advanced30 min studyLesson 404 in Mathematics

Parabolic maximum principles

Control interior values from the past and boundary while respecting forcing and reaction signs.

On this page

What you’ll understand

  • Identify the parabolic boundary of a forward cylinder.
  • Apply an upper-bound comparison with the correct forcing sign.
  • Use an exponential weight when a scalar reaction coefficient changes the simple maximum bound.

A source-free scalar heat equation cannot create a new positive interior maximum above everything supplied initially or through the boundary. This principle gives bounds without computing the solution and can compare nonlinear solutions after a suitable reduction. The word forward matters: the relevant boundary includes the initial time, not an independently prescribed terminal state. Reaction terms and forcing can alter the bound, so sign conventions must be visible in the argument.

The past and lateral boundary control a forward solution

For a bounded domain Ω and times 0 through T, the parabolic boundary is the initial slice together with the lateral spatial boundary over time. Let u be continuous on the closed cylinder and C² in space and C¹ in positive time in the interior. If uₜ−κΔu≤0 with κ>0, its supremum does not exceed the supremum on that parabolic boundary. No terminal boundary value is required to solve the forward problem, and the final time may be included among the times being bounded.

At a positive interior spatial maximum, the Hessian is negative semidefinite and the Laplacian is nonpositive. At a maximum reached at the final time, the left time derivative is nonnegative; one should not simply set it to zero. Subtracting a small positive multiple of time makes the subsolution inequality strict and rules out an interior maximum above the initial and lateral values. Letting the perturbation vanish yields the non-strict principle. This proof is for a scalar classical solution; applying it directly to an arbitrary weak representative would require a separate weak comparison argument.

ut−κΔu≤0⟹sup⁡Ω‾×[0,T]u≤sup⁡∂pQTu,∂pQT=(Ω‾×{0})∪(∂Ω×[0,T])u_t-\kappa\Delta u\le0\quad\Longrightarrow\quad \sup_{\overline\Omega\times[0,T]}u\le\sup_{\partial_pQ_T}u,\qquad \partial_pQ_T=(\overline\Omega\times\{0\})\cup(\partial\Omega\times[0,T])
Assume a bounded domain, continuous closed-cylinder data and solution, classical interior derivatives, and constant κ>0. The boundary set records initial and lateral values.↔ Scroll wide equations horizontally.

Variable coefficients and zero-order terms need their stated signs

For the scalar operator uₜ−aᵢⱼuᵢⱼ−bᵢuᵢ+cu with bounded continuous coefficients, symmetric uniformly positive definite a, and c≥0, the corresponding subsolution bound uses the maximum of zero and the parabolic-boundary supremum. At a positive maximum the drift term vanishes, the diffusion contribution has the correct sign, and cu is nonnegative. The zero term in the bound matters when all boundary values are negative. A negative zero-order coefficient can instead permit positive growth.

If a bounded coefficient gives uₜ−κΔu=a(x,t)u, choose L at least its upper bound and transform v=e⁻ᴸᵗu. The transformed equation has a nonnegative zero-order coefficient L−a in the left-side convention. This supplies a weighted comparison bound rather than an unweighted claim that every maximum decreases. For a scalar reaction F(u) locally Lipschitz on the bounded ranges of two solutions, their difference has such a bounded coefficient. An exponential weight therefore supports comparison without requiring F itself to be decreasing.

ut−κΔu=a(x,t)u,a≤L,v=e−Ltu⟹vt−κΔv+(L−a)v=0u_t-\kappa\Delta u=a(x,t)u,\quad a\le L,\quad v=e^{-Lt}u\quad\Longrightarrow\quad v_t-\kappa\Delta v+(L-a)v=0
A finite upper bound L on the scalar coefficient makes L−a nonnegative. The resulting estimate concerns the weighted solution and does not erase actual reaction growth.↔ Scroll wide equations horizontally.

Comparison creates useful barriers, but does not prove existence

A supersolution lies above the PDE's forcing after applying the same operator and lies above the initial and boundary values. Subtracting it from a solution reduces the desired inequality to a maximum principle. Spatially constant time-dependent functions are often effective barriers when the forcing has a simple bound. For homogeneous heat flow, comparison also preserves nonnegativity. Strong positivity requires further hypotheses, such as a connected region and nontrivial nonnegative data, and should not be confused with the weak upper-bound statement.

Maximum principles establish bounds for solutions in a class where comparison is valid; they do not construct those solutions by themselves. They are naturally scalar and need not hold componentwise for an arbitrary coupled parabolic system. Boundary geometry and regularity also matter when moving from interior estimates to continuous attainment of boundary data. A reliable application states the operator sign, the domain and time interval, the solution regularity, and the ordering on the entire parabolic boundary before invoking a convenient constant or time-dependent barrier.

A source-controlled upper barrier

Suppose a classical solution on (0,1) satisfies uₜ−uₓₓ=f with 0≤f≤3, initial values between 0 and 2, and endpoint values between 0 and 1. Bound u through time 2.

  1. The constant zero is a subsolution because its left-side operator is zero, at most f, and its initial and boundary values do not exceed those of u.

  2. The function v(t)=2+3t has vₜ−vₓₓ=3≥f. It starts at 2 and exceeds the endpoint values throughout the interval.

  3. Comparison yields 0≤u(x,t)≤2+3t for 0≤t≤2. At the final time the uniform upper bound is 8. This bounds any such solution without asserting that the upper barrier is attained.

The result

The data and source bounds give 0≤u≤2+3t, hence u≤8 through time 2. Positive forcing prevents the simpler source-free maximum bound.

Explore the model

Interactive lab

Interactive lab · heat

Check a source-free comparison case

Reset, set n=2 and κ=0.5, then Step once to t=0.5. Predict amplitude exp(−1), so the exact periodic mode remains between −1 and 1. Compare the readout and extrema in the drawing. This signed unforced mode illustrates an admissible bound; it does not include the worked example's positive source or establish comparison for a coupled system.

Check a source-free comparison caseThe 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 forDiffusion alone respects the initial scalar range, while forcing requires a modified barrier.

Check your understanding

Which set supplies the data in a forward parabolic maximum principle?

Try it yourself

On (0,π), solve uₜ=uₓₓ+2u with zero Dirichlet boundary and u₀=sin x. Explain why its growing maximum does not contradict the heat maximum principle.

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 sine eigenvalue is −1. Also consider the weighted variable v=e⁻²ᵗu.

Compare with the solution

The solution is u=eᵗsin x because its time coefficient satisfies c′=(2−1)c. Its maximum grows as eᵗ. The equation includes positive reaction, so u does not satisfy the source-free heat subsolution inequality. The weighted variable is v=e⁻ᵗsin x, which solves the heat equation and has decreasing maximum. Weighted comparison permits growth of u while still giving a controlled bound.

Take it with you

  • The parabolic boundary contains initial and lateral data.
  • Forcing and zero-order signs determine the valid maximum bound.
  • Comparison supplies bounds and ordering, not an automatic existence theorem or a universal system principle.

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