SYMMETRIES OF HAMILTONIAN SYSTEMS WITH TWO DEGREES ...

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abstract We classify the symmetry groups of an autonomous Hamiltonian system with two degrees of freedom. With the exception of the harmonic oscillator or a ...
SYMMETRIES OF HAMILTONIAN SYSTEMS WITH TWO DEGREES OF FREEDOM

P. A. DAMIANOU C. SOPHOCLEOUS

Department of Mathematics and Statistics University of Cyprus P. O. Box 537, Nicosia, Cyprus

abstract We classify the symmetry groups of an autonomous Hamiltonian system with two degrees of freedom. With the exception of the harmonic oscillator or a free particle where the dimension is 15, we obtain all dimensions between 1 and 7. For each system in the classification we examine integrability.

PACS numbers: 02.30.H, 02.20.T, 03.20.

1

I

Introduction

The objective of this paper is a complete classification of the symmetry groups for a Hamiltonian system with two degrees of freedom. We actually study the equations in Newtonian form since a first order system always has an infinite number of symmetries. We consider the motion of a particle of unit mass in the plane (q1 , q2 ) under the influence of a potential of the form V (q1 , q2 ). We will assume that the Hamiltonian is time independent. This is not really a restriction because a time-dependent n−dimensional system is equivalent to a time-independent (n + 1)−dimensional system by regarding the time variable as the new coordinate. For the most part we assume that the system is two-dimensional with Hamiltonian 1 1 (1) H(q1 , q2 , p1 , p2 ) = p21 + p22 + V (q1 , q2 ) . 2 2 However, in section VI we generalize some of the results to the n-dimensional case. The real valued function V (q1 , q2 ) is assumed to be smooth on some open, connected subset of R2 . Hamilton’s equations, in Newtonian form, become ∂V q¨1 = − ∂q 1 ∂V − ∂q 2

q¨2 =

(2) .

We search for point symmetries of the system (2). That is, we search for the infinitesimal transformations of the form: t0 = t +  T (t, q1 , q2 ) + O(2 ) q10 = q1 +  Q1 (t, q1 , q2 ) + O(2 ) q20 = q2 +  Q2 (t, q1 , q2 ) + O(2 ) .

(3)

The functions V (q1 , q2 ) such that the system (2) admit such transformations are completely classified. Therefore, in the following analysis we determine the functions V , T , Q 1 and Q2 . Equations (2) admit Lie transformations of the form (3) if and only if Γ(2) {¨ q 1 + V q1 } = 0 (2) Γ {¨ q 2 + V q2 } = 0 ,

(4)

where Γ(2) is the second prolongation of ∂ ∂ ∂ + Q1 + Q2 . (5) ∂t ∂q1 ∂q2 For the reader who is unfamiliar with the definition and properties of Lie point symmetries, there are a number of excellent books on the subject e.g. [1], [2], [3], [4]. Equations (4) give two identities of the form: Γ=T

E1 (t, q1 , q2 , q˙1 , q˙2 ) = 0 E2 (t, q1 , q2 , q˙1 , q˙2 ) = 0 , 2

(6)

∂V ∂V where, we have used that q¨1 = − ∂q and q¨2 = − ∂q . The functions E1 and E2 are explicit 1 2 polynomials in q˙1 and q˙2 . We impose the condition that equations (6) are identities in five variables t, q1 , q2 , q˙1 and q˙2 which are regarded as independent. These two identities enable the infinitesimal transformations to be derived and ultimately impose restrictions on the functional forms of V , T , Q1 and Q2 . After some straightforward calculations one can show, see e.g. ref. [5], that the generators necessarily have the following form:

T = a(t) + b1 (t)q1 + b2 (t)q2 Q1 = b01 (t)q12 + b02 (t)q1 q2 + c11 (t)q1 + c12 (t)q2 + d1 (t) Q2 = b01 (t)q1 q2 + b02 (t)q22 + c21 (t)q1 + c22 (t)q2 + d2 (t) .

(7)

In this paper we classify the symmetry groups of the system according to the form of the generators. Here is a preview of the various cases and the potentials that appear. Case 1. b1 6= 0, b2 6= 0. In this case the potential is of the form V =

 λ 2 q1 + q22 + λ1 q1 + λ2 q2 . 2

The symmetry group has maximum dimension. It is a 15-parameter group of transformations isomorphic to sl(4, R). Case 2. b1 = b2 = 0. In other words, T is function of time only. We consider two possibilities according to a 00 6= 0 or a00 = 0. First Subcase:

a00 6= 0

2a V =

λ 2 µ (q1 + q22 ) + 2 (q1 + κq2 )2

In this case we obtain a 6-parameter group. 2b V =

µ λ 2 (q1 + q22 ) + 2 2 q2

This is a special case of the previous one with κ = 0. 2c V =

1 λ 2 (q1 + q22 ) + 2 Φ(ξ) 2 q2

where ξ = qq12 . For Φ arbitrary we end up with a 3-parameter symmetry group. For special types of Φ we obtain a 4-parameter group. Second Subcase: a00 = 0 This is the case where T is a linear function of time. 2d V = q22 + λ1 q1 q2 + Φ(ξ) , 3

where ξ = q1 − λq2 . For Φ arbitrary we end up with a 4-parameter symmetry group. For Φ quadratic we obtain a 7-parameter group and setting λ1 = 0, for some special forms of Φ (exponential, logarithmic, nth power) results in a 5-parameter group of symmetries. 2e V = λ1 q22 + Φ(q1 ) The case λ1 = 0 is the case of a separable potential with one variable missing. We will comment on this case separately. If λ1 6= 0, we end up with a 4-parameter group.

2f The dimensions of the symmetry groups in this case are all equal to 2 except for the last three systems where the dimension is 3. Specifically, we obtain the following list of potentials: 1. V

2.

= q2N Φ( qq21 )

= λ1 log q2 + Φ( qq12 )

V 3.

V

= eµq1 Φ(q2 )

4. = eµq1 Φ(q1 − λq2 )

V 5-10 V

= λ1 f1 (q1 ) + λ2 f2 (q2 ) ,

where f1 (q1 ) = q1n , log q1 , eµq1 and f2 (q2 ) = q2m , log q2 , eµq2 . 11. V

= Φ(q12 + q22 )

12. V

= λ(q12 + q22 )n ,

n 6= −1, 0, 1

13. V

= λ log (q12 + q22 )

14. V

= λ sin−1

q12 −q22 q12 +q22

Finally, we note that the potential V (q1 , q2 ) = q1k has a 15-parameter group of symmetries for k = 0, 1, a 7-parameter group for k = 2, a 6-parameter group for k = −2 and a 5parameter group of symmetries otherwise. These dimensions generalize for arbitrary n ≥ 2 to (n + 2)2 − 1, n2 + 3, n2 + 2, n2 + 1 respectively. 4

