Hille-Yosida Theorem and some Applications

Transcription

1 Hille-Yosida Theorem and some Applications Apratim De Supervisor: Professor Gheorghe Moroșanu Submitted to: Department of Mathematics and its Applications Central European University Budapest, Hungary

2 Acknowledgements First and foremost, I would like to express my special appreciation and sincere gratitude to my supervisor, Professor Gheorghe Moroșanu for his constant support and encouragement, and his invaluable advice and guidance. He has been a tremendous mentor and I am honored to have had the opportunity of working with a mathematician of his stature. I would like to extend my special thanks to Prof. Károly Böröczky and Prof. Pál Hegedűs for their guidance, and unending support and patience. I would also like to thank Ms. Elvira Kadvany and Ms. Melinda Balazs for being the best coordinators of any department I have been a part of, always ready to extend their helping hand whenever I needed, and all my teachers and colleagues whom I came to know during my wonderful time at the Mathematics Department at CEU. I am especially thankful to my parents for everything they ve done for me and last but not the least, I would like to thank my best friends, Ella and Alfredo whose unwavering support and belief in me have helped me through my most trying times. This work is dedicated to them.

3 Contents Introduction 1 1 Preliminaries 1.1 Maximal monotone operators and their properties L p spaces Sobolev spaces Open sets of class C m Sobolev embedding theorems Green s identity for Sobolev Spaces Variational formulation of the Dirichlet boundary value problem for the Laplacian Hille-Yosida Theorem 16.1 Existence and uniqueness of solution to the evolution problem du + Au = on [, + ) with initial data u() = u Regularity of the solutions The case of self-adjoint operators Applications of Hille-Yosida Theorem Heat Equation Wave Equation Linearized equations of coupled sound and heat flow Bibliography 86 i

4 Introduction The main focus of this work is going to be the Hille-Yosida theorem, which, as we will see, is a very powerful tool in solving evolution partial differential equations. The material is organized as follows. Chapter 1 presents the background material needed for the treatment in the subsequent chapters. In Chapter, we present the Hille- Yosida Theorem and related results and offer detailed proofs. Finally, in Chapter 3, we investigate applications of the Hille-Yosida theorem to some real word phenomena. We prove existence, uniqueness and regularity of solutions of the Heat equation, the Wave equation, and the linearized equations of coupled sound and heat flow. For the proofs of most of the results established in Chapter 1, relevant references are cited wherever needed. The subsequent chapters are self-contained. 1

5 Chapter 1 Preliminaries This chapter is going to be our toolbox of important concepts and results that we will use time and again throughout the course of this Thesis. 1.1 Maximal monotone operators and their properties Here we will only consider single-valued operators but to account for the general case of multi-valued operators, a good way to define an operator A on a set X is to simply consider A as a subset of the Cartesian product X X. Then we can define the domain of A as D(A) = {u X v such that [u, v] A}. We use the notation [u, v] to denote an element of the Cartesian product so as not to confuse with a scalar product which we will denote by (, ). Note that for a single valued operator A, D(A) = {u X v X such that Au = v}. We also define the range of A as R(A) = Au. u D(A) For the case of single valued operators Au can be considered as the singleton set {Au}. Henceforth, we will only focus on operators defined on a real Hilbert space, H. Definition. An operator A on a Hilbert space H is called monotone if (y y 1, x x 1 ) [x 1, y 1 ], [x, y ] A.

6 Note that for a single valued linear operator A, it is monontone if (Au, u) u D(A). Definition. An operator A on a Hilbert space H is called maximal monotone if: (i) A is monotone. (ii) A has no proper monotone extension. That is, for any monotone operator A H H if A A, then A = A. Theorem (G. Minty). Suppose A : D(A) H H is a monotone operator. Then A is maximal monotone iff R(A + I) = H. This is a very important result used in identifying maximal monotone operators. In fact, sometimes, this is taken as the definition of maximal operators. Proposition Let A be a linear maximal monotone operator. Then we have: (i) D(A) is dense in H; (ii) A is a closed operator; (iii) (I + λa) is a bijection from D(A) onto H for every λ >. Moreover, (I + λa) 1 is a bounded operator and (I + λa) 1 L(H) 1. Remark We see from Theorem that the following are equivalent: (i) A is a maximal monotone operator. (ii) λa + I is surjective for all λ >. (iii) λa + I is surjective for some λ >. That is, A is maximal monotone iff λa is maximal monotone for some λ > ; iff λa is maximal montone for all λ >. It also holds that A + λi is surjective for some some λ > if and only if A is maximal monotone. Definition. Let A be a maximal monotone operator. For every λ >, we define the following operators: J λ = (I + λa) 1, A λ = 1 λ (I J λ). 3

7 J λ is called the resolvent, and A λ, the Yosida approximation of A. Note that D(J λ ) = D(A λ ) = H, R(J λ ) = D(A) and if A is linear, then J λ L(H) 1. The resolvent and Yosida approximation (named after Kosaku Yosida) have many useful properties that make them an invaluable tool in proving the Hille-Yosida Theorem (Theorem.1.3). We list some of their properties as follows. Proposition Let A be a linear maximal monotone operator. Then we have the following: (i) A λ v = A(J λ v) v H, λ >, (ii) A λ v = J λ (Av) v D(A), λ >, (iii) A λ v Av v D(A), λ >, (iv) lim λ J λ v = v v H, (v) lim λ A λ v = Av v D(A), (vi) (A λ v, v) v H, λ >, (vii) A λ v 1 v v H, λ >. λ For further reference regarding the material of this section, see [], [1]. 1. L p spaces Let us denote R = (, ); N = {, 1,, }. Let X be a real real Banach space with the norm X and R N for some integer N 1 be a Lebesgue measurable set. Definition. Let 1 p <. Then we define L p (; X) as the space of all equivalence classes of (strongly) measurable functions f : X such that x f(x) p X is Lebesgue integrable and the equivalence relation is equality a.e on. In general when we write, u L p (; X), we mean u is a representative of the equivalence class of functions that agree with u a.e on. L p (; X) is a real Banach space witht he norm u L p (;X) = ( u(x) p X ) 1 p. 4

8 For p =, we define L (; X) as the space of equivalence classes of measurable functions f : X such that x f(x) X is essentially bounded. L (; X) is a real Banach space with the norm u L (;X) = ess sup u(x) X. x When X = R, we use the notation L p (). If = (a, b) R, we write L p (a, b; X). For more background material regarding this section, see []. 1.3 Sobolev spaces In this section, we will define Sobolev spaces. First, we establish some notations. Here, we assume is a non-empty open subset of R N. We denote by C k () the space of functions that are continuous on and have continuous partial derivatives up to order k. We also define: C () = {φ C() φ has continuous partial derivatives of any order}, C c () = {φ C () supp φ is compact in }, where supp φ = {x φ(x) }. Let 1 p. Definition. The Sobolev space W 1,p () is defined as { } W 1,p () = u L p () g 1, g,..., g N L p () such that u ϕ x i = g iϕ, ϕ Cc (), i = 1,, N For u W 1,p (), we define u x i = g i. This definition makes sense since g i is unique a.e (this follows from the fact that fϕ = ϕ C c () f = a.e on ). Here u x i denotes the weak derivative. For p =, we write, H 1 () = W 1, (). The space W 1,p () equipped with the norm u W 1,p = ( u p L p + N u x i i=1 p L p ) 1 p for 1 p < and u W 1, = u L + N u x i i=1 L for p = 5

