Welcome to MathOverflow

Question

4

I have a very basic question in the calculus of variations:

Suppose I want to minimize the functional

$$A[r, r'] = \int_\Omega L(r, r') dx $$

When is it possible to say that extremals of $A$ agree with extremals of

$$\tilde{A}[r, r'] = \int_\Omega \left( L(r, r') \right)^2 dx $$

Assume that $L(r, r') \geq 0$. Note that $r: \mathbb{R}^{n - 1} \rightarrow \mathbb{R}^n$ (e.g. $r$ as a parameterization of a minimal surface), $r'$ is really a matrix, and $L(r, r')$ is a real number.

Any references for these types of issues would also be great.

edited Jul 10 at 18:34

asked Jul 10 at 16:58

Dan Blazevski
238●1●8

	I have only looked at the single-variable case, but combining the Euler equations for both problems seems to lead to the conclusion $dL/dx = 0$, in other words $L$ must be constant on any common extremal. In that case of course $\tilde{A}\int_\Omega dx = A^2$. – Niels Diepeveen Jul 11 at 16:00
	Sorry for the delayed response, but could you explain this further? – Dan Blazevski Jul 19 at 12:28
	Have you tried computing the Euler-Lagrange equation for each functional and comparing them? – Deane Yang Jul 28 at 14:53
	@Deane Yang, yes, though that didn't get me too far, at least at first sight. – Dan Blazevski Jul 31 at 18:18

dcs24 · Answer 1 · 2012-07-28 10:54:14Z UTC

1

I believe that the equivalence of minimisers follows if and only if when $r$ is a critical of $A$ it holds that $L(r,r') = C \in \mathbb R, \forall x \in \Omega$. The idea being inspired by the equivalence of Length/Energy minimisation in differential geometry.

answered Jul 28 at 10:54

dcs24
69●7

Thanks, I'll look into that. – Dan Blazevski Jul 31 at 18:18

Minimizing action squared versus action

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.