- Email: [email protected]
Boundedness of solutions of nonlinear p-Laplacian
Applied Mathematics and Computation 158 (2004) 397–417 www.elsevier.com/locate/amc
Boundedness of solutions of nonlinear p-Laplacian Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing 100084, PeopleÕs Republic of China
Abstract In this paper, we prove the boundedness of all the solutions for the p-Laplacian ðup ðx0 ÞÞ0 þ gðxÞ ¼ eðtÞ, where up ðuÞ ¼ jujp2 u, p P 2, g satisfies some subquasilinear conditions and eðtÞ 2 C 6 ðR=ZÞ. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Boundedness; p-Laplacian; MoserÕs theorem
1. Introduction In this paper, the boundedness of all the solutions of the following p-Laplacian-like nonlinear differential equation 0
ðup ðx0 ÞÞ þ gðxÞ ¼ eðtÞ
is discussed, where up ðuÞ ¼ juj ditions and eðtÞ 2 C 6 ðR=ZÞ. For p ¼ 2, (1.1) reduces to
0
¼
p2
d dt
ð1:1Þ
u; p P 2, g satisfies some subquasilinear con-
x00 þ gðxÞ ¼ eðtÞ:
E-mail address: [email protected] (X. Yang). 0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.08.093
ð1:2Þ
398
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
In the early 60Õs, Littlewood [1] proposed to study the boundedness of all the solutions of (1.2) in the following two cases: (i) Superlinear case: gðxÞ=x ! þ1 as jxj ! 1; (ii) Sublinear case: signðxÞgðxÞ ! þ1 and gðxÞ=x ! 0 as jxj ! 1. The first result in the superlinear case is due to Morris [2], who proved that all solutions of x00 þ 2x3 ¼ eðtÞ are bounded, where eðtÞ 2 C 0 ðR=ZÞ. Later, several authors extended MorrisÕs result to more general cases. See, for example, [3–6] and the references therein. Li [7] and the author [8] obtained boundedness result for semilinear cases. For sublinear case, however, only few results are obtained so far. Recently, the boundedness of solutions for the following simple equation a1
x00 þ jxj
x ¼ eðtÞ
ð1:3Þ
has been studied in [9,10], where 0 < a < 1 and eðtÞ 2 C 1 ðR=ZÞ. They proved that every solution is bounded. Liu [11] studied the general form Eq. (1.2) in sublinear case and obtained boundedness result under some reasonable assumptions. For the more general problem (1.1), where gðxÞ signðxÞ ! 1 and gðxÞ= up ðxÞ ! 0 as jxj ! 1, no boundedness results are available so far. Inspired by the work of [4,11], we discuss the boundedness of all the solutions of (1.1). Throughout this paper, we denote by c < 1 and C > 1, respectively, two universal positive constants without regarding their values. The main result of this paper is Theorem 1. Assume that eðtÞ 2 C 6 ðR=ZÞ, gðxÞ 2 C 6 ðRÞ satisfies that for all x 6¼ 0, (i) xgðxÞ > 0 and there exist two positive constants a1 and a2 such that 1 GðxÞg0 ðxÞ 1 < a1 6 6 a2 ; 2q g2 ðxÞ q
where q ¼
i(ii) k dk x 6 CGðxÞ; GðxÞ dxk
0 6 k 6 7;
p > 1; p1
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
399
(iii) cGðxÞ 6 GðxÞ 6 CGðxÞ: Rx where GðxÞ ¼ 0 gðsÞ ds. Then every solution of (1.1) is bounded, that is, if x ¼ xðtÞ is a solution of (1.1), then it is defined in R and supðjxðtÞj þ jx0 ðtÞjÞ < 1: t2R
Remark 1. In the proof of Theorem 1, one can easily see that the assumptions on gðxÞ can be weakened to require that they hold for jxj P d for some constant d > 0. Remark 2. From the assumptions on g, it is not difficult to prove the following facts: (I) 0 6 g0 ðxÞ 6 C, g0 ðxÞ ! 0 as jxj ! 1, (II) signðxÞ gðxÞ ! þ1, GðxÞ=jxjp ! 0, gðxÞ=jxjp2 x ! 0 as jxj ! 1, (III) xgðxÞ P GðxÞ, x2 g0 ðxÞ P cGðxÞ, x 6¼ 0. From Remarks 1 and 2 and p P 2, one can obtain the global existence and uniqueness of solutions of (1.1) with respect to initial value problem by applying the results of [12]. Therefore, the global existence and uniqueness problem in the subquasilinear case is guaranteed. We do not discuss them in the rest of this paper. Example 1. Every solution of 0
ðup ðx0 ÞÞ þ Aup ðxÞ ¼ eðtÞ is bounded, where A 6¼ 0, eðtÞ 2 C 6 ðR=ZÞ and 1 < p < p. In this cases, gðxÞ ¼ Aup ðxÞ, and 0 < GðxÞg0 ðxÞ=g2 ðxÞ ¼ 1 1p < 1 1p ¼ 1q. The proof of boundedness of all solutions for (1.1) is based on MoserÕs small twist theorem. By means of transformation, (1.1) is, outside of a large disc 2 Br ¼ fðx; x0 Þ 2 R2 , x2 þ ðx0 Þ 6 rg in the ðx; x0 Þ-plane, transformed into perturbations of integrable Hamiltonian system. The Poincare map of the transformed system is close to a so-called twist map in R2 n Br . Then MoserÕs twist theorem guarantees the existence of arbitrarily large invariant curves diffeomorphic to circles and surrounding the origin in the ðx; x0 Þ-plane. Therefore, every such invariant curves is the base of a time-periodic and flow-invariant cylinder in the extended phase space ðx; x0 ; tÞ 2 R2 R, which confines the solutions in the interior and leads to a bound of these solutions.
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X. Yang / Appl. Math. Comput. 158 (2004) 397–417
2. Action-angle variables If we introduce a new variable y ¼ uq ðx0 Þ, where q ¼ p=ðp 1Þ is the conjugate exponent of p : p1 þ q1 ¼ 1, then (1.1) is equivalent to the planar Hamiltonian system x0 ¼
oH ðx; y; tÞ; oy
y0 ¼
oH ðx; y; tÞ; ox
ð2:1Þ
where q
H ðx; y; tÞ ¼ jyj =q þ GðxÞ xeðtÞ: Consider an autonomous system x0 ¼ uq ðyÞ;
y 0 ¼ gðxÞ;
ð2:2Þ
which is an integrable Hamiltonian system with Hamiltonian H0 ðx; yÞ ¼ jyjq =q þ GðxÞ: The closed curves Ch : H0 ðx; yÞ ¼ h > 0 are the integral curves of (2.2). Now we further introduce the action and angle variable ðq; hÞ. Let q denote the area surrounded by the closed curve Ch . Then I y dx :¼ J ðhÞ: q¼ H0 ðx;yÞ¼h
Denote by Cþ1 and G1 the right and left inverse of G, respectively. Let ðx ; 0Þ and ðx ; 0Þ be the intersection points of Ch with the x-axis, that is, þ
1 þ x ¼ G1 ðhÞ < 0 < Gþ ðhÞ ¼ x :
It is easy to see that from (iii) of Theorem, we have 16
1 maxfG1 þ ðhÞ; G ðhÞg 6 C: 1 1 minfGþ ðhÞ; G ðhÞg
It is easy to see that Z xþ 1 1 J ðhÞ ¼ 2qq ðh GðxÞÞq dx:
ð2:3Þ
ð2:4Þ
x
We define now the generating function Sðx; qÞ as the area Z y dx; Sðx; qÞ ¼
ð2:5Þ
C
where C is the part of the level curve Ch : H0 ðx; yÞ ¼ h connecting the y-axis with the point H ðx; yÞ, oriented clockwise. This defines S up to an integer multiple of q ¼ Ch y dx, the area of the domain surrounded by the closed curve Ch .
