# Sturm-Liouville problem

## Homework Statement

Find the eigenvalues and the eigenfunctions of the Sturm-Liouville problem

$$\frac{d^{2}u}{dx^{2}}=\lambda u$$
$$0<x<L$$
$$\frac{du}{dx}(0) = 0$$
$$u(L) = 0$$

## The Attempt at a Solution

characteristic polynomial:
$$p^{2}=+-\lambda$$
$$u = Ae^{\sqrt{\lambda}x}+Be^{-\sqrt{\lambda}x}$$
$$u = Ccosh(\sqrt{\lambda}x)+Dsinh(-\sqrt{\lambda}x)$$

Now, i try to solve the boundaries:
$$\frac{du}{dx}(0)=-D\sqrt{\lambda}cosh(-\sqrt{\lambda}x)=0$$ ... im confused now because cosh doesn't have a root unless its translated. Can anyone help me out with this please?

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
$$u = Ccosh(\sqrt{\lambda}x)+Dsinh(-\sqrt{\lambda}x)$$

Now, i try to solve the boundaries:
$$\frac{du}{dx}(0)=-D\sqrt{\lambda}cosh(-\sqrt{\lambda}x)=0$$ ... im confused now because cosh doesn't have a root unless its translated. Can anyone help me out with this please?
Hi EngageEngage!

Isn't it just D = 0?

O yeah, thanks. not sure how i managed to screw that up. then I get
$$u = Ccosh(\sqrt{\lambda x})$$
$$u(L) = 0 = Ccosh(\sqrt{\lambda L})$$
But cosh has no root here so, i get u = 0, by setting C = 0.
I just realized it: i probably have to do this with lambda>0, <0 and = 0, is that righT?

so when: $$\lambda = 0$$
$$u = C(1)+D(0);;; u = C$$
and then i get u = 0 again.

now when $$\lambda <0$$
$$u = Ae^{i\sqrt{|\lambda|}x}+Be^{-i\sqrt{|\lambda|}x}$$
$$u = Ecos(\sqrt{\lambda}x)+Fsin(\sqrt{\lambda}x)$$
$$u'(0)=0 when F = 0$$
$$u = Ecos(\sqrt{\lambda}x)$$
$$u(L)=0=Ecos(\sqrt{\lambda}L)$$
$$\lambda = (\frac{(n-\frac{1}{2})\pi}{L})^{2}, n = 1,2,3...$$
$$u = cos(\frac{(n-\frac{1}{2})\pi x}{L})$$

Last edited:
is that right?

messed that last part up, but i got it now. Thanks!

tiny-tim
Science Advisor
Homework Helper
Hi EngageEngage!

Yes, that looks fine!

(though personally I don't like minuses, so I'd write it:

$$u = cos\left(\frac{(n+\frac{1}{2})\pi x}{L}\right)$$ )

(note the \left( and \right) in the LaTeX! )