I am looking to solve the wave equation with the following conditions:
$ \alpha^2 U_{xx}=U_{tt}, \ x>0,t>0 $
$ U(0,t)= e^{-t},\ lim_{x \to \infty} U(x,0)=0 $
$U(x,0)=0, U_{t}(x,0)=0, \ x>0$
I solved using the Laplace transform with respect to $t$ and the result was: $H(t-\frac{x}{\alpha^2})e^{-t} $ and after that I have tried to solve using the sine transform with respect to $x$ but I get a complicated integral as the solution:
$ \int_{0}^{\infty}( \frac{-cos(wat)} {a^2w(w^2+1)} -\frac{sinwat}{a^3w^2(w^2+1)} + \frac{1}{a^2(w^2+1)})\ sinwx dw $
the result of which is:
$-\dfrac{{\pi}\mathrm{e}^{-x-at}\cdot\left(2at\mathrm{e}^{x+at}+\left(-a-1\right)\mathrm{e}^{2at}+2a\mathrm{e}^{at}-a+1\right)}{4a^3}$
I am interpreting the condition $ lim_{x \to \infty} U(x,0)=0 $ as stating that the function and its derivative vanish as x goes to $\infty$ which allows for the use of the sine transform. Therefore there is no need to take the transform of the boundary condition. Is this reasoning correct? I suspect there is something wrong with taking the sine transform on this problem like I did. Thanks in advance!