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401
We define the map ðh; qÞ ! ðx; yÞ by y¼
oS ðx; qÞ; ox
h¼
oS ðx; qÞ; oq
ð2:6Þ
which is symplectic since dx ^ dy ¼ dx ^ ðSxx dx þ Sxq dqÞ ¼ Sxq dx ^ dq; dh ^ dq ¼ ðSqx dx þ Sqq dqÞ ^ dq ¼ Sqx dx ^ dq: System (2.2) in the new variables ðq; hÞ is also a Hamiltonian system system with the new Hamiltonian function hðq; hÞ ¼ J 1 ðqÞ: Under this symplectic map, (2.1) is transformed into q0 ¼
oH ðq; h; tÞ; oh
h0 ¼
oH ðq; h; tÞ; oq
ð2:7Þ
where H ðq; h; tÞ ¼ J 1 ðqÞ xðq; hÞeðtÞ :¼ J 1 ðqÞ þ H1 ðq; h; tÞ with ðx; yÞ ¼ ðxðq; hÞ; yðq; hÞÞ defined implicitly by (2.6). The following equalities give the expressions of J 0 ðhÞ and J 00 ðhÞ for h > 0. Lemma 1. For h > 0, we have 1 Z þ 1 2qq x 1 J 0 ðhÞ ¼ þ w0 ðxÞ ðh GðxÞÞq dx; q h x Z xþ 2 1 1 J 00 ðhÞ ¼ 1 w0 ðxÞ ðh GðxÞÞ p dx; p qp h x
ð2:8Þ
ð2:9Þ
where wðxÞ ¼ GðxÞ=gðxÞ for x 6¼ 0. Proof. Let þ
L ðhÞ ¼
Z
xþ
1 q
ðh GðxÞÞ dx;
L ðhÞ ¼
Z
0
0
Then 1
J ðhÞ ¼ 2qq ½Lþ ðhÞ þ L ðhÞ; 1
ðiÞ
ðiÞ
J ðiÞ ðhÞ ¼ 2qq ½ðLþ ðhÞÞ þ ðL ðhÞÞ ;
1
ðh GðxÞÞq dx: x
i ¼ 1; 2:
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X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Let GðxÞ ¼ sh, s 2 ½0; 1. Then gðxÞdx ¼ h ds and gðxÞdx=dh ¼ GðxÞ=h. Moreover 1
Lþ ðhÞ ¼ h1þq
Z
1 0
1
ð1 sÞq ds ; gðxÞ
1 Z Z 1 0 1 ðg ðxÞÞGðxÞ 1 1q 1 ð1 sÞq ds 1þ1q ðL ðhÞÞ ¼ 1 þ þh ds ð1 sÞq h q gðxÞ g3 ðxÞh 0 0 Z þ 1 þ 1q Z xþ 1 1 1 x g0 ðxÞGðxÞ q ¼ ðh GðxÞÞq dx ðh GðxÞÞ dx 2 h 0 g ðxÞ h 0 Z xþ 1 1 1 ¼ þ w0 ðxÞ ðh GðxÞÞq dx: h 0 q
þ
0
Similarly, we have ðL ðhÞÞ0 ¼
1 h
Z
0
1 1 þ w0 ðxÞ ðh GðxÞÞq dx: q
1 1 0 þ w ðxÞ ðh GðxÞÞq dx: q
x
Hence 1
2qq J ðhÞ ¼ h 0
Z
xþ x
Differentiating ðLþ ðhÞÞ0 , we obtain 1 ðL ðhÞÞ ¼ 2 h þ
00
Z
xþ
0
1 1 þ q2 q
2 2 0 00 0 1 w ðxÞ þ wðxÞw ðxÞ þ ðw ðxÞÞ q
1
ðh GðxÞÞq dx: Since Z
xþ
0
Z 0
1
ðh GðxÞÞq dx ¼
Z
xþ 0
xþ
dx ðh GðxÞÞ
1 p
Z
xþ
GðxÞdx 1
;
ðh GðxÞÞp
0
þ Z 1 1 x w0 ðxÞðh GðxÞÞq dx ¼ wðxÞðh GðxÞÞq þ 0
xþ
wðxÞgðxÞdx 1
qðh GðxÞÞp Z þ 1 x GðxÞdx ðwðxÞ ¼ GðxÞ=gðxÞÞ ¼ q 0 ðh GðxÞÞ1p 0
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
403
and Z
xþ
1
2
½wðxÞw00 ðxÞ þ ðw0 ðxÞÞ ðh GðxÞÞq dx 0
Z þ 1 x w0 ðxÞGðxÞdx ¼ ½wðxÞw ðxÞ ðh GðxÞÞ dx ¼ q 0 ðh GðxÞÞ1p 0 Z þ Z þ 1 1 x 0 h x w0 ðxÞ dx q ¼ w ðxÞðh GðxÞÞ dx þ q 0 q 0 ðh GðxÞÞ1p Z þ Z xþ 1 GðxÞ h x w0 ðxÞ dx ¼ 2 1 þ q 0 ðh GðxÞÞp q 0 ðh GðxÞÞ1p Z
xþ
1 q
0
0
which implies that 1 ðLþ ðhÞÞ00 ¼ qh
Z
xþ
w0 ðxÞ 1p dx 1
:
ðh GðxÞÞp
0
Similarly, we have 1 00 ðL ðhÞÞ ¼ qh
Z
0
x
w0 ðxÞ 1p dx 1
:
ðh GðxÞÞp
Therefore, J 00 ðhÞ ¼
2 1
qp h
Z
xþ x
w0 ðxÞ 1p dx 1
:
ðh GðxÞÞp
Lemma 2. If there exist two positive constants a1 and a2 such that for all x 6¼ 0, we have a1 6 g0 ðxÞGðxÞ=g2 ðxÞ 6
1 a2 : q
ð2:10Þ
Then the following estimates hold: 1
ch1þa2 6 J ðhÞ 6 Ch1þqa1 ; ch1 J ðhÞ 6 J 0 ðhÞ 6 Ch1 J ðhÞ; 1 0
00
1 0
ch J ðhÞ 6 J ðhÞ 6 Ch J ðhÞ: Moreover, J 0 ðhÞ ! þ1 as h ! þ1.
ð2:11Þ
404
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Proof. Since w0 ðxÞ ¼ 1 g0 ðxÞGðxÞ=g2 ðxÞ and from (2.8), (2.9), we have 1 ð1 þ a2 Þh1 J ðhÞ 6 J 0 ðhÞ 6 1 þ a1 h1 J ðhÞ q which yields that 1
ch1þa2 6 J ðhÞ 6 Ch1þqa1 and J 0 ðhÞ P ch1 J ðhÞ P cha2 ! þ1 as h ! þ1. Similarly, we have ch1 J 0 ðhÞ 6 J 00 ðhÞ 6 Ch1 J 0 ðhÞ:
Lemma 3. For 0 6 k 6 7, we have k k d o 0 k k 0 dhk J ðhÞ 6 Ch J ðhÞ; oqk J ðhÞ 6 Cq J ðhÞ
ð2:12Þ
and k o x k oqk ðq; hÞ 6 Cq jxj;
k o y k oqk ðq; hÞ 6 Cq jyj:
ð2:13Þ
The proof of Lemma 3 is similar to the proof of Lemma A 4.1 in [4], so we omit it. From (2.11) and (2.12), it is not difficult to verify that 1 1
1
Cq1þqa1 6 J 1 ðqÞ 6 Cq1þa2
ð2:14Þ
and cq1 J 1 ðqÞ 6
d 1 J ðqÞ 6 Cq1 J 1 ðqÞ: dq
ð2:15Þ
Lemma 4. Let x; y be defined by (2.5) and (2.6). Then q ox ðq; hÞ ¼ J 0 ðhÞjyjp signðyÞ; oh
oy ðq; hÞ ¼ J 0 ðhÞgðxðq; hÞÞ: oh
Proof. By the definition of S, h, for y > 0 we have Z x 1 1 Sðx; qÞ ¼ qq ðhðqÞ GðsÞÞq ds; 0
oS ðx; qÞ ¼ oq
Z 0
x
1
1
qq1 ðhðqÞ GðsÞÞq1 ds h0 ðqÞ ¼ h:
ð2:16Þ
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
405
Hence 1 o2 S ox 1 ðx; qÞ ¼ qq ðhðqÞ GðxÞÞp h0 ðqÞ ¼ 1; oq oh oh
that is, 1 ox 1 ðq; hÞ ¼ qp ðhðqÞ GðxÞÞp ðh0 ðqÞÞ1 oh 1
1
1
q
¼ J 0 ðhÞqp ðhðqÞ GðxÞÞJ 0 ðhÞqp ðhðpÞ GðxÞÞp ¼ J 0 ðhÞy p : Similarly, for y < 0, we have q ox ðq; hÞ ¼ J 0 ðhÞjyjp ; oh
therefore q ox ðq; hÞ ¼ J 0 ðhÞjyjp signðyÞ: oh
ð2:17Þ
q
From (2.17) and jyj =q þ GðxÞ ¼ h, we obtain jyjq2 y
oy ox þ gðxÞ ¼ 0; oh oh
hence oyðq; hÞ ox q2 ¼ gðxÞ =jyj y ¼ J 0 ðhÞgðxðq; hÞÞ: oh oh
Lemma 5. Let H1 ðq; h; tÞ ¼ xðq; hÞeðtÞ. Then for p 1, 0 6 k þ l 6 7, l ¼ 0; 1 and 0 6 m 6 6, we have ( 1 okþlþm k 1 p Cq ðJ ðqÞÞ ; 0 6 k 6 7; l ¼ 0; ð2:18Þ oqk ohl otm H1 ðq; h; tÞ 6 Cqk ðJ 1 ðqÞÞ1a1 ; 0 6 k 6 6; l ¼ 1: Proof. Let l ¼ 0 and q 1, we have from (2.13) and Remark 2, for 0 6 k þ m 6 7, kþm o ok x dm eðtÞ H ðq; h; tÞ ¼ oqk otm 1 oqk dtm k o x 6 C k ðq; hÞ 6 C k maxfxþ ; jx jg oq 1
k 1 ðqÞÞp : 6 Cqk G1 þ ðhÞ 6 Cq ðJ
406
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
For l ¼ 1, we have from (2.12) and (2.13) k kþ1þm o o ox ðmÞ oqk oh otm H1 ðq; h; tÞ ¼ oqk oh ðq; hÞe ðtÞ k o q ¼ k J 0 ðhÞjyjp signðyÞ eðmÞ ðtÞ oq j X di o q 0 p 6C dqi ðJ ðhÞÞ oqj jyj signðyÞ iþj¼k j o q i 0 p 6 C q jJ ðhÞj j jyj signðyÞ : oq Since j j2 i ji j X oy o q q oy 1 o2 y p pi Cji jyj 2 i opj jyj signðyÞ 6 oq oq oq 16i6j X q i j j 6C jyjp ðq1 jyjÞ 1 ðqi jyjÞ i 16i6j
