# Nonlinear PDE: regularity issues

I have this execrise I've been trying to solve

Let $B=\{x \in \mathbb R^n: |x|<1\}, \quad a: \mathbb R^n \rightarrow \mathbb R , \quad a\in C^\infty(\mathbb R^n) \cap L^\infty(\mathbb R^n)$

$$\begin{cases} -\Delta u +a(\nabla u) = 1 & \text{in B} \\ \partial_{\nu }u + u = 0 & \text{in \partial B} \end{cases}$$

I've been able to show that the solution exists and it also belong to $H^2(B)$, the next point I'm trying to show is:

Show that every solution $u \in H^2 (B)$, is also $C^\infty( \bar B)$

In the solution of my professor the answer is:

Being $a (\nabla u) − 1 \in L^\infty (B)$ we get that $\nabla u ∈ L^\infty (B)$

Since $a \in C^\infty(B)$ and $\nabla u \in H^1 (B;\mathbb R^n)$ we get $a(\nabla u) \in H^1(B)$

....

From now on is all clear to me, but how can I say that $\nabla u ∈ L^\infty (B)$ ??

Any help would be much appreciated

• Use the Definition of your pde! – Quickbeam2k1 Jun 9 '16 at 17:37
• that's the point, I don't get how can I infer $\nabla u ∈ L^\infty (B)$ from $\Delta u ∈ L^\infty (B)$ – user5609462 Jun 9 '16 at 17:38
• The weak formulation of the problem doesn't contain the Laplacian, only the gradient (and the gradient of your test function). Does that help? – Ian Jun 9 '16 at 17:41
• Ehm.... I still can't see it! – user5609462 Jun 9 '16 at 21:41
• I don't follow the sentence with "we get that", either. For all we know, $a$ could be identically zero, in which case $a(\nabla u) - 1\in L^\infty$ is not telling us anything about $\nabla u$. – user147263 Jun 10 '16 at 0:38