Numerical Functional Analysis

Transcription

1 Numerical Functional Analysis T. Betcke December 17, 28 These notes are in development and only for internal use at the University of Mancheser. For questions, comments or typos please The following books were used for the preparation of these lecture notes. C. Brenner and L.R. Scott, The mathematical theory of finite element methods, Springer C. Gasquet and P. Witomski, Fourier Analysis and Applications, Springer D. Werner, Funktionalanalysis, Springer R. Kress, Linear Integral Equations, Second Edition, Springer E. Kreyszig, Introductory functional analysis with applications, John Wiley & Sons Inc. E.H. Lieb and M. Loss, Analysis, American Mathematical Society 1

2 Contents 1 Introduction Well Posedness Discretization of operators Banach Spaces Normed vector spaces Complete Spaces Hilbert Spaces 5 4 Lebesgue Spaces 6 5 Linear Operators on Banach and Hilbert spaces 8 6 Bases of Hilbert spaces 11 7 A model diffusion problem Weak vs. strong solutions The Poisson equation in 2d Ritz-Galerkin Approximations and the Finite Element Method Finite Dimensional Approximations Approximation bases The polynomial basis Piecewise polynomial bases Spectral Galerkin Methods A Helmholtz problem Compact Operators 34 1 Spectral theory of compact operators Eigenvalues and eigenfunctions associated with the model diffusion problem The Dirac delta distribution The Green s function of the model diffusion problem Obtaining the spectrum for the operator of the model diffusion problem Linear Integral equations Definition and Existence of Solutions for equations with continuous kernel A collocation method for solving integral equations Integral equations in Hilbert spaces

3 12.4 A remark on the numerical treatment of first order and second order Fredholm integral equations Introduction 1.1 Well Posedness Mathematical Models such as PDEs should be well posed, that is: They have a solution. The solution is unique. The solution depends continously on the input paramters. This is also called well-posed in the sense of Hadamard. Not all physically relevant problems fit in this framework. One area dealing with this is the field of Inverse Problems. Example We know that Ax = b for A C n n has a unique solution if and only if det(a). In that case the continous dependence of the solution x on the input parameter b follows easily. Let Ax 1 = b 1 and Ax 2 = b 2. Then A(x 1 x 2 ) = b 1 b 2 x 1 x 2 A 1 b 1 b 2. Since A 1 exists if and only if det(a) the continuous dependence of x on b follows. We will encounter similar situations in Numerical Functional Analysis. Then x and b are functions, A is a linear operator (e.g. a partial differential operator or an integral operator) and denotes a suitable function space norm. 1.2 Discretization of operators Integral equations or partial differential equations are usually solved by a discretization process to obtain a finite dimensional problem. Example Solve Lu := u = 1 in [, 1] with u() = u(1) = 1. Let x k = k N, k =,..., N and h := 1/N. Define u k := u(x k ). Then u (x k ) 1 h 2 (u k 1 2u k + u k+1 ), k = 1,..., N 1. This leads to the linear system of equations A h x h = (1) h, where 2 1 A h = u 1 1 h R N 1 N 1, x h =., (1) h =. R N u N This problem and its discretization lead to several questions: 3

4 Is the original problem well-posed? Is the discrete problem well-posed? Does u h u as h and in what sense do we have to understand the convergence? In this module we will get to know several tools from functional analysis that will help us answer these questions. 2 Banach Spaces 2.1 Normed vector spaces Let X be a vector space. We require a way of measuring the size of elements of X and their distances. Definition (Norm) A norm on a (real or complex) vector space X is a realvalued function on X whsoe value at an x X is denoted by x and which has the following properties: (N1) x (N2) x = x = (N3) αx = α x (N4) x + y x + y (Triangle inequality) Remark A norm on X defines a distance (metric) between elements of X by d(x, y) := x y for x, y X. Example The Euclidian distance x 2 := R d. Also the following are norms on R d. x = max x j (sup-norm), x p := j ( d x j 2 ) 1/2 defines a norm on d x j p 1/p (p-norm), p 1. Let X = C([, 1], R) be the space of real continuous functions in [, 1]. Then f := sup x [,1] f(x) defines a norm. 2.2 Complete Spaces Definition (Cauchy Sequence) Consider a normed vector space (X, ). A sequence x n X is a Cauchy sequence if for all ɛ > N such that x n x m < ɛ for all n, m > N. 4

5 Definition (Convergent sequence) A sequence (x n ) in a normed space X = (X, ) is said to converge or to be convergent if there exists x X such that lim n x n x =. Theorem 2.1 Every convergent sequence is a Cauchy sequence. Proof Exercise. In the following we are interested in the converse questions. sequences have a limit in X, that is are they convergent in X? Do Cauchy Example Consider C 1 ([ 1, 1], R); the space of all continuously differentiable functions in [ 1, 1] and let f := sup t [ 1,1] f(x). Define x n (t) := (t n )1/2. We have x n (t) t uniformly in [ 1; 1]. But t is not differentiable at. Hence, the Cauchy sequence (x n ) is not convergent in the sup-norm in X. Definition (Complete Space) The normed space X = (X, ) is called complete if every Cauchy sequence in X converges with a limit in X. Definition (Banach Space) Let X = (X, ) be a normed vector space. If X is complete it is called a Banach space. Example Define l p (p 1) as the space of all infinite sequences of complex ( ) 1/p numbers x = (ξ 1, ξ 2,... ) such that for the norm x p := ξ j p we have x p < (for p = define x := sup j ξ j ). Then l p is a Banach space. Example The space C r [a; b] of r-times continuously differentiable functions in [a, b] with norm x = r j= x(j) is a Banach space. 3 Hilbert Spaces Definition (Inner Product) Let X be a vector space over K, where K denotes either C or R. A function, : X X K is called inner product if (P1) x 1 + x 2, y = x 1, y + x 2, y (P2) λx, y = λ x, y for λ K. (P3) x, y = y, x (P4) x, x x X (P5) x, x = x = Theorem 3.1 (Cauchy-Schwarz inequality) x, y x y for x, y X = (X,, ) and x := x, x. 5

