Pressure values at the boundary in Navier Stokes equations An interesting and important question: what boundary conditions should be used for the pressure field in Navier-Stokes system of equations?
I'm currently trying to find a way to derive the pressure boundary conditions from the equations themselves.
\begin{equation}
\begin{aligned}
\frac{\partial \boldsymbol{v}}{\partial t} + \boldsymbol{v} \cdot \nabla \boldsymbol{v} &= -\nabla p + \nu \nabla \cdot \nabla \boldsymbol{v} \\
\nabla \cdot \boldsymbol{v} &= 0,
\end{aligned}
\end{equation}
The impermeability condition means that the velocity component perpendicular to the wall is zero, that is flow can not penetrate the wall, mathematically:
\begin{equation}
 \label{eqn:impermeability}
  \boldsymbol{v}\cdot  \boldsymbol{n}\bigg|_{\partial \Omega}=0,
\end{equation}
where $\boldsymbol{n}$ is the normal vector to the solid wall boundary.
Recalling calculus: the gradient of any function $f$ is defined as the unique vector field whose dot product with any vector $v$ at each point $x$ is the directional derivative of $f$ along $v$, i.e. $\nabla f \cdot v=\frac{\partial f}{\partial v}$,
Applying $(\cdot \boldsymbol{n})$ to the momentum equation at the boundary, and taking into account impermeability leads to boundary condition for the pressure at the wall:
\begin{equation}\label{eqn:pressure-wall}
 \begin{aligned}
  &\left (-\frac{1}{\rho}\nabla p \cdot \boldsymbol{n} + [\nu \nabla \cdot \nabla \boldsymbol{v}]\cdot \boldsymbol{n}\right )\bigg|_{\partial \Omega}=0\implies\\
  &\frac{\partial p}{\partial \boldsymbol{n}}\bigg|_{\partial \Omega}=\rho \left[\nu \nabla \cdot \nabla \boldsymbol{v}\right]\cdot \boldsymbol{n}\bigg| _{\partial \Omega}.
 \end{aligned}
\end{equation}
The diffusive term on the right-hand side is nonzero even after the application of no-slip conditions $\boldsymbol{v}\cdot\boldsymbol{t}\bigg|_{\partial \Omega}=0$ at the wall, where $\boldsymbol{t}$ is a tangential vector to the wall, i.e. diffusion at the wall is directed in the normal direction from the boundary.
For inlets and outlets, an expression similar to the above can be obtained (differs by the addition of transient and advective terms), which leads to $\frac{\partial p}{\partial n}\bigg|_{\partial \Omega}$ as a function of diffusive, advective, transient terms at the inlet/outlet. Thus, the Poisson problem for pressure is not well-posed due to full Neumann boundary conditions.
Are the above derivation and conclusion correct or am I missing something?
 A: You have the right idea! One paper I find very helpful on the topic of the Pressure Poisson Equation (PPE) is this one here; it goes into the various formulations and boundary conditions and their relationships, as well as some numerical analysis related to boundary layers. I'll try to give useful context, and for ease of notation use $\Delta$ in place of $\nabla\cdot\nabla$ for the Laplacian.
For a moment consider a general Poisson equation $\Delta p = f$ on a bounded, sufficiently smooth domain $\Omega$ where $f$ is given and sufficiently regular. It is known that there exists
(a) a unique solution if a nonempty part of the boundary $\Gamma \subset\partial \Omega$ is given Dirichlet boundary conditions by specifying $\left.p\right|_{\Gamma}$ and the rest of the boundary is given Neumann boundary conditions by specifying $\left.\frac{\partial p}{\partial  \boldsymbol{n}}\right|_{\partial\Omega\setminus \Gamma}$
(b) a solution unique up to a constant if we are given all Neumann boundary conditions by specifying $\left.\frac{\partial p}{\partial  \boldsymbol{n}}\right|_{\partial\Omega}$, and the compatibility condition $\int_{\partial\Omega} \frac{\partial p}{\partial  \boldsymbol{n}} = \int_\Omega f$ holds, which can be derived as follows:
$$ \int_{\partial\Omega} \frac{\partial p}{\partial  \boldsymbol{n}} = \int_{\partial\Omega}\nabla p \cdot \boldsymbol{n} = \int_{\Omega}\nabla\cdot\nabla p = \int_\Omega \Delta p = \int_\Omega f;$$
the solution becomes unique as long as we fix the value of $p$ at a single point in $\Omega$.
