Normal form for the non linear Schrödinger equation

Transcription

1 Normal form for the non linear Schrödinger equation joint work with Claudio Procesi and Nguyen Bich Van Universita di Roma La Sapienza S. Etienne de Tinee 4-9 Feb. 2013

2 Nonlinear Schrödinger equation Consider the Nonlinear Schrödinger equation on the torus T n. where u := u(t, ϕ), ϕ T n, F(y) is an analytic function, F(0) = 0 no dependence on ϕ. A good model is iu t u = F( u 2 )u (1) iu t u = F( u 2 )u = u 2q u (2) with q N. I will concentrate on the CUBIC CASE namely q = 1 iu t u = u 2 u

3 The case of the cubic NLS in dimension one is integrable but - the non-cubic NLS in dim n 1 -cubic NLS in dim n > 1 should exhibit a very rich dynamics. Marcel Guardia s lectures: growth of Sobolev norms. Grebert-Thomann: quintic NLS in dimension 1. Geng-You-Xu: cubic NLS in dimension n = 2 existence of quasi-periodic solutions Wei Min Wang: analytic NLS existence of quasi-periodic solutions Procesi-Procesi: cubic NLS existence and stability of quasi-periodic solutions.

4 Linear equations This system is highly resonant in u = 0. Namely if you linearize the equation at the elliptic fixed point u=0 iu t u = F( u 2 )u0 all the bounded solutions are periodic! u(t, ϕ) = k Z n u k e i( k 2t+k ϕ)

5 Linear equations This system is highly resonant in u = 0. Namely if you linearize the equation at the elliptic fixed point u=0 iu t u = 0 all the bounded solutions are periodic! u(t, ϕ) = k Z n u k e i( k 2t+k ϕ)

6 One expects the non-linear dynamics to be much richer! In particular our aim when we started analizing this problem was to prove existence of quasi-periodic solutions with frequencies close to those of the linear NLS. We wanted to apply a KAM algorithm It is a standard procedure when applying KAM theory to resonant systems to first apply one step of Birkhoff Normal Form. The study of the NLS equation after this step will be the main topic of this lectures! It is essentially a geometric problem.

7 Quasi-periodic solutions Final goal, with C. Procesi For all m N, there exist Cantor families of small quasi periodic solutions of Equation (1) with m frequencies ω 1,..., ω m. We prove the existence of an integrable normal form close to the solution. m is arbitrarily large but finite

8 Dynamical systems approach For simplicity concentrate on the cubic NLS on T n. Passing to the Fourier representation u(t, ϕ) := k Z n u k (t)e i(k,ϕ), Eq. (1) becomes i u k (t) = k 2 u k (t)+ k 1 k 2 +k 3 =k u k1 ū k2 u k3 infinite dimensional Hamiltonian dynamical system: H = k 2 u + k u k + u + k 1 u k 2 u + k 3 u k 4 (3) k Z n k i Z n :k 1 +k 3 =k 2 +k 4 with respect to the complex symplectic form i k du + k du k. On the real subspace u + k = u k and u k = ū k.

9 If one prefers real notation u k = p k + iq k : k Z n k 2 (q 2 k + p 2 k)+o( p 2 + q 2 ) a list of harmonic oscillators coupled by a non linearity usually one writes compactly H = Ignore the order 4 term: k 2 u k 2 + u k1 ū k2 u k3 ū k4 k Z n k i Z n :k 1 +k 3 =k 2 +k 4 u k 2 = ξ k, u k (t) = ξ k e i k 2 t

10 If one prefers real notation u k = p k + iq k : k Z n k 2 (q 2 k + p 2 k)+o( p 2 + q 2 ) a list of harmonic oscillators coupled by a non linearity usually one writes compactly H = Ignore the order 4 term: k 2 u k 2 + u k1 ū k2 u k3 ū k4 k Z n k i Z n :k 1 +k 3 =k 2 +k 4 u k 2 = ξ k, u k (t) = ξ k e i k 2 t

11 The system has the constants of motion: L = u k 2, M = k u k 2 k Z n k Z n the fact that M is preserved will be crucial to the proof!

12 Birkhoff Normal Form H = K(p, q)+h (4) (p, q), K(p, q) = k λ k (p 2 k + q 2 k) where H (4) is a polynomial of degree 4 and the λ k (in our case k 2 ) are all rational. With a sympletic change of variables we reduce the Hamiltonian H to H Birk = K(p, q)+h res (4) (p, q)+h(6) where H (6) is small while H (4) res Poisson commutes with K.

13 Study the dynamics of H res = K(p, q)+h (4) res(p, q), then treat H (6) as a small perturbation. It turns out that the dynamics of H res is very rich and complex Step 1: Prove the existence of invariant subspaces on which H res is integrable (and with a simple and explicit dynamics). Step 2: Given a solution from step one, linearize H res at this solution. One gets a quadratic Hamiltonian which we call normal form N. Step 3: The structure of N depends on the choice of the invariant subspace. Look for invariant subspaces on which N is as simple as possible.

14 Functional setting work on the phase space l (a,p) := {u ± = {u ± k } k Z n u ± H = a > 0, p > n/2, k Z n k 2 u + k u k + k Z n u ± k 2 e 2a k k 2p := u ± 2 a,p < } u + k = ū k (4) k i Z n :k 1 +k 3 =k 2 +k 4 u + k 1 u k 2 u + k 3 u k 4 (5) The Hamiltonian is analytic the vector field generated by H (4) is analytic from B R l (a,p).

