DRAFT: April 28, 2008

Transcription

1 1 DRAFT: April 28,

2 Lectures on PDEs- APMA- E6301 DRAFT IN PROGRESS Michael I. Weinstein April 28, 2008 Contents 1 First order partial differential equations and the method of characteristics 4 2 The Laplacian - Laplace s and Poisson s equations The Newtonian potential in R n Boundary value problems and Green s functions Green s function for the ball in R n Green s function for the upper half plane in R 2, R n Single and Double Layer Potentials and their boundary limits Boundary integral formulation of the Dirichlet and Neumann Problems Interior and Exterior Fredholm Operators in Banach Spaces, Solvability of Fredholm Integral Equations on C 0 (S) and the Fredholm Alternative Application of Fredholm Integral Operator Theory to the solution of the Dirichlet and Neumann Problems Distributions and Fundamental Solutions 5 4 Introduction to Hilbert Space Representation of Linear Functionals on H Applications of Hilbert Space Theory to the Existence of Solutions of Poisson s Equation Generalizations of the Weak Dirichlet Problem Relation between weak and strong (classical) solutions of elliptic PDE Introduction to the Finite Element Method Department of Applied Physics and Applied Mathematics, Columbia University, New York, NY 10027; miw2103@columbia.edu 2

3 CONTENTS DRAFT: April 28, Operators on Hilbert spaces Bounded operators Compact operators Applications of compact operator theory to eigenfunction expansions for elliptic operators 21 8 Introduction to Variational Methods Variational Problem 1: Dirichlet s Principle Variational Problem 2: Smallest eigenvalue of Direct methods in the calculus of variations Ideas and general discussion Compactness and minimizing sequences The Heat / Diffusion Equation Remarks on solutions to the heat / diffusion equation Inhomogeneous Heat Equation - DuHamel s Principle Heat equation on a bounded domain, Energy inequalities and heat equation The maximum principle for the heat / diffusion equation Application: Burger s equation and Shock Waves The Wave Equation One dimensional wave equation Three dimensional wave equation D wave equation - Hadamard s method of descent Principle of causality Inhomogeneous Wave Equation - DuHamel s Principle Initial boundary value problem for the wave equation The Schrödinger equation Quantum mechanics Diffraction of classical waves Free Schrödinger - initial value problem Free Schrödinger in L p Free Schrödinger evolution of a Gaussian wave packet The Uncertainty Principle Background Linear Algebra Calculus Local existence theorem for ordinary differential equations (ODEs) Compactness and convergence results Fourier Transform Fourier inversion on S and L

4 DRAFT: April 28, First order partial differential equations and the method of characteristics Consider the first order PDE F (x, u(x), Du(x)) = 0, Du(x) = ( x1 u,.... xn u). (1.1) Here, F (x, z, p) denotes a smooth real-valued function of 2n + 1 variables x R n, p R n and z R 1, with x = (x 1,..., x n ), p = (p 1,..., p n ). Transport equation: t u(x, t) + v u(x, t) = 0, x R n, t R 1. Linear first order equations: b(x) Du(x) + c(x)u(x) = 0 Quasilinear first order equations: b(x, u(x)) Du(x) + c(x, u(x)) = 0 Fully nonlinear equations: equations which are nonlinear in Du(x), e.g. The eikonal equation, u(x) 2 = 1. 2 The Laplacian - Laplace s and Poisson s equations 2.1 The Newtonian potential in R n 2.2 Boundary value problems and Green s functions Green s function for the ball in R n Green s function for the upper half plane in R 2, R n 2.3 Single and Double Layer Potentials and their boundary limits 2.4 Boundary integral formulation of the Dirichlet and Neumann Problems Interior and Exterior 2.5 Fredholm Operators in Banach Spaces, Solvability of Fredholm Integral Equations on C 0 (S) and the Fredholm Alternative 2.6 Application of Fredholm Integral Operator Theory to the solution of the Dirichlet and Neumann Problems 4

5 DRAFT: April 28, Distributions and Fundamental Solutions 1 Definition 3.1 Let C 0 () or C c () denote the set of compactly supported C functions. C 0 () = { f : there exists K compact subset of, f(x) 0, x K } One often denotes C 0 () by D() or D 2. Functions in D are often called test functions. Definition 3.2 A linear functional, T, on D is linear mapping which associates to any φ D a number T [φ], i.e. for any λ, µ scalars and any φ, ψ D. T [ λφ + µψ ] = λ T [φ] + µ T [ψ] Examples: (a) Distributions obtained from locally integrable functions: Let f L 1 loc be any locally integrable function 3. Define T f [φ] = φ(x)f(x) dx (3.1) R n Then, T f is a linear functional on D. For simplicity, we will often write f[φ] instead of T f [φ]. (b) The Dirac delta function : δ ξ [φ] = φ(ξ) (3.2) defines a linear functional on D. Continuity of linear functionals on D, notion of distribution Definition 3.3 Consider an arbitrary sequence of functions {φ k } in D, which tend to zero in the sense of the following two conditions: (a) φ k vanish outside a fixed compact subset, K, of. and (b) For any multi-indx α, lim k α φ k (x) = 0 uniformly in x K. A linear functional T is continuous if lim k T [φ k ] = 0 for any sequence {φ k }, which tends to zero in the above sense. A continuous linear functional on D is called a distribution. The set of distributions on D() is denoted by D (). Examples: T f, defined for any locally integrable f and δ ξ are examples of distributions. Differentiation of distributions: 1 References: Chapter 3, F. John PDEs; Section E of Chapter 0 in G.B. Folland Introduction to PDEs 2 Here, is taken to be an open set. A compact set is a set which is a set which is closed and bounded. 3 f L 1 loc means that for any compact set C, f(x) dx <. C 5

6 DRAFT: April 28, 2008 Notation: A multi-index is a vector of non-negative integers α = (α 1,..., α m ) N m 0. The order of a multi-index is given by α α α m. α f(x) = α 1 x 1 αm x m f(x 1,..., x m ). Note that if f, φ D, then α f(x)φ(x) dx = ( 1) α f(x) α φ(x) dx, (3.3) which we can write as α f[φ] = ( 1) α f[ α φ ]. (3.4) Equation (3.4) motivates our definition of the derivative of a distribution. Let T D. Then, α T [φ] = ( 1) α T [ α φ ] (3.5) Distribution solutions to PDEs Let L denote the linear partial differential operator L = α I a α(x) α. Suppose we have a PDE Lu = T, (3.6) where T D. Given the above definition of derivative of a distribution, we say that the PDE has a distribution solution u, provided u[l t φ] = T [φ], φ D. (3.7) Here, L t φ = α I ( 1) α α ( a α (x) φ(x) ) denotes the adjoint of the differential operator L, which satisfies the generalization of (3.3): for φ, ψ D Lψ(x) φ(x) dx = ψ(x) L t φ(x) dx Fundamental Solutions Definition 3.4 A fundamental solution of a linear differential operator L with pole at x is a distribution solution, u, of Lu = δ x. Examples: Laplace Operator L =. L t =. We have shown that a fundamental solution for with pole at x is: Φ(x y) = 1 log x y, n = 2 2π 1 = x y 2 n, n 3 (2 n)ω n Exercise: In one space dimension, solve d 2 Φ = δ dx 2 x where δ x denotes the Dirac delta function (distribution). 6

