3
$\begingroup$

Let $B_+\subset\mathbb R^n$ be the intersection of the unit ball with the upper halfspace and $u$ be a harmonic function in $B_+$ that vanishes identically in the flat part of the boundary of $B_+$.

We can easily see that $u$ can be extended to $B$ by $-u$ in $B_-$ in a harmonic fashion by using the mean value property (Schwarz reflection principle). This allows us, for example, to use interior estimates near this part of the boundary.

I wonder whether there are other similar ideas or techniques when $u$ solves weakly a PDE like $$ \operatorname{div}(A(x)\nabla u(x))=0, \quad x\in B_+ $$ where $A$ is a symmetric, uniformly elliptic matrix with Lipschitz coefficients. To be specific, assume that $A$ is only defined on $\overline B_+$ (so that we can redefine $A$ on $B_-$ anyway we want but keeping the global Lipschitz condition on the coefficients) and $u$ also vanishes identically in the flat part of the boundary.

What I have tried: if the coefficients of $A$ have higher regularity I think that we can find a biLipschitz map $T$ from the upper half space to the lower half space and which is the identity on the hyperplane $\mathbb R^{n-1}$ such that $$ \operatorname{div}(A_T\nabla (u \circ T)) = 0, \quad x \in T^{-1}(B_+) $$ where $$ A_T = \vert JT \vert DT^{-1} A \circ T (DT^{-1})^t $$ and then we can join both domains $B_+$ and $T^{-1}(B_+)$ and both PDEs and solutions in a way that $A$ is still globally Lipschitz.

Improve this question
$\endgroup$
1
  • $\begingroup$ I should have been more clear with the question. I don't really care how weird is the reflection or the symmetry, I only mind whether I can extend $u$ (through the flat part of the boundary where $u$ vanishes) in a way such that it still is the weak solution of an elliptic PDE (uniformly elliptic, Lipschitz coefficients, ...) $\endgroup$
    – HHN
    Commented Jun 18, 2021 at 8:31

1 Answer 1

Reset to default
1
$\begingroup$

I think not. Set the dimension $n=2$ and consider the elliptic equation $$u_{,11}+u_{,12}+u_{,22}=0$$ in $B_+$. The extension being $u(x)=-u(x_1,-x_2)$, it satifies in $B_-$ the equation $$u_{,11}-u_{,12}+u_{,22}=0,$$ which means that the matrix is not even continuous accross the horizontal line.

Improve this answer
$\endgroup$
3
  • $\begingroup$ But this PDE has constant coefficients, thus we could consider a linear transformation $T$ such that $u \circ T$ is harmonic in $T^{-1}(B_+)$. We could then reflect $u \circ T$ through $T^{-1}(\{0\}\times \mathbb R^{n-1})$ and finally undo this transformation (if I'm not mistaken). $\endgroup$
    – HHN
    Commented Jun 17, 2021 at 16:46
  • $\begingroup$ @jgall. This transformation is not an isometry. Thus reflecting $u\circ T$ and then undoing the transformation yields a solution whose symmetry is uncommon. $\endgroup$
    – Denis Serre
    Commented Jun 17, 2021 at 20:39
  • $\begingroup$ I should have been more clear with the question. I don't really care how weird is the reflection or the symmetry, I only mind whether I can find an extension (through the flat part of the boundary where $u$ vanishes) that still is the weak solution of an elliptic PDE (uniformly elliptic, Lipschitz coefficients, ...) $\endgroup$
    – HHN
    Commented Jun 18, 2021 at 8:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .