# Calculus of Variations and Lagrange Multipliers

A general problem for the Calculus of Variations asks us to minimize the value of a functional $A[f]$, where $f$ is usually a differentiable function defined on $\mathbb{R}^n$.

What if, however, the domain of $A$ is not actually all differentiable functions. Suppose there is a constraint equation on $f$, such as (for example):

$L[f] = \int_{-1}^1 \sqrt{1 + f'(x)^2} dx = \pi$

and we want to minimize over functions satisfying the above and the property that $f(-1)=f(1)=0$ the functional

$A[f] = \int_{-1}^1 f(x) dx$

This sort of problem seems to me to be very similar to the problem in multivariate calculus of minimizing a function $f(x)$ with respect to a constraint equation $g(x) = 0$. In this case we are trying to minimize a functional $A[f]$ with respect to a functional constraint equation $L[f] = \pi$.

In the former, one can use Lagrange multipliers to reduce the problem to that of solving a system of equations. Is there such a technique for the variational version?

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Yes, you can do that as well. Just create the new functional $D=A-\lambda(L-\pi)$. – Raskolnikov Nov 30 '10 at 15:19

Yes. For the general theorem, see this Wikipedia page. In the particular case which you consider, you can actually consider the minimization problem

$$\int_{-1}^1 f(x) + \lambda (\sqrt{ 1 + f'(x)^2} - \pi) dx$$

which leads to the Euler-Lagrange equation

$$\left( \frac{f'}{\sqrt{1 + (f')^2}}\right)' = \lambda^{-1}$$

which leads to a somewhat unappealing looking Lagrangian with $f''$ in the denominator when you plug $\lambda$ back in.

(This is, of course, not completely unexpected, as your functional $A$ is not bounded below on the set $L[f] = \pi$. To see this, it suffice to note that $L[f + c] = L[f]$ by definition, while $A[f + c] = A[f] + c$. So for a decreasing sequence of $c\searrow -\infty$, $A[f + c]$ decreases without bound.)

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Ah, I meant to impose that $f(-1)=f(1)=0$, thanks for pointing out the problem. And also thank you for your answer! it seems to be just what I was looking for. – Eric Haengel Nov 30 '10 at 15:25
(BTW: in this new formulation, you are essentially solving the isoperimetric problem with the area bounded by the line segment from $(-1,0) \to (1,0)$ and the image of $f$, which I assume you already know the answer...) – Willie Wong Nov 30 '10 at 15:52