Exact Solutions to the Navier-Stokes Equations

The Navier-Stokes equations are notoriously nonlinear. So how can anyone solve them exactly?

Symmetry. That's the whole trick.

When the geometry of a flow is simple enough (a straight pipe, two flat plates, an infinite plane) the velocity can only point in one direction and vary along one or two coordinates. In many of these symmetric setups, the nonlinear term either vanishes or simplifies so much that the equations collapse to a linear PDE. In steady cases, you're often left with an ODE you can solve with pencil and paper.

Here's the intuition. The Navier-Stokes equations describe all possible fluid motions. But if you force the fluid into a very orderly situation, with no swirling, no chaos, everything marching in one direction, most of the equation's complexity becomes irrelevant. The hard part of Navier-Stokes is the feedback loop where the fluid pushes itself around. In these symmetric flows, there's nothing to push against. The nonlinear self-advection term just drops out.

These exact solutions aren't curiosities. They're the foundation of fluid mechanics education, the benchmarks for numerical codes, and the starting point for understanding when and how real flows go sideways.

An exact solution to the Navier-Stokes equations is a velocity field $u$ and pressure $p$ satisfying

$$\partial_t u + (u \cdot \nabla)u = -\nabla p + \nu \Delta u, \qquad \nabla \cdot u = 0$$

in closed form, without numerical approximation.

The key mechanism is the vanishing or simplification of the nonlinear advection term $(u \cdot \nabla)u$. For parallel flows (flows where $u = (u(y,t),\, 0,\, 0)$ in Cartesian coordinates or $u = (0,\, 0,\, u(r,t))$ in cylindrical) the velocity field is divergence-free automatically and the advection term vanishes identically, because the velocity has no component in, and no variation along, the streamwise direction. The momentum equation then reduces to a linear PDE, or for steady flows, an ODE.

When the flow is also steady ($\partial_t u = 0$), what remains is

$$\nu \Delta u = \nabla p,$$

a Poisson equation whose solutions are the classical exact solutions of viscous flow: Poiseuille flow, Couette flow, and their relatives.

For unsteady parallel flows ($\partial_t u \neq 0$ but advection still vanishes), the equation becomes a diffusion equation $\partial_t u = \nu\, \partial_{yy} u$, which yields Stokes' first and second problems.

If you are looking for the actual closed-form Navier-Stokes solutions, these are the classical ones to start with. Each formula below solves a reduced version of the incompressible equations under specific geometry and boundary assumptions.

The important warning is that these are not general solutions. They are exact because the geometry removes most of the nonlinear behavior that makes the three-dimensional Navier-Stokes problem difficult.

The table lists standard closed-form solutions of the incompressible Newtonian Navier-Stokes equations under no-slip boundary conditions. Here $\mu$ is dynamic viscosity, $\nu=\mu/\rho$ is kinematic viscosity, $R$ is pipe radius, $h$ is plate spacing, and $dp/dx$ is a constant streamwise pressure gradient.

These formulas are standard in graduate fluid mechanics texts such as Batchelor's An Introduction to Fluid Dynamics, Landau and Lifshitz's Fluid Mechanics, and Schlichting's Boundary-Layer Theory. The derivations below show the reductions directly rather than relying on the formulas as quoted results.

Poiseuille flow, also called Hagen–Poiseuille flow, is steady pressure-driven flow through a straight circular pipe. No slip makes the velocity zero at the wall, while the centerline moves fastest, producing a parabolic profile.

The volume flow rate is $Q=\pi R^4\Delta p/(8\mu L)$ under incompressible, Newtonian, fully developed, laminar, rigid-pipe assumptions. At fixed pressure drop, length, and viscosity, doubling the radius multiplies the predicted flow by sixteen.

Read the full Hagen–Poiseuille derivation for the Navier–Stokes reduction, boundary conditions, mean and maximum speeds, wall shear, a worked example, and the limits of the fourth-power law.