X
6C
q
jyjp qðj1 þ þiji Þ
16i6j
X
6C
q
1
jyjp qðj1 þ þji Þ 6 Chp qj ;
16i6j
where fCij g are positive constants and j1 þ þ ji ¼ j. We have kþ1þm o 0 k 1p k 1a1 6 Cqk ðJ 1 ðqÞÞ1a1 : oqk oh otm H1 ðq; h; tÞ 6 CJ ðhÞq h 6 Cq h
3. New action-angle variables We consider now the Hamiltonian system (2.7) with Hamiltonian function H ðq; h; tÞ given by H ¼ J 1 ðqÞ þ H1 ðq; h; tÞ. Note that qdh H ðq; h; tÞ dt ¼ ðH dt qðH ; t; hÞdhÞ: This implies that if one can solve q ¼ qðH ; t; hÞ from (2.7) (t and h as parameters) as a function of H ; t and h, then dH oq ¼ ðH ; t; hÞ; dh oH
dt oq ¼ ðH ; t; hÞ: dh dt
ð3:1Þ
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
407
That is, (3.1) is also a Hamiltonian system with Hamiltonian function qðH ; t; hÞ and now the action, angle and time variables are H ; t and h, respectively. This idea came from [4,11]. Since by Lemmas 3 and 5, for q 1 oH ðq; h; tÞ 6¼ 0: oq We can define qðH ; t; hÞ as the inverse function of H ðq; h; tÞ with t; h playing the role of parameters by the implicit function theorem. Now we define q1 ðH ; t; hÞ as qðH ; t; hÞ ¼ J ðH Þ þ q1 ðH ; t; hÞ: Lemma 6. The perturbation term q1 ðH ; t; hÞ possesses the following estimates 1 okþlþm CH kq J ðH Þ; if m ¼ 0; 6 q ðH ; t; hÞ ð3:2Þ oH k otl ohm 1 CH ka1 J ðH Þ; if m ¼ 1; for 0 6 k þ l 6 6 and H 1. Proof. Claim. For H 1 we have 1 1 1 H < J ðH Þ < J 1þ H : J H 6 CJ q p q
ð3:3Þ
Indeed, the first two inequalities hold since J 0 ðH Þ > 0 by (2.11). On the other hand, it follows from (2.11) that J 0 ðhÞ is increasing and from (2.8), (2.9) that 1 0 J ðhÞ 6 1 þ a1 h1 J ðhÞ; ð3:4Þ q this yields that 1 1 H J 1þ H J p q 2 0 qH 0 1 1 ¼ J ðnÞH < J 1þ H 1þ p qþ1 p q 1 qa1 1 6J 1þ J 1þ H H ; p p qþ1 1 1 H; 1 þ H n2 q p therefore we obtain the third inequality
408
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Step 1. m ¼ 0. We divide it into three cases: (I) k ¼ 0, l ¼ 0. From the definition of qðH ; t; hÞ, if we treat H as the independent variable and t; h as parameters, then we have J 1 ðqðH ÞÞ þ H1 ðqðH ÞÞ ¼ H ;
ð3:5Þ
this yields that qðH Þ ¼ J ðH H1 ðqðH ÞÞ:
ð3:6Þ
Expanding q1 ðH Þ ¼ qðH Þ J ðH Þ ¼ J ðH H1 ðqðH ÞÞÞ J ðH Þ in TaylorÕs series, q1 ðH Þ ¼ J 0 ðH ÞH1 þ
Z
H1
sJ 00 ðH H1 þ sÞ ds:
ð3:7Þ
0
From Lemma 5 and note that qðH Þ ! 1 as H ! 1, and that jH1 ðqÞj < 1 CðJ 1 ðqÞÞp < 1p J 1 ðqÞ for q 1 and all t; h; we obtain 1 1 jH1 ðqðH ÞÞj < J 1 ðqðH ÞÞ < H : p p
ð3:8Þ
By (3.4), (3.6) and (3.8) as well as the monotonicity of J in H , we obtain 1 1 J H < qðH Þ < J 1þ H : ð3:9Þ q p This, together with Claim, implies that for H 1 cJ ðH Þ < qðH Þ < CJ ðH Þ 6 J ð2H Þ:
ð3:10Þ
It follows from (2.14) and Lemma 5 that 1
1
1
1
jH1 ðqðH ÞÞj 6 CðJ 1 ðqðH ÞÞÞp 6 CðJ 1 ðCJ ðH ÞÞÞp 6 CðJ 1 ðJ ð2H ÞÞÞp 6 CH p : ð3:11Þ Now the first term in (3.7) satisfies jJ 0 ðH ÞH1 j 6 CJ ðH Þ
1þ1p
1
¼ CJ ðH ÞH q
and the second term is bounded by 2 1 2 00 e sup J ð H Þ 6 CH12 H 2 J ðH Þ 6 CJ ðH ÞH q 6 CJ ðH ÞH q ; H1 H H1 6 H~ 6 H for H 1: This completes the proof of (I).
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
409
(II) k P 1, l ¼ 0. Differentiating (3.7) k times by H , we obtain k X ok q1 ðH Þ oiþ1 J ðH Þ oki H1 ðqðH ÞÞ ok ¼ Cki þ k k iþ1 ki oH oH oH oH i¼0
Z
H1
sJ 00 ðH H1 þ sÞds;
0
where Cki is a positive integer which depends only on k and i. Note that X ok H1 ðqðH ÞÞ os ok 1 oks ¼ C H ðqðH ÞÞ qðH Þ qðH Þ k 1 s oH k oqs oH k1 oH ks k þ þks ¼k 1
and ok oH k ¼
Z
H1
0 k X i;j¼0
sJ 00 ðH H1 þ sÞ ds
Ckij
oi H1 oj H1 okijþ2 J ðH Þ oH kijþ2 oH i oH j
X
oj H1 oi H1 osþ2 J ðH Þ on1 ðIÞ ons ðIÞ oH sþ2 oH n1 oH ns oH j oH i iþjþn1 þns ¼k Z X on1 ðIÞ oni ðIÞ H oiþ2 þ Ckin s J ðH H1 þ sÞ ds; oH n1 oH ni 0 oH iþ2 n1 þ þni¼k þ
Ckijns
where I ¼ J 1 . Therefore, the proof of (II) reduces to the proof of k o 6 CH k qðH Þ; for H 1; 1 6 k 6 b: qðH Þ oH k
ð3:12Þ
Differentiating (3.6) by H yields q0 ðH Þ ¼ J 0 ðH H1 Þ ½1 H10 qðH Þ: We obtain therefore q0 ðH Þ ¼
J 0 ðH H1 Þ : 1 þ J 0 ðH H1 ÞH10
ð3:13Þ
The denominator above expression is close one for H 1, indeed, J 0 ðH H1 ÞH10 ¼ J 0 ðH H1 Þ
dIðJ ðH H1 ÞÞ H10 ðqðH ÞÞ dIðJ ðH H ÞÞ 1 dq dq
¼1
H10 ðqðH ÞÞ dIðqðH ÞÞ dq
!0
as H ! 1:
410
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
For H 1, we get q0 ðH Þ <
2dIðH H1 Þ < CðH H1 Þ1 J ðH H1 Þ < CH 1 qðH Þ dq
which proving the case k ¼ 1 in (3.12). Assuming inductively that (3.12) holds for 1 6 k 6 n 6 5, now we prove it for k ¼ n þ 1. Differentiating (3.13) n times by H yields that n X onþ1 oi J 0 oni Q qðH Þ ¼ C ðH H Þ ; ni 1 oH ni oH nþ1 oH i i¼0 1
where Q ¼ ½1 þ J 0 ðH H1 Þ H10 . The proof of (3.12) for k ¼ n þ 1 reduces to the proof of i o 0 i 0 ð3:14Þ oH i J ðH H1 Þ 6 CH J ðH H1 Þ and
i o i oH i Q 6 CH
ði 6 n; H 1Þ:
ð3:15Þ
Proof of (3.14). Note that X oi 0 oi1 oim ðmþ1Þ J ðH H Þ ¼ C J ðH H Þ ðH H Þ ðH H1 Þ 1 im 1 1 oH i oH i1 oH im i1 þ im ¼i ð3:16Þ and oj oj ðH H Þ ¼ ðIðqðH ÞÞÞ 1 oH j oH j X os oj1 ojs Cjs ðIðqðH ÞÞÞ qðH Þ qðH Þ: ¼ oH s oH j1 oH js j1 þ þJs ¼j ð3:17Þ Using inductive assumptions in (3.17), we obtain for j 6 n j o j oH j ðH H1 Þ 6 CH ðH H1 Þ; which together with (3.16) and Lemma 3 yields that for j 6 n j o 0 j 0 J ðH H Þ 1 6 CH J ðH H1 Þ oH j as desired.
