On the case of Goryachev-Chaplygin and new examples of integrable conservative systems on S^{2}

Elena N. Selivanova^{∗}

Department of Geometry, Nizhny Novgorod State Pedagogical University 603000 Russia, Nizhny Novgorod, ul. Ulyanova 1

Mathematisches Institut, Universität Tübingen Auf der Morgenstelle 10, 72076 Tübingen, Germany e-mail: lena@moebius.mathematik.uni-tuebingen.de Fax: (07071) 294322

Abstract. The aim of this paper is to describe a class of conservative systems on S^{2} possessing an integral cubic in momenta. We prove that this class of systems consists off the case of Goryachev-Chaplygin, the one-parameter family of systems which has been found by the author in the previous paper and a new two-parameter family of conservative systems on S^{2} possessing an integral cubic in momenta.

1 Introduction

Let M be an n-dimensional Riemannian manifold, and U : M → R be a smooth function on M. For the Lagrangian L : T M → R we consider

L(η) = \frac{|η|^{2}}{2} + (U ∘ π)(η)

where π : T M → M is the canonical projection, see [1]. In local coordinates q_{1}, ... , q_{n}, q̇_{1}, ... , q̇_{n} on T M we have

L(η) = \frac{1}{2} ∑ g_{ij}q̇^{i}q̇^{j} + U(q).

Identifying T M and T^{∗}M by means of the Riemannian metric, we get a Hamiltonian system with the Hamiltonian H : T^{∗}M → R which in local coordinates q_{1}, ... , q_{n}, p_{1}, ... , p_{n} on T^{∗}M has the form

∗Supported by DAAD.

H = \frac{1}{2} ∑ g^{ij}p_{i}p_{j} + U(q) = K + U.

We will call these Hamiltonian systems conservative systems on M.

A smooth function F : T^{∗}M → R which is an integral of the Hamiltonian system with the Hamiltonian H and which is independent of H we will call an integral of this system of degree m in momenta if in local coordinates F has the form

F = ∑_{k_{1}+ ... +k_{n}≤m} a_{k_{1} ... k_{n}}(q)p_{1}^{k_{1}} ... p_{n}^{k_{n}}.

We will say that two Hamiltonians H_{1} = K_{1}+U_{1} and H_{2} = K_{2}+U_{2} are equivalent if there exists a diffeomorphism ϕ of M and a diffeomorphism Φ of T^{∗}M such that the diagram

Φ : T^{∗}M → T^{∗}M
π′ ↓ ↓ π′
ϕ : M → M

is commutative, where Φ is linear for p ∈ M fixed, and if there are some nonzero constants κ_{1}, κ_{2} such that Φ^{∗}(K_{1}) = κ_{1}K_{2}, ϕ^{∗}(U_{1}) = κ_{2}U_{2}.

Clearly, if the Hamiltonians H_{1}, H_{2} are equivalent and one of the corresponding systems possesses an integral of degree m in momenta, then the other system has the same property.

In this paper we consider integrable conservative systems on S^{2} with the Hamiltonians of the form

H = λ(r^{2})(r^{2}dφ^{2} + dr^{2}) + ρ(r^{2})r cos φ = K + U  (1)

in polar coordinates r, φ where λ, ρ : R^{+} → R are some smooth functions.

There is a known example of an integrable conservative system on S^{2}, the case of Goryachev-Chaplygin in the dynamics of a rigid body, possessing an integral cubic in momenta. The total energy (the Hamiltonian) of this system has the following form

H = \frac{du_{1}^{2} + du_{2}^{2} + 4du_{3}^{2}}{4u_{1}^{2} + 4u_{2}^{2} + u_{3}^{2}} − u_{1},  (2)

where S^{2} is given by u_{1}^{2} + u_{2}^{2} + u_{3}^{2} = 1, (see [2]).

The aim of this paper is to classify (up to equivalence, see above) the Hamiltonians of the form (1) of conservative systems on S^{2}, possessing an integral

F = p_{φ}^{3} + κp_{φ}K + A  (3)

where A is a linear polynomial in momenta and κ is an arbitrary constant. We prove that these systems include the case of Goryachev-Chaplygin, a one-parameter family from [5] and a new two-parameter family of conservative systems on S^{2} possessing an integral cubic in momenta. It turns out that all these systems are due to smooth solutions of the following differential equation

x′x′′′ = xx′′ − 2x^{′′2} + x^{′2} + x^{2}.  (4)

For this reason we start with investigations of this equation. We show that there is a positive constant T such that any smooth solution of (4) which exists everywhere can be obtained from the solutions x = x(t) : R → R of the initial value problem

x′x′′′ = xx′′ − 2x^{′′2} + x^{′2} + x^{2}, x(0) = 0, x′(0) = 1, x′′(0) = τ,  (5)

where τ ∈ [0, T], or from x(t) = exp t, x(t) = cosh t by a scaling x → α_{1}x, α_{1} −const or a linear translation of time t → ±t + t_{0}, t_{0} − const.

