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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????

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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 $u \in H^2$.

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