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Consider the following Sturm-Liouville system: A²yk-1 + Ayk = 0, (3.203) (3.204) Yo = 0, YN+1 0.

Consider the following Sturm-Liouville system: A²yk-1 + Ayk = 0, (3.203) (3.204) Yo = 0, YN+1 0.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 4CEXP
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Example A Consider the following Sturm-Liouville system: A²yk-1+ Ayk = 0, (3.203) Yo = 0, YN+1 = 0. (3.204) 0, Comparison with equations (3.177) and (3.178) shows that Pk-1 rk = 1, ao = 1, a1 = 0, an = 0, and an+1= 1, qk 1. If we let 2 - X = 2 cos 0, (3.205) then equation (3.203) becomes Yk+1 – (2 cos 0)yk + Yk-1 = 0, (3.206) which has the solution Yk = c1 cos k0 + c2 sin k0, (3.207) c2 are arbitrary constants. The first boundary conditions yo = 0 0. The second boundary condition yN+1 = where C1 and gives c1 = 0 gives sin(N + 1)0 = 0 (3.208) or (N + 1)0 — пт, = 1,2, 3, .... (3.209) n = Thus, the eigenvalues A, are, from equation (3.205), given by the expression = 2 |1 COS + 1 (3.210) :4 sin? 2(N +1) ) п %3D 1,2, 3, .... 1, 2,....N, since after Note that there are only N distinct values of n, i.e., n = n = N the values of the eigenvalues repeat themselves. The N eigenfunctions associated with these eigenvalues can be determined from equations (3.207), (3.209), and (3.210); they are knT Pn,k n = 1,2, ...., N. (3.211) sin N+1
Transcribed Image Text:Example A Consider the following Sturm-Liouville system: A²yk-1+ Ayk = 0, (3.203) Yo = 0, YN+1 = 0. (3.204) 0, Comparison with equations (3.177) and (3.178) shows that Pk-1 rk = 1, ao = 1, a1 = 0, an = 0, and an+1= 1, qk 1. If we let 2 - X = 2 cos 0, (3.205) then equation (3.203) becomes Yk+1 – (2 cos 0)yk + Yk-1 = 0, (3.206) which has the solution Yk = c1 cos k0 + c2 sin k0, (3.207) c2 are arbitrary constants. The first boundary conditions yo = 0 0. The second boundary condition yN+1 = where C1 and gives c1 = 0 gives sin(N + 1)0 = 0 (3.208) or (N + 1)0 — пт, = 1,2, 3, .... (3.209) n = Thus, the eigenvalues A, are, from equation (3.205), given by the expression = 2 |1 COS + 1 (3.210) :4 sin? 2(N +1) ) п %3D 1,2, 3, .... 1, 2,....N, since after Note that there are only N distinct values of n, i.e., n = n = N the values of the eigenvalues repeat themselves. The N eigenfunctions associated with these eigenvalues can be determined from equations (3.207), (3.209), and (3.210); they are knT Pn,k n = 1,2, ...., N. (3.211) sin N+1
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