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Homework Help: Sturm-Liouville (can someone tell me if this is right?)

  1. Jun 15, 2008 #1
    Hi, I solved a simple Sturm-Liouville problem and not sure if its right or not because the answer is odd. I was hoping someone could tell me if I did it right.

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

    [tex]\lambda > 0[/tex]
    u = c_{1}cosh(\sqrt{\lambda}x)+c_{2}sinh(\sqrt{\lambda}x)
    \frac{du}{dx} = c_{1}\sqrt{\lambda}sinh(\sqrt{\lambda}x)+c_{2}\sqrt{\lambda}cosh(\sqrt{\lambda}x)
    \frac{du}{dx}(0) = c_{2}=0;
    u(0)=u(L)\Rightarrow c_{1}cosh(\sqrt{\lambda}x)=c_{1}
    It appears I cant find c1 unless i did something wrong...

    [tex] \lambda = 0[/tex]
    [tex]u = c_{1}+c_{2}x[/tex]
    [tex]\frac{du}{dx}(0) = 0\Rightarrow c_{1}=0[/tex]
    [tex]u(L) = u(0) \Rightarrow c_{2}=0[/tex]

    [tex]\lambda < 0[/tex]
    [tex] u = c_{1}cos(\sqrt{\lambda}x)+c_{2}sin(\sqrt{\lambda}x)[/tex]
    [tex] \frac{du}{dx}(0) = \sqrt{lambda}c_{2}cos(\sqrt{\lambda}0) \Rightarrow c_{2}=0[/tex]
    [tex] u(0) = u(L) \Rightarrow c_{1}=c_{1}cos(\sqrt{\lambda}L)[/tex]
    [tex] \lambda = \left(\frac{arccos(1)}{L}\right)^{2} [/tex]
    [tex]u = cos\left(\frac{arccos(1)x}{L}\right)[/tex]
    is that right? i've never seen an answer like that... Any help is appreciated!
  2. jcsd
  3. Jun 15, 2008 #2
    Hey Engage,

    Before I dive too much further into helping you out, we need to clarify some things about your initial problem:

    You have written:
    \frac{d^{2}u}{dt^{2}}=\lambda u

    However, your boundary conditions are based on x, not t. If you are to solve this ODE, you'll find a solution in terms of t, of which you've continued in terms of x. Perhaps this is just a variable mix up? Perhaps it is the case that you wish to start out with the Laplacian of u?

    Either way, it serves to clarify the situation.

  4. Jun 15, 2008 #3


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    You should be able to answer for yourself! You're trying to solve an equation, aren't you? Is your answer actually a solution to it?
  5. Jun 15, 2008 #4
    yeah, I messed up the initial problem, the independent variable is x, not t.

    \frac{d^{2}u}{dx^{2}}=\lambda u
  6. Jun 15, 2008 #5
    [tex] \lambda < 0[/tex]
    makes the diff eq:

    \frac{d^{2}u}{dx^{2}}=-\lambda u
    -cos(arcos(1)\frac{x}{L})\left(\frac{arcos(1)}{L}\right)^2=-\lambda u[/tex]
    ... So i guess it is right... Thanks for the help! So does this mean there is only one eigenfunction then?
  7. Jun 15, 2008 #6
    Hey Engage,

    I suggest you take a look at your last step, where you solve for lambda. The idea is there, but you've overstepped what you should have done.

    My suggestion is to divide out the c1's such that you have:
    c_{1}=c_{1}cos(\sqrt{\lambda}L) \rightarrow 1 = cos(\sqrt{\lambda}L)

    So then what does the argument of cosine have to be such that cos of that is always one. This is in general the way you solve these problems. You want to find the eigenvalues lambda such that this argument always holds. For finite boundary conditions, lambda is discrete.

    Hopefully this helps.
  8. Jun 15, 2008 #7
    PS. It seems you do not have enough information to solve for the constant.
  9. Jun 15, 2008 #8
    And is it u(0) = u(L) or u(0) = L ?
  10. Jun 15, 2008 #9
    its u(0) = u(L). Thanks for the help. Also, i'm not quite sure what you mean Coto in your post #6. I though thtis is what I did to find lambda. did i mess that up? Thank you all for the help! Ahh, i see so arcos(1) will be 2*pi*n where n is any integer including zero, is that correct?
  11. Jun 15, 2008 #10
    \lambda_{n} = \left(\frac{2n\pi}{L}\right)^{2}
    u_{n} = cos\left(\frac{2n\pi x}{L}\right)
    u =\sum^{\infty}_{n=0}cos\left(\frac{2n\pi x}{L}\right)
    Last edited: Jun 15, 2008
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