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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 $u\in W^{1,p}_0(\Omega)$.
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    $\begingroup$ And presumably you also know the answer when $F\in L^p$, $p>1$? $\endgroup$
    – Deane Yang
    Commented Apr 30, 2018 at 2:50
  • $\begingroup$ @DeaneYang Good point. I will modify my question. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 30, 2018 at 3:01
  • $\begingroup$ Oops. I assumed that the $p >1$ case was more straightforward using, say, elliptic regularity estimates for Sobolev spaces. Surely that’s true at least for $p=2$? $\endgroup$
    – Deane Yang
    Commented Apr 30, 2018 at 3:27
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    $\begingroup$ There is a correspondence between functionals acting on $W^{1,p'}$ and $\mathop{div} F$ for $L^p$ vector fields for $1 \leq p < \infty$, so you are essentially looking at right hand sides from $W^{-1,p}$ there (Maz'ya: Sobolev Spaces, Ch.1.1.14). For these, maximal elliptic regularity would give solutions in $W^{1,p}$, at least for $p>1$. (I think the $\mathop{div} F$ representation is classically used in the Ladyzhenskaja book(s).) I'd also be interested about $p=1$, though.. $\endgroup$
    – Hannes
    Commented Apr 30, 2018 at 8:55
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    $\begingroup$ When you use \operatorname{div} then the spacing to the right and left depends on the context, so that, for example, in $a\operatorname{div} b$ you see more space to the right of div than in $a\operatorname{div}(b),$ and you don't need to add space manually. I edited this question accordingly. $\endgroup$
    – Michael Hardy
    Commented May 1, 2018 at 17:42

1 Answer 1

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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 $$\phi(0_+)=\hat u(0)=1,\quad \phi(0_-)=-\hat u(0)=-1.$$

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  • $\begingroup$ Can you provide references pointing a particular theorem in a book? Without such references the answer is of limited use. I see that parametrix maps $W^{-1,p}$ to $W^{1,p}$ as indicated in comments in my question, but I do not see why it maps to $W_0^{1,p}(\Omega)$. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 30, 2018 at 21:59

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