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Analytic integrability of the Bianchi class A cosmological models with 0 ⩽ k < 1
Chaos, Solitons & Fractals 48 (2013) 12–21
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Chaos, Solitons & Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos
Analytic integrability of the Bianchi class A cosmological models with 0 6 k < 1 q Antoni Ferragut a, Jaume Llibre b, Chara Pantazi c,⇑ a
Departament de Matemàtica Aplicada I, Universitat Politècnica de Catalunya, ETSEIB, Av. Diagonal 647, 08028 Barcelona, Catalonia, Spain Departament de Matemàtiques, Universitat Autònoma de Barcelona, Edifici C, 08193 Bellaterra, Barcelona, Catalonia, Spain c Departament de Matemàtica Aplicada I, Universitat Politècnica de Catalunya, EPSEB, Av. Doctor Marañón 44–50, 08028 Barcelona, Catalonia, Spain b
a r t i c l e
i n f o
Article history: Received 20 July 2012 Accepted 31 December 2012 Available online 31 January 2013
a b s t r a c t Many works study the integrability of the Bianchi class A cosmologies with k = 1, where k is the ratio between the pressure and the energy density of the matter. Here we characterize the analytic integrability of the Bianchi class A cosmological models when 0 6 k < 1. We conclude that Bianchi types VI0, VII0, VIII and IX can exhibit chaos whereas Bianchi type I is not chaotic and Bianchi type II is at most partially chaotic. Ó 2013 Elsevier Ltd. All rights reserved.
1. Introduction In a cosmological model Einstein’s equations connect the geometry of the space–time with the properties of the matter. The matter occupying the space–time is determined by the stress energy tensor of the matter. Friedman in 1922 introduced the study of homogeneous cosmologies. In 1951, Taub determined the complete homogeneous solutions of the equations of the general theory of relativity. For spatially homogeneous cosmologies the Einstein field equations can be written as an autonomous system of first order differential equations, see [16] and references therein. Bianchi models describe space–times which are foliated by homogeneous (and so we have three dimensional Lie algebras) hypersurfaces of constant time. Bianchi [1,2]
q All the authors are partially supported by the MICINN/FEDER Grant MTM2008-03437. A.F. is additionally supported by grants Juan de la Cierva, 2009SGR410 and MTM2009-14163-C02-02. J.L. is additionally partially supported by an AGAUR Grant number 2009SGR410 and by ICREA Academia. C.P. is additionally partially supported by the MICINN/ FEDER Grant number MTM2009-06973 and by the AGAUR Grant number 2009SGR859. ⇑ Corresponding author. Tel.: +34934016283; fax: +34 93 401 17 13. E-mail addresses: [email protected] (A. Ferragut), jllibre@mat. uab.cat (J. Llibre), [email protected] (C. Pantazi).
0960-0779/$ - see front matter Ó 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.chaos.2012.12.007
was the first to classify three dimensional Lie algebras which are non-isomorphic. There are nine types of models according to the dimension n of the derivative subalgebra: (a) (b) (c) (d)
n = 0: n = 1: n = 2: n = 3:
type I; types II, III; types IV, V, VI, VII; types VIII, IX.
If we consider X1, X2, X3 an appropriate basis of the 3dimensional Lie Algebra, then the classification depends on a scalar a 2 R and a vector (n1, n2, n3), with ni 2 {+1, 1, 0}, such that
½X 1 ; X 2 ¼ n3 X 3 ;
½X 2 ; X 3 ¼ n1 X 1 aX 2 ;
½X 3 ; X 1 ¼ n2 X 2 þ aX 1 ; where [,] is the Lie bracket. In particular for a = 0 we obtain models of class A and for a – 0 we obtain models of class B. A good reference for the Bianchi models is Bogoyavlensky [3]. Collins in 1971 was the first to study Bianchi models as dynamical systems, see [4]. Later on Bogoyavlensky and collaborators proved that the dynamics of class A types can be reduced into a six dimensional autonomous system of differential equations of first order defined in a compact manifold with a boundary.
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In our study we follow [3] and we consider the hydrodynamical tensor of the matter. We will work with an equation of state of matter of the form p = ke, where e is the energy density of the matter, p is the pressure and 0 6 k 6 1, see also [3]. The case k = 1 is studied in [6]. Here we consider the case where 0 6 k < 1. Following [3] the Einstein equations for the homogenous cosmologies of class A without motion of matter can be formalized as a Hamiltonian system in the phase space pi, qi for i = 1, 2, 3 with the Hamiltonian function
H¼
1 1k 2
ðq1 q2 q3 Þ
1 Tðpi qi Þ þ V G ðqi Þ : 4
Table 1 The classification of Bianchi class A cosmologies. Type
I
II
VI0
VII0
VIII
IX
a n1 n2 n3
0 0 0 0
0 1 0 0
0 1 1 0
0 1 1 0
0 1 1 1
0 1 1 1
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F; 4 k1 F; x_ 5 ¼ n2 x2 ðn1 x1 þ n2 x2 n3 x3 Þ þ 4 k1 F; x_ 6 ¼ n3 x3 ðn1 x1 n2 x2 þ n3 x3 Þ þ 4 x_ 4 ¼ n1 x1 ðn1 x1 n2 x2 n3 x3 Þ þ
Here T is the kinetic energy (not positively defined) and VG is the potential. According to [3] (Section 4 of Chapter II) the kinetic and the potential energy are given, respectively, by 3 3 X X Tðpi qi Þ ¼ 2 pi pj qi qj p2i q2i ; i
i¼1
3 X
3 X
i
i¼1
V G ðqi Þ ¼ 2
n i n j qi qj
n2i q2i
We consider the Hamiltonian system
@H ; @pi
p_ i ¼
where
F ¼ n21 x21 þ n22 x22 þ n23 x23 2n1 n2 x1 x2 2n1 n3 x1 x3
for i, j 2 {1, 2, 3}.
q_ i ¼
ð1Þ
@H ; @qi
where the dot denotes derivative with respect to the time t. More precisely, the Hamiltonian system is k1
q_ 1 ¼ 2q1 ðq1 q2 q3 Þ 2 ðp1 q1 þ p2 q2 þ p3 q3 Þ; k1
q_ 2 ¼ 2q2 ðq1 q2 q3 Þ 2 ðp1 q1 p2 q2 þ p3 q3 Þ; k1
q_ 3 ¼ 2q3 ðq1 q2 q3 Þ 2 ðp1 q1 þ p2 q2 p3 q3 Þ; k1
p_ 1 ¼ ðq1 q2 q3 Þ 2 ð2p1 ðp1 q1 þ p2 q2 þ p3 q3 Þ 1 1k þ n1 ðn1 q1 þ n2 q2 þ n3 q3 Þ þ H; 2 2q1
k1
p_ 2 ¼ ðq1 q2 q3 Þ 2 ð2p2 ðp1 q1 p2 q2 þ p3 q3 Þ 1 1k H; þ n2 ðn1 q1 n2 q2 þ n3 q3 Þ þ 2 2q2 k1
p_ 3 ¼ ðq1 q2 q3 Þ 2 ð2p3 ðp1 q1 þ p2 q2 p3 q3 Þ 1 1k H: þ n3 ðn1 q1 þ n2 q2 n3 q3 Þ þ 2 2q3 Note that the constants n1, n2, n3 determine the type of the model according to Table 1. After the change of coordi1k nates ds ¼ ðq1 q2 q3 Þ 2 dt; qi ¼ xi ; pi ¼ xiþ3 =ð2xi Þ; i ¼ 1; 2; 3, we obtain the quadratic homogeneous polynomial differential system
2n2 n3 x2 x3 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 : ð2Þ Note that system (1) is a homogeneous polynomial differential system of degree 2. The Hamiltonian H becomes after the changes of variables the first integral k1
H ¼ ðx1 x2 x3 Þ 2 F ¼ ðx1 x2 x3 Þ
k1 2
n21 x21 þ n22 x22 þ n23 x23
2n1 n2 x1 x2 2n1 n3 x1 x3 2n2 n3 x2 x3 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6
ð3Þ
of system (1). Many authors have studied some models of class A for the case k = 1 considering different types of integrability, see for example [4–14]. More concretely, Szydłowski and Demaret in [15] studied the vacuum Bianchi models of class A reducing the dimension of the dynamical system. Maciejewski and Szydłowski in [12] also reduced the dimension of the dynamical system of class A. Szydłowski and Biesiada in [14] studied the algebraic integrability of class A Bianchi models first in vacuum and afterwards considering dust and radiative matter. Bianchi IX model causes a lot of interest because of its chaotic (in some sense) behavior. Many authors also studied the (different notions of) integrability of this model. More concretely Cushman and Sniatycki [5] proved that Bianchi IX is a locally integrable Hamiltonian system. Latifi et al. [7] proved that this model is not integrable since it does not satisfy the Painlevé property. Llibre and Valls in [8] proved that Bianchi IX does not admit a complete set of formal first integrals. Morales-Ruiz and Ramis [13], using tecnics of differential Galois theory, showed that the Bianchi IX model is not completely integrable with rational first integrals. The integrability of Bianchi types VIII and IX was also studied by Maciejewski and Szydłowski in [11]. Therein the authors reduced the dimension of phase space and showed that the reduced system, which is polynomial, does not admit any analytic or even formal first integral.
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In this work we study the analytic integrability of all Bianchi models of class A in the variables (x1, x2, x3, x4, x5, x6) for 0 6 k < 1. The following result is well known, see for instance [9]. P
Proposition 1. Let F be an analytic function and let F ¼ i F i be its decomposition into homogeneous polynomials of degree i. Then F is an analytic first integral of the homogeneous differential system (1) if and only if Fi is a homogeneous polynomial first integral of system (1) for all i. A differential system of n variables is completely integrable if it admits n 1 independent first integrals. According to Proposition 1 the study of the analytic first integrals of the homogeneous system (1) is reduced to the study of its polynomial homogeneous first integrals. The main result of this paper is the characterization of the polynomial first integrals of the Bianchi models of class A. Section 2 provides three technical lemmas that we will use in Section 3 to prove the following theorem. Theorem 2. For 0 6 k < 1 the following statements hold.
2. Some auxiliary lemmas In order to prove Theorem 2 we shall use the following two lemmas. Lemma 3. Let g = g(x4, x5, x6) be a homogeneous polynomial solution of the homogeneous partial differential equation
k1 @g @g @g ¼ 0; F 123 þ þ 4 @x4 @x5 @x6 ð4Þ
x25
2
0 1 0 CB C B C 1 A@ a2 A ¼ @ 0 A:
1 1
B @ 1 2 1 1
10
2
a1
1
x26
where F 123 ¼ þ þ 2ðx4 x5 þ x4 x6 þ x5 x6 Þ and a1 ; a2 ; a3 2 R are such that (a1 a2)2 + (a1 a3)2 – 0. Then g 0.
0
a3
The solution of this system is a1 = a2 = a3. This cannot happen by assumption. Therefore g is not a polynomial unless g 0. h Lemma 4. Let g = g(x4, x5, x6) and h = h(x4 x5, x4 x6) be homogeneous polynomials of respective degrees n 2 and n such that
2ðx4 x5 þ x6 Þg þ
k1 @g @g @g @h þ F 123 þ þ ¼ 0; 4 @x4 @x5 @x6 @x5 ð5Þ x25
where F 123 ¼ þ þ h = h(x4 x6) and g 0.
Statement (a) of Theorem 2 apparently is well known for people working in the area, but we cannot find a reference where it is proved.
x24
0
x24
(a) The Bianchi type I model is completely integrable. (b) The Bianchi type II model has the polynomial first integral K = x5 x6. This model does not admit any additional polynomial first integral independent from H and K. (c) The Bianchi type VI0 and VII0 models have no polynomial first integrals. (d) The Bianchi type VIII and IX models have no polynomial first integrals.