In this paper we will not consider the general case of motion in Rn , however for the case n = 3, the classification of symmetry groups is in progress; see [6]. In this case, we end up again with a maximal dimension of (n + 2)2 − 1 = 24 for the harmonic oscillator or a free particle, but the dimensions of the other groups in the classification vary from 1 to 12 (n2 + 3). We did not obtain the dimension 8 and any dimension between 13 and 23. In section II we will also consider the simplest system, with one degree of freedom, maily to illustrate the procedure we use for the two-dimensional case. The classification for a general ordinary differential equation of second order with one dependent and one independent variable goes back to Sophus Lie [7]. He showed that the dimensions of a maximal admitted algebra take only the values 1, 2, 3 and 8. Lie actually gave a group classification of all arbitrary order O.D.E.s. In this way he identified all equations that can be reduced to lowerorder equations or completely integrated by group theoretical methods [8]. The problem of classifying symmetry groups for a system of differential equations is open. This is mentioned in [9] where some known facts are presented. Some results for linear systems of second order ordinary differential equations can be found in [10]. Of course the ultimate goal in classical mechanics is to integrate explicitly the equations of motion. Such systems are called integrable. For the definition and basic concepts of Hamiltonian Systems and Symplectic Geometry there are a number of good references e.g. [11], [12], [13], [14], [15]. The key result is the following theorem of Liouville which in the 2-dimensional case translates as follows: Consider a Hamiltonian system with two degrees of freedom. If in addition to the Hamiltonian H there is a second integral of motion I, independent of H, then the system is integrable, i.e. in principle one can solve the equations by quadratures. Even though most of the well known systems of classical mechanics are Liouville integrable, the fact is, that most Hamiltonian systems are not integrable, a result demonstrated by Poincar´e. It is not surprising that most of the symmetry groups that appear in our classification correspond to integrable potentials. The non-integrable potentials appear mainly in case 2f, where the size of the symmetry group is small. The integrability of twodimensional systems has been the subject of numerous investigations; see for example [16], [17], [18], [19],[20], [21], [22]. Of course a system with symmetries should be expected to be integrable, after all this is the essence of Noether’s theorem; in this direction see for example the review [23]. On the other hand one can give a number of examples of integrable systems whose symmetry group is trivial (i.e. one dimensional). Some of the chaotic systems that do not appear in our list, for example the famous H´enon-Heiles system, are known to have only ∂ as a single symmetry. Of course, we should point out that all the systems which do not ∂t ∂ appear in our classification will have ∂t as a single symmetry. One can construct a number of systems possessing only one symmetry. For example, one can take V = q12 + 2q22 + q1k q2 ,

(8)

with k > 1. As was discussed in [24] there are integrable systems which possess only one symmetry. This situation is also investigated in [25]. We would like to point out another example: V = 4q12 + q22 +

r1 r2 . + q1 q22

(9)

In this system the associated Hamilton-Jacobi equation is separable in cartesian coordinates. We should point out that integrable systems have symmetries other than point symmetries.

5

For example, one of the integrable cases of H´enon-Heiles corresponds to the presence of a nonlocal symmetry. For the integrable systems that appear in the list we actually give the second integral (whenever it is not obvious) or a reference. Since integrability is preserved under various transformations e.g. translations, rotations, scalings, time reflections, we construct the second invariant for a representative of that class. Generally we choose the potential to be as simple as possible in order to illustrate the symmetry group and demonstrate integrability.

II

Systems with one degree of freedom

Before attacking the two-dimensional case, we classify the symmetries for a one-dimensional system, just to illustrate the techniques we use on the two-dimensional case. We consider a Hamiltonian of the form 1 H = p2 + V (q) . (10) 2 The equation of motion of the particle is q¨ = −

∂V . ∂q

(11)

We search for symmetries for equation (11) of the form t0 = t +  T (t, q) + O(2 ) q 0 = q +  Q(t, q) + O(2 ) .

(12)

Equation (11) admits symmetries of the form (12) if and only if Γ(2) {¨ q + Vq } = 0 ,

(13)

where Γ(2) is the second prolongation of

∂ ∂ +Q . (14) ∂t ∂q The definition of the second prolongation is the following: First we define the first prolongation i ∂ h Γ(1) = Γ + −Tq q˙2 + (Qq − Tt )q˙ + Qt . (15) ∂ q˙ Γ=T

The second prolongation of Γ is an extension of Γ(1) given by h

˙ q − Tqq q˙3 + (Qqq − 2Ttq )q˙2 + (2Qtq − Ttt )q˙ + Qtt Γ(2) = Γ(1) + (Qq − 2Tt − 3Tq q)¨

i ∂

∂ q¨

. (16)

Equation (13) becomes an identity of the form E(t, q, q) ˙ =0, 6

(17)

using the fact that q¨ = −Vq . The coefficient of q˙3 in (17) gives Tqq = 0. Similarly, the coefficient of q˙ 2 gives Qqq = 2Ttq . Therefore, T = a(t) + b(t)q Q = b0 (t)q 2 + c(t)q + d(t) .

(18)

Using equations (18), identity (17) becomes   ∂2V ∂V ∂V 2 0 00 00 0 3b + 3qb − a + 2c q˙ + q b + qc + d + (2a0 − c) + q 2 b000 + qc00 + d00 = 0 . (19) 2 ∂q ∂q ∂q #

"

The coefficient of q˙ in (19) gives 3b

∂V + 3qb00 − a00 + 2c0 = 0 . ∂q

(20)

We split the analysis into two exclusive cases: Case 1. b 6= 0. Case 2. b = 0. Case 1. b 6= 0. From equation (20) we obtain λ1 2 q + λ2 q + λ 3 . (21) 2 One easily calculates that the algebra of symmetries has dimension 8. It is a simple Lie algebra isomorphic to sl(3, R). V =

Case 2. b = 0. From (20) we have

1 c = a0 + c 1 2

and equation (19) becomes ∂2V ∂V [q(a + 2c1 ) + 2d] 2 + (3a0 − 2c1 ) + a000 q + 2d00 = 0 . ∂q ∂q 0

(22)

From equation (22) we deduce that V satisfies an o.d.e. of the form (λ1 q + λ2 )Vqq + λ3 Vq = λ4 q + λ5 .

(23)

In order to solve equation (23) we consider the following five possibilities: 1. λ1 = λ2 = λ3 = λ4 = λ5 = 0. In this case we get a(t) =constant, c1 = 0 and d(t) = 0. Therefore for V arbitrary we have T = c2 , Q = 0. In other words the symmetry group is trivial (one dimensional). 2. λ1 = λ3 = 0, λ2 6= 0. In this case, V is quadratic, a case already examined. 7

3. λ1 = 0, λ2 6= 0, λ3 6= 0. From (23) we get V = λeµq , where we have ignored linear terms. We obtain a 2-parameter group of symmetries with T = −2c1 t+c2 and Q = 4cµ1 . 4. λ1 6= 0, λ3 = 0. From (23) we obtain V = q log q. The symmetry group here is also trivial. 5. λ1 6= 0, λ3 6= 0. Without loss of generality we take λ2 = 0 in (23) and we obtain either i. V = λq n , n 6= 0, 1, 2, or ii. V = λ log q. i. We substitute V = λq n into (22) to obtain a(t) = c2 t2 + 2c3 t + c4 , d(t) = c1 = 0 if n = −2, and a(t) = 2(2−n) c t + c2 , d(t) = 0 if n 6= −2. To summarize, we have for n+2 1 λ V = q2 , T = c2 t2 + 2c3 t + c4 Q = (c2 t + c3 )q ,

(24)

and for V = λq n , n 6= −2, 0, 1, 2 the generators have the following form: T = Q =

2(2−n) ct n+2 1 4c1 q. n+2

+ c2

(25)

In other words, we obtain either a two-parameter or a three parameter group of symmetries. ii. When V = λ log q we get a(t) = 2c1 t + c2 and d(t) = 0. Therefore, T = 2c1 t + c2 Q = 2c1 q .

