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The complicated trajectory behaviors in the Lorenz equation
Chaos, Solitons and Fractals 19 (2004) 863–873 www.elsevier.com/locate/chaos
The complicated trajectory behaviors in the Lorenz equation Tianshou Zhou b
a,*
, Haohui Liao a, Zuohuan Zheng b, Yun Tang
a
a Department of Mathematical Science, Tsinghua University, Beijing 100084, PR China Academy of Mathematics and System Sciences, Chinese Academy of Sciences, Beijing 100080, PR China
Accepted 13 May 2003
Abstract In this paper, trajectories of the Lorenz equation are analyzed in details. In particular, it is rigorously proved that all nontrivial trajectories travel through two special Poincare projections with infinitely many times. The result would describe the structure of the Lorenz attractor from a lateral aspect. Ó 2003 Elsevier Ltd. All rights reserved.
1. Introduction The Lorenz equation proposed by Lorenz in 1963 [8,11] has been extensively studied. Much progress has been made in its study, especially in the study of the Lorenz attractor ([5,8,12–14,19,20]; for a systematic treatment of the Lorenz equation, see [6]). Most of studies, however, heavily depend on numerical simulation and computer graphics [1,4,16– 18,21,22], and the rigorous results remain few. The situation is partly because of lack of the effective mathematical tools. Until now, the real geometric structure of the Lorenz attractor has remained unknown. Just as Smale [15] pointed out, the problem ‘‘is the dynamics of the Lorenz equation described by the geometric Lorenz attraction of Williams, Guckenheimer and York?’’, is merely the tip of an iceberg. To reveal (or understand) the geometric structure of the Lorenz attractor, there have appeared many theoretical and numerical approaches [2–4,10–12]. However, little endeavor in the literature is devoted to theoretically investigating trajectory behaviors on Poincare projections. In fact, this work is of very significance since a complete understanding of the trajectory behaviors on some Poincare projections can be helpful to reveal the real geometric structure of the Lorenz attractor. In his recent paper [19], Tucker chose one usual Poincare projection, and proved the existence of the Lorenz attractor. In spite of this, its fine structure still remain unknown. Apparently, there exist many Poincare projections in the analysis of the Lorenz equation. To numerically investigate the structure of the Lorenz attractor, Xue et al. [21] selected two special Poincare projections (one of them is as it was usual). They observed some interesting phenomena, such as the switch between the trajectories of the Lorenz attractor in two distinct regions. In this paper, the trajectories of the Lorenz equation are further investigated. It is rigorously proved that all nontrivial trajectories definitely go through these two special Poincare projections with infinitely many times. This result would provide an insight for the structure of Lorenz attractor from a lateral. Moreover, it will be believed that the analytical method presented here is helpful for obtaining the finer structure of the Lorenz attractor in the theoretical analysis and numerical simulation.
* Corresponding author. Address: Department of Computational Science, Faculty of Science, Kanazawa University, KakumaMachi, Kanazawa 920-1192, Japan. E-mail address: [email protected] (T. Zhou).
0960-0779/$ - see front matter Ó 2003 Elsevier Ltd. All rights reserved. doi:10.1016/S0960-0779(03)00243-1
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2. Approximate scenarios of trajectories As is known, the Lorenz equation is described as follows: 8 0 < x ¼ aðy xÞ y 0 ¼ cx xz y : 0 z ¼ xy bz;
ð1Þ
where a > 0, b > 0, c > 1. For typical parameters: a ¼ 10, b ¼ 8=3, c ¼ 28, one has the Lorenz attractor (see Fig. 1). It has been proven [7,9] that when the time is sufficiently large, all the trajectories of (1) are contained in a ball, namely, C ¼ fðx; y; zÞ 2 R3 jx2 þ y 2 þ ðz a cÞ2 ¼ P g, where P ¼ b2 ða þ cÞ2 =4ðb 1Þ. This indicates that the trajectories of (1) are bounded. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Onpthe other hand, it is obvious that (1) has three equilibria O1 ð0; 0; 0Þ, O2 ð bðc 1Þ; bðc 1Þ; c 1Þ, and ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi O3 ð bðc 1Þ; bðc 1Þ; c 1Þ. Jacobian matrices corresponding to these points are, respectively, 0 1 a a 0 @ c 1 0 A; J1 ¼ J ðO1 Þ ¼ 0 0 b 0 1 a a 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @ 1 1 bðc 1Þ A; J2 ¼ J ðO2 Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi bðc 1Þ bðc 1Þ b 0 1 a a 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @ 1 1 bðc 1Þ A: J3 ¼ J ðO3 Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi bðc 1Þ bðc 1Þ b A simple calculation shows that matrix J1 has the following eigenvalues: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ ða 1Þ2 þ 4ac : k1 ¼ b; k2;3 ¼ 2
ð2Þ
Analogously, for matrices J2 and J3 , one has the following identity characteristic polynomials, k3 þ ða þ b þ 1Þk2 þ bða þ cÞk þ 2abc ¼ 0:
ð3Þ
It is easy to see that equilibrium point O1 is unstable, and that by Routh–Hurwitz conditions, O2 and O3 are also unstable under the condition: b < ½aða cÞ=ða þ cÞ 1 (this condition will be always assumed in the following).
Fig. 1. The typical Lorenz chaotic attractors.
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Fig. 2. The chaotic time series of zðtÞ (a ¼ 10, b ¼ 8=3, c ¼ 28), where Ô Õ represents the line z ¼ c 1 and Ô- - -Õ the line z ¼ c 1 þ ðða þ 1Þ2 =4aÞ.
For the sake of understanding, the approximate scenario of trajectories of (1) is first shown and the rigorous proof is then given in next sections. For this, consider the following system: 0 x ¼ aðy xÞ ð4Þ y 0 ¼ ðc zÞx y; where z is treated as a parameter. Note that (4) has only one equilibrium point ð0; 0Þ when z 6¼ c 1. Moreover, the Jacobian matrix of (4) at the point is a a A¼ ; c z 1 and the eigenvalues corresponding to A are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ ða þ 1Þ2 4aðz þ 1 cÞ : P1;2 ¼ 2
ð5Þ
From (5), one easily sees that if z > c 1 þ ðða þ 1Þ2 =4aÞ, then ð0; 0Þ is a stable focus of (4); if c 1 < z < c 1 þ ðða þ 1Þ2 =4aÞ, then ð0; 0Þ is a stable nodal, and if z < c 1, then ð0; 0Þ is a saddle point. Intuitively, the trajectories of (1) do not go very far on the left of the Poincare projection z ¼ c 1 because of boundedness of solutions of (1). Therefore, the trajectories will possibly cross the projection and enter its right since otherwise, they would converge to one of three equilibria of (1). Thus, at some time they possibly travel through the Poincare projection z ¼ c 1 from the right to the left and come back to the left of the projection again. Moreover, at another moment they possibly cross another Poincare projection z ¼ c 1 þ ðða þ 1Þ2 =4aÞ from the left to the right and enter the right. But, they cannot always stay on the right of z ¼ c 1 þ ðða þ 1Þ2 =4aÞ. Otherwise, they would also converge to a certain equilibrium point of (1). That is to say, they cross the Poincare projection z ¼ c 1 þ ðða þ 1Þ2 =4aÞ again and return its left. Like this, the nontrivial trajectories of (1) travel through these two Poincare projections repeatedly (see Fig. 2). In Section 4, it will be rigorously proved that all nontrivial trajectories of (1) definitely travel through the two Poincare projections z ¼ c 1 and z ¼ c 1 þ ðða þ 1Þ2 =4aÞ with infinitely many times. For it, the lemmas in the next section will be needed.
