New families of conservative systems on S^{2} possessing an integral of fourth

degree in momenta

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 e-mail: lena@moebius.mathematik.uni-tuebingen.de

Abstract. There is a well-known example of an integrable conservative system on S^{2}, the case of Kovalevskaya in the dynamics of a rigid body, possessing an integral of fourth degree in momenta. In this paper we propose new families of examples of conservative systems on S^{2} possessing an integral of fourth degree in momenta.

Key words: Integrable Hamiltonian system, Riemannian metric on S^{2}.

1991 Mathematics Subject Classification: 58F07

1 Introduction

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

L(η) = \frac{|η|^{2}}{2} + (U ∘ π)(η)

where π : T M → M is the canonical projection. 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).

The function U is reffered to as a potential function of the force field −∇U on M.

∗Supported by DAAD.

Identifying T M and T^{∗}M by means of the Riemannian metric, we get the 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

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 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}}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 exist 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 a fixed p ∈ M and Φ^{∗}(K_{1}) = κK_{2}, ϕ^{∗}(U_{1}) = κ̃U_{2} for some nonzero constants κ, κ̃.

Clearly, if the Hamiltonians H_{1} and H_{2} are equivalent and the Hamiltonian system with the Hamiltonian H_{1} possesses an integral of degree m in momenta, then the Hamiltonian system with the Hamiltonian H_{2} possesses an integral of degree m in momenta, too.

There is a well-known example of an integrable conservative system on S^{2}, the case of Kovalevskaya [10] in the dynamics of a rigid body, possessing an integral of fourth degree in momenta. The total energy (the Hamiltonian) of this system has the following form

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

where S^{2} is given by u_{1}^{2} + u_{2}^{2} + u_{3}^{2} = 1, (see [4]).

Goryachev proposed in [5] a family of examples of conservative systems on S^{2} possessing an integral of fourth degree in momenta. The Hamiltonians of these systems have the following form

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

where S^{2} is given by u_{1}^{2} + u_{2}^{2} + u_{3}^{2} = 1, B_{1}, B_{2} are some constants. Clearly, this family reduces to the case of Kovalevskaya when B_{1} = B_{2} = 0.

In this paper we propose new families of examples of conservative systems on S^{2} possessing an integral F of fourth degree in momenta, i.e. examples which are not equivalent to the cases of Kovalevskaya or Goryachev.

In our examples we use the solution ψ_{0} of the following initial value problem

ψ′′′ψ′ + 2ψ^{′′2} − 3ψ^{2} = 0, ψ(0) = 0, ψ′(0) = 1, ψ′′(0) = 0.  (3)

We set Ψ_{0}(r) = ψ_{0}(log r), Ψ_{1}(r) = ψ′_{0}(log r), Ψ_{2}(r) = ψ′′_{0}(log r).

We prove the following theorems.

Theorem 1.1 The Hamiltonian

H = \frac{1}{Ψ_{1}^{2}(r)} (dφ^{2} + \frac{dr^{2}}{r^{2}} ) − Ψ_{1}^{2}(r)(Ψ_{2}(r) − Ψ_{0}(r)) cos φ  (4)

defines a conservative system on S^{2} in polar coordinates. This system possesses an integral of fourth degree and does not possess an integral quadratic or linear in momenta.

The Hamiltonian (4) are not equivalent to (1) or (2).

Theorem 1.2 There are some constants p_{2} < p_{1} such that for any p > p_{1} or p < p_{2} the Hamiltonian H_{p} or −H_{p} correspondingly, where

H_{p} = \frac{Ψ_{1}^{2}(r) − Ψ_{0}^{2}(r) + p}{Ψ_{1}^{2}(r)} (dφ^{2} + \frac{dr^{2}}{r^{2}} ) − \frac{Ψ_{1}^{2}(r)(Ψ_{2}(r) − Ψ_{0}(r)) cos φ}{Ψ_{1}^{2}(r) − Ψ_{0}^{2}(r) + p}, (5)

defines a conservative system on S^{2}. Any system from these families possesses an integral of fourth degree in momenta and does not possess an integral quadratic or linear in momenta.

The Hamiltonians (5) for p > p_{1} or p < p_{2} are not equivalent to (1) or (2).

In [11] we proposed a new one-parameter family of conservative systems on S^{2} possessing an integral cubic in momenta.

The paper is organized as follows. In the first chapter we consider the initial value problem (3), show the existence of the solution ψ_{0} on R and investigate its asymptotic behaviour at infinity. In the second chapter we obtain a local criterion for the integrability of a geodesic flow with an integral of fourth degree and find a class of metrics with integrable geodesic flows. Finally, in the third chapter we prove Theorem 1.1 and Theorem 1.2.

2 Existence and asymptotic behaviour of the solution

In this chapter we consider the initial value problem (3), prove the existence of the solution ψ_{0} on R and investigate its asymptotic behaviour at infinity.

Theorem 2.1 The solution ψ_{0}(y) of the initial value problem (3) exists for all y ∈ R and, moreover, there are C^{∞} functions ν and μ such that

ψ′_{0}(y) = (exp y)ν(exp(−2y)) = exp(−y)ν(exp(2y)),
ψ_{0}^{′2}(y)(ψ′′_{0}(y) − ψ_{0}(y)) = exp(−y)μ(exp(−2y)) = −(exp y)μ(exp(2y))

and ν(t) > 0 for any t.

