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@Peter . Since the gradient has only radial component it is proportional to $ \hat{\mathbf{r}}=\sin\theta'\cos\varphi'\hat{\mathbf{x}}+\sin\theta'\sin\varphi'\hat{\mathbf{y}}+\cos\theta'\hat{\mathbf{z}}$. The first two terms do not contribute because the $\varphi'$ integral vanishes, and you are left with $ \int_0^{2\pi}d\theta'\:\cos\theta'\sin\theta'\Big/\sqrt{r^2+r'^2-2rr'\cos\theta'}=0 $.
Yes, the hyperbolicity and local solvability was shown by DeTurck and Yang, but we have almost no global result. For example, the $\textit{simplest}$ case of a unit disk to itself was discussed lengthly in this question* but there's no conclusive answer. I would consider an answer some criteria based on geometric properties of the domains $D_1$ and $D_2$, like an isoperimetric inequality involving areas and lengths of boundaries, that if satisfied (violated) implies that there is (no) such an $f$.
Understood, thank you. Do you $\textit{think}$ there is a way to answer if, given two (finite, connected) domains $D_1$ and $D_2$, there exists a diffeomorphism $f:D_1\rightarrow D_2$ with constant singular values ? As you mentioned, it's easy to construct $\textit{one}$ $f$ on some $D_1$, but how to incorporate the condition of being diffeomorphic to some $D_2$ ? As I stated in the question, these mappings between planar domains are currently under intense research by physicists due to newly discovered materials that deform in a similar manner.
Thank you so much. First please let me ask about the specific example you mention. Didn't we impose $\kappa_1\kappa_2\not=0$ in the aforementioned question ? In that case the domain $S$ cannot contain the $uv$-axes.
