What are the Eigenvalues and Eigenfunctions of the Sturm-Liouville Problem?

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Homework Help Overview

The discussion revolves around finding the eigenvalues and eigenfunctions of a Sturm-Liouville problem defined by a second-order differential equation with specific boundary conditions. The problem involves analyzing the behavior of the function under different values of the parameter λ.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the characteristic polynomial and the general solution to the differential equation. They raise questions about the implications of boundary conditions, particularly the behavior of the hyperbolic cosine function and its roots. There is also discussion about handling different cases for λ, including positive, negative, and zero values.

Discussion Status

Some participants have provided guidance on specific steps, such as setting certain constants to zero based on boundary conditions. Multiple interpretations of the problem are being explored, particularly regarding the implications of different values of λ on the solutions.

Contextual Notes

Participants are considering the implications of boundary conditions and the nature of the solutions for various cases of λ, which may lead to different forms of eigenfunctions. There is an acknowledgment of potential confusion regarding the roots of the hyperbolic functions involved.

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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:
<br /> \frac{du}{dx}(0)=-D\sqrt{\lambda}cosh(-\sqrt{\lambda}x)=0<br /> ... I am confused now because cosh doesn't have a root unless its translated. Can anyone help me out with this please?
 
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EngageEngage said:
u = Ccosh(\sqrt{\lambda}x)+Dsinh(-\sqrt{\lambda}x)

Now, i try to solve the boundaries:
<br /> \frac{du}{dx}(0)=-D\sqrt{\lambda}cosh(-\sqrt{\lambda}x)=0<br /> ... I am 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:
 
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
<br /> u = C(1)+D(0);;;<br /> u = C<br />
and then i get u = 0 again.

now when \lambda &lt;0
u = Ae^{i\sqrt{|\lambda|}x}+Be^{-i\sqrt{|\lambda|}x}
u = Ecos(\sqrt{\lambda}x)+Fsin(\sqrt{\lambda}x)
u&#039;(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!
 
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: )
 

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