15 One step of Birkhoff normal form produces H Birk = k Z n k 2 u k ū k + u u + H k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2 (6) (6) k 3 k 2 k 4 k 1 Even if we ignore the term H (6), this equation is still very complicated Note that H res has as constants of motion L = k u k 2, M = k k u k 2, K = k k 2 u k 2

16 One step of Birkhoff normal form produces H Birk = k Z n k 2 u k ū k + u u + H k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2 (6) (6) k 3 k 2 k 4 k 1 Even if we ignore the term H (6), this equation is still very complicated Note that H res has as constants of motion L = k u k 2, M = k k u k 2, K = k k 2 u k 2

17 The simplest rectangles are the degenerate ones k 1 = k 2 = k 3 = k 4 and k 1 = k 2 k 3 = k 4. This gives a contribution: H Res = k 2 u k u k 2 u h 2 + k Z n k k k non deg u u k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2 u k 4 + Note that L = k u k 2 is constant so we may subtract 2L 2 : H Res = k 2 u k 2 u k 4 + k Z n k non deg u u k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2

18 In dimension n = 1 there are no non-degenerate rectangles. H Res = k 2 u k 2 k Z k u k 4 u k 2 = ξ k, u k = ξ k e i( k 2 2ξ k )t Step1: Choose any finite set S = {v 1,..., v m } Z then U S := {u k = 0, k Z \ S := S c } is an invariant subspace for H res. Solution: u k = ξ k e i( k 2 2ξ k )t for k S and u k = 0 for k S c. We have parameter families of m-tori

19 H Res = k 2 u k 2 k Z k u k 4 Step 2: Consider the solution u k = ξ k e i( k 2 2ξ k )t for k S and u k = 0 for k S c. parameter families of invariant tori T m Linearize the equation (actually I work on the hamiltonian) u k = ξ k + y k e ix k for k S and u k = z k for k S c. N = (ω, y)+ k k 2 z k 2, ω i = v i 2 2ξ vi, S = (v 1, v 2,..., v m ) Step 3: N is already as simple as possible! (Kuksin-Pöschel: existence and stability of quasi-periodic solutions )

20 H Res = k 2 u k 2 k Z k u k 4 Step 2: Consider the solution u k = ξ k e i( k 2 2ξ k )t for k S and u k = 0 for k S c. Linearize the equation (actually I work on the hamiltonian) u k = ξ k + y k e ix k for k S and u k = z k for k S c. N = (ω, y)+ k k 2 z k 2, ω i = v i 2 2ξ vi, S = (v 1, v 2,..., v m ) Step 3: N is already as simple as possible! (Kuksin-Pöschel: existence and stability of quasi-periodic solutions )

21 Lemma Given a set S = {v 1,..., v m } Z n consider the subspace U S := {u = {u k } k Z n : u k = 0, if k / S} For generic choices of S the space U S is invariant for the dynamics of H Res = k 2 u k ū k u k 4 + k Z n k Z n the Hamiltonian H Res restricted on U S is non deg. k 2 u k 2 u k 4 k S k S u u k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2

22 cfr. Marcel Guardia s lectures... Definition (Completeness:) S := {v 1,..., v m } is complete if: for all triples v j1, v j2, v j3 S if then k S k = v j1 v j2 + v j3, k 2 = v j1 2 v j2 2 + v j3 2 Lemma If S is complete then U S is an invariant subspace for H res. Dim

23 tangential sites Definition (Complete and Integrable) S := {v 1,..., v m } is complete and integrable if: for all triples v j1 v j2 v j3 S we have (v j1 v j2, v j3 v j2 ) 0 Show that this set is non-empty! If S is complete and integrable then there are no non-degenerate rectangles in S so trivially k 2 u k 2 u k 4 k S k S

24 tangential sites Choice of the tangential sites Let us now partition where: S = {v 1,..., v m } is a finite set Z n = S S c, S := (v 1,..., v m ). with some constraints. Start form: S complete and integrable. The set S is called the tangential sites and the set S c the normal sites.

25 Elliptic/action-angle variables. Let us now set u k := z k for k S c, u vi := ξ i + y i e ix i for v i S, this puts the tangential sites in action angle variables y := {y 1,..., y m }, x := x 1,..., x m the ξ are parameters. ξ A ε 2 := {ξ : 1 2 ε2 ξ i ε 2 },

26 Let us now set u k := z k for k S c, u vi := ξ i + y i e ix i for v i S, For all r ε/2 this is a well known analytic and symplectic change of variables in the domain D a,p (s, r) = D(s, r) := {x, y, z : Im(x) < s, y r 2, z a,p r} T m s C m l (a,p) l (a,p). z 2 a,p := z k S c z k 2 e 2a k k 2p

27 The normal form Hamiltonian substitute in u k := z k for k S c, u vi := ξ i + y i e ix i for v i S, H res = k 2 u k 2 u k 4 + k Z n k Z n u u k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2

28 We compute the leading term of H res : N := ( v i 2 2ξ i )y i + 1 i m k 2 z k 2 (7) k S c +Q(x, z) set ω i := v i 2 2ξ i.

29 Step 3: We impose some simple constraints: 1 v i v j 0, v i + v j 0, v i + v j v l v k 0 and v i + v j v l + v k 0. 2 (v i v j, v k v j ) ( v i 2 v j 2 )+ v i v j 2 0 and 2( v i 2 + v j 2 ) v i + v j 2 0.