ða1 x4 þ a2 x5 þ a3 x6 Þg þ
arbitrary function. We note that g is a polynomial if and only if D1 2 N; D2 ¼ 0 and f is a polynomial. In particular, the relation D2 = 0 is equivalent to the linear system
x26
2ðx4 x5 þ x4 x6 þ x5 x6 Þ.
Then
P and Proof. Let h ¼ ni¼0 ai ðx4 x5 Þi ðx4 x6 Þni Pn2 g ¼ iþj¼0 bij xi4 xj5 xn2ij . Suppose that g X 0. Forcing that 6 the solution of (5) be a polynomial we obtain
gðx4 ; x5 ; x6 Þ ¼
4 1k
Z
h5 dx4 þ f ðx4 x5 ; x4 x6 Þ; F 123
@h where f is a homogeneous polynomial, h5 ¼ @x and the 5 integral is to be a polynomial. We have checked the expression of the function g with the help of an algebraic manipulator, in this case mathematica, see [17]. pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi Let A1 ¼ x5 x6 and A2 ¼ x5 þ x6 . Under this 2 2 notation F 123 ¼ ðx4 A1 Þðx4 A2 Þ. The fraction inside the above integral can be written as
! h5 1 X1 X2 ; ¼ X0 þ 2 F 123 A1 A22 x4 A21 x4 A22 where X0 = X0(x4, A1, A2), X1 = X1(A1, A2) and X2 = X2(A1, A2) are homogeneous polynomials. The integrals of the fractions in the right hand side with respect to x4 are X i log x4 A2i ; i ¼ 1; 2; hence X1 and X2 must be identically zero. X1 = 0 and X2 = 0 have the same solutions a1, . . . , an because of symmetry. Indeed X1 = 0 (or X2 = 0) reduces to Sn = 0, where
Sn ¼
n X ð3A1 A2 Þni ðA1 þ A2 Þni ð3A1 þ A2 Þi1 i¼1
ðA1 A2 Þi1 i ai : Proof. The general solution of Eq. (4) is
pffiffiffiffiD1 þD2 gðx4 ; x5 ; x6 Þ ¼ f ðx4 x5 ; x4 x6 Þx4 x5 x6 þ 2 D pffiffiffiffiD1 D2 ; x4 þ x5 þ x6 þ 2 D where D1 = 2(a1 + a2 + a3)/(3(k 1)), D2 ¼ ðð2a1 a2 a3 Þ x4ffiffiffiffi þða1 þ 2a2 a3 Þx5 þ ða1 a2 þ 2a3 Þx6 Þ=ð3ðk 1Þ p DÞ; D ¼ jx24 þ x25 þ x26 x4 x5 x4 x6 x5 x6 j and f is an
We note that we have the recursive equality
Sn ¼ ð3A1 A2 ÞðA1 þ A2 ÞSn1 þ ð3A1 þ A2 Þn1 ðA1 A2 Þn1 n an : On A1 = A2 (or equivalently on x5 = 0) we have n4n1 A22n2 an ¼ 0, and hence we have an = 0. Induction arguments prove that Sn = 0 implies a1 = = an = 0. Therefore h5 = 0, which means that Eq. (5) is a particular case of
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
Eq. (4) and then by Lemma 3 we get g 0 and then we are finished. h
In this section we prove the four statements of Theorem 2. 3.1. Proof of statement (a) of Theorem 2 According to Table 1 the Bianchi cosmological model I is obtained for n1 = n2 = n3 = 0. System (1) becomes
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; x_ 4 x_ 5 x_ 6
3ðk1Þ 2
pffiffiffiffi!x4 2xpffiffi5 þx6 x4 þ x5 þ x6 2 D D pffiffiffiffi F 1 x4 þ x 5 þ x 6 þ 2 D
and
ð6Þ
pffiffiffiffix4pxffiffi 5 1k x1 2 x4 þ x5 þ x6 2 D D pffiffiffiffi x2 x4 þ x 5 þ x 6 þ 2 D
jx3 j
3ðk1Þ 2
pffiffiffiffi!x4 þxp5ffiffi2x6 x4 þ x5 þ x6 2 D D pffiffiffiffi ; F 1 x4 þ x 5 þ x 6 þ 2 D
where D ¼ x24 þ x25 þ x26 x4 x5 x4 x6 x5 x6 . As these four first integrals of system (8) are independent and h1 is a polynomial first integral of (8), we have h1 = h1(x4 x5, x5 x6). The following lemma ends the proof of statement (b) of Theorem 2. Lemma 5. For system (7) we have that h1 = h1(x5 x6) and g1 0.
and
Proof. Suppose that g1 X 0. As h is a first integral of (7), we have
pffiffiffiffix5pxffiffi 6 1k x2 2 x4 þ x5 þ x6 2 D D pffiffiffiffi ; x x4 þ x 5 þ x 6 þ 2 D 3 with D ¼ x24 þ x25 þ x26 x4 x5 x5 x6 x4 x6 . Note that the five first integrals are independent. Statement (a) is proved. 3.2. Prove of statement (b) of Theorem 2 The Bianchi cosmological model II is obtained for n1 = 1 and n2 = n3 = 0. System (1) writes as
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F; 4 k1 F; x_ 5 ¼ 4 k1 x_ 6 ¼ F; 4
ð8Þ
where F 1 ¼ Fjx1 ¼0 . System (8) admits the two polynomial first integrals x4 x5 and x5 x6 and the two non-polynomial first integrals
jx2 j
where F ¼ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 . Straightforward computations show that system (6) has the five first integrals x4 x5, x4 x6, H defined in (3),
x_ 4 ¼ x21 þ
k1 F1; 4 k1 x_ 5 ¼ F1; 4 k1 x_ 6 ¼ F1; 4 x_ 4 ¼
3. Proof of Theorem 2
k1 F; ¼ 4 k1 F; ¼ 4 k1 F; ¼ 4
x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ;
@g xj1 jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ 1 @x1 @g 1 @g @g þ x3 ðx4 þ x5 x6 Þ 1 þ x21 1 þx2 ðx4 x5 þ x6 Þ @x2 @x3 @x4 k 1 @g 1 @g 1 @g 1 @h 1 þ x21 F þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x4
ð9Þ
We distinguish three cases depending on the value of j. If j = 1 then Eq. (9) becomes
ðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 @x1
@g @g @g þ x2 ðx4 x5 þ x6 Þ 1 þ x3 ðx4 þ x5 x6 Þ 1 þ x21 1 @x2 @x3 @x4 k 1 @g 1 @g 1 @g 1 @h1 þ x1 þ þ þ ¼ 0: F 4 @x4 @x5 @x6 @x4 ð7Þ
where F ¼ x21 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 . Let h = h(x1, x2, x3, x4, x5, x6) be a homogeneous polynomial first integral of (7). We can write h ¼ h1 ðx2 ; x3 ; x4 ; x5 ; x6 Þ þ xj1 g 1 ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ, with j 2 N and h1 and g1 homogeneous polynomials such that x1 - g1. On x1 = 0 system (7) becomes
Let g1 ¼ g 1 jx1 ¼0 X 0. Eq. (9) on x1 = 0 can be written as
@ g1 ðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ @x 2 @ g1 k 1 @ g1 @ g1 @ g1 ¼ 0: F1 þ x3 ðx4 þ x5 x6 Þ þ þ þ 4 @x3 @x4 @x5 @x6 Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
@g ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ 2 @x 2 @g 2 k 1 @g 2 @g 2 @g 2 ¼ 0: þ x3 ðx4 þ x5 x6 Þ þ þ þ F1 4 @x3 @x4 @x5 @x6
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg2 þ x3 ðx4 þ x5 x6 Þ @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0; þ þ þ F 12 4 @x3 @x4 @x5 @x6 where F 12 ¼ F 1 jx2 ¼0 . Now we write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We obtain
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g k1 @g 3 @g 3 @g 3 ¼ 0: þ x3 ðx4 þ x5 x6 Þ 3 þ þ þ F 12 4 @x3 @x4 @x5 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0; F 123 þ þ þ 4 @x4 @x5 @x6 where F 123 ¼ F 12 jx3 ¼0 . Applying Lemma 3 we get g3 0, which is a contradiction. Hence we have g1 0 and there1 fore @h 0. The lemma follows in this case. @x4 1 @h1 If j > 2 then x1 @h @x4 , and then @x4 0. Now we can proceed in a similar way as in the case j = 1 to prove that g1 0 by using Lemma 3. If j = 2 then Eq. (9) becomes
2ðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 @x1
@g @g @g þ x2 ðx4 x5 þ x6 Þ 1 þ x3 ðx4 þ x5 x6 Þ 1 þ x21 1 @x2 @x3 @x4 k 1 @g 1 @g 1 @g 1 @h1 þ F þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x4 1 ¼ g 1 jx ¼0 X 0. Eq. (9) on x1 = 0 can be written as Let g 1 @ g1 2ðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ @x 2 @ g1 k 1 @ g1 @ g1 @ g1 @h1 þ þ x3 ðx4 þ x5 x6 Þ þ þ þ ¼ 0: F1 @x3 @x4 @x5 @x6 @x4 4
Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
@g ð2ðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ 2 @x2 @g 2 k 1 @g 2 @g 2 @g 2 @h1 þ x3 ðx4 þ x5 x6 Þ þ þ þ ¼ 0: þ F1 @x3 @x4 @x5 @x6 @x4 4 1 If l > 0 then @h 0. Similar arguments to those used above @x4 lead to the desired result after applying Lemma 3. If l = 0, let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
@ g2 2ðx4 þ x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ @x3 k1 @ g2 @ g2 @ g2 @h1 þ F 12 þ þ þ ¼ 0; 4 @x4 @x5 @x6 @x4 where F 12 ¼ F 1 jx2 ¼0 . Now we write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We obtain ð2ðx4 þ x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g k1 @g 3 @g 3 @g 3 @h1 þ þ x3 ðx4 þ x5 x6 Þ 3 þ þ þ ¼ 0: F 12 @x3 @x4 @x5 @x6 @x4 4 1 If m > 0 then @h 0. Again the usual arguments lead to the @x4 desired result after applying Lemma 3. If m = 0, let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
2ðx4 þ x5 þ x6 Þg3 þ
k1 @ g3 @ g3 @ g3 @h1 F 123 þ þ ¼ 0; þ 4 @x4 @x5 @x6 @x4
where F 123 ¼ F 12 jx3 ¼0 . Applying Lemma 4 swapping x4 and x5 we get g3 0, which is a contradiction. Hence we have 1 g1 0 and therefore @h 0. The lemma follows also in this @x4 case. h After Lemma 5, h = h(x5 x6). Hence statement (b) of Theorem 2 follows. 3.3. Proof of statement (c) of Theorem 2 According to Table 1, system (1) in cases VI0 and VII0 can be written as
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F; 4 k1 F; x_ 5 ¼ n2 x2 ðx1 þ n2 x2 Þ þ 4 k1 F; x_ 6 ¼ 4 x_ 4 ¼ x1 ðx1 n2 x2 Þ þ
ð10Þ
where F ¼ ðx1 n2 x2 Þ2 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 and n22 ¼ 1. Suppose that system (10) has a homogeneous polynomial first integral h(x1, . . . , x6). We can write h ¼ h1 ðx2 ; . . . ; x6 Þ þ xj1 g 1 ðx1 ; . . . ; x6 Þ, with j 2 N and h1 and g1 homogeneous polynomials such that x1 - g1. System (10) on x1 = 0 is
x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F1; 4 k1 x_ 5 ¼ x22 þ F1; 4 k1 F1; x_ 6 ¼ 4 x_ 4 ¼
ð11Þ
where F 1 ¼ Fjx1 ¼0 . We note that h1 is a first integral of system (11). We can write h1 ¼ h2 ðx3 ; . . . ; x6 Þ þ xl2 g 2 ðx2 ; . . . ; x6 Þ, with l 2 N and h2 and g2 homogeneous polynomials such that x2 - g2. System (11) on x2 = 0 writes
x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F 12 ; 4 k1 F 12 ; x_ 5 ¼ 4 k1 F 12 ; x_ 6 ¼ 4 x_ 4 ¼
ð12Þ
where F 12 ¼ F 1 jx2 ¼0 . We note that h2 is a first integral of system (12). Straightforward computations show that system (12) has the three independent first integrals x4 x5, x5 x6 and
jx3 j
3ðk1Þ 2
pffiffiffiffi!x4 þxp5ffiffi2x6 x4 þ x5 þ x6 2 D D pffiffiffiffi ; F 12 x4 þ x5 þ x6 þ 2 D
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
where D ¼ x24 þ x25 þ x26 x4 x5 x4 x6 x5 x6 . h2 = h2(x4 x5, x4 x6).
