Maximum principles and comparison
Control solutions by their boundary and by barriers while tracking the operator sign and zero-order assumptions.
On this page
The question for today
What you’ll understand
- Derive a weak maximum principle with a positive-part test.
- Use ordered sources and boundary data to compare two solutions.
- Construct a counterexample when an essential zero-order sign condition fails.
An elliptic equation often yields useful bounds before anyone solves it exactly. Maximum and comparison principles convert the sign of an operator and the ordering of boundary values into an ordering of solutions. Their direction depends on the sign convention: here the principal operator is minus divergence, so subsolutions satisfy Lu≤0. We first use a divergence-form energy proof with nonnegative potential, then separate the stronger statement that an interior extremum forces a harmonic function to be constant. Mixing those hypotheses is a frequent source of incorrect bounds.
Test only the part that violates the bound
Let Omega be bounded Lipschitz, A bounded measurable, real symmetric and uniformly elliptic, and c bounded nonnegative. Suppose u is H-one, its boundary trace is nonpositive, and its weak inequality a(u,v)≤0 holds for every nonnegative H-zero-one test. The positive part u-plus belongs to H-zero-one. Testing with that function gives a nonnegative gradient term and a nonnegative potential term whose sum is at most zero. Poincaré then forces the positive part to vanish, proving u≤0 almost everywhere.
To obtain an upper boundary bound M, apply this reasoning to u−M. For c≥0, a nonnegative constant M contributes cM≥0, so a subsolution lies below the larger of zero and its boundary supremum. When c=0 the nonnegative restriction on M is unnecessary, and the boundary supremum alone works. These weak statements concern almost-everywhere representatives and traces; a pointwise classical maximum statement additionally assumes enough continuity to evaluate the solution and boundary values at individual points.
Comparison is a maximum principle for a difference
If Lu≤Lv weakly and u≤v on the boundary, their difference satisfies the preceding zero-boundary-order inequality and is nonpositive. This proves comparison for the same linear operator. A lower or upper barrier is a function whose operator value and boundary values bound those of the desired solution in the appropriate direction. Barriers give explicit estimates even when the source varies spatially or an exact representation would be inconvenient. They must be checked in the original equation, including every zero-order term.
The strong maximum principle contains more information and needs a compatible theorem. For a classical harmonic function on a connected open domain, an attained interior maximum or minimum forces the function to be constant. One proof uses the mean-value property to propagate equality. The weak maximum principle alone does not establish that propagation, and neither statement says a solution of a positively forced Poisson equation cannot have an interior maximum. The sign of its Laplacian decides which extremum is ruled out.
Check the assumptions before applying a slogan
A negative potential can invalidate the simple maximum principle even though the principal operator remains uniformly elliptic. On the interval from zero to pi, sin x is positive inside, vanishes at the endpoints, and solves minus u-double-prime minus u equals zero. Thus a nonpositive boundary and homogeneous equation do not force a nonpositive solution for that operator. Coercivity or an appropriate positive principal-eigenvalue condition can support other maximum-principle variants, but the missing sign cannot simply be ignored.
Boundary regularity matters differently for different conclusions. The energy proof uses traces and does not need a pointwise normal derivative. A Hopf boundary conclusion about a strict outward derivative requires additional geometric and regularity conditions, such as a suitable interior sphere condition in standard classical statements. We are not deriving that conclusion from a weak trace argument. Keeping the weak, strong, and boundary forms separate makes comparison a reliable tool rather than an all-purpose claim about every elliptic-looking equation.
Work it through
Bound a nonconstant source by two explicit barriers
Let −u″=2+2x on (0,1), with zero endpoint values. Use comparison to bound u without first solving the equation, then verify the bounds on the exact solution.
The source lies between two and four. The zero-boundary functions and satisfy −v″=2 and −w″=4. Comparison for the operator −d²/dx² therefore gives v≤u≤w.
The upper barrier has maximum one half, and the lower barrier is nonnegative. Thus 0≤u≤1/2 throughout the interval. This bound does not require locating the exact point where u reaches its maximum.
Integrating the source equation gives . Its differences are and . The direct algebra confirms the comparison bounds with the stated signs.
Ordered sources and equal boundary data give ordered solutions; simple constant-source barriers produce a useful uniform bound.
The living notebook
Explore the model
Interactive lab · function
Read which curvature supports which extremum
Pause, Reset, and set a=1 and x=0; predict a zero slope and a minimum. Repeat from Reset with both a=−1 and x=0; now the same zero slope is a maximum and −u″ is positive. This instrument displays y=ax² on the real line, with no boundary conditions or potential term. It illustrates the operator sign, not the full maximum-principle theorem.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 positive forcing for −u″ is compatible with concavity and an interior maximum; the sign of the subsolution inequality matters.
Pause & reason
Check your understanding
Your turn
Try it yourself
Check that u=sin x on (0,π) contradicts the simple zero-boundary maximum conclusion for L=−d²/dx²−1. Identify exactly which stated hypothesis 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
Calculate u″, then separate the principal second-order coefficient from the potential c.
Compare with the solution
Since u″=−sin x, one has Lu=−u″−u=0. The boundary values are zero while u is strictly positive inside and equals one at π/2. The principal coefficient is one and remains uniformly elliptic. The failed hypothesis is c≥0: here c=−1. This is a counterexample to dropping that potential sign from the stated theorem, not a counterexample to the theorem itself.
Keep the idea
Take it with you
- A positive-part test proves an ordered weak bound under precise sign assumptions.
- Comparison and barriers control solutions without requiring an explicit formula.
- Strong and boundary maximum principles need their own hypotheses and conclusions.
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