# 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

• I forgot to add that $u$ is divergence free. Sorry. Mar 1, 2013 at 4:36
• Craig, can you explain why you expect $\Delta u$ to be divergence free, to? which identity of vector calculus is in action here? Mar 1, 2013 at 7:41
• Thanks for your responce Delio. If we write $u=(u^i)$ then we $- \Delta u^i + P_{x_i}= f^i$ and then take $\partial_{x_i}$ of the $i^{th}$ equation and sum over $i$ and then reverse the Laplacian and the first order partials. We can make this rigourous by fixing $\phi$ to be smooth and compactly supported and then we can multiply the equation for $u^i$ by $\phi_{x_i}$ and integrate by parts (never using any more than the fact that $u \in H_0^1$ and $u$ is divergence free) to arrive at $\int_\Omega \nabla P \cdot \nabla \phi = \int_\Omega f \cdot \nabla \phi$. Mar 1, 2013 at 9:03
• Ok. Now about solving this weak PDE with no boundary condition for $P$? What would you do? Mar 1, 2013 at 13:02