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Many ODE's and PDE's arising in nature have a variational formulation. An example of what I mean is the following. Classical motions are solutions $q(t)$ to Lagrange's equation $$ \frac{d}{dt}\frac{\partial L(q,\dot q)}{\partial\dot q}=\frac{\partial L(q,\dot q)}{\partial q}, $$ and these are critical points of the functional $$ I(q)=\int L(q,\dot q)dt. $$ Of course one needs to be precise with what considers a solution to both equations. This amounts to specifying regularity and a domain of the functional. This example is an ODE, but many PDE examples are possible as well (for example electromagnetism, or more exotic physical theories). Once one knows a variational description of the problem, many more methods are available to solve the problem.

Now I do not expect that any PDE or ODE can be viewed (even formally) as a critical point of a suitable action functional. This is because this whole set up reminds me of De Rham cohomology: "which one-forms (the differential equations) are exact (that is, the $d$ of a functional)?". The last sentence is not correct, but the analogy maybe is? Anyway, my question is:

Are there any criteria to determine if a given differential equation admits a variational formulation?

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I'm sure that the answer is yes and that I have read it once. But where ? –  Denis Serre Jul 5 '12 at 13:33
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Ricci flow was not known to be a gradient flow until Perelman introduced his F-functional. I guess the setting was more complicated here, because there is a big gauge group of diffeomorphisms, but thats an example of where people were not sure if it could be given a variational formulation until it was done so. –  Otis Chodosh Jul 5 '12 at 15:02
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expanding a bit on the comment of Otis: It was of course know that the Ricci flow is not the gradient flow in a strict sense of a diffeo-invariant functional (reason: gradients of diffeo-invariant functionals are always divergence free, the Ricci tensor is not divergence free), so the amazing and surprising discovery of Perelman was that the Ricci flow actually is a gradient flow up to diffeomorphisms. –  Robert Haslhofer Jul 5 '12 at 16:05
    
Differential equations with a Hamiltonian formulation, automatically have a Lagrangian formulation (formally, at least). Of course, this simply moves the difficulty over to the question of when one can determine that an equation has a Hamiltonian formulation, but in practice, having a conserved integral of motion tends to be a useful clue in this regard... –  Terry Tao Jul 5 '12 at 19:49
    
Regarding Ricci flow: in retrospect, the existence of gradient shrinking Ricci solitons, and the absence of periodic-up-to-diffeomorphisms "breather" type solutions, was an indication that Ricci flow had a gradient-flow-up-to-diffeomorphism interpretation. One could also build upon this observation to try to guess the functional that gives this gradient flow, by writing down the integrals that vanish for gradient shrinking solitons, and then trying to express those integrals as a gradient-up-to-diffeomorphisms of something, though this is still far from an easy task... –  Terry Tao Jul 5 '12 at 19:53

3 Answers 3

up vote 17 down vote accepted

Others give useful references that discuss what is known about the answer, but no statement of the answer itself. The relevant algebraic setting is the variational bicomplex, which is discussed in the works of Anderson and others. In this setting, there are two differentials, the horizontal differential $d_H$ (representing derivatives with respect to independent variables like $t$) and the vertical differential $d_V$ (representing variational derivatives with respect to dependent variables like $q(t)$). Each of these differentials is "de Rham-like" and they anticommute with each other, which explains the cohomological flavor of the answer. A rough statement of the answer is as follows.

A Lagrangian $L$ density gives rise to a set of Euler-Lagrange equations $E_i=0$ as follows: $$ d_V L = E_i ~ d_V q^i - d_H\theta , $$ that is, the vertical 1-form $E_i ~ d_V q^i$ is vertically exact (up to a horizontally exact term $d_H \theta$). So, it is necessary for $E_i=0$ to be the Euler-Lagrange system of some Lagrangian that $E_i ~ d_V q^i$ is closed up to a horizontally exact term, namely $$ d_V(E_i~d_V q^i) = d_H \theta' (= -d_H d_V \theta) . $$ In fact, the same condition is also sufficient, up to obstructions related to the global topology of the manifold where the dependent variables $q$ take their values. This condition was formulated already classically by Helmholtz.

However, the above statement is restrictive in that it answers the question only when $E_i=0$ is already in Euler-Lagrange form. However, there are many transformations that one can apply to the system $E_i=0$ that gives an equivalent system $F_a=0$. Given only the system $F_a=0$, is it still possible to decide whether it is equivalent to some system $E_i=0$ in Euler-Lagrange form? This is the hard inverse problem (aka the multiplier problem). The only general result that I'm aware of in that direction is this.

If there exists a form $\omega$ of vertical degree 2 and horizontal degree $n-1$, where $n$ is the number of independent variables, such that it is both horizontally and vertically closed modulo the equations $F_a=0$ (namely $d_V \omega = A^a F_a$ and $d_H \omega = B^a F_a$), then there exists (again, up to global topological obstructions) a Lagrangian density $L$ whose Euler-Lagrange equations $E_i=0$ are equivalent to a subsystem of $F_a=0$.

To my knowledge, the above observation first appeared in Henneaux (AnnPhys, 1982) for ODEs and in Bridges, Hydon & Lawson (MathProcCPS, 2010) for PDEs. The calculation demonstrating this observation is given in a bit more detail on this nLab page. (Edit: At risk of shameless self-promotion, I'll also note that I collected these observations in a self-contained paper (arXiv; JMP, 2013).)

It reduces the solution of the hard inverse problem to classifying all such forms $\omega$ (corresponding to the so-called characteristic cohomology of the variational bicomplex restricted to $F_a=0$ in the corresponding degree) and checking that there exists a candidate that gives rise to a Lagrangian density whose Euler-Lagrange system $E_i=0$ is equivalent to the full system $F_a=0$. The calculation of the corresponding characteristic cohomology of the system $F_a=0$ is still non-trivial, but there exist ways of attacking it, which include Vinogradov's $\mathcal{C}$-spectral sequence mentioned in other responses.

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Nice answer! Since you mentioned the har inverse problem, I'll complete your answer a bit bu mentioning that the problem of determining when a system of paths---one for every tangent direction---correspond to extremals of a variational problem is quite interesting and goes back to Darboux, Hilbert (his fourth problem, for example), Douglas (3 dimensions), and has been worked on by very many people. There is a huge gap between general results (your boxed theorem) and concrete cases like Hilbert's fourth problem. –  alvarezpaiva Jul 6 '12 at 10:49
    
Thank you very much for this answer. I will ponder on it. –  Thomas Rot Jul 8 '12 at 8:25

There is a huge amount of literature on this problem. I include some works that seem "classic" and that I've consulted at some point:

Tulczyjew: http://www.springerlink.com/content/u9481124734547t6/ 105_419_0">http://www.numdam.org/numdam-bin/fitem?id=BSMF_1977_105_419_0

Takens: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jdg/1214435235

And the work of Anderson on the variational bicomplex (I think this complex was introduced by I.M. Gelfand circa 1970).

There is also quite a bit of good work by Vinogradov and his school (the C-spectral sequence). All this is mostly applications of homological algebra to the theory of PDE's.

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Thank you for the references, I'll look into it. –  Thomas Rot Jul 8 '12 at 8:26

I remember idly wondering about this once and found that Anderson's work on the variational bicomplex satisfied my idle curiosity.

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