Although this questions is very much math related, I posted it in Physics since it is related to variational (Lagrangian/Hamiltonian) principles for dynamical systems. If I should migrate this elsewhere, please tell me.

Often times, in graduate and undergraduate courses, we are told that we can only formulate the Lagrangian (and Hamiltonian) for "potential" systems, where in the dynamics satisfy the condition that: $$ m\ddot{\mathbf{x}}=-\nabla V $$ If this is true, we can formulate a functional which is stationary with respect to the system as: $$ F[\mathbf{x}]=\int^{t}_0\left(\frac{1}{2}m\dot{\mathbf{x}}(\tau)^2-V(\mathbf{x}(\tau))\right)\,\text{d}\tau $$

Taking the first variation of this functional yields the dynamics of the system, along with a condition that effectively states that the initial configuration should be similar to the final configuration (variation at the boundaries is zero).

Now, given the functional: $$ F[\mathbf{x}]=\frac{1}{2}[\mathbf{x}^{\text{T}} * D(\mathbf{x})]+\frac{1}{2}[\mathbf{x}^{\text{T}} * \mathbf{Ax}]-\frac{1}{2}\mathbf{x}'(0)\mathbf{x}(t) $$ With $\mathbf{A}$ symmetric and $\mathbf{x}(0)$ being the initial condition, and: $$ [\mathbf{f}^{\text{T}} * \mathbf{g}]=\int^{t}_0 \mathbf{f}^{\text{T}}(t-\tau)\mathbf{g}(\tau)\,\text{d}\tau $$

If we take the first variation and assume only that the initial variation is zero, the functional is stationary with respect to: $$ \frac{d\mathbf{x}(t)}{dt}= \mathbf{Ax}(t) $$

This is a functional derived by Tonti and Gurtin, it represents a variational principle for linear initial value problems with symmetric state matrices and shows, as a proof of concept, that functionals can be derived for non-potential systems, initial value or dissipative systems.

My question is, is it possible to derive these functionals for arbitrary nonlinear systems which do not have similar initial and final configurations (and cannot have similar initial and final configurations due to dissipation)?

What sorts of conditions would exists on the dynamics of these systems?

In this example, $\mathbf{A}$ must be symmetric which already implies all of it's eigenvalues are real and thus it is a non-potential system, but there is still a functional which can be derived for it.

Any related sources, information, or answers regarding specific cases would be appreciated. If anyone needs clarification, or a proof of any result I presented here, let me know.

Edit: Also, a related question anyone seeing this: I'm currently just interested in the abstract aspect of the problem (solving/investigating it for the sake of it), but why are functional representations such as these useful? I know there are some numerical application, but if I have a functional which attains a minimum for a certain system, what can I do with it?

  • 2
    $\begingroup$ Friction and dissipation are non-variational, see e.g. this post. There are "Lagrangian" formulations for dissipative forces, but they do not obey a naive principle of least action, see this post and this paper $\endgroup$
    – ACuriousMind
    Commented Feb 25, 2015 at 17:45
  • $\begingroup$ @ACuriousMind: I will review the posts and paper you linked to, I'm not sure what you mean yet, but I'll come back by the end of today to reply with more specificity. Also, another thing I want to mention: Rayleigh has a method for dealing with dissipation and external forcing, but I am specifically looking for a self contained formulation with no disspation function necessary, just one functional. $\endgroup$
    – Ron
    Commented Feb 25, 2015 at 18:05
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    $\begingroup$ The first post I linked shows a general method for deciding whether there is a Lagrangian description for a system given by differential equations. It is definite - there is no Lagrangian description of generic dissipative force (though you might be able to cheat in specific situations). The paper discusses how an "extended Lagrangian description" whose Lagrangian is not the sum of potentials and kinetic energies may be set up to model disspative forces. $\endgroup$
    – ACuriousMind
    Commented Feb 25, 2015 at 18:10
  • $\begingroup$ @ACuriousMind: I'm not necessarily looking for a Lagrangian, I'm asking a more general question about the existence of ANY functional which may be stationary with respect to ANY system. The other related question may be: when can you "cheat" and why? What sorts of conditions are there on systems where you can "cheat"? $\endgroup$
    – Ron
    Commented Feb 25, 2015 at 18:16
  • $\begingroup$ As Qmechanic writes: "This opens up a lot of possibilities, and it can be very difficult to systematically find an action principle; or conversely, to prove a no-go theorem that a given set of eoms is not variational." I think we don't have an answer to your question in such generality. $\endgroup$
    – ACuriousMind
    Commented Feb 25, 2015 at 18:19

1 Answer 1


I) The Gurtin-Tonti bi-local method [which OP mentions in an example; see also Section II below] of pairing opposite times $t\leftrightarrow (t_f-t_i)-t$ (hidden inside a convolution) is an artificial trick from a fundamental physics point of view, unless further justified. Why would such correlations into the past/future take place?

In fact, it may have non-local quantum mechanical consequences if such non-local action is supposed to be used in a path integral formalism.

