PERIODIC ORBITS FOR A GENERALIZED FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM IN DIMENSION 6 FATIMA EZZAHRA LEMBARKI AND JAUME LLIBRE

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1 This is a preprint of: Periodic orbits for a generalized Friedmann-Robertson-Walker Hamiltonian system in dimension 6, Fatima E. Lembarki, Jaume Llibre, Discrete Contin. Dyn. Syst. Ser. S, vol. 8(6, , 015. DOI: [ /dcdss ] PERIODIC ORBITS FOR A GENERALIZED FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM IN DIMENSION 6 FATIMA EZZAHRA LEMBARKI AND JAUME LLIBRE Abstract. A generalized Friedmann-Robertson-Walker Hamiltonian system is studied in dimension 6. The averaging theory is the tool used to provide sufficient conditions on the six parameters of the system which guarantee the existence of continuous families of period orbits parameterized by the energy. 1. Introduction In astrophysics the study of the dynamic of galaxies progressed considerably thanks to the discovery of important theories coming from mathematical models, see for instance the articles [], [7], [10], [13]. Calzeta and Hasi (1993 studied the simplified Friedmann-Robertson-Walker Hamiltonian in dimension 4 the Hamiltonian H = (p y p x + y x / + (bx y / which modeled a universe. They proved analytically and numerically the existence of chaotic motion of the Hamiltonian system associated to the above Hamiltonian. Although this model is too simplified to be considered realistically, it is an interesting testing ground for the implications of chaos in cosmology, see for more details [3]. Hawking [4] and Page [8] considered similar models for understanding the relation between the thermodynamic and cosmological arrow of time, in the area of quantum cosmology. We study the following generalized classical Friedmann-Robertson-Walker Hamiltonian system in dimension 6. (1 H = 1 (p y + p z p x + y + z x (ax4 + bx y + cx z + dy 4 +ey z + fz 4, Note that this Hamiltonian depends on six parameters a,b,c,d,e and f. When z = p z = 0 the previous Hamiltonian contains the planar classical Friedmann- Robertson-Walker Hamiltonian system studied by Calzeta and Hasi. Its periodic solutions were studied in [6]. 010 Mathematics Subject Classification. 37G15, 37C80, 37C30. Key words and phrases. periodic orbits, Friedmann-Robertson-Walker, averaging theory. 1

2 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, Our goal in this article is to study the periodic solutions in the different energy levels H = h of the Hamiltonian system ẋ = H = p x, p x ( ẏ = ż = ṗ x = ṗ y = H p y = p y, H p z = p z, H x = x x(ax + by + cz, H y = y y(bx + dy + ez, ṗ z = H z = z z(cx + ey + fz, associated to the Hamiltonian (1. The dot denotes derivative with respect to the independent variable t, the time. We study the existence of periodic orbits of system ( and we compute them by using the averaging theory, see for more details about this tool section. Specifically we will provide through the averaging method sufficient conditions on the six parameters a,b,c,d,e and f for ensuring the existence of periodic orbits of Friedmann- Robertson-Walker system (. The study of the periodic orbits is of special interest because they are the most simple non trivial solutions of the system after the equilibrium points, and their stability determines the kind of motion in their neighborhood. Note that the periodic orbits studied in this paper are isolated in every energy level. The averaging method provides periodic orbits of a perturbed periodic non autonomous differential system depending on a small parameter ε. Roughly speaking, the problem of finding periodic solutions of a differential system is reduced to find zeros of some convenient finite dimensional function. We check the conditions under which the averaging theory guarantees the existence of periodic orbits, and we find them as a function of the energy. In that manner we can find analytically periodic orbits in any energy level as function of the six parameters of the Friedmann-Robertson-Walker systems (. The conclusion of our main results are the following theorem and lemma. Theorem 1. At every fixed energy level H = ε h with h R the Friedmann- Robertson-Walker Hamiltonian system ( has at least (A one periodic orbit if the following conditions hold 3a + b h < 0, bc 0, (a + b(3a + b(a + c(3a + c 0, b < 1, 3a + c c < 1; (B two periodic orbits if one of the following six conditions hold (1 h > 0, eb 0, (b + 3d(b + d 0, 3d + b b < 1, e 3d e < 1;

3 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 3 ( h > 0, ec 0, (c + 3f(c + f 0, e 3f e < 1, 3f + c c < 1; (3 a + c + f 0, bc ae ce + bf 0, h(a + c + f(c + f < 0, h(a + c + f(a + c > 0, (bc c ae ce + af + bf(bc 3c ae ce + 3af + bf 0, 3c bc + ae + ce 3af bf bc ae ce + bf < 1; (4 3a+c+3f 0, (bc+3ae+ce+3bf 0, h(3a+c+3f(c+3f < 0, h(3a + c + 3f(3a+c > 0, c(c 3bc + 3ae+ce 9af 9bf(c bc + 9ae + 3ce 9af 3bf 0, bc + c + 6ae + ce 9af 6bf bc + 3ae + ce + 3bf < 1; (5 a+b+d 0, b+d 0, bc+cdaebe 0, h(b+d(a+b+d < 0, h(a + b(a + b + d > 0, b(3b bc 3ad cd + ae + be(b bc ad cd + ae + be 0, 3b bc 3ad cd + ae + be bc + cd ae be < 1; (6 3a + b + 3d 0, 3cd + 3ae + bc + be 0, h(b + 3d(3a + b + 3d < 0, h(3a + b(3a + b + 3d > 0, b(b 3bc 9ad 9cd + 3ae + be(b bc 9ad 3cd + 9ae + 3be 0 b bc 9ad 6cd + 6ae + be 3cd + 3ae + bc + be < 1. (C four periodic orbits if one of the following twenty eight conditions hold (1 cd be ce + bf 0, d e + f 0, h(d e + f(f e > 0, h(d e + f(d e > 0, e(be + bf + cdce +3df 3e (be + bf + cd ce + df e 0, cd + be + ce + 3e bf 3df < 1; cd be ce + bf ( bd cd be + bf 0, 3d + b b < 1, h(bd cd be + bf(cd bf > 0, hb(bdcdbe+bf(de > 0, be(b+d(b+3d(de+f(cebf 0; (3 3cd + be ce 3bf 0, 3d e + 3f 0, h(3f e(3d e + 3f > 0, h(3d e(3d e + 3f > 0, (9cd + be + 3ce + e 3bf 9df(3cd 3becee +9bf +9df 0, be + ce + e 6cd 6bf 9df 3cd + be ce 3bf < 1; (4 4bc+3bd+3cdbe+3bf 0, hb(4bc+3bd+3cdbe+3bf(3de > 0, h(4bc+3bd+3cdbe+3bf(4bc+3cd+3bf > 0, eb(b+d(b+3d(3d 3d + b e + 3f(4bc + ce + 3bf 0, < 1; b (5 a + c + e = 0, e c f 0, eh(e c f > 0, h(cd + df ce e (a + d(e c f > 0, ce ( ad c + cd ae ce e + af + df 0, h(ce + e + ac + af(a + d(e c f < 0; (6 a + c + e 0, c e + f = 0, c d + e 0, eh(a + c + e < 0, hc(c d + e > 0, h(ae ad cd(a + c + e(c d + e > 0, ce(c ad cd + ae + ce + e af df 0

