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Hamiltonian symmetries of the Henon-Heiles system
Volume 97A, number 1,2
PHYSICS LETTERS
8 August 1983
HAMILTONIAN SYMMETRIES OF THE HENON—HEILES SYSTEM Allan P. FORDY Mathematics Department, UMIST, P.O. Box 88, Manchester M60 I QD, UK Received 3 June 1983
Hamiltonian symmetries find all the parameter ranges for which the Hénon—Heiles hamiltonian system H 2 + c 2)/2are+ used ax2y to — by3 /3 is completely integrable. 2 + ~ + c 1x 2y
1. Introduction. The hamiltonian flow generated by: H—-(x 2 1
+y 2
+cjx 2
+c
2
2
1
2y )+axy—--by
3
(1)
There has been some speculation [3] that these are not the only integrable cases. The Painlevé method
where dot denotes the t-derivative, has been much studied in recent years. Nearly twenty years ago Hénon and Heiles [1] introduced this hamiltonian, with c1 = c2= a = b = 1. They found that for low energies this system appeared to be integrable, in so much as trajectories (numerically integrated) always lay on well defined two-dimensional surfaces. However, they also found that for large energies many of these integral surfaces were destroyedand that phase space acquired la~rgeergodic regions. Most of the recent interest in (1) has centred on determining those parameter ranges for which the system is (truly) integrable. For such cases there would exist a second hamiltonian K(x, J’, ~ 3’) which is in involution with H:
seems to leave open the integrability for certain other parameter values. In this letter it is proven that there are no values of the parameters c1, c2, a, b other than those listed above, for which the system (1) is integrable. The method used is to seek commuting hamiltonian flows. These are placed in the context of generalised symmetries, which play an important role in the investigation of partial differential equations solvable by inverse scattering techniques [7—9]. The calculation of generalised symmetries can be very long and tedious. However, if we restrict attention to those which are defined by hamiltonian flows then the whole calculation is much more efficient. The result is that all the integrable cases of(l) are very quickly determined. Unlike the Painlevé approach, the present method also gives a construction of the second ham-
{H, K} = 0
iltonian in the integrable cases.
2
,
,
(2)
,
where curly brackets denote the usual Poissoni bracket. The case = c2, a = —b has been known to be integrable for some time [2] A second integral of the motion was written down by Greene * 1 for the case a = 1, b = —6, c1, c2 arbitrary. Most recently, the Painlevé test has been applied to the general system (1) by several authors [3—5]. As well as the integrable cases enumerated above, they also isolated the case c1/c2 = —a/b = 1/16. The second integral in this case was given in refs. [5,6]. .
* 1
Greene does not seem to have published this, but his result is referred to in refs. [3,4].
0.031-9163/83/0000—0000/s 03.00 © 1983 North-Holland
2. Hamiltonian symmetries. The hamiltonian flow associated with (1) can be written as a system of second order differential equations: 2
d x/dt
2
2 =
—c1x
—
2axy
,
d
2 y/dt
2 =
—c2y
—
ax
2 + by
3 In order to define a generalised symmetry we consider the dynamical system:
dy/dr = Y(x,y, x ,y)
.
(4) 21
Volume 97A, number 1,2
PHYSICS LETTERS
The integral curves of (4) define a flow in phase space. If the flows (3) and (4) commute, then (4) is a symmetry of(3). In this case the two “time” derivatives d/dt and d/dr commute and X and Y satisfy the differential equations: 2X—(c D 1 +2ay)X— 2axY, 2Yz~~2axX+(2by—c D 2)Y, where D is the total derivative operator:
8 August 1983
nian symmetry is not only that the calculation leads directly to a secondhamiltonian, but that eqs. (5) and (8), taken together, quickly lead to the simple system of partial differential equations: K1~=0,
(9a)
Koy +~K1~ —(c1x
+
2axy)K1
=
0,
(9b)
(5) ~Ko~
—
(c1x
2axy)K0~ 2 by2)K
+
—
DX =~aX/ax÷3’ax7ay+ ~ax~a~ ÷jax/a. (6)
—
(c2y
+
ax
1
=
0,
(9c)
Whenever ii or .j occur they are replaced by the righthand sides of (3). Since (6) needs to be applied twice to each ofX andY, the equations of (5) are very long and tedious to solve. Furthermore, there may well exist solutions of(5) which are not related to a second hamiltonian. To alleviate both of these problems we restrict attention to hamiltonian symmetries. Suppose there exists a second hamiltonian K(x,y, .~,5’)in involution withH. It is assumed that K is polynomial in the derivatives and 5’. For fixed energy we can use (1) to eliminate even powers of j~.Therefore, up to equivalence, K is linear in :
where subscriptsx, y and refer to partial derivatives. These three equations can be systematically solved to find K. The first of them says that K1 is a function of just the two variables x and The second equation can then be integrated to give:
K =K0(x,y,~)+K1(x,y,)3’
of powers ofy, the following four equation arise:
.
