= Solution
In the radiative layer, take the enclosed <mass> to be $M_1$, the transported <luminosity> to be $L$, and the <mean molecular weight> to be constant. The envelope's own gravity is neglected. Divide the hydrostatic and radiative <temperature> equations, and then eliminate density using the gas equation of state:
$$
\frac{dP}{dT}=\frac{16\pi acGM_1}{3\kappa L}T^3
=B\frac{T^{3+m+n}}{P^n},\qquad
B=\frac{16\pi acGM_1}{3\kappa_0L}\left(\frac{\mathcal R}{\mu}\right)^n.
$$
Put $p=n+1$ and $b=4+m+n$. For $p,b\ne0$, integration gives
$$
P^p=\frac{pB}{b}T^b+K_P.
$$
Defining $C^p=pB/b$ and $K=K_P/C^p$ yields
$$
\boxed{P=C(T^b+K)^{1/p},\qquad
C=\left[\frac{16\pi acGM_1}{3\kappa_0L}
\left(\frac{\mathcal R}{\mu}\right)^n\frac{n+1}{4+m+n}\right]^{1/(n+1)}.}
$$
In the relevant positive-exponent case, writing $K=T_0^b$ produces the stated form whenever this integration constant is nonnegative. The exceptional values excluded above lead to logarithms: $p=0$ gives $\log P=BT^b/b+\mathrm{const}$, $b=0$ gives $P^p=pB\log T+\mathrm{const}$, and both zero give $\log P=B\log T+\mathrm{const}$.
For <radiative-envelope convection matching>, compute the logarithmic gradient of the <pressure> relation:
$$
\nabla_{\mathrm{rad}}=\frac{d\log T}{d\log P}
=\frac pb\left(1+\frac K{T^b}\right).
$$
At the marginal boundary, the <Schwarzschild criterion> equates this with the ideal-gas adiabatic gradient $\nabla_{\mathrm{ad}}=(\gamma-1)/\gamma$. Hence
$$
K=T_b^b\left[\frac bp\frac{\gamma-1}{\gamma}-1\right],
\qquad
\boxed{T_0=T_b\left[\frac{\gamma-1}{\gamma}
\frac{4+m+n}{n+1}-1\right]^{1/(4+m+n)}.}
$$
The positive real $T_0$ notation requires a nonnegative bracket; a general signed $K$ is the safer integration constant otherwise. For the later case $n=1,m=3$ and a monatomic gas, this bracket is $3/5$, so $T_0$ is of the same order as $T_b$.
Deep below the convective boundary, with the temperature-power term dominating $K$, the <pressure> law becomes $P\simeq CT^{b/p}$. Differentiating this approximation and using hydrostatic balance gives
$$
\frac bp\frac PT\frac{dT}{dr}
=-\frac{GM_1}{r^2}\frac{\mu P}{\mathcal RT},
\qquad
\frac{dT}{dr}=-\frac{\mu GM_1p}{\mathcal Rb}\frac1{r^2}.
$$
The general <temperature> is a term proportional to $1/r$ plus a boundary-dependent constant. Well inside the boundary radius, where that term dominates, it reduces to
$$
\boxed{T(r)\simeq\frac{\mu GM_1(n+1)}{\mathcal R(4+m+n)r}.}
$$
The inverse-radius profile is an inner approximation, not an assertion that the boundary term vanishes everywhere.
Now take $n=1,m=3$, so $p=2$, $b=8$. Then $P\simeq CT^4$, $\rho\simeq(\mu C/\mathcal R)T^3$, and define $A_T=\mu GM_1/(4\mathcal R)$ so that $T\simeq A_T/r$. The <luminosity> supplied by the thin burning layer is
$$
\begin{aligned}
L&\simeq4\pi\int_{R_1}^{r_b}r^2\rho\epsilon\,dr
=4\pi C^2\epsilon_0\left(\frac\mu{\mathcal R}\right)^2
A_T^{16}\int_{R_1}^{r_b}r^{-14}\,dr\\
&=\frac{4\pi}{13}C^2\epsilon_0\left(\frac\mu{\mathcal R}\right)^2
A_T^{16}(R_1^{-13}-r_b^{-13}).
\end{aligned}
$$
Because the integral is strongly concentrated near $R_1$ and $r_b\gg R_1$, the outer term is negligible. Thus
$$
\boxed{L\simeq\frac{4\pi}{13}C^2\epsilon_0
\left(\frac\mu{\mathcal R}\right)^2
\left(\frac{\mu GM_1}{4\mathcal R}\right)^{16}\frac1{R_1^{13}}.}
$$
The factor $\rho\epsilon$, rather than merely $\epsilon$, is essential since the specified energy-generation rate is per unit <mass>. This integration uses the deep radiative profile as a leading thin-shell approximation.
Crucially, $C$ is not independent of <luminosity>. Its defining relation gives
$$
C^2=\frac{4\pi acGM_1}{3\kappa_0L}\frac{\mathcal R}{\mu}
\propto\frac{M_1}{L}.
$$
Substitution into the shell integral therefore yields $L^2\propto M_1^{17}R_1^{-13}$. The degenerate-core relation gives $R_1\propto M_1^{-1/3}$, so
$$
L^2\propto M_1^{17+13/3}=M_1^{64/3},\qquad
\boxed{L\propto M_1^{32/3}.}
$$
This <degenerate-core shell-burning mass-luminosity relation> is specific to the adopted <opacity>, burning exponent and structural approximations.
Removing some envelope <mass> while retaining the hydrogen-burning shell and the deep-envelope regime leaves the leading <luminosity> approximately unchanged at fixed core <mass>. The envelope can instead change the radius and <effective temperature>. This independence is not valid for arbitrarily complete stripping: removing the fuel or the pressure-confined shell structure quenches or modifies shell burning, and the assumed giant model then fails.
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