4 4 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, (7 Σ = c ad cd + ae + ce + e af df 0, a + c + e 0, e c f 0, h(cd + df ce e Σ > 0, h(ae ad cdσ > 0, h(ce + c + ae afσ > 0, ce 0; (8 3c 3d + e 0, a + d = 0, 3a + 3c + e 0, dh(3c 3d + e < 0, h(9cd 3ce e + 9df(3c 3d + e(3a + 3c + e > 0, h(3ac + ae + 3c + 3cd + ce + 3df(3c 3d + e(3a + 3c + e > 0, cde(9c 9ad 18cd + 6ae + 6ce + e 9af 9df 0; (9 3c3d+e = 0, a+d 0, 3ce+3f 0, dh(a+d < 0, h(3ad+3cd ae(a+d(3ce+3f > 0, h(acad+af cd(a+d(3ce+3f > 0, cde(9c 9ad 18cd + 6ae + 6ce + e 9af 9df 0; (10 Σ 1 = 9c 9ad 18cd + 6ae + 6ce + e 9af 9df 0, a + d 0, 3c3d+e 0, h(9cd3cee +9dfΣ 1 > 0, h(3c +ce+ae3afσ 1 > 0, h(3ad + 3cd aeσ 1 < 0, ce 0; (11 3a+c+e = 0, ce+3f 0, he(ce+3f < 0, h(a+d(ce+3f(ce+ e 9df 3cd > 0, h(a + d(c e + 3f(ce + e + 3ac + 9af > 0, a + d 0, ce ( 9ad c + 6cd 6ae ce e + 9af + 9df 0; (1 3a + c + e 0, c e + 3f = 0, 3d c e 0, he(3a + c + e < 0, hc(3d c e > 0, h(3ad + cd ae(3d c e(3a + c + e > 0, ce ( 9ad c + 6cd 6ae ce e + 9af + 9df 0; (13 Σ = c 9ad6cd+6ae+ce+e 9af 9df 0, h(ae3adcdσ > 0, 3a + c + e 0, c e + 3f 0, h(3cd ce e + 9dfΣ > 0, h(c + ce + 3ae 9afΣ > 0, ce 0; (14 a + d = 0, c 3d + 3e 0, 3a + c + 3e 0, dh(c 3d + 3e < 0, h(cd ce 3e + 3df(c 3d + 3e(3a + c + 3e > 0, h(3ac + 9ae + c + 3cd + 3ce + 9df(c 3d + 3e(3a + c + 3e > 0, d(bc b c + 9ad + 6cd 18ae 6be 6ce 9e + 9af + 6bf + 9df(3bc d 3abce 4bc e + 9acde 3bcde + 3c de 9abe 1bce + 9cde + 9bcdf 9bdef 0; (15 a + d 0, c 3d + 3e = 0, c 3e + 3f 0, dh(a + d < 0, h(3ad + cd 3ae(a+d(c3e+3f > 0, h(ac+3af 3adcd(a+d(c3e+3f > 0, (b +bcc +9ad+6cd18ae6be6ce9e +9af+6bf+9df(bc d+ abce 3abde + 3acde 4bcde + c de 3ace + 3bcdf + 3abef 0; (16 Σ 3 = c 9ad 6cd + 18ae + 6ce + 9e 9af 9df 0, a + d 0, c3d+3e 0, h(3ae3adcdσ 3 > 0, h(cdce3e +3dfΣ 3 > 0, h(c + 3ce + 9ae 9afΣ 3 > 0, (bc b c + 9ad + 6cd 18ae 6be 6ce 9e + 9af + 6bf + 9df(3bc d 4bc e + 9acde + 3c de 9abe 9ace 1bce + 9bcdf + 9abef 0; (17 bc+de = 0, a+b+d 0, bc+ef 0, h(b+d(a+b+d < 0, h(b cd + be bc + ae ad(a + b + d(b c + e f > 0, h(cd + ab ac + ad af bf(a + b + d(b c + e f > 0,

5 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 5 (b bc + c ad cd + ae + be + ce + e af bf df(b c b 3 c b cd + bc d + ab e abce bc e + abde acde c de + ace + bce + b cf + bcdf abef b ef 0; (18 bc+de 0, a+b+d = 0, a+b+c+e 0, h(b+d(bc+de > 0, h(e cd + ce + be bf df(b c + d e(a + b + c + e > 0, h(acaecbc cdcebf df(b c+de(a+b+c+e > 0, (bcb c +ad+cdaebecee +af +bf +df(b cebc d+ bc e+acde+c deabe +bce +cde b cf bcdf b ef bdef 0; (19 ω = b bc + c ad cd + ae + be + ce + e af bf df 0, b c + d e 0, a + b + d 0, h(cd be ce e + bf + dfω > 0, h(b ad cd + ae bc + beω > 0, h(c + ce + ae af bc bfω > 0, (bc d b ce bc e acde c de + abe + ace + bce + b cf + bcdf abef b ef 0; (0 3bc+3de = 0, a+b+d 0, 3bc+e3f 0, h(b+d(a+b+d < 0, h(a + b + d(3b c + e 3f(3b cd bc + be 3ad + ae > 0, h(a + b + d(3b c + e 3f(3ab ac + 3ad + cd 3af 3bf > 0, (9b 6bc + c 9ad 6cd + 6ae + 6be + ce + e 9af 18bf 9df(3b 3 c+b c 3b cd+bc d+3ab eabce4b ce+bc e+3abde+ 3acde + c de ace bce + 3b cf + 3bcdf 3abef 3b ef 0; (1 3bc+3de 0, a+b+d = 0, 3a+3b+c+e 0, h(b+d(3bc+3d e > 0, h(3a+3b+c+e(3bc+3de(3be+ce+e 3cd9bf9df > 0, h(3a+3b+c+e(3bc+3de(3ac+3ae+3bc+c +3cd+ce+9bf+9df > 0, (6bc 9b c + 9ad + 6cd 6ae 6be ce e + 9af + 18bf + 9df(bc d + 4abce + 3b ce + bc e + 3acde + 4bcde + c de + abe + bce + cde + 3b cf + 3bcdf + 3b ef + 3bdef 0; ( ω 1 = 9b 6bc+c 9ad6cd+6ae+6be+ce+e 9af18bf9df 0, 3bc+3de 0, a+b+d 0, h(3cd3becee +9bf +9dfω 1 > 0, h(3b 3ad cd + ae bc + beω 1 > 0, h(c 3bc + 3ae + ce 9af 9bfω 1 > 0, (bc d 5b ce + bc e + 3acde + c de + abe ace bce + 3b cf + 3bcdf 3abef 3b ef 0; (3 b3c+3de = 0, 3a+b+3d 0, b3c+e3f 0, h(b 9ad9cd+ 3ae3bc+be(3a+b+3d(b3c+e3f > 0, h(b+3d(3a+b+3d < 0, h(3cdbf +ab3ac+3ad3af(3a+b+3d(b3c+e3f > 0, (b 6bc+9c 9ad18cd+6ae+be+6ce+e 9af 6bf 9df(b 3 c 3b c + 3b cd 9bc d + 3ab e 9abce + 3bc e + 9abde + 9acde + 9c de + 1bcde 3ace bce 3b cf 9bcdf 9abef 3b ef 0; (4 b 3c + 3d e 0, 3a + b + 3d = 0, 3a + b + 3c + e 0, h(b+3d(b3c+3de > 0, h(b3c+3de(3a+b+3c+e(e 9cd+ be+3ce3bf 9df > 0, h(b3c+3de(3a+b+3c+e(3ac+ae+bc+ 3c +3cd+ce+bf+3df < 0, (b 6bc+9c 9ad18cd+6ae+be+6ce+ e 9af 6bf 9df(1abce9bc d+5b ce+15bc e+9acde+1bcde+

6 6 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, 9c de + 3abe + 5bce + 3cde 3b cf 9bcdf + 3b ef + 9bdef 0; (5 b 3c + 3d e 0, 3a + b + 3d 0, ω = b + 9c 9ad + 6ae + e 18cd + 6ce 6bc + be 6bf 9af 9df 0, h(9cd be 3ce e + 3bf + 9dfω > 0, h(b 9ad 9cd + 3aeb 3bc + beω > 0, h(3c + ce + ae 3af bc bfω > 0, (5bc e 3bc d b ce + 3acde + 3c de + abe ace + bce b cf 3bcdf 3abef b ef 0; (6 cb3d+3e = 0, 3a+b+3d 0, bc+3e3f 0, h(b+3d(3a+b+ 3d < 0, h(3a+b+3d(bc+3e3f(b bc9ad3cd+9ae+3be > 0, h(3a + b + 3d(b c + 3e 3f(cd + ab ac + 3ad 3af bf > 0, (b bc+c 9ad6cd+18ae+6be+6ce+9e 9af 6bf 9df(b 3 c b c + 3b cd 3bc d + 3ab e 3abce + 4b ce bc e + 9abde 9acde + 1bcde 3c de + 9ace + 3bce 3b cf 9bcdf 9abef 3b ef 0; (7 c b 3d + 3e 0, 3a + b + 3d = 0, h(b + 3d(b c + 3d 3e > 0, h(b c + 3d 3e(3a + b + c + 3e(be bf + ce cd + 3e 3df > 0, h(bc+3d3e(3a+b+c+3e(3ac+bc+c +3cd+9ae+3ce+3bf+9df < 0, (bc b c + 9ad + 6cd 18ae 6be 6ce 9e + 9af + 6bf + 9df(b ce bc d + bc e 3acde c de + 3abe + 3bce 3cde b cf 3bcdf + b ef + 3bdef 0; (8 c b 3d + 3e 0, 3a + b + 3d 0, ω 3 = b bc + c 9ad 6cd + 18ae + 6be + 6ce + 9e 9af 6bf 9df 0 h(cd be ce 3e + bf + 3dfω 3 > 0, h(b bc 9ad 3cd + 9ae + 3beω 3 > 0, h(c bc + 9ae + 3ce 9af 3bfω 3 > 0, (bc d b ce bc e + 3acde + c de 3abe 3ace 5bce + b cf + 3bcdf + 3abef + b ef 0. Theorem 1 is proved in section 3. Lemma. The family of periodic orbits of the system ( generated by the periodic solutions (r, α, R, β of differential system (15 at every fixed energy level H = εh with h R is (3 x(t, ε = ε r cos t + O(ε 3/, y(t, ε = ε h r R cos(α t + O(ε 3/, z(t, ε = ε R cos(β t + O(ε 3/, p x (t, ε = ε r sin t + O(ε 3/, p y (t, ε = ε h + r R sin(α t + O(ε 3/, p z (t, ε = ε R sin(β t + O(ε 3/. Lemma 3 is proved in the end of section 3.. The averaging theory of First Order In this section we recall the basic results from averaging theory that we shall need for proving Theorem 1.

