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On the stability of double homoclinic and heteroclinic cycles
Nonlinear Analysis 53 (2003) 701 – 713
www.elsevier.com/locate/na
On the stability of double homoclinic and heteroclinic cycles Maoan Hana , Shouchuan Hub;∗ , Xingbo Liuc a Department
of Mathematics, Shanghai Jiao Tong University, 200030 Shanghai, China of Mathematics, Southwest Missouri State University, Spring eld, MO 65804, USA c Department of Mathematics, East China Normal University, 200062 Shanghai, China
b Department
Received 21 September 2001; accepted 2 October 2001
Abstract In this paper we give a criterion for the stability of planar double homoclinic and heteroclinic cycles with one or two saddles in some degenerate case. c 2003 Published by Elsevier Science Ltd. Keywords: Stability; Double homoclinic cycle; Heteroclinic cycle
1. Main results Consider a planar system x˙ = f(x; y);
y˙ = g(x; y);
(1.1)
5
where f; g are at least C functions on the plane. Suppose (1.1) has a heteroclinic cycle L = L1 ∪ L2 with two di3erent hyperbolic saddles Si (xi ; yi ); i = 1; 2. Let i1 ¿ 0 and i2 ¡ 0 be the eigenvalues of the matrix (@(f; g)=@(x; y))(Si ) ≡ Di ; i = 1; 2. Then the hyperbolicity ratio of Si is given by ri = − i2 = i1 . As we know [9], Cherkas proved that if r1 r2 ¿ 1(¡ 1) then L is stable (unstable). Han et al. [5] proved further that if r1 = r2 = 1 and 2 1 = (fx + fy ) dt = (fx + fy ) dt ≡ 11 + 21 ¡ 0(¿ 0); L
i=1
Li
then L is stable (unstable). In this paper, we suppose r1 = r2 = 1 and 1 = 0, and give a criterion for the stability of L in this more degenerate case. ∗
Corresponding author. E-mail address: [email protected] (S. Hu).
c 2003 Published by Elsevier Science Ltd. 0362-546X/03/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 3 0 1 - 2
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Let us Arst recall the deAnition of the Arst saddle value. Let Ti be an inversible matrix such that det Ti = 1;
Ti Di Ti−1 = diag( i1 ; i2 );
i = 1; 2:
(1.2)
By making a linear transformation of the form x − xi u ; = Ti y − yi v we obtain from (1.1) (i) ajl uj vl + O(|u; v|4 ) ; u˙ = i1 u + k=2;3 j+l=k
v˙ = − i2 −v +
k=2;3 j+l=k
(i) j l bjl u v + O(|u; v|4 )
(1.3)
for (u; v) near the origin. Then according to [6] when ri = 1 the Arst saddle value of (1.1) at Si is given by (i) (i) (i) (i) (i) (i) + b12 − a20 a11 + b02 b11 : Ri1 = a21
(1.4)
By [2,6] when ri = 1, there exists a C 5 transformation of the form x − xi u + O(|x − xi ; y − yi |2 ) ≡ Ti∗ (x − xi ; y − yi ); = Ti y − yi v
(1.5)
which carries (1.1) into the normal form u˙ = i1 u[1 + a∗i1 uv + O(u2 v2 )]; v˙ = i1 v[ − 1 + b∗i1 uv + O(u2 v2 )]:
(1.6)
In this case we have Ri1 = a∗i1 + b∗i1 : If
(1.7)
f(u + xi ; v + yi ) = i1 v +
k=2;3 j+l=k
g(u + xi ; v + yi ) = i1 u +
k=2;3 j+l=k
(i) j l ajl u v + O(|u; v|4 ) ;
(i) j l bjl u v + O(|u; v|4 ) ;
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
703
Fig. 1.
then instead of (1.4) and (1.7) we have (i) (i) (i) (i) (i) (i) (i) (i) (i) (i) (i) − b03 ) + (b21 − a12 ) + 2(a02 b02 − a20 b20 ) + a11 (a02 − a20 ) Ri1 = 3(a30 (i) (i) (i) (b02 − b20 ): + b11
(1.8)
We note that the sign of time is retained when we change (1.1) into the form of (1.3) or (1.6). Now for deAniteness we assume (L1 ) = S2
and
L is oriented clockwise:
(1.9)
There are two types of L, as shown in Fig. 1, according to the side of L where the Poincare map is well deAned. Our main result is the following. Theorem 1.1. Suppose (1.9) holds. Let r1 = r2 = 1;
1 (=11 + 21 ) = 0;
2 ≡ R11 + R21
11 11 e = 0:
21
(1.10)
(i) If the Poincare map of (1.1) is well de ned in an inner neighborhood of L, then it is stable (unstable) when 2 ¿ 0(¡ 0). (ii) If the Poincare map of (1.1) is well de ned in an outer neighborhood of L, then it is stable (unstable) when 2 ¡ 0(¿ 0). For case (ii) in Fig. 1 L becomes a double homoclinic cycle if S1 = S2 . In this case it was proved in [4] that if r1 ¿ 1(¡ 1) or =1 and 1 ¡ 0(¿ 0), then L is stable (unstable). By the proof of Theorem 1.1 we have Theorem 1.2. Suppose L = L1 ∪ L2 is a clockwise-oriented double homoclinic cycle with a hyperbolic saddle S1 . Let r1 = 1, 1 = 0 and R11 = 0. If R11 ¡ 0(¿ 0) then L is outer stable (unstable), and, if in addition 11 = 0, L1 and L2 are inner unstable (stable). Remark 1.1. If L is oriented counter clockwise, one can determine its stability by applying Theorems 1.1 and 1.2 to the system x˙ = −f(x; y);
y˙ = −g(x; y):
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Fig. 2.
