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The one-dimensional Euler-Lagrange equation for an energy with a potential

Example

Example. Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let V∈C1(R), let a<b be real numbers and let I(u)=∫ab(12u′(x)2+V(u(x)))dx. On the admissible class C2([a,b]) with fixed endpoint values u(a)=u0, u(b)=u1, a local minimiser in the C2 norm satisfies the boundary value problem u′′=V′(u)  in (a,b),u(a)=u0,u(b)=u1, the classical Euler-Lagrange equation of the Lagrangian f(x,s,ξ)=12ξ2+V(s) (The classical Euler-Lagrange equation under regularity). In the special case V=0 this is the one-dimensional Laplace equation u′′=0 with the affine solution u(x)=u0+u1−u0b−a(x−a); for V(s)=12ω2s2 it is the equation of an inverted harmonic oscillator u′′=ω2u.

Facts & Assumptions

Given: Countable Choice; a function V∈C1(R), a compact interval [a,b]⊂R with a<b, the functional I(u)=∫ab(12u′(x)2+V(u(x)))dx on the admissible class of u∈C2([a,b]) with fixed endpoint values u(a)=u0, u(b)=u1, and the Lagrangian f(x,s,ξ)=12ξ2+V(s).

[F1]

The conclusion has the shape of the classical Euler-Lagrange equation of The classical Euler-Lagrange equation under regularity for the Lagrangian f(x,s,ξ)=12ξ2+V(s), whose partials are fξ=ξ and fs=V′(s). That corollary also assumes f∈C2 and the growth hypotheses of the weak Euler-Lagrange theorem (The weak Euler-Lagrange equation for integral functionals with fixed trace), which need not hold for a general V∈C1; the verification below therefore computes the equation directly from the minimality of u.

[F2]

One-variable mean value theorem and Fermat's interior-extremum theorem: a differentiable function on an interval with an interior local extremum has vanishing derivative there, and the mean value theorem identifies (V(u+εφ)−V(u))/ε with V′(u+θεφ)φ for some θ∈(0,1) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a), Fermat's interior extremum theorem: if f has a local extremum at a point c interior to its domain and is differentiable at c, then f′(c)=0).

[F3]

Integration by parts on a compactly supported test function: ∫abu′φ′ dx=−∫abu′′φ dx for φ∈Cc∞((a,b)) and u∈C2([a,b]). This is Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives with F=u′ and G=φ; its continuous integrands have equal Riemann and Lebesgue integrals by A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral.

[F4]

Fundamental lemma: a continuous function on an open set orthogonal to every compactly supported smooth test function vanishes identically (The fundamental lemma of the calculus of variations).

[F5]

Verification

technique · direct, by testing the minimality of $u$ against compactly supported variations
1.1F1F2givenalgebra

The first variation. Let φ∈Cc∞((a,b)). Since φ vanishes near the endpoints, u+εφ belongs to the admissible class for every ε, and for ∣ε∣ small it is close to u in C2([a,b]), so Φ(ε):=I(u+εφ) has a local minimum at ε=0. Computing the difference quotient, ε−1(Φ(ε)−Φ(0))=∫ab(u′φ′+ε2φ′2+ε−1(V(u+εφ)−V(u)))dx, and by the mean value theorem [F2] the last term equals φ(x)V′(u(x)+θxεφ(x)) with θx∈(0,1); as ε→0 this converges to φV′(u) uniformly on [a,b], because V′ is uniformly continuous by Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous on a compact interval containing the values u(x)+θxεφ(x) and ∣u+θεφ−u∣≤∣ε∣∥φ∥∞. Hence Φ is differentiable at 0 with Φ′(0)=∫ab(u′φ′+V′(u)φ) dx. This is the direct computation announced in [F1].

2.1F2step 1.1

Fermat's theorem. The point 0 is an interior local minimum of the differentiable function Φ, so [F2] gives Φ′(0)=0, that is ∫ab(u′φ′+V′(u)φ) dx=0 for every φ∈Cc∞((a,b)).

3.1F3F4step 2.1

The differential equation. For φ∈Cc∞((a,b)) integration by parts [F3] gives ∫abu′φ′ dx=−∫abu′′φ dx, so the identity of step 2.1 reads ∫ab(−u′′+V′(u))φ dx=0 for every such φ. The function −u′′+V′(u) is continuous on (a,b), being a sum of continuous functions, so the fundamental lemma [F4] gives −u′′+V′(u)=0 on (a,b), that is u′′=V′(u).

4.1F5step 3.1givenalgebra∎

Endpoint conditions and the two instances. The admissible class fixes u(a)=u0 and u(b)=u1, so u solves the boundary value problem of the statement. For V=0 the equation is u′′=0, and [F5] makes u affine, with values determined by the endpoints: u(x)=u0+u1−u0b−a(x−a). For V(s)=12ω2s2 one has V′(s)=ω2s, so the equation reads u′′=ω2u, the equation of an inverted harmonic oscillator.

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