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Couette-Poiseuille flow superposes wall-driven shear and pressure-driven parabolic flow between no-slip parallel plates. With one plate moving at $U$, separation $a$, and $dp/dx=-\rho \nu G$, the displayed profile has volume flux $Ua/2+G{a}^3/12$ per unit width. Requiring zero flux gives $G=-6U/a^2$ and $u=U[3(y/a{)}^2-2y/a]$, with a pressure gradient induced by return-flow constraints.

For a periodic thin mucus gap $h(x)$ translating as a backward wave relative to a force-free foot, put $I_j={\int }_0^L{h}^{-j}dx$. In the wave frame, the plane and foot have horizontal velocities $V-U$ and $V$. A Couette-Poiseuille flow with periodic pressure has flux $q=(V-U/2)I_2/I_3$. Zero net plane force gives the displayed crawling speed. The leading viscous dissipation per wavelength is $\Phi =2\mu UV{I}_1$.

A generalized Newtonian fluid can approximate the shear response of a fluid with viscoelasticity when the flow varies slowly relative to its stress-memory time scales and elastic stresses are negligible, or in locally steady unidirectional flow where a fitted steady shear viscosity captures the relevant stress and elastic normal stresses do not significantly influence the motion. It is not a general replacement in rapidly changing flows or in flows controlled by elastic stresses. Two specific failures are stress relaxation after deformation stops, since this law has no memory, and nonzero normal-stress differences in steady simple shear flow, since the only nonzero nonpressure components are off diagonal. A rate-dependent viscosity alone cannot repair either limitation.

Take the usual physical power-law fluid range $k>0$, $n>0$, and $G>0$. The steady unidirectional Couette-Poiseuille flow satisfies $0=G+{\partial }_y{\sigma }_{xy}$. The stated upper-wall stress thus givesPut $Y=y/h$. Inverting this monotone stress law gives $u_y=(Gh/k{)}^{1/n}\mathrm{sgn}(\alpha -Y)|\alpha -Y{|}^{1/n}$. Integration from the lower no-slip boundary condition $u(-h)=-U$ gives the power-law Couette-Poiseuille flow profileThe upper no-slip boundary condition requiresThe bracket is a strictly increasing function of $\alpha$ with value zero at $\alpha =0$, so the equation determines a unique positive $\alpha$. The absolute values are essential: for sufficiently strong pressure driving, the shear can reverse inside the channel. For a Newtonian fluid, $n=1$, $\mu =k$, and the bracket reduces to $4\alpha$, yieldingFor zero pressure gradient the limit is pure Couette flow; the parameterization of wall stress by $Gh\alpha$ then becomes singular rather than indicating zero Couette stress.

For weak pressure forcing, let $s_0=U/h$ and $s_c=k{s}_0^n$. Write the center-plane shear stress as $s_c+\delta s$. Linearization of the power-law relation about the positive base shear givesIntegrating $u_y$ across the channel must still give $2U$. The Gy term has zero mean, so $\delta s=0$ at first order. Hence the weak-pressure Couette-Poiseuille expansion isIt is a small positive parabolic displacement of the straight Couette profile, preserving both wall velocities. For fixed positive n the stated $Gh\ll k(U/h{)}^n$ controls the expansion; when n is very small one needs the stronger incremental condition $Gh\ll nk(U/h{)}^n$.

The volume flux per unit spanwise width is $Q={\int }_{-h}^{h}u(y)dy$. The odd Couette part contributes zero and the parabola contributes $4h^3/3$ times its coefficient, soThis correctly reduces to the exact Newtonian flux $2G{h}^3/(3k)$ at $n=1$.

The extension to a general smooth generalized Newtonian fluid uses its differential shear viscosity, rather than its secant viscosity. Let $\mathcal{T}(s)=s\mu (s)$ for positive shear rate, and assumeLinearizing the stress law repeats the preceding calculation with $nk{s}_0^{n-1}$ replaced by ${\eta }_d$. ThusThe expansion requires small rate changes, $Gh\ll s_0|{\eta }_d|$, and a smooth constitutive law on the perturbed rates. At zero differential viscosity the regular first-order expansion fails. A negative differential viscosity is still locally invertible mathematically, so the formula is formally applicable, but belongs to a decreasing, mechanically unstable constitutive branch. The usual stable-flow interpretation requires positive differential viscosity. Thus “any” generalized-Newtonian law needs these local qualifications.