Also the Gurtin-Tonti convolution method does not work for a non-compact time interval $[t_i,t_f]$, i.e. if $t_i=-\infty$ or $t_f=\infty$.

Most fundamental physics models typically obey locality, but there are various non-local proposals on the market.

The question of whether a certain set of equations of motions $E_i(t)$ has an action principle (or not!) can be very difficult to answer, and is often an active research area, cf. e.g. this Phys.SE post.

Also what constitutes an acceptable action principle? E.g. can we just introduce some Lagrange multipliers $\lambda^{i}(t)$ and an action $S=\int\! dt ~\lambda^i(t) E_i(t)$ so that $\delta S/ \delta\lambda^i(t) = E_i(t)$, and call it a day? Or are we not allowed to introduce auxiliary variables or non-locality? Should it satisfy a minimum principle rather than a stationary principle? And so forth.

II) Example. Let us for simplicity consider the unit time interval $[t_i,t_f]=[0,1]$. A symmetrized version of the Gurtin-Tonti model is the following bi-local action

$$S[q]~:=~ \frac{1}{4}\iint_{[0,1]^2} \!dt~du~\left\{ q^i(t) \left(\frac{dq^i(u)}{du}- A_{ij}(t,u) q^j(u)\right)+(t\leftrightarrow u) \right\}\delta(t+u-1) $$ $$~=~\frac{1}{2}\int_{[0,1]} \!dt~\left\{\frac{1}{2} q^i(1\!-\!t) \frac{dq^i(t)}{dt}-\frac{1}{2}q^i(t) \frac{dq^i(1\!-\!t)}{dt}- q^i(1\!-\!t)A_{ij}(1-t,t) q^j(t) \right\} $$ $$~=~\frac{1}{2}\int_{[0,1]} \!dt~\left\{ q^i(1\!-\!t) \frac{dq^i(t)}{dt}- q^i(1\!-\!t)A_{ij}(1-t,t) q^j(t) \right\} \tag{1}$$

with symmetric matrix

$$ A_{ij}(t,u) ~=~A_{ji}(u,t) .\tag{2}$$

Interestingly, the boundary contributions in the variation $\delta S$ cancel without imposing any boundary conditions (BC). In other words, as far as the finding stationary solutions, we may assume that the variables $q^i$ are free at both end points. (However, there might be other reasons to impose BCs.)

The functional derivative

$$\frac{\delta S[q]}{\delta q^i(t)}~=~\left.\left\{\frac{dq^i(u)}{du}- A_{ij}(t,u) q^j(u)\right\}\right|_{u=1-t}. \tag{3} $$

Hence the equations of motion become

$$ \frac{dq^i(t)}{dt}~\approx~A_{ij}(1\!-\!t,t) q^j(t). \tag{4}$$

III) Let us for completeness mention that a related method is the classical Schwinger/Keldysh "in-in" formalism, cf. e.g my Phys.SE answer here.


  1. V. Berdichevsky, Variational Principles of Continuum Mechanics: I. Fundamentals, 2009; Appendix B.
  • $\begingroup$ I think I see what you mean. Effectively, you're making the argument that mixing these opposite times doesn't make sense from a physical standpoint? What kind of justification would make if more reasonable? Take for example this paper: arxiv.org/abs/1112.2286 here, the author uses fractional derivatives to formulate a variational principle. Fractional derivatives are also non-local quantities, and there is something to be said about processes with friction being non-local in a sense, as they are path dependent. $\endgroup$
    – Ron
    Commented Feb 26, 2015 at 16:49
  • 1
    $\begingroup$ I see what you mean about the issue of locality. It's interesting to note that in the frequency domain, the convolution takes on a local form and the inner product does the opposite. $\endgroup$
    – Ron
    Commented Mar 6, 2015 at 23:33
  • $\begingroup$ The convolution could also be useful for describing dissipative processes specifically because of it's non-local nature. Sort of a way of saying "I'm currently at time $t$, now let me convolve all of the information about my state $t$ time in the past with the information going back to that time.". If you look at the requirements for a system to be able to work in the convolutive framework with path independence, for linear systems only systems with symmetric state matrices work (which only have real eigenvalues, which can be dissipative). $\endgroup$
    – Ron
    Commented Mar 6, 2015 at 23:34
  • $\begingroup$ Also, there are ways to formulate local variational principles for dissipative systems, my only worry would be that the configuration of the system isn't truly the same at the beginning and the end, so how can one justify saying the variation is zero at both endpoints? $\endgroup$
    – Ron
    Commented Mar 6, 2015 at 23:36
  • $\begingroup$ Further references: C.R. Galley, arxiv.org/abs/1210.2745 The trick here is briefly 1. double up $q$s. 2. $L=L_1-L_2+\ldots$. 3. $q_-$ acts like a Lagrange multiplier that imposes the eom for $q_+$. $\endgroup$
    – Qmechanic
    Commented Dec 9, 2018 at 23:24

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