We know that the Incompressible Navier Stokes equation is as follows,

$ρ[\frac{∂U}{∂t}+(\bar{U}∙∇U)]=-∇P-ρg+[μ∇^2 \bar{U}+\frac{μ}{3} ∇(∇∙\bar{U})]$

For the Steady State flow,

$\frac{∂U}{∂t}=0and∇∙\bar{U}$=0 (Continuity equation for incompressible flows)

$ρ[(U ̅∙∇U)]=-∇P-ρg+[μ∇^2 U ̅ ]$

Couette flow is a viscous laminar flow of a Newtonian fluid between 2 parallel plates separated by gap, in which the lower plate is stationary or moving and the upper plate moves with a constant relative velocity.

The Assumptions made are as follows:

Width of the plates are infinite (implying no transverse flow, perpendicular to the plane of the paper)

The flow is steady and laminar

The Flow is thus 2 Dimensional

Therefore the velocity vector in terms of its components,

$U ̅=ui ̅+vj ̅$

Now, v=0,$\frac{∂v}{∂x}=0,\frac{∂v}{∂y}$=0

and from continuity equation ∇∙$U ̅=0⇒\frac{∂u}{∂x}+\frac{∂v}{∂y}$=0

∴$\frac{∂u}{∂x}=0 and (∂^2 u)/∂x^2 =0$

Hence, the Y-direction momentum equation becomes 0, while the X-direction momentum equation is,

$ρu \frac{∂u}{∂x}=-\frac{∂P}{∂x}+μ[\frac{(∂^2 u)}{∂y^{2}}]$=0

∴μ $\frac{(∂^2 u)}{dy^2} = \frac{∂P}{∂x}$

For a constant pressure differential, the partial differential equation gets converted second order ordinary differential equation,

$\frac{∂P}{∂x}=\frac{∆P}{∆x}$=constant

∴$μ \frac{(d^2 u)}{dy}^2 = \frac{∆P}{∆x}$

Now, for Couette flow as can be seen in the figure, u=0 at y=0 and u=U at y=h

Integrating the differential eq, we get,

$μ \frac{du}{dy}=\frac{∆P}{∆x} y+c_1$

Integrating again,

$μu=\frac{∆P}{∆x} (\frac{y^2}{2})+yc_1+c_2$

Applying the boundary conditions,

u=0 at y=0 ⇒$c_2$=0 and u=U at y=h

$μU=\frac{∆P}{∆x}(\frac{h^2}{2})+hc_1⇒c_1=\frac{μU}{h}-\frac{∆P}{∆x}\frac{h}{2}$

Substituting the integration constants, we get

$μu=\frac{∆P}{∆x} (\frac{y^2}{2})+y[\frac{μU}{h}-\frac{∆P}{∆x }(\frac{h}{2})]$

Rearranging the terms, the velocity profile for couette flow is,

u=$\frac{1}{2}μ \frac{∆P}{∆x} y(y-h)+(\frac{Uy}{h})$