7 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 7 Consider the differential system (4 ẋ = εf 1 (t, x + ε F (t, x, ε, x(0 = x 0 with x D, where D is an open subset of R n, t 0. Furthermore we suppose that the functions F 1 (t, x and F (t, x, ε are T periodic in t. We define in D the averaged differential system (5 ẏ = εf 1 (y, y(0 = x 0, where (6 f 1 (y = 1 T T 0 F 1 (t, ydt. As we shall see under convenient assumptions, the equilibria solutions of the averaged system will provide T periodic solutions of system (4. Theorem 3. Consider the two initial value problems (4 and (5. Assume that (i the functions F 1, F 1 / x, F 1 / x, F and F / x are defined, continuous and bounded by a constant independent of ε in [0, D and ε (0, ε 0 ]; (ii the functions F 1 and F are T periodic in t (T independent of ε. Then the following statements hold. (a If p is an equilibrium point of the averaged system (5 satisfy ( f1 (7 det 0, y y=p then there is a T periodic solution φ(t, ε of system (4 such that φ(0, ε p as ε 0. (b The kind of stability or instability of the periodic solution φ(t, ε coincides with the kind of stability or instability of the equilibrium point p of the averaged system (5. The equilibrium point p has the kind of stability behavior of the Poincaré map associated to the periodic solution φ(t, ε. For a proof of theorem 3, see theorems 11.5, 11.6 and sections 6.3, 11.8 of Verhulst [1]. In order to apply the averaging theory we need to do some changes of variables in the Hamiltonian differential system ( in order to write it in the normal form of the averaging theory, i.e. to write it as system (4. Therefore first we do a scaling by a factor ε in order to introduce a small parameter ε > 0 in the Hamiltonian system. Second, using some generalized polar coordinates in R 6, and after taking as new independent variable an angle coordinate instead of the time, we get a π periodic differential system. Third, fixing the energy level and omitting a redundant variable in every energy level, we will obtain a differential system written in the normal form for applying the averaging theorem (Theorem 3, and finally we shall prove the existence of some isolated periodic solutions in every positive energy level. 3. Proof of theorem 1 We do a scaling using a small parameter ε > 0. In fact, we change in the Hamiltonian system ( the variables (x, y, z, p x, p y, p z by (X, Y, Z p X, p Y, p Z where

8 8 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, x = ε X, y = ε Y, z = ε Z, p x = ε p X, p y = ε p Y new variables, system ( becomes and p z = ε p Z. In the (8 Ẋ = p X, Ẏ = p Y, Ż = p Z, ṗ X = X εx(ax + by + cz, ṗ Y = Y εy (bx + dy + ez, ṗ Z = Z εz(cx + ey + fz. This differential system again is Hamiltonian with Hamiltonian (9 H = ε ( ( p X + p Y + p Z X + Y + Z + ε ax 4 + bx Y 4 +cx Z + dy 4 + ey Z + fz. 4 The original and the transformed systems ( and (8 have the same topological phase portrait because the change of variables is only a scale transformation for all ε > 0, and also system (8 for ε sufficiently small is close to an integrable one. The periodicity in the independent variable of the differential system is needed to apply the averaging theory, so we change the Hamiltonian (9 and its equations of motion (8 to a kind of generalized polar coordinates (r, θ, ρ, α, R, β in R 6. Defined by (10 X = r cos θ, Y = ρ cos(α θ, Z = R cos(β θ, p X = r sin θ, p Y = ρ sin(α θ, p Z = R sin(β θ. Of course in this change of variables r 0, ρ 0 and R 0. The first integral H in the new variables becomes (11 H = ε (ρ + R r + ε [ ar 4 cos 4 θ + br ρ cos θ 4 cos (α θ + dρ 4 cos 4 (α θ + R ( cr cos θ +eρ cos (α θ cos (β θ + fr 4 cos 4 (β θ ],

9 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 9 and the equations of motion (8 become ṙ = εr cos θ sin θ [ a r cos θ + b ρ cos (α θ + c R cos (β θ ], (1 θ = 1 ε cos θ [ a r cos θ + bρ cos (α θ +cr cos (β θ ], ρ = ερ cos(α θ sin(α θ [ b r cos θ +dρ cos (α θ + er cos (β θ ], [ ( α = ε a r cos 4 θ cos θ b(r + ρ cos (α θ ( +cr cos (β θ cos (α θ dρ cos (α θ ] +er cos (β θ, Ṙ = εr cos(β θ sin(β θ [ c r cos θ +eρ cos (α θ + fr cos (β θ ], β = [ ε a r cos 4 θ bρ cos θ cos (α θ ( c(r + R cos θ eρ cos (α θ ] cos (β θ fr cos 4 (β θ. We note that in this system if we take the variable θ as the new independent variable instead of t, we obtain the necessary periodicity for writing the system in the normal form of the averaging theory. In what follows the independent variable will be θ. This means that the new differential system will have only five equations. We denote by a prime the derivative with respect to θ and we expand system (1 in Taylor series in ε. Thus, system (1 becomes (13 r = ε r sin θ cos θ [ a r cos θ + bρ cos (α θ +cr cos (β θ ] + O(ε, ρ = ερ cos(α θ sin(α θ [ b r cos θ + dρ cos (α θ +er cos (β θ ] + O(ε, α = ε [ ar cos 4 θ dρ cos 4 (α θ er cos (α θ cos (β θ cos θ ( b(r + ρ cos (α θ + cr cos (β θ ] + O(ε, R = ε R sin(β θ cos(β θ [ c r cos θ +eρ cos (α θ + fr cos (β θ ] + O(ε, β = ε [ fr cos 4 (β θ cos θ ( ar cos θ +bρ cos (α θ cos (β θ ( c(r + R cos θ +eρ cos (α θ ] + O(ε. System (13 is π-periodic respect to the variable θ, i.e. it is written as the normal form (4 but it is not ready for applying the averaging theory, we must fix the value of the first integral H = εh with h R, otherwise the Jacobian (7 will be zero because the periodic orbits are non-isolated leaving on cylinders parameterized by the energy, see for more details [1].

10 10 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, Solving ρ from H = εh, we get two positive solutions, but the unique with physical meaning expanded in Taylor series in ε is (14 ρ = h + r R + O(ε. Since ρ 0 we need that h + r R 0. Substituting ρ in system (13 we get the differential system (15 r = ε r sin θ cos θ [ a r cos θ + b(h + r R cos (α θ +cr cos (β θ ] + O(ε, [ α = ε ar cos 4 θ + d(r h r cos 4 (α θ ( er cos (α θ cos (β θ + cos θ ] b(r h r cos (α θ cr cos (β θ + O(ε, R = ε R sin(β θ cos(β θ [ c r cos θ + e(h + r R cos (α θ + fr cos (β θ ] + O(ε, [ β = ε ar cos 4 θ + cos (β θ(e(h + r R ( cos (α θ fr cos (β θ cos θ ] b(h + r R cos (α θ + c(r + r cos (β θ + O(ε. Following the notation of the averaging theory given in section, the function F 1 = (F 11, F 1, F 13, F 14 of (4 is (16 F 11 = r sin θ cos θ [ a r cos θ + b(h + r R cos (α θ + cr cos (β θ ], F 1 = ar cos 4 θ d(h + r R cos 4 (α θ er cos (α θ cos (β θ cos θ ( b(h + r R cos (α θ + cr cos (β θ, F 13 = R sin(β θ cos(β θ [ c r cos θ + e(h + r R cos (α θ + fr cos (β θ ], F 14 = ar cos 4 θ e(h + r R cos (α θ cos (β θ fr cos 4 (β θ cos θ ( b(h + r R cos (α θ +c(r + r cos (β θ, where F 1j = F 1j (θ, r, α, R, β for j = 1,, 3, 4.