(7)
Comment. Since H is invariant under the discrete symmetry t —t we can assume K to have the same symmetry. Thus K must be an even function of the velocities and 5’. This means that K0 is an even function of~while K1 is odd. This observation saves work in the next section. Because of the involution condition (2), the hamiltonian flow generated by K commutes with (3), so is indeed a symmetry in the sense of(4) and (5) above. Such a symmetry is called hamiltonian symmetry. If (4) are to be the first half of Hamilton’s equations, then: X=aK0/a~÷(aK1/a,~)~, Y=K1,
~.
K0(x,y, ~)= K~(x,~)+ (c1xK1~ 2K + axy 1~ (10) where K00 (x, ~)is the “constant” of integration. We now have the complete dependence of K upony. By subtituting (10) into (9c) and matching coefficients —
,
K1~ = 0
(lla)
,
—~
DX = —aK0/ax DY
—
(aK1/ax))
aK0/~y (~K1J3y) —
2axK00
=
0,
(8)
(lIb)
c1(~K1+ xKi~+ 2x~K1~~) —
2K J~Koo~ c1xK00~ ax —
1
—
(lic) (lId)
0.
ForK odd mx, (1 la) gives K1 (x, ~)= K11 (x),
(12)
while (1 Ib) gives SaxKiix + (a + b)K11 0. Defining X = a/b, there are two cases:
,
,
a(~Kj~ + 2xKi~+ 3xK1~~) + bK1
X
—1,
(13)
K11 constant,
where D is the total derivative operator (6).
X = —I/(5m
3. Results. The eqs. (5), taken alone are quite cornplicated to solve. The advantage of seeking a hamilto-
a constant. (14b) The remainder of the calculation is very simple and
22
+ I)
,
m
(14a) 0
,
K11
=
ei.xm
Volume 97A, number 1 ,2
PHYSICS LETTERS
splits into three cases: Case A —1. This is the well-known example [2] which separates in the coordinatesx ±y. Case (14b) m = 1. This is the case shown to be integrable by Greene (see footnote). With a = 1 and a = 4 the second hamiltonian is 4+4x2y2+4c
Kx
8 August 1983
iltonian symmetry. Finally, the Painlevd method can only isolate an integrable case; it does not construct the second hamiltonian in the process. The method of this letter can be applied to much more general hamiltonians than (1). For instance, it is possible to characterise the potentials V(x,y) for which the conservative system with 2 degrees of free-
2y—4~2y
dom
1x
+(4c
2+c 2)+4x~, (15) 1 —c2)( 1x as written down by Greene. Case(14b) m ~‘2. For the consistency of(l lc—d) we need m = 3, so that A = —1/16, and c 2 = 16c1 With a = —4a the second hamiltonian is 4 +6c 22+12ax2y—4ac 4y K=3 1x 1x —4a2x4y2 + 3c?x4 —~ 2 6 4ax3~3’, (16)
H’~~(~2+3’2)+V(x,y),
(17)
is integrable [10] I thank the SERC for financial support during this work. References
.
[11 M. Hénon and C. Heiles, Astron. J. 69 (1964) 73.
—
3a x
as written down in refs. [5,6]. There are no other solutions of eqs. (9).
[2] Y. Aizawa and N. Saito, J. Phys. Soc. Japan 32 (1972) 1636. [3] Y.F. Chang, M. Tabor and J. Weiss, J. Math. Phys. 23
4. Conclusions. This note has used the Hdnon— Heiles system as a vehicle to exhibit a quick and efficient way of testing a hamiltonian system for complete integrability.It has isolated the three values A = 1, as does the Painlevé method. However, the
(1982) 531. [4] T. Bountis, H. Segur and F. Vivaldi, Phys. Rev. A25 (1982) 1257. [51B. Grammaticos, 89A (1982) 111. B. Dorizzi and R. Padjen, Phys. Lett. [6] L.S. Hall, Preprint UC1D - 18980 (1982). [71 R.L. Anderson, S. Kumei and C.E. Wulfman, Phys. Rev.
latter approach also conjectures other values of A for which the Hénon—Heiles system may be integrable [3] whereas (14a) and the formula (14b) with m = 1 or 3 show that these are the only values corresponding to integrabiity. It should be understood that the existence of a second hamiltonian implies the existence of a ham-
.
[8] [9] [10]
Lett. 28 (1972) 988. AS. Fokas, J. Math. Phys. 21(1980)1318. N.K. Ibragimov and A.B. Shabat, Funct. Anal. AppI. 14 (1980) 25 (Russian); 14 (1980) 19 (English). A.P. Fordy, On the integrability of the system with hamiltonian H = (~2 + ~2) + V(x, y), preprint.
