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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.

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1 Answer 1

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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.

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  • $\begingroup$ 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

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