, for every test function ϕ C c ( ]0, [, one has } uϕ t + u(a ij ϕ xj xi + u(b i ϕ xi c uϕ dx dt = 0 (218 To prove (218, consider a sequence of initial data f m Dom(A such that f m f L 2 0 Then, for any fixed time interval [0, T ], the trajectories t u m (t = S t f m converge to the continuous trajectory t u(t = S t f in C 0 ([0, T ] ; L 2 ( Since each u m is clearly a solution in distributional sense, writing } u m ϕ t + u m (a ij ϕ xj xi + u m (b i ϕ xi c u m ϕ dx dt = 0, and letting m we obtain (218 19

20 3 Hyperbolic equations Consider a linear hyperbolic initial-boundary value problem, of the form u tt + L 0 u = 0 t IR, x, u(t, x = 0 t IR, x, u(0, x = f(x, u t (0, x = g(x x, (319 where L 0 is the second order elliptic operator L 0 u = (a ij (xu xi xi (320 We assume that the coefficients a ij satisfy a ij = a ji W 1, (, a ij (xξ i ξ j θ ξ 2 for all x, ξ IR n (321 According to Lemma 82, the space L 2 ( admits an orthonormal basis φ k ; k 1} consisting of eigenfunctions of L 0, so that for a sequence of eigenvalues µ k + as k φ k Dom(L 0, L 0 φ k = µ k φ k, (322 It is convenient to reformulate (319 as a first order system, setting v = u t On the product space X = H 1 0 ( L 2 (, (323 we thus consider the evolution problem ( ( ( d u 0 I u =, dt v L 0 0 v ( u v (0 = ( f g (324 A semigroup of solutions of (324 will be constructed using the eigenfunctions φ k In the special case where f = a k φ k, g = b k φ k, (325 for a given k 1 and a k, b k IR, an explicit solution of (324 is found in the form where the coefficient a(t satisfies u(t = a(tφ k, v(t = u t (t = a (tφ k, a (t + µ k a(t = 0, a(0 = a k, a (0 = b k An elementary computation yields a(t = a k cos( µ k t + b k µk sin( µ k t Hence u(t = v(t cos( µ k t 1 µk sin( µ k t a k φ k µ k sin( µ k t cos( (326 µ k t b k φ k 20

21 Observe that t ( u(t, v(t is a continuously differentiable map from IR into H 1 0 ( L2 ( which satisfies the initial conditions and the differential equation in (324 By linearity, more general solutions can be obtained by taking linear combinations of solutions of the form (326 This motivates the definition f S t = g cos( µ k t 1 µk sin( µ k t (f, φ k L 2 φ k µ k sin( µ k t cos( (327 µ k t (g, φ k L 2 φ k Theorem 31 (solutions of a linear hyperbolic problem In the above setting, the formula (327 defines a strongly continuous group of bounded linear operators S t ; t IR} on the space X = H0 1( L2 ( Each operator S t : X X is an isometry wrt the equivalent norm (u, v X = (B 0 [u, u] + v 2 L 2 1/2 (328 Remark 31 Consider an elastic membrane which occupies a region in the plane Let u(t, x denote the vertical displacement of a point on this membrane, at time t Then the square of the norm (u, u t 2 X can be interpreted as the total energy of the vibrating membrane Indeed, the term B 0 [u, u] describes an elastic potential energy, while u t 2 L yields the kinetic energy 2 Proof of Theorem 31 1 The equivalence between the norm (328 and the standard product norm is an immediate consequence of (123 (u, v H 1 L 2 = ( u 2 H 1 + v 2 L 2 1/2 2 Let the functions f, g be given by f = a k φ k, g = b k φ k (a k = (f, φ k L 2, b k = (g, φ k L 2 Then B 0 [f, f] = (µ k a k φ k, a k φ k ( f Therefore X if and only if g L 2 = µ k a 2 k, g 2 L 2 = b 2 k µ k a 2 k <, b 2 k < (329 21

