Functional Analytic Methods for PDE s. G. Seregin

Transcription

1 Functional Analytic Methods for PDE s G. Seregin November 28, 2017

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3 Contents 1 Introduction Why Functional Analysis Methods are important for PDE s?. 5 2 Lebesgue and Sobolev Spaces Lebesgue s Spaces Spaces L p (E) and L p (E) Sets that are dense in L p (E) Linear Functionals and Weak Convergence in L p (E) Mollification in L p () Distributions Spaces of Differentiable Functions Distributions Distributional Derivatives Sobolev Spaces Weak Derivatives Mollification of Functions with Weak Derivatives Sobolev Spaces Extension of Functions with Weak Derivatives The First Emedding Theorem Sobolev embedding of W 1,p () into L q () Traces of functions with weak derivatives Surface Integral Traces of functions from Sobolev Spaces Functional Methods for PDE s Methods of Functional Analysis for PDE s

4 4 CONTENTS Notation Dirichlet Boundary Value Problems for Elliptic Equation Variational Method Spectrum of Elliptic Differential Operators under Dirichlet Boundary Condition Smoothness of Weak Solutions The Second Embedding Theorem Solvability in H 2 H Smoothness of Distributional Solutions A Functional Analysis Background 69 A.1 Normed Spaces A.2 Completeness A.3 Separability A.4 Compactness A.5 Linear Operators A.6 Duality A.7 Fredholm Alternative B Lebesgue s Integration 79 B.1 Lebesgue s Measure B.2 Measurable Functions B.3 Lebesgue s Integral B.4 Sequences of Integrable Functions

5 Chapter 1 Introduction 1.1 Why Functional Analysis Methods are important for PDE s? Let us consider, as a simple but important example, the general linear PDE (partial differential equation) of elliptic type Lu := (a ij u,j ),i +b i u,i +cu = f +g i,i in. (1.1.1) Here, is a domain in R n, n 2, u is unknown function, u,i = u/ x i, a = (a ij ) is a given symmetric matrix field, b = (b i ) and g = (g i ) are given vector valued functions, c and f are given scalar functions, and summation over repeated indices running from 1 to n is adopted. It is assumed that the matrix a satisfies the ellipticity condition νi a ν 1 I ( ν ξ 2 ξ aξ ν 1 ξ 2 ξ R n ) with a positive ν. Here, I is the identity matrix. Equations (1.1.1) can be written also in the following invariant form div(a u)+b u+cu = f +divg in. In general, equation (1.1.1) may have infinitely many solutions. In order to select a particular one, describing a given situation, one should prescribe a boundary condition u = u 0 on. (1.1.2) 5

6 6 CHAPTER 1. INTRODUCTION Problem (1.1.1), (1.1.2) is called the Dirichlet boundary value problem. A classicalsolutionto(1.1.1), (1.1.2)isasolutionu, belongingtoc 2 () C(). The existence and the multiplicity of solutions to boundary value problem (1.1.1), (1.1.2) are important questions in the theory of PDE s. Answers to them are not very easy. Moreover, there might be no classical solution at all, for example, if a, b, c, f, and g are not smooth enough. Several warning messages, indicating that the classical approach to the aforesaid problems does not work, came from physics and the Calculus of Variations in the first part of 20th century. In particular, to prove the existence of a minimizer, say, of the multiple integral F( u)dx, one has to extend it to non-smooth functions and thus to assume that solutions to the Euler-Lagrange equation for the integrand F are not necessary smooth. On the other hand, δ-function (or Dirac function), introduced by physicists, suggests that the notion of functions as well as solutions should be revised. That time, the main trend was to include PDE problems into the functional analysis framework. The principal objects in powerful functional analysis schemes are function spaces and operators, acting there. Suitable spaces (Lebesgue and Sobolev spaces), in which differential operators have reasonable properties, were discovered in the first part of 20th century. Our course can be regarded an introduction to the theory of spaces of functions with so-called weak derivatives and includes the celebrated Sobolev embedding theorems. Having in hands good function spaces, we shall define the notion of weak solutions, re-discovering old ideas of mechanics and the Calculus of Variations. Namely, we replace our differential equation (1.1.1) with the integral identity L(u,v) = (a ij u,j v,i +b i u,i v +cuv)dx = (fv g i v,i )dx, being valid for any test function v that is sufficiently smooth and vanishes in a neighborhood of the boundary of. If everything is sufficiently smooth,

7 1.1. WHY FUNCTIONAL ANALYSIS METHODS ARE IMPORTANT FOR PDE S?7 we may integrate by parts and and derive the following identity (Lu f divg)vdx = 0, showing that all classical solutions are of course weak solutions as well. It turns out that the existence of weak solutions is a relatively simple consequence of well-known theorems of functional analysis. Methods of functional analysis give modern and powerful tools to treat problems related to PDE s and, nowadays, it is difficult to imagine modern mathematics of PDE s without them.

8 8 CHAPTER 1. INTRODUCTION

9 Chapter 2 Lebesgue and Sobolev Spaces 2.1 Lebesgue s Spaces Spaces L p (E) and L p (E) Let 1 p and E R n be measurable. For simplicity, we always assume that Lebesgue s measure of E is finite, i.e., E := µ(e) <, although many statements hold true without this restriction. Define for a measurable function f : E R ( )1 f p,e := f(x) p p dx E if p < and if p =. Here, f,e := esssup f(x) x E ess sup f(x) := inf{c > 0 : f c a.e. ine}. x E We let then L p (E) := {f : f is measurable and f p,e < }. L p (E) is a vector space. Indeed, if f,g are measurable in E and λ R, then f +λg is measurable. In addition, f +λg p 2 p 1 ( f p + λ p g p ) for 1 p <. So, f +λg L p (E). 9

10 10 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES Lemma 1.1. (Hölder inequality) Let f L p (E), and g L p (E) with 1/p+ 1/p = 1. Then fg L 1 (E) and E ( f(x)g(x)dx E )1 ( f(x) p p dx E ) 1 g(x) p p dx. Proof Sheet 1. With the help of Hölder inequality, we can easily prove the following version of Theorem 4.1 of Appendix B on dominated convergence. Theorem 1.2. Let f m, m = 1,2,..., be a sequence of measurable functions in E. Suppose that (i) f m f a.e. in E; (ii) sup m f m p,e < for some p > 1. Then f L p (E) and f f m q,e 0 as m and for any 1 q < p. Proof By Fatou s lemma, f L p (E). Let γ > 0 and E m = {x E : f m (x) f(x) γ}. By Theorem 2.3 (Lebesgue) of Appendix B, E m 0 as m 0. By Hölder inequality, f f m q q,e = f f m q q,e\e m + f f m q q,e m γ q E \E m + E m (1 q p ) f f m q p,e m γ q E + E m (1 q p ) c(p,q)sup f m q p,e. m Passing to the limit as m, we find lim sup f f m q,e γ E 1 q. m Letting γ tend to 0, we complete the proof. Lemma 1.3. (Minkovski inequality) ( )1 f(x)+g(x) p p dx ( )1 ( f(x) p p dx + )1 g(x) p p dx. E E E