Proof. The initial value problem (3) has a unique solution ψ(y) = ψ_{0}(y) which is positive on (0, ε) and negative on (−ε, 0) for sufficiently small ε.

Let us consider the case y > 0.

Let R(y) = ln ψ(y) and q(y) = R′(y). Then the differential equation from (3) is equivalent to the following system of differential equations of the first order:

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

The system (6) has two singular points: the nodes p = 0, q = ±1.

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

Show that orbit Γ of (6), related to ψ_{0}(y) for y ≥ 0, converges to the point q = 1, p = 0 as y → +∞.

For ψ_{0}(y) we have R(y) → −∞ as y → 0+ and then q(y) = R′(y) → +∞ as y → 0+.

Introduce the function g such that Γ = {(q, p)|p = −q^{2} + qg(q^{−1})} and denote s = q^{−1}.

So, we may rewrite (3) as the following initial value problem

g′(1 − gs) + 3g^{2} − 3s^{2} = 0, g(0) = lim_{q→+∞} \frac{p + q^{2}}{q} = \frac{ψ′′_{0}(0)}{ψ′_{0}(0)} = 0  (7)

The solution of (7) is g(s) = s^{3}. So,

Γ = \{(q, p)|p = −q^{2} + \frac{1}{q^{2}} \}  (8)

Since p(1) = 0 in (8), the orbit Γ converges to the point q = 1, p = 0.

The Jacobian of (6) at (q, p) = (1, 0) has the eigenvalues −3, −4. Therefore (6) is C^{∞}-conjugate to their linear system in a neighborhood of q = 1, p = 0, see [8]. Taking into account (8), we conclude that there exists C^{∞}-functions ξ_{1}, ξ_{2} such that the solution of (6), related to ψ_{0}(y), y > 0, can be written as

q = 1 + ξ_{1}(exp(−4y)),
p = ξ_{2}(exp(−4y))

where ξ_{1}(0) = ξ_{2}(0) = 0.

So, we get

ψ_{0}(y) = (exp y)β(exp(−2y))

where

β(u) = exp(−\frac{1}{2} ∫ ^{u} \frac{ξ_{1}(t^{2})}{t}dt).

Thus, β is of class C^{∞}.

Since

ψ′_{0}(y) = ψ_{0}(y)q = (exp y)β(exp(−2y))(1 + ξ_{1}(exp(−4y)) = (exp y)ν(exp(−2y)),

ν is of class C^{∞} and ν(t) > 0 for any t ≥ 0.

Then we compute

ψ_{0}^{′2}(y)(ψ′′_{0}(y) − ψ_{0}(y)) = ψ_{0}^{′2}(y)ψ_{0}(y)(p + q^{2} − 1) =

exp(−y)ν^{2}(exp(−2y))β(exp(−2y))\frac{ξ_{2}(exp(−4y)) + 2ξ_{1}(exp(−4y)) + ξ_{1}^{2}(exp(−4y))}{exp(−4y)}

= exp(−y)μ(exp(−2y))

and, therefore, μ is of class C^{∞}.

Since ψ_{0}(y) = −ψ_{0}(−y), we get

ψ′_{0}(y) = (exp y)ν(exp(−2y)) = exp(−y)ν(exp(2y))

and

ψ_{0}^{′2}(y)(ψ′′_{0}(y) − ψ_{0}(y)) = exp(−y)μ(exp(−2y)) = −(exp y)μ(exp(2y)).
□

3 A criterion for the 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  (9)

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

H = \frac{p_{w}pw̄}{4θ(w, w̄)}.  (10)

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 (10), 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,  (11)

where k = 0, ..., n + 1 and b_{−1} = b_{n+1} = 0. Substituting k = 0 and k = n + 1 in (11) we get immediately

\frac{∂}{∂w̄}b_{0} ≡ 0

and

\frac{∂}{∂w}b_{n+1} ≡ 0.

Note that in another conform coordinate system

b̂_{0}(ŵ) = b_{0}(w) (\frac{dŵ}{dw})^{n} .  (12)

Proposition 3.1 The geodesic flow of a Riemannian metric on S^{2}

ds^{2} = λ(r^{2})(r^{2}dφ^{2} + dr^{2})  (13)

does not possess a nontrivial integral quadratic in momenta (which does not depend on H and linear integrals).

Proof. Let us introduce the conform coordinate system u = r cos φ, v = r sin φ. Denote w = u + iv and consider ŵ = w^{−1} for ds^{2}.

Let F = b_{0}(w, w̄)p_{w}^{2} +b_{1}(w, w̄)p_{w}pw̄ +b_{2}(w, w̄)p2 ; w̄ be an integral of the geodesic of (13). As above, b_{0} = b_{0}(w), b_{2} = b_{2}(w̄) = b̄_{0}(w) and Im b_{1}(w, w̄) = 0. Since (13) is a Riemannian metric on S^{2}, from (12) it follows that b_{0}(w) is a polynomial of degree n where n ≤ 4. If F does not depend on the Hamiltonian and linear integrals, then b_{0}(w) does not equal zero identically.

