advanced29 min studyLesson 403 in Mathematics

Energy dissipation and uniqueness

Use the correct boundary terms and coercive norm to prove stability and distinguish decay from conservation.

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What you’ll understand

  • Derive the heat-flow energy identity with declared boundary conditions.
  • Prove uniqueness by applying an estimate to a difference.
  • Identify which mean or boundary condition permits an exponential decay rate.

A temperature profile becomes flatter because diffusion dissipates spatial variation. An energy calculation turns this statement into a quantitative inequality and often a uniqueness proof. The calculation is sensitive to boundary terms: an insulated domain preserves its mean, while an absorbing boundary permits decay to zero. Choosing the wrong reference state or suppressing a boundary contribution can turn a valid identity into an incorrect long-time conclusion.

Multiplication by the solution exposes dissipation

Consider uₜ=κΔu with κ>0 on a bounded smooth domain and homogeneous Dirichlet data. For a sufficiently regular solution, multiply by u and integrate by parts. The boundary term vanishes because u has zero trace. The derivative of half the squared L² norm equals minus κ times the squared gradient norm. With homogeneous Neumann data, the same boundary term vanishes because the normal derivative is zero. Periodic data also cancel paired boundary contributions. Nonhomogeneous conditions generally contribute work or heat exchange and must be retained.

The identity extends to the weak energy class through time regularization or the Hilbert-space chain rule. In that setting the squared L² norm is absolutely continuous and its derivative is the dual pairing with uₜ almost everywhere. It is not necessary to assert that u is classically differentiable at every time. For bounded measurable symmetric uniformly elliptic divergence coefficients, the dissipation becomes the integral of ∇u dotted with A∇u, still bounded below by the ellipticity constant times gradient energy.

E(t)=12∫Ωu2 dx,E′(t)=−κ∫Ω∣∇u∣2 dxE(t)=\frac12\int_\Omega u^2\,dx,\qquad E'(t)=-\kappa\int_\Omega|\nabla u|^2\,dx
This unforced heat identity assumes homogeneous Dirichlet, homogeneous Neumann, or periodic conditions eliminating the boundary term. In the weak energy class it holds almost everywhere in time.↔ Scroll wide equations horizontally.

A difference estimate proves uniqueness without explicit formulas

If two weak solutions share the same forcing, boundary data, and initial trace, their difference w solves the homogeneous equation with zero data. Its energy is nonincreasing and initially zero. Since energy is nonnegative, it remains zero, so the solutions agree in the energy space. This argument does not require a heat-kernel expression or an eigenfunction expansion. It also explains continuous dependence: a nonzero initial difference cannot grow in L² for this homogeneous linear diffusion problem.

If the forcing differs, pair the forcing difference with w and use a dual norm. With the H⁻¹ norm dual to the H₀¹ gradient norm, Young's inequality absorbs half the gradient term and leaves a controlled forcing contribution. Integrating yields a stability bound. For equations with lower-order terms, additional estimates and a Gronwall factor may appear. An estimate for one particular equation should not be generalized unchanged to a reaction term capable of amplifying solutions. Uniqueness concerns a specified solution class and prescribed data.

12ddt∥w∥22+κ2∥∇w∥22≤12κ∥f1−f2∥H−12\frac12\frac{d}{dt}\lVert w\rVert_2^2+\frac{\kappa}{2}\lVert\nabla w\rVert_2^2\le\frac{1}{2\kappa}\lVert f_1-f_2\rVert_{H^{-1}}^2
For the difference of two homogeneous-Dirichlet heat solutions on the same domain, this uses the gradient-dual H⁻¹ norm. Equal forcing and initial trace imply uniqueness.↔ Scroll wide equations horizontally.

Dissipation becomes a decay rate only after a coercive inequality

On a bounded domain with homogeneous Dirichlet data, Poincaré's inequality bounds the squared gradient norm below by λ₁ times the squared solution norm, where λ₁>0 is the first Dirichlet Laplacian eigenvalue. The energy identity then gives E(t)≤E(0)exp(−2κλ₁t). The L² norm itself decays with exponent κλ₁. Confusing squared norm and norm creates a factor-of-two error. This rate is sharp when the initial profile is a first eigenfunction.

For Neumann or periodic flow, a constant profile has zero gradient and does not decay. The mean is conserved, so apply Poincaré only after subtracting that mean, on a connected bounded domain or a connected periodic torus. A disconnected insulated domain conserves a separate mean on each component; one global average is insufficient to remove every zero mode. Energy dissipation alone says that variation costs energy. Identifying the equilibrium and proving a positive spectral gap are the extra steps needed to quantify convergence toward it.

Check the norm-versus-energy exponent

On (0,π) with zero Dirichlet boundary and κ=1/2, take u₀=sin x. Compute energy, dissipation, and energy half-life.

  1. The solution is u=exp(−t/2)sin x. Since the integral of sin²x over the interval is π/2, E(t)=πe⁻ᵗ/4.

  2. The gradient is exp(−t/2)cos x, so κ times its squared L² norm is (1/2)(π/2)e⁻ᵗ=πe⁻ᵗ/4, exactly −E′(t).

  3. Solving E(t)/E(0)=1/2 gives t=log2. The solution norm decays as exp(−t/2), whose own half-life is 2log2.

The result

The energy identity and the first-eigenvalue Poincaré estimate agree exactly. Energy and norm half-lives differ because energy is quadratic.

Explore the model

Interactive lab

Interactive lab · heat

Recover energy from the displayed spatial norm

Reset, set n=1 and κ=0.5, then Step once to t=0.5. Predict L² norm √π exp(−0.25) and energy π exp(−0.5)/2 on the instrument's periodic interval. Square the norm readout and divide by two to check the energy prediction. This exact sine mode has zero mean; it does not demonstrate decay of a nonzero constant Neumann mode.

Recover energy from the displayed spatial normThe 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 forQuadratic energy decays twice as fast in its exponent as the underlying amplitude.

Check your understanding

Why can a nonzero constant solution of insulated heat flow fail to decay while energy dissipation remains valid?

Try it yourself

On a periodic interval of length 2π, let u₀=3. Compute E(t), and explain why the conclusion E(t)≤E(0)e⁻²κᵗ would be wrong.

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

Give me a hint

A spatial constant has zero Laplacian and zero gradient at every time.

Compare with the solution

The solution stays u=3, so E(t)=(1/2)·9·2π=9π. Its dissipation is zero and the energy identity is satisfied. The proposed exponential estimate incorrectly applies a positive spectral gap to the constant mode. Subtracting the conserved mean gives u−3=0, whose variation energy is zero from the start. The periodic gap controls mean-zero functions, not the full density.

Take it with you

  • Boundary terms determine whether a stated energy identity is valid.
  • A difference-energy estimate proves uniqueness and data stability in a declared class.
  • Exponential relaxation requires removing zero modes and applying an appropriate spectral gap.

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