11 2.1. LEBESGUE S SPACES 11 Proof Sheet 1. Note f p,e = 0 implies f = 0 a.e. in E and thus, from Lemma 1.3, it follows that f p,e is a semi-norm in L p (E). To work with semi-norms is inconvenient and, in order to avoid this, we introduce equivalence classes in the following natural way. Two measurable functions f and g are equivalent in E (f g) if f = g a.e. in E. An equivalence class generated by a measurable function f is denoted by [f]. The space of all equivalence classes whose representatives are integrable with power p is denoted by L p (E). However, in what follows, we shall denote an equivalence class [f] by the function that generates it, i.e., simply by f. Null element of L p (E) consists of all functions that are equal to zero a.e. in E and, hence, L p (E) is a normed space. Theorem 1.4. L p (E) is a Banach space. Proof (for 1 p <, p = is an exercise). Let f m, m = 1,2,..., be a Cauchy sequence in L p (E). In particular, this implies ε m = sup f k f m p,e 0 k>m as m. One can find a subsequence ε ms such that ε ms < 2 s, s = 1,2,... Then, by Hölder inequality, f ms+1 f ms 1,E E 1 p f ms+1 f ms p,e ε ms E 1 p and thus the series E f m1 dx+ s=1 E f ms+1 f ms dx converges. By Beppo Levi theorem (see Theorem 4.2) of Appendix B, the series f m1 + f ms+1 f ms s=1 converges a.e. in E and the series f m1 + (f ms+1 f ms ) s=1

12 12 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES convergence a.e. in E absolutely. The later means that S j = f m1 +f m2 f m f mj+1 f mj = f mj+1 converges a.e. in E to a measurable function f. So, for sufficiently large j, we have ε p m f mj f m p dx E and f mj f m p f f m p a.e. in E. By Fatou s lemma (see Lemma 4.3) of Appendix B, ε p m f f m p dx for any m. This completes the proof. E Sets that are dense in L p (E) Let T = {E j } m j=1 be a partition of E, i.e., E = m j=1 E j, E j E k = if j k. f : E R is a simple function if there exists a partition T such that f(x) = c i for x E i. Theorem 1.5. Let 1 p. The set of all simple functions in E is dense in L p (E). Proof The proof is based on Lebesgue s partition, on Sheet 1. Theorem 1.6. Let 1 p <. C(R n ) is dense in L p (E). We start with two auxiliary lemmata. Lemma 1.7. Let A R n and (x,a) = inf x z. Then (x,a) z A (y,a) x y. Proof We have for z A (x,a) x z x y + y z, which implies (x,a) x y + (y,a). Replacing x with y and y with x, we complete the proof.

13 2.1. LEBESGUE S SPACES 13 Lemma 1.8. Let 1 p < and E 0 E be two measurable sets in R n. Given ε > 0, there exists a function g C(R n ) such that χ E0 g p,e < ε, where χ E0 (x) = 1 if x E 0 and χ E0 (x) = 0 if x E \E 0. Proof Since E 0 is measurable, there exist a closed set F E 0 and an open set O E 0 such that O\F < (ε/2) p. We let 0 g(x) := (x,r n \O) (x,f)+ (x,r n \O) 1 for x R n. Obviously, g is continuous function in R n. And χ E0 g p,e 2 O \F 1 p ε. Proof of Theorem 1.6 (Sheet 1. Hint: First approximate a function by simple functions and then approximate simple functions by continuous functions) Corollary 1.9. L p (E) is separable if 1 p <. Indeed, let Q be an open cube such that E Q. Since C(Q) is separable, we find a countable set {f k } k=1 C(Q) that is dense in C(Q) with respect to L -norm. By Theorem 1.6, given ε > 0 and given f L p (E), there exist a function g C(Q) such that f g p,e ε/2 and a function f i such that g f i p,c(q) g f i,c(q) Q 1 p < ε/2. So, f fi p,e < ε. However, L (E) is not separable (on problem sheet). Arguments are as follows. Let T be a countable partition of E, i.e., T = {E j } j=1, E = j=1 E j, E i E j = if i j. We fix it. Define X 0 L (E) so that f X 0 if and only if f L (E) and f(x) = c i for x E i, i = 1,2,... The mapping π : X 0 l defined by πf = c = (c i ) is an isometric isomorphism. If L (E) is separable, then X 0 is separable as well (Explain why, see remark below). But this implies separability of l, which is wrong. Remark Let (X, ) be a separable normed spaced and X 0 be a subset of X. Then X 0 is separable. Indeed, there exists a countable set {x k } k=1 that is dense in X. Let ε m > 0 tend to zero as m goes to. We can find z km X 0 such that x k z km < ε m /3+ (x k,x 0 ).

14 14 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES To show that {z km } is dense in X 0, take an arbitrary ε and sufficiently large m so that ε m < ε. Now, let x X 0. First, we can find x k such that x x k < ε/3. We have x z km x x k + x k z km < ε/3+ε m /3+ (x k,x 0 ) 2ε/3+ x k x < ε. Theorem (integral continuity or continuity of translations) Let E be a measurable bounded set of R n and 1 p <. Let f L p (E) be extended by zero from E to the whole R n. Then for any ε > 0, there exists δ > 0 such that ( )1 f( +h) f( ) p,e := f(x+h) f(x) p p dx < ε whenever h < δ. E Proof We fix a large cube Q such that E + h Q for any h R n such that h 1. Since f L p (Q), given ε > 0, there exists a function g C(Q) such that f g p,q < ε. Since g in uniformly continuous in Q, there exists 0 < δ < 1 such that g(x+h) g(x) < ε E 1 p as long as x,x+h Q and h < δ. So, f( +h) f( ) p,e f g p,e+h + g( +h) g( ) p,e + g f p,e 2 f g p,q + g( +h) g( ),E E 1 p < 3ε Linear Functionals and Weak Convergence in L p (E) Lemma Let f L p (E). Then f p,e = I, where { I := sup E } f(x)g(x)dx : g p,e = 1. Proof By Hölder inequality, I(g) := E f(x)g(x)dx f p,e if g p,e = 1 and thus I f p,e.

15 2.1. LEBESGUE S SPACES 15 Consider first the case 1 < p. Define g 0 (x) = signf(x) f(x) p 1 p p / f p,e (g 0 (x) = signf(x) if p = ). It is easy check that g 0 p,e = 1 and I I(g 0 ) = f p,e. This completes the proof for the case 1 < p. If p = 1, then, given ε > 0, define a set E ε = {x E : f(x) f,e ε} and a function g 0 (x) = χ Eε (x)signf(x)/ E ε. Simple calculations show that g 0 1,E = 1 and I I(g 0 ) f,e ε. Passing ε 0, we get the statement of the lemma for p = 1. Theorem (Riesz) Let 1 p <. There exists isometric isomorphism π : (L p (E)) L p (E), p = p, so that πt = f with p 1 T(g) = f(x)g(x)dx g L p (E). (2.1.1) So, we have (L p (E)) = L p (E). E From the Riesz representation theorem, it follows that the space L p (E) is reflexive provided 1 < p <. Proposition Assume that sup f m p,e <. m If 1 < p <, there exists a subsequence f mk such that f mk f as k, i.e., for any g L p (E), f mk gdx as k. If p = there exists a subsequence f mk such that E E f mk f, i.e., (2.1.2) holds true for any g L 1 (E). f gdx (2.1.2) Proof It is a direct consequence of Theorems 6.4 and 6.5 of Appendix A and Theorem 1.13.

