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The weak maximum principle needs the zero-order sign condition

Statement refuted

Statement refuted. Let $L$ be a uniformly elliptic divergence-form operator on a bounded smooth domain with bounded coefficients. Without a sign condition on its zero-order coefficient, every weak subsolution of $L u = 0$ satisfies ${ess sup}_{Ω} u \leq \sup_{\partial Ω} u^{+}$ , where boundary order means $(u - k)^{+} \in H_{0}^{1} (Ω)$ .

Counterexample. Assume the Axiom of Choice and Countable Choice. Let $Ω = B_{π} (0) \subset R^{3}$ , $a^{i j} = δ^{i j}$ , $b = 0$ , $c = - 1$ , and $L = - Δ - 1$ in the convention of Uniformly elliptic divergence-form operators and their sesquilinear forms. Set $u (x) = \sin ∣ x ∣ / ∣ x ∣$ for $x \neq 0$ and $u (0) = 1$ . Then $u \in C^{\infty} (\overline{Ω}) \cap H_{0}^{1} (Ω)$ , $L u = 0$ , $u > 0$ in $Ω$ and $u = 0$ on $\partial Ω$ . Thus $u$ is a weak solution and a weak subsolution, but ${ess sup}_{Ω} u = 1 > 0 = \sup_{\partial Ω} u^{+}$ . The sign condition in Weak maximum principle for coercive divergence-form equations fails: for every nonzero nonnegative $ζ \in C_{c}^{\infty} (Ω)$ , $\int_{Ω} c ζ d x < 0$ .

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the ball $Ω = B_{π} (0) \subset R^{3}$ ; the coefficients $a^{i j} = δ^{i j}$ , $b^{i} = 0$ , $c = - 1$ ; and the function $u (x) = \sin ∣ x ∣ / ∣ x ∣$ for $x \neq 0$ , $u (0) = 1$ .

[F1]

The operator $L = - Δ - 1$ is the divergence-form operator with $a^{i j} = δ^{i j}$ , $b = 0$ , $c = - 1$ in the convention of Uniformly elliptic divergence-form operators and their sesquilinear forms; the ball $B_{π} (0)$ is a bounded $C^{\infty}$ domain with trace $T$ (Bounded C^k domains and boundary charts, Weak subsolutions and supersolutions of a divergence-form equation).

[F2]

Assume the Axiom of Choice and Countable Choice. Classical-to-weak consistency: if $Ω$ is a bounded $C^{1}$ domain, $u \in C^{2} (\overline{Ω}) \cap H_{0}^{1} (Ω)$ and $L u = f$ with $f \in C (\overline{Ω})$ , then $u$ is a weak solution of $L u = f$ in the sense of Weak Dirichlet solutions for a divergence-form operator, i.e. $a (u, v) = \int_{Ω} f \overline{v} d x$ for every $v \in H_{0}^{1} (Ω)$ (Classical solutions satisfy the weak formulation).

[F3]

Assume the Axiom of Choice. ${w \in H^{1} (Ω) : T w = 0} = H_{0}^{1} (Ω)$ (The kernel of the trace is the closure of the test functions); consequently a class in $C (\overline{Ω}) \cap H^{1} (Ω)$ vanishing on $\partial Ω$ has zero trace and belongs to $H_{0}^{1} (Ω)$ .

[F4]

The smooth radial function $r \mapsto \sin r / r$ on $(0, \infty)$ has the convergent power series $\sum_{j \geq 0} (- 1)^{j} r^{2 j} / (2 j + 1)!$ , so $u$ extends to a $C^{\infty}$ function on $R^{3}$ with $u (0) = 1$ (The notation $H^{k}$ and the reserved zero-boundary symbol for the Sobolev class notation).

Counterexample

1.1givenF4algebra

The function $u$ is smooth and solves the equation classically. Because the power series $\sum_{j \geq 0} (- 1)^{j} r^{2 j} / (2 j + 1)!$ converges everywhere, $u$ is the $C^{\infty}$ function $x \mapsto \sum_{j \geq 0} (- 1)^{j} ∣ x ∣^{2 j} / (2 j + 1)!$ on $R^{3}$ , with $u (x) > 0$ for $∣ x ∣ < π$ and $u = 0$ on the sphere $∣ x ∣ = π$ . On $r > 0$ one computes $u^{'} (r) = \cos r / r - \sin r / r^{2}$ and $u^{''} (r) = - \sin r / r - 2 \cos r / r^{2} + 2 \sin r / r^{3}$ , hence the radial Laplacian satisfies $u^{''} (r) + 2 u^{'} (r) / r = - \sin r / r = - u (r)$ ; by continuity this identity $Δ u = - u$ holds on all of $Ω$ , and $L u = - Δ u - u = 0$ there.

2.1step 1.1F1F2F3

The boundary conditions and the weak equation. Since $u$ is smooth on the closed ball and $u = 0$ on $\partial Ω$ , its trace vanishes; by [F3] $u \in H_{0}^{1} (Ω)$ , and with $u \in C^{2} (\overline{Ω})$ and $f = 0 \in C (\overline{Ω})$ , [F2] exhibits $u$ as a weak solution of $L u = 0$ in the sense of Weak subsolutions and supersolutions of a divergence-form equation.

3.1step 2.1F1algebra

The supremum is larger than the boundary supremum. As a weak solution $u$ is also a weak subsolution; since $u (x) > 0$ for $∣ x ∣ < π$ and $u (0) = 1$ , while $u (x) < 1$ for every $x \neq 0$ (because $\sin r < r$ on $(0, π)$ ), the essential supremum is ${ess sup}_{Ω} u = 1$ . On the boundary $u = 0 \geq 0$ is nonnegative, so $u^{+} = u$ and $(u - 0)^{+} = u \in H_{0}^{1} (Ω)$ ; thus $\sup_{\partial Ω} u^{+} = 0$ in the boundary-order convention of [F1]. Hence ${ess sup}_{Ω} u = 1 > 0 = \sup_{\partial Ω} u^{+}$ , and the refuted statement fails for this weak subsolution.

4.1step 1.1step 3.1F1∎

The sign condition fails. For every nonnegative $ζ \in C_{c}^{\infty} (Ω)$ with $ζ \equiv̸ 0$ , the sign functional is $\int_{Ω} (c ζ + b^{i} D_{i} ζ) d x = - \int_{Ω} ζ d x < 0$ , so the weak sign condition required by Weak maximum principle for coercive divergence-form equations does not hold. This is a direct adaptation of the adverse-zero-order obstruction in [S] Section II.2; the radial eigenfunction and its weak verification are computed here, and the example uses only the explicit function, the trace theorem and the classical-to-weak consistency, so no choice principle beyond the declared Axiom of Choice and Countable Choice is used.

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Sources

Armin Schikorra, Partial Differential Equations (University of Pittsburgh, version 4 December 2019; complete 185-page lecture notes) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (author manuscript, version 11 February 2025; complete 392-page archived text) (standard reference, not scraped)