Now suppose instead that we start from a vector equation $\nabla p = \boldsymbol F$ in $\Omega$ and want to solve for $p$. Note that this equation only specifies $p$ up to a constant, so let's assume that there is some point $x_0 \in \partial\Omega$ where $p$ is specified. By taking the divergence of both sides, we get the Poisson equation $\Delta p = f$ for $f = \nabla\cdot\boldsymbol{F}$. What is a 'natural' boundary condition for this Poisson equation given the vector equation we started from? Since the boundary condition is a scalar, there is some freedom in how we do this. One way as you noticed is to take the dot product with the normal vector $\boldsymbol{n}$ to get the Neumann condition
$\left.\frac{\partial p}{\partial  \boldsymbol{n}}\right|_{\partial\Omega} = \left. \boldsymbol{F}\cdot \boldsymbol{n}\right|_{\partial\Omega};$ if this is the only condition, notice that the compatibility condition is automatically satisfied since
$$ \int_{\partial\Omega} \frac{\partial p}{\partial  \boldsymbol{n}} = \int_{\partial\Omega}\boldsymbol{F} \cdot \boldsymbol{n} = \int_{\Omega}\nabla\cdot\boldsymbol{F}  = \int_\Omega f.$$
Therefore we have a well-posed Poisson problem with Neumann boundary conditions. However, this is not necessarily the only condition we could derive. For example, in two dimensions say one could instead project the vector equation onto the tangent boundary vector $\boldsymbol \tau$ to give $\left.\frac{\partial p}{\partial  \boldsymbol{\tau}}\right|_{\partial\Omega} = \left. \boldsymbol{F}\cdot \boldsymbol{\tau}\right|_{\partial\Omega}$ which could be integrated along $\partial\Omega$ to give Dirichlet boundary conditions for $p$ (where we also need the value of $p$ at $x_0$ to uniquely determine the boundary values). This condition is obviously a bit more complex so not seen as often, but in the paper I cited they show that the derived Dirichlet and Neumann conditions for the PPE are in fact equivalent.
Now what about the PPE derived from Navier Stokes? It has the exact setup as before just with $\boldsymbol F = \rho\left[\nu \Delta \boldsymbol{v} - \frac{\partial \boldsymbol{v}}{\partial t} - \boldsymbol{v} \cdot \nabla \boldsymbol{v}\right]$ and the additional condition $\nabla\cdot \boldsymbol{v} = 0$. This divergence-free condition yields $\nabla\cdot \boldsymbol{F} = -\rho\nabla\cdot(\boldsymbol{v} \cdot \nabla \boldsymbol{v})$
and so the PPE with 'natural' Neumann BCs is
\begin{equation}
\begin{aligned}
\Delta p &= -\rho\nabla\cdot(\boldsymbol{v} \cdot \nabla \boldsymbol{v}) \\
\\
\left.\frac{\partial p}{\partial  \boldsymbol{n}}\right|_{\partial\Omega} &= \left. \rho\left[\nu \Delta \boldsymbol{v} - \frac{\partial \boldsymbol{v}}{\partial t} - \boldsymbol{v} \cdot \nabla \boldsymbol{v}\right]\cdot \boldsymbol{n}\right|_{\partial\Omega}
\end{aligned}
\end{equation}
Notice we did not use any boundary conditions for $\boldsymbol v$ to come up with the PPE; it is well posed regardless of the possible choices for boundary conditions on the velocity since $\boldsymbol v$ is assumed to be a known fixed quantity. However, the impermeability condition $\boldsymbol v\cdot \boldsymbol n = 0$ leads to a slightly less complicated Neumann condition for $p$ since the middle term drops out.
So you basically have all the right ideas, but hopefully with this additional context you can see that the PPE is in fact well-posed even with all Neumann boundary conditions
