1
$\begingroup$

In the derivation of Euler-Lagrange equation, as given here (under the "Statement" section, with the name "Derivation of one-dimensional Euler–Lagrange equation"), at the end of the derivation, it is assumed that the arbitrary pertubation path $\eta(t)$ has compact support in order to use the fundamental lemma of calculus of variations.

However, is there any justification for this assumption? Can't a path have non-compact support? If so, why?

$\endgroup$
  • $\begingroup$ Your link is not to a derivation of the Euler-Lagrange equation. There is one above it for the one-dimensional version, but the $\eta$ there is an arbitrary perturbation, and the fact that $\eta(a) = \eta(b) = 0$ simply comes from the fact that $f + \eta$ has to satisfy the same boundary value requirements as $f$. $\endgroup$ – Paul Sinclair Jun 26 '18 at 4:03
  • $\begingroup$ @PaulSinclair I did not understand anything from your comment expect the first sentence. $\endgroup$ – onurcanbektas Jun 26 '18 at 5:32
1
$\begingroup$

The derivation takes place on the compact interval $[a,b]$. Any function whose domain is a compact space automatically has compact support.

Since the support of a function is the closure of the set where the function is not $0$, it is a closed subset of a compact space, and is therefore compact itself. The function doesn't even need to be continuous for this to be true.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.