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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?

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    The inverse Laplace transform of the terms proportional to $T_0$ is quite simple, and Mathematica can handle it. The term proportional to $T_1$ is a different story, but at least you can get the $T_0$ part out of the way. – 2017-01-10
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    Yes, when $T_1 = 0$ the answer is well known. My question is about when $T_1$ is not vanished. I am looking for a solution in the form of a infinite series – 2017-01-10
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    So really all you need is the inverse Laplace transform of $\exp(-a\sqrt{s})/(s+\beta)$. Not trivial, but a little more clear what is being asked. It's probably also worth noting your boundary conditions are inconsistent. – 2017-01-10
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    Do either of you know how to employ contour integration techniques to compute the ILT directly without the use of black-box software? Neither of your solutions provides any insight into how one computes such an ILT of a function having both a branch point at the origin and a pole on the negative real axis.. – 2017-01-11

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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.

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Using Maple I am obtaining the following closed solution

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Where "CylinderD" is the Whittaker's Parabolic Function (https://en.wikipedia.org/wiki/Parabolic_cylinder_function)

The trick is to use

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Using the identity

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it is possible to rewrite the closed form solution in terms of Hermite functions as

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From other side, using Mathematica and the same trick we obtain the closed solution in terms of hypergeometric functions as follows

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This last solution coincides with the solution given by Maple in terms of parabolic cylinder functions when the following identity is used

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The universal closed solution is obtained using Heun functions and it takes the form

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Also it is possible to write the closed solution in terms of the MeijerG function as

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The closed solution can be obtained using Mathematica with Raspberry pi. It is wonderful.

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    Also, we use MathJax here, which is a lot easier to write and a lot easier on the eyes than this. – 2017-01-11
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Other closed solution is given by

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