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Let $\Omega\subset\mathbb{R}^d$, $d>1$ be an open and bounded Lipschitz domain. It is well-known, that for a given function $f\in L^q(\Omega)$ with $1<q<\infty$ and and such that $\int_\Omega f=0$, there exists a solution $u\in W^{1,q}_0(\Omega;\mathbb{R}^d)$ of the underdetermined PDE: $$\mathrm{div} u=f$$ and the estimate $$\lVert \nabla u \rVert_q\leq C\lVert f\rVert_q$$ holds for a scaling and translation invariant constant $C=C(\Omega,q)>0$. The solution to this PDE is $u:=\mathcal{B}f$, where $\mathcal{B}:L^q(\Omega)\to W^{1,q}(\Omega;\mathbb{R}^d)$ is so-called Bogovskii's operator.

I'm interested if there is a version of this theorem, preferably with the similar estimate, where the divergence operator is replaced by an arbitrary first order PDE operator $\mathcal{A}:=a^i\partial_i$ with some restrictions on coefficients $a^i$. Unfortunately, it seems that the original proof of Bogovskii does not carry over to the general case.

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    $\begingroup$ It might help if you think of $u$ as an $(n-1)$-form, and then the equation is $du=f\Omega$ where $\Omega$ is a volume form. Since all volume forms of equal volume are isomorphic (a theorem of Moser), this gives you some freedom to change variables. $\endgroup$
    – Ben McKay
    Commented Jun 11, 2018 at 17:48
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    $\begingroup$ Here is a place of some interest: projecteuclid.org/download/pdf_1/euclid.pja/1195511890 $\endgroup$ Commented Jun 12, 2018 at 0:12
  • $\begingroup$ @BenMcKay Unfortunately, the language of differential forms is still a black magic to me. Could that be rephrased more elementary? $\endgroup$
    – user114232
    Commented Jun 18, 2018 at 16:04

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