advanced36 min studyLesson 417 in Mathematics

Entropy conditions and Kružkov solutions

Select bounded scalar weak solutions with a full entropy family and state the assumptions behind uniqueness and L¹ contraction.

On this page

What you’ll understand

  • Construct a convex entropy and its compatible entropy flux.
  • Apply the signed entropy jump inequality to Burgers discontinuities.
  • State bounded scalar Kružkov uniqueness with the initial trace and integrability qualifications.

Weak conservation admits more continuations than the intended physical evolution. An entropy inequality removes some of that ambiguity by requiring a compatible convex quantity to dissipate at shocks. For a scalar conservation law, a full family of Kružkov entropies gives a strong uniqueness and stability theory under precise boundedness and trace assumptions. This is a mathematical selection principle. It must not be confused with a claim that any one convenient quadratic quantity guarantees uniqueness for every nonlinear vector system.

Compatible entropies turn viscosity into dissipation

For a scalar flux f, a smooth convex function η\eta has a compatible entropy flux q determined up to a constant by q′=η′f′q'=\eta'f'. Smooth inviscid solutions satisfy an entropy equality by the chain rule. Adding positive viscosity produces an extra divergence term and a nonpositive term involving η′′ux2\eta''u_x^2. This calculation explains the sign of the entropy inequality in a sufficiently controlled vanishing-viscosity limit; establishing that limit still requires bounds and compactness rather than a formal deletion of terms.

The mathematical entropy is taken convex, so the selected distributional inequality has nonpositive production. A physical thermodynamic entropy may use the opposite concavity or sign convention. At a moving jump, the distributional coefficient is the entropy-flux jump minus shock speed times entropy jump, using the same right-minus-left convention as conservation. This coefficient must be nonpositive. Checking it is different from checking the state-flux Rankine–Hugoniot equality.

q′(u)=η′(u)f′(u),∂tη(u)+∂xq(u)≤0,[q]−s[η]≤0q'(u)=\eta'(u)f'(u),\qquad\partial_t\eta(u)+\partial_xq(u)\le0,\qquad [q]-s[\eta]\le0
The last condition is the entropy production at an isolated constant-state jump. All jumps use right minus left, and the inequality is distributional.↔ Scroll wide equations horizontally.

The full Kružkov family gives scalar selection

For every real constant k, use absolute distance from k as entropy and the signed difference of fluxes as its entropy flux. The family detects how a candidate solution compares with every constant state, not only with zero. A bounded weak solution is a Kružkov entropy solution when all these inequalities hold and it attains its initial datum strongly in local L¹ as time decreases to zero. The sign at equality can be set to zero; its value there does not alter the flux difference.

For the autonomous scalar Cauchy problem on the whole line, a twice continuously differentiable flux and essentially bounded initial data admit a unique bounded entropy solution, represented continuously in time with values in local L¹. Boundedness on each finite time strip and the initial trace are part of the class in which uniqueness is asserted. Neither strict convexity of f nor bounded variation of the initial datum is required for this scalar theorem. Boundary data, spatially discontinuous fluxes, unbounded solution classes, and systems require additional or different theory.

ηk(u)=∣u−k∣,qk(u)=sgn⁡(u−k)(f(u)−f(k)),∂tηk(u)+∂xqk(u)≤0for every k∈R\eta_k(u)=|u-k|,\quad q_k(u)=\operatorname{sgn}(u-k)(f(u)-f(k)),\qquad\partial_t\eta_k(u)+\partial_xq_k(u)\le0\quad\text{for every }k\in\mathbb R
The theorem here concerns a bounded scalar solution of u_t+f(u)_x=0 on R with f∈C² and a strong local-L¹ initial trace. It is not an unrestricted systems uniqueness theorem.↔ Scroll wide equations horizontally.

