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Exponential dichotomies and Melnikov functions for singularly perturbed systems1
Nonlinear Analysis 36 (1999) 401 – 422
Exponential dichotomies and Melnikov functions for singularly perturbed systems 1 Zeng Weiyao∗ , Luo Jiaowan Institute of Sciences, Changsha Railway University, Changsha, Hunan 410075, People’s Republic of China Received 17 November 1995; received in revised form 8 December 1996; accepted 13 January 1998
Keywords: Exponential dichotomies; Melnikov functions; Singularly perturbed systems
1. Introduction Recently, many authors have worked on the existence of homoclinic solutions in singularly perturbed systems. Some of them worked on systems where the homoclinic solution possesses a transition layer connecting two equilibria on two slow manifolds. Others worked on systems where the homoclinic solution does not have a transition layer and is con ned to one slow manifold. This paper belongs to the second category and we mainly consider the problem of the existence of transversal homoclinic orbits of systems of singularly perturbed ODE of the form x˙ = f(x; y) + h1 (t; x; y; );
x ∈ Rn;
(1)
y˙ = g(x; y) + h2 (t; x; y; );
y ∈ Rm;
(2)
where ≥ 0 is a small parameter, f; g; h1 and h2 are C 2 -smooth. However, rst we recall some results for general singularly perturbed systems of the form
∗ 1
x˙ = f(t; x; y; );
(3)
y˙ = g(t; x; y; );
(4)
Corresponding author. E-mail: [email protected]. The project supported by NSF of China.
0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 6 1 4 - 7
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where x ∈ R n , y ∈ R m , is a small parameter. Setting = 0 in the system (3) and (4), we obtain the equations x˙ = f(t; x; y; 0);
(5)
0 = g(t; x; y; 0);
(6)
which are called the degenerate equations of Eqs. (3) and (4). We assume that C1. The degenerate equations (5) and (6) have a nontrivial bounded (or periodic or almost periodic) solution x = u(t), y = v(t). We call the equation ˙ = (fx (t) − fy (t)gy−1 (t)gx (t));
(7)
the rst variational equations of Eqs. (3) and (4) along the bounded solution u(t); v(t), where fx (t) = fx (t; u(t); v(t); 0) and fy (t); gx (t) and gy (t) have similar meanings. For the nonautonomous singularly perturbed equations (3) and (4), if condition C1 and conditions C2. There exist a constant ¿0 such that the absolute values of the real parts of the eigenvalues of the m ×m matrix gy (u(t); v(t); 0) are greater than , C3. The rst variational Eq. (7) has an exponential dichotomy on R are satis ed, then Chang [4] showed that for suciently small the singularly perturbed equations (3) and (4) have a unique bounded solution x(t; ); y(t; ) satisfying lim {|x(t; ) − u(t)| + |y(t; ) − v(t)|} = 0;
→0
(∗)
uniformly in t ∈ R. If f(t; x; y; ); g(t; x; y; ); u(t); v(t) are T -periodic (or almost periodic in t) then for 6= 0 suciently small Eqs. (3) and (4) have a unique T -periodic (or almost periodic) solution x(t; ); y(t; ) satisfying Eq. (∗). However, for the special singularly perturbed system (1) and (2) with nonautonomous perturbed terms, condition C3 cannot be satis ed. In fact, degenerate equations of Eqs. (1) and (2) are x˙ = f(x; y); 0 = g(x; y); which have a nontrivial bounded solution x = u(t), y = v(t). It is easy to show ˙ u(t) = [fx (t) − fy (t)gy−1 (t)gx (t)] u(t): Hence, the rst variational equation of singularly perturbed equations (1) and (2) along u(t); v(t) have a nontrivial bounded solution u(t) ˙ on R, and the rst variational equation does not admit an exponential dichotomy on R (refer to [1]). Thus, the above result on bounded solutions for Eqs. (3) and (4) cannot be applied to Eqs. (1) and (2). Recently, Feckan [8], and Battelli and Lazzari [2] investigated the existence of homoclinic orbits of singularly perturbed system (1) and (2) using a Melnikov function. In their papers they want Eqs. (1) and (2) to satisfy the following conditions: (a) g(x; y) = 0 ⇔ y = 0.
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403
(b) The degenerate equation x˙ = f(x; 0) has a nontrivial bounded solution u(t), and the variational equation along u(t) x˙ = A(t)x;
(A(t) =: fx (u(t); 0))
(8)
has an exponential dichotomy on both R+ and R− and the sum of the dimensions of the stable and unstable subspaces of Eq. (8) is n. Moreover, u(t) ˙ is the unique (up to a scalar multiple) bounded solution of variational equation (8). (c) C(t) =: gy (u(t); 0) = diag(C1 (t); C2 (t)), and the real parts of the eigenvalues (t) of C1 (t) (resp. C2 (t)) satisfy Re (t) ≤ −2 ¡0 (resp. Re (t) ≥ 2 ¿0) for any t ∈ R. Under conditions (a)–(c), Feckan [8], and Battelli and Lazzari [2] obtained the following Melnikov-type function: Z +∞ ∗ (t){h1 (t + ; u(t); 0; 0) − fy (t)C −1 (t)h2 (t + ; u(t); 0; 0)} dt M ( ) = −∞
and proved that if M ( ) has a simple zero then for 6= 0 suciently small singularly perturbed system (1) and (2) has a transversal homoclinic orbit. It seems to us that the method used in [2] is very complex, and Feckan [8] did not prove the transversality of homoclinic orbits. Moreover, it should be noted, from the conditions in known results on periodic solutions and bounded solutions of singularly perturbed equations, that conditions (a) and (c) in their papers are not natural (compared with conditions C1, C2). The main purpose of this paper is to drop these conditions. We need to emphasize that the results in this paper cannot be obtained by simply making a change in the variable y = y + v(t) of Eq. (2). Battelli and Lazzari [2] deduced from the Melnikov function obtained in [2] that under conditions (a) – (c) the contribution of the singular part of the system is not important for the existence of transversal homoclinic orbits in the case where h2 (t; ; 0; 0) = 0. From the Melnikov function M ( ) obtained in this paper we see, under general conditions H1–H4, that even when h2 (t; x; y; 0) = 0, we cannot neglect the role of the singular part of the system for the existence of transversal homoclinic orbits. We obtain a very general Melnikov-type function and show that if the Melnikov-type of function has a simple zero then for 6= 0 suciently small the singularly perturbed system has a transversal homoclinic orbit. It seems to us that our proof in this paper is simpler than the one in Battelli and Lazzari [2], and we also give a new method for proving the transversality of homoclinic orbits of the singularly perturbed equations. The main tool used in this paper is the theory of exponential dichotomies and the functional analystic method. We should note that the paper lacks geometric insight. Geometric singular perturbation has become an essential tool for understanding singular perturbation problems, we therefore refer to Fenichel [9]. This paper is organized as follows: Section 1 is an introduction, in Section 2 we give the main result of this paper; Section 3 is devoted to the proof of the main results and a example. The notations for function spaces in this paper are the same as those in Battelli and Lazzari [2].
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2. Main results We assume H1. There exists a compact set S1 ×S2 ⊂ R n × R m such that the degenerate system x˙ = f(x; y);
(9)
0 = g(x; y)
(10)
admits a nontrivial bounded solution x = u(t), y = v(t) in S1 × S2 . H2. There exists a constant ¿0 such that the absolute values of the real parts of the eigenvalues of the m×m matrix gy (u(t); v(t)) are greater than . Remark. If conditions H1 and H2 hold, we can easily prove that u(t) = [fx (t) − fy (t)gy−1 (t)gx (t)] u(t); ˙ where fx (t) =: fx (u(t); v(t)) and fy (t); gx (t) and gy (t) have similar meanings, thus u(t) ˙ is a nontrivial bounded solution of the rst variational equation x˙ = [fx (t) − fy (t)gy−1 (t)gx (t)]x:
(11)
It follows from Coppel [7] that Eq. (11) does not have an exponential dichotomy on R. H3. The rst variational Eq. (11) admits an exponential dichotomy on both R+ and R− and the sum of the dimensions of the stable and unstable subspaces is n. Moreover, Eq. (11) has a unique (up to a scalar multiple) nontrivial bounded solution u(t). ˙ Remark. If H3 is satis ed, then Palmer [12] has proved that the adjoint equation x˙ = −[fx (t) − fy (t)gy−1 (t)gx (t)]∗ x
(12)
also has, up to a scalar multiple, a unique nontrivial bounded solution on R; denoted by (t). The main results of this paper are as follows: Theorem 1. We assume H1, H2 and H3 are satis ed and we de ne Z +∞ ∗ (t){fy (t)gy−1 (t)[v(t) ˙ − h2 (t + ; u(t); v(t); 0)] M ( ) = −∞
+ h1 (t + ; u(t); v(t); 0)} dt: If there exists an 0 ∈ R satisfying M ( 0 ) = 0;
M 0 ( 0 ) 6= 0;
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then there exists a continuous function () with (0) = 0 ; such that for 6= 0 suciently small singularly perturbed system (1) and (2) has a unique bounded solution x(t; ); y(t; ) satisfying u˙∗ (0)[x( (); ) − u(0)] = 0; sup |x(t + (); ) − u(t)| = O(); t∈R
sup |y(t + (); ) − v(t)| = O(): t∈R
Moreover, for suciently small 6= 0 variational equations of Eqs. (1) and (2) along the bounded solution x(t; ); y(t; ) admits an exponential dichotomy on R. A remarkable situation where Theorem 1 applies, occurs when H3 is replaced by H30 . The degenerate equations (9) and (10) have a nontrivial bounded solution u(t); v(t) in S1 × S2 and a trivial solution x0 ; y0 satisfying lim u(t) = x0 ;
t→±∞
lim v(t) = y0
t→±∞
and the real parts of the eigenvalues of the matrix fx (x0 ; y0 ) − fy (x0 ; y0 )gy−1 (x0 ; y0 ): ˙ is a unique (up to a scalar multiple) nongx (x0 ; y0 ) are di erent from zero, and u(t) trivial bounded solution of Eq. (11) on R. By making use of Theorem 1 and the method for proving Corollary 5.2 in Palmer [12], we can show the following result. Theorem 2. Suppose H1, H2 and H30 are satis ed and hi (t + T; x; y; ) = hi (t; x; y; ); i = 1; 2. Then for 6= 0 suciently small Eqs. (1) and (2) admit a unique T -periodic solution x0 (t; ); y0 (t; ) satisfying lim {kx0 (t; ) − x0 k + ky0 (t; ) − y0 k} = 0
→0
and variational equations of Eqs. (1) and (2) along x0 (t; ); y0 (t; ) have an exponential dichotomy on R. Moreover if there exists 0 ∈ R such that M ( 0 ) = 0;
M 0 ( 0 ) 6= 0;
then there is a continuous function () with (0) = 0 ; such that for 6= 0 suciently small Eqs. (1) and (2) have a unique bounded solution x(t; ); y(t; ) satisfying u˙∗ (0)[x( (); ) − u(0)] = 0; sup |x(t + (); ) − u(t)| = O(); t∈R
sup |y(t + (); ) − v(t)| = O(); t∈R
lim {|x(t; ) − x0 (t; )| + |y(t; ) − y0 (t; )|} = 0:
t→±∞
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Moreover; for 6= 0 suciently small variational equations Eqs. (1) and (2) along the bounded solution x(t; ); y(t; ) admits an exponential dichotomy on R. Remark. From the result in Palmer [12] and Theorem 2 of this paper; we see that if the conditions of Theorem 2 are satis ed then the Poincare map P of Eqs. (1) and (2) have a transversal homoclinic point x(0; ); y(0; ) biasymptotic to the hyperbolic T -periodic point x0 (0; ); y0 (0; ). 3. The proofs of the theorems For simplicity, we let U (t) =: gy−1 (t)gx (t). For singularly perturbed system (1) and (2) we make the change of variable as in [5] x(t + ) = (t) + u(t); y(t + ) = (t) + v(t) − U (t)(t); then system (1) and (2) becomes ˙ = f( + u(t); + v(t) − U (t)) + h1 (t + ; + u(t); + v(t) − U (t); ) −f(u(t); v(t));
(13)
˙ = g( + u(t); + v(t) − U (t)) + h2 (t + ; + u(t); + v(t) − U (t); ) ˙ − v(t) ˙ + U˙ (t) + U (t):
(14)
We write Eqs. (13) and (14) in the following form: ˙ = [fx (t) − fy (t)gy−1 (t)gx (t)] + fy (t) + h1 (t + ; + u(t); + v(t) − U (t); ) + F1 (t; ; ; ); ˙ = gy (t) + F2 (t; ; ; ; );
(15) (16)
where F1 (t; ; ; ) = f( + u(t); + v(t) − U (t)) − fy (t) −f(u(t); v(t)) − [fx (t) − fy (t)gy−1 (t)gx (t)];
(17)
F2 (t; ; ; ; ) = −v(t) ˙ + U˙ (t) + U (t)˙ + [g( + u(t); + v(t) − U (t)) − gy (t)] + h2 (t + ; + u(t); + v(t) − U (t); ):
(18)
For simplicity, we now give some notations. For suciently small 0 ¿0; we let I0 = (0; 0 ]; I1 = [0; 0 ] and I2 be a small interval containing 0 .