30 The normal form Hamiltonian 2 1 i<j m h,k S c Q(x, z) = 4 1 i j m h,k S c ξ i ξ j e i(x i+x j ) z h z k + 2 ξ i ξ j e i(x i x j ) z h z k + (8) 1 i<j m h,k S c ξ i ξ j e i(x i+x j) z h z k. This is a non-integrable potentially very complicated infinite dimensional quadratic form!

31 The constraints, mean that the terms are resonant with the quadratic part K, that is: Definition Here denotes that (h, k, v i, v j ) give a rectangle: {(h, k, v i, v j ) h + v i = k + v j, h 2 + v i 2 = k 2 + v j 2 }. We say h H i,j, k H j,i. means that (h, v i, k, v j ) give a rectangle: {(h, v i, k, v j ) h + k = v i + v j, h 2 + k 2 = v i 2 + v j 2 }. We say h, k S i,j

32 H i,j v j v i + v j h k S i,j v i v j h v i k Figure: The plane H i,j and the sphere S i,j. The points h, k, v i, v j form the vertices of a rectangle. Same for the points h, v i, k, v j

33 Examples: h 1, h 2 belong to one and only one resonant figure: h 2 h 1 + v 2 v 1 = 0, h 2 2 h v 2 2 v 1 2 = 0 we have the Hamiltonian 4 ξ 1 ξ 2 e i(x 1 x 2 ) z h1 z h2 + 4 ξ 1 ξ 2 e i(x 1 x 2) z h1 z h2 h 2 + h 1 v 2 v 1 = 0, h h 1 2 v 2 2 v 1 2 = 0 we have the Hamiltonian 4 ξ 1 ξ 2 e i(x 1+x 2 ) z h1 z h2 + 4 ξ 1 ξ 2 e i(x 1+x 2) z h1 z h2

34 a more complicated block: A k1 = k 4, (9) v 2 +v 1 k 1 v 1 v 3 k 2 v 2 v 3 k 3 this means that k 2 k 1 + v 3 v 1 = 0 k 2 2 k v 3 2 v 1 2 = 0 k 3 k 2 + v 3 v 2 = 0 k 3 2 k v 3 2 v 2 2 = 0 k 4 + k 2 v 2 v 1 = 0 k k 2 2 v 2 2 v 1 2 = 0 we have the Hamiltonian: 4 ξ 1 ξ 3 e i(x 1 x 3 ) z k1 z k2 +4 ξ 2 ξ 3 e i(x 2 x 3 ) z k2 z k3 +4 ξ 1 ξ 2 e i(x 1+x 2 ) z k2 z k4 + 4 ξ 1 ξ 3 e i(x 1 x 3) z k1 z k2 +4 ξ 2 ξ 3 e i(x 2 x 3) z k2 z k3 +4 ξ 1 ξ 2 e i(x 1+x 2) z k2 z k4

35 In dimension n = 2 you may strongly simplify by applying arithmetic conditions. Proposition (Geng-You-Xu, for the cubic NLS) For n = 2, and for every m there exist infinitely many choices of generic tangential sites S = {v 1,..., v m } such that, if A is a connect component of the geometric graph, then A is either a vertex or a single edge. This is the simplest form for the normal form N! (ω, y)+ k k 2 z k 2 + i<j 4 ξ i ξ i e i(x i x j ) z h1 z h2 +4 ξ i ξ j e i(x i x j) z h1 z h2 + i<j 4 ξ i ξ j e i(x i+x j ) z k1 z k2 + 4 ξ i ξ j e i(x i+x j) z k1 z k2

36 What to do in dimension n > 2? How big can the resonant blocks be?

37 What to do in dimension n > 2? How big can the resonant blocks be? GO ASK DADDY!

38 A first theorem for generic v i s. Theorem For generic v i s the quadratic Hamiltonian Q(x, w) is an infinite sum of independent (decoupled) terms each depending on a finite number of variables (at most n+1 variables z j together with their conjugates z j ). One can exhibit an explicit symplectic change of variables which integrates N, namely makes all the angles disappear from Q(x, w).

39 Theorem There exists a map S c k L(k) Z m, L(k) < 2n such that the analytic symplectic change of variables: z k = e il(k).x z k, y = y + L(k) z k 2, x = x. k S c Φ : (y, x) (z, z ) (y, x) (z, z) from D(s, r) D(s, r/2) such that N Φ = (ω(ξ), y )+ ( k 2 + L i (k) v i 2 ) z k 2 + Q(w ), (11) k S c i

40 Generiticity condition: Resonance polynomials Definition Given a list R := {P 1 (y),..., P N (y)} of non zero polynomials in k vector variables y i, we say that a list of vectors S = {v 1,..., v m }, v i C n is GENERIC relative to R if, for any list A = {u 1,..., u k } such that u i S, i, the evaluation of the resonance polynomials at y i = u i is non zero. If m is finite this condition is equivalent to requiring that S (considered as a point in C nm ) does not belong to the algebraic variety where at least one of the resonance polynomials is zero.

