Stokes problem; naive question on the  regularity of pressure term I am attempting to recall some basic knowledge related to Stokes' problem.
In particular I am following along in Evans PDE book in section 8.4.
So lets assume that 
$-\Delta u + \nabla P = f $ in $ \Omega$ 
with $ u=0 $ on $ \partial \Omega$ where $ \Omega$ some domain in $ R^N$.   Here $u$ and $f$ are vector valued and $P$ is scalar valued.  $ f \in L^2$ is given.  
In Evans the regularity for the $P$ is only a local $L^2$.    
My question is,  at least formally, can't one take a divergence of the equation to see that 
$\Delta P = div(f)$ and hence we'd expect  $P \in H^1_{loc}$ ?   
thanks
Craig
 A: You are looking at 
$$
-\Delta u + \nabla p = f \mbox{ in }\Omega
$$
with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.
Now what you said is that $P$ is more than $L^2$ by doing the following. Take  a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. 
$$
\int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi  
$$
you integrate by parts in $j$ in the first term to obtain
$$
\int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega  u_i  \partial_{jij}\phi
$$
which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain 
$$
\int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi
$$
Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of 
$$
-\Delta p = -\textrm{div}(f).
$$ 
in the sense that
$$
- \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. 
$$
Once you are here, since your identity does not require more than two derivatives, you can lower it to $W^{2,2}$, and write it as
$$
- \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{2,2}(\Omega)\textrm{ s.t. } \phi= \partial_n\phi=0\textrm{ on } \partial\Omega. 
$$
So your question is: what does it say on the regularity of $p$ ? Does it imply in particular that $p$ has a weak derivative in $L^2$ ?
The problem I see is that you need to cancel also the gradient. If you did not need the gradient to be also zero, the answer would be yes (just as outlined below), but that's not the case. 
You would like to know if this implies that
for  dense family $\psi\in H^1(\Omega)$, there holds
$$
\int_\Omega p \partial_i \psi \leq C \|\psi\|_{H^1(\Omega)},
$$
so that indeed the weak derivative exists. The natural way to use the equation is to find a function $\phi$ such that
$$
\Delta \phi = \partial_i \psi \textrm{ in }\Omega, \textrm{ with } \phi=\partial_n\phi=0 \textrm{ on }
\partial \Omega.
$$
But of course, that's very unlikely that such a $\phi$ exists... 
A: In fact, under reasonable assumptions on $\Omega$, you do get $u\in H^2$ and $p\in H^1$.
However, this needs the regularity theory for elliptic systems, it does not follow from the calculus of variations method discussed in the relevant section of Evans' book.
