6
$\begingroup$

I have a question which looks like some sort of inverse problem.
Let $B$ denote the unit ball centered at the origin in $R^N$ (take $N \ge 2$).
Given any $h:\partial B \rightarrow (0,\infty)$ (smooth) we would like to find some $ f \ge 0$ in $B$ (sufficiently regular, say $ f \in L^q(B) $ for some $q>N$) such that $v$ satisfies the following:

$\Delta v(x)=f(x)$ in $B$ with $ v=0$ on $ \partial B$ and $ x \cdot \nabla v(x)=h(x)$ on $ |x|=1$.

In the case of $h=C>0$ (constant) one sees they can just explicitly write out a solution. In the case of $h(x)=1 +\epsilon g(x)$ ($g$ fixed $ \epsilon $ small they can also do it).

Any comments would be greatly appreciated.

$\endgroup$

1 Answer 1

3
$\begingroup$

It is possible to do this. Here is a sketch of the construction:

1) Let $w = (|x|-1)h(x/|x|)$. Then $w$ satisfies the desired boundary conditions, and is smooth away from the origin with $\Delta w = (n-1)h > 0$ on $\partial B_1$. The idea is to find an appropriate extension of $w$ from a neighborhood of the boundary to the interior.

2) Let $v_0 = \max\{w, \, c_0(|x|^2 - 1) - c_0\}$. For $c_0 > 0$ small, $v_0 = w$ near the boundary and $v_0$ is the quadratic on a set $E$ whose boundary is a radial graph just inside of $\partial B_1$. Furthermore, $\Delta v_0 > c_1 > 0$ in the distributional sense.

3) Now let $\eta$ be a smooth cutoff function that is $1$ near $\partial B_1$ and $0$ in a neighborhood of $E$. Take $$v = \eta v_0 + (1-\eta) v_{\epsilon},$$ where $v_{\epsilon}$ is a mollification of $v_0$ (so $\Delta v_{\epsilon} > c_1$). Then $v = w$ near $\partial B_1$. Since $v_0 = w$ is smooth where the derivatives of $\eta$ are supported, it is straightforward to check that $\Delta v$ is smooth and positive in $B_1$ for $\epsilon$ small, completing the construction.

$\endgroup$
1
  • $\begingroup$ Thank you very much for the great answer (I will attempt to see if i can understand the proof fully,...) Thanks again. $\endgroup$
    – Math604
    Jun 15, 2016 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.