= Solution
An isotropic <simple fluid> has local <stress> determined by the deformation history of the material element. A <viscometric flow> is one whose relative deformation history is that of steady <simple shear flow>, up to rigid changes of frame. At a given element the <shear rate> is constant along its history, although it can differ between elements. This includes the locally simple-shear kinematics used in standard viscometric measurements.
Use the component-first <velocity gradient> $K_{ij}=\partial v_i/\partial x_j$. The first two <Rivlin-Ericksen tensors> are
$$
A_1=K+K^T,\qquad A_2=D_tA_1+A_1K+K^TA_1.
$$
In the axes of a steady shear $v_x=sy$, the relative <deformation gradient> over a lag $q$ is $F=I+qK$, since $K^2=0$. The material history is determined by $s$, and its tensor symmetries exclude $xz$ and $yz$ <stress> components. The remaining symmetric <stress> consists of an isotropic part, one shear component, and two independent <normal-stress differences>. The relevant matrices are
$$
A_1=\begin{pmatrix}0&s&0\\s&0&0\\0&0&0\end{pmatrix},\qquad
A_1^2=\begin{pmatrix}s^2&0&0\\0&s^2&0\\0&0&0\end{pmatrix},\qquad
A_2=\begin{pmatrix}0&0&0\\0&2s^2&0\\0&0&0\end{pmatrix}.
$$
Thus $I,A_1,A_1^2,A_2$ span the allowed <stress> components. <Material isotropy> and reversal of the shear show that their scalar coefficients depend on $|s|=(A_1:A_1/2)^{1/2}$. Define the <viscometric functions> by
$$
\mu=\frac{\sigma_{xy}}{s},\qquad
\psi_1=\frac{\sigma_{xx}-\sigma_{yy}}{s^2},\qquad
\psi_2=\frac{\sigma_{yy}-\sigma_{zz}}{s^2}.
$$
Solving for the matrix coefficients gives
$$
\boxed{\sigma=-pI+\mu A_1+(\psi_1+\psi_2)A_1^2-\frac12\psi_1A_2.}
$$
This is an exact steady-viscometric representation, rather than an assumption that the fluid is everywhere a <second-order fluid>. The <shear viscosity> $\mu$ measures the shear resistance, while $\psi_1$ and $\psi_2$ measure the first and second <normal-stress differences> generated by shear. Their zero-rate values are understood as limits when these exist. An isotropic <pressure> changes neither normal difference.
For a flow in the <lubrication approximation> with gap $H$, longitudinal scale $\ell\gg H$ and speed $U$, the shear-rate scale is $U/H$. The viscous shear force per unit volume scales as $\mu U/H^2$. Quadratic viscoelastic <normal stresses> scale as $\psi_iU^2/H^2$; their longitudinal gradients, and the comparable mixed-component corrections, contribute forces of scale $\psi_iU^2/(H^2\ell)$. Therefore the <normal-stress smallness in lubrication> criterion is
$$
\boxed{\max_{i=1,2}\frac{|\psi_i|U}{\mu\ell}\ll1.}
$$
Equivalently, the thin-gap aspect ratio must multiply the shear-based elastic parameter to give a small quantity: $(H/\ell)|\psi_i|\dot\gamma/\mu\ll1$. Normal stresses need not themselves be smaller than the <shear stress>; it is their effect on the leading force balance that matters. If the flow changes on an additional short time scale $T$, the $D_tA_1$ term also requires $|\psi_1|/(\mu T)\ll1$ for its neglect. Material coefficients should be evaluated at the characteristic <shear rate>.
Now take positive cylinder rotation, measure $\theta$ counterclockwise from the rightward horizontal, and let $y=r-a$ increase outward through the film. The tangent has upward component $\cos\theta$, so the tangential gravitational acceleration is $-g\cos\theta$. To leading order in $h/a$, neglect inertia and higher-order hydrostatic or capillary pressure gradients, and write $U=\Omega a$. Tangential <force balance> and the shear-free <free surface> give
$$
\frac{\partial\sigma_{\theta y}}{\partial y}=\rho g\cos\theta,\qquad \sigma_{\theta y}(h)=0,
$$
hence
$$
\boxed{\sigma_{\theta y}(y)=\rho g\cos\theta\,(y-h),\qquad 0\le y\le h.}
$$
The wall <shear stress> is $-\rho gh\cos\theta$. It opposes rotation on the uphill side and assists it on the downhill side.
For a <power-law fluid>, $\sigma_{\theta y}=k|u_y|^{n-1}u_y$. Its inversion must retain the sign:
$$
u_y=-\operatorname{sgn}(\cos\theta)\left(\frac{\rho g|\cos\theta|}{k}\right)^{1/n}(h-y)^{1/n}.
$$
Integrating with the <no-slip boundary condition> $u(0)=U$ yields
$$
u(y)=U-\operatorname{sgn}(\cos\theta)\frac{n}{n+1}\left(\frac{\rho g|\cos\theta|}{k}\right)^{1/n}\left[h^{(n+1)/n}-(h-y)^{(n+1)/n}\right].
$$
The <volume flux> per unit axial length is $Q=\int_0^h u(y)dy$. The integrated bracket is $[(n+1)/(2n+1)]h^{(2n+1)/n}$, so the <power-law coating on a rotating cylinder> has
$$
\boxed{Q=\Omega ah-\operatorname{sgn}(\cos\theta)\frac{n}{2n+1}\left(\frac{\rho g|\cos\theta|}{k}\right)^{1/n}h^{(2n+1)/n}.}
$$
At $\cos\theta=0$ the gravitational term vanishes by continuity. Steady <mass conservation> makes $Q$ independent of $\theta$.
On the uphill half, put $D=(\rho g\cos\theta/k)^{1/n}$ and $m=(2n+1)/n$. The local flux is $Q(h)=Uh-(D/m)h^m$, with
$$
Q'(h)=U-Dh^{(n+1)/n},\qquad Q''(h)<0\quad(h>0).
$$
It has a unique maximum at
$$
h_*(\theta)=\left(\frac{kU^n}{\rho g\cos\theta}\right)^{1/(n+1)},\qquad
Q_*(\theta)=\frac{n+1}{2n+1}Uh_*(\theta).
$$
A positive coating must carry the same steady flux at every angle. The tightest local bound occurs at $\cos\theta=1$, not on the downhill side, where the flux increases monotonically with thickness. Consequently
$$
\boxed{Q_{\max}=\frac{n+1}{2n+1}\left(\frac{k}{\rho g}\right)^{1/(n+1)}(\Omega a)^{(2n+1)/(n+1)}.}
$$
For $0<Q<Q_{\max}$ the thinner positive root on the uphill side connects through $\cos\theta=0$ to the unique downhill root, giving a periodic leading-order coating profile. At the limiting flux the two uphill roots meet at $\theta=0$; it is the maximum flux of this lubrication model, with a possible limiting corner resolved by neglected higher-order effects. For $n=1$ the check is $Q_{\max}=(2/3)U\sqrt{kU/(\rho g)}$. All these conclusions require the resulting film to remain thin enough for the <lubrication approximation> and for the neglected <normal stresses> and inertia to satisfy their scaling restrictions.
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