In the next section we obtain a criterion for integrability for geodesic flows of Riemannian metrics as a partial differential equation. In the last section we consider a class of solutions of this equation and reduce the problem of classification of conservative systems on S^{2} to the classification (up to a scaling or linear translations) of smooth solutions of (4). We will show that the known case of Goryachev-Chaplygin is due to the solution of (5) if τ = T (if τ = 0 in (5) we get the metric of constant positive curvature). Then we prove the main theorem of classification and find new integrable cases on S^{2}.

2 Smooth solutions

Theorem 2.1 There is a positive constant T such that any smooth solution of (4) which exists everywhere can be obtained from x(t) = exp t, x(t) = cosh t or from the solutions of the initial value problem (5), where τ ∈ [0, T], by a scaling x → α_{1}x, α_{1} − const or a linear translation of time t → ±t + α_{0}, α_{0} − const.

For any solution x_{τ} of (5) with τ ∈ [0, T] there are smooth functions ξ_{τ}, ζ_{τ}, μ_{τ}, ν_{τ} such that

x′_{τ}(t) = (exp(−t))ξ_{τ}(exp(2t)) = (exp t)ζ_{τ}(exp(−2t)),
(x′′_{τ} − x_{τ})x_{τ}^{′2}(t) = (exp t)μ_{τ}(exp(2t)) = (exp(−t))ν_{τ}(exp(−2t))

where x′_{τ}(t) is positive everywhere if and only if τ ∈ [0, T) and there is one value of the parameter t = t_{0} such that x′_{T} (t_{0}) = 0.

Proof. The initial value problem (4) has a unique solution x(t) = Θ_{τ}(t) which is positive on (0, ε) and negative on (−ε, 0) for a sufficiently small ε.

Let us consider the case t > 0.

The differential equation from (4) can be replaced by the following differential equation of the second order

q̈ = \frac{1}{q} (1 + 2q^{2} − 3q^{4} + q̇ − 7q^{2}q̇ − 2q̇^{2})

with q(t) = R′(t) where R(t) = log x(t). We may rewrite this equation as a system of differential equations of the first order:

q̇ = p, ṗ = \frac{1}{q} (1 + 2q^{2} − 3q^{4} + p − 7q^{2}p − 2p^{2}) .  (6)

Since system (6) is symmetric with respect to q ↦ −q, t ↦ −t, it suffices to consider the case q > 0.

In order to obtain the phase portrait of (6) we may consider the following smooth system

q̇ = qp, ṗ = 1 + 2q^{2} − 3q^{4} + p − 7q^{2}p − 2p^{2}.  (7)

The solutions of (6) are obtained from the solutions of (7) by a reparametrization. The system (7) has four singular points: two saddle points p = 1, q = 0, p = −\frac{1}{2}, q = 0 and two nodes p = 0, q = ±1.

So, if a solution of (4) is smooth everywhere, then it corresponds to a smooth solution of (6) and, therefore, it corresponds to p = 0, q = ±1 or an orbit p = −q^{2} + 1, |q| < 1 or maybe two orbits of (6) where q → −∞ as t → 0− and q → +∞ as t → 0+. In the cases p = 0, q = ±1 and p = −q^{2} + 1, |q| < 1 a solution of (4) is obtained from x(t) = exp t and from x(t) = cosh t correspondingly by a scaling or a linear translation of time.

The aim of our further investigations is to show that the orbits of (6), corresponding to the solutions of (4), when τ belongs to a certain interval, converge to the singular point q = 1, p = 0.

For any solution x(t) = Θ_{τ}(t), t ∈ (0, ε) of (4) there is an orbit {Γ_{τ} : p = p_{τ}(q)} in the phase spase of (6).

For any solution of (4) it holds R(t) → −∞ as t → 0+ and, therefore, q(t) = R′(t) → +∞, p(t) = q′(t) → −∞ as t → 0+. Thus p_{τ}(q) → −∞ as q → +∞.

We may now consider only the orbits of (6) where p → −∞ as q → +∞.

Show that if τ_{1} > τ_{2}, then p_{τ_{1}}(q) > p_{τ_{2}}(q). Indeed, for a solution x(t) = Θ_{τ_{1}}(t) of (4) it holds

τ_{1} = lim_{t→0+} \frac{Θ′′_{τ_{1}}(t)}{Θ′_{τ_{1}}(t)} = lim_{q→+∞} \frac{q^{2} + p_{τ_{1}}(q)}{q}.  (8)

Note that the function x(t) = sinh t satisfies (4) when τ = 0. The related orbit of (6) has then the following form {Γ_{0} : p = p_{0}(q) = −q^{2} + 1}.

So, the orbits of (6), corresponding to the solutions of (4), for τ ≥ 0 converge to the singular point q = 1, p = 0.

We prove below that orbit (∗), where p → −\frac{1}{2} as q → 0+, corresponds to the solution of (4) when τ is equal to a negative constant −T and all orbits of (6), lying between (∗) and Γ_{0} correspond to the solutions of (4) for τ ∈ (−T, 0).

Assume first that there exists a constant τ_{0} < 0 such that orbit Γ_{τ_{0}} does not converge to the point q = 1, p = 0.

Consider the set W of the orbits of (6), lying between Γ_{τ_{0}} and Γ_{0}. Show that for any orbit in W the value q becomes infinite in a finite time interval.