16 16 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES Mollification in L p () Define a function for non-negative t as h(t) = 0 if t 1 and h(t) = exp{ 1 } t 1 if 0 t < 1. We also let ω 1 (x) = c 1 (n)h( x 2 ), x R n, with constant c 1 chosen so that ω 1 (x)dx = 1. R n For positive, we define a mollifier as ω (x) = nω 1 1 ( x ) that is an infinitely differentiable function and equal to 0 outside B( ). By scaling and by shift, we have for any x R n : ω (x y)dy = ω (x y)dy = 1. R n B(x, ) Let be a domain in R n and f L 1 () be extended by zero to the whole R n. A mollification of f is f (x) := (ω f)(x) = ω (x y)f(y)dy. It is an infinitely differentiable function in R n (explain why) and vanishes outside := {x R n : dist(x,) < }. Our aim is to show that we can approximate functions from L p () with the help of mollification. Lemma Let f L p () and 1 p. f p, f p,. Proof Let 1 < p <. We apply Hölder inequality in the following way 1 p f (x) ω + 1 p (x y) f(y) dy = 1 p ω (x y)ω p (x y) f(y) dy 1

17 2.1. LEBESGUE S SPACES 17 ( ) 1 p ω (x y)dy ( )1 ω (x y) f(y) p p dy. We know ω (x y)dy 1. It remains to apply Tonelli s theorem and conclude f p p, = f (x) p dx dx ω (x y) f(y) p dy = f(y) p dy ω (x y)dx f(y) p dy = f p p,. Cases p = 1 and p = are considered in the same way (explain why). Theorem Let f L p () and 1 p <. Then f f in L p () as 0. Proof For any x, we have (f is extended by zero outside ) f (x) f(x) = ω (x y)f(y)dy f(x) = = ω (x y)f(y)dy f(x) = ω (x y)(f(y) f(x))dy B(x, ) B(x, ) and thus f (x) f(x) ω (x y) f(y) f(x) dy z=y x = B(x, ) ω (z) f(x+z) f(x) dz. B( ) Repeating arguments used in the proof of Lemma 1.15, we find f (x) f(x) p dx dx ω (z) f(x+z) f(x) p dz B( )

18 18 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES = ω (z)dz f(x+z) f(x) p dx and, therefore, B( ) f f p, sup f( +z) f( ) p, (2.1.3) z < By Theorem 1.11, the right hand side of (2.1.3) tends to 0 as 0. Theorem (Riesz) Let F be a subset in L p (). Let 1 p <. F is a precompact in L p () if and only if (i) sup f p, =: M < f F (ii) sup sup f( +z) f( ) p, =: δ( ) 0 as 0. f F z < Proof We start with a proof of sufficient conditions of compactness. Let us denote by F the set of all f with f F and show that for each fixed > 0 this family satisfies all conditions of Ascoli-Arzela theorem, see Theorem 4.7 of Appendix A. First, this family is uniformly bounded, i.e., It is equi-continuous, since f (x 1 ) f (x 2 ) sup f (x) c(, ) f p, = cm. x ω (x 1 y) ω (x 2 y) f(y) dy c(, ) x 1 x 2 f p, cm x 1 x 2 for any x 1 and x 2 from. We take an arbitrary ε > 0 and choose > 0 so that δ( ) < ε/2 and fix it. For this > 0, the family F is precompact in C(). By Hausdorff s theorem, see Theorem 4.5 of Appendix A, there exists a finite ε/(2 1 p )-net {h j C()} m j=1 and we are going to show that it is ε-net for F. Indeed, for (f) F, there exists h j such that f h j p, 1 p f h j, < ε/2 and thus, by (2.1.3) and by our choice of > 0, f h j p, f f p, + f h j p, < ε.

19 2.2. DISTRIBUTIONS 19 Now, we are in a position to prove necessary conditions. Let Q be a large cube so that + h Q for any h 1. Let F Q consists of all functions f L p (Q) such that f(x) = g(x) if x for some g F and f(x) = 0 if x Q\. Obviously, if F is precompact in L p (), then F Q is precompact in L p (Q). By Hausdorff s theorem, there exists a finite 1-net of F Q, say, f j, j = 1,2,...,m. Then by the definition, given f F Q, there exists f j from this 1-net such that f f j p,q 1 and thus we have f p, < f j p,q +1 sup f j p,q +1 =: M 1 j m for any f F. So, uniform boundedness is proved. Next, for an arbitrary ε > 0, we have ε-net of F Q, say, f j, j = 1,2,...,m. So, for f F Q, there exists f j from ε-net of F Q such that f f j p,q < ε. Then, for h <. f( +h) f( ) p, f( +h) f j ( +h) p, + f j ( +h) f j ( ) p, + f j f p, = f f j p,+h + f j ( +h) f j ( ) p, + f j f p, 2 f j f p,q + sup 2ε+ sup 1 j m h < 1 j m h < sup f j ( +h) f j ( ) p, sup f j ( +h) f j ( ) p,. It follows from Theorem 1.11, that the second term on the right hand side of the latter inequality tends to 0 as 0. So, lim sup 0 By arbitrariness of ε, sup sup f( +h) f( ) p, 2ε. f F h < lim sup sup 0 f F h < 2.2 Distributions f( +h) f( ) p, = Spaces of Differentiable Functions Definition 2.1. A n-dimensional vector α = (α 1,α 2,...,α n ) = (α i ) with nonnegative integer components is a multi-index of order n. α = α 1 +α α n is a length of α.

20 20 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES For x R n, we let x α := x α 1 1 x α x αn n multi-indices is denoted as = n i=1 xα i i. Summation over a α := α k k a α = i=0 α =i k a α1,α 2,...,α n. i=0 α 1 +α α n=i In this notation, a polynomial of order k of n variables can be written as P k (x) := p α x α. α k Denote D i = / x i, D i f = f/ x i, D k i = k / x k i and introduce a formal n-dimensional vector D = (D 1,D 2,...,D n ) so that D α f = D α 1 1 D α Dn αn f = α 1+α α n x α 1 1 x α x αn Exercise Prove that D α D β f = D β D α f. Let be a domain (open connected set) in R n. For bounded, C k () is a B-space with norm f C k := α k max D α f(x). x For f : R n R, define suppf := {x R n : f(x) > 0}. We say that f is compactly supported in if suppf is compact and contained in. By definition, C k 0() consists of all f : R n R being continuously differentiable up to order k and having a compact support in. The important case is a linear space C 0 () consisting of all infinitely differentiable functions compactly supported in. Fore example, the function h(t) = 0 if t 1 and h(t) = exp{ 1 t 2 1 } if t < 1 is of C 0 (R) Distributions Definition 2.2. D() is the space of test functions consisting of all functions from C 0 (). It is endowed with the following notion of convergence. Let ϕ m C 0 (), m = 1,2,..., and ϕ C 0 (), we say that ϕ m ϕ in D() as m if there exists a compact K such that suppϕ m K ( m = 1,2,...), suppϕ K, and D α ϕ m D α ϕ uniformly in K for any multi-indices α. n f.

