advanced34 min studyLesson 416 in Mathematics

Shock formation and Rankine–Hugoniot

Continue conservation across a moving jump using a signed distributional balance, without confusing that balance with entropy admissibility.

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What you’ll understand

  • Write the weak conservation identity with its initial term.
  • Derive the jump speed using a consistent right-minus-left convention.
  • Explain why a mass-balanced jump still needs an admissibility test.

When characteristics collide, a smooth density description breaks down while the underlying conservation principle remains meaningful. A moving discontinuity can replace the overlapping profile, provided its motion balances the flux entering and leaving it. Rankine–Hugoniot gives that balance exactly. Its signs are easiest to understand by fixing one jump convention and checking a control-volume calculation. The condition determines whether a jump is a weak solution; a further entropy principle will decide which weak continuation is selected.

Move derivatives onto a test function

A bounded discontinuous function can still define a weak scalar conservation law because both the state and its flux are locally integrable. Multiply the conservative equation by a smooth compactly supported test function and integrate by parts in space and time. The derivatives then act on the test function, and the prescribed initial state contributes a boundary term at time zero. A weak solution satisfies this identity for every such test, rather than only at points where a classical derivative happens to exist.

The conservative form is crucial. Replacing the flux derivative by a discontinuous coefficient times a distributional state derivative is not an ordinary chain-rule operation at a jump. The weak flux identity instead provides a well-defined distribution without assigning an arbitrary product at the discontinuity. Its local integrals also allow constant left and right Riemann states, even though those states may not have an integrable total mass over the entire line. Local conservation remains meaningful on every finite control interval.

∫0∞ ⁣∫R(uϕt+f(u)ϕx) dx dt+∫Ru0(x)ϕ(0,x) dx=0\int_0^\infty\!\int_{\mathbb R}\left(u\phi_t+f(u)\phi_x\right)\,dx\,dt+\int_{\mathbb R}u_0(x)\phi(0,x)\,dx=0
The identity holds for every smooth test φ compactly supported in space and in nonnegative time. The initial term has the plus sign after integration by parts.↔ Scroll wide equations horizontally.

A moving jump supplies two delta contributions

Write [a]=aR−aL[a]=a_R-a_L throughout this lesson. For constant states separated by x=st, represent the state with a Heaviside jump. Its time derivative contributes minus s times the state jump on the moving line, while the spatial flux derivative contributes the flux jump. Their distributional sum vanishes exactly when speed times state jump equals flux jump. Reversing both jump conventions would give the same equation; mixing conventions between numerator and denominator would reverse the physical speed.

For distinct scalar states, the speed is the secant slope of the flux graph. With Burgers' quadratic flux it is the arithmetic mean of the two states, a result derived from conservation rather than guessed from characteristic drawings. For a vector system, Rankine–Hugoniot is a vector equation: one common speed must balance every conserved component. Componentwise scalar averages generally cannot satisfy that requirement, so nonlinear wave curves and family structure enter the system problem.

−s[u]+[f]=0,s=f(uR)−f(uL)uR−uL,sBurgers=uL+uR2-s[u]+[f]=0,\qquad s=\frac{f(u_R)-f(u_L)}{u_R-u_L},\qquad s_{\mathrm{Burgers}}=\frac{u_L+u_R}{2}
The scalar quotient assumes u_R≠u_L. The first relation is the right-minus-left jump balance and also makes sense componentwise as a vector equation.↔ Scroll wide equations horizontally.

Conservation does not choose every weak continuation

For a strictly convex flux, a downward jump can gather characteristics arriving from both sides. In the scalar compressive case, the shock speed lies between the left and right characteristic speeds. An upward jump may also satisfy Rankine–Hugoniot algebraically, but its neighboring characteristics spread away and leave a region whose values are not determined by that jump picture. Both can obey the weak conservation identity, which demonstrates why weak balance alone does not establish uniqueness.

An entropy or vanishing-viscosity selection rule excludes the inappropriate expansion jump for the convex scalar problem and replaces it by a rarefaction fan. One should therefore distinguish three checks: whether a classical solution has broken, whether a proposed discontinuity conserves the quantity, and whether it satisfies the appropriate admissibility principle. Numerical schemes have an analogous distinction: a shared interface flux can conserve mass while still converging to an incorrectly selected discontinuity if no suitable entropy mechanism is present.

f′(uL)>s>f′(uR)f'(u_L)>s>f'(u_R)
This is the strict scalar compressive inequality for a shock between distinct states of a strictly convex flux with u_L>u_R. It is an admissibility statement beyond the bare jump equation.↔ Scroll wide equations horizontally.

A control volume verifies the sign

For Burgers' equation take left state three and right state minus one. Find the shock speed and verify the mass balance on [−2,2] while the moving jump stays inside.

  1. The right-minus-left state jump is −1−3=−4. The fluxes are 1/2 and 9/2, so the flux jump is −4 and the speed is s=1.

  2. The characteristic speeds are three on the left and minus one on the right. Thus three is greater than one, which is greater than minus one: characteristics approach the shock from both sides.

  3. With shock location st, mass inside the interval is 3(st+2)−(2−st)=4st+4. Its derivative is 4s=4, matching the boundary inflow minus outflow, 9/2−1/2=4.

s=1/2−9/2−1−3=1,ddt∫−22u dx=f(3)−f(−1)=4s=\frac{1/2-9/2}{-1-3}=1,\qquad\frac{d}{dt}\int_{-2}^{2}u\,dx=f(3)-f(-1)=4
The control-volume calculation applies during the time interval in which the jump lies strictly between the two endpoints.↔ Scroll wide equations horizontally.
The result

The speed is positive one, and the finite-interval mass balance confirms the jump sign convention.

Explore the model

Interactive lab

Interactive lab · function

Secant speed versus tangent speeds

Pause and Reset, then set coefficient a=0.5 and position x=2, treating the horizontal coordinate as state u. Record the Burgers flux value and tangent slope. For a fresh comparison, Reset and set a=0.5 and x=0. Compute the secant slope between these two flux points on paper.

Secant speed versus tangent speedsThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe tangent speeds are two and zero, while the secant shock speed is one. The graph displays local flux information; it does not draw a secant automatically or evolve the shock in space-time.

Check your understanding

With [a]=a_R−a_L and a jump at x=st, which distributional coefficient must vanish for weak conservation?

Try it yourself

For Burgers left state four and right state two, calculate the Rankine–Hugoniot speed and compare it with the two characteristic speeds. Would reversing these states still satisfy some Rankine–Hugoniot speed?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the flux secant or factor the difference of squares. Then distinguish weak balance from the compressive ordering.

Compare with the solution

The shock speed is (4+2)/2=3, between the left characteristic speed four and the right speed two. Reversing the states gives the same secant speed three because numerator and denominator both reverse sign, so weak conservation still holds. The reversed jump is expansive, however: the left speed is two and the right four. It needs the entropy test and is replaced by a rarefaction in the convex scalar entropy solution.

Take it with you

  • Weak conservation uses the flux divergence and a consistently signed initial trace.
  • Rankine–Hugoniot equates speed times state jump to flux jump.
  • A weakly balanced jump is not automatically the admissible continuation.

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