For steady, fully developed, axisymmetric flow $\mathbf{u}=u_z(r)\mathbf{e}_z$ in a circular pipe, incompressibility holds automatically and $(\mathbf{u}\cdot\nabla)\mathbf{u}=0$. With $G=-dp/dz>0$, the axial equation is $r^{-1}(ru_z')'=-G/\mu$.

Centerline regularity and no slip at $r=R$ give $u_z(r)=G(R^2-r^2)/(4\mu)$, $u_{\max}=2\bar u$, and $Q=\pi GR^4/(8\mu)$. These formulas remain in the parent catalog; the dedicated Poiseuille-flow page owns the complete derivation, resistance and shear relations, Reynolds-number caveats, and source boundaries.

Couette flow is the exact solution for fluid trapped between two parallel plates when one plate moves and the other stays still. It is one of the simplest exact solutions in fluid mechanics.

Imagine a deck of cards lying flat on a table. Drag the top card sideways and the cards underneath shift too, each one a little less than the one above. That's it. The velocity varies linearly from zero at the bottom plate to the speed of the top plate, and there's nothing more to it than that.

A straight-line velocity profile. Bottom plate: stationary. Top plate: moving. Everything in between just interpolates linearly, no pressure gradient needed, no complicated setup, the motion driven purely by the moving boundary dragging the fluid along.

Things get more interesting when you also apply a pressure gradient along the channel, because then you're combining shear-driven and pressure-driven flow in the same gap. The velocity profile warps into a parabola superimposed on the linear profile, sometimes called plane Poiseuille-Couette flow, and depending on the pressure gradient's strength relative to the plate speed, you can even get backflow near one wall.

Strip away the moving plate entirely, keep both walls stationary, and let a pressure difference do all the work. That's plane Poiseuille flow, the flat-plate analogue of pipe flow. Parabolic. Fastest in the middle. Zero at both walls.

Consider steady flow between two infinite parallel plates separated by a gap $h$. Let $y$ be the coordinate perpendicular to the plates, with the bottom plate at $y = 0$ and the top plate at $y = h$. The flow is parallel: $u = (u(y),\, 0,\, 0)$.

Simple Couette flow. The top plate moves at velocity $U$ in the $x$-direction; the bottom plate is stationary; $dp/dx = 0$. The momentum equation reduces to $d^2u/dy^2 = 0$, yielding

$$u(y) = U\frac{y}{h}.$$

This is the simplest nontrivial exact solution to the Navier-Stokes equations: a linear velocity profile driven entirely by the boundary condition.

Plane Poiseuille flow. Both plates are stationary; a constant pressure gradient $dp/dx < 0$ drives the flow. The momentum equation becomes

$$\mu \frac{d^2 u}{dy^2} = \frac{dp}{dx},$$

with $u(0) = u(h) = 0$. The solution is

$$u(y) = \frac{1}{2\mu}\left(-\frac{dp}{dx}\right)y(h - y),$$

a parabolic profile symmetric about the centerline $y = h/2$.

General Couette-Poiseuille flow. Combining a moving top plate with a pressure gradient gives the superposition

$$u(y) = U\frac{y}{h} + \frac{1}{2\mu}\left(-\frac{dp}{dx}\right)y(h - y).$$

The first term is the shear-driven contribution; the second is the pressure-driven contribution. Depending on the sign and magnitude of $dp/dx$ relative to $\mu U / h^2$, the profile can be monotone, have an interior maximum, or even exhibit backflow near one wall.

Poiseuille and Couette flows are steady. Nothing changes in time. Stokes' problems are the simplest unsteady exact solutions, and they reveal something beautiful: viscosity makes momentum diffuse through a fluid, just as heat diffuses through a solid.

Stokes' first problem (also called the Rayleigh problem): imagine a vast, still pool of fluid resting above a flat plate. At time zero, the plate suddenly starts sliding sideways at constant speed. The fluid right next to the plate gets dragged along immediately, but fluid farther away takes time to notice. A smooth boundary layer grows outward from the plate, getting thicker as time passes.

