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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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    Yes, it's common in the inhomogeneous p-Laplace literature to write the RHS as the divergence of a field. Usually without a reference to why it's possible; perhaps because the fact was internalized by the collective mind of the researchers in the area. I gave a recent reference because they are easier to find online. The book by Brezis is a natural source for this sort of results.2012-12-22

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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 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.2012-12-27