(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] orthogonal eigenfunctions

From Sturm-Louisville eigenvalue theory we know that eigenfunctions corresponding to different eigenvalues are orthogonal. For example,

[tex]\Phi_{xx} + \lambda \Phi = 0[/tex]

would be of Sturm-Louisville form (note: [itex]\Phi_{xx}[/itex] represents the second derivative of Phi with respect to x)

[tex]\frac{d}{dx}[p(x) \frac{d \Phi}{dx}] + (q(x) + \sigma(x) \lambda)\Phi = 0[/tex]

where q = 0, p = 1, and [itex]\sigma[/itex] = 1. The boundary conditions are

BC1: [tex]\Phi(0) = 0[/tex]

and

BC2: [tex]\Phi_x(L) + h \Phi(L,t) = 0 [/tex]

which means that there can be one eigenfunction corresponding to a negative eigenvalue and the rest are sines relating to the positive eigenvalues, specifically that the eigenfunctions would be

[tex]\Phi_n(x) = \left\{ {\begin{array}{*{20}c}

{sinh \sqrt{s_1}x} & {n=1} \\

{sin\sqrt{\lambda_n}x} & {n > 1} \\

\end{array}} \right.[/tex]

Basically all I need to do is show that

[tex]\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = 0[/tex]

with integration by parts I have gotten to the point

[tex]\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = \frac{\sqrt{s_e}cosh(\sqrt{s_e}L)sin(L\sqrt{\lambda}) - \sqrt{\lambda_n}cos(\sqrt{\lambda_n}L)sinh(L\sqrt{s_e})}{s_e+\lambda}[/tex]

and mathematica gives the same thing. I don't see this being zero any time soon. Any suggestions?

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# Homework Help: Orthogonal eigenfunctions

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