Solving Sturn-Liousville problem

[JasonZ]

My problem is:

\frac{d^2\phi}{dx^2} + 6\frac{d\phi}{dx} + \lambda\phi = 0; \\ \frac{d\phi}{dx}(0) = 0, \frac{d\phi}{dx}(L) = 0.

I am told to begin by finding every eigenvalue and corresponding eigenfunction. I missed the last class where we went over this and the book is not giving me much advice on how to begin. Can someone just provide some help on how I begin solving this? I would show what I have but I have nothing yet ;).

Thanks,
-Jason

 

[cookiemonster]

Try a solution of the form

\phi = e^{rx}

then consider cases for lambda, i.e. positive, zero, negative, and match boundary conditions.

cookiemonster

 

[JasonZ]

Try a solution of the form

\phi = e^{rx}

then consider cases for lambda, i.e. positive, zero, negative, and match boundary conditions.

cookiemonster

Sorry for seeming so dense, but I understand how to apply that to normal O.D.E's, but the eigenvalue is throwing me off. What I am doing is:

<br /> Let \ \phi = e^{rx}<br />

<br /> \frac{d\phi}{dx} = re^{rx}<br />

<br /> \frac{d^2\phi}{dx^2} = r^2e^{rx}<br />

After substituting this back in for \phi(x) I get:

<br /> r^2 + 6r + \lambda = 0<br />

This is easily solved for \lambda = 0, but how do I go about solving it for non-zero eigenvalues? I am pretty embarrased for having to ask questions on such a simple problem, but thank you for helping!

 

[cookiemonster]

How about the quadratic formula? =]

cookiemonster

 

[JasonZ]

Yeah I did that originally, and I think i was looking for some sort of Fourier series like lambda to come about, and getting confused by that. I got it now though, thanks man.

-Jason