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Existence of analytic solutions of an iterative functional equation
Applied Mathematics and Computation 217 (2011) 7245–7257
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Existence of analytic solutions of an iterative functional equation q Lingxia Liu Department of Mathematics, Weifang University, Weifang, Shandong 261061, PR China
a r t i c l e
i n f o
Keywords: Iterative functional equation Analytic solution Auxiliary equation Convergent power series
a b s t r a c t This paper is concern analytic solutions of an iterative functional equation of the form
f ðpðzÞ þ qðf ðzÞÞÞ ¼ hðf ðzÞÞ; z 2 C: Byconstructing a convergent power series solution of an auxiliary equation
gð/ða2 zÞ pð/ðazÞÞÞ ¼ hðgð/ðazÞ pð/ðzÞÞÞÞ; z 2 C; analytic solutions of the original equation are obtained. We discuss not only these a appeared in the auxiliary equation at the hyperbolic case 0 < jaj – 1 and resonance, i.e., at a root of the unity, but also those a near resonance (i.e., near a root of the unity) under Brjuno condition. Ó 2011 Elsevier Inc. All rights reserved.
1. Introduction Invariant curves of the area preserving maps play an important role in the theory of periodic stability of discrete dynamical systems. A common and useful method to understand behaviors of a discrete dynamical system generated by the iteration of a self-mapping is to find a simple structures in its phase space and to describe the dynamics in terms of the effect caused by the presence of these structures. In particular, invariant curve, in dimension two, is one of such structures. Diamond [1] researched the existence of analytic invariant curves for two-dimensional maps of the form
Tðx; yÞ ¼ ðx þ y; yð1 þ bxk Þ þ Fðx; yÞÞ: For an investigation of other related problems the interested reader is referred to [2–12] and the monograph [13]. In particular, Wen Rong Li and Sui Sun Cheng [4] discussed the analytic solutions of functional equation
f ðpðzÞ þ bf ðzÞÞ ¼ hðzÞ
ð1:1Þ
by using the method of majorant series. The purpose of this paper is to find invertible and linearizable invariant analytic curves of the 2-D complex map T : C2 ! C2 ; ðz; wÞ # ðz1 ; w1 Þ, defined by
z1 ¼ pðzÞ þ qðwÞ; w1 ¼ hðwÞ:
ð1:2Þ
Throughout this paper, we assume that p(z), q(z) and h(z) are analytic in a neighborhood of the origin, and p(0) = 0,h(0) = 0,q(0) = 0,p0 (0) = n – 0,h0 (0) = g – 0 and q0 (0) = f – 0. Clearly, map T has a fixed point O = (0,0) with the Jacobian matrix
q
Supported by the Natural Science Foundation of Shandong Province (2006ZRB01066). E-mail address: [email protected]
0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.02.015
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L. Liu / Applied Mathematics and Computation 217 (2011) 7245–7257
A ¼ DTð0Þ ¼
n
f
0 g
at O. Its characteristic polynomial is
PA ðkÞ ¼ a2 ðn þ gÞa þ ng: As is well known that a curve w = f(z) is said to be an invariant curve of T if w1 = f(z1). Thus, we see that map T has an invariant curve w = f(z) if and only if f satisfies the functional equation
f ðpðzÞ þ qðf ðzÞÞÞ ¼ hðf ðzÞÞ; z 2 C:
ð1:3Þ
The results of this paper can be regarded as a generalization of the results obtained in [4]. From the above assumptions, it is easy to see that the inverse function of q(z) exists and is analytic in a neighborhood of the q(0) = 0. We denote the inverse function of q(z) by g(z) and consider the following equation
gð/ða2 zÞ pð/ðazÞÞÞ ¼ hðgð/ðazÞ pð/ðzÞÞÞÞ; z 2 C;
ð1:4Þ
