elliptic PDE with discontinuous coefficients Consider $ \nabla \cdot (\sigma \nabla u)=0  $ in the unit ball $B\subset\mathbb{R}^3$ with some nice boundary conditions. Further assume that $\sigma\in (L^\infty)^{3\times3 } $ is symmetric positive definite at each point and piecewise constant on some smooth subdomains e.g. inbetween the radii $0<r_1<...<r_N<1$.
In physical applications I often see the comment that $u$ and $(\sigma \nabla u) \cdot n$ are continuous across these boundaries.
Does this follow by itself or are these further assumptions?
Edit: I think it won't make much difference, but we can consider the case where $\sigma$ is a scalar valued function.
Edit: Let $\Gamma$ be the smooth surface where $\sigma$ jumps. For simplicity we assume, that $\Gamma $ is a hypersurface, otherwise we have to locally straighten it. On each side $u$ is smooth and $\sigma$ constant. We will denote the restrictions to each side with subscripts $i=1,2$. Let $\phi\in C_c^\infty(B_\epsilon(x))$ be a testfunction, where $x\in \Gamma$. We get $$0=\int_B (\sigma \nabla u)\cdot\nabla \phi =\int_{\Omega_1}(\sigma_1 \nabla u_1)\cdot\nabla \phi+\int_{\Omega_2}(\sigma_2 \nabla u_2)\cdot\nabla \phi=-\int_{\Omega_1}\nabla\cdot(\sigma_1 \nabla u_1) \phi-\int_{\Omega_2}\nabla\cdot(\sigma_2 \nabla u_2)\phi+\int_\Gamma\phi(\sigma_1\nabla u_1-\sigma_2\nabla u_2)\cdot n.$$ Since on each side $u$ is a classical solution we get $0=\int_\Gamma\phi(\sigma_1\nabla u_1-\sigma_2\nabla u_2)\cdot n.$ Therefore we have $(\sigma_1\nabla u_1)\cdot n=(\sigma_2\nabla u_2)\cdot n$ a.e. on $\Gamma$ and since both are continuous we get equality on $\Gamma$.
Is this correct?
 A: First of all $u$ is not differentiable so we need to define what it means for $u$ to satisfy $\mathrm{div} (\sigma \nabla u) = 0$.

Definition: A function $u \in H^1(B_1)$ satisfies $\mathrm{div} (\sigma \nabla u) = 0$ if $$\int_{B_1}\nabla \varphi \cdot \sigma \nabla u \, dx =0 $$ for all $\varphi \in C^1_0(B_1)$.

The regularity theory for elliptic PDE with coefficients with low regularity is addressed in a lot of detail in Chapter 8 of Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger. For example, an application of Theorem 8.24 to your problem is:

If $u \in H^1(B_1)$ is weak solution of $\mathrm{div} (\sigma \nabla u) = 0$ with $\sigma_{ij} \in L^\infty(B_1)$ then for any $\Omega' \subset \subset B_1$ $$\| u\|_{C^\alpha (\overline{\Omega'})} \leqslant C\| u \|_{L^2(B_1)}$$ where $C=C(n,\Lambda/\lambda)$ and $\alpha=\alpha(n,\Lambda/\lambda)>0$.

Here $$ \Lambda = \sum_{i,j} \vert \sigma_{ij} \vert^2 \leqslant C(n) \|\sigma\|_{L^\infty(B_1)}$$ and $$\sum_{i,j} \sigma_{ij} \xi_i \xi_j \geqslant \lambda \vert \xi \vert^2 $$ for all $\xi\in \mathbb R^n$. This in particular implies $u$ is continuous in $B_1$.
As for $\sigma \nabla u$ I couldn't find a counter-example but I would have thought you would need to assume a little more regularity of $\sigma$. I'll let you know if I think of anything.