41 Generiticity condition In our specific case the required list of the resonances, P 1 (y),..., P N (y), are non zero polynomials with integer coefficients depending on k = 2n vector variables y = (y 1,..., y 2n ) with y i = (yi 1,..., y i n ). The explicit list of these resonances is generated so to exclude a finite list of undesired graphs. Thus it depends on some non trivial combinatorics of graphs. Example: means that we require for all i j k P 1 (y 1, y 2, y 3 ) = (y 1 y 2, y 1 y 3 ) (v i v j, v i v k ) 0

42 Generiticity condition In our specific case the required list of the resonances, P 1 (y),..., P N (y), are non zero polynomials with integer coefficients depending on k = 2n vector variables y = (y 1,..., y 2n ) with y i = (yi 1,..., y i n ). Example: P 1 (y 1, y 2, y 3 ) = (y 1 y 2, y 1 y 3 ) means that we require for all i j k (v i v j, v i v k ) 0

43 Some remarks There is no a-priori reason why this change of variables should exist. If one does not impose good genericity conditions then this is false. This change of variables exists for all analytic NLS iu t u = F( u 2 )u provided that F does not explicitly depends on ϕ. We can proceed in the same way also when S is an infinite set.

44 Some remarks There is no a-priori reason why this change of variables should exist. If one does not impose good genericity conditions then this is false. This change of variables exists for all analytic NLS iu t u = F( u 2 )u provided that F does not explicitly depends on ϕ. We can proceed in the same way also when S is an infinite set.

45 The final Theorem and goal for the normal form We will show that Theorem for generic values of the parameters (outside some algebraic hyper surface) we can find a further symplectic change of coordinates so that N is diagonal (possibly with some complex terms) N is non degenerate namely it has distinct eigenvalues.. there exists a positive measure region of the parameters ξ in which N is elliptic (all real eigenvalues).

46 H fin = (ω(ξ), y)+ k S c Ω k z k 2 + P(ξ, x, y, z, z) (12) Ω k = k 2 + i ω i = v i 2 2ξ i L i (k) v i 2 + θ k (ξ), k S c The L i (k) are integers θ k (ξ) {θ (1) (ξ),..., θ (K) (ξ)}, K := K(n, m), (13) list different analytic homoeogeneous functions of ξ.

47 Why restrict to the cubic NLS? In the cubic NLS we are able to prove NON DEGENERACY conditions: (ω(ξ), ν) = 0, (ω(ξ), ν)+ω k (ξ) = 0 (ω(ξ), ν)+ω k (ξ) Ω h (ξ) = 0, holds true on a proper algebraic hypersurface for all non trivial choices of ν Z m h, k S c (recall that Z n = S S c ) compatible with momentum conservation. Recall ω i = v i 2 2ξ i Ω k = k 2 + i L (i) k v i 2 + θ k (ξ) For ν small it is NOT obvious that the second Melnikov condition is NOT identically zero!

48 Why restrict to the cubic NLS? In the cubic NLS we are able to prove NON DEGENERACY conditions: (ω(ξ), ν)+ω k (ξ) Ω h (ξ) = 0, holds true on a proper algebraic hypersurface for all non trivial choices of ν Z m h, k S c (recall that Z n = S S c ) compatible with momentum conservation. Recall ω i = v i 2 2ξ i Ω k = k 2 + i L (i) k v i 2 + θ k (ξ) For ν small it is NOT obvious that the second Melnikov condition is NOT identically zero!

49 H = k 2 u k 2 + u k1 ū k2 u k3 ū k4 k Z n k i Z n :k 1 +k 3 =k 2 +k 4 Do 1 step of Birkhoff normal form: H res = k 2 u k 2 u k 4 + k Z n k Z n Non deg. u u +H k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2 (6) Choose a set S = {v 1, v 2,..., v m } so that (v i v j, v l v j ) 0 for all triples (v i v j v l ). Divide Z n = S S c and substitute on the board ANSATZ: the value of u k 2 with k S c is much smaller than u v 2 with v S

50 H = k 2 u k 2 + u k1 ū k2 u k3 ū k4 k Z n k i Z n :k 1 +k 3 =k 2 +k 4 Do 1 step of Birkhoff normal form: H res = k 2 u k 2 u k 4 + k Z n k Z n Non deg. u u +H k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2 (6) Choose a set S = {v 1, v 2,..., v m } so that (v i v j, v l v j ) 0 for all triples (v i v j v l ). Divide Z n = S S c and substitute on the board ANSATZ: the value of u k 2 with k S c is much smaller than u v 2 with v S

51 H = k 2 u k 2 + u k1 ū k2 u k3 ū k4 k Z n k i Z n :k 1 +k 3 =k 2 +k 4 Do 1 step of Birkhoff normal form: H res = k 2 u k 2 u k 4 + k Z n k Z n Non deg. u u +H k1ūk2 k3ūk4 k i Z n :k 1 +k 3 =k 2 +k 4 k k 3 2 = k k 4 2 (6) Choose a set S = {v 1, v 2,..., v m } so that (v i v j, v l v j ) 0 for all triples (v i v j v l ). Divide Z n = S S c and substitute on the board ANSATZ: the value of u k 2 with k S c is much smaller than u v 2 with v S

52 Show: dynamics on the invariant subspace Substitute u k := z k for k S c, u vi := ξ i + y i e ix i for v i S, with u vi 2 ξ i > 0, y i ξ i and z k 2 ξ H = N + higher order terms N = 1 i m ( v i 2 2ξ i )y i + Write the Hamilton equations Recall the goal k S c k 2 z k 2 + Q(ξ; x, z)

53 4 4 N = 1 i<j m h,k S c 1 i<j m h,k S c 1 i m explain the building blocks! ( v i 2 2ξ i )y i + k S c k 2 z k 2 + ( ξ i ξ j e i(x i x j ) z h z k + ξ i ξ j e i(x i x j) z h z k ) + ( ξ i ξ j e i(x i+x j ) z h z k + ξ i ξ j e i(x i+x j) z h z k ).