Take positive $x$ in the snail's direction of travel, and let $y$ measure height above the rigid plane. The wave moves at laboratory velocity $U-V$. In the wave frame the plane's horizontal velocity is $V-U$ and the foot's is $V$. The foot is stationary as a shape, so its material velocity is tangent to $y=h(x)$: to leading lubrication theory order its components are $(V,V{h}_x)$. Apply the no-slip boundary condition to both surfaces.

The leading Stokes flow equations are $p_y=0$ and $\mu u_{yy}=p_x$. Their solution is the local Couette-Poiseuille flowThe normal-velocity condition on each steady surface and incompressibility make the wave-frame flux independent of $x$. Integrating the velocity givesPut $C=V-U/2$. Periodicity of pressure gives $0={\int }_0^L{p}_{x}dx=12\mu (C{I}_2-q{I}_3)$, and henceThe shear traction exerted by the fluid on the plane, positive in the $x$ direction, is

The snail is force-free horizontally: its gait consists of internal motion and it has no externally applied horizontal drive. Uniform locomotion therefore requires zero net horizontal fluid force on the foot per wavelength. The stresses on opposite periodic fluid faces cancel, so horizontal momentum balance also makes the net force on the plane zero. ThusSubstitution of $q$ proves the lubrication model of a crawling snail:The Cauchy-Schwarz inequality applied to $h^{-1/2}$ and $h^{-3/2}$ gives $0<\alpha \le 1$, with equality exactly for a uniform gap. Thus the denominator is positive, and a nonuniform waveform gives forward motion; the flat gap gives $U=0$.

For the sinusoidal shape, the given integrals yieldTherefore $\alpha =1/(1+A^2/2)$ andFor small amplitude, $U\sim 3A^{2}V$: the leading locomotion is quadratic, and there is no propulsion from a flat foot. As $A\to 1^{-}$, $U\to V$, so the wave becomes nearly stationary in the laboratory frame while the snail advances relative to it. However, the minimum gap tends to zero; finite speed in this limit does not imply finite dissipation, and $A=1$ itself is excluded by the noncontact model.

For the general waveform, use the mechanical work identity for viscous dissipation. Pressure does no boundary work because the boundary motion is tangential in the wave frame. The plane's velocity is constant, and its total force is zero, so it contributes zero net work. The upper boundary supplies, at leading lubrication order,The zero plane-force relation is $\mu U{I}_1=\frac{1}{2}\int h{p}_{x}dx$. ConsequentlyAs a direct check, integrating the squared shear rate gives $\Phi ={\int }_0^L[\mu U^2/h+h^3{p}_x^2/(12\mu )]dx=\mu I_1[U^2+12C^2(1-\alpha )]$, which reduces to the same result using the force-free speed relation. Zero horizontal force on the foot does not mean zero gait work: the tangential material velocity along its sloping surface includes vertical motion.

For prescribed speed $U>0$ and sinusoidal shape, eliminate $V=U(1+2A^2)/(3A^2)$ to obtainWriting $y=A^2$, the logarithmic derivative of the amplitude-dependent factor isThe only root in $0<y<1$ is $y=1/2$. The dissipation diverges at both endpoints, so this is the global minimum and gives the optimal sinusoidal waveform for viscous snail locomotion:The magically driven flat glider has simple Couette flow $u=Uy/h_0$ in the laboratory frame, and dissipation ${\Phi }_{\mathrm{glide}}=\mu L{U}^2/h_0$ per wavelength. HenceThe comparison is at the same $U$, mean thickness $h_0$ and wavelength $L$; the flat glider is externally propelled, whereas the snail must power its internal waveform.

The baffles create an induced pressure gradient even though no external gradient is applied. Away from them the same Couette-Poiseuille flow has flux per unit widthZero flux therefore gives $G=-6U/a^2$ andThe positive pressure gradient drives the return flow near the stationary lower plate, balancing the forward flow near the moving plate. The rectilinear approximation excludes the short end regions next to the baffles.

The steady rectilinear equation is $u^{\prime \prime }(y)=-G$, with no-slip conditions $u(0)=0$ and $u(a)=U$. Two integrations give the Couette-Poiseuille flowThe first term is driven by the moving wall and the second by the specified pressure gradient.