My system of PDEs is for $\vec{V}$ and $R$ constants: $$\mathrm{div}\vec{v}=0 \tag1$$ $$(\vec{V}\cdot\vec{\mathrm{grad}})\vec{v}-\frac{1}{R}\triangle \vec{v}=-\vec{\mathrm{grad}}p\tag2$$ with the following BCs $$\lim_{|x|\to\infty}\vec{v}=\vec{0} \qquad \lim_{|x|\to\infty}\vec{\mathrm{grad}}\vec{v}=\widetilde{0} \qquad \lim_{|x|\to\infty}p=0$$
$\textbf{PROBLEM:}$
I would like to find out if I can replace equation $(1)$ by: $$\triangle p=0\tag3$$ Which is obtained taking the divergence of $(2)$, taking $(1)$ into account. With an aditional BC ($p$ must have a jump over some surface, but its derivative must be continuous).
$\textbf{WHAT I'VE TRIED SO FAR:}$
I know:
1) Thanks to $(3)$ and its additional BC I know that $\vec{\mathrm{grad}}p$ is continuous except possibly the tangential component of the gradient through the boundary (surface).
2) Thanks to $(2)$ that the velocity and its derivatives are continuous (because of the Laplacian).
3) Due to points 1) and 2) from $(2)$ that also the second derivatives of $\vec{v}$ are also continuous (because they are equal to functions that are also continuous).
Therefore, given equation $(3)$ and $(2)$, taking the divergence of $(2)$ we have an additional equation for $\phi=\mathrm{div}\vec{v}$
$$\mathrm{div}\left(\vec{V}\phi -\frac{1}{R}\vec{\mathrm{grad}}\phi\right)=0\tag4$$
Integrating equation $(4)$ once we obtain that what is within brackets must be equal to a divergence-free vector field $A$ that is finite everywhere and vanishes when $|x|\to\infty$ [applying 1), 2) and 3)]: $$\vec{V}\phi -\frac{1}{R}\vec{\mathrm{grad}}\phi=\vec{A}$$
According to Helmholtz $\vec{A}$ must be a curl of some vector potential $\vec{A}=\vec{\mathrm{curl}\vec{\Psi}}$, where
$$\vec{\Psi} = \frac{1}{4\pi}\int_{V}{\frac{\vec{curl}'\vec{A}(\vec{x}')}{|\vec{x}-\vec{x}'|}\,dV'}-\frac{1}{4\pi}\oint_{\partial V}{\vec{n}'\times\frac{\vec{A}(\vec{x}')}{|\vec{x}-\vec{x}'|}\,d\sigma'} \tag{H} $$
Since the potential part of the decomposition (which is a solution of Laplace equation) must vanish at infinity and must be continuous everywhere this part must be $0$ everywhere.
Considering an unbounded domain, $\textit{i.e.}$ $V=\mathbb{R}^3$, for the second integral of $(H)$ to vanish the vector function $\vec{A}$ must, at least decay as $\vec{A}\propto |\vec{x}'|^{-p}$ with $p>1$.
As a result $\vec{curl}'\vec{A}\propto |\vec{x}'|^{-p-1}$ and the integrand of the first addend behaves as $|\vec{x}|^{-p-2}$. Since $p>1$ this integrand is singular for some point within $V$. Therefore $V$ can be split into $V=B(\epsilon,\vec{x}) \cup V\backslash B(\epsilon,\vec{x})$ with $\epsilon\to 0$.
but to fulfill the smoothness requirement of the derivatives of the velocity (variable $\phi$) this integral must be $0$ setting $\vec{curl}'\vec{A}=\vec{0}$. The integral within the ball $B$ if it is solvable ($2>p>1 $) would vanish and if it is not, the variable $\phi$ would not be defined and therefore $\vec{A}=\vec{0}$.
I have if $p>2$: $$\vec{V}\phi -\frac{1}{R}\vec{\mathrm{grad}}\phi=0\tag5$$
The general solution of $(5)$ is: $$\phi = c\exp{\left(R\vec{V}\cdot\vec{x}\right)}$$ We see that $c=0$ because we need a solution for $\vec{v}$ in which $\vec{v}$ itself and its derivatives vanish at infinity. And therefore we have: $$\phi = 0$$
$ \textbf{I can not prove the case for which $2>p>1$}$.
$\textbf{Thank you}$ for your time and interest.