Singularities of bihamiltonian systems and the multidimensional rigid body (792482), страница 14
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Byanalyticity, it is non-resonant everywhere.6.18Proof of stability theoremsProof of Theorem 12. Take a regular symplectic leaf O passing through M .The conditions of the theorem imply that the point M has pure elliptic typeon O. Therefore, there exists an integral f such thatf (M ) = 0,d(f |O )(M ) = 0and the Hessian of f |O is positive definite at M.Since O is regular there exists a coordinate system x1 , . . .
xN in the neighborhood of M such thatO = {xi = 0, i = 1, . . . , k}.Now it is easy to see that2f +kXx2ii=1is a Lyapunov function.To prove the instability theorem, we will need the followingLemma 6.18.1 (About unstable cone). Let a vector field v on a manifoldM n vanish at point x0 . Suppose that the linearization of v at x0 has aneigenvalue with a positive real part.
Then there is an open subset K ⊂ M nsuch that1. There exists δ > 0 such that all trajectories starting at K leave Uδ (x0 ).1102. Intersection of K with any open neighbourhood of x0 has non-emptyinterior.Proof of lemma. First suppose that the linearization of v has a real eigenvalue λ > 0. Then we can find a coordinate system x1 , . .
. , xn in the neighbourhood V of x0 such that v has the formẋ1 = λx1 + f (x)...,where f (x) = o(||x||).Let K = {x ∈ V : x1 > 0, ||x|| < ε, |xi | < x1 for all i > 1}, where ε > 0.Obviously, the intersection of K with any open neighbourhood of x0 hasnon-empty interior. We claim that for sufficiently small ε there exists δ > 0such that all trajectories starting at K leave Uδ (x0 ). Indeed, letM = supx∈K|f (x)|.||x||√√Then ẋ1 > λx1 − M nx1 = (λ − M n)x1 .If ε is small, then M is also small and x1 will have exponential growth,q.e.d.The case of a complex eigenvalue is analogous.Corollary 6.18.1.
Let v = sgrad H be a non-resonant integrable Hamiltonian system and x be an equilibrium point of it. Suppose that there exists anintegral f such that sgrad f (x) = 0 and the linearization of sgrad f at x hasan eigenvalue with non-zero real part. Then x is an unstable equilibrium forsgrad H.Proof.
Since the linearization of sgrad f at x has an eigenvalue with nonzero real part, it has an eigenvalue with positive real part. Therefore, we111can find a set K from the lemma. For any ε the intersection Uε (x) ∩ Khas non-empty interior. Therefore, we can find a non-resonant torus passingthrough Uε (x) ∩ K.
The trajectory of sgrad f lies on this torus, therefore thistorus will leave Uδ (x). But since the torus is non-resonant, all trajectoriesof sgrad H are dense on it and will leave Uδ (x) as well. Therefore, x is anunstable equilibrium, q.e.d.Proof of Theorem 13. By Proposition 4.2.3 the following sets of operatorsare equal{Df Pα |Ker Pλ , f ∈ F, df ∈ Ker Pα } = {adξ , ξ ∈ gλ ∩ L},where adξ is the adjoint operator in gλ .Conditions of the theorem imply that for some λ there is ξ ∈ gλ ∩ Lsuch that the operator adξ has an eigenvalue with non-zero real part (see theformulas for the eigenvalues given by Propositions 6.15.2, 6.15.3). Therefore,there exists an integral f such that Df Pα also has such an eigenvalue.
ButDf Pα is dual to the linearization of sgrad f . Now it suffices to apply corollary6.18.1.Now we shall prove that all exotic equilibria are unstable.Proposition 6.18.1. Let M be an exotic equilibrium. Then there exists λsuch that sgrad Hλ 6= 0, where√√1Hλ = − h(J + λE)−1 Ω(J + λE)−1 , M i.2Proof. The vector field sgrad Hλ has the formṀ = [(J + νE)−1 Ω(J + νE)−1 , M ],where ν =√λ.112Suppose that M is an equilibrium point for sgrad Hλ for all λ.
Then0 = [(J + νE)−1 Ω(J + νE)−1 , JΩ + ΩJ] == (J + νE)−1 Ω2 + (J + νE)−1 Ω(J + νE)−1 Ω(J − νE)−− (J − νE)Ω(J + νE)−1 Ω(J + νE)−1 − Ω2 (J + νE)−1 .Since M is an equilibrium point of the body, Ω2 commutes with J. Therefore,it also commutes with (J + νE)−1 . Consequently,(J + νE)−1 Ω(J + νE)−1 Ω(J − νE) == (J − νE)Ω(J + νE)−1 Ω(J + νE)−1 .But this equality means that the matrix(J + νE)−1 Ω(J + νE)−1 Ω(J − νE)is symmetric. Let us denote by ωij the entries of the matrix Ω. λi are, asusual, diagonal entries of J. Then the symmetry condition can be writtenas:λk − ν X ωij ωjkλi − ν X ωij ωjk=λi + ν j λj + νλk + ν j λj + νfor all i, k.Since all eigenvalues of J are distinct, this impliesX ωij ωjk=0λj + νjfor all i 6= k.This equation should be satisfied for all ν.
But this means thatωij ωjk = 0for all j and all distinct i, k. Therefore, we can bring Ω to the block-diagonalform with two-by-two blocks on the diagonal by permuting basis vectors.113Such a permutation will preserve diagonal form of J. Consequently, M isnot exotic, but regular. Contradiction.Proof of Theorem 14. By the previous Proposition we can find an integral fsuch that sgrad f (M ) 6= 0 for a given exotic equilbrium M . Obviously, thetrajectories of sgrad f leave sufficiently small neighbourhood of M . Therefore, Liouville tori leave this neighbourhood as well. Since our system isnon-resonant, it’s trajectories are dense on most Liouville tori and will alsoleave the neighbourhood, q.e.d.Remark 6.18.1.
We used Proposition 6.18.1 to show that exotic equilibriaare not rank zero singular points. For equilbria not belonging to the set Badthis follows automatically from Theorem 16.Remark 6.18.2. Since sgrad Hλ (M ) 6= 0 for any exotic equilibrium M andsome value λ, exotic equilibria are not isolated on the symplectic leaves ofthe so(n)-bracket, but form smooth families. The dimension of such a familyessentially depends on the sizes of the blocks Ai , entering formula (6.2).114Bibliography[1] V. I. Arnold. Conditions for Nonlinear Stability of the Stationary PlaneCurvilinear Flows of an Ideal Fluid.
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