Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Extremal condition calculus of variations

  1. Apr 2, 2013 #1
    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?
  2. jcsd
  3. Apr 5, 2013 #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 John Baez's exemplary notes for more information.
    Last edited: Apr 5, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook