Incompressible Fluid form of the Navier Stokes Equations - Is pressure given?

Question

I am trying to understand the Navier-Stokes equation for incompressible fluid flow. $\newcommand{\vect}[1]{{\bf #1}}$ $$ \frac{\partial \vect{u}}{\partial t} + (\vect{u} \cdot \nabla)\vect{u} - \nu\nabla^2\vect{u} = -\nabla \frac{p}{\rho_0} + \vect{g} \tag{1} $$

where

• $\vect{u}$ is the flow vector field
• $t$ is time
• $\nu$ is the viscosity of the fluid
• $p$ is the pressure (a map from $\mathbb{R}^3 \mapsto \mathbb{R})$
• $\vect{g}$ is some external force (a vector)
• $\rho$ is the density of the fluid.

Additionally the mass conservation condition $$ \nabla \cdot \vect{u} = 0\label{2}\tag{2} $$ holds.

According to wikipedia,

The solution of the equations is a flow velocity. It is a vector field - to every point in a fluid, at any moment in a time interval, it gives a vector whose direction and magnitude are those of the velocity of the fluid at that point in space and at that moment in time.

Is the pressure given? Or are we also solving for the pressure?

Answer 1

The pressure is solved for. That said, it is common to apply the Leray projector $\mathbb P$ (projection to the space of divergence free vector fields; recall the Helmholtz decomposition) to the Navier-Stokes equation to obtain an equation without explicit mention of $p$,

$$ \partial_t u + \mathbb P ((u\cdot\nabla) u) - \nu \Delta u = \mathbb P g$$ If you manage to find the solution $u$ to this system, then $p$ can be recovered (assuming everything is $C^\infty$ and integrable enough, say) by taking the divergence of the Navier-Stokes equation, leaving you with a Poisson equation for the pressure $p$: $$ -\frac1{\rho_0}\Delta p = \nabla\cdot ((u\cdot\nabla u)) - \nabla \cdot g $$ (the density $\rho \equiv \rho_0$ is constant for incompressible N-S) so you can use the solution theory for Poisson equations to find $p$.

I learned this from the book of Majda and Bertozzi "Vorticity and Incompressible Flow", and I believe it is also in this more recent book "The Three-Dimensional Navier–Stokes Equations" by Robinson, Rodrigo, & Sadowski. Its got to be in most books on the incompresible Navier-Stokes equation.

Answer 2

Pressure is usually obtained by using the auxiliary incompressibility condition: $\nabla \cdot \bf u=0$

Answer 3

Sorry if the following is bullshit, but my Navier-Stokes notes has the following pattern, which looks like solving the pressure field (but with a warning that I haven't properly tried if the pattern works):

Let's consider Navier-Stokes equation as the following funcion: $$ f(u,p,x_1,x_2,x_3,t)=0, $$ with $$ f(u,p,x_1,x_2,x_3,t) = \frac{\partial}{\partial t}u_i + \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} - v \Delta u_i + \frac{\partial p}{\partial x_i} - g_i(x,t) = 0. $$ Pressure can be solved by implicit functions technique: $$ f(x,y) = 0 \Rightarrow f\big(x,h(x)\big) =0 \implies h(x)\text{ can be calculated.} $$ Same way, for the Navier-Stokes equation we have: $$ f\big(u,p(u,x_1,x_2,x_3,t),x_1,x_2,x_3,t\big) = 0 $$ $$ \implies \frac{\partial}{\partial t}u_i + \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} - v \Delta u_i + \frac{\partial p(u,x_1,x_2,x_4,t)}{\partial x_i} - g_i(x,t) =0 $$ This only works in time 0, because initially we only know $u$ at time 0.
Moving to other side: $$ \begin{split} \implies \frac{\partial p(u,x_1,x_2,x_4,t)}{\partial x_i} &= -\frac{\partial}{\partial t}u_i - \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} + v \Delta u_i + g_i(x,t) \\ \partial p(u,x_1,x_2,x_4,t) &= \bigg( -\frac{\partial}{\partial t}u_i - \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} + v \Delta u_i + g_i(x,t)\bigg) \partial x_i \end{split}$$

then integrate on both sides: $$ p(u,x_1,x_2,x_4,t) = \int \bigg(-\frac{\partial}{\partial t}u_i - \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} + v \Delta u_i + g_i(x,t)\bigg) \partial x_i$$