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
411
Proof of (3.15). It is not difficult to see that ij X oi oi j 1j o Q¼ Cij ½1 þ J 0 ðH H1 Þ H10 ðQ1 Þ ij ðQ1 Þ; I i oH oHi1 oH 06j6i
ð3:18Þ where i1 þ þ ij ¼ i. From (3.14) and Lemma 5, it is easy to see that for 1 6 i 6 n i i o o 1 0 0 i 0 0 ðQ Þ ¼ ½1 þ J ðH H Þ H 1 1 6 CH jJ ðH H1 Þ H1 j: oH i oH i Since J 0 ðH H1 Þ H10 ! 0
as H ! 1
we have for 1 6 i 6 n and H 1 j1 þ J 0 ðH H1 Þ H10 j 6 C and i o 0 0 i oH i ½1 þ J ðH H1 Þ H1 6 C H : It follows from (3.18) that for i 6 n and H 1 i 1 o ðQ Þ i oH i 6 C H : This proves (3.15) and (II) is proved. (III) k P 0, l > 1. Similar to the above calculation, the proof of (III) reduces to the proof of kþl o k ð3:19Þ oH k otl qðH Þ 6 C H qðH Þ; for k P 0; l > 1; H 1: But, respect bound. Step
according to (3.13) and Lemma 5, the differentiation of qðkÞ ðH Þ with to the variables t does not increase the order of growth of the upper Therefore (III) is proved. 2. m ¼ 1. Differentiating the equality
qðH ðq; h; t; t; hÞ ¼ q by h we obtain from q1 ðH Þ ¼ qðH Þ J ðH Þ and H ¼ IðqÞ þ H1 ðqÞ that o o o q1 ðH Þ ¼ qðH Þ H1 ðqðH ÞÞ: oh oH oh
ð3:20Þ
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X. Yang / Appl. Math. Comput. 158 (2004) 397–417
It is not difficult to obtain
X okþ1 os os H H ðqðH ÞÞ ¼ C ðqðH ÞÞ 1 ks 1 oH k oh oqs oh k þ þks ¼k 1
ok 1 oks qðH Þ ks qðH Þ: k oH 1 oH
Therefore, by applying the results of (3.12), (3.19), (3.20) and Lemma 5, we obtain for H 1 kþlþ1 o 6 CH ka1 J ðH Þ: q ðH ; t; hÞ oH k otl oh 1
4. Proof of Theorem 1 Lemma 7. There exists a canonical transformation T : ðr; sÞ ! ðH ; tÞ of the form T : H ¼ r þ U ðr; s; hÞ;
t ¼ s þ V ðr; s; hÞ;
ð4:1Þ
where the functions U and V are 1-periodic in h and satisfy U ðr; s; hÞ ; r
V ðr; s; hÞ ! 0
as r ! þ1
uniformly for ðs; hÞ 2 S 1 S 1 such that under this transformation, the system (3.1) is changed into the form dr oR ¼ ðr; s; hÞ; dh os
ds oR ¼ ðr; s; hÞ; dh or
ð4:2Þ
Rðr; s; hÞ ¼ J ðrÞ þ q1 ðr; hÞ; þR1 ðr; s; hÞ
ð4:3Þ
where
with q1 ðr; hÞ ¼
Z
1
q1 ðr; t; hÞ dt:
0
Moreover the new perturbation R1 possesses the estimate: kþl o 6 C rkþ1a1 ; 0 6 k þ l 6 5: R ðr; s; hÞ ork osl 1
ð4:4Þ
Proof. We choose T :H ¼rþ
oF ðr; s; hÞ; ot
s¼tþ
oF ðr; t; hÞ; or
ð4:5Þ
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
413
where F ðr; t; hÞ ¼
Z
t
0
1 ½q1 ðr; t; hÞ q1 ðr; hÞ dt: J 0 ðrÞ
Then the transformed system (3.1) is of the form dr oR ¼ ðr; s; hÞ; dh os
ds oR ¼ ðr; s; hÞ; dh or
where oF oF oF ; t; h þ Rðr; s; hÞ ¼ J r þ þ q1 r þ ot ot oh oF ¼ J ðrÞ þ J 0 ðrÞ þ q1 ðr; t; hÞ þ R1 ðr; s; hÞ ot with 2 Z 1 oF oF oF 00 þ R1 ðr; s; hÞ ¼ ð1 sÞJ r þ s ds oh ot ot 0 Z 1 oq1 oF oF ; t; h ds: rþs þ ot ot 0 oH Then the transformation T satisfies desired conditions. It remains to prove (4.4). It follows from (2.11) and (3.2) that for 0 6 k þ l 6 6 and m ¼ 0; 1, kþlþm 1 o Crkþp ; m ¼ 0; 6 F ðr; t; hÞ ork otl ohm Crkþ1a1 ; m ¼ 1: In particular, 2 o 6 Cra1 6 1 F ðr; t; hÞ or ot 2 if r 1. So one can solve the second equation of (4.5) for t, t ¼ s þ V ðr; s; hÞ; where V satisfies V ðr; s; hÞ ¼
oF ðr; s þ V ; hÞ: or
Let U ðr; s; hÞ ¼
oF ðr; s þ V ; hÞ: ot
ð4:6Þ
414
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Then we are done. Moreover, similar to the proof of [3, Lemma 2], one can verify that kþl kþl o 6 C rkþ1p ; o 6 C rk1q U ðr; s; hÞ V ðr; s; hÞ ð4:7Þ ork osl ork osl for 0 6 k þ l 6 5 and U =r; V ! 0 as r ! þ1. Denote oF ðr; s; þV ; hÞ; oh Z 1 f2 ðr; s; hÞ ¼ ð1 sÞJ 00 ðr þ sU Þ U 2 ds; 0 Z 1 oq1 ðr þ sU ; s þ V ; hÞ U ds: f3 ðr; s; hÞ ¼ 0 oH f1 ðr; s; hÞ ¼
Then one can verify that (The proof is similar to the proof of (3.14)) for 0 6 k þ l 6 5, r 1, kþ1 o kþ1a1 ; ork osl f1 ðr; s; hÞ 6 Cr kþ1 o 6 Crk1 ; f ðr; s; hÞ ork osl 2 kþ1 o 6 Crk1 : f ðr; s; hÞ ork osl 3 1
Note that by (2.11), J ðrÞ 6 Cr1þqa1 , for r 1, we have finally kþl 3 o X okþl kþ1a1 ork osl R1 ðr; s; hÞ 6 ork osl fi ðr; s; hÞ 6 Cr i¼1 for 0 6 k þ l 6 5; r 1:
For r0 > 0, denote by Dr0 by the domain Dr0 ¼ fðr; s; hÞjr P r0 ; ðh; tÞ 2 S 1 S 1 g: The following lemma will be used for the application of MoserÕs twist theorem. Lemma 8 [3]. The Poincare mapping P of (4.2) has the intersection property on Dr0 ; that is, if C is an embedded circle in Dr0 homotopic to a circle r ¼constant in Dr0 , then P ðCÞ \ C 6¼ ;.
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
415
If we define k ¼ J 0 ðrÞ;
s ¼ s;
h ¼ h;
ð4:8Þ
then the transformed system (4.2) is of the form dk ¼ g1 ðk; s; hÞ; dh
ds ¼ k þ g2 ðk; s; hÞ; dh
ð4:9Þ
where oR1 ðr; s; hÞ; os o q oR1 ðr; s; hÞ; g2 ðk; s; hÞ ¼ 1 ðr; hÞ þ or or g1 ðk; s; hÞ ¼ J 00 ðrÞ
with r ¼ rðkÞ implicitly given by (4.8). From (2.11) and (3.2), we have 1
cra2 6 J 0 ðrÞ 6 Crqa1 ; o q1 ðr; hÞ 6 Cra1 : or Therefore q
1
ð4:10Þ
ck1qa1 6 rðkÞ 6 Cka2 ;
and k 1 if and only if r 1. Moreover, from (2.11) and (2.12), we obtain k o 6 Ckk rðkÞ 6 CrðkÞ; 0 6 k 6 4: rðkÞ ð4:11Þ okk From (2.12), (3.2), (4.4) and (4.11), we have for 0 6 k þ l 6 4, kþl o 6 Ckk r2 J ðrÞr1a1 6 Crð1þa1 Þ J ðrÞ 6 Cr1q2a1 6 Cke1 ; g ðk; s; hÞ okk osl 1 and kþl o k a1 e2 2 a1 okk osl g2 ðk; s; hÞ 6 Ck ðr þ r J ðrÞÞ 6 Cr 6 Ck ; where e1 ¼
2qa1 1 > 0; 1 qa1
e2 ¼
a1 q > 0: 1 qa1
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X. Yang / Appl. Math. Comput. 158 (2004) 397–417
We are now in a position to prove Theorem 1. Proof of Theorem 1. Since the functions g1 and g2 are sufficiently small if k 1, one can verify that the solutions of (4.9) do exist for 0 6 h 6 1 if the initial value kð0Þ ¼ k is sufficiently large. Integrating (4.9) from h ¼ 0 to h ¼ 1, we obtain that the Poincare mapping P of (4.9) is of the form P : s1 ¼ s0 þ k0 þ D1 ðk0 ; s0 Þ;
k1 ¼ k0 þ D2 ðk0 ; s0 Þ;
where D1 and D2 possess the same estimates as g1 and g2 , that is, for 0 6 k þ l 6 4, okþl k l Di 6 Cke0 ; ok0 os0 where e0 ¼ minðe1 ; e2 Þ. Since T is a diffeomorphism, P possesses the intersection property on Dk0 . Therefore P satisfies all the assumptions of MoserÕs twist theorem [5,13]. From this theorem, for any x 1 satisfying m 5 x P c0 jnj 2 ; m; n 2 N ; ðm; nÞ ¼ 1 n there exists an invariant curve C of P and on which P is the form s1 ¼ s0 þ x: Therefore there exist invariant curves of the Poincare mapping of the system (2.1) which surrounding the origin ðx; yÞ ¼ ð0; 0Þ and are arbitrarily far from the origin. So every solution of (1.1) is bounded. h
References [1] J. Littlewood, Some Problems in Real and Complex Analysis, Heath, Lexington, MA, 1968. [2] G. Morris, A case of boundedness of LittlewoodÕs problem on oscillatory differential equations, Bull. Austral. Math. Soc. 14 (1976) 71–93. [3] R. Dieckerhoff, E. Zehnder, Boundedness of solutions via the twist theorem, Ann. Scuola. Norm. Sup. Pisa Cl. Sci. 14 (1987) 79–95. [4] M. Levi, Quasi-periodic motions in superquadratic time-periodic potentials, Commun. Math. Phys. 143 (1991) 43–85. [5] H. R€ ussmann, On the existence of invariant curves of twist mapping of an annulus, in: Lecture Notes in Mathematics, vol. 1007, Springer-Verlag, Berlin, Heidelberg, New York, 1981, pp. 677–718. [6] J. You, Boundedness for solutions of superlinear DuffingÕs equatiions via twist curves theorems, Sci. Sin. 35 (1992) 399–412. [7] X. Li, Boundedness of solutions for DuffingÕs equations with semilinear potentials, J. Differ. Equat. 176 (2001) 248–268. [8] X. Yang, Boundedness in nonlinear asymmetric oscillations, J. Differ. Equat. 183 (2002) 108– 131.