The proof of the above theorems is given in Section 2 and some application examples are presented in Section 3 to show the way to And limit cycles by using the main results. 2. Proof We Arst consider case (i) in Fig. 1. Choose points A2 ; A3 ∈ L1 near S1 and A1 ; A4 ∈ L2 near S2 . Let li be a cross section of (1.1) normal to L at Ai ∈ Li with the directional vector ni = (−g(Ai ); f(Ai ))=|f(Ai ); g(Ai )|: DeAne a map Pij : li → lj by using the orbits of (1.1). Let B = A1 + P(a)n1 be the Arst intersection point of the positive orbit of (1.1) starting at A = A1 + an1 with l1 . Then the Poincare map P : l1 → l1 satisAes (see Fig. 2) P(a) = (P41 ◦ P34 ◦ P23 ◦ P12 )(a)
for 0 ¡ − a 1:
The maps P12 and P34 are called Dulac maps and P23 and P41 regular maps. To study the Dulac maps we use the normal form (1.6). For i = 1 we get from (1.6) dv v = [ − 1 + R11 uv + O(|uv|2 )]; du u
(2.1)
where R11 is given by (1.7). Let l1 = {v = −p; 0 6 u 6 p};
l2 = {u = p; −p 6 v 6 0}
for p ¿ 0 being small. Let B (p; −q1 (r)) be the Arst intersection point of the positive orbit of (1.6) (i = 1) starting at A (r; −p) ∈ l1 with l2 for 0 ¡ r ¡ p (see Fig. 3 (i)). Set z = uv. We have from (2.1) d z R11 + O(|uv|) = du u z2
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
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Fig. 3.
and
−pq1
−pr
dz = z2
p
r
R11 + O(v) du; u
which yields
1 1 1 = −R11 ln r + O(1) − r p q1 or q1 (r) = r + R11 pr 2 ln r + O(r 2 ):
(2.2)
Lemma 2.1. If we choose A1 and A2 such that T1∗ (A1 − S1 ) = (0; −p)T ;
T1∗ (A2 − S1 ) = (p; 0)T ;
(2.3)
where T1∗ is given by (1.5), then there exist C 5 functions Wi (r)=−Mi (p)r +O(r 2 ); i= 1; 2; such that P12 ◦ W1 = W2 ◦ q, where lim Mi (p) = )i sin *1 ;
p→0
)1 = |T1−1 (1; 0)T |;
)1 )2 sin *1 = 1;
)2 = |T1−1 (0; 1)T |;
and *1 being the inner angle of L1 and L2 at S1 . Proof. We relate A and A such that the orbit AˆB of (1.6) is the image of the orbit of (1.1) passing through A. In this way a is a function of r. We want to prove a = W1 (r). For this purpose let T1∗ (A∗ − S1 ) = A (=(r; −p)T ):
(2.4)
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Then A∗ and A are on the same orbit of (1.1). Let a∗ = |A1 A∗ |. Applying the sine theorem to the triangle with vertices A1 ; A and A∗ , we obtain sin “A∗ |a| = a∗ : (2.5) sin “A Let K1 denote the tangent line of L1 at S1 , and K2 the tangent line of L2 at A1 . Then K2 is the limit position of the line passing through A and A∗ as r → 0. By (2.3) and (2.4) we have A1 − A∗ = (T1∗ )−1 (0; −p) − (T1∗ )−1 (r; −p) = T1−1 (r; o)T + O(r 2 );
(2.6)
which implies that (A1 − A∗ )=r = T1−1 (1; 0)T + O(r) → T1−1 (1; 0)T as r → 0. Note that the above limit vector is parallel to K1 . Thus, K1 is the limit position of the line passing through A1 and A∗ . Hence, noting that l1 is orthogonal to K2 , we obtain “A = -=2 + O(r);
“A∗ = *(p) + O(r);
(2.7)
where *(p) denotes the angle between K1 and K2 satisfying limp→0 *(p) = *1 . By (2.6) we also have a∗ = |A1 − A∗ | = r)1 + O(r 2 ):
(2.8)
Substituting (2.7) and (2.8) into (2.5) we have a = −|a| = −r)1 sin *(p) + O(r 2 ) ≡ W1 (r): In the same way we can prove P12 (a) = W2 (q1 (r)). Hence, it follows that P12 ◦ W1 = W2 ◦ q. Finally, from det T1 = 1 it implies that the area of the region T1−1 ([0; 1] × [0; 1]) ≡ G is equal to 1. Note that G is a parallelogram with sides T1−1 (1; 0)T and T1−1 (0; 1)T . The two sides have the angle -−*1 . Therefore, )1 )2 sin *1 =1. The proof is completed. By (2.2) and Lemma 2.1 we have immediately that if A1 and A2 satisfy (2.3), then )2 )2 pR11 P12 (a) = (1 + op (1))a − 2 (1 + op (1))a2 ln|a| + O(a2 ); (2.9) )1 )1 sin *1 where limp→0 Op (1) = 0: Lemma 2.2. (Han [3]). Let (2.3) hold and r1 = 1. Then for p ¿ 0 small |f(A1 ); g(A1 )| = )2 11 p + O(p2 );
|f(A2 ); g(A2 )| = )1 11 p + O(p2 ):
Proof. Note that f(S1 ) = g(S1 ) = 0. We have (f(A1 ); g(A1 ))T = D1 (A1 − S1 ) + O(|A1 − S1 |2 ); T1∗ (A1 − S1 ) = T1 (A1 − S1 ) + O(|A1 − S1 |2 ):
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
707
Hence, by (1.2) and (2.3), we have (f(A1 ); g(A1 ))T = T1−1 diag( 11 ; − 11 )T1 (A1 − S1 ) + O(p2 ) = T1−1 diag( 11 ; − 11 )(0; −p)T + O(p2 ) = p 11 T1−1 (0; 1)T + O(p2 ); which implies the Arst formula. The second one can be proved similarly. The proof is completed. Similar to (2.9) and Lemma 2.2, we can also prove that if r2 = 1 one can choose A3 ; A4 = S2 + O(p) suitably such that |f(A3 ); g(A3 )| = )4 21 p + O(p2 );