54 Let us go back to our Hamiltonian and study the blocks of the graph explain the graph! H i,j v j v i + v j h k S i,j v i v j h v i k

55 Example block: A k1 = k 4, (14) v 2 +v 1 k 1 v 1 v 3 k 2 v 2 v 3 k 3 this means that k 2 k 1 + v 3 v 1 = 0 k 2 2 k v 3 2 v 1 2 = 0 k 3 k 2 + v 3 v 2 = 0 k 3 2 k v 3 2 v 2 2 = 0 k 4 + k 2 v 2 v 1 = 0 k k 2 2 v 2 2 v 1 2 = 0 we have the Hamiltonian: 4 ξ 1 ξ 3 e i(x 1 x 3 ) z k1 z k2 +4 ξ 2 ξ 3 e i(x 2 x 3 ) z k2 z k3 +4 ξ 1 ξ 2 e i(x 1+x 2 ) z k2 z k4 + 4 ξ 1 ξ 3 e i(x 1 x 3) z k1 z k2 +4 ξ 2 ξ 3 e i(x 2 x 3) z k2 z k3 +4 ξ 1 ξ 2 e i(x 1+x 2) z k2 z k4

56 write the equations in a simpler form k 2 k 1 + v 3 v 1 = 0 k 2 2 k v 3 2 v 1 2 = 0 k 3 k 2 + v 3 v 2 = 0 k 3 2 k v 3 2 v 2 2 = 0 k 4 + k 2 v 2 v 1 = 0 k k 2 2 v 2 2 v 1 2 = 0 k 2 = k 1 v 3 + v 1 k 1 v 3 + v 1 2 k v 3 2 v 1 2 = 0 k 3 = k 1 + v 1 + v 2 2v 3 k 1 + v 1 + v 2 2v 3 2 k 1 v 3 + v v 3 2 v 2 2 = 0 k 4 = k 1 + v 3 + v 2 k 1 + v 3 + v k 1 v 3 + v 1 2 v 2 2 v 1 2 = 0

57 k 2 = k 1 v 3 + v 1 k 3 = k 1 + v 1 + v 2 2v 3 k 4 = k 1 + v 3 + v 2 k 1 + v 1 + v 2 2v 3 2 k 1 v 3 + v v 3 2 v 2 2 = 0 k 1 v 3 + v 1 2 k v 3 2 v 1 2 = 0 k 1 + v 3 + v k 1 v 3 + v 1 2 v 2 2 v 1 2 = 0 k 2 = k 1 v 3 + v 1 k 3 = k 1 + v 1 + v 2 2v 3 k 4 = k 1 + v 3 + v 2 2(k 1, v 3 + v 1 )+ v 3 v v 3 2 v 1 2 = 0 2(k 1, v 2 v 3 )+ v 1 + v 2 2v 3 2 v 3 + v v 3 2 v 2 2 = 0 2 k (k 1, v 1 2v 3 v 2 )+ v 3 + v v 3 v 1 2 v 2 2 v 1 2 =

58 2(k 1, v 3 + v 1 )+ v 3 v v 3 2 v 1 2 = 0 2(k 1, v 2 v 3 )+ v 1 + v 2 2v 3 2 v 3 + v v 3 2 v 2 2 = 0 2 k (k 1, v 1 v 2 2v 3 )+ v 3 + v v 1 v 3 2 v 2 2 v 1 2 = A k1 = k 4, (16) v 2 +v 1 k 1 v 1 v 3 k 2 v 2 v 3 k 3 corresponds to two linear and one quadratic equation on k 1 the graph appears if this equations have solutions in S c.

59 A tree with e edges corresponds to e equations ( for x S c C n ): 2(x, l 1 (v 1, v 2,..., v m )) = Q 1 (v 1, v 2,..., v m ) 2(x, l 2 (v 1, v 2,..., v m )) = Q 2 (v 1, v 2,..., v m ). 2(x, l K (v 1, v 2,..., v m )) = Q K (v 1, v 2,..., v m ) 2 x 2 + 2(x, l K+1 (v 1, v 2,..., v m )) = Q K+1 (v 1, v 2,..., v m ) 2 x 2 + 2(x, l K+2 (v 1, v 2,..., v m )) = Q K+2 (v 1, v 2,..., v m ). 2 x 2 + 2(x, l e (v 1, v 2,..., v m )) = Q e (v 1, v 2,..., v m ) so if the equations are independent we may have only graphs with at most n+1 vertices. Unfortunately it is not clear why the equations should be independent!

