advanced36 min studyLesson 420 in Mathematics

Finite-volume fluxes and discrete entropy

Separate conservation, flux consistency, monotonicity, and entropy dissipation in a fully discrete finite-volume update.

Before you begin

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The question for today

What you’ll understand

Explain why shared interface fluxes telescope to conservation.
Prove a convex entropy inequality for scalar upwind transport under its CFL condition.
Identify which additional properties are needed beyond consistency or a semidiscrete entropy claim.

A finite-volume method updates cell averages by balancing fluxes through their faces. Using one shared flux at an interface guarantees that what leaves one cell enters the next. This is conservation, but it does not by itself guarantee the correct weak solution, positivity, or entropy dissipation. A simple linear upwind example lets us prove a fully discrete entropy inequality directly. It also shows how violating the time-step condition can preserve total mass while producing an inadmissible oscillation.

Conservation and consistency are different checks

On a uniform one-dimensional mesh, let U represent cell averages and let a numerical flux depend on the left and right neighboring values. The same interface flux enters the two adjacent updates with opposite signs. Summing over a periodic grid therefore cancels every internal face and preserves the total discrete mass. On a physical bounded domain, boundary fluxes remain, just as in the continuous control-volume balance. Equal face values do not justify dropping those boundary terms.

Consistency requires the numerical flux at two equal states to match the physical flux at that state. It ensures constant states are represented correctly and supports weak-limit arguments when combined with appropriate boundedness and convergence assumptions. It does not prove that a sequence of approximate solutions actually converges, nor that its limit is entropy admissible. A centered flux can be conservative and consistent while lacking enough dissipation for a discontinuous problem. Reconstruction and the time update are also part of the complete method.

U_j^{n+1}=U_j^n-\frac{\Delta t}{\Delta x}\left(\widehat f(U_j^n,U_{j+1}^n)-\widehat f(U_{j-1}^n,U_j^n)\right),\qquad\widehat f(v,v)=f(v)

Shared fluxes provide conservation; the equality at identical states provides consistency. These properties have separate roles and do not alone establish entropy convergence.↔ Scroll wide equations horizontally.

A convex combination gives a discrete entropy proof

For linear scalar transport with positive speed a, the upwind flux is a times its left argument. Define the Courant number $\nu=a\Delta t/\Delta x$ . When it lies between zero and one, the new average is a convex combination of its old value and its left neighbor. Jensen's inequality then gives a local inequality for every convex scalar entropy. The numerical entropy flux is a times the upstream entropy, which is consistent with the continuous entropy flux a times eta.

Summing that cell inequality over a periodic mesh makes entropy fluxes telescope and proves nonincrease of the entropy sum. The proof uses the fully discrete update, including the step restriction; it is not merely an entropy property of a spatial flux formula. At Courant number one, the method shifts cell values exactly by one cell and may preserve the entropy sum rather than strictly decrease it. At intermediate values it generally mixes unequal neighboring states and dissipates strictly convex entropies.

U_j^{n+1}=(1-\nu)U_j^n+\nu U_{j-1}^n,\qquad\eta(U_j^{n+1})\le(1-\nu)\eta(U_j^n)+\nu\eta(U_{j-1}^n)\quad(0\le\nu\le1)

This elementary fully discrete entropy inequality follows from convexity. It also yields a cellwise maximum principle for this linear scalar method.↔ Scroll wide equations horizontally.

Entropy stability must match the entire discretization

For nonlinear scalar laws, Godunov fluxes and suitable monotone updates provide a route to entropy-stable approximation under appropriate CFL restrictions. For systems, a flux may be designed to satisfy one compatible convex entropy relation, while positivity of density or pressure and convergence require additional conditions. An entropy-conservative spatial flux has zero designed entropy production; shock computations typically need an admissible dissipative mechanism rather than a visual expectation that conservation will select the correct jump automatically.