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407
We have the following Lemma 1. Lemma 1. Suppose H1 and H2 hold. Then for any constant 0 ¿0 and for any (t) ∈ {x | x ∈ Cb1 ; |x(t)| ≤ 0 } the equation ˙ = gy (t) + F2 (t; (t); ; ; )
(19)
has; for suciently small; a unique small bounded solution = (t; (·); ; ) satisfying (see remark following Lemma 1): ˙ − h2 (t + ; u(t); v(t); 0)]; (t; (·); ; 0) = gy−1 (t)[v(t)
(20)
(t; (·); ; 0) = 0;
(21)
(t; (·); ; 0) = 0;
(22)
for (t; (·); ) ∈ R × Cb1 × I2 . Moreover; (t; ; ; ) is C 2 in (t; (·); ; ) ∈ R × Cb1 × I2 × 1 (·); ; ) is C in (t; ; ; ) ∈ R × Cb1 × I2 × I1 . I0 ; (t; (·); ; ) are de ned only for 6= 0. Remark. Actually; (t; (·); ; ); (t; (·); ; ); (t; Since we will see that as → 0 their limits exist; we de ned their values at = 0 to be their limits as → 0. Proof. Since H2 holds, it follows from Chang and Coppel [5] that for ¿0 suciently small the linear equation ˙ = gy (t) possesses an exponential dichotomy on R with |X (t; )P()X −1 (s; )| ≤ Ke−=(t−s) ;
t ≥ s;
|X (t; ) (I − P()X −1 (s; )| ≤ Ke−=(s−t) ;
s≥t
(23)
for t; s ∈ R. Using a standard perturbation method (contraction xed point theorem and theory of exponential dichotomies, refer to Coppel [6]), we can prove that for ¿0 suciently small Eq. (19) has a unique bounded solution = (t; (·); ; ) satisfying (·); ; ) = 0 lim (t;
→0
(24)
uniformly in (t; (·); ) ∈ R × Cb1 × I2 and (t; (·); ; ) Z t 1 X (t; )P()X −1 (s; )F2 (t; (t); (t; (t); ; ) ds = −∞ Z +∞ −1 X (t; )(I − P())X (s; )F2 (t; (t); (t; (t); ; ); ; ) ds : − t
So we can de ne (t; (·); ; 0) = 0. First of all, if Eq. (19) were ˙ = gy (t) + F2 (t; (t); ; ; );
(25)
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Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
where the fundamental matrix X (t) of ˙ = gy (t) satis ed |X (t)PX −1 (s)| ≤ Ke−(t−s) ; |X (t) (I − P)X
−1
(s)| ≤ Ke
−(s−t)
t ≥ s; ;
s ≥ t;
then it is well-known that the solution (t; ; ; ) is C 2 in (; ; ). Now, we deal with Eq. (19). We want to prove that (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . We rst prove that (t; (·); ; ) is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Set (·); ; ); 2 (t) = (t + t; (·) + (·); + ; + h); (t) = 2 (t) − 1 (t). 1 (t) = (t; Then (t) is a solution of equation ˙ = gy (t) + F2 (t + t; ( + h)((t) + (t); 2 (t); + ; + h) − F2 (t; (t); 1 (t); ; ) − h · ˙2 (t) − [gy (t) − gy (t + t)]2 (t) and so, using Eqs. (23)–(25) |(t)| ≤ K−1
sup
[|F2 (t + t; ( + h)((t) + (t)); 2 (t); + ; + h)
−∞¡t¡+∞
− F2 (t; (t); 1 (t); ; ) − h · ˙2 (t) − [gy (t) − gy (t + t)]2 (t)|] 1 sup |(t)| + constant · |h| + |t| + |(h + )| sup |(t)| + | | ; ≤ 2 −∞¡t¡∞ t∈R provided the parameters are suciently small. Hence, sup |(t)| ≤ constant · |h| + |t| + |(h + )| sup |(t)| + | | −∞¡t¡∞
t∈R
and so |(t + t; (·) + (·); + ; + h) − (t; ; ; )| ≤ constant · |h| + |t| + |(h + )| sup |(t)| + | | : t∈R
Hence, (t; (·); ; ) is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . To show that (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 ; let z(t; (·); ; ) be the unique bounded solution of equation z˙ = gy (t)z + +
@F2 @F2 (t; (t); (t; (·); ; ); ; )z + (t; (t); (t; (·); ; ); ; )(t) @ @
@F2 (t; (t); (t; (·); ; ); ; ) − (t; ˙ (·); ; ): @
Then, by considering (t) = (t; (·); ; + h) − (t; (·); ; ) − z(t)h and reasoning as above, we can show that (t) is o(h) and hence conclude that (t; (·); ; ) = z(t; (·); ; ). By the same way as proving that (t; (·); ; ) is continuous in (t; (·); ; )
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
409
∈ R × Cb1 × I2 × I0 , we can show that (t; (·); ; ) = z(t; (·); ; ) is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Similarly, we can prove that t (t; (·); ; ); (t; (·); ; ) and (t; (·); ; ) exist and are continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Thus, (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Similarly, we can prove that (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 , for simplicity, we omit the proof. Now, we show that (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 . Obviously, we only need to prove (t; (·); ; ) is continuously di erentiable at any point (t; (·); ; 0). We rst prove that (t; (·); ; 0) exists and (t; (·); ; ) is continuous at (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . For simplicity, we let m(t) = m(t; (·); ) = gy−1 (t) [v(t) ˙ − h2 (t + ; u(t); v(t); 0)]. Di erentiating both sides of the equation (·); ; ) + F2 (t; (t); (t; (·); ; ); ; ) (t; ˙ (·); ; ) = gy (t)(t; with respect to ; we obtain ˙ (·); ; ) ˙ (t; (·); ; ) + (t; ˙ + h2 (t + ; u(t) + (t); (t; (·); ; ) + v(t); ) = gy (t) (t; (·); ; ) − v(t) + O() (t; (·); ; ) + O(): Thus, ˙ (·); )} {˙ (t; (·); ; ) − m(t; ˙ (·); ; )+ O() (t; (·); ; ) = gy (t){ (t; (·); ; ) − m(t; (·); )}+ O() − (t; (·); ; ) − m(t; ˙ (·); ) + h2 (t + ; u(t) + (t); (t; + v(t); ) − h2 (t + ; u(t); v(t); 0): Since for ∈ I0 the linear part of the above equation y˙ = gy (t)y admits an exponential dichotomy (23) on R, and (t; (·); ; ) is bounded for t ∈ R and xed ∈ I0 , (t; (·); ; ) has the following form: (t; (·); ; ) − m(t) Z t 1 = X (t; )P()X −1 (s; ){O() − (t; ˙ (t); ; ) −∞ + O(){ (t; (t); ; ) − m(t)} (·); ; ) + v(t); ) + O()m(t) − m(t) ˙ + h2 (t + ; u(t) + (t); (t; − h2 (t + ; u(t); v(t); 0) ds Z −
t
+∞
X (t; )(I − P())X −1 (s; ){O() − (t; ˙ (t); ; )
+ O(){ (t; (t); ; ) − m(t)}
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Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
+ O()m(t) − m(t) ˙ + h2 (t + ; u(t) + (t); (t; (·); ; ) + v(t); ) − h2 (t + ; u(t); v(t); 0)}ds : Thus, there exists a positive constant L0 such that | (t; (·); ; ) − m(t)| ˙ (·); ; ) + O() − O()m(t) − m(t) ˙ ≤ L0 · sup |(t; t∈R
(·); ; ) + v(t); ) − h2 (t + ; u(t); v(t); 0)|: (26) + h2 (t + ; u(t) + (t); (t; Noting that (t; (·); ; 0) = 0 and is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 , we obtain ˙ − h2 (t + ; u(t); v(t); 0)]: lim (t; (·); ; ) = m(t) = gy−1 (t)[v(t)
→0
So we may de ne ˙ − h2 (t + ; u(t); v(t); 0)]: (t; (·); ; 0) = gy−1 (t)[v(t) Following almost the same method used to prove Eq. (26), we can prove | (t + t; (·) + (·); + ; h) − (t; (·); ; 0)| = | (t + t; (·) + (·); + ; h) − m(t; (·); )| ≤ constant |t| + sup |(t)| + | | + |h| : t∈R
Thus, (t; (·); ; ) exists and is continuous at (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . In the same way, we can prove that t (t; (·); ; 0) = 0; (t; (·); ; 0) = 0 and (t; (·); ; 0) = 0 exists, and t (t; (·); ; ); (t; (·); ; ) and (t; (·); ; ) are con (·); ; ) is continuously di erentinuous at (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . Hence, (t; (·); ; ) is C 1 in (t; (·); ; ) tiable at point (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . Thus, (t; 1 ∈ R × C b × I2 × I1 . We need a result in Palmer [12]. Lemma 2. We consider the linear equation x˙ = A(t)x; n
(27)
x ∈ R . Assume that Eq. (27) admits an exponential dichotomy on both R+ and R− . Then for f ∈ Cb0 equation x˙ = A(t)x + f(t) has bounded solutions on R if and only if Z +∞ ∗ (t)f(t) dt = 0 −∞
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
holds for all bounded solutions
411
(t) of the equation adjoint to Eq. (27)
∗
x˙ = −A (t)x: We make a change of variables (t) ∼ (t) in Eqs. (15) and (16) and obtain a new system of equations ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + fy (t) + F1 (t; (t); ; ; ) (t) y + h1 (t + ; (t) + u(t); + v(t) − U (t)(t); );
(28)
(t) ˙ = gy (t)(t) + F2 (t; (t); (t); ; ):
(29)
From Lemma 1, we see for sucient small, Eq. (14) has a bounded solution = (t; (·); ; ). We put it in Eq. (28) and have ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) (t) y fy (t)(t; (·); ; ) + F1 (t; (t); (t; (·); ; ); ; ) + h1 (t + ; (t) + u(t); + v(t) − U (t)(t); ): +
(30)
From Lemma 1 we see that if Eq. (30) has a bounded solution = (t; ), then Eqs. (13) and (14) have a bounded solution = (t; ); = (t; (t; ); ; ). Hence, we only need to solve Eq. (30). We de ne a function f (t)(t; (·); ; ) + F1 (t; (·); ; ; ) y 6= 0; F3 (t; (·); ; ) = ˙ − h2 (t + ; u(t); v(t); 0)] = 0: fy (t)gy−1 (t)[v(t) Then F3 (t; (·); ; 0) = fy (t)gy−1 (t)[v(t) ˙ − h2 (t + ; u(t); v(t); 0)] fy (t)(t; (·); ; ) + F1 (t; (·); ; ; ) →0 d = [fy (t)(t; (·); ; ) + F1 (t; (·); ; ; )|=0 F3 (t; (·); ; 0): d = lim
Since (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 from Lemma 1, we have fy (t)(t; (·); ; ) + F1 (t; (·); ; ; ) is also C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 . Thus, F3 (t; (·); ; ) is C 0 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 . F3 (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R×Cb1 ×I2 ×I0 is obvious because (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R× Cb1 ×I2 ×I0 . Similarly, we can prove that (@=@ )F3 (t; (·); ; ) is C 0 in (t; (·); ; ) ∈ R× Cb1 × I2 × I1 . It is obvious that when 6= 0 Eq. (30) is equivalent to the equation ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + F3 (t; (·); ; ) (t) y + h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); ): We have the following lemma.