The speed at any height above the plate depends on the ratio of that height to a characteristic diffusion length $\sqrt{\nu t}$, where $\nu$ is the viscosity and $t$ is the elapsed time. More viscous fluid? The motion spreads upward faster.

Stokes' second problem: same setup, but now the plate oscillates back and forth sinusoidally instead of moving at constant speed. The oscillation only penetrates a finite distance into the fluid. Farther up, the fluid barely notices. The amplitude of the motion decays exponentially with height, creating a thin oscillatory boundary layer. This is the mechanism behind oscillatory boundary layers: an oscillating plate sets the nearby fluid in motion, but the disturbance dies off exponentially with distance from the plate.

Both Stokes problems involve a semi-infinite fluid ($y > 0$) above a flat plate at $y = 0$. The flow is parallel: $u = (u(y,t),\, 0,\, 0)$. The advection term vanishes, and the governing equation is the one-dimensional diffusion equation:

$$\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}.$$

Stokes' first problem (Rayleigh problem). Initial condition: $u(y,0) = 0$ for $y > 0$. Boundary condition: $u(0,t) = U$ for $t > 0$. Far-field: $u \to 0$ as $y \to \infty$.

The similarity variable $\eta = y / (2\sqrt{\nu t})$ reduces the PDE to an ODE. The solution is

$$u(y,t) = U\,\operatorname{erfc}\!\left(\frac{y}{2\sqrt{\nu t}}\right),$$

where $\operatorname{erfc}$ is the complementary error function. The boundary layer thickness grows as $\delta \sim \sqrt{\nu t}$, the hallmark of diffusive spreading.

Stokes' second problem. The plate oscillates: $u(0,t) = U\cos(\omega t)$. Seeking a solution of the form $u(y,t) = \operatorname{Re}[\hat{u}(y)\, e^{i\omega t}]$ gives

$$\hat{u}(y) = U\exp\!\left(-(1+i)\frac{y}{\delta_s}\right), \qquad \delta_s = \sqrt{\frac{2\nu}{\omega}},$$

so that

$$u(y,t) = U\exp\!\left(-\frac{y}{\delta_s}\right)\cos\!\left(\omega t - \frac{y}{\delta_s}\right).$$

The amplitude decays exponentially with penetration depth $\delta_s$, and the phase lags linearly: a transverse wave whose energy is entirely dissipated by viscosity. Higher frequencies penetrate less deeply.

Poiseuille, Couette, and Stokes flows get most of the attention. They aren't the only exact solutions, though. Not by a long shot.

• Taylor-Green vortex: a decaying pattern of swirling vortices in two dimensions with genuine vortical structure. Tested a CFD code? You've probably run it against this. It's the benchmark everyone reaches for first, and it's been that way for decades.
• Jeffery-Hamel flow: flow in a wedge-shaped channel that converges or diverges. It captures how fluid accelerates into a narrowing gap or decelerates into an expanding one.
• Hiemenz stagnation-point flow: fluid slamming head-on into a flat wall, slowing to zero at the surface and diverting sideways. Wind hitting a building. A jet striking a plate.

Symmetry. Every one of these exploits a specific geometric symmetry to make the equations tractable, and they matter in specialized contexts, but for everyday introductory fluid mechanics Poiseuille and Couette still do all the heavy lifting.

Beyond parallel flows, several other families of exact solutions exploit specific symmetries or self-similar structures:

• Taylor-Green vortex. In 2D, $u = (A\cos(ax)\sin(by),\, -B\sin(ax)\cos(by))$ with $aA = bB$ (incompressibility) and exponential temporal decay $e^{-\nu(a^2+b^2)t}$. This is an exact solution to the full 2D Navier-Stokes equations, including the nonlinear term (which turns out to be a gradient and gets absorbed into the pressure). It's a standard validation case for DNS codes.
• Jeffery-Hamel flow. Steady, purely radial flow $u_r(r,\theta)$ in a wedge of half-angle $\alpha$. The stream function $\psi = \nu f(\theta)$ satisfies a third-order nonlinear ODE in $\theta$. Solutions exist for both converging and diverging channels, with rich bifurcation structure at higher Reynolds numbers.
• Hiemenz stagnation-point flow. A 2D similarity solution for flow impinging on a plane wall. The stream function $\psi = \sqrt{a\nu}\, x\, f(\eta)$, $\eta = y\sqrt{a/\nu}$, reduces Navier-Stokes to the Hiemenz ODE: $f''' + ff'' - f'^2 + 1 = 0$ with $f(0)=f'(0)=0$, $f'(\infty)=1$.