6 Proof Let λ K. Assume x, y (otherwise the proof is trivial). We have Choose λ = x,y y,y x λy 2 = x, x λ x, y λ y, x + λ 2 y, y. to obtain x, x x, y 2 y, y. This is equivalent to x, y 2 x, x y, y from which the proof follows. Theorem 3.2 Let X = (X,, ) be an inner product space. Then x := x, x defines a norm on X. Proof Exercise. Definition (Hilbert Space) Let X = (X,, ) be an inner product space and let x := x, x be the induced norm. X is called a Hilbert space if X is complete with respect to the norm. Example C d with the Euclidian inner product x, y = d x iy i is a Hilbert space. Also l 2 with inner product x, y = x iy i is a Hilbert space. 4 Lebesgue Spaces Definition (Lebesgue Spaces) L p (Ω) (p 1) is the normed space of all functions f : Ω K, where K is either C or R such that For p = define ( 1/p f LP (Ω) := f(x) dx) p <, 1 p < Ω f L (Ω) := inf{k : f(x) K for almost every x Ω}. Remark Let f = f a.e. Then f f LP (Ω) = even though f and f might not be identical. One avoids this technical problem formally by introducing equivalence classes of functions that are identical a.e. Hence, two functions f, f L P (Ω) are identified if they agree a.e. With this assumption L P (Ω) can be shown to be a norm for all 1 p. Theorem 4.1 The spaces L P (Ω) (1 p ) are complete and therefore Banach spaces. Proof See for example [5, Theorem 2.7] Theorem 4.2 L 2 (Ω) with the inner product f, g := f(x)g(x)dx defines a Ω Hilbert space. 6

7 Proof Exercise. The following Lemma introduces the concept of dual indices and is fundamental for the subsequent theorem. Lemma 4.3 (Young s inequality) Let a, b, p > 1 and 1/p + 1/q = 1. Then Proof Exercise. ab ap p + bq q. In the following we call two indices p and q dual if if 1/p + 1/q = 1 (1 p ). Theorem 4.4 (Hölder s inequality) Assume that f L p (Ω) and g L q (Ω), where 1/p + 1/q = 1. Then fg L 1 (Ω), and fg L1 (Ω) f Lp (Ω) g Lq (Ω) Proof p = 1 and p = are trivial. For 1 < p < apply Young s inequality with and integrate on both sides. a = αf, b = g, α = g q/p L q (Ω) f p The following theorem describes inclusion of L p spaces. Theorem 4.5 Assume that Ω K d is bounded (K is either R or C). Then 1. L (Ω) L P (Ω), p L q (Ω) L p (Ω), q p There exists a constant c = c(p, q) such that h p c h q for all h L q (Ω), 1 p q Proof 1. Let f L (Ω). Then Ω f(x) p dx Ω 1dx f p L (Ω) result follows. and the 2. Let p < q and f L q (Ω) Let S := {x Ω : f(x) 1}. For x S, f(x) p f(x) q and therefore f(x) p dx f(x) q dx+ f(x) p dx f(x) q dx+ 1dx <. Ω S Ω\S Ω Ω\S Hence, f Lp (Ω) <, from which the proof follows. 7

8 3. Let h L q (Ω). Apply Hölder s inequality with Then, 1/r + 1/s = 1 and f(x) = h(x) p, g(x) = 1, r = q p, s = r r 1. Ω from which the result follows. ( 1/r h(x) p dx f(x) dx) r ( Ω Ω 1dx) 1/s, 5 Linear Operators on Banach and Hilbert spaces Let T : X Y be a linear operator between normed spaces X and Y. Definition (Boundedness) We say that T is bounded if there exists c > such that T x c x for all x X. Denote by B(X, Y ) the space of bounded linear operators between X and Y. One can define a norm on B(X, Y ) by defining T := sup T x x X, x =1 It is easy to show that T indeed satisfies the norm axioms. It follows immediately that T x T x. Theorem 5.1 If Y is a Banach space, then B(X, Y ) is a Banach space. Proof See [4, Theorem 2.1-1] Example Define T : C[; 1] C[; 1] by y = T x, where y(t) = 1 k(t, τ)x(τ)dτ. Theorem 5.2 Assume that C[, 1] is equipped with the usual sup-norm. Then T is bounded if k is continuous on the unit square [, 1] [, 1]. Proof Exercise. The following theorem gives an important connection between boundedness and continuity of linear operators. Theorem 5.3 Let T : X Y be a linear operator between the normed spaces X and Y. The following holds. 1. T is bounded if and only if it is continuous. 8

9 2. If T is continuous at x = it is continuous everywhere. Proof 1. Assume that T is bounded and let x X. Then from T x T x T x x for every x X it follows that T is Lipschitz continuous. Now assume that T is continuous. Then for every ɛ > there exists δ > such that Now let x X and choose x = x + T x T x ɛ for x x δ. T x T x = δ x x. Hence, x x = δ and δ T x ɛ. x Since this holds for any x X we obtain that T is bounded with T x ɛ x, x X. δ 2. Choose x = in the proof of 1. to see that T is bounded. Continuity everywhere then follows from boundedness of T. An important special case of linear operators are linear functionals. These are linear operators whose range lies in R. Definition (Dual Space) Let X be a normed space. Then the space X of all bounded linear functionals on X is called the dual space of X. Theorem 5.4 The dual space X of a normed space X is a Banach space. Proof Since R is a complete space this is an immediate consequence from Theorem 5.1. The following theorem gives a description of the dual of L p spaces. Theorem 5.5 Let 1 p <, 1/p + 1/q = 1 and Ω K d. The dual of L p (Ω) is L q (Ω) in the sense that every L L p (Ω) has the form Lg = v(x)g(x)dx for some unique v L q (Ω) and L = v. Proof The proof of this result can be found in [5, Theorem 2.14]. Ω Remark This result is also known as Riesz representation theorem. We will mostly use this theorem in the special case p = q = 2, which we will discuss below. The dual space of L is harder to describe. We will not do this here. 9

10 For L 2 spaces the following theorem is a consequence of the previous theorem. For general Hilbert spaces a direct prove is possible but not given here. Theorem 5.6 (Riesz Representation Theorem for Hilbert Spaces) Let H be a Hilbert space. We have H = H in the sense that every bounded linear functional f H can be represented as where z is unique and f = z. f(x) = x, z, x H, A consequence is a representation of sesquilinear forms on Hilbert spaces. Sesquilinear forms will play an important role in the analysis of PDEs on Hilbert spaces and the design of finite element methods. Definition (Sesquilinear Form) Let X and Y be vector spaces over K. Then a sesquilinear form h on X Y is a mapping such that h : X Y K (S1) h(x 1 + x 2, y) = h(x 1, y) + h(x 2, y) (S2) h(λx, y) = λh(x, y), λ K (S3) h(x, y) = h(y, x) If there exists c > such that for all x, y h(x, y) c x y, then h is said to be bounded and the number h = sup x =1 y =1 h(x, y) Theorem 5.7 Let H 1, H 2 be Hilbert spaces and h : H 1 H 2 K a bounded sesquilinear form. Then h has a representation h(x, y) = Sx, y, where S : H 1 H 2 is a bounded linear operator. S is uniquely determined by h and has norm S = h. Proof We have to show that 1. There exists a linear operator S such that h(x, y) = Sx, y. 2. S is bounded with S = h. 3. S is unique. The details are left as exercise. 1

