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Integrability of Hamiltonian Systems with Two Degrees of Freedom and Homogenous Potential of Degree Zero

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DOI: 10.4236/jamp.2018.611184    99 Downloads   155 Views  

ABSTRACT

We provide necessary conditions in order that the Hamiltonian systems with Hamiltonian and one of the following potentials  are integrable in the Liouville sense.

1. Introduction

In this paper we consider classical Hamiltonian systems with two degrees of freedom for which the Hamiltonian is of the form

H = 1 2 ( p 1 2 + p 2 2 ) + V ( q 1 , q 2 ) , (1)

where V is a homogeneous function of degree 0. Although these systems arising from physics and applied sciences are generally understood to involve only real variables, we will assume that the Hamiltonian system associated to (1) is defined on the complex symplectic manifold M = 4 \ Σ equipped with the canonical symplectic form

ω = j = 1 2 d q j d p j ,

where Σ is a set of zero Lebesgue measure.

We will assume that the system of equations

grad V ( q 1 , q 2 ) = V ( q 1 , q 2 ) = ( q 1 , q 2 ) ,

has a nonzero solution d = ( d 1 , d 2 ) 2 , called a Darboux point of the potential V.

A Hamiltonian system of two degrees of freedom is integrable in the Liouville sense if there exists a first integral F, defined in 4 except perhaps in a set of zero Lebesgue measure, which is independent with the Hamiltonian, i.e. the rank of the matrix

( H q 1 H q 2 H p 1 H p 2 F q 1 F q 2 F p 1 F p 2 )

is 2, except perhaps in a zero Lebesgue measure set of 4 .

The aim of this paper is to find necessary conditions for the integrability of some homogeneous potentials of degree of homogeneity 0. More precisely, our main results are the following.

Proposition 1 Assume that a Hamiltonian system of two degrees of freedom with Hamiltonian of the form (1) and potential

V = a 0 q 1 + a 1 q 2 a 2 q 1 + a 3 q 2 , (2)

where a j = b j , 1 + i b j , 2 for j = 0 , 1 , 2 , 3 , is integrable in the Liouville sense, and that it has Darboux points. If a 3 0 , then

b 01 = b 11 ( b 21 b 31 + b 22 b 32 ) b 31 2 + b 32 2 b 12 ( b 22 b 31 b 21 b 32 ) b 31 2 + b 32 2 , b 02 = b 11 ( b 21 b 32 b 22 b 31 ) b 31 2 + b 32 2 + b 12 ( b 21 b 31 + b 22 b 32 ) b 31 2 + b 32 2 . (3)

If a 3 = 0 is a particular case of Proposition 3.

Proposition 2 Assume that a Hamiltonian system of two degrees of freedom with Hamiltonian of the form (1) and potential

V = a 0 q 1 2 + a 1 q 1 q 2 + a 2 q 2 2 a 3 q 1 2 + a 4 q 1 q 2 + a 5 q 2 2 , (4)

where a j = b i , 1 + i b j , 2 for j = 0 , , 5 , is integrable in the Liouville sense, that it has Darboux points. If a 5 0 , then

b 01 = b 21 b 31 b 51 + b 21 b 32 b 52 + b 22 b 31 b 52 b 22 b 32 b 51 b 51 2 + b 52 2 , b 02 = b 21 b 31 b 52 b 21 b 32 b 51 b 22 b 31 b 51 b 22 b 32 b 52 b 51 2 + b 52 2 , b 11 = b 21 b 41 b 51 + b 21 b 42 b 52 + b 22 b 41 b 52 b 22 b 42 b 51 b 51 2 + b 52 2 , b 12 = b 21 b 41 b 52 + b 21 b 42 b 51 + b 22 b 41 b 51 + b 22 b 42 b 52 b 51 2 + b 52 2 . (5)

If a 5 = 0 , then

b 01 = b 21 , b 02 = b 22 and b 11 = b 12 = 0. (6)