9 is a real Banach space. The space H 1 () is equipped with the scalar product (u, v) H 1 = (u, v) L + = = = uv + N i=1 N i=1 uv + uv + ( u, v ) x i x i L N i=1 u v x i x i u v x i x i u v. H 1 () is a real Hilbert space with this scalar product and the associated norm is ( u H 1 = u + Remark If u C 1 () L p (), and if u u x i x i u ) 1. L p () for all i = 1,, N (here denotes the partial derivative in the usual sense), then u W 1,p () and the usual partial derivatives coincide with their Sobolev counterparts. consistent. So our notations are Definition. For integers m and 1 p, we define the Sobolev space W m,p () inductively as follows: { } W m,p () = u W m 1,p () u W m 1,p (), i = 1,, N. x i Equivalently, we can define: { W m,p () = u L p () α, with α m, g α L p () such that udα ϕ = ( 1) α g αϕ, ϕ Cc (). where we use the standard multi-index notation α = (α 1, α,, α N ) with α i, an integer and α = N i=1 α i and D α ϕ = α ϕ x α 1 1 x α. N N We denote D α u = g α, and as before, this is well-defined. The space W m,p () equipped with the norm u W m,p = D α u p L p α m 1 p for 1 p < }, 6

10 and u W m,p = D α u L for p = α m is a real Banach space. We write H m () = W m, (). The space H m () equipped with the scalar product (u, v) H m = (D α u, D α v) L is a real Hilbert space. α m Definition. Let 1 p. The Sobolev space W m.p () is defined as the closure of Cc () in W m,p (). More precisely, u W m,p () iff a sequence of functions, u n Cc () such that u u n W m,p. As before, we write H m () = W m, (). Remark 1.3. Consider = R N +. Then Γ = = R N 1 {}. In this case it can be shown that the map u u Γ defined from Cc (R N ) into L p (Γ) extends by density to a bounded linear operator from W 1,p () into L p (Γ). The operator is defined to be the trace of u on Γ and it is also denoted by u Γ. Now if is a regular open set in R N, e.g., if is of class C 1 with Γ = bounded, then it is possible to define the trace of a function u W 1,p () on Γ =. In this case, u Γ L p (Γ) (for the surface measure dσ). One important result regarding the trace is as follow: the kernel of the trace operator is W 1,p (), that is, For further reference, see [11], []. W 1,p () = { u W 1,p () u Γ = }. 7

11 1.4 Open sets of class C m First we denote the following sets: R N + = {x = (x 1,..., x N ) R N x N > }, N 1 Q = {x = (x 1,..., x N ) R N ( x i ) 1/ < 1, x N < 1}, Q + = R N + Q, i=1 N 1 Q = {(x 1,..., x N 1, ) R N ( x i ) 1/ < 1}. Definition. An open set is said to be of class C 1 if for every x Γ =, there exists a neighborhood U x of x in R N and a bijective map H : Q U x such that H C 1 (Q), i=1 H 1 C 1 (U x ), H(Q + ) = U x Q, H(Q ) = U x Γ. Definition. Similarly, an open set is said to be of class C m for an integer m 1 if for every x Γ =, there exists a neighborhood U x of x in R N and a bijection H : Q U x such that H C m (Q), H 1 C m (U x ), H(Q + ) = U x, H(Q ) = U x Γ. is said to be of class C if it is of class C m m N. Remark For an open set R N, we say that the boundary Γ = is C k, if for every x Γ, r > and a C k function γ : R N 1 R such that we have (after possibly relabeling and reorienting the coordinate axes) B(x, r) = {x B(x, r) x N > γ(x 1,, x N 1 )}. Likewise Γ is class C, if it is C k, k N. This definition for Γ to be of class C k and that for an open set, of class C k are equivalent. See [5] for further reference. 8

12 1.5 Sobolev embedding theorems In this section we will state some useful results regarding continuous injections of Sobolev spaces to some L p spaces. We will need these results later on. We suppose here that R N is an open set of class C 1 and Γ = is bounded, or else that = R N +. Theorem Let 1 p <. Then the following hold: W 1,p () L q (), W 1,p () L q (), W 1,p () L (), where 1 q = 1 p 1 N q [p, + ) if p < N, if p = N, if p > N, where all the above injections are continuous. u W 1,p (), Furthermore, if p > N, we have, u(x) u(y) C u W 1,p x y α for a.e x, y where α = 1 N and C depends only on, p and N. In particular, W 1,p () p C()(this inclusion is modulo the choice of a continuous representative). Theorem Let m 1 be an integer and 1 p <. Then the following hold: W m,p () L q (), W m,p () L q (), W m,p () L (), where 1 q = 1 p m N q [p, + ) if 1 p m N >, if 1 p m N =, if 1 p m N <, where all the above injections are continuous. Furthermore, if m N > is not an [ ] p integer, set k = m N and θ = m N k ( < θ < 1). Then u W m,p (), p p and D α u L C u W m,p α with α k D α u(x) D α u(y) C u W m,p x y θ for a.e x, y R N, α with α = k. In particular, if m N/p > is not an integer, then [ W m,p () C k (), where k = m N ] p and C k () = { u C k () D α u has a continuous extension on, α with α k }. 9

13 Proofs. (c.f. [], [5]) Corollary Given any k N, there exists an integer m > k such that W m, () C k () and this inclusion is a continuous injection. Proof. if R N, where N is odd, then for sufficiently large m > N, we have that > is not an integer and it follows from Theorem 1.5. that W m, () C k (). m N that Now if N is even, choose m 1 = N, an integer. Then it follows from Theorem 1.5., W m1, () L q () is a continuous injection (1.5.1) [ ] for some irrational q >. Now let s choose an integer m = k + N + 1. Then q ] m [m N > is not an integer and by Theorem 1.5. with k = q N, q W m,q () C k () (1.5.) is a continuous injection. Now, since W m1, () L q (), { u W m1+1, () = u W m1, () u } W m1, (), i = 1,..., N x i u L q (), u x i L q (), i = 1,..., N. Therefore, u W 1,q (), that is, W m1+1, W 1,q (). Also note that W m1+1, () W m1, () L q (), where each injection is continuous. Now, u W = m 1 +1, u W + m 1 D α u, L (1.5.3) α =m 1 +1 N ( ) = u L + u Dα x i=1 i α m 1 L N = u L + u x i (1.5.4) W m 1, i=1 Therefore, u W m 1 +1, u W m 1, (from (1.5.3)) u L q (from (1.5.1)) 1

14 Again, u W m 1 +1, u x i (from (1.5.4)) W m 1, u x i (from (1.5.1)) L q for i = 1 to N. But u L q and u L x i i = 1 to N together imply q u W 1,q. Thus, W m1+1, () W 1,q () is a continuous injection. Proceeding inductively, and using a similar argument as above, we can show that W m 1+l, () W l,q () for any integer l 1. Then taking l = m, we have, W m 1+m,q () W m,q () is a continuous injection. And from (1.5.), it follows that W m 1+m, () C k () is [ a continuous ] injection.thus given any k N, if we take m m 1 + m = N + k + N + 1, q W m, () C k () is a continuous injection. 1.6 Green s identity for Sobolev Spaces We define the gradient and the Laplacian for Sobolev functions as follows: ( ) u u u = grad u =,,, x 1 x N u = N i=1 u, x i where all the partial derivatives are in the Sobolev sense. For example, u = N i=1 u x i = N i=1 g i, where u ϕ x i = g i ϕ ϕ C c (). Clearly, u makes sense for u H 1 () and u makes sense for u H (). If in addition u C 1 (), by Remark 1.3.1, u (in the usual classical) sense coincides with its Sobolev counterpart. It s important to note here that Sobolev functions 11

