- #1

Markov2

- 149

- 0

$\begin{eqnarray*}

{{u}_{t}}&=&{{u}_{xx}},\text{ }0<x<\infty ,\text{ }0<t<\infty \\

u(0,t)&=&\sin t,\text{ }0<t<\infty \\

u(x,0)&=&0,\text{ }0\le x<\infty .

\end{eqnarray*}$

I need to apply the Laplace transform to solve this, so by applying it I get $su(x,s)-u(x,0)=u_{xx}(x,s)$ and $u(0,s)=\frac1{1+s^2}.$ I need to treat the first equation as a second order ODE right? Then the solution would be $u(x,s)=c_1e^{x\sqrt s}+c_2e^{-x\sqrt s},$ here's where I'm getting stuck, I've been told that I must found bounded solutions, so in order to get it bounded I should consider $u(x,s)=c_2e^{-x\sqrt s},$ by applying the initial conditions I get $u(0,s)=c_2=\frac1{1+s^2}$ so we have $u(x,s)=\frac1{1+s^2}e^{-x\sqrt s},$ now I must apply the inverse Laplace transform, so first $u(x,s)=\frac s{1+s^2}\frac{e^{-x\sqrt s}}s=\mathcal L(\cos t)\mathcal L^{-1}\left(\frac{e^{-x\sqrt s}}s\right)$

I can't find the inverse of the remaining function.

Is this correct?

Thanks!