I am looking for a rigorous way to deal with discontinuous thermal conductivity $K(x)$ in the heat equation (e.g. on the boundary of two different materials)

$$ u' = \nabla_x\cdot (K(x)\nabla_x u) + \text{\{possible other terms\}},$$

where $u(t,x)$ is the temperature. Clearly the equation breaks down at discontinuity since the derivative of $K$ does not exist, what becomes the boundary condition of the non-discretized equation above to replace the discontinuity?

When the equation is discretized, for basic finite difference scheme at boundary it is possible to in a loose sense use the "average value" (is there a derivation somewhere?) $$K(x)\approx (1/2) (K(x+1)+K(x)).$$

For finite volume scheme not at the boundary $$Au' = \int\int\int (\nabla_x\cdot K\nabla_x u)dA$$ $$=\int\int (K\nabla_x u) \cdot \bar ndS$$ $$=K\int \int\nabla_xu \cdot \bar n dS$$ Let's say a finite difference scheme is applicable, although there could be more generality $$\approx K \sum_k (u(x_1,...,x_k+1,...,x_d) + u(x,...,x_k-1,...,x_d)-2u(x))$$

The average value approximation then yields $$\approx \sum_k \frac{1}{2}(K(x)+K(x_1,...,x_k+1,...,x_d)) \cdot (u(x_1,...,x_k+1,...,x_d) -u(x)) +...,$$ for the finite volume method, is it valid on the boundary? References to books are welcome.

  • 1
    $\begingroup$ The first who studied and found solutions for discontinuities in the heat equation were John Forbes Nash and Ennio De Giorgihttps : karlin.mff.cuni.cz/~kaplicky/pages/pages/2011z/Nash1958.pdf $\endgroup$
    – The Tiler
    Jun 15, 2023 at 8:49
  • $\begingroup$ @HVAC I had no idea Nash worked in this area of mathematics... Anyway, since the equation is not defined at the boundary should not the boundary condition be defined rather than derived? $\endgroup$ Jun 15, 2023 at 10:39
  • $\begingroup$ Related, if not dupe of, physics.stackexchange.com/q/107761/25301 $\endgroup$
    – Kyle Kanos
    Jun 15, 2023 at 11:42
  • $\begingroup$ @KyleKanos It is referenced in the question. Particularly, whether the same "averaging" works on finite volume is being asked. $\endgroup$ Jun 15, 2023 at 11:46
  • $\begingroup$ Magemathician: Sorry, I'm not a specialist in the field, However, the first notable use of his work was the solution of Hilbert's 19th problem on the analyticity of minimizers of functionals with analytic integrand. $\endgroup$
    – The Tiler
    Jun 15, 2023 at 13:23

2 Answers 2


Put a grid point $x_j$ at the interface between the two regions. Then $$k^+\frac{T_{j+1}-T_j}{\Delta x}=k^-\frac{T_j-T_{j-1}}{\Delta x}$$That is, the heat flux is continuous at the interface.

  • $\begingroup$ So in the original equation $$-K(x)\nabla_x u,$$ is continuous? It's not odd that then the gradient is discontinuous? Where does the condition come from? $\endgroup$ Jun 15, 2023 at 11:38
  • $\begingroup$ Like I said, the heat flux has to be continuous. Where else is the heat going to go? $\endgroup$ Jun 15, 2023 at 11:44
  • $\begingroup$ I have not thought about the meaning of the flux. So it's continuous because conservation of energy? $\endgroup$ Jun 15, 2023 at 11:50
  • $\begingroup$ Sure. It’s helpful to understand this stuff at the gut level. $\endgroup$ Jun 15, 2023 at 12:24
  • $\begingroup$ @Chef Miller Yes, divergence theorem needs even continuous differentiability and should hold. $\endgroup$ Jun 16, 2023 at 9:03

The scheme appears applicable, as discussed here. First, from Chet Miller's answer and comments the divergence theorem must hold and the flux be cotinuously differentiable, so at boundary the quantity of interest is $$\int\int (K(x)\nabla_x u) \cdot \bar ndS.$$ Now make the approximation the quantities vary linearly within the cell and there are finite $N$ number of faces. Apparently there is a theorem that then the integral is equal to the value found at the face centroid $$\approx \sum_{f=1}^N K_f (\nabla u)_f \cdot \bar n_f S_f,$$

where the subscript denotes the value at the face centre (and $S_f$ the surface area of the face). To estimate the face-value central differencing can be used for $K_f$, which gives the weighted average, weighted by the relative distance to the face.


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