Then from (11) when k = 1 we obtain

λ\frac{∂b_{0}}{∂w} + 2b_{0} \frac{∂λ}{∂w} + λ\frac{∂b_{1}}{∂w̄} + b_{1} \frac{∂λ}{∂w̄} = 0.

This is equivalent to the following condition

Im \frac{∂}{∂w} (λ\frac{∂b_{0}}{∂w} + 2b_{0} \frac{∂λ}{∂w}) = 0

where b_{0}(w) = B_{0}w^{4} + B_{1}w^{3} + B_{2}w^{2} + B_{3}w + B_{4} and B_{0}, B_{1}, B_{2}, B_{3}, B_{4} are some constants such that at least one of them does not equal zero.

Denote ww̄ as t. So, we get the following conditions for λ(t)

Im (w^{2}(12B_{0}λ + 12B_{0}λ′t + 2B_{0}λ′′t^{2}) + w̄^{2}(2B_{4}λ′′))+
Im (w(6B_{1}λ + 9B_{1}λ′t + 2B_{1}λ′′t^{2}) + w̄(3B_{3}λ′ + 2B_{3}λ′′t))+

Im (B_{2}(2λ + 6λ′t + 2λ′′t^{2})) ≡ 0.

So, we have to consider the following linear differential equations

ϵ_{1}(6λ + 6λ′t + λ′′t^{2}) = ϵ_{2}(λ′′),  (14)
ϵ_{1}(6λ + 9λ′t + 2λ′′t^{2}) = ϵ_{2}(3λ′ + λ′′t),  (15)
λ + 3λ′t + 2λ′′t^{2} = 0  (16)

where ϵ_{1}, ϵ_{2} are constants equal to −1, 0 or 1. (In another case consider the rescaling t ↦ Dt, D − const, of a solution of (14) or (15).)

It is easy to see that for any smooth solution of (16) λ(0) = 0 and then ds^{2} is not a Riemannian metric on S^{2}. We get Im B_{2} = 0.

It is also easy to see that any solution of the equations (14), (15) if ϵ_{1} = 0 or ϵ_{2} = 0 does not define by (13) a Riemannian metric on S^{2}.

Solutions of (14) if ϵ_{1} = 1 and ϵ_{2} = 1 have the following form

λ(t) = C_{1} \frac{1 + t^{2}}{(−1 + t^{2})^{2}} + C_{2} \frac{t}{(−1 + t^{2})^{2}} , C_{1}, C_{2} − const.

Solutions of (14) if ϵ_{1} = 1 and ϵ_{2} = −1 have the following form

λ(t) = C_{1} \frac{1 − t^{2}}{(1 + t^{2})^{2}} + C_{2} \frac{t}{(1 + t^{2})^{2}} , C_{1}, C_{2} − const.

We get it noting that the solutions of (14) if ϵ_{1} = 1 and ϵ_{2} = 1 can be obtained from the solutions of (14) if ϵ_{1} = 1 and ϵ_{2} = −1 by complex transformation t → it.

It is easy to note that if λ(0) = 1 there is only one smooth solution of (15) if ϵ_{1} = 1 or ϵ_{2} = ±1. These solutions correspond to the metrics of curvature −1 if ϵ_{2} = 1 and to the metrics of curvature −1 if ϵ_{2} = −1. So, if ϵ_{1} = 1 and ϵ_{2} = 1 any solution of (15) which is smooth in zero has the following form

λ(t) = \frac{C_{1}}{(−1 + t)^{2}} , C_{1} − const

and if ϵ_{1} = 1 and ϵ_{2} = −1 any solution of (15) which is smooth in zero has the following form

λ(t) = \frac{C_{1}}{(1 + t)^{2}} , C_{1} − const.

Thus, we may see that either Im B_{2} = 0 and B_{i} = 0, i ≠ 2 or

λ(t) = \frac{C_{1}}{(1 + Dt)^{2}} , C_{1}, D − const

and, therefore,

ds^{2} = \frac{C_{1}}{(1 + Dr^{2})^{2}} (r^{2}dφ^{2} + dr^{2}) ,  (17)

where C_{1}, D − const, i.e. ds^{2} is a metric of constant positive curvature.

Remember that the geodesic flow of a metric (13) possesses the linear integral F_{1} = iwp_{w} − iw̄pw̄. If Im B_{2} = 0 and B_{i} = 0, i ≠ 2, we will consider the integral F_{2} = F + B_{2}F_{1}^{2} which is equal, obviously, to ĈH where Ĉ is a constant. Therefore, F depends on H and F_{1}.

So, if the geodesic flow of a metric (13) on S^{2} possesses a nontrivial quadratic integral, it can be only a metric of constant positive curvature. But it is well known that the geodesic flows of metrics of constant curvature possess two independent linear integrals. (Indeed, the geodesic flows with the Hamiltonians H = l^{′2}(y)(p_{φ}^{2} + p_{y}^{2}) where l′′(y) = l(y) possess the integrals F_{1} = p_{φ} and F̂_{1} = l(y) cos φp_{φ} − l′(y) sin φp_{y}.) So, in this case any quadratic integral is also trivial, i.e. depends on the Hamiltonian and linear integrals. □

Proposition 3.2 The geodesic flow of a metric (13) on S^{2} possesses two independent linear integrals if and only if it has the form (17) in polar coordinates, i.e. if it is a metric of constant positive curvature.

