advanced32 min studyLesson 409 in Mathematics

Entropy methods and convergence to equilibrium

Choose a nonnegative functional, calculate its dissipation, and obtain a rate only from a valid functional inequality.

On this page

What you’ll understand

  • Derive mass-relative entropy dissipation for positive insulated heat flow.
  • Separate monotonicity of a functional from a quantitative convergence rate.
  • Use Poincaré or a stated logarithmic Sobolev inequality with consistent normalization.

An evolution can approach equilibrium even when no explicit solution formula is available. Entropy methods seek a nonnegative quantity measuring deviation from equilibrium and calculate how the PDE dissipates it. The sign of that derivative is only the first step. A convergence rate requires an inequality connecting the functional to its dissipation, and the equilibrium itself depends on conserved quantities and boundary conditions.

Conserved mass chooses the reference state

Take heat flow with κ>0 on a connected bounded smooth insulated domain, or on a connected periodic torus. Its mass M is conserved, so the equilibrium candidate is the mean ū=M/|Ω|. For a positive sufficiently smooth solution bounded away from zero on the time interval of calculation, define relative entropy as the integral of u log(u/ū)−u+ū. The integrand is nonnegative by convexity and vanishes only at u=ū. The linear terms integrate to zero because mass matches the reference.

Differentiate the functional and use the PDE. Integrating by parts removes the boundary term under zero normal flux or periodicity, leaving minus κ times the integral of |∇u|²/u. This is nonnegative dissipation with a negative sign. For solutions that touch zero, the logarithm and division require an approximation or a justified extended formulation; the classical calculation should not be applied pointwise without that step. The reference mean must be positive for this logarithmic functional, so an identically zero mass solution is handled separately.

H(u∣uˉ)=∫Ω[ulog⁡uuˉ−u+uˉ]dx,dHdt=−κI(u),I(u)=∫Ω∣∇u∣2u dxH(u\mid\bar u)=\int_\Omega\left[u\log\frac{u}{\bar u}-u+\bar u\right]dx,\qquad \frac{dH}{dt}=-\kappa I(u),\quad I(u)=\int_\Omega\frac{|\nabla u|^2}{u}\,dx
Assume a positive smooth heat solution, positive conserved mean, and periodic or homogeneous Neumann boundary conditions. Lower positivity bounds justify this classical differentiation; weak extensions require a limiting argument.↔ Scroll wide equations horizontally.

A decreasing functional does not supply its own rate

Knowing H′≤0 says entropy cannot increase, but it does not identify an exponential rate or rule out slow relaxation. If a valid logarithmic Sobolev inequality in the chosen normalization gives H≤C I, then H′≤−(κ/C)H and Gronwall gives exponential decay. The value and meaning of C depend on domain, reference measure, and normalization. An inequality written for probability densities cannot be transplanted to an unnormalized mass convention without checking how all factors transform.

A simpler quadratic functional is half the squared L² distance to the mean. Its dissipation is κ times the squared gradient norm, and a mean-zero Poincaré inequality gives exponential decay with rate 2κλ₁ for this quadratic energy. Logarithmic entropy and squared L² distance are different functionals. They may control related notions of closeness under additional positivity or boundedness assumptions, but their numerical values and sharp constants should not be conflated. Even identifying the correct zero modes is necessary before claiming a positive gap.

H(u∣uˉ)≤CI(u)⟹H(t)≤H(0)e−(κ/C)t,E(t)=12∥u−uˉ∥22≤E(0)e−2κλ1tH(u\mid\bar u)\le C I(u)\quad\Longrightarrow\quad H(t)\le H(0)e^{-(\kappa/C)t},\qquad E(t)=\frac12\lVert u-\bar u\rVert_2^2\le E(0)e^{-2\kappa\lambda_1t}
The entropy implication assumes the displayed functional inequality is valid for the evolving positive densities. The quadratic estimate assumes a positive mean-zero Poincaré gap λ₁; its normalization and exponent are separate.↔ Scroll wide equations horizontally.

Boundary conditions and nonlinear mechanisms determine the entropy

On an insulated disconnected domain, mass is conserved separately on each component, so the equilibrium is componentwise constant. One global mean can leave a persistent mismatch and invalidate a claimed global gap. Under Dirichlet conditions, a nonzero prescribed boundary or forcing changes the equilibrium and can create boundary contributions in the entropy balance. On the whole space, a finite-mass heat solution spreads rather than approaching a nonzero spatial constant; self-similar coordinates and a corresponding reference profile are needed for a meaningful equilibrium comparison.

For nonlinear diffusion, a power-type energy may be more natural than the logarithm. For sufficiently regular whole-space porous-medium flow with vanishing boundary flux, differentiating the integral of uᵐ⁺¹/(m+1) gives minus the integral of |∇uᵐ|². Turning such dissipation into a sharp convergence theorem requires mass constraints, scaling, compactness or functional inequalities, and an appropriate stationary or self-similar reference. A formal negative derivative motivates the analysis; it is not by itself a proof of existence, uniqueness, or a universal relaxation rate.

An exact quadratic energy and an entropy upper bound

On the periodic interval [0,2π], let u=1+(1/2)exp(−0.4t)cos x. Compute E=‖u−1‖²/2 and use log s≤s−1 to bound H(u|1).

  1. The mean is one and u≥1/2 for t≥0, so the logarithmic calculation is well defined. Orthogonality gives E(t)=πexp(−0.8t)/8.

  2. Multiplying log u≤u−1 by positive u and integrating gives H=∫u logu≤∫u(u−1). Since ∫(u−1)=0, the last expression is ∫(u−1)²=2E.

  3. Therefore 0≤H(t)≤πexp(−0.8t)/4. The exact energy derivative is −0.8E, matching κ=0.4 and the first mean-zero periodic eigenvalue one.

The result

The quadratic energy is known exactly, while πexp(−0.8t)/4 is an entropy bound, not an asserted exact logarithmic entropy formula.

Explore the model

Interactive lab

Interactive lab · heat

Translate mode damping into a functional decay

Reset, set n=1 and κ=0.5, then Step once to t=0.5. Predict the mode L² norm ratio exp(−0.25), hence squared-distance energy ratio exp(−0.5). Compare with a fresh n=2, κ=0.5 run after Reset and one Step, predicting energy ratio exp(−2). The instrument is a signed zero-mean sine; logarithmic density entropy would require adding a sufficiently large positive mean outside this model.

Translate mode damping into a functional decayThe 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 forA decay exponent for a norm must be converted carefully before it is used for a quadratic functional.

Check your understanding

What extra ingredient turns H′=−κI into a quantitative exponential entropy rate?

Try it yourself

Suppose positive insulated heat densities obey H≤2I in the exact normalization above, and κ=1/2. Derive an entropy bound and its guaranteed half-life. Is the factor 2 an exact value of H for every density?

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

Give me a hint

Insert the functional inequality into H′=−κI before solving the scalar differential inequality.

Compare with the solution

Since I≥H/2, H′≤−H/4. Therefore H(t)≤H(0)exp(−t/4), and time 4log2 guarantees H has fallen to at most half its initial value. The assumed constant 2 belongs to an inequality; it does not assert equality H=2I, an exact trajectory, or a universally sharp rate. The conclusion is conditional on that normalized inequality holding throughout the solution class.

Take it with you

  • Conserved quantities determine the reference equilibrium.
  • Entropy dissipation and a functional inequality play separate roles in proving a rate.
  • Norms, mass normalization, geometry, and boundary terms must be tracked consistently.

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