I am trying to solve the following problem
$$\rho\,C_{{p}}{\frac {\partial }{\partial t}}T \left( x,t \right) = \kappa\,{\frac {\partial ^{2}}{\partial {x}^{2}}}T \left( x,t \right) $$
$$T(x,0)=T_0$$ $$T(0,t)=T_1e^{-\beta t}$$ $$T \left( \infty ,t \right) =T_0$$
where the initial condition is for $x >0$.
I am using the laplace transform method. In the laplace domain I am obtaining the following solution
$$T \left( x \right) ={{\rm e}^{-{\frac {\sqrt {\rho}\sqrt {C_{{p}}} \sqrt {s}x}{\sqrt {\kappa}}}}} \left( {\frac {T_{{1}}}{s+\beta}}-{ \frac {T_{{0}}}{s}} \right) +{\frac {T_{{0}}}{s}} $$
I am implementing the procedure both in Mathematica and Maple respectively but none of them is able to compute the inverse laplace transform of the last expression.
My question is: it is possible to obtain a closed form for the solution?
OK, it is possible to find a closed form for this. Mathematica informs me the term proportional to $T_0$ is $\mathcal{L}\left[T_0\mathrm{erf}\left(\frac{x}{2}\sqrt{\frac{\rho C_p}{\kappa t}}\right)\right]$. For the term muliplying $T_1$, we use $$ \mathcal{L}\left[e^{-\beta t}f(\beta t)\right] = \frac{1}{\beta}F\left(\frac{s}{\beta}+1\right) $$ Mathematica informs me that $$ \mathcal{L}^{-1}\left[\frac{\exp\left(-a \sqrt{u - 1}\right)}{u}\right] = \mathrm{Re}\left[e^{i a}\mathrm{erfc}\left(\frac{a + 2iv}{2\sqrt{v}}\right)\right] $$ So we have $$ \mathcal{L}^{-1}\left[T_1 \frac{\exp\left(-\sqrt{\frac{\rho C_p}{\kappa}} x\sqrt{s}\right)}{s+\beta}\right] = T_1 e^{-\beta t}\mathrm{Re}\left[e^{i\alpha x}\mathrm{erfc}\left(\frac{\alpha x + 2i\beta t}{2\sqrt{\beta t}}\right)\right] $$ where $\alpha = \sqrt{\rho C_p \beta/\kappa}$. I don't know of any way to simplify that real part term, but this should give you your answer.