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For a regular material constraint $F(A)=0$, admissible rates of the deformation gradient lie in the kernel of $F_{,A}$. The difference between the nominal stress and the derivative of the strain energy density annihilates that tangent space by virtual work, so it is a Lagrange multiplier times $F_{,A}$. For incompressibility, the derivative of the determinant is its cofactor matrix, and the material-first reaction is $q{A}^{-1}$ when $\det A=1$.

Differentiate the advection equation componentwise: $D_t({\partial }_i\theta )=-({\partial }_i{u}_j){\partial }_j\theta$. Along a Lagrangian trajectory, the scalar gradient therefore evolves by the negative transpose of the velocity gradient tensor. If $F$ is the deformation gradient, this says $\mathbf{k}(t)=F(t{)}^{-T}\mathbf{k}(0)$, as expected for a covector. Fluid material vectors and scalar gradients transform oppositely under stretching.

A deformation map sends each particle's reference configuration position to its current configuration position. Its spatial derivative with respect to the Lagrangian coordinates is the deformation gradient.

A hyperelastic material derives its elastic stress from a strain energy density. In reference-volume units its first Piola-Kirchhoff stress tensor is the derivative of that density with respect to the deformation gradient.

Let $e=(\cos \theta ,\sin \theta )$ be the current fibre direction and $n=(-\sin \theta ,\cos \theta )$. Inextensibility fixes the first column of the deformation gradient to $e$, and incompressibility fixes the second to $\gamma e+n$. Commuting mixed derivatives gives ${\gamma }_{,1}={\theta }_{,1}$ and ${\theta }_{,2}=\gamma {\theta }_{,1}$. Thus $\gamma =\theta +f(X_2)$, and curves tangent to $(-\gamma ,1)$ map to straight transverse curves with constant $\theta$.

A unit reference fibre direction $m_0$ retains its length under this constraint on the deformation gradient. Its reaction contributes an undetermined tension along the current fibre direction, independently of the pressure reaction enforcing incompressibility.

An internal variable records a material state beyond its instantaneous deformation gradient and temperature, such as a relaxing microstructural state. A constitutive equation specifies its evolution and its contribution to the free energy.

The left Cauchy-Green tensor measures stretch on the current configuration. It is a positive-definite symmetric matrix for an invertible deformation gradient, with eigenvalues equal to the squares of the principal stretches.

For an orientation-preserving deformation gradient $F$, oriented area transforms by its cofactor matrix: $(Fa)\times (Fb)=(\det F)F^{-T}(a\times b)$. Taking the dot product with $Fc$ proves this from the determinant identity for the scalar triple product. Applying it to tangent vectors of a surface proves Nanson's formula. Equating current and reference tractions then gives the material-first nominal stress tensor $P=J{F}^{-1}\sigma$ for symmetric Cauchy stress tensor $\sigma$.

For axial stretch $\lambda$ and twist $\alpha$ per reference length, an incompressible neo-Hookean solid has current radius $a=R/\sqrt{\lambda }$. In local cylindrical orthonormal bases the deformation gradient is $A=\left(\begin{array}{ccc}{\lambda }^{-1/2}& 0& 0\\ 0& {\lambda }^{-1/2}& \alpha r\\ 0& 0& \lambda \end{array}\right)$. The Cauchy stress tensor is $\mu A{A}^T+qI$. Radial equilibrium gives $q^{\prime }=\mu {\alpha }^{2}r$, and a traction-free curved surface gives $q=-\mu /\lambda +\mu {\alpha }^2(r^2-a^2)/2$. The nominal end traction is the displayed vector, measured per reference area. Its moment about the axis is $\int (x_1{t}_2-x_2{t}_1)d{S}_0=\mu \alpha \int r^{2}d{S}_0=\mu \pi R^4\alpha /(2\lambda )$. This calculation uses current moment arms with reference-area tractions.

Write $a=\Lambda T$. The deformation gradient of one interval of shear flow is $S(a)=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An initial unit material line at angle $\theta$ therefore becomes $(\cos \theta +a\sin \theta ,\sin \theta )$. Its orientation, including the correct quadrant, isFor an unoriented line the angle is taken modulo $\pi$. If $\Lambda =0$, there is no shear flow: every ${\lambda }_N=1$ and $\mu =0$. The fluctuation and maximizing calculations below concern nonzero shear.

