Differential holonomic constraint is an integrable homogeneous first order differential equation: $$\sum_{i}\mathcal{E}_{i}(q)\frac{dq_{i}}{d\tau}=0;$$ in which $\sum_{i}\mathcal{E}_{i}(q)dq_{i}$ is a complete differential, that is the conditions $$\partial\mathcal{E}_{i}(q)/\partial q_{j}= \partial\mathcal{E}_{j}(q)/\partial q_{i}$$ are met for all $i, j$.
In case when the potential function $E(q)$ for the corresponding form, such that: $$\mathcal{E}_{i}(q)=\partial E(q)/\partial q_{i};$$ is known, the general one-parametric solution of the above differential equation is obviously: $$E(q_{1},...q_{n})=Const;$$

  1. What can we say about the solution of this differential equation when the potential function is not known, can not be given in a closed explicit form (does not exist)?
    Can we still expect in this case that the dimension of the configuration space is effectively reduced by this constraint, or maybe even there still exist in some implicit sense a general solution, as a one-parametric family of relations between variables $q_{i}$?

  2. If so, than can we find / build those solutions?

  3. How to deal with such a constraint? We can not just add it in its differential form to the Lagrangian as a Lagrangian multiplier, as we can do for both explicit holonomic and non-holonomic constraints.

I appreciate your suggestions, specific references.


1 Answer 1


The case that $E$ does not exist cannot happen on topologically trivial configuration spaces, and even then, it exists locally by the Poincaré lemma.

Your "complete conditions" say nothing but that the form $\mathcal{E}=\mathcal{E}_i\mathrm{d}q^i$ is closed, i.e. $\mathrm{d}\mathcal{E} = 0$. By the Poincaré lemma, every closed form is locally exact, i.e. there is some function $E$ such that $\mathcal{E} = \mathrm{d}E$.

So, on a topologically non-trivial configuration space, it can happen that a constraint given by a closed form is not exact, so it will not be holonomic.

This doesn't say much about your constraint, it says something about the configuration space - it has topological defects for which the system will change when it travels "around" them. If your configuration space is topologically non-trivial, many of the standard methods only work locally. However, finding local solutions to the equations of motion is enough, since you can glue them together. So, just restrict to a contractible neighbourhood around your initial conditions, and write the Lagrangian with the constraint added by a Lagrange multiplier as is standard. Solve those local equations of motion. When you near the boundary of that neighbourhood, take the endpoint of this local solution to the e.o.m., and choose another contractible neighbourhood around that endpoint to propagate the solution. Repeat.

  • $\begingroup$ Thank you for your response. It looks like it answers my first question - the potential function $E(q)$ exists (in topologically trivial neighborhood of configuration space). This gives ground to the expectation that a differential holonomic constraint always reduces the dimension of the configuration space. Though I afraid I still do not understand quite well what kind of existence it presumes. If there is no way to represent this potential function in any explicit (integrated) form, what can we learn about its characteristics, like its smoothness, area of existence, ... ? Thank you. $\endgroup$
    – MirOdin
    Jan 23, 2016 at 23:33
  • $\begingroup$ Still my third question remains open for me. === The problem is: If no explicit form for $E(q)$ is known, or can even exist, is there any way to actually take this constraint into account in Lagrange formalism, say any form in which it could actually be added to the Lagrangian with a Lagrange multiplier... ? $\endgroup$
    – MirOdin
    Jan 23, 2016 at 23:59
  • $\begingroup$ @peth: The proof of the Poincarè lemma is constructive and contains your answers. For a closed 1-form $\omega$, you pick any point $x_0$, and then, for every other point $x$, define $f(x) = \int_\gamma \omega$ where $\gamma$ is any path from $x_0$ to $x$. Since the form is closed, this will not depend on the path chosen if the space is contractible. Now, $\mathrm{d}f = \omega$. So given the 1-form, one can write down an expression for $f$. It's smooth (since forms are generally smooth), and it exists on every contractible subset of your space. $\endgroup$
    – ACuriousMind
    Jan 24, 2016 at 13:51
  • $\begingroup$ Great, this will do! Thank you very much for your time. Sorry, do not have enough reputation to upvote. @:) $\endgroup$
    – MirOdin
    Jan 24, 2016 at 17:53

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