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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.