To quantify the renovating shear flow, use the intended isotropic model: each shear orientation is independent and uniform modulo $\pi$. Independence by itself does not specify a probability law and cannot fix the answer. Under the uniform law, conditional on every earlier deformation, the relative angle to the next shear is again uniform. Thus the logarithmic increments ${\xi }_n=\log \ell ({\theta }_n)$ are independent and identically distributed, even though the absolute orientation of the stretched material line need not be uniform. Since $\log {\lambda }_N={\sum }_{n=1}^N{\xi }_n$, only the one-step angular mean is needed.

Write ${\ell }^2=A+B\cos (2\theta -\delta )$, where $A=1+a^2/2$, $B=|a|\sqrt{1+a^2/4}$ and $A^2-B^2=1$. The supplied angular logarithmic integral, with parameter $B/A$, givesTherefore the mean logarithmic stretching rate isThis is the mean logarithmic stretching in an isotropic renovating shear, and the strong law of large numbers gives the same almost-sure Lyapunov exponent. Taking $\Lambda >0$ without loss for this statistic puts it in the requested form $\Lambda F(\Lambda T)$.

For $b=|\Lambda |T$, $F(b)=\log (1+b^2/4)/(2b)$. At small $b$, $\mu ={\Lambda }^{2}T/8+O({\Lambda }^4{T}^3)$; at large $b$, $\mu \sim \log (b/2)/T$, tending to zero when $T$ increases at fixed shear rate. The maximizing nonzero $b$ solvesThe difference $2(1-z^{-1})-\log z$ rises from zero up to $z=2$ and then strictly decreases to $-\infty$, so there is exactly one root beyond $2$. Numerically,Very rapid renovation gives little deformation per interval and cancels the first-order mean stretch. Long intervals provide strong deformation but progressively align a material line with the shear direction, so one interval stretches it only algebraically in $T$. Intermediate persistence combines substantial deformation with frequent reorientation, producing the finite maximum.

For the distribution, let $\gamma =\mathrm{arsinh}(|a|/2)$, so the singular stretches of $S(a)$ are $e^{\pm \gamma }$. In the right-singular-vector basis, ${\ell }^2=e^{2\gamma }{\cos }^2\phi +e^{-2\gamma }{\sin }^2\phi$. Put $q=\tanh \gamma =|a|/\sqrt{a^2+4}$. Factoring this expression and expanding its logarithm givesThe variance follows from angular orthogonality. These bounded independent increments implyAccordingly the central part of the large-$N$ stretch distribution is approximately described by a log-normal distribution, with density $[\lambda \sqrt{2\pi Nv}{]}^{-1}\exp [-(\log \lambda -Nr{)}^2/(2Nv)]$. This central limit theorem is not an assertion about arbitrarily extreme tails: the exact stretch has support inside $e^{-N\gamma }\le {\lambda }_N\le e^{N\gamma }$.

For $|a|\ll 1$, the log-stretch fluctuations in an isotropic renovating shear haveTo leading order the mean and variance of $\log {\lambda }_N$ both equal $N{a}^2/8$. Typical stretch is $\exp (N{a}^2/8)$, while its mean grows as $\exp (3N{a}^2/16)$. These moment statements can be justified directly rather than extrapolating the central Gaussian approximation: expansion of ${\ell }^p$ givesfor fixed $p$. In particular $⟨{\lambda }_N^2⟩=(1+a^2/2{)}^N$ exactly. The leading small-$a$ log-normal approximation is especially precise in the many-small-renovations scaling with $N{a}^2$ fixed and $N{a}^4\to 0$.

If the orientation law is not isotropic these formulas need not hold. For example, repeating a fixed shear direction gives $S(a{)}^N=S(Na)$, only algebraic stretch, and zero asymptotic logarithmic rate. The associated degenerate direction variables are independent, illustrating why the stated independence condition alone is insufficient to select the isotropic result.