21 2.2. DISTRIBUTIONS 21 Definition 2.3. A distribution T on is a linear contnuous functional T : D() R. The latter means: ϕ m ϕ in D() T(ϕ m ) T(ϕ) in R. The set of all distributions is denoted D (). Very often, the action T on ϕ D() is also denoted by < T,ϕ >:= T(ϕ). Examples: 1. T is a regular distribution if there exists a function f L 1 loc () such that T(ϕ) = fϕdx. A regular distribution is denoted also as T = T f. Lemma 2.4. T f = T g if and only if f = g a.e. in. Proof T f = T g (f g)ϕdx = 0 for any ϕ C 0 (). The result follows from Lemma 2.5 below. Lemma 2.5. Let f L 1 loc () and in. fϕdx = 0 for ϕ C 0 (). Then f = 0 Proof Without loss of generality, we may assume that is bounded and f L 1 (). Define = {x : dist(x, ) > }. Fix 0 > 0. Then, for 0 < < 0, the function y ω (x y) C 0 () for x 0. By the assumption, f (x) = 0 for all x 0. From Theorem 1.16, it follows that f = 0 a.e. in 0 and thus in. 2. A bounded Radon measure T is a distribution satisfying T(ϕ) M ϕ, for any ϕ D(). Given a, the Dirac δ-function is a distribution T(ϕ) = ϕ(a), ϕ D(). Let us show that the Dirac δ-function is not a regulardistribution. Ifyes,thereexistsf L 1 loc ()suchthat fϕdx = ϕ(a) for any ϕ C0 (). By Lemma 2.5, f = 0 a.e. in \{a} and thus f = 0 a.e. in, which implies ϕ(a) = 0 for any ϕ C0 (). This is a contradiction. Nevertheless, physicists often use the formal notation δ a (x) for density so that ϕ(a) = δ a(x)ϕ(x)dx. However, this is just formal notation since the right hand side of the last identity makes no sense. Remark 2.6. We say that a sequence {T j } of D () converges to T in the sense of distributions if T j (ϕ) T(ϕ) for any ϕ C 0 (). Obviously, T is a linear functional in D() and, moreover, T D (), which is a bit more difficult to prove.

22 22 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES Distributional Derivatives Any distribution has partial derivatives of any order in the following way. Let T be a distribution on and consider the linear functional S(ϕ) = ( 1) α T(D α ϕ), ϕ D(). It is easy to check that S is continuous on D(). Definition 2.7. A distributional derivative of a distribution T is a distribution S denoted by D α T, i.e., S = D α T. Definition 2.7 is in accordance with the classical notion of partial derivatives. Indeed, suppose that f C α (), then integration by parts gives ( 1) α T f (D α ϕ) = ( 1) α fd α ϕdx = D α fϕdx for any ϕ C 0 (). So, we conclude that in this case D α T f = T D α f. Example: Fundamental solution to Laplace s equation. Let, for x 0, f(x) = 1 x n 2 for n 3. Direct calculations shows that f(x) = 0 (2.2.1) if x R n \{0}. Let us find T f on R n. By the definition, T f (ϕ) = T f ( ϕ) for all ϕ D(). Then T f (ϕ) = f ϕdx = lim ε 0 R n So, after double integration by parts, we show T f (ϕ) = lim ε 0 B(ε) R n \B(ε) ( ϕ ν f f ϕ)ds +lim ν ε 0 f ϕdx. R n \B(ε) ϕ f dx,

23 2.3. SOBOLEV SPACES 23 where ν is the outward unit normal to the surface B(ε). The last term on the right hand side vanishes by (2.2.1) and thus 1 ϕ 1 T f (ϕ) = lim ν ε 0 ε n 2 i ds (n 2)lim ϕds. x i ε 0 ε n 1 B(ε) B(ε) Since the surface area of the ball B(ε) is equal to S n 1 ε n 1, where S n 1 is the surface area of the unit ball in R n and since ϕ is a smooth function, we find after taking the limit as ε 0 the following identity T f (ϕ) = (n 2)S n 1 ϕ(0). (2.2.2) However, very often, physicists use the classical notation for (2.2.2) f(x) = (n 2)S n 1 δ 0 (x), x R n, just mentioning that the latter relation is understood in the sense of distributions. 2.3 Sobolev Spaces Weak Derivatives Let be a domain in R n. Definition 3.1. Let u L 1 loc (). A regular distribution T v is a weak (or Sobolev) derivative of u in if T v = D α T u and classical notation is used v := D α u. In other words, v L 1 loc () is a weak derivative of u in if vϕdx = ( 1) α ud α ϕdx, ϕ C0 (). If u C α (), then the integration by parts gives: D α uϕdx = ( 1) α ud α ϕdx, ϕ C0 (). So, by the uniqueness lemma, see Lemma 2.4, u has a weak derivative D α u that coincides with the corresponding usual (classical) derivative.

24 24 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES Let us list some elementary properties of weak derivatives: 1. By Lemma 2.4, u has the only weak derivative D α u (if it does exist). 2. Let u L 1 loc () have a weak derivative Dα u in and. Then u has a weak derivative D α u in and this derivative coincides with the restriction of the original weak derivative D α u to. 3. Let u 1,u 2 L 1 loc () and Dα u 1,D α u 1 be the corresponding weak derivatives of u 1, u 2 in, respectively. Then, for any c 1,c 2 R, c 1 u 1 +c 2 u 2 has a weak derivative in and it is equal to c 1 D α u 1 +c 2 D α u 2. Proof of 1 3 Exercise. Let u m L 1 loc (), m = 1,2,..., and u L1 loc (). We say that u m u in L 1 loc () as m if u m u in L 1 (K) for each compact K. Lemma 3.2. Let u m u in L 1 loc () and Dα u m v in L 1 loc (). Then v = D α u in. Proof By definition, for each m, we have D α u m ϕdx = ( 1) α u m D α ϕdx, ϕ C 0 (). Since ϕ is compactly supported in, we can take a limit for each fixed ϕ and show vϕdx = ( 1) α ud α ϕdx Mollification of Functions with Weak Derivatives Lemma 3.3. Let u L 1 () have a weak derivative D α u L 1 (). Let x and 0 < < dist(x, ). Then D α u (x) = (D α u) (x). Proof For simplicity only, consider the case α = 1. By assumptions, B(x, ) and thus the function ω (x ) C0 (). Next, u (x) = ω (x y)u(y)dy and u (x) = x k ω x k (x y)u(y)dy.

