Partial Differential Equations and Sobolev Spaces MATINF4300 autumn Snorre H. Christiansen November 10, 2016


 Alaina Williams
 3 months ago
 Views:
Transcription
1 Partial Differential Equations and Sobolev Spaces MATINF4300 autumn 2016 Snorre H. Christiansen November 10,
2 2
3 Contents 1 Introduction 5 2 Notations for differential operators 7 3 Generalities about differential operators 9 4 Exercises on partial derivatives 11 5 Continuous linear or bilinear maps 13 6 Completion of a normed vector space 15 7 Laplace equation 17 8 More on the Laplace equation 19 9 Integration and mollification Some references for background material Measures and integration The standard bump Convolution Liouville s theorem and convolution Smoothing by convolution Extension to L 1 (U) Epsilon neighborhoods and epsilon interiors Two results Exercises and L p spaces Exercises on Hölder s inequality L p spaces and weak derivatives L p spaces Integral operators Weak derivatives Weak derivatives and Sobolev spaces Exercises on Hölder Partitions of unity Integration on submanifolds Stokes theorem Density theorems 47 3
4 21 Trace theorem The trace operator W 1,p (U) L p ( U) On W 1,p (U) Characterizations of W 1,p 0 (U) Poincaré inequality Some exercises The Dirichlet principle Inner products LaxMilgram Second order elliptic PDEs Variational formulation Galerkin method Some exercises The space W 1,1 of an interval The space W 1,1 of an interval The Cantor  Lebesgue function Some exercises 61 4
5 1 Introduction 22 August 2016 Practical information. Prerequisites: Basic real analysis, e.g. MAT2400 (say compact subsets of R n, uniform convergence of sequences of functions). Aquaintance with PDEs, e.g. MATINF3360 (some exposure to partial derivatives including GaussGreenStokes theorem, examples from physics, chemistry, biology, social sciences, finance or whatever). Strongly recommended (either before or in parallel): Basic measure theory and functional analysis: MAT4400. Curriculum: Lawrence C. Evans Partial Differential Equations, Amer. Math. Soc., Graduate Studies in Mathematics. Selected topics in the following chapters: Chap. 2: Some explicit formulas, for the heat and Poisson equations. Chap. 5: Sobolev spaces, definition and some properties. Chap. 6: Applications of Sobolev spaces to linear elliptic PDEs. Time permitting: some theory for finite element methods and/or nonlinear elliptic equations. Language. A comparison of formulations. Let I = [a, b] be a (closed and bounded) interval in R (i.e. < a < b < + ). Consider the following statements: Suppose f, g : I R are two functions. Pick x I. Suppose f and g are both differentiable at x. Then f +g is differentiable at x and (f +g) (x) = f (x) + g (x). Let F denote the set of (real valued) functions on I and D denote the set of differentiable functions. Then F has a natural structure of vector space, and D is then a subspace. Moreover differentiation defines a linear map from D to F. Denote by C k (I) the space of functions which are k times differentiable on I, the derivatives up to order k being continuous. It is a vector space, and the following defines a norm: u k = sup{ u (l) (x) : l k, x I}. (1) Here u (l) denotes the derivative of order l of u. Then derivation defines a bounded linear map from C k (I) to C k 1 (I) (k 1), by construction. Moreover C k (I) equipped with this norm, is complete. 5
6 For this course I am going to assume that the students think the first statement is a distant memory, that they are familiar with the second statement and that they can work through the last. This is mostly about reformulating key results from a second calculus course in the language of functional analysis. This defines some function spaces. Sobolev spaces are other function spaces, that turn out to be more adapted to the study of partial differential equations, for instance because Hilbert space techniques can be applied to them. Substance. For most PDEs there is no explicit formula for the solution. One then studies questions such as: does the PDE have a solution, is it unique, and what are its properties. Different types of properties are of interest, an important one is how smooth the solution is... Let U be an open bounded subset of R n. Let f : U R be a function and g : U R be another one. In the course we will mainly be interested in the equation: u = f on U, (2) u = g on U. (3) In dimension n = 1 this equation can be solved explicitely, by two integrations and a two by two system (do it!); and if f is C 0 (U) then u is C 2 (U). In higher dimensions (n 2), there are explicit solution techniques only for special choices of domain U (e.g. with Fourier series); not when, say, U has the shape of the human body. Moreover one is lead to seek solutions that are not of class C 2 (U). We will give meaning to a notion of generalized solution, involving functions whose derivatives are not defined in the classical sense (i.e. by pointwise everywhere differentiation). This framework is important for numerical computations, because approximate solutions will typically be piecewise linear functions, which are not differentiable in the classical sense. For a numerical method applied to some PDE, one wants to estimate the distance from the computed (approximate) solution to the exact solution, with respect to some norm. It turns out that this is easier in Sobolev space norms than in those, more elementary, defined by (1). Reading assigment. Introduction in Evans book. Welcome! 6