11 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 11 From (6 and (16 we compute the averaged function f 1 = (f 11, f 1, f 13, f 14 and we obtain f 11 = 1 8 r[ b(h + r R sin α + cr sin β ], f 1 = 1 8[ 6dh 3ar 3dr + 3dR + (bh br + br ( + cos α er ( + cos (α β cr ( + cos β ], f 13 = 1 8 R[ e(h + r R sin (α β cr sin β ], f 14 = 1 [ ( 3ar 3fR (h + r R (b + e + b cos α 8 ] +e cos (α β c(r + R ( + cos β, where f 1j = f 1j (r, α, R, β for j = 1,, 3, 4. According to theorem 3 our aim is to find the zeros (r, α, R, β of (17 f 1i (r, α, R, β = 0 for i = 1,, 3, 4, and after we must check that the Jacobian determinant (7 evaluated in these zeros are different from zero. Then from f 11 (r, α, R, β = 0 we obtain either r = 0, or b sin α 0 and r = R h cr sin β, or b sin α = 0. b sin α Case 1: r = 0. Substituting r in f 1i, for i =, 3, 4, we get f 1 (0, α, R, β = 1 8[ 6dh + 3dR + (bh + br ( + cos α er ( + cos (α β cr ( + cos β ], f 13 (0, α, R, β = 1 8 er(h + R sin (α β, f 14 (0, α, R, β = 1 4(b + eh + (b (c e 3fR 8[ ( (h R b cos α + e cos (α β cr cos β ]. From f 13 (0, α, R, β = 0 we have the following four subcases: e = 0, R = 0, R = h, α = β + kπ with k Z. Subcase 1.1: e = 0. The Jacobian is zero in this subcase because f 13 vanish and the averaging theory cannot provide information about the periodic orbits. Hence, in what follows in case 1 we assume that e 0. Subcase 1.: When R = 0, f 1 and f 14 become f 1 (0, α, 0, β = 1 4 h[ b + 3d + b cos α ], f 14 (0, α, 0, β = 1 4 h[ b + e + b cos α + e cos (α β ].

12 1 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, If b 0 then f 1 (0, α, 0, β = 0 implies α ± = ± 1 in f 14 (0, α, R, β we get f 14 (0, α, 0, β = 1 ( [e 4 h cos β + arccos Hence f 14 (0, α, 0, β = 0 implies β +± = 1 e 3d] arccos if e 0. e Substituting α in f 14 (0, α, 0, β we get f 14 (0, α, 0, β = 1 4 h [e cos + b arccos 3d. Substituting α + b 3d + b b [ 3d + b arccos b ( β arccos 3d + b b ] + 3d e. ± ] + 3d e. Therefore f 14 (0, α, 0, β = 0 implies β ± = 1 [ 3d + b arccos ± b e 3d] arccos if e 0. e Supposing that (18 h > 0, eb 0, 3d + b b < 1 and e 3d e < 1. System (17 has four zeros S = (r, α, R, β with ρ = h ( 0, 1 + b arccos 3d, 0, 1 [ 3d + b e 3d ] arccos ± arccos, b b e (19 ( 0, 1 + b arccos 3d, 0, 1 [ 3d + b e 3d ] arccos ± arccos, b b e which reduce to two solutions if either 3d + b b = 1, e 3d e < 1, h > 0 and eb 0, or (3d + b/b < 1, (e 3d/e = 1, h > 0 and eb 0. We have one solution if (3d + b/b = (e 3d/e = 1, h > 0 and eb 0. Now we check if the Jacobian of f 1 evaluated in these solutions is different from zero. By definition the Jacobian is f 11 f 11 f 11 f 11 r α R β f 1 f 1 f 1 f 1 r α R β J f1 = D rαrβ f 1 (S = f 13 f 13 f 13 f 13. r α R β f 14 f 14 f 14 f 14 r α R β [ So J f1 (S = 3 ( ] e 3d 64 h4 e (b + 3d(b + d 1. e (r,α,r,β=s Assuming that (b + d(b + 3d 0 and (18 hold. This supposition is not empty because the value b = 1, d = /3, e = 4/3, h = 1 satisfy it. Then J f1(s 0 and the four solutions (19 of system (17 provide only two periodic solutions of

13 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 13 differential system (15 because when R = 0 the two solutions of β provide the same initial conditions in (10. In the remaining subcases in what follows of all this paper we have two possibilities and in both of them we do not have results. The first one when the J f1(s = 0, therefore the averaging theory does not provide information about the periodic solution. The second one when the set of conditions on the parameters which guarantee the existence of the solutions is empty. If b = 0 then f 1 = 3dh/4= constant. The Jacobian will be zero and the averaging theory does not give results. If b 0 and e = 0 then f 14 = 3dh/4= constant. The Jacobian will be zero and the averaging theory does not give results. Subcase 1.3: R = h so h > 0. Then f 1 and f 14 become f 1 (0, α, h, β = 1 4 h[ e + c + c cos β + e cos (α β ], f 14 (0, α, h, β = 1 4 h[ c + 3f + c cos β ]. If c 0, solving f 14 = 0 we get β ± = ± 1 respectively in f 1 we have f 1 (0, α, h, β ± = h 4 [ ( e + 3f + e cos α ± arccos + c arccos 3f. Substituting β ± c ] 3f + c. c Solving f 1 = 0 with respect to α, if e 0 we obtain the four solutions α ± = 1 [ ] e 3f 3f + c ± arccos arccos. e c Assuming (0 h > 0, ec 0, (1 0, ± 1 ( e 3f e < 1 and 3f + c c < 1. System (17 has four solutions S = (r, α, R, β with ρ = 0 ( e 3f arccos 1 e 0, ± 1 arccos e 3f e + 1 3f + c arccos, h, 1 + c arccos 3f c c 3f + c arccos, h, 1 + c arccos 3f c c which reduce to two solutions if either h > 0, (e 3f/e < 1, ec 0 and (3f + c/c = 1 or (e 3f/e = 1 and (3f + c/c < 1. We have one solution if h > 0, ec 0 and (e 3f/e = (3f + c/c = 1. Computing the Jacobian on these solutions [ we get J f1(s = 3 ( ] e 3f 3 h4 e (c + 3f(c + f 1. e Supposing that (c + 3f(c + f 0 and (0 hold. This assumption is not empty because it is satisfied for the value c = 1, f = 3/4, e = 7/4, h = 1. Therefore J f1 (S 0 and the four zeros (1 of system (17 provide only two periodic solutions of differential system (15 because since ρ = 0 the two solutions of α provide the same initial conditions in (10. If c = 0 then f 14 = 3fh/4= constant.,,

14 14 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, If c 0 and e = 0 then f 1 = 3fh/4= constant. Subcase 1.4: α = β + kπ. We consider four subcases k = 0, 1,, 3. Due to the periodicity of the cosinus we study the cases k = 0 and k =, and the cases k = 1 and k = 3 together. Subcase 1.4.1: Assume that either k = 0 or k =, i.e. α = β + π. Then either α = β or f 1 (0, α, R, β = 1 8[ h(b + 3d + b cos β ( b c + 3d 3e+ (b c cos β R ], f 14 (0, α, R, β = 1 8[ 3fR + cr ( + cos β + (h R (b + 3e + b cos β ]. Subcase : D 1 = b c + 3d 3e + (b c cos β 0. Solving f 1 = 0 we get R = h (b + 3d + b cos β /D 1. Substituting R in f 14 and solving f 14 = 0 with respect to the variable β we get, if cd be ce + bf 0, β = ± 1 arccos[(cd + be + ce + 3e bf 3df/(cd be ce + bf]. d e + f 0, R = h(d e/(d e + f and ρ = h(f e/(d e + f. With the condition that ( cd be ce + bf 0, (d e + f 0, h(d e + f(f e > 0, h(d e + f(d e > 0 and 1 < 1, where 1 = (cd + be + ce + 3e bf 3df/(cd be ce + bf, system (17 for k = 0 and k = has four solutions S h(f e = (r, α, R, β with ρ = (d e + f ( h(d e 0, β, (d e + f, β = ±1 arccos 1, (3 ( h(d e 0, β + π, (d e + f, β = ±1 arccos 1, which reduce to two solutions if cd be ce + bf 0, (d e + f 0, h(d e + f(f e > 0, h(d e + f(d e > 0 and 1 = 1. Its Jacobian J f1 (S = 9eh 4 3(d e + f 3 (f e(d e(be + bf + cd ce + 3df 3e (be + bf + cd ce + df e. Assuming that e(be + bf + cd ce + 3df 3e (be + bf + cd ce + df e 0 and ( hold. This supposition is satisfied when b = 5, c = 0, d = 0, f = 1/, e = 1, h = 1. Then J f1 (S 0 and the four solutions (3 of system (17 provide four periodic solutions of differential system (15. If cd be ce + bf = 0 we get f 14 = 3h(be + 3e cd + ce bf 3df/4D 1. So either f 14 never is zero, or f 14 is identically zero. In both cases the averaging theory does not provide information. Let

15 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 15 Subcase : D 1 = 0. We have f 1 (0, α, R, β = f 14 (0, α, R, β = 1 4 h[ b + 3d + b cos β], 1 4 h[ b + 3e + b cos β + R ( b c + 3e 3f +(b c cos β ]. If b 0, solving f 1 = 0 we have β = ± 1 + b arccos 3d. Substituting β in f 14 b hb(d e and solving f 14 = 0 we get if bd cd be + bf 0, R = bd cd be + bf. Then h(cd bf ρ = bd cd be + bf. In case that (4 bd cd be + bf 0, b 0, 3d + b b < 1, h(bd cd be + bf(cd bf > 0 and hb(bd cd be + bf(d e > 0, system (17 for k = 0 and k = has four solutions S = (r, α, R, β with h(cd bf ρ = (bd cd be + bf (5 ( hb(d e 0, β, bd cd be + bf, β = ±1 + b arccos 3d, b ( hb(d e 0, β + π, bd cd be + bf, β = ±1 + b arccos 3d, b which reduce to two solutions if bd cd be + bf 0, b 0, 3d + b b = 1, h(bd cd be + bf(cd bf > 0 and hb(bd cd be + bf(d e > 0. 9beh 4 [ J f1(s = (b + d(b + 3d(d e(d e + f(bf cd(ce 3(bd cd be + bf 4 bf ]. Assuming that e(b + d(b + 3d(d e + f(ce bf 0 and (4 hold. This supposition is not empty because the value b = 3, c = 5, d =, f = 3, e = 1, h = 1 satisfy it. Then J f1 (S 0 and the four solutions (5 of system (17 provide four periodic solutions of differential system (15. If b = 0 we have f 1 = 3dh/4=constant. If b 0 and bd cd be + bf = 0 we get f 14 = 3(d eh/4=constant.

