2.4 Eigenvalue problems

Transcription

1 2.4 Eigenvalue problems Associated with the boundary problem (2.1) (Poisson eq.), we call λ an eigenvalue if Lu = λu (2.36) for a nonzero function u C 2 0 ((0, 1)). Recall Lu = u. Then u is called an eigenfunction. Multiply both sides with u and integrate, and this gives Lu, u = λ u, u By positive definiteness, λ > 0, so we define β = λ and get from the eigenvalue equation u (x) + β 2 u(x) = 0 with solutions u(x) = c 1 cos(βx) + c 2 sin(βx). The boundary conditions give c 1 = 0 and c 2 sin(β) = 0 so (as eigenfunctions are nonzero) β = kπ for k = 1, 2, / 14

2 Lemma 2.7 The eigenvalues and eigenfunctions of the problem (2.36) are given by and λ k = (kπ) 2 for k = 1, 2,... sin(kπx) for k = 1, 2,.... Note that (i) there is an infinite number of eigenvalues (and eigenfunctions), and distinct of these are orthogonal; this follows from basic trigonometric identities or the symmetry of L: λ k u k, u m = Lu k, u m = u k, Lu m = λ m u k, u m so (λ k λ m ) u k, u m = 0 and as λ k λ m when k m, orthogonality follows. Lemma 2.8 The functions {sin(knx)} k 1 satisfy that sin(kπx), sin(mπx) equals 0 when k m and 1/2 when k = m, so distinct eigenfunctions here are orthogonal in this inner product. 2 / 14

3 We call µ an eigenvalue of difference method (2.13) if L h v = µv (2.44) for a nonzero function v in D h,0. Then µ must be an eigenvalue of the matrix A = (1/h 2 ) (2.18) One can solve the second-order difference equation (2.44) and find eigenvectors and eigenvalues: Lemma 2.9 The eigenvalues µ k of the problem (2.44) are given by µ k = (4/h 2 ) sin 2 (kπh/2) for k = 1, 2,..., n and the corresponding discrete eigenfunctions in D h,0 k = 1, 2..., n are given by v k (x j ) = sin(kπx j ) for k = 1, 2,..., n 3 / 14

4 Furthermore, the discrete eigenfunctions are orthogonal and satisfy that v k, v m h equals 0 when k m and (1/2) otherwise. The n eigenfunctions ν k are orthogonal and nonzero, so they are linearly independent, and span a subspace of dimension n in D h,0. But this space has dimension n, so the ν k s gives an orthogonal basis for D h,0. Any function g D h,0 may therefore be written n g(x j ) = 2 g, v k h sin(kπx j ) (j = 1, 2,..., n). k=1 where the factor 2 is because the eigenfunctions have length 2. This representation is called a finite Fourier series. If we let the number n of mesh points go to infinite, i.e., h goes to zero, then we may, under certain assumptions, for given f, get convergence so f (x) = c k u k = c k sin(kπx j ) (j = 1, 2,..., n). k=1 k=1 where c k = 2 f, u k = f (x)u k(x)dx. 4 / 14

5 3 The Heat Equation We consider Fourier s method for solving PDEs. 3.1 A Brief Overview Consider the heat equation. u t (x, t) = u xx (x, t), x (0, 1), t > 0. (3.1) subject to boundary conditions u(0, t) = u(1, t) = 0, t > 0. (3.2) and the initial condition u(x, 0) = f (x) x (0, 1). (3.3) For systems of ODEs the approach (from MAT1120) is to decouple the system by diagonalizing the matrix A, using eigenvalues/eigenvectors. Fourier s method extends this idea to PDEs. 5 / 14