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[9] T. K€ upper, J. You, Existence of quasiperiodic solutions and LittlewoodÕs boundedness problem of Duffing equations with subquadratic potentials, Nonlinear Anal. 35 (1999) 549– 559. [10] B. Liu, F. Zanolin, Boundedness of solutions for second order quasilinear ODEs, preprint. [11] B. Liu, On LittlewoodÕs boundedness problem for sublinear Duffing equations, Trans. Amer. Math. Soc. 353 (2001) 1567–1585. [12] M. del Pino, R. Manasevich, Multiple solutions for the p-Laplacian under global nonresonance, Proc. Amer. Math. Soc. 112 (1991) 131–138. [13] J. Moser, On invariant curves of area-preserving mappings of annulus, Nachr. Akad. Wiss. Gotingen Math-Phys. Math. Kl. II (1962) 1–20.
Boundedness of solutions of nonlinear p-Laplacian Xiaojing Yang Department of Mathematics, Tsinghua University, Beijing 100084, PeopleÕs Republic of China
Abstract In this paper, we prove the boundedness of all the solutions for the p-Laplacian ðup ðx0 ÞÞ0 þ gðxÞ ¼ eðtÞ, where up ðuÞ ¼ jujp2 u, p P 2, g satisfies some subquasilinear conditions and eðtÞ 2 C 6 ðR=ZÞ. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Boundedness; p-Laplacian; MoserÕs theorem
1. Introduction In this paper, the boundedness of all the solutions of the following p-Laplacian-like nonlinear differential equation 0
ðup ðx0 ÞÞ þ gðxÞ ¼ eðtÞ
is discussed, where up ðuÞ ¼ juj ditions and eðtÞ 2 C 6 ðR=ZÞ. For p ¼ 2, (1.1) reduces to
0
¼
p2
d dt
ð1:1Þ
u; p P 2, g satisfies some subquasilinear con-
x00 þ gðxÞ ¼ eðtÞ:
E-mail address: [email protected] (X. Yang). 0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.08.093
ð1:2Þ
398
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
In the early 60Õs, Littlewood [1] proposed to study the boundedness of all the solutions of (1.2) in the following two cases: (i) Superlinear case: gðxÞ=x ! þ1 as jxj ! 1; (ii) Sublinear case: signðxÞgðxÞ ! þ1 and gðxÞ=x ! 0 as jxj ! 1. The first result in the superlinear case is due to Morris [2], who proved that all solutions of x00 þ 2x3 ¼ eðtÞ are bounded, where eðtÞ 2 C 0 ðR=ZÞ. Later, several authors extended MorrisÕs result to more general cases. See, for example, [3–6] and the references therein. Li [7] and the author [8] obtained boundedness result for semilinear cases. For sublinear case, however, only few results are obtained so far. Recently, the boundedness of solutions for the following simple equation a1
x00 þ jxj
x ¼ eðtÞ
ð1:3Þ
has been studied in [9,10], where 0 < a < 1 and eðtÞ 2 C 1 ðR=ZÞ. They proved that every solution is bounded. Liu [11] studied the general form Eq. (1.2) in sublinear case and obtained boundedness result under some reasonable assumptions. For the more general problem (1.1), where gðxÞ signðxÞ ! 1 and gðxÞ= up ðxÞ ! 0 as jxj ! 1, no boundedness results are available so far. Inspired by the work of [4,11], we discuss the boundedness of all the solutions of (1.1). Throughout this paper, we denote by c < 1 and C > 1, respectively, two universal positive constants without regarding their values. The main result of this paper is Theorem 1. Assume that eðtÞ 2 C 6 ðR=ZÞ, gðxÞ 2 C 6 ðRÞ satisfies that for all x 6¼ 0, (i) xgðxÞ > 0 and there exist two positive constants a1 and a2 such that 1 GðxÞg0 ðxÞ 1 < a1 6 6 a2 ; 2q g2 ðxÞ q
where q ¼
i(ii) k dk x 6 CGðxÞ; GðxÞ dxk
0 6 k 6 7;
p > 1; p1
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
399
(iii) cGðxÞ 6 GðxÞ 6 CGðxÞ: Rx where GðxÞ ¼ 0 gðsÞ ds. Then every solution of (1.1) is bounded, that is, if x ¼ xðtÞ is a solution of (1.1), then it is defined in R and supðjxðtÞj þ jx0 ðtÞjÞ < 1: t2R
Remark 1. In the proof of Theorem 1, one can easily see that the assumptions on gðxÞ can be weakened to require that they hold for jxj P d for some constant d > 0. Remark 2. From the assumptions on g, it is not difficult to prove the following facts: (I) 0 6 g0 ðxÞ 6 C, g0 ðxÞ ! 0 as jxj ! 1, (II) signðxÞ gðxÞ ! þ1, GðxÞ=jxjp ! 0, gðxÞ=jxjp2 x ! 0 as jxj ! 1, (III) xgðxÞ P GðxÞ, x2 g0 ðxÞ P cGðxÞ, x 6¼ 0. From Remarks 1 and 2 and p P 2, one can obtain the global existence and uniqueness of solutions of (1.1) with respect to initial value problem by applying the results of [12]. Therefore, the global existence and uniqueness problem in the subquasilinear case is guaranteed. We do not discuss them in the rest of this paper. Example 1. Every solution of 0
ðup ðx0 ÞÞ þ Aup ðxÞ ¼ eðtÞ is bounded, where A 6¼ 0, eðtÞ 2 C 6 ðR=ZÞ and 1 < p < p. In this cases, gðxÞ ¼ Aup ðxÞ, and 0 < GðxÞg0 ðxÞ=g2 ðxÞ ¼ 1 1p < 1 1p ¼ 1q. The proof of boundedness of all solutions for (1.1) is based on MoserÕs small twist theorem. By means of transformation, (1.1) is, outside of a large disc 2 Br ¼ fðx; x0 Þ 2 R2 , x2 þ ðx0 Þ 6 rg in the ðx; x0 Þ-plane, transformed into perturbations of integrable Hamiltonian system. The Poincare map of the transformed system is close to a so-called twist map in R2 n Br . Then MoserÕs twist theorem guarantees the existence of arbitrarily large invariant curves diffeomorphic to circles and surrounding the origin in the ðx; x0 Þ-plane. Therefore, every such invariant curves is the base of a time-periodic and flow-invariant cylinder in the extended phase space ðx; x0 ; tÞ 2 R2 R, which confines the solutions in the interior and leads to a bound of these solutions.
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X. Yang / Appl. Math. Comput. 158 (2004) 397–417
2. Action-angle variables If we introduce a new variable y ¼ uq ðx0 Þ, where q ¼ p=ðp 1Þ is the conjugate exponent of p : p1 þ q1 ¼ 1, then (1.1) is equivalent to the planar Hamiltonian system x0 ¼
oH ðx; y; tÞ; oy
y0 ¼
oH ðx; y; tÞ; ox
ð2:1Þ
where q
H ðx; y; tÞ ¼ jyj =q þ GðxÞ xeðtÞ: Consider an autonomous system x0 ¼ uq ðyÞ;
y 0 ¼ gðxÞ;
ð2:2Þ
which is an integrable Hamiltonian system with Hamiltonian H0 ðx; yÞ ¼ jyjq =q þ GðxÞ: The closed curves Ch : H0 ðx; yÞ ¼ h > 0 are the integral curves of (2.2). Now we further introduce the action and angle variable ðq; hÞ. Let q denote the area surrounded by the closed curve Ch . Then I y dx :¼ J ðhÞ: q¼ H0 ðx;yÞ¼h
Denote by Cþ1 and G1 the right and left inverse of G, respectively. Let ðx ; 0Þ and ðx ; 0Þ be the intersection points of Ch with the x-axis, that is, þ
1 þ x ¼ G1 ðhÞ < 0 < Gþ ðhÞ ¼ x :
It is easy to see that from (iii) of Theorem, we have 16
1 maxfG1 þ ðhÞ; G ðhÞg 6 C: 1 1 minfGþ ðhÞ; G ðhÞg
It is easy to see that Z xþ 1 1 J ðhÞ ¼ 2qq ðh GðxÞÞq dx:
ð2:3Þ
ð2:4Þ
x
We define now the generating function Sðx; qÞ as the area Z y dx; Sðx; qÞ ¼
ð2:5Þ
C
where C is the part of the level curve Ch : H0 ðx; yÞ ¼ h connecting the y-axis with the point H ðx; yÞ, oriented clockwise. This defines S up to an integer multiple of q ¼ Ch y dx, the area of the domain surrounded by the closed curve Ch .
