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110 Difference Equations 3.6.1 Example A Consider the following Sturm-Liouville system: A?yk-1+ Ayk = 0, (3.203) Yo = 0, YN+1 = 0. (3.204) Comparison with equations (3.177) and (3.178) shows that pk-1 = 1, qk = 0, rk = 1, ao = 1, a1 = 0, an = 0, and an+1 = 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) where ci and c2 are arbitrary constants. The first boundary conditions yo = 0 gives c1 = 0. The second boundary condition YN+1 = 0 gives

110 Difference Equations 3.6.1 Example A Consider the following Sturm-Liouville system: A?yk-1+ Ayk = 0, (3.203) Yo = 0, YN+1 = 0. (3.204) Comparison with equations (3.177) and (3.178) shows that pk-1 = 1, qk = 0, rk = 1, ao = 1, a1 = 0, an = 0, and an+1 = 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) where ci and c2 are arbitrary constants. The first boundary conditions yo = 0 gives c1 = 0. The second boundary condition YN+1 = 0 gives

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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110 Difference Equations 3.6.1 Example A Consider the following Sturm-Liouville system: A²yk-1+ Ayk = 0, (3.203) (3.204) Yo = 0, YN+1 = 0. Comparison with equations (3.177) and (3.178) shows that pk–1=1, qk = 0, rk = 1, ao = 1, a1 = 0, aN = 0, and aN+1 = 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) where c1 and c2 are arbitrary constants. The first boundary conditions yo = 0 gives c1 = 0. The second boundary condition YN+1 = 0 gives sin(N +1)0 = 0 (3.208) or (N+ 1)0 = na, n = 1, 2, 3, .... (3.209) Thus, the eigenvalues A, are, from equation (3.205), given by the expression An = 2 |1 – cos N +1 (3.210) = 4 sin? п %3 1,2, 3, .... 2(N + 1), Note that there are only N distinct values of n, i.e., n = 1, 2, ....N, since after 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 On,k = sin n = 1,2, ...., N. (3.211) N+1
Transcribed Image Text:110 Difference Equations 3.6.1 Example A Consider the following Sturm-Liouville system: A²yk-1+ Ayk = 0, (3.203) (3.204) Yo = 0, YN+1 = 0. Comparison with equations (3.177) and (3.178) shows that pk–1=1, qk = 0, rk = 1, ao = 1, a1 = 0, aN = 0, and aN+1 = 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) where c1 and c2 are arbitrary constants. The first boundary conditions yo = 0 gives c1 = 0. The second boundary condition YN+1 = 0 gives sin(N +1)0 = 0 (3.208) or (N+ 1)0 = na, n = 1, 2, 3, .... (3.209) Thus, the eigenvalues A, are, from equation (3.205), given by the expression An = 2 |1 – cos N +1 (3.210) = 4 sin? п %3 1,2, 3, .... 2(N + 1), Note that there are only N distinct values of n, i.e., n = 1, 2, ....N, since after 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 On,k = sin n = 1,2, ...., N. (3.211) N+1
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