15 are considered equal if they agree a.e. Similarly, if u C () then u (in the usual classical sense) coincides with its Sobolev counterpart. So our notations are consistent. Also note that both and are linear operators. Green s identity for Sobolev functions is stated as follows: Theorem For any u H () and v H 1 (), we have u v + v u = v( u n)dσ (1.6.1) where n is the outward pointing unit normal on the surface element dσ. Proof. We know that (1.6.1) holds for u C () and v C 1 () in which case u, v, u coincide with their usual classical counterparts. This identity can be extended to Sobolev spaces if both sides are continuous wrt the Sobolev norm. We will make use of the fact that if two continuous functions agree on a dense subset, then they agree everywhere. Note that for linear expressions like u ϕ u or bilinear like (u, v) u v, continuity and boundedness are equivalent. Note that v u v L u L C v H 1 u H. (by C-S inequality) Similarly, it can be shown that u v is bounded. Thus the left hand side of is continuous on H () H 1 (). For a regular open set R N (for example, is of class C 1 and Γ = is bounded), the trace operator is bounded from H 1 () L (Γ), and thus, That takes care of the RHS. Γ v( u n)dσ v L (Γ) u L (Γ) C 1 v H 1 u H. Note that in the case when v H() 1 and u H (), Green s identity reads u v + v u = (1.6.) because v H() 1 v Γ =. 1

16 1.7 Variational formulation of the Dirichlet boundary value problem for the Laplacian In this section we set up the variational formulation of the Dirichlet boundary value problem for the Laplacian and state some important results that we will use time and again later. Let R N be an open set. We are looking for a solution u : R of the problem { u + u = f in u = on Γ = (1.7.1) for a given function f. The condition u = on Γ is called the (homogeneous) Dirichlet condition. Here u = N u i=1. x i A classical solution of the above problem is a function u C () that satisfies (1.7.1). A weak solution of the problem is defined to be a function u H 1 () such that u v + uv = fv v H 1 (). Note that if u H 1 () then u Γ = and the boundary condition is incorporated in the definition. It can be shown that a classical solution is also a weak solution. We have the following theorem that deals with the existence and the uniqueness of a weak solution. Theorem (Dirichlet, Riemann, Poincare-Hilbert). Given any f L (), a unique weak solution u H() 1 of (1.7.1). The unique solution is given by { } 1 min ( u + v ) fv. v H 1() Proof. (cf. [], [5]) It can be further shown that if the weak solution u H 1 () is also in C () and is of class C 1, then the weak solution actually turns out to be a classical solution. (A similar treatment can be done for the Neumann boundary problem u + u = f in Γ, u n =, 13

17 where u n = u n, n being the outward pointing unit normal vector to Γ). The next theorem deals with the regularity of the weak solution for the Dirichlet problem. Theorem Suppose is of class C and Γ = is bounded or else = R N +. Let f L () and u H() 1 satisfy u ϕ + uϕ = fϕ ϕ H(). 1 Then u H () and u H C f L where the constant C depends only on. Moreover, if is of class C m+ and f H m (), then u H m+ () and u H m+ C f H m. Furthermore, if m > N and f Hm (), then u C (). And if is of class C, and f C (), then u C (). Proof. (cf. []) Corollary Consider the unbounded linear operator A : D(A) H H, where H = L (), D(A) = H () H(), 1 Au = u. Then A is a maximal montone operator.(here u = N u i=1 x i in the Sobolev sense.) Proof. Note that by Theorem and Theorem 1.7., it follows that given any f L (), a unique u H () H 1 () such that u + u = f, that is, (A + I)u = f. So, R(A + I) = H = L (). A is also monotone and therefore, by Minty s Theorem A is maximal monotone. Remark We saw that : H () H 1 () L () is a maximal monotone operator. Then by Remark 1.1.1, we have that + λi is a bijection from H () 14

18 H 1 () onto L (), λ >. Then for any f L (), a unique u H () H 1 () such that ( + λi)u = f, that is, u + λu = f. Remark On the other hand, suppose u H () H 1 (). Then a unique f = f u L () such that u + u = f u. And by Theorem 1.7., we have u H C f u L for some constant C that depends only on. Putting f u = u + u, we have u H C u + u L C ( u L + u ) L. Furthermore if, u H k+ (), we have u H k (). In this case, f u = u + u H k (). Suppose is of class C k+. Then by Theorem 1.7., we have u H k+ C f u H k. Therefore, putting f u = u + u, u H C ( u k+ H + u ) k H. k 15

19 Chapter Hille-Yosida Theorem In this chapter, we will state and prove the central theorem of the thesis, viz. the Hille- Yosida Theorem, named after the mathematicians Einar Hille and Kosaku Yosida who independently discovered the result around We will establish some results first along our way to the proof. Many of the results will be applied in the final chapter. We will adapt the proofs presented in [] and offer detailed explanations wherever necessary..1 Existence and uniqueness of solution to the evolution problem du + Au = on [, + ) with initial data u() = u We begin with the following general result: Theorem.1.1 (Cauchy-Lipschitz-Picard). Let E be a Banach space. Let F : E E be a Lipschitz map with Lipschitz constant L, that is, F (u) F (v)) L u v u, v E Then for every u E, there exists a unique solution, u to the problem: du = F (u(t)), on [, + ) (..1) u() = u such that u C 1 ([, + ); E). Proof. Note that, finding a solution u C 1 ([, + ); E) to the above problem is equivalent to finding a solution u C([, + ); E) of the following equation: u(t) = u + t F (u(s))ds, t. (..) 16

20 First we will define an appropriate Banach space of functions, X. Let { } X = u C([, + ); E) sup e tk u(t) < t [,+ ) where k is a positive constant that we will fix later. We will now check that X is indeed a Banach space for the norm u X = sup e tk u(t) t [,+ ) Consider any Cauchy sequence (u n ) X. That means, given any ɛ >, N ɛ such that m, n > N ɛ, u n u m X < ɛ. For some fixed t, take ɛe tk >. Since {u n } is a Cauchy sequence, N = N ɛe tk that m, n > N, such u n u m X < ɛe tk sup e tk u n (t) u m (t) < ɛe tk t e tk u n (t) u m (t) < ɛe tk, t u n (t) u m (t) < ɛ, t. This holds for any ɛ >. That is, {u n (t)} is a Cauchy sequence in E. Since E is a Banach space, this means {u n (t)} converges to some point in E. Let s denote it by u(t) That is, u(t) := lim n u n (t). Now we show that the function u defined as above is in X. Since {u n (t)} converges to u(t), there exists N 1 such that for all n > N 1, u(t) u n (t) < ɛ/3. Similarly, since {u n (t )} converges to u(t ), there exists N such that for all n > N, u(t ) u n (t ) < ɛ/3. Note that u n is continuous. Therefore, given ɛ >, there exists δ > such that u n (t) u n (t ) < ɛ/3. 17

21 whenever t t < δ. Now take n > max(n 1, N ). Then, whenever t t < δ, we have, u(t) u(t ) u(t) u n (t) + u n (t) u n (t ) + u n (t ) u(t ) u(t) u n (t) + u n (t) u n (t ) + u n (t ) u(t ) ɛ/3 + ɛ/3 + ɛ/3 = ɛ. Thus we showed that u : [, + ) E is continuous, that is, u C([, + ); E). Now, for a fixed t, and for any ɛ >, choose n large enough such that u(t) u n (t) < ɛ. Note that u n X implies that sup t e tk u n (t) <. That means M > such that e tk u n (t) < M, t. That is, u n (t) < Me tk. Therefore, u(t) = u(t) u n (t) + u n (t) u(t) u n (t) + u n (t) < ɛ + Me tk, ɛ >. This implies u(t) < Me tk. The choice of t was arbitrary, so it holds for all t. Thus, for all t, e tk u(t) < M, that is, sup t e tk u(t) < Thus, we have shown that u X. Next we show that u n u X. Since u n is a Cauchy sequence in X, given ɛ >, there exists N ɛ such that m, n > N ɛ, u n u m X < ɛ sup e tk u n (t) u m (t) < ɛ t u n (t) u m (t) < ɛe tk, t. 18