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3. Lemmas Denote by ðxðtÞ; yðtÞ; zðtÞÞ an orbit of (1). Lemma 1. For any interval ða; bÞ, if xðtÞ const. as t 2 ða; bÞ, then ðxðtÞ; yðtÞ; zðtÞÞ is merely the trivial solution of (1), namely, one of O1 , O2 and O3 . Similar conclusion holds for yðtÞ and zðtÞ. In other words, if x0 ðtÞ 6 0, t 2 ð1; þ1Þ, then for any ða; bÞ, one has x0 ðtÞ 6 0, y 0 ðtÞ 6 0 z0 ðtÞ 6 0, for t 2 ða; bÞ. Proof. In the case that xðtÞ const. for t 2 ða; bÞ, one has x0 ðtÞ 0 as t 2 ða; bÞ. By (1), one sees xðtÞ ¼ yðtÞ const. for t 2 ða; bÞ. Therefore, y 0 ðtÞ 0 as t 2 ða; bÞ. Also, by the second equation of (1), one has zðtÞ ¼ const. which implies that z0 ðtÞ 0 as t 2 ða; bÞ. Thus, in the interval ða; bÞ, ðxðtÞ; yðtÞ; zðtÞÞ ¼ O1 , or O2 , or O3 . However, O1 , O2 and O3 are three trivial solutions of (1). By a fundamental property of ordinary differential equations, one can determine that ðxðtÞ; yðtÞ; zðtÞÞ must be one of O1 , O2 and O3 as t 2 ð1; 1Þ. It is impossible since the lemma assumes x0 ðtÞ 6 0 for t 2 ð1; 1Þ. The lemma is thus proved. h Lemma 2. If limt!þ1 xðtÞ, limt!þ1 yðtÞ, limt!þ1 zðtÞ all exist, then ðxðtÞ; yðtÞ; zðtÞÞ must be a trivial orbit of (1). In fact, assume limt!þ1 xðtÞ ¼ n, limt!þ1 yðtÞ ¼ g, limt!þ1 zðtÞ ¼ f. By (1), limt!þ1 x0 ðtÞ, limt!þ1 y 0 ðtÞ, limt!þ1 z0 ðtÞ all exist. But, ðxðtÞ; yðtÞ; zðtÞÞ are all boundary. So, limt!þ1 x0 ðtÞ ¼ limt!þ1 y 0 ðtÞ ¼ limt!þ1 z0 ðtÞ ¼ 0. Thus, n, g and f must satisfy 8 0 for t P T nor x0 ðtÞ < 0 for t P T . Similar conclusion applies to yðtÞ and zðtÞ. Proof. First, consider the case of xðtÞ. Without loss of generality, assume x0 ðtÞ > 0 for t P T . Then, xðtÞ keeps the constant sign as t P T . Therefore, limt!þ1 xðtÞ exist, denoted by n. By (1), one has for any fixed Dt, Z Dtþt yðsÞeas ds: ð7Þ xðtÞ ¼ xðtÞeaDt þ aeat t
Note that lim lim
Dt!0 t!þ1
xðtÞð1 eaDt Þ xðtÞð1 eaDt Þ ¼ lim lim ¼ an; t!þ1 Dt!0 Dt Dt
which means that the above two limits are exchangeable. Therefore, R tþDt R tþDt eat t yðsÞeas ds eat t yðsÞeas ds ¼ lim lim ¼ n: lim lim t!þ1 Dt!0 Dt!0 t!þ1 Dt Dt This implies that limt!1 yðtÞ not only exists but also equates to n. Also by (1), one has limt!þ1 x0 ðtÞ ¼ 0. If n ¼ 0, then limt!þ1 xðtÞ ¼ limt!þ1 yðtÞ ¼ 0. By y 0 ðtÞ ¼ ðc zðtÞÞxðtÞ yðtÞ and boundedness of zðtÞ, limt!þ1 y 0 ðtÞ ¼ 0. In addition, it follows from z0 ðtÞ ¼ xðtÞyðtÞ bðtÞ that for any T1 > T , Z t xðsÞyðsÞebs ds: ð8Þ zðtÞ ¼ zðT1 ÞebðT1 tÞ þ ebt T1
So, limt!þ1 zðtÞ exists and equates to zero. This is a contradiction by Lemma 2. If n 6 0, then since xðtÞ keeps the constant sign for t P R t T , and limt!þ1 xðtÞ ¼ limt!þ1 yðtÞ, there exists a T2 > T , such that yðtÞ keeps the constant sign for t P T2 . So limt!1 ð T3 xðsÞyðsÞebs dsÞebt exists for any T3 P T2 . By (8), one sees that limt!1 zðtÞ exists. It is also a contradiction by Lemma 2. In a word, x0 ðtÞ > 0 for t P T is impossible. Secondly, assume y 0 ðtÞ > 0 for t P T . Then yðtÞ keeps the constant sign as t P T . By the boundedness of yðtÞ, one Rt knows that limt!þ1 yðtÞ exists. Furthermore, limt!þ1 eat b yðsÞeas ds exists for any b P T . It follows from x0 ðtÞ ¼ aðyðtÞ xðtÞÞ that Z t yðsÞeas ds: ð9Þ xðtÞ ¼ xðbÞeaðbtÞ þ aeat b
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So, limt!þ1 x0 ðtÞ exists. Moreover, limt!þ1 x0 ðtÞ must be zero because of the boundedness of xðtÞ. Thus, limt!þ1 xðtÞ ¼ limt!þ1 yðtÞ. In addition, yðtÞ keeps the constant sign as t is sufficiently large since xðtÞ keeps the constant sign. Therefore, xðtÞyðtÞ keeps the constant sign as t is large enough. By (8), limt!þ1 zðtÞ exists. By Lemma 2, one then encounters a contradiction. Finally, assume z0 ðtÞ keeps the constant sign for t P T . Then zðtÞ also keeps the constant sign for t P T . By boundedness of zðtÞ, limt!þ1 zðtÞ must exist. Furthermore, xðtÞyðtÞ keeps the constant sign since xðtÞyðtÞ ¼ z0 ðtÞ þ bzðtÞ. That is to say, xðtÞ and yðtÞ all keep the constant sign as t is sufficiently large. By (1), one sees that x0 ðtÞ and y 0 ðtÞ all keep the constant sign as t is large enough. Therefore, limt!1 xðtÞ, limt!þ1 yðtÞ all exist. By Lemma 2 and the assumed condition, it is impossible. In conclusion, Lemma 3 holds. h Corollary 1. Assume x0 ðtÞ 6 0 for t 2 ðþ1; þ1Þ. Then, for any T < þ1, x0 ðtÞ, y 0 ðtÞ and z0 ðtÞ all change the sign with infinitely many times in the interval ðT ; þ1Þ. Corollary 2. Assume that x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, for any T < þ1, xðtÞ, yðtÞ and zðtÞ all have infinitely local extreme value points in the interval ðT ; þ1Þ. Lemma 4. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. If there exists b such that x0 ðbÞ ¼ x00 ðbÞ ¼ 0, then t ¼ b is the inflection point of xðtÞ, i.e. t ¼ b is not the extreme value point of xðtÞ. The similar conclusion holds for yðtÞ and zðtÞ. Proof. For such a b, one has by (1) yðbÞ ¼ ð1=aÞx0 ðbÞ þ xðbÞ ¼ xðbÞ and y 0 ðbÞ ¼ ð1=aÞx00 ðbÞ þ x0 ðbÞ ¼ 0. Furthermore, by (1), ½c 1 zðbÞxðbÞ ¼ 0:
ð10Þ
If xðbÞ ¼ 0, then y 00 ðbÞ ¼ ½czðbÞx0 ðbÞ z0 ðbÞxðbÞ y 0 ðbÞ ¼ 0. Therefore, xð3Þ ðbÞ ¼ aðy 00 ðbÞ x00 ðbÞÞ ¼ 0. Inductively, one can obtain that xðnÞ ðbÞ ¼ 0;
n ¼ 1; 2; . . .
ð11Þ
Consider the Taylor expansion of xðtÞ at t ¼ b, xðtÞ ¼
1 X xðnÞ ðbÞ ðt bÞn : n! n¼0
ð12Þ
So, xðtÞ 0 at a neighborhood of t ¼ b. By Lemma 1, it is impossible. Thus, it follows from (10) that zðbÞ ¼ c 1:
ð13Þ
A direct calculation yields xð3Þ ðbÞ ¼ az0 ðbÞxðbÞ:
ð14Þ
Note a fact that x0 ðbÞ, y 0 ðbÞ and z0 ðbÞ cannot equate to zero at the same time. Thus, z0 ðbÞxðbÞ 6¼ 0, which implies that xð3Þ ðbÞ 6¼ 0. This indicates that t ¼ b could not be the extreme value point of xðtÞ. Similarly, one can obtain the same conclusion for yðtÞ and zðtÞ. The lemma is hence proved. h Corollary 3. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. At every extreme value point of xðtÞ, x00 ðtÞ 6¼ 0 holds. The same conclusion applies to yðtÞ and zðtÞ. Lemma 5. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, for any T < þ1, it is impossible to have zðtÞ < 0 for all t P T . Proof. Conversely, there would exist a T < þ1 such that zðtÞ < 0 for t P T . By Corollary 2, zðtÞ has infinitely locally extreme value point as t P T . Take a locally maximum value point, denoted by b, of zðtÞ. Then zðbÞ < 0;
z0 ðbÞ ¼ 0;
z00 ðbÞ < 0:
ð15Þ
By z0 ðbÞ ¼ 0 and (1), one has xðbÞyðbÞ ¼ bzðbÞ < 0. Let V ðtÞ ¼ xðtÞyðtÞ. Then V 0 ðbÞ ¼ z00 ðbÞ < 0. However, V 0 ðbÞ ¼ x0 ðbÞyðbÞ þ xðbÞy 0 ðbÞ ¼ ay 2 ðbÞ ða þ 1ÞxðbÞyðbÞ þ ðc zðbÞÞx2 ðbÞ > 0. This is a contradiction, completing the proof of the lemma. h
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a
z=0
a1 a0
t
b
Fig. 3. Sketch diagram of zðtÞ corresponding to the proof of Lemma 6.
