Sturm-Liouville problem

[EngageEngage]

1. The problem statement, all variables and given/known data
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

3. 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?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[tiny-tim]

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? :smile:

[EngageEngage]

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?

[EngageEngage]

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})

[EngageEngage]

is that right?

[EngageEngage]

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

[tiny-tim]

Hi EngageEngage! :smile:

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! :smile: )