(26)

This is a two parameter group of symmetries. To summarize the results: In the case of one degree of freedom we obtain a maximal dimension of 8 for the harmonic oscillator or a free particle, but the dimensions in the other groups in the classification vary from 1 to 3. We do not obtain any dimension between 4 and 7.

III

Systems with two degrees of freedom

We return now to the case of two-degrees of freedom. The analysis is analogous to the one used in the case of one-degree of freedom. We substitute the forms of T , Q1 , Q2 in (7) into equations (6). The coefficient of q˙1 in equation (6) [E2 = 0] gives

8

∂V ∂ 2 b1 ∂c21 b1 + 2 q2 + 2 ∂q2 ∂t ∂t

!

=0.

(27)

On the other hand, the coefficient of q˙2 in equation (6) [E1 = 0] implies ∂c12 ∂ 2 b2 ∂V b2 + 2 q1 + 2 ∂q1 ∂t ∂t

!

=0.

(28)

Similarly, the coefficient of q˙1 in equation (6) [E1 = 0] gives 3b1

∂c11 ∂ 2 b2 ∂ 2 a ∂V ∂ 2 b1 ∂V + b2 + 3q1 2 + q2 2 − 2 + 2 =0, ∂q1 ∂q2 ∂t ∂t ∂t ∂t

(29)

while the coefficient of q˙2 in equation (6)[E2 = 0] implies 3b2

∂V ∂V ∂ 2 b2 ∂ 2 b1 ∂ 2 a ∂c22 + b1 + 3q2 2 + q1 2 − 2 + 2 =0. ∂q2 ∂q1 ∂t ∂t ∂t ∂t

(30)

If b1 (t) 6= 0 and b2 (t) 6= 0, then from equations (27) and (28) we deduce that V is quadratic in q1 and q2 . We also note that if b1 = 0, b2 6= 0 (or b1 6= 0, b2 = 0), then from equations (28) and (29) V has again a quadratic form. We therefore split the analysis into two exclusive cases: Case 1. b1 6= 0, b2 6= 0.

Case 2. b1 = b2 = 0.

IV

Case 1

From equations (27) and (28) we deduce that V = λ1 q12 + λ2 q22 + λ3 q1 + λ4 q2 + λ5 .

(31)

Now, substituting (31) into (6) the coefficients of q1 q˙1 in E1 and q2 q˙1 in E2 give respectively: 3(b001 + 2λ1 b1 ) = 0 3(b001 + 2λ2 b1 ) = 0 .

(32)

Hence, it follows that λ1 = λ2 = 0 or λ1 = λ2 6= 0. In the case λ1 = λ2 = 0, V is linear. We shall present the symmetries for this case, but in the remaining part of the analysis we shall ignore linear terms in the form of V . Adding a constant to equations (2) has no effect on the symmetry groups. V = λ 1 q1 + λ 2 q2 + λ 3

Subcase 1a

9

Note that we have taken λ1 = λ2 = 0 in (31) and then renamed the constants. Without presenting any calculations, we state that the system (2) with V linear has the following 15 symmetries: Γ1 = Γ2 = Γ3 = Γ4 = Γ5 = Γ6 = Γ7 Γ8 Γ9 Γ10 Γ11 Γ12 Γ13 Γ14 Γ15

= = = = = = = = =

∂ ∂t ∂ ∂q1 ∂ ∂q2 λ2 ∂ − ( 41 t3 + 23 λ1 tq1 ) ∂q∂1 − ( λ14λ2 t3 + 12 λ1 tq2 + λ2 tq1 ) ∂q∂2 q1 ∂t ∂ t ∂t + ( 12 q1 − 34 λ1 t2 ) ∂q∂1 + ( 12 q2 − 34 λ2 t2 ) ∂q∂2 ∂ + (q1 t − λ21 t3 ∂q∂1 + (q2 t − λ22 t3 ) ∂q∂2 t2 ∂t λ2 ∂ − ( λ14λ2 t3 + λ1 tq2 + 12 λ2 tq1 ) ∂q∂1 − ( 42 t3 + 32 λ2 tq2 ) ∂q∂2 q2 ∂t ∂ (2tq2 + λ2 t3 ) ∂t + (2q1 q2 + λ2 t2 q1 − λ1 t2 q2 − 12 λ1 λ2 t4 ) ∂q∂1 + (2q22 − 12 λ22 t4 ) ∂q∂2 ∂ + (2q12 − 12 λ21 t4 ) ∂q∂1 + (2q1 q2 + λ1 t2 q2 − λ2 t2 q1 − 12 λ1 λ2 t4 ) ∂q∂2 (2tq1 + λ1 t3 ) ∂t (q2 + 12 λ2 t2 ) ∂q∂1 (q1 + 12 λ1 t2 ) ∂q∂2 t ∂q∂1 t ∂q∂2 (q1 + 12 λ1 t2 ) ∂q∂1 (q2 + 12 λ2 t2 ) ∂q∂2 .

Remark: This system is of course integrable. The second integral is I = 12 p21 + λ1 q1 or I = 21 p22 + λ2 q2 . It also has constants of motion linear in the momenta, for example I = −λ2 p1 + λ1 p2 . V = λ2 (q12 + q22 )

Subcase 1b

We will choose λ = 1. We substitute the form of V into equations (6) and equate coefficients. In E1 , q˙1 q1 = 0 implies d 2 b1 + b1 = 0 . dt2 In E1 , q˙1 q2 = 0 implies d 2 b2 + b2 = 0 . dt2 In E1 , q˙2 = 0 implies dc12 =0, dt therefore c12 is constant. Similarly by examining the coefficient of q˙1 in E2 we see that c21 is also constant. In E1 , q˙1 = 0 and in E2 q˙2 = 0, imply that dc11 d2 a − 2 =0, dt2 dt 10

and dc22 d2 a −2 =0. 2 dt dt Similarly, using the coefficient of q1 = 0 in E1 and q2 = 0 in E2 , we obtain da d2 c11 +2 =0, 2 dt dt and

d2 c22 da +2 =0. 2 dt dt Finally, E1 = 0 and E2 = 0 imply that the functions d1 (t) and d2 (t) are solutions of the equation d2 x +x=0 . dt2 Therefore the form of the generators in this case is the following: T = k1 + k2 cos 2t + k3 sin 2t + (k4 cos t + k5 sin t)q1 + (k6 cos t + k7 sin t)q2 Q1 = (−k4 sin t + k5 cos t)q12 + (−k6 sin t + k7 cos t)q1 q2 + (−k2 sin 2t + k3 cos 2t + c11 )q1 + c12 q2 + k8 cos t + k9 sin t Q2 = (−k4 sin t + k5 cos t)q1 q2 + (−k6 sin t + k7 cos t)q22 + c21 q1 + (−k2 sin 2t + k3 cos 2t + c22 )q2 + +k10 cos t + k11 sin t . We note that the system (2) with V = λ2 (q12 + q22 ) admit a 15-parameter group of transformations isomorphic to sl(4, R). Remark: This system is the 2-dimensional isotropic oscillator. A second integral is I 1 = 1 2 p + 12 q12 or I2 = 21 p22 + 12 q22 . We also have constants of motion linear in the momenta, for 2 1 example I3 = q2 p1 − q1 p2 . Remark: Cases 1a and 1b give the most general form of Hamiltonian for which the second invariant is linear in the momenta [21].