Therefore
Lemma 6. For system (11) we have that h2 = h2(x4 x6) and g2 0. Proof. Suppose that g2 X 0. As h1 ¼ h2 þ xl2 g 2 is a first integral of system (11), we can write
@g xl2 lðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 @x2 @g 2 k1 @g 2 @g 2 @g 2 2 @g 2 F1 þ x2 þ þ þ þ x3 ðx4 þ x5 x6 Þ 4 @x3 @x5 @x4 @x5 @x6 2 @h2 ¼ 0: ð13Þ þ x2 @x5 We distinguish three cases depending on the value of l. If l = 1 then Eq. (13) becomes
@g ðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g 2 k 1 @g 2 @g 2 @g 2 @h2 @g þ x2 F1 þ þ þ þ x22 2 ¼ 0: 4 @x3 @x4 @x5 @x6 @x5 @x5 Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
ðx4 x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0: F 12 þ þ þ 4 @x3 @x4 @x5 @x6 Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. Then
ððx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g k1 @g 3 @g 3 @g 3 ¼ 0: F 12 þ þ 3þ 4 @x3 @x4 @x5 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we get
ððx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0; F 123 þ þ þ 4 @x4 @x5 @x6 3 0, where F 123 ¼ F 12 jx3 ¼0 . Applying Lemma 3 we obtain g which is a contradiction. Hence g2 0. Back to Eq. (13) we 2 have @h 0. Then the lemma follows. @x5 2 If l > 2, then from Eq. (13) we have that x2 @h @x5 and thus @h2 @x5 0. Therefore h2 = h2(x4 x6). Now we can proceed in a similar way as in the case l = 1 to obtain the equation
ðlðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0: þ þ þ F 123 4 @x4 @x5 @x6 Applying again Lemma 3 we arrive to contradiction and hence g2 0. If l = 2, then Eq. (13) writes as
@g 2ðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g 2 k 1 @g 2 @g 2 @g 2 @g @h2 þ x22 2 þ F1 þ þ þ ¼ 0: 4 @x3 @x4 @x5 @x6 @x5 @x5 Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
@ g2 2ðx4 x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ @x 3 k1 @ g2 @ g2 @ g2 @h2 þ þ þ þ ¼ 0: F 12 4 @x4 @x5 @x6 @x5 Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. Then
xm 3 ½ð2ðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g k1 @g 3 @g 3 @g 3 @h2 3þ þ F 12 þ þ ¼ 0: 4 @x3 @x4 @x5 @x6 @x5 We distinguish two cases depending on the value of m. 2 2 . Hence @h If m > 0 then x3 @h 0 and h2 = h2(x4 x6). Now @x5 @x5 let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we obtain
ð2ðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0: þ þ þ F 123 4 @x4 @x5 @x6 Applying Lemma 3 we get a contradiction and hence g2 0. If m = 0, let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we obtain k1 @ g3 @ g3 @ g3 @h2 2ðx4 x5 þ x6 Þg3 þ þ þ ¼ 0: þ F 123 @x4 @x5 @x6 @x5 4 2 Applying Lemma 4 we get @h 0 and g3 0, hence the @x5 lemma follows. All the subcases are finished and therefore the lemma is proved. h After Lemma 6 we have that h ¼ h2 ðx4 x6 Þþ xj1 g 1 ðx1 ; . . . ; x6 Þ, with j 2 N and x1 - g1. We recall that h is a first integral of system (10). Thus
xj1 ½jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ @g @g 1 þ x2 ðx4 x5 þ x6 Þ 1 þ x3 ðx4 þ x5 x6 Þ @x1 @x2 @g 1 @g 1 þ x1 ðx1 n2 x2 Þ n2 x2 ðx1 n2 x2 Þ @x3 @x4 @g k 1 @g 1 @g 1 @g 1 F þ þ 1þ 4 @x5 @x4 @x5 @x6 @h2 ¼ 0: þ x1 ðx1 n2 x2 Þ @x4
ð14Þ
Lemma 7. For system (10) we have that h2 0 and g1 0. Proof. Suppose that g1 X 0. We distinguish two cases depending on the value of j. If j > 1 then from Eq. (14) we 2 have that x1 @h , and hence h2 0. Therefore Eq. (14) can @x4 be written as
jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 þ x2 ðx4 x5 þ x6 Þ @x1
@g 1 @g þ x3 ðx4 þ x5 x6 Þ 1 þ x1 ðx1 n2 x2 Þ @x2 @x3 @g 1 @g k 1 @g 1 @g 1 @g 1 ¼ 0: F n2 x2 ðx1 n2 x2 Þ 1 þ þ þ 4 @x4 @x5 @x4 @x5 @x6
Let g1 ¼ g 1 jx1 ¼0 X 0. Eq. (14) on x1 = 0 becomes
@ g1 jðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ þ x3 ðx4 þ x5 x6 Þ @x2 @ g1 @ g1 k 1 @ g1 @ g1 @ g1 ¼ 0: F1 þ x22 þ þ þ 4 @x3 @x5 @x4 @x5 @x6
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Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g 2 @g @g þ x3 ðx4 þ x5 x6 Þ 2 þ x22 2 @x2 @x3 @x5 k1 @g 2 @g 2 @g 2 ¼ 0: F1 þ þ þ 4 @x4 @x5 @x6
Let g2 ¼ g 2 jx2 ¼0 X 0. Then, on x2 = 0 we have
2 ¼ It only remains to consider the case l = 1. Let g g 2 jx2 ¼0 X 0. From Eq. (15) on x2 = 0 we have
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg2 þ x3 ðx4 þ x5 x6 Þ @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0: F 12 þ þ þ 4 @x3 @x4 @x5 @x6
2x6 g2 þ x3 ðx4 þ x5 x6 Þ
Now write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We get
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g k1 @g 3 @g 3 @g 3 ¼ 0: F 12 þ x3 ðx4 þ x5 x6 Þ 3 þ þ þ 4 @x3 @x4 @x5 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. Then, on x3 = 0 we have
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0: F 123 þ þ þ 4 @x4 @x5 @x6
@g 1 þ x2 ðx4 x5 þ x6 Þ @x1
@g 1 @g þ x3 ðx4 þ x5 x6 Þ 1 þ x1 ðx1 n2 x2 Þ @x2 @x3 @g @g k 1 @g 1 @g 1 @g 1 F þ þ 1 n2 x2 ðx1 n2 x2 Þ 1 þ 4 @x4 @x5 @x4 @x5 @x6
þ ðx1 n2 x2 Þ
n2
@ g2 k 1 @ g2 @ g2 @ g2 F 12 þ þ þ 4 @x3 @x4 @x5 @x6
@h2 ¼ 0: @x4
Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We get:
@g 3 xm 3 ð2x6 þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @x3 k1 @g 3 @g 3 @g 3 @h2 n2 F 12 þ þ ¼ 0: þ 4 @x4 @x5 @x6 @x4
2 , and hence h2 0. Therefore we obtain If m > 0 then x3 @h @x4 an equation of type (4) and hence by Lemma 3 we get g1 0. If m = 0, let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0, we have
3 0, a contradiction. Applying Lemma 3 we obtain g Hence g1 0 and the lemma follows in this case. If j = 1 then Eq. (14) becomes
ðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
Similar arguments to those used before lead to an equation of type (4) and hence applying Lemma 3 we get a contradiction. Therefore g1 0 and the lemma follows. If l = 0 then we can use the same arguments to arrive from Eq. (15) to an equation of type (4), and hence applying Lemma 3 we get a contradiction. Therefore g1 0 and h2 0, so the lemma follows.
@h2 ¼ 0: @x4
Let g1 ¼ g 1 jx1 ¼0 X 0. On x1 = 0 we have
2x6 g3 þ
k1 @ g3 @ g3 @ g3 @h2 n2 F 123 þ þ ¼ 0: 4 @x4 @x5 @x6 @x4
As h2 = h2(x4 x6) is a homogeneous polynomial of degree 2 n, we have h2 = a0(x4 x6)n. Thus @h ¼ a0 nðx4 x6 Þn1 . On @x4 x6 = 0 Eq. (16) writes
k1 @ g3 @ g3 @ g3 ðx4 x5 Þ2 þ þ n2 a0 nxn1 ¼ 0: 4 4 @x4 @x5 @x6 x6 ¼0 Therefore we must take a0 = 0 and hence h2 0. Now Eq. (16) is of type (4) and hence by Lemma 3 we get g1 0 and the lemma follows. h After Lemma 7 statement (c) of Theorem 2 is proved, as it follows that h 0.
