Full wall Stokes pressure requires a divergence inverse

Assume full AC. Let $d\ge 2$, $R>0$, and let $\mathrm{\Omega }\subset (-R,R{)}^d$ be a nonempty bounded open set. For each $\mu >0$ and $f\in L^2(\mathrm{\Omega };{\mathbb{R}}^d)$, the projected stationary Stokes theorem gives a unique $u$ in the distributionally divergence-free subspace $V$ of the closure-based $H_0^1(\mathrm{\Omega };{\mathbb{R}}^d)$ space, satisfying $\mu {\int }_{\mathrm{\Omega }}\mathrm{\nabla }u:\mathrm{\nabla }vdx={\int }_{\mathrm{\Omega }}f\cdot vdx(v\in V),$ with $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\le 2R\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{/}\mu$; this conclusion contains no pressure. If $\mathrm{\Omega }={\prod }_{i=1}^d(a_i,b_i)\subset (-R,R{)}^d$ is a nonempty finite open rectangle, the separate rectangular theorem additionally gives $p\in H^{-1}(\mathrm{\Omega })$ such that $-\mu \mathrm{\Delta }u+\mathrm{\nabla }p=f$ as vector distributions, with $p$ unique modulo constant distributions. The rectangular result does not assert an $L^2$ pressure estimate or smoothness at rectangle corners. Neither theorem claims pressure recovery for arbitrary bounded open sets; such a claim needs a domain-appropriate divergence right inverse or inf-sup argument.

Martin Vohralík, A posteriori error estimates for efficiency and error control in numerical simulations, Course NM497 lecture notes (2010–2011).

Theo Bühler and Dietmar Salamon, Functional Analysis.