Non-unique pressure for Navier-Stokes for incompressible fluid

Question

The Navier-Stokes system for incompressible fluid says $$ \begin{aligned} \dot{\mathbf{u}} + (\mathbf{u}\cdot\nabla)\mathbf{u}-\nu\Delta\mathbf{u}+\nabla{p} &= 0,\\ \nabla\cdot\mathbf{u}&=0. \end{aligned} $$ The non-slip boundary condition for the velocity $\mathbf{u}$ says $\mathbf{u}|_{\partial\Omega}=0.$ But the pressure boundary condition is unclear. Most references I found use Neumann boundary condition which is $\nabla{p}\cdot\mathbf{n}=0$ on $\partial\Omega$ ($\mathbf{n}$ is normal vector of $\partial\Omega$). As consequence, $p$ can be determined up to an arbitrary additive constant! Does this make any sense in physics? How can I determine the constant?

There is also an initial value condition for $p$. We can add an arbitrary function $q(t)$ satisfying $q(0)=0$ to $p(\mathbf{x},t).$ That is, if $p(\mathbf{x},t)$ is a solution, so is $p(\mathbf{x},t)+q(t).$

Answer 1

The motion of the fluid parcel only depends on the gradient of the pressure, not the pressure itself. Think about it this way: we know that pressure increases the deeper you go into water. But if I place a cylinder in the water, the flow will always look like (the colors represent pressure with red being high pressure and blue low pressure; the arrows are velocity vectors)

no matter how deep it is in the water because it is the change in the pressure that determines the flow. So yes, $p$ being defined to an additive constant makes complete sense.

I don't know that you can solve for the constant because there are a multitude of gauges, $\phi$, that can satisfy the $\nabla\phi=0$ requirement in the NS equations.

Answer 2

As long as we are far from compressible effects (low Mach) and the 0 pressure under which cavitation may appear, yes, pressure is defined up to a constant if boundary conditions are given in terms of velocity. If you provide boundary condition in terms of normal stress (e.g. at an inlet), then you provide a pressure value with this boundary condition. But you can add any constant to this boundary pressure and you will obtain the same flow.

Note that the B.C. in $\nabla p\cdot n$ you mention is only an artifact of some numerical techniques where the pressure equation is not solved directly (Chorin method).