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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 ()). Let , let be real numbers and let On the admissible class with fixed endpoint values , , a local minimiser in the norm satisfies the boundary value problem the classical Euler-Lagrange equation of the Lagrangian (The classical Euler-Lagrange equation under regularity). In the special case this is the one-dimensional Laplace equation with the affine solution ; for it is the equation of an inverted harmonic oscillator .
Facts & Assumptions
Given: Countable Choice; a function , a compact interval with , the functional on the admissible class of with fixed endpoint values , , and the Lagrangian .
The conclusion has the shape of the classical Euler-Lagrange equation of The classical Euler-Lagrange equation under regularity for the Lagrangian , whose partials are and . That corollary also assumes 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 ; the verification below therefore computes the equation directly from the minimality of .
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 with for some (The mean value theorem, as the case of Cauchy's: for continuous on with and differentiable on there is with , Fermat's interior extremum theorem: if has a local extremum at a point interior to its domain and is differentiable at , then ).
Integration by parts on a compactly supported test function: for and . This is Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives with and ; 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.
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).
A function on with on is affine, since has vanishing derivative and is therefore constant (The mean value theorem, as the case of Cauchy's: for continuous on with and differentiable on there is with ).
Verification
The first variation. Let . Since vanishes near the endpoints, belongs to the admissible class for every , and for small it is close to in , so has a local minimum at . Computing the difference quotient, , and by the mean value theorem [F2] the last term equals with ; as this converges to uniformly on , because 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 and . Hence is differentiable at with . This is the direct computation announced in [F1].
Fermat's theorem. The point is an interior local minimum of the differentiable function , so [F2] gives , that is for every .
The differential equation. For integration by parts [F3] gives , so the identity of step 2.1 reads for every such . The function is continuous on , being a sum of continuous functions, so the fundamental lemma [F4] gives on , that is .
Endpoint conditions and the two instances. The admissible class fixes and , so solves the boundary value problem of the statement. For the equation is , and [F5] makes affine, with values determined by the endpoints: . For one has , so the equation reads , the equation of an inverted harmonic oscillator.
Depends on
- The classical Euler-Lagrange equation under regularity
- The weak Euler-Lagrange equation for integral functionals with fixed trace
- 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 $f'(c) = 0$
- The fundamental lemma of the calculus of variations
- 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 $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives
- A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral
- Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous
Used by
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Sources
- Sung-Jin Oh, Lecture Notes for Math 222A, UC Berkeley, 19 March 2024 (complete 179-page author PDF) (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)