The Navier–Stokes Equations

Honey oozing off a spoon, air peeling over a wing, blood pulsing through an artery, the smoke curling off a candle, the weather over an entire hemisphere — all of them, every last one, obey the same equations. Write them once and you have written down the law of motion for practically every fluid on Earth. They are the Navier–Stokes equations, and they are simply Euler's equation — Newton's second law for a fluid — with one extra ingredient bolted on: viscosity, the internal friction that makes honey thick and water thin.

That one extra term changes everything. It is what makes a fluid stick to a wall, what turns smooth flow into roaring turbulence, and what makes the equations so ferociously hard that nobody on the planet knows whether their solutions in three dimensions always stay smooth or can blow up into a singularity. The Clay Mathematics Institute will pay one million dollars to whoever settles the question. This page builds the equations term by term, reads each one out loud, meets the no-slip condition that ties a fluid to a solid surface, and solves them exactly for flow down a channel — before pointing at the million-dollar hole in our understanding.

Starting point: Euler, then add friction

For an ideal (frictionless) fluid, Newton's law \mathbf{F} = m\mathbf{a} applied to a fluid parcel gives Euler's equation. Using the material derivative D/Dt = \partial/\partial t + (\mathbf{u}\cdot\nabla) for the acceleration a parcel actually feels, it reads

\rho\,\frac{D\mathbf{u}}{Dt} = -\nabla p + \rho\mathbf{g}.

The left side is mass-per-volume times acceleration; the right side is the two forces per unit volume acting on the parcel — the pressure pushing it from high pressure toward low, and gravity (or any body force). This is exact and elegant, but it tells a lie: it says a fluid feels no friction. Real fluids drag on themselves. A fast-moving layer pulls its slower neighbour along and gets held back in return. That internal drag is viscosity, and for a Newtonian fluid — one whose internal stress is simply proportional to the rate at which layers shear past each other — it adds exactly one term:

\rho\,\frac{D\mathbf{u}}{Dt} = -\nabla p + \mu\,\nabla^2\mathbf{u} + \rho\mathbf{g}.

Here \mu is the dynamic viscosity (units \text{Pa·s}) — big for honey, tiny for air. The new term \mu\nabla^2\mathbf{u} is a diffusion of momentum: the Laplacian \nabla^2 compares the velocity at a point with the average of its neighbours, so wherever a parcel is moving differently from the fluid around it, this term nudges it back toward the local average — smearing sharp velocity differences out, exactly as heat conduction smears out temperature.

The equations, read out loud

A fluid also has to conserve mass. For an incompressible fluid (constant density — an excellent approximation for liquids and for gases well below the speed of sound) that conservation law takes the beautifully simple form \nabla\cdot\mathbf{u} = 0: whatever flows into any little region must flow back out, so the velocity field has no sources or sinks. Put the momentum balance and the mass balance together and you have the whole system.

Term by term, the momentum equation says: the mass of a parcel times its acceleration (\rho\,D\mathbf{u}/Dt, split into the local change \rho\,\partial\mathbf{u}/\partial t and the convective change \rho(\mathbf{u}\cdot\nabla)\mathbf{u}) equals the pressure force -\nabla p, plus the viscous force \mu\nabla^2\mathbf{u}, plus the body force \rho\mathbf{g}. Delete the viscous term and Navier–Stokes collapses back into Euler. Everything hard and everything real lives in that one Laplacian.

Two viscosities: dynamic μ and kinematic ν

Divide the whole momentum equation by the density \rho and the viscous term becomes (\mu/\rho)\nabla^2\mathbf{u}. That combination is so common it earns its own name and symbol, the kinematic viscosity:

\nu = \frac{\mu}{\rho} \qquad (\text{units } \text{m}^2/\text{s}).

The two are genuinely different animals. \mu (dynamic) measures the force a fluid's shearing generates; \nu (kinematic) measures how quickly momentum diffuses through it, and it is \nu that governs whether a flow stays orderly or goes turbulent. A famous surprise: air is far runnier than water to the touch, yet air's kinematic viscosity is about fifteen times larger than water's, because air is so much less dense. Low \mu does not guarantee low \nu — always check which one a formula wants.

No-slip and viscous shear stress

The no-slip condition is one of the most counter-intuitive facts in all of physics: at a solid surface, the fluid touching it moves with the surface — velocity zero at a stationary wall, no matter how fast the flow rushes by just a hair's breadth away. It is why a fan blade or a car keeps a stubborn film of dust: the air right at the surface is not moving to blow it off. Between the stuck-still fluid at the wall and the fast free stream lies a thin boundary layer where the speed ramps up, and inside it the fluid layers slide across one another.

That sliding is where viscosity earns its keep. In a simple shear where the speed u varies in the cross-stream direction y, a Newtonian fluid resists with a shear stress (force per unit area) proportional to the velocity gradient:

\tau = \mu\,\frac{du}{dy}.