|f(A4 ); g(A4 )| = )3 21 p + O(p2 )
(2.10)
and P34 (a) =
)4 )4 pR21 (1 + op (1)) − 2 (1 + op (1))a2 ln |a| + O(a2 ); )3 )3 sin *2
(2.11)
where )3 )4 sin *2 = 1. Let X1 (t) be a representation of L1 and let t2 ¡ t3 be such that X1 (tj ) = Aj ;
j = 2; 3:
(2.12)
By Lemmas 4.3.2 and 4.3.3 of [8] we can introduce the change of variables (x; y)T = X1 (*) − JV (*)r; where X (*) V (*) = 1 ; |X1 (*)|
J=
t2 6 * 6 t3 ; 0
1
−1
0
(2.13)
;
to (1.1) to obtain dr=d* = A(*)r + O(r 2 );
(2.14)
where A(*) = (fx + gy )(X1 (*)) −
d ln|f(X1 (*)); g(X1 (*))|: d*
Let r(*) be the solution of (2.14) with r(t2 ) = a. The formula of constant variation yields that
t3 |f(A2 ); g(A2 )| r(t3 ) = (fx + gy ) dt a + O(a2 ): (2.15) exp |f(A3 ); g(A3 )| t2 Note that −JV (tj ) = nj ;
j = 2; 3:
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
By (2.12) and (2.13) and the deAnition of P23 , we obtain P23 (a)=r(t3 ). Since r1 =r2 =1 it follows that ([7] or [8, Lemma 2.3.5]) t3 (fx + gy ) dt → 11 (as p → 0): t2
Hence by Lemma 2.2, (2.10) and (2.15) we have P23 (a) =
)1 11 11 e (1 + op (1))a + O(a2 ): )4 21
(2.16)
An analogue procedure yields that P41 (a) =
)3 21 21 e (1 + op (1))a + O(a2 ): )2 11
(2.17)
Then by (2.9), (2.11), (2.16) and (2.17) we obtain
pR11 )2 11 11 2 (1+op (1))a− (1+op (1))a ln|a| +O(a2 ); e (P23 ◦ P12 )(a)= )1 sin *1 )4 21
)4 21 21 pR21 (P41 ◦ P34 )(a)= (1+op (1))a− e (1+op (1))a2 ln|a| +O(a2 ) )2 11 )3 sin *2 and
P(a) = e1 N1 (p)a − p)2 N2 (p)a2 ln|a| + O(a2 );
(2.18)
where N1 (p) = (1 + op (1));
N2 (p) = R11 +
11 R21 e11 + op (1):
21
Here we have used )1 )2 sin *1 = )3 )4 sin *2 = 1 for the expression of N2 . On the other hand, by [1,7] we know that 1 P (a) = e AB (fx +gy ) dt → e1 1 + oa (1) as a → 0, which gives that N1 = 1. Therefore, if 1 = 0, (2.18) becomes P(a) = a − p)2 (2 + op (1))a2 ln|a| + O(a2 ): Now conclusion (i) of Theorem 1.1 is clear. For case (ii) in Fig. 1, we can deAne regular maps P23 near L1 and P41 near L2 , and Dulac maps P12 near S1 and P34 near S2 as before. We can also prove that P23 and P41 have the same form as in the Arst case, and P12 and P34 can be obtained by replacing R11 and R21 by −R11 and −R21 , respectively, in (2.9) and (2.11). This is because the coeIcients a∗i1 and b∗i1 in (1.6) change sign if u is replaced by −u. Thus, in this case we have P(a) = a + p)2 (2 + op (1))a2 ln|a| + O(a2 ); which implies conclusion (ii). Hence, the proof of theorem 1.1 is Anished.
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
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Now suppose L is a homoclinic cycle with a hyperbolic saddle S1 . By [7] we know that if 0 ≡ (fx + gy )(S1 ) ¡ 0(¿ 0) or 0 = 0 and 1 ≡ L (fx + gy ) dt ¡ 0(¿ 0) then L is stable (unstable). Further, by the proof of Theorem 1.1 we have Corollary 2.1. Suppose the homoclinic cycle L is oriented clockwise. Let 0 = 1 = 0 and R11 = 0. Then (i) If the Poincare map of (1.1) is well de ned in an inner neighborhood of L, then it is stable (unstable) when R11 ¿ 0(¡ 0). (ii) If the Poincare map of (1.1) is well de ned in an outer neighborhood of L, then it is stable (unstable) when R11 ¡ 0(¿ 0). For case (ii) in Fig. 1 and S1 = S2 , we have from Theorem 1.1 (ii) Corollary 2.2. Let L be a double homoclinic cycle with a hyperbolic saddle S1 and let L be oriented clockwise. If 0 = 1 = 0 and R11 ¡ 0(¿ 0), then L is stable (unstable). Obviously, Theorem 1.2 follows from the above two corollaries.
3. Application examples In this section, we consider a system of the form x˙ = Hy + aHx H ≡ f(x; y); y˙ = −Hx − Hy H ≡ g(x; y);
(3.1)
where H is a polynomial in (x; y). For (3.1) we have fx + gy = aHx2 − Hy2 + H (aHxx − Hyy ):
(3.2)
We Arst suppose H = 12 (y2 − x2 ) + 14 x4 :
(3.3)
Then (3.1) has a homoclinic cycle Li : H (x; y)=0; (−1)i x ¿ 0 (i=1; 2). Let L=L1 ∪L2 . Proposition 3.1. Suppose (3.3) holds. Then (i) If a ¡ 57 (¿ 57 ), then L, L1 and L2 are all stable (unstable); if a = 57 then L is unstable, and L1 and L2 are stable. (ii) There exists b ¿ 0. such that (3.1) has two “small ” limit cycles Li1 ⊂ Li (i = 1; 2) for a = 57 , has four “small” limit cycles Lij ⊂ Li (i; j = 1; 2) for 57 ¡ a ¡ 57 + b and has two “small ” limit cycles Li1 ⊂ Li (i = 1; 2), and one “large” limit cycle for 57 − b ¡ a ¡ 57 (see Fig. 4).