60 Suppose we only have linear equations 2(x, l 1 (v 1, v 2,..., v m )) = Q 1 (v 1, v 2,..., v m ) 2(x, l 2 (v 1, v 2,..., v m )) = Q 2 (v 1, v 2,..., v m ). 2(x, l K (v 1, v 2,..., v m )) = Q K (v 1, v 2,..., v m ) If K < n It may be that the equations are resonant because ai l i (v 1,..., v m ) vanishes for some choice of the v i choose the v i s to avoid this resonance Example

61 2(x, l 1 (v 1, v 2,..., v m )) = Q 1 (v 1, v 2,..., v m ) 2(x, l 2 (v 1, v 2,..., v m )) = Q 2 (v 1, v 2,..., v m ). 2(x, l K (v 1, v 2,..., v m )) = Q K (v 1, v 2,..., v m ) Even when K > n it may be that the equations are resonant because a i l i (v 1,..., v m ) is identically zero. Then if a i Q 1 (v 1, v 2,..., v m ) is not identically zero choose the v i s to avoid this resonance Example

62 2(x, l 1 (v 1, v 2,..., v m )) = Q 1 (v 1, v 2,..., v m ) 2(x, l 2 (v 1, v 2,..., v m )) = Q 2 (v 1, v 2,..., v m ). 2(x, l K (v 1, v 2,..., v m )) = Q K (v 1, v 2,..., v m ) ai l i (v 1,..., v m ) = 0, a i Q 1 (v 1, v 2,..., v m ) = 0 identically then the resonant equations are unavoidable! Lemma: the graphs in our construction cannot generate unavoidable resonances made only with vertices of the same colour! EXPLAIN!

63 2(x, l 1 (v 1, v 2,..., v m )) = Q 1 (v 1, v 2,..., v m ) 2(x, l 2 (v 1, v 2,..., v m )) = Q 2 (v 1, v 2,..., v m ). 2(x, l K (v 1, v 2,..., v m )) = Q K (v 1, v 2,..., v m ) If K = n+1 then surely the vectors l 1,..., l N (even if they are formally independent) cannot be independent. If possible: choose the v i s to avoid this resonance Then if they produce an unavoidable resonance EXPLAIN! Lemma: the unique solution is x = v i (a point in S)

64 Proposition Our graphs may have at most 2n+1 vertices! 2(x, l 1 (v 1, v 2,..., v m )) = Q 1 (v 1, v 2,..., v m ) 2(x, l 2 (v 1, v 2,..., v m )) = Q 2 (v 1, v 2,..., v m ). 2(x, l K (v 1, v 2,..., v m )) = Q K (v 1, v 2,..., v m ) 2 x 2 + 2(x, l K+1 (v 1, v 2,..., v m )) = Q K+1 (v 1, v 2,..., v m ) 2 x 2 + 2(x, l K+2 (v 1, v 2,..., v m )) = Q K+2 (v 1, v 2,..., v m ). 2 x 2 + 2(x, l K+M (v 1, v 2,..., v m )) = Q K+M (v 1, v 2,..., v m ) Lemma1: the linear equations are formally independent Lemma1: the quadratic equations are formally independent Lemma2: if either M or K are n+1 the only solution is in S

65 The normal form Hamiltonian as matrix Since Q(x, w) and N are quadratic Hamiltonians we study their matrix representation: acting by Poisson bracket on two spaces V 0,1, F 0,1 V 0,1 is the space with basis the elements z k, z k, k S c (and coefficients trigonometric poly. in x). (compute ({Q, z k }, {Q, z k }) F 0,1 is the subspace of V 0,1 satisfying conservation of momentum and mass, so it has as basis e iν.x z k with ν i v i + k = 0, i i and the conjugates e iν.x z k (compute ({Q, e iν.x z k }, {Q, e iν.x z k }). Explain ν i + 1 = 0, k S c

66 One term Suppose that h+v i = k + v j, h 2 + v i 2 = k 2 + v j 2 and Q(x, w) contains the term 4( ξ i ξ j e i(x i x j ) z h z k + ξ i ξ j e i(x i x j) z h z k ) we describe this by two matrices compute: 0 ξi ξ 4i j e i(x i x j ) ξi ξ j e i(x i x j ) 0, 4i 0 ξi ξ j ξi ξ j 0 respectively in the bases: z h, z k, resp. (ν,+) := e iν x z h, (µ,+) := e iµ x z k, µ i v i + k = 0, ν i v i + h = 0, ν = µ+e i e j i i e i are the basis vectors in Z m compute the Poisson bracket.

67 A k1 = k 4 v 2 +v 1 k 1 v 1 v 3 k 2 v 2 v 3 k 3 S c (17) Let now a = (a 1, a 2,..., a m ) Z m be any vector such that a i v i = k 1, draw a graph in Z m {+, } A (a,+) = ( e 2 e 3 a, ) e 2 e 1 (a,+) e 3 e 1 (a+e 3 e 1,+) e 3 e 2 (a e 1 e 2 + 2e 3,+) (18)

68 The edges of the graph Λ S, We construct a graph Λ S having as vertices Z m {+, }) we join two basis elements with an edge (with suitable marking) if and only if the corresponding matrix element (of {Q, }) is non zero. The marking tells us what is the matrix entry.