11 6 Bases of Hilbert spaces In the following we will deal with Hilbert spaces that are separable. We are not going into details about the theory of separable spaces. For our purpose we will only need that separable Hilbert spaces have a countable basis. Let H be a Hilbert space. We call two elements x, y H orthogonal if x, y =. If also x = y = 1 then x and y are called orthonormal. A set M H is called orthogonal if all elements of M are pairwise orthogonal. If in addition all elements of M are of unit norm then the set is called orthonormal. Definition Let H be a Hilbert space. An orthonormal set S H is called an orthonormal basis if H = spans. Remark spans denotes the closure of the set of all linear combinations of elements of S. Example Let H = l 2. An orthonormal basis for H is given by the set S := {e 1, e 2,... }, where e j is the sequence consisting of 1 at position j and everywhere else. Let S := {e 1, e 2,... } be an orthonormal basis of a Hilbert space H. Then for every x H x = e j x, e j. The coefficients x, e j are called Fourier coefficients. By taking norms and using orthonormality one obtains e j x, e j 2 = x 2. This is called Parseval identity. Remark If a Hilbert space H is separable and S is an orthonormal basis of H one can define an isomorphism between H and l 2 by going over from elements in H to the coefficients of the basis vectors. The following result describes best approximations in Hilbert spaces. Theorem 6.1 Let H be a Hilbert space with orthonormal basis S := {e j }. Let x H. Then the best approximation x n to x from the set S n := {e 1,..., e n } is given by x n = n x, e j e j in the sense that minimizes x y over all y S n. Proof Let x = c je j the expansion of x in terms of the basis {e j } with c j = x, e j and let y = n a je j S n. We want to choose the a j such that x y is minimized. A short calculation shows that n n x y 2 = x 2 + a j 2 c j a j a j c j = x 2 + a j c j 2 c j 2. 11

12 This is minimzed by choosing a j = c j = x, e j. Let us now consider the space L 2 [, 2π] of square-integrable functions in the interval [, 2π]. A particularly important basis are the functions e k (t) = e ikt for t [, 2π], k Z. We state the following important result without proof. Theorem 6.2 The functions e k (t) := e ikt, k Z form an orthogonal basis in L 2 [, 2π]. It follows that every function f L 2 [, 2π] can be decomposed as f(t) = k= c k e ikt, (1) where convergence is to be understood in the L 2 norm. The coefficients c k are obtained by 2πc k = f, e ik = 2π fe ikt dt. The factor 2π accommodates for the fact that e ik, e ik = 2π. The series (1) is called Fourier series. It is an important tool in a vast amount of applications with a wealth of literature. If the function is smooth enough then one can establish convergence even in a pointwise sense. However, we will not go into this fascinating topic here and refer to the literature. The importance of the basis functions e ikt lie in the fact that they are eigenfunctions of differentiable operators, that is d dt eikt = ike ikt. Hence, the corresponding eigenvalues are (ik) and the differentiation of a Fourier series is identical to a multiplication with ik. Example Find u = u(r, θ) such that u = in the unit disk and u = f on the boundary. The solution is given by u(r, θ) = k= where the c k are the Fourier coefficients of f. c k r k e ikθ, Fourier series are suitable to obtain the components of functions in frequency space at integer frequencies. However, we would like to go over to arbitrary frequencies. This is done by the Fourier transform. Let x, ξ R n. We use the notation 1/2 n n n xξ = x j ξ j, x 2 = x 2 j, x =. x 2 j 12

13 Definition Let f L 1 (R n ). Define (Ff)(ξ) = f(x)e 2πixξ dx, ξ R n. R n Since for f L 1 (R n ) f(x)e 2πixξ dx f(x) dx < R n R n the Fourier transform is well defined. We will from now on use the standard notation ˆf := (Ff) for the Fourier transform of f. The Fourier transform ˆf of f L 1 (R n ) can be shown to be a continuous function. Furthermore, one obtains easily ˆf L (R n ), ˆf f 1 For nonnegative f we have ˆf() = f 1 and therefore even ˆf = f 1. By (f g)(x) = f(x y)g(y)dy R n we denote the convolution of two functions f and g. By Hölder s inequality the convolution is well defined if f L p (R n ), g L q (R n ) with 1/p + 1/q = 1. However, it also turns out to be well defined if p=q=1. The Fourier transform of convolutions is particularly simple. (f g)(ξ) = e 2πixξ f(x y)g(y)dydx R n R n = e 2πiyξ g(y) e 2πiξ(x y) f(x y)dxdy R n R n = ˆf(ξ)ĝ(ξ) (2) The following theorem gives the extension of the Fourier transform to functions in L 2 (R n ). Theorem 6.3 (Plancherel s theorem) If f L 1 (R n ) L 2 (R n ) then ˆf is in L 2 (R n ) and the following formula of Plancherel holds. ˆf 2 = f 2. (3) The map f ˆf has a unique extension to a continuous, linear map from L 2 (R n ) which is an isometry, that is (3) holds for this extension. Proof see [5, Theorem 5.3] For the following corollary we need the polarization identity f, g = 1 2 { f + g 2 2 i if + g 2 2 (1 i) f 2 2 (1 i) g 2 2}. (4) 13

14 Corollary 6.4 Let f, g L 2 (R n ). Then f, g = ˆf, ĝ Proof The result follows from (3) and the polarization identity. For the following results we need the notion of adjoints of operators. Definition Let T : H 1 H 2 be a linear operator between the Hilbert spaces H 1 and H 2. Then the adjoint operator T : H 2 H 1 is defined by T x, y = x, T y for all x H 1, y H 2. Theorem 6.5 The adjoint F of F is given by (F f)(ξ) = f(x)e 2πixξ dx R n Proof Let f, g L 2 (R n ). Then Ff, g = f(x)e 2πixξ dxg(ξ)dξ = f(x) g(ξ)e 2πixξ dξdx = f, F g. R n R n R n R n We can now proof the following inversion theorem. Theorem 6.6 Let f L 2 (R n ). Then Hence, F 1 = F. f = F Ff. Proof From Corollary 6.4 and Theorem 6.5 it follows that for all f, g L 2 (Ω). Hence, F = F 1. f, g = Ff, Fg = F Ff, g In the following we will define the notion of generalized derivatives. This will allow us to define subspaces of L 2 for which differentiation makes sense in a generalized way. Furthermore, these spaces will have a very simple characterization in terms of the Fourier transform. Definition (Space of test functions) Let Ω R n be an open nonempty set. The space of test functions D(Ω) is defined as the space of all infinitely differentiable functions in Ω who are compactly supported in Ω. We now introduce the notion of a multiindex to have a simple way of writing down derivatives in D(Ω). Let Ω R n and φ r-times continuously differentiable in Ω. Define D α φ := α1 αn x α1 1 φ, xαn n where α := (α 1,..., α n ) with α j and α := α 1 + α α n r. α is called a multiindex. Furthermore, for x R n let x α := x α xαn n. 14