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Fig. 4.
Proof. By (3.2) we have 11 = 21 = (aHx2 − Hy2 ) dt = a(x − x3 ) dy − y d x: L2
L2
By integration by parts we have
L2
(x − x3 ) dy =
that 11 = 21 = 2
√
2
2
x(3ax − a − 1)
0
L2
(3x2 − 1)y d x. Hence it is direct
1 4 1 − x2 d x = 2 3
7 a−1 : 5
Thus, L, L1 and L2 are all stable (unstable) if a ¡ 57 (¿ 57 ). For a = 57 , by (1.8) we have R11 = 2(a + 1) = 24 7 . Note that L is oriented clockwise. It follows from Theorem 1.2 that L is unstable and L1 and L2 are both stable. It is easy to see that (3.1) has three critical points (0,0) and (±1; 0) in the region H (x; y) 6 0 for a ¡ 16. By (3.2) (fx + gy )(±1; 0) =
1 − 2a : 4
Thus the points (±1; 0) are stable for a = 57 . Hence there is an unstable limit cycle Li1 inside Li (i = 1; 2). For 0 ¡ a − 57 1, Li has changed its stability from stable into unstable. Hence a stable limit cycle Li2 between Li and Li1 has been bifurcated from Li . For 0 ¡ 57 − a 1, L has changed its stability and a large unstable limit cycle surrounding L has appeared. The proof is completed. Now we suppose H = x(y2 + x − 1): Then (3.1) has a heteroclinic cycle L : H (x; y) = 0;
x ¿ 0; |y| 6 1;
(3.4)
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
711
with L1 : x = 0; |y| 6 1 and L2 : y2 + x − 1 = 0; 0 6 x 6 1. Note that Hx = 2x + y2 − 1; Hy = 2xy and d x = −2y dy along L2 . We have by (3.2) 1 = −aHx dy − Hy d x L
and
11 =
−1
21 =
=
− a(y2 − 1) dy = −2a
−1
1
=
1
1
−1
− a(1 − y2 ) dy −
−a(y2 − 1) dy +
−1
1 1
−1
4a 1 −1 = ; 3 3 2y(1 − y2 )(−2y) dy
4(y4 − y2 ) dy
4a 16 4a 1 1 = +4×2 − − : 3 5 3 3 15
Hence, 1 =
8 15 [5a
− 2]:
Thus, L is stable (unstable) if a ¡ 25 (¿ 25 ). For a = 25 , removing the saddle S2 (1; 0) to the origin we obtain x˙ = 2[x + xy + ax3 + 3ax2 y + 2xy2 ] + O(|x; y|4 ); y˙ = 2[ − x − y − x3 − 12 y2 − 2x2 y] + O(|x; y|4 ): Introducing a new variable v = 12 x + y, the above system becomes
1 1−a 3 x˙ = 2 x − x2 + xv + x + (3a − 2)x2 v + 2xv2 + O(|x; v|4 ); 2 2
3 2 1 2 1−a 3 3 2 2 v˙ = 2 −v − x + xv − v + x + (a − 2)x v + xv + O(|x; v|4 ): 8 2 4 2 Then (1.4) gives that R21 = 3a − 1 = 15 . Similarly, removing S1 (−1; 0) to the origin and introducing a new variable v = − x=2 + y, we obtain x˙ = 2[ − x + xy + ax3 − 3ax2 y + 2xy2 ] + O(|x; y|4 ); y˙ = −2x − y2 + 2y + 2x3 − 4x2 y + O(|x; y|4 ) and
1 2 3 2 3 1−a 3 2 2 x + O(|x; v|4 ); v˙ = 2 v − v − xv − x − v x + (a − 2)vx − 2 8 2 4
1 2 1−a 3 2 2 x˙ = 2 −x + vx + x + 2v x + (2 − 3a)vx + x + O(|x; v|4 ): 2 2
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Fig. 5.
Thus, 1 R11 = 1 − 3a = − : 5 Therefore, 1 2 = (e8=15 − 1) ¿ 0: 5 It follows from Theorem 1.1 that L is stable for a = 25 . By (3.2) and (3.4) we have fx + gy 12 ; 0 = − 14 (2a − 1): Thus the critical point ( 12 ; 0) is unstable for a ¡ 12 . It is not hard to see that ( 21 ; 0) is stable for a ¿ 12 . Then similar to Proposition 3.1 we have Proposition 3.2. Let (3.4) hold. Then (i) The heteroclinic cycle L is stable (unstable) if a 6 25 (a ¿ 25 ). (ii) There exist a stable limit cycle L1 ⊂ L for 25 ¡ a ¡ 12 (see Fig. 5).
References [1] S.N. Chow, J.K. Hale, Methods of Bifurcation Theory, Springer, New York, 1982. [2] M. Han, On the number of limit cycles bifurcated from a homoclinic or heteroclinic loop, Sci. China (Ser. A) 2 (1993) 113–122. [3] M. Han, Bifurcations of limit cycles from a heteroclinic cycle of Hamiltonian systems, Chin. Ann. Math 19B (2) (1998) 189–196. [4] M. Han, D. Luo, D. Zhu, The uniqueness of limit cycles bifurcated from a separatrix cycle (II), Acta Math. Sin. 4 (1992) 541–548. [5] M. Han, D. Luo, D. Zhu, The uniqueness of limit cycles bifurcated from a separatrix cycle (III), Acta Math. Sin. 5 (1992) 673–684.
