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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 isHere $\tau$ is the constant shear stress magnitude. The lower wall's Navier slip boundary condition holds exactly because $b{u}^{\prime }(0)=bU/(H+b)=u(0)$. 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 $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $U=1\mathrm{mm}{\mathrm{s}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. The bottom velocity is $0.167\mathrm{mm}{\mathrm{s}}^{-1}$ 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 $z=-2\mu \mathrm{m}$.

For fixed $H$, $U$ and $\mu$, $b\to 0$ recovers $u=Uz/H$. As $b\to \infty$, $u\to U$ throughout the gap and $\tau \to 0$: 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 $\mu u^{\prime \prime }=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.

Linearize about a relaxed state $\mathsf{T}=0$ with constant mass density ${\rho }_0$. In the unforced momentum conservation equation ${\rho }_0{D}_t\mathbf{u}=-\nabla p+\nabla \cdot \mathsf{T}$, the $x$ component of the pressure gradient is zero for a freely evolving unidirectional flow independent of $x$. A prescribed external pressure gradient would add a forcing term.

For $\mathbf{u}=u(y,t)\hat{\mathbf{x}}$, incompressibility is automatic and the only nonzero entries of the rate-of-strain tensor are $E_{xy}=E_{yx}=\frac{1}{2}{\partial }_{y}u$. Products of stress and velocity gradient are quadratic in the small amplitude. The constitutive equation consequently becomes $\mathsf{T}+\tau {\partial }_t\mathsf{T}=2\mu \mathsf{E}$ to first order. With $T=T_{xy}=T_{yx}$, its shear component and the $x$ component of momentum conservation areThe diagonal stress components have only homogeneous relaxation at this order and vanish if initially relaxed. In particular, $\alpha$ does not enter this linear response.

For this steady unidirectional flow, the convective acceleration $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanishes because $u$ depends only on $y$. The axial Navier-Stokes equation reduces to $0=-p_x+\mu u^{\prime \prime }$, with $p_x=-G$. The no-slip boundary conditions therefore giveThis is a superposition of Couette flow and pressure-driven Poiseuille flow.

For the unidirectional flow $\mathbf{u}=(0,0,w)$, incompressibility gives ${\partial }_{z}w=0$. Hence $w=w(x,y,t)$, and the convective acceleration $w{\partial }_{z}w$ vanishes. The transverse components of the Navier-Stokes equation give $p_x=p_y=0$, while the axial component givesA steady-flow assumption is implicit in the question. Neither unidirectional flow nor a time-independent pressure gradient alone eliminates ${\partial }_{t}w$: a decaying transient can be superposed on the final pipe profile. The native PDF does not explicitly say steady flow in this part. For the intended steady flow, $w_t=0$, and thereforeTaking the pipe walls to be stationary, the no-slip boundary condition isThe no-penetration boundary condition is already satisfied by the assumed velocity field. This is the Poisson equation defining rectangular duct flow.

Rectangular duct flow is steady flow of a Newtonian fluid down a straight rectangular pipe under a constant axial pressure gradient $-G$. For stationary walls and unidirectional flow, incompressibility eliminates axial dependence and the Navier-Stokes equation becomes a two-dimensional Poisson equation with zero Dirichlet boundary conditions. Subtracting a plane Poiseuille flow profile leaves a harmonic function that repairs the side-wall values using separation of variables. Uniqueness follows from Green's first identity: the difference of two solutions has zero boundary values and zero integral of the squared gradient, and hence vanishes.