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Thread: Nonhomogeneous wave equation

  1. #1
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    Nonhomogeneous wave equation

    Hi, how should I start to solve the following PDE?

    $\displaystyle u_{tt} - u_{xx} = sin(x) cos(5x) sin(w t)$ 0<x<pi, t>0

    u(0,t)=u(pi, t)=0
    u(x,0)=$\displaystyle u_{t}(x,0)$=0.

    My aim is actually to find the values for w, for which the solution does not diverge. Is there a way to find it without completeley solving the problem.
    Thanks for the help.
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  2. #2
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    Quote Originally Posted by Gaudium View Post
    Hi, how should I start to solve the following PDE?

    $\displaystyle u_{tt} - u_{xx} = sin(x) cos(5x) sin(w t)$ 0<x<pi, t>0

    u(0,t)=u(pi, t)=0
    u(x,0)=$\displaystyle u_{t}(x,0)$=0.

    My aim is actually to find the values for w, for which the solution does not diverge. Is there a way to find it without completeley solving the problem.
    Thanks for the help.
    Yes. If you look for solutions in the form

    $\displaystyle
    u = \sum_{n=1}^{\infty} T_n(t) \sin n \pi x
    $
    and sub into you PDE noting that $\displaystyle \sin x \cos 5x = \frac{1}{2} \sin 6x - \frac{1}{2} \sin 4x$, then comparing terms of $\displaystyle \sin n \pi x$ you get

    $\displaystyle
    T_n'' + n^2 \pi^2 T_n = 0\;\; \text{for}\; n \ne 4 \; \text{and}\;6
    $ and

    $\displaystyle
    T_4'' + 16 \pi^2 T_4 = - \frac{1}{2} \sin \omega t
    $
    and

    $\displaystyle
    T_6'' + 36 \pi^2 T_6 = \frac{1}{2} \sin \omega t
    $

    Now the solution of the first is bounded and then you ask, for what values of $\displaystyle \omega $ are the solutions of the second and third unbounded?
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