This is a cross-post. Let $D \subseteq \mathbb{R}^2$ be the **closed** unit disk, and let $f:D \to D$ be a diffeomorphism.

Does there exist a smooth $h \in C^{\infty}(D)$ such that $h\cdot f$ is an

area-preservingdiffeomophism of $D$?

(*Clarification:* by $h\cdot f$ I mean multiplication of a scalar by a vector, i.e. $\big( h\cdot f\big) (x):=h(x)\cdot f(x)$).

Clearly, such an $h$ must send the entire boundary $\partial D$ either to $1$ or to $-1$. Another necessary condition is $\det\big(d (h\cdot f)\big) = 1$. Since

$$ d (h\cdot f)=h df+dh \otimes f=h df+f \cdot (\nabla h)^T, $$

the matrix determinant lemma implies that $$ 1=\det\big(d (h\cdot f)\big)=h^2\det(df)+(\nabla h)^T \operatorname{adj}( hdf ) f. \tag{1} $$

In fact, I am not even sure whether the PDE $(1)$ always has a solution for every diffeomorphism $f$. (that is even when omitting the requirement that $h\cdot f$ would be a diffeomorphism, or even a map from $D$ into $D$-when looking only at the PDE with no other restrictions on the result-does there always exist a solution?)