= Couette flow with a slipping lower wall
{title2=$u(z)=U(z+b)/(H+b)$}
Consider <steady flow> of a <Newtonian fluid> in <incompressible flow> with constant <dynamic viscosity> $\mu>0$ between parallel plates at $z=0$ and $z=H$. The bottom plate is stationary and obeys the <Navier slip boundary condition> with <slip length> $b$. The top plate moves at <velocity> $U>0$ and obeys the <no-slip boundary condition>. There is no streamwise <pressure gradient> or body <force>. The <Couette flow> solution is
$$
u(z)=U\frac{z+b}{H+b},\qquad
u(0)=\frac{Ub}{H+b},\qquad
\tau=\mu u'(z)=\frac{\mu U}{H+b}.
$$
Here $\tau$ is the constant <shear stress> magnitude. The lower wall's <Navier slip boundary condition> holds exactly because $bu'(0)=bU/(H+b)=u(0)$. At the top, $u(H)=U$ is exactly the prescribed <no-slip boundary condition>; no finite upper-wall <slip length> is imposed in this example.
For a numerical example, stipulate $H=10\,\mathrm{\mu m}$, $b=2\,\mathrm{\mu m}$, $U=1\,\mathrm{mm\,s^{-1}}$ and $\mu=10^{-3}\,\mathrm{Pa\,s}$. The bottom <velocity> is $0.167\,\mathrm{mm\,s^{-1}}$ and the <shear stress> is $0.0833\,\mathrm{Pa}$, compared with $0.100\,\mathrm{Pa}$ when both plates obey the <no-slip boundary condition>. The extrapolated profile reaches zero at $z=-2\,\mathrm{\mu m}$.
\Image[/navier-slip-couette.png]
{title=Couette profiles reveal slip length by extrapolation beneath the lower wall}
{description=The bottom wall is stationary with <slip length> $b$; the upper wall at $z=H$ moves at <velocity> $U$ with a <no-slip boundary condition>. Solid lines show <Couette flow> for $b/H=0$, $0.25$ and $0.75$. Dashed continuations inside the shaded solid reach zero at $z=-b$. Increasing <slip length> reduces the <shear rate> required for the same upper-wall <velocity>.}
For fixed $H$, $U$ and $\mu$, $b\to0$ recovers $u=Uz/H$. As $b\to\infty$, $u\to U$ throughout the gap and $\tau\to0$: the moving upper wall selects the limiting uniform <velocity>, while the lower wall approaches a <stress-free boundary condition>. The <dynamic viscosity> stays positive; the disappearing <shear stress> comes from the disappearing <velocity gradient>.
A short derivation uses <unidirectional flow>: the <convective acceleration> vanishes and the <Navier-Stokes equation> reduces to $\mu u''=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.
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