I was trying to solve this 1D diffusion problem \begin{equation} \dfrac{\partial^2 T}{\partial \xi^2} = \dfrac{1}{\kappa_S}\dfrac{\partial T}{\partial t}\, , \label{eq_diff_xi} \end{equation} with the boundary conditions \begin{align} &T(\xi = 2Bt^{1/2},t) = A t^{1/2}\, ,\\ &T(\xi=\infty,t) = 0\, ,\\ &T(\xi,0) = 0\, , \end{align} where $A$ is a constant.

I know that the solution is $T = A \sqrt{t}\, \rm{erfc}(\xi/2\sqrt{kt})/\rm{erfc}(B\sqrt{k})$

I tried by using the Laplace transformation, but I found a problem since I have conditions on $\xi = 2Bt^{1/2}$ instead of $\xi = 0$.

More precisely, if the Laplace function of $T(\xi,t)$ is $\Theta(\xi,s)$, after apply the Laplace transformation plus $T(\xi=\infty,t) = 0$ and $T(\xi,0) = 0$, I got

\begin{equation} \Theta(\xi,s) = C_1(s)\exp{\left(-\sqrt{\dfrac{s}{\kappa_T}}\xi\right)}\, . \end{equation}

So now, to find $C_1(s)$ and use the convolution property of the Laplace transformation, I need a condition on $\xi = 0$, but I only now that $T(\xi = 2Bt^{1/2},t) = A t^{1/2}$.

Does any of you know if the Laplace transform has some other properties that allow me to solve the problem?

  • 1
    $\begingroup$ Are you allowed to solve by separation of variables? That way you just get 2 ODEs that are easy to solve. $\endgroup$
    – Ballanzor
    Commented Mar 30, 2019 at 0:52
  • $\begingroup$ I know that the solution is $T = A \sqrt{t}\, \rm{erfc}(\xi/2\sqrt{kt})/\rm{erfc}(B\sqrt{k})$, so it is clearly not separable :/ $\endgroup$
    – jorafb
    Commented Mar 30, 2019 at 3:08
  • $\begingroup$ I have no idea why that happens... Sorry. It definitely looks separable at first glance $\endgroup$
    – Ballanzor
    Commented Mar 30, 2019 at 9:27

1 Answer 1


The boundary condition hints to try a change of variables. Let's will be looking for the solution in the form $$ T(\xi,t) = A\sqrt{t}\ \tau(\xi/2B\sqrt{t},t) $$ Then boundary conditions and equation for $\tau(x,t)$ are respectivly $$ \tau(1,t) = 1,\qquad \tau(\infty,t) = 0 $$ and $$ t\frac{\partial\tau}{\partial t}(x,t) = \frac12\left(x\frac{\partial\tau}{\partial x}(x,t) - \tau(x,t)\right)+\frac{k}{4B^2}\frac{\partial^2\tau}{\partial x^2}(x,t) $$ This problem is solvable by the separation of varfiables method. If we will look for the solution in the form $\tau(x,t) = f(x)g(t)$, then we'll get $$ t\frac{\dot{g}(t)}{g(t)} = \frac1{2f(x)}\left(xf'(x)-f(x)+\frac{k}{2B^2}f''(x)\right) = \lambda = const $$ Boundary condition looks like $\tau(1,t) = f(1)g(t) = 1$. It follows that $\lambda = 0$. Then the equation for $f(x)$ is $$ \frac{k}{2B^2}f''(x)+xf'(x)-f(x) = 0. $$ If we choose $g(t) = 1$, then boundary conditions for $f(x)$ are $$ f(1) = 1,\qquad f(\infty) = 0. $$ It's up to you to check if a solution to this problem gives the known solution.


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