7 2 Notations for differential operators 25 August 2016 Notation for partial derivatives. Divergence, gradient and Laplace operator. Linear partial differential operators. Linearity of partial derivation, Leibniz rule for partial derivative of product, commutation of partial derivatives. Polynomials and linear partial differential operators with constant coefficients, correspond naturally. C 0 (I) is a complete normed vector space, when I is a compact interval. 7
8 8
9 3 Generalities about differential operators 29 August 2016 For k N, C k (I) is a complete normed vector space, when I is a compact interval. If a α are continuous functions and the partial differential operator: a α α (4) α is zero, as an operator from smooth functions to continuous functions, then each a α is zero. Classification of some constant coefficient PDEs in the plane: transport, elliptic, parabolic and hyperbolic equations. Explicit solutions to transport and heat equations. Reading assigment. equation). [Evans] 2.1 (Transport equation) and 2.3.1, a & b (Heat 9
10 10
11 4 Exercises on partial derivatives Let U be an open subset of R n. 1 September 2016 Exercise 4.1. Check that if u C 1 (U) (scalar field) and v C 1 (U) n (vector field): div(uv) = (grad u) v + u div v. (5) Exercise 4.2. Choose reals a i < b i for integer i such that 1 i n. Suppose U =]a 1, b 1 [ ]a n, b n [. You may start with n = 2. What is the boundary of U? What is the unit outward pointing normal on U? We denote it by ν. Give a direct proof of Stokes theorem in this case, namely that if u C 1 (R n ) n then : div u = u(x) ν(x)dx. (6) U U Deduce (informally) the formula for a finite union of such boxes, possibly sharing faces. Exercise 4.3. For a function u C 1 (U), we define for any vector a = (a 1,, a n ) R n : a u(x) = lim ɛ 0 ɛ 1 (u(x + ɛa) u(x)). (7) Check: a u(x) = n a i i u(x). (8) i=1 Let (e i ) 1 i n denote the canonical basis of R n : Compared with standard notation we have: e 1 = (1, 0, 0,..., 0), (9) e 2 = (0, 1, 0,..., 0), (10)... (11) e n = (0, 0, 0,..., 1). (12) i u = ei u. (13) Let (f i ) 1 i n be another orthonormal basis of R n. Prove that for u C 1 (U): ( i u)e i = ( fi u)f i. (14) i i and that for u C 1 (U) n : i u i = i i fi (u f i ). (15) Prove also that for u C 2 (U): n n i 2 u = f 2 i u. (16) i=1 i=1 11
12 Exercise 4.4. Let A be an invertible real n n matrix, considered also as a map R n R n, x Ax. Check that for u C 1 (U) we have: and that for v C 1 (U) n : Deduce that if A is an orthogonal matrix: grad(u A) = (A T grad u) A, (17) div(a 1 v A) = (div v) A. (18) (u A) = ( u) A. (19) Whether A is orthogonal or not, find a positive definite matrix B such that: (u A) = (div B grad u) A. (20) 12
13 5 Continuous linear or bilinear maps 5 September 2016 Not very much functional analysis is required for this course. Essentially we need the following background material. In the following all vectorspaces are vectorspaces over a field K which is either R or C. Proposition 5.1. E and F two normed vectorspaces. A : E F a linear map. The following are equivalent: A is continuous. A is continuous at 0. There is C > 0 such that for all u E, Au C u. Such linear maps are also said to be bounded. Proof. See Lemma 4.1 p. 88 in [5]. For a continuous linear map A : E F one defines: A = inf{c 0 : u E Au C u }, (21) = sup{ Au : u E u 1}. (22) The continuous linear maps E F form a vector space denoted L(E, F ) and the above expression defines a norm on this space. In the special case F = K, the space L(E, K) is called the dual space of E and denoted E. Very similarly one proves: Proposition 5.2. E, F and G three normed vectorspaces. a : E F G a bilinear map. The following are equivalent: a is continuous. a is continuous at 0. There is C > 0 such that for all u E, all v F, a(u, v) C u v. Such bilinear maps are also said to be bounded. For a continuous bilinear map a : E F G one defines: a = inf{c 0 : u E v F a(u, v) C u v }, (23) = sup{ a(u, v) : u E u 1, v F v 1}. (24) The continuous bilinear maps constitute a normed vector space. Notice that a continous linear map is uniformly continuous, whereas a (non zero) continuous bilinear map is not. 13
14 Proposition 5.3. Suppose that E is a normed vector space and that E 0 is a dense subspace. Suppose that F is a complete normed vector space. Suppose that A : E 0 F is a linear map and that for some C 0 we have: u E 0 Au C u. (25) Then there is a unique continuous map Ã : E F which coincides with A on E 0. It is linear and satisfies: u E Ãu C u. (26) Proposition 5.4. Suppose E 0 is a normed vector space. Then there exists a complete normed vector space E equipped with an isometry i : E 0 E such that i(e 0 ) is dense in E. If F is another complete normed vector space equipped with an isometry j : E 0 F such that j(e 0 ) is dense in F, there exists a unique continuous map J : E F such that j = J i ; and it is a linear isometric bijection. Proof. See [3] Chapter VII, Section 4. 14
15 6 Completion of a normed vector space 8 September 2016 Completion of a normed vector space. Exercises on partial derivatives. Exercise for next week: [1] problem 13 p
16 16
17 7 Laplace equation 12 September 2016 Uses of Stokes theorem. Integration by parts. A uniqueness result. Support of a function. The space C k c (U). Continuous functions with compact support are bounded. Linear differential operators preserve the support. Radial solutions to the Laplace equation. Harmonic functions. Weakly harmonic functions. 17