22 Taking the limit as m, we have u n (t) u(t) < ɛe tk, sup e tk u n (t) u(t) < ɛ t u n u X < ɛ n > N ɛ. t n > N ɛ n > N ɛ That is, u n u X This concludes the proof that X is Banach with the norm X. Now let us define a function Φ : X X by (Φu)(t) = u + We claim that Φu X. Note that t (Φu)(t) = u + F (u(s))ds u + u + t t u + L t u + Lt u + L Now we derive some inequalities. L t t F (u(s))ds F (u(s)) F (u ) + F (u ) ds F (u(s)) F (u ) ds + u(s) ds = L t u(s) u ds + t F (u ) t t F (u ) ds u(s) ds + t F (u ). (..3) e sk e sk u(s) ds t L u X e sk ds ( since e sk u(s) sup t e tk u(t) = u X ) (e tk 1) L u X. (..4) k 19

23 From (..4) we can derive, e tk L t u(s) ds = e tk L u X (e tk 1) k = L u X (1 e tk ) k L k u X. (..5) We also have, e tk u u. (..6) Note that e tk > tk. So, e tk < 1/tk. Therefore, e tk Lt u 1 tk Lt u = L k u. (..7) Also, we have e tk t F (u ) 1 tk t F (u ) = 1 k F (u ). (..8) Finally, combining (..3), (..5), (..6), (..7), (..8), we get e tk (Φu)(t) u + L k u + L k u X + 1 k F (u ) ( t ) sup e tk (Φu)(t) <. (..9) t Next we show that (Φu) C([, + ); E). Let g(t) = t F (u(s))ds. Note that g(t) is continuous. That is, Therefore, lim g(t) g(t ) =. (for any t ) t t lim (Φu)(t) (Φu)(t ) t t = lim g(t) g(t ) t t =. This together with (..9) helps us conclude that Φu X. Furthermore, we claim that Φu Φv X L k u v X.

24 Indeed, note that t e tk (Φu)(t) (Φv)(t) = e tk (F (u(s) F (v(s)))ds Taking sup t of the left side, we obtain e tk L = e tk L t t u(s) v(s) ds e sk e sk u(s) v(s) ds t e tk L u v X e sk ds [ ] (e = L u v X e tk tk 1) k (1 e tk ) = L u v X k L k u v X. Φu Φv X L k u v X. (..1) Taking any k > L, we see from (..1) that Φ : X X is a contraction map. Thus from Banach s Fixed Point Theorem, we get that Φ has a unique fixed point. Denote it by u. Then u = Φu. That is, u(t) = (Φu)(t) = u + t F (u(s))ds. But then clearly, u is a solution of (..1) with u() = u. This proves the existence part. Now we prove uniqueness. Suppose u and ū are two solutions to (..1). Then using (..), we have, u(t) ū(t) = That is, φ(t) L t L t t t (F (u(s)) F (ū(s))) ds (F (u(s)) F (ū(s))) ds (u(s) ū(s)) ds. φ(s), where φ(t) = u(t) ū(t). (..11) 1

25 Let f(t) = t φ(s)ds. Then f (t) = φ(t). Therefore, from (..11), we have Let h(t) = e Lt f(t). Then f (t) Lf(t) f (t) Lf(t). (..1) h (t) = e Lt f (t) Le Lt f(t) = e Lt (f (t) Lf(t)). Thus h(t) is a non-increasing function. So we have h(t), for any t e Lt f(t). But e Lt >, so it must be that f(t). Again, f(t) = t φ(s)ds = Therefore, f(t) = for all t. That is, t u(s) u(s) ds. φ(t) Lf(t) = φ(t) =, t φ. Thus, we have shown uniqueness of the solution. This concludes our proof. Now we will go on to state and prove the Hille-Yosida Theorem. Henceforth, by H, we will denote a real Hilbert space. First we will state and prove the following lemmas which we will use during the course of the proof. Lemma.1.. Let w C 1 ([, + ); H) be a function satisfying dw + A λw = on [, + ) (.1.a) where A λ = 1 λ (I J λ) is the Yosida approximation, and J λ is the resolvent of a maximal monotone operator A as defined in Section 1.1. Then the functions t w(t) and t dw (t) are non-increasing on [, + )

26 Proof. First note that [ ] d w(t) w(t + h) w(t) = lim h h 1 = lim [(w(t + h), w(t + h)) (w(t), w(t))] h h 1 = lim [(w(t + h), w(t + h)) (w(t), w(t + h)) + (w(t + h), w(t)) (w(t), w(t))] h h = lim [(w(t + h) w(t), w(t + h)) + (w(t + h) w(t), w(t))] ( h ) ( ) w(t + h) w(t) w(t + h) w(t) = lim, w(t + h) + lim, w(t) h h h h ( ) dw(t) =, w(t). (.1.b) h 1 ( ) Since w satisfies (.1.a), we have dw(t), w(t) = ( A λ w(t), w(t)) = (A λ w(t), w(t)). Recall that (A λ v, v) for all v H. Therefore, d w(t) = (A λ w(t), w(t)). That is, w(t) is an non-increasing function on [, + ). Then, w(t) = ( w(t) ) 1/ is also non-increasing on [, + ). Since A λ is a linear bounded operator we have ( ) ( ) d dw dw + A λ =, t. (.1.e) So, dw/ satisfies (.1.a). Using the same argument as that used in showing w(t) is non-increasing, we conclude that dw/ is non-increasing. Moreover, d ( ) d k w k ( ) d k w + A λ =, t. (for any order k) k This is turn implies dk w k is non-increasing for any order k. 3

27 Lemma.1.3. Consider a sequence of functions f n C 1 ([a, b]; H). Suppose the sequence of functions dfn converges uniformly to a function g : [a, b] H. If f n (t ) converges for some point t on [a, b], then f n converges uniformly to some function, f(say). Then f is differentiable and df = g. Proof. A similar result holds for real valued functions (see Theorem 7.17 in [16]).For our case, let s fix some v H. Denote h n (t) = (f n (t), v); h(t) = (f(t), v); h 1 (t) = (g(t), v). Note that h n(t) = (f n(t), v) and these are all scalar functions. So we can apply Theorem 7.17 ( [16]) to conclude that the sequence of function h n converges uniformly to the function h(t) = (f(t), v) and h (t) = h 1 (t) = (g(t), v), t [a, b]. That is, (f (t), v) = (g(t), v). The choice of v H was arbitrary. Therefore, f (t) = g(t). Lemma.1.4. Let u D(A). Then for any ɛ >, ū D(A ) such that u ū < ɛ and Au Aū < ɛ. That is, D(A ) is dense in D(A) for the graph norm. Proof. Take u D(A). Recall that D(J λ ) = H, R(J λ ) = D(A). Denote ū = J λ u = (I + λa) 1 u. Then ū D(A) and u = ū + λaū. Therefore, λaū = u ū D(A) Aū D(A). But ū D(A), Aū D(A) ū D(A ). Now recall that J λ (Av) = A λ v v D(A), λ > and A(J λ v) = A λ v v H, λ > Therefore, v D(A), λ >, we have J λ (Av) = A(J λ v) Also, recall that lim λ J λ v = v, v H. Therefore, (.1.h) lim J λu u = λ and lim J λ(au ) Au = λ 4