We can solve the Navier-Stokes equations exactly in all these cases. So why is there still a million-dollar open problem?

Tricks. Every single one relies on a trick. The geometry is chosen so carefully that the hardest part of the equation (the nonlinear term) either vanishes entirely or reduces to something manageable, and the problem becomes solvable precisely because it's been drained of everything that makes Navier-Stokes hard. Pipe flow? One-dimensional. Couette? A line. The Taylor-Green vortex hides its nonlinearity inside the pressure.

The Millennium Prize problem asks about general three-dimensional flows. No symmetry tricks. No simplifying geometry. Smooth divergence-free initial data, full nonlinear interaction across all scales. In that setting, nobody's proved solutions always stay smooth, and nobody's proved they blow up. We genuinely don't know.

So these exact solutions tell us something, but nowhere near enough. They prove the equations can produce explicit smooth solutions when you hand them strong symmetry to lean on. The million-dollar question is whether smoothness holds always, for arbitrary smooth divergence-free initial data in the standard 3D formulations, or whether somewhere in the full violence of turbulence something goes catastrophically and irreversibly wrong.

Every exact solution discussed above achieves tractability by eliminating or trivializing the nonlinear advection term $(u \cdot \nabla)u$. In parallel flows it vanishes identically. In the Taylor-Green vortex it's a gradient absorbed into pressure. In similarity solutions it reduces via a change of variables to a lower-dimensional problem.

The Clay Millennium Prize problem concerns the initial value problem for the 3D incompressible Navier-Stokes equations with arbitrary smooth, rapidly decaying initial data, precisely the regime where none of these simplifications apply.

The question is whether solutions to

$$\partial_t u + (u \cdot \nabla)u = -\nabla p + \nu \Delta u, \quad \nabla \cdot u = 0, \quad u(x,0)=u_0(x)$$

on $\mathbb{R}^3$ remain in $C^\infty(\mathbb{R}^3 \times [0,\infty))$ for all time, given smooth, divergence-free, rapidly decaying initial data. Exact solutions don't address this because they inhabit subspaces of the solution space where the full nonlinear dynamics never engage.

In short: exact solutions provide explicit smooth solutions in highly symmetric regimes. The open problem is whether well-posedness extends to unrestricted 3D flows. For the precise formulation, see Navier-Stokes Existence and Smoothness.

To understand the equations themselves and what each term means, start with What Are the Navier-Stokes Equations?

To see how these equations are built up from first principles, read How the Navier-Stokes Equations Are Derived.

To understand when laminar flows like Poiseuille and Couette break down into turbulence, read Reynolds Number, Turbulence, and Why Small Scales Matter.

To understand the million-dollar question that exact solutions can't answer, read The Millennium Prize Problem: Existence and Smoothness.

For the bigger picture of what "solving" Navier-Stokes means, from these closed forms through CFD to the open problem, read Solving the Navier-Stokes Equations.

Recommended next pages on this site:

• What Are the Navier-Stokes Equations? (the governing system and its terms)
• How the Navier-Stokes Equations Are Derived (from momentum balance and the Newtonian constitutive law to the PDE)
• Reynolds Number, Turbulence, and Why Small Scales Matter (the transition from the laminar regimes described here to fully turbulent flow)
• The Millennium Prize Problem: Existence and Smoothness (the precise formulation of the open problem that exact solutions don't resolve)
• Solving the Navier-Stokes Equations (the exact, numerical, and well-posedness senses of solution, and the gaps between them)