Proposition 3 If a Hamiltonian system of two degrees of freedom with Hamiltonian of the form (1) and potential

V = a 0 q 2 n + a 1 q 2 n 1 q 1 + + a n q 1 n q 1 n , (7)

where a j = b j , 1 + i b j , 2 for j = 0 , , n , is integrable in the Liouville sense and has Darboux points, then it satisfies

b 1 , 1 = k = 1 [ ( n 1 ) / 2 ] b 2 k + 1 , 1 ( 2 k + 1 ) 2 ( 1 ) k , b 2 , 1 = k = 2 [ n / 2 ] b 2 k , 1 k 2 ( 1 ) k , b 1 , 2 = k = 1 [ ( n 1 ) / 2 ] b 2 k + 1 , 2 ( 2 k + 1 ) 2 ( 1 ) k , b 2 , 2 = k = 2 [ n / 2 ] b 2 k , 2 k 2 ( 1 ) k . (8)

These three propositions are proved in Section 3. The proof of Proposition 2 needs the help of an algebraic manipulator like mathematica for doing it, and a such manipulator also helps in the proof of Proposition 1 but there is not strictly necessary.

The integrability of other Hamiltonian systems with two degrees of freedom and different potentials have been studied in [1] - [30] .

2. Preliminary Results

It was proved in Theorem 1.2 of [31] the following result for Hamiltonian systems with n degrees of freedom of the form (1) with homogeneous potentials of degree 0.

Theorem 4 Assume that V [ q 1 , , q n ] is homogeneous of degree 0 and that the following conditions are satisfied:

1) there exists a non-zero d = ( d 1 , , d n ) n such that V ( d 1 , , d n ) = ( d 1 , , d n ) , and

2) the system is integrable in the Liouville sense with rational first integrals.

Then

1) all eigenvalues of the Hessian matrix V ( d 1 , , d n ) are integers; and

2) the matrix V ( d 1 , , d n ) is diagonalizable.

When n = 2 Theorem 4 has the following easier formulation given in section 5 of [31]

Theorem 5 Assume that V [ q 1 , q 2 ] is homogeneous of degree 0 and that the following conditions are satisfied:

1) there exists a non-zero d = ( d 1 , d 2 ) 2 such that V ( d 1 , d 2 ) = ( d 1 , d n ) , and

2) the system is integrable in the Liouville sense with rational first integrals.

Let z = q 2 / q 1 with q 1 0 , be the affine coordinate on P 1 and set v ( z ) : = V ( 1, z ) . Then the Darboux points are ± i , and the function v satisfies

v ( z * ) + z * v ( z * ) = 0. (9)

Proof. Darboux points of V are non-zero solutions of equations

V q 1 = q 1 , V q 2 = q 2 . (10)

It is convenient to consider Darboux points in the projective line P 1 . Let z = q 2 / q 1 with q 1 0 , be the affine coordinate on P 1 . Then we can rewrite system (10) in the form

v ( z ) z = q 1 2 , v ( z ) = z q 1 2 , (11)

where v ( z ) : = V ( 1 , z ) . From the above formulae it follows that z * is a Darboux point of V if and only if z * { i , i } , and v ( z * ) 0 . Thus the location of the Darboux points does not depend on the potential.

If z * is the affine coordinate of the Darboux point d = ( d 1 , d 2 ) of V, then the Hessian matrix V ( d ) expressed in this coordinate has the form

V ( d ) = ( v ( z * ) x * 2 2 [ v ( z * ) + z * v ( z * ) ] x * 2 [ v ( z * ) + z * v ( z * ) ] x * 2 v ( z * ) x * 2 ) ,

where

x * = v ( z * ) z * = v ( z * ) / z * .