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
401
We define the map ðh; qÞ ! ðx; yÞ by y¼
oS ðx; qÞ; ox
h¼
oS ðx; qÞ; oq
ð2:6Þ
which is symplectic since dx ^ dy ¼ dx ^ ðSxx dx þ Sxq dqÞ ¼ Sxq dx ^ dq; dh ^ dq ¼ ðSqx dx þ Sqq dqÞ ^ dq ¼ Sqx dx ^ dq: System (2.2) in the new variables ðq; hÞ is also a Hamiltonian system system with the new Hamiltonian function hðq; hÞ ¼ J 1 ðqÞ: Under this symplectic map, (2.1) is transformed into q0 ¼
oH ðq; h; tÞ; oh
h0 ¼
oH ðq; h; tÞ; oq
ð2:7Þ
where H ðq; h; tÞ ¼ J 1 ðqÞ xðq; hÞeðtÞ :¼ J 1 ðqÞ þ H1 ðq; h; tÞ with ðx; yÞ ¼ ðxðq; hÞ; yðq; hÞÞ defined implicitly by (2.6). The following equalities give the expressions of J 0 ðhÞ and J 00 ðhÞ for h > 0. Lemma 1. For h > 0, we have 1 Z þ 1 2qq x 1 J 0 ðhÞ ¼ þ w0 ðxÞ ðh GðxÞÞq dx; q h x Z xþ 2 1 1 J 00 ðhÞ ¼ 1 w0 ðxÞ ðh GðxÞÞ p dx; p qp h x
ð2:8Þ
ð2:9Þ
where wðxÞ ¼ GðxÞ=gðxÞ for x 6¼ 0. Proof. Let þ
L ðhÞ ¼
Z
xþ
1 q
ðh GðxÞÞ dx;
L ðhÞ ¼
Z
0
0
Then 1
J ðhÞ ¼ 2qq ½Lþ ðhÞ þ L ðhÞ; 1
ðiÞ
ðiÞ
J ðiÞ ðhÞ ¼ 2qq ½ðLþ ðhÞÞ þ ðL ðhÞÞ ;
1
ðh GðxÞÞq dx: x
i ¼ 1; 2:
402
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Let GðxÞ ¼ sh, s 2 ½0; 1. Then gðxÞdx ¼ h ds and gðxÞdx=dh ¼ GðxÞ=h. Moreover 1
Lþ ðhÞ ¼ h1þq
Z
1 0
1
ð1 sÞq ds ; gðxÞ
1 Z Z 1 0 1 ðg ðxÞÞGðxÞ 1 1q 1 ð1 sÞq ds 1þ1q ðL ðhÞÞ ¼ 1 þ þh ds ð1 sÞq h q gðxÞ g3 ðxÞh 0 0 Z þ 1 þ 1q Z xþ 1 1 1 x g0 ðxÞGðxÞ q ¼ ðh GðxÞÞq dx ðh GðxÞÞ dx 2 h 0 g ðxÞ h 0 Z xþ 1 1 1 ¼ þ w0 ðxÞ ðh GðxÞÞq dx: h 0 q
þ
0
Similarly, we have ðL ðhÞÞ0 ¼
1 h
Z
0
1 1 þ w0 ðxÞ ðh GðxÞÞq dx: q
1 1 0 þ w ðxÞ ðh GðxÞÞq dx: q
x
Hence 1
2qq J ðhÞ ¼ h 0
Z
xþ x
Differentiating ðLþ ðhÞÞ0 , we obtain 1 ðL ðhÞÞ ¼ 2 h þ
00
Z
xþ
0
1 1 þ q2 q
2 2 0 00 0 1 w ðxÞ þ wðxÞw ðxÞ þ ðw ðxÞÞ q
1
ðh GðxÞÞq dx: Since Z
xþ
0
Z 0
1
ðh GðxÞÞq dx ¼
Z
xþ 0
xþ
dx ðh GðxÞÞ
1 p
Z
xþ
GðxÞdx 1
;
ðh GðxÞÞp
0
þ Z 1 1 x w0 ðxÞðh GðxÞÞq dx ¼ wðxÞðh GðxÞÞq þ 0
xþ
wðxÞgðxÞdx 1
qðh GðxÞÞp Z þ 1 x GðxÞdx ðwðxÞ ¼ GðxÞ=gðxÞÞ ¼ q 0 ðh GðxÞÞ1p 0
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
403
and Z
xþ
1
2
½wðxÞw00 ðxÞ þ ðw0 ðxÞÞ ðh GðxÞÞq dx 0
Z þ 1 x w0 ðxÞGðxÞdx ¼ ½wðxÞw ðxÞ ðh GðxÞÞ dx ¼ q 0 ðh GðxÞÞ1p 0 Z þ Z þ 1 1 x 0 h x w0 ðxÞ dx q ¼ w ðxÞðh GðxÞÞ dx þ q 0 q 0 ðh GðxÞÞ1p Z þ Z xþ 1 GðxÞ h x w0 ðxÞ dx ¼ 2 1 þ q 0 ðh GðxÞÞp q 0 ðh GðxÞÞ1p Z
xþ
1 q
0
0
which implies that 1 ðLþ ðhÞÞ00 ¼ qh
Z
xþ
w0 ðxÞ 1p dx 1
:
ðh GðxÞÞp
0
Similarly, we have 1 00 ðL ðhÞÞ ¼ qh
Z
0
x
w0 ðxÞ 1p dx 1
:
ðh GðxÞÞp
Therefore, J 00 ðhÞ ¼
2 1
qp h
Z
xþ x
w0 ðxÞ 1p dx 1
:
ðh GðxÞÞp
Lemma 2. If there exist two positive constants a1 and a2 such that for all x 6¼ 0, we have a1 6 g0 ðxÞGðxÞ=g2 ðxÞ 6
1 a2 : q
ð2:10Þ
Then the following estimates hold: 1
ch1þa2 6 J ðhÞ 6 Ch1þqa1 ; ch1 J ðhÞ 6 J 0 ðhÞ 6 Ch1 J ðhÞ; 1 0
00
1 0
ch J ðhÞ 6 J ðhÞ 6 Ch J ðhÞ: Moreover, J 0 ðhÞ ! þ1 as h ! þ1.
ð2:11Þ
404
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Proof. Since w0 ðxÞ ¼ 1 g0 ðxÞGðxÞ=g2 ðxÞ and from (2.8), (2.9), we have 1 ð1 þ a2 Þh1 J ðhÞ 6 J 0 ðhÞ 6 1 þ a1 h1 J ðhÞ q which yields that 1
ch1þa2 6 J ðhÞ 6 Ch1þqa1 and J 0 ðhÞ P ch1 J ðhÞ P cha2 ! þ1 as h ! þ1. Similarly, we have ch1 J 0 ðhÞ 6 J 00 ðhÞ 6 Ch1 J 0 ðhÞ:
Lemma 3. For 0 6 k 6 7, we have k k d o 0 k k 0 dhk J ðhÞ 6 Ch J ðhÞ; oqk J ðhÞ 6 Cq J ðhÞ
ð2:12Þ
and k o x k oqk ðq; hÞ 6 Cq jxj;
k o y k oqk ðq; hÞ 6 Cq jyj:
ð2:13Þ
The proof of Lemma 3 is similar to the proof of Lemma A 4.1 in [4], so we omit it. From (2.11) and (2.12), it is not difficult to verify that 1 1
1
Cq1þqa1 6 J 1 ðqÞ 6 Cq1þa2
ð2:14Þ
and cq1 J 1 ðqÞ 6
d 1 J ðqÞ 6 Cq1 J 1 ðqÞ: dq
ð2:15Þ
Lemma 4. Let x; y be defined by (2.5) and (2.6). Then q ox ðq; hÞ ¼ J 0 ðhÞjyjp signðyÞ; oh
oy ðq; hÞ ¼ J 0 ðhÞgðxðq; hÞÞ: oh
Proof. By the definition of S, h, for y > 0 we have Z x 1 1 Sðx; qÞ ¼ qq ðhðqÞ GðsÞÞq ds; 0
oS ðx; qÞ ¼ oq
Z 0
x
1
1
qq1 ðhðqÞ GðsÞÞq1 ds h0 ðqÞ ¼ h:
ð2:16Þ
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
405
Hence 1 o2 S ox 1 ðx; qÞ ¼ qq ðhðqÞ GðxÞÞp h0 ðqÞ ¼ 1; oq oh oh
that is, 1 ox 1 ðq; hÞ ¼ qp ðhðqÞ GðxÞÞp ðh0 ðqÞÞ1 oh 1
1
1
q
¼ J 0 ðhÞqp ðhðqÞ GðxÞÞJ 0 ðhÞqp ðhðpÞ GðxÞÞp ¼ J 0 ðhÞy p : Similarly, for y < 0, we have q ox ðq; hÞ ¼ J 0 ðhÞjyjp ; oh
therefore q ox ðq; hÞ ¼ J 0 ðhÞjyjp signðyÞ: oh
ð2:17Þ
q
From (2.17) and jyj =q þ GðxÞ ¼ h, we obtain jyjq2 y
oy ox þ gðxÞ ¼ 0; oh oh
hence oyðq; hÞ ox q2 ¼ gðxÞ =jyj y ¼ J 0 ðhÞgðxðq; hÞÞ: oh oh
Lemma 5. Let H1 ðq; h; tÞ ¼ xðq; hÞeðtÞ. Then for p 1, 0 6 k þ l 6 7, l ¼ 0; 1 and 0 6 m 6 6, we have ( 1 okþlþm k 1 p Cq ðJ ðqÞÞ ; 0 6 k 6 7; l ¼ 0; ð2:18Þ oqk ohl otm H1 ðq; h; tÞ 6 Cqk ðJ 1 ðqÞÞ1a1 ; 0 6 k 6 6; l ¼ 1: Proof. Let l ¼ 0 and q 1, we have from (2.13) and Remark 2, for 0 6 k þ m 6 7, kþm o ok x dm eðtÞ H ðq; h; tÞ ¼ oqk otm 1 oqk dtm k o x 6 C k ðq; hÞ 6 C k maxfxþ ; jx jg oq 1
k 1 ðqÞÞp : 6 Cqk G1 þ ðhÞ 6 Cq ðJ
406
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
For l ¼ 1, we have from (2.12) and (2.13) k kþ1þm o o ox ðmÞ oqk oh otm H1 ðq; h; tÞ ¼ oqk oh ðq; hÞe ðtÞ k o q ¼ k J 0 ðhÞjyjp signðyÞ eðmÞ ðtÞ oq j X di o q 0 p 6C dqi ðJ ðhÞÞ oqj jyj signðyÞ iþj¼k j o q i 0 p 6 C q jJ ðhÞj j jyj signðyÞ : oq Since j j2 i ji j X oy o q q oy 1 o2 y p pi Cji jyj 2 i opj jyj signðyÞ 6 oq oq oq 16i6j X q i j j 6C jyjp ðq1 jyjÞ 1 ðqi jyjÞ i 16i6j
