= Solution
Write $a=\Lambda T$. The <deformation gradient> of one interval of <shear flow> is $S(a)=\begin{pmatrix}1&a\\0&1\end{pmatrix}$. An initial unit <material line> at angle $\theta$ therefore becomes $(\cos\theta+a\sin\theta,\sin\theta)$. Its orientation, including the correct quadrant, is
$$
\boxed{\theta_T=\operatorname{atan2}(\sin\theta,\cos\theta+a\sin\theta),\quad
\ell(\theta)=\sqrt{1+a\sin2\theta+a^2\sin^2\theta}.}
$$
For 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 $B/A$, gives
$$
r(a):=\langle\xi\rangle=\frac12\log\frac{A+\sqrt{A^2-B^2}}2
=\frac12\log(1+a^2/4).
$$
Therefore \b[the mean logarithmic stretching rate] is
$$
\boxed{\mu=\frac{\log(1+\Lambda^2T^2/4)}{2T}
=|\Lambda|\frac{\log(1+a^2/4)}{2|a|}.}
$$
This 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 $b$, $\mu=\Lambda^2T/8+O(\Lambda^4T^3)$; at large $b$, $\mu\sim\log(b/2)/T$, tending to zero when $T$ increases at fixed shear rate. The maximizing nonzero $b$ solves
$$
\log z=2(1-z^{-1}),\qquad z=1+b^2/4.
$$
The 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,
$$
\boxed{|\Lambda|T\simeq3.9605826,\qquad \mu_{\max}\simeq0.2012|\Lambda|.}
$$
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 $\gamma=\operatorname{arsinh}(|a|/2)$, so the singular stretches of $S(a)$ are $e^{\pm\gamma}$. In the right-singular-vector basis, $\ell^2=e^{2\gamma}\cos^2\varphi+e^{-2\gamma}\sin^2\varphi$. Put $q=\tanh\gamma=|a|/\sqrt{a^2+4}$. Factoring this expression and expanding its logarithm gives
$$
\xi=r(a)+\sum_{j=1}^{\infty}\frac{(-1)^{j+1}q^j}{j}\cos(2j\varphi),\qquad
v(a):=\operatorname{Var}\xi=\frac12\sum_{j=1}^{\infty}\frac{q^{2j}}{j^2}.
$$
The variance follows from angular orthogonality. These bounded independent increments imply
$$
\boxed{\frac{\log\lambda_N-Nr(a)}{\sqrt{Nv(a)}}\ \Longrightarrow\ N(0,1).}
$$
Accordingly 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 $e^{-N\gamma}\leq\lambda_N\leq e^{N\gamma}$.
For $|a|\ll1$, the <log-stretch fluctuations in an isotropic renovating shear> have
$$
r(a)=\frac{a^2}{8}-\frac{a^4}{64}+O(a^6),\qquad
v(a)=\frac{a^2}{8}-\frac{3a^4}{128}+O(a^6).
$$
To leading order the mean and variance of $\log\lambda_N$ both equal $Na^2/8$. Typical stretch is $\exp(Na^2/8)$, while its mean grows as $\exp(3Na^2/16)$. These moment statements can be justified directly rather than extrapolating the central Gaussian approximation: expansion of $\ell^p$ gives
$$
\langle\ell^p\rangle=1+\frac{p(p+2)}{16}a^2+O(a^4),\qquad
\boxed{\langle\lambda_N^p\rangle=\exp\left[\frac{Np(p+2)a^2}{16}+O(Na^4)\right]}
$$
for fixed $p$. In particular $\langle\lambda_N^2\rangle=(1+a^2/2)^N$ exactly. The leading small-$a$ log-normal approximation is especially precise in the many-small-renovations scaling with $Na^2$ fixed and $Na^4\to0$.
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.
Back to article page