The steeper the velocity gradient — the harder neighbouring layers are forced to slide past each other — the bigger the drag. This single relation is the definition of a Newtonian fluid, and integrating it over a surface is how you turn a velocity profile into the actual force felt by a wall (skin-friction drag). Drag the top plate below and watch the no-slip condition pin the fluid to both walls while a straight-line velocity profile builds across the gap.

An exact solution: flow down a channel

Navier–Stokes is usually unsolvable by hand, but in a few clean geometries the nonlinear convective term (\mathbf{u}\cdot\nabla)\mathbf{u} vanishes and the equations become gentle. The classic case is steady flow between two parallel plates a distance 2h apart, with the walls at y = \pm h. The flow is purely horizontal, \mathbf{u} = (u(y), 0), so (\mathbf{u}\cdot\nabla)\mathbf{u} = u\,\partial u/\partial x = 0 and the horizontal momentum equation reduces to a simple balance between the pressure push and the viscous term:

0 = -\frac{dp}{dx} + \mu\frac{d^2u}{dy^2}\quad\Longrightarrow\quad \frac{d^2u}{dy^2} = \frac{1}{\mu}\frac{dp}{dx}.

Integrate twice and impose no-slip at both walls (u = 0 at y = \pm h). With a constant pressure gradient driving the flow you get plane Poiseuille flow — a parabola:

u(y) = u_{\max}\left(1 - \frac{y^2}{h^2}\right), \qquad u_{\max} = -\frac{h^2}{2\mu}\frac{dp}{dx}.

Zero at both walls (no-slip), fastest right down the middle. If instead there is no pressure gradient and the top wall is simply dragged at speed U, the parabola straightens into a line — plane Couette flow, u(y) = U(y+h)/(2h). Toggle the two profiles below and drag the sliders to feel how the peak speed and wall speed reshape them; notice both are pinned to zero at any stationary wall.

Worked examples

Example 1 — kinematic from dynamic viscosity. Water has dynamic viscosity \mu \approx 1.0\times 10^{-3}\ \text{Pa·s} and density \rho \approx 1000\ \text{kg/m}^3. Its kinematic viscosity is

\nu = \frac{\mu}{\rho} = \frac{1.0\times 10^{-3}}{1000} = 1.0\times 10^{-6}\ \text{m}^2/\text{s}.

Air, by contrast, has \mu \approx 1.8\times 10^{-5}\ \text{Pa·s} but \rho \approx 1.2\ \text{kg/m}^3, giving \nu \approx 1.5\times 10^{-5}\ \text{m}^2/\text{s} — about fifteen times larger than water's, even though air's \mu is fifty times smaller.

Example 2 — shear stress in Couette flow. Oil of viscosity \mu = 0.2\ \text{Pa·s} fills a gap h = 2\ \text{mm} = 0.002\ \text{m} between plates; the top plate slides at U = 0.5\ \text{m/s}. The profile is linear, so du/dy = U/h and the shear stress on the plate is

\tau = \mu\frac{du}{dy} = \mu\frac{U}{h} = 0.2\times\frac{0.5}{0.002} = 50\ \text{Pa}.

Example 3 — reading the Poiseuille parabola. A channel flow has centreline speed u_{\max} = 3\ \text{m/s} and half-gap h = 1\ \text{cm}. Halfway to the wall, at y = h/2, the speed is

u = u_{\max}\left(1 - \frac{(h/2)^2}{h^2}\right) = 3\left(1 - \tfrac{1}{4}\right) = 3\times 0.75 = 2.25\ \text{m/s}.

Right at the wall y = h it gives 3(1-1) = 0 — no-slip, exactly as it must.

A tempting picture is that viscosity is "friction", so \mu\nabla^2\mathbf{u} must be a brake that everywhere reduces the speed. Not so. The term is a diffusion of momentum, not a drag force: it drives velocity toward the local average of its surroundings. Where a parcel is moving slower than its neighbours, the Laplacian is positive and the term speeds it up — it is how the fast free stream drags the near-wall fluid along and how the moving wall in Couette flow accelerates the layers above it. Viscosity smears velocity differences out in both directions; only the fluid near a faster region gets a push, and the fluid near a slower region gets held back. Whether the net effect on a given parcel is to speed it up or slow it down depends entirely on the shape of the profile around it.

Because being able to write an equation is a world away from being able to solve it — or even to prove a solution exists. The trouble is the innocent-looking convective term (\mathbf{u}\cdot\nabla)\mathbf{u}: it is nonlinear, velocity multiplying its own gradient, and nonlinearity is what breeds turbulence and mathematical chaos. The Clay Millennium Prize Problem asks a very basic question about the 3-D incompressible equations: given smooth initial data, do the solutions stay smooth and finite for all time, or can they spontaneously develop a singularity — a point where velocity becomes infinite — in finite time? Nobody knows. We use these equations to design aircraft and forecast weather with spectacular success, yet we cannot prove they never break. The viscous term \mu\nabla^2\mathbf{u} is thought to be the peacemaker that keeps things smooth, but proving it always wins against the nonlinearity is the open problem. Solve it, either way, and the million dollars is yours.