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Separation of Variables

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Hey guys,

    I have this problem which I am having a hard time solving.

    $$u_{tt} -x^2u_{xx} = 0$$
    $$1<x<2 \hspace{4mm} t>0$$
    $$u(x,0)=0$$
    $$u_t(x,0)=g(x)$$
    $$u(1,t)=0=u(2,t)$$


    2. Relevant equations

    $$u_{tt} -x^2u_{xx} = 0$$
    $$1<x<2 \hspace{4mm} t>0$$
    $$u(x,0)=0$$
    $$u_t(x,0)=g(x)$$
    $$u(1,t)=0=u(2,t)$$



    3. The attempt at a solution

    I used separation of variables to reach:

    $$x^2X'' - \lambda X=0$$
    $$T''- \lambda T = 0$$

    Solving for $$X$$ since the determinant $$a^2-4b > 0$$ (since $$x$$ is between 1 and 2)
    and the roots are both real, we have to use the form of a solution:
    $$X(x) = Ae^{rx} + Be^{rx}$$

    solving for the roots yields $$r = \pm \frac{\lambda}{x}$$

    Plugging the roots into the solution gives:
    $$X(x) = Ae^{\lambda} + Be^{\lambda}$$

    but this solution does not $$x$$ at all!

    Where did I go wrong?!
     
  2. jcsd
  3. Nov 4, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    I assume you're thinking of [itex]aX'' + bX = 0[/itex], but assuming a solution of the form [itex]e^{rx}[/itex] doesn't work here because [itex]a = x^2[/itex] is not a constant.

    Try a solution of the form [itex]x^{\alpha}[/itex] instead.

    Or set [itex]X(x) = Z(\log x) = Z(z)[/itex], which does give you a second-order linear ODE with constant coefficients for [itex]Z(z)[/itex].
     
    Last edited: Nov 4, 2013
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