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Lagrange multipliers

  1. Sep 12, 2012 #1
    Normally lagrange multipliers are used in the following sense.

    Suppose we are given a function f(x,y.z..,) and the constraint g(x,y,z,...,) = c
    Define a lagrange function:
    L = f - λ(g-c)

    And find the partial derivatives with respect to all variables and λ. This gives you the extrema since for an extrema ∇f = λ∇g

    However, I find that in my mechanics book this is used differently. I have attached a picture of the place where lagrange multipliers are used. I don't see how they in this case are used to maximize a quantity (which should be L) under a constraint like the above. Can anyone show me how they are and which gradients are to be parallel?
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2012 #2
    I'm not sure if I understand; how are they used differently? You have a set of constraints [itex] f_\alpha(\mathbf{q},\dot{\mathbf{q}},t) = 0 [/itex] and then you just write [itex] L = L + \sum_\alpha \lambda_\alpha f_\alpha [/itex]
     
  4. Sep 12, 2012 #3
    maybe I just don't see it. But my point is that you usually want to make the gradients of f and g=y parallel. How is that done in the example? There are no gradients being taken...
     
  5. Sep 12, 2012 #4
    Aah sorry, I somehow missed the whole point of your question :) They are not maximizing L. They are minimizing (or maximizing) the integral of L. This is entirely different.

    So do you understand how you get from [tex]S = \int dt L [/tex] by varying the action to
    [tex]\delta S = \int dt \left( \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \right) \delta q = 0 \rightarrow \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0 [/tex]?
     
  6. Sep 12, 2012 #5
    Yes, I know that the above implies the Euler-lagrange equation.
    So they want to make the lagrangian stationary. Normally if the qi's are independent you that leads to n independent lagrange equations. But now they are instead using lagrange multipliers. But how exactly are they used? In analogy to my original example, what is f and what is g and how does the above correspond to the problem of making the gradient of f and g parallel. Maybe it is completely obvious, but I can't quite see it.
     
  7. Sep 12, 2012 #6

    Ray Vickson

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    Imagine that you approximate the integral by a finite sum of a large number of terms (which we actually do in numerical work), and you replace the continuum of constraints ##f(q,\dot{q},t) = 0## by a large number of constraints ##f(q_i,v_i ,t_i) = 0, \; i=1,2, \ldots, N.## Here, the ##q_i## are you estimate of ##q(t_i)## and the ##v_i## are your estimate of ##\dot{q}(t_i).## (Again, this is often how such problems ARE dealt with using numerical methods---we replace a continuous problem by a disctete problem involving a large number of small time steps.) Now your solution to
    [tex] \min \int_a^b L(q,\dot{q},t)\, dt, \: \text{ subject to } f(q,\dot{q},t) = 0 \text{ for all } t \in [a,b][/tex] would be replaced by that of the problem
    [tex] \min \sum_{i=1}^N \Delta t_i f(q_i,v_i,t_i),\\
    \text{subject to}\\
    v_i = (q_i - q_{i-1})/ \Delta_i \;(\text{ or } v_i = (q_{i+1} - q_i)/ \Delta_i), i=1, \ldots N\\
    f(q_i,v_i ,t_i) = 0, \; i=1, \ldots, N.
    [/tex]
    Now, of course, you would have a Lagrange multiplier ##\lambda_i## for each of the f = 0 constraints, so your Lagrangian would contain
    [tex] \sum_{i=1}^N \lambda_i f(q_i,v_i,t_i). [/tex]
    and this would go over into an integral in the limit of infinite N.

    Well, that is the intuition, anyway; rigorous justification is another matter.

    RGV
     
  8. Sep 12, 2012 #7
    Thanks, just the type of argument I was looking for.
     
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