Exact solutions: Couette, Poiseuille and Stokes' first problem

The Navier-Stokes equations have few exact solutions, and most of them share a trick: the nonlinear term vanishes identically. For a unidirectional flow \(\mathbf{u} = (u(y,t),0,0)\), incompressibility holds automatically and \((\mathbf{u}\cdot\nabla)\mathbf{u} = u\,\partial_x u\,\mathbf{e}_x = 0\). The \(y\) and \(z\) components of the momentum equation give \(\partial_y p = \partial_z p = 0\), and the \(x\) component \[ \frac{\partial u}{\partial t} = -\frac{1}{\rho}\frac{\partial p}{\partial x} + \nu\frac{\partial^2 u}{\partial y^2} \] shows \(\partial_x p\) depends only on time. So \(\partial_x p = -G(t)\), and what is left is a linear heat equation.

Couette flow (Couette, 1890). Fluid between a fixed plate at \(y = 0\) and a plate at \(y = h\) moving at speed \(U\), with no pressure gradient. Steady flow needs \(u'' = 0\), and no-slip gives \(u = Uy/h\). The shear stress \(\mu U/h\) is the same at every height. Plane Poiseuille flow. Both plates fixed, a pressure gradient \(G\) driving the fluid: \(\mu u'' = -G\), \(u(0) = u(h) = 0\), so \[ u = \frac{G}{2\mu}\,y(h-y), \qquad Q = \int_0^h u\,dy = \frac{Gh^3}{12\mu}. \] A parabola, fastest in the middle, with mean speed two thirds of the maximum. In a round pipe of radius \(a\) the same reasoning in cylindrical coordinates gives the Hagen-Poiseuille profile (Hagen 1839, Poiseuille 1840) \(u = \frac{G}{4\mu}(a^2 - r^2)\) and flow rate \(Q = \frac{\pi G a^4}{8\mu}\). The fourth power of the radius is why narrowing an artery by a fifth cuts the flow by more than half.

Stokes' first problem (Stokes, 1851; often called Rayleigh's problem). Fluid at rest above a plate that suddenly starts moving at speed \(U\): solve \(u_t = \nu u_{yy}\), \(u(0,t) = U\), \(u(y,0) = 0\). There is no length scale in the problem except \(\sqrt{\nu t}\), so the solution depends on \(\eta = y/(2\sqrt{\nu t})\) alone, and \[ u = U\operatorname{erfc}\!\left(\frac{y}{2\sqrt{\nu t}}\right). \] The disturbance spreads a distance of order \(\sqrt{\nu t}\): viscosity diffuses momentum, slowly. If the plate oscillates instead (Stokes' second problem), \(u = Ue^{-ky}\cos(\omega t - ky)\) with \(k = \sqrt{\omega/(2\nu)}\).

A genuinely two-dimensional example is the decaying Taylor-Green vortex (1937), \[ \mathbf{u} = (\sin x\cos y,\ -\cos x\sin y)\,e^{-2\nu t}, \qquad p = \tfrac14(\cos 2x + \cos 2y)\,e^{-4\nu t}. \] Here the nonlinear term does not vanish, but it is a gradient and is balanced exactly by the pressure. Checking such a solution is a matter of differentiating and substituting, and it is worth doing once by hand.

Picture it: the velocity profile across a channel is a straight line when a wall drags the fluid and a parabola when pressure pushes it; they add, because the equation for them is linear. Think it: every exact solution here lives in a subspace where \((\mathbf{u}\cdot\nabla)\mathbf{u}\) is zero or a gradient, so Navier-Stokes collapses to a heat equation. Such solutions say nothing about the hard nonlinear questions, which is precisely why they can be solved.