V

Case 2

We use the identities E1 = 0 and E2 = 0 in (6) to obtain: E1 E2 E1 E2

: q˙2 : q˙1 : q˙1 : q˙2

=0 =0 =0 =0

⇔ ⇔ ⇔ ⇔ 11

c012 = 0 c021 = 0 2c011 − a00 = 0 2c022 − a00 = 0 .

(33)

Therefore, c12 = c1 , c21 = c2 , c11 = 12 a0 + c3 and c22 = 12 a0 + c4 . Here and elsewhere the ci are constants. Using these results, equations (6) take the form

and

(a0 q2 + 2c2 q1 + 2c4 q2 + 2d2 )Vq1 q2 + (a0 q1 + 2c3 q1 + 2c1 q2 + 2d1 )Vq1 q1 + 0 (3a − 2c3 )Vq1 − 2c1 Vq2 + a000 q1 + 2d001 = 0 ,

(34)

(a0 q1 + 2c3 q1 + 2c1 q2 + 2d1 )Vq1 q2 + (a0 q2 + 2c2 q1 + 2c4 q2 + 2d2 )Vq2 q2 + (3a0 − 2c4 )Vq2 − 2c2 Vq1 + a000 q2 + 2d002 = 0 .

(35)

We differentiate equations (34) and (35) with respect to t to obtain respectively: (a00 q2 + 2d02 )Vq1 q2 + (a00 q1 + 2d01 )Vq1 q1 + 3a00 Vq1 + a0000 q1 + 2d000 1 = 0 ,

(36)

(a00 q1 + 2d01 )Vq1 q2 + (a00 q2 + 2d02 )Vq2 q2 + 3a00 Vq2 + a0000 q2 + 2d000 2 = 0 .

(37)

We now split the analysis into two parts: a00 6= 0 or a00 = 0. a00 6= 0

Non-linear T

We divide equations (36) and (37) by a00 and then differentiate with respect to t to obtain respectively, d0 2 002 a

!0

d0 2 001 a

!0

V q1 q2

d0 + 2 001 a

!0

V q1 q2

d0 + 2 002 a

!0

V q1 q1

a0000 + a00

!0

d000 q1 + 2 100 a

!0

=0,

(38)

V q2 q2

a0000 + a00

!0

d000 q2 + 2 200 a

!0

=0.

(39)

From equations (38) and (39) we deduce that the function V (q1 , q2 ) satisfies two partial differential equations of the form λ 1 V q1 q2 + λ 2 V q1 q1 + λ 3 q 1 + λ 4 = 0 ,

(40)

λ 2 V q1 q2 + λ 1 V q2 q2 + λ 3 q 2 + λ 5 = 0 .

(41)

In order to solve equations (40) and (41) we consider the following cases: • λ1 6= 0, λ2 6= 0 • λ1 = 0, λ2 6= 0 (or λ1 6= 0, λ2 = 0) • λ1 = λ2 = λ3 = λ4 = 0

12

In the following three subcases we determine the form of V from equations (40) and (41). The corresponding generators may be obtained with the employment of equations (34)–(39). Subcase 2a In this case, V takes the form V =

µ λ 2 , (q1 + q22 ) + 2 (q1 + κq2 )2

(42)

with the corresponding generators : T = a(t) Q1 = 12 a0 (t)q1 + κ2 c4 q1 − κc4 q2 + κd2 (t) Q2 = 21 a0 (t)q2 − κc4 q1 + c4 q2 − d2 (t) ,

(43)

where a00 + 4λa = c8 and d2 is a solution of d002 + λd2 = 0. ( A 6-parameter group). Subcase 2b Setting λ1 = 0 in (40) and (41) we deduce that V has the form: V = k1 q13 + k2 q12 + k3 q1 q22 + k4 q1 q2 + k5 q1 + Φ(q2 ) .

(44)

Using equations (36) and (37) we get k1 = k3 = k4 = 0 and Φ = k2 q22 + k6 q2 + k7 + Therefore, ignoring the linear terms, V takes the form V =

λ 2 µ (q1 + q22 ) + 2 . 2 q2

µ . q22

(45)

Finally, using equations (34) and (35) we obtain the forms of the group generators : T = a(t) Q1 = c3 q1 + 21 a0 (t)q1 + d1 (t) Q2 = 21 a0 (t)q2 .

(46)

where a00 + 4λa = c8 and d001 + λd1 = 0. ( A 6-parameter group). The case λ = 0 is equivalent to the system V (q1 , q2 ) = q12 with generators: 1

T = c1 t2 + 2c2 t + c3 Q1 = (c1 t + c2 )q1 Q2 = (c1 t + c2 + c4 )q2 + c5 t + c6 . We choose the following basis for the Lie algebra of symmetries: X1 X2 X3 X4 X5 X6

= = = = = =

∂ ∂t

∂ + q1 ∂q∂1 + q2 ∂q∂2 2t ∂t ∂ t2 ∂t + q1 t ∂q∂1 + q2 t ∂q∂2 q2 ∂q∂2 t ∂q∂2 ∂ , ∂q2

13

(47)

with (non-zero) bracket relations [X1 , X2 ] [X1 , X3 ] [X1 , X5 ] [X2 , X3 ] [X2 , X5 ] [X2 , X6 ] [X3 , X6 ] [X4 , X5 ] [X4 , X6 ]

= = = = = = = = =

2X1 X2 X6 2X3 X5 −X6 −X5 −X5 −X6 .

(48)

This algebra is not semi-simple since the ideal generated by X5 , X6 is abelian. It is not solvable either because L(1) = {X1 , X2 , X3 , X5 , X6 }, and L(2) = L(1) . Remark: It is clear that Subcase 2b is a special case of Subcase 2a by setting κ = 0. Subcase 2c Since all the coefficients of the terms in equations (38) and (39) vanish, the functions a(t), d1 (t) and d2 (t) may be determined. From equations (36) and (37) we deduce that V = where ξ =

q1 . q2

λ 2 1 (q1 + q22 ) + 2 Φ(ξ) 2 q2

(49)

We now use equations (34) and (35) to determine the forms of the generators.

Without presenting any more calculations we state the following results: Φ arbitrary: T = a(t) Q1 = 21 a0 (t)q1 Q2 = 12 a0 (t)q2 , where a00 + 4λa = c8 . ( A 3-parameter group). For λ = 0 the lie algebra has a basis given by ∂ X1 = ∂t ∂ + q1 ∂q∂1 + q2 ∂q∂2 X2 = 2t ∂t ∂ X3 = t2 ∂t + tq1 ∂q∂1 + tq2 ∂q∂2

with bracket relations of a simple Lie algebra: [X1 , X2 ] = 2X1 [X1 , X3 ] = X2 [X2 , X3 ] = 2X3 . Φ=

µ ξ2

q2

= µ q22 : 1

14

(50)

It follows from (49) that V =

µ λ 2 (q1 + q22 ) + 2 . 2 q1

(51)

λ 2 µ (q1 + q22 ) + 2 . 2 q2

(52)

This potential already appeared in 2b. Φ = µ = constant: It follows from (49) that V =

This system is not different from the previous one. Φ = ξ2µ+1 ec tan ξ : It follows from (49) that −1

V =

q λ 2 µ c tan−1 q1 2 . (q1 + q22 ) + 2 e 2 q1 + q22

(53)

The generators take the form: T = a(t) Q1 = 41 cc1 q1 + c1 q2 + 12 a0 (t)q1 Q2 = 41 cc1 q2 − c1 q1 + 12 a0 (t)q2 ,

(54)

where a00 + 4λa = c8 . ( A 4-parameter group). The Lie algebra for this system is a direct sum of an sl(2, R) and a one dimensional Lie algebra. It has a basis consisting of the vectors X1 X2 X3 X4 with bracket relations

∂ = ∂t ∂ = 2t ∂t + q1 ∂q∂1 + q2 ∂q∂2 ∂ + tq1 ∂q∂1 + tq2 ∂q∂2 = t2 ∂t     = 14 cq1 + q2 ∂q∂1 + 41 cq2 − q1 ∂q∂2 ,

[X1 , X2 ] [X1 , X3 ] [X2 , X3 ] [Xi , X4 ]

= = = =

2X1 X2 2X3 0 i = 1, 2, 3 .