@ g1 ðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ þ x3 ðx4 þ x5 x6 Þ @x2 @ g1 @ g1 k 1 @ g1 @ g1 @ g1 @h2 n2 x2 F1 þ x22 þ þ þ ¼ 0: 4 @x3 @x5 @x4 @x5 @x6 @x4
3.4. Proof of statement (d) of Theorem 2
Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ;
xl2 ½ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g 2 @g @g þ x3 ðx4 þ x5 x6 Þ 2 þ x22 2 @x2 @x3 @x5 k1 @g 2 @g 2 @g 2 @h2 n2 x2 F1 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x4
We distinguish three cases depending on the value of l. 2 If l > 1 then x2 @h . Hence h2 0. Thus, from (15), @x4
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g @g @g 2 þ x3 ðx4 þ x5 x6 Þ 2 þ x22 2 @x2 @x3 @x5 k1 @g 2 @g 2 @g 2 ¼ 0: F1 þ þ þ 4 @x4 @x5 @x6
According to Table 1, Bianchi cases VIII and IX correspond to n1 ¼ n2 ¼ n23 ¼ 1 and can be written into the form
k1 F; 4 k1 F; x_ 5 ¼ x2 ðx1 þ x2 n3 x3 Þ þ 4 k1 F; x_ 6 ¼ n3 x3 ðx1 x2 þ n3 x3 Þ þ 4 x_ 4 ¼ x1 ðx1 x2 n3 x3 Þ þ
ð15Þ
ð16Þ
ð17Þ
where F ¼ x21 þ x22 þ x23 2x1 x2 2n3 x1 x3 2n3 x2 x3 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 and n23 ¼ 1. Let h = h(x1, . . . , x6) be a homogeneous polynomial first integral of degree n of system (17). Write h ¼ h1 ðx2 ; . . . ; x6 Þ þ xj1 g 1 ðx1 ; . . . ; x6 Þ, with j 2 N, h1 and g1 homogeneous polynomials and x1 - g1. System (17) on x1 = 0 becomes
A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; x_ 4 ¼
k1 F1; 4
k1 x_ 5 ¼ x2 ðx2 n3 x3 Þ þ F1; 4 k1 x_ 6 ¼ n3 x3 ðx2 n3 x3 Þ þ F1; 4
ð18Þ
where F 1 ¼ Fjx1 ¼0 . System (18) admits h1 = h1(x2, . . . , x6) as first integral. Write h1 ¼ h2 ðx3 ; . . . ; x6 Þ þ xl2 g 2 ðx2 ; . . . ; x6 Þ, with l 2 N, h2 and g2 homogeneous polynomials and x2 - g2. System (18) on x2 = 0 becomes
x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F 12 ; 4 k1 F 12 ; x_ 5 ¼ 4 k1 F 12 ; x_ 6 ¼ x23 þ 4
ð19Þ
where F 12 ¼ F 1 jx2 ¼0 . We note that h2 = h2(x3, . . . , x6) is a first integral of system (19). Write h2 ¼ h3 ðx4 ; x5 ; x6 Þ þ xm 3 g3 ðx3 ; x4 ; x5 ; x6 Þ, with m 2 N, h3 and g3 homogeneous polynomials and x3 - g3. System (19) on x3 = 0 is
k1 F 123 ; 4 k1 x_ 5 ¼ F 123 ; 4 k1 F 123 ; x_ 6 ¼ 4
x_ 4 ¼
ð20Þ
where F 123 ¼ F 12 jx3 ¼0 . Note that h3 is a polynomial first integral of system (20). Since system (20) admits the two independent first integrals x4 x5 and x5 x6, any polynomial first integral of (20) must be a polynomial in the variables x4 x5 and x5 x6. Therefore h3 = h3(x4 x5, x5 x6). The next three lemmas end the proof of statement (d) of Theorem 2. The first one shows that h2 = h2(x4 x5). Lemma 8. For system (19) we have that g3 0 and h3 = h3(x4 x5). We recall that Proof. Suppose that g3 X 0. h2 ¼ h3 ðx4 x5 ; x5 x6 Þ þ xm 3 g 3 ðx3 ; x4 ; x5 ; x6 Þ, where m 2 N and x3 - g3. As h2 is a first integral of system (19), we have
@g 3 @g xm þ x23 3 3 mðx4 þ x5 x6 Þg 3 þ x3 ðx4 þ x5 x6 Þ @x3 @x6 k1 @g 3 @g 3 @g 3 @h 3 þ x23 F 12 þ þ ¼ 0: þ 4 @x4 @x5 @x6 @x6
ð21Þ
We distinguish three cases depending on the value of m. If m = 1 then
@g @g ðx4 þ x5 x6 Þg 3 þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 @x3 @x6 k1 @g 3 @g 3 @g 3 @h3 þ x3 F 12 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
ðx4 þ x5 x6 Þg3 þ
k1 @ g3 @ g3 @ g3 ¼ 0: F 123 þ þ 4 @x4 @x5 @x6
3 0 and hence g3 0. ConseApplying Lemma 3 we get g 3 quently @h ¼ 0 and the lemma followsin this case. @x6 3 If m > 2 then from (21) we have x3 @h @x6 and so h3 = h3(x4 x5). Now from Eq. (21) on x3 = 0 we get an equation of type (4), hence applying Lemma 3 we get g3 0 and the lemma follows in this case. If m = 2 then from (21) we have
@g @g 2ðx4 þ x5 x6 Þg 3 þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 @x3 @x6 k1 @g 3 @g 3 @g 3 @h3 þ F 12 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we obtain
2ðx4 þ x5 x6 Þg3 þ
x_ 4 ¼
19
k1 @ g3 @ g3 @ g3 @h3 þ F 123 þ þ ¼ 0: 4 @x4 @x5 @x6 @x6
3 Applying Lemma 4 swapping x5 and x6 we get @h ¼ 0 and @x6 g3 0. Hence h3 = h3(x4 x5), g3 0 and the lemma follows in this case. h The second lemma shows that h1 0.
Lemma 9. For system (18) we have that g2 0 and h2 0. Proof. Suppose that g2 X 0. We recall that h1 ¼ h2 ðx4 x5 Þ þxl2 g 2 , with l 2 N and x2 - g2. As h1 is a first integral of system (18), we have
@g xl2 lðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g 2 @g 2 @g þ x2 ðx2 n3 x3 Þ n3 x3 ðx2 n3 x3 Þ 2 @x3 @x5 @x6 k1 @g 2 @g 2 @g 2 F1 þ þ þ 4 @x4 @x5 @x6 We distinguish two cases depending on the value of l. If 2 2 and hence @h l > 1 then x2 @h 0, which means that @x5 @x5 h2 0. Substituting in the equation above we have
@g 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g @g @g 2 þ x2 ðx2 n3 x3 Þ 2 n3 x3 ðx2 n3 x3 Þ 2 @x3 @x5 @x6 k1 @g 2 @g 2 @g 2 ¼ 0: þ þ þ F1 4 @x4 @x5 @x6
lðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ
The usual arguments lead to Eq. (4), hence we obtain g2 0 by Lemma 3. If l = 1 then we have
@g 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g @g @g 2 þ x2 ðx2 n3 x3 Þ 2 n3 x3 ðx2 n3 x3 Þ 2 @x3 @x5 @x6 k1 @g 2 @g 2 @g 2 @h2 þ ðx2 n3 x3 Þ F1 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x5
ðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ
Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
@ g2 @ g2 ðx4 x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ þ x23 @x3 @x6 k1 @ g2 @ g2 @ g2 @h2 n 3 x3 þ þ þ ¼ 0: F 12 4 @x4 @x5 @x6 @x5
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. Then
xm 3 ½ððx4
x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g 3 @g k1 @g 3 @g 3 @g 3 þ x23 3 þ þ þ F 12 4 @x3 @x6 @x4 @x5 @x6
Now we distinguish three cases depending on the value of m. If m = 0 then we are in (22) again and the usual arguments lead to g2 0 and h2 0. 2 @h2 If m > 1 then x3 @h @x5 and hence @x5 0, which means that h2 0. Then we have
ððx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g @g k1 @g 3 @g 3 @g 3 3 þ x23 3 þ ¼ 0: F 12 þ þ 4 @x3 @x6 @x4 @x5 @x6 The usual arguments finish the proof in this case. Finally if m = 1 then we have
jðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ
@ g1 þ x3 ðx4 þ x5 x6 Þ @x2
@ g1 @ g1 þ x2 ðx2 n3 x3 Þ þ n3 x3 ðx2 þ n3 x3 Þ @x3 @x5 @ g1 k 1 @ g1 @ g1 @ g1 ¼ 0: þ þ þ F1 4 @x6 @x4 @x5 @x6
1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get Write g
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g @g 2 þ x3 ðx4 þ x5 x6 Þ 2 þ x2 ðx2 n3 x3 Þ @x2 @x3 @g 2 @g þ n3 x3 ðx2 þ n3 x3 Þ 2 @x5 @x6 k1 @g 2 @g 2 @g 2 ¼ 0: þ þ þ F1 4 @x4 @x5 @x6 2 ¼ g 2 jx ¼0 X 0. On x2 = 0 we have Let g 2
@g @g k1 2x4 g 3 þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 þ 4 @x3 @x6 @g 3 @g 3 @g 3 @h2 n3 þ þ ¼ 0: F 12 @x4 @x5 @x6 @x5 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
2x4 g3 þ
k1 @ g3 @ g3 @ g3 @h2 n3 F 123 þ þ ¼ 0: 4 @x4 @x5 @x6 @x5
As h2 = h2(x4 x5) is a homogeneous polynomial of degree n, we have h2 = a0(x4 x5)n. Hence
2x4 g3 þ
k1 @ g3 @ g3 @ g3 þ n3 a0 nðx4 x5 Þn1 ¼ 0: F 123 þ þ 4 @x4 @x5 @x6
On x4 = 0 we have
k1 @ g3 @ g3 @ g3 þ þ þ n3 a0 nðx5 Þn1 ¼ 0; ðx5 x6 Þ2 4 @x4 @x5 @x6 x4 ¼0 which means that a0 = 0. Therefore h2 0. The equation is now of type (4) and leads to g2 0 by Lemma 3. All the subcases are considered and the proof of the lemma is finished. h The last lemma shows that g1 0 and therefore that h 0. Lemma 10. For system (17) we have that g1 0. Proof. Suppose that g1 X 0. We recall that h ¼ xj1 g 1 , with j 2 N and x1 - g1, is a first integral of system (17). Then
jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 þ x2 ðx4 x5 þ x6 Þ @x1
@g @g 1 þ x3 ðx4 þ x5 x6 Þ 1 þ x1 ðx1 x2 n3 x3 Þ @x2 @x3 @g 1 @g þ x2 ðx1 þ x2 n3 x3 Þ 1 þ n3 x3 ðx1 x2 þ n3 x3 Þ @x4 @x5 @g 1 k 1 @g 1 @g 1 @g 1 ¼ 0: þ þ þ F 4 @x6 @x4 @x5 @x6 Let g1 ¼ g 1 jx1 ¼0 X 0. On x1 = 0 we have
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg2 þ x3 ðx4 þ x5 x6 Þ @ g2 @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0: þ x23 þ þ þ F 12 4 @x3 @x6 @x4 @x5 @x6 2 ¼ xm Write g 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We get
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g @g þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 @x3 @x6 k1 @g 3 @g 3 @g 3 ¼ 0: þ þ þ F 12 4 @x4 @x5 @x6 3 ¼ g 3 jx ¼0 X 0. On x3 = 0 we have Let g 3
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 F 123 þ þ þ 4 @x4 @x5 @x6 ¼ 0: We can apply Lemma 3. Hence the lemma follows. h After Lemma 10, we get h 0. Thus the proof of statement (d) of Theorem 2 is finished. 4. Conclusions According to Proposition 1 the study of analytic first integrals of systems (1) is reduced to the study of polynomial first integrals. Hence the non-existence of polynomial first integrals implies the non-existence of analytic first integrals. From Theorem 2 we have that Bianchi type I model is not chaotic whereas Bianchi type II is at most partially chaotic. The rest of the models of class A, the Bianchi types VI0, VII0, VIII and IX, can exhibit chaos. References [1] Bianchi L. Sugli spazi a tre dimensioni che ammettono un gruppo continuo di movimenti (On the spaces of three dimensions that admit a continuous group of movements). Mem Math Fis Soc Ital Sci 1898;11:267–352. [2] Bianchi L. Lezioni sulla teoria dei gruppi continui finiti di trasformazioni (Lectures on the theory of finite continuous transformation grups), Spoerri, Pisa, 1918, pp. 550–557. [3] Bogoyavlensky OI. Qualitative theory of dynamical systems in astrophysics and gas dynamics. Springer-Verlag; 1985.
A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21 [4] Collins CB. More qualitative cosmology. Commun Math Phys 1971;23:137–58. [5] Cushman R, Sniatycki J. Local integrability of the Mixmaster model. Rep Math Phys 1995;36:75–89. [6] Ferragut A, Llibre J, Pantazi C. Analytic integrability of Bianchi class A cosmological models with k = 1. J Geom Phys 2012;62:381–6. [7] Latifi A, Musette M, Conte R. The Bianchi IX cosmological model is not integrable. Phys Lect A 1994;194:83–92. [8] Llibre J, Valls C. Formal and analytical integrability of Bianchi IX system. J Math Phys 2006;47:022704–15. [9] Llibre J, Zhang X. Polynomial first integrals for quasi-homogeneous polynomial differential systems. Nonlinearity 2002;15:1269–80. [10] Maciejewski AJ, Strelcyn J, Szydłowski M. Nonintegrability of Bianchi VIII Hamiltonian systems. J Math Phys 2001;42:1728. 13pp.
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[11] Maciejewski AJ, Szydłowski M. On the integrability of Bianchi cosmological models. J Phys A Math Gen 1998;31:2031–43. [12] Maciejewski AJ, Szydłowski M. Bianchi cosmologies as dynamical systems. Celestial Mech Dyn Astron 1999;73:17–24. [13] Morales-Ruiz JJ, Ramis JP. Galoisian obstructions to integrability of Hamiltonian systems II. Methods Appl Anal 2001;8:97–112. [14] Szydłowski M, Biesiada M. Kovalevski exponents and integrability properties in class A homogeneous cosmological models. J Nonlinear Math Phys 2002;9:1–10. [15] Szydłowski M, Demaret J. Bianchi A cosmological models as the simplest dynamical system in R4 . Gen Rel Grav 1999;31:897–911. [16] Wainwright J, Ellis GFR. Dynamical systems in cosmology. Cambridge University Press; 1997. [17] Wolfram Research, Inc., Mathematica, V.8.0, Champaign, IL, 2010.
