Consider steady incompressible flow of a Newtonian fluid between two large parallel plates at

$$y=0\qquad\text{and}\qquad y=H.$$

Let the lower plate move in the $x$ direction with speed $U_0$ and the upper plate with speed $U_H$. Allow a constant axial pressure gradient $dp/dx$.

For fully developed unidirectional flow,

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

and the Navier-Stokes equations reduce to

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

No slip at the two walls gives

$$u(0)=U_0, \qquad u(H)=U_H.$$

Solving the velocity profile

Integrating twice,

$$u(y)=\frac{1}{2\mu}\frac{dp}{dx}y^2+C_1y+C_2.$$

From $u(0)=U_0$,

$$C_2=U_0.$$

Applying $u(H)=U_H$ gives

$$C_1=\frac{U_H-U_0}{H}-\frac{H}{2\mu}\frac{dp}{dx}.$$

Therefore

$$\boxed{u(y)=U_0+\frac{U_H-U_0}{H}y -\frac{1}{2\mu}\frac{dp}{dx},y(H-y)}.$$

This is plane Couette-Poiseuille flow. The profile is the sum of two independent contributions:

$$\boxed{u=u_{\rm wall}+u_{\rm pressure}},$$

with

$$u_{\rm wall}=U_0+\frac{U_H-U_0}{H}y$$

and

$$u_{\rm pressure}=-\frac{1}{2\mu}\frac{dp}{dx}y(H-y).$$

The wall-driven part is linear; the pressure-driven part is parabolic.

Pure Couette flow

If there is no pressure gradient,

$$\frac{dp}{dx}=0,$$

then

$$\boxed{u(y)=U_0+\frac{U_H-U_0}{H}y}.$$

For a stationary lower wall and an upper wall moving at speed $U$,

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

Viscosity transmits the motion of the wall through the fluid. The velocity gradient is constant,

$$\frac{du}{dy}=\frac{U}{H},$$

so the Newtonian shear stress is also constant:

$$\boxed{\tau_{xy}=\mu\frac{U}{H}}.$$

Pure plane Poiseuille flow

If both walls are stationary,

$$U_0=U_H=0,$$

then

$$\boxed{u(y)=-\frac{1}{2\mu}\frac{dp}{dx}y(H-y)}.$$

For flow driven toward increasing $x$, pressure decreases downstream, so

$$\frac{dp}{dx}<0,$$

and the velocity is positive between the plates.

The maximum speed occurs at the midplane $y=H/2$:

$$\boxed{u_{\max}=-\frac{H^2}{8\mu}\frac{dp}{dx}}.$$

The mean speed is

$$\bar u=\frac1H\int_0^H u(y),dy =\boxed{-\frac{H^2}{12\mu}\frac{dp}{dx}},$$

so

$$\boxed{u_{\max}=\frac32\bar u}.$$

Combined flow and superposition

With both wall motion and pressure forcing present,

$$\boxed{\bar u=\frac{U_0+U_H}{2} -\frac{H^2}{12\mu}\frac{dp}{dx}}.$$

The pressure gradient can reinforce the wall-driven flow or oppose it. If the two effects oppose strongly enough, the velocity can change sign within the gap even while the net flow rate remains in one direction.

The shear stress is

$$\tau_{xy}=\mu\frac{du}{dy},$$

so

$$\boxed{\tau_{xy}(y)= \mu\frac{U_H-U_0}{H} -\frac12\frac{dp}{dx}(H-2y)}.$$

Unlike pure Couette flow, the combined flow generally has shear stress that varies across the gap.

Worked example

A fluid with viscosity

$$\mu=0.10,\mathrm{Pa,s}$$

fills a gap

$$H=4.0,\mathrm{mm}=0.0040,\mathrm m.$$

The lower plate is stationary and the upper plate moves at

$$U_H=0.40,\mathrm{m/s}.$$

The pressure gradient is

$$\frac{dp}{dx}=-5.0\times10^3,\mathrm{Pa/m}.$$

The wall-motion contribution to the mean speed is

$$\frac{U_0+U_H}{2}=0.20,\mathrm{m/s}.$$

The pressure-driven contribution is

$$-\frac{H^2}{12\mu}\frac{dp}{dx} =-\frac{(0.0040)^2}{12(0.10)}(-5.0\times10^3) \approx0.0667,\mathrm{m/s}.$$

Therefore

$$\boxed{\bar u\approx0.267,\mathrm{m/s}}.$$

The moving wall and the favorable pressure gradient both drive the fluid in the positive $x$ direction.

Why this solution matters

Couette-Poiseuille flow is one of the simplest exact demonstrations of how different physical mechanisms appear as separate terms in a velocity field:

• moving boundaries drive shear;
• pressure gradients drive curvature in the profile;
• viscosity controls how strongly either forcing changes velocity across the gap.

It is also a foundation for lubrication theory, rheometry, thin-film flow, and many microfluidic systems.