From the proof procedure of the lemma, one sees that Corollary 4. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, there exists no maximum point for zðtÞ in the side of zðtÞ < 0. Lemma 6. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, there exists a T < þ1 such that zðtÞ > 0 for all t P T . Proof. Conversely, for any T < þ1, there would exist an instant t0 P T such that zðt0 Þ < 0. By Corollaries 2 and 4, there exist two instants a and b such that zðaÞ < 0;
z0 ðaÞ ¼ 0;
zðbÞ > 0;
z0 ðbÞ ¼ 0;
z00 ðbÞ < 0;
z0 ðtÞ > 0;
z00 ðtÞ > 0
for tða; bÞ:
ð16Þ
And zðtÞ has no extreme value point as t 2 ½a; bÞ (see Fig. 3). Moreover, there exists a unique a0 2 ða; bÞ such that zða0 Þ ¼ 0, and zðtÞ < 0 as t 2 ða; a0 Þ while zðtÞ > 0 as t 2 ða0 ; bÞ. Now, one can determine both x0 ðtÞ 6¼ 0 and y 0 ðtÞ 6¼ 0 as t 2 ða; bÞ. In fact, by z0 ðaÞ ¼ 0 and z0 ðbÞ ¼ 0, one has xðaÞyðaÞ ¼ bzðaÞ < 0;
xðbÞyðbÞ ¼ bzðbÞ > 0:
ð17Þ
Without loss of generality, assume xðaÞ < 0;
yðaÞ > 0;
xðbÞ > 0;
yðbÞ > 0:
ð18Þ
0
If there exists an a1 such that x ða1 Þ ¼ 0 and a1 2 ða; bÞ, then a < a1 6 a0 < b (see Fig. 3). Also, without loss of generality, assume x00 ða1 Þ < 0. Then, in a left small interval of t ¼ a1 , one has x0 ðtÞ > 0 while in a right small interval of t ¼ a1 , x0 ðtÞ < 0. Therefore a1 ¼ a0 . Furthermore, by (1), yða0 Þ ¼ ð1=aÞx0 ða0 Þ þ xða0 Þ ¼ 0. Also by (1), one has y 0 ða0 Þ ¼ ðc zða0 ÞÞxða0 Þ yða0 Þ ¼ 0. So, x00 ða0 Þ ¼ 0. Inductively, one can obtain xðnÞ ða0 Þ ¼ 0, n ¼ 1; 2; . . . Considering Taylor expansions of xðtÞ in a neighborhood of t ¼ a0 , one will encounter a contradiction. Similarly, one can prove y 0 ðtÞ 6¼ 0 for all t 2 ða; bÞ. Because of this and by (18), there exists a unique instant, still denoted by a1 , such that xða1 Þ ¼ 0 and a < a1 < a0 < b. Since xðaÞ < 0 and yðaÞ > 0, x0 ðaÞ ¼ aðyðaÞ xðaÞÞ > 0. Thus, x0 ðtÞ > 0 for t 2 ða; bÞ. Furthermore, one can show yðbÞ > xðbÞ. It follows from z00 ðbÞ < 0 that ay 2 ðbÞ ða þ 1ÞxðbÞyðbÞ þ ðc zðbÞÞx2 ðbÞ < 0, which yields qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ 1 þ ða þ 1Þ2 þ 4aðzðbÞ þ 1 cÞ xðbÞ: ð19Þ yðbÞ < 2a Combining (19) with yðbÞ > xðbÞ, one has zðbÞ > c 1:
ð20Þ
Therefore, y 0 ðbÞ ¼ ðc zðbÞÞxðbÞ yðbÞ < ðc 1 zðbÞÞxðbÞ < 0. Moreover, yðtÞ < 0 as t 2 ða; bÞ. Summarizing the above discussions, one sees z0 ðtÞ > 0, x0 ðtÞ > 0, y 0 ðtÞ < 0 and yðtÞ > 0 as t 2 ða; bÞ. In addition, xðtÞ < 0 as t 2 ða; a1 Þ while xðtÞ > 0 as t 2 ða1 ; bÞ. Let V ðtÞ ¼ xðtÞyðtÞ. Then, V 0 ðtÞ > 0 as t 2 ða; a1 Þ and V 0 ðtÞ < 0 as t 2 ða1 ; bÞ. Since z00 ðaÞ > 0 and z00 ðbÞ < 0, V 0 ða1 Þ ¼ 0. Note that xða1 Þ ¼ 0 and V 0 ða1 Þ ¼ ay 2 ða1 Þ ða þ 1Þxða1 Þyða1 Þ þ ðc zða1 ÞÞx2 ða1 Þ. So, yða1 Þ ¼ 0. By (1), one inductively has xðnÞ ða1 Þ ¼ 0, n ¼ 1; 2; . . . This is impossible. In conclusion, Lemma 6 holds. h Lemma 7. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, for any T < þ1, neither yðtÞ < 0 for all t P T nor yðtÞ > 0 for all t P T , i.e. yðtÞ definitely changes the sign in the interval ðT ; þ1Þ. The similar conclusion applies to xðtÞ.
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Proof. First, consider the R t case for yðtÞ. Conversely, there would exist a T < þ1 such that yðtÞ > 0 for t P T . By (1) one has (9). But,R limt!þ1 b yðsÞeas ds ¼ þ1 since yðtÞ > 0 for t P b P T . According to boundedness of xðtÞ and (9), t limt!þ1 eat b yðsÞeas ds must exist. limt!þ1 xðtÞ then exists. Furthermore, by (9), there exists a t0 > b such that xðtÞ keeps the positive sign as t P b. Therefore, xðtÞyðtÞ keeps the constant sign as t P b. By (8), one knows that limt!þ1 zðtÞ must exist. According to Lemma 2, this is impossible. Secondly, assume xðtÞ > 0 for t P T . By Lemma 6, without loss of generality, one may assume zðtÞ > 0 as t P T . By (1), one has for any fixed b P T , Z t yðtÞ ¼ yðbÞebt þ et ðc zðsÞÞxðsÞet ds: ð21Þ b
Since xðtÞ and zðtÞ all have the positive sign, limt!þ1 yðtÞ must exist by (9). Besides, by (9), limt!þ1 xðtÞ also exists. Furthermore, by (8), limt!þ1 zðtÞ exists. A collection of the above discussion contracts to the conclusion of the Lemma 2. Lemma 7 is then proved. h Corollary 5. Assume x0 ðtÞ 6 for t 2 ð1; þ1Þ. Then, for any T < þ1, xðtÞ and yðtÞ all change the sign with infinitely many times in the interval ðT ; þ1Þ. After having the above preparations, now one may prove the main results of this paper.
4. Main results Theorem 1. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ, a > 1 and b > 1. For any T < þ1, then the trajectories of (1) definitely travel through the Poincare z ¼ c 1 from the left to the right at some moments t: t P T . Proof. Conversely, there would exist a T < þ1 such that zðtÞ < c 1 as t P T . According to Corollary 2, one sees that zðtÞ has at least one maximum value point, denoted by b. Furthermore, by Corollary 3 one has that there exists a small neighborhood of t ¼ b, denoted by ða; cÞ (where a < b < c) such that z0 ðbÞ ¼ 0, z00 ðbÞ < 0, zðtÞ < zðbÞ as t 6¼ b and t 2 ða; cÞ. Note that z0 ðbÞ ¼ 0 yields xðbÞyðbÞ ¼ bzðbÞ:
ð22Þ
By Lemma 6, without loss of generality, assume xðbÞ > 0, yðbÞ > 0. Now, one can determine x0 ðbÞ 6¼ 0. Otherwise, it would have yðbÞ ¼ xðbÞ. By (1), y 0 ðbÞ ¼ ðc 1 zðbÞÞxðbÞ > 0. On the other hand, 0 > z00 ðbÞ ¼ x0 ðbÞyðbÞ þ xðbÞy 0 ðbÞ ¼ xðbÞy 0 ðbÞ > 0. This is a contradiction. In addition, one can determine y 0 ðbÞ 6¼ 0. Conversely, y 0 ðbÞ ¼ 0 will yield ðc zðbÞÞxðbÞ ¼ yðbÞ:
ð23Þ
Thus, by (22) and (23) one obtains x2 ðbÞ ¼
bzðbÞ ; c zðbÞ
y 2 ðbÞ ¼ bðc zðbÞÞzðbÞ:
ð24Þ
However, z00 ðbÞ ¼ ay 2 ðbÞ ða þ 1ÞxðbÞyðbÞ þ ðc zðbÞÞx2 ðbÞ ¼ zðbÞ½abðc 1 zðbÞÞ þ ab a þ 1 þ b > 0 as b > 1. It also produces a contradiction. So y 0 ðbÞ 6¼ 0. Next, one needs to distinguish the following four cases: (1) (2) (3) (4)
x0 ðbÞ > 0, x0 ðbÞ < 0, x0 ðbÞ < 0, x0 ðbÞ > 0,
y 0 ðbÞ > 0; y 0 ðbÞ < 0; y 0 ðbÞ > 0; y 0 ðbÞ < 0.
In the following discussions, note that xðbÞ > 0;
yðbÞ > 0;
zðbÞ < c 1;
z0 ðbÞ ¼ 0;
z00 ðbÞ < 0:
Case 1: x0 ðbÞ > 0, y 0 ðbÞ > 0. This case is impossible since 0 > z00 ðbÞ ¼ x0 ðbÞyðbÞ þ xðbÞy 0 ðbÞ > 0. Case 2: x0 ðbÞ < 0, y 0 ðbÞ < 0.