25 2.3. SOBOLEV SPACES 25 We notice that ω (x y) = ω (x y) x k y k and, using the definition of weak derivatives, find u x k (x) = ω y k (x y)u(y)dy = ω (x y) u y k (y)dy = ( u x k ) (x). Lemma 3.4. Let 1 p < and u and D α u be in L p (). Then D α u D α u in L p loc (). Proof We know that u u and (D α u) D α u in L p (). By previous lemma, for any compact K and sufficiently small, D α u (x) = (D α u) (x) for any x K. This completes our proof. Proposition 3.5. Let u L 1 loc () and all the weak derivatives of the first order vanish. Then u is a constant in. Proof Suppose first that is a ball of radius r and u L 1 (). Let us show that u is a constant there. Let B be a ball of radius r ε with the same center as. For 0 < < ε, by Lemma 3.3, u x k (x) = ( u x k ) (x) = 0, x B, k = 1,2,...,n, and thus u (x) = c for x B. We know that u u in L 1 (B), which implies that u is a constant in B. This constant is in fact independent of ε (explain why). Tending ε 0, we get that u is constant in. From this particular case and from the fact that is connected, we can deduce the statement, noticing that if two balls containing in have an intersection that u is a constant in the union of these balls.. Theorem 3.6. Let φ : be diffeomorphism of class C 1 and let a locally integrable function x u(x) have all weak derivatives of the first order in. Then the function y v(y) = u(φ(y)) also has all weak derivatives of the first order calculated according to the classical chain rule, i.e., v n u x i (y) = (x) (y). y k x i x=φ(y) y k Proof On sheet 2. i=1

26 26 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES Sobolev Spaces Definition 3.7. Sobolev space W l,p () is a vector space of functions that are integrable with power p and have all weak derivatives up to order l also being integrable with power p. This is a normed space with respect to the norm ( )1 ( u W l,p () = u p,l, = D α u p p p, = α l α l )1 D α u p p dx or equivalently u W l,p () = α l D α u p,. Theorem 3.8. W l,p () is a B-space. Proof Let u k u m p,l, 0 as k,m. Then D α u k D α u m p, 0 for any α l. Since L p () is a Banach space, there exist functions w α L p () such that D α u m w α p, 0. We let u = w 0. By Lemma 3.2, w α = D α u for all 0 α l and thus u u m p,l, 0 as m. For the space W l,2 (), we introduce a special notation setting H l () = W l,2 (). It becomes a Hilbert space with a scalar product (u,v) H l () = D α ud α vdx = α l α l(d α u,d α v) L 2 (). Theorem 3.9. Let φ : be diffeomorphism of class C l so that the mapping φ and all its derivatives up to order l are continuous in the closure. Moreover, its Jacobian does not change the sign there. Then if x u(x) W l,p () then y v(y) = u(φ(y)) W l,p ( ) and there exist positive constants c 1 and c 2 depending only on φ and its derivatives such that c 1 u p,l, v p,l, c 2 u p,l,. Proof Easy consequence of chain rule. Definition For finite p, W l,p 0 () is the closure of C0 () in W l,p (). It is a B-space. We also define H0() l = W l,2 0 ().

27 2.3. SOBOLEV SPACES 27 Lemma Let and u W l,p 0 (). Define ũ(x) = u(x) if x and ũ(x) = 0 if x \. Then ũ W l,p 0 ( ) and ũ l,p, = u l,p,. Proof By the definition, there exists a sequence u m C0 () such that u m u l,p, 0asm. Letusdenotebyũ m theextensionofu m byzero to. Obviously, ũ m C0 ( ) and of course ũ m ũ k l,p, = u m u k l,p, and ũ m l,p, = u m l,p,. From the first identity, it follows that ũ m is a Cauchy sequence in W l,p ( ) and thus ũ W l,p 0 ( ). The second identity yields the statement of the lemma. Lemma Let u W l,p 0 (). Then u u in W l,p () as 0. Proof See Problem Sheet 2. Proposition (integration by parts) Let u W l,p () and v W l,p 0 () so that 1/p+1/p = 1 with 1 p <. Then, for any α l, vd α udx = ( 1) α ud α vdx. (2.3.1) Proof Let v m C0 () is an approximating sequence for v. Then, by definition of weak derivatives, we have v m D α udx = ( 1) α ud α v m dx. We find (2.3.1) by tending m to. Lemma (Friedrichs inequality) Let u W l,p 0 (). Then where d = diam, u p,l, = u p, d l u p,l,, (2.3.2) ( )1 α =l Dα u p p p,. Proof Obviously, it is sufficient to prove (2.3.2) for u C 0 (). Without loss of generality, we may assume that Q n = {x = (x i ) : 0 < x i < d}. We extend u by zero to the cube Q n. We let x = (y,x n ), where y = (x 1,x 2,...,x n 1 ). Then u(y,x n ) = x n 0 u t (y,t)dt.

28 28 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES By Hölder inequality, u(y,x n ) ( xn 0 t (y,t) p )1 ( xn p dt u 0 ) 1 ( d 1 p p dt d 1 p 0 u t (y,t) p dt )1 p Integrating the latter inequality over Q n and applying Tonelli s theorem, we find d u(y,x n ) p dx = dx n u(y,x n ) p dy Q n 0 Q n 1 d p p d 0 dx n Q n 1 dy Since 1+ p p = p, we have ( d 0 Q n u p dx u (y,t) p dt = d 1+ p p x n )1 p ( d Q n Q n u p )1 p dx. x n u (y,x n ) p dx. x n. Proceeding in the same way, we show that ( u p )1 ( p dx d x n Q n Q n 2 u p dx x 2 n )1 p and so on. As a result, ( Q n u p dx )1 p d l ( Qn l u p dx x l n )1 p d l u p,l,. Corollary p,l, and p,l, are equivalent in W l,p 0 (). Our next question is about density of smooth functions in Sobolev spaces. We denote by W l,p () the closure of C () in W l,p (). Obviously, W l,p () is a subspace in W l,p (). Very often, these two spaces coincide and this depends on how good or bad domain is. Definition is a star-shaped domain if there exists a point x 0 such that, for any x, the line segment, joining x and x 0, is in.

29 2.3. SOBOLEV SPACES 29 Theorem W l,p () = W l,p () for bounded star-shaped domains. Proof The idea of the proof is as follows. We extend a function u W l,p () outside of so that the extension slightly differs from the original one and then we approximate the extended function by mollification. We may assume that x 0 = 0. Lemma Let f L p () with 1 p < be extended by zero to the whole R n. Then ( )1 f( /λ) f( ) p, := f(x/λ) f(x) p p dx 0 as λ 1. Proof On problem sheet 3, similar to the proof of the integral continuity, see Theorem Let us proceed with a proof of Theorem We restrict ourselves to the case l = 1. For λ > 1, we let λ = φ λ () where φ λ (x) = λx. Define u λ (x) = u(y), setting x = λy λ for y. By Theorem 3.9, u λ W 1,p ( λ ) and u λ (x) = 1 u (y), x λ. x k λ y k y=x/λ By the lemma above, u λ u p,1, 0 as λ 1. Now, consider mollification of u λ, i.e., (u λ ) (x) = ω (x y)u λ (y)dy. λ Obviously, (u λ ) C () and we know, see Lemma 3.4, that (u λ ) u λ p,1, 0 as 0. Given k N, we first find λ k > 1 such that u λk u p,1, < 1/k and find k > 0 such that (u λk ) k u λk p,1, < 1/k. Letting u k := (u λk ) k, we deduce from the triangle inequality that u k u p,1, < 2/k 0 as k. So, sequence u k is required. Definition A domain is locally star-shaped if for any x, there exists a neighborhood O x such that the domain O x is star-shaped. Theorem Let be a bounded locally star-shaped domain. Then W l,p () = W l,p () provided 1 p <.

