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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.

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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 '12 at 16:00
Sorry for the delayed response, but could you explain this further? –  Dan Blazevski Jul 19 '12 at 12:28
Have you tried computing the Euler-Lagrange equation for each functional and comparing them? –  Deane Yang Jul 28 '12 at 14:53
@Deane Yang, yes, though that didn't get me too far, at least at first sight. –  Dan Blazevski Jul 31 '12 at 18:18
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1 Answer

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.

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Thanks, I'll look into that. –  Dan Blazevski Jul 31 '12 at 18:18
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