2.2 Weak solutions and the Rankine–Hugoniot condition
After breaking, the conservation law still makes sense in its integral form: for every interval \([\alpha , \beta ]\), \[ \frac {d}{dt}\int _\alpha ^\beta u\,dx = f\bigl (u(\alpha , t)\bigr ) - f\bigl (u(\beta , t)\bigr ) , \] which does not require derivatives of \(u\). Equivalently, \(u\) is a weak solution if for every smooth test function \(\phi \) with compact support in \(\mathbb R \times [0, \infty )\), \[ \int _0^\infty \!\!\int _{-\infty }^\infty \bigl (u\phi _t + f(u)\phi _x\bigr )\,dx\,dt + \int _{-\infty }^\infty u_0(x)\phi (x, 0)\,dx = 0 , \] obtained by multiplying the equation by \(\phi \) and integrating by parts. A classical solution is a weak solution, and a weak solution that is \(C^1\) is classical. Weak solutions may have jumps.
Theorem 2.2.1 (Rankine–Hugoniot). Let \(u\) be a weak solution that is \(C^1\) on either side of a curve \(x = s(t)\), with limits \(u_l\) and \(u_r\) on the left and right. Then the speed of the discontinuity satisfies \[ s'(t) = \frac {f(u_r) - f(u_l)}{u_r - u_l} = \frac {[f]}{[u]} . \] For Burgers’ equation, \(s' = \frac 12(u_l + u_r)\).
Proof. Take \(\alpha < s(t) < \beta \) with \([\alpha , \beta ]\) small. Splitting the integral at the shock and differentiating, \[ \frac {d}{dt}\Bigl (\int _\alpha ^{s(t)}u\,dx + \int _{s(t)}^\beta u\,dx\Bigr ) = \int _\alpha ^{s}u_t\,dx + \int _{s}^\beta u_t\,dx + s'u_l - s'u_r . \] On each side \(u_t = -f(u)_x\), so the integrals equal \(f(u(\alpha )) - f(u_l) + f(u_r) - f(u(\beta ))\). Comparing with the integral form of the law, \(f(u(\alpha )) - f(u(\beta ))\), gives \(s'(u_l - u_r) = f(u_l) - f(u_r)\). □
The condition says that the amount of \(u\) swept up as the shock moves is supplied by the difference of the fluxes on its two sides. It applies to any conservation law, and is the jump condition across a shock wave in a gas.
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