which is called the auxiliary equation of (1.3). We first construct analytic solutions of (1.4) in the cases: (H1) 0 < jaj – 1. P log qkþ1 (H2) a = e2pih, where h 2 R n Q is a Brjuno number ([14] and [15]), i.e., BðhÞ ¼ 1 < 1, where {pk/qk} denotes the k¼0 qk sequence of partial fraction of the continued fraction expansion of h, said to satisfy the Brjuno condition. (H3) a = e2pi q/p for some integers p 2 N with p P 2 and q 2 Z n f0g, and a – e2pil/k for all 1 6 k 6 p 1 and l 2 Z n f0g. Observe that a is the inside or outside of the unit circle S1 in the case of (H1) but on S1 in the rest cases. More difficulties are encountered for a on S1, as mentioned in the so-called ‘‘small-divisor problem’’ (seen in [16] p. 22 and p. 146 and [17]). Under Diophantine condition: a = e2pih, where h 2 R n Q and there exist constants f > 0 and r > 0 such that jan 1j P f1nr for all n P 1, the number a 2 S1 is ‘‘far’’ from all roots of the unity and was considered in different settings [10–12]. In [10] the case of (H3), where a is a root of the unity, was also discussed for a general class of iterative equations. Since then, we have been striving to give a result of analytic solutions for those a ‘‘near’’ a root of the unity, i.e., neither being roots of the unity nor satisfying the Diophantine condition. The Brjuno condition in (H2) provides such a chance for us. As stated in [18], for a real number h, we let [h] denote its integer part and {h} = h [h] its fractional part. Then every irrational number h has a unique expression of the Gauss’ continued fraction
h ¼ a0 þ h0 ¼ a0 þ
1 ¼ ; a1 þ h1
denoted simply by h = [a0,a1, . . ., an, . . .], where aj’s and hj’s are calculated by the algorithm: (a) a0 = [h], h0 = {h}, and (b) 1 1 an ¼ ½hn1 ; hn ¼ fhn1 g for all n P 1. Define the sequences ðpn Þn2N and ðqn Þn2N as follows:
q2 ¼ 1; q1 ¼ 0; qn ¼ an qn1 þ qn2 ; p2 ¼ 0; p1 ¼ 1; pn ¼ an pn1 þ pn2 : It is easy to show that pn/qn = [a0,a1, . . ., an]. Thus, For every h 2 R n Q we associate, using its convergence, an arithmetical P log q function BðhÞ ¼ nP0 qnnþ1 . We say that h is a Brjuno number or that it satisfies Brjuno condition if B(h) <+1. The Brjuno condition is weaker than the Diophantine condition. For example, if anþ1 6 cean for all n P 0, where c > 0 is a constant, then h = [a0,a1, . . ., an, . . .] is a Brjuno number but is not a Diophantine number. So, the case (H2) contains both Diophantine condition and a part of a ‘‘near’’ resonance. In this paper, considering the Brjuno condition instead of the Diophantine one, we discuss not only the cases (H1) and (H3) but also (H2) for analytic invariant curves of the mapping T defined in (1.2).
2. Substituting power series in Eq. (1.4) Let
pðzÞ ¼
1 X n¼1
an zn ;
hðzÞ ¼
1 X
c n zn ;
gðzÞ ¼
n¼1
1 X n¼1
g n zn ;
a1 ¼ n; c1 ¼ g; g 1 ¼ s ¼
1 : f
ð2:1Þ
Without loss of generality , we can assume that
jan j 6 1; jcn j 6 1; jg n j 6 1; n ¼ 2; 3; . . . :
ð2:2Þ
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In fact, since the series 1 X
nþ
an zn1 ;
gþ
n¼2
1 X
cn zn1 ;
sþ
1 X
n¼2
g n zn1
n¼2
are uniformly convergent in a neighborhood of O, there exists a constant q > 0 such that
jan j 6 qn1 ; jcn j 6 qn1 ; jg n j 6 qn1 ; n ¼ 2; 3; . . . :
ð2:3Þ
Introducing new functions
~ ~ðzÞ ¼ qpðq1 zÞ; /ðzÞ ¼ q/ðq1 zÞ; p ~ hðzÞ ¼ qhðq1 zÞ; g~ðzÞ ¼ qgðq1 zÞ; ~ we see that /ðzÞ satisfies
~ g~ð/ð ~ azÞÞÞ ¼ hð ~ azÞ p ~ ~ a2 zÞ p ~ð/ð ~ð/ðzÞÞÞÞ; g~ð/ð which is the same form of (1.4). From (2.3) and the following expansions 1 X
~ðzÞ ¼ qpðq1 zÞ ¼ nz þ p ~ hðzÞ ¼ qhðq1 zÞ ¼ gz þ
n¼2 1 X
an q1n zn ; cn q1n zn ;
n¼2
and
g~ðzÞ ¼ qgðq1 zÞ ¼ sz þ
1 X
g n q1n zn ;
n¼2
we get janq1nj 6 1, jcnq1nj 6 1, jgnq1nj 6 1, n = 2, 3, . . .. Let 1 X
/ðzÞ ¼
bn zn
ð2:4Þ
n¼1
be the expansion of a formal solution of (1.4). Substituting to /, p, h, g, their power series (2.4) and (2.1) respectively in (1.4), we have
1 3n 9n > > > > 6 > C 7 B > 7 6 = 1 1 C B X X X 7 6 C B n n7 6 cn gn6 at bl1 bl2 . . . blt Cz 7 Bbn a > C 7 > > > 6 n¼1 B n¼1 n¼1 > A 5 > @ l1 þ l2 þ þ lt ¼ n > > 4 > > > > ; : t ¼ 1; 2; . . . ; n 3n 2 1 0 8 > > > > > >
2
0
7 6 C B 7 6X C X 7 61 B C B n n n7 ¼ gn6 b a b b . . . b a z a B n t l1 l2 lt C 7 : 6 C B 7 6 n¼1 @ n¼1 A l1 þ l2 þ þ lt ¼ n 5 4 t ¼ 1; 2; . . . ; n 1 X
