TITLE: Solution form for Stokes flows
QUESTION [6 upvotes]: If 
$p:\mathbb{R^3} \rightarrow \mathbb{R} $
and
$u: \mathbb{R}^3 \rightarrow \mathbb{R}^3$
satisfy:
$$\nabla p-\nabla^2u=0$$
$$\nabla\cdot u=0$$
How can we prove that every solution is of the form:
$$u=\nabla \phi+v$$ where 
$\phi:\mathbb{R^3} \rightarrow \mathbb{R} $
and
$v: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ satisfy
$$\nabla^2\phi=p $$
$$\nabla^2v=0$$
$$\nabla\cdot v=-p$$
$$$$

REPLY [2 votes]: Assume that $u$ solves your problem.
Take $\phi$ such that $$\Delta\phi=p\tag{1}$$
Define $v=u-\nabla\phi$ and note that $$\operatorname{div}(w)=\operatorname{div}(u)-\operatorname{div}(-\nabla \phi)=0-\Delta\phi=-p\tag{2}$$
On the other hand $$\Delta v=\Delta (u-\nabla \phi)=\Delta u-\nabla(\Delta \phi)=\nabla p-\nabla p=0\tag{3}$$
It follows from $(1)$-$(3)$ that $u=v+\nabla\phi$ satisfies the precribed requisites.