Spinodal decomposition: a) The formation of spinodal - Gibbs free energy $G$ vs composition $x_B$. b) solvus line ($\Delta G' = 0$, solid line) and chemical spinodal ($\Delta G'' = 0$, dashed line) vs $T$
Assume
$$ ⁍ $$
We have
$$ f_V = f_V^0 + K[c']^2 $$
We can further assume that the distribution $c(x) - c_0 = A \cos \beta x$ where $\beta = 2\pi/\lambda$.
$f'' = \mathrm{d}f_V / \mathrm{d}c^2$
$\beta$: modulation wavelength
Analysis 1:
$$ \frac{\Delta F}{V} = \frac{A^2}{4} \left[ f'' + 2K\beta^2\right] $$
where $f'' = \partial^2 f / \partial c^2$
$$ f'' < -2K\beta^2 $$
and we have $\beta_\text{c}$, where $\Delta F / V = 0$
$$ \beta_\text{c} = \sqrt{-\frac{f''}{2K}} $$