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I have a question regarding a decomposition of a vector field. So fix $ 1<p<\infty$ and let $ \Omega$ denote a smooth bounded domain in $ R^N$. Now let $ F $ denote a smooth vector field $F:\Omega \rightarrow R^N$ with $ \|F\|_{L^p}=1$. Now consider $\Delta \phi =div(F)$ in $\Omega$ with $ \phi=0$ on $ \partial \Omega$.

My question is: does there exist some $C_p$ (independent of $F$ and $ \phi$) such that $ \| \nabla \phi \|_{L^p} \le C_p$?

When $p=2$ this is fairly obvious. Can one prove it for another range of $p$ by some interpolation or ???

My real question here is given $ F \in L^p(\Omega, R^N)$ can we decompose $F= \nabla \phi +g$ where $ \phi$ is as above and $div(g)=0$ and where the projections define by $ F \mapsto \nabla \phi$ and $ F \mapsto g$ are continuous in $L^p$.

I was convinced at one point this was always true and I had easy proof...but now I not so sure. I have tried googling Helmholtz and Hodge decomposition but i don't appear to be getting what i want.

thanks.

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  • $\begingroup$ I think i managed to prove this result by first proving an interior regularity result and then on $B_2^+$ by using an odd reflection of $ \phi$ (say $ \psi$ and then $ \Delta \psi=div(G)$) where $G^i$ for $1 \le i \le N-1$ is the odd reflection of $F^i$ and then $G^N(x)= G(x', - x_N)$ for $ x_N<0$ ... does seem like a possible valid approach (I am more interested in whether the result is true as compared to whether by approach is correct) thanks $\endgroup$
    – Math604
    Mar 5, 2017 at 14:19
  • $\begingroup$ Look at the Theorem 1 in sciencedirect.com/science/article/pii/S0022123685710671 When the boundary is $C^1$ it says that your $\phi\in W^{1,p}$. The boundedness of another projection will follow by the triangle inequality. $\endgroup$
    – ABMath
    Mar 5, 2017 at 18:30
  • $\begingroup$ @almaz. Thanks a bunch for your comment. It really helps me. $\endgroup$
    – Math604
    Mar 5, 2017 at 19:05

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