18 18
19 8 More on the Laplace equation 15 September 2016 Convolution. Nonhomogeneous Laplace equation on R n. Mean value property. Reading assigment In Evans. Section and Get used to the notations in Appendix A. Appendix C.1 C.3. We will soon be considering C.5. 19
20 20
21 9 Integration and mollification 19 September Some references for background material PDE theory draws on many (all?) sources. I hope you will find references that please you. Here are some of my favorites. Differential Intrinsic definition of the differential of a function from one normed vector space to another, compared with partial derivatives: [8] Chapter 17: Several variable differential calculus. [3] Chapter XV: Functions on nspace, and Chapter XVII: Derivatives in vector spaces. Lebesgue measure and integral [7] Chapter 1: Measure theory. [2] Chapter 2.6: The ndimensional Lebesgue integral. Remark 9.1. An approach to the Lebesgue integral through completion. Let S denote the space of step functions [a, b] X. Equip S with the seminorm defined by: f = b a f(x) dx. (27) It turns out that the completion of this space with respect to this seminorm, has all the properties we would like for a nice space L 1 ([a, b], X). In particular its elements can be identified with functions (defined up to a subset of measure zero). Realizing this program requires some work of course, see [3] Chapter X.4 p , and [4] Appendix A, p Remark that completing S with respect to the supnorm yields also yields a space of functions on which the integral is well defined. It is comparatively easy, but unfortunately the space obtained is complete with respect to the supnorm, not the L 1 norm. Convolution Convolution and Dirac sequences: [3] Chapter XI: Approximation with convolution. [7] Chapter 3.2: Good kernels and approximations to the identity. See also: [6] Chapter 2.3: Convolutions, Chapter 2.4: Good kernels. [2] Chapter 8.2: Convolutions. 21
22 9.2 Measures and integration Definition 9.1. Let E be a set. A σalgebra on E is a subset M of P(E) such that: M, For all F M we have E \ F M, For any sequence (F n ) n N in M we have n N F n M. When E is a topological space, there is a smallest σalgebra containing the open subsets of E, called the Borel σalgebra on E. Definition 9.2. Let E be a set and M a σalgebra on E. A measure on M is a function µ : M [0, ] such that for any sequence (F n ) n N in M of two by two disjoint sets: µ( n N F n ) = n N µ(f n ). (28) Let E be a set, M a σalgebra on E and µ a measure on M. Then the following is also a σalgebra: M = {F G : F M, and G M µ(g ) = 0 and G G }. (29) The measure µ has a unique extension to a measure µ on M. For any subset F of E, if there is F M such that µ(f ) = 0 and F F, we have F M. One says that (M, µ) is the completion of (M, µ). Once a σalgebra on E has been fixed, as well as a measure, the elements of the σalgebra will be referred to as the measurable subsets of E. A property P(x) defined for x E is said to hold almost everywhere, if there is a set F of measure 0 such that P(x) is true for x E \ F. Definition 9.3. A function f : E R is measurable if for any open interval I in R, f 1 (I) is a measurable subset of E. Definition 9.4. For a function f : E R the following are equivalent: f is measurable, has finite range and for a 0, f 1 ({a}) has finite measure. f is a (finite) linear combination of characteristic functions of measurable subsets with finite measure. Such functions are called simple. They form a vector space. Lemma 9.1. Suppose f is a simple function represented as: f = n a i χ Fi, (30) i=1 with F i measurable, with finite measure. If f(x) = 0 for almost every x E, then: n a i µ(f i ) = 0. (31) i=1 22
23 As a consequence, if a simple function f is represented as in (30), the number: n a i µ(f i ), (32) i=1 does not depend on the choice of representation. It is called the integral of f and denoted f. For a measurable function f : E [0, ] we define: f = sup{ φ : φ is simple and φ f} [0, ]. (33) We define L 1 (E) to consist of measurable functions f : E R such that f = f <. It is a normed vector space and the simple functions are dense in it. The integral, defined a priori for simple functions, has a unique extension to a continuous linear form on L 1 (E) called the integral and still denoted. Theorem 9.2. (dominated convergence) Suppose that f : E R is some function, (f n ) n N is a sequence in L 1 (E) and g : E [0, ] satisfies: g is measurable and g <, for almost every x E, for all n N, f n (x) g(x), for almost every x E, f n (x) f(x) as n. Then f L 1 (E) and: lim n Proposition 9.3. L 1 (E) is complete. f n = f. (34) Actually the norm on L 1 (E) is just a seminorm, since the L 1 (E) norm of a function is 0 iff the function is zero almost everywhere. One usually overlooks this problem and considers that such functions are 0. Proposition 9.4. On the Borel σalgebra of R n there is a unique measure which: is translation invariant, and is such that the measure of the open unit cube is 1. The completion of this measure is called the Lebesgue measure and the associated integral is called the Lebesgue integral. By default this is the measure and integral we will always consider. If S is an open subset of R n we denote by L 1 (S) the space of functions on S whose extension by 0 to R n is in L 1 (R n ). This is also a Banach space. Proposition 9.5. For any open subset S of R n, the space C c (S) is dense in L 1 (S). We define: L 1 (S) loc = {u : R n R : K S K compact χ K u L 1 (S)}. (35) For instance the fundamental solution to the Laplace operator is in L 1 (R n ) loc but not L 1 (R n ). 23