Proof. As mentioned above the geodesic flow of a metric (13) on S^{2} possesses the linear integral F_{1} = p_{φ}.

In the coordinates w, w̄, p_{w}, pw̄ where w = r cos φ + ir sin φ it can be written as follows F_{1} = iwp_{w} − iw̄pw̄. (Indeed, in the coordinates z = φ + iy, y = log r we have F_{1} = p_{z} + pz̄ and, clearly, w = exp iz.)

Assume that a metric ds^{2} of the form (13) on S^{2} possesses another linear integral F_{2} = d_{0}(w)p_{w} + d̄_{0}(w)pw̄ which is independent of F_{1} and, therefore, the function d_{0}(w) does not equal iwC_{3} where C_{3} is a real constant.

Let us consider then the function F_{2}^{2} = d_{0}^{2}(w)P_{w} + 2|d_{0}|^{2}p_{w}pw̄ + d̄_{0}(w)p2 ; w̄ which is also an integral of the geodesic flow of ds^{2}.

Compare now d_{0}^{2}(w) with b_{0}(w) from Proposition 3.1. Since d_{0}(w) ≠ iwC_{3} in our case, then at least one of the coefficients B_{i}, i ≠ 2 (see Proposition 3.1) does not equal zero. In the same way as in Proposition 3.1 we may show that ds^{2} has the form (17) in polar coordinates, i.e. ds^{2} is a metric of constant positive curvature.

□

Remember that conform coordinates x, y of a metric ds^{2} are Liouville if ds^{2} = (f(x) + h(y))(dx^{2} + dy^{2}) for some functions f, h. Due to Darboux [6] Liouville coordinates exist if and only if the geodesic flow of ds^{2} possesses an integral quadratic in momenta and moreover if one of the function f or h is constant then the geodesic flow possesses an integral linear in momenta. So, we may formulate the following

Corollary 3.3 Liouville coordinates φ, y = log r related to polar coordinates φ, r of a metric (13) on S^{2} are unique up to shifts and the transform y → −y.

Proof. Assume that a metric ds^{2} on S^{2} can be written in two different forms (13)

ds^{2} = λ_{1}(r^{2})(r^{2}dφ^{2} + dr^{2})

and

ds^{2} = λ_{2}(r̃^{2})(r̃^{2}dφ̃^{2} + dr̃^{2})

where r̃ ≠ Dr^{±1}, D − const. This means that the geodesic flow of ds^{2} possesses two independent linear integrals. So, from Proposition 3.2 it follows that ds^{2} has the form (17), and therefore, r̃ = Dr^{±1}, D − const. Thus, the Liouville coordinates φ, y = log r related to polar coordinates φ, r of ds^{2} on S^{2} are unique up to shifts and the transform y → −y.

□

We must note that Proposition 3.1 follows also from the results of Kolokol’tsov published in his Ph.D. thesis (in Russian) but we give here a complete another proof.

One may show that if a polynomial in momenta integral F of the geodesic flow of (9) 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 3.4 Let ds^{2} = λ(z, z̄)dzdz̄ be a metric such that there exists a function f : R^{2} ↦ R, satisfying the following conditions

λ(z, z̄) = \frac{∂^{2}f}{∂z∂z̄}, Im (\frac{∂^{4}f}{∂z^{4}} \frac{∂^{2}f}{∂z∂z̄} + 3\frac{∂^{3}f}{∂z^{3}} \frac{∂^{3}f}{∂z^{2}∂z̄} + 2\frac{∂^{2}f}{∂z^{2}} \frac{∂^{4}f}{∂z^{3}∂z̄}) = 0.  (18)

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

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

Proof. Suppose that (18) holds. We will now construct an integral

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 (11) has then the following form

4\frac{∂λ}{∂z} = −\frac{∂(a_{1}λ)}{∂z̄},  (19)

\frac{∂(a_{1}λ^{3})}{∂z} = −λ\frac{∂(a_{2}λ^{2})}{∂z̄},  (20)
a_{2} = \overline{a_{2}}.  (21)

Consider

a_{1} = −4\frac{∂^{2}f}{∂z^{2}} (\frac{∂^{2}f}{∂z∂z̄})^{−1} .

Then (19) holds.

Show that for

g = \frac{∂(a_{1}λ^{3})}{∂z}λ^{−1}

it holds

Im \frac{∂g}{∂z} = 0.

Indeed, by computation we obtain

Im \frac{∂g}{∂z} = −4 Im (\frac{∂^{4}f}{∂z^{4}} \frac{∂^{2}f}{∂z∂z̄} + 3\frac{∂^{3}f}{∂z^{3}} \frac{∂^{3}f}{∂z^{2}∂z̄} + 2\frac{∂^{2}f}{∂z^{2}} \frac{∂^{4}f}{∂z^{3}∂z̄}) = 0.