69 I impose conditions in order to ensure that the graphs A (a,+) and and A k1 are isomorphic. A k1 = k 4 v 2 +v 1 k 1 v 1 v 3 k 2 v 2 v 3 k 3 A (a,+) = ( e 2 e 3 a, ) e 2 e 1 (a,+) e 3 e 1 (a+e 3 e 1,+) e 3 e 2 (a e 1 e 2 + 2e 3,+)

70 Why do I like to work in this setting: We have a group structure: (a,+)(b,+) = (a+b,+), (a, )(b,+) = (a b, ), (a,+)(b, ) = (a+b, ), (a, )(b, ) = (a b,+). We use this to traslate to (0,+) our graphs: A (a,+) = ( e 2 e 3 a, ) e 2 e 1 e 3 e 1 (a,+) (a+e 3 e 1,+) goes to A = ( e 2 e 3, ) e 3 e 2 (a e 1 e 2 + 2e 3,+) e 2 e 1 (0,+) e 3 e 1 (e 3 e 1,+) e 3 e 2 ( e 1 e 2 + 2e 3,+) (19)

71 By our previous argument the trees A k1 = k 4, v 2 +v 1 k 1 v 1 v 3 k 2 v 2 v 3 k 3 and A = ( e 2 e 3, ) e 2 e 1 (0,+) e 3 e 1 (e 3 e 1,+) e 3 e 2 ( e 1 e 2 + 2e 3,+) are isomorphic. To each k S c associate the corresponding vertex L(k) Z m.

72 Theorem The map S c k L(k) Z m, L(k) < 2n is such that the analytic symplectic change of variables: z k = e il(k).x z k, y = y + k S c L(k) z k 2, x = x. Φ : (y, x) (z, z ) (y, x) (z, z) from D(s, r) D(s, r/2) bring you to N Φ = (ω(ξ), y )+ ( k 2 + L i (k) v i 2 ) z k 2 + Q(w ), (20) k S c i

73 To each graph A = ( e 2 e 3, ) e 2 e 1 (0,+) e 3 e 1 (e 3 e 1,+) e 3 e 2 ( e 1 e 2 + 2e 3,+) associate the equations: { (x, i a iv i ) = 1 2 ( i a iv i 2 + i a i v i 2 ) for (a = i a ie i,+) x 2 +(x, i a iv i ) = 1 2 ( i a iv i 2 + i a i v i 2. ) for (a, ) (21) (a, ±) (0,+) are the vertices of the graph! Proposition A combinatorial graph A contributes to Q if and only if the equations (21) have solutions in S c.

74 Take a graph with only black lines, we have the equations: (x, i a(1) i v i ) = 1 2 ( i a(1) i v i 2 + i a(1) i v i 2 ) (x, i a(2) i v i ) = 1 2 ( i a(2) i v i 2 + i a(2) i v i 2 ). (x, i a(n) i v i ) = 1 2 ( i a(n) i v i 2 + i a(n) i v i 2 ) suppose that the left hand side is resonant for all choices of the v i s ( this means that the vertices a of the graph are linearly dependent) then the right hand side cannot cancel identically we have an avoidable resonance Example If there are both linear and quadratic equations this is not true!

75 Compatible and incompatible equations The main difficulty is when edges are linearly dependent, we need to exclude certain resonant graphs which are always formally compatible but produce no real solutions. e 2 e 3. e 2 e 3 e 1 e 4 e 1 e 3e 1 + e 2 2e 1 e 2 e 4 4 e e1 e 2 1 e 2 e 2 e 3 0 e 2 e 3

76 Compatible and incompatible equations One of the main steps is thus to show that only special combinatorial graphs produce equations which have real solutions for generic values of the tangential sites v i. Theorem (Main) Only graphs which have at most n linearly independent edges appear. The proof is quite complex it requires some algebraic geometry and a long combinatorial analysis.

77 The reduced Hamiltonian : N = (ω, y)+ m ( k 2 + L i (k) v i 2 ) z k 2 + Q A (ξ, z, z) k S c i=1 A Γ S Represent the action as a block diagonal matrix with blocks described by the connected graphs. one block αi 0 0 αi + Q A (ξ) 0 0 Q A (ξ) to diagonalize I need to compute the eigenvalues

78 The combinatorial matrices: example d 4,3 c 4,1 2,3 a 1,2 b 0 2 ξ 2 ξ ξ 1 ξ 4 2 ξ 2 ξ 1 ξ 2 ξ 1 2 ξ 2 ξ ξ 2 ξ 3 ξ 1 + ξ 3 2. ξ 4 ξ 3 2 ξ 1 ξ ξ 4 ξ 3 ξ 4 ξ 1

79 With characteristic polynomial 4 ξ1 3 ξ ξ1 2 ξ 2 ξ 3 4 ξ1 3 ξ ξ1 2 ξ 2 ξ ξ1 2 ξ 3 ξ 4 8 ξ 1 ξ 2 ξ 3 ξ 4 +(ξ1 3 9 ξ2 1 ξ 2 ξ1 2 ξ 3+ξ 1 ξ 2 ξ 3 9 ξ1 2 ξ 4+9 ξ 1 ξ 2 ξ 4 +ξ 1 ξ 3 ξ 4 +7 ξ 2 ξ 3 ξ 4 ) t +(3 ξ1 2 6 ξ 1 ξ 2 2 ξ 1 ξ 3 3 ξ 2 ξ 3 6 ξ 1 ξ 4 + ξ 2 ξ 4 3 ξ 3 ξ 4 ) t 2 +(3 ξ 1 ξ 2 ξ 3 ξ 4 ) t 3 + t 4

80 The combinatorial matrices: example a (1,2) b (i,j) c (h,k) d, 0 2 ξ 2 ξ ξ 2 ξ 1 ξ 2 + ξ 1 2 ξ i ξ j ξ i ξ j ξ i + ξ j + ξ 2 + ξ 1 2 ξ k ξ h ξ h ξ h ξ k ξ h ξ i + ξ j + ξ 2 + ξ 1

81 Non degeneracy: STEP 2 Non degeneracy Non degeneracy

82 Our main goal is to prove that Theorem For generic values of the parameters ξ all eigenvalues of all these infinitely many matrices are distinct. We can change symplectic coordinates and make the matrix diagonal with distinct eigenvalues (as functions of ξ). In order to prove this fact we cannot proceed directly but we can prove it as a consequence of a stronger algebraic Theorem.