24 9.3 The standard bump Define a function φ : R R by, for x > 0 : φ(x) = exp( 1/x), (36) and for x 0, φ(x) = 0. Exercise 9.1. By induction on k, show that for any k N there exists a polynomial P and l N such that the kth order derivative of φ takes the form, for x > 0: φ (k) (x) = P (x) x l exp( 1/x). (37) Deduce that φ C (R). Define Φ : R n R by, for x R n : Φ(x) = φ(1 x 2 ). (38) What is the support of Φ? Why is its integral strictly positive? Define then: Ψ(x) = Φ(x)/ Φ, (39) and: What is the integral of Ψ ɛ? mollifiers. 9.4 Convolution Ψ ɛ (x) = ɛ n Ψ(ɛ 1 x). (40) These functions will be called the standard Given two functions u, v : R n R one defines u v : R n R by: u v(x) = u(y)v(x y)dy, (41) whenever this makes sense. Proposition 9.6. For u, v L 1 (R n ) we get a well defined u v L 1 (R n ) and: Exercise 9.2. Check that u v = v u. u v L 1 (R n ) u L 1 (R n ) v L 1 (R n ). (42) Lemma 9.7. If u C c (R n ), then u is uniformly continuous. Proposition 9.8. If u L 1 loc (Rn ) and v C c (R n ) then u v C(R n ). Proposition 9.9. If u L 1 loc (Rn ) and v C 1 c(r n ) then u v C 1 (R n ) and: i (u v) = u i v. (43) Proposition If u C c (R n ) then as ɛ 0, u Ψ ɛ converges to u, uniformly on R n. Proposition If u L 1 (R n ) then as ɛ 0, u Ψ ɛ converges to u in L 1 (R n ). Corollary If u L 1 loc (Rn ) satisfies, for all v C c (R n ): uv = 0, (44) then u = 0. 24
25 10 Liouville s theorem and convolution 22 September 2016 Smoothing effect of convolution by the standard bump. C 2 (R n ) functions that are harmonic are smooth. Liouville s theorem. An existence and uniqueness theorem of solutions to the Laplace equation on R n, n 3. Generalities about Lebesgue measure. Convolution in L 1 (R n ) and Fubini Tonelli. Exercise 3 page 85 in Evans. 25
26 26
27 11 Smoothing by convolution 26 september 2016 Convolution of L 1 (R n ) functions with C 0 c(r n ), C 1 c(r n ), and C c (R n ) functions : continuity and differentiability. Convolution of C 0 c(r n ) and L 1 (R n ) functions with Ψ ɛ (standard mollifier) : convergence as ɛ 0. 27
28 28
29 12 Extension to L 1 (U) 29 september Epsilon neighborhoods and epsilon interiors For a subset E of R n and a point x R n one defines: d(x, E) = inf{ x y : y E}. (45) Exercise Check: and: and: d(x, E) = 0 x E, (46) d(x, E) d(y, E) x y, (47) d(x, E) = d(x, E). (48) Definition Let E be a subset of R n : The ɛ neighborhood of E is: The ɛ interior of a set E is: Lemma If ɛ < ɛ then: N ɛ (E) = {x R n : d(x, E) < ɛ}. (49) I ɛ (E) = {x E : d(x, R n \ E) > ɛ}. (50) N ɛ (E) N ɛ (E). (51) Lemma If K U R n, with K compact and U open, there exists ɛ > 0 such that N ɛ (K) U. Exercise supp(v Ψ ɛ ) N ɛ (supp(v)). Can one have equality? Exercise N ɛ (I ɛ (E)) E Two results Proposition Suppose U is open in R n and u L 1 loc (U) satisfies: φ C c (U) uφ = 0. (52) Then u = 0. Proposition Suppose U is open in R n. Then C c (U) is dense in L 1 (U). 29
30 30
31 13 Exercises and L p spaces 3 October 2016 Correction of Exercises. L p spaces. Hölder and Minkowski inequalities Exercises on Hölder s inequality Exercise Suppose U R n is open. Suppose we have reals p, q 1 such that: 1 p + 1 = 1. (53) q Given u L p (U), find a nonzero v L q (U) such that: uv = u L p (U) v L q (U). (54) Hint: find v such that uv = u p. Exercise Suppose U R n is open and has finite measure (e.g. bounded). Suppose we have reals 1 p < q. Suppose u L q (U). Show that u L p (U) and: u Lp (U) (meas U) ( 1 p 1 q ) u Lq (U). (55) Hint: write u p = 1 u p and find a Hölder inequality that can be applied to this product. Exercise Suppose U R n is open and that reals p, q, r 1 satisfy: 1 p + 1 q = 1 r. (56) Suppose u L p (U) and v L q (U). Show that uv L r (U) and: uv L r (U) u L p (U) v L q (U). (57) Exercise Suppose U R n is open and that reals p, q, r 1 satisfy: 1 p + 1 q + 1 = 1. (58) r Suppose u L p (U), v L q (U) and w L r (U). Show that uvw L 1 (U) and: uvw L 1 (U) u L p (U) v L q (U) w L r (U). (59) Exercise Suppose U R n is open and that we have reals p, q 1, as well as θ [0, 1]. Define r by: 1 r = θ 1 p + (1 θ)1 q. (60) Suppose u L p (U) L q (U). Show that u L r (U) and: u Lr (U) u θ L p (U) u (1 θ) L q (U). (61) 31
32 Definition Suppose U R n is open. For u : U R measurable, one defines One also defines: u L (U) = inf{c [0, ] : for a.e. x U u(x) C}. (62) L (U) = {u : U R : u is measurable and u L (U) < }. (63) One checks that this is a vectorspace and that (62) defines a norm on this space. Moreover this normed vectorspace is complete. Exercise Try to extend the preceding exercises to the case where at least one of p, q and r is infinite. 32
33 14 L p spaces and weak derivatives 6 October L p spaces For proofs of the following results we refer to [2] Chapter 6. L p (U) is complete. C c (U) is dense in L p (U) for p <. These two properties characterize L p (U) up to unique isomorphism (see notes on completion) Integral operators In the following (as always) we only consider integration with respect to Lebesgue measure. Theorem Suppose U R n and V R m are open. Suppose K : U V R is measurable, and that we have two constants 0 C 0, C 1 < such that: for a.e. x U K(x, y) dy C 0. (64) and for a.e. y V K(x, y) dx C 1. (65) Pick p [1, ] and define q [1, ] by: 1 p + 1 = 1. (66) q Pick u L p (V ). Then the following holds. For a.e. x U the function y K(x, y)u(y) is in L 1 (V ). Moreover the expression: (Ku)(x) = K(x, y)u(y)dy, (67) defines a function Ku L p (U) which satifies: V Ku Lp (U) C 1 q 0 C 1 p 1 u L p (V ). (68) If one does not worry about which functions are measurable, this is just a clever application of Hölder s inequality. The measuretheoretic details of the proof, which involves an appeal to the Fubini and Tonelli theorems, are beyond the scope of this course. Reference: [2] Theorem Application to mollification: Uniform boundedness of mollification: C c (U) is dense in L p (U) for p <. u Ψ ɛ Lp (I ɛ(u)) u Lp (U). (69) 33
