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?

2019-12-02 02:52:45
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Answers: 1

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.)

2019-12-03 04:17:44