Let
X
sn ¼ bn an
at bl1 bl2 . . . blt ;
un ¼
X
l1 þ l2 þ þ lt ¼ n
l1 þ l2 þ þ lt ¼ n
t ¼ 1; 2 . . . ; n
t ¼ 1; 2; . . . ; n
g t sl1 sl2 . . . slt ; ð2:5Þ
then we obtain 1 X n¼1
un an zn ¼
1 X
X
n¼1
l1 þ l2 þ þ lt ¼ n t ¼ 1; 2; . . . ; n
ct ul1 ul2 . . . ult zn :
ð2:6Þ
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L. Liu / Applied Mathematics and Computation 217 (2011) 7245–7257
Comparing coefficients we obtain
8 ða gÞu1 ¼ 0; > > > P < ðan gÞu ¼ ct ul1 ul2 . . . ult ; n ¼ 2; 3; . . . ; n l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð2:7Þ
From (2.5), it follows that s1 = b1(a n) and u1 = b1s(a n). And then from the first equation of (2.7) we have b1s(a n)(a g) = 0. In view of s – 0 and (a n)(a g) = 0, we can choose b1 = s – 0. 3. Auxiliary equation in case (H1) In this section, we discuss Eq. (1.4) in the case 0 < jaj – 1. Theorem 3.1. Assume that a 2 (H1), then for any s 2 C, the auxiliary Eq. (1.4) has an analytic solution /(z) in a neighborhood of the origin such that /(0) = 0 and /0 (0) = s. Proof. If a – g, then a = n. Thus, we have s1 = 0 and u1 = 0. From (2.5) and (2.7) it is easily seen that sn = 0 and un = 0 for all n P 1. Consequently,
X
bn an ¼ nbn þ
at bl1 bl2 . . . blt ;
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n that is,
8 ða nÞb1 ¼ 0; > > > n P < at bl1 bl2 . . . blt ; n ¼ 2; 3; . . . : ða nÞbn ¼ l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð3:1Þ
In view of a = n and the condition (H1), the second equality of (3.1) can be written as
bn ¼
X
1
an a
at bl1 bl2 . . . blt ; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n
ð3:2Þ
Then for b1 = s – 0, we can uniquely determine the sequence fbn g1 n¼2 by (3.2), recursively. In what follows we prove the convergence of series (2.4) in a neighborhood of the origin. Since
lim
n!1
1 ¼ an a
(
a1 ; 0 < jaj < 1; 0;
jaj > 1;
there exists a positive number M such that
1 an a 6 M;
n P 2:
ð3:3Þ
From (2.2), (3.2) and (3.3), we see
X
jbn j 6 M
jbl1 j:jbl2 j jblt j; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n
ð3:4Þ
t ¼ 2; 3; . . . ; n To construct a governing series, we consider the implicit functional equation
ðM þ 1ÞðGðzÞÞ2 ðjsjz þ 1ÞGðzÞ þ jsjz ¼ 0; i.e.,
GðzÞ ¼ jsjz þ M
ðGðzÞÞ2 : 1 GðzÞ
ð3:5Þ
Define the function
RðM; s; z; GÞ ¼ ðM þ 1ÞG2 ðjsjz þ 1ÞG þ jsjz
ð3:6Þ
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for (z,G) in a neighborhood of (0,0). Then R(M,s,0,0) = 0, R0G ðM; s; 0; 0Þ ¼ 1 – 0. Thus, there exists a unique function G(M,s,z), 0 s;0;0Þ analytic in a neighborhood of zero, such that G(M,s,0) = 0, G0 ðM; s; 0Þ ¼ RR0z ðM; ¼ jsj and R(M,s,z,G(M,s,z)) = 0. G(M,s,z) can G ðM;s;0;0Þ be expanded into a convergent series
GðM; s; zÞ ¼
1 X
An z n :
ð3:7Þ
n¼1
Replacing (3.7) into (3.5) and comparing coefficients, we obtain that
8 A1 ¼ jsj; > > > P < Al1 Al2 . . . ; Alt ; n ¼ 2; 3; . . . : An ¼ M þ l þ þ l ¼ n l > 1 2 t > > : t ¼ 2; 3; . . . ; n
ð3:8Þ
Furthermore,
jbn j 6 An ; n ¼ 1; 2; . . . :
ð3:9Þ
In fact, jb1j = jsj = A1. For inductive proof we assume that jbjj 6 Aj, j 6 n 1. Observe that in (3.4), jblj j 6 Alj ; j ¼ 1; 2; . . . ; k, because 1 6 l1, . . ., lk 6 n 1. From (3.8) we know jbnj 6 An and (3.9) is proved. By the convergence of (3.7) and the inequality (3.9), we see that the series (2.4) converges uniformly in a neighborhood of the origin. If a – n, then a = g. It is easy to see that s1 = b1(a n) – 0 and u1 = b1s(a n) – 0. The second equality of (2.7) can be written as
un ¼
X
1
an a
ct ul1 ul2 . . . ult ; n ¼ 2; 3; . . . : ð3:10Þ
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n
As the proof of convergence of the power series (2.4), we can show that power series
UðzÞ ¼
1 X
un zn
ð3:11Þ
n¼1
converges in a neighborhood of the origin. Thus, there is a 0 < q1 < 1 such that
jun j 6 qn1 :
ð3:12Þ
Now we show the convergence of the power series From (2.5), we get
P1
n¼1 sn z
n
.
