Question:
Let $\Omega\subset\mathbb{R}^d$ be convex and open with compact closure, $\varphi\in W^{1,p}(\Omega,\mathbb{R})$ for $2<p<\infty$ and $\Phi:\Omega\to\Omega$ a diffeomorphism. Can I find a function $\psi\in W^{1,p}(\Omega,\mathbb{R})$ such that
$$ \nabla \psi(y) = \nabla\varphi(\Phi(y)) \cdot |\det D\Phi(y)| $$
holds?
IISC the naive choice $\psi(y) = \varphi(\Phi(y))$ does not do the trick (the chain rule gives $D\Phi$ rather than $|\det D\Phi|$).
Background:
I have some function $q\in L^{p'}(\Omega,\mathbb{R}^d)$ satisfying $$ \int_\Omega \nabla \varphi \cdot q = - \int_\Omega \varphi\mu \quad \forall \varphi\in W^{1,p}(\Omega,\mathbb{R}) $$
for some scalar valued $\mu$ satisfying $\int_\Omega \mu = 0$ and I'd like this to also hold true for $q\circ\Phi^{-1}$. So the determinant comes from the transformation formula.
Actually, for now I only consider the case $\Omega=B_r(0)$ and $\Phi(y) = \eta(|y|) y$ where $\eta:[0,r] \to [0,r]$ is a diffeomorphism, but if it works out, I'd like to extend the result to more general sets and transformations.