16 16 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, Subcase 1.4.: Assume that either k = 1 or k = 3, i.e. either α = β + π or α = β + 3π then f 1 (0, α, R, β = 1 8[ h(b + 3d + b cos β + (c b 3d + e+ (b + c cos βr ], f 14 (0, α, R, β = 1 8[ (b + eh + (b c + e 3fR + ( bh (b + cr cos β ]. Subcase : D = c b 3d + e + b cos β + c cos β 0. So solving f 1 (0, α, R, β = 0 we get R = h (b 3d b cos β /D. Substituting R in f 14 and solving f 14 = 0 with respect to the variable β we obtain, if 3cd + be ce 3bf 0, β = ± 1 arccos where = be + ce + e 6cd 6bf 9df. Then 3cd + be ce 3bf h(3d e h(3f e R = 3d e + 3f and ρ = 3d e + 3f. Conceding that (6 3cd + be ce 3bf 0, 3d e + 3f 0, h(3f e(3d e + 3f > 0, h(3d e(3d e + 3f > 0 and < 1. System (17 for k = 1 or k = 3 has four solutions S = (r, α, R, β with h(3f e ρ = 3d e + 3f ( 0, β + π h(3d e, 3d e + 3f, β = ±1 arccos, (7 ( 0, β + 3π h(3d e, 3d e + 3f, β = ±1 arccos, which reduce to two solutions if 3cd + be ce 3bf 0, 3d e + 3f 0, h(3f e(3d e + 3f > 0, h(3d e(3d e + 3f > 0 and = 1. The Jacobian is J f1 (S = eh4 (3f e(3d e 3(3d e + 3f 3 (9cd + be + 3ce + e 3bf 9df (3cd 3be ce e + 9bf + 9df. Assuming that (9cd+be+3ce+e 3bf 9df(3cd3becee +9bf +9df 0 and (6 hold. This supposition is not empty because the value b = 5/8, c = 1, d = 1, f = 1, e = 0, h = 1 satisfy it. Therefore J f1(s 0 and the four solutions (7 of system (17 provide four periodic solutions of differential system (15. When 3d e + 3f and 3cd + be ce 3bf = 0 we obtain f 14 = h(6cd be ce e + 6bf + 9df/(4D. So either f 14 never is zero, or f 14 is identically zero. In both cases the averaging theory does not provide information. Subcase 1.4..: D = 0. Then f 1 = h 4 (b cos β b 3d. Solving f 1 = 0 when b 0 we get β = ± 1 arccos(3d + b/b. Substituting β in f 14 and solving

17 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 17 f 14 = 0 if 4bc+3bd+3cdbe+3bf 0 we obtain R = h(4bc + 3cd + 3bf and ρ = 4bc + 3bd + 3cd be + 3bf. Assuming that hb(3d e 4bc + 3bd + 3cd be + 3bf (8 4bc + 3bd + 3cd be + 3bf 0, b 0, hb(4bc + 3bd + 3cd be + 3bf(3d e > 0, h(4bc + 3bd + 3cd be + 3bf(4bc + 3cd + 3bf > 0, 3d + b and < 1. b System (17 for k = 1 and k = 3 has four solutions S = (r, α, R, β with h(4bc + 3cd + 3bf ρ = 4bc + 3bd + 3cd be + 3bf (9 (0, β + π, hb(3d e 4bc + 3bd + 3cd be + 3bf, β = ±1 (0, β + 3π, b + 3d arccos, b hb(3d e 4bc + 3bd + 3cd be + 3bf, β = ±1 b + 3d arccos b, which reduce to two solutions when 4bc + 3bd + 3cd be + 3bf 0, b 0, h(4bc + 3bd+3cdbe+3bf(4bc+3cd+3bf > 0, h(4bc+3bd+3cdbe+3bfhb(3de > 0 and b + 3d b = 1. We have J f1 (S = 3beh 4 (b + d(b + 3d (3d e(3d e + 3f(4bc + 3cd 3(4bc + 3bd + 3cd be + 3bf 4 +3bf(4bc + ce + 3bf. In case that e(b + d(b + 3d(3d e + 3f(4bc + ce + 3bf 0 and (8 hold, this supposition is true for the value b =, c = 8, d = 1, f = 6, e = 9, h = 1. Then we have J f1 (S 0 and the four solutions (9 of system (17 provide four periodic solutions of differential system (15. If b = 0, f 1 = 3dh/4=constant. If b 0 and 4bc+3bd+3cdbe+3bf = 0 we get f 14 = h(3de/(4b=constant.

18 18 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, Case : b sin α 0, r = R h cr sin β b sin α. Then 1 [ f 1 (r, α, R, β = 4b(6ah + 4bh (3a + (b + c + er 3b +b(h R cos α 4bR (c + e cos α cos β + R sin β (6ac + 8bc + 6cd be + 4bc cos α sin α cos α +be cos 4α ], f 13 (r, α, R, β = f 14 (r, α, R, β = cr sin β [ R h [b sin α c sin β 4 b sin α +e sin (α β] ], [ 1 8b(3a + ch R (6ab + 8bc ce + 6bf 3b [ +8bc(h R cos β + cr e cos 4β 1 [ + (c + e cos α sin 4β + sin β sin α cos α +(b cos α + 6a + 4b + 4c + 4e ]]]. So if D 3 = b sin α c sin β + e sin (α β = 0, c = 0, R = 0, f 13 (r, α, R, β = 0 β = kπ with k Z. and if D 3 0 we get f 13 (r, α, R, β = 0 R = bh sin α D 3. Subcase.1: D 3 = 0 and c = 0. No information as in subcase 1.1. Hence, in what follows in the rest of case we assume that c 0. Subcase.: D 3 = 0 and R = 0. Then r = h, h < 0 and ρ = 0. f 1 (r, α, R, β = 1 4[ h(3a + b + b cos α ], f 14 (r, α, R, β = 1 4[ h(3a + c + c cos β ]. If b 0 and c 0, solving f 1 = f 14 = 0 we get α = ± 1 + b arccos 3a and β = ± 1 b With the condition that (30 h < 0, bc 0, 3a + b b + c arccos 3a. c < 1 and 3a + c c < 1,

19 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 19 system (17 has four solutions S = (r, α, R, β with ρ = 0 (31 ( h, ± 1 + b arccos 3a, 0, ± 1 b + c arccos 3a, c 3a + b which reduce to two solutions if either h < 0, bc 0, b < 1 and 3a + c c = 3a + b 1 or h < 0, bc 0, b = 1 and 3a + c < 1, and to one solution if h < 0, c 3a + b bc 0, b = 1 and 3a + c c = 1. The Jacobian J f1(s = 9h4 (a + b(3a + b(a + c(3a + c. 3 Assuming that D 3 = 0, (a + b(3a + b(a + c(3a + c 0 and (30 hold. This supposition is true for the value a = (/3, b = 1, c = 4/3, h = 1. Then J f1(s 0 and the four solutions (31 of system (17 provide only one periodic solution of differential system (15 because when R = 0 and ρ = 0 the two solutions of both α and β provide the same initial conditions in (10. If b = 0, f 1 = 3ah/4=constant. If c = 0, f 14 = 3ah/4=constant. Subcase.3: D 3 = 0 and β = kπ. Due to the periodicity of the cosinus we study the cases k = 0 and k =, and the cases k = 1 and k = 3, together. Subcase.3.1: Assume that either k = 0 or k =, i.e. either β = 0 or β = π. Then substituting β in r we obtain r = R h. f 1 and f 14 become f 1 (r, α, R, β = 1 8[ 6ah + 4bh (3a + b + 3c + er +(bh (b + er cos α ], f 14 (r, α, R, β = 3 8[ (a + ch (a + c + fr ]. If a + c + f 0, solving f 14 = 0 with respect to R we obtain R = h(a + c/a + c + f. Substituting R in f 1 and assuming that bc ae ce + bf 0 we obtain solving f 1 = 0 with respect to α, α = ± 1 arccos 3, where 3 = 3c bc + ae + ce 3af bf. Substituting R in r and in ρ we obtain bc ae ce + bf h(c + f r = and ρ = 0. a + c + f Supposing that (3 a + c + f 0, (bc ae ce + bf 0, h(a + c + f(c + f < 0, h(a + c + f(a + c > 0 and 3 < 1.