15 We can now motivate weak derivates. Let φ D(Ω) and f L 2 (Ω). Then we can formally write D α f, φ = D α f(x)φ(x)dx = ( 1) α f(x)d α φ(x)dx. Ω If f is sufficiently often differentiable this is valid by partial integration. Otherwise, we use it to define the notion of a weak derivative. Definition Let Ω R n be open, α a multiindex and f L 2 (Ω). The function g L 2 (Ω) is called weak derivative of f if Ω g, φ = ( 1) α f, D α φ, φ D(Ω). Example Let Ω = ( 1, 1) and f(x) = x, x Ω. The weak derivative of f is given by { 1 x g(x) = 1 x > This follows by partial integration from 1 1 g, φ = g(x)φ(x)dx = x φ (x)dx = f, φ, φ D(Ω) 1 1 We are now ready to define Sobolev spaces. These will turn out to be the right spaces for the treatment of a large class of partial differential equations. Definition (Sobolev spaces) Let Ω R n be open. The Sobolev space H m (Ω) is defined as H m (Ω) = {f L 2 (Ω) : D α f L 2 (Ω) α m} with associated inner product f, g H m = D α f, D α g L 2 α m Theorem 6.7 H m (Ω) is a Hilbert space. Proof The result follows easily from the definition of weak derivatives and the completeness of L 2 spaces. Theorem 6.8 Let f H m (R n ). Then for all α m. F(D α f) = (2πi) α ξ α Ff (5) Proof A formal verification of (5) is given as exercise. The full proof requires an additional closure argument. We conclude this section with the following important theorem that states under which conditions functions in Sobolev spaces can be represented by continuously differentiable functions. 15

16 Theorem 6.9 Let Ω R n be an open set and let m, k N with m > k + n/2. If f H m (Ω) there exists a k times continuously differentiable function f such that f = f a.e. Hence, we can represent f by a function in C k (Ω). Example Let n = 1. Then it follows that every function in H 2 (R) has a continuously differentiable representation. 7 A model diffusion problem All spaces in this section are assumed to be spaces over the real numbers R. In this section we will start using our previous tools to analyze certain partial differential equations. We start with the model diffusion problem u = f in (, 1), u() = u(1) = (6) for a given function f L 2 (, 1). In the following we will reformulate this problem as a variational problem in a Sobolev space. Let v be a sufficiently smooth test function satisfying v() = v(1) =. We can multiply (6) on both sides with v and apply integration by parts to obtain 1 u (x)v(x)dx = 1 u (x)v (x)dx = 1 f(x)v(x)dx. (7) Define a(u, v) := 1 u (x)v (x)dx and let (f, v) := 1 f(x)v(x)dx be the standard L 2 inner product. We haven t yet fixed from what function space we choose v and in what space we look for u. Since the first derivative of v and u appears under an integral sign and there is no second derivative it is suggestive to use the space H 1 (, 1) := {v H 1 (, 1) : v() = v(1) = }. We obtain the variational problem Find u H 1 (, 1) such that a(u, v) = (f, v) v H 1 (, 1). (8) This is called the weak formulation of (6). If u is a solution of (6) then u H 1 (, 1) and by integration by parts it follows that u satisfies (8). We will now show that solutions of (8) are unique. Assume that u 1 and u 2 are both solutions of (8). Then a(u 1, v) = (f, v), a(u 2, v) = (f, v) v H 1 (, 1). Hence, a(u 1 u 2, v) = for all v H 1 (, 1). The difficulty now is to conclude that this implies u 1 u 2 =. The needed result is provided by the following lemma Lemma 7.1 (Poincaré-Friedrich) If w H 1 (, 1) then (w, w) (Dw, Dw). 16

17 Proof If w H 1 (, 1) then w(x) = w() + Since w() = it follows that w 2 = x 2 Dw(ξ)dξ ( x ) ( x 1 2 dx ( 1 ) ( dx x ) (Dw(ξ)) 2 dξ Dw(ξ)dξ. (Cauchy-Schwarz) ) ( 1 ) (Dw(ξ)) 2 dξ = (Dw(ξ)) 2 dξ. Hence, w 2 (x) (Dw, Dw) for x (, 1). Integrating on both sides then leads to (w, w) = 1 w 2 (x)dx 1 (Dw, Dw)dx = (Dw, Dw). From the Poincaré-Friedrichs inequality it now follows that a(u 1 u 2, v) = v H 1 (, 1) u 1 = u 2 since for the special choice v = u 1 u 2 we have (u 1 u 2, u 1 u 2 ) a(u 1 u 2, u 1 u 2 ) = and therefore u 1 u 2 L 2 (,1) =. We have now shown that if the original diffusion problem (6) has a solution u then this is the unique solution of the weak formulation (8). However, the converse needs not be true. If u is a solution of the weak formulation (8) then it needs not necessarily be a twice continuously differentiable function. Hence, a larger class of problems is covered by the weak formulation. It is now left to proof existence of a solution of (8). We need the following two theorems. Theorem 7.2 Let a(u, v) := 1 DwDvdx for u, v H1 (, 1). Then there exists C, α > such that and α u 2 H 1 (,1) a(u, u) u a(u, v) C u H 1 (,1) v H 1 (,1) H1 (, 1) (Coercivity) u, v H 1 (, 1) (Continuity) Proof Coercivity follows from the Poincaré-Friedrichs inequality by u 2 H 1(,1) = u 2 L 2 (,1) + Du 2 L 2 (,1) 2 Du 2 L 2 (,1) = 2a(u, u). Continuity follows from 1 a(u, v) = DuDv Du L 2 (,1) Dv L2 (,1) u H1 (,1) v H1 (,1). 17