28 By the first inequality above, we have, for sufficiently small λ, ū u = J λ u u < ɛ (.1.i) And by the second inequality. we have for sufficiently small λ, J λ (Au ) Au < ɛ A(J λ u ) Au < ɛ Aū Au < ɛ The equations (.1.i) and (.1.j) conclude the proof of the Lemma. (using (.1.h)) (.1.j) Theorem.1.5 (Hille-Yosida). Let A : D(A) H H be a maximal monotone operator, where H is a real Hilbert space. Then given any u D(A), there exists a unique solution, u C 1 ([, + ); H) C([, + ); D(A)) to the following problem: du + Au = on [, + ) u() = u. (.1.1) Furthermore, u(t) u, du (t) = Au(t) Au, t. Remark. Note that the space D(A) is equipped with the norm ( v + Av ) 1/ or with the equivalent norm v + Av. Proof. First we prove the uniqueness part. Suppose u and ū be two solutions to the system (.1.1). Then, we have d(u ū) = A(u ū). Therefore, ( ) d(u ū), u ū = ( A(u ū), (u ū))) (since A is monotone, (Av, v), v H.) Let φ(t) = u(t) ū(t). Then the above inequality reads ( ) dφ(t), φ(t). 5

29 Note that φ = (φ, φ). We have d φ(t) [ ] φ(t + h) φ(t) = lim h h 1 = lim [(φ(t + h), φ(t + h)) (φ(t), φ(t))] h h 1 = lim [(φ(t + h), φ(t + h)) (φ(t), φ(t + h)) + (φ(t + h), φ(t)) (φ(t), φ(t))] h h = lim [(φ(t + h) φ(t), φ(t + h)) + (φ(t + h) φ(t), φ(t))] ( h ) ( ) φ(t + h) φ(t) φ(t + h) φ(t) = lim, φ(t + h) + lim, φ(t) h h h h ( ) dφ(t) =, φ(t). h 1 (.1. ) This means φ(t) is an non-increasing function on [, + ). Then for any t, φ(t) φ() = u() ū() = u u =. Therefore, φ(t) =, for any t. That is, φ. This proves uniqueness. Now we will prove the Existence part. We will use the following strategy. We replace A by A λ in (.1.1) and we use Theorem.1.1(Cauchy-Picard-Lipschitz) on the approximate problem. Then using the fact that lim λ A λ v = Av and a number of estimates that are independent of λ, we pass on to the limit as λ +. Recall that A λ v 1 v v H and λ >. λ That is, A λ is Lipschitz with Lipschitz constant 1/λ. Using Theorem.1.1, we have that for any λ > there exists a solution (say u λ ) of the problem: dw (t) = A λw(t) on [, + ), w() = u D(A), 6

30 such that u λ C 1 ([, + ); H). That is, for any λ >, du λ (t) = A λu λ (t) on [, + ), u λ () = u. (.1.) Now using Lemma.1., we have u λ (t) and du λ (t) are non-increasing on [, + ). Therefore, for any t, u λ (t) u() = u and du λ (t) du λ () A λ u λ (t) A λ u λ () = A λ u Au. The last inequality holds because A λ v Av v D(A), following estimates: λ >. So we have the u λ (t) u t, λ >, (.1.3) A λ u λ (t) Au t, λ >. (.1.4) Now, note that for any λ, µ >, we have du λ du µ + A λu λ A µ u µ = d(u λ u µ ) + A λ u λ A µ u µ = d(u λ u µ ) = (A λ u λ A µ u µ ). Using the same argument as (.1.b), we have ( ) 1 d u λ u µ d(uλ u µ ) =, u λ u µ = (A λ u λ A µ u µ, u λ u µ ). (.1.5) Now, (A λ u λ A µ u µ, u λ u µ ) = (A λ u λ A µ u µ, u λ J λ u λ + J λ u λ J µ u µ + J µ u µ u µ ) = (A λ u λ A µ u µ, (u λ J λ u λ ) (u µ J µ u µ )) + (A λ u λ A µ u µ, J λ u λ J µ u µ ). 7

31 Recall that A λ = 1 λ (I J λ). Therefore, u λ J λ u λ = (I J λ )u λ = λa λ u λ. Again recall that A λ v = A(J λ v) v H and λ >. Then we have, (A λ u λ A µ u µ, u λ u µ ) = (A λ u λ A µ u µ, λa λ u λ µa µ u µ ) + (A(J λ u λ J µ u µ ), J λ u λ J µ u µ ) (A λ u λ A µ u µ, λa λ u λ µa µ u µ ). (.1.6) The last inequality holds because A is maximal monotone and hence (Av, v) v D(A). From (.1.5), we have, 1 d u λ u µ = (A λ u λ A µ u µ, u λ u µ ) (A λ u λ A µ u µ, λa λ u λ µa µ u µ ) (from.1.6) = λ(a λ u λ, A λ u λ ) + µ(a λ u λ, A µ u µ ) + λ(a µ u µ, A λ u λ ) µ(a µ u µ, A µ u µ ) = (λ + µ)(a λ u λ, A µ u µ ) λ A λ u λ µ A µ u µ (λ + µ) (A λ u λ, A µ u µ ) + λ A λ u λ + µ A µ u µ (λ + µ) A λ u λ A µ u µ + λ A λ u λ + µ A µ u µ (λ + µ) Au. For the last inequality, we used the estimate (.1.4). So we have d u λ u µ 4(λ + µ) Au. Integrating the inequality with respect to t, we get (using Cauchy-Schwarz inequality) u λ u µ 4t(λ + µ) Au u λ u µ t(λ + µ). ( ) It is obvious from that u λ (t) converges uniformly as λ + on every bounded interval [, T ]. Note that each u λ (t) C([ + ); H). It follows from the uniform limit theorem that the limit function u C([, + ); H). By Lemma.1., if u λ C 1 ([, + ); H) satisfies dw + A λw = on [, + ), then ( ) ( ) d duλ duλ + A λ =, t. 8

32 Also from the proof of Lemma.1., we have u λ C ([, + ); H). Let s denote v λ = du λ, then v λ C ([, + ); H) and dv λ + A λ v λ = Then d(v λ v µ ) = (A λ v λ A µ v µ ). Using the same argument as (.1.b), we have ( ) 1 d v λ v µ d(vλ v µ ) =, v λ v µ Now, Therefore, (A λ v λ A µ v µ, v λ v µ ) = (A λ v λ A µ v µ, v λ v µ ). = (A λ v λ A µ v µ, v λ J λ v λ + J λ v λ J µ v µ + J µ v µ v µ ) = (A λ v λ A µ v µ, λa λ v λ µa µ v µ ) + (A λ v λ A µ v µ, J λ v λ J µ v µ ) (since A λ = 1/λ(I J λ )) = (A λ v λ A µ v µ, λa λ v λ µa µ v µ ) + (A(J λ v λ J µ v µ ), J λ v λ J µ v µ ) (A λ v λ A µ v µ, λa λ v λ µa µ v µ ). (since (Av, v), A being monotone) 1 d v λ v µ = (A λ v λ A µ v µ, λa λ v λ µa µ v µ ) (A λ v λ A µ v µ, λa λ v λ µa µ v µ ) A λ v λ A µ v µ λa λ v λ µa µ v µ (by Cauchy-Schwarz inequality) ( A λ v λ + A µ v µ )(λ A λ v λ + µ A µ v µ ). (.1.7) Again from the proof of Lemma.1., dv λ = d u λ is non-increasing on [, + ). Therefore, dv λ (t) A λv λ () = A du λ λ () = A λ A λ u = A λu A λ v λ (t) A λu λ >. (.1.8) Now we assume that u D(A ), so, Au D(A). Since A λ v = J λ (Av), v D(A), λ >, we have A λu = A λ (A λ u ) = A λ (J λ (Au )) = J λ A(J λ (Au )) = J λ J λ A(Au ) = JλA u. 9

