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I Inverse laplace PDE

  1. Sep 28, 2016 #1
    I am trying to solve with Laplace Transforms in an attempt to prove duhamels principle but cant find the Laplace transform inverse at the end. The book I am reading just says "from tables"...

    The problem :
    $$
    U_t = U_{xx}\\\\
    U(0,t)=0 \quad 0<t< \infty\\\\
    U(1,t)=1\\\\
    U(x,0)=0 \quad 0<x<1\\\\
    $$

    The solution attempt :
    $$
    SU(x,s) = U_{xx}(x,s)\\\\
    U(1,s) = \frac{1}{S}\\\\
    U = \frac{1}{S} \frac{e^{\sqrt{S}x}-e^{-\sqrt{S}x}}{e^{\sqrt{S}}-e^{-\sqrt{S}}} = \frac{1}{S} \frac{sinh(\sqrt(S)x)}{sinh(\sqrt{S}})\\\\
    $$
    The inverse transform is the convolution $$1 \ast
    \mathcal{L}^{-1}(\frac{sinh(\sqrt(S)x)}{sinh(\sqrt{S}}) $$

    Does anyone know of a table where I can find this... The integral to actually compute it myself is... terrifying. Do I have to use the integral... if so... can someone show me how...
     
  2. jcsd
  3. Oct 4, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Oct 4, 2016 #3
    I dunno but I found a good inverse Laplace and inverse Fourier calculator on wolfram. Maybe I will try the integral again after I study residue theory or something.
     
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