7 4 Introduction to Hilbert Space DRAFT: April 28, 2008 Definition (a) S is a vector space or linear space: u, v S implies, for all λ, µ scalars that λu+µv S, plus vector spaces axioms. (b) S is an inner product space if it is a linear space and there is a function (, ) : S S R u, v S (u, v) R (4.1) with the following properties (P1) (u, v) = (v, u) (P2) (λu + µv, w) = λ(u, v) + µ(v, w), λ, µ R. (P3) (u, u) 0 and (u, u) = 0 = u = 0. Example 1a: S = R n. u, v R n. (u, v) = n i=1 u iv i. Example 2a: S = C (), R n bounded open, (f, g) = u(x) v(x) dx. Definition 4.2 A normed linear space is a vector space equipped with a notion of length, called the norm,, satisfying the following properties (n1) u = 0 implies u = 0 (n2) λu = λ u, λ R (n3) u + v u + v Given an inner product space, S, there is a natural norm, or measure of size of a vector in S u = (u, u) (4.2) 4 References: We seek to give the required working knowledge of the basic functional analysis, e.g. Hilbert space, linear operators, spectral theory,..., with a view toward its application to PDEs. See, for example, (a) L.C. Evans, PDEs, Appendix D on Linear Functional Analysis and Appendix E on Measure and Integration; (b) F. John, PDEs, Springer 4th Ed., Chapter 4, section 5; G.B. Folland, Introduction to PDEs, Princeton University Press, Chapter 0. A good text on elementary notions in analysis (convergence, uniform convergence, integration,... is the text: W. Rudin, Principles of Mathematical Analysis, McGraw Hill. a more advanced text is that of E. Lieb and M. Loss, Analysis, Am. Math. Soc. 5 Throughout we will construct spaces of functions, with notions of size and distance within these spaces defined in terms of integrals. These integrals are assumed throughout to be taken in the Lebesgue sense, an important extension notion of Riemann integral. Note that if f is integrable in the sense of Riemann, it is in the sense of Lebesque and the values of these integrals is the same. Appendix E of L.C. Evans PDEs contains a very terse treatment of measure theory and the Lebesgue integral. 7

8 DRAFT: April 28, 2008 Any norm satisfies the Cauchy Schwarz inequality (u, v) u v (4.3) Example 1b: S = R n. u, v R n. u = k=1 u k 2. Example 2b: S = C (), R n bounded open. u 2 = ( u(x) 2 dx ) 1 2. Convergence: The sequence {u j } j 1 in S converges to a point u S provided lim u j u = 0. (4.4) j In this case we say that the sequence {u j } j 1 converges strongly to u or converges to u in norm. Later, we shall introduce the notion of weak convergence. Cauchy sequence: The sequence {u j } j 1 in S is called a Cauchy sequence if u j u k 0 as j, k. Completeness: The normed linear space S is complete if every Cauchy sequence in S converges to some element of S. A complete normed linear space is called a Banach space. A complete normed linear space, whose norm derives from an inner product, as in (4.2), is called a Hilbert space. Completion Theorem: Every normed linear space can be completed. Specifically, if S is a normed linear space, then it is possible to naturally extend S to a possibly larger set, S, and the norm to be defined on all S, in such a way that the set S, equipped with the norm. is complete. The completion is constructed by identifying all Cauchy sequences with points in the extended space S. If v S but v / S, then there is a sequence {v k } in S, such that v k v 0 as k. We say that the space S is dense in S. Example 1c: Let Q denote the set of rational numbers, numbers of the form p/q, where p and q are integers and q 0. Consider the inner product space S = Q n. u, v Q n. u = k=1 u k 2, the Euclidean norm. While Q n equipped with this norm is not complete, Q n has a completion, namely R n equipped with the norm u. Example 2c: C () is not complete with respect to the norm u 2 = u(x) 2 dx. There are many important completions of C () with respect to different norms: (1) The Hilbert Space L 2 (): The completion of C () with respect to the norm 2 is called L 2 (). Since the norm on L 2 derives from an inner product, f, g fg, L2 is a Hilbert space. 8

9 DRAFT: April 28, 2008 (2) Sobolev Spaces - H k (), k 0: Define the inner product of two functions f, g H 1 by (f, g) H 1 = f g + fg dx (4.5) The H 1 Sobolev norm is defined by: u 2 H 1 = (u, u) H 1 = u(x) 2 + u(x) 2 dx (4.6) More generally, for any integer k 0, define H k () to be the completion of C () with respect to the norm u 2 H = α u(x) 2 dx (4.7) k Note that H 0 () = L 2 (). α k (3) Sobolev Spaces - H k (R n ), k 0: Replace the bounded open set,, with = R n. First consider C0 (R n ) and define H k (R n ) to be the completion of C0 (R n ) with respect to the H k (R n ) norm: u 2 H = α u(x) 2 dx (4.8) k R n α k Remark 4.1 Sobolev spaces are important in analysis of functions largely due to their role in relating global (integral) properties and local (pointwise) properties. For example, here is the most elementary Sobolev inequality: There is a constant K > 0 such that for any f C0 (R) we have ( ) f(x) 2 K f(x) 2 + f (x) 2 dx = K f 2 H (4.9) 1 Proof: Exercise Weak convergence: A sequence {u j } j 1 in H converges to a limit u in H if for all w H (u j, w) H (u, w) H (4.10) Compactness A closed subset, C, of H is compact if any sequence {u j } j 1 in C has a convergent subsequence. Remark 4.2 Any closed and bounded subset of C n is compact. However, in infinite dimensional spaces, e.g. L 2, this is not the case. See section

