Diffusion Equation on the One Half Plane How do i solve the Diffusion Equation with Fourier Series solution below?
$u_t=ku_{xx}\quad,0\lt x,0\lt t \\ u(0,t)=T_0\, (constant) \\
u(x,0)=e^{-2x^2}$
It's not a homework actually. Please help with the step by step solution and i want to learn it. Thanks
 A: Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=kX''(x)T(t)$
$\dfrac{T'(t)}{kT(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{kT(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-kts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-kts^2}\cos xs~ds$
$u(0,t)=T_0$ :
$\int_0^\infty C_2(s)e^{-kts^2}~ds=T_0$
$C_2(s)=T_0\delta(s)$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty T_0\delta(s)e^{-kts^2}\cos xs~ds$
$u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+T_0$
$u(x,0)=e^{-2x^2}$ :
$\int_0^\infty C_1(s)\sin xs~ds+T_0=e^{-2x^2}$
$\mathcal{F}_{s,s\to x}\{C_1(s)\}=e^{-2x^2}-T_0$
$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{e^{-2x^2}-T_0\}=\dfrac{1}{\sqrt2}F\left(\dfrac{s}{2\sqrt2}\right)-\dfrac{2T_0}{\pi s}$ (according to https://www.wolframalpha.com/input/?i=int+e%5E(-2x%5E2)sin(sx),x,0,inf and http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf)