22 At any time t IR, we have ( f 2 S t g X = µ k a 2 k(t + b k (t 2, (330 where, according to (327, a k (t = cos( µ k ta k + 1 µk sin( µ k t b k, b k (t = a k(t = µ k sin( µ k t a k + cos( µ k t b k (331 ( ( f f If X, then for every t IR the series defining S g t in (327 is convergent Using g (331 inside (330, we obtain ( f 2 S t g X = µ k a 2 k + b 2 k = f 2 ( (332 g X This shows that each linear operator S t is an isometry wrt the equivalent norm X 3 It is easy to check that the family of linear operators S t ; t IR} satisfies the group properties S 0 ( f g = ( ( f f, S g t S s g = S t+s ( f g t, s IR ( f To complete the proof, we need to show that, for f, g fixed, the map t S t is continuous g from IR into X But this is clear, because the above map is the uniform limit of the continuous maps ( fm m m t S t, f g m = (f, φ k L 2 φ k, g m = (g, φ k L 2 φ k, (333 m as m Having constructed the group of operators S t ; t IR}, one needs to understand in which sense the trajectories ( ( u(t f t = S v(t t (334 g provide a solution to the hyperbolic initial-boundary value problem (319 ( f (1 - Initial and boundary conditions Consider an arbitrary initial data X = g H 1 0 ( L2 ( By the continuity of the map (334, it follows that u(t f H 1 0, v(t g L 2 0 as t 0 22

23 Hence the initial conditions in (319 are satisfied Moreover, by the definition of the space X, we have u(t H0 1 ( for all t 0 Hence the boundary conditions in (319 are also satisfied (2 - Regular solutions If both functions f and g are finite linear combinations of the eigenfunctions φ k, then the corresponding trajectory (334 is a continuously differentiable map from [0, [ into X, and it satisfies the ODE (324 at all times t > 0 (3 - Distributional solutions Given general initial data f H0 1( and g L2 (, one can construct a sequence of approximations f m, g m as in (333, so that f m f H 1 ( 0, ( g m um (t fm g L 2 0 as m The corresponding semigroup trajectories t = S v m (t t g m converge to the trajectory (334, uniformly for t IR Relying on these approximations, we now show that the function u = u(t, x provides a solution to the parabolic equation u tt = (a ij (xu xi xj in distributional sense Indeed, consider any test function ϕ C c ( ]0, [ Since each function u m = u m (t, x is a distributional solution of (319, there holds u m ϕ tt + a ij (x (u m xi ϕ xj }dx dt = 0 Letting m and using the convergence u m (t u(t H 1 0, uniformly for t IR, we obtain } u ϕ tt + a ij (x u xi ϕ xj dx = 0 Example 84 As in Example 81, let = ]0, π[ IR and L 0 u = u xx Given f H 1 0 ( and g L 2 (]0, π[, consider the hyperbolic initial-boundary value problem u tt = u xx t > 0, 0 < x < π, u(0, x = f(x, u t (0, x = g(x 0 < x < π, u(t, 0 = u(t, π = 0 In this special case, the formula (327 yields the solution as the sum of a Fourier sine series: u(t, x = [ ( 2 π ( sin kt π ] cos kt f(y sin ky dy + g(y sin ky dy sin kx π 0 k 0 23

24 4 Problems 1 Let = (x, y ; x 2 + y 2 < 1} be the open unit disc in IR 2 Prove that, for every bounded measurable function f = f(x, y, the problem uxx + x u xy + u yy = f on has a unique weak solution u = 0 on 2 Let = (x, y ; x 2 + y 2 < 1} be the unit disc in IR 2 On the space X = H0 1 (, consider the inner product u, v = [ ] u x v x + 2u y v y + y(u x v y + u y v x dx (i Prove that, is indeed an inner product on X, which makes X into a Hilbert space (ii Given f L 2 (, show that there exists a unique u X such that u, v = f v dx for all v X = H0 1 ( What elliptic equation does u solve? 3 On a bounded open set IR n consider an elliptic operator of the form Lu = (a ij (xu xi xj + c(xu, (435 assuming that the coefficients a ij, c satisfy (13 and (14 (i Prove that the space L 2 ( admits an orthonormal basis φ k ; eigenfunctions of L, so that k 1} consisting of φ k H 1 0 (, Lφ k = µ k φ k for all k 1 The corresponding eigenvalues satisfy µ k + as k (ii Show that the general solution to the parabolic initial-boundary value problem (22 can be written as S t g = e µ kt (g, φ k L 2 φ k 4 Prove that a representation similar to (327 continues to hold, for the solution to the more general hyperbolic initial-boundary value problem u tt + Lu = 0 t IR, x, u(t, x = 0 t IR, x, u(0, x = f(x, u t (0, x = g(x x, 24