10 4.1 Representation of Linear Functionals on H DRAFT: April 28, Representation of Linear Functionals on H (1) A linear functional on S is a mapping φ : S R, u φ(u), such that for all u, v S and all scalars λ, µ we have φ(λu + µv) = λφ(u) + µφ(v). One sometimes represents the action of φ on a vector u with notation φ(u) = φ, u. (2) A bounded linear functional is a linear functional for which there is a constant, M > 0, such that for any u S φ(u) M u (4.11) The set of bounded linear functionals on a Hilbert Space, H, is denoted H and is called the dual space of H. The smallest constant, M, for which the inequality (4.11) holds is called the norm of the linear functional φ, φ H. Example 1d: For any vector v R n, φ(u) = (u, v) = u v = n k=1 u kv k is a bounded linear functional with norm φ (R n ) = v Rn. Moreover, it is straightforward to see that given any linear functional on R n, φ, there is a unique vector v φ R n such that φ(u) = (u, v φ ) for all u R n. What about bounded linear functionals on Hilbert space? Let H denote any Hilbert space and let v be a fixed vector in H. Then, in analogy with the construction in Example 1d, φ( ) = (, v) defines a bounded linear functional on H with norm v. The joke is that every bounded linear functional on a Hilbert space is obtained in this way: Theorem 4.1 Riesz Representation Theorem: Let H denote a Hilbert space and H the set of all bounded linear functionals on H. For any φ H, there is a unique vector v φ H, such that φ(u) = (u, v φ ), for any u H. (4.12) Furthermore 6, φ H = v φ H. The proof of this theorem relies on some elementary geometry of Hilbert space, in particular, the projection theorem. Proof of the Riesz Representation Theorem: If φ(u) = 0 for every u H, then we take v φ = 0. Thus we suppose that φ(q) 0 for some q H. By linearity, φ(q φ(q) 1 ) = 1. The proof can be broken down into a sequence of claims. Claim 1: Let m = inf{ u : φ(u) = 1 }. The infimum, m, is attained, i.e. there exists w H such that φ(w) = 1 and w = m. We begin with the identity (exercise), which holds for all a, b H: Parallelogram Law : 6 H and H are isometrically isomorphic 1 4 a + b a b 2 = 1 2 a b 2, (4.13) 10

11 4.1 Representation of Linear Functionals on H DRAFT: April 28, 2008 which follows by direct computation, using that a 2 = (a, a). Now let w k denote a minimizing sequence. Thus, w k m, as k, φ(w k ) = 1. We use (4.13) to show that {w k } is a Cauchy sequence. Set a = w k and b = w k. Then, (4.13) 1 4 w k w j 2 = 1 2 w k w j 2 w k + w j w k w j 2 m 2 The latter inequality holds because φ( w k+w j ) = 1, and therefore w k+w j 2 2 m. Now let k, j tend to infinity and we have, since w k,j m, that w k w j 0, i.e. the sequence {w k } is Cauchy. It therefore has a limit w H. Claim 2: The minimizer, w, is unique. Again this follows from the parallelogram law (4.13). If w A and w B both belong to H with w A = w B = m, we have w A+w B 2 m 2. Substitution into (4.13) gives: 2 m w A w B 2 w A + w B w A w B 2 = m 2. This gives a contradiction. Claim 3: w is orthogonal to the null space of φ, i.e. for all u such that φ(u) = 0, we have (w, u) = 0. If φ(u) = 0 then Since w has minimal norm, ( φ w ) (w, u) (u.u) u w (w, u) (u, u) u 2 = φ(w) = 1 (4.14) w 2. Expanding and cancelling terms gives (w, u) 2 0 and therefore (w, u) = 0. Claim 4: w 0. 1 = φ(w k ) φ(w) by continuity of φ. If w = 0 then φ(w) = 0, a contradiction. Claim 5 and Completion of the Proof: Define v φ = w 2 w. Then, for all u H, φ(u) = (u, v φ ). Let u H be arbitrary. Recall that by Claims 3 and 4, w 0 and {x : φ(x) = 0} w. Now express u as the sum of two orthogonal components, one in the direction of w and one lying in the null space of φ. u = [ u wφ(u) ] + wφ(u) (u, w) = (u wφ(u), w) + φ(u) (w, w) Note: (u wφ(u), w) = 0 since φ(u wφ(u)) = 0. Therefore, (u, w) = φ(u) (w, w) (u, v φ ) = φ(u), v φ w w 2 This completes the proof of the Riesz Representation Theorem. 11

12 DRAFT: April 28, Applications of Hilbert Space Theory to the Existence of Solutions of Poisson s Equation We begin with Poisson s equation for a function u defined on an open bounded subset of R n : v = f, v = 0 (5.1) Here, we begin by taking f C( ). We seek a weak formulation of (5.1). That is, we seek a formulation which is satisfied by classical solutions, i.e. u of class C 2 ( ) satisfying (5.1). To derive a weak formulation, let u C0 () and suppose that v is a classical solution. Now multiply (5.1) by u and integrate over. Integration by parts, using the Gauss divergence theorem and that u vanishes on gives uf dx = u ( v) = (u v) dx + u v dx = u v dx. (5.2) Define φ(u) = (u, v) D = u(x)f(x)dx u(x) v(x) dx The result (5.2) can be expressed as follows: If v is a classical solution to Poisson s equation with zero (Dirichlet) boundary conditions (v(x) = 0, x ), then for all u C 0 (), we have φ(u) = (u, v) D. Note, on the other hand, that φ(u) is well-defined for any u, v L 2 () and (u, v) D is well-defined if u and v are both in L 2 (). With the Riesz Representation Theorem in mind, it is natural to regard (u, v) D as defining an inner product with respect to which we shall represent the bounded linear functional φ(u). Toward the correct formulation, we define the Dirichlet norm on C 0 () u D = (u, u) D. (5.3) The key to showing that D is indeed a norm is to observe that if u D = 0, then u is identically constant, and because u vanishes on, we must have u 0. Next, to put our problem in a Hilbert space setting we define the Sobolev space H 1 0(), which encodes zero boundary conditions, to be the completion of the space C 0 () with respect to the Dirichlet norm, D. We can now formulate the Weak Dirichlet Problem: Let f L 2 (). Find v H0() 1 such that for all u H0(), 1 we have (u, v) D = u(x) v(x) dx = u(x) f(x) dx = φ f (u). (5.4) 12

13 DRAFT: April 28, 2008 Theorem 5.1 The weak Dirichlet problem has a unique H 1 0() solution. Proof: We have set things up so that the result follows directly from the Riesz Representation Theorem. The main thing to check is that φ(u) is a bounded linear functional on H0(), 1 i.e. there exists a constant M > 0, such that φ(u) uf dx M f u D, (5.5) for some M f, depending on f but independent of u. By the Cauchy Schwarz inequality 7 uf dx f L 2 u L 2 Therefore, to prove that φ(u) is a bounded linear functional and therewith the theorem, we must prove the: Poincaré Inequality: Let be an open connected subset in R n, which lies between two planes, a distance 2a apart 8. There exists a positive constant, p, depending on, such that for any u H 1 0() u L 2 p u D (5.6) Note that by the Poincaré inequality, φ(u) is a bounded linear functional on H 1 0(), with norm bound M f = p f 2. Proof of the Poincaré inequality: After possibly rotating the region, we may assume it lies between the two planes x 1 = a and x 1 = a. Note that it suffices to prove the inequality for u belonging to a dense subset of H0(), 1 u C0 (). Let x = (x 1, x ), with x = (x 2,..., x n ). For fixed u(x), we have u(x 1, x ) = x1 a x1 u(ξ 1, x ) dξ 1 Squaring and applying the Cauchy-Schwarz inequality, we have u 2 (x 1, x ) = ( x1 ) 2 1 x1 u(ξ 1, x ) dξ 1 a ( x1 ) ( x1 ) 1 2 dξ 1 x1 u(ξ 1, x ) 2 dξ 1 a (x 1 + a) a a a x1 u(ξ 1, x ) 2 dξ 1 7 fg f 2 g 2 8 Thus we allow domains, more general than bounded domains. 13

