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Using undetermined multipliers in Lagrangian Mechanics

  1. Mar 25, 2015 #1

    gulfcoastfella

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    An example problem in Chapter 7 of "Classical Dynamics of Particles and Systems" by Marion, Thornton uses Lagrangian equations with undetermined multipliers to solve for the motion of a disc rolling down an incline. The resulting Lagrangian equations are:

    Mg sin α - M d2y/dt2 + λ = 0

    MR2 d2θ/dt2 - λR = 0

    y = R θ (equation of constraint)

    I understand how these equations were developed, and I realize that three unknowns (y, θ, λ) require three equations. I don't feel comfortable, though, with including the equation of constraint as one of the equations when it's already incorporated into the other two equations as a partial derivative. Can anyone make me feel better about this?
     
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  3. Mar 26, 2015 #2

    vanhees71

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    The idea (for the most simple case of holonomous constraints) is as follows: You have a set of generalized coordinates ##q^j## and a set of constraints
    $$f_k(q^1,\ldots,q^n;t)=0, \quad k \in \{1,2,\ldots,m \}.$$
    One way to describe the dynamics is to introduce a new set of independent variables ##Q^k##, ##k \in \{1,2,\ldots,n-m\}##, write the Lagrangian (action) in terms of these independent variables and then write down the equations of motion, following from the stationarity of the corresponding action.

    Another way is to introduce auxiliary degrees of freedom ##\lambda^k## (Lagrange multipliers) and make up an action, depending on the ##q^j## and ##\lambda^k## as generalized coordinates and define the action as
    $$\tilde{L}=L(q,\dot{q};t)-\sum_{k=1}^m \lambda^k f_k(q;t).$$
    Then you minimize the corresponding action
    $$A[q,\lambda]=\int_{t_1}^{t_2} \mathrm{d} t \tilde{L}(q,\dot{q}\lambda;t).$$
    The variation with respect to the ##q## gives ##n## 2nd-order differential equations, and the variation with respect to the ##\lambda## gives you back the ##m## constraint equations. All together you have ##(n+m)## equations for ##(n+m)## unknown functions ##q## and ##\lambda##, which lead to unique solutions for the initial-value problem, if no pathologies occur.

    Try to solve your example, and it should become very clear immediately that, of course, you need the constraint equation to have a complete set of equations to be solved for the three independent degrees of freedom ##y##, ##\theta##, and ##\lambda##.

    In this case, it's of course also trivial to solve the constraint to eliminate one degree of freedom instead of introducing the Lagrange multiplier. You'll see that you get, of course, the same solutions and describe the same physics, using both methods.
     
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