# Diffusion equation with time-dependent boundary condition

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

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

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?

• Are you allowed to solve by separation of variables? That way you just get 2 ODEs that are easy to solve. – Ballanzor Mar 30 at 0:52
• 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 :/ – jorafb Mar 30 at 3:08
• I have no idea why that happens... Sorry. It definitely looks separable at first glance – Ballanzor Mar 30 at 9:27

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.