Singularities of bihamiltonian systems and the multidimensional rigid body, страница 10

Moreover, let all eigenvalues of Ω (eigenfrequenciesof rotation) be pairwise distinct. Then there exists an orthonormal basis suchthat J is diagonal, while Ω and M are block-diagonal with two-by-two blockson the diagonal.In other words, stationary rotation with pairwise distinct eigenfrequenciesis rotation in main axes of inertia.Definition 18. We will say that an equilibrium M is regular if there existsan orthonormal basis such that J is diagonal and Ω is block-diagonal withtwo-by-two blocks on the diagonal (i.e.

M is rotation in main axes of inertia).Otherwise, we will say that M is exotic.Corollary 6.3.1 says that all stationary rotations with pairwise distincteigenfrequencies are regular.We will see later that regular equilibria are exactly rank zero singularpoints of the system in the sense of integrable systems theory, which meansthat these points are critical points for all integrals. Exotic equilibria are, onthe contrary, non-zero rank singular points, which means that there existsan integral the differential of which doesn’t vanish in the equilibrium point.This implies that exotic equilibria are not isolated on the orbits of so(n)72bracket, but form whole smooth submanifolds of equilibrium points, whileregular equilibria are always isolated (on a coadjoint orbit).We will also show that exotic equilibria are always unstable.6.4Parabolic diagram of a regular relativeequilibriumLet M be a regular relative equilibrium.

Then there exists an orthonormalbasis such that J is diagonal and Ω is block-diagonal with two-by-two blockson the diagonal. In other words, there exists a decomposition!mMnR =Πi ⊕ Π0 ,(6.3)i=1where• Each Πi , i > 0 is spanned by two main axes of inertia. There is arotation in each of these planes with angular velocity ωi .• Π0 is spanned by n − 2m axes of inertia and is fixed.Therefore, to define a regular relative equilibrium, we need to choose aninteger m such that 0 ≤ m ≤ [n/2] and pick m pairs out of the set ofprincipal axes of inertia. For each pair we need to define angular velocity.Choice of pairs and values of angular velocities uniquely define a regularrelative equilibrium.

Knowing this data, we want to understand whether anequilibrium is stable or not.We are going to define an object called the parabolic diagram of a relativeequilibria of a multidimensional rigid body. This object will allow us toexpress stability conditions in geometric terms. For each plane Πi , i > 0 let us73denote the corresponding eigenvalues of the mass tensor J by λ1 (Πi ), λ2 (Πi )and the corresponding angular velocity by ω(Πi ).Definition 19. The parabolic diagram of a relative equilibrium is the following set of parabolas and vertical lines drawn on the same coordinate plane:• For each Πi draw a parabola given byχi (x) =(x − λ1 (Πi )2 )(x − λ2 (Πi )2 ).ω(Πi )2 (λ1 (Πi ) + λ2 (Πi ))2• For all fixed axes draw verticals lines through the squares of corresponding eigenvalues.Remark 6.4.1. Each χi is a quadratic function the roots of which are squaresof the eigenvalues of J.

Therefore what we do is we simply draw parabolasthrough the squares of eigenvalues corresponding to moving axes and verticallines through the squares of eigenvalues corresponding to fixed axes.Remark 6.4.2. The leading coefficient of χi is inverse proportional to thesquare of the angular momentum m(Πi ) = ω(Πi )(λ1 (Πi ) + λ2 (Πi )).Let us accept the following formal agreement:• Two parabolas intersect at infinity, if their leading coefficients are equaland non-zero, i.e. if they have only one point of intersection (of multiplicity one).• Two parabolas are tangent at infinity if they can be obtained fromeach other by a vertical shift, i.e.

have no points of intersection (realor complex).Definition 20. We will say that a parabolic diagram is generic if all intersections on it are simple, i.e. it contains no multiple intersections and nopoints of tangency.746.5Stability theoremsTheorem 12. Let M be a regular equilibrium. Suppose that• M has no more than two fixed axes (dim Π0 ≤ 2).• The parabolic diagram of M is generic.• All intersections on the parabolic diagram are either real and belong tothe upper half-plane or infinite.Then M is stable.Theorem 13.