30 30 CHAPTER 2. LEBESGUE AND SOBOLEV SPACES Extension of Functions with Weak Derivatives Let y be a Cartesian system of coordinates in R n. C y (R,h) = {y = (y,y n ) R n : y = (y 1,y 2,...,y n 1 ), y < R, y n < h} is a right circle cylinder. Definition Let x be a Cartesian system of coordinates in R n. A Lipschitz domain (or domain with Lipschitz boundary) is a domain with the boundary satisfying the following property. For any x 0, there exist positive numbers L, R, a Cartesian system of coordinates y centred at the point x 0, and a function φ : { y R} R such that: (i) C y (R,2LR) = {y R n : y n = φ(y ), y R} (ii) C y (R,2LR) = {y R n : y R, φ(y ) y n 2LR} (iii) function φ is Lipschitz continuous with the Lipschitz constant L, i.e., φ(y ) φ(z ) L y z for y,z R n 1, y R, z R. Remark Numbers R, L, and function φ may depend on x 0. Relationship between old (global) and new (local) Cartesian coordinates x and y is given by y = Q(x x 0 ) with an orthogonal matrix Q. Remark Any Lipschitz domain is a locally star-shaped. Remark A Lipschitz domain is of class C k if the function φ of Definition 3.21 belongs to C k { y R}. Examples Theorem Let be a Lipschitz domain and 0 be a domain such that 0. For any u W 1,p (), there exists a function v W 1,p 0 ( 0 ) with the following properties: (i) v(x) = u(x), x (ii) v p,1,0 c u p,1, with a constant depending only on n, p,, and 0.

31 Chapter 3 The First Emedding Theorem 3.1 Sobolev embedding of W 1,p () into L q () Let B 1 and B 2 be two B-spaces. We say that B 2 is embedded into B 1 if B 2 B 1. The embedding of B 2 into B 1 is continuous if there exists a constant c such that u B1 c u B2 for any u B 2. When talking about embedding, we always keep in mind continuous embedding. Trivial examples are: (i) B 1 = L p 1 (), B 2 = L p 2 () for p 1 < p 2 if is a bounded domain (ii) B 1 = L p (), B 2 = W 1,p () Embedding is called compact if any set bounded in B 2 is precompact in B 1. Lemma 1.1. Let n > 1. For any u C0(R 1 n ), ( )n 1 n n ( ) 1 n n u n 1 dx D i u dx, (3.1.1) R n i=1 R n where D i = / x i. Proof Proof is by induction on n. Let n = 2. Then and thus u(x 1,x 2 ) x 1 u(x 1,x 2 ) = u t (t,x 2) dt x 1 31 u t (t,x 2)dt u (x 1,x 2 ) dx 1 x 1 D 1 u dx 1.

32 32 CHAPTER 3. THE FIRST EMEDDING THEOREM The same arguments give u(x 1,x 2 ) D 2 u dx 2. From the latter bounds, it follows that u 2 dx dx R 2 R 2 D 1 u dx 1 D 2 u dx 2. By Tonelli s theorem the right hand side of the last inequality is D 1 u dx D 2 u dx. R 2 R 2 Now, assume that our statement is valid for n 1 and let us show its validity for n. In the way as in 2D case, one can prove that u(x) D i u dx i, i = 1,2,...,n. (3.1.2) Next, letting x = (x 1,x 2,...,x n 1 ), we have, by Tonelli s theorem, n 1 u n 1 dx = dx n u u n 1 dx. (3.1.3) R n R n 1 By Hölder inequality and by induction ( ) 1 ( )n 2 1 u u n 1 dx 1 ( u n 1 ) n 1 dx n 1 n 1 u n 2 dx n 1 R n 1 R n 1 R n 1 ( ) 1 n 1 ( ) 1 u dx n 1 D i u dx n 1. R n 1 i=1 R n 1 Using (3.1.3) and (3.1.2), we derive from the last estimate ( n u n 1 dx dx n R n R n 1 D n u dx dx n ) 1 n 1 n 1 ( i=1 R n 1 D i u dx ) 1 n 1 =

33 3.1. SOBOLEV EMBEDDING OF W 1,P () INTO L Q () 33 ( = D n u dx R n ) 1 n 1 dx n n 1 ( i=1 R n 1 D i u dx ) 1 n 1. Applying Hólder inequality one more time to the second multiplier on the right hand side of the latter inequality, we complete the proof of the lemma. Theorem 1.2. (Gagliardo-Nirenberg inequality) Let be an arbitrary domain in R n and let 1 p < n. For any u W 1,p 0 (), where p = np n p. u p, p(n 1) n p u p,1,, (3.1.4) Proof We shall prove the theorem for the case p > 1. The case p = 1 is an exercise. We let κ = n p. It is easy to check that 1/κ > 1. For an arbitrary (n 1)p u C0 () extended by zero to the whole R n, define v = u κ. 1 Since D i v = 1 u 1 κ 1 sign(u)d κ i u, v C0(R 1 n ). By Lemma 1.1, v n n 1, After direct calculations, we see n ( i=1 D i v dx v n n 1, = u 1 κ p,. For the right hand side of (3.1.5), we apply Hölder inequality D i v dx = 1 u 1 κ 1 D i u dx κ 1 κ ( )1 ( D i u p p dx ( u 1 κ 1 ) p )p 1 p p 1 dx ) 1 n. (3.1.5) 1 κ u p,1, u 1 κ 1 p,. Now, from (3.1.5) and from two latter bounds, we get required inequality (3.1.4) for u C 0 ().

34 34 CHAPTER 3. THE FIRST EMEDDING THEOREM If u W 1,p 0 (), then, by the definition, there exists a sequence u m C 0 ()suchthat u u m p,1, 0asm. So, itisacauchysequencein W 1,p ()and,byinequality(3.1.4)forsmoothcompactlysupportedfunctions, is a Cauchy sequence in L p (). Then we finish our proof by taking the limit in (3.1.4) with u = u m. Corollary 1.3. (Poincarè inequality) For any u W 1,2 0 (), u 2, c(n) 1 n u 2,1,. Proof Let us find p for which p = 2. It is p = 2n n+2 u 2, 2(n 1) u 2n n n+2,1, = 2(n 1) ( n n i=1 < 2. So, by (3.1.4), D i u 2n )n+2 2n n+2 dx It remains to apply Hölder inequality for sums and integrals and complete the proof. Theorem 1.4. (Sobolev) Let be a bounded domain with Lipschitz boundary and let 1 p n. Then: (i) if 1 p < n then Sobolev space W 1,p () is embedded (continuously) into pn n p ]; Lebesgue space L q () for any q [1, (ii) if p = n then Sobolev space W 1,p () is embedded (continuously) into Lebesgue space L q () for any 1 q <. Proof. Let us fix a bounded domain 0 so that 0. Then for any u W 1,p (), there exists a function v W 1,p 0 ( 0 ) such that (i) v = u in (ii) v p,1,0 c 1 (n,p, 0,) u p,1,. Obviously, by (3.1.4), we have u p, v p,0 (n 1)p n p v p,1, 0 (n 1)p n p v p,1, 0 c 1 (n 1)p n p u p,1,. To finish the proof of the first part of the theorem, it is sufficient to note that u q, 1 q 1 p u p,. The second part follows from the first one and obvious continuous embedding W 1,p () into W 1,q () provided 1 q < p. Our next question is under which assumptions the above continuous embeddings are compact. We start with the following theorem.