This example generalizes in n dimension to a Lie algebra which is a direct sum sl(2, R) ⊕ so(n, R). Remark: In polar coordinates this system is V =

µ λ 2 (r ) + 2 ecθ . 2 r

It is integrable, by taking B(θ) = µecθ and I = 12 l2 + B(θ), where l = q1 p2 − p1 q2 .

15

Taking Φ(ξ) =

r1 ξ2

+ r2 we end up with the system V =

r1 r2 λ 2 (q1 + q22 ) + 2 + 2 . 2 q1 q2

The associated Hamilton-Jacobi equation is separable in cartesian and polar coordinates. This system is an example of a system with closed trajectories under the influence of Noncentral field [26]. For λ = 1, the generators take the form: T = c1 + c2 cos 2t + c3 sin 2t Q1 = (−c2 sin 2t + c3 cos 2t)q1 Q2 = (−c2 sin 2t + c3 cos 2t)q2 . They form a 3-dimensional Lie algebra ∂ X1 = ∂t ∂ − q1 sin 2t ∂q∂1 − q2 sin 2t ∂q∂2 X2 = cos 2t ∂t ∂ X3 = sin 2t ∂t + q1 cos 2t ∂q∂1 + q2 cos 2t ∂q∂2 ,

with bracket relations [X1 , X2 ] = −2X3 [X1 , X3 ] = 2X2 [X2 , X3 ] = 2X1 .

(55)

In other words, it is a simple Lie algebra of type A1 isomorphic to so(3, R). For λ = 0 we obtain a Lie algebra isomorphic to sl(2, R). In [27] the most general form of a differential equation invariant under the action of the generators of sl(2, R) is determined. We note that the similar system V =

r1 r2 λ 2 (4q1 + q22 ) + + 2 2 q1 q2

(56)

∂ has ∂t as the only symmetry. In general, the system V = λ2 (q12 + q22 ) + q12 Φ( qq12 ) is integrable: Changing to polar coordinates 2 q1 = r cos θ, q2 = r sin θ we find that

V =

λ B(θ) λ 2 Φ(cot θ) r + 2 2 = r2 + 2 . 2 2 r r sin θ

Letting l = q1 p2 − p1 q2 , the second integral is I = 12 l2 + B(θ). The system V = λ2 (q12 + q22 ) + µ . (ξ+κ)2

µ (q1 +κq2 )2

is integrable. It is a special case of (49) with Φ(ξ) =

a00 = 0

T linear

In this case a(t) is a linear function of time. From equations (36) and (37) we deduce that the function V (q1 , q2 ) satisfies two partial differential equations of the form 16

λ 1 V q1 q1 + λ 2 V q1 q2 + λ 3 = 0 ,

(57)

λ 1 V q1 q2 + λ 2 V q2 q2 + λ 4 = 0 .

(58)

In order to solve these equations, we consider the following cases: • λ1 6= 0, λ2 6= 0 • λ1 = 0, λ2 6= 0 (or λ1 6= 0, λ2 = 0) • λ1 = λ2 = λ3 = λ4 = 0 Without giving any more details, using equations (34) and (35), we are led to the following results: V = λ1 q22 + λ2 q1 q2 + Φ(ξ)

Subcase 2d

where ξ = q1 − λq2 . We obtain various forms of the generators depending on the form of the function Φ. Φ = λ3 ξ 2 : That is, V is quadratic of the form V = λ1 q22 + λ2 q1 q2 + λ3 q12 .

(59)

We have followed the common practice of renaming the constants. The corresponding generators are: T = c6 Q1 = c1 q2 + c3 q1 + d1 (t)  Q2 = c1 q1 + 2 λλ12 − λλ23 c1 + c3 q2 + d2 (t) ,

(60)

where d1 (t) and d2 (t) satisfy the o.d.e.’s d001 + 2λ3 d1 (t) + λ2 d2 (t) = 0, and d002 + 2λ1 d2 (t) + λ2 d1 (t) = 0. ( A 7-parameter group). If λ2 = 0 then λ1 6= λ3 , Q1 = c3 q1 + d1 (t), Q2 = c4 q2 + d2 (t) and d1 (t), d2 (t) satisfy the same o.d.e.s with λ2 = 0. We describe explicitly the Lie algebra for the potential V (q1 , q2 ) = − 12 q12 + 12 q22 . The Lie algebra is 7-dimensional with generators: X1 X2 X3 X4 X5 X6 X7

= = = = = = =

∂ ∂t

q1 ∂q∂1 et ∂q∂1 e−t ∂q∂1 q2 ∂q∂2 cos t ∂q∂2 sin t ∂q∂2 .

with (non-zero) bracket relations 17

[X1 , X3 ] [X1 , X4 ] [X1 , X6 ] [X1 , X7 ] [X2 , X3 ] [X2 , X4 ] [X5 , X6 ] [X5 , X7 ]

= = = = = = = =

X3 −X4 −X7 X6 −X3 −X4 −X6 −X7 .

This Lie algebra L is solvable with L(1) = [L, L] = {X3 , X4 , X6 , X7 } and L(2) = {0}. Remark: The system with Hamiltonian 1 1 (61) H = p21 + p22 + λ1 q12 + λ2 q1 q2 + λ3 q22 . 2 2 is integrable. We can actually rotate the Hamiltonian to a separable one, obtain the second integral and then rotate back to obtain the invariant in the original coordinates. So, we set q1 q2 p1 p2

= = = =

cos θ Qx + sin θ Qy − sin θ Qx + cos θ Qy cos θ px + sin θ py − sin θ px + cos θ py .

The Hamiltonian H will be transformed to a new Hamiltonian which is a function of Q x , Qy , px and py . The coefficient of Qx Qy in the rotated Hamiltonian is (λ1 − λ3 ) sin 2θ + λ2 cos 2θ . If λ1 = λ3 , we choose θ = π4 . If λ1 6= λ3 , then we choose θ to satisfy tan 2θ =

λ2 . λ3 − λ 1

Therefore, in the new coordinates the Hamiltonian is separable of the form 1 2 1 2 px + py + µ1 Q2x + µ2 Q2y . 2 2 We may choose the second integral to be p2x + µ1 Q2x . The second integral for the original system is I = (cos θ p1 − sin θ p2 )2 + µ1 (cos θ q1 − sin θ q2 )2 . Φ arbitrary: In this case, V has the form 1 − λ2 2 q ) + Φ(q1 − λq2 ) , V = λ2 (q1 q2 + 2λ 2 The corresponding generators are: 18

(62)

T = c6 Q1 = λc2 q1 + c2 q2 + λd2 (t) Q2 = c2 q1 + λ1 c2 q2 + d2 (t) ,

(63)

where d2 (t) satisfies the o.d.e. d002 + λλ2 d2 (t) = 0. (A 4-parameter group). Remark: Assume λ = 1. The system with Hamiltonian 1 1 H = p21 + p22 + λ2 q1 q2 + Φ(q1 − q2 ) (64) 2 2 is integrable. We can actually transform the Hamiltonian to a separable one, obtain the second integral and then rotate back to obtain the invariant in the original coordinates. So, we set q1 q2 p1 p2

= √12 Qx + √12 Qy = − √12 Qx + √12 Qy = √12 px + √12 py = − √12 px + √12 py .

