Heteroclinic, homoclinic and closed orbits in the Chen system

Bounded orbits such as closed, homoclinic and heteroclinic orbits are discussed in this work for a Lorenzlike 3D nonlinear system. For a large spectrum of the parameters the system has neither closed nor homoclinic orbits but has exactly two heteroclinic orbits, while under other constraints the system has symmetrical homoclinic orbits.

In this paper we first refine some results reported in [7] and present a different proof of these results.Secondly, we consider an important case not treated in [7] and show the system has homoclinic orbits for a large spectrum of the parameters, by transforming the system into a new form and using results reported in [1].Proving existence of homoclinic or heteroclinic orbits in nonlinear ode systems is in general a difficult task [2], [5], [6], [12], [15], [16].
Recall that, if x 0 is an equilibrium hyperbolic point such that the stable manifold W s (x 0 ) intersected with the unstable manifold W u (x 0 ) is not empty, then the orbits belonging to W s (x 0 ) ∩ W u (x 0 ) ̸ = Φ are called homoclinic orbits.They are doubly asymptotic to an equilibrium point.Similarly, if x 1 , x 2 are two hyperbolic equilibrium points such there exists an orbit Γ ⊂ W s (x 1 ) ∩W u (x 2 ) or inversely Γ ⊂ W u (x 1 ) ∩W s (x 2 ), then Γ is called a hyperbolic orbit.
The system (1) has the origin O as an equilibrium point for any a, b, c > 0 and it has two more equilibrium points Assume also further a > c.As the transformation (x, y, z) → (−x, −y, z) leaves the system unchanged, the orbits of the system are symmetrical with respect to the z-axis.