34 14.3 Weak derivatives Definition and first properties. 34
35 15 Weak derivatives and Sobolev spaces Evans Chapter October
36 36
37 16 Exercises on Hölder 13 October 2016 Correction of the exercises on Hölder s inequality. 37
38 38
39 17 Partitions of unity 17 October
40 40
41 18 Integration on submanifolds 20 October 41
42 42
43 Compulsory exercise To be handed in on 27 October 2016 Some notation: For x R n : x = (x x 2 n) 1/2. (70) Exercise We define: R + = {x R : x > 0}. (71) For any continuous function w : R n R + define the space: X w = {u : R n R : u is measurable and u 2 w < }. (72) For u X w we define: u w = ( u 2 w) 1/2. (73) (a) Fix a continuous function w : R n R +. Check that X w is a vector space and that w defines a norm on this space. In the following we always consider X w as equipped with this norm. Show that for any open bounded subset U of R n we have a continous linear surjection: X w L 2 (U), (74) defined by u u U. (b) Pick two functions w 0 and w 1, which are continuous R n R +. For θ ]0, 1[ define: w θ = w0 1 θ w1. θ (75) Show that for any θ ]0, 1[ and any u X w0 X w1 we have u X wθ and: u wθ u 1 θ w 0 u θ w 1. (76) (c) Consider the setup of question (b). Suppose we have, for each n N a function u n X w0 X w1. Suppose that the sequence (u n ) converges in X w0 to u X w0. Suppose also that (u n ) is bounded in X w1. Check that there exists a constant C 0 such that for any open bounded subset U of R n : u 2 w 1 C. (77) U Use the monotone convergence theorem to deduce that u X w1. Show that (u n ) converges to u in X wθ for any θ ]0, 1[. Exercise (a) Show that there exists a function θ C c (R n ) such that: You may use smoothing by convolution. x R n x 1 θ(x) = 1. (78) 43
44 (b) Let θ be a function as in (a). For nonzero m N define θ m : R n R by : θ m (x) = θ(x/m). (79) Fix p [1, + [.  Show that for any u L p (R n ), the functions θ m u are in L p (R n ) and converge to u in L p (R n ), as m. You may want to prove it first for u C c (R n ).  Show that for any u W 1,p (R n ), the functions θ m u are in W 1,p (R n ) and converge to u in W 1,p (R n ), as m. (c) Use known results on smoothing by convolution to deduce that C c (R n ) is dense in W 1,p (R n ) for all p [1, + [. Exercise Let u and v be two continuous maps R n R. Suppose supp u E and supp v F, with E and F closed subsets of R n, F being, in addition, bounded. (a) Show that E + F is closed, where we denote: (b) Show that supp(u v) E + F. E + F = {x + y : x E, y F }. (80) Exercise In this exercise we identify C with R 2 in the standard way. We say that f : C C is an entire function if, for each x C, the following limit exists in C: f (x) = lim h 0,h C\{0} f(x + h) f(x), (81) h and moreover f : C C is a continuous function. (a) Show that f is entire iff f is of class C 1 and satisfies the partial differential equation: 1 f + i 2 f = 0. (82) (b) Show that complex polynomials are entire functions. (c) Show that if f is entire and does not take the value 0, then 1/f is entire and (1/f) = f /f 2. (d) Show that if f is entire and of class C 2, then the real and imaginary parts of f are harmonic functions. (e) Use Liouville s theorem and the above, to show that any non constant complex polynomial has a root. Exercise Choose a < b in R and let I be the interval I =]a, b[. Show that there exists a constant C such that for all smooth functions on I = [a, b]: u L (I) C u 1/2 L 2 (I) u 1/2 H 1 (I). (83) Hint: use the identity: u(y) 2 = y x 2u(s)u (s)ds + u(x) 2. (84) 44
45 19 Stokes theorem 31 October
46 46
47 20 Density theorems 3. November 2016 Reference: [1] 5.3. Evans presents three theorems on the approximation of elements in Sobolev spaces by smooth functions. Theorem 1 says that as long as one stays away from the boundary, smoothing by convolution will do the job. Theorem 2 shows how one can get global approximation by smooth functions defined on the whole of U. Theorem 3 shows how one can get global approximation by smooth funcitons defined on a neighborhood of U. Notice that contrary to Theorem 2, this uses some smoothness assumption on U. Of these three, the most useful one is the last, because it shows that there are dense subspaces of W 1,p (U) for which integration by parts makes sense. This is used to define traces of functions on the boundary (See Evans 5.5). 47
48 48
49 21 Trace theorem 7. November 2016 Reference: [1] The trace operator W 1,p (U) L p ( U) Notice that there are essentially three ingredients in its definition: A lemma on the construction of linear operators (extension by continuity from dense subspaces). A socalled estimate on functions, proved for smooth enough functions defined on a halfspace. In this setting integration by parts, Leibniz rules and Hölder s inequality can all be safely applied. The technique of partitions of unity and straightening of the boundary. 49
50 50
