 Solution

For $\alpha /\lambda \ll 1$, follow the unstable root near the clean-film value $L=\lambda s(1-s)$. Write $\beta =L+\Delta$ in $(\beta -L)(\beta +S)+3LS/4=0$. At first order in $S$, $L\Delta +3LS/4=0$, givingThis expansion is uniform near the maximizing wavenumber. Differentiation gives $s_{\ast }=1/2-3\alpha /(8\lambda )+O((\alpha /\lambda {)}^2)$. Evaluating the growth there yieldsThe change in maximizing wavenumber only affects the maximum at second order. The first-order decrease therefore has exactly the requested coefficient, relative to the clean-film rate with its surface tension replaced by $\bar{\sigma }$.