(31)
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Lemma 3. Suppose H1–H3 are satis ed. Then there exist constants 0 ¿0 and 0 ¿0 such that for ||¡0 the equation ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + F3 (t; (·); ; ) (t) y + h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); ) Z +∞ ∗ (t){F3 (t; (·); (t; (·); ; ); ; ) − −∞
+ h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); )} dt· (t)
(32)
has a unique bounded solution = (t; ; ) satisfying |(t; ; )| ≤ 0 sup | + F3 (t; (t; ; ); ; ) + h1 (t + ; (t; ; ) t∈R
+ u(t); (t; (t; ; ); ; ) + v(t) − U (t)(t); )|; u˙∗ (0)(0; ; ) = 0: Moreover; (t; ; ) and (@=@ )(t; ; ) are Cb1 in (t; ; ) ∈ R × I2 × I0 ; and C 0 in (t; ; ) ∈ R × I2 × I1 . Proof. Since (t; (·); ; ) is Cb1 in (t; ; ) ∈ R×I2 ×I0 , and is Cb0 in (t; ; ) ∈ R×I2 ×I1 , the functions on the right-hand side of Eq. (32) are also Cb1 in (t; ; ) ∈ R × I2 × I0 , and is Cb0 in (t; ; ) ∈ R × I2 × I1 . The proof of Lemma 3 is almost the same as that in Battelli and Palmer [3] (noting (t; (·); ; 0) = 0), so we omit the proof. R +∞ Proof of Theorem 1. Without loss of generality, we may assume −∞ ∗ (t) (t) dt = 1. Using the Liapunov–Schmidt method, we know that Eq. (31) is equivalent to the following equation: ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + F3 (t; (·); ; ) (t) y + h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); ) − B((·); ; )· (t); Z B((·); ; ) =
+∞
−∞
∗
(33)
(t){fy (t)(t; (·); ; ) + F3 (t; (·); (t; (·); ; ); )
+ h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); )} dt: (34) It follows from Lemma 3 that for suciently small Eq. (33) has a unique bounded solution = (t; ; ) which is Cb1 in (t; ; ) ∈ R×I2 ×I0 and is Cb0 in (t; ; ) ∈ R×I2 ×I1 and satis es u˙∗ (0)(0; ; ) = 0;
|(t; ; )| ≤ 0 ; (t; ; ) ∈ R × I2 × I1 :
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413
Moreover, (@=@ )(t; ; ) is Cb0 in (t; ; ) ∈ R × I2 × I1 . Substituting = (t; ; ) into Eq. (34), we obtain the bifurcation equation for Eq. (31) Z +∞ ∗ (t){F3 (t; (t; ; ); ; ) B( ; ) =: ∞
+ h1 (t + ; (t; ; ) + u(t); (t; (t; ; ); ; ) + v(t) − U (t)(t); )} = 0:
(35)
Since (t; (·); ; ) and (t; ; ) are Cb1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 , and are C 0 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 , and (@=@ )(t; ; ; ) and (@=@ )(t; ; ) are Cb1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 , and are C 0 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 , B( ; ) and B ( ; ) are Cb1 in ( ; ) ∈ I2 × I0 , and are Cb0 in ( ; ) ∈ I2 × I1 . From Eq. (35), we obtain Z +∞ ∗ (t){F3 (t; (t; ; 0); ; 0) + h1 (t + ; u(t); v(t); 0)} dt B( ; 0) = −∞
= M ( ): Hence, B( 0 ; 0) = M ( 0 ) = 0 and B ( 0 ; 0) = M 0 ( 0 ) 6= 0. Since B ( ; ) and B( ; ) are continuous in ( ; ) ∈ I2 × I1 , it follows from the implicit function theorem that for ∈ I1 suciently small there exists a unique function () satisfying (0) = 0 and B( (); ) = 0: Moreover, () is C 0 in ∈ I1 and C 1 in ∈ I0 . Hence, for suciently small Eqs. (13) and (14) with = () have a unique bounded solution (t; ) = (t; (); );
(t; ) = (t; (t; (); ); (); )
satisfying u˙∗ (0)(0; ) = 0;
|(t; )| = O();
|(t; )| = O():
Moreover, (t; ); (t; ) are C 0 in ∈ I1 and C 1 in ∈ I0 . Hence, for suciently small Eqs. (1) and (2) have a unique bounded solution x(t; ) = (t − (); ) + u(t − ()); y(t; ) = (t − (); ) + v(t − ()) − U (t − ())(t − (); ) satisfying the rst part of Theorem 1. Now, we want to prove an important fact which is needed in the following proofs. From the above discussion we see that we have shown that the solution (t; ); (t; ) is continuous in ∈ I1 and continuously di erentiable in ∈ I0 . Now, we want to prove that (t; ); (t; ) is continuously di erentiable in ∈ I1 . This implies that we want to prove (t; ); (t; ) is di erentiable at 6= 0 suciently small. We have
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Lemma 4. We have (1) (t; ) is continuously di erentiable at = 0 and (t; 0) = (t; 0 ; 0) is the unique bounded solution; satisfying u˙∗ (0)x(0) = 0 on R; of equation x˙ = [fx (t) − fy (t)gy−1 (t)gx (t)]x + fy (t) (t; 0) + h1 (t + 0 ; u(t); v(t); 0): (2) (t; ) is continuously di erentiable at 6= 0 suciently small and ˙ − h2 (t + 0 ; u(t); v(t); 0)]: (t; 0) = gy−1 (t)[v(t) (3) ˙ ˙ (t; 0) = −gy−1 (t)[gxy (t) − gyy (t)U (t)] (t; 0)u(t) −gy−1 (t){[U˙ (t) + U (t)(fx (t) − fy (t)gy−1 (t)gx (t)) + h2x (t; 0) ˙ + h2t (t + 0 ; u(t); v(t); 0)}: −h2y (t; 0)U (t)]u(t) Proof of Lemma 4. Noting that (t; ; ) is C 0 in (t; ; ) ∈ R × I2 × I1 from Lemma 3 and (t; ; 0) = 0, we have (t; 0) = lim
→0
= lim
→0
(t; (); ) (t; (); ) − 0
= lim (t; (); ) = (t; 0 ; 0); →0
which is bounded on R. The proof of (1) is complete. Noting that (t; ; ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 and (t; 0; ; 0) = 0; and (t; 0) = 0, we have (t; 0) = lim
→0
= lim
→0
(t; (t; ); (); ) (t; (t; ); (); ) − (t; (t; 0); (); 0)
= (t; 0; 0 ; 0) + (t; 0; 0 ; 0) (t; 0) ˙ − h2 (t + 0 ; u(t); v(t); 0)]: = gy−1 (t)[v(t) The proof of (2) is complete. The proof of (3) is a direct computation from (2) and we omit its proof. Now, we will show that variational equations of Eqs. (1) and (2) along x(t; ); y(t; ) admits an exponential dichotomy on R, which are equivalent to proving that variational
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415
equations of Eqs. (13) and (14) along (t; ); (t; ) have an exponential dichotomy on R. We let fx (t; ) = fx ((t; ) + u(t); (t; ) + v(t) − U (t)(t; )); fy (t; ); gx (t; ) and gy (t; ) have similar meanings, and let h1x (t; ) = h1x (t + (); (t; ) + u(t); (t; ) + v(t) − U (t)(t; ); ); h1y (t; ); h2x (t; ), and h2y (t; ) have similar meanings. Let Fx (t; ) = fx (t; ) − fy (t; )U (t) + h1x (t; ) − h1y (t; )U (t); Fy (t; ) = fy (t; ) + h1y (t; ): With the above notations, variational equations of Eqs. (13) and (14) along (t; ); (t; ) can be written in the following form: fx (t; ) − fy (t; )U (t) + h1x (t; ) ! ! −h (t; )U (t); fy (t; ) + h1y (t; ) x˙ 1y x : = gx (t; ) − gy (t; )U (t) + h2x (t; ) gy (t; ) + h2y (t; ) y y˙ +U (t)Fy (t; ) −h2y (t; )U (t) + U˙ (t) + U (t)Fx (t; ); (36) We denote the coecient matrix of the linear Eq. (36) by A(t; ). Using the same method in Battelli and Lazzari [1], we can prove that for 6= 0 suciently small Eq. (36) has an exponential dichotomy on both R+ and R− and the sum of the dimensions of the stable and unstable subspaces is n + m. For any f(t) = (f1 (t); f2 (t))∗ ∈ Cb0 , we will prove that the equation ! ! ! x˙ x 2 f1 (t) (37) = A(t; ) + f2 (t) y˙ y has bounded solutions. If the result is true for 6= 0 suciently small then Eq. (36) admits an exponential dichotomy on R. In fact, otherwise, Eq. (36) has at least one bounded solution [7], so the adjoint equation of Eq. (36) has at least one bounded solution on R too. We denote it by (t; ). On the other hand, since the equation ! ! x˙ x (38) = A(t; ) + 2 (t; ) y˙ y has bounded solutions on R, it follows from Lemma 2 that Z +∞ ∗ 2 (t; ) (t; ) dt = 0; −∞
which is a contradiction. Hence, if for any f(t) Eq. (37) has bounded solutions then for 6= 0 suciently small then Eq. (36) admits an exponential dichotomy on R. For
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any x(t) ∈ Cb0 (R; R n ) we want to solve the second equation of Eq. (37) y˙ = [gy (t; ) + h2y (t; ) + U (t)Fy (t; )]y + [gy (t; ) − gy (t; )U (t) + h2x (t; ) − h2y (t; )U (t) + U˙ (t) + U (t)Fx (t; )]x(t) + 2 f2 (t):
(39)
From the proof of Lemma 1, we see that the equation y˙ = gy (t; 0)y
(gy (t) = gy (t; 0))
possesses an exponential dichotomy (23) and (24) on R for 6= 0 suciently small with constants independent of , hence the equation y˙ = [gy (t; ) + h2y (t; ) + U (t)Fy (t; )]y admits an exponential dichotomy on R too when 6= 0 is suciently small. Moreover, since the second term on the right-hand side of Eq. (39) is O(), using a standard perturbation method (refer to [6]), we can prove for 6= 0 suciently small Eq. (39) has a unique bounded solution y = y(t; x(·); ). The bounded solution y = y(t; x(·); ) has the following properties. Lemma 5. The bounded solution y = y(t; x(·); ) satis es (1) y(t; x(·); 0) = 0 and yx (t; x(·); 0) = 0; (2) y (t; x(·); 0) = y (t; x(·); 0) for ∈ R; that is, y (t; x(·); 0) is linear in x. Moreover, y(t; x(·); ) is Cb1 in (t; x(·); ) ∈ R × Cb1 × I1 . Proof. From the proof of the rst part of Theorem 1, we see that (t; ); (t; ) are continuously di erentiable in ∈ I1 . Thus, the functions on the right-hand side of the equation (39) are continuously di erentiable in ∈ I1 too. Using the method similar to that used to prove Lemma 1, we can prove Lemma 5. So we omit the proof of Lemma 5. Now, we substitute y = y(t; x(·); ) in the rst equation of the Eq. (37) and obtain x(t) ˙ = [ fx (t; ) − fy (t; )U (t) + h1x (t; ) − h1y (t; )U (t)]x(t) + [ fy (t; ) + h1y (t; )]y(t; x(·); ) + 2 f1 (t):
(40)
We should note that the functions on the right-hand side of Eq. (40) are C 1 in (t; x(·); ) ∈ R × Cb1 × I1 because y(t; x(·); ) is C 1 in (t; x(·); ) ∈ R × Cb1 × I1 . For Eq. (40), we make a change of variable x = x + u(t), ˙ and if x is still written as x after the change is made, then Eq. (40) becomes x˙ = [ fx (t) − fy (t)U (t)]x + [ fx (t; ) − fy (t; )U (t) + h1x (t; ) ˙ − h1y (t; )U (t) − (fx (t) − fy (t)U (t))][x + u(t)] ˙ ) + 2 f1 (t): + [ fy (t; ) + h1y (t; )]y(t; x + u(t);
(41)
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417