In fact, for any solution of (6) it holds

∫ ^{q(t)} \frac{dq}{p} = t + const.  (9)

For orbits in W we have p < −q^{2} + 1. Hence, the left hand side of (9) is bounded for any orbit in W as q → +∞. Without loss of generality we assume that for these solutions of (6) t vanishes as q becomes infinite.

We conclude that for any orbit from W it holds by (8)

(τ_{0} + o(1))q < p + q^{2} < 1

and for a corresponding solution of (4) we get for t → 0+

τ_{0} + o(1) < \frac{x′′(t)}{x′(t)} < 1.

Thus, for a corresponding solution of (4) the function (log x′(t))′ is bounded in an interval, containing 0+. Hence, lim_{t→0+} x′(t) is bounded for a solution x(t) of (4), corresponding to an orbit in W and, therefore, x(t) = \frac{x′(t)}{q} → 0 as t → 0+.

Let us consider the orbit (∗) where p = p^{∗}(q). If (∗) is the same as Γ_{τ_{0}} then T = −τ_{0}.

If (∗) is not the same as Γ_{τ_{0}} then it belongs to W and, hence, corresponds to the solution of (4) for τ = −T, where T is equal to a positive constant and, moreover, T < −τ_{0}.

Thus we have shown that for any τ ∈ (−T, T) where T is a positive constant the system (4) has a solution for all t ≥ 0.

Let us assume now that there is no such orbit Γ_{τ_{0}} and, so, all orbits of (6), corresponding to the solutions of (4), converge to the singular point q = 1, p = 0.

Clearly, the orbit (∗) corresponds to a solution x(t) = η(t) of the differential equation in (4). As mentioned above, the function q(t) = \frac{η′(t)}{η(t)} becomes infinite in a finite time interval of t, say, as t → 0+. Since p^{∗}(q) < −q^{2} − kq for any k and for sufficiently large q, for η(t) it holds

lim_{t→0+} \frac{η′′(t)}{η′(t)} = −∞.  (10)

Note that η(0) is finite because R(0) − R(t) = ∫_{t}^{0} q(t)dt < 0 for t > 0 and η(t) = exp R(t). So, there are two cases: η(0) = 0 or η(0) ≠ 0.

Assume that η(0) ≠ 0. It follows that η′(0+) = +∞ and from (10) we get η′′(0+) = −∞.

Rewrite now the differential equation in (4) in the following form

\frac{x′′′}{x} = −3\frac{x′′ − x}{x′} − 2\frac{(x′′ − x)^{2}}{xx′} + \frac{x′}{x} .

Therefore, it holds

\frac{η′′′}{η} = −3\frac{p^{∗}(q) + q^{2} − 1}{q} − 2\frac{(p^{∗}(q) + q^{2} − 1)^{2}}{q} + q.

We obtain η′′′(0+) = −∞. This is a contradiction η′′(0+) = −∞.

Assume that η(0) = 0 and η′(0) ≠ 0. From (10) we get η′′(0+) = −∞. Rewriting the differential equation in (4) in the following form

x′′′ = x′ − 3\frac{(x′′ − x)x}{x′} − 2\frac{(x′′ − x)^{2}}{x′} ,

we obtain again η′′′(0+) = −∞.

Therefore, we only have to consider one case η(0) = 0 and η′(0) = 0. Taking into account that η′(t) = η(t)q(t) and q(t) > 0 as t → 0+, we conclude that η′(t) → 0+ as t → 0+ but on the other hand from (10) it follows that η′′(t) < 0.

These contradictions finally show that there is an orbit Γ_{τ_{0}} which does not converge to the point q = 1, p = 0 and, therefore, for any τ ∈ (−T, T), where

T = − lim_{q→+∞} \frac{p^{∗}(q) + q^{2}}{q} < ∞,  (11)

solutions of (4) exist on [0, +∞).

Since a solution x(t) = Θ_{τ}(t) of (4) equals −Θ_{−τ}(−t) if t ≤ 0, solutions of (4) exist on (−∞, +∞).

We show now the asymptotic behaviour of the solutions of (5) with τ ∈ [0, T] where T is defined by (11).

Put s = exp(−2y) and

g(s) = \sqrt{s}x (−\frac{1}{2} log s) .

Then the initial value problem (4) can be rewritten as

g′′′ = 3\frac{g′′g′}{g − 2g′s} + s\frac{g^{′′2}}{g − 2g′s}, g(1) = 0, g′(1) = −\frac{1}{2}, g′′(1) = τ  (12)

We will consider τ ∈ (−T, T].

Compute now for a solution of (4):

x′(t) = (exp t)(g − 2g′s),  (13)
x′′(t) − x(t) = 4 exp (−3t)g′′(s).  (14)

Let us consider the system (6). Since the eigenvalues of the Jacobian of (6) at (q, p) = (1, 0) are equal to −2, −4, there exist functions P and Q of class C^{1} such that

q = Q(s) = 1 + Cs + o_{1}(s),

p = P (s) = −2Cs + o_{2}(s),

where C is a constant, see [4].

We may write:

R(t) = ∫ ^{t} q = t + ψ_{1}(s).

