# Orthogonal eigenfunctions

1. Dec 2, 2007

### 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

$$\Phi_n(x) = \left\{ {\begin{array}{*{20}c} {sinh \sqrt{s_1}x} & {n=1} \\ {sin\sqrt{\lambda_n}x} & {n > 1} \\ \end{array}} \right.$$

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?

2. Dec 2, 2007

### Avodyne

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

3. Dec 2, 2007

### 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.

4. Dec 2, 2007

### Avodyne

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

5. Dec 3, 2007

### 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.

6. Dec 3, 2007

### Avodyne

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

Last edited: Dec 3, 2007
7. Dec 3, 2007

### Mindscrape

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

8. Dec 3, 2007

### Avodyne

You're welcome!