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Let $D=u_1>0$ denote the front speed in the stationary upstream frame. In the shock frame, the upstream and downstream velocities are $-D$ and $-D/r$. Taking $M_1\to \infty$ in the Rankine-Hugoniot conditions for a perfect gas givesThe downstream laboratory velocity is $D-D/r=2D/(\gamma +1)$, which distinguishes the gas speed from the front speed.

For the planar blast-wave energy scaling of a self-similar blast wave, integration of the total energy density over the shocked interval $0<z<Z$ gives the energy per unit area on this side:The similarity solution makes $I$ time-independent; a finite positive explosion energy requires $0<I<\infty$. Conservation of energy therefore gives $\dot{Z}=(E_{+}/({\rho }_{1}I){)}^{1/2}{Z}^{-1/2}$ for the expanding front. Integrating from $Z(0)=0$,If $E$ denotes the one-sided energy, $C=(9/(4I){)}^{1/3}$. If the released energy $E$ feeds two symmetric fronts, $E_{+}=E/2$ and $C=(9/(8I){)}^{1/3}$. In either convention the requested scaling is $Z=C(E/{\rho }_1{)}^{1/3}{t}^{2/3}$. The constant depends on the similarity profiles and the energy convention; energy conservation determines the exponent without solving those profiles. The Strong-shock Rankine-Hugoniot conditions additionally fix $f(1)=(\gamma +1)/(\gamma -1)$ and $g(1)=h(1)=2/(\gamma +1)$.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

The original PDF supplies the intended profiles $\rho ={\rho }_{1}f(\eta )$, $p={\rho }_1{\dot{R}}^{2}g(\eta )$ and $u=\dot{R}h(\eta )$; the TeX has corrupted the first two. At fixed $r$, ${\partial }_t\eta =-\eta \dot{R}/R$, and at fixed $t$, ${\partial }_r\eta =1/R$. The spherical continuity equation is thereforeSince $\dot{R}/R=3/(5t)$ is nonzero, the continuity equation for a superbubble similarity solution isThe ambient gas is at rest. In the locally stationary shock frame, its velocity is $-\dot{R}$; the downstream velocity is $u-\dot{R}$. Using the shock compression ratio from part (a) converts the downstream speed back to the laboratory frame and gives the inner shock boundary conditionsImmediately outside the shock, the corresponding unperturbed profiles are $f=1$, $g=0$ and $h=0$, with $g=0$ understood in the negligible-ambient-pressure limit.

The inner boundary must supply the injected power. If the source is idealized as a point and the profiles extend all the way inward, central energy input in a similarity solution requiresor equivalently ${\lim }_{\eta \downarrow 0}{\eta }^{2}h(f{h}^2/2+\gamma g/(\gamma -1))=125/(108\pi {\alpha }^5)$. A finite source region or an inner wind/contact region supplies a different matching description. One cannot impose a regular zero-flux centre on a source-free interior and still inject nonzero power.

Work in the shock frame. The undisturbed upstream gas enters the stationary discontinuity at speedwhile the downstream gas, which moves with the piston in the laboratory frame, leaves atThe Rankine-Hugoniot conditions for a perfect gas areandEliminating the velocities between these equations gives the perfect-gas density jump

The mass and momentum equations also implysoSubstitution of the density ratio into mass conservation, followed by this momentum relation, givesUsing the adiabatic sound speed $c_1^2=\gamma p_1/{\rho }_1$ therefore yieldsFinally, eliminating $U$ from $V=\beta p_1/({\rho }_{1}U)$ gives the piston-driven normal shock relationFor a compressive shock $\beta >0$, the first boxed equation has $U>c_1$: the shock propagates supersonically into the undisturbed gas, approaching the sound speed in the weak shock limit $\beta \to 0$.

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

Work in the shock frame, and let the positive upstream and downstream speeds be $w_0$ and $w_1$. The Rankine-Hugoniot conditions for a perfect gas give conservation of mass and momentum:PutMass conservation gives $w_1=w_0/r$, and momentum conservation then gives

Applying the energy integral from part (a) on the two sides givesSubstitution of (1) into (2), followed by cancellation of $p_0/{\rho }_0$, yieldsSolving this linear equation for the density ratio gives