The Hamiltonian H will be transformed to a new Hamiltonian which is a function of Q x , Qy , px and py . Therefore, in the new coordinates the Hamiltonian is separable of the form √ 1 2 1 2 λ2 px + py + Φ( 2Qx ) + (Q2y − Q2x ) . 2 2 2 √ We may choose the second integral to be 12 p2x + Φ( 2Qx ) − λ22 Q2x . The second integral for the original system is 1 λ2 I = (p1 − p2 )2 + Φ(q1 − q2 ) − (q1 − q2 )2 . 4 4 Φ = λ3 ξ n , n 6= 0, 1, 2 : In this case, V has the form V = λ2 (q1 q2 +

1 − λ2 2 q ) + λ3 (q1 − λq2 )n , 2λ 2

(65)

with n 6= 0, 1, 2. The generators are: T = 2c5 t + c6 4 Q1 = c2 (λq1 + q2 ) − n−2 c5 q1 + λd2 (t) 4 1 Q2 = c2 (q1 + λ q2 ) − n−2 c5 q2 + d2 (t) ,

(66)

where d2 (t) satisfies the o.d.e. d002 + λλ2 d2 (t) = 0. If λ2 = 0 then d2 (t) = c3 + c4 t and we end-up with a 5-parameter group. Note that for n = −2 we are in subcase 2a with a 6-parameter group. If λ2 6= 0, then we set c5 = 0 and we end up with a 4-parameter group (the same as Φ arbitrary). For example, if V (q1 , q2 ) = q1 q2 + (q1 − q2 )3 then the Lie algebra is generated by 19

∂ X1 = ∂t   X2 = (q1 + q2 ) ∂q∂1 + ∂q∂2   X3 = cos t ∂q∂1 + ∂q∂2

X4 = sin t



∂ ∂q1

+

∂ ∂q2



with (non-zero) bracket relations [X1 , X3 ] [X1 , X4 ] [X2 , X3 ] [X2 , X4 ]

= = = =

−X4 X3 −2X3 −2X4 .

This Lie algebra L is solvable with L(1) = [L, L] = {X3 , X4 } and L(2) = {0}. On the other hand, the potential V (q1 , q2 ) = (q1 − q2 )3 gives a five dimensional Lie algebra. This Lie algebra is isomorphic with the symmetry Lie algebra for the potential V (q 1 , q2 ) = q13 which we examine later. Φ = λ3 eµξ : In this case, V has the form V = λ2 (q1 q2 +

1 − λ2 2 q2 ) + λ3 eµ(q1 −λq2 ) , 2λ

(67)

The generators are: T = 2c5 t + c6 Q1 = c2 (λq1 + q2 ) − µ4 c5 + λd2 (t) Q2 = c2 (q1 + λ1 q2 ) + d2 (t) ,

(68)

where d2 (t) satisfies the o.d.e. d002 + λλ2 d2 = 0. If λ2 = 0 then d2 (t) = c3 + c4 t and we end-up with a 5-parameter group. If λ2 6= 0, then we set c5 = 0 and we end up with a 4-parameter group (the same as Φ arbitrary). The case λ = µ = 1 and λ2 = 0 is the Toda Lattice, a well-known integrable system [28], [29]. We will calculate the Lie algebra of symmetries for the potential of the Toda lattice V (q1 , q2 ) = eq1 −q2 . We obtain a five dimensional Lie algebra with generators X1 X2 X3 X4

∂ = ∂t ∂ = 2t ∂t − 4 ∂q∂1  = (q1 + q2 ) ∂q∂1 +   = ∂q∂1 + ∂q∂2

X5 = t



∂ ∂q1

+

with (non-zero) bracket relations 20

∂ ∂q2



,

∂ ∂q2



[X1 , X2 ] [X1 , X5 ] [X2 , X3 ] [X2 , X5 ] [X3 , X4 ] [X3 , X5 ]

= = = = = =

2X1 X4 −4X4 2X5 −2X4 −2X5 .

This Lie algebra L is solvable with L(1) = [L, L] = {X1 , X4 , X5 }, L(2) = {X4 } and L(3) = {0}. For the case λ2 6= 0 we obtain a Lie algebra which is identical with the one in Φ arbitrary. Φ = λ3 log ξ : Setting λ2 = 0, V takes the form V = λ3 log (q1 − λq2 ) ,

(69)

T = 2c5 t + c6 Q1 = λc2 q1 + c2 q2 + 2c5 q1 + λd2 (t) Q2 = c2 q1 + λ1 c2 q2 + 2c5 q2 + d2 (t) ,

(70)

where d2 (t) = c3 t + c4 ( A 5-parameter group). If λ2 6= 0 the result again is the same as in Φ arbitrary. V = λ1 q22 + Φ(q1 )

Subcase 2e

We obtain various forms of the generators depending on the form of the function Φ. Φ arbitrary: The generators take the form: T = c6 Q1 = 0 Q2 = c4 q2 + d2 (t) ,

(71)

where d2 (t) satisfies the o.d.e. d002 + 2λ1 d2 (t) = 0. (A 4-parameter group). Φ = λ2 q1n , n 6= −2, 0, 1, 2 : If λ1 = 0, then T = 2c5 t + c6 4 c 5 q1 Q1 = 2−n Q2 = c 4 q2 + c 1 t + c 2 ,

(72)

(A 5-parameter group). If λ1 6= 0, we set c5 = 0. We end up with a 4-parameter group. It is the same as in Φ arbitrary. We will calculate explicitly the Lie algebra for the potential V (q1 , q2 ) = q13 . For a basis we choose the following five vector fields:

21

X1 X2 X3 X4 X5

∂ = ∂t ∂ − 2q1 ∂q∂1 = t ∂t = q2 ∂q∂2 = t ∂q∂2 = ∂q∂2 ,

with (non-zero) bracket relations [X1 , X2 ] [X1 , X4 ] [X2 , X4 ] [X3 , X4 ] [X3 , X5 ]

= = = = =

X1 X5 X4 −X4 −X5 .

This Lie algebra L is solvable with L(1) = [L, L] = {X1 , X4 , X5 }, L(2) = {X5 } and L(3) = {0}. On the other hand, for the potential V (q1 , q2 ) = 12 q22 + q13 we obtain a 4-parameter group with basis X1 X2 X3 X4

∂ = ∂t = q2 ∂q∂2 = cos t ∂q∂2 = sin t ∂q∂2 .

