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!

Homework Help: 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

    User Avatar
    Science Advisor
    Homework Helper

    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.
    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.

  8. Sep 12, 2012 #7
    Thanks, just the type of argument I was looking for.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook