How to prove the orthogonality of eigenfunctions in Sturm-Louisville problems?

[Mindscrape]

[SOLVED] orthogonal eigenfunctions

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

\Phi_{xx} + \lambda \Phi = 0

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

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

where q = 0, p = 1, and \sigma = 1. The boundary conditions are

BC1: \Phi(0) = 0

and

BC2: \Phi_x(L) + h \Phi(L,t) = 0

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

\Phi_n(x) = \left\{ {\begin{array}{*{20}c}<br /> {sinh \sqrt{s_1}x} &amp; {n=1} \\<br /> {sin\sqrt{\lambda_n}x} &amp; {n &gt; 1} \\<br /> \end{array}} \right.

Basically all I need to do is show that

\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = 0

with integration by parts I have gotten to the point

\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}

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

 

[Avodyne]

I think it is zero. Use the information on lambda and s that you get by imposing the second BC.

 

[Mindscrape]

The second BC gives transcendental equations though.

tan(\sqrt{\lambda}x) = -\sqrt{\lambda}/h

and

tanh(\sqrt{s}L) = -\sqrt{s}/h

which require numerical methods. I'm pretty sure this can be done analytically.

 

[Avodyne]

You don't have to solve them, you have to USE them to show that your overlap integral is zero.

 

[Mindscrape]

I'm probably going to feel stupid when I finally figure it out, but I still don't really see what to do. The second BC will tell us that the eigenvalues are different, which is why Sturm-Louisville would say that the integral must be zero. Beyond that though, I can't see how it directly influences the integration.

 

[Avodyne]

I can write your overlap integral as
\frac{\sinh(\sqrt{s_e}L)\sin(\sqrt{\lambda_n}L)}{s_e+\lambda_n}\left[<br /> \sqrt{s_e}\coth(\sqrt{s_e}L) - \sqrt{\lambda_n}\cot(\sqrt{\lambda_n}L)}\right]
Now look at the equations you get from the boundary conditions, and use them to simplify each term in the square brackets ...

 

[Mindscrape]

Shoot dang Avodyne, nice algebra trick and nice foresight. Thanks for the help.

 

[Avodyne]

You're welcome!