To make it plain and simple, if I have a holonomic constraint, that I want to treat using a lagrange multiplier, in any textbook I concern, they are just expressed as "$\lambda$" (omitting possible arguments). What I'd like to know is whether a lagrange multiplier is something like "$\lambda(t)$", that would be $\lambda$ to be an additional degree of freedom, whose time dependence is not known yet. Or instead, is the lagrange multiplier a (yet to exactly determine) extension to the lagrange function, and as such "$\lambda(q, \dot{q}, t)$"?

I will expand my thoughts here on why I think that both versions do work out: Starting with "$\lambda(q, \dot{q}, t)$", The complete Lagrangian of the System would be $$ \mathcal L = L(q, \dot{q},t) - \lambda(q, \dot{q},t) f(q,t) $$ Variations in the paths $\delta q$ and the requirement that $f(q(t),t)=0$ holds, will yield the correct EOM: $$ \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot{q}} (q(t),\dot{q}(t),t) - \frac{\partial \mathcal L}{\partial q} (q(t),\dot{q}(t),t) = \\ \frac{d}{dt} \frac{\partial \lambda }{\partial \dot{q}} (q(t),\dot{q}(t),t) - \frac{\partial \lambda}{\partial q} (q(t),\dot{q}(t),t))f(q(t), t) + \frac{\partial \lambda}{\partial \dot{q}}(q(t),\dot{q}(t),t) \frac{d}{dt}f(q(t),t) + \lambda(q(t), \dot{q}(t), t) \frac{\partial f}{\partial q} $$

Applying the constraints, $f = 0$ and $\frac{d}{dt} f = 0$, then we get:

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} (q(t),\dot{q}(t),t) - \frac{\partial L}{\partial q} (q(t),\dot{q}(t),t) = + \lambda(q(t), \dot{q}(t), t) \frac{\partial f}{\partial q} $$ And $$ f(q(t), t) = 0 $$ which are just the equations that describe the motion of the system.

Alternatively, (as said), I can treat the multipliers as additional degrees of freedom in the configuration-space. The total lagrangian, dependent on $q$, $\dot{q}$, $\lambda$, $\dot{\lambda}$ (this dependence is just listed here for completeness, the total lagrangian won't depend on $\dot{\lambda}$), and t, is: $$ L(q, \dot{q},t) - \lambda(t) f(q,t) $$ Lagrange's equations then will be: $$ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}}(q(t), \dot{q}(t),t) - \frac{\partial L}{\partial q} (q(t), \dot{q}(t), t) = \lambda(t) \frac{\partial f}{\partial q}(q(t), t) $$ And $$ f(q(t), t) = 0 $$

Both methods, although using different assumptions, yield the same equations of motion. Which one is more feasible? Are there cases where my reasoning doesn't work, which favor one of the two options I have given?


2 Answers 2


What you noticed is that there can be a dual description of lagrangian systems under holonomic constraints.

Consider a lagrangian system with $n$ variables and $m$ holonomic constraints. Lagrange multipliers can actually be viewed simply as undetermined functions chosen so that $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}-\sum_a\lambda_a\frac{\partial f_a}{\partial q_i}=0,$$ for $m$ of the $q_i$. The remaining $n-m$ variables can be varied independently and the problem has $n-m$ degrees of freedom.

On the other hand, lagrange multipliers can also be seen as new independent variables, $\lambda_a(t)$. Given a lagrangian $L(q,\dot q,t)$ and $m$ holonomic constraints $f_a(q,t)=0$ then we can formulate a dual problem which consists on the unconstrained dual lagrangian $$\tilde L(q,\dot q,\lambda,t)=L(q,\dot q,t)+\sum_a \lambda_a(t)f_a(q,t).\tag 1$$ The lagrange multipliers $\lambda_a(t)$ are then considered new independent variables in a variational problem with $n+m$ degrees of freedom. Variation with respect to $q_i$ gives the dynamic equations, $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=\sum_a\lambda_a\frac{\partial f_a}{\partial q_i},$$ while the variation with respect to $\lambda_a$ give the constraint equations, $$f_a(q,t)=0.$$

Note that this dual description does not hold for non-holonomic constraints since we cannot write a dual lagrangian such as (1) due to the lack of the constraint equations of the form $f_a=0$

  • $\begingroup$ So in principle, you say that as long as one of the versions is possible, it isn't favored over the other one? $\endgroup$ May 22, 2017 at 6:51
  • $\begingroup$ They are completely equivalent even from the practical point of view. I just prefer the second because it is easier to remember the procedure. It may also be slightly more elegant. $\endgroup$
    – Diracology
    May 22, 2017 at 13:50

Conventionally, the counting goes as follows: There are as many Lagrange multipliers $\lambda_i(t)$ as there are constraints $f^i(t)\approx 0$. Since a constraint $f^i(t)\approx 0$ should be satisfied at each instant $t$, there is one Lagrange multiplier $\lambda_i(t)$ for each instant $t$. Equivalently: the Lagrange multipliers $\lambda_i(t)$ depends on time $t$.


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