- #1
fahraynk
- 185
- 5
I am trying to solve with Laplace Transforms in an attempt to prove duhamels principle but can't 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...
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...