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In this case, it follows from y 0 ðtÞ ¼ ðc 1 zðtÞÞxðtÞ þ xðtÞ yðtÞ that there exist a and c, a < b < c, such that y ðtÞ > xðtÞ yðtÞ as t 2 ða; cÞ. So, when t 2 ðb; cÞ, 0
yðtÞ > yðbÞebt þ xðtÞ xðtÞebt :
ð25Þ
0
Furthermore, by x ¼ ay ax, one has xðtÞ > xðbÞ þ aðyðbÞ xðbÞÞð1 ebt Þ
ð26Þ
for t 2 ðb; cÞ. (25) multiplying (26) will yield for t 2 ðb; cÞ ð27Þ
xðtÞyðtÞ > gðtÞ; where gðtÞ ¼ xðbÞyðbÞebt þ axðtÞðyðbÞ xðbÞÞð1 2ebt þ e2ðbtÞ Þ þ ½ayðbÞðyðbÞ xðbÞÞ þ xðtÞxðbÞð1 ebt Þ. It follows from z0 ¼ xy bz that one has for t 2 ðb; cÞ, Z t gðsÞebs ds: zðtÞ > zðbÞebt þ ebt
ð28Þ
b
The direct calculation will lead to zðtÞ zðbÞebðbtÞ ¼ z0 ðbÞ þ bzðbÞ ¼ bzðbÞ; tb t!b Rt ebt b gðsÞebs ds ¼ xðbÞyðbÞ: lim tb t!bþ Rt Let GðtÞ ¼ zðtÞ zðbÞebðbtÞ ebt b gðsÞebs ds. Then GðtÞ > 0 as t 2 ðb; cÞ, and 2 limt!bþ GðtÞ=ðt bÞ2 ¼ z00 ðbÞ ay 2 ðbÞ þ ða þ 1ÞxðbÞyðbÞ x2 ðbÞ P 0. By z00 ðbÞ < 0, one has ðyðbÞ xðbÞÞðayðbÞ xðbÞÞ < 0. This is impossible since x0 ðbÞ > 0 and a > 1. Case 3: x0 ðbÞ < 0, y 0 ðbÞ > 0. Similar to the proof in the case 2, one will encounter a contradiction. Case 4: x0 ðbÞ > 0, y 0 ðbÞ < 0. In this case, if considers the left small interval ða; bÞ of t ¼ b in the proof procedure in the case 2, one will also obtain a contradiction. Theorem 1 is thus proved. h limþ
Corollary 6. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, there exists no locally maximum value point for both xðtÞ and yðtÞ within the side of z < c 1. Similarly, one can prove Theorem 2. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, all trajectories of (1) definitely cross the Poincare projection z ¼ c 1 þ ðða þ 1Þ2 =4aÞ from the right to the left at some moments. Corollary 7. Under the assumed conditions of Theorem 2, the exists no locally minimal value point for xðtÞ and yðtÞ within the side of z > c 1 þ ðða þ 1Þ2 =4aÞ. Theorem 3. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, for any T < þ1 all trajectories of (1) definitely travel through the Poincare z ¼ c 1 from the right to the left at some instances. Proof. Otherwise, there would have zðtÞ > c 1 for t P T . By Corollaries 2 and 6, one may assume that there exists a b > T such that c 1 < zðbÞ < c 1 þ ðða þ 1Þ2 =4aÞ, z0 ðbÞ ¼ 0, z00 ðbÞ > 0. By Lemma 6 and z0 ðbÞ ¼ 0, without loss of generality, assume xðbÞ > 0, yðbÞ > 0. There exists a neighborhood, denoted by ða; cÞ, a < b < c, of t ¼ b, such that c 1 < zðtÞ < c 1 þ ðða þ 1Þ2 =4aÞ as t 2 ða; cÞ. In the neighborhood, by y 0 ¼ ðc zÞx y one has ! ða þ 1Þ2 0 xðtÞ yðtÞ > y ðtÞ > 1 xðtÞ yðtÞ: ð29Þ 4a It follows from z0 ðbÞ ¼ 0, z00 ðbÞ > 0 that
T. Zhou et al. / Chaos, Solitons and Fractals 19 (2004) 863–873
yðbÞ <
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ ða þ 1Þ2 þ 4aðzðbÞ cÞ 2a
871
xðbÞ
ð30Þ
xðbÞ > xðbÞ:
ð31Þ
or
yðbÞ >
ða þ 1Þ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ2 þ 4aðzðbÞ cÞ 2a
However, (30) does not appear since zðbÞ > c 1, xðbÞ > 0 and yðbÞ > 0. Moreover, by (1), one sees x0 ðbÞ > 0. Furthermore, 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 a þ 1 þ ða þ 1Þ2 þ 4aðzðbÞ cÞ 7 6 0 y ðbÞ ¼ ðc zðbÞÞxðbÞ yðbÞ < 4c zðbÞ 5xðbÞ < 0 2a as c 1 < zðbÞ < c 1 þ ðða þ 1Þ2 =4aÞ. So, y 0 ðbÞ < 0. Without loss of generality, it is assumed that x0 ðtÞ > 0, y 0 ðtÞ < 0 as t 2 ða; cÞ. When t 2 ðb; cÞ, by (29) one knows that ! ða 1Þ2 ða 1Þ2 xðtÞ þ xðtÞ þ yðbÞ ebt < yðtÞ < xðtÞ þ ðyðbÞ xðtÞÞebt : ð32Þ 4a 4a Therefore, by x0 ðtÞ ¼ aðyðtÞ xðtÞÞ, one has " # ða 1Þ2 ða þ 1Þ2 a xðtÞ þ yðbÞ ebt xðtÞ < x0 ðtÞ < a½yðbÞ xðtÞebt : 4a 4
ð33Þ
Integrating both sides of (33) from b to t for t 2 ½b; cÞ will lead to " # 2 4a ½ða1Þ2 =4ðbtÞ þ xðbÞ þ yðbÞ ebt e½ðaþ1Þ =4ðbtÞÞ < xðtÞ < xðbÞ þ a½yðbÞ xðbÞð1 ebt Þ: xðbÞe 2 ða 1Þ
ð34Þ
Thus, xðtÞyðtÞ > g1 ðtÞ
ð35Þ
as t 2 ½b; mÞ, where " ða 1Þ2 g1 ðtÞ ¼ xðtÞ þ 4a (
! # ða 1Þ2 xðtÞ þ yðbÞ ebt 4a " # ) 4a ½ða1Þ2 =4aðbtÞ bt ½ðaþ1Þ2 =4ðbtÞ þ xðbÞ þ yðbÞ ðe e Þ : xðbÞe ða 1Þ2
Also, it follows from z0 ¼ xy bt that Z t g1 ðsÞebs ds zðtÞ > zðbÞebðbtÞ þ ebt
ð36Þ
b
as t 2 ½b; mÞ. Combined (36) with z0 ðbÞ ¼ 0, z00 ðbÞ > 0, a direct calculation will yield ð37Þ
xðbÞyðbÞ 6 0: So, either xðbÞ 6 0 or yðbÞ 6 0 which contradict xðbÞ > 0 and yðbÞ > 0. The lemma is then proved.
h
Corollary 8. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. For any T < þ1, then zðtÞ has no minimal value within the side of zðtÞ > c 1 for t P T . Similarly, one can prove
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Theorem 4. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, for any T < þ1, all trajectories of (1) definitely travel through the Poincare z ¼ c 1 þ ðða þ 1Þ2 =4aÞ from the left to the right at some moments. Corollary 9. Under assumed conditions of Theorem 4, for any T > þ1, zðtÞ has no maximum value within the side of zðtÞ < c 1 þ ðða þ 1Þ2 =4aÞ for t P T . A combination of Theorems 1–4 gives Theorem 5 (Main result). Assume a > 1, b > 1, c > 1, b < ½aða 1Þ=ða þ 1Þ 1. Then, all nontrivial half-positive trajectories of (1) definitely travel through two Poincare projections z ¼ c 1 and zðtÞ < c 1 þ ðða þ 1Þ2 =4aÞ with infinitely many times.
5. Conclusion This paper has proved that all nontrivial trajectories of the Lorenz system cross two special Poincare projections under some conditions. By writing down instants at which these orbits travel through the Poincare projections, one may be able to capture some periodic orbits embedded in the Lorenz attractor. In addition, if z is viewed as a time-varying parameter, then the projected trajectories of the system on the ðx; yÞ planes can be divided into a sequence of small projections, each of which has different characteristics from its two other projections. The fine structure of the Lorenz attractor can be thus described. In spite of this, further studies remain to be carried out, for example, how to distinguish quasi-periodic orbits from chaotic orbits.
Acknowledgements The first author thanks the hospitality of the Kanazawa University of Japan, where he conducted part of this research under the fellowship awarded by the Japanese Society for the Promotion of Science. This paper is partly supported by National Key Basic Research Special Fund (no. G1998020309), National Nature Science Foundation of China (no. 10272059).
