Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B / 6

A weak solution is a field $\mathbf{u}$ for which the composed densities $\mathbf{P}(x,t,\mathbf{u})$ and fluxes $\mathbf{Q}(x,t,\mathbf{u})$ are locally integrable andfor every smooth compactly supported vector test function $\mathbf{\phi }$ in the spacetime domain. This is the distributional form of the conservation law, obtained by moving derivatives onto the test function. If tests are allowed to meet an initial boundary $t=0$, the specified initial density contributes the additional term $\int \mathbf{P}(x,0,{\mathbf{u}}_0)\cdot \mathbf{\phi }(x,0)dx$ in the usual initial-value formulation.

If the fields are continuously differentiable and obey the pointwise conservation law, integration by parts against a compactly supported test function gives the weak identity, with no boundary terms. Conversely, integrate the weak identity by parts for a continuously differentiable field:for every test function. The fundamental lemma of the calculus of variations, applied to each component, makes the continuous integrand vanish at every point. Hence the classical and weak formulations are equivalent for continuously differentiable fields.

Write the shock as $x=s(t)$ and let the plus trace be on $x>s(t)$. Locally its discontinuous part can be represented with the Heaviside factor $H(x-s(t))$. Distributional differentiation gives ${\partial }_{t}H=-s^{\prime }(t)\delta (x-s(t))$ and ${\partial }_{x}H=\delta (x-s(t))$. The singular part of the conservation law is thereforeIt must vanish, so the Rankine-Hugoniot conditions areThe smooth parts of the fields still satisfy the classical conservation law on each side. The jump condition ensures that the moving discontinuity creates no extra source of the conserved quantities.

Put $v=u-V$, the velocity in the shock frame. The mass jump condition gives a common mass flux $j=\rho v$. Subtracting $V$ times the mass balance from the momentum balance givesFor energy, transform the conserved energy to the shock frame by subtracting $V$ times momentum and adding $V^2/2$ times mass. The resulting energy flux isFor a genuine shock, $j\ne 0$; dividing its continuous energy flux by the same nonzero $j$ on both sides givesThe second term is the ideal-gas specific enthalpy. A contact discontinuity has zero relative mass flux and does not in general require equal specific enthalpies; the division here is appropriate for the shock specified in the question.

Let the undisturbed state ahead have $({\rho }_0,p_0,u_0)=({\rho }_0,p_0,0)$, and the uniform compressed state behind have velocity $u_1=U$. The piston boundary fixes that downstream velocity. Mass and momentum jumps giveThus ${\rho }_1/{\rho }_0=V/(V-U)$. Equality of specific stagnation enthalpy in the shock frame isSubstitute $p_1=p_0+{\rho }_{0}UV$ and $1/{\rho }_1=(V-U)/({\rho }_{0}V)$, then multiply by $2(\gamma -1)V$. The resulting difference factors as $U[-2V^2+(\gamma +1)UV+2\gamma p_0/{\rho }_0]=0$. Since the piston moves into the gas with $U>0$, division givesThis determines the constant speed of the compression shock ahead of the piston.

For a piston-driven normal shock with $U>0$, $c_0>0$ and $\gamma >1$, the quadratic has one positive and one negative root. The forward-moving shock usesSince the square root is greater than $(\gamma +1)U$,Therefore there is always a forward root with shock speed greater than the gas and piston speed under the stated compression assumptions. This also makes the density ratio $V/(V-U)$ positive and greater than one. Evaluating the quadratic at $V=c_0$ gives a negative value, so the positive root additionally has $V>c_0$: it is supersonic relative to the undisturbed gas. The negative root does not describe the shock driven into the region $x>0$.