I've been given the following problem:
Find the solution to the half-line problem for the heat equation
$u_t = u_{xx}$, $0<x<\infty$, $t>0$
$u_x(0,t) = 0$, $t > 0$,
$u(x,0)= e^{−x}$, $0<x<\infty$
This is Neumann's Problem so I know it's solved with the following integral:
$$\frac{1}{\sqrt{4\pi t}}\int_0^\infty e^{\frac{-(x-y)^2}{4t}-y} + e^{\frac{-(x+y)^2}{4t} -y}dy$$
My problem is simplifying this integral. I did it through once by completing the square and got the answer $u(x,t) = e^{t-x} + e^{t+x}$ but this does not satisfy the condition $u(x,0) = e^{-x}$
Could someone walk me through this nasty integration? Thanks!
Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=X''(x)T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds$
$u_x(x,t)=\int_0^\infty sC_1(s)e^{-ts^2}\cos xs~ds-\int_0^\infty sC_2(s)e^{-ts^2}\sin xs~ds$
$u_x(0,t)=0$ :
$\int_0^\infty sC_1(s)e^{-ts^2}~ds=0$
$C_1(s)=0$
$\therefore u(x,t)=\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds$
$u(x,0)=e^{−x}$ :
$\int_0^\infty C_2(s)\cos xs~ds=e^{−x}$
$\mathcal{F}_{c,s\to x}\{C_2(s)\}=e^{−x}$
$C_2(s)=\mathcal{F}^{-1}_{c,x\to s}\{e^{−x}\}=\dfrac{2}{\pi(s^2+1)}$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourCos2.pdf)
$\therefore u(x,t)=\dfrac{2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\cos xs}{s^2+1}~ds$