Introduction to Mathematical Physics/Some mathematical problems and their solution/Use of change of variables

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Normal forms[edit]

As written in ([ma:equad:Arnold83]) it is very powerfull not to solve differential equations but to tranform them into a simpler differential equation. ([ma:equad:Arnold83],[ma:equad:Guckenheimer83],[ma:equad:Berry78]) Let the system:

and a fixed point of the system: . Without lack of generality, we can assume . Assume that application can be develloped around :

where the dots represent polynomial terms in of degree . There exists the following lema:



Let be a vectorial polynom of order and . The change of variable transforms the differential equation into the equation:

where and where the dots represent terms of order .

Note that is the Poisson crochet between and . We note and we call the following equation:

the homological equation associated to the linear operator .

We are now interested in the reverse step of theorem lemplo: We have a nonlinear system and want to find a change of variable that transforms it into a linear system. For this we need to solve the homological equation, {\it i.e.} to express as a function of associated to the dynamics.

Let us call , the basis of eigenvectors of , the associated eigenvalues, and the coordinates of the system is this basis. Let us write where contains the monoms of degree , that is the terms , being a set of positive integers such that . It can be easily checked (see [ma:equad:Arnold83]) that the monoms are eigenvectors of with eigenvalue where :

One can thus invert the homological equation to get a change of variable that eliminate the nonom considered. Note however, that one needs to invert previous equation. If there exists a with and such that , then the set of eigenvalues is called resonnant. If the set of eigenvalues is resonant, since there exist such , then monoms can not be eliminated by a change of variable. This leads to the normal form theory ([ma:equad:Arnold83]).

KAM theorem[edit]

An hamiltonian system is called integrable if there exist coordinates such that the Hamiltonian doesn't depend on the .


Variables are called action and variables are called angles. Integration of equation eqbasimom is thus immediate and leads to:

and where and are the initial conditions.

Let an integrable system described by an Hamiltonian in the space phase of the action-angle variables . Let us perturb this system with a perturbation .

where is periodic in .

If tori exist in this new system, there must exist new action-angle variable such that:


Change of variables in Hamiltonian system can be characterized ([ph:mecac:Goldstein80]) by a function called generating function that satisfies:

If admits an expension in powers of it must be:

Equation eqdefHip thus becomes:


Calling the frequencies of the unperturbed Hamiltionan :

Because and are periodic in , they can be decomposed in Fourier:

Projecting on the Fourier basis equation equatfondKAM one gets the expression of the new Hamiltonian:

and the relations:

Inverting formally previous equation leads to the generating function:

The problem of the convergence of the sum and the expansion in has been solved by KAM. Clearly, if the are resonnant (or commensurable), the serie diverges and the torus is destroyed. However for non resonant frequencies, the denominator term can be very large and the expansion in may diverge. This is the {\bf small denominator problem}.

In fact, the KAM theorem states that tori with ``sufficiently incommensurable frequencies[1]

are not destroyed: The series converges[2].

  1. In the case two dimensional case the KAM theorem proves that the tori that are not destroyed are those with two frequencies and whose ratio is sufficiently irrational for the following relation to hold:

    where is a number that tends to zero with the .

  2. To prove the convergence, KAM use an accelerated convergence method that, to calculate the torus at order uses the torus calculated at order instead of the torus at order zero like an classical Taylor expansion. See ([ma:equad:Berry78]) for a good analogy with the relative speed of the Taylor expansion and the Newton's method to calculate zeros of functions.