Let us show that ψ_{1} is of class C^{1}. By differentiation we obtain

\frac{dψ_{1}(s)}{ds} = −\frac{q − 1}{2s} = −\frac{Cs + o_{1}(s)}{2s} ∈ C^{0}.

Write now for a solution of (4)

x(t) = exp R(t) = (exp t) exp ψ_{1}(s) = (exp t)g(s),

where we have used that g = r^{−1}u = exp(−t)u = exp(−t)x(t), and, therefore, g(0) = exp ψ_{1}(0) ≠ 0.

For a solution of (4) which can be extended into infinity it holds

x′(t) = x(t)q(t) = (exp t)(exp ψ_{1}(s))(1 + Cs + o_{1}(s)) =
= (exp t)g(s)(1 + Cs + o_{1}(s)), s = exp(−2t).

On the other hand, from (13) we obtain

g(s)(1 + Cs + o_{1}(s)) = g(s) − 2sg′(s).

Thus, for any solution of (12) where τ ∈ (−T, T] it holds g′(0) < ∞.

As mentioned above, the solution of (4) for τ = 0 has the form x(t) = sinh t.

Let us show that for any solution of (12) where τ ∈ (−T, T] and τ ≠ 0 it holds 0 < |g′′(0)| < ∞.

We compute

x′′(t) − x(t) = x(t)(p + q^{2} − 1) = (exp t)(exp ψ_{1}(s))(P (s) + Q^{2}(s) − 1) =

= (exp t)(exp ψ_{1}(s))(−2Cs + o_{2}(s) + (1 + Cs + o_{1}(s))^{2} − 1).

Thus,

x′′(t) − x(t) = (exp t)ψ_{2}(s),

where ψ_{2} ∈ C^{1}, ψ_{2}(s) = o_{3}(s). Rewrite the differential equation in (4) in the following form

(x′′′ − x′)x′ = −3(x′′ − x)x − 2(x′′ − x)^{2}.

We obtain either x′′(t) − x(t) ≡ 0 (then τ = 0) or

(log |x′′(t) − x(t)|)′ = −3\frac{1}{q} − 2|x′′(t) − x(t)|(exp t exp ψ_{1}(s)(1 + Cs + o_{1}(s)))^{−1} =

= −3\frac{1}{q} − 2|ψ_{2}(s)|(exp ψ_{1}(s)(1 + Cs + o_{1}(s)))^{−1}.

So,

(log |x′′(t) − x(t)|)′ = −3(1 − Cs) + o_{4}(s).

Then it follows

(log ((exp t)|ψ_{2}(s)|))′ = −3(1 − Cs) + o_{4}(s).

Thus,

(t + log |ψ_{2}(s)|)′ = −3(1 − Cs) + o_{4}(s).

By integrating we get

(log |ψ_{2}(s)|) = −4t − \frac{3C}{2} s + ψ_{3}(s),

where ψ_{3} ∈ C^{1}.

Thus,

|ψ_{2}(s)| = (exp(−4t)) exp(−\frac{3C}{2} s + ψ_{3}(s)) = s^{2}ψ_{4}(s),

since s = exp(−2t). So, ψ_{4} ∈ C^{1} and ψ_{4}(0) = const ≠ 0. Taking into account (14), we obtain |g′′(0)| = \frac{1}{4}ψ_{4}(0) and therefore 0 < |g′′(0)| < ∞.

So, we conclude that the solutions of (12) for τ ∈ (−T, T] are of class C^{∞} in zero.

Now the functions ζ_{τ}, ν_{τ} can be calculated in terms of g. From (13) ζ_{τ}(0) > 0 in view of g(0) > 0. Therefore ζ_{τ} > 0 everywhere on [0, +∞).

Since a solution x(t) = x_{τ}(t) of (4) equals −x_{−τ}(−t) for τ ∈ (−T, T) and x_{T} (t + t_{0}) = x_{T} (−t + t_{0}), the theorem follows.

□

3 A criterion for integrability

Consider a metric ds^{2} = Θ(u, v)(du^{2} + dv^{2}) in conformal coordinates u, v. It can also be written as

ds^{2} = θ(w, w̄)dwdw̄  (15)

where w = u+iv. The geodesic flow of ds^{2} is a Hamiltonian system with Hamiltonian

H = \frac{p_{w}pw̄}{4θ(w, w̄)}.  (16)

A polynomial F in momenta p_{u}, p_{v} can be also written as

F = ∑_{k=0}^{n} b_{k}(w, w̄)p_{w}^{k}pn−k ; w̄

where

b_{k} = \overline{b_{n−k}}, k = 0, ..., n.

If the polynomial F is an additional integral of the geodesic flow with the Hamiltonian (16), then {F, H} = 0 and the following holds

θ\frac{∂b_{k−1}}{∂w} + (n − (k − 1))b_{k−1} \frac{∂θ}{∂w} + θ\frac{∂b_{k}}{∂w̄} + kb_{k} \frac{∂θ}{∂w̄} = 0,  (17)

where k = 0, ..., n + 1 and b_{−1} = b_{n+1} = 0. Substituting k = 0 and k = n + 1 in (17) we get immediately

\frac{∂}{∂w̄}b_{0} ≡ 0

and

\frac{∂}{∂w}b_{n+1} ≡ 0.

