Use of Lagrange's equations in classical mechanics

[PrathameshR]

I have been studying classical mechanics for a while from Goldstein book and can't go ahead of the following derivation. I understand the method of Lagrange's multipliers for getting extrima of a function subjected to equality constraints but can't understand it's relevance here because in that method we find "points" which give extremum value but here we want to find "function" which extrimizes a perticular integral.

In the 4rth line of 2nd paragraph it says that delta a 'subscript I ' may not be consistent with contraints , how is that possible?

In the title I ment use of Lagrange's "multipliers"

 

[vanhees71]

Skip this chapter from Goldstein, at least if it comes to non-holonomous constraints since this is awfully flawed. Otherwise the idea of Lagrange multipliers is precisely the same in variational calculus (finding stationary points of functionals) as in finding stationary points of functions under constraints.

 

[PrathameshR]

Skip this chapter from Goldstein, at least if it comes to non-holonomous constraints since this is awfully flawed. Otherwise the idea of Lagrange multipliers is precisely the same in variational calculus (finding stationary points of functionals) as in finding stationary points of functions under constraints.

Can you suggest any good place where I can read a proper treatment of this topic?

 

[vanhees71]

It's in Landau&Lifshitz vol. I.

 

[Orodruin]

but can't understand it's relevance here because in that method we find "points" which give extremum value but here we want to find "function" which extrimizes a perticular integral.

As @vanhees71 mentioned, it is just the same idea. In fact, the only difference is that your function space is infinite-dimensional whereas you have likely only seen the Lagrange multiplier method applied to finite-dimensional vector spaces before. As a heuristic argument, consider a discretisation of your integral
$$
\mathcal F = \int f(\phi(x),\phi'(x)) dx \to \Delta x \sum_{i = 1}^N f(\phi(x_i),[\phi(x_{i+1})-\phi(x_i)]/\Delta x) \equiv F(\vec \phi)
$$
where $F$ is some function of the function values $\phi_i = \phi(x_i)$. Now think of the $\phi_i$ as the coordinates in $\mathbb R^N$. It is rather easy to convince yourself that the partial derivative $\partial F/\partial\phi_i$ is just the discretisation of the functional derivative $\delta \mathcal F/\delta\phi(x_i)$ at $x_i$. Furthermore, you therefore also have that the discretisation of the variation
$$
\delta \mathcal F = \int \frac{\delta F}{\delta \phi(x)} \delta\phi(x) dx
$$
is on the form
$$
\Delta x \sum_i \delta\phi_i \frac{\partial F}{\partial\phi_i} \propto \delta\vec \phi \cdot \nabla F,
$$
where $\nabla$ is the gradient in the space with coordinates $\phi_i$ and $\delta\vec \phi$ is a variation in that space.

If you are considering a vector space of functions, the entire argument for the Lagrange multiplier method that is used in a finite-dimensional vector space goes through without modification to the function space.