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[6] P. Joyal, Generalized Hopf bifurcation and its dual generalized homoclinic bifurcation, SIAM J. Math. 48 (1988) 481–496. [7] D. Luo, M. Han, D. Zhu, The uniqueness of limit cycles bifurcated from a separatrix cycle (I), Acta Math. Sin. 3 (1992) 407–417. [8] D. Luo, X. Wang, D. Zhu, M. Han, Bifurcation Theory and Methods of Dynamical Systems, in: Advanced Series in Dynamical Systems, Vol. 15, World ScientiAc, Singapore, 1997. [9] Y. Ye, et al., Theory of Limit Cycles, Translations of Mathematical Monographs, Vol. 66, American Mathematical Society, Providence, RI, 1986.
www.elsevier.com/locate/na
On the stability of double homoclinic and heteroclinic cycles Maoan Hana , Shouchuan Hub;∗ , Xingbo Liuc a Department
of Mathematics, Shanghai Jiao Tong University, 200030 Shanghai, China of Mathematics, Southwest Missouri State University, Spring eld, MO 65804, USA c Department of Mathematics, East China Normal University, 200062 Shanghai, China
b Department
Received 21 September 2001; accepted 2 October 2001
Abstract In this paper we give a criterion for the stability of planar double homoclinic and heteroclinic cycles with one or two saddles in some degenerate case. c 2003 Published by Elsevier Science Ltd. Keywords: Stability; Double homoclinic cycle; Heteroclinic cycle
1. Main results Consider a planar system x˙ = f(x; y);
y˙ = g(x; y);
(1.1)
5
where f; g are at least C functions on the plane. Suppose (1.1) has a heteroclinic cycle L = L1 ∪ L2 with two di3erent hyperbolic saddles Si (xi ; yi ); i = 1; 2. Let i1 ¿ 0 and i2 ¡ 0 be the eigenvalues of the matrix (@(f; g)=@(x; y))(Si ) ≡ Di ; i = 1; 2. Then the hyperbolicity ratio of Si is given by ri = − i2 = i1 . As we know [9], Cherkas proved that if r1 r2 ¿ 1(¡ 1) then L is stable (unstable). Han et al. [5] proved further that if r1 = r2 = 1 and 2 1 = (fx + fy ) dt = (fx + fy ) dt ≡ 11 + 21 ¡ 0(¿ 0); L
i=1
Li
then L is stable (unstable). In this paper, we suppose r1 = r2 = 1 and 1 = 0, and give a criterion for the stability of L in this more degenerate case. ∗
Corresponding author. E-mail address: [email protected] (S. Hu).
c 2003 Published by Elsevier Science Ltd. 0362-546X/03/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 3 0 1 - 2
702
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Let us Arst recall the deAnition of the Arst saddle value. Let Ti be an inversible matrix such that det Ti = 1;
Ti Di Ti−1 = diag( i1 ; i2 );
i = 1; 2:
(1.2)
By making a linear transformation of the form x − xi u ; = Ti y − yi v we obtain from (1.1) (i) ajl uj vl + O(|u; v|4 ) ; u˙ = i1 u + k=2;3 j+l=k
v˙ = − i2 −v +
k=2;3 j+l=k
(i) j l bjl u v + O(|u; v|4 )
(1.3)
for (u; v) near the origin. Then according to [6] when ri = 1 the Arst saddle value of (1.1) at Si is given by (i) (i) (i) (i) (i) (i) + b12 − a20 a11 + b02 b11 : Ri1 = a21
(1.4)
By [2,6] when ri = 1, there exists a C 5 transformation of the form x − xi u + O(|x − xi ; y − yi |2 ) ≡ Ti∗ (x − xi ; y − yi ); = Ti y − yi v
(1.5)
which carries (1.1) into the normal form u˙ = i1 u[1 + a∗i1 uv + O(u2 v2 )]; v˙ = i1 v[ − 1 + b∗i1 uv + O(u2 v2 )]:
(1.6)
In this case we have Ri1 = a∗i1 + b∗i1 : If
(1.7)
f(u + xi ; v + yi ) = i1 v +
k=2;3 j+l=k
g(u + xi ; v + yi ) = i1 u +
k=2;3 j+l=k
(i) j l ajl u v + O(|u; v|4 ) ;
(i) j l bjl u v + O(|u; v|4 ) ;
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
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Fig. 1.
then instead of (1.4) and (1.7) we have (i) (i) (i) (i) (i) (i) (i) (i) (i) (i) (i) − b03 ) + (b21 − a12 ) + 2(a02 b02 − a20 b20 ) + a11 (a02 − a20 ) Ri1 = 3(a30 (i) (i) (i) (b02 − b20 ): + b11
(1.8)
We note that the sign of time is retained when we change (1.1) into the form of (1.3) or (1.6). Now for deAniteness we assume (L1 ) = S2
and
L is oriented clockwise:
(1.9)
There are two types of L, as shown in Fig. 1, according to the side of L where the Poincare map is well deAned. Our main result is the following. Theorem 1.1. Suppose (1.9) holds. Let r1 = r2 = 1;
1 (=11 + 21 ) = 0;
2 ≡ R11 + R21
11 11 e = 0:
21
(1.10)
(i) If the Poincare map of (1.1) is well de ned in an inner neighborhood of L, then it is stable (unstable) when 2 ¿ 0(¡ 0). (ii) If the Poincare map of (1.1) is well de ned in an outer neighborhood of L, then it is stable (unstable) when 2 ¡ 0(¿ 0). For case (ii) in Fig. 1 L becomes a double homoclinic cycle if S1 = S2 . In this case it was proved in [4] that if r1 ¿ 1(¡ 1) or =1 and 1 ¡ 0(¿ 0), then L is stable (unstable). By the proof of Theorem 1.1 we have Theorem 1.2. Suppose L = L1 ∪ L2 is a clockwise-oriented double homoclinic cycle with a hyperbolic saddle S1 . Let r1 = 1, 1 = 0 and R11 = 0. If R11 ¡ 0(¿ 0) then L is outer stable (unstable), and, if in addition 11 = 0, L1 and L2 are inner unstable (stable). Remark 1.1. If L is oriented counter clockwise, one can determine its stability by applying Theorems 1.1 and 1.2 to the system x˙ = −f(x; y);
y˙ = −g(x; y):
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Fig. 2.