One may show that if a polynomial in momenta integral F of the geodesic flow of (15) is independent of the Hamiltonian and an integral of smaller degree, then there is a conformal coordinate system z = z(w) of this metric such that the coefficients of F for p_{z} and pz̄ are equal to 1 identically.

Theorem 3.1 Let ds^{2} = λ(z, z̄)dzdz̄ (z = φ + iy) be a metric such that there exists a function f : R^{2} ↦ R, satisfying the following conditions

λ = \frac{1}{4} (\frac{∂^{2}f}{∂φ^{2}} + \frac{∂^{2}f}{∂y^{2}} ) ,

and

\frac{∂}{∂φ} ((\frac{∂^{2}f}{∂φ^{2}} − \frac{∂^{2}f}{∂y^{2}} ) (\frac{∂^{2}f}{∂φ^{2}} + \frac{∂^{2}f}{∂y^{2}} )) = 2\frac{∂}{∂y} ((\frac{∂^{2}f}{∂φ∂y} ) (\frac{∂^{2}f}{∂φ^{2}} + \frac{∂^{2}f}{∂y^{2}} )) (18)

Then the geodesic flow of ds^{2} possesses an integral cubic in momenta.

If the geodesic flow of a metric ds^{2} possesses an integral which is cubic in momenta and it does not depend on the Hamiltonian and an integral of smaller degree then there exist conformal coordinates φ, y and a function f : R^{2} ↦ R such that ds^{2} = λ(z, z̄)dzdz̄ where z = φ + iy and (18) holds.

Proof. We will consider integrals

F = ∑_{k=0}^{n} a_{k}(z, z̄)p_{z}^{k}pn−k ; z̄

of the geodesic flow of ds^{2} where a_{0} = a_{n} ≡ 1.

The system (17) has then the following form

3\frac{∂λ}{∂z} + \frac{∂(a_{1}λ)}{∂z̄} = 0,  (19)

\frac{∂(a_{1}λ^{2})}{∂z} + \frac{∂(a_{2}λ)}{∂z̄} = 0.  (20)

(19) holds if and only if there exists a function h such that

λ = \frac{∂h}{∂z̄} and a_{1}λ = −3\frac{∂h}{∂z}.

(20) can be also rewritten in the following form

\frac{∂(a_{1}λ^{2})}{∂z} = −\frac{∂(\overline{a_{1}}λ^{2})}{∂z̄} = −\overline{\frac{∂(a_{1}λ^{2})}{∂z}}.

Therefore, the system (19), (20) is equivalent to the following condition

Re \frac{∂(a_{1}λ^{2})}{∂z} = Re \frac{∂}{∂z} (\frac{∂h}{∂z} \frac{∂h}{∂z̄}) = 0

where

Im \frac{∂h}{∂z̄} = Im λ(z, z̄) = 0,

i.e. there is a real function f such that \frac{∂f}{∂z} = h and (18) holds.

□

The equation (18) has been obtained first in [3]. We need to give hier another proof because further we need the following corollaries which are due to the above proof of Theorem 3.1.

Corollary 3.2 If in a conform coordinate system φ, y of ds^{2} = λ(φ, y)(dφ^{2} + dy^{2}) there is an integral of the geodesic flow of ds^{2} which has the form

F = α_{3}(p_{z}^{3} + a_{1}p_{z}^{2}pz̄ + \overline{a_{1}}p_{z}p2 ; z̄ + p3 ; z̄), z = φ + iy, α_{3} − const ≠ 0,  (21)

then there is a function f such that (18) holds.

Corollary 3.3 If in a conform coordinate system φ, y of ds^{2} = λ(φ, y)(dφ^{2} + dy^{2}) there is a function f such that (18) holds, then there is an integral F which has the form

F = p_{z}^{3} + a_{1}p_{z}^{2}pz̄ + \overline{a_{1}}p_{z}p2 ; z̄ + p3 ; z̄, z = φ + iy.

4 The Classification

Theorem 4.1 Let

ds^{2} = (Ψ_{1}(r^{2})r cos φ + Ψ_{2}(r^{2}))(r^{2}dφ^{2} + dr^{2})  (22)

be a metric on S^{2} in polar coordinates. Then the geodesic flow of ds^{2} possesses an integral of the form

F = iw^{3}p_{w}^{3} + b_{1}p_{w}^{2}pw̄ + \overline{b_{1}}p_{w}p2 ; w̄ − iw̄^{3}p3 ; w̄, w = r cos φ + ir sin φ  (23)

if and only if

Ψ_{1}(r^{2})r = ((ψ′′ − ψ)(log r))r^{−2}

and

Ψ_{2}(r^{2}) = cr^{−2}ψ^{′−2}(log r)

or

Ψ_{2}(r^{2}) = ar^{−2} \frac{ψ^{′2} − ψ^{2} + b}{ψ^{′2}}(log r)

where ψ is a smooth solution of (4) which exist everywhere on (−∞, +∞) and a, c, p are appropriate constants.

Proof. Assume that the geodesic flow of ds^{2} possesses an integral of the form (23). Let us consider z = φ + iy = −i ln w. Then (22) can be rewritten in the following form

ds^{2} = (Ψ_{1}(exp(2y))(exp y) cos φ + Ψ_{2}(exp(2y)))(exp(2y))(dφ^{2} + dy^{2}).