6 First, one guesses that there is a solution of the form u k (x, t) = X k (x)t k (t) (3.4) i.e., that the variables can be separated. This is called separation of variables. This leads to ODEs that can be solved. Step 1: Find a family u k (x, t) of solutions satisfying the differential equation (3.1) and the boundary condition (3.2). Consider linear combinations u(x, t) = k c k u k (x, t) (3.5) Then we use the principle of superposition: Step 2: Any linear combination of the form (3.5) is a new solution of (3.1) and (3.2). Then we determine the unknown coefficients. This is the heart of Fourier series. Step 3: Find the coefficients {c k } such that the initial condition (3.3) is satisfied. 6 / 14

7 Step 4: (a) Verify that the series in (3.5) converges towards a well-defined function u = u(x, t). (b) Verify that the limit u solves the differential equation (3.1). (c) Verify that the limit u satisfies the boundary condition (3.2). (d) Verify that the limit u satisfies the initial condition (3.3). 7 / 14

8 3.2 Separation of Variables We shall find particular solutions {u k (x, t)} of the form (3.4), i.e., separable. So they must satisfy: (u k ) t (x, t) = (u k ) xx (x, t), x R, t > 0. (3.7) subject to boundary conditions We start with u k (0, t) = u k (1, t) = 0, t > 0. (3.8) Inserting into the equation gives Divide by X k (x)t k (t), so u k (x, t) = X k (x)t k (t) X k (x)t k (t) = X k (x)t k(t) 8 / 14

9 T k (t) T k (t) = X k (x) X k (x) The left side only depends on t, the right only on x, so they must both be equal to some constant, say λ k. From this we get the ODEs X k (x) + λ kx k (x) = 0 (3.12) and T k (t) + λ kt k (t) = 0 (3.13) The first equation has boundary conditions X k (0) = X k (1) = 0, and we have looked at it before, so λ k = (kπ) 2 (k = 1, 2,...) (3.15) and X k (x) = sin(kπx) (k = 1, 2,...) (3.15) 9 / 14

10 The first-order equation (3.13) has solution T k (t) = e λ kt = e (kπ)2 t (k = 1, 2,...) (3.17) So, our final solutions are u k (x, t) = e (kπ)2t sin(kπx) (k = 1, 2,...) (3.18) which are the particular solutions we looked for. 10 / 14

11 3.3 The Principle of Superposition The functions u k we found in Step 1 satisfy: (u k ) t (x, t) = (u k ) xx, x (0, 1), t > 0, u k (0, t) = u k (1, t) = 0, t > 0. (3.19) u k (x, 0) = sin(kπx) for k = 1, 2,.... But we want to solve u t (x, t) = u xx, x (0, 1), t > 0, u(0, t) = u(1, t) = 0, t > 0 (3.20) u(x, 0) = f (x) If f can be written as a finite linear combination of the eigenfunctions {sin(kπx)}, so f (x) = n c k sin(kπx) (3.21) k=1 11 / 14

12 Then, by linearity, u(x, t) = n c k e (kπ)2t sin(kπx) (3.22) k=1 is a solution of (3.20). Later we will se that we can find a solution also for other given functions f, by letting n go to infinity in the sum. 12 / 14

13 Example 3.2 Consider a uniform rod of length 1 with initial temperature u of the entire rod equal to 1. Then, at t = 0, we start cooling the rod at the endpoints x = 0 and x = 1. By an appropriate choice of scales, the heat equation (3.20) with f (x) = 1 models the temperature distribution in the rod for t > 0. In order to find the temperature by following the steps outlined above, we have to represent the function f (x) = 1 as a finite sum of sine functions. However, this is impossible and the procedure fails at the simplest possible initial condition! On the other hand, if we allow infinite linear combinations, it can be shown that 1 = 4 π for x in the unit interval. k=1 1 sin((2k 1)πx) 2k 1 13 / 14

14 See Fig. 3.3, which shows the Nth partial sum of this series for N = 3, 10, and 100. The series converge towards f (x) = 1 within the unit interval, and we notice that the convergence is very slow near the boundaries. The solution if the PDE is u(x, k) = 4 π k=1 1 2k 1 e ((2k 1)π)2 t sin((2k 1)πx) (3.24) 14 / 14