= Solution
Put $U=b\xi$ and $D=d\xi$. The <shock wave> conditions fix $b=2/(\gamma+1)$ and $d=(\gamma+1)/(\gamma-1)$. The <continuity equation> then reduces to
$$
(b-1)+3b-\beta=0,
$$
so a linear <mass density> profile requires
$$
\boxed{\beta=4b-1=\frac{7-\gamma}{\gamma+1}}.
$$
Momentum requires $P'=-db(b-1+q)\xi^2$. At this exponent $q=-2(\gamma-1)/(\gamma+1)$, hence integration gives $P=2\xi^3/(\gamma+1)+P_0$. Substitution in the <pressure> equation makes the cubic coefficient vanish because $(\gamma+1)b=2$, while its constant term requires $(3\gamma b-3)P_0=0$. For $\gamma>1$ this implies $P_0=0$. Therefore all three equations and all <shock wave> conditions hold for the <linear-profile spherical blast wave>,
$$
U=\frac{2\xi}{\gamma+1},\qquad
D=\frac{\gamma+1}{\gamma-1}\xi,\qquad
P=\frac{2\xi^3}{\gamma+1}.
$$
The prescribed range $0<\beta<3$ restricts this family to $1<\gamma<7$.
For $\gamma=5/3$ one has $\beta=2$, $a=2/3$, $U=3\xi/4$, $D=4\xi$ and $P=3\xi^3/4$. The <energy> integral is
$$
\int_0^1\left[\frac12DU^2+\frac{P}{\gamma-1}\right]\xi^2\,d\xi
=\int_0^1\frac94\xi^5\,d\xi=\frac38.
$$
Consequently $E=(3\pi/2)CR\dot R^2=(2\pi/3)CR^3/t^2$, which fixes the radius and dimensional fields completely:
$$
\boxed{R(t)=\left(\frac{3E}{2\pi C}\right)^{1/3}t^{2/3},\qquad
u(r,t)=\frac{r}{2t},}
$$
$$
\boxed{\rho(r,t)=\frac{4Cr}{R^3(t)}=\frac{8\pi C^2r}{3Et^2},\qquad
p(r,t)=\frac{Cr^3}{3R^3(t)t^2}=\frac{2\pi C^2r^3}{9Et^4},\quad 0<r<R(t)}.
$$
The <kinetic energy> and <internal energy> integrands are each $9\xi^5/8$, so each accounts for half the explosion <energy>. Integrating the <mass density> also gives the correct swept-up mass $4\pi CR$.
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