By almost the same method used to obtain the bifurcation function H ( ; ) for Eq. (31), we can obtain the bifurcation function H1 ( ; ) for Eq. (41) B1 ( ; ) ; 6= 0; (42) H1 ( ; ) = B1 ( ; 0); = 0 where B1 ( ; 0) Z +∞ =
∗
−∞
(t){[fx (t; 0) − fy (t; 0)U (t) + h1x (t; 0) − h1y (t; 0)U (t)] u(t) ˙
+ fy (t; 0)y (t; u(t); ˙ 0)} dt and H1 ( ; ) and H1 ( ; ) are C 0 in ( ; ) ∈ R × I1 . We let fxx (t) = fxx (u(t); v(t)); fxy (t) = fxy (u(t); v(t)), and fyy (t) = fyy (u(t); v(t)). By calculating fx (t; 0) and fy (t; 0), and using the fact that y(t; x; ) is linear in x, we obtain B1 ( ; 0) Z +∞ = −∞
∗
(t){[(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) + (fxy (t)
− fyy (t)U (t)) (t; 0) + h1x (t; 0) − h1y (t; 0)U (t)] u(t) ˙ + fy (t; 0)y (t; u(·); ˙ 0) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt − fxy (t)U (t) (t; 0)u(t) Z +∞ ∗ (t){(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) = −∞
+ (fxy (t) (t; 0) − fyy (t)U (t)) (t; 0) ˙ + fy (t; 0)y (t; u(t); ˙ 0) + h1x (t; 0) − h1y (t; 0)U (t)]u(t) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt: − fxy (t)U (t) (t; 0)u(t)
(43)
In order to calculate Eq. (43), we need some lemmas. Lemma 6. We have y (t; u(·); ˙ 0) ˙ ˙ (t) (t; 0) − gxy (t)U (t)u(t) ˙ = −gy−1 (t){[gxx (t)u(t) (t; 0) − gxy (t)u(t)U (t; 0) ˙ (t) (t; 0) + [gxy (t) − gyy (t)U (t)] (t; 0)u(t) ˙ + gyy (t)U (t)u(t)U ˙ + U˙ (t)[ fx (t) − fy (t)gy−1 (t)gx (t)]u(t)}: ˙ + [h2x (t; 0) − h2y (t; 0)U (t) + U˙ (t)]u(t) (44)
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Proof. Since the proof of Lemma 6 is almost the same as that of Lemma 1, we omit the detail. Di erentiating both sides of the equation y(t; ˙ u(t); ˙ ) ˙ ) = [gy (t; ) + h2y (t; ) + U (t)Fy (t; )]y(t; u(t); + [gx (t; ) − gy (t; )U (t) + h2x (t; ) − h2y (t; )U (t) ˙ + 2 f2 (t); + U˙ (t) + U (t)Fx (t; )]u(t) with respect to and letting → 0 and noting y(t; x; 0) = 0 and v(t) ˙ = −U (t)u(t), ˙ we have 0 = y (t; u(t); ˙ 0) + = −gy−1 (t){[gxx (t)u(t) ˙ ˙ (t) (t; 0) (t; 0) − gxy (t)u(t)U ˙ ˙ (t) (t; 0) − gxy (t)U (t)u(t) (t; 0 + gyy (t)U (t)u(t)U ˙ + h2x (t; 0) − h2y (t; 0)U (t) + U˙ (t)]u(t) ˙ + [gxy (t) − gyy (t)U (t)] (t; 0)u(t) ˙ + U˙ (t)[fx (t) − fy (t)gy−1 (t)gx (t)]u(t)}: Hence, we have y (t; u(·); ˙ 0) ˙ ˙ (t) (t; 0) − gxy (t)U (t)u(t) ˙ = −gy−1 (t){[gxx (t)u(t) (t; 0) − gxy (t)u(t)U (t; 0) ˙ (t) (t; 0) + [gxy (t) − gyy (t)U (t)] (t; 0)u(t) ˙ + gyy (t)U (t)u(t)U ˙ + U˙ (t)[fx (t) − fy (t)gy−1 (t)gx (t)]u(t)}: ˙ + h2x (t; 0) − h2y (t; 0)U (t) + U˙ (t)]u(t) The proof of Lemma 6 is complete. Lemma 7. We have U˙ (t) = gx−1 (t)[gxx (t)u(t) ˙ − gxy (t)u(t)U ˙ (t) − gxy (t)U (t)u(t) ˙ + gyy (t)U (t)u(t)U ˙ (t)]: (45) The proof of lemma 7 is a direct computation and so we omit the proof. Lemma 8. We have Z +∞ ∗ (t){[(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) −∞
+ (fxy (t) − fyy (t)U (t)) (t; 0) + (h1x (t; 0) − h1y (t; 0)U (t))]u(t) ˙ + h1t (t; 0) + fy (t)˙ (t; 0) − fy (t)U˙ (t) (t; 0) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt = 0: − fxy (t)U (t) (t; 0)u(t)
(46)
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419
Proof. Di erentiating both sides of the equation ˙ ) = f((t; ) + u(t); (t; ) + v(t) − U (t)(t; )) − f(u(t); v(t)) (t; + h1 (t + (); (t; ) + u(t); (t; ) + v(t) − U (t)(t; ); ) with respect to and letting = 0 and noting (t; 0) = 0; (t; 0) = 0, we have ˙ (t; 0) = [fx (t) − fy (t)gy−1 (t)gx (t)] (t; 0) + fy (t) (t; 0) + h1 (t + 0 ; u(t); v(t); 0): Di erentiating the above equation with respect to t and noting v(t) ˙ = −U (t)u(t), ˙ we have (t; 0) = [fx (t) − fy (t)gy−1 (t)gx (t)]˙ (t; 0) ˙ + [ fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t)]u(t) ˙ − fy (t)U˙ (t) (t; 0) + [fxy (t) − fyy (t)U (t)] (t; 0)u(t) ˙ + fy (t)˙ (t; 0) + h1t (t + 0 ; u(t); v(t); 0) + [h1x (t) − h1y (t)U (t)]u(t) ˙ − fxy (t) (t; 0)U (t)u(t): ˙ − fxy (t)U (t) (t; 0)u(t) From Lemma 4 we see that (t; 0) is bounded on R, hence ˙ (t; 0) is also bounded on R. It follows from Lemma 2 that Z +∞ ∗ (t){[(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) −∞
+ (fxy (t) − fyy (t)U (t)) (t; 0) + (h1x (t; 0) − h1y (t; 0)U (t))]u(t) ˙ + h1t (t; 0) + fy (t)˙ (t; 0) − fy (t)U˙ (t) (t; 0) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt = 0: − fxy (t)U (t) (t; 0)u(t) Lemma 8 is proved. Now, we calculate B1 ( ; 0). From Eqs. (43), (45) and (46) we have Z +∞ ∗ (t){fy (t)y (t; u(·); ˙ 0) + fy (t)U˙ (t) (t; 0) B1 ( ; 0) = −∞
− fy (t)˙ (t; 0) − h1t (t + 0 ; u(t); v(t); 0)} dt: Substituting Eqs. (44) and (46) into Eq. (47), we have Z +∞ ∗ (t){−fy (t)gy−1 (t)h2t (t + 0 ; u(t); v(t); 0) B1 ( ; 0) = − −∞
+ h1t (t + 0 ; u(t); v(t); 0)} dt = − M 0 ( 0 ): Hence H1 (0; 0) = 0;
H1 (0; 0) = −M 0 ( 0 ) 6= 0:
(47)
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It follows from the implicit function theorem that there exists a continuous function = (); (0) = 0, such that H1 ( (); ) = 0 for || suciently small. So for suciently small Eq. (39) has bounded solutions, and hence Eq. (37) has bounded solutions. From the previous discussion we see that for 6= 0 suciently small, the variational Eq. (36) admits an exponential dichotomy on R. The proof of Theorem 1 is complete. Remark. Battelli and Lazzari [2] stated that under conditions (a)– (c) the singular part of the system makes no contribution to the existence of transversal homoclinic orbits in the case where h2 (t; ; 0; 0) = 0. From the Melnikov function M ( ) obtained in the paper we see; under general conditions H1–H4, even though h2 (t; x; y; 0) ≡ 0; we cannot neglect the contribution of the singular part of the system for the existence of transversal homoclinic orbits. Now, we give an example to illustrate out theory. We consider the four-dimensional di erential equation x˙1 = y1 − y2 ; ˙ + r(t)); x˙2 = x1 − 2x13 + ( 75 r(t) y˙ 1 = y1 + 21 y22 − 12 x2 − 18 x22 + r(t);
(48)
y2 = −y2 − 12 x2 ; where r(t) = sech t. From Gruendler [12], we easily calculate Z +∞ Z +∞ 2 r˙2 (t) dt = ; r(t) ˙ r2 (t) dt = 0; 3 −∞ −∞ Z +∞ Z +∞ 14 2 r (t) dt = ; r(t)r(t) ˙ dt = 0: 15 −∞ −∞
(49)
Moreover, r(t) = r(t) − 2r 3 (t):
(50)
The degenerate equation corresponding to the singularly perturbed Eq. (48) is x˙1 = y1 − y2 ; x˙2 = x1 − 2x13 ; 1 0 = y1 + 12 y22 − x2 − 18 x22 ; 2 0 = −y2 − 12 x2 ; which has a bounded solution (using the notations of Section 2) ! ! 1 r(t) ˙ 2 r(t) : u= ; v= r(t) ˙ ˙ − 12 r(t)
(51)
(52)
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
connecting (0; 0)∗ to itself. Moreover, ! ! 1 −1 −r(t) ; (t) = ; fy (t) = 0 0 r(t) ˙ ! ! ˙ ˙ 1 − 12 r(t) 1 − 12 r(t) −1 ; gy (t) = : gy (t) = 0 −1 0 −1
421
(53)
(54)
The Melnikov function for the singularly perturbed Eq. (48) is Z +∞ 7 1 M ( ) = ˙ r2 (t) + r(t) ˙ + ) dt: (55) ˙ r(t + ) + r(t −r(t) r(t + ) + r(t) 4 5 −∞ It is easy to verify M (0) = 0;
M 0 (0) = 23 6= 0:
It follows from Theorem 1 that for 6= 0 suciently small the singularly perturbed system (48) has a transversal homoclinic orbit x(t; ) = (x1 (t; ); x2 (t; )∗ ; y(t; ) = (y1 (t; ); y2 (t; ))∗ . Remark. Since gy (t) does not satisfy condition (c) of Feckan [8] and Battelli and Lazzari [2], the results in Feckan [8] and Battelli and Lazzari [2] cannot be applied to Eq. (48), so this paper generalizes their results. Acknowledgements I would like to thank Professor Ken Palmer and the referees for carefully reading the paper and also for their invaluable suggestions on the paper. References [1] F. Battelli, C. Lazzari, Dichotomies and stability in the singularly perturbed systems of O.D.E., Nonlinear Anal. 11 (1987) 259 – 273. [2] F. Battelli, C. Lazzari, Bounded solutions to the singularly perturbed systems of ODE, J. Di erential Equations 100 (1992) 49 – 81. [3] F. Battelli, K.J. Palmer, Chaos in the Dung equations, J. Di erential Equations 102 (1993) 276 – 301. [4] K.W. Chang, Almost periodic solutions of singularly perturbed systems of di erential equations, J. Di erential Equations 4 (1968) 300 – 307. [5] K.W. Chang, W.A. Coppel, Singular perturbations of initial value problems over a nite interval, Arch. Rational Mech. Anal. 32 (1969) 269 – 280. [6] W.A. Coppel, Stability and Asymptotic Behavior of Di erential Equations, Heath, Boston, 1965. [7] W.A. Coppel, Dichotomies in Stability Theory, Lecture Notes in Mathematics, vol. 629, Springer, Berlin, 1978. [8] M. Feckan, Melnikov functions for singularly perturbed di erential equations, Nonlinear Anal. 19 (4) (1992) 393 – 401. [9] N. Fenichel, Geometric singular perturbation theory, J. Di erential Equations 17 (1979) 308 – 328.
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[10] L. Flatto, N. Levinson, Periodic solutions of singularly perturbed systems, J. Math. Mech. 4 (1955) 943 – 950. [11] J. Gruendler, Homoclinic solutions for autonomous dynamical system in arbitrary dimension, SIAM J. Math. Anal. 23 (3) (1992) 702–721. [12] K.J. Palmer, Exponential dichotomies and transversal homoclinic points, J. Di erential Equations 55 (1984) 225 – 256. [13] K.J. Palmer, Transversal heteroclinic orbits and Cherry’s example of a nonintegrable Hamiltonian system, J. Di erential Equations 65 (1986) 321– 360. [14] P. Szmolyan, Heteroclinic orbits in singularly perturbed di erential equations. preprint. [15] Z. Weiyao, Exponential dichotomies of linear systems depending on small parameter, Rocky Mountain J. Math. 25 (1995) 1565 –1576. [16] S. Wiggins, Introduction to Applied Nonlinear Dynamical Systems and Chaos, Texts in Appl. Math., No. 2, Springer, Berlin. [17] Z. Weiyao, Exponential dichotomies and Melnikov vector in degenerate case, J. Dyn. Di erential Equations 8 (1995) 521– 548.
