I have been reading through "The Variational Principles of Mechanics" by Lanczos (if anyone is familiar with this text), and I am currently reading through the section discussing holonomic constraints. I think that I understand the concept pretty well, but when I encountered the example problem, I was a little confused. I think that I figured it out mostly, but I still have a few questions regarding the problem. The problem is:

Investigate the integrability of the following differential relation: $$ x dz + (y^2-x^2-z)dx + (z-y^2-xy)dy=0 \tag{16.9}$$

He says that the differential is indeed integrable with constraint $$z=x^2-xy+y^2.\tag{16.10}$$ It is not too hard to see that if one plugs in the given $z$, divides by $x$, and multiplies by a negative (although this step is unnecessary if the signs are consistent), then the differential will be of the form:

$$ -dz + (2x-y)dx + (2y-x)dy = 0 $$

Which is the correct differential given by taking the differential of the given $z$. We can also see that each

$$\frac{\partial B_i}{\partial q_k} = \frac{\partial B_k}{\partial q_i}.\tag{16.8}$$

Holds true in the final form of the differential. Given all of this information, my question is this: If we do indeed have a differential that is holonomic then the aforementioned requirement should hold true for each of the $B_i$'s and $q_k$'s. In the final form of the differential, as previously mentioned, we see that this is indeed true. However, with the initial form of the differential, we see that this is not the case. Why is this? I found this to be very confusing. Even attempting a solution with considering $z$ to depend upon $x$ and $y$, beginning with the initial differential we would have the equation:

$$ \frac{\partial}{\partial y} (y^2-x^2-z) = \frac{\partial}{\partial x} (z-y^2-xy)$$

Which leads to the equation:

$$ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 3y. $$

Which is clearly not true with the given solution for $z$. So what then is the proper way to evaluate whether a differential constraint is holonomic? Because it seems that, based on this example, if the solution is indeed correct, that we can't purely rely on the mechanical process of evaluating and comparing the partial derivatives of the $B_i$'s. If there is any way that I have formulated the problem incorrectly or misstepped at all, I would love to know!


1 Answer 1

  1. The type of non-holonomic constraint, that Ref. 1 is discussing at this point, is a so-called semi-holonomic constraint, which is a non-holonomic constraint given by a one-form $$ \omega~\equiv~\sum_{j=1}^na_j(q,t)~\mathrm{d}q^j+a_0(q,t)\mathrm{d}t~=~0. \tag{S}$$

  2. If there exist (i) a holonomic constraint $$f(q,t)~=~0,\tag{H}$$ (ii) an integrating factor $\lambda(q,t)\neq 0$ and (iii) a one-form $\eta$ such that $$ \lambda\omega+ f\eta~\equiv~\mathrm{d}f , \tag{I}$$ then the constraint (S) is equivalent to the holonomic constraint (H).

  3. Example. From the problem of Ref. 1, we have $$ \omega~\equiv~x\mathrm{d}z + (y^2-x^2-z)\mathrm{d}x + (z-y^2-xy)\mathrm{d}y,\tag{16.9} $$ $$ f~\equiv~z-x^2+xy-y^2,\tag{16.10} $$ $$ \lambda~\equiv~x^{-1}, $$ $$\eta~\equiv~x^{-1}\mathrm{d}(x-y).$$

  4. A closed 1-form satisfy closedness relations/Maxwell relations. It is locally exact, cf. the Poincare lemma. Returning to OP's specific question, the one-form (S) needs not satisfy Maxwell relations.

  5. Given a one-form (S) how do we detemine an underlying holonomic constraint (H)? Ref. 1 suggests to assume the constraint (H) can be put on graph form $$f(x,z)\equiv z-g(x),\tag{G}$$ where $g$ is the sought-for function. Similarly, normalize the one-form (S) so that it becomes of the form $$\omega\equiv \mathrm{d}z-B_i(x,z) \mathrm{d}x^i.\tag{N}$$ It follows from eq. (I) that the integrating factor is $$\lambda~\equiv ~1+ {\cal O}(f)$$ and that $$\mathrm{d}g(x)~\equiv~B_i(x,g(x))\mathrm{d}x^i\tag{O}$$ on-shell. Each Maxwell relations $$(i\leftrightarrow j)~=~\frac{d B_i(x,g(x))}{d x^j}~\equiv~\frac{\partial B_i(x,g(x))}{\partial x^j}+\frac{\partial B_i(x,g(x))}{\partial z}\frac{\partial g(x)}{\partial x^j} \tag{M}$$ is a first-order quasi-linear PDE for the sought-for function $g$. $\Box$

    Concretely OP is unable to conclude any Maxwell relations in their example because they skipped the normaliation step (N).


  1. C. Lanczos, The variational principles of mechanics, 4th eds, 1970; p. 26.
  • $\begingroup$ So then, based on what you're saying, we can't just mechanically evaluate the partial derivatives and equate them? Because some instances might require an integrating factor? $\endgroup$
    – user132849
    Feb 24, 2020 at 7:41
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Feb 24, 2020 at 7:42
  • $\begingroup$ Haha so then using only the partial derivatives, it is then impossible to tell apart the necessity of an integrating factor from that of a totally unintegrable differential? $\endgroup$
    – user132849
    Feb 24, 2020 at 8:13
  • $\begingroup$ Great! Thanks for the clarification! However, I am still curious about the partial derivative question. $\endgroup$
    – user132849
    Feb 25, 2020 at 0:57
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Feb 27, 2020 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.