Contents lists available at SciVerse ScienceDirect
Chaos, Solitons & Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos
Analytic integrability of the Bianchi class A cosmological models with 0 6 k < 1 q Antoni Ferragut a, Jaume Llibre b, Chara Pantazi c,⇑ a
Departament de Matemàtica Aplicada I, Universitat Politècnica de Catalunya, ETSEIB, Av. Diagonal 647, 08028 Barcelona, Catalonia, Spain Departament de Matemàtiques, Universitat Autònoma de Barcelona, Edifici C, 08193 Bellaterra, Barcelona, Catalonia, Spain c Departament de Matemàtica Aplicada I, Universitat Politècnica de Catalunya, EPSEB, Av. Doctor Marañón 44–50, 08028 Barcelona, Catalonia, Spain b
a r t i c l e
i n f o
Article history: Received 20 July 2012 Accepted 31 December 2012 Available online 31 January 2013
a b s t r a c t Many works study the integrability of the Bianchi class A cosmologies with k = 1, where k is the ratio between the pressure and the energy density of the matter. Here we characterize the analytic integrability of the Bianchi class A cosmological models when 0 6 k < 1. We conclude that Bianchi types VI0, VII0, VIII and IX can exhibit chaos whereas Bianchi type I is not chaotic and Bianchi type II is at most partially chaotic. Ó 2013 Elsevier Ltd. All rights reserved.
1. Introduction In a cosmological model Einstein’s equations connect the geometry of the space–time with the properties of the matter. The matter occupying the space–time is determined by the stress energy tensor of the matter. Friedman in 1922 introduced the study of homogeneous cosmologies. In 1951, Taub determined the complete homogeneous solutions of the equations of the general theory of relativity. For spatially homogeneous cosmologies the Einstein field equations can be written as an autonomous system of first order differential equations, see [16] and references therein. Bianchi models describe space–times which are foliated by homogeneous (and so we have three dimensional Lie algebras) hypersurfaces of constant time. Bianchi [1,2]
q All the authors are partially supported by the MICINN/FEDER Grant MTM2008-03437. A.F. is additionally supported by grants Juan de la Cierva, 2009SGR410 and MTM2009-14163-C02-02. J.L. is additionally partially supported by an AGAUR Grant number 2009SGR410 and by ICREA Academia. C.P. is additionally partially supported by the MICINN/ FEDER Grant number MTM2009-06973 and by the AGAUR Grant number 2009SGR859. ⇑ Corresponding author. Tel.: +34934016283; fax: +34 93 401 17 13. E-mail addresses: [email protected] (A. Ferragut), jllibre@mat. uab.cat (J. Llibre), [email protected] (C. Pantazi).
0960-0779/$ - see front matter Ó 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.chaos.2012.12.007
was the first to classify three dimensional Lie algebras which are non-isomorphic. There are nine types of models according to the dimension n of the derivative subalgebra: (a) (b) (c) (d)
n = 0: n = 1: n = 2: n = 3:
type I; types II, III; types IV, V, VI, VII; types VIII, IX.
If we consider X1, X2, X3 an appropriate basis of the 3dimensional Lie Algebra, then the classification depends on a scalar a 2 R and a vector (n1, n2, n3), with ni 2 {+1, 1, 0}, such that
½X 1 ; X 2 ¼ n3 X 3 ;
½X 2 ; X 3 ¼ n1 X 1 aX 2 ;
½X 3 ; X 1 ¼ n2 X 2 þ aX 1 ; where [,] is the Lie bracket. In particular for a = 0 we obtain models of class A and for a – 0 we obtain models of class B. A good reference for the Bianchi models is Bogoyavlensky [3]. Collins in 1971 was the first to study Bianchi models as dynamical systems, see [4]. Later on Bogoyavlensky and collaborators proved that the dynamics of class A types can be reduced into a six dimensional autonomous system of differential equations of first order defined in a compact manifold with a boundary.
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
In our study we follow [3] and we consider the hydrodynamical tensor of the matter. We will work with an equation of state of matter of the form p = ke, where e is the energy density of the matter, p is the pressure and 0 6 k 6 1, see also [3]. The case k = 1 is studied in [6]. Here we consider the case where 0 6 k < 1. Following [3] the Einstein equations for the homogenous cosmologies of class A without motion of matter can be formalized as a Hamiltonian system in the phase space pi, qi for i = 1, 2, 3 with the Hamiltonian function
H¼
1 1k 2
ðq1 q2 q3 Þ
1 Tðpi qi Þ þ V G ðqi Þ : 4
Table 1 The classification of Bianchi class A cosmologies. Type
I
II
VI0
VII0
VIII
IX
a n1 n2 n3
0 0 0 0
0 1 0 0
0 1 1 0
0 1 1 0
0 1 1 1
0 1 1 1
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F; 4 k1 F; x_ 5 ¼ n2 x2 ðn1 x1 þ n2 x2 n3 x3 Þ þ 4 k1 F; x_ 6 ¼ n3 x3 ðn1 x1 n2 x2 þ n3 x3 Þ þ 4 x_ 4 ¼ n1 x1 ðn1 x1 n2 x2 n3 x3 Þ þ
Here T is the kinetic energy (not positively defined) and VG is the potential. According to [3] (Section 4 of Chapter II) the kinetic and the potential energy are given, respectively, by 3 3 X X Tðpi qi Þ ¼ 2 pi pj qi qj p2i q2i ; i
i¼1
3 X
3 X
i
i¼1
V G ðqi Þ ¼ 2
n i n j qi qj
n2i q2i
We consider the Hamiltonian system
@H ; @pi
p_ i ¼
where
F ¼ n21 x21 þ n22 x22 þ n23 x23 2n1 n2 x1 x2 2n1 n3 x1 x3
for i, j 2 {1, 2, 3}.
q_ i ¼
ð1Þ
@H ; @qi
where the dot denotes derivative with respect to the time t. More precisely, the Hamiltonian system is k1
q_ 1 ¼ 2q1 ðq1 q2 q3 Þ 2 ðp1 q1 þ p2 q2 þ p3 q3 Þ; k1
q_ 2 ¼ 2q2 ðq1 q2 q3 Þ 2 ðp1 q1 p2 q2 þ p3 q3 Þ; k1
q_ 3 ¼ 2q3 ðq1 q2 q3 Þ 2 ðp1 q1 þ p2 q2 p3 q3 Þ; k1
p_ 1 ¼ ðq1 q2 q3 Þ 2 ð2p1 ðp1 q1 þ p2 q2 þ p3 q3 Þ 1 1k þ n1 ðn1 q1 þ n2 q2 þ n3 q3 Þ þ H; 2 2q1
k1
p_ 2 ¼ ðq1 q2 q3 Þ 2 ð2p2 ðp1 q1 p2 q2 þ p3 q3 Þ 1 1k H; þ n2 ðn1 q1 n2 q2 þ n3 q3 Þ þ 2 2q2 k1
p_ 3 ¼ ðq1 q2 q3 Þ 2 ð2p3 ðp1 q1 þ p2 q2 p3 q3 Þ 1 1k H: þ n3 ðn1 q1 þ n2 q2 n3 q3 Þ þ 2 2q3 Note that the constants n1, n2, n3 determine the type of the model according to Table 1. After the change of coordi1k nates ds ¼ ðq1 q2 q3 Þ 2 dt; qi ¼ xi ; pi ¼ xiþ3 =ð2xi Þ; i ¼ 1; 2; 3, we obtain the quadratic homogeneous polynomial differential system
2n2 n3 x2 x3 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 : ð2Þ Note that system (1) is a homogeneous polynomial differential system of degree 2. The Hamiltonian H becomes after the changes of variables the first integral k1
H ¼ ðx1 x2 x3 Þ 2 F ¼ ðx1 x2 x3 Þ
k1 2
n21 x21 þ n22 x22 þ n23 x23
2n1 n2 x1 x2 2n1 n3 x1 x3 2n2 n3 x2 x3 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6
ð3Þ
of system (1). Many authors have studied some models of class A for the case k = 1 considering different types of integrability, see for example [4–14]. More concretely, Szydłowski and Demaret in [15] studied the vacuum Bianchi models of class A reducing the dimension of the dynamical system. Maciejewski and Szydłowski in [12] also reduced the dimension of the dynamical system of class A. Szydłowski and Biesiada in [14] studied the algebraic integrability of class A Bianchi models first in vacuum and afterwards considering dust and radiative matter. Bianchi IX model causes a lot of interest because of its chaotic (in some sense) behavior. Many authors also studied the (different notions of) integrability of this model. More concretely Cushman and Sniatycki [5] proved that Bianchi IX is a locally integrable Hamiltonian system. Latifi et al. [7] proved that this model is not integrable since it does not satisfy the Painlevé property. Llibre and Valls in [8] proved that Bianchi IX does not admit a complete set of formal first integrals. Morales-Ruiz and Ramis [13], using tecnics of differential Galois theory, showed that the Bianchi IX model is not completely integrable with rational first integrals. The integrability of Bianchi types VIII and IX was also studied by Maciejewski and Szydłowski in [11]. Therein the authors reduced the dimension of phase space and showed that the reduced system, which is polynomial, does not admit any analytic or even formal first integral.
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In this work we study the analytic integrability of all Bianchi models of class A in the variables (x1, x2, x3, x4, x5, x6) for 0 6 k < 1. The following result is well known, see for instance [9]. P
Proposition 1. Let F be an analytic function and let F ¼ i F i be its decomposition into homogeneous polynomials of degree i. Then F is an analytic first integral of the homogeneous differential system (1) if and only if Fi is a homogeneous polynomial first integral of system (1) for all i. A differential system of n variables is completely integrable if it admits n 1 independent first integrals. According to Proposition 1 the study of the analytic first integrals of the homogeneous system (1) is reduced to the study of its polynomial homogeneous first integrals. The main result of this paper is the characterization of the polynomial first integrals of the Bianchi models of class A. Section 2 provides three technical lemmas that we will use in Section 3 to prove the following theorem. Theorem 2. For 0 6 k < 1 the following statements hold.
2. Some auxiliary lemmas In order to prove Theorem 2 we shall use the following two lemmas. Lemma 3. Let g = g(x4, x5, x6) be a homogeneous polynomial solution of the homogeneous partial differential equation
k1 @g @g @g ¼ 0; F 123 þ þ 4 @x4 @x5 @x6 ð4Þ
x25
2
0 1 0 CB C B C 1 A@ a2 A ¼ @ 0 A:
1 1
B @ 1 2 1 1
10
2
a1
1
x26
where F 123 ¼ þ þ 2ðx4 x5 þ x4 x6 þ x5 x6 Þ and a1 ; a2 ; a3 2 R are such that (a1 a2)2 + (a1 a3)2 – 0. Then g 0.
0
a3
The solution of this system is a1 = a2 = a3. This cannot happen by assumption. Therefore g is not a polynomial unless g 0. h Lemma 4. Let g = g(x4, x5, x6) and h = h(x4 x5, x4 x6) be homogeneous polynomials of respective degrees n 2 and n such that
2ðx4 x5 þ x6 Þg þ
k1 @g @g @g @h þ F 123 þ þ ¼ 0; 4 @x4 @x5 @x6 @x5 ð5Þ x25
where F 123 ¼ þ þ h = h(x4 x6) and g 0.
Statement (a) of Theorem 2 apparently is well known for people working in the area, but we cannot find a reference where it is proved.
x24
0
x24
(a) The Bianchi type I model is completely integrable. (b) The Bianchi type II model has the polynomial first integral K = x5 x6. This model does not admit any additional polynomial first integral independent from H and K. (c) The Bianchi type VI0 and VII0 models have no polynomial first integrals. (d) The Bianchi type VIII and IX models have no polynomial first integrals.