35 3.1. SOBOLEV EMBEDDING OF W 1,P () INTO L Q () 35 Theorem 1.5. (Reillich-Kondrachov) Let R n be a bounded Lipschitz domain. Then W 1,p () is compactly embedded into L p (). Proof We shall prove the theorem for finite p. Let u C 0 (R n ). Fix an arbitrary vector z R n and define the function ϑ(t) = u(x+tz). Then u(x+z) u(x) = ϑ(1) ϑ(0) = 1 0 dϑ dt (t)dt, where Next, we have dϑ n dt (t) = D i u(y) y=x+tz z i. i=1 I := u(x+z) u(x) p dx = dx R n R n 1 0 n D i u(x+tz)z i dt p. i=1 Then we apply Hölder inequality for integrals and sums: I dx R n 1 0 n i=1 D i u(x+tz)z i p dt n p p R n dx 1 0 n D i u(x+tz) p z i p i=1 n p 1 z p 1 0 R n n D i u(x+tz) p dx. i=1 After change of variables y = x+tz, we find u( +z) u( ) p,r n c 2 (n,p) z u p,1,r n (3.1.6) for any u C0 (R n ). Now, let U be a bounded set in W 1,p (), i.e., u p,1, M < for any u U. Wefixaboundeddomain 0 suchthat 0. DefineV W 1,p 0 ( 0 ) as follows: v V if and only if there exists u U such that v = u in and v p,1,0 c 3 (n,p,, 0 ) u p,1,. So, V is also bounded in W 1,p ( 0 ) and therefore in L p ( 0 ). Next, by the definition of W 1,p 0 ( 0 ), there exists a

36 36 CHAPTER 3. THE FIRST EMEDDING THEOREM sequence v m C 0 ( 0 ) is converging to v in W 1,p ( 0 ) and thus in L p ( 0 ) and even in L p (R n ). Since (3.1.6) gives: v m ( +z) v m ( ) p,0 c 2 (n,p) z v m p,1,0, (3.1.6) is valid for any function v W 1,p 0 ( 0 ) and, therefore, v( +z) v( ) p,0 v( +z) v( ) p,r n c 2 (n,p) z v p,1,0 c 2 c 3 z M for any v V and for any z R n. Hence, by Theorem 1.17, the set V is precompact in L p ( 0 ) and therefore the set U is precompact in L p (). Theorem 1.6. (Sobolev-Kondrachov) Let R n be a bounded Lipschitz domain. Then W 1,p () is compactlyembeddedinto L q () with any 1 q < p, where p = np if 1 p < n and p = if p = n. n p Proof Consider the case 1 p < n only. The case p = n is an exercise. Let U is a bounded set in W 1,p () and u m is an arbitrary sequence in U. By Theorem 1.5, there exists a subsequence u mk such that u mk u p, 0 as k for some u W 1,p () and thus u mk u in measure in. On the other hand, by Theorem 1.4, this sequence is bounded in L p (). Then, by Theorem 1.2, we show that u mk u in L q () with q < p. 3.2 Traces of functions with weak derivatives Surface Integral Let be a bounded domain of class C 1. Going back to Definition 3.21, note that cylinders C y (R(x),2L(x)R(x)) for x are an open cover of the compact. By the Heine-Borel lemma, there exists a finite subcover, i.e., m C y(k) (R k,2l k R k ), k=1 where R k = R(x (k) ), L k = L(x (k) ), y (k) = Q k (x x (k) ), and Q k is an orthogonal matrix. Denote that S k = { y(k) < R k} R n 1. Under our assumptions, the function φ k, the graph of which is a part of, belongs to C 1 (S k ). For this subcover, there exists a finite partition of unity ϑ k C0 (R n ) such that 0 ϑ k 1 and suppϑ C y(k) (R k,2l k R k ), k = 1,2,...,m, and m ϑ k (x) = 1 for all x. k=1

37 3.2. TRACES OF FUNCTIONS WITH WEAK DERIVATIVES 37 Definition 2.1. f : R is Lebesgue integrable in if the value m I := ϑ k (x(y(k),φ k (y(k))))f(x(y (k),φ k (y(k)))) k=1 S k n 1 1+ φ k (y(k) ) 2 dy (k). is finite i=1 y i(k) All the integrals in the above definition are taken with respect to Lebesgue measure in R n 1. If f is integrable in, we shall write fds := I. One can show that the integral is well-define, i.e., independent of the choice of subcover. We also can introduce a surface Lebesgue measure and measurable (with respect to this measure) sets as follows. Let Γ and χ Γ be its characteristicfunction. Ifχ Γ isintegrable, thenthesetγiscalledmeasurable and the corresponding integral is called its surface measure. It is a natural generalisation of the surface area. Wesaythatf : Γ RisintegrableinΓifitsextension f : R by zero to the whole is integrable in and we let fds = fds. Γ For functions integrable in Γ, the same statements as for functions integrable in R n are valid as well. We also can introduce Lebesgue space L p (Γ). Remark 2.2. All the statements and constructions remains to be true for domains with Lipschitz boundary Traces of functions from Sobolev Spaces Lemma 2.3. Let be a domain of class C 1 and let 1 p <. There exists a constant c(n,p,) such that u p, c u p,1,, u C 1 (). (3.2.1)

38 38 CHAPTER 3. THE FIRST EMEDDING THEOREM Proof We shall consider the case p > 1. The case p = 1 is an exercise. To avoid technical difficulties and demonstrate the essence of the matter, consider the following particular case. Assume that contains a flat part Γ = {x = (x,x n ) : x < r, x n = 0} and prove instead of (3.2.1) a simpler inequality u p,γ c u p,1,, u C 1 (). (3.2.2) Let h > 0 be so small that h := {x = (x,x n ) : x < r, 0 < x n < h}. Fix a function η C 0 (R) so that η(t) = 1 for t h/3 and η(t) = 0 for t > 2h/3. Then we have h p 0 h u(x,0) p = 0 t η(t)u(x,t) p dt = η(t)u(x,t) p 1 sign(η(t)u(x,t)) ηu t (x,t)dt. After applying Hölder inequality, we find ( h u(x,0) p c 1 0 ) 1 u(x,x n ) p dx ( h p n +c 1 h 0 0 u(x,x n ) p dx n. u (x,x n ) p )1 p dx n + x n To complete the proof, it is sufficient to integrate the latter inequality over Γ, then apply consequently Hölder inequality and Young inequality (ab a p /p+b p /p ). Let us define a linear operator γ : C 1 () W 1,p () L p ( ) so that γu = u. By (3.2.1), this operator can be considered as a linear bounded operator from W 1,p () into L p ( ) with domain of definition D(γ) = C 1 (), which is dense in W 1,p (). Its operator norm γ D(γ) = { γu p, : u D(γ), u p,1, 1} is finite. It is known from the Functional Analysis that such an operator admits unique continuation γ to the whole W 1,p (). The operator γ : W 1,p () L p ( ) has the following important properties:

39 3.2. TRACES OF FUNCTIONS WITH WEAK DERIVATIVES 39 (i) γ = γ D(γ) c(n,p,) (ii) γu = γu for all u C 1 (). γ isthetrace operator, actingonw 1,p ()intol p ( ), and γuisthetrace of a function u W 1,p (). For the trace of u, we use the classical notation, i.e., γu = u. It should be understood in the sense described above. Lemma 2.4. Let be a bounded domain of class C 1 and let 1 < p <. Then we have ud i vdx = uvν i ds vd i udx (3.2.3) for all u W 1,p () and for all v W 1,p (). Here, ν is the unit outward normal to the surface. Proof(3.2.3)isvalidforallu,v fromc 1 (). Therefore, wecanwrite(3.2.3) for sequences of smooth functions approximating functions u W 1,p () and v W 1,p () and then take the limit using continuity of the trace operator with respect to strong convergence in Sobolev spaces. Corollary 2.5. Let be a bounded domain of class C 1 and 1 p <. Suppose that u W 1,p () with u = 0. Let 0 and ũ is an extension of u by zero from to 0. Then ũ W 1,p ( 0 ). Proof On problem sheet 3. 1,p Remark 2.6. Let W 0 () = {u W 1,p () : u = 0}. In fact, W 1,p 0 () W 1,p 0 () (explain why). However, under our assumptions on (it is of class C 1 ), W 1,p 1,p 0 () = W 0 (). Remark 2.7. All above statements can be extended to bounded domains with Lipschitz boundaries.