Exponential dichotomies and Melnikov functions for singularly perturbed systems 1 Zeng Weiyao∗ , Luo Jiaowan Institute of Sciences, Changsha Railway University, Changsha, Hunan 410075, People’s Republic of China Received 17 November 1995; received in revised form 8 December 1996; accepted 13 January 1998
Keywords: Exponential dichotomies; Melnikov functions; Singularly perturbed systems
1. Introduction Recently, many authors have worked on the existence of homoclinic solutions in singularly perturbed systems. Some of them worked on systems where the homoclinic solution possesses a transition layer connecting two equilibria on two slow manifolds. Others worked on systems where the homoclinic solution does not have a transition layer and is con ned to one slow manifold. This paper belongs to the second category and we mainly consider the problem of the existence of transversal homoclinic orbits of systems of singularly perturbed ODE of the form x˙ = f(x; y) + h1 (t; x; y; );
x ∈ Rn;
(1)
y˙ = g(x; y) + h2 (t; x; y; );
y ∈ Rm;
(2)
where ≥ 0 is a small parameter, f; g; h1 and h2 are C 2 -smooth. However, rst we recall some results for general singularly perturbed systems of the form
∗ 1
x˙ = f(t; x; y; );
(3)
y˙ = g(t; x; y; );
(4)
Corresponding author. E-mail: [email protected]. The project supported by NSF of China.
0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 6 1 4 - 7
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where x ∈ R n , y ∈ R m , is a small parameter. Setting = 0 in the system (3) and (4), we obtain the equations x˙ = f(t; x; y; 0);
(5)
0 = g(t; x; y; 0);
(6)
which are called the degenerate equations of Eqs. (3) and (4). We assume that C1. The degenerate equations (5) and (6) have a nontrivial bounded (or periodic or almost periodic) solution x = u(t), y = v(t). We call the equation ˙ = (fx (t) − fy (t)gy−1 (t)gx (t));
(7)
the rst variational equations of Eqs. (3) and (4) along the bounded solution u(t); v(t), where fx (t) = fx (t; u(t); v(t); 0) and fy (t); gx (t) and gy (t) have similar meanings. For the nonautonomous singularly perturbed equations (3) and (4), if condition C1 and conditions C2. There exist a constant ¿0 such that the absolute values of the real parts of the eigenvalues of the m ×m matrix gy (u(t); v(t); 0) are greater than , C3. The rst variational Eq. (7) has an exponential dichotomy on R are satis ed, then Chang [4] showed that for suciently small the singularly perturbed equations (3) and (4) have a unique bounded solution x(t; ); y(t; ) satisfying lim {|x(t; ) − u(t)| + |y(t; ) − v(t)|} = 0;
→0
(∗)
uniformly in t ∈ R. If f(t; x; y; ); g(t; x; y; ); u(t); v(t) are T -periodic (or almost periodic in t) then for 6= 0 suciently small Eqs. (3) and (4) have a unique T -periodic (or almost periodic) solution x(t; ); y(t; ) satisfying Eq. (∗). However, for the special singularly perturbed system (1) and (2) with nonautonomous perturbed terms, condition C3 cannot be satis ed. In fact, degenerate equations of Eqs. (1) and (2) are x˙ = f(x; y); 0 = g(x; y); which have a nontrivial bounded solution x = u(t), y = v(t). It is easy to show ˙ u(t) = [fx (t) − fy (t)gy−1 (t)gx (t)] u(t): Hence, the rst variational equation of singularly perturbed equations (1) and (2) along u(t); v(t) have a nontrivial bounded solution u(t) ˙ on R, and the rst variational equation does not admit an exponential dichotomy on R (refer to [1]). Thus, the above result on bounded solutions for Eqs. (3) and (4) cannot be applied to Eqs. (1) and (2). Recently, Feckan [8], and Battelli and Lazzari [2] investigated the existence of homoclinic orbits of singularly perturbed system (1) and (2) using a Melnikov function. In their papers they want Eqs. (1) and (2) to satisfy the following conditions: (a) g(x; y) = 0 ⇔ y = 0.
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403
(b) The degenerate equation x˙ = f(x; 0) has a nontrivial bounded solution u(t), and the variational equation along u(t) x˙ = A(t)x;
(A(t) =: fx (u(t); 0))
(8)
has an exponential dichotomy on both R+ and R− and the sum of the dimensions of the stable and unstable subspaces of Eq. (8) is n. Moreover, u(t) ˙ is the unique (up to a scalar multiple) bounded solution of variational equation (8). (c) C(t) =: gy (u(t); 0) = diag(C1 (t); C2 (t)), and the real parts of the eigenvalues (t) of C1 (t) (resp. C2 (t)) satisfy Re (t) ≤ −2 ¡0 (resp. Re (t) ≥ 2 ¿0) for any t ∈ R. Under conditions (a)–(c), Feckan [8], and Battelli and Lazzari [2] obtained the following Melnikov-type function: Z +∞ ∗ (t){h1 (t + ; u(t); 0; 0) − fy (t)C −1 (t)h2 (t + ; u(t); 0; 0)} dt M ( ) = −∞
and proved that if M ( ) has a simple zero then for 6= 0 suciently small singularly perturbed system (1) and (2) has a transversal homoclinic orbit. It seems to us that the method used in [2] is very complex, and Feckan [8] did not prove the transversality of homoclinic orbits. Moreover, it should be noted, from the conditions in known results on periodic solutions and bounded solutions of singularly perturbed equations, that conditions (a) and (c) in their papers are not natural (compared with conditions C1, C2). The main purpose of this paper is to drop these conditions. We need to emphasize that the results in this paper cannot be obtained by simply making a change in the variable y = y + v(t) of Eq. (2). Battelli and Lazzari [2] deduced from the Melnikov function obtained in [2] that under conditions (a) – (c) the contribution of the singular part of the system is not important for the existence of transversal homoclinic orbits in the case where h2 (t; ; 0; 0) = 0. From the Melnikov function M ( ) obtained in this paper we see, under general conditions H1–H4, that even when h2 (t; x; y; 0) = 0, we cannot neglect the role of the singular part of the system for the existence of transversal homoclinic orbits. We obtain a very general Melnikov-type function and show that if the Melnikov-type of function has a simple zero then for 6= 0 suciently small the singularly perturbed system has a transversal homoclinic orbit. It seems to us that our proof in this paper is simpler than the one in Battelli and Lazzari [2], and we also give a new method for proving the transversality of homoclinic orbits of the singularly perturbed equations. The main tool used in this paper is the theory of exponential dichotomies and the functional analystic method. We should note that the paper lacks geometric insight. Geometric singular perturbation has become an essential tool for understanding singular perturbation problems, we therefore refer to Fenichel [9]. This paper is organized as follows: Section 1 is an introduction, in Section 2 we give the main result of this paper; Section 3 is devoted to the proof of the main results and a example. The notations for function spaces in this paper are the same as those in Battelli and Lazzari [2].
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2. Main results We assume H1. There exists a compact set S1 ×S2 ⊂ R n × R m such that the degenerate system x˙ = f(x; y);
(9)
0 = g(x; y)
(10)
admits a nontrivial bounded solution x = u(t), y = v(t) in S1 × S2 . H2. There exists a constant ¿0 such that the absolute values of the real parts of the eigenvalues of the m×m matrix gy (u(t); v(t)) are greater than . Remark. If conditions H1 and H2 hold, we can easily prove that u(t) = [fx (t) − fy (t)gy−1 (t)gx (t)] u(t); ˙ where fx (t) =: fx (u(t); v(t)) and fy (t); gx (t) and gy (t) have similar meanings, thus u(t) ˙ is a nontrivial bounded solution of the rst variational equation x˙ = [fx (t) − fy (t)gy−1 (t)gx (t)]x:
(11)
It follows from Coppel [7] that Eq. (11) does not have an exponential dichotomy on R. H3. The rst variational Eq. (11) admits an exponential dichotomy on both R+ and R− and the sum of the dimensions of the stable and unstable subspaces is n. Moreover, Eq. (11) has a unique (up to a scalar multiple) nontrivial bounded solution u(t). ˙ Remark. If H3 is satis ed, then Palmer [12] has proved that the adjoint equation x˙ = −[fx (t) − fy (t)gy−1 (t)gx (t)]∗ x
(12)
also has, up to a scalar multiple, a unique nontrivial bounded solution on R; denoted by (t). The main results of this paper are as follows: Theorem 1. We assume H1, H2 and H3 are satis ed and we de ne Z +∞ ∗ (t){fy (t)gy−1 (t)[v(t) ˙ − h2 (t + ; u(t); v(t); 0)] M ( ) = −∞
+ h1 (t + ; u(t); v(t); 0)} dt: If there exists an 0 ∈ R satisfying M ( 0 ) = 0;
M 0 ( 0 ) 6= 0;
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405
then there exists a continuous function () with (0) = 0 ; such that for 6= 0 suciently small singularly perturbed system (1) and (2) has a unique bounded solution x(t; ); y(t; ) satisfying u˙∗ (0)[x( (); ) − u(0)] = 0; sup |x(t + (); ) − u(t)| = O(); t∈R
sup |y(t + (); ) − v(t)| = O(): t∈R
Moreover, for suciently small 6= 0 variational equations of Eqs. (1) and (2) along the bounded solution x(t; ); y(t; ) admits an exponential dichotomy on R. A remarkable situation where Theorem 1 applies, occurs when H3 is replaced by H30 . The degenerate equations (9) and (10) have a nontrivial bounded solution u(t); v(t) in S1 × S2 and a trivial solution x0 ; y0 satisfying lim u(t) = x0 ;
t→±∞
lim v(t) = y0
t→±∞
and the real parts of the eigenvalues of the matrix fx (x0 ; y0 ) − fy (x0 ; y0 )gy−1 (x0 ; y0 ): ˙ is a unique (up to a scalar multiple) nongx (x0 ; y0 ) are di erent from zero, and u(t) trivial bounded solution of Eq. (11) on R. By making use of Theorem 1 and the method for proving Corollary 5.2 in Palmer [12], we can show the following result. Theorem 2. Suppose H1, H2 and H30 are satis ed and hi (t + T; x; y; ) = hi (t; x; y; ); i = 1; 2. Then for 6= 0 suciently small Eqs. (1) and (2) admit a unique T -periodic solution x0 (t; ); y0 (t; ) satisfying lim {kx0 (t; ) − x0 k + ky0 (t; ) − y0 k} = 0
→0
and variational equations of Eqs. (1) and (2) along x0 (t; ); y0 (t; ) have an exponential dichotomy on R. Moreover if there exists 0 ∈ R such that M ( 0 ) = 0;
M 0 ( 0 ) 6= 0;
then there is a continuous function () with (0) = 0 ; such that for 6= 0 suciently small Eqs. (1) and (2) have a unique bounded solution x(t; ); y(t; ) satisfying u˙∗ (0)[x( (); ) − u(0)] = 0; sup |x(t + (); ) − u(t)| = O(); t∈R
sup |y(t + (); ) − v(t)| = O(); t∈R
lim {|x(t; ) − x0 (t; )| + |y(t; ) − y0 (t; )|} = 0:
t→±∞
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Moreover; for 6= 0 suciently small variational equations Eqs. (1) and (2) along the bounded solution x(t; ); y(t; ) admits an exponential dichotomy on R. Remark. From the result in Palmer [12] and Theorem 2 of this paper; we see that if the conditions of Theorem 2 are satis ed then the Poincare map P of Eqs. (1) and (2) have a transversal homoclinic point x(0; ); y(0; ) biasymptotic to the hyperbolic T -periodic point x0 (0; ); y0 (0; ). 3. The proofs of the theorems For simplicity, we let U (t) =: gy−1 (t)gx (t). For singularly perturbed system (1) and (2) we make the change of variable as in [5] x(t + ) = (t) + u(t); y(t + ) = (t) + v(t) − U (t)(t); then system (1) and (2) becomes ˙ = f( + u(t); + v(t) − U (t)) + h1 (t + ; + u(t); + v(t) − U (t); ) −f(u(t); v(t));
(13)
˙ = g( + u(t); + v(t) − U (t)) + h2 (t + ; + u(t); + v(t) − U (t); ) ˙ − v(t) ˙ + U˙ (t) + U (t):
(14)
We write Eqs. (13) and (14) in the following form: ˙ = [fx (t) − fy (t)gy−1 (t)gx (t)] + fy (t) + h1 (t + ; + u(t); + v(t) − U (t); ) + F1 (t; ; ; ); ˙ = gy (t) + F2 (t; ; ; ; );
(15) (16)
where F1 (t; ; ; ) = f( + u(t); + v(t) − U (t)) − fy (t) −f(u(t); v(t)) − [fx (t) − fy (t)gy−1 (t)gx (t)];
(17)
F2 (t; ; ; ; ) = −v(t) ˙ + U˙ (t) + U (t)˙ + [g( + u(t); + v(t) − U (t)) − gy (t)] + h2 (t + ; + u(t); + v(t) − U (t); ):
(18)
For simplicity, we now give some notations. For suciently small 0 ¿0; we let I0 = (0; 0 ]; I1 = [0; 0 ] and I2 be a small interval containing 0 .
