 Solution

Use the interior $\Omega =\{1/2<|x|<2\}$ for the Sobolev space notation; the closed shell in the PDF specifies its boundary. Put $A(x)=\mathrm{diag}(1,1,g(|x|))$. A weak solution has zero boundary trace in the sense of the Sobolev trace theorem, equivalently $u\in H_0^1(\Omega )$, and satisfiesThe minus sign is necessary because the operator in the paper is positive divergence, not its negative. One can first test with $C_c^{\infty }$ functions and then use density in the zero-boundary Sobolev space. Smoothness of $g$ makes its restriction to the compact radial interval bounded, so both sides define continuous functionals on this space. Merely belonging to $H^1$ without the zero trace would omit the boundary condition.