Existence of heteroclinic orbits
The matrix of the linearized system to (1) at the origin has the eigenvalues and the corresponding eigenvectors given by: Considering 2c > a > c > 0, it follows that d 1 > 0 and d 2,3 < 0, that is, the origin is a saddle point having a onedimensional unstable manifold W u 0 and a two-dimensional stable manifold W s 0 .The tangent unstable subspace TW u 0 is This is a preprint of: "Heteroclinic, homoclinic and closed orbits in the Chen system", Gheorghe given by x, z = 0 } .The unstable manifold W u 0 contains O(0, 0, 0) and is tangent to TW u 0 at the origin.Using the method of undetermined coefficients in a small neighborhood of the origin we get where |x| ≪ 1 and 0 is indeed tangent to TW u 0 since z ′ (0) = 0, y ′ (0) = a 1 and the vector (1, y ′ (0), z ′ (0)) T is collinear to the direction vector u 1 of the line TW u 0 .Note also that the z-axis is included in the stable manifold W s 0 .Denote in the following by ϕ t u 0 = (x(t, u 0 ), y(t, u 0 ), z(t, u 0 )) a solution of the system (1) through the initial point u 0 = (x 0 , y 0 , z 0 ) and by W u Remark 1 Different from [7], the Lyapunov function U is defined both for b > 2a and b = 2a.
PROPOSITION 1 If 2c > a > c > 0, b ≥ 2a the following assertions are true: a) If there exist t 1 and t 2 such that t 1 < t 2 and U satisfies U(ϕ t 1 u 0 ) = U(ϕ t 2 u 0 ), then either • u 0 is an equilibrium point of system (1), or Proof a) From ( 2) and from the hypothesis of a), one gets for all t ∈ (t 1 ,t 2 ), i.e u 0 is one of the equilibria of (1), or b > 2a and the orbit ϕ t u 0 is contained in the intersection of the plane x = y with the parabolic cylinder bz = x 2 .But this latter case leads again to (3), i.e u 0 is one of the equilibria of system (1), because from ϕ t u 0 ∈ {x = y} ∩ {bz = x 2 } for all t, we get ẋ(t, u 0 ) = 0, ż(t, u 0 ) = 0. Hence x(t) = x 0 , but y(t) = x(t) for all t, i.e ẏ(t, u 0 ) = 0. We notice that the three equilibrium points lie on the non-invariant curve {x = y} ∩ {bz = x 2 }.Finally, (2) and the hypothesis of a) lead also to b = 2a and the orbit ϕ t u 0 is contained in the parabolic cylinder bz = x 2 (for these values of the parameters this cylinder is invariant by the flow of system (1), i.e. if an orbit has a point in it the whole orbit is contained in the cylinder).b) We prove first U(O) > U(ϕ t u 0 ) for all t ∈ R; U(O) > 0. To this end, assume by contrary that there exists a t 0 ∈ R such that 0 < U(O) ≤ U(ϕ t 0 u 0 ).From ϕ t u 0 → O as t → −∞ and U continuous on t, it follows that there exists a sequence t n → −∞ and an integer positive number n 1 such that |U(ϕ t n u 0 ) − U(O)| < ε for all ε > 0 and n > n 1 .Since t n → −∞ and t 0 ∈ R, there is an integer positive number n 2 such that t n < t 0 for all n > n 2 .Denote further by n 0 = max{n 1 , n 2 } and take ε On the other hand U(t) is decreasing with respect to t, which, by definition, leads to U(ϕ t n u 0 ) ≥ U(ϕ t 0 u 0 ) for all t n < t 0 and n > n 0 .Therefore, U(ϕ t n u 0 ) = U(ϕ t 0 u 0 ) and by virtue of a) we get that u 0 is an equilibrium point of system (1).Since ϕ t u 0 → O we get u 0 ≡ O and x(t, u 0 ) = 0 for all t.But this contradicts the hypothesis x(t, u 0 ) > 0 for some t.Hence, U(O) > U(ϕ t u 0 ) for all t ∈ R.
Let us prove now that x(t, u 0 ) > 0 for all t ∈ R. Assuming that there exists a t ′ ∈ R such that x(t ′ , u 0 ) ≤ 0 and using x(t ′′ , u 0 ) > 0 for some t ′′ ∈ R from the hypothesis of b), one gets that there exists a τ ∈ R such that x(τ, u 0 ) = 0.As U(O) > U(ϕ t u 0 ) for all t ∈ R, it follows that ϕ τ u 0 ∈ Ω ∩ P, where Ω = {(x, y, z) : U(O) > U(x, y, z)} and P is the plane x = 0. On the other hand, Ω ∩ P is given by Ay 2 + Bz 2 +Cb 2 (2c − a) 2 < Cb 2 (2c − a) 2 , i.e.Ay 2 +Bz 2 < 0 with A, B ≥ 0. It leads to Ω ∩P = Φ which is a contradiction.Therefore x(t, u 0 ) > 0 for all t ∈ R.This completes the proof of the proposition.
Theorem 1 Consider 2c > a > c > 0, b > 2a and the above function U. Then the following assertions are true: a) The ω-limit of any trajectory of system ( 1) is an equilibrium point.In particular system (1) has no closed trajectories.b) System (1) has no homoclinic trajectories.c) System (1) has exactly two heteroclinic trajectories.
Proof While the function U is different, the proof is similar to the one presented in [7].However, we choose to present it here as it may be a useful handy exercise for some readers.a) If a > c > 0, the function U is decreasing along trajectories of the system both for b > 2a and b = 2a, except perhaps for the orbits on the cylinder bz = x 2 if b = 2a, and the orbits contained in {x = y} ∩ {bz = x 2 } if b > 2a, where the function U is constant and equal to zero.This implies that for all t ≥ 0 0 ≤ U(ϕ t u 0 ) ≤ U(u 0 ), where ϕ t u 0 is a trajectory of the system through the initial point u 0 .It implies that, the limit lim t→∞ U(ϕ t u 0 ) exists.Denote it by U * (u 0 ).From (4) one gets that U(ϕ t u 0 ) is bounded for t ≥ 0, which implies further that x(t, u 0 ), y(t, u 0 ), z(t, u 0 ) are bounded for t ≥ 0, i.e. ϕ t u 0 is bounded for t ≥ 0. De- note by Ω (u 0 ) the ω-limit set of the orbit ϕ t u 0 .It is known that, if u ∈ Ω (u 0 ), then all points of the orbit through u belong to Ω (u 0 ), i.e. ϕ t u ∈ Ω (u 0 ).Therefore, for any point ϕ t u, t ≥ 0, there exists a sequence t n → ∞ for n → ∞ such that lim n→∞ ϕ t n u 0 = ϕ t u which leads to for all t ≥ 0. So there exists t 1 < t 2 such that U(ϕ t 1 u) = U(ϕ t 2 u) for all t ∈ (t 1 ,t 2 ), and by Proposition 1 either u is one of the equilibria of the system, or b = 2a and the point u is contained in the invariant cylinder bz = x 2 , or b > 2a and the point u is contained in the curve {x = y} ∩ {bz = x 2 }.Assume b = 2a and the point u is contained in the invariant cylinder bz = x 2 .Then the function U takes the value zero on this cylinder, and eventually the system can have periodic orbits on the cylinder.Assume b > 2a and u ∈ {x = y} ∩ {bz = x 2 }.Since in the connected curve {x = y}∩{bz = x 2 } there are the three equilibria of the system, and u is an ω-limit set, u is one of the equilibria of the system by the Bendixson-Poincaré Theorem (see for instance Corollary 1.30 of [4]).b) Assume that the system has a homoclinic orbit γ(t) at one of the equilibria O, S 1 or S 2 , that is, lim t→±∞ γ(t) = q where q ∈ {O, S 1 , S 2 }.Since U is decreasing along the trajectories of the system, it follows that ) By statement a) every one-dimensional branch of the unstable manifold W u has ω-limit an equilibrium point p.Assume b > 2a.Since U(O) > U(S i ) = 0 for i = 1, 2, the equilibrium point p must be either S 1 or S 2 , and by the symmetry of the system with respect to the z-axis, one of the two branches of W u must go to S 1 and the other to S 2 , obtaining in this way two heteroclinic orbits.A numerical case with two heteroclinic orbits is illustrated in Fig. 1 a).This completes the proof of the theorem.
The case b = 2a has been studied in [7] using a Lyapunovlike function in the form In particular, the following results have been reported: a) If the negative orbit from a point u 0 is bounded, then the solution ϕ t u 0 approaches one of the equilibria of the system as t → −∞.Consequently, the system has no closed orbits.b) The system (1) has no homoclinic orbits.c) The system (1) has exactly two heteroclinic orbits: one linking O to S 1 and the other O to S 2 .
It implies that for any a, b, c with 2c > a > c > 0 and 2a > b > 0, system (1) has two symmetrical homoclinic orbits to the equilibrium point O.A particular numerical case in this regard is presented in Fig. 1 b) where the homoclinic orbits are depicted.

Conclusions
In this paper we have investigated closed, homoclinic and heteroclinic orbits in a three-dimensional autonomous system, known as the Chen system.Using a convenient Lyapunovlike function, we proved that the system under some constraints of its parameters has no homoclinic orbits and no closed orbits but it has exactly two heteroclinic orbits, symmetrically with respect to the z-axis.Moreover, transforming the system to a new form and using some known results from [1], we obtained results on cases not covered in [7] concerning homoclinic orbits.More exactly, we have proved that the system has two symmetrical homoclinic orbits to the equilibrium point O.