= Solution
Write the <ideal gas> law as $P=\mathcal R\rho T/\mu$. Since $P=g\Sigma$,
$$
\rho=B\frac{\Sigma}{T},
\qquad
B=\frac{\mu g}{\mathcal R}.
$$
The prescribed opacity and burning laws become
$$
\kappa=\kappa_0B\Sigma T^{-3},
\qquad
\epsilon=\epsilon_0B\Sigma T^{14}.
$$
The radiative equation is consequently
$$
\frac{dT}{d\Sigma}=K\Sigma F T^{-6},
\qquad
K=\frac{3\kappa_0B}{4ac}.
$$
With $y=T^7$ and $x=\Sigma^2/2$,
$$
\frac{dy}{dx}=7KF\equiv AF.
$$
The energy equation similarly gives
$$
\frac{dF}{dx}=-\epsilon_0B y^2.
$$
Differentiating the first relation therefore produces the <nonlinear ordinary differential equation>
$$
\boxed{\frac{d^2y}{dx^2}=-\omega^2y^2},
\qquad
\boxed{\omega^2=A\epsilon_0B\gt0}.
$$
At the idealized zero-temperature surface, $x=0$ and $y=0$. The outward flux there is $F_s=L/(4\pi R^2)$, so
$$
\boxed{y(0)=0,
\qquad
y'(0)=\frac{AL}{4\pi R^2}}.
$$
At the base, $x_0=\Sigma_0^2/2$ and $y=T_0^7$. The core supplies no luminosity in this model, so all flux has been generated in the overlying hydrogen envelope and $F(x_0)=0$. Hence
$$
\boxed{y(x_0)=T_0^7,
\qquad
y'(x_0)=0}.
$$
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