For an arrested domain size $\Lambda$, make the prescribed replacement
$$
\dot L\longrightarrow\frac{\Lambda}{\tau},
\qquad
\ddot L\longrightarrow\frac{\Lambda}{\tau^2}.
$$
Writing
$$
\Lambda=L_0g(w),
\qquad
w=\frac{\tau}{t_0},
$$
gives the schematic algebraic equation
$$
(\alpha+\beta)\frac{g}{w^2}
=\gamma\frac1{wg}+\frac{\delta}{g^2},
$$
or
$$
(\alpha+\beta)g^3=\gamma wg+\delta w^2.
$$

For $w\ll1$, viscous-capillary balance gives
$$
g(w)\sim\frac{\delta}{\gamma}w,
\qquad
\Lambda\sim\frac{\sigma\tau}{\eta},
$$
which is independent of $\rho$. For $w\gg1$, inertial-capillary balance gives
$$
g(w)\sim
\left(\frac{\delta}{\alpha+\beta}\right)^{1/3}w^{2/3},
\qquad
\Lambda\sim
\left(\frac{\sigma\tau^2}{\rho}\right)^{1/3},
$$
which is independent of $\eta$. These are the two limits of the <stirring-arrested binary-fluid domain size>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 2 / e / Solution

If stirring imposes a time scale $\tau$, replacing time derivatives by $1/\tau$ predicts a steady domain size. Viscous-capillary balance gives $\Lambda\sim\sigma\tau/\eta$, while inertial-capillary balance gives $\Lambda\sim(\sigma\tau^2/\rho)^{1/3}$.


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