20 0 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, System (17 for k = 0 and k = has four zeros solutions S = (r, α, R, β with ρ = 0 (33 ( h(c + f h(a + c a + c + f, ±1 arccos 3, a + c + f, 0, ( h(c + f h(a + c (a + c + f, ±1 arccos 3, a + c + f, π, which reduce to two solutions if (bc ae ce + bf 0, a + c + f 0, 3 = 1, h(a + c + f(c + f < 0 and h(a + c + f(a + c > 0. The Jacobian is J f1(s = 9ch 4 16(a + c + f 3 (a + c(c + f( bc c ae ce + af + bf ( bc 3c ae ce + 3af + bf. In the case that (bc c ae ce + af +bf ( bc 3c ae ce + 3af + bf 0 and (3 hold, this supposition is not empty because the value a =, b = 3, c = 1, f = 1, e = 0, h = 1 satisfy it. Then J f1 (S 0 and the four zeros (33 of system (17 provide only two periodic solutions of differential system (15 because since ρ = 0 the two solutions of α provide the same initial conditions in (10. If a + c + f = 0, f 14 = 3h(a + c/4=constant. If a + c + f 0 and bc + bf ae ce = 0 we get f 1 = h(3af + b(c + f 3c ae ce/4(a + c + f=constant. Subcase.3.: Assume that either k = 1 or k = 3, i.e. either β = π or β = 3π. Then substituting β in r we get r = R h. f 1 and f 14 become f 1 (r, α, R, β = 1 8[ 6ah + 4bh (3a + b + c + er +(bh br + er cos α ], f 14 (r, α, R, β = 1 8[ h(3a + c (3a + c + 3fR ]. Solving f 14 = 0 we get if 3a + c + 3f 0, R = h(3a + c 3a + c + 3f. Substituting R in f 1 and solving f 1 = 0, we obtain if bc + 3ae + ce + 3bf 0, α = ± 1 arccos 4 where 4 = (bc + c + 6ae + ce 9af 6bf/bc + 3ae + ce + 3bf. Substituting h(c + 3f R in r and in ρ we have r = and ρ = 0. 3a + c + 3f Assuming that (34 3a + c + 3f 0, bc + 3ae + ce + 3bf 0, 4 < 1, h(3a + c + 3f(c + 3f < 0, and h(3a + c + 3f(3a + c > 0.

21 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 1 System (17 for k = 1 and k = 3 has four solutions S = (r, α, R, β with ρ = 0 ( h(c + 3f h(3a + c 3a + c + 3f, ±1 arccos 4, 3a + c + 3f, π, (35 ( h(c + 3f h(3a + c 3a + c + 3f, ±1 arccos 4, 3a + c + 3f, 3π, which reduce to two solutions if 3a + c + 3f 0, bc + 3ae + ce + 3bf 0, 4 = 1, h(3a + c + 3f(c + 3f < 0 and h(3a + c + 3f(3a + c > 0. The Jacobian is J f1(s = ch4 (3a + c(c + 3f (163a + c + 3f 3 ( 3bc + c + 3ae + ce 9af 9bf ( bc + c + 9ae + 3ce 9af 3bf. Conceding that c ( 3bc + c + 3ae + ce 9af 9bf ( bc + c + 9ae + 3ce 9af 3bf 0 and (34 hold. This assumption is not empty because the value a = 1, b =, c = 1, f = 1, e = 0, h = 1 satisfy it. Therefore J f1 (S 0 and the four zeros (35 of system (17 provide only two periodic solutions of differential system (15 because since ρ = 0 the two solutions of α provide the same initial conditions in (10. If 3a + c + 3f = 0, we have f 14 = h(3a + c/4=constant. If 3a + c + 3f 0 and bc + 3ae + ce + 3bf = 0, then f 1 = h(9af + b(c + 3f c 6ae ce/4(3a + c + 3f=constant. bh sin α eh sin (β α Subcase.4: D 3 0 and R =. Then r =. D 3 D 3 f 1, f 14 become f 1 (r, α, R, β = f 14 (r, α, R, β = h 4D 3 [ (3ae bc + be sin (α β + (bc be +3cd sin β b(c + e sin α ], h 4D 3 [ (bc + ce 3bf sin α + (bc + 3ae + ce sin (α β + c(b + e sin β ]. To calculate the zeros of these two last functions we need their numerators h [ (3ae bc + be sin (α β + (bc be + 3cd sin β b(c + e sin α ], h [ (bc + ce 3bf sin α + (bc + 3ae + ce sin (α β + c(b + e sin β ]. Expanding the trigonometrical terms of these numerators and using the notation sin α = s; cos α = ± 1 s ; sin β = S; cos β = ± 1 S we obtain using the sign + for cos α and cos β P 1 (s, S = hs 1 s ( 6aeS + 3ae + bcs 3bc 4beS hs 1 S ( 6aes + 3ae + bcs 3bc 4bes +3be 3cd, P 14 (s, S = hs 1 s ( 6aeS + 3ae + bcs 3bc 3bf 4ceS + 3ce hs 1 S ( 6aes + 3ae + bcs 3bc 4ces.

22 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, Note that the other three subcases provide, taking into account the different signs, the same zeros than the system P 1 (s, S = P 14 (s, S = 0. This last system is equivalent to the system Q 1 (s, S = Q 14 (s, S = 0 where (36 Q 1 (s, S = 4h S ( 1 S ( 6aes + 3ae + bcs 3bc 4bes + 3be 3cd 4h s ( 1 s ( 6aeS + 3ae + bcs 3bc 4beS, Q 14 (s, S = 4h S ( 1 S ( 6aes + 3ae + bcs 3bc 4ces 4h s ( 1 s ( 6aeS +3ae + bcs 3bc 3bf 4ceS + 3ce. Calculating the resultant of Q 1 and Q 14 with respect to s and S, we obtain (37 R 1 (S R 14 (s = h 16 (1 + S 4 S 8 (1 + S 4 T (SU (S, = h 16 (1 + s 4 s 8 (1 + s 4 V (sw (s, with T(S and V(s two polynomials of the form AS + B and Cs + D respectively with A, B, C, D constants, and U(S and W(s two polynomials of the form ES 4 + F S + G and Hs 4 + Is + J respectively with E, F, G, H, I, J constants. Solving (37 we obtain 81 pairs (s, S. Only 9 of these pairs are solutions of (36. When we calculate (α,β corresponding to an (s, S solution we find the zeros S = (r, ρ, α, R, β of (17 ( eh ch S1,±,± = c e, c e, ±π, 0, ±π ; ( eh ch S,± = c e, c e, ±π, 0, 0 ; ( eh h(c e S3,± = c e,, 0, 0, ± π ; c e ( eh h(c e S4 = c e,, 0, 0, 0. c e The values of r and R are not well defined in these solutions. So the averaging theory in subcase.4 does not give information. Case 3: b sin α = 0. Then b = 0 or sin α = 0. Subcase 3.1: b = 0. Then f 11 = crr sin(β/8. Solving f 11 = 0 we obtain the following four subcases: c = 0, r = 0 (studied in case 1, R = 0, β = mπ with m Z. Subcase 3.1.1: c = 0. No information as in subcase 1.1. Hence in what follows in subcase 3.1 we assume that c 0. Subcase 3.1.: R = 0. This subcase does not give results because f 13 and f 11 will be zero. Subcase β = mπ with m Z. Due to the periodicity of the sinus we study the subcases m = 0 and m =, and the subcases m = 1 and m = 3 together.

23 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 3 Subcase : Assume that either β = 0 or β = π. So f 13 = er 8 (h + r R sin α. e = 0, R = 0, f 13 (r, α, R, β = 0 R = h + r, α = nπ with n Z. Subcase : e = 0. No information as in subcase 1.1. Hence, in what follows in subcase we assume that e 0. Subcase : R = 0. Then we have f 1 = 3 8 [dh + (a + dr ], f 14 = 1 8 [(3a + 3c + e + e cos αr + eh( + cos α]. dh ah If a + d 0, f 1 = 0 r = a + d. So ρ = a + d. Substituting r in f 14 and solving f 14 = 0 if ae 0, we get α = ± 1 3ad + 3cd ae arccos. ae In the case that 3ad + 3cd ae ae < 1, ae(a + d 0, hd(a + d < 0 and ah(a + d > 0, system (17 for m = 0 and m = has four zeros S = (r, α, R, β ah with ρ = a + d ( dh ( 3ad + 3cd ae a + d, ±1 arccos, 0, 0, ae ( dh ( 3ad + 3cd ae a + d, ±1 arccos, 0, π, ae which reduce to two zeros when ae(a + d 0, hd(a + d < 0, ah(a + d > 0 and 3ad + 3cd ae ae = 1. But when the Jacobian is evaluated on these solutions it becomes zero so the averaging theory does not give information in this subcase. If a + d = 0, we have f 1 = 3dh/4=constant. If a + d 0 and ae = 0, f 14 = 3hd(a + c 4(a + d =constant. Subcase : R = h + r. Studied in the subcase.4.1. Subcase : α = nπ. Due to the periodicity of the sinus we study the cases n = 0 and n =, and the cases n = 1 and n = 3 together. Subcase : Assume that either n = 0 or n =, i.e. either α = 0 or α = π. f 1 = 1 [ 6dh + 3(a + dr + 3(c d + er ], 8 f 14 = 1 8 [ 6eh + 3(a + c + er 3(e c fr ].

