# Comments needed on the doubts of PDEs in moving boundary problems

We know that in classical two-phase Stefan problems, let's say in the temperature distribution of ice-water problem here, the governing PDEs are: \left. \begin{aligned} C_1\frac{\partial T_1}{\partial t}=\nabla\cdot(k_1\nabla T_1),\ in\ \Omega_1\\ C_2\frac{\partial T_2}{\partial t}=\nabla\cdot(k_2\nabla T_2),\ in\ \Omega_2\\ \end{aligned} \right\} which is also usually written in the following form: $$C_i\frac{\partial T}{\partial t}=\nabla\cdot(k_i\nabla T),\ in\ \Omega,\\$$ where k_i (C_i)=\left\{ \begin{aligned} k_1 (C_1)\ in\ \Omega_1\\ k_2 (C_2)\ in\ \Omega_2\\ \end{aligned} \right. so that it becomes a PDE with space-dependent coefficient, which seems to be reasonable. But after looking into this equation, I found that in this way I am enforcing additional constraints on temperature derivatives at the phase-change interface. I mean, the temperature derivatives with respect to both time and space are discontinuous and thus is not defined at the interface in the original problem. But when combining them together in a single form, I think we are implicitly violating the above conditions. Therefore, I don't think they can be written in the second form (the combined form).

But I am not sure if there is something inappropriate with my point of view. Perhaps it's just that I am thinking too much? Could you please leave your comments about it? Any comment is welcome! I would greatly appreciate it!

• In continuum mechanics, the so-called jump condition for a stationary interface is usually given by $$\mathbf{n}\cdot\left[\![\mathbf{q}]\!\right]=0\,.$$ This indeed indicates a jump in the derivatives. But the temperature is continuous of course. For a moving interface it is more complicated. – sebastian_g Jan 14 '16 at 8:33
• What is capital $K$? Should that be lower case? – James Jan 18 '16 at 3:31

Your statement suggests that you assume that because you have a term $\nabla\cdot k \nabla T$ in the PDE, that the derivatives $\nabla T$ need to be continuous. But that's not true. You take derivatives not of $\nabla T$ but of $k\nabla T$ (i.e., the heat flux), and this quantity is continuous [1]. So it is differentiable almost everywhere, and that's all you need.
Where it gets a bit more complicated is if you have a moving interface. In that case, if you are sitting at a fixed point $x$ then the time evolution of the temperature at this point, $T(\cdot,x)$ will be continuous, but will probably have a kink at that time where the interface moves across $x$. That's not a problem: it's just at one time instant, and the temperature remains constant, so the time derivative is not defined at that point, but at least it's not a delta function.
• Thanks for your answer, Prof. @Bangerth. So, is the $k \nabla T$ (i.e., the heat flux) also needed to be continuous at the moving interface? I mean, it is already enforced with Stefan condition. – user123 Jan 15 '16 at 11:16