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407
We have the following Lemma 1. Lemma 1. Suppose H1 and H2 hold. Then for any constant 0 ¿0 and for any (t) ∈ {x | x ∈ Cb1 ; |x(t)| ≤ 0 } the equation ˙ = gy (t) + F2 (t; (t); ; ; )
(19)
has; for suciently small; a unique small bounded solution = (t; (·); ; ) satisfying (see remark following Lemma 1): ˙ − h2 (t + ; u(t); v(t); 0)]; (t; (·); ; 0) = gy−1 (t)[v(t)
(20)
(t; (·); ; 0) = 0;
(21)
(t; (·); ; 0) = 0;
(22)
for (t; (·); ) ∈ R × Cb1 × I2 . Moreover; (t; ; ; ) is C 2 in (t; (·); ; ) ∈ R × Cb1 × I2 × 1 (·); ; ) is C in (t; ; ; ) ∈ R × Cb1 × I2 × I1 . I0 ; (t; (·); ; ) are de ned only for 6= 0. Remark. Actually; (t; (·); ; ); (t; (·); ; ); (t; Since we will see that as → 0 their limits exist; we de ned their values at = 0 to be their limits as → 0. Proof. Since H2 holds, it follows from Chang and Coppel [5] that for ¿0 suciently small the linear equation ˙ = gy (t) possesses an exponential dichotomy on R with |X (t; )P()X −1 (s; )| ≤ Ke−=(t−s) ;
t ≥ s;
|X (t; ) (I − P()X −1 (s; )| ≤ Ke−=(s−t) ;
s≥t
(23)
for t; s ∈ R. Using a standard perturbation method (contraction xed point theorem and theory of exponential dichotomies, refer to Coppel [6]), we can prove that for ¿0 suciently small Eq. (19) has a unique bounded solution = (t; (·); ; ) satisfying (·); ; ) = 0 lim (t;
→0
(24)
uniformly in (t; (·); ) ∈ R × Cb1 × I2 and (t; (·); ; ) Z t 1 X (t; )P()X −1 (s; )F2 (t; (t); (t; (t); ; ) ds = −∞ Z +∞ −1 X (t; )(I − P())X (s; )F2 (t; (t); (t; (t); ; ); ; ) ds : − t
So we can de ne (t; (·); ; 0) = 0. First of all, if Eq. (19) were ˙ = gy (t) + F2 (t; (t); ; ; );
(25)
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Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
where the fundamental matrix X (t) of ˙ = gy (t) satis ed |X (t)PX −1 (s)| ≤ Ke−(t−s) ; |X (t) (I − P)X
−1
(s)| ≤ Ke
−(s−t)
t ≥ s; ;
s ≥ t;
then it is well-known that the solution (t; ; ; ) is C 2 in (; ; ). Now, we deal with Eq. (19). We want to prove that (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . We rst prove that (t; (·); ; ) is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Set (·); ; ); 2 (t) = (t + t; (·) + (·); + ; + h); (t) = 2 (t) − 1 (t). 1 (t) = (t; Then (t) is a solution of equation ˙ = gy (t) + F2 (t + t; ( + h)((t) + (t); 2 (t); + ; + h) − F2 (t; (t); 1 (t); ; ) − h · ˙2 (t) − [gy (t) − gy (t + t)]2 (t) and so, using Eqs. (23)–(25) |(t)| ≤ K−1
sup
[|F2 (t + t; ( + h)((t) + (t)); 2 (t); + ; + h)
−∞¡t¡+∞
− F2 (t; (t); 1 (t); ; ) − h · ˙2 (t) − [gy (t) − gy (t + t)]2 (t)|] 1 sup |(t)| + constant · |h| + |t| + |(h + )| sup |(t)| + | | ; ≤ 2 −∞¡t¡∞ t∈R provided the parameters are suciently small. Hence, sup |(t)| ≤ constant · |h| + |t| + |(h + )| sup |(t)| + | | −∞¡t¡∞
t∈R
and so |(t + t; (·) + (·); + ; + h) − (t; ; ; )| ≤ constant · |h| + |t| + |(h + )| sup |(t)| + | | : t∈R
Hence, (t; (·); ; ) is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . To show that (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 ; let z(t; (·); ; ) be the unique bounded solution of equation z˙ = gy (t)z + +
@F2 @F2 (t; (t); (t; (·); ; ); ; )z + (t; (t); (t; (·); ; ); ; )(t) @ @
@F2 (t; (t); (t; (·); ; ); ; ) − (t; ˙ (·); ; ): @
Then, by considering (t) = (t; (·); ; + h) − (t; (·); ; ) − z(t)h and reasoning as above, we can show that (t) is o(h) and hence conclude that (t; (·); ; ) = z(t; (·); ; ). By the same way as proving that (t; (·); ; ) is continuous in (t; (·); ; )
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
409
∈ R × Cb1 × I2 × I0 , we can show that (t; (·); ; ) = z(t; (·); ; ) is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Similarly, we can prove that t (t; (·); ; ); (t; (·); ; ) and (t; (·); ; ) exist and are continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Thus, (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 . Similarly, we can prove that (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 , for simplicity, we omit the proof. Now, we show that (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 . Obviously, we only need to prove (t; (·); ; ) is continuously di erentiable at any point (t; (·); ; 0). We rst prove that (t; (·); ; 0) exists and (t; (·); ; ) is continuous at (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . For simplicity, we let m(t) = m(t; (·); ) = gy−1 (t) [v(t) ˙ − h2 (t + ; u(t); v(t); 0)]. Di erentiating both sides of the equation (·); ; ) + F2 (t; (t); (t; (·); ; ); ; ) (t; ˙ (·); ; ) = gy (t)(t; with respect to ; we obtain ˙ (·); ; ) ˙ (t; (·); ; ) + (t; ˙ + h2 (t + ; u(t) + (t); (t; (·); ; ) + v(t); ) = gy (t) (t; (·); ; ) − v(t) + O() (t; (·); ; ) + O(): Thus, ˙ (·); )} {˙ (t; (·); ; ) − m(t; ˙ (·); ; )+ O() (t; (·); ; ) = gy (t){ (t; (·); ; ) − m(t; (·); )}+ O() − (t; (·); ; ) − m(t; ˙ (·); ) + h2 (t + ; u(t) + (t); (t; + v(t); ) − h2 (t + ; u(t); v(t); 0): Since for ∈ I0 the linear part of the above equation y˙ = gy (t)y admits an exponential dichotomy (23) on R, and (t; (·); ; ) is bounded for t ∈ R and xed ∈ I0 , (t; (·); ; ) has the following form: (t; (·); ; ) − m(t) Z t 1 = X (t; )P()X −1 (s; ){O() − (t; ˙ (t); ; ) −∞ + O(){ (t; (t); ; ) − m(t)} (·); ; ) + v(t); ) + O()m(t) − m(t) ˙ + h2 (t + ; u(t) + (t); (t; − h2 (t + ; u(t); v(t); 0) ds Z −
t
+∞
X (t; )(I − P())X −1 (s; ){O() − (t; ˙ (t); ; )
+ O(){ (t; (t); ; ) − m(t)}
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Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
+ O()m(t) − m(t) ˙ + h2 (t + ; u(t) + (t); (t; (·); ; ) + v(t); ) − h2 (t + ; u(t); v(t); 0)}ds : Thus, there exists a positive constant L0 such that | (t; (·); ; ) − m(t)| ˙ (·); ; ) + O() − O()m(t) − m(t) ˙ ≤ L0 · sup |(t; t∈R
(·); ; ) + v(t); ) − h2 (t + ; u(t); v(t); 0)|: (26) + h2 (t + ; u(t) + (t); (t; Noting that (t; (·); ; 0) = 0 and is continuous in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 , we obtain ˙ − h2 (t + ; u(t); v(t); 0)]: lim (t; (·); ; ) = m(t) = gy−1 (t)[v(t)
→0
So we may de ne ˙ − h2 (t + ; u(t); v(t); 0)]: (t; (·); ; 0) = gy−1 (t)[v(t) Following almost the same method used to prove Eq. (26), we can prove | (t + t; (·) + (·); + ; h) − (t; (·); ; 0)| = | (t + t; (·) + (·); + ; h) − m(t; (·); )| ≤ constant |t| + sup |(t)| + | | + |h| : t∈R
Thus, (t; (·); ; ) exists and is continuous at (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . In the same way, we can prove that t (t; (·); ; 0) = 0; (t; (·); ; 0) = 0 and (t; (·); ; 0) = 0 exists, and t (t; (·); ; ); (t; (·); ; ) and (t; (·); ; ) are con (·); ; ) is continuously di erentinuous at (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . Hence, (t; (·); ; ) is C 1 in (t; (·); ; ) tiable at point (t; (·); ; 0) ∈ R × Cb1 × I2 × I1 . Thus, (t; 1 ∈ R × C b × I2 × I1 . We need a result in Palmer [12]. Lemma 2. We consider the linear equation x˙ = A(t)x; n
(27)
x ∈ R . Assume that Eq. (27) admits an exponential dichotomy on both R+ and R− . Then for f ∈ Cb0 equation x˙ = A(t)x + f(t) has bounded solutions on R if and only if Z +∞ ∗ (t)f(t) dt = 0 −∞
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
holds for all bounded solutions
411
(t) of the equation adjoint to Eq. (27)
∗
x˙ = −A (t)x: We make a change of variables (t) ∼ (t) in Eqs. (15) and (16) and obtain a new system of equations ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + fy (t) + F1 (t; (t); ; ; ) (t) y + h1 (t + ; (t) + u(t); + v(t) − U (t)(t); );
(28)
(t) ˙ = gy (t)(t) + F2 (t; (t); (t); ; ):
(29)
From Lemma 1, we see for sucient small, Eq. (14) has a bounded solution = (t; (·); ; ). We put it in Eq. (28) and have ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) (t) y fy (t)(t; (·); ; ) + F1 (t; (t); (t; (·); ; ); ; ) + h1 (t + ; (t) + u(t); + v(t) − U (t)(t); ): +
(30)
From Lemma 1 we see that if Eq. (30) has a bounded solution = (t; ), then Eqs. (13) and (14) have a bounded solution = (t; ); = (t; (t; ); ; ). Hence, we only need to solve Eq. (30). We de ne a function f (t)(t; (·); ; ) + F1 (t; (·); ; ; ) y 6= 0; F3 (t; (·); ; ) = ˙ − h2 (t + ; u(t); v(t); 0)] = 0: fy (t)gy−1 (t)[v(t) Then F3 (t; (·); ; 0) = fy (t)gy−1 (t)[v(t) ˙ − h2 (t + ; u(t); v(t); 0)] fy (t)(t; (·); ; ) + F1 (t; (·); ; ; ) →0 d = [fy (t)(t; (·); ; ) + F1 (t; (·); ; ; )|=0 F3 (t; (·); ; 0): d = lim
Since (t; (·); ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 from Lemma 1, we have fy (t)(t; (·); ; ) + F1 (t; (·); ; ; ) is also C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 . Thus, F3 (t; (·); ; ) is C 0 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 . F3 (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R×Cb1 ×I2 ×I0 is obvious because (t; (·); ; ) is C 2 in (t; (·); ; ) ∈ R× Cb1 ×I2 ×I0 . Similarly, we can prove that (@=@ )F3 (t; (·); ; ) is C 0 in (t; (·); ; ) ∈ R× Cb1 × I2 × I1 . It is obvious that when 6= 0 Eq. (30) is equivalent to the equation ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + F3 (t; (·); ; ) (t) y + h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); ): We have the following lemma.