So, it follows that there is a function h : R^{2} ↦ R such that

\frac{∂h}{∂z̄} = g.

Consider a_{2} = −\frac{h}{λ^{2}} . Then a_{2} : R^{2} ↦ R and, moreover,

\frac{∂(a_{1}λ^{3})}{∂z}λ^{−1} = −\frac{∂(a_{2}λ^{2})}{∂z̄},

i.e. conditions (20) and (21) hold also.

Put then a_{3} = ā_{1}. So, we have an integral of fourth degree in momenta of the geodesic flow of ds^{2}.

Prove now the second statement of the theorem. As mentioned above, in this case there exists a conformal coordinate system of ds^{2} where a_{0} = a_{4} ≡ 1. Note that this system is unique up to a shift. So, in this coordinate system (19)-(21) hold. From (19) it follows that there exists a function V where

\frac{∂V}{∂z̄} = λ

and

a_{1}λ = −4\frac{∂V}{∂z} .

From (20) and (21) it follows immediately that

Im \frac{∂}{∂z} (\frac{∂(a_{1}λ^{3})}{∂z}λ^{−1}) = 0.

On the other hand

Im \frac{∂}{∂z} (\frac{∂(a_{1}λ^{3})}{∂z}λ^{−1}) = −4 Im \frac{∂}{∂z} (\frac{∂^{2}V}{∂z^{2}} \frac{∂V}{∂z̄} + 2\frac{∂^{2}V}{∂z∂z̄} \frac{∂V}{∂z} ) = 0.

Taking into account that Im \frac{∂V}{∂z̄} = Im λ = 0, we conclude that there exists a function f : R^{2} ↦ R such that

\frac{∂f}{∂z} = V.

Thus,

λ = \frac{∂^{2}f}{∂z∂z̄}

and

0 = Im \frac{∂}{∂z} (\frac{∂^{3}f}{∂z^{3}} \frac{∂^{2}f}{∂z∂z̄} + 2\frac{∂^{3}f}{∂z^{2}∂z̄} \frac{∂^{2}f}{∂z^{2}} ) =

= Im (\frac{∂^{4}f}{∂z^{4}} \frac{∂^{2}f}{∂z∂z̄} + 3\frac{∂^{3}f}{∂z^{3}} \frac{∂^{3}f}{∂z^{2}∂z̄} + 2\frac{∂^{2}f}{∂z^{2}} \frac{∂^{4}f}{∂z^{3}∂z̄}) .

□

Equation (18) in some other form has been obtained in [7] but in order to show when the corresponding integral is nontrivial we needed to give here a complete proof.

Using this criterion we may find families of metrics with integrable geodesic flows.

We will look for a class of solutions of (18) and show that in this class of solutions the problem of integrability can be reduced to an ordinary differential equation.

Denote

\frac{∂^{2}f}{∂z^{2}} = (A(x, y) + iB(x, y)) .

By computation we obtain

Im \frac{∂^{4}f}{∂z^{4}} = \frac{1}{4} (A_{xx} − A_{yy} − 2A_{xy}) ,

Im \frac{∂^{3}f}{∂z^{3}} \frac{∂^{3}f}{∂z^{2}∂z̄} = \frac{1}{2} (A_{x}A_{x} + A_{y}A_{y}) ,

Im \frac{∂^{2}f}{∂z^{2}} \frac{∂^{4}f}{∂z^{3}∂z̄} = \frac{1}{4} (B(A_{xx} + A_{yy}) + A(A_{xx} + A_{yy})) .

Thus, (18) can be written as follows

6(A_{y}A_{y} +A_{x}A_{x})+2(B(A_{xx} +A_{yy})+A(A_{xx} +A_{yy}))+(A_{xx} −A_{yy} −2A_{xy})λ = 0. (22)

Consider now the solutions of (18) of the following form:

f(x, y) = ψ(y) cos x + ξ(y) + a(x^{2} − y^{2})  (23)

where ψ, ξ are some smooth functions and a is a constant.

Then we obtain

λ = \frac{1}{4} ((ψ′′(y) − ψ(y)) cos x + ξ′′(y))

and

A = −\frac{1}{4} ((ψ′′(y) + ψ(y)) cos x + ξ′′(y) − 4a) , B = \frac{1}{2} (ψ′(y) sin x) .  (24)

Substituting (24) in (22) we get the following condition of integrability

6(((ψ′′′ + ψ′) cos x + ξ′′′)2ψ′′ sin x − (ψ′′ + ψ)2ψ′ sin x cos x+

2(2ψ′ sin x((ψ^{(4)} − ψ) cos x + ξ^{(4)}) + ((ψ′′ + ψ) cos x + ξ′′ − 4a)2(ψ′′′ − ψ′) sin x))+

4(ψ′′′ + ψ′) sin x((ψ′′ − ψ) cos x + ξ′′) ≡ 0.

Therefore,

3ψ′′ξ′′′ + ψ′ξ^{(4)} + 2ψ′′′ξ′′ = 4a(ψ′′′ − ψ′),  (25)
5ψ′′′ψ′′ − 6ψ′ψ + ψ′ψ^{(4)} = 0.  (26)

Equation (25) can be written as

(ψ′ξ′′′)′ + 2(ψ′′ξ′′)′ = 4a(ψ′′ − ψ)′.