Vector d is an eigenvector of V ( d ) with corresponding eigenvalue λ = 1 . As the trace of V ( d ) is −2, λ = 1 is the only eigenvalue of V ( d ) . Thus the first hypothesis of Theorem 4 is satisfied, and the second also by our assumptions, then the matrix V ( d ) is diagonalizable, and since its eigenvalues are −1 and −1, we get that V ( d ) is diagonal. Hence the second condition of Theorem 4 is satisfied if and only if (9) holds.

3. Proof of the Propositions

Proof of Proposition 1. We will apply Theorem 5 to potential (2). Then it has the Darboux points ± i . Since

v ( z ) = a 0 + a 1 z a 2 + a 3 z ,

we have

v ( z ) + z v ( z ) = ( a 1 a 2 a 0 a 3 ) ( a 2 a 3 z ) ( a 2 + a 3 z ) 3 .

Taking a j = b j , 1 + i b j , 2 for j = 0 , , 3 , the condition (9) implies

( b 21 + i b 22 i ( b 31 + i b 32 ) ) ( ( b 11 + i b 12 ) ( b 21 + i b 22 ) ( b 01 + i b 02 ) ( b 31 + i b 32 ) ) ( b 21 + i b 22 + i ( b 31 + i b 32 ) ) 3 = 0 (12)

and

( b 21 + i b 22 + i ( b 31 + i b 32 ) ) ( ( b 11 + i b 12 ) ( b 21 + i b 22 ) ( b 01 + i b 02 ) ( b 31 + i b 32 ) ) ( b 21 + i b 22 i ( b 31 + i b 32 ) ) 3 = 0. (13)

Taking the real part and the imaginary part of (12) we get

b 01 b 21 2 b 31 2 b 01 b 21 b 31 b 32 b 01 b 22 2 b 31 2 b 01 b 22 b 32 2 + b 01 b 31 3 + b 01 b 31 b 32 2 + b 02 b 21 2 b 32 2 b 02 b 21 b 31 2 + b 02 b 22 2 b 32 2 b 02 b 22 b 31 b 32 b 02 b 31 2 b 32 b 02 b 32 3 + b 11 b 21 3 + b 11 b 21 b 22 2 + 2 b 11 b 21 b 22 b 31 b 11 b 21 b 31 2 b 11 b 21 b 32 2 + 2 b 11 b 22 2 b 32 b 12 b 21 2 b 22 + 2 b 12 b 21 2 b 31 + 2 b 12 b 21 b 22 b 32 b 12 b 22 3 + b 12 b 22 b 31 2 + b 12 b 22 b 32 2 = 0

and

b 01 b 21 2 b 32 + 2 b 01 b 21 b 31 2 b 01 b 22 2 b 32 + 2 b 01 b 22 b 31 b 32 + b 01 b 31 2 b 32 + b 01 b 32 3 b 02 b 21 2 b 31 2 b 02 b 21 b 31 b 32 b 02 b 22 2 b 31 2 b 02 b 22 b 32 2 + b 02 b 31 3 + b 02 b 31 b 32 2 + b 11 b 21 2 b 22 2 b 11 b 21 2 b 31 2 b 11 b 21 b 22 b 32 + b 11 b 22 3 b 11 b 22 b 31 2 b 11 b 22 b 32 2 + b 12 b 21 3 + b 12 b 21 b 22 2 + 2 b 12 b 21 b 22 b 31 b 12 b 21 b 31 2 b 12 b 21 b 32 2 + 2 b 12 b 22 2 b 32 = 0.

