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Lagrange Multipliers in Calculus of Variations

  1. Nov 13, 2011 #1
    In Lagrangian mechanics, can anyone show how to find the extrema of he action functional if you have more constraints than degrees of freedom (for example if the constraints are nonholonomic) using Lagrange Multipliers?

    I've looked everywhere for this (books, papers, websites etc.) but none of them use the Calc. of Variations approach, they simply say something like "well assume a multiplier exists that makes this term zero" and go from there. I've never seen this derived from Hamilton's principle, which is what I think this is all about. I'll bet Lagrange looked at the problem in terms of finding a stationary point of a function, then applied the same idea to functionals. I'd like to see how this works, so I'd appreciate if someone could show me.

    Thanks
     
  2. jcsd
  3. Nov 18, 2011 #2
    Here's what I mean:

    http://en.wikibooks.org/wiki/Classi...#Lagrange_multipliers_and_constraining_forces

    They define a new Lagrangian,
    [tex]\tilde L = L - \lambda(t)G(\boldsymbol x)[/tex]

    Where they subtract the lagrange multiplier times the constraint function from the usual Lagrangian. Why? Can anyone show why this is?

    From what I know, we have an action function defined:
    [tex]S=\int_{t1}^{t2}f(q,q',t)dt[/tex]
    and this function must be stationary for the correct path of the particle(s). For the unconstrained case, I know that S is stationary if the function integrated is the Lagrangian. But if it is constrained, can we do this?
    [tex]\nabla S = \lambda \nabla G[/tex] for a constraint function G(q,q',t)
    so then can we do this:
    [tex]\frac{\partial S}{\partial q}=\lambda \frac{\partial G}{\partial q} , \frac{\partial S}{\partial q'}=\lambda \frac{\partial G}{\partial q'},\frac{\partial S}{\partial t}=\lambda \frac{\partial G}{\partial t} ?[/tex]

    I'm stuck here...
     
    Last edited: Nov 18, 2011
  4. Nov 20, 2011 #3

    Matterwave

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    1) You can't have more (non-redundant) constraints than degrees of freedom.
    2) If you had the same number of constraints as degrees of freedom, the constraints would determine the entire motion.
    3) You are free to add G because G is defined to be 0. You are adding zero to your Lagrangian.
     
  5. Nov 21, 2011 #4
    Then can you show that the [itex]\lambda(t)G(\boldsymbol x)[/itex] is equal to the constraint forces?

    And can you show that the multiplier is unique? And under what conditions it is unique?
     
  6. Nov 21, 2011 #5

    I like Serena

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    What you would have is the standard Lagrangian formula:
    [tex]{d \over dt}{\partial\tilde{L} \over \partial \dot q} - {\partial\tilde{L} \over \partial q} = 0[/tex]
     
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