A large molecular Schmidt number $\mathrm{Sc}=\nu /\kappa$ separates the velocity's smallest smooth length scale from the scalar's smaller diffusive length scale. Follow a Lagrangian trajectory $X(t)$ and measure displacement $x$ from it. Within a sufficiently small scalar packet, Taylor expansion of the smooth velocity field givesThe moving frame removes uniform sweeping, leaving a linear random velocity gradient tensor. Thus the local scalar equation is the advection-diffusion equation with velocity $Cx$, and averaging over realizations of $C$ gives the corresponding ensemble average. The approximation requires the packet to lie within the velocity's locally smooth region. Its thin, diffusive direction can satisfy this condition even when its long direction eventually becomes too large for one global linearization; applications to a spatially varying flow then use local packet descriptions. The statistical conclusions below refer to the linear-flow model and the usual mixing assumptions, not to every conceivable trace-free random matrix process.

Write $Q=\int \chi dA$ and assume sufficiently rapid decay to discard boundary terms. Integration by parts givesFor $J_{ij}=\int x_i{x}_j\chi dA$, integrate the advective term once and the diffusive term twice. Since ${\partial }_k(x_i{x}_j)={\delta }_{ki}{x}_j+{\delta }_{kj}{x}_i$ and $\Delta (x_i{x}_j)=2{\delta }_{ij}$, one obtainsThese are mass conservation and the evolution of the unnormalized second moment tensor. The first moment obeys $\dot{M}=CM$, so a centered packet stays centered.

For a positive Gaussian scalar packet in a linear incompressible flow, $B$ must be a positive-definite matrix. Diagonalizing its symmetric matrix and applying the Gaussian integral givesThe factor $Q^2$ in the last identity matters because $J$ has not been divided by the scalar mass. Substitution into the advection-diffusion equation also givesThese identities agree with the second moment equation and prove that the Gaussian family is preserved. For an initially negative packet the same analysis applies after replacing $\chi$ by $-\chi$ and taking its positive mass.

To see the roles of stretching and diffusion explicitly, let the deformation gradient $F$ solve $\dot{F}=CF$, $F(0)=I$. Incompressibility gives $\det F=1$. Differentiating $F^{-1}J{F}^{-T}$ integrates the moment equation:Without diffusion, $J=FJ(0)F^T$ and $\det J$ is constant. In a statistically stationary, mixing flow with positive Lyapunov exponent $\lambda$, the two stretching exponents are $\lambda$ and $-\lambda$. Writing the ordered eigenvalues of $J$ as $e^{2{\rho }_1},e^{2{\rho }_2}$, their typical initially nondiffusive evolution isThe fluctuation scale presumes the usual central limit theorem for accumulated strain; it is not implied by incompressibility alone. During this stage the packet becomes long and thin but keeps its area and peak concentration.

In the instantaneous principal-axis basis, write $\tilde{C}$ for the rotated velocity gradient tensor. Differentiation of an eigenvalue givesThe formulas hold through a repeated eigenvalue by continuity of their sum. The compressed physical variance is $e^{2{\rho }_2}/Q$. Its contraction is arrested when it is of order $\kappa /\lambda$, the square of the Batchelor scalar microscale ${\ell }_B\sim \sqrt{\kappa /\lambda }$. If its initial length is $L\gg {\ell }_B$, the crossover occurs atThereafter ${\rho }_2$ has stationary fluctuations associated with the strain and the diffusive floor, while ${\rho }_1$ continues to grow typically at rate $\lambda$. Under sufficiently fast mixing, the central part of the distribution of ${\rho }_1$ is approximately Gaussian with mean $\lambda t+O(1)$ and variance $D_{\rho }t+o(t)$ for a strain-dependent coefficient $D_{\rho }$. This central approximation does not determine far tails, which control high concentration moments; those require a full large deviation principle. The two logarithmic widths need not be independent. The area now increases and $f$ decreases typically as $e^{-\lambda t}$. This is diffusion arrest of Gaussian packet compression.

For example, a constant principal strain $C=\mathrm{diag}(\lambda ,-\lambda )$ gives the exact physical variancesThe thin variance approaches its diffusive floor, the long variance grows exponentially, and the mass-to-area relation gives the quoted typical amplitude decay. A random strain additionally makes rare histories important for concentration moments.

For a precise statistical formulation define the logarithmic area increaseThe Gaussian identities give the exact relation $f(t)=f(0)e^{-H_t}$; henceAssume the long-time strain statistics give a large deviation principle for $H_t/t$ on $s\ge 0$, with convex good rate function $I(s)$, $I(\lambda )=0$. At exponential accuracy the probability of $H_t/t$ near $s$ is $e^{-tI(s)}$. Combining that cost with the amplitude factor and applying the Laplace principle for probability measures gives the large-deviation decay rates of a Gaussian scalar packetThe initial amplitude only changes a time-independent prefactor. One can see the minimization directly by dividing the possible area-growth rates into small intervals: each contributes its probability cost times $e^{-\mu ts}$, and the interval with least total exponential cost dominates. An interval around a minimizing $s$ gives the matching lower bound. The nonnegative area increase bounds the exponential weight by one, so unusually large positive growth cannot produce a divergent weighted tail. Using the area rather than only ${\rho }_1$ also avoids discarding possibly important fluctuations of the compressed direction.

Suppose $I$ is strictly convex and differentiable, with its minimum at $\lambda >0$ and finite negative right derivative at zero. DefineFor $0<\mu <{\mu }_{\ast }$ the minimizing rate $s_{\mu }$ is positive and uniquely solvesFor $\mu \ge {\mu }_{\ast }$, convexity puts the minimum at the endpoint $s=0$, givingHigh moments are dominated by rare histories with little net area growth, in which a packet keeps a substantial concentration. Their probability cost is $I(0)$, independent of the moment order. A finite threshold is conditional: if $I^{\prime }(0+)=-\infty$, the minimizer can approach zero without reaching it at any finite $\mu$.

Below the threshold, differentiation of the minimizing expression cancels the terms involving $s_{\mu }^{\prime }$ because $I^{\prime }(s_{\mu })+\mu =0$. Thus ${\gamma }_{\mu }^{\prime }=s_{\mu }$, andStrict negativity uses a nondegenerate rate function with a unique zero at $\lambda$: for positive $\mu$, $s_{\mu }<\lambda$ and $I(s_{\mu })>0$. For deterministic strain the ratio can instead be constant. Near the typical rate, if $I(s)\simeq (s-\lambda {)}^2/(2D)$ with $D>0$, then ${\gamma }_{\mu }\simeq \lambda \mu -D{\mu }^2/2$ for small $\mu$. If this quadratic expression holds all the way to zero, it gives the illustrative full resultThe quadratic form is an example, not a universal law of turbulent strain.

A thin scalar packet is stretched until molecular diffusion broadens its compressed direction; conservation of its scalar mass then converts area growth into dilution. Ensemble averages sample different stretching histories, producing different decay rates for different moments and saturation when weakly stretched histories dominate. The present moment calculation is specifically for the centered Gaussian packet and averaging over the flow; signed random initial fields with overlapping packets can also have cancellations and require their initial-field statistics.

Finally, positive exponential decay is a statistical conclusion requiring the strain assumptions above, not a consequence of the local differential equation by itself. For $C\equiv 0$ and initially isotropic physical variance $L^2$, the exact solution has $J=Q(L^2+2\kappa t)I$ and $f=Q/[2\pi (L^2+2\kappa t)]$. Its moments decay algebraically as $t^{-\mu }$. Thus the general equations for the exponential rates are the rate-function minimization above; numerical rates and a finite saturation threshold require the long-time statistics of the flow.

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 areIn 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 areThus $I,A_1,A_1^2,A_2$ 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 bySolving for the matrix coefficients givesThis 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 ${\psi }_i{U}^2/H^2$; their longitudinal gradients, and the comparable mixed-component corrections, contribute forces of scale ${\psi }_i{U}^2/(H^2\ell )$. Therefore the normal-stress smallness in lubrication criterion isEquivalently, the thin-gap aspect ratio must multiply the shear-based elastic parameter to give a small quantity: $(H/\ell )|{\psi }_i|\dot{\gamma }/\mu \ll 1$. 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 $D_t{A}_1$ term also requires $|{\psi }_1|/(\mu T)\ll 1$ 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 givehenceThe 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:Integrating with the no-slip boundary condition $u(0)=U$ yieldsThe 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 hasAt $\cos \theta =0$ 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 $Q(h)=Uh-(D/m)h^m$, withIt has a unique maximum atA 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. ConsequentlyFor $0<Q<Q_{\max }$ 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.