14 5.1 Generalizations of the Weak Dirichlet Problem DRAFT: April 28, 2008 Integration with respect to x 1 gives a a u(x 1, x ) 2 dx 1 2a 2 a a x1 u(ξ 1, x ) 2 dξ 1 Finally, integation with respect to the remaining n 1 variables, x gives: u(x 1, x ) 2 dx 1 dξ 1 dx 2a 2 x1 u(ξ 1, x ) 2 dξ 1 dx This proves Poincaré s inequality with p = 2a. 5.1 Generalizations of the Weak Dirichlet Problem We briefly discussed how the forgoing discussion can be generalized from the Laplacian,, to a treatment of the Dirichlet problem for for more general linear elliptic partial differential operators, L: Lv = f, x, v = 0 (a) L = a(x), where a(x) is a smooth function satisfying the strong ellipticity condition: min x a(x) θ. Define the Dirichlet inner produce (u, v) a = a(x) u(x) v(x) dx. We then introduce the Dirichlet norm is then u H 1 a. Let Ha 1 denote the completion of C0 () with respect to u H 1 a. Then, we can prove (Exercise using Poincaré s inequality!) that φ(u) = (u, f) L 2 is a bounded linear functional on Ha. 1 We can then apply the Riesz Representation Theorem to show that there exists a unique v φ Ha, 1 such that for all u Ha, 1 φ(u) = (u, v φ ) a. (b) More general strongly elliptic operators: L = jk x j a jk (x) xk, where A = (a jk (x)) is a n by n symmetric and positive definite matrix function, whose minimum eigenvalue, λ min (x) satisfies min x λ min (x) θ > 0, for some θ. Note that example (a) is a special case with A = δ jk a(x). Introduce the Dirichlet norm: u H 1 A and HA 1, the completion of C0 () with respect to u H 1 A. Then, (Exercise using Poincaré s inequality) that φ(u) = (u, f) L 2 is a bounded linear functional on HA 1. We can then apply the Riesz Representation Theorem to show that there exists a unique v φ HA 1, such that for all u HA 1, φ(u) = (u, v φ) A. (c) Yet more general elliptic operators - elliptic operators with non-symmetric lower order terms: Garding s inequality and the Lax-Milgram Lemma (representation theorem, which generalizes Riesz Rep Th m, which is applicable to non-symmetric forms). 14

15 5.2 Relation between weak and strong (classical) solutions of elliptic DRAFT: PDEApril 28, Relation between weak and strong (classical) solutions of elliptic PDE For strongly elliptic linear partial differential operators (see 5.1 one has the following important theorem, showing that weak implies strong. Theorem 5.2 (Elliptic Regularity Theorem) Consider the Dirichlet problem Lv(x) = f(x), x, v(x) = 0, x, (5.7) where is an open and bounded subset of R n with smooth boundary and f L 2 (). Let v H0() 1 be a weak solution of Lv = f, where f C (). Then, v C and lim x, x v(x) = 0. The proof is very technical; see L.C. Evans, PDE, section Introduction to the Finite Element Method 9 The notion of weak solution of the Dirichlet problem for Poisson s equation v(x) = f(x), v(x) = 0, x (see (5.4) ) motivates The Finite Element Method - (FEM), a method of central importance in the numerical solution of PDEs. We introduce FEM via a simple example. Consider the two-point boundary value problem: 2 xv(x) = f(x), v(0) = v(1) = 0. (5.8) Above we saw that this can be formulated, in terms of the Weak Dirichlet Problem (5.4), which we present in modified form as follows. Define U = {u : u C 0 [0, 1], u C 1 piecewise[0, 1] }. (5.9) Here, f C 1 piecewise[0, 1] if there are finitely many points 0 < ξ 1 < < ξ r < 1 such that f C 1 (ξ j, ξ j+1 ), j = 1,..., r 1 Our modified form of the weak Dirichlet problem is: Find u U such that for all v U ( x u, x v) = (f, v) (5.10) To obtain the Galerkin or Finite Element Method for the (5.8) we will replace U, by a finite-dimensional approximation, U h, which we now construct. 9 Reference: Numerical solution of partial differential equations by the finite element method, Claes Johnson, Cambridge

16 5.3 Introduction to the Finite Element Method DRAFT: April 28, 2008 Partition the interval [0, 1] into subintervals: [0, 1] = [x 0, x 1 ) (x 1, x 2 ) (x M 1, x M ) (x M, x M+1 ] = M+1 j=1 I j 0 = x 0 < x 1 < x 2 < < x M 1 < x M < x M+1 = 1 The interval lengths in the partition are h j = x j+1 x j. If the points, x j, are equally spaced than h j = h = (M + 1) 1. Definition of U h : U h = {u : u linear on I j, j = 1,..., M + 1, v continuous, u(0) = u(1) = 0 } (5.11) Proposition: U h has dimension M and has the basis (spanning set which is linearly independent) consisting of the triangular tent functions defined by: ϕ j U h ϕ j (x i ) = δ ij, (1, i = j; 0, i j) j = 1, 2,..., M proof: Let u U h. Then, u(x) = M j=1 η jϕ j (x), where η j = u(x j ). Therefore, the functions ϕ i (x), i = 1,..., M span U h. It is simple to check they are linearly independent. The Galerkin / Finite Element Method Find u h U h such that ( x v h, x u) = (f, u), for all u U h or equivalently ( x v h, x ϕ j ) = (f, ϕ j ), for all j = 1,..., M (5.12) Exercise: Using the ideas from our discussion of the Dirichlet s principle, show that (5.12) is equivalent to finding u h U h, such that I[v h ] = min u U h I[u] (5.13) Implementation of the finite element method Since v h U h, we have that v h (x) = M k=1 ξ kϕ k (x), where ξ k = v h (x k ) is to be determined. Substitution into (5.12) gives ( M ) ξ k x ϕ k, x ϕ j = (f, ϕ j ), j = 1,..., M. k=1 Equivalently, A ξ = b, (5.14) 16

