1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrange multipliers and variation of functions

  1. Apr 25, 2009 #1
    Let F and f be functions of the same n variables where F describes a mechanical system and f defines a constraint. When considering the variation of these functions why does eliminating the nth term (for example using the Lagrange multiplier method) result in a free variation problem where it wasn't necessarily so before the elimination?
  2. jcsd
  3. Apr 26, 2009 #2
    I didn't state my question very well. Anyway, I think I may be getting some insight from a Dover booklet - "Calculus of Variations", Gelfand and Fomim. The portion that is bearing some fruit in my mind says:
    "As is familiar from elementary analysis, in finding an extremum of a function of n variables subject to k constraints (k<n), we can use the constraints to express k variables in terms of the other n - k variables. In this way, the problem is reduced to that of finding an unconstrained extremum of a function of n - k variables; i.e. an extremum subject to no subsidiary conditions."​
    I need to work out some simple examples to get the feel for why this works but it seems to me that this is a very amazing little mathematical "trick". Lagrange was pretty smart, huh? :wink:
  4. Apr 27, 2009 #3
    I guess the gist of it is that when you eliminate variables using constraint equations, you are incorporating information about your system into the problem. When extremizing the result, the information is utilized, consequently, during differentiation. i.e. You no longer regard certain variables as having independent variations with respect to the varying parameter, rather, you acknowledge the constraint, substitute in for the variable in question and observe the differential changes in the substituted expression.

    A physical example of this would be to take the Lagrangian of a 2d pendulum system, written in Cartesian (not polar!) coordinates and substitute length = x^2 + y^2 for x in the Lagrangian. We know that this system has 1 degree of freedom and, now, the Lagrangian is expressed using one coordinate, y. When we look at variations of L with respect to some parameter, we will also be looking at variations in the expression sqrt(length - y^2), wherever x had appeared. Thus we are asserting the truth of both the constraint equation and L when we extremize.

    The method of Lagrange multipliers is a little different, procedurally and conceptually, yet with similar results. Returning to the 2D pendulum, this method suggests that we set the variational derivative of L - \lambda * (x^2 + y^2) to 0. The reason for this is often explained geometrically, where, at a stationary point, the gradient of the two terms should be parallel (hence the lambda and the 0).

    A good way to get a sense for these equations is to break them, in a controlled way. So, for example, what happens when you don't include any constraint information into the 2D Lagrangian? You get a constant momentum in the x-direction and the point mass either hits the V=0 level with a thud (if we asserted there is a ground) or oscillates about it in this odd mathematical construction. It is as if we snipped the string connecting the point mass to the pendulum pivot point. The first point to make is that we were successful in extremizing the Lagrangian without the constraint equation. We, in return, were given the EOMs (because these are the extremums) of a system that had no constraints. The second point is that this result is not consistent with the pendulum system we had in mind when we began, as it should be, for there was no way our Lagrangian could acquire the knowledge that "there is a string."
  5. Apr 27, 2009 #4
    Yes - and what mystifies me is the actual "mechanics" (in the mathematical sense) of how the lambda method accomplishes a reduction in the number of resulting differential equations from n to n-m if when we have n variables for the functional:
    and and m constraints,
    Gj(x1,...,xn), {j=1,2,...,m},m<n.​
    Actually, the variation of F (or the variation of the definite integral of F) under the constraining condition(s) given by G. The lambda (Lagrange multiplier) method seems to accomplish a result similar to what it does when using ordinary functions in multi-variable calculus, as you mentioned. In ordinary calculus, the lamda method works because the constraining function is tangent to the function under examination so that their respective gradients are parallel (or anti-parallel) and therefore are a real number multiple of each other. For some reason I'm having a mental block in seeing the similarity when dealing with the variation of functionals. I understand that in principle it should be the same. It's the mechanics of how variation(F+lamda G)=0 reduces the number of "admissible curves" that F can take as input from m to m-n. (If that's even an allowable way to say it?)

    Hey! I see what you mean here! I'm going to play with this today. Thanks a lot.
    [I'm kind of playing in my head with the geometrical notion of how constraints (as functionals) accomplish dimensional reduction and therefore path limitation in configuration space and your example is a perfect playground]
    Last edited: Apr 27, 2009
  6. Apr 27, 2009 #5
    It's probably best to split our peas and carrots. The method of Lagrange multipliers does not reduce the number of variables or the number of equations. What it does is incorporate the constraint information by adding a term to the constrained equation, then it observes a parallel gradient condition for singular points. In the 2D pendulum problem, one usually extremizes both the x and y coordinates, separately. On the other hand, solving the constraint for x and substituting this into the Lagrangian gives an expression free of x that can be normally extremized. The resulting y-EOM then gives the x-EOM without any further extremization. This sounds messy and I haven't tried it. The method of Lagrange multipliers is accessible in this problem, however.

    There is a simple example of this problem in the mechanics book, by Hand and Finch, that I had used in school and I just picked it up and read that the authors suggest using the constraint length = sqrt(x^2 + y^2), rather than the length squared, because this gives a more interesting interpretation of the Lagrange multiplier.

    https://www.amazon.com/Mechanics-Ne...sr_1_1?ie=UTF8&s=books&qid=1240857299&sr=1-1" has a good section on the geometry of mechanics, as well. I also remember some Cal Tech notes that were floating around their physics site that handled this particular topic nicely. They were supplements (more like a step-by-step walk-through, lol) of Hand and Finch. Unfortunately, I do not remember the professor's name (I went to Berkeley).
    Last edited by a moderator: May 4, 2017
  7. Apr 28, 2009 #6
    Yes. I was wrong about the reduction. Not sure why I thought that. In fact, in going back and re-studying what I have on this, now it seems that when finding the stationary value of a functional with n degrees of freedom, F(u1,...,un), with m constraints, gi(u1,...,un)=0 {i=1,...,m}, we actually end up with n+m simultaneous equations. The n equations are from the modified functions (modified to incorporate the constraints into F by using separate Lagrange multipliers for each constraint g), F+lambda1g1+...+lambdamgm. The m additional equations would be the m constraining functions gi=0. So this would be n+m equations in m+n unknowns. The unknowns are u1,..., un, lambda1,..., lambdam.
    At least, that's what I think today... :rolleyes:

    That looks like a great book! I wish I could afford it :cry:
    Communist :rofl:
    Last edited by a moderator: May 4, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook