Extremal condition calculus of variations

In summary, the conversation discusses the process of solving a minimization problem for a functional with a Lagrangian. This involves finding necessary conditions through the Euler-Lagrange equations, and determining if the minimum found is an actual minimum. To do this, one can check if the Lagrangian is convex or calculate a second derivative. However, differentiating the Lagrangian twice only gives the local minimum, while the goal is to find the global minimum of the action. Therefore, finding the minimum of the Lagrangian itself is not useful, but finding the minimum action reveals the equations the particle will follow. Additionally, it is noted that some phenomena may have maximum action in nature. Further information and resources are recommended for a deeper understanding of
  • #1
Gavroy
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if I have a functional with a Lagrangian L(t,x(t),y(t),x'(t),y'(t)), meaning two functions x and y of one parameter t. And want to solve the minimization problem $$ \int_0^t L dt $$ . Then I get necessary conditions to find extrema by getting the two Euler Lagrange equation $$ \frac{\partial L}{\partial x}- \frac{d}{dt} \frac {\partial L}{\partial x'}=0$$ and $$ \frac{\partial L}{\partial y}- \frac{d}{dt} \frac {\partial L}{\partial y'}=0$$

now, if i solved these functions. how do i find out, that it is an actual minimum? are there methods to show this in general? i know, that in case of one variable it would be sufficient to show somehow that the lagrangian is convex. but is there a way to do this in this case too? or do i need to calculate a second derivative? if this is necessary, can someone give me a referece, where this is done for functionals of several functions or show me a way to do this?
 
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  • #2
Look at the proof of Euler-Lagrange equation here (look at the 'one dimensional case').

You wouldn't differentiate L twice, as you're not trying to minimise [itex]L[/itex] itself, but the action [itex]A=\int L dt[/itex], so differentiating it twice will just give you locally where [itex]L[/itex] happens to be a minimum, whereas you really want the global minimum of the action, [itex]\int L dt[/itex]. Finding the minimum of the Lagrangian will give you a point in time when the Lagrangian is least, not really useful, but given the Lagrangian in the first place, finding the minimum action tells you the equations the particle will follow (e.g. the path it will take). There is no principle of least Lagrangian, only of least action.

The equations [tex]\frac{\partial L}{\partial x}=\frac{d}{dt}\frac{\partial L}{\partial \frac{dx}{dt}}[/tex] have actually done all the work for you: they are true as a result of [itex]\int L dt[/itex] being minimised, and are only true when [itex]\int L dt[/itex] is at a minimum (e.g. in nature) for all initial conditions.

If you want to do something to check whether it is a minimum, choose a slightly different function [itex]L[/itex], and calculate [itex]\int L dt[/itex] for that function (I think the notion of 'slightly different' is formalised in norm spaces, but I'm no expert on that).

Also note that (apparently) Hamilton showed that in nature some phenomena have maximum action (as the proof of the Euler-Lagrange equations only posit that the action is stationary, not minimum).

Have a look at [URL='https://www.physicsforums.com/insights/author/john-baez/']John Baez's exemplary notes[/URL] for more information.
 
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1. What is the Extremal Condition Calculus of Variations?

The Extremal Condition Calculus of Variations is a mathematical tool used to find the optimal solution to a problem involving an unknown function. It involves setting up an integral and finding the function that minimizes or maximizes it, subject to certain constraints.

2. How is the Extremal Condition Calculus of Variations used in real-life applications?

The Extremal Condition Calculus of Variations is used in a variety of fields such as physics, engineering, economics, and biology. It is used to optimize systems and processes, to find the most efficient path or shape, and to solve problems involving energy minimization or cost optimization.

3. What are some common techniques used in the Extremal Condition Calculus of Variations?

Some common techniques used in the Extremal Condition Calculus of Variations include the Euler-Lagrange equation, the method of variation of parameters, and the fundamental lemma. These techniques help to find the critical points of the integral and ultimately the optimal solution.

4. How does the Extremal Condition Calculus of Variations differ from other optimization methods?

The Extremal Condition Calculus of Variations is different from other optimization methods in that it deals with unknown functions, rather than known variables. It also considers the entire function, rather than just a few points, making it a more accurate and powerful tool for optimization.

5. What are some potential challenges when using the Extremal Condition Calculus of Variations?

One potential challenge when using the Extremal Condition Calculus of Variations is the complexity of the calculations involved. It can also be difficult to set up the integral and determine the appropriate boundary conditions. Additionally, finding the global minimum or maximum may be challenging, as the function may have multiple extrema or be non-convex.

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