Timeline for Bernstein's corollary for the case of half space
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2022 at 4:36 | vote | accept | ling | ||
| May 15, 2022 at 4:30 | comment | added | ling | Oh, I understand what you mean. In your example, $a_{ij}$ could be discontinuous and we could guling coefficients to construct a locally uniformly elliptic equation. Then there comes a question is that if we can assume some conditions on $a_{ij}$ to make this problem still true, is it must be uniformly elliptic? Thank you very much for your warm answer. | |
| May 15, 2022 at 3:26 | comment | added | Connor Mooney | To clarify, the positive and negative Hessian eigenvalues of $u$ are locally bounded away from $0$ and $\pm \infty$ outside $B_{1/4}$ (where the equation is already uniformly elliptic), so one can cook up locally uniformly elliptic coefficients on the plane such that $u$ solves the corresponding equation. | |
| May 15, 2022 at 2:55 | comment | added | Connor Mooney | Since $\det D^2u < 0$ outside of $B_{1/4}$, $u$ solves a locally uniformly elliptic equation on the whole plane. | |
| May 15, 2022 at 2:47 | comment | added | ling | Thanks for your answer, it’s so excited for me to have your answer. I am using your notes to study Monge-ampere equation. As for the question, I want to know if it is right for $a_{ij}$ just strictly elliptic but not uniformly elliptic. In your counterexample, the equation is degenerate at $x=\pm\sqrt{2}/2$. Is it could be true if we allow the degeneration only at $\infty$? I think it maybe right, but I have no idea how to prove it. Thank you again for your answer! | |
| May 14, 2022 at 14:29 | history | answered | Connor Mooney | CC BY-SA 4.0 |