This Lie algebra L is also solvable with L(1) = [L, L] = {X3 , X4 } and L(2) = {0}. It is isomorphic with the algebra of symmetries of potential (65) which we already examined. Φ = λ2 eµq1 : If λ1 = 0, then T = 2c5 t + c6 Q1 = −4 c µ 5 Q2 = c4 q2 + d2 (t) ,

(73)

where d2 (t) satisfies the o.d.e. d002 + 2λ1 d2 = 0. (A 5-parameter group). If λ1 6= 0, we set c5 = 0. It is the same case as in Φ arbitrary. Φ = λ2 log q1 : We set λ1 = 0. Then V = λ2 log q1 and the generators take the form: T = 2c5 t + c6 Q1 = 2c5 q1 Q2 = c 4 q2 + c 7 t + c 8 ,

(74)

(A 5-parameter group). Remark: The potentials that appear in this case are clearly integrable, being separable potentials. At this point we have completed the analysis of a separable potential with one variable missing. The potential q12 was considered in subcase 2b. The potentials q1n for n = 0, 1 are 1

22

covered by Case 1. The potential q12 was considered in subcase 2d. The potential f (q1 ) falls under subcase 2e. Subcase 2f : Equations (36) and (37) are satisfied ( d1 (t) = constant , d2 (t) = constant). From equations (34) and (35) we obtain the following results: 1.

V T Q1 Q2

= q2N Φ( qq12 ) = 21 c3 (2 − N )t + c6 = c 3 q1 = c 3 q2 ,

(75)

a 2-parameter group of transformations. The Lie algebra in this case is the twodimensional non-abelian Lie algebra with bracket [X1 , X2 ] = 21 (2 − N )X1 if N 6= 2 and an abelian 2-dimensional Lie algebra if N = 2. We should mention that for certain choices of Φ we may obtain a larger symmetry group, e.g. for Φ(x) = xN , but generically the Lie algebra is 2-dimensional. Some values of N will also give different results. For example, N = −2 falls under subcase 2c.

Remark: In general, this system is not integrable. However, there are some integrable examples. We mention the Holt potentials [19], [20], [30] −2

V (q1 , q2 ) = q2 3 (cq22 + q12 ) , where c = 34 , c =

9 2

(76)

and c = 12.

Also the Fokas-Lagerstr¨om potential [31], V (q1 , q2 ) =

1 (q12

2

− q22 )− 3

.

(77)

Case 2f includes H´enon-Heiles type potentials of the form cq23 +q12 q2 . They are integrable for the following values of c: c = 13 , c = 2, and c = 16 [32], [33], [34]. 3 Finally we mention the potential V (q1 , q2 ) =

q1 . q2

(78)

It was shown by Hietarinta [35] that the second integral for this potential is a transcendental function. It can be expressed as a combination of W+ and W− , the standard Whittaker functions, i.e. the solutions of the equation 1 2 x −a y =0 . y + 4 00





23

(79)

2.

V T Q1 Q2

λ1 log q2 + Φ( qq12 ) c3 t + c6 c 3 q1 c 3 q2 ,

= = = =

(80)

a 2-parameter group of transformations. The Lie algebra in this case is the twodimensional non-abelian Lie algebra with bracket [X1 , X2 ] = X1 . 3.

= eµq1 Φ(q2 ) = 12 c5 t + c6 = − cµ5 = 0,

V T Q1 Q2

(81)

a 2-parameter group of transformations. Remark: Taking Φ(q2 ) = e−µq2 we obtain again the Toda lattice. However, we already have seen that this system has a 5-parameter group of transformations. Therefore, for the specific potential we do not obtain a maximal admitted algebra. Generically, the symmetry group is 2-dimensional. For example, taking V (q1 , q2 ) = eq1 q23 gives a ∂ ∂ two-dimensional non-abelian algebra with basis X1 = ∂t and X2 = t ∂t − 2 ∂q∂1 . 4.

V T Q1 Q2

= = = =

eµq1 Φ(q1 − λq2 ) − 12 λµc8 t + c6 λc8 c8 ,

(82)

a 2-parameter group of transformations. The Lie algebra is again the two-dimensional non-abelian Lie algebra with bracket [X1 , X2 ] = − 21 λµX1 .

Remark: We should mention that because of symmetry we do not list potentials of the form V (q1 , q2 ) = eµq2 Φ(q2 − λq1 ). We can also replace q1 − λq2 with αq1 + βq2 . Taking µ = 1, α = 1 and β = −2 we obtain the potential V (q1 , q2 ) = eq1 −q2 + eq2 . This is a generalized Toda lattice associated with a Lie algebra of type B2 , first considered by Bogoyavlensky in [36]. 5.

V T Q1 Q2

= λ1 q1n + λ2 q2m = 12 c5 t + c6 c5 = 2−n q1 c5 = 2−m q2 ,

(83)

a 2-parameter group of transformations. Here, n 6= 0, 1, 2 and m 6= 0, 1, 2 and m, n not both equal to −2. The Lie algebra in this case is the two-dimensional non-abelian Lie algebra with bracket [X1 , X2 ] = 12 X1 . The symmetry Lie algebra for the potentials 6-10 satisfies precisely the same bracket relation.

24

6.

7.

V T Q1 Q2

V T Q1 Q2

8.

= λ1 q1n + λ2 log q2 , = 21 c5 t + c6 c5 = 2−n q1 = c25 q2 .

= λ1 q1n + λ2 eµq2 , = 21 c5 t + c6 c5 = 2−n q1 = − cµ5 .

V T Q1 Q2

9.

10.

n 6= 0, 1, 2 (84)

n 6= 0, 1, 2 (85)

= λ1 log q1 + λ2 log q2 = 12 c5 t + c6 = c25 q1 = c25 q2 .

(86)

V T Q1 Q2

= λ1 log q1 + λ2 eµq2 = 12 c5 t + c6 = c25 q1 = − cµ5 .

(87)

V T Q1 Q2

= λ 1 e µ1 q 1 + λ 2 e µ2 q 2 = 12 c5 t + c6 = − µc51 = − µc52 .

(88)

11.

Φ(q12 + q22 ) c6 c 1 q2 −c1 q1 ,

(89)

λ(q12 + q22 )n , n 6= −1, 0, 1 2c5 t + c6 2 c 5 q1 c1 q2 − n−1 2 c 5 q2 , −c1 q1 − n−1

(90)

V T Q1 Q2

= = = =

a 2-parameter group of transformations. This is a unit mass in 2-dimensional space moving in a central field, i.e. a potential which is a function of r only. The function q1 p2 − q2 p1 is a second integral. Note that in this case the Lie algebra is abelian. 12.

V T Q1 Q2

= = = =

25

a 3-parameter group of transformations. The Lie algebra is 3-dimensional with only non-zero bracket [X1 , X2 ] = 2X1 . The case n = − 12 is a Kepler problem. For n = −1 the Lie algebra is 4-dimensional, it falls under subcase 2c. See (53) with λ = c = 0 and µ = 1. Remark: This potential is a special case of system 1 with N = 2n. Taking n = 2, we have a system of the form aq14 +bq12 q22 +cq24 . In general (for a, b, c non-zero) this system has a 2-dimensional group of symmetries unless b = 2a = 2c. Generically the potential V (q1 , q2 ) = aq14 + bq12 q22 + cq24 is not integrable, but for certain values of the parameters it becomes integrable. That is the case when b = 6a = 6c or a = 16c, b = 12c or b = 6a, c = 8a [19], [21], [37]. 13.