By integrating we get

ψ′ξ′′′ + 2ψ′′ξ′′ = 4a(ψ′′ − ψ) + a_{1}

where a_{1} is a constant. So, if a = 0

ξ′′ = \frac{a_{1}ψ(y) + c}{(ψ′(y))^{2}}.  (27)

or if a ≠ 0

ξ′′ = 2a\frac{ψ^{′2}(y) − ψ^{2}(y) + a_{1}(2d)^{−1}ψ(y) + p}{ψ′(y))^{2}}.  (28)

where p, c are some constants.

On the other hand there is an integral of (26):

2ψ^{′′2} − 3ψ^{2} + ψ′ψ′′′ = const.  (29)

So, we may now formulate the following proposition.

Proposition 3.5 For any solution ψ of (29) there are two families of metrics ds_{1}^{2} = Λ(x, y)(dx^{2} + dy^{2}) where

Λ(x, y) = (ψ′′(y) − ψ(y)) cos x + \frac{a_{1}ψ(y) + c}{ψ^{′2}(y)}  (30)

or

Λ(x, y) = (ψ′′(y) − ψ(y)) cos x + c\frac{ψ^{′2}(y) − ψ^{2}(y) + a_{1}ψ(y) + p}{ψ^{′2}(y)}  (31)

where c, a_{1}, p are constants and the corresponding geodesic flows possess an integral of fourth degree in momenta. If \frac{∂^{2}Λ}{∂x∂y} and \frac{∂^{2}Λ}{∂x^{2}} − \frac{∂^{2}Λ}{∂y^{2}} are not equal to zero identically, then this integral is nontrivial, i.e. it does not depend on the Hamiltonian and an integral of smaller degree.

Proof. As we have shown, for every metric in one of these families there exists a function f of the form (23) satisfying (18). In fact, compare (30) with (27) and (31) with (28) (in the last case we denote 4d in (28) as c).

Since every metric in one of these families satisfies the conditions of theorem 3.4, the integrability follows.

So, we have only to prove that the integrals of the geodesic flows of these geodesic flows are nontrivial.

According to our construction the integral F of the geodesic flow with Hamiltonian

H = \frac{p_{z}pz̄}{4Λ}

where Λ is given by (30), (31) has the following form

F = p_{z}^{4} + D(x, y)p_{z}^{3}pz̄ + E(x, y)p_{z}^{2}p2 ; z̄ + D̄(x, y)p_{z}p3 ; z̄ + p4 ; z̄

where D(x, y) : R^{2} ↦ C, E(x, y) : R^{2} ↦ R.

Thus, if F depends on H and another integral polynomial of smaller degree, then there is an integral F̃ quadratic in momenta which has the following form

F̃ = p_{z}^{2} + Ẽ_{1}(x, y)p_{z}^{2}p2 ; z̄ + p2 ; z̄

or

F̃ = ip_{z}^{2} + Ẽ_{2}(x, y)p_{z}^{2}p2 ; z̄ − ip2 ; z̄

where Ẽ_{1}(x, y), Ẽ_{2}(x, y) : R^{2} ↦ R.

Therefore, due to Darboux [6] the following holds

\frac{∂^{2}Λ}{∂x∂y} ≡ 0

or

\frac{∂^{2}Λ}{∂x^{2}} − \frac{∂^{2}Λ}{∂y^{2}} ≡ 0.

□

4 Smooth metrics on S^{2}

In this chapter we will prove Theorem 1.1 and Theorem 1.2.

Proof of Theorem 1.1. Prove that the Hamiltonian system with the Hamiltonian (4) is a smooth system on S^{2}. Using theorem 2.1, we may write

H = \frac{1}{ν^{2}(r^{2})}(r^{2}dφ^{2} + dr^{2}) + μ(r^{2})r cos φ =

\frac{1}{ν^{2}(r̃^{2})}(r̃^{2}dφ̃^{2} + dr̃^{2}) − μ(r̃^{2})r̃cos φ̃

where r̃ = \frac{1}{r}, φ̃ = −φ. Since ν, μ are of class C^{∞} and ν ≠ 0, see Theorem 2.1, then this system is a smooth system on S^{2}.

Prove that (4) possesses an integral of fourth degree in momenta. Consider the metrics (30) from Proposition 3.5 and put in (30) ψ(y) = ψ_{0}(y) and d_{1} = 0. Then we obtain the following one-parameter family of metrics:

((ψ′′_{0}(y) − ψ_{0}(y)) cos x + c\frac{1}{ψ_{0}^{′2}(y)}) (dx^{2} + dy^{2})

or in polar coordinates φ = x, r = exp y:

((Ψ_{2}(r) − Ψ_{0}(r)) cos φ + c\frac{1}{Ψ_{1}^{2}(r)}) (dφ^{2} + \frac{1}{r^{2}} dr^{2})

where c is an arbitrary constant. From Proposition 3.5 we know that any metric in this family possesses an integral of fourth degree in momenta. Using the well-known Maupertuis’s principle, see [1], we conclude that (4) possesses an integral of fourth degree in momenta, see also [3], [4].

