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Let $\Omega\subset \mathbb R^N$ be an open, smooth and bounded subset.

Given a $N\times N$, bounded and elliptic matrix of Hölder continuous functions. That is, $A(x)= \{a_{ij}(x)\}_{N\times N}$, $a_{ij}(x)\in C^{0,\alpha}(\Omega)$ and for some fixed $C>0$, $$ \frac{1}{C} |\xi|^2 \leq \langle A(x)\xi,\xi \rangle\leq C|\xi|^2. $$ for every $\xi \in \mathbb R^N$

$\textbf{Question:}$

Does there exist a vector field $\Phi$ (expected to be $C^{1,\alpha}$ smooth) such that the following system of PDEs is solved: $$ |det(D\Phi)(x)|^{1/2} |D\Phi (x)\xi|= |A(x)\xi| $$ for every $\xi \in \mathbb R^N$

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In general, No. Your equation is overdetermined. In components, $\partial_j \Phi_i = B_{ji}$ (defining $B$ in terms of $A$), implying $\partial_k B_{ji} - \partial_j B_{ki} = 0$ (you need to interpret the derivative on $C^{0,\alpha}$ in a suitable weak/distributional sense). If $B_{ji}$ does satisfy this integrability condition, then $\Phi_i$ exists locally, using a Poincaré lemma at the right level of regularity. Globally on $\Omega$, $\Phi_i$ may be forced to be multivalued if $\Omega$ is not simply connected.

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  • $\begingroup$ thank you for your answer, since I am not very familiar with differential forms, if yiu could clarify what "integrability condition" on $B_{ij}$ you are referring to? moreover, is the claim true if instead of equality, we have the following condition :$ |D\Phi (x) \xi| = \frac {|A(x)\xi|}{|det(A(x))|^{1/N+2}}$? this condition is more relaxed than one in question, instead of matrices being equal we only need to preserve distanced from otigin. $\endgroup$
    – Harish
    Commented Feb 23, 2021 at 0:50
  • $\begingroup$ @Harish By "this integrability condition" I referred to "\partial_k B_{ji} - \partial_j B_{ki} = 0". Since the $\xi$ in your new condition is arbitrary, equivalently $(D\Phi)^T (D\Phi) = B$ (collect the coefficients of $\xi_i\xi_j$) with $B$ a symmetric matrix proportional to $A^T A$ with the corresponding determinant normalization factor. This equation is no longer linear in $\Phi$ (it is now "fully nonlinear") and honestly I don't know how to approach it. $\endgroup$
    – Igor Khavkine
    Commented Feb 23, 2021 at 10:10
  • $\begingroup$ I saw thhe problem bit differently, if the solution $\Phi$ is such that $D\Phi$ only need to preserve distances, then $D\Phi(x)= U(x) B (x) U(x)^{-1} $, for some distance preserving matric $U(x)$. where $B(x)= |det (A(x))|^{1/N+2} A(x)$. Now that we have more flexibility in solving the problem by choosing any $U$, does it not make the problem easier? (though I still not know the answer) $\endgroup$
    – Harish
    Commented Feb 23, 2021 at 13:59
  • $\begingroup$ @Harish The two generalizations that you proposed are not equivalent. The second one reduces to $D\Phi = B'$, where $B' = U B U^{-1}$. Now it is $B'$ that needs to satisfy the integrability condition, which becomes a condition on $U$. Unfortunately, the condition on $U$ is again non-linear and I also don't see how to approach it. Sorry! $\endgroup$
    – Igor Khavkine
    Commented Feb 23, 2021 at 14:36

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