17 5.3 Introduction to the Finite Element Method DRAFT: April 28, 2008 where A = (a ij ) denotes the stiffness matrix and b = (b i ) denotes the load vector given by: Clearly, A is symmetric. a ij = ( x ϕ i, x ϕ j ), i, j = 1,..., M b i = (f, ϕ i ). Computation of and properties of the stiffness matrix (1) Note that ( x ϕ i, x ϕ j ) = 0, i j > 1. Therefore, A is a tridiagonal matrix. The diagonal and off-diagonal elements are given by: a ii = ( x ϕ i, x ϕ i ) = h 1 i + h 1 i+1, a i,i+1 = a i+1,i = ( x ϕ i, x ϕ i+1 ) = h 1 i (2) A is symmetric and positive definite. Symmetry was noted above. Positive definiteness is seen as follows: For any η R M η T A η = = M η i ( x ϕ i, x ϕ j )η j i,j=1 ( M ) M η i x ϕ i, η j x ϕ j i=1 j=1 = v 2 0. where v = M i=1 η i x ϕ i. Moreover, η T Aη = 0 only if η = 0 because {ϕ i : i = 1,..., M} is a linearly independent set. Theorem: The finite element method reduces to the solution of the system of M nonhomogeneous linear algebraic equations Aξ = b, (5.14), with stiffness matrix, A, and load vector, b. A is symmetric and positive definite. It is therefore invertible and the system can be solved for any load vector b. Special case: Uniform partition of [0, 1]: Here, h j = x j+1 x j = h = (M +1) 1. Therefore, a ii = 2/h and a i,i+1 = a i+1,i = 1/h. Thus, A = 1 h (5.15) 17

18 6 Operators on Hilbert spaces DRAFT: April 28, 2008 The setting: complex Hilbert space with inner product, (, ) H. 6.1 Bounded operators Definition 6.1 Bounded linear transformation or bounded linear operator A linear transformation on H is bounded if there is a constant, C 0, such that for all x H, T (x) C x (6.1) The smallest constant, C, for which (6.1) holds, is called the norm of T and is denoted by T, T (x) T = sup = sup T (x) (6.2) x 0 x x =1 We sometime write T L(H,H) to explicitly denote the operator norm of T, which maps H to H, and more generally write, T L(H1,H 2 ) = sup x H1 =1 T (x) H2 when speaking of T as an operator between different spaces H 1 and H 2. Definition 6.2 (Self-adjoint operator) T : H H is self-adjoint if (T x, y) = (x, T y), for all x, y H (6.3) Example 1: H = C n, (x, y) C n = j x jy j. Linear transformation on C n : T (x) = T x, T = (t ij ), n by n matrix. T is self-adjoint if T = (t ij ) = (t ji ) = T. T = T = T = max{ λ : λ σ(t ) }, where σ(t ) denotes the set of eigenvalues, the spectrum, of the matrix T. Example 2: H = L 2 (R n ). Fix τ R n. Define T τ f = f(x + τ). By change of variables T τ f = f. Thus, T τ is unitary on L 2 ; T τ = 1. Example 3: H = L 2 (R n ). Define F[f](ξ) = ˆf(ξ) = e 2πix ξ f(x) dx, the Fourier transform of f; see section The Plancherel Theorem states: F[f] = f. Thus, F is unitary on L 2 ; F = 1. Example 4: H = the set of 1 periodic L 2 functions: L 2 (S 1 ) = {f : f(x + 1) = f(x), x R, Define, for n Z, the Fourier coefficients 1 0 f 2 < } F[f](n) = ˆf(n) = (f, φ n ) = f(z)e 2πinz dz

19 6.2 Compact operators DRAFT: April 28, 2008 Parseval - Plancherel Theorem states f 2 L 2 (S 1 ) = 1 0 f(z) 2 dz = n Z ˆf(n) 2 Thus, the mapping F : L 2 (S 1 ) l 2 (Z), where l 2 (Z) is the set of sequences, which are square summable, is unitary in L 2 ; F L(L 2 (S 1 );l 2 (Z)) = 1. Definition 6.3 An operator T : H H has finite rank if its range is finite dimensional. Example 5: Let φ H, and φ H = 1. Define the projection operator P φ f = (f, φ) H φ An important example is the family of projection operators, P j, defined by P j f(x) = ˆf(j) e 2πijx for any f L 2 (S 1 ). Note that the family of operators {P j } are orthogonal projections. That is, P j P j = P 2 j = P j and P j P k = 0 if j k. Example 6: Let P (N) = span{ e 2πijx : j N }. Let P (N) = j N P j; see the previous example. The operator d P (N) maps P (N) to itself. What is its matrix representation with dx the respect to the basis { e 2πijx : j N }? 6.2 Compact operators This would be a good time to review the background material on compactness and convergence in section Definition 6.4 Let T : H 1 H 2 denote a bounded linear transformation between Hilbert spaces. T is compact if for any bounded sequence, {u k } in H 1, the sequence {T (u k )} has a convergent subsequence in H 2. That is, if for all k, u k M, then there exists a subsequence {T (u kj )} and a v H 2 such that T (u kj ) v H2 0 as j. Example 6.1 Finite rank operators are compact. The proof is an exercise, using the compactness of closed and bounded subsets of C n. The importance of finite rank operators in the general theory is that any compact operator in a separable Hilbert space can be approximated arbitrarily well by finite rank operators. Theorem 6.1 If H is a separable Hilbert space and T is compact, then T is the norm limit of operators of finite rank. That is, there exists T j compact and finite rank such that T j T L(H,H) 0. A very important class of operators is the class of Hilbert Schmidt operators: 19

20 6.2 Compact operators DRAFT: April 28, 2008 Definition 6.5 K(x, y) : S S R is a called a Hilbert-Schmidt kernel if K(x, y) 2 dx dy < S S We use K(x, y) to generate an operator on the Hilbert space L 2 (S). T K [f](x) = K(x, y) f(y) dy (6.4) Theorem 6.2 If K(x, y) is a Hilbert-Schmidt kernel, then T K is a compact operator. S Proof: Let {φ j (x)} j 1 denote an orthonormal basis for L 2 (S). Then, ψ jk (x, y) = φ j (x)φ k (y) is an orthonormal basis for L 2 (S S). Define K N (x, y) to be the partial sum approximation to K(x, y) given by: K N (x, y) = a jk ψ jk (x, y) (6.5) Here, a jk = S j+k N S K(x, y)φ j(x) dx φ k (y)dy is the (j, k) Fourier coefficient of K. Then, T K T KN K K N L 2 (S S) 0, N Definition 6.6 Let T denote a bounded operator on H. The resolvent set of T, denoted ρ(t ), given by ρ(t ) = {z C : (T zi) 1 exists and is a bounded operator on H }. (6.6) The spectrum of T, denoted by σ(t ), is the complement of ρ(t ): σ(t ) = C ρ(t ) Theorem 6.3 (Variant of the Riesz-Schauder Theorem) Let T denote a compact operator on H. (1) σ(t ) is discrete (2) If H is infinite dimensional, then σ(t ) is infinite and consists of eigenvalues accumulating at zero. Thus, 0 σ(t ). (3) The eigenvalues of T have finite multiplicity. If λ 0 is an eigenvalue of T, then dim{x : T x = λx } is finite. The following result generalizes the well-known fact that Hermitian symmetric or self-adjoint n n matrices have a complete set of eigenvectors that span C n. Theorem 6.4 (Hilbert Schmidt Theorem) Let T : H H be a compact and selfadjoint operator. Then, there exists an orthonormal basis for H consisting of eigenvectors of T. That is, there exists {x j } j 1, T x j = λ j x j, (x j, x k ) = δ jk, such that for any x H N x 0 j=1 (x, x j )x j H 20