The proof of the above theorems is given in Section 2 and some application examples are presented in Section 3 to show the way to And limit cycles by using the main results. 2. Proof We Arst consider case (i) in Fig. 1. Choose points A2 ; A3 ∈ L1 near S1 and A1 ; A4 ∈ L2 near S2 . Let li be a cross section of (1.1) normal to L at Ai ∈ Li with the directional vector ni = (−g(Ai ); f(Ai ))=|f(Ai ); g(Ai )|: DeAne a map Pij : li → lj by using the orbits of (1.1). Let B = A1 + P(a)n1 be the Arst intersection point of the positive orbit of (1.1) starting at A = A1 + an1 with l1 . Then the Poincare map P : l1 → l1 satisAes (see Fig. 2) P(a) = (P41 ◦ P34 ◦ P23 ◦ P12 )(a)
for 0 ¡ − a 1:
The maps P12 and P34 are called Dulac maps and P23 and P41 regular maps. To study the Dulac maps we use the normal form (1.6). For i = 1 we get from (1.6) dv v = [ − 1 + R11 uv + O(|uv|2 )]; du u
(2.1)
where R11 is given by (1.7). Let l1 = {v = −p; 0 6 u 6 p};
l2 = {u = p; −p 6 v 6 0}
for p ¿ 0 being small. Let B (p; −q1 (r)) be the Arst intersection point of the positive orbit of (1.6) (i = 1) starting at A (r; −p) ∈ l1 with l2 for 0 ¡ r ¡ p (see Fig. 3 (i)). Set z = uv. We have from (2.1) d z R11 + O(|uv|) = du u z2
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
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Fig. 3.
and
−pq1
−pr
dz = z2
p
r
R11 + O(v) du; u
which yields
1 1 1 = −R11 ln r + O(1) − r p q1 or q1 (r) = r + R11 pr 2 ln r + O(r 2 ):
(2.2)
Lemma 2.1. If we choose A1 and A2 such that T1∗ (A1 − S1 ) = (0; −p)T ;
T1∗ (A2 − S1 ) = (p; 0)T ;
(2.3)
where T1∗ is given by (1.5), then there exist C 5 functions Wi (r)=−Mi (p)r +O(r 2 ); i= 1; 2; such that P12 ◦ W1 = W2 ◦ q, where lim Mi (p) = )i sin *1 ;
p→0
)1 = |T1−1 (1; 0)T |;
)1 )2 sin *1 = 1;
)2 = |T1−1 (0; 1)T |;
and *1 being the inner angle of L1 and L2 at S1 . Proof. We relate A and A such that the orbit AˆB of (1.6) is the image of the orbit of (1.1) passing through A. In this way a is a function of r. We want to prove a = W1 (r). For this purpose let T1∗ (A∗ − S1 ) = A (=(r; −p)T ):
(2.4)
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Then A∗ and A are on the same orbit of (1.1). Let a∗ = |A1 A∗ |. Applying the sine theorem to the triangle with vertices A1 ; A and A∗ , we obtain sin “A∗ |a| = a∗ : (2.5) sin “A Let K1 denote the tangent line of L1 at S1 , and K2 the tangent line of L2 at A1 . Then K2 is the limit position of the line passing through A and A∗ as r → 0. By (2.3) and (2.4) we have A1 − A∗ = (T1∗ )−1 (0; −p) − (T1∗ )−1 (r; −p) = T1−1 (r; o)T + O(r 2 );
(2.6)
which implies that (A1 − A∗ )=r = T1−1 (1; 0)T + O(r) → T1−1 (1; 0)T as r → 0. Note that the above limit vector is parallel to K1 . Thus, K1 is the limit position of the line passing through A1 and A∗ . Hence, noting that l1 is orthogonal to K2 , we obtain “A = -=2 + O(r);
“A∗ = *(p) + O(r);
(2.7)
where *(p) denotes the angle between K1 and K2 satisfying limp→0 *(p) = *1 . By (2.6) we also have a∗ = |A1 − A∗ | = r)1 + O(r 2 ):
(2.8)
Substituting (2.7) and (2.8) into (2.5) we have a = −|a| = −r)1 sin *(p) + O(r 2 ) ≡ W1 (r): In the same way we can prove P12 (a) = W2 (q1 (r)). Hence, it follows that P12 ◦ W1 = W2 ◦ q. Finally, from det T1 = 1 it implies that the area of the region T1−1 ([0; 1] × [0; 1]) ≡ G is equal to 1. Note that G is a parallelogram with sides T1−1 (1; 0)T and T1−1 (0; 1)T . The two sides have the angle -−*1 . Therefore, )1 )2 sin *1 =1. The proof is completed. By (2.2) and Lemma 2.1 we have immediately that if A1 and A2 satisfy (2.3), then )2 )2 pR11 P12 (a) = (1 + op (1))a − 2 (1 + op (1))a2 ln|a| + O(a2 ); (2.9) )1 )1 sin *1 where limp→0 Op (1) = 0: Lemma 2.2. (Han [3]). Let (2.3) hold and r1 = 1. Then for p ¿ 0 small |f(A1 ); g(A1 )| = )2 11 p + O(p2 );
|f(A2 ); g(A2 )| = )1 11 p + O(p2 ):
Proof. Note that f(S1 ) = g(S1 ) = 0. We have (f(A1 ); g(A1 ))T = D1 (A1 − S1 ) + O(|A1 − S1 |2 ); T1∗ (A1 − S1 ) = T1 (A1 − S1 ) + O(|A1 − S1 |2 ):
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
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Hence, by (1.2) and (2.3), we have (f(A1 ); g(A1 ))T = T1−1 diag( 11 ; − 11 )T1 (A1 − S1 ) + O(p2 ) = T1−1 diag( 11 ; − 11 )(0; −p)T + O(p2 ) = p 11 T1−1 (0; 1)T + O(p2 ); which implies the Arst formula. The second one can be proved similarly. The proof is completed. Similar to (2.9) and Lemma 2.2, we can also prove that if r2 = 1 one can choose A3 ; A4 = S2 + O(p) suitably such that |f(A3 ); g(A3 )| = )4 21 p + O(p2 );