(31)
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Lemma 3. Suppose H1–H3 are satis ed. Then there exist constants 0 ¿0 and 0 ¿0 such that for ||¡0 the equation ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + F3 (t; (·); ; ) (t) y + h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); ) Z +∞ ∗ (t){F3 (t; (·); (t; (·); ; ); ; ) − −∞
+ h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); )} dt· (t)
(32)
has a unique bounded solution = (t; ; ) satisfying |(t; ; )| ≤ 0 sup | + F3 (t; (t; ; ); ; ) + h1 (t + ; (t; ; ) t∈R
+ u(t); (t; (t; ; ); ; ) + v(t) − U (t)(t); )|; u˙∗ (0)(0; ; ) = 0: Moreover; (t; ; ) and (@=@ )(t; ; ) are Cb1 in (t; ; ) ∈ R × I2 × I0 ; and C 0 in (t; ; ) ∈ R × I2 × I1 . Proof. Since (t; (·); ; ) is Cb1 in (t; ; ) ∈ R×I2 ×I0 , and is Cb0 in (t; ; ) ∈ R×I2 ×I1 , the functions on the right-hand side of Eq. (32) are also Cb1 in (t; ; ) ∈ R × I2 × I0 , and is Cb0 in (t; ; ) ∈ R × I2 × I1 . The proof of Lemma 3 is almost the same as that in Battelli and Palmer [3] (noting (t; (·); ; 0) = 0), so we omit the proof. R +∞ Proof of Theorem 1. Without loss of generality, we may assume −∞ ∗ (t) (t) dt = 1. Using the Liapunov–Schmidt method, we know that Eq. (31) is equivalent to the following equation: ˙ = [fx (t) − fy (t)g−1 (t)gx (t)](t) + F3 (t; (·); ; ) (t) y + h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); ) − B((·); ; )· (t); Z B((·); ; ) =
+∞
−∞
∗
(33)
(t){fy (t)(t; (·); ; ) + F3 (t; (·); (t; (·); ; ); )
+ h1 (t + ; (t) + u(t); (t; (·); ; ) + v(t) − U (t)(t); )} dt: (34) It follows from Lemma 3 that for suciently small Eq. (33) has a unique bounded solution = (t; ; ) which is Cb1 in (t; ; ) ∈ R×I2 ×I0 and is Cb0 in (t; ; ) ∈ R×I2 ×I1 and satis es u˙∗ (0)(0; ; ) = 0;
|(t; ; )| ≤ 0 ; (t; ; ) ∈ R × I2 × I1 :
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
413
Moreover, (@=@ )(t; ; ) is Cb0 in (t; ; ) ∈ R × I2 × I1 . Substituting = (t; ; ) into Eq. (34), we obtain the bifurcation equation for Eq. (31) Z +∞ ∗ (t){F3 (t; (t; ; ); ; ) B( ; ) =: ∞
+ h1 (t + ; (t; ; ) + u(t); (t; (t; ; ); ; ) + v(t) − U (t)(t); )} = 0:
(35)
Since (t; (·); ; ) and (t; ; ) are Cb1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 , and are C 0 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 , and (@=@ )(t; ; ; ) and (@=@ )(t; ; ) are Cb1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I0 , and are C 0 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 , B( ; ) and B ( ; ) are Cb1 in ( ; ) ∈ I2 × I0 , and are Cb0 in ( ; ) ∈ I2 × I1 . From Eq. (35), we obtain Z +∞ ∗ (t){F3 (t; (t; ; 0); ; 0) + h1 (t + ; u(t); v(t); 0)} dt B( ; 0) = −∞
= M ( ): Hence, B( 0 ; 0) = M ( 0 ) = 0 and B ( 0 ; 0) = M 0 ( 0 ) 6= 0. Since B ( ; ) and B( ; ) are continuous in ( ; ) ∈ I2 × I1 , it follows from the implicit function theorem that for ∈ I1 suciently small there exists a unique function () satisfying (0) = 0 and B( (); ) = 0: Moreover, () is C 0 in ∈ I1 and C 1 in ∈ I0 . Hence, for suciently small Eqs. (13) and (14) with = () have a unique bounded solution (t; ) = (t; (); );
(t; ) = (t; (t; (); ); (); )
satisfying u˙∗ (0)(0; ) = 0;
|(t; )| = O();
|(t; )| = O():
Moreover, (t; ); (t; ) are C 0 in ∈ I1 and C 1 in ∈ I0 . Hence, for suciently small Eqs. (1) and (2) have a unique bounded solution x(t; ) = (t − (); ) + u(t − ()); y(t; ) = (t − (); ) + v(t − ()) − U (t − ())(t − (); ) satisfying the rst part of Theorem 1. Now, we want to prove an important fact which is needed in the following proofs. From the above discussion we see that we have shown that the solution (t; ); (t; ) is continuous in ∈ I1 and continuously di erentiable in ∈ I0 . Now, we want to prove that (t; ); (t; ) is continuously di erentiable in ∈ I1 . This implies that we want to prove (t; ); (t; ) is di erentiable at 6= 0 suciently small. We have
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Lemma 4. We have (1) (t; ) is continuously di erentiable at = 0 and (t; 0) = (t; 0 ; 0) is the unique bounded solution; satisfying u˙∗ (0)x(0) = 0 on R; of equation x˙ = [fx (t) − fy (t)gy−1 (t)gx (t)]x + fy (t) (t; 0) + h1 (t + 0 ; u(t); v(t); 0): (2) (t; ) is continuously di erentiable at 6= 0 suciently small and ˙ − h2 (t + 0 ; u(t); v(t); 0)]: (t; 0) = gy−1 (t)[v(t) (3) ˙ ˙ (t; 0) = −gy−1 (t)[gxy (t) − gyy (t)U (t)] (t; 0)u(t) −gy−1 (t){[U˙ (t) + U (t)(fx (t) − fy (t)gy−1 (t)gx (t)) + h2x (t; 0) ˙ + h2t (t + 0 ; u(t); v(t); 0)}: −h2y (t; 0)U (t)]u(t) Proof of Lemma 4. Noting that (t; ; ) is C 0 in (t; ; ) ∈ R × I2 × I1 from Lemma 3 and (t; ; 0) = 0, we have (t; 0) = lim
→0
= lim
→0
(t; (); ) (t; (); ) − 0
= lim (t; (); ) = (t; 0 ; 0); →0
which is bounded on R. The proof of (1) is complete. Noting that (t; ; ; ) is C 1 in (t; (·); ; ) ∈ R × Cb1 × I2 × I1 and (t; 0; ; 0) = 0; and (t; 0) = 0, we have (t; 0) = lim
→0
= lim
→0
(t; (t; ); (); ) (t; (t; ); (); ) − (t; (t; 0); (); 0)
= (t; 0; 0 ; 0) + (t; 0; 0 ; 0) (t; 0) ˙ − h2 (t + 0 ; u(t); v(t); 0)]: = gy−1 (t)[v(t) The proof of (2) is complete. The proof of (3) is a direct computation from (2) and we omit its proof. Now, we will show that variational equations of Eqs. (1) and (2) along x(t; ); y(t; ) admits an exponential dichotomy on R, which are equivalent to proving that variational
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
415
equations of Eqs. (13) and (14) along (t; ); (t; ) have an exponential dichotomy on R. We let fx (t; ) = fx ((t; ) + u(t); (t; ) + v(t) − U (t)(t; )); fy (t; ); gx (t; ) and gy (t; ) have similar meanings, and let h1x (t; ) = h1x (t + (); (t; ) + u(t); (t; ) + v(t) − U (t)(t; ); ); h1y (t; ); h2x (t; ), and h2y (t; ) have similar meanings. Let Fx (t; ) = fx (t; ) − fy (t; )U (t) + h1x (t; ) − h1y (t; )U (t); Fy (t; ) = fy (t; ) + h1y (t; ): With the above notations, variational equations of Eqs. (13) and (14) along (t; ); (t; ) can be written in the following form: fx (t; ) − fy (t; )U (t) + h1x (t; ) ! ! −h (t; )U (t); fy (t; ) + h1y (t; ) x˙ 1y x : = gx (t; ) − gy (t; )U (t) + h2x (t; ) gy (t; ) + h2y (t; ) y y˙ +U (t)Fy (t; ) −h2y (t; )U (t) + U˙ (t) + U (t)Fx (t; ); (36) We denote the coecient matrix of the linear Eq. (36) by A(t; ). Using the same method in Battelli and Lazzari [1], we can prove that for 6= 0 suciently small Eq. (36) has an exponential dichotomy on both R+ and R− and the sum of the dimensions of the stable and unstable subspaces is n + m. For any f(t) = (f1 (t); f2 (t))∗ ∈ Cb0 , we will prove that the equation ! ! ! x˙ x 2 f1 (t) (37) = A(t; ) + f2 (t) y˙ y has bounded solutions. If the result is true for 6= 0 suciently small then Eq. (36) admits an exponential dichotomy on R. In fact, otherwise, Eq. (36) has at least one bounded solution [7], so the adjoint equation of Eq. (36) has at least one bounded solution on R too. We denote it by (t; ). On the other hand, since the equation ! ! x˙ x (38) = A(t; ) + 2 (t; ) y˙ y has bounded solutions on R, it follows from Lemma 2 that Z +∞ ∗ 2 (t; ) (t; ) dt = 0; −∞
which is a contradiction. Hence, if for any f(t) Eq. (37) has bounded solutions then for 6= 0 suciently small then Eq. (36) admits an exponential dichotomy on R. For
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any x(t) ∈ Cb0 (R; R n ) we want to solve the second equation of Eq. (37) y˙ = [gy (t; ) + h2y (t; ) + U (t)Fy (t; )]y + [gy (t; ) − gy (t; )U (t) + h2x (t; ) − h2y (t; )U (t) + U˙ (t) + U (t)Fx (t; )]x(t) + 2 f2 (t):
(39)
From the proof of Lemma 1, we see that the equation y˙ = gy (t; 0)y
(gy (t) = gy (t; 0))
possesses an exponential dichotomy (23) and (24) on R for 6= 0 suciently small with constants independent of , hence the equation y˙ = [gy (t; ) + h2y (t; ) + U (t)Fy (t; )]y admits an exponential dichotomy on R too when 6= 0 is suciently small. Moreover, since the second term on the right-hand side of Eq. (39) is O(), using a standard perturbation method (refer to [6]), we can prove for 6= 0 suciently small Eq. (39) has a unique bounded solution y = y(t; x(·); ). The bounded solution y = y(t; x(·); ) has the following properties. Lemma 5. The bounded solution y = y(t; x(·); ) satis es (1) y(t; x(·); 0) = 0 and yx (t; x(·); 0) = 0; (2) y (t; x(·); 0) = y (t; x(·); 0) for ∈ R; that is, y (t; x(·); 0) is linear in x. Moreover, y(t; x(·); ) is Cb1 in (t; x(·); ) ∈ R × Cb1 × I1 . Proof. From the proof of the rst part of Theorem 1, we see that (t; ); (t; ) are continuously di erentiable in ∈ I1 . Thus, the functions on the right-hand side of the equation (39) are continuously di erentiable in ∈ I1 too. Using the method similar to that used to prove Lemma 1, we can prove Lemma 5. So we omit the proof of Lemma 5. Now, we substitute y = y(t; x(·); ) in the rst equation of the Eq. (37) and obtain x(t) ˙ = [ fx (t; ) − fy (t; )U (t) + h1x (t; ) − h1y (t; )U (t)]x(t) + [ fy (t; ) + h1y (t; )]y(t; x(·); ) + 2 f1 (t):
(40)
