# Calculus of Variations and also Lagrange Multipliers

A basic trouble for the Calculus of Variations asks us to decrease the value of an useful $A[f]$, where $f$ is generally a differentiable function specified on $\mathbb{R}^n$.

What happens if, nonetheless, the domain name of $A$ is not in fact all differentiable features. Intend there is a restraint formula on $f$, such as (as an example) :

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

and also we intend to decrease over features pleasing the above and also the building that $f(-1)=f(1)=0$ the useful

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

This type of trouble appears to me to be really comparable to the trouble in multivariate calculus of decreasing a function $f(x)$ relative to a restraint formula $g(x) = 0$. In this instance we are attempting to decrease an useful $A[f]$ relative to an useful restraint formula $L[f] = \pi$.

In the previous, one can make use of Lagrange multipliers to lower the trouble to that of addressing a system of formulas. Exists such a strategy for the variational variation?

Yes. For the basic theory, see this Wikipedia page. In the certain instance which you take into consideration, you can in fact take into consideration the reduction trouble

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

which brings about the Euler - Lagrange formula

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

which brings about a rather uninviting looking Lagrangian with $f''$ in the when you connect $\lambda$ back in.

(This is, certainly, not entirely unanticipated, as your useful $A$ is not bounded listed below on the set $L[f] = \pi$. To see this, it are adequate to keep in mind that $L[f + c] = L[f]$ necessarily, while $A[f + c] = A[f] + c$. So for a lowering series of $c\searrow -\infty$, $A[f + c]$ lowers without bound.)