|f(A4 ); g(A4 )| = )3 21 p + O(p2 )
(2.10)
and P34 (a) =
)4 )4 pR21 (1 + op (1)) − 2 (1 + op (1))a2 ln |a| + O(a2 ); )3 )3 sin *2
(2.11)
where )3 )4 sin *2 = 1. Let X1 (t) be a representation of L1 and let t2 ¡ t3 be such that X1 (tj ) = Aj ;
j = 2; 3:
(2.12)
By Lemmas 4.3.2 and 4.3.3 of [8] we can introduce the change of variables (x; y)T = X1 (*) − JV (*)r; where X (*) V (*) = 1 ; |X1 (*)|
J=
t2 6 * 6 t3 ; 0
1
−1
0
(2.13)
;
to (1.1) to obtain dr=d* = A(*)r + O(r 2 );
(2.14)
where A(*) = (fx + gy )(X1 (*)) −
d ln|f(X1 (*)); g(X1 (*))|: d*
Let r(*) be the solution of (2.14) with r(t2 ) = a. The formula of constant variation yields that
t3 |f(A2 ); g(A2 )| r(t3 ) = (fx + gy ) dt a + O(a2 ): (2.15) exp |f(A3 ); g(A3 )| t2 Note that −JV (tj ) = nj ;
j = 2; 3:
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By (2.12) and (2.13) and the deAnition of P23 , we obtain P23 (a)=r(t3 ). Since r1 =r2 =1 it follows that ([7] or [8, Lemma 2.3.5]) t3 (fx + gy ) dt → 11 (as p → 0): t2
Hence by Lemma 2.2, (2.10) and (2.15) we have P23 (a) =
)1 11 11 e (1 + op (1))a + O(a2 ): )4 21
(2.16)
An analogue procedure yields that P41 (a) =
)3 21 21 e (1 + op (1))a + O(a2 ): )2 11
(2.17)
Then by (2.9), (2.11), (2.16) and (2.17) we obtain
pR11 )2 11 11 2 (1+op (1))a− (1+op (1))a ln|a| +O(a2 ); e (P23 ◦ P12 )(a)= )1 sin *1 )4 21
)4 21 21 pR21 (P41 ◦ P34 )(a)= (1+op (1))a− e (1+op (1))a2 ln|a| +O(a2 ) )2 11 )3 sin *2 and
P(a) = e1 N1 (p)a − p)2 N2 (p)a2 ln|a| + O(a2 );
(2.18)
where N1 (p) = (1 + op (1));
N2 (p) = R11 +
11 R21 e11 + op (1):
21
Here we have used )1 )2 sin *1 = )3 )4 sin *2 = 1 for the expression of N2 . On the other hand, by [1,7] we know that 1 P (a) = e AB (fx +gy ) dt → e1 1 + oa (1) as a → 0, which gives that N1 = 1. Therefore, if 1 = 0, (2.18) becomes P(a) = a − p)2 (2 + op (1))a2 ln|a| + O(a2 ): Now conclusion (i) of Theorem 1.1 is clear. For case (ii) in Fig. 1, we can deAne regular maps P23 near L1 and P41 near L2 , and Dulac maps P12 near S1 and P34 near S2 as before. We can also prove that P23 and P41 have the same form as in the Arst case, and P12 and P34 can be obtained by replacing R11 and R21 by −R11 and −R21 , respectively, in (2.9) and (2.11). This is because the coeIcients a∗i1 and b∗i1 in (1.6) change sign if u is replaced by −u. Thus, in this case we have P(a) = a + p)2 (2 + op (1))a2 ln|a| + O(a2 ); which implies conclusion (ii). Hence, the proof of theorem 1.1 is Anished.
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Now suppose L is a homoclinic cycle with a hyperbolic saddle S1 . By [7] we know that if 0 ≡ (fx + gy )(S1 ) ¡ 0(¿ 0) or 0 = 0 and 1 ≡ L (fx + gy ) dt ¡ 0(¿ 0) then L is stable (unstable). Further, by the proof of Theorem 1.1 we have Corollary 2.1. Suppose the homoclinic cycle L is oriented clockwise. Let 0 = 1 = 0 and R11 = 0. Then (i) If the Poincare map of (1.1) is well de ned in an inner neighborhood of L, then it is stable (unstable) when R11 ¿ 0(¡ 0). (ii) If the Poincare map of (1.1) is well de ned in an outer neighborhood of L, then it is stable (unstable) when R11 ¡ 0(¿ 0). For case (ii) in Fig. 1 and S1 = S2 , we have from Theorem 1.1 (ii) Corollary 2.2. Let L be a double homoclinic cycle with a hyperbolic saddle S1 and let L be oriented clockwise. If 0 = 1 = 0 and R11 ¡ 0(¿ 0), then L is stable (unstable). Obviously, Theorem 1.2 follows from the above two corollaries.
3. Application examples In this section, we consider a system of the form x˙ = Hy + aHx H ≡ f(x; y); y˙ = −Hx − Hy H ≡ g(x; y);
(3.1)
where H is a polynomial in (x; y). For (3.1) we have fx + gy = aHx2 − Hy2 + H (aHxx − Hyy ):
(3.2)
We Arst suppose H = 12 (y2 − x2 ) + 14 x4 :
(3.3)
Then (3.1) has a homoclinic cycle Li : H (x; y)=0; (−1)i x ¿ 0 (i=1; 2). Let L=L1 ∪L2 . Proposition 3.1. Suppose (3.3) holds. Then (i) If a ¡ 57 (¿ 57 ), then L, L1 and L2 are all stable (unstable); if a = 57 then L is unstable, and L1 and L2 are stable. (ii) There exists b ¿ 0. such that (3.1) has two “small ” limit cycles Li1 ⊂ Li (i = 1; 2) for a = 57 , has four “small” limit cycles Lij ⊂ Li (i; j = 1; 2) for 57 ¡ a ¡ 57 + b and has two “small ” limit cycles Li1 ⊂ Li (i = 1; 2), and one “large” limit cycle for 57 − b ¡ a ¡ 57 (see Fig. 4).