24 4 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, eh If a+c+e = 0 and (ecf 0, solving f 14 = 0 we get R = e c f. Substituting R in f 1 and solving f 1 = 0 if a+d 0, we have r = h(cd + df ce e (a + d(e c f h(ce + e + ac + af and ρ =. (a + d(e c f Supposing that (38 a + c + e = 0, e c f 0, eh(e c f > 0 h(cd + df ce e (a + d(e c f > 0 and h(ce + e + ac + af(a + d(e c f < 0. System (17 for n = 0 and n = has the following four zeros S = (r, α, R, β with ρ = h(ce + e + ac + af (a + d(e c f (39 ( h(cd + df ce e eh, 0, (a + d(e c f e c f, 0, ( h(cd + df ce e eh, π, (a + d(e c f e c f, 0, ( h(cd + df ce e eh, 0, (a + d(e c f e c f, π, ( h(cd + df ce e eh, π, (a + d(e c f e c f, π. The Jacobian J f1 (s = 9ce 3 h 4 16(a + d (e c f 4 (ce + e + ac + af(cd + df ce e ( ad c + cd ae ce e + af + df. With the condition ce ( ad c + cd ae ce e + af + df 0 and (38 we have J f1 (S 0. The condition is not empty because the value a =, c = 1, d = 3, f = 1, e = 1, h = 1 satisfy it. Therefore the four zeros (39 of (17 provide four periodic solutions of (15. If a + c + e = 0 and (e c f = 0, f 14 = 3eh/4. If a + c + e = 0, (e c f 0 and a + d = 0, f 1 = 3h(ce + e dc df/(4(e c f=constant. If a + c + e 0 and (e c f = 0, solving f 14 = 0 we get r = Substituting r in f 1 and solving f 1 = 0 if c d + e 0, we have h(ae ad cd hc R = (a + c + e(c d + e and ρ = c d + e. Assuming that (40 a + c + e 0, c e + f = 0, c d + e 0, eh(a + c + e < 0, hc(c d + e > 0 and h(ae ad cd(a + c + e(c d + e > 0. eh a + c + e.

25 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 5 System (17 for n = 0 and n = has the following four zeros S = (r, α, R, β with hc ρ = c d + e (41 ( eh a + c + e, 0, h(ae ad cd (a + c + e(c d + e, 0, ( eh a + c + e, π, h(ae ad cd (a + c + e(c d + e, 0, ( eh a + c + e, 0, h(ae ad cd (a + c + e(c d + e, π, ( eh a + c + e, π, h(ae ad cd (a + c + e(c d + e, π. The Jacobian J f1(s = 9c e h 4 16(a + c + e 3 (c d + e 3 (ad + cd ae (c ad cd +ae + ce + e af df. With the condition ce ( c ad cd + ae + ce + e af df 0 and (40, we have J f1(s 0. The set of conditions is not empty because it is satisfied for the value a = 9/4, c = 1, d = 4, e =, f = 1, h = 1. Then the four zeros (41 of (17 provide four periodic solutions of (15. If a + c + e 0, (e c f = 0 and c d + e = 0, f 1 = 3h(ad + cd ae/(4(a + c + e=constant. If a + c + e 0 and e c f 0. Then solving f 14 = 0 we get r = (e c fr he. Substituting r in f 1 and solving f 1 = 0 we have, if a + c + e Σ = c ad cd + ae + ce + e af df 0, h(ae ad cd h(cd + df ce e R = then r = and Σ Σ h(ce + c + ae af ρ =. Σ Wherever (4 Σ = c ad cd + ae + ce + e af df 0, a + c + e 0, e c f 0, h(cd + df ce e Σ > 0, h(ae ad cdσ > 0 and h(ce + c + ae afσ > 0,

26 6 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, system (17 for n = 0 and n = has four zeros h(ce + c S + ae af = (r, α, R, β with ρ = Σ ( h(cd + df ce e h(ae ad cd, 0,, 0, Σ Σ ( h(cd + df ce e h(ae ad cd, π,, 0, Σ Σ (43 ( h(cd + df ce e h(ae ad cd, 0,, π, Σ Σ ( h(cd + df ce e h(ae ad cd, π,, π. Σ Σ The Jacobian is J f1 (s = 9ceh4 16Σ 3 (cd + df ce e (ce + c + ae af(ae ad cd. Assuming that ce 0 and (4 hold. This supposition is not empty because the value a = 5, c =, d =, f =, e = 1, h = 1, satisfy it. Then J f1 (S 0 and the four zeros (43 of (17 provide four periodic solutions of (15. If a + c + e 0, e c f 0 and Σ = 0 we have f 1 = 3h(ad + cd ae/(4(a + c + e=constant. Subcase : Assume that either n = 1 or n = 3, i.e. either α = π or 3π. Then f 1 = 1 8[ 6dh + 3(a + dr + (3c 3d + er ], f 14 = 1 8[ eh + (3a + 3c + er + (3c + 3f er ]. If 3c 3d + e 0 and a + d = 0, solving f 1 = 0 we get 6dh R = 3c 3d + e. Substituting R in f 14 and solving f 14 = 0 with respect to R h(9cd 3ce e + 9df if 3a + 3c + e 0, we obtain r = (3c 3d + e(3a + 3c + e, 6h(3ac + ae + 3c + 3cd + ce + 3df ρ =. (3c 3d + e(3a + 3c + e Whenever (44 3c 3d + e 0, a + d = 0, 3a + 3c + e 0, dh(3c 3d + e < 0, h(9cd 3ce e + 9df(3c 3d + e (3a + 3c + e > 0 and h(3ac + ae + 3c + 3cd + ce + 3df (3c 3d + e(3a + 3c + e > 0, system (17 for n = 1 and n = 3 has four zeros S = (r, α, R, β with

27 ρ = FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 7 6h(3ac + ae + 3c + 3cd + ce + 3df (3c 3d + e(3a + 3c + e (45 ( h(9cd 3ce e + 9df (3c 3d + e(3a + 3c + e, π, 6dh (3c 3d + e, 0, ( h(9cd 3ce e + 9df (3c 3d + e(3a + 3c + e, 3π, 6dh (3c 3d + e, 0, ( h(9cd 3ce e + 9df (3c 3d + e(3a + 3c + e, π, 6dh (3c 3d + e, π, ( h(9cd 3ce e + 9df (3c 3d + e(3a + 3c + e, 3π, 6dh (3c 3d + e, π. Its Jacobian is J f1(s = 7cd eh 4 16(3a + 3c + e (3c 3d + e 4 (9cd 3ce e + 9df(3ac +ae + 3c + 3cd + ce + 3df(9c 9ad 18cd + 6ae + 6ce +e 9af 9df. In case that cde(9c 9ad18cd+6ae+6ce+e 9af 9df 0 and (44 hold, the set of these conditions is not empty because the value a =, c = 1, d =, f = 0, e = 1, h = 1, satisfy it. Therefore we have J f1 (S 0 and the four zeros (45 of (17 provide four periodic solutions of differential system (15. If 3c 3d + e = 0 and a + d = 0 we get f 1 = 3dh/4=constant. If 3c 3d + e 0, a + d = 0 and 3a + 3c + e = 0 we have f 14 = h(9cd 3ce e + 9df/(4(3c 3d + e=constant. dh If 3c 3d + e = 0 and a + d 0, f 1 = 0 r = a + d. Substituting r in f 14 and solving f 14 = 0 with respect to R if 3c e + 3f 0 we obtain h(3ad + 3cd ae R = (a + d(3c e + 3f, ρ = 6h(ac ad + af cd (a + d(3c e + 3f. Assuming that (46 3c 3d + e = 0, a + d 0, 3c e + 3f 0, dh(a + d < 0, h(3ad + 3cd ae(a + d(3c e + 3f > 0 and h(ac ad + af cd(a + d(3c e + 3f > 0. System (17 for n = 1 and n = 3 has four zeros S = (r, α, R, β with