A semidiscrete inequality controls the spatially discretized differential equation in continuous time. An arbitrary time integrator or excessive step can destroy the fully discrete version. High-order reconstructions, limiters, boundary treatments, and changing meshes can likewise alter the argument. State which entropy, which time update, and which step restrictions were actually verified. A numerical entropy sum is a useful diagnostic, but matching one sum on one mesh does not prove all entropy inequalities or a continuum convergence theorem.

\eta(U_j^{n+1})\le\eta(U_j^n)-\frac{\Delta t}{\Delta x}\left(\widehat q_{j+1/2}^n-\widehat q_{j-1/2}^n\right),\qquad\widehat q(v,v)=q(v)

A fully discrete cell entropy inequality includes an entropy-consistent numerical flux and the actual time update. Its validity must be established for the chosen scheme and restrictions.↔ Scroll wide equations horizontally.

Work it through

Mass stays fixed while quadratic entropy falls

Use positive transport speed a=2, cell width 0.5, and step 0.1 on a periodic three-cell grid with averages (1,3,2). Compute one upwind update and the unweighted quadratic entropy sum.

The Courant number is 2·0.1/0.5=0.4. Periodic upstream neighbors give new values 0.6·1+0.4·2=1.4, 0.6·3+0.4·1=2.2, and 0.6·2+0.4·3=2.4.
Both old and new averages sum to six, so the actual discrete mass, cell width times that sum, stays at three.
For eta(U)=U²/2, the old unweighted entropy sum is seven. The new sum is (1.4²+2.2²+2.4²)/2=6.28. In the middle cell, the local inequality reads 2.42≤0.6·4.5+0.4·0.5=2.9.

U^{n+1}=(1.4,2.2,2.4),\qquad\sum_jU_j^{n+1}=6,\qquad\sum_j\frac{(U_j^{n+1})^2}{2}=6.28\le7

Conservation follows by telescoping; the entropy inequality follows separately from the convex-combination structure.↔ Scroll wide equations horizontally.

The result

The conservative update preserves mass and satisfies the convex entropy bound under its valid CFL number. Multiplying both entropy sums by the common cell width gives the corresponding spatial entropy totals.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

A two-cell conservative mixing update

Pause and Reset, then set p=0.4 and q=0.4. Starting from (1,0), predict the two probabilities displayed inside the state circles after one Step. Interpret this symmetric transition as the same update matrix as two periodic upwind cells with ν=0.4. For a fresh comparison, Reset and set p=1 and q=1, then Step twice.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. 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 first update gives (0.6,0.4), with quadratic sum 0.26 instead of 0.5. The ν=1 comparison alternates the two entries and keeps that sum fixed. This exact two-state model illustrates the matching two-cell operator, not a general PDE solver or nonlinear entropy theorem.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The proof uses both the upwind spatial flux and the actual forward time step under its CFL bound. Conservation and consistency alone do not supply the convex-combination inequality.

Your turn

Try it yourself

On a periodic two-cell grid with old values (0,1), use the same upwind formula but ν=1.2. Find the new values, their sum, and the quadratic entropy sum. Which earlier guarantee fails?

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 coefficient 1−ν is now negative. Compute the two updates before applying any intuition about conservation.

Compare with the solution

The new first value is (−0.2)·0+1.2·1=1.2, and the second is (−0.2)·1+1.2·0=−0.2. Their sum remains one, so conservation survives. The quadratic entropy sum becomes (1.44+0.04)/2=0.74, larger than the old 0.5. The convex-combination and maximum-principle argument fails because ν>1; the method creates a negative value and an overshoot despite preserving total mass.

Keep the idea

Take it with you

Shared fluxes conserve and equal-state matching ensures consistency, but neither alone establishes entropy selection.
The linear upwind CFL bound makes a fully discrete Jensen entropy proof possible.
Entropy claims must include reconstruction, boundaries, time integration, and the relevant restrictions.

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