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.