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Consider the 1D heat equation

$$\frac{\partial u}{\partial t}(x,t) = \frac{\partial}{\partial x}\left(k(x)\frac{\partial u}{\partial x}(x,t)\right) \qquad x \in \mathbb{R}$$

where the thermal diffusivity is a piecewise constant function modelling two semi-infinite rods with different diffusivities coming in contact at $x=0$, i.e. $k(x) = k_1$ for $x<0$ and $k(x) = k_2$ for $x\geq 0$. The boundary conditions are well known boundary conditions for heat transfer across an interface: 1) temperature continuity $u(0^-,t) = u (0^+ , t)$, and 2) heat flux continuity $k_1 u_x(0^-,t) = k_2 u_x(0^+,t)$.

There's one limiting case of this problem that seems inconsistent to me. Say $k_2 = 0$. From the heat flux condition, we get that $u_x(0^-,t) = 0$. So in this case, for the $x<0$ region, we just have the problem of a semi-infinite rod with insulating boundary condition. This should have a unique solution, which I'll call $u^*(x,t)$.

Now for the $x\geq 0$ region, we have

$$\frac{\partial u}{\partial t}(x,t) =0 \qquad x\geq 0,$$

i.e. $u$ is constant in time. But how can $u$ be constant for $x\geq 0$ while $u^*$ is potentially time-dependent? This contradicts the continuity of temperature across the boundary $x = 0$.

It would be great if someone could clear up my confusion.

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1 Answer 1

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While the continuity you mention should hold for $k$ continuous, it need not hold in general.

For example, if $k(x) = \operatorname{Heaviside}(x)$, the heat distribution takes the following form:

If you think about it from the perspective of a diffusion process, the behavior becomes more clear. Consider a particle $X$ whose motion is given by the SDE $$ dX_{t}=\sqrt{2\operatorname{Heaviside}(X_{t})}dW_{t}. $$ By the Feynman–Kac formula, the heat distribution is $$ u(x,t)=\mathbb{E}\left[\phi(X_{0})\middle|X_{-t}=x\right]. $$

Remark. In the notation above, the initial time is $-t$ and time moves forward to the final time $0$.

If the particle is at position $x < 0$ at the initial time, it does not move (i.e., there is no heat transfer). In other words, its position at the final time is also $x$. By virtue of this, the distribution in the left rod does not change. However, there is clearly heat transfer in the right rod, and hence we arrive at a discontinuity.

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  • $\begingroup$ Thank you for the answer! I was taught that you "prove" the continuity by the following argument (admittedly not rigorous by any standard): If $u(x,t)$ was discontinuous, its derivative would have a $\delta(x)$, and its second derivative would contain a term like $\delta'(x)$. But none of the other terms in the PDE contain the derivative of a Dirac delta, which is a contradiction. I was wondering what part of this breaks down in this example? What's special about the case where $k(x)$ is not continuous? $\endgroup$ Commented Nov 16, 2023 at 4:53
  • $\begingroup$ I don't know about this argument but as you yourself pointed out, in this case, the PDE on the left hand side of the equation is $\partial u / \partial t = 0$ (i.e., no change of the solution over time). $\endgroup$
    – parsiad
    Commented Nov 16, 2023 at 5:24
  • $\begingroup$ If $k$ is not continuous, then its generalized derivatives do contain various $\delta$ terms. You could also look what happens if $k_2$ is a very small but still positive number, or if $k$ goes continuously between $k_1$ and $0$ in a very short $x$ distance. $\endgroup$
    – aschepler
    Commented Nov 16, 2023 at 5:26
  • $\begingroup$ @parsiad yeah I'm totally convinced of that. I just want to know when exactly I'm allowed to invoke the continuity condition when I'm solving problems like this. Because it is a common condition people tend to use when solving heat transfer problems. $\endgroup$ Commented Nov 16, 2023 at 5:41
  • $\begingroup$ @aschepler don't you just get a $\delta$ term from $k$? You have $\frac{\partial}{\partial x} [k(x) u_x(x,t)] = k'(x) u_x(x,t) + k(x) u_{xx}(x,t) = -k_1 \delta(x) u_x(x,t) + k(x) u_{xx}(x,t)$. It just gives a Dirac delta, not its derivative. $\endgroup$ Commented Nov 16, 2023 at 5:44

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