Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 3 / iii / Solution

Multiplying $y^{\prime \prime }=-{\omega }^2{y}^2$ by $y^{\prime }$ and integrating giveswhere the base conditions fixed the constant. At the surface, $y=0$ and $y^{\prime }=A{F}_s$, with $F_s=L/(4\pi R^2)$. Thereforeand

Separating variables in the first integral givesSet $y=T_0^7\eta$. ThenSince $x_0={\Sigma }_0^2/2$ and ${\Sigma }_0=M_{\mathrm{env}}/(4\pi R^2)$,Combining this with $F_s\propto T_0^{21/2}$ yields

This inverse relation signals thin-shell instability. Hydrogen burning has extreme temperature sensitivity, $ϵ\propto \rho T^{15}$, while the weight of the thin envelope fixes its pressure and limits thermostatic expansion. As burning consumes envelope mass, the equilibrium luminosity rises sharply, accelerating consumption rather than restoring the original state. A thermal runaway can therefore lead to a shell flash or classical nova rather than steady stable burning.