18 From the coercivity of a(u, v) it follows that a(v, v) for all v H 1 (, 1) and a(v, v) = if and only if v =. Hence a(u, v) defines an inner product on H 1 (, 1) (the axioms (P1) to (P5) are satisfied). Together with the continuity of a(u, v) it follows that H 1 (, 1) equipped with the inner product a(u, v) is even a Hilbert space as the following theorem shows. Theorem 7.3 Let H be a Hilbertspace and a(, ) a symmetric linear form that is coercive and continuous on a closed subspace V of H. Then (V, a(, )) is a Hilbert space. Proof From the coercivity it follows that a(, ) is an inner product on V. It is left to show that V is complete in the norm u E := a(u, u). Let {u n } be a Cauchy sequence in (V, E ) Then by coercivity {u n } is also a Cauchy sequence in H with its associated norm. Since V is closed in H there exists u V such that lim n u u n =. From the continuity of a(, ) it follows that u u n E C u u n for some C >. Hence, u n u in the norm E showing the completeness of (V, E ). We have now everything in place to prove the existence of a solution of the variational problem (8). Theorem 7.4 Let H be a Hilbert space and V a closed subspace of H. Let a(, ) : V V R be a coercive and continuous symmetric linear form on V and l : V R a bounded linear functional on V. Then the variational problem Find u V such that a(u, v) = l(v) v V has a unique solution. Proof From Theorem 7.3 it follows that a(, ) is an inner product in V. The result follows now by an application of the Riesz representation theorem. From this theorem it now follows that there exists a unique solution of (8). Let us summarize the previous steps. Suppose the following holds: (H,, ) is a Hilbert space V is a closed subspace of H a(, ) is a bounded symmetric linear form that is coercive on V. Under these conditions the following variational problem has a unique solution. Remark Note, we have implicitly assumed that we are working in spaces over the real numbers. For Hilbert spaces over C we need to require instead of symmetry that a(u, v) = a(v, u). Then the previous proofs carry also through for the complex case. 18

19 Given F V, find u V such that a(u, v) = F (v) for all v V (9) An important condition for the previous results was that a(u, v) is symmetric. Hence, a(u, v) = a(v, u) for all u, v V. Otherwise, a(, ) does not define an inner product. In order to treat the nonsymmetric case we need the following lemma. Lemma 7.5 Given a Banach space V and a mapping T : V V satisfying T v 1 T v 2 M v 1 v 2 for all v 1, v 2 V and fixed M, M < 1, there exists a unique u V such that u = T u Proof see [1, Lemma 2.7.2] We are now ready to prove the following famous theorem. Theorem 7.6 (Lax-Milgram) Given a Hilbert space (V, (, ), a continuous, coercive bilinear form a(, ) and a continuous linear functional F V, there exists a unique u V such that a(u, v) = F (v) v V. Proof For any u V define the linear functional l u by l u (v) = a(u, v) for all v V. l u is linear. Continuity of l u follows from the continuity of a(, ) by l u (v) = a(u, v) C u v. Hence, l u C u < and therefore l u V. Furthermore, it follows that the map l : V V defined by u l u is continuous with l C. We are finished if we can find u such that l u (v) = F (v) for all v V. From the Riesz representation theorem it follows that for every u V there exists a bijective linear map τ : V V with τ = 1 such that l u ( ) =, τ(l u ). Furthermore, there exists z V such that F ( ) =, z. Hence, we need to find u such that τ(l u ) = z since then a(u, v) = l u (v) = v, τ(l u ) = v, z = F (v) v V. In order to solve τ(l u ) = z we define the map T v := v ρ(τ(l v ) z) for some ρ. If T is a contraction map then from Lemma 7.5 it follows that there exists a unique u with T u = u and hence τ(l u ) z =. We therefore 19

20 have to show that there exists ρ such that T is a contraction map. Let v = v 1 v 2. Then T v 1 T v 2 2 = v 1 v 2 ρ(τ(l v1 ) τ(l v2 )) 2 = v ρτ(l v ) 2 = v 2 2ρ τ(l v ), v + ρ 2 τ(l v ), τ(l v ) = v 2 ρa(v, v) + ρ 2 a(v, τ(l v )) = v 2 2ρα v 2 + ρ 2 C v τ(l v ) (Coercivity and continuity) (1 2ρα + ρ 2 C 2 ) v 2 := M v 1 v 2 2 with M = 1 2ρα + ρ 2 C 2. To obtain M < 1 we need that 1 2ρα + ρ 2 C 2 < 1 or equivalently ρ(ρc 2 2α) <. But this is achieved for ρ (, 2α/C 2 ). 7.1 Weak vs. strong solutions Consider again the model diffusion problem u = f in (, 1), u() = u(1) = (1) for a given function f L 2 (, 1). We have transformed it into the variational problem Find u H 1 (, 1) such that a(u, v) = (f, v) v H 1 (, 1). (11) and shown that there exists a unique solution to this problem. If a solution u of the variational problem is twice continuously differentiable and f is continuous then u is also a solution of the original problem. From a(u, v) = 1 we obtain by partial integration 1 u v = DuDv = 1 fv 1 fv, v H 1 (, 1) 1 (f u )v = for all v H 1 (, 1). Since f and u are continuous it follows easily that f = u. Furthermore, since u H 1 (, 1) also u() = u(1) = and u is a solution of the original problem. We now discuss an example, where the solution u of the variational problem is not a solution of the original strong formulation. Consider the problem u = H(x) in ( 1, 1), u( 1) = u(1) =, (12) { 1, x where H(x) = This function is not continuous. Therefore, no 1, x > twice continuously differentiable function u exists, which solves this differential 2

21 equation. However, in a variational sense a solution exists. We can obtain it by integrating H(x) twice. This leads to the function ũ(x) = 1 2 x x. Indeed, we have ũ (x) = x and ũ (x) = H(x) (in the weak sense). ũ does not satisfy the boundary conditions. However, the linear function w(x) = 1 x 2 ũ( 1) x ũ(1) 2 satisfies w (x) = and ũ(1) w(1) = ũ( 1) w( 1) =. Hence, a candidate for a weak solution of (12) (that is a solution of the corresponding variational problem) is u(x) = ũ(x) w(x) = 1 x(1 x ). 2 Let us now show that u is indeed a solution of the associated variational problem. First of all, it is obvious that u H 1 ( 1, 1). Furthermore, since u = f in a weak sense we have for every test function v D( 1, 1) (the set of infinitely differentiable functions with compact support in ( 1, 1). 1 1 u (x)v(x)dx = 1 1 f(x)v(x)dx and therefore by using the definition of weak derivatives u (x)v(x)dx = u (x)v (x)dx = f(x)v(x)dx Hence, u satisfies a(u, v) = (u, v) for every v D( 1, 1). It follows then also that a(u, v) = (u, v) for every v H 1 ( 1, 1). This is a direct consequence of the following theorem. Theorem 7.7 Let Ω R n. Then D(Ω) is dense in H 1 (Ω). Assume now that we approximate f L 2 ( 1, 1) by continuous functions f n such that lim n f n f L2 ( 1,1) =. For every f n there exists a strong solution u n with u n = f in ( 1, 1) and u n ( 1) = u n (1) =. Let u be the weak solution of the variational problem with f. Using the variational formulation we have a(u n, v) = (f n, v) a(u, v) = (f, v) (13) 21

