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Stationarity of the Euler-Lagrange equation does not imply a minimum

Statement refuted

Counterexample. Assume the Axiom of Choice (The Axiom of Choice). Let $Ω \subseteq R^{n}$ be a bounded $C^{1}$ domain and let $J (u) = - \frac{1}{2} \int_{Ω} ∣ D u ∣^{2} d x (u \in H_{0}^{1} (Ω)) .$ Then $u = 0$ is the only stationary point: its weak Euler-Lagrange (stationarity) equation reads $- \int_{Ω} D u \cdot D φ d x = 0$ for every $φ \in H_{0}^{1} (Ω)$ , which forces $D u = 0$ and then $u = 0$ almost everywhere by the Poincare inequality (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction). But $J$ is not bounded below: for any fixed nonzero $φ \in H_{0}^{1} (Ω)$ one has $J (t φ) = - \frac{t^{2}}{2} ∥ D φ ∥_{2}^{2} \to - \infty$ as $∣ t ∣ \to \infty$ . So the Euler-Lagrange equation is a necessary condition only; the functional is concave, not convex, and Stationarity is sufficient for a global minimum of a convex differentiable functional does not apply.

Facts & Assumptions

Given: The Axiom of Choice; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , the functional $J (u) = - \frac{1}{2} \int_{Ω} ∣ D u ∣^{2} d x$ on $H_{0}^{1} (Ω)$ , and the Lagrangian $f (x, s, ξ) = - \frac{1}{2} ∣ ξ ∣^{2}$ .

[F1]

A stationary point of $J$ is a point $u \in H_{0}^{1} (Ω)$ whose first variation vanishes in every direction $φ \in H_{0}^{1} (Ω)$ . The quadratic expansion below computes this variation directly for every $n \geq 1$ . For $n \geq 2$ , its vanishing is also the fixed-zero-trace weak Euler-Lagrange formula of The weak Euler-Lagrange equation for integral functionals with fixed trace, since $f_{s} = 0$ , $f_{ξ} = - ξ$ and the $p = 2$ differentiation growth bounds hold.

[F2]

On the full linear space $H_{0}^{1} (Ω)$ , at a local minimiser the first variation vanishes (The first variation vanishes at an interior minimiser); the converse requires convexity and stationarity in the sense $δ I (u; v - u) \geq 0$ for all competitors, by Stationarity is sufficient for a global minimum of a convex differentiable functional.

[F3]

Poincare's inequality controls the $L^{2}$ norm by the Dirichlet energy on $H_{0}^{1} (Ω)$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction); in particular $\int_{Ω} ∣ D u ∣^{2} = 0$ forces $u = 0$ almost everywhere.

[F4]

The Lagrangian $f (x, s, ξ) = - \frac{1}{2} ∣ ξ ∣^{2}$ is concave, not convex, in $ξ$ (Convex and strictly convex functionals on a convex subset of a real vector space).

Counterexample

technique · direct computation of the stationary equation and along the line $t\varphi$

1.1F1algebra

The stationary equation. For $f (x, s, ξ) = - \frac{1}{2} ∣ ξ ∣^{2}$ one has $f_{ξ} (x, s, ξ) = - ξ$ and $f_{s} = 0$ , so the exact expansion $J (u + ε φ) - J (u) = - ε \int D u \cdot D φ - \frac{1}{2} ε^{2} \int ∣ D φ ∣^{2}$ gives the bounded first variation $δ J (u; φ) = - \int D u \cdot D φ$ by Holder (Holder's inequality for integrals, including the endpoint cases). Thus a point $u \in H_{0}^{1} (Ω)$ satisfies the weak Euler-Lagrange equation of [F1] exactly when $\int_{Ω} D u \cdot D φ d x = 0$ for every $φ \in H_{0}^{1} (Ω)$ .

1.2F3algebra

$0$ is not a minimiser, not even locally. Fix any nonzero $φ \in H_{0}^{1} (Ω)$ , which exists: choose a ball $B_{R} (a) \subseteq Ω$ and a nonzero smooth bump supported inside it (Compactly supported scaled Euclidean bumps). Poincare [F3] gives $∥ D φ ∥_{2} > 0$ , and consider $J (t φ) = - \frac{t^{2}}{2} \int_{Ω} ∣ D φ ∣^{2} d x$ . As $∣ t ∣ \to \infty$ this tends to $- \infty$ , so $J$ is not bounded below on $H_{0}^{1} (Ω)$ ; and for every $t \neq 0$ one has $J (t φ) < 0 = J (0)$ , and $∥ t φ ∥_{H^{1}} = ∣ t ∣ ∥ φ ∥_{H^{1}} \to 0$ as $t \to 0$ , so $0$ is not a local minimiser either.

2.1F3step 1.1

The only stationary point. If $u$ is stationary, step 1.1 applies with the admissible test function $φ = u \in H_{0}^{1} (Ω)$ , giving $\int_{Ω} ∣ D u ∣^{2} = 0$ ; by [F3] this forces $D u = 0$ and hence $u = 0$ almost everywhere. Conversely $u = 0$ satisfies the equation because $D 0 = 0$ . So $0$ is the only stationary point of $J$ .

3.1F2F4step 2.1step 1.2∎

Conclusion. The only stationary point of $J$ fails to be a minimiser by step 1.2, so the Euler-Lagrange equation is a necessary condition only; the failure is consistent with [F2], since $J$ is concave in the gradient by [F4] and the convex stationarity-sufficiency theorem therefore does not apply.

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Sources

Riccardo Cristoferi, Calculus of Variations: Lecture Notes, Carnegie Mellon University 2016 (complete 133-page notes) (standard reference, not scraped)
Francesco Paolo Maiale (course by Giovanni Alberti), Lecture Notes Calculus of Variations A, University of Pisa (last update 21 August 2019; complete 149-page notes) (standard reference, not scraped)