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?
Best Answer
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.)