23
PHYSICS LETTERS
8 August 1983
HAMILTONIAN SYMMETRIES OF THE HENON—HEILES SYSTEM Allan P. FORDY Mathematics Department, UMIST, P.O. Box 88, Manchester M60 I QD, UK Received 3 June 1983
Hamiltonian symmetries find all the parameter ranges for which the Hénon—Heiles hamiltonian system H 2 + c 2)/2are+ used ax2y to — by3 /3 is completely integrable. 2 + ~ + c 1x 2y
1. Introduction. The hamiltonian flow generated by: H—-(x 2 1
+y 2
+cjx 2
+c
2
2
1
2y )+axy—--by
3
(1)
There has been some speculation [3] that these are not the only integrable cases. The Painlevé method
where dot denotes the t-derivative, has been much studied in recent years. Nearly twenty years ago Hénon and Heiles [1] introduced this hamiltonian, with c1 = c2= a = b = 1. They found that for low energies this system appeared to be integrable, in so much as trajectories (numerically integrated) always lay on well defined two-dimensional surfaces. However, they also found that for large energies many of these integral surfaces were destroyedand that phase space acquired la~rgeergodic regions. Most of the recent interest in (1) has centred on determining those parameter ranges for which the system is (truly) integrable. For such cases there would exist a second hamiltonian K(x, J’, ~ 3’) which is in involution with H:
seems to leave open the integrability for certain other parameter values. In this letter it is proven that there are no values of the parameters c1, c2, a, b other than those listed above, for which the system (1) is integrable. The method used is to seek commuting hamiltonian flows. These are placed in the context of generalised symmetries, which play an important role in the investigation of partial differential equations solvable by inverse scattering techniques [7—9]. The calculation of generalised symmetries can be very long and tedious. However, if we restrict attention to those which are defined by hamiltonian flows then the whole calculation is much more efficient. The result is that all the integrable cases of(l) are very quickly determined. Unlike the Painlevé approach, the present method also gives a construction of the second ham-
{H, K} = 0
iltonian in the integrable cases.
2
,
,
(2)
,
where curly brackets denote the usual Poissoni bracket. The case = c2, a = —b has been known to be integrable for some time [2] A second integral of the motion was written down by Greene * 1 for the case a = 1, b = —6, c1, c2 arbitrary. Most recently, the Painlevé test has been applied to the general system (1) by several authors [3—5]. As well as the integrable cases enumerated above, they also isolated the case c1/c2 = —a/b = 1/16. The second integral in this case was given in refs. [5,6]. .
* 1
Greene does not seem to have published this, but his result is referred to in refs. [3,4].
0.031-9163/83/0000—0000/s 03.00 © 1983 North-Holland
2. Hamiltonian symmetries. The hamiltonian flow associated with (1) can be written as a system of second order differential equations: 2
d x/dt
2
2 =
—c1x
—
2axy
,
d
2 y/dt
2 =
—c2y
—
ax
2 + by
3 In order to define a generalised symmetry we consider the dynamical system:
dy/dr = Y(x,y, x ,y)
.
(4) 21
Volume 97A, number 1,2
PHYSICS LETTERS
The integral curves of (4) define a flow in phase space. If the flows (3) and (4) commute, then (4) is a symmetry of(3). In this case the two “time” derivatives d/dt and d/dr commute and X and Y satisfy the differential equations: 2X—(c D 1 +2ay)X— 2axY, 2Yz~~2axX+(2by—c D 2)Y, where D is the total derivative operator:
8 August 1983
nian symmetry is not only that the calculation leads directly to a secondhamiltonian, but that eqs. (5) and (8), taken together, quickly lead to the simple system of partial differential equations: K1~=0,
(9a)
Koy +~K1~ —(c1x
+
2axy)K1
=
0,
(9b)
(5) ~Ko~
—
(c1x
2axy)K0~ 2 by2)K
+
—
DX =~aX/ax÷3’ax7ay+ ~ax~a~ ÷jax/a. (6)
—
(c2y
+
ax
1
=
0,
(9c)
Whenever ii or .j occur they are replaced by the righthand sides of (3). Since (6) needs to be applied twice to each ofX andY, the equations of (5) are very long and tedious to solve. Furthermore, there may well exist solutions of(5) which are not related to a second hamiltonian. To alleviate both of these problems we restrict attention to hamiltonian symmetries. Suppose there exists a second hamiltonian K(x,y, .~,5’)in involution withH. It is assumed that K is polynomial in the derivatives and 5’. For fixed energy we can use (1) to eliminate even powers of j~.Therefore, up to equivalence, K is linear in :
where subscriptsx, y and refer to partial derivatives. These three equations can be systematically solved to find K. The first of them says that K1 is a function of just the two variables x and The second equation can then be integrated to give:
K =K0(x,y,~)+K1(x,y,)3’
of powers ofy, the following four equation arise:
.