Taking the real part and the imaginary part of (13) we get

b 01 b 21 2 b 31 + 2 b 01 b 21 b 31 b 32 b 01 b 22 2 b 31 + 2 b 01 b 22 b 32 2 + b 01 b 31 3 + b 01 b 31 b 32 2 + b 02 b 21 2 b 32 + 2 b 02 b 21 b 31 2 + b 02 b 22 2 b 32 + 2 b 02 b 22 b 31 b 32 b 02 b 31 2 b 32 b 02 b 32 3 + b 11 b 21 3 + b 11 b 21 b 22 2 2 b 11 b 21 b 22 b 31 b 11 b 21 b 31 2 b 11 b 21 b 32 2 2 b 11 b 22 2 b 32 b 12 b 21 2 b 22 2 b 12 b 21 2 b 31 2 b 12 b 21 b 22 b 32 b 12 b 22 3 + b 12 b 22 b 31 2 + b 12 b 22 b 32 2 = 0

and

b 01 b 21 2 b 32 2 b 01 b 21 b 31 2 b 01 b 22 2 b 32 2 b 01 b 22 b 31 b 32 + b 01 b 31 2 b 32 + b 01 b 32 3 b 02 b 21 2 b 31 + 2 b 02 b 21 b 31 b 32 b 02 b 22 2 b 31 + 2 b 02 b 22 b 32 2 + b 02 b 31 3 + b 02 b 31 b 32 2 + b 11 b 21 2 b 22 + 2 b 11 b 21 2 b 31 + 2 b 11 b 21 b 22 b 32 + b 11 b 22 3 b 11 b 22 b 31 2 b 11 b 22 b 32 2 + b 12 b 21 3 + b 12 b 21 b 22 2 2 b 12 b 21 b 22 b 31 b 12 b 21 b 31 2 b 12 b 21 b 32 2 2 b 12 b 22 2 b 32 = 0.

Solving these four equations we obtain that the condition (3) given in the statement of Proposition 1.

Proof of Proposition 2. We will apply Theorem 5 to potential (4). First assume that a 5 0 . Therefore this potential has the Darboux points ± i . Since

v ( z ) = a 0 + a 1 z + a 2 z 2 a 3 + a 4 z + a 5 z 2 ,

we obtain

v ( z ) + z v ( z ) = 1 ( a 3 + z ( a 4 + a 5 z ) ) 3 ( a 0 ( z ( a 4 2 + 3 a 4 a 5 z + 4 a 5 2 z 2 ) a 3 ( a 4 + 4 a 5 x ) ) + a 1 ( a 3 2 a 3 z ( a 4 + 6 a 5 z ) + a 5 z 3 ( a 5 z a 4 ) ) + a 2 z ( 4 a 3 2 + a 3 z ( 3 a 4 4 a 5 x ) + a 4 z 2 ( a 4 a 5 z ) ) ) = 0.

By condition (9) on v and since a j = b j , 1 + i b j , 2 for j = 0 , , 5 , we get

(14)

and

(15)

Taking the real part and the imaginary parts of (14) and (15) and solving the corresponding equations with the help of an algebraic manipulator such as Mathematica we get the unique solution given in (5).

Now we assume that. In this case

Then

By condition (9) on v and since for, we get

(16)

and

(17)

Taking the real part and the imaginary parts of (16) and (17) and solving the corresponding equations with the help of an algebraic manipulator such as Mathematica we get the unique solution given in (6).

Proof of Proposition 3. We will apply Theorem 5 to potential (7). Then its Darboux points are. Since

and consequently

Now applying condition (9) on v and taking into account that for, we get

which yields

and

Therefore

Hence

This yields (8).

4. Conclusions

We have characterized the Liouville integrability of the Hamiltonian systems with Hamiltonian

and one of the following potentials

For doing this we have used the Darboux theory of integrability.

Acknowledgements

The first author is partially supported by the Ministerio de Economía, Industria y Competitividad, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER) and MDM-2014-0445, the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. The second author is partially supported by FCT/Portugal through UID/MAT/ 04459/2013.

Conflicts of Interest

The authors declare no conflicts of interest.

Cite this paper

Llibre, J. and Valls, C. (2018) Integrability of Hamiltonian Systems with Two Degrees of Freedom and Homogenous Potential of Degree Zero. Journal of Applied Mathematics and Physics, 6, 2192-2201. doi: 10.4236/jamp.2018.611184.

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