Let M be a regular equilibrium. Suppose that• M has no more than two fixed axes (dim Π0 ≤ 2).• There is at least one intersection on the parabolic diagram of M whichis either complex or belong to the lower half-plane.Then M is unstable.Theorem 14. Let M be an exotic equilibrium.

Then M is unstable.Question 1. Can we omit the condition dim Π0 ≤ 2 in Theorems 12, 13 4 ?Question 2. Can we omit the condition that the parabolic diagram is genericin Theorem 12?Whether or not the answer to these questions is positive, Theorems 12,13, 14 are already enough to solve the stability problem for an open dense4The problem is that if dim Π0 > 2, then M ∈ Bad in terms of the first part of thisthesis (i.e.

the rank of all brackets of the pencil drops at M ) and we don’t know how towork with such points. However, it is quite natural to consider rotations with lots of fixedaxes. The answer to the question is probably positive.75Figure 6.2: Rotation of a 3d-body around the shortest principal axis of inertia. The intersection point is above the X axis ⇒ stableFigure 6.3: Rotation of a 3d-body around the middle principal axis of inertia.The intersection point is below the X axis ⇒ unstablesubset of relative equilibria.

Proof of these three theorems can be found inSection 6.18.Now we shall discuss some examples. First let us recover the classicalthree-dimensional result.Example 6.5.1 (Three-dimensional body). Parabolic diagrams for a 3dimensional body are illustrated on figures 6.2, 6.3, 6.4. We see from theparabolic diagrams that the rotation around the middle axis is unstable,while the rotations about the shortest and the longest axes are stable.76Figure 6.4: Rotation of a 3d-body around the longest principal axis of inertia.The intersection point is above the X axis ⇒ stableExample 6.5.2 (Four-dimensional body). There are three different cases:1.

The first plane is spanned by two short axes of inertia, the secondplane is spanned by two long axes of inertia. We look at the parabolicdiagram (see figure 6.5) and see that the rotation is stable.2. The first plane is spanned by the shortest and the second longest axes,the second plane is spanned by the longest and the third shortest axes.We look at the parabolic diagram (see figure 6.6) and see that therotation is unstable.3.

The first plane is spanned by the longest and the shortest axes, andthe second plane is spanned by two middle axes. See figures 6.7, 6.8,6.9, 6.10, 6.11. We see that everything depends on the ratio of angularvelocities here. If the rotation in the “inner” (i.e. spanned by themiddle axes) plane is fast enough, we have stability. If it is slow, wehave instability.The situation is similar to the rotation of a 3d body with a gyroscopeinside around the middle axis of inertia.

If the gyroscope is rotating77Figure 6.5: Rotation of a 4d-body. Parabolic diagram is generic, all intersection points are above the X axis ⇒ stableFigure 6.6: Rotation of a 4d-body. One intersection point is above the Xaxis, second is below ⇒ unstable78Figure 6.7: Rotation of a 4d-body. Parabolic diagram is generic, all intersection points are above the X axis ⇒ stablefast enough, it stabilises the rotation of the body. The “inner” planein a four-dimensional body plays the role of a gyroscope.In the non-generic case illustrated on figure 6.10 our theorems do notgive an answer. However, it is possible to show that the correspondingrotation is stable (and admits a Lyapunov function of degree four). Thisimplies that stability loss in four-dimensional body is always “soft”, i.e.the set of stable equilibria is closed.79Figure 6.8: Rotation of a 4d-body.