21 DRAFT: April 28, Applications of compact operator theory to eigenfunction expansions for elliptic operators By Theorem 5.1, the Dirichlet problem for Poisson s equation v = f, x ; u = 0 has a weak H0() 1 solution, i.e. There exists a unique v f H0() 1 such that u v f dx = uf dx, u H0() 1 (7.1) The solution operator T : f v f, L 2 () H0() 1 is a bounded operator. Think of this mapping as ( D ) 1 : f ( D ) 1 f, where D denotes the Dirichlet Laplacian, the Laplace operator acting in the space of functions with zero boundary conditions. We next view T = ( D ) 1 as an operator from L 2 () to itself and can estimate its norm. Setting u = v f in the weak formulation (7.1) we obtain v f 2 dx = v f fdx. By the Cauchy Schwarz inequality, v f 2 dx ( v f 2 dx ) 1 2 ( f 2 dx ) 1 2. By the Poincaré inequality (5.6), ( g L 2 p g L 2, for g H0()), 1 we have p 2 v f 2 L v 2 f 2 dx v f L 2 f L 2 In other words, T is bounded on L 2 and T p 2. We now claim that T is compact and self-adjoint. T f L 2 p 2 f L 2 (); (7.2) Theorem 7.1 The solution operator T : f T f = v f, the solution of the Dirichlet problem for Poisson s equation is compact and self-adjoint operator for L 2 () to L 2 (). Therefore, by the Hilbert-Schmidt Theorem 6.4 there exists an orthonormal set of eigenfunctions of T which is complete in L 2 (). 21

22 DRAFT: April 28, 2008 Proof: To show that T is compact, we must show that if f j is a bounded sequence in L 2 (), then T f j has a convergent subsequence. The key tool here is the Rellich compactness Lemma, Theorem Suppose that f j is a sequence for which there is a constant M such that f j L 2 M. By (7.2) T f j D p 2 f L 2 () p 2 M (7.3) Thus, T f j is a bounded sequence in H0(). 1 By Theorem 12.11, this sequence has an L 2 () convergent subsequence, {T f jk }. Next, we verify self-adjointness of T. By the weak formulation of the Dirichlet problem and the definition of T : T f u dx = f u dx, u H0() 1 (7.4) Setting u = T g, where g L 2 (), we have T f T g dx = f T g dx, (7.5) Similarly, T g T f dx = g T f dx, u H 1 0() (7.6) Therefore, T f g dx = f T g dx. (7.7) That is, (T f, g) L 2 = (f, T g) L 2; T is self-adjoint. The existence of a complete set of eigenfunctions of T is now a consequence of the Hilbert Schmidt Theorem, Theorem 6.4. Corollary 7.1 L 2 () has an orthonormal basis of eigenfunctions of the, satisfying Dirichlet boundary conditions. Proof: Let φ j H 1 0() denote the orthonormal sequence of eigenfunctions of T. By the definition of T (T φ j, u) D = (φ j, u) L 2 (α j φ j, u) D = (φ j, u) L 2 α j (φ j, u) D = (φ j, u) L 2, α j real 1 φ j u dx = φ j u dx α j Therefore, the eigenfunctions of T are weak solutions of φ j = λ j φ j, λ j = 1 α j, φ j H 1 0(). Interior elliptic regularity results ensures that these are, in fact, C (). It can 10 Theorem: ( Rellich compactness Lemma) Suppose {u j } is a sequence in H 1 0 () such that u j H 1 0 () C. Then, there exists a subsequence u jk and an element u H 1 0 such that u jk u L 2 0 as j k. 22

23 DRAFT: April 28, 2008 also be shown, using boundary elliptic regularity results that if is sufficiently smooth, then lim x φ j (x) = Moreover, setting u = φ j above yields positivity of the eigenvalues λ j = 1 = φ j 2 dx. α j FInally, since α j 0 as j, λ j. 11 See L.C. Evans, PDE, section 6.3 for a discussion of interior and boundary elliptic regularity 23

24 DRAFT: April 28, Introduction to Variational Methods In these notes we introduce Variational Methods or The Calculus of Variations via two classical problems: (1) Dirichlet s principle, which characterizes the solution of the Poisson equation as the solution of an optimization (minimization) problem 12. (2) The Rayleigh-Ritz, characterization of the smallest eigenvalue of the Dirichlet problem for the Laplacian as the solution of a minimization problem Variational Problem 1: Dirichlet s Principle We introduce a class of admissible functions, A 2 A {w C (Ū) : w = g on U } (8.1) For w A, define the Dirichlet energy: 1 I[w] = 2 w(x) 2 w(x)f(x)dx U (8.2) Theorem 8.1 (Dirichlet s principle 14 ) (1) If u A and u = f in U, then I[u] = min I[w]. (8.3) w A (2) Conversely, if u A and I[u] = min w A I[w], then u = f and u = g for x U. We have studied, by Hilbert Space - functional analytic methods, the solution of the Dirichlet problem for Poisson s equation, for the case g = 0. The above Theorem gives an alternative characterization of the solution. 8.2 Variational Problem 2: Smallest eigenvalue of Poincaré inequality 15 : Let denote a domain in R n which is open, connected and can be bounded between two planes. There exists a positive constant, p, depending on the domain, such that for any u H0() 1 u 2 dx p u 2 dx. (8.4) 12 Section of L.C. Evans PDEs, pages Section 6.5.1,Theorem 2, We follow an approach to Theorem 2, using the methods of Chapter 8 of L.C. Evans, PDEs. 14 Theorem 17, page 42 of L.C. Evans PDEs 15 We established Poincaré s inequality in our study of weak solutions to Poisson s equation. 24

