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Seeking the solution of an integro-differential equation

  1. Aug 10, 2015 #1
    I'm seeking the solution to the following integro-differential equation:
    $$ \frac{\partial c(x,t)}{\partial t}=-xc(x,t)+2\int_{x}^{\infty} c(y,t)dy$$

    I know that to solve this, the Laplace transform must be taken term by term. Let
    $$ \mathcal{L}\{c(x,t)\}=\int_{0}^{\infty}e^{-st}c(x,s) $$
    Then, $$ \mathcal{L}\{ \frac{\partial c(x,t)}{\partial t} \}=s\mathcal{L}\{c(x,t)\}-c(x,0)$$
    and, $$ \mathcal{L}\{xc(x,t)\}=x\mathcal{L}\{c(x,t)\} $$

    The problem arises with the Laplace transform of the integral.
    \begin{align*}
    \mathcal{L}\{ \int_{x}^{\infty}c(y,t)dt \} &= \mathcal{L}\{ \int_{0}^{\infty}c(y,t)dt \}-\mathcal{L}\{ \int_{0}^{x}c(y,t)dt \}
    \\&=\mathcal{L}\{ \int_{0}^{\infty}c(y,t)dt \}-\frac{\mathcal{L}\{ c(x,t) \}}{s}
    \end{align*}
    My question is what do I do with the first integral of the last equation?
     
  2. jcsd
  3. Aug 10, 2015 #2

    fzero

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    Science Advisor
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    Gold Member

    This should be
    $$ \mathcal{L}\{c(x,t)\}=\int_{0}^{\infty}e^{-st}c(x,t) dt.$$

    You've made a mistake in swapping the original integration over the position domain with one over the time domain. If we let ##\mathcal{L}\{c(x,t)\} = C(x,s)##, then
    $$\mathcal{L}\{ \int_{x}^{\infty}c(y,t)dy \} = \int_{x}^{\infty}C(y,s)dy.$$
    You'll still end up with an integral equation, but without the time derivative. You should be able to differentiate the equation with respect to ##x## to get a first-order differential equation for ##C(x,s)##.
     
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