ða1 x4 þ a2 x5 þ a3 x6 Þg þ
arbitrary function. We note that g is a polynomial if and only if D1 2 N; D2 ¼ 0 and f is a polynomial. In particular, the relation D2 = 0 is equivalent to the linear system
x26
2ðx4 x5 þ x4 x6 þ x5 x6 Þ.
Then
P and Proof. Let h ¼ ni¼0 ai ðx4 x5 Þi ðx4 x6 Þni Pn2 g ¼ iþj¼0 bij xi4 xj5 xn2ij . Suppose that g X 0. Forcing that 6 the solution of (5) be a polynomial we obtain
gðx4 ; x5 ; x6 Þ ¼
4 1k
Z
h5 dx4 þ f ðx4 x5 ; x4 x6 Þ; F 123
@h where f is a homogeneous polynomial, h5 ¼ @x and the 5 integral is to be a polynomial. We have checked the expression of the function g with the help of an algebraic manipulator, in this case mathematica, see [17]. pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi Let A1 ¼ x5 x6 and A2 ¼ x5 þ x6 . Under this 2 2 notation F 123 ¼ ðx4 A1 Þðx4 A2 Þ. The fraction inside the above integral can be written as
! h5 1 X1 X2 ; ¼ X0 þ 2 F 123 A1 A22 x4 A21 x4 A22 where X0 = X0(x4, A1, A2), X1 = X1(A1, A2) and X2 = X2(A1, A2) are homogeneous polynomials. The integrals of the fractions in the right hand side with respect to x4 are X i log x4 A2i ; i ¼ 1; 2; hence X1 and X2 must be identically zero. X1 = 0 and X2 = 0 have the same solutions a1, . . . , an because of symmetry. Indeed X1 = 0 (or X2 = 0) reduces to Sn = 0, where
Sn ¼
n X ð3A1 A2 Þni ðA1 þ A2 Þni ð3A1 þ A2 Þi1 i¼1
ðA1 A2 Þi1 i ai : Proof. The general solution of Eq. (4) is
pffiffiffiffiD1 þD2 gðx4 ; x5 ; x6 Þ ¼ f ðx4 x5 ; x4 x6 Þx4 x5 x6 þ 2 D pffiffiffiffiD1 D2 ; x4 þ x5 þ x6 þ 2 D where D1 = 2(a1 + a2 + a3)/(3(k 1)), D2 ¼ ðð2a1 a2 a3 Þ x4ffiffiffiffi þða1 þ 2a2 a3 Þx5 þ ða1 a2 þ 2a3 Þx6 Þ=ð3ðk 1Þ p DÞ; D ¼ jx24 þ x25 þ x26 x4 x5 x4 x6 x5 x6 j and f is an
We note that we have the recursive equality
Sn ¼ ð3A1 A2 ÞðA1 þ A2 ÞSn1 þ ð3A1 þ A2 Þn1 ðA1 A2 Þn1 n an : On A1 = A2 (or equivalently on x5 = 0) we have n4n1 A22n2 an ¼ 0, and hence we have an = 0. Induction arguments prove that Sn = 0 implies a1 = = an = 0. Therefore h5 = 0, which means that Eq. (5) is a particular case of
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
Eq. (4) and then by Lemma 3 we get g 0 and then we are finished. h
In this section we prove the four statements of Theorem 2. 3.1. Proof of statement (a) of Theorem 2 According to Table 1 the Bianchi cosmological model I is obtained for n1 = n2 = n3 = 0. System (1) becomes
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; x_ 4 x_ 5 x_ 6
3ðk1Þ 2
pffiffiffiffi!x4 2xpffiffi5 þx6 x4 þ x5 þ x6 2 D D pffiffiffiffi F 1 x4 þ x 5 þ x 6 þ 2 D
and
ð6Þ
pffiffiffiffix4pxffiffi 5 1k x1 2 x4 þ x5 þ x6 2 D D pffiffiffiffi x2 x4 þ x 5 þ x 6 þ 2 D
jx3 j
3ðk1Þ 2
pffiffiffiffi!x4 þxp5ffiffi2x6 x4 þ x5 þ x6 2 D D pffiffiffiffi ; F 1 x4 þ x 5 þ x 6 þ 2 D
where D ¼ x24 þ x25 þ x26 x4 x5 x4 x6 x5 x6 . As these four first integrals of system (8) are independent and h1 is a polynomial first integral of (8), we have h1 = h1(x4 x5, x5 x6). The following lemma ends the proof of statement (b) of Theorem 2. Lemma 5. For system (7) we have that h1 = h1(x5 x6) and g1 0.
and
Proof. Suppose that g1 X 0. As h is a first integral of (7), we have
pffiffiffiffix5pxffiffi 6 1k x2 2 x4 þ x5 þ x6 2 D D pffiffiffiffi ; x x4 þ x 5 þ x 6 þ 2 D 3 with D ¼ x24 þ x25 þ x26 x4 x5 x5 x6 x4 x6 . Note that the five first integrals are independent. Statement (a) is proved. 3.2. Prove of statement (b) of Theorem 2 The Bianchi cosmological model II is obtained for n1 = 1 and n2 = n3 = 0. System (1) writes as
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F; 4 k1 F; x_ 5 ¼ 4 k1 x_ 6 ¼ F; 4
ð8Þ
where F 1 ¼ Fjx1 ¼0 . System (8) admits the two polynomial first integrals x4 x5 and x5 x6 and the two non-polynomial first integrals
jx2 j
where F ¼ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 . Straightforward computations show that system (6) has the five first integrals x4 x5, x4 x6, H defined in (3),
x_ 4 ¼ x21 þ
k1 F1; 4 k1 x_ 5 ¼ F1; 4 k1 x_ 6 ¼ F1; 4 x_ 4 ¼
3. Proof of Theorem 2
k1 F; ¼ 4 k1 F; ¼ 4 k1 F; ¼ 4
x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ;
@g xj1 jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ 1 @x1 @g 1 @g @g þ x3 ðx4 þ x5 x6 Þ 1 þ x21 1 þx2 ðx4 x5 þ x6 Þ @x2 @x3 @x4 k 1 @g 1 @g 1 @g 1 @h 1 þ x21 F þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x4
ð9Þ
We distinguish three cases depending on the value of j. If j = 1 then Eq. (9) becomes
ðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 @x1
@g @g @g þ x2 ðx4 x5 þ x6 Þ 1 þ x3 ðx4 þ x5 x6 Þ 1 þ x21 1 @x2 @x3 @x4 k 1 @g 1 @g 1 @g 1 @h1 þ x1 þ þ þ ¼ 0: F 4 @x4 @x5 @x6 @x4 ð7Þ
where F ¼ x21 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 . Let h = h(x1, x2, x3, x4, x5, x6) be a homogeneous polynomial first integral of (7). We can write h ¼ h1 ðx2 ; x3 ; x4 ; x5 ; x6 Þ þ xj1 g 1 ðx1 ; x2 ; x3 ; x4 ; x5 ; x6 Þ, with j 2 N and h1 and g1 homogeneous polynomials such that x1 - g1. On x1 = 0 system (7) becomes
Let g1 ¼ g 1 jx1 ¼0 X 0. Eq. (9) on x1 = 0 can be written as
@ g1 ðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ @x 2 @ g1 k 1 @ g1 @ g1 @ g1 ¼ 0: F1 þ x3 ðx4 þ x5 x6 Þ þ þ þ 4 @x3 @x4 @x5 @x6 Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
@g ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ 2 @x 2 @g 2 k 1 @g 2 @g 2 @g 2 ¼ 0: þ x3 ðx4 þ x5 x6 Þ þ þ þ F1 4 @x3 @x4 @x5 @x6
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A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg2 þ x3 ðx4 þ x5 x6 Þ @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0; þ þ þ F 12 4 @x3 @x4 @x5 @x6 where F 12 ¼ F 1 jx2 ¼0 . Now we write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We obtain
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g k1 @g 3 @g 3 @g 3 ¼ 0: þ x3 ðx4 þ x5 x6 Þ 3 þ þ þ F 12 4 @x3 @x4 @x5 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0; F 123 þ þ þ 4 @x4 @x5 @x6 where F 123 ¼ F 12 jx3 ¼0 . Applying Lemma 3 we get g3 0, which is a contradiction. Hence we have g1 0 and there1 fore @h 0. The lemma follows in this case. @x4 1 @h1 If j > 2 then x1 @h @x4 , and then @x4 0. Now we can proceed in a similar way as in the case j = 1 to prove that g1 0 by using Lemma 3. If j = 2 then Eq. (9) becomes
2ðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 @x1
@g @g @g þ x2 ðx4 x5 þ x6 Þ 1 þ x3 ðx4 þ x5 x6 Þ 1 þ x21 1 @x2 @x3 @x4 k 1 @g 1 @g 1 @g 1 @h1 þ F þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x4 1 ¼ g 1 jx ¼0 X 0. Eq. (9) on x1 = 0 can be written as Let g 1 @ g1 2ðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ @x 2 @ g1 k 1 @ g1 @ g1 @ g1 @h1 þ þ x3 ðx4 þ x5 x6 Þ þ þ þ ¼ 0: F1 @x3 @x4 @x5 @x6 @x4 4
Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
@g ð2ðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ 2 @x2 @g 2 k 1 @g 2 @g 2 @g 2 @h1 þ x3 ðx4 þ x5 x6 Þ þ þ þ ¼ 0: þ F1 @x3 @x4 @x5 @x6 @x4 4 1 If l > 0 then @h 0. Similar arguments to those used above @x4 lead to the desired result after applying Lemma 3. If l = 0, let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
@ g2 2ðx4 þ x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ @x3 k1 @ g2 @ g2 @ g2 @h1 þ F 12 þ þ þ ¼ 0; 4 @x4 @x5 @x6 @x4 where F 12 ¼ F 1 jx2 ¼0 . Now we write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We obtain ð2ðx4 þ x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g k1 @g 3 @g 3 @g 3 @h1 þ þ x3 ðx4 þ x5 x6 Þ 3 þ þ þ ¼ 0: F 12 @x3 @x4 @x5 @x6 @x4 4 1 If m > 0 then @h 0. Again the usual arguments lead to the @x4 desired result after applying Lemma 3. If m = 0, let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
2ðx4 þ x5 þ x6 Þg3 þ
k1 @ g3 @ g3 @ g3 @h1 F 123 þ þ ¼ 0; þ 4 @x4 @x5 @x6 @x4
where F 123 ¼ F 12 jx3 ¼0 . Applying Lemma 4 swapping x4 and x5 we get g3 0, which is a contradiction. Hence we have 1 g1 0 and therefore @h 0. The lemma follows also in this @x4 case. h After Lemma 5, h = h(x5 x6). Hence statement (b) of Theorem 2 follows. 3.3. Proof of statement (c) of Theorem 2 According to Table 1, system (1) in cases VI0 and VII0 can be written as
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F; 4 k1 F; x_ 5 ¼ n2 x2 ðx1 þ n2 x2 Þ þ 4 k1 F; x_ 6 ¼ 4 x_ 4 ¼ x1 ðx1 n2 x2 Þ þ
ð10Þ
where F ¼ ðx1 n2 x2 Þ2 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 and n22 ¼ 1. Suppose that system (10) has a homogeneous polynomial first integral h(x1, . . . , x6). We can write h ¼ h1 ðx2 ; . . . ; x6 Þ þ xj1 g 1 ðx1 ; . . . ; x6 Þ, with j 2 N and h1 and g1 homogeneous polynomials such that x1 - g1. System (10) on x1 = 0 is
x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F1; 4 k1 x_ 5 ¼ x22 þ F1; 4 k1 F1; x_ 6 ¼ 4 x_ 4 ¼
ð11Þ
where F 1 ¼ Fjx1 ¼0 . We note that h1 is a first integral of system (11). We can write h1 ¼ h2 ðx3 ; . . . ; x6 Þ þ xl2 g 2 ðx2 ; . . . ; x6 Þ, with l 2 N and h2 and g2 homogeneous polynomials such that x2 - g2. System (11) on x2 = 0 writes
x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F 12 ; 4 k1 F 12 ; x_ 5 ¼ 4 k1 F 12 ; x_ 6 ¼ 4 x_ 4 ¼
ð12Þ
where F 12 ¼ F 1 jx2 ¼0 . We note that h2 is a first integral of system (12). Straightforward computations show that system (12) has the three independent first integrals x4 x5, x5 x6 and
jx3 j
3ðk1Þ 2
pffiffiffiffi!x4 þxp5ffiffi2x6 x4 þ x5 þ x6 2 D D pffiffiffiffi ; F 12 x4 þ x5 þ x6 þ 2 D
17
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where D ¼ x24 þ x25 þ x26 x4 x5 x4 x6 x5 x6 . h2 = h2(x4 x5, x4 x6).