51 22 On W 1,p (U) 22.1 Characterizations of W 1,p 0 (U) 10. November 2016 Theorem Let U be bounded, open and of class C 1. We suppose p [1, [. Then for any u W 1,p (U) the following are equivalent: u can be approximated by elements of C c The trace of u on U (defined in L p ( U)) is zero. The extension of u by zero outside U is in W 1,p (R n ) Poincaré inequality (U) in the W 1,p (U) norm. Proposition Let U R n be open and bounded. There exists C > 0 such that for all u H 1 0(U) we have: u 2 C grad u 2. (85) 22.3 Some exercises Evans Chapter 5, exercises 7 and 8. An exercise using density of smooth functions. U Exercise For any function u : R n R and x R n we define τ x u : R n R by: y R n (τ x u)(y) = u(y x). (86) Show that for any u H 1 0(R n ) and any x R n : U u τ x u L 2 (R n ) x grad u L 2 (R n ). (87) You may start by noticing that for u C c (R n ): u(y) u(z) = 1 0 grad u(z + t(y z)) (y z)dt. (88) An exercise to look back at the first few lectures. Exercise Pick a real c > 0. Given a function v : R 3 R we consider the equation, on R 3 : u + c 2 u = v. (89) Given v C(R 3 ), a strong solution to the equation (89) is a function u C 2 (R 3 ) such that (89) holds in the sense of pointwise differentiation. Given v L 1 loc (R3 ), a weak solution to the equation (89) is a function u L 1 loc (R3 ) such that for all φ Cc (R 3 ): u( φ + c 2 φ) = vφ. (90) 51
52 (a) Show that if v C(R 3 ), a function u C 2 (R n ) is a strong solution to (89) if and only if it is a weak solution. Also show that the homogeneous equation (that is, with v = 0) always has infinitely many strong solutions. (b) Find all functions f C 2 (]0, + [) such that the function u f : R 3 \{0} R defined by, x R 3 \ {0} u f (x) = f( x ), (91) satisfies (in the sense of pointwise differentiation): (c) Define the function E : R 3 R by: u f + c 2 u f = 0 on R 3 \ {0}. (92) E(x) = 1 exp( c x ). (93) 4π x Check that E is locally integrable and show that for all φ C c (R 3 ): E( φ + c 2 φ) = φ(0). (94) Deduce that for any v Cc (R 3 ), E v is a strong solution to (89). (d) Show that E W 1,1 (R 3 ). Deduce that for v L 2 (R 3 ), E v H 1 (R 3 ) and is a weak solution to (89). (e) For this question you may use the result of question (c) in Exercise Pick v L 2 (R 3 ) and u H 1 (R 3 ). Show that u is a weak solution to (89) if and only if for any w H 1 (R 3 ): grad u grad w + c 2 uw = vw. (95) Deduce that, given v L 2 (R 3 ), equation (89) has a unique weak solution in H 1 (R 3 ). Other exercises in Evans, Chapter 5: 9,11,13,14. 52
53 23 The Dirichlet principle Already known to Gauss, but used and extended by Riemann under this name. Reading assignment: in Evans Inner products In the following all vector spaces are real. The following is the CauchySchwarz inequality (we check that it doesn t require positive definiteness). Proposition Let X be a vector space and a a symmetric bilinear form on X with the property: u X a(u, u) 0. (96) Then: and the function: defines a seminorm on X. u, v X a(u, v) a(u, u) 1/2 a(v, v) 1/2, (97) { X R, u a(u, u) 1/2. Proof. (i) Pick u, v X, for which we want to prove (97). We first write for all λ R:, (98) 0 a(u + λv, u + λv) = a(u, u) + 2λa(u, v) + λ 2 a(v, v). (99) With λ = sign a(u, v) we get: From this we deduce, for all t > 0: 2 a(u, v) a(u, u) + a(v, v). (100) 2 a(u, v) = 2 a(tu, t 1 v) t 2 a(u, u) + t 2 a(v, v). (101) If both a(u, u) = 0 and a(v, v) = 0, then a(u, v) = 0 and we are done. Suppose then, that a(u, u) 0. If a(v, v) = 0 we get a(u, v) = 0 by letting t 0, so we are also done. If a(v, v) 0, choosing t so that: gives the result. (ii) We have: t 2 = a(v, v) 1/2 /a(u, u) 1/2, (102) a(u + v, u + v) 1/2 = (a(u, u) + 2a(u, v) + a(v, v)) 1/2, (103) (a(u, u) + 2a(u, u) 1/2 a(v, v) 1/2 + a(v, v)) 1/2, (104) a(u, u) 1/2 + a(v, v) 1/2. (105) This is the main ingredient in showing that (98) defines a seminorm. 53
54 In the above circumstaces, the seminorm is a norm iff: a(u, u) = 0 u = 0. (106) In this case we call a an inner product on X and we say that (X, a) is an inner product space. When X is equipped with the norm defined by a, a is a continuous bilinear form on X. Definition A Hilbert space is an inner product space which is complete. Recall that for any normed vector space X, its dual is the space of continuous linear forms on X. We denote the dual by X. It is equipped with the norm: where it is understood that only nonzero u are used. l(u) l = sup u X u, (107) Proposition Let (X, a) be an innerproduct space and l X. Let F : X R denote the functional defined by: For any u X, the following are equivalent: There is at most one such u. F (u) = 1 a(u, u) l(u). (108) 2 v X a(u, v) = l(v). (109) v X F (u) F (v). (110) Remark Let (X, a) be an inner product space. Then, for u X: a(u, v) u = sup = a(u, ) X. (111) u X v In other words the linear map: { X X, u a(u, ). (112) is an isometry. The following result is called the Riesz representation theorem. Theorem Let (X, a) be a Hilbert space. Then for any l X there exists a unique u X such that: Proof. Let F be the functional defined in (108). For any v X we have: v X a(u, v) = l(v). (113) F (v) 1 2 v 2 l v, (114) 1 2 ( v l )2 1 2 l l 2. (115) 54
