# Inconsistency in heat equation over two semi-infinite rods

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.

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.

• 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? Commented Nov 16, 2023 at 4:53
• 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). Commented Nov 16, 2023 at 5:24
• 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. Commented Nov 16, 2023 at 5:26
• @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. Commented Nov 16, 2023 at 5:41
• @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. Commented Nov 16, 2023 at 5:44