All intersection points are complex ⇒unstableFigure 6.9: Rotation of a 4d-body. All intersection points are below the Xaxis ⇒ stable80Figure 6.10: Rotation of a 4d-body. Parabolic diagram is not generic, allintersections point are above the X axis ⇒ we don’t know!Figure 6.11: Rotation of a 4d-body. Parabolic diagram is not generic, butthere is an intersection point below the X axis ⇒ unstable816.6The bihamiltonian structureWe will denote the Lie-Poisson bracket on so(n)∗ by { , }∞ . It is given by{f, g}∞ (M ) = hM, [df, dg]i, where M ∈ so(n)∗ .Proposition 6.6.1. The equations (6.1) are hamiltonian with respect to theLie-Poisson bracket on so(n)∗ with the hamiltonian given by the kinetic energy1H = hΩ, M i2Proof. Consider arbitrary function f .

The derivative of f with respect to theequations will behdf, [M, Ω]i = hM, [Ω, df ]i = hM, [dH, df ]i = {H, f },q.e.d.Now we introduce a second operation on so(n) defined by[X, Y ]J 2 = XJ 2 Y − Y J 2 X.Proposition 6.6.2. [ , ]J 2 is a Lie bracket compatible with the standard Liebracket. In other words, any linear combinations of these brackets define aLie algebra structure on so(n).Proof. It suffices to show that all operations on so(n) having the form[X, Y ]A = XAY − Y AXare Lie brackets, i.e.

they satisfy the Jacobi identity. But this is just astraightforward computation.82Now introduce an operation { , }0 on so(n)∗ given by{f, g}0 = hM, (df )J 2 (dg) − (dg)J 2 (df )i.This is a Lie-Poisson brackets associated with the Lie bracket [ , ]J 2 . Proposition 6.6.2 implies the followingProposition 6.6.3. { , }0 is a Poisson bracket compatible with the LiePoisson bracket.We will write down the corresponding pencil in the form{f, g}λ = {f, g}0 − λ{f, g}∞ = hM, df (J 2 − λE)dg − dg(J 2 − λE)df i.Theorem 15 (A.Bolsinov, [5, 6]). The system (6.1) is Hamiltonian withrespect to any bracket { , }λ , i.e.

it is bihamiltonian. The hamiltonian isgiven by√√1Hλ = − h(J + λE)−1 Ω(J + λE)−1 , M i.2√Remark 6.6.1. Since J is posisitve-definite, the matrix J + λE is invertiblefor any λ.Proof. Let us write down the condition that a system is hamiltonian withrespect to { , }λ . Let Hλ be the corresponding hamiltonian and f be anarbitrary function. Thendf= {Hλ , f }λ = hM, dHλ (J 2 − λE)df − df (J 2 − λE)dHλ i =dt= hdf, M dHλ (J 2 − λE) − (J 2 − λE)dHλ M i,where df /dt is the derivative with respect to our system. On the other hand,df= hdf, Ṁ i,dt83thereforeṀ = M dHλ (J 2 − λE) − (J 2 − λE)dHλ M.But we know thatṀ = [ΩJ + JΩ, Ω] = JΩ2 − Ω2 J.Consequently, we get an equation on Hλ of the form(ΩJ + JΩ)dHλ (J 2 − λE) − (J 2 − λE)dHλ (ΩJ + JΩ) = JΩ2 − Ω2 J.Denote X = dHλ , ν =√λ.

Now our equation can be rewritten asJΩ2 − Ω2 J = (Ω(J + νE) + (J − νE)Ω)X(J + νE)(J − νE)−− (J − νE)(J + νE)X(Ω(J − νE) + (J + νE)Ω) == Ω(J + νE)X(J + νE)(J − νE) − (J − νE)(J + νE)X(J + νE)Ω++ (J − νE)(ΩX(J + νE) − (J + νE)XΩ)(J − νE)Now denoteY = (J + νE)X(J + νE).We haveJΩ2 − Ω2 J = ΩY (J − νE) − (J − νE)Y Ω++ (J − νE)(Ω(J + νE)−1 Y − Y (J + νE)−1 Ω)(J − νE)Obvoiusly, Y = −Ω is a solution. Therefore,dHλ = −(J + νE)−1 Ω(J + νE)−1and√√1Hλ = − h(J + λE)−1 Ω(J + λE)−1 , M i,2q.e.d.84(6.4)Remark 6.6.2.