33 Therefore, A λu = J λa u J λ L(H) A u A u. (.1.9) The last inequality holds because J λ L(H) 1 for all λ >. From (.1.7) and (.1.8), we have 1 d v λ v µ A λu (λ + µ) A λu = (λ + µ) A λu (λ + µ) A u. (from (.1.9)) Integrating wrt t, we have v λ (t) v µ (t) t(λ + µ) A u. (.1.9a) Therefore v λ v uniformly on every bounded interval [, T ] as λ +. Since v λ C([, + ); H), v being the uniform limit of v λ, satisfies v C([, T ]; H) for every T >, and hence v C([, + ); H). So far we have, as λ + u λ (t) u(t) uniformly on [, T ] du λ (t) v(t) uniformly on [, T ], T >. (.1.1) By Lemma.1.3, we have u has a derivative du du λ C([, + ); H), and v(t) = du C([, + ); H). And hence, u C 1 ([, + ); H). and v(t) = du (t). Besides, since (t) is the uniform limit, it also holds that du Now note that we can write du λ (t) + A λu λ (t) = as du λ (t) + A(J λu λ (t)) =. (since A λ x = A(J λ x) x H, λ > ) 3

34 Recall that lim λ J λ x = x. Therefore J λ u λ (t) u(t) J λ u λ (t) J λ u(t) + J λ u(t) u(t) J λ L(H) u λ (t) u(t) + J λ u(t) u(t) as λ + That is, J λ u λ (t) u(t) as λ +. (.1.11) Again, from (.1.1) and (.1.11), A(J λ u λ (t)) = du λ (t) du as λ +. Since A is closed, this means u(t) D(A) and du (t) = Au(t) du (t) + Au(t) =. Furthermore, note that since u C 1 ([, + ); H), Au(t) = du follows that u(t) u(t ) D(A) = u(t) u(t ) + Au(t) Au(t ) as t t for all t C([, + ); H). It Thus, u C([, + ); D(A)). Moreover, recall that from the derived estimates (.1.3) and (.1.4), we have u λ (t) u t, λ >, A λ u λ (t) Au t, λ >. And passing on to the limit as λ +, we have u(t) u t, Au(t) Au t. Now note that by Lemma.1.4, D(A ) is dense in D(A). Therefore, for any u D(A), we can find a sequence (u n ) D(A ) such that u n u D(A). That is, u n u + Au n Au. 31

35 That is, u n u and Au n Au. By the preceding analysis, for every u n D(A ), we know that the problem: has a solution, say, u n which satisfies dw + Aw = on [, + ), w() = u n, u n (t) u du n (t) = Au n(t) Au t. Note that u n u m is a solution to the problem dw + Aw = on [, + ), w() = u n u m, therefore, for all t, we have u n (t) u m (t) u n u m, (.1.1) du n (t) du m (t) Au n Au m. (.1.13) Since u n u in H, (u n ) is a Cauchy sequence in H. Therefore given any ɛ > for sufficiently large m, n, we have u n (t) u m (t) u m u n < ɛ That is, (u n (t)) is a Cauchy sequence in H. Thus (u n (t)) has a pontwise limit in H. Let s denote it as u(t) := lim n u n (t). Then from (.1.1), lim u n(t) u m (t) lim u n u m m m u n (t) u(t) u n u Therefore, for any T sup u n (t) u(t) u n u t [,T ] as n. Thus, u n (t) u(t) uniformly on [, T ] T. (.1.14) 3

36 By a similar argument, we have from (.1.13), du n (t) φ(t) uniformly on [, T ] T (.1.15) where φ(t) is the pointwise limit of dun (t). Therefore by Lemma.1.3, it follows that u(t) is differentiable and φ(t) = du(t) on [, T ] for any T. Since u n are continuous on [, T ], so is the uniform limit u on [, T ] T. That is, u C([, + ); H). Again, since u n C 1 ([, + ); H), it follow that dun (t) C([, + ); H), and thus, du du (t) being the uniform limit, it follows that (t) C([, + ); H). Hence, That is, du Now, we note that u C 1 ([, + ); H). Au n (t) = du n (t) du (t) and u n (t) u(t). (t) is continuous on [, T ] T. But since A is a closed operator, it follows that u(t) D(A) and Au(t) = du(t), that is, du (t) + Au(t) =. Moreover, note that Au = du C([, + ); H). Also since u C1 ([, + ); H), we have u(t) u(t ) D(A) = u(t) u(t ) + Au(t) Au(t ) as t t for any t. Thus, u C([, + ); D(A)). This concludes the proof of the Theorem.. Regularity of the solutions It turns out that if one were to make additional assumptions on the initial data u, the solution to the system (.1.1) in Theorem.1.5 (Hille-Yosida Theorem) is 33

37 more regular than just C 1 ([, + ); H) C([, + ); D(A)). For that purpose we inductively define the space D(A k ) = {v D(A k 1 ); Av D(A k 1 )}, where k is an integer. Claim : D(A k ) is a Hilbert space for the scalar product the corresponding norm is (u, v) D(A k ) = u D(A k ) = k (A j u, A j v); j= ( k j= A j u ) 1. Proof. Note that (u, v) D(A k ) being a finite sum of scalar products is indeed a scalar product itself. Let (u n ) D(A k ) be a Cauchy sequence. That is, given ɛ >, N ɛ such that m, n > N ɛ, u n u m D(A k ) < ɛ ( k ) 1 A j (u n u m ) < ɛ j= u n u m + Au n Au m + + A k u n A k u m < ɛ u n u m < ɛ u n u m < ɛ m, n > N ɛ This means u n is also Cauchy in H. Then u n converges to some limit u H. Similarly, (Au n ), (A u n ),, (A k u n ) are Cauchy sequences in H and hence they converge to the limits, say u 1, u,, u k respectively. Recall that A is maximal monotone and hence a closed operator. Note that (u n ) D(A k ) implies (u n ) D(A j ) for j = to k. Now (u n ) D(A), u n u H, Au n u 1 H. Since A is closed, this means u D(A) and u 1 = Au. Again, (u n ) D(A ) and we just saw that Au n Au. Also, A(Au n ) = A u n u Since A is closed, it follows that Au D(A) and u = A u. Repeating this argument inductively, it is clear that u j = A j u and A j u D(A) for j = to k. This in turn implies that u D(A k ). Thus the Cauchy sequence (u n ) u in D(A k ) and hence D(A k ) is complete wrt the distance function induced by the scalar product. 34