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[18] Sachdev PL, Sarathy R. Periodic and chaotic solutions for a nonlinear system arising from a nuclear spin generator. Chaos, Solitons & Fractals 1994;4(11):2015–41. [19] Tucker W. The Lorenz attractor exists. CR Acad Sci Paris Ser I Math 1999;328(12):1197–202. [20] Williams R. The structure of Lorenz attractors. In: Bermard P, Ratiu T, editors. Turbulence Seminar Berkeley 1996/97. Berlin– Heidelberg–New York: Springer-Verlag; 1997. p. 94–112. [21] Xue YS, Wu QH et al. The fine structure of Lorenz attractor, preprint 2002. [22] Yang SK, Chen CL, Yau HT. Control of chaos in Lorenz systenz. Chaos, Solitons & Fractals 2002;13(4):767–80.
The complicated trajectory behaviors in the Lorenz equation Tianshou Zhou b
a,*
, Haohui Liao a, Zuohuan Zheng b, Yun Tang
a
a Department of Mathematical Science, Tsinghua University, Beijing 100084, PR China Academy of Mathematics and System Sciences, Chinese Academy of Sciences, Beijing 100080, PR China
Accepted 13 May 2003
Abstract In this paper, trajectories of the Lorenz equation are analyzed in details. In particular, it is rigorously proved that all nontrivial trajectories travel through two special Poincare projections with infinitely many times. The result would describe the structure of the Lorenz attractor from a lateral aspect. Ó 2003 Elsevier Ltd. All rights reserved.
1. Introduction The Lorenz equation proposed by Lorenz in 1963 [8,11] has been extensively studied. Much progress has been made in its study, especially in the study of the Lorenz attractor ([5,8,12–14,19,20]; for a systematic treatment of the Lorenz equation, see [6]). Most of studies, however, heavily depend on numerical simulation and computer graphics [1,4,16– 18,21,22], and the rigorous results remain few. The situation is partly because of lack of the effective mathematical tools. Until now, the real geometric structure of the Lorenz attractor has remained unknown. Just as Smale [15] pointed out, the problem ‘‘is the dynamics of the Lorenz equation described by the geometric Lorenz attraction of Williams, Guckenheimer and York?’’, is merely the tip of an iceberg. To reveal (or understand) the geometric structure of the Lorenz attractor, there have appeared many theoretical and numerical approaches [2–4,10–12]. However, little endeavor in the literature is devoted to theoretically investigating trajectory behaviors on Poincare projections. In fact, this work is of very significance since a complete understanding of the trajectory behaviors on some Poincare projections can be helpful to reveal the real geometric structure of the Lorenz attractor. In his recent paper [19], Tucker chose one usual Poincare projection, and proved the existence of the Lorenz attractor. In spite of this, its fine structure still remain unknown. Apparently, there exist many Poincare projections in the analysis of the Lorenz equation. To numerically investigate the structure of the Lorenz attractor, Xue et al. [21] selected two special Poincare projections (one of them is as it was usual). They observed some interesting phenomena, such as the switch between the trajectories of the Lorenz attractor in two distinct regions. In this paper, the trajectories of the Lorenz equation are further investigated. It is rigorously proved that all nontrivial trajectories definitely go through these two special Poincare projections with infinitely many times. This result would provide an insight for the structure of Lorenz attractor from a lateral. Moreover, it will be believed that the analytical method presented here is helpful for obtaining the finer structure of the Lorenz attractor in the theoretical analysis and numerical simulation.
* Corresponding author. Address: Department of Computational Science, Faculty of Science, Kanazawa University, KakumaMachi, Kanazawa 920-1192, Japan. E-mail address: [email protected] (T. Zhou).
0960-0779/$ - see front matter Ó 2003 Elsevier Ltd. All rights reserved. doi:10.1016/S0960-0779(03)00243-1
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2. Approximate scenarios of trajectories As is known, the Lorenz equation is described as follows: 8 0 < x ¼ aðy xÞ y 0 ¼ cx xz y : 0 z ¼ xy bz;
ð1Þ
where a > 0, b > 0, c > 1. For typical parameters: a ¼ 10, b ¼ 8=3, c ¼ 28, one has the Lorenz attractor (see Fig. 1). It has been proven [7,9] that when the time is sufficiently large, all the trajectories of (1) are contained in a ball, namely, C ¼ fðx; y; zÞ 2 R3 jx2 þ y 2 þ ðz a cÞ2 ¼ P g, where P ¼ b2 ða þ cÞ2 =4ðb 1Þ. This indicates that the trajectories of (1) are bounded. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Onpthe other hand, it is obvious that (1) has three equilibria O1 ð0; 0; 0Þ, O2 ð bðc 1Þ; bðc 1Þ; c 1Þ, and ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi O3 ð bðc 1Þ; bðc 1Þ; c 1Þ. Jacobian matrices corresponding to these points are, respectively, 0 1 a a 0 @ c 1 0 A; J1 ¼ J ðO1 Þ ¼ 0 0 b 0 1 a a 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @ 1 1 bðc 1Þ A; J2 ¼ J ðO2 Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi bðc 1Þ bðc 1Þ b 0 1 a a 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @ 1 1 bðc 1Þ A: J3 ¼ J ðO3 Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi bðc 1Þ bðc 1Þ b A simple calculation shows that matrix J1 has the following eigenvalues: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ ða 1Þ2 þ 4ac : k1 ¼ b; k2;3 ¼ 2
ð2Þ
Analogously, for matrices J2 and J3 , one has the following identity characteristic polynomials, k3 þ ða þ b þ 1Þk2 þ bða þ cÞk þ 2abc ¼ 0:
ð3Þ
It is easy to see that equilibrium point O1 is unstable, and that by Routh–Hurwitz conditions, O2 and O3 are also unstable under the condition: b < ½aða cÞ=ða þ cÞ 1 (this condition will be always assumed in the following).
Fig. 1. The typical Lorenz chaotic attractors.
T. Zhou et al. / Chaos, Solitons and Fractals 19 (2004) 863–873
865
Fig. 2. The chaotic time series of zðtÞ (a ¼ 10, b ¼ 8=3, c ¼ 28), where Ô Õ represents the line z ¼ c 1 and Ô- - -Õ the line z ¼ c 1 þ ðða þ 1Þ2 =4aÞ.
For the sake of understanding, the approximate scenario of trajectories of (1) is first shown and the rigorous proof is then given in next sections. For this, consider the following system: 0 x ¼ aðy xÞ ð4Þ y 0 ¼ ðc zÞx y; where z is treated as a parameter. Note that (4) has only one equilibrium point ð0; 0Þ when z 6¼ c 1. Moreover, the Jacobian matrix of (4) at the point is a a A¼ ; c z 1 and the eigenvalues corresponding to A are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ ða þ 1Þ2 4aðz þ 1 cÞ : P1;2 ¼ 2
ð5Þ
From (5), one easily sees that if z > c 1 þ ðða þ 1Þ2 =4aÞ, then ð0; 0Þ is a stable focus of (4); if c 1 < z < c 1 þ ðða þ 1Þ2 =4aÞ, then ð0; 0Þ is a stable nodal, and if z < c 1, then ð0; 0Þ is a saddle point. Intuitively, the trajectories of (1) do not go very far on the left of the Poincare projection z ¼ c 1 because of boundedness of solutions of (1). Therefore, the trajectories will possibly cross the projection and enter its right since otherwise, they would converge to one of three equilibria of (1). Thus, at some time they possibly travel through the Poincare projection z ¼ c 1 from the right to the left and come back to the left of the projection again. Moreover, at another moment they possibly cross another Poincare projection z ¼ c 1 þ ðða þ 1Þ2 =4aÞ from the left to the right and enter the right. But, they cannot always stay on the right of z ¼ c 1 þ ðða þ 1Þ2 =4aÞ. Otherwise, they would also converge to a certain equilibrium point of (1). That is to say, they cross the Poincare projection z ¼ c 1 þ ðða þ 1Þ2 =4aÞ again and return its left. Like this, the nontrivial trajectories of (1) travel through these two Poincare projections repeatedly (see Fig. 2). In Section 4, it will be rigorously proved that all nontrivial trajectories of (1) definitely travel through the two Poincare projections z ¼ c 1 and z ¼ c 1 þ ðða þ 1Þ2 =4aÞ with infinitely many times. For it, the lemmas in the next section will be needed.
