= Solution
At fixed <cylindrical radius>, let $w=u_z$ and use the <continuity equation> $\partial_t\rho+\partial_z(\rho w)=0$. For the <similarity solution> $z=\xi\eta(t)$ and $\rho=\widetilde\rho(\xi)/\eta$, the <continuity equation> reduces to
$$
\partial_\xi\left[\widetilde\rho(w-\dot\eta\xi)\right]=0.
$$
Here $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a <homologous vertical motion of an astrophysical disk>:
$$
w=\dot\eta\xi=\frac{\dot\eta}{\eta}z,\qquad
\partial_z=\eta^{-1}\partial_\xi,\qquad
\left.\partial_t\right|_z=\left.\partial_t\right|_\xi-\frac{\dot\eta}{\eta}\xi\partial_\xi,\qquad
D_t=\left.\partial_t\right|_\xi.
$$
The <hydrostatic approximation> and <ideal gas> law become
$$
\frac{d\widetilde P}{d\xi}=-\Omega^2\xi\widetilde\rho,\qquad
\widetilde P=\mathcal R\widetilde\rho\widetilde T.
$$
The <material derivatives> are $D_tP=\widetilde P\dot\eta$ and $D_t\rho=-\widetilde\rho\dot\eta/\eta^2$, so the energy equation gives
$$
\frac{\gamma+1}{\gamma-1}\widetilde P\dot\eta=-A\widetilde T^{1+\beta}\eta^{2+2\beta}.
$$
<Separation of variables> requires a positive constant $K$ such that
$$
\dot\eta=-K\eta^{2+2\beta},\qquad
\widetilde P=\frac{(\gamma-1)A}{(\gamma+1)K}\widetilde T^{1+\beta}.
$$
Integrating with $\eta(0)=1$ gives the <slow cooling of an astrophysical disk> similarity factor
$$
\boxed{\eta(t)=(1+Ct)^{-1/(1+2\beta)},\qquad C=(1+2\beta)K>0.}
$$
A closed set of profile equations is therefore
$$
\boxed{\widetilde P'=-\Omega^2\xi\widetilde\rho,\qquad
\widetilde P=\mathcal R\widetilde\rho\widetilde T,\qquad
\widetilde P=\frac{(\gamma-1)(1+2\beta)A}{(\gamma+1)C}\widetilde T^{1+\beta}.}
$$
These have the same <polytropic vertical structure in stellar gravity> as part (b). Writing
$$
H_0^2=\frac{2(1+\beta)\mathcal R T_0}{\Omega^2},\qquad
\widetilde T=T_0\left(1-\frac{\xi^2}{H_0^2}\right),\qquad
\widetilde\rho=\rho_0\left(1-\frac{\xi^2}{H_0^2}\right)^\beta,\qquad
\widetilde P=P_0\left(1-\frac{\xi^2}{H_0^2}\right)^{1+\beta},
$$
with $P_0=\mathcal R\rho_0T_0$, fixes
$$
C=\frac{(\gamma-1)(1+2\beta)A T_0^{1+\beta}}{(\gamma+1)P_0}.
$$
Returning to physical height, the explicit profiles and evolving semi-thickness for $|z|\leq H(t)$ are
$$
\boxed{H(t)=H_0(1+Ct)^{-1/(1+2\beta)},\qquad
T(z,t)=\frac{\Omega^2}{2(1+\beta)\mathcal R}\left[H(t)^2-z^2\right],}
$$
and
$$
\rho(z,t)=\frac{\rho_0}{\eta(t)}\left[1-\frac{z^2}{H(t)^2}\right]^\beta,\qquad
P(z,t)=P_0\eta(t)\left[1-\frac{z^2}{H(t)^2}\right]^{1+\beta}.
$$
Outside this moving surface $\rho=P=0$. The <surface density of a disk> is conserved because $\rho\,dz=\widetilde\rho\,d\xi$. The column \b[cools and contracts without changing its scaled profile]: its midplane <temperature> falls as $\eta^2$, its <mass density> rises as $\eta^{-1}$ and its <pressure> falls as $\eta$.
The value $T_0$ determines $H_0$, but the <mass density> normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating gives
$$
C=\frac{9\alpha\Omega}{4}\frac{(\gamma-1)(1+2\beta)}{\gamma+1}.
$$
Otherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed <vertical dynamical timescale of a disk>. This is an exact solution of the reduced <hydrostatic approximation>, rather than the full momentum equation: its omitted vertical acceleration is $\ddot\eta\xi$, with $\ddot\eta/(\eta\Omega^2)=2(1+\beta)C^2/[(1+2\beta)^2\Omega^2(1+Ct)^2]$, which remains small and decreases if $C\ll\Omega$.
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