38 Theorem..1. Suppose u D(A k ) for some integer k. Then the solution u to the problem (.1.1): satisfies du + Au = on [, + ), u() = u, u C k j ([, + ); D(A j )) j =, 1,, k. Proof. Assume first k =. Denote by H 1 the Hilbert space D(A) equipped with the scalar product (u, v) D(A) = (u, v) + (Au, Av) and the norm u D(A) = ( u + Au ) 1/. Denote by A 1 the operator A 1 : D(A 1 ) H 1 H 1 such that D(A 1 ) = D(A ), A 1 v = Av for v D(A 1 ) = D(A ). We will show that A 1 is maximal monotone in H 1 = D(A). Note that for v D(A 1 ) = D(A ) D(A), (A 1 v, v) = (Av, v). Now, take any f H 1 = D(A). Since A is maximal monotone, this means u D(A) = H 1 such that u + Au = f. That is, Au = f u. Now f, u H 1 Au = f u H 1. But u D(A) and Au D(A) imply u D(A ). Thus for any f H 1, we have found u D(A 1 ) such that u + A 1 u = u + Au = f. So A 1 is maximal monotone in H 1. Consider the system du + A 1u = on [, + ), u() = u, (..1) with maximal monotone operator A 1 in the space H 1. So we can apply Theorem.1.5 to the above system to obtain a solution u C 1 ([, + ); H 1 ) C([, + ); D(A 1 )). Note that u(t) D(A 1 ) = D(A ) D(A). So A 1 u(t) = Au(t). But then u satisfies du + Au = on [, + ), u() = u. By the uniqueness of the solution obtained in Theorem.1.5, this u is the unique solution to the system (.1.1). 35

39 So we have that u C 1 ([, + ); D(A)) C([, + ); D(A )). Now we will show that u C ([, + ); H). For any x H 1 with x H1 = 1, we have x H1 = (x, x) H1 = 1 x + Ax = 1 x = 1 Ax 1. This is true for all x H 1 such that x H1 = 1. So, Ax 1, x H1 = 1 and therefore sup Ax 1. x H 1 x H1 = 1 x H 1 such that That is, when A is considered as an operator from H 1 (as a Hilbert space with (, ) H1 ), it is a a bounded linear operator(hence continuous). In other words, A L(H 1 ; H) Again, we have from before that u C([, + ); H 1 ). Therefore Au C([, + ); H). We had previously shown that u C([, + ); D(A 1 )). Therefore, u C([, + ); D(A 1 )) u(t) D(A 1 ) = D(A ) Au(t) D(A) du (t) D(A) = H 1. 36

40 Now, Au(t + h) Au(t) lim h ( h ) u(t + h) u(t) = lim A h h ( ) u(t + h) u(t) =A lim h h ( ) du =A d ( ) du (Au) = A ( ) du = d ( Au) d = A ( du ). (since A is linear bdd, hence continuous) (..) Therefore, d ( ) du = A du C1 (, + ]; H) u C ([, + ]; H). ( ) du C([, + ); H) Also, ( ) ( ) d du du = A d ( ) ( ) du du + A = on [, + ). (..3) Now we use induction for the general case k 3. Assume the Theorem holds up to order k 1. Suppose u D(A k ). We have shown above that the solution u of the problem du + Au = on [, + ), u() = u. 37

41 satisfies : u C ([, + ); H) C 1 ([, + ); D(A)), ( ) ( ) d du du + A = on [, + ). Let s denote v = du. Note that v satisfies the system: dv + Av = on [, + ), v() = du () = Au() = Au. (..4) By assumption, u D(A k ). So Au D(A k 1 ). By Induction hypothesis the theorem holds for the system (..4) with v() = Au D(A k 1 ). Then v C k 1 j ([, + ); D(A j )) for j =, 1,, k 1 du Ck 1 j ([, + ); D(A j )) for j =, 1,, k 1 u C k j ([, + ); D(A j )) for j =, 1,, k 1. Now we only need to verify that u C([, + ); D(A k )). For j = k 1, we have u C 1 ([, + ); D(A k 1 ) du C([, + ); D(Ak 1 )) Au C([, + ); D(A k 1 )). So, we have, u(t) D(A k 1 ) and Au(t) D(A k 1 ). Therefore, u(t) D(A k ) for all t. Note that u(t) u(t ) D(A k ) = And, k A j (u(t) u(t )) for any t. j= k 1 Au(t) Au(t ) D(A k 1 ) = A j (Au(t) Au(t )) = j= k A j (u(t) u(t )) for any t. j=1 38

42 Therefore, ) 1/ u(t) u(t ) D(A k ) = ( u(t) u(t ) + Au(t) Au(t ) D(A k 1 lim t t u(t) u(t ) D(A k ) = ( ) 1/ lim u(t) u(t ) + lim Au(t) Au(t ) t t t t = (since u C([, + ); H) and Au C([, + ); D(A k 1 )).) Therefore, u C([, + ); D(A k )). This concludes the proof..3 The case of self-adjoint operators Suppose A : D(A) H H is an unbounded linear operator and D(A) is dense in H, that is, D(A) = H. Define D(A ) to be the set of all f H such that the linear functional g (f, Ag) extends to a bounded linear functional on all of H. Since D(A) is dense in H, by Riesz representation theorem it follows that there exists a unique h H such that (f, Ag) = (h, g). We define the adjoint operator, A of A as A f = h. Clearly, A is also a linear operator. An operator A is called symmetric if (u, Av) = (Au, v) A is called self-adjoint if D(A) = D(A ) and A = A. u, v D(A). An operator Claim 1: A is a symmetric operator if and only if D(A) D(A ), A = A on D(A). Proof. ( :) A is symmetric. That is, (u, Av) = (Au, v) for all u, v D(A). Take f D(A) and the functional g (f, Ag). Now, since A is symmetric, (f, Ag) = (Af, g) for all g D(A). Denote by T the functional: T (g) = (Af, g) g H. 39

43 Note that for g D(A), T (g) = (Af, g) = (g, Af). And thus T is an extension of the functional g (f, Ag) on all of H. Also, note that (Af, g) Af g by Cauchy- Schwarz inequality. Hence, T is a bounded linear functional that is an extension of the functional g (f, Ag) on all of H. Therefore by definition of D(A ), f D(A ). The choice of f D(A) was arbitrary. So, we have D(A) D(A ). Again, by Riesz s Representation Theorem, h H such that T (g) = (h, g) for all g H. By definition, A f = h. But T (g) = (Af, g) g H. Therefore, (Af, g) = (A f, g) Af = A f. g H This is true for any f D(A). Therefore, A = A on D(A). ( : ) We have D(A) D(A ), A = A on D(A). Take any u D(A). Then u D(A ). Then by definition of D(A ), the functional g (u, Ag) can be extended to a bounded linear functional T on all of H such that T (g) = (A u, g) g H. Now A = A T (g) = (Au, g) g H. So, for g D(A), T (g) = (u, Ag) (Au, g) = (u, Ag). (since T is an extension of the functional) Therefore, A is symmetric. Claim : If A is a self-adjoint operator, then it is symmetric. Proof. A is self-adjoint. Therefore, D(A ) = D(A) and A = A. Take any u D(A) = D(A ). By definition of D(A ), the functional g (u, Ag) extends to bounded linear functional, say T on all of H. Then T (g) = (A u, g) = (Au, g) for all g H. But since T is an extension, for all g D(A), (u, Ag) = T (g) = (Au, g). Thus for all u, g D(A), (u, Ag) = (Au, g). That is, A is symmetric. 4