25 8.2 Variational Problem 2: Smallest eigenvalue of DRAFT: April 28, 2008 We proved, in particular, that if is bounded between two planes a distance 2a apart, then p can be taken to be 2a 2. Question: What is the smallest constant, p for which the Poincaré s inequality holds? To answer this question, we need to compute 1 p inf R[u] 0 u H0 1() = inf u 2 dx 0 u H0 1() u 2 dx (8.5) Remark 8.1 As in Dirichlet s principle, we have an admissible class of functions,, A = H 1 0() in this case, and a functional R[u], defined on A which we seek to minimize. Suppose the minimium in (8.5) is attained a function u H 1 0(). Let η C 0 () be fixed. Then, we have for any ɛ, u + ɛη H 1 0() and R[u + ɛη] R[u ] Thus the function r(ɛ) = R[u + ɛη] is minimized at ɛ = 0. In particular, we must have Equation (8.6) can be rewritten equivalently as u η dx R(u ) Thus, for any η C 0 () d dɛ R[u + ɛη] ɛ=0 = 0. (8.6) uη = 0 ( u η R(u ) uη ) dx = 0 (8.7) In other words, (u, R(u )) is a weak solution of the eigenvalue problem ϕ = λϕ, ϕ = 0, x. (8.8) u is an eigenfunction of with corresponding eigenvalue R(u ). We have previously seen that the eigenvalue problem (8.8) has an infinite sequence of Dirichlet eigenvalues λ j, with λ 0 < λ 1 λ 2..., and corresponding eigenfunctions ϕ j H 1 0() 16. Note that R(ϕ λj ) = λ j and therefore we have λ 1 = R(u ). Borrowing from terminology in quantum physics, λ 1 () is sometimes called the ground state energy and u the ground state eigenfunction. We therefore have the following equivalence 16 This we shown by proving that the mapping ( ) 1 : f u f = ( ) 1 f which maps f L 2 () to the weak H 1 0 () solution of the Dirichlet problem is a compact linear operator from L 2 () to itself. We then applied the general spectral theorem for compact self-adjoint operators in Hilbert space. 25

26 8.3 Direct methods in the calculus of variations DRAFT: April 28, 2008 Theorem 8.2 (a) λ 0 () > 0 is the smallest Dirichlet eigenvalue of the Laplacian,. (b) (c) λ 0 () = min R(u), the minimum of the Rayleigh quotient over u H 1 0(). (λ 0 ()) 1 is the best (smallest) constant for which the Poincaré inequality holds, i.e. for any u H0() 1 u(x) 2 1 dx u(x) 2 dx. (8.9) λ 0 () 8.3 Direct methods in the calculus of variations In both variational problems, we assumed that the minimum of of the Dirichlet energy, (8.3), and the Rayleigh quotient, R(u) in (8.5) are attained at an admissible function. The question of whether this is in fact the case is quite subtle. In both problems we have the following set up: (a) a functional J[u] defined on an admissible class of functions, A (J[u] = I[u] in variational problem 1 and J[u] = R[u] in variational problem 2). (b) J[u] > for u A. Thus, inf u A J[u] = J exists. u j A such that J[u j ] J. In particular, there is a minimizing sequence of functions Exercise 1: Prove that for both Dirichlet s and Rayleigh-Ritz principles that J[u] has a lower bound, when u varies over the appropriate admissible set of functions, A. THE QUESTION: Does the minimizing sequence {u j } converge to some u which is (a) admissible, i.e. u A and (b) a minimizer, i.e. J[u ] = lim j J[u j ] = J? The property that {u j } is a minimizing sequence can often be used to conclude some kind of (c) uniform upper bound on norms of the functions u j. The goal is to deduce from (a), (b) and (c) that (d) the sequence of {u j }, or possibly some subsequence of it, has good convergence properties to a limiting admissible function u A, for which J[u ] = j. Assuming this and if we set aside the issue of whether minimizers are C 2 functions 17, the arguments of the previous section apply to give variational characterizations to solutions of PDEs. For the remainder of these notes, we shall focus primarily on the solution of Variational Problem 2: Lowest eigenvalue of and Poincaré s inequality. 17 This is an important and very technical issue. See the discussion in L.C. Evans PDEs, Chapter

27 8.4 Ideas and general discussion DRAFT: April 28, Ideas and general discussion We now turn to the following questions: (Q1) How do we deduce a uniform bound on norms of the function u j from (a) and (b)?, and (Q2) How do we show that, perhaps by passing to some subsequence of the u js that we can find a sequence which converges to a function u in the admissible set, at which the functional attains its minimum. The approach we take to the existence of solutions of a PDE via the proof that a functional is minimized is called the direct method in the calculus of variations 18. Exercise 2: Concerning (Q1), prove that if u j is a minimizing sequence for either of the variational problems in section 8, then there is a constant C, such that u j H 1 0 () C for all j. Concerning (Q2), we are interested in when bounds on a sequence of functions imply the existence of a convergent subsequence. This is the subject of compactness; see section Compactness and minimizing sequences Exercise 5: Prove that minimizing sequences for the variational problems in section 8 have subsequences which converge weakly in H 1. To show that subsequence of any minimizing sequence for the problems of section 8 converge to an admissible minimizer, we use the following key properties: Property A Weak compactness of minimizing sequences: There is a subsequence {u j } such u j u H 1 0() weakly. Property B (Strong) compactness in L 2 () of minimizing sequences: There is a subsequence for which u j u 2 0. Property C Weak lower semicontinuity of a functional with respect to weak convergence 19 : u 2 dx lim inf u j 2 dx (8.10) j Here s how Properties A, B and C are used to established the minimum is attained. Consider variational problem 2 to minimize R(u). Recall that J = 1/p > 0 denotes the positive infimum. By scaling, we can assume that {u j } is such that u j 2 = 1. Moreover, it is clear since R(u j ) approaches its positive infimum, that u j 2 is bounded. Thus u j is a bounded sequence in H0(). 1 By weak compactness, there s a subsequence which we ll, for simplicity call u j, which converges weakly to some u H0(). 1 Furthermore, Rellich s Lemma implies there is a subsequence, which we ll call once again u j which converges strongly to some u. (Note: the weak limit is unique.). By strong convergence in L 2 we u j 2 u 2 u j u An excellent introductory reference is the book of Gelfand and Fomin: Calculus of Variations 19 This is a special case of the lower semi-continuity of any convex functional with respect to weak convergence. Compare also, with Fatou s Lemma from basic real analysis. 27

28 8.5 Compactness and minimizing sequences DRAFT: April 28, 2008 Thus, u is admissible, i.e. u H 1 0() and u 2 = 1. By weak lower semicontinuity and the fact that {u j } is a minimizing sequence J = lim j R[u j ] = lim inf j R[u j ] R[u ]. But u is admissible, so R[u ] J and the minimum is attained R[u ] = j. What s left is the Justification of Properties A, B and C: Property A follows from the above theorem on weak compactness. Property B is a consequence of the following theorem, which was at the heart of the Hilbert space approach to the Dirichlet problem and eigenfunction expansions. Proof of Property C - weak lower semicontinuity: To prove (8.10) we begin with the following observation: u k 2 u 2 = (u k u ) u (u k u ), (8.11) which is proved by simply expanding (u k u ) (u k u ). Dropping the positive term on the right hand side gives u k 2 u 2 2 u (u k u ), (8.12) Letting k tend to infinity and using that u k tends weakly to u in H 1 we see that the term on the right hand side approaches zero as k. This implies u 2 lim inf u k 2. k 28

