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given a symmetric matrix of Holder continuous functions $A(x)$ such that $$ \frac{1}{C} |\xi|^2 \leq \langle A(x)\xi,\xi \rangle \leq C |\xi|^2 $$

find a vector field $\Phi$ such that $$ D \Phi(x)^t D \Phi (x) = A(x)^tA(x) $$

as a simplified version of the problem, if someone could help me with the following fully nonlinear first order PDE

given a Holder continuous function $a(x)>\lambda >0$, find a function $\Psi$ such that $$ |\nabla \Psi(x)|^2= |a(x)|^2 $$

do any of the above problems have a solution?

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  • $\begingroup$ Your simplified equation is an eikonal equation, which is a special case of a Hamilton-Jacobi equation. There is a ton of literature on that, some with very limited regularity solution. However, my understanding is that for a lot of the theory there the fact that $\Psi$ is scalar is used strongly. I am not sure if the theory carries to the vector case you asked about. $\endgroup$
    – Willie Wong
    Commented Feb 24, 2021 at 1:41

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I think, at least if A is twice differentiable, then this has a solution if and only if the curvature of the metric $g$, where $g_x(U,V) := U^t A(x)^t A(x) V$, vanishes. If so, then $\Phi$ corresponds to the derivative matrix of a geodesic normal co-ordinate system for $g$.

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  • $\begingroup$ 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? $\endgroup$
    – Harish
    Commented Feb 24, 2021 at 1:22
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    $\begingroup$ 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. $\endgroup$
    – David Hughes
    Commented Feb 24, 2021 at 12:00
  • $\begingroup$ @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. $\endgroup$
    – Igor Khavkine
    Commented Feb 24, 2021 at 12:07
  • $\begingroup$ @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. $\endgroup$
    – Igor Khavkine
    Commented Feb 24, 2021 at 12:08

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