Contraction controls dependence on initial data

If two bounded scalar entropy solutions have an integrable initial difference, their L¹ distance cannot increase. Individual states need not both have finite L¹ norms: their difference is what must be integrable for the global inequality to be finite. More generally a local contraction estimate enlarges the initial observation interval by the maximal characteristic speed over the relevant bounded state range. This combines admissibility with finite propagation in a way that yields uniqueness for identical initial data.

The proof uses a doubling-of-variables comparison and the whole entropy family. A calculation for one shock or one quadratic entropy illustrates the mechanism but does not replace that proof for arbitrary bounded data. A numerical entropy diagnostic is useful for the same reason: a violated necessary inequality exposes an error, while passing one scalar test does not establish every hypothesis of a convergence or uniqueness theorem. Keep the solution class, entropy family, initial trace, and comparison norm visible when using the result.

∥u(t,⋅)−v(t,⋅)∥L1(R)≤∥u0−v0∥L1(R)\|u(t,\cdot)-v(t,\cdot)\|_{L^1(\mathbb R)}\le\|u_0-v_0\|_{L^1(\mathbb R)}
The global bound assumes the same smooth autonomous scalar flux, bounded entropy solutions in the stated class, and u₀−v₀∈L¹. Use the standard time-continuous representatives.↔ Scroll wide equations horizontally.

The two orientations of one Burgers jump

For Burgers flux f=u²/2, use η=u²/2 and q=u³/3 to compare a jump from zero to two with its reverse. Both have Rankine–Hugoniot speed one.

  1. For the expansion jump u_L=0,u_R=2, the entropy jump is two and entropy-flux jump is 8/3. Thus [q]−s[η]=8/3−2=2/3>0.

  2. For the reversed jump u_L=2,u_R=0, the entropy jump is −2 and entropy-flux jump is −8/3. Its production is −8/3+2=−2/3≤0.

  3. The first orientation violates a necessary convex entropy inequality despite conserving the state. The reversed compressive Burgers shock passes this test and is the entropy-admissible orientation for these Riemann states.

D0→2=23,D2→0=−23,D=[q]−s[η]\mathcal D_{0\to2}=\frac23,\qquad\mathcal D_{2\to0}=-\frac23,\qquad\mathcal D=[q]-s[\eta]
The state jump equality holds in both directions, while admissibility requires nonpositive entropy production.↔ Scroll wide equations horizontally.
The result

The sign of entropy production distinguishes compression from expansion here. Passing this single test is an illustration, not the full Kružkov uniqueness proof.

Explore the model

Interactive lab

Interactive lab · heat

A controlled picture of viscous smoothing

Pause and Reset, then set spatial mode n=2 and diffusivity κ=0.5. Predict its amplitude after one Step to t=0.5. For a fresh comparison, Reset and set n=2 and κ=1, then Step once. Compare how positive diffusion suppresses the same spatial oscillation.

A controlled picture of viscous smoothingThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. 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 amplitudes are exp(−1) and exp(−2). The exact linear heat mode illustrates dissipation from positive viscosity; it does not compute viscous Burgers shocks or prove the Kružkov limit theorem.

Check your understanding

Which qualification belongs to the scalar L¹ contraction statement in this lesson?

Try it yourself

For Burgers left state one and right state minus one, compute the speed and entropy production using both η=u²/2,q=u³/3 and the Kružkov pair with k=0.

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

Give me a hint

The speed is zero. Both absolute-value entropies and quadratic entropies have equal values on these two states, but their fluxes jump.

Compare with the solution

The speed is (1−1)/2=0. For the quadratic pair, [η]=0 and [q]=−1/3−1/3=−2/3, so production is −2/3. For k=0, η₀=|u| has zero jump, while q₀=sgn(u)u²/2 changes from 1/2 to −1/2, giving production −1. Both inequalities hold for this compressive shock. These two checks do not assert that only two entropy functions suffice for general data.

Take it with you

  • Entropy-flux compatibility fixes the distributional dissipation sign.
  • Kružkov selection uses every constant-state entropy and a strong initial trace in a bounded scalar class.
  • L¹ contraction concerns an integrable difference and does not automatically transfer to nonlinear systems.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.