22 for every v H 1 ( 1, 1). It follows that a(u n u, v) = (f n f, v). By setting v = u n u and using coercivity we obtain α u n u 2 H 1 ( 1,1) a(u n u, u n u) = (f n f, u n u). We can now use Cauchy-Schwarz to obtain α u n u 2 H 1 ( 1,1) f n f L 2 ( 1,1) u n u L 2 ( 1,1) f n f L 2 ( 1,1) u n u H 1 ( 1,1). It follows that u n u H1 ( 1,1) 1 α f n f L2 ( 1,1). Hence, if we approximate f L 2 by a sequence of continuous functions. Then the corresponding strong solutions u n of the boundary value problem converge in the H 1 norm against the weak solution u. 7.2 The Poisson equation in 2d In this section we investigate the weak formulation of the 2d Poisson problem u = 2 u x 2 2 u y 2 = f (14) for u Ω R 2 with zero Dirichlet boundary conditions, that is u = on Ω. We often use the abbreviation Γ := Ω. Without going into too much technical detail we assume that Ω is open and Ω is sufficiently smooth for the results in this section to hold. The space, in which we look for solutions of this equation will be H 1 (Ω) := { v H 1 (Ω) : v Γ = }. There is a technical problem with this definition. In the 1d case we could conclude from Theorem 6.9 that every function in H 1 (, 1) has a representation as continuous function. Therefore, it made sense to say v() = v(1) =. In two dimensions we do not have this property any more and it does not directly make sense to say that v Γ = since Γ is only a set of measure zero. One way to define H 1 (Ω) is to use traces. For a brief introduction we refer to [1]. To derive a weak formulation from (14) we need the following result, which is proved in [1, Chapter 5]. Theorem 7.8 Let u H 2 (Ω) and v H 1 (Ω). Then u ( u)vdx = u vdx ν vds, where u ν Ω Ω is the derivative in the direction of the unit outward normal ν at Γ. 22 Γ

23 Now assume that in (14) u H 2 (Ω). It follows that f must be in L 2 (Ω) since it is a sum of second derivatives of u. Together with Theorem 7.8 it follows for v H 1 (Ω) that (f, v) = ( u)vdx = u vdx v u Ω Ω Ω ν ds = u vdx := a(u, v), (15) Ω where (f, v) denotes the standard L 2 inner product between f and v. We obtain the following variational problem Find u H 1 (Ω) such that a(u, v) = (f, v) v H 1 (Ω). (16) From the derivation of this problem it is clear that if u H 2 (Ω) solves (14) then u also solves (16). To show that a unique solution of (16) always exists we need to prove continuity and corcivity of a(u, v). Continuity follows simply by a(u, v) u L2 (Ω) v L2 (Ω) u H1 (Ω) v H1 (Ω). For coercivity we refer to [1]. Let us now briefly consider the Poisson problem with zero Neumann boundary conditions. That means, instead of u Γ = we require u ν = on Ω. There are two essential differences to the Poisson problem with Dirichlet boundary conditions. Solutions are only unique up to a constant. If u is a solution of the Neumann problem. Then any function u + C, where C is an arbitrary constant, is also a solution. A necessary condition for the existence of solutions is f(x)dx =. This follows from Ω Ω f(x)dx = u(x)dx Ω = u 1dx Ω Γ u ds =. ν A suitable space for the solution of the variational form of the Neumann problem turns out to be { } V = v H 1 (Ω) : v(x)dx =. For further details see [1, Section 5.2]. Ω 23

24 8 Ritz-Galerkin Approximations and the Finite Element Method 8.1 Finite Dimensional Approximations Let V be a real Hilbert space with inner product,. Let a(u, v) be a continuous bilinear form. Initially, we will assume that a(, ) is symmetric. For F V we want to find numerical approximations to the following variational problem: Find u V such that a(u, v) = F (v), v V. If V is infinite dimensional we cannot solve this problem directly. The idea is to introduce finite dimensional approximations. Let V h V be a finite dimensional subspace of V. The Ritz-Galerkin approximation u h to u is defined as the solution of Find u h V such that a(u h, v) = F (v), v V h. (17) This problem is finite-dimensional and equivalent to solving a linear system of equations. Let {φ j }, j = 1,..., N be a basis of V h. Then u h = N γ j φ j for some coefficient vector x = [ γ 1,..., ] T γ N R N. Similarly, any v V h can be represented as N v = µ j φ j for some coefficient vector m = [ µ 1... µ n ] R N. It follows that i=1 n n a(u, v) = a( γ i φ i, µ j φ j ) = n γ i µ j a(φ i, φ j ) = m T Kx, i, where the matrix A R n n is defined by K ij = a(φ j, φ i ). Similarly, for v V h the functional F (v) has the form F (φ 1 ) F (v) = m T. =: m T b. F (φ N ) Hence, the variational problem (17) is equivalent to Find x R N such that m T Kx = m T b m R N. (18) It can be easily seen that this is equivalent to the linear system of equations Kx = b. 24

25 Lemma 8.1 The linear system Kx = b defined above has a unique solution. Proof Let c. Since the set of functions {φ j } forms a basis of V h we therefore also have v := N c jφ j. From the definition of K and the coercivity of a(, ) it follows that c T Kc = a(v, v) α v 2 H 1 (,1) >. Hence, the matrix K is nonsingular and there exists a unique solution of Kx = b for any right-hand side b. The matrix K is often called stiffness-matrix. This term comes from structural mechanics. We have now established that the finite dimensional variational problem (17) has a unique solution u h, which is obtained by solving a linear system of equations Kx = b. We now want to investigate the properties of the solution u h. We have a(u, v) = F (v), v V a(u h, v) = F (v), v V h Since V h V we can subtract both equations for v V h to obtain a(u u h, v) = v V h. (19) This is called Galerkin-Orthogonality. In the inner product defined by a(, ) the error u u h is orthogonal to the space V h. It is one of the fundamental relationships of finite element methods and the key to their successful understanding. In the following we denote by v E := a(v, v) the energy norm of v V defined by the inner product a(, ). Theorem 8.2 u u h E = min{ u v E : v V }. Proof By Galerkin-orthogonality we have for any v v h that a(u u h, u h ) = a(u u h, v) =. It follows that u u h 2 E = a(u u h, u u h ) = a(u u h, u) a(u u h, u h ) = a(u u h, u) a(u u h, v) = a(u u h, u v) u u h E u v E (Cauchy-Schwarz inequality) If u u h E = the theorem is trivial. Assume that u u h E. Divide the above equation by u u h E to obtain u u h E u v E v V. Hence, u u h E inf{ u v E : v V }. Since u h V the infimimum is attained and u u h E = min{ u v E : v V }. 25

