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Solving Partial Differential Equations with Laplace Transform

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data


    \dfrac{\partial^2 \varphi }{ \partial x^2} - \dfrac{\partial ^2 \varphi }{\partial t^2} = 1


    Initial Conditions:

    [itex] \varphi (x, 0) = 1; \varphi_t (x, 0) = 1 [/itex]

    Boundary Condition:

    [itex] \varphi (0, t) = 1 [/itex]

    On [itex] 0 \leq x < \infty, 0 \leq t < \infty [/itex]

    2. Relevant equations

    Let [itex] {\Phi} [/itex] denote the Laplace transform from t to s.

    3. The attempt at a solution

    Apply Laplace Transform to PDE:


    \dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + s\varphi (x, 0) + \varphi _t (x, 0) = \dfrac{1}{s}


    Apply Initial Conditions:


    \dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + s + 1 - \dfrac{1}{s} = 0



    \dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + \dfrac{s^2 + s - 1}{s} = 0


    This leads to the eigenvalue being something like:


    r = \dfrac{s^3 \pm \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}


    which is always positive for both cases.

    The solution in Laplace space should be something like:


    \Phi (x, s) = A(s) e ^{\dfrac{s^3 + \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}x} + B(s) e ^{\dfrac{s^3 - \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}x}


    and since the eigenvalue is always positive, as x goes to infinity the exponential terms will explode, requiring both A(s) and B(s) to be equal to 0.

    What am I doing wrong? Any ideas?
  2. jcsd
  3. May 2, 2013 #2
    I would check your final calculation for the solution Φ in Laplace space. It seems to not depend on x, which is weird. If you're still working on this, that is.
  4. May 2, 2013 #3


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    Homework Helper

    I would double-check this step. It looks like you assumed a solution ##\Phi(x,s) = e^{rx}##, and then tried to solve the characteristic equation ##r^2 - s^2 r + (s^2 + s -1)/s = 0##, but this equation for r is not correct. Your purely ordinary differential equation in x is not homogeneous - only two of your terms depend on ##\Phi##. If you plug in ##\Phi(x,s) = e^{rx}##, what happens when you try to divide out the ##e^{rx}## to get the characteristic equation for r?
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