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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.