25 where L is the elliptic operator in (435 5 Let IR n be a bounded open set Let β > 0 be the constant in Poincaré s inequality: u L 2 ( β u L 2 ( (i Establish a lower bound on the eigenvalues of the operator Lu = u More precisely, if φ H0 1 ( provides a weak solution to φ = µ φ x, φ = 0 x, prove that µ 1/β 2 (ii Prove that the solution of the parabolic initial-boundary value problem u t = u t > 0, x, u(t, x = 0 t > 0, x, u(0, x = g(x x, decays to zero as t Indeed, u(t L 2 e t/β2 g L 2 for every t 0 6 On the interval [0, T ], consider the Sturm-Liouville eigenvalue problem (p(tu + q(tu = µ u 0 < t < T, u(0 = u(t = 0 (436 Assume that p C 1 (]0, T [, q C 0 ([0, T ], p(t θ > 0 for all t Prove that the space L 2 ([0, T ] admits an orthonormal basis φ k ; k 1} where each φ k satisfies (436, for a suitable eigenvalue µ k Moreover, µ k as k 7 In Lemma 82, take L 0 u = u Let φ k ; k 1} be an orthonormal basis of L 2 ( consisting of eigenfunctions of L 1 0 Show that in this special case the eigenfunctions φ k are mutually orthogonal also wrt the inner product in H 1, namely (φ j, φ k H 1 = 0 whenever j k 8 On a bounded open set IR n, let 0 < µ 1 µ 2 be the eigenvalues of the operator Lu = u (with zero boundary conditions (i Given any ϕ Cc (, prove that ϕ 2 dx µ 1 25 ϕ 2

26 (ii Prove that the best possible constant in Poincare s inequality is C = 1/µ 1 u 2 L 2 C u 2 L 2 u H1 0 ( (437 9 Let IR n be bounded open sets Let 0 < µ 1 µ 2 be the eigenvalues of the operator on the domain, and let 0 < µ 1 µ 2 be the eigenvalues of the operator on Prove that µ 1 µ 1 10 Consider the open rectangle Q = (x, y ; 0 < x < a, 0 < y < b} Define the functions φ m,n (x, y = 2 ab sin mπ x a sin nπ y b m, n 1 (i Check that φ m,n H0 1 (Q Moreover, prove that the countable set of functions φ m,n ; m, n 1} is an orthonormal basis of L 2 (Q consisting of eigenfunctions of the elliptic operator Lu = u Compute the corresponding eigenvalues µ m,n (ii If IR 2 is an open domain contained inside a rectangle Q with sides a, b, prove that u L 2 ( ab π a 2 + b 2 u L 2 ( for all u H 1 0 ( 11 (Galerkin approximations Let IR n be a bounded open set Given f L 2 (, consider the elliptic problem φ = f x, φ = 0 x, Let ϕ 1,, ϕ m } be any set of linearly independent functions in H0 1 ( Construct an approximate solution by setting m U = c k ϕ k where the coefficients c 1,, c m are chosen so that B[U, ϕ j ] = U ϕ j dx = (f, ϕ j L 2 j = 1,, m (438 Show that (438 can be written as an algebraic system of m linear equations for the m variables c 1,, c m Prove that this system has a unique solution 12 Let = (x, y ; x 2 + y 2 < 1} be the open unit disc in IR 2, and let u be a smooth solution to the equation u tt = u xx + x u xy + 3u yy on [0, T ], (439 26

27 u = 0 on [0, T ] (i Write the equation (439 in the form u tt +Lu = 0, proving that the operator L is uniformly elliptic on the domain (ii Define a suitable energy e(t =[kinetic energy] + [elastic potential energy], and check that it is constant in time 13 Consider the differential operator on IR 2 Lu = (xu x x (yu y y + 2u xy + 3(u x + u y 6u (i At which points (x, y is L elliptic? (ii Determine for which open bounded subsets IR 2 one can say that operator L is uniformly elliptic on 14 Let = (x, y ; x 2 + y 2 < 1} be the open unit disc in IR 2, and let u be a smooth solution to the equation u tt = u xx + 2x u xy + 3u yy + u y on [0, T ], (440 u = 0 on [0, T ] (i Write the equation (440 in the form u tt +Lu = 0, proving that the operator L is uniformly elliptic on the domain (ii Define a suitable energy E(t = [kinetic energy] + [elastic potential energy], and check that it is constant in time 15 Consider the homogenous linear elliptic operator L 0 in (?? assuming that (14 holds, together with a ij = a ji L ( Work out the following alternative proof of Theorem 11 (i Show that the bilinear functional B 0 in (122 is an inner product on H0 1( corresponding norm 1/2 u = a ij (xu xi u xj is equivalent to the H 1 norm Namely The 1 C u H 1 u C u H 1 for all u H

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