# Elliptic regularity for $-\nabla \cdot(a(x)\nabla u)$

Define $$Du= -\nabla \cdot (a(x)\nabla u)$$ where $$a$$ is a smooth function which is away from $$0$$ and bounded.

Suppose I have $$u \in H^1(\Omega)$$ on a smooth bounded domain and I also know that $$Du \in L^2(\Omega)$$.

Does this imply that $$\Delta u \in L^2(\Omega)$$ and $$u \in H^2(\Omega)$$?

For the Laplacian case: we get immediately that the Laplacian is in $$L^2$$ and then elliptic regularity theory gives $$H^2$$ for $$u$$. But this case, I don't know????

If you expand the divergence in $$Du$$, you get $$- \nabla a \cdot \nabla u - a \Delta u = Du$$ Rearranging yields $$\Delta u(x) = - \frac{1}{a(x)} \nabla a(x) \cdot \nabla u(x) - \frac{1}{a(x)} Du(x)$$ Since $$a$$ is smooth and bounded away from $$0$$, we conclude using Holder's inequality that both terms on the right are in $$L^2$$. In particular, their norms in $$L^2$$ are controlled by $$\left\lVert \frac{1}{a(x)} \nabla a(x) \cdot \nabla u(x) \right\rVert_{L^2} \leq \left\lVert \frac{1}{a} \right\rVert_{L^\infty} \lVert \nabla a \rVert_{L^\infty} \lVert u \rVert_{H^1},$$ $$\left\lVert \frac{1}{a(x)} Du(x) \right\rVert_{L^2} \leq \left\lVert \frac{1}{a} \right\rVert_{L^\infty} \lVert Du \rVert_{L^2}$$ Thus, $$\Delta u \in L^2(\Omega)$$, and elliptic regularity puts $$u \in H^2$$.