28 8 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, ρ = 6h(ac ad + af cd (a + d(3c e + 3f (47 ( dh a + d, π, h(3ad + 3cd ae (a + d(3c e + 3f, 0, ( dh a + d, 3π, h(3ad + 3cd ae (a + d(3c e + 3f, 0, ( dh a + d, π, h(3ad + 3cd ae (a + d(3c e + 3f, π, ( dh a + d, 3π, h(3ad + 3cd ae (a + d(3c e + 3f, π. The Jacobian is J f1(s = 3cdeh 4 16(3c e + 3f 3 (a + d 4 (3ad + 3cd ae (ac ad + af cd(9c 9ad 18cd + 6ae + 6ce + e 9af 9df. With the condition that cde(9c 9ad 18cd + 6ae + 6ce + e 9af 9df 0 and (46 hold, we have J f1 (S 0 and the four zeros (47 of (17 provide four periodic solutions of differential system (15. The set of conditions is not empty because it is satisfied for the value a =, c = /3, d = 1, f = 1, e = 1, h = 1. If a + d 0, 3c 3d + e = 0 and 3c e + 3f = 0 we get f 14 = (3ad + 3cd aeh/(4(a + d=constant. 6dh + (3c 3d + er If a + d 0 and 3c 3d + e 0, we have r =. 3(a + d Substituting r in f 14 and solving f 14 = 0 if Σ 1 = 9c 9ad 18cd + 6ae + 6ce + e 9af 9df 0, 6h(3ad + 3cd ae we have R =, r = Σ 1 6h(3c + ce + ae 3af. Σ 1 Conceding that h(9cd 3ce e + 9df Σ 1 and ρ = (48 Σ 1 = 9c 9ad 18cd + 6ae + 6ce + e 9af 9df 0, a + d 0, 3c 3d + e 0, h(9cd 3ce e + 9dfΣ 1 > 0, h(3c + ce + ae 3afΣ 1 > 0 and h(3ad + 3cd aeσ 1 < 0.

29 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 9 System (17 for n = 1 and n = 3 has four zeros S = (r, α, R, β with ρ = 6h(3c + ce + ae 3af/Σ 1 (49 ( h(9cd 3ce e + 9df, π 6h(3ad + 3cd ae Σ 1,, 0, Σ 1 ( h(9cd 3ce e + 9df, 3π 6h(3ad + 3cd ae Σ 1,, 0, Σ 1 ( h(9cd 3ce e + 9df, π 6h(3ad + 3cd ae Σ 1,, π, Σ 1 ( h(9cd 3ce e + 9df, 3π 6h(3ad + 3cd ae Σ 1,, π. Σ 1 Its Jacobian is J f1 (s = 7ceh4 16Σ 1 3 (3ad + 3cd ae (3c + ae + ce 3af(9cd 3ce e + 9df. Supposing ce 0 and (48 hold. This assumption is not empty because the value a = 1/, c = 1, d = 1, f = 3, e = 1, h = 1 satisfy it. Then J f1 (S 0 and the four zeros (49 of system (17 provide four periodic solutions of differential system (15. If a + d 0, 3c 3d + e 0 and Σ 1 = 0 we get f 14 = h(3ad + 3cd ae/(4(a + d=constant. Subcase : Assume that either β = π or β = 3π. So if f 13 = er 8 (R h r sin α = 0, then consequently one of the following four subcases holds: e = 0, R = 0, R = h + r and α = lπ/ with l Z. Subcase : e = 0. No information as in subcase 1.1. So in what follows in this subcase we assume that e 0. Subcase : R = 0. Then f 1 = 3 8 [dh + (a + dr ], f 14 = 1 8 [4eh + (3a + c + er e(h + r cos α]. If a+d 0 solving f 1 = 0, we obtain r = dh/(a + d and ρ = ah/(a + d. Substituting r in f 14 and solving f 14 = 0 if ae 0, we get α = ± 1 ae 3ad cd arccos. Therefore when hd(a + d < 0, ah(a + d > 0 and ae ae 3ad cd ae < 1, system (17 has four zeros S = (r, α, R, β with ρ = ah/(a + d ( dh a + d, ±1 ae 3ad cd arccos, 0, π ae,

30 30 FATIMA EZZAHRA LEMBARKI AND J. LLIBRE, ( dh a + d, ±1 ae 3ad cd arccos, 0, 3π, ae ae 3ad cd which reduces to two zeros if hd(a+d < 0, ah(a+d > 0 and ae = 1. Evaluating the Jacobian on these zeros we get zero so the averaging theory does not give information in this subcase. If a + d = 0, we have f 1 = 3dh/4=constant. If a + d 0 and ae = 0, we get f 14 = hd(3a + c/(4(a + d=constant. Subcase : R = h + r. Studied in the subcase.4.. Subcase : α = lπ with l Z. Due to the periodicity of the sinus we study the cases l = 0 and l =, and the cases l = 1 and l = 3 together. Subcase : Assume that either l = 0 or l =, i.e. either α = 0 or α = π. Then f 1 = 1 [ 6dh + 3(a + dr + (c 3d + er ], 8 f 14 = 1 [ he + (3a + c + er + (c e + 3fR ]. 8 he If 3a + c + e = 0 and c e + 3f 0, solving f 14 = 0, we obtain R = c e + 3f. Substituting R in f 1 and solving f 1 = 0 with respect to r if a + d 0 we get h(ce + e 9df 3cd h(ce + e + 3ac + 9af r = and ρ =. 3(a + d(c e + 3f 3(a + d(c e + 3f Supposing that (50 3a + c + e = 0, c e + 3f 0, he(c e + 3f < 0, h(a + d(c e + 3f(ce + e 9df 3cd > 0, h(a + d (c e + 3f(ce + e + 3ac + 9af > 0 and a + d 0. System (17 for l = 0 and l = has four zeros S = (r, α, R, β with ρ = h(ce + e + 3ac + 9af 3(a + d(c e + 3f (51 ( h(ce + e 9df 3cd he, 0, 3(a + d(c e + 3f c e + 3f, π, ( h(ce + e 9df 3cd he, π, 3(a + d(c e + 3f c e + 3f, π, ( h(ce + e 9df 3cd he, 0, 3(a + d(c e + 3f c e + 3f, 3π, ( h(ce + e 9df 3cd he, π, 3(a + d(c e + 3f c e + 3f, 3π.

31 FRIEDMANN-ROBERTSON-WALKER HAMILTONIAN SYSTEM 31 The Jacobian is J f1 (s = ce 3 h 4 144(a + d (c e + 3f 4 (3ac + ce + e + 9af(3cd ce e + 9df ( 9ad c + 6cd 6ae ce e + 9af + 9df. Conceding that ce ( 9ad c + 6cd 6ae ce e + 9af + 9df 0 and (50 hold. This assumption is not empty because it is satisfied for the value a = /3, d = 1, c = 1, f = 1, e = 1, h = 1. Then J f1(s 0 and the four zeros (51 of system (17 provide four periodic solutions of differential system (15. If 3a + c + e = 0 and c e + 3f = 0 we have f 14 = eh/4=constant. If 3a + c + e = 0, c e + 3f 0 and a + d = 0, we get f 1 = h(ec + e 6dc 18df/(8(c e + 3f=constant. If 3a + c + e 0 and c e + 3f = 0, solving f 14 = 0, we obtain r = he/(3a + c + e. Substituting r in f1 and solving f 1 = 0 with respect to 6h(3ad + cd ae ch R if 3d c e 0 we get R = (3d c e(3a + c + e and ρ = 3d c e. Supposing that (5 3a + c + e 0, c e + 3f = 0, 3d c e 0, he(3a + c + e < 0, hc(3d c e < 0. and h(3ad + cd ae(3d c e(3a + c + e > 0. System (17 for l = 0 and l = has four zeros S = (r, α, R, β with ρ = ch 3d c e ( he 3a + c + e, 0, 6h(3ad + cd ae (3d c e(3a + c + e, π, ( he 3a + c + e, π, 6h(3ad + cd ae (3d c e(3a + c + e, π, (53 ( he 3a + c + e, 0, 6h(3ad + cd ae (3d c e(3a + c + e, 3π, ( he 3a + c + e, π, 6h(3ad + cd ae (3d c e(3a + c + e, 3π. The Jacobian is J f1 (s = 9c e h 4 (3ad + cd ae (c 9ad 6cd + 6ae + ce + e 9af 9df 16(3d c e 3 (3a + c + e 3. Supposing that ce ( 9ad c + 6cd 6ae ce e + 9af + 9df 0 and (5 hold. This assumption is not empty because it is satisfied for the value a = 3/, d =, c = 1, f = 1, e = 4, h = 1. Then J f1 (S 0 and the four zeros (53 of system (17 provide four periodic solutions of differential system (15. If 3a + c + e 0, c e + 3f = 0 and 3d c e = 0, we get f 1 = 3h(3ad + cd ae/(4(3a + c + e=constant.

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