We should note that the functions on the right-hand side of Eq. (40) are C 1 in (t; x(·); ) ∈ R × Cb1 × I1 because y(t; x(·); ) is C 1 in (t; x(·); ) ∈ R × Cb1 × I1 . For Eq. (40), we make a change of variable x = x + u(t), ˙ and if x is still written as x after the change is made, then Eq. (40) becomes x˙ = [ fx (t) − fy (t)U (t)]x + [ fx (t; ) − fy (t; )U (t) + h1x (t; ) ˙ − h1y (t; )U (t) − (fx (t) − fy (t)U (t))][x + u(t)] ˙ ) + 2 f1 (t): + [ fy (t; ) + h1y (t; )]y(t; x + u(t);
(41)
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417
By almost the same method used to obtain the bifurcation function H ( ; ) for Eq. (31), we can obtain the bifurcation function H1 ( ; ) for Eq. (41) B1 ( ; ) ; 6= 0; (42) H1 ( ; ) = B1 ( ; 0); = 0 where B1 ( ; 0) Z +∞ =
∗
−∞
(t){[fx (t; 0) − fy (t; 0)U (t) + h1x (t; 0) − h1y (t; 0)U (t)] u(t) ˙
+ fy (t; 0)y (t; u(t); ˙ 0)} dt and H1 ( ; ) and H1 ( ; ) are C 0 in ( ; ) ∈ R × I1 . We let fxx (t) = fxx (u(t); v(t)); fxy (t) = fxy (u(t); v(t)), and fyy (t) = fyy (u(t); v(t)). By calculating fx (t; 0) and fy (t; 0), and using the fact that y(t; x; ) is linear in x, we obtain B1 ( ; 0) Z +∞ = −∞
∗
(t){[(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) + (fxy (t)
− fyy (t)U (t)) (t; 0) + h1x (t; 0) − h1y (t; 0)U (t)] u(t) ˙ + fy (t; 0)y (t; u(·); ˙ 0) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt − fxy (t)U (t) (t; 0)u(t) Z +∞ ∗ (t){(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) = −∞
+ (fxy (t) (t; 0) − fyy (t)U (t)) (t; 0) ˙ + fy (t; 0)y (t; u(t); ˙ 0) + h1x (t; 0) − h1y (t; 0)U (t)]u(t) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt: − fxy (t)U (t) (t; 0)u(t)
(43)
In order to calculate Eq. (43), we need some lemmas. Lemma 6. We have y (t; u(·); ˙ 0) ˙ ˙ (t) (t; 0) − gxy (t)U (t)u(t) ˙ = −gy−1 (t){[gxx (t)u(t) (t; 0) − gxy (t)u(t)U (t; 0) ˙ (t) (t; 0) + [gxy (t) − gyy (t)U (t)] (t; 0)u(t) ˙ + gyy (t)U (t)u(t)U ˙ + U˙ (t)[ fx (t) − fy (t)gy−1 (t)gx (t)]u(t)}: ˙ + [h2x (t; 0) − h2y (t; 0)U (t) + U˙ (t)]u(t) (44)
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Proof. Since the proof of Lemma 6 is almost the same as that of Lemma 1, we omit the detail. Di erentiating both sides of the equation y(t; ˙ u(t); ˙ ) ˙ ) = [gy (t; ) + h2y (t; ) + U (t)Fy (t; )]y(t; u(t); + [gx (t; ) − gy (t; )U (t) + h2x (t; ) − h2y (t; )U (t) ˙ + 2 f2 (t); + U˙ (t) + U (t)Fx (t; )]u(t) with respect to and letting → 0 and noting y(t; x; 0) = 0 and v(t) ˙ = −U (t)u(t), ˙ we have 0 = y (t; u(t); ˙ 0) + = −gy−1 (t){[gxx (t)u(t) ˙ ˙ (t) (t; 0) (t; 0) − gxy (t)u(t)U ˙ ˙ (t) (t; 0) − gxy (t)U (t)u(t) (t; 0 + gyy (t)U (t)u(t)U ˙ + h2x (t; 0) − h2y (t; 0)U (t) + U˙ (t)]u(t) ˙ + [gxy (t) − gyy (t)U (t)] (t; 0)u(t) ˙ + U˙ (t)[fx (t) − fy (t)gy−1 (t)gx (t)]u(t)}: Hence, we have y (t; u(·); ˙ 0) ˙ ˙ (t) (t; 0) − gxy (t)U (t)u(t) ˙ = −gy−1 (t){[gxx (t)u(t) (t; 0) − gxy (t)u(t)U (t; 0) ˙ (t) (t; 0) + [gxy (t) − gyy (t)U (t)] (t; 0)u(t) ˙ + gyy (t)U (t)u(t)U ˙ + U˙ (t)[fx (t) − fy (t)gy−1 (t)gx (t)]u(t)}: ˙ + h2x (t; 0) − h2y (t; 0)U (t) + U˙ (t)]u(t) The proof of Lemma 6 is complete. Lemma 7. We have U˙ (t) = gx−1 (t)[gxx (t)u(t) ˙ − gxy (t)u(t)U ˙ (t) − gxy (t)U (t)u(t) ˙ + gyy (t)U (t)u(t)U ˙ (t)]: (45) The proof of lemma 7 is a direct computation and so we omit the proof. Lemma 8. We have Z +∞ ∗ (t){[(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) −∞
+ (fxy (t) − fyy (t)U (t)) (t; 0) + (h1x (t; 0) − h1y (t; 0)U (t))]u(t) ˙ + h1t (t; 0) + fy (t)˙ (t; 0) − fy (t)U˙ (t) (t; 0) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt = 0: − fxy (t)U (t) (t; 0)u(t)
(46)
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419
Proof. Di erentiating both sides of the equation ˙ ) = f((t; ) + u(t); (t; ) + v(t) − U (t)(t; )) − f(u(t); v(t)) (t; + h1 (t + (); (t; ) + u(t); (t; ) + v(t) − U (t)(t; ); ) with respect to and letting = 0 and noting (t; 0) = 0; (t; 0) = 0, we have ˙ (t; 0) = [fx (t) − fy (t)gy−1 (t)gx (t)] (t; 0) + fy (t) (t; 0) + h1 (t + 0 ; u(t); v(t); 0): Di erentiating the above equation with respect to t and noting v(t) ˙ = −U (t)u(t), ˙ we have (t; 0) = [fx (t) − fy (t)gy−1 (t)gx (t)]˙ (t; 0) ˙ + [ fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t)]u(t) ˙ − fy (t)U˙ (t) (t; 0) + [fxy (t) − fyy (t)U (t)] (t; 0)u(t) ˙ + fy (t)˙ (t; 0) + h1t (t + 0 ; u(t); v(t); 0) + [h1x (t) − h1y (t)U (t)]u(t) ˙ − fxy (t) (t; 0)U (t)u(t): ˙ − fxy (t)U (t) (t; 0)u(t) From Lemma 4 we see that (t; 0) is bounded on R, hence ˙ (t; 0) is also bounded on R. It follows from Lemma 2 that Z +∞ ∗ (t){[(fxx (t) (t; 0) + fyy (t)U (t) (t; 0)U (t) −∞
+ (fxy (t) − fyy (t)U (t)) (t; 0) + (h1x (t; 0) − h1y (t; 0)U (t))]u(t) ˙ + h1t (t; 0) + fy (t)˙ (t; 0) − fy (t)U˙ (t) (t; 0) ˙ − fxy (t) (t; 0)U (t)u(t)} ˙ dt = 0: − fxy (t)U (t) (t; 0)u(t) Lemma 8 is proved. Now, we calculate B1 ( ; 0). From Eqs. (43), (45) and (46) we have Z +∞ ∗ (t){fy (t)y (t; u(·); ˙ 0) + fy (t)U˙ (t) (t; 0) B1 ( ; 0) = −∞
− fy (t)˙ (t; 0) − h1t (t + 0 ; u(t); v(t); 0)} dt: Substituting Eqs. (44) and (46) into Eq. (47), we have Z +∞ ∗ (t){−fy (t)gy−1 (t)h2t (t + 0 ; u(t); v(t); 0) B1 ( ; 0) = − −∞
+ h1t (t + 0 ; u(t); v(t); 0)} dt = − M 0 ( 0 ): Hence H1 (0; 0) = 0;
H1 (0; 0) = −M 0 ( 0 ) 6= 0:
(47)
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It follows from the implicit function theorem that there exists a continuous function = (); (0) = 0, such that H1 ( (); ) = 0 for || suciently small. So for suciently small Eq. (39) has bounded solutions, and hence Eq. (37) has bounded solutions. From the previous discussion we see that for 6= 0 suciently small, the variational Eq. (36) admits an exponential dichotomy on R. The proof of Theorem 1 is complete. Remark. Battelli and Lazzari [2] stated that under conditions (a)– (c) the singular part of the system makes no contribution to the existence of transversal homoclinic orbits in the case where h2 (t; ; 0; 0) = 0. From the Melnikov function M ( ) obtained in the paper we see; under general conditions H1–H4, even though h2 (t; x; y; 0) ≡ 0; we cannot neglect the contribution of the singular part of the system for the existence of transversal homoclinic orbits. Now, we give an example to illustrate out theory. We consider the four-dimensional di erential equation x˙1 = y1 − y2 ; ˙ + r(t)); x˙2 = x1 − 2x13 + ( 75 r(t) y˙ 1 = y1 + 21 y22 − 12 x2 − 18 x22 + r(t);
(48)
y2 = −y2 − 12 x2 ; where r(t) = sech t. From Gruendler [12], we easily calculate Z +∞ Z +∞ 2 r˙2 (t) dt = ; r(t) ˙ r2 (t) dt = 0; 3 −∞ −∞ Z +∞ Z +∞ 14 2 r (t) dt = ; r(t)r(t) ˙ dt = 0: 15 −∞ −∞
(49)
Moreover, r(t) = r(t) − 2r 3 (t):
(50)
The degenerate equation corresponding to the singularly perturbed Eq. (48) is x˙1 = y1 − y2 ; x˙2 = x1 − 2x13 ; 1 0 = y1 + 12 y22 − x2 − 18 x22 ; 2 0 = −y2 − 12 x2 ; which has a bounded solution (using the notations of Section 2) ! ! 1 r(t) ˙ 2 r(t) : u= ; v= r(t) ˙ ˙ − 12 r(t)
(51)
(52)
Z. Weiyao, L. Jiaowan / Nonlinear Analysis 36 (1999) 401 – 422
connecting (0; 0)∗ to itself. Moreover, ! ! 1 −1 −r(t) ; (t) = ; fy (t) = 0 0 r(t) ˙ ! ! ˙ ˙ 1 − 12 r(t) 1 − 12 r(t) −1 ; gy (t) = : gy (t) = 0 −1 0 −1
421
(53)
(54)
The Melnikov function for the singularly perturbed Eq. (48) is Z +∞ 7 1 M ( ) = ˙ r2 (t) + r(t) ˙ + ) dt: (55) ˙ r(t + ) + r(t −r(t) r(t + ) + r(t) 4 5 −∞ It is easy to verify M (0) = 0;
M 0 (0) = 23 6= 0:
It follows from Theorem 1 that for 6= 0 suciently small the singularly perturbed system (48) has a transversal homoclinic orbit x(t; ) = (x1 (t; ); x2 (t; )∗ ; y(t; ) = (y1 (t; ); y2 (t; ))∗ . Remark. Since gy (t) does not satisfy condition (c) of Feckan [8] and Battelli and Lazzari [2], the results in Feckan [8] and Battelli and Lazzari [2] cannot be applied to Eq. (48), so this paper generalizes their results. Acknowledgements I would like to thank Professor Ken Palmer and the referees for carefully reading the paper and also for their invaluable suggestions on the paper. References [1] F. Battelli, C. Lazzari, Dichotomies and stability in the singularly perturbed systems of O.D.E., Nonlinear Anal. 11 (1987) 259 – 273. [2] F. Battelli, C. Lazzari, Bounded solutions to the singularly perturbed systems of ODE, J. Di erential Equations 100 (1992) 49 – 81. [3] F. Battelli, K.J. Palmer, Chaos in the Dung equations, J. Di erential Equations 102 (1993) 276 – 301. [4] K.W. Chang, Almost periodic solutions of singularly perturbed systems of di erential equations, J. Di erential Equations 4 (1968) 300 – 307. [5] K.W. Chang, W.A. Coppel, Singular perturbations of initial value problems over a nite interval, Arch. Rational Mech. Anal. 32 (1969) 269 – 280. [6] W.A. Coppel, Stability and Asymptotic Behavior of Di erential Equations, Heath, Boston, 1965. [7] W.A. Coppel, Dichotomies in Stability Theory, Lecture Notes in Mathematics, vol. 629, Springer, Berlin, 1978. [8] M. Feckan, Melnikov functions for singularly perturbed di erential equations, Nonlinear Anal. 19 (4) (1992) 393 – 401. [9] N. Fenichel, Geometric singular perturbation theory, J. Di erential Equations 17 (1979) 308 – 328.
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