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3. Lemmas Denote by ðxðtÞ; yðtÞ; zðtÞÞ an orbit of (1). Lemma 1. For any interval ða; bÞ, if xðtÞ const. as t 2 ða; bÞ, then ðxðtÞ; yðtÞ; zðtÞÞ is merely the trivial solution of (1), namely, one of O1 , O2 and O3 . Similar conclusion holds for yðtÞ and zðtÞ. In other words, if x0 ðtÞ 6 0, t 2 ð1; þ1Þ, then for any ða; bÞ, one has x0 ðtÞ 6 0, y 0 ðtÞ 6 0 z0 ðtÞ 6 0, for t 2 ða; bÞ. Proof. In the case that xðtÞ const. for t 2 ða; bÞ, one has x0 ðtÞ 0 as t 2 ða; bÞ. By (1), one sees xðtÞ ¼ yðtÞ const. for t 2 ða; bÞ. Therefore, y 0 ðtÞ 0 as t 2 ða; bÞ. Also, by the second equation of (1), one has zðtÞ ¼ const. which implies that z0 ðtÞ 0 as t 2 ða; bÞ. Thus, in the interval ða; bÞ, ðxðtÞ; yðtÞ; zðtÞÞ ¼ O1 , or O2 , or O3 . However, O1 , O2 and O3 are three trivial solutions of (1). By a fundamental property of ordinary differential equations, one can determine that ðxðtÞ; yðtÞ; zðtÞÞ must be one of O1 , O2 and O3 as t 2 ð1; 1Þ. It is impossible since the lemma assumes x0 ðtÞ 6 0 for t 2 ð1; 1Þ. The lemma is thus proved. h Lemma 2. If limt!þ1 xðtÞ, limt!þ1 yðtÞ, limt!þ1 zðtÞ all exist, then ðxðtÞ; yðtÞ; zðtÞÞ must be a trivial orbit of (1). In fact, assume limt!þ1 xðtÞ ¼ n, limt!þ1 yðtÞ ¼ g, limt!þ1 zðtÞ ¼ f. By (1), limt!þ1 x0 ðtÞ, limt!þ1 y 0 ðtÞ, limt!þ1 z0 ðtÞ all exist. But, ðxðtÞ; yðtÞ; zðtÞÞ are all boundary. So, limt!þ1 x0 ðtÞ ¼ limt!þ1 y 0 ðtÞ ¼ limt!þ1 z0 ðtÞ ¼ 0. Thus, n, g and f must satisfy 8
Note that lim lim
Dt!0 t!þ1
xðtÞð1 eaDt Þ xðtÞð1 eaDt Þ ¼ lim lim ¼ an; t!þ1 Dt!0 Dt Dt
which means that the above two limits are exchangeable. Therefore, R tþDt R tþDt eat t yðsÞeas ds eat t yðsÞeas ds ¼ lim lim ¼ n: lim lim t!þ1 Dt!0 Dt!0 t!þ1 Dt Dt This implies that limt!1 yðtÞ not only exists but also equates to n. Also by (1), one has limt!þ1 x0 ðtÞ ¼ 0. If n ¼ 0, then limt!þ1 xðtÞ ¼ limt!þ1 yðtÞ ¼ 0. By y 0 ðtÞ ¼ ðc zðtÞÞxðtÞ yðtÞ and boundedness of zðtÞ, limt!þ1 y 0 ðtÞ ¼ 0. In addition, it follows from z0 ðtÞ ¼ xðtÞyðtÞ bðtÞ that for any T1 > T , Z t xðsÞyðsÞebs ds: ð8Þ zðtÞ ¼ zðT1 ÞebðT1 tÞ þ ebt T1
So, limt!þ1 zðtÞ exists and equates to zero. This is a contradiction by Lemma 2. If n 6 0, then since xðtÞ keeps the constant sign for t P R t T , and limt!þ1 xðtÞ ¼ limt!þ1 yðtÞ, there exists a T2 > T , such that yðtÞ keeps the constant sign for t P T2 . So limt!1 ð T3 xðsÞyðsÞebs dsÞebt exists for any T3 P T2 . By (8), one sees that limt!1 zðtÞ exists. It is also a contradiction by Lemma 2. In a word, x0 ðtÞ > 0 for t P T is impossible. Secondly, assume y 0 ðtÞ > 0 for t P T . Then yðtÞ keeps the constant sign as t P T . By the boundedness of yðtÞ, one Rt knows that limt!þ1 yðtÞ exists. Furthermore, limt!þ1 eat b yðsÞeas ds exists for any b P T . It follows from x0 ðtÞ ¼ aðyðtÞ xðtÞÞ that Z t yðsÞeas ds: ð9Þ xðtÞ ¼ xðbÞeaðbtÞ þ aeat b
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So, limt!þ1 x0 ðtÞ exists. Moreover, limt!þ1 x0 ðtÞ must be zero because of the boundedness of xðtÞ. Thus, limt!þ1 xðtÞ ¼ limt!þ1 yðtÞ. In addition, yðtÞ keeps the constant sign as t is sufficiently large since xðtÞ keeps the constant sign. Therefore, xðtÞyðtÞ keeps the constant sign as t is large enough. By (8), limt!þ1 zðtÞ exists. By Lemma 2, one then encounters a contradiction. Finally, assume z0 ðtÞ keeps the constant sign for t P T . Then zðtÞ also keeps the constant sign for t P T . By boundedness of zðtÞ, limt!þ1 zðtÞ must exist. Furthermore, xðtÞyðtÞ keeps the constant sign since xðtÞyðtÞ ¼ z0 ðtÞ þ bzðtÞ. That is to say, xðtÞ and yðtÞ all keep the constant sign as t is sufficiently large. By (1), one sees that x0 ðtÞ and y 0 ðtÞ all keep the constant sign as t is large enough. Therefore, limt!1 xðtÞ, limt!þ1 yðtÞ all exist. By Lemma 2 and the assumed condition, it is impossible. In conclusion, Lemma 3 holds. h Corollary 1. Assume x0 ðtÞ 6 0 for t 2 ðþ1; þ1Þ. Then, for any T < þ1, x0 ðtÞ, y 0 ðtÞ and z0 ðtÞ all change the sign with infinitely many times in the interval ðT ; þ1Þ. Corollary 2. Assume that x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, for any T < þ1, xðtÞ, yðtÞ and zðtÞ all have infinitely local extreme value points in the interval ðT ; þ1Þ. Lemma 4. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. If there exists b such that x0 ðbÞ ¼ x00 ðbÞ ¼ 0, then t ¼ b is the inflection point of xðtÞ, i.e. t ¼ b is not the extreme value point of xðtÞ. The similar conclusion holds for yðtÞ and zðtÞ. Proof. For such a b, one has by (1) yðbÞ ¼ ð1=aÞx0 ðbÞ þ xðbÞ ¼ xðbÞ and y 0 ðbÞ ¼ ð1=aÞx00 ðbÞ þ x0 ðbÞ ¼ 0. Furthermore, by (1), ½c 1 zðbÞxðbÞ ¼ 0:
ð10Þ
If xðbÞ ¼ 0, then y 00 ðbÞ ¼ ½czðbÞx0 ðbÞ z0 ðbÞxðbÞ y 0 ðbÞ ¼ 0. Therefore, xð3Þ ðbÞ ¼ aðy 00 ðbÞ x00 ðbÞÞ ¼ 0. Inductively, one can obtain that xðnÞ ðbÞ ¼ 0;
n ¼ 1; 2; . . .
ð11Þ
Consider the Taylor expansion of xðtÞ at t ¼ b, xðtÞ ¼
1 X xðnÞ ðbÞ ðt bÞn : n! n¼0
ð12Þ
So, xðtÞ 0 at a neighborhood of t ¼ b. By Lemma 1, it is impossible. Thus, it follows from (10) that zðbÞ ¼ c 1:
ð13Þ
A direct calculation yields xð3Þ ðbÞ ¼ az0 ðbÞxðbÞ:
ð14Þ
Note a fact that x0 ðbÞ, y 0 ðbÞ and z0 ðbÞ cannot equate to zero at the same time. Thus, z0 ðbÞxðbÞ 6¼ 0, which implies that xð3Þ ðbÞ 6¼ 0. This indicates that t ¼ b could not be the extreme value point of xðtÞ. Similarly, one can obtain the same conclusion for yðtÞ and zðtÞ. The lemma is hence proved. h Corollary 3. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. At every extreme value point of xðtÞ, x00 ðtÞ 6¼ 0 holds. The same conclusion applies to yðtÞ and zðtÞ. Lemma 5. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, for any T < þ1, it is impossible to have zðtÞ < 0 for all t P T . Proof. Conversely, there would exist a T < þ1 such that zðtÞ < 0 for t P T . By Corollary 2, zðtÞ has infinitely locally extreme value point as t P T . Take a locally maximum value point, denoted by b, of zðtÞ. Then zðbÞ < 0;
z0 ðbÞ ¼ 0;
z00 ðbÞ < 0:
ð15Þ
By z0 ðbÞ ¼ 0 and (1), one has xðbÞyðbÞ ¼ bzðbÞ < 0. Let V ðtÞ ¼ xðtÞyðtÞ. Then V 0 ðbÞ ¼ z00 ðbÞ < 0. However, V 0 ðbÞ ¼ x0 ðbÞyðbÞ þ xðbÞy 0 ðbÞ ¼ ay 2 ðbÞ ða þ 1ÞxðbÞyðbÞ þ ðc zðbÞÞx2 ðbÞ > 0. This is a contradiction, completing the proof of the lemma. h
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T. Zhou et al. / Chaos, Solitons and Fractals 19 (2004) 863–873 z
a
z=0
a1 a0
t
b
Fig. 3. Sketch diagram of zðtÞ corresponding to the proof of Lemma 6.
