How to solve: \begin{cases} u_t - ku_{xx} = 0 \hspace{2cm} k>0, x>0, t>0 \\ u(x,0) = 0 \hspace{2.8cm} x>0\\ u(0,t) = h(t) \hspace{3cm} t>0 \end{cases} I am trying to solve diffusion equation in two methods, Laplace transform and reflection method. I have solved it by Laplace transform. Here is the solution $$ \int^{t}_{0} \frac{1}{\sqrt {4\pi k(t-s)^3}}xe^\frac{-x^2}{4k(t-s)} \quad dx $$
I want to solve it by reflection method and make sure they are same. The following is the solution of reflection method. $$ \int^{\infty}_{0} \int^{t}_{0} \frac{\partial}{\partial s} [S(x-y, t-s) - S(x+y, t-s)] h(s)dsdy + h(t) \\ ,where \hspace{1cm} S(x,t) = \frac{1}{\sqrt {4 \pi kt}}e^\frac{-x^2}{4kt} $$
However, I can't simplify them. Any help in solving this is much appreciated!
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\mbox{With}\ k > 0,\ x >0\ \mbox{and}\ t>0;\quad \left\vert\begin{array}{l} \ds{\ \partiald{\on{u}\pars{x,t}}{t} - k\partiald[2]{\on{u}\pars{x,t}}{x} = 0} \\[1mm] \left\{\begin{array}{rcl} \ds{\on{u}\pars{x,0}} & \ds{=} & \ds{0} \\ \ds{\on{u}\pars{0,t}} & \ds{=} & \ds{\on{h}\pars{t}} \end{array}\right. \\[3mm] \mbox{Lets}\ \ds{\hat{\on{u}}\pars{x,s} \equiv \int_{0}^{\infty}\on{u}\pars{x,t}\expo{-st}\dd t} \end{array}\right.$
I'll assume that $\ds{\underline{\lim_{x \to \infty}\on{u}\pars{x,t} = 0\ \mbox{and}\ \lim_{x \to \infty}\hat{\on{u}}\pars{x,s} = 0}}$.
\begin{align} & \mbox{Therefore,}\ -\ \overbrace{\on{u}\pars{x,0}}^{\ds{0}}\ +\ s\hat{\on{u}}\pars{x,s} - k\,\partiald[2]{\hat{\on{u}}\pars{x,s}}{x} = 0 \\ & \mbox{where}\ \hat{\on{u}}\pars{0,s} = \hat{\on{h}}\pars{s} \equiv \int_{0}^{\infty}\on{h}\pars{t}\expo{-st}\dd t. \\ & \mbox{Then,}\ {\large \hat{\on{u}}\pars{x,s} = \hat{\on{h}}\pars{s}\expo{-\root{s/k}x}}. \\[5mm] & \phantom{\ =\ }\mbox{Henceforth,} \\ \color{#44f}{\large\on{u}\pars{x,t}} & = \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} \hat{\on{h}}\pars{s}\expo{-\root{s/k}x}\expo{ts}{\dd s \over 2\pi\ic} \\[5mm] & = \int_{0}^{\infty}\on{h}\pars{\tau}\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} \expo{-\root{s/k}x}\expo{\pars{t - \tau}s}\ {\dd s \over 2\pi\ic}\,\dd\tau \\[5mm] & = \int_{0}^{\infty}\on{h}\pars{\tau}\bracks{t > \tau}\ \times \\ & \bracks{% -\int_{-\infty}^{0} \expo{-\ic\root{-s/k}x}\expo{\pars{t - \tau}s}\ {\dd s \over 2\pi\ic} - \int_{0}^{-\infty} \expo{\ic\root{-s/k}x}\expo{\pars{t - \tau}s}\ {\dd s \over 2\pi\ic}}\dd\tau \\[5mm] & = \int_{0}^{t}\on{h}\pars{\tau}\ \times \\ & \bracks{% -\int_{0}^{\infty} \expo{-\ic\root{s/k}x}\expo{-\pars{t - \tau}s}\ {\dd s \over 2\pi\ic} + \int_{0}^{\infty} \expo{\ic\root{s/k}x}\expo{-\pars{t - \tau}s}\ {\dd s \over 2\pi\ic}}\dd\tau \\[5mm] & = {1 \over \pi}\int_{0}^{t}\on{h}\pars{\tau}\ \overbrace{\int_{0}^{\infty} \sin\pars{\root{s \over k}x}\expo{-\tau s}\dd s} ^{\ds{{1 \over 2}\root{\pi \over k}x\ \tau^{-3/2}\,\exp\pars{-\,{x^{2} \over 4k\tau}}}}\dd\tau \\[5mm] & = \bbx{\color{#44f}{\large{x \over 2\root{\pi k}}\int_{0}^{t}{\expo{-x^{2}/\pars{4k\tau}} \over \tau^{3/2}}\on{h}\pars{\tau}\dd\tau}} \\ & \end{align} A further reduction requires a $\ds{particular\ \on{h}\pars{\tau}\ expression}$.