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Thread: Separation of variables

  1. #1
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    Separation of variables

    Hi, I have this problem, but I am stuck in one thing.

    Assuming that solutions to $\displaystyle y_{tt}=g(zy_z)_z$ are of the form $\displaystyle y(z,t)=Z(z)T(t)$, and that the separation constant is negative, written as $\displaystyle -v^2$ with $\displaystyle v>0$, I need to show that $\displaystyle Z(z)$ and $\displaystyle T(t)$ must satisfy the differential equations

    $\displaystyle T''+v^2gT=0$, and
    $\displaystyle zZ''+Z'+v^2Z=0$

    So well, I kind know what to do once I have substituted $\displaystyle y(z,t)$ into $\displaystyle y_{tt}=g(zy_z)_z$... However, I keep on getting the wrong results:

    I get this:

    $\displaystyle Z(z)T''(t)=g(Z(z)Z'(z)T(t))_z$
    $\displaystyle Z(z)T''(t)=g_z(T(t)(Z'(z)Z'(z)+Z(z)Z''(z)))$

    But this is clearly not what should be on both sides :/

    I would really appreciate it if any of you could give me a hand. Once I get both results on each side, I will be able to separate the variables.

    Thanks a lot!
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  2. #2
    MHF Contributor
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    Re: Separation of variables

    $\displaystyle zy_z \neq Z(z)Z'(z)T(t)$ unless $\displaystyle Z(z) = z$.

    I think you should have:

    $\displaystyle \begin{align*}Z(z)T''(t) & = g\left(zZ'(z)T(t)\right)_z \\ & = g'\left(zZ'(z)T(t)\right) \left[zZ''(t)T(t) + Z'(t)T(t)\right]\end{align*}$

    Also, I am assuming that $\displaystyle g$ is a function. If it is a constant, that changes things significantly.
    Last edited by SlipEternal; Dec 1st 2013 at 02:03 PM.
    Thanks from mathstudent1992
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