Question

42 CHAPTER 2. BASICS Example 2.15 We consider the one-dimensional Sturm-Liouville eigenvalue problem (2.24) - u"(x) = \u()0<<<, (0) = u(T) = 0, that models the vibration of a homogeneous string of length that is fired at both ends. The eigenvalues and eigenvectors or eigenfunctions of (2.24) are x = k?, ux() = sin ka, KEN Let u" denote the approximation of an (eigen)function u at the grid point Ii, uiuti), Di=ih, 0<i<n +1, h = n+1 We approximate the second derivative of u at the interior grid points by finite differ- ences [3, 7] 1 (2.25) 22 (-2-1 + 2ui – Wi+1) = dui, 1<i<n. Collecting these equations and taking into account the boundary conditions, uo = 0 and Un+1 = 0, we get a (matrix) eigenvalue problem (2.26) Tx= x where 2 2 (n+1) -1 2 -1 Tn := ERXn -1 - 1 -1 -1 2 -1 -1 2 The matrix eigenvalue problem (2.26) can be solved explicitly (9, p.229]. Eigenvalues and eigenvectors are given by 4(n + 1)2 km (2 - 2 cos Ox) = #2 72 (2.27) 2(n+1) 2 1/2 ka [sin øk, sin 20k, ..., sin nox], n+1 OR = n+1 1 = (n +12 sin? Clearly, 1) converges to Ak as n + 20. (Note that sing + as & +0.) When we identify u" with the piecewise linear function that takes on the values given by ") at the grid points r; then this function evidently converges to sin kr. Let p(x) be a polynomial of degree d, p(x) = 00 + a11 + 0212 + ... + ad. As A = (UAU*)) = UNU* we can define a matrix polynomial as d (2.28) P(A) = Ža;4° = { a;UN'U* = U Q;A' U* j=0 This equation shows that p(A) has the same eigenvectors as the original matrix A. The eigenvalues are modified though, I becomes p(x). Similarly, more complicated functions of A can be computed if the function is defined on spectrum of A.

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   0 are Ergen vector A; = (jm)? of 4; (x) = sin(arx) eigen function u eun hal J N # Consider the 10 Strum Liwille eigenvaluen problem - L"(x) = luxe) 0<x< 1 46-4(1)=0 - The Eigen value and and Let denote the approx, el at the geid point x; 021 <N-1 Hi ñ u(x;) ; ; = inh we approx. the second dezivative af u at the interior geid points by finite differences 1212N-1 + du; - Hitl] = id; he these equations into account the B.G 40=0 į UN=0 we get. Tnx =AR whese 2 -1 1 I Fui- and taking Collecting o Tn a 0 -1 -1 =A st - Q 0 1 A is A; = a th hal - Cosją] +2 Therefore the eigenvalue [I - Casjit eigen vector V; - f [sin ja , Sinagre, anal of is A IT Sin (N-1 ja]