X
6C
q
jyjp qðj1 þ þiji Þ
16i6j
X
6C
q
1
jyjp qðj1 þ þji Þ 6 Chp qj ;
16i6j
where fCij g are positive constants and j1 þ þ ji ¼ j. We have kþ1þm o 0 k 1p k 1a1 6 Cqk ðJ 1 ðqÞÞ1a1 : oqk oh otm H1 ðq; h; tÞ 6 CJ ðhÞq h 6 Cq h
3. New action-angle variables We consider now the Hamiltonian system (2.7) with Hamiltonian function H ðq; h; tÞ given by H ¼ J 1 ðqÞ þ H1 ðq; h; tÞ. Note that qdh H ðq; h; tÞ dt ¼ ðH dt qðH ; t; hÞdhÞ: This implies that if one can solve q ¼ qðH ; t; hÞ from (2.7) (t and h as parameters) as a function of H ; t and h, then dH oq ¼ ðH ; t; hÞ; dh oH
dt oq ¼ ðH ; t; hÞ: dh dt
ð3:1Þ
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
407
That is, (3.1) is also a Hamiltonian system with Hamiltonian function qðH ; t; hÞ and now the action, angle and time variables are H ; t and h, respectively. This idea came from [4,11]. Since by Lemmas 3 and 5, for q 1 oH ðq; h; tÞ 6¼ 0: oq We can define qðH ; t; hÞ as the inverse function of H ðq; h; tÞ with t; h playing the role of parameters by the implicit function theorem. Now we define q1 ðH ; t; hÞ as qðH ; t; hÞ ¼ J ðH Þ þ q1 ðH ; t; hÞ: Lemma 6. The perturbation term q1 ðH ; t; hÞ possesses the following estimates 1 okþlþm CH kq J ðH Þ; if m ¼ 0; 6 q ðH ; t; hÞ ð3:2Þ oH k otl ohm 1 CH ka1 J ðH Þ; if m ¼ 1; for 0 6 k þ l 6 6 and H 1. Proof. Claim. For H 1 we have 1 1 1 H < J ðH Þ < J 1þ H : J H 6 CJ q p q
ð3:3Þ
Indeed, the first two inequalities hold since J 0 ðH Þ > 0 by (2.11). On the other hand, it follows from (2.11) that J 0 ðhÞ is increasing and from (2.8), (2.9) that 1 0 J ðhÞ 6 1 þ a1 h1 J ðhÞ; ð3:4Þ q this yields that 1 1 H J 1þ H J p q 2 0 qH 0 1 1 ¼ J ðnÞH < J 1þ H 1þ p qþ1 p q 1 qa1 1 6J 1þ J 1þ H H ; p p qþ1 1 1 H; 1 þ H n2 q p therefore we obtain the third inequality
408
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Step 1. m ¼ 0. We divide it into three cases: (I) k ¼ 0, l ¼ 0. From the definition of qðH ; t; hÞ, if we treat H as the independent variable and t; h as parameters, then we have J 1 ðqðH ÞÞ þ H1 ðqðH ÞÞ ¼ H ;
ð3:5Þ
this yields that qðH Þ ¼ J ðH H1 ðqðH ÞÞ:
ð3:6Þ
Expanding q1 ðH Þ ¼ qðH Þ J ðH Þ ¼ J ðH H1 ðqðH ÞÞÞ J ðH Þ in TaylorÕs series, q1 ðH Þ ¼ J 0 ðH ÞH1 þ
Z
H1
sJ 00 ðH H1 þ sÞ ds:
ð3:7Þ
0
From Lemma 5 and note that qðH Þ ! 1 as H ! 1, and that jH1 ðqÞj < 1 CðJ 1 ðqÞÞp < 1p J 1 ðqÞ for q 1 and all t; h; we obtain 1 1 jH1 ðqðH ÞÞj < J 1 ðqðH ÞÞ < H : p p
ð3:8Þ
By (3.4), (3.6) and (3.8) as well as the monotonicity of J in H , we obtain 1 1 J H < qðH Þ < J 1þ H : ð3:9Þ q p This, together with Claim, implies that for H 1 cJ ðH Þ < qðH Þ < CJ ðH Þ 6 J ð2H Þ:
ð3:10Þ
It follows from (2.14) and Lemma 5 that 1
1
1
1
jH1 ðqðH ÞÞj 6 CðJ 1 ðqðH ÞÞÞp 6 CðJ 1 ðCJ ðH ÞÞÞp 6 CðJ 1 ðJ ð2H ÞÞÞp 6 CH p : ð3:11Þ Now the first term in (3.7) satisfies jJ 0 ðH ÞH1 j 6 CJ ðH Þ
1þ1p
1
¼ CJ ðH ÞH q
and the second term is bounded by 2 1 2 00 e sup J ð H Þ 6 CH12 H 2 J ðH Þ 6 CJ ðH ÞH q 6 CJ ðH ÞH q ; H1 H H1 6 H~ 6 H for H 1: This completes the proof of (I).
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
409
(II) k P 1, l ¼ 0. Differentiating (3.7) k times by H , we obtain k X ok q1 ðH Þ oiþ1 J ðH Þ oki H1 ðqðH ÞÞ ok ¼ Cki þ k k iþ1 ki oH oH oH oH i¼0
Z
H1
sJ 00 ðH H1 þ sÞds;
0
where Cki is a positive integer which depends only on k and i. Note that X ok H1 ðqðH ÞÞ os ok 1 oks ¼ C H ðqðH ÞÞ qðH Þ qðH Þ k 1 s oH k oqs oH k1 oH ks k þ þks ¼k 1
and ok oH k ¼
Z
H1
0 k X i;j¼0
sJ 00 ðH H1 þ sÞ ds
Ckij
oi H1 oj H1 okijþ2 J ðH Þ oH kijþ2 oH i oH j
X
oj H1 oi H1 osþ2 J ðH Þ on1 ðIÞ ons ðIÞ oH sþ2 oH n1 oH ns oH j oH i iþjþn1 þns ¼k Z X on1 ðIÞ oni ðIÞ H oiþ2 þ Ckin s J ðH H1 þ sÞ ds; oH n1 oH ni 0 oH iþ2 n1 þ þni¼k þ
Ckijns
where I ¼ J 1 . Therefore, the proof of (II) reduces to the proof of k o 6 CH k qðH Þ; for H 1; 1 6 k 6 b: qðH Þ oH k
ð3:12Þ
Differentiating (3.6) by H yields q0 ðH Þ ¼ J 0 ðH H1 Þ ½1 H10 qðH Þ: We obtain therefore q0 ðH Þ ¼
J 0 ðH H1 Þ : 1 þ J 0 ðH H1 ÞH10
ð3:13Þ
The denominator above expression is close one for H 1, indeed, J 0 ðH H1 ÞH10 ¼ J 0 ðH H1 Þ
dIðJ ðH H1 ÞÞ H10 ðqðH ÞÞ dIðJ ðH H ÞÞ 1 dq dq
¼1
H10 ðqðH ÞÞ dIðqðH ÞÞ dq
!0
as H ! 1:
410
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
For H 1, we get q0 ðH Þ <
2dIðH H1 Þ < CðH H1 Þ1 J ðH H1 Þ < CH 1 qðH Þ dq
which proving the case k ¼ 1 in (3.12). Assuming inductively that (3.12) holds for 1 6 k 6 n 6 5, now we prove it for k ¼ n þ 1. Differentiating (3.13) n times by H yields that n X onþ1 oi J 0 oni Q qðH Þ ¼ C ðH H Þ ; ni 1 oH ni oH nþ1 oH i i¼0 1
where Q ¼ ½1 þ J 0 ðH H1 Þ H10 . The proof of (3.12) for k ¼ n þ 1 reduces to the proof of i o 0 i 0 ð3:14Þ oH i J ðH H1 Þ 6 CH J ðH H1 Þ and
i o i oH i Q 6 CH
ði 6 n; H 1Þ:
ð3:15Þ
Proof of (3.14). Note that X oi 0 oi1 oim ðmþ1Þ J ðH H Þ ¼ C J ðH H Þ ðH H Þ ðH H1 Þ 1 im 1 1 oH i oH i1 oH im i1 þ im ¼i ð3:16Þ and oj oj ðH H Þ ¼ ðIðqðH ÞÞÞ 1 oH j oH j X os oj1 ojs Cjs ðIðqðH ÞÞÞ qðH Þ qðH Þ: ¼ oH s oH j1 oH js j1 þ þJs ¼j ð3:17Þ Using inductive assumptions in (3.17), we obtain for j 6 n j o j oH j ðH H1 Þ 6 CH ðH H1 Þ; which together with (3.16) and Lemma 3 yields that for j 6 n j o 0 j 0 J ðH H Þ 1 6 CH J ðH H1 Þ oH j as desired.