So, we have to prove only that this integral nontrivial, i.e. depends on the Hamiltonian and an integral of smaller degree. Denote the kinetik energy in (4) as K = λ(y)(dφ^{2} + dy^{2}) and the potential as V. We will denote the Hamiltonian of the geodesic flow of the metric K as K also.

Let us assume that a system of this family has an integral which is independent of the Hamiltonian (total energy (4)) and which is a polynomial of second degree in momenta (clearly, this assumption includes the case of linear integrals). So, there is an integral F̃ of (4) which is quadratic in momenta p_{φ}, p_{y}. Thus, F̃ = A(p_{φ}, p_{y}, φ, y) + B(φ, y) where A(p_{φ}, p_{y}, φ, y) is a polynomial of second degree in momenta. We may write {F̃, H} = {A(p_{φ}, p_{y}, φ, y) + B(φ, y), K + V } ≡ 0 and, therefore, {A(p_{φ}, p_{y}, φ, y), K} ≡ 0.

Thus, the geodesic flow of K has an integral which is a polynomial of second degree in momenta.

Since a function ψ such that ψ′(y) = c_{1} exp(−y)(1 + exp(2y)), c_{1} − const does not satisfy (3) and, therefore, Ψ_{1}(r) ≠ r^{−1}(1 + Dr^{2}), D is a positive constant, then K is not a metric of constant positive curvature. So, the quadratic integral A of the geodesic flow of K depends on the Hamiltonian K and the integral p_{φ}. Since F does not depend on H, w.l.o.g. we may put A = p_{φ}^{2}.

So, we may write

{A, V } + {B, K} = {p_{φ}^{2}, V } + {B, K} ≡ 0.

By computation we obtain

\frac{∂B}{∂y} ≡ 0,

and

\frac{∂B}{∂φ} = \frac{∂V}{∂φ} λ(y).

So, we get

V = \frac{B(φ) + α(y)}{λ(y)}

for a smooth function α(y).

In our case

λ(y) = \frac{1}{ψ_{0}^{′2}(y)}

and

V (x, y) = (ψ′′_{0}(y) − ψ_{0}(y))ψ_{0}^{′2}(y) cos φ

and we get then ψ′′_{0}(y) − ψ_{0}(y) ≡ const, that is not true. So, there is no quadratic integral of the system given by (4).

In oder to prove that the system given by (4) is really a new example of the integrability by a polynomial of fourth degree in momenta we have to compare this case with the known cases of Kovalevskaya and Goryachev. First of all, we note that from Corollary 3.3 it follows that in the family (2) we have to consider only the case B_{1} = B_{2} = 0. As mentioned above, for B_{1} = B_{2} = 0 the system given by (2) is simply the case of Kovalevskaya. So, we may consider only the case of Kovalevskaya. Let us write the Hamiltonian (1) of the case of Kovalevskaya in polar coordinates. We obtain

H = γ_{1}(r^{2})(r^{2}dφ^{2} + dr^{2}) − γ_{2}(r^{2}) cos φ  (32)

where γ_{2}(r^{2}) ≠ 0 for all 0 < r < +∞.

From Corollary 3.3 we have that if the Hamiltonians (32) (or (1)) and (4) are equivalent then the potentials U in (32) and in (4) are proportional, see the Introduction. Then comparing the potential in (32) with the potential in (4) we see that they cannot be proportional, because γ_{2}(r) > 0, r ∈ (0, +∞) in (32) but Ψ_{2}(1) − Ψ_{0}(1) = 0.

□

Proof of Theorem 1.2. Using theorem 2.1, we may rewrite H_{p} in (5) as

H_{p} = \frac{−Φ(r^{2}) + p + 1}{ν^{2}(r^{2})}(r^{2}dφ^{2} + dr^{2}) + \frac{μ(r^{2})}{−Φ(r^{2}) + p + 1}r cos φ =

\frac{−Φ(r̃^{2}) + p + 1}{ν^{2}(r̃^{2})}(r̃^{2}dφ̃^{2} + dr̃^{2}) − \frac{μ(r̃^{2})}{−Φ(r̃^{2}) + p + 1}r̃cos φ̃

where r̃ = \frac{1}{r}, φ̃ = −φ and

Φ(t) = ∫_{1}^{t} μ(s)ν^{−1}(s)ds.

Since ν, μ are of class C^{∞} and ν ≠ 0, see Theorem 2.1, there are the constants

M_{1} = max_{[0,1]} Φ(t) and M_{2} = min_{[0,1]} Φ(t),

where M_{2} < M_{1}. Thus, for p > p_{1} = M_{1} − 1 or p < p_{2} = M_{2} − 1 a system with the Hamiltonian H_{p} or −H_{p} correspondingly, where H_{p} has the form (5), is a conservative system on S^{2}.