V T Q1 Q2

= = = =

λ log (q12 + q22 ) 2c5 t + c6 c1 q2 + 2c5 q1 −c1 q1 + 2c5 q2 ,

(91)

a 3-parameter group of transformations. The Lie algebra, which is the same as in the previous case, may not seem interesting, but in n dimensions it is a direct sum of a 2-dimensional Lie algebra with so(n, R). 14. V T Q1 Q2

= = = =

q 2 −q 2

λ sin−1 q12 +q22 2 1 2c5 t + c6 c1 q2 + 2c5 q1 −c1 q1 + 2c5 q2 ,

(92)

a 3-parameter group of transformations. This potential can be written in the form λ1 + λ2 θ, in polar coordinates. In other words, it is a linear function of θ.

VI

Generalizations, Questions

Case 1 generalizes in n dimensions. Consider a Hamiltonian with n degrees of freedom, H=

n 1X p2 + V (q1 , q2 , . . . , qn ) , 2 i=1 i

and the associated Lagrange-Newton equations q¨i + Vqi = 0 ,

i = 1, 2, . . . , n .

(93)

As in the case of two degrees of freedom we seek point symmetries of equations (93). We consider the equations Γ(2) {¨ qi + Vqi } = 0 i = 1, 2, . . . , n , (94) 26

where Γ(2) is the second prolongation of Γ=T

n X ∂ ∂ Qi + . ∂t i=1 ∂qi

(95)

Equations (94) give n identities of the form: Ei (t, q1 , q2 , . . . , qn , q˙1 , q˙2 , . . . , q˙n ) = 0

i = 1, 2, . . . , n ,

(96)

∂V . The functions Ei are explicit polynomials in q˙1 , q˙2 , . . . , q˙n . where, we have used that q¨i = − ∂q i We impose the condition that equations (96) are identities in the variables t, q i , q˙i which are regarded as independent. Again, the functions T and Qi must be of the form

T = a(t) + ni=1 bi (t)qi Pn Pn 0 Qi = k=1 bk (t)qi qk + k=1 cik (t)qk + di (t) P

i = 1, 2, . . . , n .

(97)

We substitute (97) into (96). By considering the coefficient of q˙k in Ej we obtain the following n2 equations: For j 6= k, ∂V bk = 0 , (98) b00k (t)qj + c0jk (t) + ∂qj and for j = k, 2c0jj (t)

00

− a (t) +

3b00j (t)qj

!

X ∂V ∂V bj + bi (t) = 0 . +3 b00i (t)qi + ∂qj ∂qi i6=j

(99)

It follows from equations (98) that V is quadratic of the form n X

V =

λi qi2 +

i=1

n X

µi qi ,

i=1

unless bi (t) = 0 for i = 1, 2, . . . , n. Substituting (100) into (99), we obtain b00i + 2λi bi = 0

i = 2, 3, . . . , n

b00i + 2λ1 bi = 0

i = 2, 3, . . . , n .

and Therefore for non-zero bi (t), we necessarily have λ1 = λ 2 = . . . = λ n . Hence, V is of the form V =

n λX q2 , 2 i=1 i

where the linear terms are ignored. One can easily deduce the form of the generators: a(t) is a solution of a second order equation of the form a00 + 4λa = c. (3 parameters) 27

(100)

bi (t) is a solution of b00i + λbi = 0. (2n parameters) di (t) is a solution of d00i + λdi = 0. (2n parameters) cij (t) are constant for i 6= j and ckk = ct + ck − 2λ a(t)dt. (n2 parameters) Therefore, the dimension of the symmetry algebra is 3 + 2n + 2n + n2 = (n + 2)2 − 1. The case of the harmonic oscillator has been studied in [38] where it is shown that the symmetry group for a time-dependent harmonic oscillator is SL(n + 2, R). R

When λ = 0, the potential energy is zero and we have a free particle moving in Rn . In this case the generators take the following simple form: a(t) bi (t) di (t) cii cij

= = = = =

c1 + c2 t + ct2 α i + βi t γ i + δi t i + ct κij i 6= j .

(101)

The dimension is again (n + 2)2 − 1. This dimension is in agreement with the results in [39], where upper bounds for the dimension of symmetry groups are obtained. In case 2, bi (t) = 0 for i = 1, 2, . . . , n and equations (99) and (100) imply that cjk (t) = cjk f or j 6= k cjj (t) = 12 a0 (t) + cjj , where cjk are constants. Equations (96) now become n n X X 1 000 3 0 ∂V ∂2V ∂V 00 a (t)qi + di (t) + γk + a (t) − =0, cik 2 2 ∂qi k=1 ∂qk k=1 ∂qi ∂qk

for i = 1, 2, . . . , n, where

n X 1 0 γk = qk a (t) + ckj qj + dk (t) . 2 j=1

(102)

(103)

Using equations (102) we can prove the following: 1. The potential q12 has an n2 + 3 parameter group of symmetries. 2. The potential

1 q12

has an n2 + 2 parameter group of symmetries.

3. The potential q1k , k ∈ / {−2, 0, 1, 2} has an n2 + 1 parameter group of symmetries.

4. The potential f (q1 ) where f is arbitrary but not exponential, logarithmic or a power has an n2 parameter group of symmetries. We give the proof for the potential 1 V (q1 , q2 , . . . , qn ) = q12 . 2 The proof for the other three cases is similar. Since the variables q2 , q3 , . . . , qn are missing, equations (102) for k = 2, 3, . . . , n become 1 000 a qk + d00k − ck1 q1 = 0 . 2 28

Therefore, a000 = 0, dk (t) = ξk + ηk t, and ck1 = 0 for k = 2, 3, . . . , n. On the other hand, the first equation in (102) gives d001

3 0 + γ1 + a − c11 q1 = 0 , 2 



where n 1 0 X c1j qj + d1 . γ1 = q 1 a + 2 j=1

Therefore, a0 (t) = 0, d1 (t) = c2 cos t + c3 sin t, and c1j = 0 for j = 2, 3, . . . , n. We obtain the following form for the generators: T = c1 Q1 = cP11 q1 + c2 cos t + c3 sin t n Qk = j=2 ckj qj + ξk + ηk t

(104) k = 2, . . . , n .

Therefore, the dimension of the algebra of symmetries is 4 + (n − 1)2 + 2(n − 1) = n2 + 3.

We would like to close with some questions:

1. Is every dimension between 1 and n2 + 3 obtained? In the case n = 2 it is true. In the case n = 3 we did not obtain the dimension 8 so far. In n = 4 the dimensions 13 and 14 are probably missing. Can one show that no dimension in the range {n2 + 4, . . . , n2 + 4n + 2} is obtained? 2. It seems that linear combinations of the qik , i = 1, 2, . . . , n and k ∈ {−2, 0, 1, 2, 3} are enough for the classification of symmetry groups. One should investigate the connection with singularity theory. 3. Equations (93) are not the most general second order ordinary differential equations in n variables. However, using a Laguerre type transformation one can transform the general system of second order equations closer to the ones in (93). It is natural to ask what is the classification of symmetry groups for a general system of second order equations. 4. Is every Lie algebra the symmetry algebra of an equation of the form (93)? The same question for simple Lie algebras. Aknowledgement: We would like to thank Peter Leach for sending us preprints of his work and also for some valuable remarks on the transcript. We would also like to thank Rui Fernandes and Peter Olver for some helpful comments.

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