3
$\begingroup$

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: \begin{equation} \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\} \end{equation} which is also usually written in the following form: \begin{equation} C_i\frac{\partial T}{\partial t}=\nabla\cdot(k_i\nabla T),\ in\ \Omega,\\ \end{equation} where \begin{equation} k_i (C_i)=\left\{ \begin{aligned} k_1 (C_1)\ in\ \Omega_1\\ k_2 (C_2)\ in\ \Omega_2\\ \end{aligned} \right. \end{equation} 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!

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Commented Jan 14, 2016 at 8:33
  • $\begingroup$ What is capital $K$? Should that be lower case? $\endgroup$
    – James
    Commented Jan 18, 2016 at 3:31

1 Answer 1

3
$\begingroup$

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.

Secondly, the temperature is continuous both in time and in space. So you can take derivatives in time.

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.

[1] All of the above assumes that the strong form of the PDE makes sense. This may not be the case with reentrant corners, etc, but the fundamental points of the statements above remain true.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – user123
    Commented Jan 15, 2016 at 11:16
  • 1
    $\begingroup$ Yes, energy conservation requires it to be continuous: if energy flow from the right to the left into an interface, it needs to flow out of the interface at the other side. So it needs to be continuous. The exception is if you consider phase changes -- in that case, some of the energy is put into the phase change and the rest flows out on the other side. The difference in energy flux is then what is used to affect the phase change, and determines the speed with which the interface moves. $\endgroup$ Commented Jan 15, 2016 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.