Solution to the heat equation with mixed boundary conditions and step function. PDE with the given intial and boundary conditions
$\gamma \frac{\partial^{2}p}{\partial x^{2}}=\frac{\partial p}{\partial t}$
Initial condition: $p(x,t=0)=0$
Outer Boundary Condition: $p(x\rightarrow\infty,t)=0$
Inner Boundary Condition: $\left.\frac{\partial p}{\partial x}\right|_{x=0}=c\left.\frac{\partial p}{\partial t}\right|_{x=0}+\left(\alpha b\left.\frac{\partial p}{\partial t}\right|_{x=0}-t^{-\alpha}\right)\left(1-\theta\left(t-t_{i}\right)\right)$
Where $\theta$ is the unit step function
I am only interested in the solution at the inner boundary. i.e.
$p(x=0,t>0)=?$
Other information:
$0\leq\alpha\leq0.5$
$\gamma>0$
$t>0$
$c\geq0$
$b\geq0$
My Skill level:
I have solved this equation for simplified cases using Laplace transforms for $t<t_i$. However I do not know how to proceed when $t>t_i$
Any help is greatly appreciated!
I have also checked the solution numerically using COMSOL for both $t<t_i$ and $t>t_i$
 A: Hint:
Of course use separation of variables:
Let $p(x,t)=X(x)T(t)$ ,
Then $\gamma X''(x)T(t)=X(x)T'(t)$
$\dfrac{T'(t)}{\gamma T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-\gamma s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-\gamma ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore p(x,t)=\int_0^\infty C_1(s)e^{-\gamma ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-\gamma ts^2}\cos xs~ds$
$\left.\dfrac{\partial p}{\partial x}\right|_{x=0}=c\left.\dfrac{\partial p}{\partial t}\right|_{x=0}$ when $t\geq t_i$ :
$\int_0^\infty sC_1(s)e^{-\gamma ts^2}~ds=-c\gamma\int_0^\infty s^2C_2(s)e^{-\gamma ts^2}~ds$
$\int_0^\infty C_1(s)e^{-\gamma ts^2}~ds=-\int_0^\infty c\gamma sC_2(s)e^{-\gamma ts^2}~ds$
$C_1(s)=-c\gamma sC_2(s)$
$\therefore p(x,t)=-\int_0^\infty c\gamma sC_2(s)e^{-\gamma ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-\gamma ts^2}\cos xs~ds=\int_0^\infty C_2(s)e^{-\gamma ts^2}(\cos xs-c\gamma s\sin xs)~ds~\text{when}~t\geq t_i$
$p(x,0)=0$ :
$\int_0^\infty C_2(s)(\cos xs-c\gamma s\sin xs)~ds=0$
$C_2(s)=0$
$\therefore p(x,t)=0$ when $t\geq t_i$
$\left.\dfrac{\partial p}{\partial x}\right|_{x=0}=(c+\alpha b)\left.\dfrac{\partial p}{\partial t}\right|_{x=0}-t^{-\alpha}$ when $0<t<t_i$ :
$\int_0^\infty sC_1(s)e^{-\gamma ts^2}~ds=-(c+\alpha b)\gamma\int_0^\infty s^2C_2(s)e^{-\gamma ts^2}~ds-t^{-\alpha}$
$\int_0^\infty(sC_1(s)+(c+\alpha b)\gamma s^2C_2(s))e^{-\gamma ts^2}~ds=-t^{-\alpha}$
$sC_1(s)+(c+\alpha b)\gamma s^2C_2(s)=-\dfrac{2\gamma^\alpha s^{2\alpha-1}}{\Gamma(\alpha)}$ (according to http://en.wikipedia.org/wiki/Gaussian_integral#Integrals_of_similar_form)
$C_1(s)=-(c+\alpha b)\gamma sC_2(s)-\dfrac{2\gamma^\alpha s^{2\alpha-2}}{\Gamma(\alpha)}$
$\therefore p(x,t)=\int_0^\infty C_2(s)e^{-\gamma ts^2}(\cos xs-(c+\alpha b)\gamma s\sin xs)~ds-\dfrac{2\gamma^\alpha}{\Gamma(\alpha)}\int_0^\infty s^{2\alpha-2}e^{-\gamma ts^2}\sin xs~ds$
$p(x,0)=0$ :
$\int_0^\infty C_2(s)(\cos xs-(c+\alpha b)\gamma s\sin xs)~ds=\dfrac{2\gamma^\alpha}{\Gamma(\alpha)}\int_0^\infty s^{2\alpha-2}\sin xs~ds$