From the proof procedure of the lemma, one sees that Corollary 4. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, there exists no maximum point for zðtÞ in the side of zðtÞ < 0. Lemma 6. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, there exists a T < þ1 such that zðtÞ > 0 for all t P T . Proof. Conversely, for any T < þ1, there would exist an instant t0 P T such that zðt0 Þ < 0. By Corollaries 2 and 4, there exist two instants a and b such that zðaÞ < 0;
z0 ðaÞ ¼ 0;
zðbÞ > 0;
z0 ðbÞ ¼ 0;
z00 ðbÞ < 0;
z0 ðtÞ > 0;
z00 ðtÞ > 0
for tða; bÞ:
ð16Þ
And zðtÞ has no extreme value point as t 2 ½a; bÞ (see Fig. 3). Moreover, there exists a unique a0 2 ða; bÞ such that zða0 Þ ¼ 0, and zðtÞ < 0 as t 2 ða; a0 Þ while zðtÞ > 0 as t 2 ða0 ; bÞ. Now, one can determine both x0 ðtÞ 6¼ 0 and y 0 ðtÞ 6¼ 0 as t 2 ða; bÞ. In fact, by z0 ðaÞ ¼ 0 and z0 ðbÞ ¼ 0, one has xðaÞyðaÞ ¼ bzðaÞ < 0;
xðbÞyðbÞ ¼ bzðbÞ > 0:
ð17Þ
Without loss of generality, assume xðaÞ < 0;
yðaÞ > 0;
xðbÞ > 0;
yðbÞ > 0:
ð18Þ
0
If there exists an a1 such that x ða1 Þ ¼ 0 and a1 2 ða; bÞ, then a < a1 6 a0 < b (see Fig. 3). Also, without loss of generality, assume x00 ða1 Þ < 0. Then, in a left small interval of t ¼ a1 , one has x0 ðtÞ > 0 while in a right small interval of t ¼ a1 , x0 ðtÞ < 0. Therefore a1 ¼ a0 . Furthermore, by (1), yða0 Þ ¼ ð1=aÞx0 ða0 Þ þ xða0 Þ ¼ 0. Also by (1), one has y 0 ða0 Þ ¼ ðc zða0 ÞÞxða0 Þ yða0 Þ ¼ 0. So, x00 ða0 Þ ¼ 0. Inductively, one can obtain xðnÞ ða0 Þ ¼ 0, n ¼ 1; 2; . . . Considering Taylor expansions of xðtÞ in a neighborhood of t ¼ a0 , one will encounter a contradiction. Similarly, one can prove y 0 ðtÞ 6¼ 0 for all t 2 ða; bÞ. Because of this and by (18), there exists a unique instant, still denoted by a1 , such that xða1 Þ ¼ 0 and a < a1 < a0 < b. Since xðaÞ < 0 and yðaÞ > 0, x0 ðaÞ ¼ aðyðaÞ xðaÞÞ > 0. Thus, x0 ðtÞ > 0 for t 2 ða; bÞ. Furthermore, one can show yðbÞ > xðbÞ. It follows from z00 ðbÞ < 0 that ay 2 ðbÞ ða þ 1ÞxðbÞyðbÞ þ ðc zðbÞÞx2 ðbÞ < 0, which yields qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ 1 þ ða þ 1Þ2 þ 4aðzðbÞ þ 1 cÞ xðbÞ: ð19Þ yðbÞ < 2a Combining (19) with yðbÞ > xðbÞ, one has zðbÞ > c 1:
ð20Þ
Therefore, y 0 ðbÞ ¼ ðc zðbÞÞxðbÞ yðbÞ < ðc 1 zðbÞÞxðbÞ < 0. Moreover, yðtÞ < 0 as t 2 ða; bÞ. Summarizing the above discussions, one sees z0 ðtÞ > 0, x0 ðtÞ > 0, y 0 ðtÞ < 0 and yðtÞ > 0 as t 2 ða; bÞ. In addition, xðtÞ < 0 as t 2 ða; a1 Þ while xðtÞ > 0 as t 2 ða1 ; bÞ. Let V ðtÞ ¼ xðtÞyðtÞ. Then, V 0 ðtÞ > 0 as t 2 ða; a1 Þ and V 0 ðtÞ < 0 as t 2 ða1 ; bÞ. Since z00 ðaÞ > 0 and z00 ðbÞ < 0, V 0 ða1 Þ ¼ 0. Note that xða1 Þ ¼ 0 and V 0 ða1 Þ ¼ ay 2 ða1 Þ ða þ 1Þxða1 Þyða1 Þ þ ðc zða1 ÞÞx2 ða1 Þ. So, yða1 Þ ¼ 0. By (1), one inductively has xðnÞ ða1 Þ ¼ 0, n ¼ 1; 2; . . . This is impossible. In conclusion, Lemma 6 holds. h Lemma 7. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ. Then, for any T < þ1, neither yðtÞ < 0 for all t P T nor yðtÞ > 0 for all t P T , i.e. yðtÞ definitely changes the sign in the interval ðT ; þ1Þ. The similar conclusion applies to xðtÞ.
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Proof. First, consider the R t case for yðtÞ. Conversely, there would exist a T < þ1 such that yðtÞ > 0 for t P T . By (1) one has (9). But,R limt!þ1 b yðsÞeas ds ¼ þ1 since yðtÞ > 0 for t P b P T . According to boundedness of xðtÞ and (9), t limt!þ1 eat b yðsÞeas ds must exist. limt!þ1 xðtÞ then exists. Furthermore, by (9), there exists a t0 > b such that xðtÞ keeps the positive sign as t P b. Therefore, xðtÞyðtÞ keeps the constant sign as t P b. By (8), one knows that limt!þ1 zðtÞ must exist. According to Lemma 2, this is impossible. Secondly, assume xðtÞ > 0 for t P T . By Lemma 6, without loss of generality, one may assume zðtÞ > 0 as t P T . By (1), one has for any fixed b P T , Z t yðtÞ ¼ yðbÞebt þ et ðc zðsÞÞxðsÞet ds: ð21Þ b
Since xðtÞ and zðtÞ all have the positive sign, limt!þ1 yðtÞ must exist by (9). Besides, by (9), limt!þ1 xðtÞ also exists. Furthermore, by (8), limt!þ1 zðtÞ exists. A collection of the above discussion contracts to the conclusion of the Lemma 2. Lemma 7 is then proved. h Corollary 5. Assume x0 ðtÞ 6 for t 2 ð1; þ1Þ. Then, for any T < þ1, xðtÞ and yðtÞ all change the sign with infinitely many times in the interval ðT ; þ1Þ. After having the above preparations, now one may prove the main results of this paper.
4. Main results Theorem 1. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ, a > 1 and b > 1. For any T < þ1, then the trajectories of (1) definitely travel through the Poincare z ¼ c 1 from the left to the right at some moments t: t P T . Proof. Conversely, there would exist a T < þ1 such that zðtÞ < c 1 as t P T . According to Corollary 2, one sees that zðtÞ has at least one maximum value point, denoted by b. Furthermore, by Corollary 3 one has that there exists a small neighborhood of t ¼ b, denoted by ða; cÞ (where a < b < c) such that z0 ðbÞ ¼ 0, z00 ðbÞ < 0, zðtÞ < zðbÞ as t 6¼ b and t 2 ða; cÞ. Note that z0 ðbÞ ¼ 0 yields xðbÞyðbÞ ¼ bzðbÞ:
ð22Þ
By Lemma 6, without loss of generality, assume xðbÞ > 0, yðbÞ > 0. Now, one can determine x0 ðbÞ 6¼ 0. Otherwise, it would have yðbÞ ¼ xðbÞ. By (1), y 0 ðbÞ ¼ ðc 1 zðbÞÞxðbÞ > 0. On the other hand, 0 > z00 ðbÞ ¼ x0 ðbÞyðbÞ þ xðbÞy 0 ðbÞ ¼ xðbÞy 0 ðbÞ > 0. This is a contradiction. In addition, one can determine y 0 ðbÞ 6¼ 0. Conversely, y 0 ðbÞ ¼ 0 will yield ðc zðbÞÞxðbÞ ¼ yðbÞ:
ð23Þ
Thus, by (22) and (23) one obtains x2 ðbÞ ¼
bzðbÞ ; c zðbÞ
y 2 ðbÞ ¼ bðc zðbÞÞzðbÞ:
ð24Þ
However, z00 ðbÞ ¼ ay 2 ðbÞ ða þ 1ÞxðbÞyðbÞ þ ðc zðbÞÞx2 ðbÞ ¼ zðbÞ½abðc 1 zðbÞÞ þ ab a þ 1 þ b > 0 as b > 1. It also produces a contradiction. So y 0 ðbÞ 6¼ 0. Next, one needs to distinguish the following four cases: (1) (2) (3) (4)
x0 ðbÞ > 0, x0 ðbÞ < 0, x0 ðbÞ < 0, x0 ðbÞ > 0,
y 0 ðbÞ > 0; y 0 ðbÞ < 0; y 0 ðbÞ > 0; y 0 ðbÞ < 0.
In the following discussions, note that xðbÞ > 0;
yðbÞ > 0;
zðbÞ < c 1;
z0 ðbÞ ¼ 0;
z00 ðbÞ < 0:
Case 1: x0 ðbÞ > 0, y 0 ðbÞ > 0. This case is impossible since 0 > z00 ðbÞ ¼ x0 ðbÞyðbÞ þ xðbÞy 0 ðbÞ > 0. Case 2: x0 ðbÞ < 0, y 0 ðbÞ < 0.
