Consider a semi-infinite incompressible Newtonian fluid occupying

$$y>0.$$

Initially the fluid is at rest. At $t=0$, an infinite flat plate at $y=0$ is suddenly set into motion in the $x$ direction at constant speed $U$.

This is Stokes' first problem, also called Rayleigh's problem. It is a canonical example of how wall momentum diffuses into a viscous fluid.

Governing equation and conditions

For parallel flow

$$\mathbf v=u(y,t)\hat{\mathbf x},$$

with no axial pressure gradient, the momentum equation is

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

The conditions are

$$u(y,0)=0\qquad(y>0),$$

$$u(0,t)=U\qquad(t>0),$$

and

$$u(y\to\infty,t)\to0.$$

The no-slip condition imposes the plate velocity at the wall, while the semi-infinite fluid remains undisturbed infinitely far away.

Map the problem onto the diffusion similarity solution

The equation has exactly the form

$$\phi_t=D\phi_{xx}$$

for diffusion into a semi-infinite domain. The mapping is

$$\phi\rightarrow u,$$

$$D\rightarrow\nu,$$

$$x\rightarrow y,$$

$$\phi_i=0,$$

$$\phi_s=U.$$

Therefore

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

Thus

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

Check the physical limits

At the plate,

$$y=0,$$

so

$$u(0,t)=U\operatorname{erfc}(0)=U.$$

Far from the plate,

$$y\to\infty,$$

so

$$u\to0.$$

For any fixed $y>0$, as

$$t\to0^+,$$

the similarity variable tends to infinity, so

$$u\to0.$$

The solution therefore satisfies the wall condition, far-field condition, and initial condition.

Growth of the viscous layer

The velocity profile depends on position and time only through

$$\eta=\frac{y}{2\sqrt{\nu t}}.$$

A fixed fractional velocity therefore occurs at a distance proportional to

$$\boxed{\delta\sim\sqrt{\nu t}}.$$

The wall does not instantly set the entire fluid into motion. Its influence penetrates progressively farther from the surface through viscous momentum diffusion.

The profile is self-similar: plotted against $y/\sqrt{\nu t}$, profiles at all times collapse onto the same complementary-error-function curve.

Transient wall shear stress

For a Newtonian fluid,

$$\tau_{xy}=\mu\frac{\partial u}{\partial y}.$$

Using

$$\frac{d}{d\eta}\operatorname{erfc}(\eta) =-\frac{2}{\sqrt\pi}e^{-\eta^2},$$

and

$$\eta=\frac{y}{2\sqrt{\nu t}},$$

we obtain

$$\frac{\partial u}{\partial y} =-\frac{U}{\sqrt{\pi\nu t}} \exp!\left(-\frac{y^2}{4\nu t}\right).$$

At the wall,

$$\boxed{ \tau_{xy}(0,t) =-\frac{\mu U}{\sqrt{\pi\nu t}} }.$$

The negative sign means that the shear stress exerted within the fluid opposes increasing $u$ with distance away from the moving plate. The magnitude of the viscous traction associated with the wall is

$$\boxed{|\tau_w|=\frac{\mu U}{\sqrt{\pi\nu t}}}.$$

Since

$$\nu=\frac{\mu}{\rho},$$

this can also be written

$$\boxed{|\tau_w|=U\sqrt{\frac{\rho\mu}{\pi t}}}.$$

The required shear decays as

$$|\tau_w|\propto t^{-1/2}.$$

As momentum spreads through a thicker layer, the wall velocity gradient becomes less steep.

The ideal impulsive-start singularity

The formula predicts

$$|\tau_w|\to\infty$$

as

$$t\to0^+.$$

This does not mean a real plate can produce infinite stress. It reflects the idealization that the wall jumps instantaneously from zero speed to $U$. Any real plate accelerates over a finite time, which regularizes the initial response.

The singularity is useful because it reveals how strongly an ideal diffusion model reacts to a discontinuous initial-boundary condition.

Worked example

Take water-like properties

$$\nu=1.0\times10^{-6},\mathrm{m^2/s},$$

$$\mu=1.0\times10^{-3},\mathrm{Pa,s},$$

and let the plate move at

$$U=0.50,\mathrm{m/s}.$$

After

$$t=1.0,\mathrm s,$$

consider the fluid at

$$y=1.0,\mathrm{mm}=1.0\times10^{-3},\mathrm m.$$

The similarity variable is

$$\eta=\frac{1.0\times10^{-3}} {2\sqrt{(1.0\times10^{-6})(1.0)}}=0.50.$$

Using

$$\operatorname{erfc}(0.50)\approx0.480,$$

we obtain

$$u\approx(0.50)(0.480) =\boxed{0.240,\mathrm{m/s}}.$$

The wall-shear magnitude at the same time is

$$|\tau_w| =\frac{(1.0\times10^{-3})(0.50)} {\sqrt{\pi(1.0\times10^{-6})(1.0)}}$$

$$\approx\boxed{0.282,\mathrm{Pa}}.$$

At $t=4.0,\mathrm s$, the shear magnitude has fallen by a factor of two because of the $t^{-1/2}$ scaling.

Why this problem matters

Stokes' first problem is more than a special plate calculation. It shows, in an exact solution, how

• a boundary velocity generates viscous shear;
• momentum diffuses away from a wall;
• the affected thickness grows like $\sqrt{\nu t}$;
• the wall stress decays like $1/\sqrt t$;
• the same similarity mathematics used for heat and mass diffusion also governs transient momentum transport.

It is a basic model for start-up shear layers and a conceptual bridge from viscosity to transient boundary-layer behavior.