Thus,

ds^{2} = (Ψ_{3}(y) cos φ + Ψ_{4}(y))(dφ^{2} + dy^{2}) = λ(φ, y)(dφ^{2} + dy^{2}).  (24)

for some functions Ψ_{3}, Ψ_{4}. An integral of the form (23) has then the form (21). Now we can apply Corollary 3.2. So, there is a function f such that (18) holds.

Let us consider a function h(φ, y) = ψ(y) cos φ + ξ(y) where ψ′′ − ψ = Ψ_{3} and ξ′′ = Ψ_{4}. Then a function f from (18) has the form f = h + δ where δ_{φφ} + δ_{yy} = 0. Then for the function a_{1} from (21) we get

a_{1} = −3\frac{f_{φφ} − f_{yy}}{f_{φφ} + f_{yy}} + 6i\frac{f_{φy}}{f_{φφ} + f_{yy}} = −3\frac{h_{φφ} − h_{yy} + δ_{φφ} − δ_{yy}}{λ(φ, y)} + 6i\frac{h_{φy} + δ_{φy}}{λ(φ, y)} . (25)

Thus, the functions δ_{φy} and δ_{φφ} − δ_{yy} are periodic in φ. Therefore,

δ(φ, y) = P_{1}(φ, y) + a(φ^{2} − y^{2}) + bφy + ∑_{k∈N} (C_{k}(exp(−ky)) + D_{k}(exp(ky))) sin kφ

+ ∑_{k∈N} (E_{k}(exp(−ky)) + G_{k}(exp(ky))) cos kφ  (26)

where a, b, C_{k}, D_{k}, E_{k}, G_{k} are some constant and P_{1}(φ, y) is a linear polynomial in φ, y.

For the function b_{1} from (23) we have b_{1}(w, w̄) = a_{1}|w|^{2}w is bounded if |w| = 0, since ds^{2} is a metric on S^{2}. Then from (25) and (26) we obtain C_{k}, D_{k}, E_{k}, G_{k} are equal to zero when k ≥ 2.

Thus, we have to consider functions f of the form

f(φ, y) = ψ(y) cos φ + ξ(y) + a(φ^{2} − y^{2}) + (C_{1}(exp(−y)) + D_{1}(exp y)) sin φ (27)

satisfying (18). Taking into account that ds^{2} is a metric on S^{2} from (18) for (27) we obtain b = 0 and C_{1} = D_{1} = 0.

So, we must consider solutions of (18) of the following form

f(φ, y) = ψ(y) cos φ + ξ(y) + a(φ^{2} − y^{2}), a − const.

Then (18) is equivalent to the following conditions

λ = (ψ′′ − ψ) cos x + ξ′′

where

ξ′′ψ′′ + (ξ′′ψ′)′ = 2a(ψ′′ − ψ)  (28)

and

ψ′ψ′′′ = ψψ′′ − 2ψ^{′′2} + ψ^{′2} + ψ^{2}.  (29)

Multiply (28) by ψ′:

2ξ′′ψ′ψ′′ + ξ′′′ψ^{′2} = 2a(ψ′′ − ψ)ψ′

and integrate

ξ′′(y) = a\frac{ψ^{′2}(y) − ψ^{2}(y) + b}{ψ^{′2}(y)}, b − const if a ≠ 0

or

ξ′′(y) = cψ^{′−2}(y), c − const if a = 0.

Conversely integrability of the corresponding geodesic flows follows immediately from Corollary 3.3 and above calculations.

□

Thus, due to the Maupertuis’s principle we obtain the following corollary from Theorem 4.1.

Corollary 4.2 Hamiltonian systems with the Hamiltonians of the form

H = \frac{dφ^{2} + dy^{2}}{ψ^{′2}(y)} − (ψ′′(y) − ψ(y))ψ^{′2}(y) cos φ  (30)

and

H_{b} = \frac{ψ^{′2}(y) − ψ^{2}(y) + b}{ψ^{′2}(y)} (dφ^{2} + dy^{2}) − \frac{ψ^{′2}(y)(ψ′′(y) − ψ(y))}{ψ^{′2}(y) − ψ^{2}(y) + b} cos φ, (31)

where ψ is a solution of (4), possessing an integral cubic in momenta of the form (3).

Now we prove the main theorem.

Theorem 4.3 A Hamiltonian system with the Hamiltonian of the form (1) is a conservative system on S^{2}, possessing an integral of the form (3) and not possessing an integral quadratic or linear in momenta if and only if the corresponding Hamiltonian is equivalent to one of the following

1. (30) where ψ is a solution of (5) for τ ∈ (0, T).

2. (31) where ψ is a solution of (5) for τ ∈ (0, T) and

b > b_{∗}(τ) = max_{y∈R} (ψ^{2} − ψ^{′2}) or b < b^{∗}(τ) = min_{y∈R} (ψ^{2} − ψ^{′2})

where b_{∗}(τ) < +∞ and b^{∗}(τ) > −∞ for τ ∈ (0, T).