44 Claim 3: Suppose T L(H; H). Then T is symmetric if and only if it s self-adjoint. Proof. From Claim, we know that a self-adjoint operator is always symmetric. So we only need to prove ( :) T is symmetric. Therefore, (u, T v) = (T u, v) for all u, v H. Then from Claim 1, we have, D(T ) D(T ). That is, H = D(T ) = D(T ) H. Therefore, D(T ) = D(T ) = H and T = T on D(T ) = H. Therefore, T is self-adjoint. Theorem.3.1. Suppose A is a maximal monotone operator that is symmetric. Then A is self-adjoint. Proof. Denote J 1 = (I + A) 1. Recall that D(J 1 ) = H, R(J 1 ) = D(A), and J 1 L(H) 1. We will first show that J 1 is self-adjoint. Since J 1 is linear and bounded, by Claim 3, it suffices to show that J 1 is symmetric. Take any u, v H. Denote J 1 u = u 1, J 1 v = v 1.Note that u 1, v 1 D(A). Then u = u 1 + Au 1, v = v 1 + Av 1. So, Au 1 = u u 1, Av 1 = v v 1. A is symmetric. Therefore, (u 1, Av 1 ) = (Au 1, v 1 ) (u 1, v v 1 ) = (u u 1, v 1 ) (u 1, v) (u 1, v 1 ) = (u, v 1 ) (u 1, v 1 ) (J 1 u, v) = (u, J 1 v) Thus J 1 is symmetric and hence, self-adjoint. Take any u D(A ). Let f = u + A u, that is, f u = A u. Recall that from the definition of adjoint operator, we have (u, Ag) = (A u, g) for all g D(A). Therefore, (u, Ag) = (f u, g) = (f, g) (u, g), that is (u, g + Ag) = (f, g) g D(A). (.3.1) 41

45 Recall that D(J 1 ) = H, R(J 1 ) = D(A). Take any w H. Then J 1 w D(A). Denote v = J 1 w. Then v + Av = w. Therefore, by (.3.1), (u, v + Av) = (f, v) (u, w) = (f, J 1 w) w H. Since we showed that J 1 is symmetric, we have (f, J 1 w) = (J 1 f, w) = (u, w) w H. Taking w = J 1 f u, (J 1 f, J 1 f u) = (u, J 1 f u) (J 1 f u, J 1 f u) = J 1 f = u. Therefore, u R(J 1 ) = D(A). Since the choice of u D(A ) was arbitrary, this means, D(A ) D(A). Again, since A is symmetric, by Claim 1, D(A) D(A ), A = A on D(A). Therefore, D(A) = D(A ) and A = A on D(A) = D(A ). Hence, A is self-adjoint. Theorem.3.. Suppose A is a maximal monotone operator that is self-adjoint. Then for any u H, a unique solution u C([, + ); H) C 1 ((, + ); H) C((, + ); D(A)) to the problem: du + Au = on (, + ), u() = u. (.3.) Furthermore, we have the following estimates: u(t) u t >, du (t) = Au(t) 1 t u t >, u C k ((, + ); D(A l )) for all non-negative integers k, l. 4

46 Proof. First we show uniqueness. Suppose u and ū are two solutions to the system (.3.). We have du dū + Au Aū = on (, + ) d(u ū) = A(u ū) ( ) d(u ū), u ū = (A(u ū), u ū). (since A is monotone, (Av, v), v D(A)) Denote φ(t) = u(t) ū(t). Then similar to (.1. ), we get ( ) d dφ(t) φ(t) =, φ(t) ( ) d(u ū) =, u ū on (, + ). Therefore, φ is non-increasing on (, + ). Since u, ū C([, + ); H), φ = u ū is continuous on [, + ). Therefore, φ(t + h) φ(h) for h, t φ(t) φ() =, t φ on [, + ). Thus the solution to (.3.) is unique. Now we prove the existence part. Let us first assume that u D(A ) (that is u D(A), Au D(A)). Let u be the solution to the system: du on [, + ), u() = u, obtained in Theorem.1.5. We will show that du 1 u t t >. Recall that J λ = (I + λa) 1 has D(J λ ) = H and A λ = 1 (I J λ λ) has D(A λ ) = D(J λ ) = H. 43

47 We showed in the proof of Theorem.3.1 that J 1 is symmetric. By a similar argument it follows that J λ is symmetric and hence by Claim 3, that J λ is self-adjoint, that is, J λ = J λ. Now, (A λ u, v) = ( 1 λ (I J λ)u, v) = 1 λ (u, v) 1 λ (J λu, v) = 1 λ (u, v) 1 λ (u, J λv) (since J λ is symmetric) ( = u, 1 ) λ (v J λv) = (u, A λ v). Thus A λ is symmetric and hence by Claim 3, it is self-adjoint as well, that is, A λ = A λ. Consider the following approximate problem that was used in the proof of Theorem.1.5: du λ + A λu λ = on [, + ), u() = u. (.3.3) Recall that A λ is Lipschitz, since A λ v 1 v, v H, λ >. Then by λ Theorem.1.1(Cauchy-Lipschitz-Picard), the unique solution u λ to (.3.3) satisfies u λ C 1 ([, + ); H). Now from (.3.3), du λ = A λu λ ( ) duλ (t), u λ(t) = (A λ u λ (t), u λ (t)). Again, similar to (.1. ), we have, ( ) 1 d u λ(t) duλ = (t), u λ(t) = (A λ u λ (t), u λ (t)). Integrating wrt to t over [, T ], we get 1 ( u λ(t ) u λ () ) = 1 u λ(t ) + T T (A λ u λ (t), u λ (t)) (A λ u λ (t), u λ (t)) = u. (.3.4) Taking the scalar product of (.3.3) with t du t du λ, we have ( + t A λ u λ (t), du λ (t) 44 ) =.

48 Integrating wrt to t over [, T ], we get Now, d (A λu λ, u λ ) T du λ (t) t + T ( A λ u λ (t), du ) λ (t) t =. (.3.5) (A λ u λ (t + h), u λ (t + h)) (A λ u λ (t), u λ (t)) = lim h h (A λ u λ (t + h) A λ u λ (t), u λ (t + h)) (A λ u λ (t), u λ (t + h) u λ (t)) = lim h ( ( ) h ) ( uλ (t + h) u λ (t) = lim A λ, u λ (t + h) + lim A λ u λ (t), u ) λ(t + h) u λ (t) h h h h ( ) ( du λ = A λ, u λ + A λ u λ, du ) λ ( ) ( duλ =, A λu λ + A λ u λ, du ) λ (since A λ is self-adjoint) ( = A λ u λ, du ) λ. (.3.6) From (.3.6), = 1 = 1 T T [ t ( A λ u λ, du λ ) t [ d (A λu λ, u λ ) ] t d (A λu λ, u λ ) ] T = T (A λu λ (T ), u λ (T )) 1 1 T T ( ) d (A λu λ, u λ ) (using integration by parts) (A λ u λ, u λ ). (.3.7) By Lemma.1.1, t du λ (t) is non-increasing. Therefore, T du λ (t) t du λ (T ) T t = du λ (T ) T. (.3.8) 45

49 From (.3.7), we have So, T (A λ u λ, u λ ) = T (A λ u λ (T ), u λ (T )) T T ( A λ u λ, du λ ) t = T (A λ u λ (T ), u λ (T )) + du λ t (using (.3.5)) T (A λ u λ (T ), u λ (T )) + T du (T ). (using (.3.8)) T Now, from (.3.4), (A λ u λ, u λ ) T (A λ u λ (T ), u λ (T )) + T du (T ). (.3.9) u T = 1 u λ(t ) + (A λ u λ, u λ ) 1 u λ(t ) + T (A λ u λ (T ), u λ (T )) + T du λ (T ). (using (.3.9)) That is, u λ (T ) + T (A λ u λ (T ), u λ (T )) + T du λ (T ) u u λ (T ) + T (A λ u λ (T ), u λ (T )) + T du λ (T ) + T du λ (T ) u u λ (T ) + T (A λ u λ (T ), u λ (T )) + T A λ u λ (T ) + T du λ (T ) u (using du λ / = A λ u λ ) u λ(t ) + T du λ (T ) + T du λ (T ) u T du λ (T ) u du λ (T ) 1 T u, T >, λ >. (.3.1) Recall that in the proof of Theorem.1.5, we showed that du λ du uniformly as λ +. 46