55 Define: Let (u n ) be a sequence in X such that: (one says that (u n ) is a minimizing sequence ). We have (from the parallelogram identity ): I = inf{f (v) : v X}. (116) F (u n ) I. (117) 1 4 u n u m 2 = 1 2 u n u m 2 u n + u m 2, 2 (118) = F (u n ) l(u n ) + F (u n ) l(u n ) (119) 2(F ( u n + u m 2 ) l( u n + u m )), (120) 2 = F (u n ) + F (u m ) 2F ( u n + u m ), 2 (121) F (u n ) I + F (u m ) I. (122) It follows that (u n ) is Cauchy (notice how the fact that I > enters the argument). Let u be its limit. By continuity of F, F (u) = I, so F achieves its minimum at u. Remark Let (X, a) be an inner product space. It follows from the completeness of R, that X is complete 1, so in order for the map (112) to be onto, it is necessary that (X, a) is complete. The theorem says that this condition is also sufficient LaxMilgram The following defines orthogonal projection in a Hilbert space. Proposition Let X be a Hilbert space and Y a closed subspace. Then for any u X, there is a unique v Y such that u v Y. Proof. Let, denote the scalar product of X. The vector space Y, equipped with the restriction of, is a Hilbert space. Also, u, is a continuous linear form on Y. We may apply the Riesz representation theorem. The following theorem, proposition or lemma (according to taste) is known as LaxMilgram. Proposition Let X be a Hilbert space, and let a be continuous bilinear form on X, which is coercive in the sense that for some α > 0: u X a(u, u) α u 2. (123) Then for any l X there exists a unique u X such that: v X a(u, v) = l(v). (124) 1 More generally, if X is a normed vector space any Y is a Banach space, then the space L(X, Y ) of continuous linear maps X Y, equipped with the operator norm, is a Banach space. 55
56 Proof. (sketch) Let, denote the scalar product of X. For any u X denote by Au X the unique solution to the problem: v X Au, v = a(u, v), (125) given by the Riesz representation theorem. Then A : X X is linear and continuous. Moreover: Au α u. (126) One deduces that Ran A is closed. If Ran A X we may find, by the above corollary, a nonzero u X such that u Ran A. We would get: Therefore A is an isomorphism. α u 2 a(u, u) = Au, u = 0. (127) Remark Consider the hypotheses of the proposition and suppose in addition that a is symmetric. Then a is an inner product on X. It defines a norm on X which is equivalent to the original one. Therefore (X, a) is a Hilbert space and LaxMilgram reduces to the Riesz representation theorem. 56
57 24 Second order elliptic PDEs 24.1 Variational formulation [1] 6.1, 6.2.1, Theorem Let U R n be open and bounded. For any f L 2 (U) there is a unique u H 1 0(U) such that: u H 1 0(U) grad u grad u = fu. (128) Remark Equation (128) is called (by definition) the weak formulation of the equation: u = f. (129) The boundary condition u U = 0 is encoded by the condition u H 1 0(U). u is the unique point where the Dirichlet functional reaches its minimum on H 1 0(U). Any minimizing sequence converges to u in H 1 0(U) Galerkin method Céa s lemma Some exercises Evans Chapter 6: Exercises 1,2,3,4. In Exercises 3 and 4 give explanations for the why these can be called weak solutions. 57
58 25 The space W 1,1 of an interval 25.1 The space W 1,1 of an interval Proposition Let a < b be two reals and consider the interval I =]a, b[. (i) Suppose v L 1 (I) and define u : I R by: Then: u(x) = x a v(y)dy. (130) a. u is continuous. b. u is differentiable almost everywhere, (in the classical sense) with u (x) = v(x). c. v is the weak derivative of u. (ii) We have: a. There exists C > 0 such that for all u W 1,1 (I): u L (I) C u W 1,1 (I). (131) b. Any u W 1,1 (I) coincides a.e. with an element of C(I). c. If u W 1,1 (I) C(I) and v L 1 (I) denotes the weak derivative of u, we have: x I u(x) = u(a) + x Proof. (i) a: by the Lebesgue dominated convergence theorem b: by the Lebesgue differentiation theorem. c: We write, for any φ C 0 (I): b where χ is defined by: and Fubini was used. (ii) a u(x)φ (x)dx = = = = χ(x, y) = b x ( a a b b a a b b a b a ( y a v(y)dy. (132) v(y)φ (x)dy)dx, (133) v(y)φ (x)χ(x, y)dydx, (134) v(y)φ (x)dx)dy, (135) v(y)φ(y)dy, (136) { 1 for y x, 0 for x < y, (137) 58
59 a and b : For u C 1 (I) we have: Hence: So: (b a) u(x) u(x) = u(y) + u(y) + b a y x b a u(y) dy + (b a) u (z)dz, (138) u (z) dz. (139) b a u (z) dz. (140) u L (I) 1 b a u L 1 (I) + u L 1 (I). (141) For u W 1,1 (I), we may find a sequence u n C 1 (I), such that u n u in W 1,1 (I). By the inequality (141) it is Cauchy in C(I). The limit in C(I) must be equal a.e. to the limit in W 1,1 (I) (why?). So we may consider that u C(I). Then (141) holds for each u n and we may pass to the limit n. c: For u W 1,1 (I) C(I), we may find a sequence u n C 1 (I), such that u n u in W 1,1 (I). By the preceding results it converges uniformly to u. For each x I: u n (x) = u n (a) + x a u n(y)dy. (142) Since u n converges pointwise to u and u n converges in L 1 (I) to v, we may let n in this identity. Exercise Let I be a bounded open interval. Show that if p ]1, [, then for any u W 1,p (I) we have for a.e. x, y I: u(x) u(y) x y 1 1/p u Lp (I). (143) Exercise Let U R n be open. Pick u L 1 (U). Show that for any ɛ > 0 there exists δ > 0 such that for any open V U: meas(v ) δ u ɛ. (144) Hint: You may check it first when u is a step function, or a continuous function with compact support. Definition A function u : [a, b] R is said to be absolutely continuous if for any ɛ > 0 there exists δ > 0 such that for any n N, any family of two by two disjoint intervals ]a i, b i [ [a, b] for i [0, n], we have: n b i a i δ i=0 V n u(b i ) u(a i ) ɛ. (145) i=0 Exercise Check that functions in W 1,1 (I) are absolutely continuous, and that absolutely continuous functions are continuous. 59
