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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 toso a linear mass density profile requiresMomentum requires $P^{\prime }=-db(b-1+q){\xi }^2$. 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,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 isConsequently $E=(3\pi /2)CR{\dot{R}}^2=(2\pi /3)C{R}^3/t^2$, which fixes the radius and dimensional fields completely: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 $4\pi CR$.