26 For Theorem 8.2 it is essential that a(, ) is symmetric and thereby defines an inner product. Otherwise, the energy norm E is not defined. If a(, ) is unsymmetric we therefore have to proceed differently. This is done in the following theorem. Theorem 8.3 Suppose that a(, ) is a coercive and continuous bilinear form on a closed subspace V of a Hilbert space H with associated Hilbert space norm. Then u u h C α min v V h u v, where C is the continuity constant and α is the coercivity constant of a(, ) on V. Proof By (19) we have a(u u h, v) = for every v V h. We obtain α u u h 2 a(u u h, u u h ) (Coercivity) = a(u u h, u v) ( since a(u u h, u h ) = a(u u h, v) = ) For u u h it follows that Since V h is closed we obtain C u u h u v (Continuity) u u h C α u v v V h. u u h C α min v V h u v. Assume that a(, ) is symmetric. Then both Theorems 8.2 and 8.3 can be applied leading to the two error estimates u u h E = min u V h u v E, u u h C α min u V h u v. Let û and û E be the best approximations of u in the original Hilbert space norm and in the energy norm, that is u û = min u v v V h u û E E = min u v E v V h The Theorems 8.2 and 8.3 lead to the following observations. For the FEM solution u h we have u u h E = u û E from Theorem 8.2. Hence, u h is optimal in the energy norm. In the original Hilbert space norm u h is optimal only up to the constanct C α. If C α is large then the error u u h can be much larger than the optimal error u û. 26

27 8.2 Approximation bases In this section we focus on the 1d model problem Find u H 1 (, 1) such that a(u, v) = F (v) v H 1 (, 1), where a(u, v) = 1 Du(x)Dv(x)dx and F (v) = 1 f(x)v(x)dx for f L2 (, 1). In order to find a Ritz-Galerkin approximation u h to u we need to define a finite dimensional subspace V h of H 1 (, 1). In this section we discuss different choices of basis functions The polynomial basis Let Ṽk := span{1, x,..., x k 1 }. To ensure that Ṽk H 1 (, 1) two conditions need to be satisfied. 1. Every v Ṽk must be weakly differentiable in (, 1). 2. v() = v(1) = for every v Ṽk. The first condition is obviously satisfied as any polynomial is infinitely often differentiable. The second condition is not satisfied for all functions in Ṽk. One subspace of Ṽk which also satisfies the second condition is given by V k := span{x(x 1), x 2 (x 1),..., x k 2 (x 1)}. Example: Let f = 1 in the definition of the variational problem and let our finite dimensional approximation space be given by V 2 = span{ψ 1, ψ 2 }, where ψ 1 (x) = x(x 1) and ψ 2 (x) = x 2 (x 1). The linear system Kx = b is given by K 11 = a(ψ 1, ψ 2 ) = 1 3 K 12 = K 21 = a(ψ 1, ψ 2 ) = 1 6 K 22 = a(ψ 2, ψ 2 ) = 2 15 b 1 = b 2 = 1 1 ψ 1 (x)f(x)dx = 1 6, ψ 2 (x)f(x)dx = The solution of the system Kx = b is given by x 1 = 1 2, x 2 =. It follows that the Ritz-Galerkin approximation u k is given by u k (x) = x 1 ψ 1 (x) + x 2 ψ 2 (x) = 1 2 (x x2 ). 27

28 In this example this is identical to the exact solution of the variational problem. However, in general this will not be true. In general, the stiffness matrix K arising from a polynomial basis is full. Furthermore, even for small matrix sizes it is badly conditioned. These two properties make it essentially infeasible to solve Kx = b as the dimension k of V k increases. A solution to this problem is the introduction of a piecewise polynomial basis Piecewise polynomial bases Partition the interval [, 1] into points = x < x 1 < < x n+1 = 1. Let h = max i=1,...,n+1 (x i x i 1 ) and let the space V h consist of piecewise linear functions v who have the following properties: v C ([, 1]) v [xi 1,x i] is linear for i = 1,..., n + 1. v() = v(1) = A basis of the space V h is given by the piecewise linear functions φ i, i = 1,..., n defined by { 1, i = j φ i (x j ) = (2), i j It can be easily seen that V h = span{φ 1,..., φ n }. Since every piecewise linear function is weakly differentiable with derivative in L 2 and v() = v(1) = for any v V h it follows that V h H 1 (, 1). Let us compute the stiffness matrix K in this basis. For simplicity we assume that the points x i are equally spaced, that is h = x i x i 1 for all i = 1,..., n + 1. We need to evaluate K ij = a(φ i, φ j ). We distinguish the following cases. i j > 1 In this case the basis functions φ i and φ j do not overlap and we immediately have a(φ i, φ j ) =. i j = 1. Assume that j = i + 1. Then i = j In this case a(φ i, φ j ) = xj x i a(φ i, φ i ) = Dφ i (x)dφ j (x)dx = 1/h. xi+1 x i 1 Dφ(x) 2 dx = 2/h. It follows that the matrix K is given by K = h R n n

29 To analyze the rate of convergence in this basis in dependence on h we will proceed in two steps. First, we show that for every w V there is a function w I V h such that w w I E Ch w L 2 (,1) for some C > independent of h. Based on this approximation property of V h we use an argument involving a dual problem to show that u u h L 2 (,1) (Ch) 2 f L 2 (,1). Theorem 8.4 Let h = max x i x i 1. i=1,...,n+1 H 2 (, 1) there exists w I V h such that Then for every w H 1 (, 1) w w I E Ch w L2 (,1). Proof Let w H 1 (, 1) H 2 (, 1). We define w I to be the interpolant of w at the points x i, that is n w I = w(x i )φ i. Then, w I (x i ) = w(x i ) for i = 1,..., n. We want to show that xi i=1 x i 1 (w w I ) (x)dx c(x i x i 1 ) 2 xi x i 1 w (x) 2 dx (21) for i = 1,..., n + 1. Then by summing over i the proof follows with C = c. With e = u u I (21) is equivalent to xi xi e (x) 2 dx c(x i x i 1 ) 2 e (x) 2 dx. x i 1 x i 1 By a change of variables this inequality is equivalent to 1 ẽ (x) 2 d x c 1 ẽ ( x) 2 d x for ẽ( x) = e(x j 1 + x(x i x i 1 )). The proof of this inequality is an exercise. Theorem 8.4 is the basis for the following result. Theorem 8.5 u u h L 2 (,1) Ch u u h E (Ch) 2 f L 2 (,1) Proof The proof makes use of a duality argument. Let w be the solution of w = u u h in (, 1) with w() = w(1) =. 29