3. (31) where ψ is a solution of (5) for τ = T and p = ψ^{2}(y_{0}), when ψ′(y_{0}) = 0, which is in fact the case of Goryachev-Chaplygin in the dynamics of a rigid body.

Proof. From Theorem 4.1 and the Maupertuis’s principle we know that if a Hamiltonian system with the Hamiltonian of the form (1) defines a conservative system on S^{2}, possessing an integral of the form (3), then (1) is equivalent to (30) or (31) where ψ is a smooth solution of (4) which exists everywhere on (−∞, +∞). So, we must only describe when in this case (30), (31) define a conservative system on S^{2}. From Theorem 2.1 it follows that we must consider only solutions of (5) for τ ∈ [0, T] and ψ(y) = exp y, ψ(y) = cosh y. We note that if ψ is a solution of (5) for τ = 0 (then ψ(y) = sinh y) or ψ(y) = exp y, ψ(y) = cosh y, then the Hamiltonians (30),

(31) define simply a metric of constant curvature where there are two independent linear integrals.

So, we must only consider solutions ψ of (5) when τ ∈ (0, T].

Let us write then (30), (31) in polar coordinates φ, r = log y and φ̃ = −φ, r̃ = − log y. Using Theorem 2.1 we compute

H = \frac{1}{ξ_{τ}^{2}(r^{2})}(r^{2}dφ^{2} + dr^{2}) − μ_{τ}(r^{2})r cos φ

= \frac{1}{ζ_{τ}^{2}(r̃^{2})}(r̃^{2}dφ̃^{2} + dr̃^{2}) − ν_{τ}(r̃^{2})r̃cos φ̃

and

H_{b} = \frac{Φ_{τ}(r^{2}) + b + 1}{ξ_{τ}^{2}(r^{2})}(r^{2}dφ^{2} + dr^{2}) − \frac{μ_{τ}(r^{2})r cos φ}{Φ_{τ}(r^{2}) + b + 1}

= \frac{−Φ̃_{τ}(r̃^{2}) + b + 1}{ζ_{τ}^{2}(r̃^{2})}(r̃^{2}dφ̃^{2} + dr̃^{2}) − \frac{ν_{τ}(r̃^{2})r̃cos φ̃}{−Φ̃_{τ}(r̃^{2}) + b + 1}

where

Φ_{τ}(t) = ∫_{1}^{t} μ_{τ}(s)ξ_{τ}^{−1}(s)ds,
Φ̃_{τ}(t) = ∫_{1}^{t} ν_{τ}(s)ζ_{τ}^{−1}(s)ds.

Consider τ = T. In Theorem 2.1 it has been proved that for the corresponding solution ψ of (5) there is y = y_{0} such that ψ′(y_{0}) = 0, and therefore there is r = r_{0} > 0 such that ξ_{T} (r_{0}^{2}) = 0. Thus, in the case τ = T (30) does not define a system on S^{2} and (31) in this case defines a system on S^{2} if and only if p = ψ^{2}(y_{0}).

Consider 0 < τ < T. Then (30) defines a family of conservative systems on S^{2} which has been found in [5]. In [5] it has been proved that for τ_{1} ≠ τ_{2} the corresponding Hamiltonians are not equivalent, the systems possess nontrivial cubic integrals (there is no quadratic or linear integral) and no Hamiltonian from this family is equivalent to the case of Goryachev-Chaplygin.

Let us find the admissible values of the parameter b in this case. From the above expressions for H_{b} in polar coordinates we have

b + 1 > max_{[0,1]} −Φ_{τ} = M_{1}(τ) and b + 1 > max_{[0,1]} Φ̃_{τ} = M̃_{1}(τ)

or

b + 1 < min_{[0,1]} −Φ_{τ} = M_{2}(τ) and b + 1 < min_{[0,1]} Φ̃_{τ} = M̃_{2}(τ).

So, b > b_{∗}(τ) = max {M_{1}(τ), M̃_{1}(τ)} − 1 or b < b^{∗}(τ) = min {M_{2}(τ), M̃_{2}(τ)} − 1 where b_{∗}(τ) < +∞ and b^{∗}(τ) > −∞ for τ ∈ (0, T).

With the same arguments as in [5] we can prove that the corresponding systems do not possess integrals linear or quadratic in momenta.

Let us show that the Hamiltonians (31) where ψ is a solution of (5) for τ ∈ (0, T) are not equivalent for different values of the parameters τ and p.

Assume that the Hamiltoians of the form (31) for τ_{1}, b_{1} and τ_{2}, b_{2} are equivalent. We will use the following lemma which is in fact a particular case of some re-

sults of Kolokol’tsov, published in his Ph.D. Dissertation, (Moscow State University, 1984), another proof one can find in [5].

Lemma 4.4 Let

ds^{2} = λ(r^{2})(r^{2}dφ^{2} + dr^{2})  (32)

be a metric on S^{2} in polar coordinates r, φ. Then ds^{2} can be written in the form (32) in polar coordinates r̃, φ̃ if and only if r̃ = Dr^{±1}, D − const and φ̃ = ±φ + φ_{0}, φ_{0} − const.

