---
tag: K76033B5G1
kind: definition
name: Navier–Stokes equations
msc: [35Q30, 76D05]
defines: 18KXZHVF4M
uses: [GC5BDZ6QBC, 0299VCW1TH, 0ABK9V9Y64, 0MWV6ZTK6S, 0E0B6RA8K0]
sources:
  - url: https://en.wikipedia.org/wiki/Navier–Stokes_equations
---

## Definition

Let $\Omega \subseteq \R^3$ be an open set, let $T > 0$, let $\nu > 0$ be a constant, and let $f \colon \Omega \times (0, T) \to \R^3$ be a given force.
The *incompressible Navier–Stokes equations* are the system
$$\partial_t u + (u \cdot \nabla) u = \nu \Delta u - \nabla p + f, \qquad \nabla \cdot u = 0,$$
for an unknown [[vector field|0299VCW1TH]] $u \colon \Omega \times (0, T) \to \R^3$, the velocity, and an unknown function $p \colon \Omega \times (0, T) \to \R$, the pressure.
Here $\nabla p$ is the [[gradient|0ABK9V9Y64]] of $p$ in $x$, and $\nabla \cdot u$ is the [[divergence|0MWV6ZTK6S]] of $u$.
The [[Laplacian|0E0B6RA8K0]] $\Delta$ acts on each component, and $(u \cdot \nabla) u$ has components $\sum_{j=1}^3 u_j \partial_j u_i$.
The constant $\nu$ is the kinematic viscosity, and the density is normalised to $1$.

## Remarks

Example: on $\Omega = \R^3$ with $f = 0$, the shear flow $u(x, y, z, t) = (y, 0, 0)$ with constant pressure is a solution.
Indeed $\partial_t u = 0$, $(u \cdot \nabla) u = y \, \partial_x u = 0$, $\Delta u = 0$, $\nabla p = 0$ and $\nabla \cdot u = 0$.

Non-example: $u = (x, 0, 0)$ with any pressure is not a solution, because $\nabla \cdot u = 1 \ne 0$.

The same equations are studied in two dimensions and on the torus.
For a compressible fluid the density $\rho$ becomes an unknown, with the continuity equation $\partial_t \rho + \nabla \cdot (\rho u) = 0$ in place of $\nabla \cdot u = 0$.
Setting $\nu = 0$ gives the Euler equations of an ideal fluid.