30 Integrating by parts yields u u h 2 L 2 (,1) = (u u h, u u h ) = (u u h, w ) = a(u u h, w) = a(u u h, w v), for any v V h by Galerkin orthogonality. By the Cauchy-Schwarz inequality we obtain u u h L2 (,1) u u h E w v E / u u h L2 (,1) = u u h E w v E / w L2 (,1). Now take the infimum over all v V h to obtain u u h L2 (,1) u u h E From Theorem 8.4 it follows that Hence, we obtain inf v V h w v E / w L2 (,1) inf v V h w v E Ch w L 2 (,1). u u h L 2 (,1) Ch u u h E We can now use Theorem 8.4 again with w replaced by u since u h is the optimal approximation in the space V h in the energy norm. This yields u u h L 2 (,1) Ch u u h E (Ch) 2 u L 2 (,1) = (Ch) 2 f L 2 (,1). Theorem 8.5 uses L 2 norms and therefore does not give any information about the error u u h at specific points in the interval [, 1]. Usually, local pointwise estimates of the error are hard to obtain. However, in this 1d model problem the corresponding analysis is not too hard and one can show that u u h Ch 2 u, where is the usual sup-norm [1, Section.7] Spectral Galerkin Methods We now give a simple example of a spectral Galerkin Method. The idea is to use basis functions for the Ritz-Galerkin method that are specially adapted to the the PDE that one wants to solve. Consider the following eigenvalue problem. Find ψ L 2 (, 1) and λ > such that ψ (x) = λψ, x (, 1), ψ() = ψ(1) =. (22) 3

31 We will later investigate eigenvalue problems more in detail. Usually, eigenvalue problems cannot be solved directly. But in this special case we can easily write down the eigenvalues and corresponding eigenfunctions as ψ j (x) = sin πjx with corresponding eigenvalues λ j = π 2 j 2. We define the subspace V j as V k := span{ψ 1, ψ 2,..., ψ k } = span{sin πx, sin 2πx,..., sin kπx} The stiffness matrix K now has the entries K ij = a(ψ i, ψ j ) = = 1 1 ψ i(x)ψ j(x)dx π 2 ij cos(jπx) cos(iπx)dx Using the trigonometric identity cos A cos B = 1 2 (cos(a + B) + cos(a B)) we obtain 1 a(ψ i, ψ j ) = ijπ2 cos(i + j)πx + cos(i j)πxdx 2 [ ] 1 = ijπ2 1 2 π(i + j) sin(i + j)πx + 1 sin(i j)πx π(i j) {, i j = i 2 π 2 2, i = j It follows that the stiffness matrix K is diagonal. This is one of the big advantages of spectral Galerkin methods. It makes the resulting linear system Kx = b, where b j = 1 f(x) sin jπxdx, trivial to solve. The coefficients x j of x are just given by x j = 2 1 π 2 j 2 f(x) sin jπxdx. It follows that u h = k x j sin jπx. The disadvantage of this scheme is that we need to know a representation of the eigenvalues and eigenfunctions of our operator. Nevertheless, there are efficient implementations of spectral Galerkin methods for a variety of problems. 8.3 A Helmholtz problem In this section we want to discuss some aspects of the Helmholtz equation u k 2 u = f (23) 31

32 in (, 1) with boundary conditions u() = u(1) =. There are some important differences to the model diffusion problem. The Helmholtz equation describes time-harmonic waves with frequency k/(2π). Especially, if k is large there are still many open questions and the efficient computation of solutions in 2d and 3d in this case is still an open problem. To arrive at a variational formulation we proceed as in the model diffusion problem. Let u H 1 (, 1). Multiplying (23) and integrating by parts leads to 1 Hence, the variational problem is where a(u, v) = Du(x)Dv(x)dx k 2 u(x)v(x) = 1 f(x)v(x)dx. Find u H 1 (Ω) such that a(u, v) = (f, v) v H 1 (Ω), (24) 1 Du(x)Dv(x)dx k 2 u(x)v(x). The bilinear form is obviously symmetric. Continuity follows easily from a(u, v) Du L 2 (,1) Dv L 2 (,1)+k 2 u L 2 (,1) v L 2 (,1) C u H 1 (,1) v H 1 (,1) (25) for some C > that only depends on k. However, for large enough k the bilinar form a(u, v) needs not be coercive. Consider u(x) = sin(πx). This function satisfies u() = u(1) =. We have 1 1 a(u, u) = π 2 cos 2 (πx)dx k 2 sin 2 (πx)dx = 1 [ π 2 k 2]. 2 Hence, if k > π then a(u, u) < and a(, ) is not coercive. It is therefore not guaranteed that a unique solution of the variational problem (24) exists. Before we go more into details of the existence of solutions let us consider Ritz-Galerkin approximations to the problem (24). Let V h be the usual space of piecewise linear functions with basis functions φ j defined in (2). The matrix K associated with the operator 1 Du(x)Dv(x)dx for u, v V h was already computed. For 1 u(x)v(x)dx, u, v V h we obtain after a short calculation the matrix representation B := h R n n. Hence, the matrix representation of a(u, v) is given by A := K k 2 B = h k 2 h

33 Let us now consider the questions of existence of solutions of the variational problem (24). Consider again the eigenvalue problem u = λu in (, 1) with boundary conditions u() = u(1) =. The eigenvalues are given as λ j = π 2 j 2 with corresponding eigenfunctions φ j (x) = 2 sin(jπx). The φ j are normalized such that φ j L 2 (,1) = 1. One can show that the {φ j } form a basis of L 2 (, 1). Hence, we can represent f and the unknown solution u of the Helmholtz equation u k 2 u = f with zero boundary conditions u() = u(1) = in the form f = γ j φ j, u = µ j φ j. The coefficients γ j are obtained as γ j = f, φ j. The = is to be understand in the sense of L 2 convergence. We want to determine the unknown coefficients µ j of the solution u such that u k 2 u = µ j φ k 2 µ j φ j = f = γ j φ j Together with the relationship φ j = λ jφ j we obtain (µ j λ j k 2 µ j )φ j = γ j φ j. Taking on both sides the inner product with φ j and using orthonormality of the basis functions we arrive at Hence, µ i = µ j λ j k 2 µ j = γ j. γ j λ j k 2 and u = γ j λ j k 2 φ j. Now consider that k 2 λ i = i 2 π 2 γ i for some λ i. Then λ i k 2 1 and for k 2 λ i the solution u becomes unbounded in L 2. Indeed, one can prove the following theorem. Theorem 8.6 The variational problem (24) has a unique solution as long as k jπ for j N\{}. Proof The arguments given here are not yet a complete proof as several technical details are left out. A more complete presentation of the existence theory of weak solutions of elliptic PDEs is given for example in [2]. 33