version of the fundamental lemma of calculus of variations
Lemma. If a real function f is continuous on the interval [a,b] and if
∫baf(x)φ(x)𝑑x= 0 |
for all functions φ continuously differentiable on the interval and vanishing at its end points, then
f(x)≡ 0 on the whole interval.
Proof. We make the antithesis that f does not vanish identically. Then there exists a point x0 of the open interval (a,b) such that f(x0)≠0; for example f(x0)>0. The continuity of f implies that there are the numbers α and β such that a<α<x0<β<b and f(x)>0 for all x∈[α,β]. Now the function φ0 defined by
φ0(x):= |
fulfils the requirements for the functions . Since both and are positive on the open interval , we however have
Thus the antithesis causes a contradiction. Consequently, we must have .
Title | version of the fundamental lemma of calculus of variations![]() |
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Canonical name | VersionOfTheFundamentalLemmaOfCalculusOfVariations |
Date of creation | 2013-03-22 19:12:04 |
Last modified on | 2013-03-22 19:12:04 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 4 |
Author | pahio (2872) |
Entry type | Theorem |
Classification | msc 26A15 |
Classification | msc 26A42 |
Synonym | fundamental lemma of calculus of variations |
Related topic | EulerLagrangeDifferentialEquation |