83 In fact the eigenvalues are solutions of complicated polynomial equations as graph G := e 1 e 2 0 e 1 e 2 ξ 1 ξ 2 2 ξ 1 ξ 2 0 matrix C G = 2 ξ 1 ξ ξ 1 ξ ξ 1 ξ 2 ξ 2 ξ 1 χ G (t) = det(ti C G ) = t 3 + 2ξ 1 t 2 +(ξ 2 1 ξ 2 2)t 8ξ 1 ξ 2 2 = 0. (22)

84 graph G := d 4,3 c 4,1 2,3 a 1,2 b 4 ξ1 3 ξ ξ1 2 ξ 2 ξ 3 4 ξ1 3 ξ ξ1 2 ξ 2 ξ ξ1 2 ξ 3 ξ 4 8 ξ 1 ξ 2 ξ 3 ξ 4 +(ξ1 9 3 ξ1 2 ξ 2 ξ1 2 ξ 3 +ξ 1 ξ 2 ξ 3 9 ξ1 2 ξ 4 +9 ξ 1 ξ 2 ξ 4 +ξ 1 ξ 3 ξ 4 +7 ξ 2 ξ 3 ξ 4 ) t +(3 ξ1 2 6 ξ 1 ξ 2 2 ξ 1 ξ 3 3 ξ 2 ξ 3 6 ξ 1 ξ 4 + ξ 2 ξ 4 3 ξ 3 ξ 4 ) t 2 +(3 ξ 1 ξ 2 ξ 3 ξ 4 ) t 3 + t 4 = 0

85 A direct method one should compute discriminants and resultants, which are polynomials in ξ and show that they are not identically zero. This can be done by direct computations only for very small cases but in general it is out of question. So we prove that all these discriminants and resultants are not identically zero by showing a stronger property of all characteristic polynomials!

86 MAIN THEOREM of Step 2 Main Theorem The characteristic polynomials of the combinatorial matrices are all irreducible and different from each other. Implications 1 Eigenvalue separation 2 Validity of the 3 Melnikov conditions. 3 Symplectic coordinates which diagonalize the Hamiltonian.

87 The proof of this Theorem is very complicated, the main idea is to apply induction on the size of the graph and on the number of variables ξ i. The starting point is the remark that, if we set a variable ξ i = 0 in the matrix associated to a graph G we obtain the matrix of the graph obtained from G by removing all edges where i appears in the marking. So the characteristic polynomial specializes to the product of characteristic polynomials of the connected components of the obtained graph.

88 The combinatorial matrices: example In a (1,2) b (i,j) c (h,k) d, set ξ 1 = 0 get a b (i,j) c (h,k) d, ξ 2 2 ξ i ξ j ξ i ξ j ξ i + ξ j + ξ 2 2 ξ k ξ h ξ h ξ h ξ k ξ h ξ i + ξ j + ξ 2

89 The combinatorial matrices: example In set ξ i = 0 get a (1,2) b (i,j) c (h,k) d, a (1,2) b c (h,k) d, 0 2 ξ 2 ξ ξ 2 ξ 1 ξ 2 + ξ ξ j + ξ 2 + ξ 1 2 ξ k ξ h ξ h ξ h ξ k ξ h + ξ j + ξ 2 + ξ 1

90 Example By induction in the first case one has that the characteristic polynomial specializes to a product of a linear and a cubic irreducible factor, in the second to the product of two irreducible quadratic factors. This can happen only if we start from an irreducible polynomial.

91 Example: this argument does not work for graph G := d 4,3 c 4,1 2,3 a 1,2 b whichever variable you set equal zero, you always get a linear and a cubic term. You need a subtler combinatorial argument. Many special cases thus arise!!

92 KAM scheme: STEP 3 KAM scheme KAM scheme

93 The new Hamiltonian H fin = (ω(ξ), y)+ Ω k z k 2 + P(ξ, x, y, z, z) (23) k S c \H ω i = v i 2 2ξ i Ω k = k 2 i L i (k) v i 2 + θ k (ξ), k S c The L i (k) are integers ( L(k) < 2n) and θ k (ξ) {θ (1) (ξ),..., θ (K) (ξ)}, K := K(n, m), (24) list different analytic homoeogeneous functions of ξ. ( θ k (ξ) are the eigenvalues of the matrices C A which are a finite list)

94 The Perturbation P is small in some appropriate norm

95 Melnikov conditions Theorem (ω(ξ), ν) = 0, (ω(ξ), ν)+ω k (ξ) = 0, (25) (ω(ξ), ν)+ω k (ξ)+σω h (ξ) = 0 1. The set of parameter values ξ for which the three Melnikov resonances in (25) occur has zero measure (and for each condition it is algebraic).

96 KAM scheme. 1. A regularity/smallness condition on the perturbation P namely that X P r α. 2. A regularity condition, namely ω(ξ) must be a diffeomorphism and Ω k (ξ) k 2 must be a bounded Lipschitz function. 3. A non degeneracy condition, that is the three Melnikov resonances 25 hold in a set of measure A Quasi-Töplitz condition to control the measure estimates in the second Melnikov condition.