Thus, from Lemma 4.4 it follows that there are some constants C_{0} ≠ 0, C_{3} ≠ 0, y_{1} such that

ψ′′_{1}(y) − ψ_{1}(y) = C_{0}(ψ′′_{2}(y + y_{1}) − ψ_{2}(y + y_{1})),  (33)
\frac{ψ_{1}^{′2}(y) − ψ_{1}^{2}(y) + b_{1}}{ψ_{1}^{′2}(y)} = C_{3} \frac{ψ_{2}^{′2}(y + y_{1}) − ψ_{2}^{2}(y + y_{1}) + b_{2}}{ψ_{2}^{′2}(y + y_{1})}  (34)

where ψ_{1}, ψ_{2} are solutions of (5) for τ = τ_{1} and τ = τ_{2} correspondingly. So, we get from (33)

ψ_{1}(y) = C_{0}ψ_{2}(y + y_{1}) + C_{1} exp y + C_{2} exp(−y)

for some constants C_{1}, C_{2}. As in the proof of Theorem 2.1 let us write for a solution ψ of (4): ψ(y) = (exp y)g(exp(−2y)). It has been shown that g satisfies then the differential equation from (12). So, from our assupmtion it follows that there are two solutions of the differential equation from (12) g_{1} and g_{2} such that g_{1}(s) = g_{2}(s) + C_{1} + C_{2}s. Then by substituting g_{1} in the differential equation from (12) we obtain C_{1} = C_{2} = 0. Thus,

ψ_{1}(y) = C_{0}ψ_{2}(y + y_{1}).

From the initial conditions of (5) we get 0 = ψ_{1}(0) = C_{0}ψ_{2}(y_{1}) and therefore ψ_{2}(y_{1}) = 0 but ψ_{2}(0) = 0 and ψ′_{2}(y) is positive everywhere. So, y_{1} = 0 and from ψ′_{1}(0) = ψ′_{2}(0) = 1 we obtain C_{0} = 1. Thus, τ_{1} = τ_{2}.

Prove now that b_{1} = b_{2}. So, we have from (34)

ψ_{1}^{′2}(y) − ψ_{1}^{2}(y) + b_{1} = C_{3}(ψ_{1}^{′2}(y) − ψ_{1}^{2}(y) + b_{2}),

and therefore, either C_{3} = 1 and b_{1} = b_{2} or ψ_{1}^{′2}(y) − ψ_{1}^{2}(y) ≡ const that is not true for τ ≠ 0. So we get b_{1} = b_{2}.

Assume that a Hamiltonian from (1.) is equivalent to a Hamiltonian from (2.). Then we obtain in the same way as above that for a solution ψ_{τ} of (5) it holds ψ_{τ}^{′2}(y) − ψ^{2}(y) ≡ const and therefore τ = 0. Thus, no Hamiltonian from (1.) is equivalent to a Hamiltonian from (2.).

Let us show that no Hamiltonian from this family is equivalent to the case of Gorychev-Chaplygin. There are polar coordinates r, φ such that (2) can be rewritten as

H = γ_{1}(r^{2})(r^{2}dφ^{2} + dr^{2}) − γ_{2}(r^{2})r cos φ = γ_{1}(r̃^{2})(r̃^{2}dφ^{2} + dr̃^{2}) − γ_{2}(r̃^{2})r̃cos φ (35)

where r̃ = r^{−1}.

Assume that the Hamiltonian (31) where ψ is a solution of (5) for τ = τ_{0} ∈ (0, T) and b = b_{1} is equivalent to the Hamiltonian of the case of Gorychev-Chaplygin. This means also from the symmetry of (35) that (33) and (34) hold for b_{1} = b_{2} and τ_{1} = τ_{0}, τ_{2} = −τ_{0}. In the same way as above we obtain τ_{0} = 0 but then the Hamiltonian of the case of Gorychev-Chaplygin is equivalent to the Hamiltonian of a metric of constant positive curvature.

So, we described all conservative systems on S^{2} with the Hamiltonians of the form (1) possessing an integral of the form (3) and not possessing an integral quadratic or linear in momenta. We proved also that no Hamiltonian from (1.) or (2.) is equivalent to the Hamiltonian of the case of Goryachev-Chaplygin. On the other hand the case of Goryachev-Chaplygin belongs to the class of conservative systems on S^{2} with the Hamiltonians of the form (1) possessing an integral of the form (3) and not possessing an integral quadratic or linear in momenta, see [5]. So, the Hamiltonian of the case of Goryachev-Chaplygin is equivalent to the Hamiltonian (31) where ψ is a solution of (5) for τ = T and b = ψ^{2}(y_{0}), when ψ′(y_{0}) = 0.

□

References

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[2] Bolsinov A.V., Kozlov V.V., Fomenko, A.T. The Maupertuis principle and geodesic flows on S^{2} arising from integrable cases in the dynamics of a rigid body, Russ. Math. Surv. 50, n.3, 473-501 (1995)

[3] Hall, L.S. A theory of exact and approximate configurational invariants, Phys 8D, 90-116 (1983)

[4] Hartman, P. Ordinary differential equations. New York-London-Sydney, John Wiley and Sons, 1964

[5] Selivanova E.N. New examples of conservative systems on S^{2} possessing an integral cubic in momenta, Preprint dg-ga/9711005 (1997)