40 40 CHAPTER 3. THE FIRST EMEDDING THEOREM

41 Chapter 4 Functional Methods for PDE s 4.1 Methods of Functional Analysis for PDE s Notation There are several ways to denote partial derivatives. Here, it is a list of some of them u x k = D k u = xk u = u xk = u,k. In the rest of the course, the last notation will be used mostly both for classical and weak derivatives. Another important thing is the so-called nabla-operator. So, the action of this operator on u is the gradient of u and denoted by u. This makes our notation closer to physical notation, in which fundamental equations of the physics are invariant (independent of a coordinate system). In particular, in Cartesian coordinates x = (x k ), u = (u,1,u,2,...,u,n ) = (u,k ). We also let A : B = spa T B for two n n matrices A and B. So that A : B = A ij B ij, where summation over repeated indices running from 1 to n is adopted. And for a given vector-valued field a = (a k ), we denote diva = a k,k. Now, we can consider two types of differential operators in. The fist one has a divergence form: Lu := div(a u)+b u+cu, where a is a symmetric matrix-valued field, b is a vector-valued field, c is a scalar field in. The second type has a non-divergence form: Nu := a : 2 u+b u+c. 41

42 42 CHAPTER 4. FUNCTIONAL METHODS FOR PDE S In this course, we are going to deal with differential operators in the divergence form only. Example 1. Laplace operator: here, a = I := (δ ij ), b = 0, and c = 0. Then Lu = div u = u = u,ii. 2. Helmholtz operator: Lu = u+k 2 u Dirichlet Boundary Value Problems for Elliptic Equation We always assume that a symmetric matrix a L (;R n n ) satisfies the ellipticity condition a.e. in with a positive constant ν. ν ξ 2 ξ a(x)ξ 1 ν ξ 2, ξ R n (4.1.1) Definition 1.1. A function u C 2 () C() satisfying Lu = f +divg in (4.1.2) u = u 0 on (4.1.3) with given f, g, and u 0 is called a classical solution to boundary value problem (4.1.2), (4.1.3). Of course, a necessary condition for the existence of a classical solution to boundary value problem (4.1.2), (4.1.3) is sufficient smoothness of given functions a, b, c, f, g, and u 0. However, even under these conditions on the data of the problem, it is not so easy to prove the existence of classical solutions especially in the case of variable coefficients a, b, and c. On the other hand, there are quite a number of interesting and physically relevant cases, in which those coefficients are not smooth enough. The modern way to tackle the solvability issue, which is, by the way, closely connected with modern ways to approximate solutions, is as follows. The classical set-up of boundary value problems is replaced with a weak setting based on integral identities rather than point-wise equations. This allows us to use powerful methods of functional analysis in order to prove existence theorems and develop a rigorous foundation of solving problems

43 4.1. METHODS OF FUNCTIONAL ANALYSIS FOR PDE S 43 approximately, for example, by the finite-difference method or the finite element method. In what follows, we always assume that a := a : a L (), b := b b L (), c L (), (4.1.4) and ellipticity condition (4.1.1) holds. f L 2 (), g L 2 (), (4.1.5) u 0 H 1 () := W 1,2 (), (4.1.6) Definition 1.2. A function u H0() 1 + u 0 is a weak (or generalized) solution to boundary value problem (4.1.2), (4.1.3) if it satisfies the integral (variational) identity L(u,w) = (fw g w)dx, w C0 (), (4.1.7) where L(u,w) := ((a u) w +b uw +cuw)dx. Variational identity (4.1.7) is motivated by the following formal identity (Lu f divg)wdx = 0, w C0 (), which can be obtained by means of a single integration by parts involving the terms wdiv(a u) and wdiv g. Boundary condition (4.1.3) is satisfied in the sense of traces, i.e., u u 0 H 1 0(). Remark 1.3. If u is a weak solution, then variational identity (4.1.7) holds true for any test functions w H 1 0(). (Explain why) Theorem 1.4. (uniqueness implies existence) Let given functions a, b, and c satisfy conditions (4.1.1) and (4.1.4). Suppose, in addition, that any function v H 1 0() subject to the identity L(v,w) = 0, w C 0 (), must be equal to zero. Then, for any f, g, and u 0, satisfying conditions (4.1.5) and (4.1.6), boundary value problem (4.1.2), (4.1.3) has a unique weak solution.

44 44 CHAPTER 4. FUNCTIONAL METHODS FOR PDE S Proof We are going to reduce the problem in question to Theorem 7.2. Our Hilbert space is going to be U = H0(). 1 By Poincaré inequality, 2,1, is a norm on U that is equivalent to the standard norm 2,1,, see Corollary The ellipticity condition gives to following bounds ( ν u 2,1, IuI 2,1, := )1 2 u a u 1 ν u 2,1,. They imply that I I 2,1, is a norm in U as well and it is equivalent to the norm 2,1,. The norm I I 2,1, is generated by the scalar product [u,v] = (a u) vdx = u a vdx, u,v U. Then our bilinear form L can be presented as follows L(u,v) = [u,v] L 1 (u,v), where L 1 (u,v) = (vb u+cuv)dx, u,v U. By Cauchy-Schwartz inequality, by Poincaré inequality, see Corollary 1.3, and by the above equivalence of norms, we L 1 (u,v) C 1 ( v 2, u 2, + v 2, u 2, ) C 2 IuI 2,1, IvI 2,1,, u,v U. So, givenu U, thelinearfunctionalv L 1 (u,v)isboundedinu. ByRiesz theorem on representation of linear functional in Hilbert space, there exist a unique K(u) U such that L 1 (u,v) = [K(u),v] for any u,v U. It is easy to check that K : U U is a bounded linear operator (indeed, IKuI 2,1, C 2 IuI 2,1, ). Our aim is to show that K is a compact operator. To this end, we should show that, for any bounded sequence u m, the sequence Ku m is precompact. WLOG,wemayassumethatu m uinu andthusku m Ku in U. If not, using a boundedness of u m and a sufficient condition of weak compactness in Hilbert spaces, select a required subsequence, which could be denoted again by u m. By Reillich-Kondrachov theorem, the embedding of U into L 2 () is compact and we also may assume that u m u and Ku m Ku in L 2 (). Then, denoting w m = Ku m and w = Ku, we show IK(u m u)i 2 2,1, = [K(u m u),w m w] = L 1 (u m u,w m w) =