Therefore
Lemma 6. For system (11) we have that h2 = h2(x4 x6) and g2 0. Proof. Suppose that g2 X 0. As h1 ¼ h2 þ xl2 g 2 is a first integral of system (11), we can write
@g xl2 lðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 @x2 @g 2 k1 @g 2 @g 2 @g 2 2 @g 2 F1 þ x2 þ þ þ þ x3 ðx4 þ x5 x6 Þ 4 @x3 @x5 @x4 @x5 @x6 2 @h2 ¼ 0: ð13Þ þ x2 @x5 We distinguish three cases depending on the value of l. If l = 1 then Eq. (13) becomes
@g ðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g 2 k 1 @g 2 @g 2 @g 2 @h2 @g þ x2 F1 þ þ þ þ x22 2 ¼ 0: 4 @x3 @x4 @x5 @x6 @x5 @x5 Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
ðx4 x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0: F 12 þ þ þ 4 @x3 @x4 @x5 @x6 Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. Then
ððx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g k1 @g 3 @g 3 @g 3 ¼ 0: F 12 þ þ 3þ 4 @x3 @x4 @x5 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we get
ððx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0; F 123 þ þ þ 4 @x4 @x5 @x6 3 0, where F 123 ¼ F 12 jx3 ¼0 . Applying Lemma 3 we obtain g which is a contradiction. Hence g2 0. Back to Eq. (13) we 2 have @h 0. Then the lemma follows. @x5 2 If l > 2, then from Eq. (13) we have that x2 @h @x5 and thus @h2 @x5 0. Therefore h2 = h2(x4 x6). Now we can proceed in a similar way as in the case l = 1 to obtain the equation
ðlðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0: þ þ þ F 123 4 @x4 @x5 @x6 Applying again Lemma 3 we arrive to contradiction and hence g2 0. If l = 2, then Eq. (13) writes as
@g 2ðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g 2 k 1 @g 2 @g 2 @g 2 @g @h2 þ x22 2 þ F1 þ þ þ ¼ 0: 4 @x3 @x4 @x5 @x6 @x5 @x5 Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
@ g2 2ðx4 x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ @x 3 k1 @ g2 @ g2 @ g2 @h2 þ þ þ þ ¼ 0: F 12 4 @x4 @x5 @x6 @x5 Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. Then
xm 3 ½ð2ðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g k1 @g 3 @g 3 @g 3 @h2 3þ þ F 12 þ þ ¼ 0: 4 @x3 @x4 @x5 @x6 @x5 We distinguish two cases depending on the value of m. 2 2 . Hence @h If m > 0 then x3 @h 0 and h2 = h2(x4 x6). Now @x5 @x5 let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we obtain
ð2ðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0: þ þ þ F 123 4 @x4 @x5 @x6 Applying Lemma 3 we get a contradiction and hence g2 0. If m = 0, let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we obtain k1 @ g3 @ g3 @ g3 @h2 2ðx4 x5 þ x6 Þg3 þ þ þ ¼ 0: þ F 123 @x4 @x5 @x6 @x5 4 2 Applying Lemma 4 we get @h 0 and g3 0, hence the @x5 lemma follows. All the subcases are finished and therefore the lemma is proved. h After Lemma 6 we have that h ¼ h2 ðx4 x6 Þþ xj1 g 1 ðx1 ; . . . ; x6 Þ, with j 2 N and x1 - g1. We recall that h is a first integral of system (10). Thus
xj1 ½jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ @g @g 1 þ x2 ðx4 x5 þ x6 Þ 1 þ x3 ðx4 þ x5 x6 Þ @x1 @x2 @g 1 @g 1 þ x1 ðx1 n2 x2 Þ n2 x2 ðx1 n2 x2 Þ @x3 @x4 @g k 1 @g 1 @g 1 @g 1 F þ þ 1þ 4 @x5 @x4 @x5 @x6 @h2 ¼ 0: þ x1 ðx1 n2 x2 Þ @x4
ð14Þ
Lemma 7. For system (10) we have that h2 0 and g1 0. Proof. Suppose that g1 X 0. We distinguish two cases depending on the value of j. If j > 1 then from Eq. (14) we 2 have that x1 @h , and hence h2 0. Therefore Eq. (14) can @x4 be written as
jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 þ x2 ðx4 x5 þ x6 Þ @x1
@g 1 @g þ x3 ðx4 þ x5 x6 Þ 1 þ x1 ðx1 n2 x2 Þ @x2 @x3 @g 1 @g k 1 @g 1 @g 1 @g 1 ¼ 0: F n2 x2 ðx1 n2 x2 Þ 1 þ þ þ 4 @x4 @x5 @x4 @x5 @x6
Let g1 ¼ g 1 jx1 ¼0 X 0. Eq. (14) on x1 = 0 becomes
@ g1 jðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ þ x3 ðx4 þ x5 x6 Þ @x2 @ g1 @ g1 k 1 @ g1 @ g1 @ g1 ¼ 0: F1 þ x22 þ þ þ 4 @x3 @x5 @x4 @x5 @x6
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Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g 2 @g @g þ x3 ðx4 þ x5 x6 Þ 2 þ x22 2 @x2 @x3 @x5 k1 @g 2 @g 2 @g 2 ¼ 0: F1 þ þ þ 4 @x4 @x5 @x6
Let g2 ¼ g 2 jx2 ¼0 X 0. Then, on x2 = 0 we have
2 ¼ It only remains to consider the case l = 1. Let g g 2 jx2 ¼0 X 0. From Eq. (15) on x2 = 0 we have
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg2 þ x3 ðx4 þ x5 x6 Þ @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0: F 12 þ þ þ 4 @x3 @x4 @x5 @x6
2x6 g2 þ x3 ðx4 þ x5 x6 Þ
Now write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We get
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g k1 @g 3 @g 3 @g 3 ¼ 0: F 12 þ x3 ðx4 þ x5 x6 Þ 3 þ þ þ 4 @x3 @x4 @x5 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. Then, on x3 = 0 we have
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 ¼ 0: F 123 þ þ þ 4 @x4 @x5 @x6
@g 1 þ x2 ðx4 x5 þ x6 Þ @x1
@g 1 @g þ x3 ðx4 þ x5 x6 Þ 1 þ x1 ðx1 n2 x2 Þ @x2 @x3 @g @g k 1 @g 1 @g 1 @g 1 F þ þ 1 n2 x2 ðx1 n2 x2 Þ 1 þ 4 @x4 @x5 @x4 @x5 @x6
þ ðx1 n2 x2 Þ
n2
@ g2 k 1 @ g2 @ g2 @ g2 F 12 þ þ þ 4 @x3 @x4 @x5 @x6
@h2 ¼ 0: @x4
Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We get:
@g 3 xm 3 ð2x6 þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @x3 k1 @g 3 @g 3 @g 3 @h2 n2 F 12 þ þ ¼ 0: þ 4 @x4 @x5 @x6 @x4
2 , and hence h2 0. Therefore we obtain If m > 0 then x3 @h @x4 an equation of type (4) and hence by Lemma 3 we get g1 0. If m = 0, let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0, we have
3 0, a contradiction. Applying Lemma 3 we obtain g Hence g1 0 and the lemma follows in this case. If j = 1 then Eq. (14) becomes
ðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
Similar arguments to those used before lead to an equation of type (4) and hence applying Lemma 3 we get a contradiction. Therefore g1 0 and the lemma follows. If l = 0 then we can use the same arguments to arrive from Eq. (15) to an equation of type (4), and hence applying Lemma 3 we get a contradiction. Therefore g1 0 and h2 0, so the lemma follows.
@h2 ¼ 0: @x4
Let g1 ¼ g 1 jx1 ¼0 X 0. On x1 = 0 we have
2x6 g3 þ
k1 @ g3 @ g3 @ g3 @h2 n2 F 123 þ þ ¼ 0: 4 @x4 @x5 @x6 @x4
As h2 = h2(x4 x6) is a homogeneous polynomial of degree 2 n, we have h2 = a0(x4 x6)n. Thus @h ¼ a0 nðx4 x6 Þn1 . On @x4 x6 = 0 Eq. (16) writes
k1 @ g3 @ g3 @ g3 ðx4 x5 Þ2 þ þ n2 a0 nxn1 ¼ 0: 4 4 @x4 @x5 @x6 x6 ¼0 Therefore we must take a0 = 0 and hence h2 0. Now Eq. (16) is of type (4) and hence by Lemma 3 we get g1 0 and the lemma follows. h After Lemma 7 statement (c) of Theorem 2 is proved, as it follows that h 0.
