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Let $\Omega\subset\mathbb{R}^N$ be a bounded domain, $p\in (1,\infty)$, $p< N$, $q>\frac{N}{p-1}$, $F=(f_1,...,f_n)$ with $f_i\in L^q(\Omega)$ and $g\in W^{-1,q}(\Omega)$. How can i solve this equation (I want to find $F$ and $g$ is fixed): $$\int_{\Omega}F\cdot\nabla v=\int_{\Omega}gv,\ \forall\ v\in W^{1,p}_0(\Omega)$$

Edt: I have added a hypothesis that was missing

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Normally, one inverts the divergence operator by convolution with the Newtonian potential: $g*|x|^{2-n}$. Looks like you don't have any boundary conditions on $F$? – user53153 Dec 21 '12 at 23:57
No boundary conditions @PavelM – Tomás Dec 22 '12 at 0:19
Then the domain does not matter much. You can extend $g$ to a larger ball and invert the divergence operator there via integration along segments (Bogovskii operator). See Proposition 2.1 here, or in published version. – user53153 Dec 22 '12 at 1:29
Thank you @PavelM. Im gonna ask you another thing, maybe you can help me. Take a look here ( page 98 Theorem E.0.19. This notes are from 1997 and in this theorem Peral uses the fact that this problem posses a solution. Is there a solution to this problem in more old literature? "Sorry about the english" – Tomás Dec 22 '12 at 11:01
Just to make it clear @PavelM, take a look in (E.4) – Tomás Dec 22 '12 at 11:18
up vote 1 down vote accepted

This is tiny adaptation of demonstration of the theorem 9.20 from Brezis book: Brezis, Haim Functional analysis, Sobolev spaces and partial differential equations. Universitext. Springer, New York, 2011. xiv+599 pp.

Let $L^p(\Omega)^N=L^p(\Omega)\times...\times L^p(\Omega)$ with the norm $\|u\|^p=\int|u|^p$ where $u=(u_1,...,u_n)$. Because $\Omega$ is bounded, we can use the norm $\|u\|_{1,p}=\|\nabla u\|_p$ in $W_0^{1,p}(\Omega)$. Define $T:W^{1,p}\rightarrow L^p(\Omega)^N$ by $$Tu=\nabla u$$

Because $T$ is a isometry we have that $T$ is a bijection on $T(W^{1,p})$. Let $S=T^{-1}:T(W^{1,p}_0)\rightarrow W_0^{1,p}$

Note that by using the hypothesis the functional $Gv=\int_\Omega gv$ is a well defined bounded linear functional in $W^{1,p}_0$. Define $P:T(W^{1,p}_0)\rightarrow\mathbb{R}$ by $$Pv=GSv$$

Because $S$ is continuous and linear, we have that $P$ is a bounded linear functional defined in $T(W^{1,p}_0)$. By using Hahn-Banach we can extend $P$ to a bounded linear functional $\tilde{P}$ defined in $L^{q'}(\Omega)^N$ (because $p>q'$, where $\frac{1}{q}+\frac{1}{q'}=1$).

Therefore we can find $F\in L^{q}(\Omega)^N$ such that $$\tilde{P}v=\int Fv$$

where $Fv=F_1v_1+...+f_nv_n$. If we take $u\in W_0^{1,p}(\Omega)$ then we have $$\int gu=\int F\nabla u$$

Please verify if it is correct.

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I don't see why you consider $\tilde P$ on $(L^q)^N$. We have $P$ bounded on a closed subspace of $(L^p)^N$. Therefore, $P$ can be extended to all of $(L^p)^N$. This extension is represented by an element of the dual space $(L^q)^N$. This is the $F$ that we wanted. – user53153 Dec 22 '12 at 18:57
It is not on $q$, it is on $q'$, i correct it. I did it to guarantee that $F\in L^q$. – Tomás Dec 22 '12 at 19:21
I see, I forgot for the moment that $p$ and $q$ were not conjugate exponents. – user53153 Dec 22 '12 at 19:25
I think the proof would be neater without $p$ in it. That is, define $T:W_0^{1,q'}\to (L^{q'})^N$, then use the fact that $g$ defines a linear functional on $W_0^{1,q'}$ to get a linear functional on $(L^{q'})^N$. By duality you get $F\in (L^q)^N$, which satisfies the desired equation with test functions from $W_0^{1,q'}$. Now that the problem has been solved, you can restrict the space of test functions to the smaller space $W_0^{1,p}$ if you wish.... Whatever you do, don't forget to accept your answer so that SE marks it as resolved. – user53153 Dec 27 '12 at 20:18

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