Solution form for Stokes flows

Question

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 $$\nabla^2\phi=p $$ $$\nabla^2v=0$$ $$\nabla\cdot v=-p$$ $$$$

Answer 1

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.