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Fig. 4.
Proof. By (3.2) we have 11 = 21 = (aHx2 − Hy2 ) dt = a(x − x3 ) dy − y d x: L2
L2
By integration by parts we have
L2
(x − x3 ) dy =
that 11 = 21 = 2
√
2
2
x(3ax − a − 1)
0
L2
(3x2 − 1)y d x. Hence it is direct
1 4 1 − x2 d x = 2 3
7 a−1 : 5
Thus, L, L1 and L2 are all stable (unstable) if a ¡ 57 (¿ 57 ). For a = 57 , by (1.8) we have R11 = 2(a + 1) = 24 7 . Note that L is oriented clockwise. It follows from Theorem 1.2 that L is unstable and L1 and L2 are both stable. It is easy to see that (3.1) has three critical points (0,0) and (±1; 0) in the region H (x; y) 6 0 for a ¡ 16. By (3.2) (fx + gy )(±1; 0) =
1 − 2a : 4
Thus the points (±1; 0) are stable for a = 57 . Hence there is an unstable limit cycle Li1 inside Li (i = 1; 2). For 0 ¡ a − 57 1, Li has changed its stability from stable into unstable. Hence a stable limit cycle Li2 between Li and Li1 has been bifurcated from Li . For 0 ¡ 57 − a 1, L has changed its stability and a large unstable limit cycle surrounding L has appeared. The proof is completed. Now we suppose H = x(y2 + x − 1): Then (3.1) has a heteroclinic cycle L : H (x; y) = 0;
x ¿ 0; |y| 6 1;
(3.4)
M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
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with L1 : x = 0; |y| 6 1 and L2 : y2 + x − 1 = 0; 0 6 x 6 1. Note that Hx = 2x + y2 − 1; Hy = 2xy and d x = −2y dy along L2 . We have by (3.2) 1 = −aHx dy − Hy d x L
and
11 =
−1
21 =
=
− a(y2 − 1) dy = −2a
−1
1
=
1
1
−1
− a(1 − y2 ) dy −
−a(y2 − 1) dy +
−1
1 1
−1
4a 1 −1 = ; 3 3 2y(1 − y2 )(−2y) dy
4(y4 − y2 ) dy
4a 16 4a 1 1 = +4×2 − − : 3 5 3 3 15
Hence, 1 =
8 15 [5a
− 2]:
Thus, L is stable (unstable) if a ¡ 25 (¿ 25 ). For a = 25 , removing the saddle S2 (1; 0) to the origin we obtain x˙ = 2[x + xy + ax3 + 3ax2 y + 2xy2 ] + O(|x; y|4 ); y˙ = 2[ − x − y − x3 − 12 y2 − 2x2 y] + O(|x; y|4 ): Introducing a new variable v = 12 x + y, the above system becomes
1 1−a 3 x˙ = 2 x − x2 + xv + x + (3a − 2)x2 v + 2xv2 + O(|x; v|4 ); 2 2
3 2 1 2 1−a 3 3 2 2 v˙ = 2 −v − x + xv − v + x + (a − 2)x v + xv + O(|x; v|4 ): 8 2 4 2 Then (1.4) gives that R21 = 3a − 1 = 15 . Similarly, removing S1 (−1; 0) to the origin and introducing a new variable v = − x=2 + y, we obtain x˙ = 2[ − x + xy + ax3 − 3ax2 y + 2xy2 ] + O(|x; y|4 ); y˙ = −2x − y2 + 2y + 2x3 − 4x2 y + O(|x; y|4 ) and
1 2 3 2 3 1−a 3 2 2 x + O(|x; v|4 ); v˙ = 2 v − v − xv − x − v x + (a − 2)vx − 2 8 2 4
1 2 1−a 3 2 2 x˙ = 2 −x + vx + x + 2v x + (2 − 3a)vx + x + O(|x; v|4 ): 2 2
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M. Han et al. / Nonlinear Analysis 53 (2003) 701 – 713
Fig. 5.
Thus, 1 R11 = 1 − 3a = − : 5 Therefore, 1 2 = (e8=15 − 1) ¿ 0: 5 It follows from Theorem 1.1 that L is stable for a = 25 . By (3.2) and (3.4) we have fx + gy 12 ; 0 = − 14 (2a − 1): Thus the critical point ( 12 ; 0) is unstable for a ¡ 12 . It is not hard to see that ( 21 ; 0) is stable for a ¿ 12 . Then similar to Proposition 3.1 we have Proposition 3.2. Let (3.4) hold. Then (i) The heteroclinic cycle L is stable (unstable) if a 6 25 (a ¿ 25 ). (ii) There exist a stable limit cycle L1 ⊂ L for 25 ¡ a ¡ 12 (see Fig. 5).
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[6] P. Joyal, Generalized Hopf bifurcation and its dual generalized homoclinic bifurcation, SIAM J. Math. 48 (1988) 481–496. [7] D. Luo, M. Han, D. Zhu, The uniqueness of limit cycles bifurcated from a separatrix cycle (I), Acta Math. Sin. 3 (1992) 407–417. [8] D. Luo, X. Wang, D. Zhu, M. Han, Bifurcation Theory and Methods of Dynamical Systems, in: Advanced Series in Dynamical Systems, Vol. 15, World ScientiAc, Singapore, 1997. [9] Y. Ye, et al., Theory of Limit Cycles, Translations of Mathematical Monographs, Vol. 66, American Mathematical Society, Providence, RI, 1986.