I have PDE (diffusion equation) on the half plane and just have one BC and one IC only. says,
$u(0,t)=T_0\quad \text{Constant} \\ u(x,0)=e^{-2x^2}$
I'm try to solve this with Fourier Series and i don't get any conclusion about this. Cause of the three cases makes me confused. I mean three cases is $\lambda>0\, ,\lambda=0,\text{and}\, ,\lambda<0$.
When i'm try to use this formula:
$\upsilon(x,t)=\dfrac{1}{\sqrt{4\pi kt}}\displaystyle\int_{0}^{\infty}\left[e^{\frac{-(x-y)^2}{4kt}}+e^{\frac{-(x+y)^2}{4kt}}\right]\phi_{\text{even}}(y)\,dy\\ \phi_{\text{even}}(y)=e^{-2y^2}\\$
But it's too complicated to compute even though i've use Error Function. And actually i don't want to use this formula.
Then, my question is with my BC and IC and each of them just 1, is it possible to compute with fourier series solution (because the command questions in the book suggest that method)?
And what if i do with Fourier Transform?
Please help me to understand this and sorry for my mistakes.
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)