60 25.2 The Cantor  Lebesgue function The Cantor set C is a certain compact subset of [0, 1] with measure 0. The Cantor  Lebesgue function f is a certain continuous function [0, 1] [0, 1] with the following two rather paradoxical properties: f C : C [0, 1] is onto. on the complement of C, f is differentiable with differential 0. In particular the Cantor  Lebesgue function is a continuous and a.e. differentiable function, such that the a.e. differential is in L 1 (]0, 1[), but such that, for x > 0: f(x) f(0) + x 0 f (y)dy, (146) where we denote by f the a.e. differential of f (which is 0). Notice that f is not in W 1,1 (]0, 1[). References: page in [2], or page in [7]. 60
61 26 Some exercises Exercise Choose a < b in R and let I be the interval I =]a, b[. Show that there exists a C > 0 such that for all u H 1 (I): u 1 u L2 (I) C u L2 (I). (147) b a Hint: Use that smooth functions on an interval attain their average. I Exercise Let I be the interval ]a, b[. Define for any smooth function u C (I), the affine function P I u : I R by: x I (P I u)(x) = u(a) + x a (u(b) u(a)). (148) b a Check that P I is a projection. Show that P I has a unique extension to a continuous operator H 1 (I) H 1 (I). 2. Deduce from Exercise 26.1 that there exists a C > 0 such that for all u H 1 (I): u P I u L 2 (I) C u L 2 (I). (149) 3. In the preceding question a and b were considered given we now study the dependence of the constant C on a and b. Show that there exists a C > 0 such that for all a < b and all u H 1 (I) (with I =]a, b[) : u P I u L 2 (I) C(b a) u L 2 (I). (150) Hint: Use the preceding result in the case I =]0, 1[ and make a change a variable from ]0, 1[ to ]a, b[. 4. (Optional) By similar arguments show that there exists a C > 0 such that for any h > 0, any interval I of length h, we have for any u H 2 (I): and for any u H 1 (I): (u P I u) L 2 (I) Ch u L 2 (I), (151) (P I u) L 2 (I) C u L 2 (I), (152) Exercise Let I be the interval ]0, 1[. 1. Pick n > 0 an integer. Define for integer i [0, n]: x n i = i/n I. (153) Define the function λ n i : I R by: 0 for x x n i 1, λ n n(x x i (x) = n i 1 ) for xn i 1 < x xn i, 1 n(x x n i ) for xn i < x xn i+1, (154) 0 for x n i+1 < x. 61
62 Check that λ i is continuous on I, affine on every subinterval [x n j, xn j+1 ] and that: λ n i (x n j ) = δ ij. (155) Define an operator P n : H 1 (I) H 1 (I) by: P n u = n u(x i )λ n i. (156) i=0 Show that P n is a continuous projector in H 1 (I). Show, using (150) and (151) on each subinterval [x n j, xn j+1 ], that for any u H 2 (I) there exists C > 0 such that for all n: 2. Define X n by: u P n u H 1 (I) C/n. (157) X n = span{λ n i : 0 < i < n}. (158) Check that for any u H 1 0(I) we have that P n u X n. Show using (150) and (152) that P n is uniformly bounded in H 1 0 (I) H 1 (I), that is, there exists C > 0 such that for all u H 1 0(I) and all n: P n u H 1 (I) C u H 1 (I). (159) Then show that for any u in H 1 0(I), P n u converges to u in H 1 (I), as n. 3. Pick δ > 0 and γ R. For a given f L 2 (I) we want to solve the equation: δu + γu = f. (160) with boundary conditions u(0) = u(1) = 0. Define the bilinear form a on H 1 0(I) by, for all u, v H 1 0(I): a(u, v) = δ u v + γ u v. (161) Show that there exists C > 0 such that for all u H 1 0(I): a(u, u) 1/C u 2 H 1 (I). (162) Apply the LaxMilgram theorem to show that there is a unique solution u H 1 0(I) to the weak formulation of (160). Check that u H 2 (I). 4. Check that there is a unique u n X n solving: v X n a(u n, v) = fv. (163) 5. Show using (162) that there is a constant C such that for all n: u u n H1 (I) C u P n u H1 (I). (164) Deduce that u n converges to u in H 1 (I), at least at the rate O(1/n). 62
63 Exercise Fix an integer n > 0. Let φ : R n R be a smooth function such that: For any ɛ > 0 we define φ ɛ by: supp φ B(0, 1), (165) φ = 1, (166) x R n φ(x) 0. (167) x R n φ(x) = φ( x). (168) φ ɛ (x) = ɛ n φ(x/ɛ). (169) a) Pick ψ Cc (R n ) and define ˆψ by ˆψ(x) = ψ( x). Show that for all u, v L 2 (R n ) we have: u(v ψ) = (u ˆψ)v. (170) b) We suppose that we have a function u L 2 (R n ) for which there exists a M > 0 such that for each i {1,, n} and all ɛ > 0: i (φ ɛ u) L2 (R n ) M. (171) Show that there exists a C > 0 such that for all ψ Cc (R n ): u i ψ C ψ L2 (R n ). (172) c) Deduce that u H 1 (R n ). 63
64 64
65 References [1] L. C. Evans. Partial differential equations, volume 19 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, second edition, [2] G. B. Folland. Real analysis. Pure and Applied Mathematics (New York). John Wiley & Sons Inc., New York, second edition, Modern techniques and their applications, A WileyInterscience Publication. [3] S. Lang. Undergraduate analysis. Undergraduate Texts in Mathematics. SpringerVerlag, New York, second edition, [4] P. D. Lax. Functional analysis. Pure and Applied Mathematics (New York). WileyInterscience [John Wiley & Sons], New York, [5] B. P. Rynne and M. A. Youngson. Linear functional analysis. Springer Undergraduate Mathematics Series. SpringerVerlag London, Ltd., London, second edition, [6] E. M. Stein and R. Shakarchi. Fourier analysis. Princeton Lectures in Analysis, I. Princeton University Press, Princeton, NJ, An introduction. [7] E. M. Stein and R. Shakarchi. Real analysis. Princeton Lectures in Analysis, III. Princeton University Press, Princeton, NJ, Measure theory, integration, and Hilbert spaces. [8] T. Tao. Analysis. II, volume 38 of Texts and Readings in Mathematics. Hindustan Book Agency, New Delhi, second edition,