Regularity of solutions to $-\Delta u = \operatorname{div} F$, $F\in L^1$ Let $\Omega\subset\mathbb{R}^n$ be a bounded domain with smooth boundary. 


*

*What are the regularity results for solutions to
$$
-\Delta u= \operatorname{div} F,
\qquad
F\in L^1(\Omega,\mathbb{R}^n)?
$$
Can one find a solution in $W^{1,1}_0(\Omega)$? (I think the answer is no, but I do not know a counterexample). I know that $-\Delta u=f\in L^1$ has a unique solution in $W^{1,p}_0(\Omega)$ for all $1\leq p<n/(n-1)$, (see https://mathoverflow.net/a/298962/121665) but now the right hand side is much worse.

*What are the regularity results for solutions to
$$
-\Delta u= {\rm div }\, F,
\qquad
F\in L^p(\Omega,\mathbb{R}^n), 1<p<\infty?
$$
If $1<p<\infty$ we can solve the quation $-\Delta U=F$, $U\in W^{2,p}(\Omega)$. Then $u={\rm div}\, U\in W^{1,p}$ is a solution to $-\Delta u=\operatorname{div} F$, but I do not know if we can take $u\in W^{1,p}_0(\Omega)$.

 A: About your second question. If I understand things correctly, you want to solve the Dirichlet problem
$$
\Delta u = \operatorname{div} F, \quad u_{\vert \partial \Omega}=0.
$$
Since your open set is smooth, the parametrix for this problem is a classical pseudo-differential operator with order $-2$ on the the manifold $\Omega$ (with boundary $\partial \Omega$) and that operator has $L^p$ continuity properties for $p\in (1,+\infty).$ You could say in particular that since $\operatorname{div} F$ belongs to $W^{-1, p}(\Omega)$, you can solve the Dirichlet problem and get a solution in $W^{1, p}_0(\Omega)$.
For your first question, there is a standard difficulty with the space $L^1$ which is poorly behaved with respect to singular integrals. For instance the Hilbert transform, i.e. the convolution with $\operatorname{pv}(i/π x)$, or the Fourier multiplier $\operatorname{sign}\xi$ is not bounded on $L^1(\mathbb R)$. Take for instance  a function $u$ in $L^1$ with integral 1. The Fourier transform $\hat u$ is a continuous function such that $\hat u(0)=1$. Now consider
$$
\phi(\xi)=\operatorname{sign}\xi\times \hat u(\xi).
$$
The inverse Fourier transform  of $\phi$ is the Hilbert transform of $u$ and does not belong to $L^1$, otherwise it would be a continuous function, which is not the case since
$$\phi(0_+)=\hat u(0)=1,\quad \phi(0_-)=-\hat u(0)=-1.$$