Prove that the system with energy (5) possesses an integral of fourth degree in momenta. Consider the metrics (31) from Proposition 3.5 and put in (31) ψ(y) = ψ_{0}(y) and a_{1} = 0. We get the following two-parameter family of metrics:

((ψ′′_{0}(y) − ψ_{0}(y)) cos x + c\frac{ψ_{0}^{′2}(y) − ψ_{0}^{2}(y) + p}{ψ_{0}^{′2}(y)}) (dx^{2} + dy^{2})

which can be written also in polar coordinates φ = x, r = exp y as

((Ψ_{2}(r) − Ψ_{0}(r)) cos φ + c\frac{Ψ_{1}^{2}(r) − Ψ_{0}^{2}(r) + p}{Ψ_{1}^{2}(r)}) (dφ^{2} + \frac{1}{r^{2}} dr^{2})

where c is an arbitrary constant. It was shown in Proposition 3.5 that any metric in this family possesses an integral of fourth degree in momenta. From the well-known Maupertuis’s principle, see [1], it follows that a system with the Hamiltonian (5) possesses an integral of fourth degree in momenta.

In order to prove that for any p > p_{0} the system with the Hamiltonian (5) does not possess an integral quadratic or linear in momenta and in oder to compare these cases with the above cases of Kovalevskaya and Goryachev we can apply the same arguments as for the system with the Hamiltonian (4), see the proof of Theorem 1.1.

We must prove that for different values of parameter p the Hamiltonians (5) are not equivalent.

Let us introduce the coordinates φ, y = log r. From Corollary 3.3 it follows that we need to consider only the transformations y ↦ ±y + κ_{1} where κ_{1} is a constant.

Let us rewrite H_{p} in the coordinates φ, y:

H_{p} = \frac{ψ_{0}^{′2}(y) − ψ_{0}^{2}(y) + p}{ψ_{0}^{′2}(y)} (dφ^{2} + dy^{2}) −

(ψ′′_{0}(y) − ψ_{0}(y)) (\frac{ψ_{0}^{′2}(y) − ψ_{0}^{2}(y) + p}{ψ_{0}^{′2}(y)})^{−1} cos φ  (33)

From Corollary 3.3 it follows that if the Hamiltonians (33) for p = p_{1} and p = p_{2} are equivalent, then there are some constants K_{1} ≠ 0, K_{2} ≠ 0 and κ_{1} such that the following holds

\frac{ψ_{0}^{′2}(y) − ψ_{0}^{2}(y) + p}{ψ_{0}^{′2}(y)} =

K_{1} \frac{ψ_{0}^{′2}(±y + κ_{1}) − ψ_{0}^{2}(±y + κ_{1}) + p}{ψ_{0}^{′2}(±y + κ_{1})}  (34)

and

(ψ′′_{0}(y) − ψ_{0}(y)) (\frac{ψ_{0}^{′2}(y) − ψ_{0}^{2}(y) + p}{ψ_{0}^{′2}(y)})^{−1} =

K_{2}(ψ′′_{0}(±y + κ_{1}) − ψ_{0}(±y + κ_{1})) (\frac{ψ_{0}^{′2}(±y + κ_{1}) − ψ_{0}^{2}(±y + κ_{1}) + p}{ψ_{0}^{′2}(±y + κ_{1})})^{−1} . (35)

Thus, we get

ψ′′_{0}(y) − ψ_{0}(y) = K_{1}K_{2}(ψ′′_{0}(±y + κ_{1}) − ψ_{0}(±y + κ_{1})).  (36)

Put y = 0 in (36) and we get ψ′′_{0}(κ_{1}) − ψ_{0}(κ_{1}) = 0. W.l.o.g. we assume that κ_{1} ≥ 0. From the proof of Theorem 2.1 we know that the function ψ′′_{0}(y) − ψ_{0}(y), y > 0 can be written as follows

ψ′′_{0}(y) − ψ_{0}(y) = (exp R(y))(p(q) + q^{2} − q), y > 0

where p(q) = −q^{2} + \frac{1}{q^{2}} and q > 1.

So, we obtain ψ′′_{0}(y) − ψ_{0}(y) = exp R(y)(q^{−2} − q) and, therefore, if ψ′′_{0}(κ_{1}) − ψ_{0}(κ_{1}) = 0 and κ_{1} ≠ 0, then q(κ_{1}) = 1 but as mentioned above q(y) > 1 for finite values of y. Thus, κ_{1} = 0. Then taking into account that ψ_{0}(y) = −ψ_{0}(−y) from (34) we get K_{1} = 1 and p_{1} = p_{2} and then from (35) we get K_{2} = 1.

Thus, for p_{1} ≠ p_{2} the Hamiltonians H_{p_{1}} and H_{p_{2}} of the form (5) are not equivalent.

□

Acknowledgement. The author would like to thank Professor Gerhard Huisken and the Arbeitsbereich ”Analysis” of the University of Tübingen for their hospitality.

After the paper was finished, K. P. Hadeler informed the author that the differential equation in (3) can be solved by successive integration and the inverse function of the solution ψ_{0}(y) to the initial value problem (3) can be given as:

y = y(ψ_{0}) = \frac{1}{4} log \frac{(ψ_{0}^{4} + 1)^{\frac{1}{4}} + ψ_{0}}{(ψ_{0}^{4} + 1)^{\frac{1}{4}} − ψ_{0}} − \frac{1}{2} arctan \frac{(ψ_{0}^{4} + 1)^{\frac{1}{4}}}{ψ_{0}} .

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