@ g1 ðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ þ x3 ðx4 þ x5 x6 Þ @x2 @ g1 @ g1 k 1 @ g1 @ g1 @ g1 @h2 n2 x2 F1 þ x22 þ þ þ ¼ 0: 4 @x3 @x5 @x4 @x5 @x6 @x4
3.4. Proof of statement (d) of Theorem 2
Write g1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get
x_ 1 ¼ x1 ðx4 þ x5 þ x6 Þ; x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ;
xl2 ½ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g 2 @g @g þ x3 ðx4 þ x5 x6 Þ 2 þ x22 2 @x2 @x3 @x5 k1 @g 2 @g 2 @g 2 @h2 n2 x2 F1 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x4
We distinguish three cases depending on the value of l. 2 If l > 1 then x2 @h . Hence h2 0. Thus, from (15), @x4
ððx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g @g @g 2 þ x3 ðx4 þ x5 x6 Þ 2 þ x22 2 @x2 @x3 @x5 k1 @g 2 @g 2 @g 2 ¼ 0: F1 þ þ þ 4 @x4 @x5 @x6
According to Table 1, Bianchi cases VIII and IX correspond to n1 ¼ n2 ¼ n23 ¼ 1 and can be written into the form
k1 F; 4 k1 F; x_ 5 ¼ x2 ðx1 þ x2 n3 x3 Þ þ 4 k1 F; x_ 6 ¼ n3 x3 ðx1 x2 þ n3 x3 Þ þ 4 x_ 4 ¼ x1 ðx1 x2 n3 x3 Þ þ
ð15Þ
ð16Þ
ð17Þ
where F ¼ x21 þ x22 þ x23 2x1 x2 2n3 x1 x3 2n3 x2 x3 þ x24 þ x25 þ x26 2x4 x5 2x5 x6 2x4 x6 and n23 ¼ 1. Let h = h(x1, . . . , x6) be a homogeneous polynomial first integral of degree n of system (17). Write h ¼ h1 ðx2 ; . . . ; x6 Þ þ xj1 g 1 ðx1 ; . . . ; x6 Þ, with j 2 N, h1 and g1 homogeneous polynomials and x1 - g1. System (17) on x1 = 0 becomes
A. Ferragut et al. / Chaos, Solitons & Fractals 48 (2013) 12–21
x_ 2 ¼ x2 ðx4 x5 þ x6 Þ; x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; x_ 4 ¼
k1 F1; 4
k1 x_ 5 ¼ x2 ðx2 n3 x3 Þ þ F1; 4 k1 x_ 6 ¼ n3 x3 ðx2 n3 x3 Þ þ F1; 4
ð18Þ
where F 1 ¼ Fjx1 ¼0 . System (18) admits h1 = h1(x2, . . . , x6) as first integral. Write h1 ¼ h2 ðx3 ; . . . ; x6 Þ þ xl2 g 2 ðx2 ; . . . ; x6 Þ, with l 2 N, h2 and g2 homogeneous polynomials and x2 - g2. System (18) on x2 = 0 becomes
x_ 3 ¼ x3 ðx4 þ x5 x6 Þ; k1 F 12 ; 4 k1 F 12 ; x_ 5 ¼ 4 k1 F 12 ; x_ 6 ¼ x23 þ 4
ð19Þ
where F 12 ¼ F 1 jx2 ¼0 . We note that h2 = h2(x3, . . . , x6) is a first integral of system (19). Write h2 ¼ h3 ðx4 ; x5 ; x6 Þ þ xm 3 g3 ðx3 ; x4 ; x5 ; x6 Þ, with m 2 N, h3 and g3 homogeneous polynomials and x3 - g3. System (19) on x3 = 0 is
k1 F 123 ; 4 k1 x_ 5 ¼ F 123 ; 4 k1 F 123 ; x_ 6 ¼ 4
x_ 4 ¼
ð20Þ
where F 123 ¼ F 12 jx3 ¼0 . Note that h3 is a polynomial first integral of system (20). Since system (20) admits the two independent first integrals x4 x5 and x5 x6, any polynomial first integral of (20) must be a polynomial in the variables x4 x5 and x5 x6. Therefore h3 = h3(x4 x5, x5 x6). The next three lemmas end the proof of statement (d) of Theorem 2. The first one shows that h2 = h2(x4 x5). Lemma 8. For system (19) we have that g3 0 and h3 = h3(x4 x5). We recall that Proof. Suppose that g3 X 0. h2 ¼ h3 ðx4 x5 ; x5 x6 Þ þ xm 3 g 3 ðx3 ; x4 ; x5 ; x6 Þ, where m 2 N and x3 - g3. As h2 is a first integral of system (19), we have
@g 3 @g xm þ x23 3 3 mðx4 þ x5 x6 Þg 3 þ x3 ðx4 þ x5 x6 Þ @x3 @x6 k1 @g 3 @g 3 @g 3 @h 3 þ x23 F 12 þ þ ¼ 0: þ 4 @x4 @x5 @x6 @x6
ð21Þ
We distinguish three cases depending on the value of m. If m = 1 then
@g @g ðx4 þ x5 x6 Þg 3 þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 @x3 @x6 k1 @g 3 @g 3 @g 3 @h3 þ x3 F 12 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
ðx4 þ x5 x6 Þg3 þ
k1 @ g3 @ g3 @ g3 ¼ 0: F 123 þ þ 4 @x4 @x5 @x6
3 0 and hence g3 0. ConseApplying Lemma 3 we get g 3 quently @h ¼ 0 and the lemma followsin this case. @x6 3 If m > 2 then from (21) we have x3 @h @x6 and so h3 = h3(x4 x5). Now from Eq. (21) on x3 = 0 we get an equation of type (4), hence applying Lemma 3 we get g3 0 and the lemma follows in this case. If m = 2 then from (21) we have
@g @g 2ðx4 þ x5 x6 Þg 3 þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 @x3 @x6 k1 @g 3 @g 3 @g 3 @h3 þ F 12 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x6 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we obtain
2ðx4 þ x5 x6 Þg3 þ
x_ 4 ¼
19
k1 @ g3 @ g3 @ g3 @h3 þ F 123 þ þ ¼ 0: 4 @x4 @x5 @x6 @x6
3 Applying Lemma 4 swapping x5 and x6 we get @h ¼ 0 and @x6 g3 0. Hence h3 = h3(x4 x5), g3 0 and the lemma follows in this case. h The second lemma shows that h1 0.
Lemma 9. For system (18) we have that g2 0 and h2 0. Proof. Suppose that g2 X 0. We recall that h1 ¼ h2 ðx4 x5 Þ þxl2 g 2 , with l 2 N and x2 - g2. As h1 is a first integral of system (18), we have
@g xl2 lðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g 2 @g 2 @g þ x2 ðx2 n3 x3 Þ n3 x3 ðx2 n3 x3 Þ 2 @x3 @x5 @x6 k1 @g 2 @g 2 @g 2 F1 þ þ þ 4 @x4 @x5 @x6 We distinguish two cases depending on the value of l. If 2 2 and hence @h l > 1 then x2 @h 0, which means that @x5 @x5 h2 0. Substituting in the equation above we have
@g 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g @g @g 2 þ x2 ðx2 n3 x3 Þ 2 n3 x3 ðx2 n3 x3 Þ 2 @x3 @x5 @x6 k1 @g 2 @g 2 @g 2 ¼ 0: þ þ þ F1 4 @x4 @x5 @x6
lðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ
The usual arguments lead to Eq. (4), hence we obtain g2 0 by Lemma 3. If l = 1 then we have
@g 2 þ x3 ðx4 þ x5 x6 Þ @x2 @g @g @g 2 þ x2 ðx2 n3 x3 Þ 2 n3 x3 ðx2 n3 x3 Þ 2 @x3 @x5 @x6 k1 @g 2 @g 2 @g 2 @h2 þ ðx2 n3 x3 Þ F1 þ þ þ ¼ 0: 4 @x4 @x5 @x6 @x5
ðx4 x5 þ x6 Þg 2 þ x2 ðx4 x5 þ x6 Þ
Let g2 ¼ g 2 jx2 ¼0 X 0. On x2 = 0 we have
@ g2 @ g2 ðx4 x5 þ x6 Þg2 þ x3 ðx4 þ x5 x6 Þ þ x23 @x3 @x6 k1 @ g2 @ g2 @ g2 @h2 n 3 x3 þ þ þ ¼ 0: F 12 4 @x4 @x5 @x6 @x5
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Write g2 ¼ xm 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. Then
xm 3 ½ððx4
x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g 3 @g k1 @g 3 @g 3 @g 3 þ x23 3 þ þ þ F 12 4 @x3 @x6 @x4 @x5 @x6
Now we distinguish three cases depending on the value of m. If m = 0 then we are in (22) again and the usual arguments lead to g2 0 and h2 0. 2 @h2 If m > 1 then x3 @h @x5 and hence @x5 0, which means that h2 0. Then we have
ððx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 þ x3 ðx4 þ x5 x6 Þ @g @g k1 @g 3 @g 3 @g 3 3 þ x23 3 þ ¼ 0: F 12 þ þ 4 @x3 @x6 @x4 @x5 @x6 The usual arguments finish the proof in this case. Finally if m = 1 then we have
jðx4 þ x5 þ x6 Þg1 þ x2 ðx4 x5 þ x6 Þ
@ g1 þ x3 ðx4 þ x5 x6 Þ @x2
@ g1 @ g1 þ x2 ðx2 n3 x3 Þ þ n3 x3 ðx2 þ n3 x3 Þ @x3 @x5 @ g1 k 1 @ g1 @ g1 @ g1 ¼ 0: þ þ þ F1 4 @x6 @x4 @x5 @x6
1 ¼ xl2 g 2 X 0, with l 2 N [ f0g and x2 - g2. We get Write g
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg 2 þ x2 ðx4 x5 þ x6 Þ @g @g 2 þ x3 ðx4 þ x5 x6 Þ 2 þ x2 ðx2 n3 x3 Þ @x2 @x3 @g 2 @g þ n3 x3 ðx2 þ n3 x3 Þ 2 @x5 @x6 k1 @g 2 @g 2 @g 2 ¼ 0: þ þ þ F1 4 @x4 @x5 @x6 2 ¼ g 2 jx ¼0 X 0. On x2 = 0 we have Let g 2
@g @g k1 2x4 g 3 þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 þ 4 @x3 @x6 @g 3 @g 3 @g 3 @h2 n3 þ þ ¼ 0: F 12 @x4 @x5 @x6 @x5 Let g3 ¼ g 3 jx3 ¼0 X 0. On x3 = 0 we have
2x4 g3 þ
k1 @ g3 @ g3 @ g3 @h2 n3 F 123 þ þ ¼ 0: 4 @x4 @x5 @x6 @x5
As h2 = h2(x4 x5) is a homogeneous polynomial of degree n, we have h2 = a0(x4 x5)n. Hence
2x4 g3 þ
k1 @ g3 @ g3 @ g3 þ n3 a0 nðx4 x5 Þn1 ¼ 0: F 123 þ þ 4 @x4 @x5 @x6
On x4 = 0 we have
k1 @ g3 @ g3 @ g3 þ þ þ n3 a0 nðx5 Þn1 ¼ 0; ðx5 x6 Þ2 4 @x4 @x5 @x6 x4 ¼0 which means that a0 = 0. Therefore h2 0. The equation is now of type (4) and leads to g2 0 by Lemma 3. All the subcases are considered and the proof of the lemma is finished. h The last lemma shows that g1 0 and therefore that h 0. Lemma 10. For system (17) we have that g1 0. Proof. Suppose that g1 X 0. We recall that h ¼ xj1 g 1 , with j 2 N and x1 - g1, is a first integral of system (17). Then
jðx4 þ x5 þ x6 Þg 1 þ x1 ðx4 þ x5 þ x6 Þ
@g 1 þ x2 ðx4 x5 þ x6 Þ @x1
@g @g 1 þ x3 ðx4 þ x5 x6 Þ 1 þ x1 ðx1 x2 n3 x3 Þ @x2 @x3 @g 1 @g þ x2 ðx1 þ x2 n3 x3 Þ 1 þ n3 x3 ðx1 x2 þ n3 x3 Þ @x4 @x5 @g 1 k 1 @g 1 @g 1 @g 1 ¼ 0: þ þ þ F 4 @x6 @x4 @x5 @x6 Let g1 ¼ g 1 jx1 ¼0 X 0. On x1 = 0 we have
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 ÞÞg2 þ x3 ðx4 þ x5 x6 Þ @ g2 @ g2 k 1 @ g2 @ g2 @ g2 ¼ 0: þ x23 þ þ þ F 12 4 @x3 @x6 @x4 @x5 @x6 2 ¼ xm Write g 3 g 3 X 0, with m 2 N [ f0g and x3 - g3. We get
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg 3 @g @g þ x3 ðx4 þ x5 x6 Þ 3 þ x23 3 @x3 @x6 k1 @g 3 @g 3 @g 3 ¼ 0: þ þ þ F 12 4 @x4 @x5 @x6 3 ¼ g 3 jx ¼0 X 0. On x3 = 0 we have Let g 3
ðjðx4 þ x5 þ x6 Þ þ lðx4 x5 þ x6 Þ þ mðx4 þ x5 x6 ÞÞg3 k1 @ g3 @ g3 @ g3 F 123 þ þ þ 4 @x4 @x5 @x6 ¼ 0: We can apply Lemma 3. Hence the lemma follows. h After Lemma 10, we get h 0. Thus the proof of statement (d) of Theorem 2 is finished. 4. Conclusions According to Proposition 1 the study of analytic first integrals of systems (1) is reduced to the study of polynomial first integrals. Hence the non-existence of polynomial first integrals implies the non-existence of analytic first integrals. From Theorem 2 we have that Bianchi type I model is not chaotic whereas Bianchi type II is at most partially chaotic. The rest of the models of class A, the Bianchi types VI0, VII0, VIII and IX, can exhibit chaos. References [1] Bianchi L. Sugli spazi a tre dimensioni che ammettono un gruppo continuo di movimenti (On the spaces of three dimensions that admit a continuous group of movements). Mem Math Fis Soc Ital Sci 1898;11:267–352. [2] Bianchi L. Lezioni sulla teoria dei gruppi continui finiti di trasformazioni (Lectures on the theory of finite continuous transformation grups), Spoerri, Pisa, 1918, pp. 550–557. [3] Bogoyavlensky OI. Qualitative theory of dynamical systems in astrophysics and gas dynamics. Springer-Verlag; 1985.
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