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

  • $\begingroup$ Use the Definition of your pde! $\endgroup$ – Quickbeam2k1 Jun 9 '16 at 17:37
  • $\begingroup$ that's the point, I don't get how can I infer $\nabla u ∈ L^\infty (B)$ from $\Delta u ∈ L^\infty (B)$ $\endgroup$ – user5609462 Jun 9 '16 at 17:38
  • $\begingroup$ The weak formulation of the problem doesn't contain the Laplacian, only the gradient (and the gradient of your test function). Does that help? $\endgroup$ – Ian Jun 9 '16 at 17:41
  • $\begingroup$ Ehm.... I still can't see it! $\endgroup$ – user5609462 Jun 9 '16 at 21:41
  • $\begingroup$ 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 $\nabla u$. $\endgroup$ – user147263 Jun 10 '16 at 0:38

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