29 9 The Heat / Diffusion Equation DRAFT: April 28, 2008 Initial value problem t u = u u(x, 0) = f(x) Solution by Fourier transform t û(ξ, t) = 4π 2 ξ 2 û(ξ, t), û(ξ, 0) = ˆf(ξ) û(ξ, t) = e 4π2 ξ 2 t ˆf(ξ) u(x, t) = e 2πiξ x e 4π2 ξ 2 t ˆf(ξ) dξ = f(y) dy e 2πiξ (x y) e 4π2 ξ 2t dξ = 1 K t (x y) f(y) dy, K t (z) = (4πt) n 2 e z2 /4t Theorem 9.1 Let f be bounded and continuous on R n. For t > 0, define u(x, t) = K t (x y) f(y) dy e t f. (9.1) R n Then, (a) u C for any t > 0, (b) u(x, t) satisfies the heat equation, and (c) lim t 0 u(x, t) = f(x). K t (x) is called the fundamental solution for the operator L = t, i.e. ( t ) K t (x) = 0, t > 0 lim t 0 K t (x) = δ 0, where δ 0 denotes the Dirac delta distribution. Proof: The explicit construction guarantees that (a) and (b) hold. To prove part (c) we note from the following properties of K t (x). (K1) K t (x) > 0, t > 0 (K2) Kt (x) dx = 1, t > 0. (K3) Let δ > 0 be fixed. Then, lim 0<t 0 x y δ K t (y) dy = 0, uniformly in x. 29

30 9.1 Remarks on solutions to the heat / diffusion equation DRAFT: April 28, 2008 We need to show u(x, t) f(ξ) 0 as (x, t) (ξ, 0). u(x, t) f(ξ) = K t (x y) ( f(y) f(ξ) ) dy R n = K t (x y) f(y) f(ξ) dy, (K t > 0) R n = K t (x y) f(y) f(ξ) dy + K t (x y) f(y) f(ξ) dy x y δ = I 1 + I 2 x y >δ Here, we have used property (K1). Estimation of I 1 : Here, x y δ: Let ε > 0 be arbitrary. Using the continuity of f, we can choose δ = δ(ε) so that if y ξ < 2δ then f(y) f(ξ) < ε. By taking x ξ < δ, we have y ξ y x + x ξ 2δ. Thus, I 1 ε K t (x y) dy K t (x y) dy = ε, x y δ R n where we have used property (K2). Estimation of I 2 : Here, x y δ. I 2 2 f L x y δ(ε) K t (x y) dy, (9.2) which tends to zero, as t 0 uniformly in x, by property (K3). Thus we have shown u(x, t) f(ξ) ε + o(1) as t 0, where ε is arbitrary. This proves the theorem. 9.1 Remarks on solutions to the heat / diffusion equation Let u(x, t) = e t f, the solution of the heat equation with initial data f(x). (1) Instantaneous smoothing and infinite propagation speed f 0, bounded and of compact support = for all t > 0, e t f C, e t f > 0 for all x. Contrast this with the transport (wave) equation ( t + c x )u(x, t) = 0 with initial data u(x, 0) = f(x). The solution,u(x, t) = f(x ct) is clearly no smoother than the initial data. (2) f 0, u(x, 0) = f L 1 = u(x, t)dx = u(x, 0)dx, t > 0. (3) e t f L 2 f L 2, t 0. More precisely, d u(x, t) 2 dx = 2 dt u(x, t) 2 dx (9.3) (4) u(x, t) C t n 2 f L 1, t > 0. 30

31 9.2 Inhomogeneous Heat Equation - DuHamel s Principle DRAFT: April 28, 2008 (5) inf f e t f sup f, t > 0. (6) The basic properies of e t can be proved for e Lt, where L is a strictly elliptic divergence form operator: L = a jk (x), (9.4) x j x k j,k where j,k a jk(x)ξ j ξ k θ ξ 2, where θ > 0 is independent of x and ξ R n is arbitrary. Theorem 9.2 ( Nash (1958) and Aronson (1967) ) e Lt f = Kt L (x, y) f(y) dy and there exist constants a, a, b, b > 0 depending on θ and such that for all x, y, R n, Kt L satisfies Gaussian upper and lower bounds: a t n x y 2 2 exp( b ) Kt L (x, y) a t n x y 2 2 exp( b ) t t 9.2 Inhomogeneous Heat Equation - DuHamel s Principle Consider the scalar ODE du = au + f(t), u(0) = u 0 dt where a denotes a constant. Multiplying by the integrating factor e at, the resulting equation can be rewritten as d dt (e at u(t)) = e at f(t) which we can now integrate and conclude: u(t) = e at u 0 + t 0 e a(t s) f(s) ds The formula generalizes immediately to the case u(t) is the solution of the system of ordinary differential equations: du dt = Au + f(t), u(0) = u 0 where A is a constant n n matrix and f(t) is a n 1 vector function of t: u(t) = e At u 0 + t 0 e A(t s) f(s) ds (9.5) Under suitable hypotheses this can be generalized to the case where A is an operator acting on a Hilbert space and f(t) takes values in a Hilbert space A. Pazy, Semigroups of linear operators and application to partial differential equations, Springer

32 9.3 Heat equation on a bounded domain, DRAFT: April 28, 2008 Equation (9.5) is often called DuHamel s formula or DuHamel s principle. Applying it with A =, we obtain that the solution to the inhomogeneous heat equation: t u(t, x) = u(t, x) + f(t, x), u(0, x) = u 0 (x) is given by u(t, x) = K t u 0 + t where e t g = K t g and K t denotes the heat-kernel 0 K t s f(s) ds, (9.6) K t (z) = 1 (4πt) n 2 e z 2 4t (9.7) 9.3 Heat equation on a bounded domain, t u = u, u(t = 0, x) = f(x) (t, x) (0, ) u(t, x) = 0, x, t > 0 (9.8) Solution: From section 7, we have that L 2 () is spanned by the complete orthonormal set of eigenfunctions of, F j (x), j = 0, 1, 2,... with eigenvalues λ j > 0: F j (x) = λ j F j (x), x F j (x) = 0, x F j (x) F k (x) dx = δ jk We write the solution u(x, t) as a weighted sum of solutions of the form e λ jt F j (x): u(x, t) = a j F j (x) e λ jt (9.9) j=0 The formal sum (9.9), by construction, solves the heat equation and satisfies the Dirichlet boundary condition. To complete the solution, it suffices to choose the constants a j so that the initial condition u(0, x) = f(x) is satisfied: f(x) = u(0, x) = a j F j (x) j=0 Since the eigenfunctions F j form a complete orthonormal set we have a j = (f, F j ) L 2 (9.10) Therefore, u(x, t) = (f, F j ) L 2 F j (x) e λ jt (9.11) j=0 32