= Solution
\b[The local conservation laws.] The first equation is the radial <Euler equation>: the <material acceleration> $Du/Dt=\partial_tu+u\partial_ru$ is caused by the <pressure gradient> force per unit mass, $-\partial_rp/\rho$. It neglects gravity and viscosity. The second equation is <mass conservation> in spherical coordinates,
$$
\partial_t\rho+\frac1{r^2}\partial_r(r^2\rho u)=0.
$$
The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.
To obtain the third equation, let $e=p/[(\gamma-1)\rho]$ be the <specific internal energy> of an <ideal gas> with constant <heat capacity ratio> $\gamma$. The <first law of thermodynamics> for a smooth adiabatically moving element is
$$
\frac{De}{Dt}+p\frac{D(1/\rho)}{Dt}=0,
\qquad
\frac{De}{Dt}=\frac p{\rho^2}\frac{D\rho}{Dt}.
$$
Differentiating $e$ and collecting terms gives
$$
\frac{1}{(\gamma-1)\rho}\frac{Dp}{Dt}
-\frac{p}{(\gamma-1)\rho^2}\frac{D\rho}{Dt}
=\frac p{\rho^2}\frac{D\rho}{Dt},
$$
and hence
$$
\frac{Dp}{Dt}=\frac{\gamma p}{\rho}\frac{D\rho}{Dt},\qquad
\boxed{\left(\partial_t+u\partial_r\right)(p\rho^{-\gamma})=0.}
$$
Thus the entropy parameter $p/\rho^\gamma$ is conserved along each smooth <pathline>; for fixed composition, <specific entropy> differs from $c_v\ln(p/\rho^\gamma)$ only by a constant. This is local <isentropic flow> behind the shock. A physical <shock wave> itself generates <entropy> and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.
\b[The front-radius scaling.] In the <self-similar blast wave> approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a\rho_0^bt^d$. Using $[E]=\mathsf M\mathsf L^2\mathsf T^{-2}$ and $[\rho_0]=\mathsf M\mathsf L^{-3}$, <dimensional analysis> gives
$$
a+b=0,\qquad 2a-3b=1,\qquad -2a+d=0.
$$
Thus $a=1/5$, $b=-1/5$, $d=2/5$ and
$$
\boxed{R=\xi_0\left(\frac E{\rho_0}\right)^{1/5}t^{2/5}.}
$$
The dimensionless coefficient $\xi_0$ depends on $\gamma$ and the resolved <similarity profiles>; dimensions alone do not fix it. Define the length scale $\ell(t)=(Et^2/\rho_0)^{1/5}$, so $\xi=r/\ell$ and the front is at the fixed coordinate $\xi=\xi_0$.
\b[Kinetic energy.] Substitute $\rho=\rho_0f_\rho(\xi)$ and $u=(r/t)f_u(\xi)$ into the volume integral. Since $r=\ell\xi$ and $dr=\ell\,d\xi$,
$$
E_{\rm kin}=\frac{2\pi\rho_0}{t^2}\int_0^Rr^4f_\rho f_u^2\,dr
=\frac{2\pi\rho_0\ell^5}{t^2}\int_0^{\xi_0}\xi^4f_\rho f_u^2\,d\xi.
$$
But $\rho_0\ell^5/t^2=E$, so the first of the <spherical blast-wave similarity energy integrals> is
$$
\boxed{E_{\rm kin}=2\pi E\int_0^{\xi_0}\xi^4f_\rho(\xi)f_u(\xi)^2\,d\xi
=\text{constant}.}
$$
The conclusion requires a fixed, integrable <similarity profile>, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.
\b[Pressure and internal energy.] The derivatives of the <similarity variable> are
$$
\left.\partial_t\xi\right|_r=-\frac{2\xi}{5t},\qquad
\left.\partial_r\xi\right|_t=\frac1\ell.
$$
Writing a prime for differentiation with respect to $\xi$, the <material acceleration> becomes
$$
\partial_tu=\frac r{t^2}\left[-f_u-\frac25\xi f_u'\right],\qquad
u\partial_ru=\frac r{t^2}\left[f_u^2+\xi f_uf_u'\right],
$$
$$
\frac{Du}{Dt}=\frac r{t^2}\left[f_u^2-f_u+\xi\left(f_u-\frac25\right)f_u'\right].
$$
The radial momentum equation therefore requires a <pressure gradient> on the scale $\rho_0\ell/t^2$. Integrating over radius gives <pressure similarity in a spherical blast wave>:
$$
\boxed{p(r,t)=\rho_0\frac{\ell^2}{t^2}f_p(\xi)
=E^{2/5}\rho_0^{3/5}t^{-6/5}f_p(\xi),}
$$
where
$$
\boxed{f_p'=-\xi f_\rho\left[f_u^2-f_u+\xi\left(f_u-\frac25\right)f_u'\right].}
$$
The prefactor has units $\mathrm{kg\,m^{-1}\,s^{-2}}$, as required for <pressure>. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.
For constant $\gamma>1$, the <internal energy> is consequently
$$
E_{\rm int}=\frac{4\pi}{\gamma-1}\frac{\rho_0\ell^5}{t^2}
\int_0^{\xi_0}\xi^2f_p(\xi)\,d\xi,
$$
$$
\boxed{E_{\rm int}=\frac{4\pi E}{\gamma-1}\int_0^{\xi_0}\xi^2f_p(\xi)\,d\xi
=\text{constant}.}
$$
If $E$ is the total excess explosion <energy>, its normalization is
$$
1=2\pi\int_0^{\xi_0}\xi^4f_\rho f_u^2\,d\xi
+\frac{4\pi}{\gamma-1}\int_0^{\xi_0}\xi^2f_p\,d\xi.
$$
The two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.
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