in our finite element class we were talking about interface problems our teacher came up with the following, where $K_i$ are two given functions and $u_i$ is the restriction of the solution $u$ to $\Omega_i$.

$$-\nabla \cdot (K_i \nabla u_i) =f_i \text{ in } \Omega_i$$ $$u_i=0 \text{ on } \partial \Omega \cap \Omega_i$$ $$n_1 \cdot K_1 \nabla u_1 + n_2 \cdot \nabla u_2 = 0 \text{ on } \Gamma:=\partial \bar{\Omega_1} \cap \partial \bar{\Omega_2}$$ $$[u]=0 \text{ on } \Gamma$$

We assumed that $\Omega_2 \subset \Omega_1$, i.e. one domain is included in another.

Question: what is the physical meaning of the condition $$n_1 \cdot K_1 \nabla u_1 + n_2 \cdot K_2 \nabla u_2 = 0 \text{ on } \Gamma$$ ? Can you provide a simple example where this condition means something from the physical standpoint?

For sure it is a continuity condition, but I cannot imagine a physical situation where this occurs. Examples from elasticity/solid mechanics are of course welcome.


2 Answers 2


$\vec\Phi = K_i \nabla u_i$ is the flux across the interface. For example, if $u$ is the thermal energy density and $K$ the thermal conductivity, then $\vec\Phi$ is the thermal energy flux. Energy conservation then dictates that whatever flows into the interface on one side ($\vec n_1 \cdot K_1 \nabla \vec\Phi_1$) better be equal to what flows out on the other side ($-\vec n_2 \cdot K_2 \nabla \vec\Phi_2$). This then results in the condition you show.

Similarly, if $u$ is the concentration of a substance diffusing in a medium, where the diffusion constant is $K$, you end up with the same condition by considering the conservation of mass of the substance.

  • $\begingroup$ Thanks @WolfgangBangerth, that was exactly what I was looking for. One side question. If I have instead a problem like this (where $\Omega_2 \subset \Omega_1$, so $\Gamma = \partial \Omega_2$): $$- \nabla \cdot ( K_i \nabla u_i) = f_i \text{ in } \Omega_1 \cup \Omega_2$$ $$u= 0 \text{ on } \partial \Omega_1$$ $$[u]=0 \text{ on } \Gamma$$ $$n_1 \cdot K_1 \nabla u_1=0 \text{ on } \Gamma$$ I'd say that that Neumann boundary condition is not physical, meaning that energy is not conserved. Is this correct? $\endgroup$
    – FEGirl
    Commented Nov 28, 2021 at 20:20
  • $\begingroup$ It just means that from that one side no energy flows. That's fine -- the interface might be insulated. But this only makes sense if no energy flows from the other side either. $\endgroup$ Commented Nov 29, 2021 at 1:28
  • $\begingroup$ Apparently the question implies that n1 and n2 are in general two different vectors. On the other hand, in this answer n1 and n2 would not be different (other than the sign) because it is the same interface. $\endgroup$ Commented Nov 29, 2021 at 7:41
  • 1
    $\begingroup$ @MaximUmansky I think that's a mistake in the question. $\endgroup$ Commented Nov 29, 2021 at 13:34
  • $\begingroup$ @MaximUmansky Of course, indeed what I meant is $n_1 = -n_2$. $\endgroup$
    – FEGirl
    Commented Nov 29, 2021 at 19:52

In addition to Wolfgang Bangerth's explanation of temperature and concentration, let me give an other application where such interface conditions arise: (linear) elasticity, which has a similar structure to the elliptic equation in your example. Consider the situation with no interface first. Then the equilibrium, constitutive and kinematic equations of linear elasticity read \begin{gather} \sigma\cdot\nabla = \mathbf{0}, \quad \mathbf{x}\in\Omega \\ \sigma = \mathcal{C}:\varepsilon, \quad \mathbf{x}\in\Omega \\ \varepsilon = \nabla^s\mathbf{u}, \quad \mathbf{x}\in\Omega \end{gather} with properly defined boundary conditions, where $\Omega = \Omega_1 \cup \Omega_2$, $\mathcal{C}$ is the Hooke tensor, $\sigma$ is the Cauchy stress tensor, $\varepsilon$ is the linearized strain tensor and $\mathbf{u}$ is the displacement vector.

A physically equivalent formulation is when these equations are written for each subdomain and the subdomains are tied together: \begin{gather} \sigma_i\cdot\nabla = \mathbf{0}, \quad \mathbf{x}\in\Omega_i \\ \sigma_i = \mathcal{C}_i:\varepsilon_i, \quad \mathbf{x}\in\Omega_i \\ \varepsilon_i = \nabla^s\mathbf{u}_i, \quad \mathbf{x}\in\Omega_i \end{gather} with the continuity of the displacement field (no interface separation) and the traction field (Newton's third axiom): \begin{gather} \mathbf{u}_1| = \mathbf{u}_2, \quad \mathbf{x} \in \Gamma \\ \mathbf{t}_i = -\mathbf{t}_j, \quad \mathbf{x} \in \Gamma. \end{gather} By Cauchy's theorem, the traction vector can be expressed with the stress tensor as \begin{equation} \sigma_i|_\Gamma\cdot\mathbf{n}_i = -\sigma_j|_\Gamma\cdot\mathbf{n}_j. \end{equation} If you substitute the constitutive and kinematics equations into this equation, you get the structurally same condition as what your teacher wrote. Here, the fourth-order tensor $\mathcal{C}_i$ corresponds to the scalar $K_i$ and $\mathbf{u}_i$ corresponds to $u_i$.

Remark 1: The above formulae hold even if $\Omega_2 \not\subset \Omega_1$ (i.e. $\Gamma$ is not closed), but $\Omega_1$ and $\Omega_2$ have a common boundary (i.e. $\Gamma \neq \emptyset$).

Remark 2: The same considerations hold for multiple subdomains, in which two neighbouring subdomains $\Omega_i$ and $\Omega_j$ are separated by the interface $\Gamma_{ij}$.

Remark 3: Note that writing the PDEs for each subdomain independently and then tying them together resembles a domain decomposition method. Indeed, if the continuity conditions are imposed by a Lagrange multiplier (which turns out to be equivalent to the interface traction), the resulting mixed formulation is the continuous analogue of the FETI domain decomposition method.

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    $\begingroup$ Yes, nice example! The (normal) fluxes in this case correspond to tractions. $\endgroup$ Commented Nov 29, 2021 at 1:29

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