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In this case, it follows from y 0 ðtÞ ¼ ðc 1 zðtÞÞxðtÞ þ xðtÞ yðtÞ that there exist a and c, a < b < c, such that y ðtÞ > xðtÞ yðtÞ as t 2 ða; cÞ. So, when t 2 ðb; cÞ, 0
yðtÞ > yðbÞebt þ xðtÞ xðtÞebt :
ð25Þ
0
Furthermore, by x ¼ ay ax, one has xðtÞ > xðbÞ þ aðyðbÞ xðbÞÞð1 ebt Þ
ð26Þ
for t 2 ðb; cÞ. (25) multiplying (26) will yield for t 2 ðb; cÞ ð27Þ
xðtÞyðtÞ > gðtÞ; where gðtÞ ¼ xðbÞyðbÞebt þ axðtÞðyðbÞ xðbÞÞð1 2ebt þ e2ðbtÞ Þ þ ½ayðbÞðyðbÞ xðbÞÞ þ xðtÞxðbÞð1 ebt Þ. It follows from z0 ¼ xy bz that one has for t 2 ðb; cÞ, Z t gðsÞebs ds: zðtÞ > zðbÞebt þ ebt
ð28Þ
b
The direct calculation will lead to zðtÞ zðbÞebðbtÞ ¼ z0 ðbÞ þ bzðbÞ ¼ bzðbÞ; tb t!b Rt ebt b gðsÞebs ds ¼ xðbÞyðbÞ: lim tb t!bþ Rt Let GðtÞ ¼ zðtÞ zðbÞebðbtÞ ebt b gðsÞebs ds. Then GðtÞ > 0 as t 2 ðb; cÞ, and 2 limt!bþ GðtÞ=ðt bÞ2 ¼ z00 ðbÞ ay 2 ðbÞ þ ða þ 1ÞxðbÞyðbÞ x2 ðbÞ P 0. By z00 ðbÞ < 0, one has ðyðbÞ xðbÞÞðayðbÞ xðbÞÞ < 0. This is impossible since x0 ðbÞ > 0 and a > 1. Case 3: x0 ðbÞ < 0, y 0 ðbÞ > 0. Similar to the proof in the case 2, one will encounter a contradiction. Case 4: x0 ðbÞ > 0, y 0 ðbÞ < 0. In this case, if considers the left small interval ða; bÞ of t ¼ b in the proof procedure in the case 2, one will also obtain a contradiction. Theorem 1 is thus proved. h limþ
Corollary 6. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, there exists no locally maximum value point for both xðtÞ and yðtÞ within the side of z < c 1. Similarly, one can prove Theorem 2. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, all trajectories of (1) definitely cross the Poincare projection z ¼ c 1 þ ðða þ 1Þ2 =4aÞ from the right to the left at some moments. Corollary 7. Under the assumed conditions of Theorem 2, the exists no locally minimal value point for xðtÞ and yðtÞ within the side of z > c 1 þ ðða þ 1Þ2 =4aÞ. Theorem 3. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, for any T < þ1 all trajectories of (1) definitely travel through the Poincare z ¼ c 1 from the right to the left at some instances. Proof. Otherwise, there would have zðtÞ > c 1 for t P T . By Corollaries 2 and 6, one may assume that there exists a b > T such that c 1 < zðbÞ < c 1 þ ðða þ 1Þ2 =4aÞ, z0 ðbÞ ¼ 0, z00 ðbÞ > 0. By Lemma 6 and z0 ðbÞ ¼ 0, without loss of generality, assume xðbÞ > 0, yðbÞ > 0. There exists a neighborhood, denoted by ða; cÞ, a < b < c, of t ¼ b, such that c 1 < zðtÞ < c 1 þ ðða þ 1Þ2 =4aÞ as t 2 ða; cÞ. In the neighborhood, by y 0 ¼ ðc zÞx y one has ! ða þ 1Þ2 0 xðtÞ yðtÞ > y ðtÞ > 1 xðtÞ yðtÞ: ð29Þ 4a It follows from z0 ðbÞ ¼ 0, z00 ðbÞ > 0 that
T. Zhou et al. / Chaos, Solitons and Fractals 19 (2004) 863–873
yðbÞ <
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ ða þ 1Þ2 þ 4aðzðbÞ cÞ 2a
871
xðbÞ
ð30Þ
xðbÞ > xðbÞ:
ð31Þ
or
yðbÞ >
ða þ 1Þ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða þ 1Þ2 þ 4aðzðbÞ cÞ 2a
However, (30) does not appear since zðbÞ > c 1, xðbÞ > 0 and yðbÞ > 0. Moreover, by (1), one sees x0 ðbÞ > 0. Furthermore, 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 a þ 1 þ ða þ 1Þ2 þ 4aðzðbÞ cÞ 7 6 0 y ðbÞ ¼ ðc zðbÞÞxðbÞ yðbÞ < 4c zðbÞ 5xðbÞ < 0 2a as c 1 < zðbÞ < c 1 þ ðða þ 1Þ2 =4aÞ. So, y 0 ðbÞ < 0. Without loss of generality, it is assumed that x0 ðtÞ > 0, y 0 ðtÞ < 0 as t 2 ða; cÞ. When t 2 ðb; cÞ, by (29) one knows that ! ða 1Þ2 ða 1Þ2 xðtÞ þ xðtÞ þ yðbÞ ebt < yðtÞ < xðtÞ þ ðyðbÞ xðtÞÞebt : ð32Þ 4a 4a Therefore, by x0 ðtÞ ¼ aðyðtÞ xðtÞÞ, one has " # ða 1Þ2 ða þ 1Þ2 a xðtÞ þ yðbÞ ebt xðtÞ < x0 ðtÞ < a½yðbÞ xðtÞebt : 4a 4
ð33Þ
Integrating both sides of (33) from b to t for t 2 ½b; cÞ will lead to " # 2 4a ½ða1Þ2 =4ðbtÞ þ xðbÞ þ yðbÞ ebt e½ðaþ1Þ =4ðbtÞÞ < xðtÞ < xðbÞ þ a½yðbÞ xðbÞð1 ebt Þ: xðbÞe 2 ða 1Þ
ð34Þ
Thus, xðtÞyðtÞ > g1 ðtÞ
ð35Þ
as t 2 ½b; mÞ, where " ða 1Þ2 g1 ðtÞ ¼ xðtÞ þ 4a (
! # ða 1Þ2 xðtÞ þ yðbÞ ebt 4a " # ) 4a ½ða1Þ2 =4aðbtÞ bt ½ðaþ1Þ2 =4ðbtÞ þ xðbÞ þ yðbÞ ðe e Þ : xðbÞe ða 1Þ2
Also, it follows from z0 ¼ xy bt that Z t g1 ðsÞebs ds zðtÞ > zðbÞebðbtÞ þ ebt
ð36Þ
b
as t 2 ½b; mÞ. Combined (36) with z0 ðbÞ ¼ 0, z00 ðbÞ > 0, a direct calculation will yield ð37Þ
xðbÞyðbÞ 6 0: So, either xðbÞ 6 0 or yðbÞ 6 0 which contradict xðbÞ > 0 and yðbÞ > 0. The lemma is then proved.
h
Corollary 8. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. For any T < þ1, then zðtÞ has no minimal value within the side of zðtÞ > c 1 for t P T . Similarly, one can prove
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Theorem 4. Assume x0 ðtÞ 6 0 for t 2 ð1; þ1Þ and a > 1, b > 1. Then, for any T < þ1, all trajectories of (1) definitely travel through the Poincare z ¼ c 1 þ ðða þ 1Þ2 =4aÞ from the left to the right at some moments. Corollary 9. Under assumed conditions of Theorem 4, for any T > þ1, zðtÞ has no maximum value within the side of zðtÞ < c 1 þ ðða þ 1Þ2 =4aÞ for t P T . A combination of Theorems 1–4 gives Theorem 5 (Main result). Assume a > 1, b > 1, c > 1, b < ½aða 1Þ=ða þ 1Þ 1. Then, all nontrivial half-positive trajectories of (1) definitely travel through two Poincare projections z ¼ c 1 and zðtÞ < c 1 þ ðða þ 1Þ2 =4aÞ with infinitely many times.
5. Conclusion This paper has proved that all nontrivial trajectories of the Lorenz system cross two special Poincare projections under some conditions. By writing down instants at which these orbits travel through the Poincare projections, one may be able to capture some periodic orbits embedded in the Lorenz attractor. In addition, if z is viewed as a time-varying parameter, then the projected trajectories of the system on the ðx; yÞ planes can be divided into a sequence of small projections, each of which has different characteristics from its two other projections. The fine structure of the Lorenz attractor can be thus described. In spite of this, further studies remain to be carried out, for example, how to distinguish quasi-periodic orbits from chaotic orbits.
Acknowledgements The first author thanks the hospitality of the Kanazawa University of Japan, where he conducted part of this research under the fellowship awarded by the Japanese Society for the Promotion of Science. This paper is partly supported by National Key Basic Research Special Fund (no. G1998020309), National Nature Science Foundation of China (no. 10272059).
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