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
411
Proof of (3.15). It is not difficult to see that ij X oi oi j 1j o Q¼ Cij ½1 þ J 0 ðH H1 Þ H10 ðQ1 Þ ij ðQ1 Þ; I i oH oHi1 oH 06j6i
ð3:18Þ where i1 þ þ ij ¼ i. From (3.14) and Lemma 5, it is easy to see that for 1 6 i 6 n i i o o 1 0 0 i 0 0 ðQ Þ ¼ ½1 þ J ðH H Þ H 1 1 6 CH jJ ðH H1 Þ H1 j: oH i oH i Since J 0 ðH H1 Þ H10 ! 0
as H ! 1
we have for 1 6 i 6 n and H 1 j1 þ J 0 ðH H1 Þ H10 j 6 C and i o 0 0 i oH i ½1 þ J ðH H1 Þ H1 6 C H : It follows from (3.18) that for i 6 n and H 1 i 1 o ðQ Þ i oH i 6 C H : This proves (3.15) and (II) is proved. (III) k P 0, l > 1. Similar to the above calculation, the proof of (III) reduces to the proof of kþl o k ð3:19Þ oH k otl qðH Þ 6 C H qðH Þ; for k P 0; l > 1; H 1: But, respect bound. Step
according to (3.13) and Lemma 5, the differentiation of qðkÞ ðH Þ with to the variables t does not increase the order of growth of the upper Therefore (III) is proved. 2. m ¼ 1. Differentiating the equality
qðH ðq; h; t; t; hÞ ¼ q by h we obtain from q1 ðH Þ ¼ qðH Þ J ðH Þ and H ¼ IðqÞ þ H1 ðqÞ that o o o q1 ðH Þ ¼ qðH Þ H1 ðqðH ÞÞ: oh oH oh
ð3:20Þ
412
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
It is not difficult to obtain
X okþ1 os os H H ðqðH ÞÞ ¼ C ðqðH ÞÞ 1 ks 1 oH k oh oqs oh k þ þks ¼k 1
ok 1 oks qðH Þ ks qðH Þ: k oH 1 oH
Therefore, by applying the results of (3.12), (3.19), (3.20) and Lemma 5, we obtain for H 1 kþlþ1 o 6 CH ka1 J ðH Þ: q ðH ; t; hÞ oH k otl oh 1
4. Proof of Theorem 1 Lemma 7. There exists a canonical transformation T : ðr; sÞ ! ðH ; tÞ of the form T : H ¼ r þ U ðr; s; hÞ;
t ¼ s þ V ðr; s; hÞ;
ð4:1Þ
where the functions U and V are 1-periodic in h and satisfy U ðr; s; hÞ ; r
V ðr; s; hÞ ! 0
as r ! þ1
uniformly for ðs; hÞ 2 S 1 S 1 such that under this transformation, the system (3.1) is changed into the form dr oR ¼ ðr; s; hÞ; dh os
ds oR ¼ ðr; s; hÞ; dh or
ð4:2Þ
Rðr; s; hÞ ¼ J ðrÞ þ q1 ðr; hÞ; þR1 ðr; s; hÞ
ð4:3Þ
where
with q1 ðr; hÞ ¼
Z
1
q1 ðr; t; hÞ dt:
0
Moreover the new perturbation R1 possesses the estimate: kþl o 6 C rkþ1a1 ; 0 6 k þ l 6 5: R ðr; s; hÞ ork osl 1
ð4:4Þ
Proof. We choose T :H ¼rþ
oF ðr; s; hÞ; ot
s¼tþ
oF ðr; t; hÞ; or
ð4:5Þ
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
413
where F ðr; t; hÞ ¼
Z
t
0
1 ½q1 ðr; t; hÞ q1 ðr; hÞ dt: J 0 ðrÞ
Then the transformed system (3.1) is of the form dr oR ¼ ðr; s; hÞ; dh os
ds oR ¼ ðr; s; hÞ; dh or
where oF oF oF ; t; h þ Rðr; s; hÞ ¼ J r þ þ q1 r þ ot ot oh oF ¼ J ðrÞ þ J 0 ðrÞ þ q1 ðr; t; hÞ þ R1 ðr; s; hÞ ot with 2 Z 1 oF oF oF 00 þ R1 ðr; s; hÞ ¼ ð1 sÞJ r þ s ds oh ot ot 0 Z 1 oq1 oF oF ; t; h ds: rþs þ ot ot 0 oH Then the transformation T satisfies desired conditions. It remains to prove (4.4). It follows from (2.11) and (3.2) that for 0 6 k þ l 6 6 and m ¼ 0; 1, kþlþm 1 o Crkþp ; m ¼ 0; 6 F ðr; t; hÞ ork otl ohm Crkþ1a1 ; m ¼ 1: In particular, 2 o 6 Cra1 6 1 F ðr; t; hÞ or ot 2 if r 1. So one can solve the second equation of (4.5) for t, t ¼ s þ V ðr; s; hÞ; where V satisfies V ðr; s; hÞ ¼
oF ðr; s þ V ; hÞ: or
Let U ðr; s; hÞ ¼
oF ðr; s þ V ; hÞ: ot
ð4:6Þ
414
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
Then we are done. Moreover, similar to the proof of [3, Lemma 2], one can verify that kþl kþl o 6 C rkþ1p ; o 6 C rk1q U ðr; s; hÞ V ðr; s; hÞ ð4:7Þ ork osl ork osl for 0 6 k þ l 6 5 and U =r; V ! 0 as r ! þ1. Denote oF ðr; s; þV ; hÞ; oh Z 1 f2 ðr; s; hÞ ¼ ð1 sÞJ 00 ðr þ sU Þ U 2 ds; 0 Z 1 oq1 ðr þ sU ; s þ V ; hÞ U ds: f3 ðr; s; hÞ ¼ 0 oH f1 ðr; s; hÞ ¼
Then one can verify that (The proof is similar to the proof of (3.14)) for 0 6 k þ l 6 5, r 1, kþ1 o kþ1a1 ; ork osl f1 ðr; s; hÞ 6 Cr kþ1 o 6 Crk1 ; f ðr; s; hÞ ork osl 2 kþ1 o 6 Crk1 : f ðr; s; hÞ ork osl 3 1
Note that by (2.11), J ðrÞ 6 Cr1þqa1 , for r 1, we have finally kþl 3 o X okþl kþ1a1 ork osl R1 ðr; s; hÞ 6 ork osl fi ðr; s; hÞ 6 Cr i¼1 for 0 6 k þ l 6 5; r 1:
For r0 > 0, denote by Dr0 by the domain Dr0 ¼ fðr; s; hÞjr P r0 ; ðh; tÞ 2 S 1 S 1 g: The following lemma will be used for the application of MoserÕs twist theorem. Lemma 8 [3]. The Poincare mapping P of (4.2) has the intersection property on Dr0 ; that is, if C is an embedded circle in Dr0 homotopic to a circle r ¼constant in Dr0 , then P ðCÞ \ C 6¼ ;.
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
415
If we define k ¼ J 0 ðrÞ;
s ¼ s;
h ¼ h;
ð4:8Þ
then the transformed system (4.2) is of the form dk ¼ g1 ðk; s; hÞ; dh
ds ¼ k þ g2 ðk; s; hÞ; dh
ð4:9Þ
where oR1 ðr; s; hÞ; os o q oR1 ðr; s; hÞ; g2 ðk; s; hÞ ¼ 1 ðr; hÞ þ or or g1 ðk; s; hÞ ¼ J 00 ðrÞ
with r ¼ rðkÞ implicitly given by (4.8). From (2.11) and (3.2), we have 1
cra2 6 J 0 ðrÞ 6 Crqa1 ; o q1 ðr; hÞ 6 Cra1 : or Therefore q
1
ð4:10Þ
ck1qa1 6 rðkÞ 6 Cka2 ;
and k 1 if and only if r 1. Moreover, from (2.11) and (2.12), we obtain k o 6 Ckk rðkÞ 6 CrðkÞ; 0 6 k 6 4: rðkÞ ð4:11Þ okk From (2.12), (3.2), (4.4) and (4.11), we have for 0 6 k þ l 6 4, kþl o 6 Ckk r2 J ðrÞr1a1 6 Crð1þa1 Þ J ðrÞ 6 Cr1q2a1 6 Cke1 ; g ðk; s; hÞ okk osl 1 and kþl o k a1 e2 2 a1 okk osl g2 ðk; s; hÞ 6 Ck ðr þ r J ðrÞÞ 6 Cr 6 Ck ; where e1 ¼
2qa1 1 > 0; 1 qa1
e2 ¼
a1 q > 0: 1 qa1
416
X. Yang / Appl. Math. Comput. 158 (2004) 397–417
We are now in a position to prove Theorem 1. Proof of Theorem 1. Since the functions g1 and g2 are sufficiently small if k 1, one can verify that the solutions of (4.9) do exist for 0 6 h 6 1 if the initial value kð0Þ ¼ k is sufficiently large. Integrating (4.9) from h ¼ 0 to h ¼ 1, we obtain that the Poincare mapping P of (4.9) is of the form P : s1 ¼ s0 þ k0 þ D1 ðk0 ; s0 Þ;
k1 ¼ k0 þ D2 ðk0 ; s0 Þ;
where D1 and D2 possess the same estimates as g1 and g2 , that is, for 0 6 k þ l 6 4, okþl k l Di 6 Cke0 ; ok0 os0 where e0 ¼ minðe1 ; e2 Þ. Since T is a diffeomorphism, P possesses the intersection property on Dk0 . Therefore P satisfies all the assumptions of MoserÕs twist theorem [5,13]. From this theorem, for any x 1 satisfying m 5 x P c0 jnj 2 ; m; n 2 N ; ðm; nÞ ¼ 1 n there exists an invariant curve C of P and on which P is the form s1 ¼ s0 þ x: Therefore there exist invariant curves of the Poincare mapping of the system (2.1) which surrounding the origin ðx; yÞ ¼ ð0; 0Þ and are arbitrarily far from the origin. So every solution of (1.1) is bounded. h
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