# Problem2.4.3 - „x H f L = ï 2 C 2 è l Sin I è l p M =...

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Problem 2.4.3 If we let x = x - p , then the problem is defined over a symmetric domain -p < x < p . The ODE in terms of x becomes (1) 2 f ÅÅÅÅÅÅÅÅÅ „x 2 = -lf with BCs (2) f H -p L = f H p L „f ÅÅÅÅÅÅÅ „x H -p L = „f ÅÅÅÅÅÅÅ „x H f L The general solution to this eigenvalue problem is (3) f H x L = C 1 Sin I è!!! l x M + C 2 Cos I è!!! l x M Applying the BCs give (4) f H p L = C 1 Sin I è!!! l p M + C 2 Cos I è!!! l p M = f H -p L = - C 1 Sin I è!!! l p M + C 2 Cos I è!!! l p M Thus (5) f H -p L - f H p L = 0 ï 2 C 1 Sin I è!!! l p M = 0 In a analogous calculation we find that (6) „f ÅÅÅÅÅÅÅ „x H -p L - „f
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Unformatted text preview: „x H f L = ï 2 C 2 è!!! l Sin I è!!! l p M = Hence the only way we can satisfy both requirements (55) and (6) and not have a trivial solution is (7) l = n 2 , n = 1, 2, 3, … Thus the solution is (8) f H x L = f H x- p L = C 1 Sin @ n H x- p LD + C 2 Cos @ n H x- p LD = C 1 H-1 L n Sin H n x L + C 2 H-1 L 2 Cos H n x L which can be expressed as (9) f H x L = C 1 ' Sin H n x L + C 2 ' Cos H n x L...
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