I have question about Navier-Stokes equation, but firstly some details:

given Navier-Stokes equations:

  1. $\nabla \cdot u = 0$
  2. $\dfrac{\partial u}{\partial t}=-(u\cdot \nabla )u - \dfrac{1}{\rho }\nabla \rho +\nu \nabla ^{2}u+f$

the Helmholtz-Hodge decomposition states that any vector field $w$ can be uniquely decomposed into the form, $w=u+\nabla q$ provided that: $\nabla \cdot u = 0$.

Let's define the operator $P$ as the one which projects any vector field $w$ onto its divergence free part $u=Pw$.

I want to apply $P$ on Eq. 2, but I am not sure how to do it. My solution is the following:

$P\dfrac{\partial u}{\partial t}=-P(u\cdot \nabla )u - P\dfrac{1}{\rho }\nabla \rho +P\nu \nabla^{2} u+Pf$

but how can I apply the operator on the equation? I cannot understand it.


1 Answer 1


The role of pressure field in incompressible Navier-Stokes equations

Before starting the answer, let's recap the role of the pressure field in incompressible Navier-Stokes equations.

Replacing the mass equation $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{u}) = 0$ with the incompressibility constraint $\nabla \cdot \mathbf{u} = 0$, we're dropping thermodynamics from our problem, and applying the kinematic constraint of incompressibility: namely, we are assuming that each material particle keeps constant volume, and thus undergoes an isochoric transformation.

In order to satisfy this kinematic constraint, a scalar field called "pressure" $P(\mathbf{r},t)$ (even if we could discuss a lot about the meaning of calling it pressure...) acts as a Lagrange multiplier needed to prescribe the constraint in each point of the domain. If the pressure field was not there in the incompressible Navier-Stokes equations,

$\dfrac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} - \nu \Delta \mathbf{u} = \mathbf{0}$,

we would have the momentum equation that is not compatible with the kinematic constraint since it's not divergence free in general, since the kinematic constraint sets the divergence of time derivative and the Laplacian to zero, while the term $(\mathbf{u} \cdot \nabla) \mathbf{u}$ has divergence different from zero in general, $\nabla \cdot \left[ (\mathbf{u} \cdot \nabla) \mathbf{u} \right] \ne 0$.

Projection in physical space

Assuming constant and uniform density, and hiding it into the pressure field $\frac{P}{\rho} \rightarrow P$, momentum equation of the Navier-Stokes as we know reads

$\dfrac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} - \nu \Delta \mathbf{u} + \nabla P = \mathbf{0}$

The role of this field is to project the non-linear term, into a divergence free term, namely $(\mathbf{u} \cdot \nabla) \mathbf{u} + \nabla P$. Taking the divergence of the momentum equation

$\dfrac{\partial }{\partial t}\underbrace{(\nabla \cdot\mathbf{u})}_{=0} + \nabla \cdot(\mathbf{u} \cdot \nabla) \mathbf{u} - \nu \Delta \underbrace{\nabla \cdot \mathbf{u}}_{=0} + \underbrace{\nabla \cdot \nabla}_{=\Delta} P = \mathbf{0}$

we can find a Poisson's equation for the pressure field

$-\Delta P = \nabla \cdot [ (\mathbf{u} \cdot \nabla) \mathbf{u} ]$.

Projection in Fourier space

We can have a more clear picture in the domain of the wave numbers, transforming the space domain equation using Fourier transform in space,

$\displaystyle \mathbf{\hat{u}}(\mathbf{k},t) = \mathscr{F}\{\mathbf{u}(\mathbf{r},t)\} = \int \mathbf{u}(\mathbf{r},t) e^{i\mathbf{k} \cdot \mathbf{r}} d^3\mathbf{r}$.

Transform of space derivative reads $\mathscr{F}\{\nabla\} = i \mathbf{k}$ and the transformed Navier-Stokes equations become

$i \mathbf{k} \cdot \mathbf{\hat{u}} = 0$
$\partial_t \mathbf{\hat{u}} + \mathscr{F}\{ (\mathbf{u} \cdot \nabla) \mathbf{u}\} + \nu k^2 \mathbf{\hat{u}} + i \mathbf{k} \hat{P} = \mathbf{0}$

In this domain, a projector naturally arises. Taking the divergence of the momentum equation, i.e. taking the scalar product with $i\mathbf{k}$, we get

$\partial_t \underbrace{i \mathbf{k} \cdot \mathbf{\hat{u}}}_{=0} + \mathbf{k} \cdot \mathscr{F}\{ (\mathbf{u} \cdot \nabla) \mathbf{u}\} + i \nu k^2 \underbrace{\mathbf{k} \cdot \mathbf{\hat{u}}}_{=0} + \underbrace{i^2 \mathbf{k} \cdot \mathbf{k} \hat{P}}_{=-k^2 \hat{P}} = \mathbf{0}$$\qquad \rightarrow \qquad$$\hat{P} = \dfrac{\mathbf{k}}{k^2} \cdot \mathscr{F}\{ (\mathbf{u} \cdot \nabla) \mathbf{u}\}$,

and replacing this expression for $\hat{P}$ in the momentum equation, a orthogonal projector $\mathbb{P}$ that acts on the non-linear term naturally arise to get

$\partial_t \mathbf{\hat{u}} + \nu k^2 \mathbf{\hat{u}} + \underbrace{\left( \mathbb{I} - \dfrac{\mathbf{k} \cdot \mathbf{k}}{k^2} \right)}_{\mathbb{P}} \cdot \mathscr{F}\{ (\mathbf{u} \cdot \nabla) \mathbf{u}\} = \mathbf{0}$.

The projector $\mathbb{P}$ project the Fourier transform of the non-linear term in the direction orthogonal to the wave vector $\mathbf{k}$, and this operation is equivalent in the physical space to projecting this term into a divergence-free space.

Numerical applications for solving Navier-Stokes equations

You can use the projection methods for the numerical solution of Navier-Stokes equations with fractional-step algorithms, like Chorin's method, to solve Navier-Stokes equations as a sequence of:

  • elliptical equations for the pressure field
  • parabolic equations for the velocity field,

without the need for treating Navier-Stokes equations as a nonlinear system of DAEs, formed by the discrete counterparts of the dynamical momentum equation and the algebraic incompressibility constraint.

  • $\begingroup$ Thank you for your help, you answered in a very detailed way. With your permission, I should please go through your answer again. This is a new topic for me. $\endgroup$
    – hch
    Nov 4, 2022 at 17:09
  • $\begingroup$ Tell me what you need, and what you're doing, because the topic could be quite challenging and it can be approached in many different ways. If I know what you need, we can talk not in general, but focusing on the required details $\endgroup$
    – basics
    Nov 4, 2022 at 17:13

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