(7)
Comment. Since H is invariant under the discrete symmetry t —t we can assume K to have the same symmetry. Thus K must be an even function of the velocities and 5’. This means that K0 is an even function of~while K1 is odd. This observation saves work in the next section. Because of the involution condition (2), the hamiltonian flow generated by K commutes with (3), so is indeed a symmetry in the sense of(4) and (5) above. Such a symmetry is called hamiltonian symmetry. If (4) are to be the first half of Hamilton’s equations, then: X=aK0/a~÷(aK1/a,~)~, Y=K1,
~.
K0(x,y, ~)= K~(x,~)+ (c1xK1~ 2K + axy 1~ (10) where K00 (x, ~)is the “constant” of integration. We now have the complete dependence of K upony. By subtituting (10) into (9c) and matching coefficients —
,
K1~ = 0
(lla)
,
—~
DX = —aK0/ax DY
—
(aK1/ax))
aK0/~y (~K1J3y) —
2axK00
=
0,
(8)
(lIb)
c1(~K1+ xKi~+ 2x~K1~~) —
2K J~Koo~ c1xK00~ ax —
1
—
(lic) (lId)
0.
ForK odd mx, (1 la) gives K1 (x, ~)= K11 (x),
(12)
while (1 Ib) gives SaxKiix + (a + b)K11 0. Defining X = a/b, there are two cases:
,
,
a(~Kj~ + 2xKi~+ 3xK1~~) + bK1
X
—1,
(13)
K11 constant,
where D is the total derivative operator (6).
X = —I/(5m
3. Results. The eqs. (5), taken alone are quite cornplicated to solve. The advantage of seeking a hamilto-
a constant. (14b) The remainder of the calculation is very simple and
22
+ I)
,
m
(14a) 0
,
K11
=
ei.xm
Volume 97A, number 1 ,2
PHYSICS LETTERS
splits into three cases: Case A —1. This is the well-known example [2] which separates in the coordinatesx ±y. Case (14b) m = 1. This is the case shown to be integrable by Greene (see footnote). With a = 1 and a = 4 the second hamiltonian is 4+4x2y2+4c
Kx
8 August 1983
iltonian symmetry. Finally, the Painlevd method can only isolate an integrable case; it does not construct the second hamiltonian in the process. The method of this letter can be applied to much more general hamiltonians than (1). For instance, it is possible to characterise the potentials V(x,y) for which the conservative system with 2 degrees of free-
2y—4~2y
dom
1x
+(4c
2+c 2)+4x~, (15) 1 —c2)( 1x as written down by Greene. Case(14b) m ~‘2. For the consistency of(l lc—d) we need m = 3, so that A = —1/16, and c 2 = 16c1 With a = —4a the second hamiltonian is 4 +6c 22+12ax2y—4ac 4y K=3 1x 1x —4a2x4y2 + 3c?x4 —~ 2 6 4ax3~3’, (16)
H’~~(~2+3’2)+V(x,y),
(17)
is integrable [10] I thank the SERC for financial support during this work. References
.
[11 M. Hénon and C. Heiles, Astron. J. 69 (1964) 73.
—
3a x
as written down in refs. [5,6]. There are no other solutions of eqs. (9).
[2] Y. Aizawa and N. Saito, J. Phys. Soc. Japan 32 (1972) 1636. [3] Y.F. Chang, M. Tabor and J. Weiss, J. Math. Phys. 23
4. Conclusions. This note has used the Hdnon— Heiles system as a vehicle to exhibit a quick and efficient way of testing a hamiltonian system for complete integrability.It has isolated the three values A = 1, as does the Painlevé method. However, the
(1982) 531. [4] T. Bountis, H. Segur and F. Vivaldi, Phys. Rev. A25 (1982) 1257. [51B. Grammaticos, 89A (1982) 111. B. Dorizzi and R. Padjen, Phys. Lett. [6] L.S. Hall, Preprint UC1D - 18980 (1982). [71 R.L. Anderson, S. Kumei and C.E. Wulfman, Phys. Rev.
latter approach also conjectures other values of A for which the Hénon—Heiles system may be integrable [3] whereas (14a) and the formula (14b) with m = 1 or 3 show that these are the only values corresponding to integrabiity. It should be understood that the existence of a second hamiltonian implies the existence of a ham-
.
[8] [9] [10]
Lett. 28 (1972) 988. AS. Fokas, J. Math. Phys. 21(1980)1318. N.K. Ibragimov and A.B. Shabat, Funct. Anal. AppI. 14 (1980) 25 (Russian); 14 (1980) 19 (English). A.P. Fordy, On the integrability of the system with hamiltonian H = (~2 + ~2) + V(x, y), preprint.
23