17
$\begingroup$

I am looking for an explicit (preferably simple) example of an ODE with time-independent coefficients in $\mathbb{R}^3$ such that there does not exist an Euler-Lagrange equation $$\frac{\partial L}{\partial q^i}=\frac{d}{dt}\frac{\partial L}{\partial \dot q^i}, \, \, i=1,2,3,$$ with the same solutions.

I prefer to distinguish two cases: either $L$ depends explicitly on time or not.

$\endgroup$
3
  • 4
    $\begingroup$ Isn't Euler Lagrange equation equivalent to Hamiltonian's and therefore implies conservation laws ? So I guess anything with a dissipative term, for example $\frac{d^2}{dt^2}q = -\frac{d}{dt}q$ should be a counter example. $\endgroup$ – RaphaelB4 Dec 29 '20 at 10:29
  • 7
    $\begingroup$ @RaphaelB4: That argument only works for time-independent Lagrangians. For example, the Euler-Lagrange equation for the functional $\int \mathrm{e}^t \dot q^2\,\mathrm{d}t$ is $\ddot q + \dot q = 0$. $\endgroup$ – Robert Bryant Dec 29 '20 at 10:37
  • 1
    $\begingroup$ The article "Generalized Classical Dynamics, and the 'Classical Analogue' of a Fermi Oscillator", J. L. Martin, Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences Vol. 251, No. 1267 (Jun. 23, 1959), pp. 536-542, discusses examples of classical Hamiltonian systems possessing no Lagrangian formulation. $\endgroup$ – Konstantinos Kanakoglou Dec 29 '20 at 21:15
22
$\begingroup$

Note: I'm updating my answer to give a better (i.e., simpler) example plus a little more information about how to derive the example from Douglas' results (which may not be entirely clear upon first reading of his paper). This also addresses the question of time-dependent Lagrangians originally raised by the OP.

Have a look at Jesse Douglas' paper Solution of the inverse problem of the calculus of variations, Trans. Amer. Math. Soc. 50 (1941), 71–128. In this paper, Douglas derives necessary conditions for a system of second order equations in any number of dependent variables to be the Euler-Lagrange equations of a first-order functional. He shows how to reduce the problem to an overdetermined linear system whose compatability can be checked by differentiation.

For example, it's a consequence of his results that the system $$ \ddot x = 0,\quad\ddot y = 0,\quad \ddot z = (y^2+z^2)\tag1 $$ is not equivalent (in the sense of having the same solutions) to the Euler-Lagrange equations for any nondegenerate first-order functional $$ \int \phi(t,x,y,z,\dot x, \dot y,\dot z)\,\mathrm{d}t,\tag2 $$ where 'nondegenerate' has the usual meaning that the Hessian of $\phi$ with respect to the variables $(\dot x, \dot y,\dot z)$ is invertible.

What Douglas does is show that the problem of finding such a $\phi$ is equivalent to finding a nondegenerate solution of an overdetermined linear system of equations for the components of the Hessian of $\phi$ with respect to the variables $(\dot x, \dot y,\dot z)$. This is his system (4.7–10) together with the nondegeneracy condition (4.11).

For the given right-hand sides in the above system (1), one easily finds that Douglas' system (4.7–10) implies $$ \frac{\partial^2\phi}{\partial\dot x\,\partial\dot z} =\frac{\partial^2\phi}{\partial\dot y\,\partial\dot z} =\frac{\partial^2\phi}{\partial\dot z\,\partial\dot z} = 0, $$ which implies that $\phi$ cannot be nondegenerate.

Remark: Most of Douglas' paper concerns explicitly working out, in the case of two dependent variables, the consequences of the criterion he derives in Part II for an arbitrary number of dependent variables. Part II is quite short and readable while the later parts are much more technical.

$\endgroup$
5
  • $\begingroup$ Thanks very much. To make sure: do I understand correctly that $L$ does not depend on time here? $\endgroup$ – makt Dec 29 '20 at 11:29
  • 1
    $\begingroup$ @makt: Yes, I left out time because I wasn't sure that this particular example (which I happened to remember) is not equivalent to the E-L equation of some time-dependent Lagrangian. However, Douglas' methods do extend to the time-dependent case, and I imagine that, if Douglas didn't give such an example explicitly in that paper, it wouldn't be hard to construct one using his methods. $\endgroup$ – Robert Bryant Dec 29 '20 at 11:38
  • 1
    $\begingroup$ @makt: I have revised my answer so that it covers the case of a time-dependent Lagrangian as well. Also, I have given some indications of what parts of Douglas' paper need to be understood in order to verify the above example. (The bulk of the paper is technical and concentrates on the case of 2 dependent variables, but you don't need to read that to understand the method and to verify the above example.) $\endgroup$ – Robert Bryant Dec 30 '20 at 13:15
  • $\begingroup$ Many thanks. By the way, is it local statement, i.e. true in any open subset of $\mathbb{R}^3$? $\endgroup$ – makt Dec 30 '20 at 13:53
  • 2
    $\begingroup$ @makt: Yes, it is a purely local statement, i.e., there is no non-degenerate first-order Lagrangian for system (1) on any open subset of $\mathbb{R}\times T\mathbb{R}^3$. Note by the way, that Douglas claims that his examples are the first known explicit examples of non-E-L second order systems, though, as he points out, it had been supposed for a long time that such examples existed, on the basis of counting 'arbitrary functions' in the equations versus 'arbitrary functions' in the Lagrangian. $\endgroup$ – Robert Bryant Dec 30 '20 at 14:02
2
$\begingroup$

If you're willing to consider PDEs, I coauthored an explicit example of a PDE in a gauge field $A$ that fails to be the Euler-Lagrange equation arising from any gauge-invariant lagrangian involving $A$ alone in arXiv:1501.07548.

The PDE in question is formula (12), which is a deformation of the (topologically massive, for $\mu\neq 0$) 3D Yang-Mills equation of motion. The relevant term is the one involving the mass parameter $m$; for $m\to\infty$, we find (topologically massive) Yang-Mills. To avoid reproducing most of that (short) paper here, I will only sketch the proof that this is not an EL equation for any gauge-invariant lagrangian: the PDE is gauge invariant under standard Yang-Mills type gauge transformations, but fails to satisfy a Noether identity. However, gauge-invariant PDEs arising as EL equations always satisfy Noether identities; contradiction.

$\endgroup$
3
  • 1
    $\begingroup$ The equation could still come from a Lagrangian that is not gauge invariant, or not? $\endgroup$ – Tobias Diez Dec 29 '20 at 22:47
  • $\begingroup$ @TobiasDiez: indeed! We were careful to point that out at the time, but I was not so careful just now. I'm editing accordingly $\endgroup$ – AlexArvanitakis Dec 29 '20 at 23:33
  • 2
    $\begingroup$ For PDE, it's easier to give examples of non-Euler-Lagrange equations. For example, in $n\ge 2$ independent variables $x = (x^i)$ and one dependent variable $z$, the equation $$\Delta z = f(x,z,\nabla z)$$ is (equivalent to) the E-L equation of some 1st order Lagrangian iff $f$ can be written in the form $$f(x,z,\nabla z)=b_z(x,z)\,|\nabla z|^2 + 2 \nabla_x b(x,z)\cdot \nabla z + a(x,z)$$ for some functions $a$ and $b$ of $x$ and $z$. See Exterior differential systems and Euler-Lagrange partial differential equations, by RL Bryant, PA Griffiths, DA Grossman, CUP, 2003 or arXiv:math/0207039. $\endgroup$ – Robert Bryant Dec 31 '20 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.