Timeline for Solving a fully nonlinear first order PDE
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 24, 2021 at 12:08 | comment | added | Igor Khavkine | @Harish The scalar version of your equation has no integrability conditions, so you could concentrate purely on the regularity aspect, if you so wished. Willie Wong gave some relevant keywords in his comment if you're interested in this direction. | |
| Feb 24, 2021 at 12:07 | comment | added | Igor Khavkine | @Harish There are two orthogonal difficulties to solving your matrix equation, one has to do with regularity (which is what you seem to be concerned about) and one having to do with inegrability conditions (as rightly pointed out by David here and by me for your previous question). As soon as curvature can be defined at a point $x$ for the metric $g=A^t A$ and it is non-vanishing, then your equation cannot be solved on any neighborhood of $x$, independent of the regularity elsewhere. | |
| Feb 24, 2021 at 12:00 | comment | added | David Hughes | Actually, $A$ only needs to be once differentiable. Also, even if $A$ is only continuous, the zero curvature condition could be generalized using the distance function defined by $g$ (and hence $A$), and then comparing triangles in $g$ with those in spaces of constant curvature $\kappa$, getting Toponogov-style bounds and letting $\kappa \to 0$. That would be a necessary condition for $\Phi$ I think, and might be sufficient. | |
| Feb 24, 2021 at 1:22 | comment | added | Harish | nothing can be said when $A$ is just Holder? what about the second case, cant any holder continuous function be modulus of gradient of a $C^{1,\alpha}$ function? | |
| Feb 24, 2021 at 0:58 | history | answered | David Hughes | CC BY-SA 4.0 |