= Solution
Take a positive initial separation $0<l_0\ll\eta$. In the <dissipation range>, the <velocity field> is smooth, so the difference of the two <Lagrangian trajectories> obeys the linearized separation equation $\dot{\boldsymbol l}\simeq(\nabla\mathbf u)\boldsymbol l$. Chaotic stretching gives a positive <Lyapunov exponent> of order $\tau_\eta^{-1}$ and a typical separation
$$
l(t)\sim l_0e^{\lambda t},\qquad t_1\sim\lambda^{-1}\log(\eta/l_0).
$$
This needs a nonzero initial separation: exactly coincident particles remain coincident in a smooth flow.
Once $l$ lies in the <inertial range>, the <Kolmogorov 1941 theory> gives a relative speed of order $(\epsilon l)^{1/3}$. A scale-local typical-separation estimate $dl/dt\sim(\epsilon l)^{1/3}$ integrates to
$$
l(t)^{2/3}\sim\eta^{2/3}+C\epsilon^{1/3}(t-t_1).
$$
After loss of the entry-scale memory, the statistical <Richardson pair dispersion> law is $\mathbb E[l^2]\sim g_R\epsilon(t-t_1)^3$. Thus the characteristic separation grows as $(t-t_1)^{3/2}$ while $\eta\ll l\ll L$. This is a statistical scaling closure, rather than a deterministic equation for every pair.
For $l\gg L$, the particle velocities become approximately independent. If their <Lagrangian velocity autocorrelations> have a finite <Lagrangian integral time> $T_L\sim L/U$, the <diffusive large-scale pair dispersion> has effective relative <diffusivity> of order $UL$. After a further velocity-decorrelation transient,
$$
\mathbb E[l^2(t)]\sim L^2+C_DUL(t-t_2),\qquad l_{\mathrm{rms}}\propto(t-t_2)^{1/2}.
$$
This is the long-time mechanism of <Taylor turbulent dispersion> applied to the difference of two decorrelated trajectories. The <root mean square> separation eventually loses memory of the $L^2$ term.
Reaching $l=L$ at the end of the <inertial range> takes
$$
\boxed{t_2-t_1\sim\epsilon^{-1/3}\left(L^{2/3}-\eta^{2/3}\right)\sim\left(\frac{L^2}{\epsilon}\right)^{1/3}\sim\frac LU.}
$$
Order-one coefficients depend on the dispersion closure. The initial exponential stage can take a long time if $l_0$ is extremely small, but the inertial-range stage is of order one outer <eddy turnover time>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-69-pair-dispersion.png]
{title=Exponential, Richardson and diffusive regimes of turbulent pair separation}
The <three-regime turbulent pair-separation model> in the sketch matches the regimes continuously for illustration. Its coefficients are schematic; the universal claims here are the scaling powers under the stated stretching, locality and decorrelation assumptions.
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