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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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  • $\begingroup$ 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 $\tilde{A}\int_\Omega dx = A^2$. $\endgroup$ Jul 11, 2012 at 16:00
  • $\begingroup$ Sorry for the delayed response, but could you explain this further? $\endgroup$
    – user7807
    Jul 19, 2012 at 12:28
  • $\begingroup$ Have you tried computing the Euler-Lagrange equation for each functional and comparing them? $\endgroup$
    – Deane Yang
    Jul 28, 2012 at 14:53
  • $\begingroup$ @Deane Yang, yes, though that didn't get me too far, at least at first sight. $\endgroup$
    – user7807
    Jul 31, 2012 at 18:18

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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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