8 s1 ¼ ða nÞs; > > > P < g t sl1 sl2 . . . slt ; n ¼ 2; 3; . . . : ssn ¼ un l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð3:13Þ
For given un and s1 = (a n)s – 0, we can uniquely determine the sequence fsn g1 n¼2 by (3.13), recursively, and
1
0
C B C X 1B C B g t sl1 sl2 . . . slt C; n ¼ 2; 3; . . . : sn ¼ Bun C sB A @ l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n In view of (3.12) and (2.2) and f ¼ 1s – 0,
0
1
C B C B X C B n jsl1 j:jsl2 j . . . jslt jC; jsn j 6 jfjBq1 þ C B A @ l1 þ l2 þ þ lt ¼ n
n ¼ 2; 3; . . . :
t ¼ 2; 3; . . . n To construct a governing series we consider the implicit functional equation
fq21 z2 fq21 z2 ð1 þ fÞðWðzÞÞ2 1 þ þ jða nÞsjz WðzÞ þ jða nÞsjz þ ¼ 0; 1 q1 z 1 q1 z
ð3:14Þ
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L. Liu / Applied Mathematics and Computation 217 (2011) 7245–7257
i.e.,
WðzÞ ¼ jða nÞsjz þ
fq21 z2 ðWðzÞÞ2 þf : 1 q1 z 1 WðzÞ
ð3:15Þ
Now let us define the function
2 2 2 2 e WÞ ¼ ð1 þ fÞW 2 1 þ fq1 z þ jða nÞsjz W þ jða nÞsjz þ fq1 z Rðz; 1 q1 z 1 q1 z e e 0 ð0; 0Þ ¼ 1 – 0, there exists a unique function W(z), anafor (z, W) in a neighborhood of the zero. Since Rð0; 0Þ ¼ 0 and R W eR 0z ð0; 0Þ 0 ¼j ða nÞs j – 0. So W(z) can be expanded into a conlytic in a neighborhood of zero, such that Wð0Þ ¼ 0; W ð0Þ ¼ 0 eR W ð0; 0Þ vergent power series
WðzÞ ¼
1 X
Bn zn
ð3:16Þ
n¼1
uniformly in a neighborhood of zero. Replacing (3.16) into (3.15) and comparing coefficients, we obtain
8 B1 ¼ jða nÞsj; > > 1 0 > > > > > < C B C B P C B n > ; n ¼ 2; 3; . . . B ¼ f q þ B . . . B B B n l l l > tC 1 1 2 > C B > > l1 þ l2 þ þ lt ¼ n A @ > > : t ¼ 2; 3; . . . n then by induction and using (3.14) it is not difficult to show that
jsn j 6 Bn ; n ¼ 1; 2; . . . :
ð3:17Þ
By the convergence of (3.16) and the inequality (3.17) we see that the series
SðxÞ ¼
1 X
sn zn
ð3:18Þ
n¼1
converges uniformly in a neighborhood of the origin. Thus, there is a 0 < q2 < 1 such that
jsn j 6 qn2 :
ð3:19Þ
Now we prove the convergence of the power series (2.4) in the case a – n,a = g. From (2.5) we see
8 ða nÞb1 ¼ s1 ; > > > P < n at bl1 bl2 . . . blt ; n ¼ 2; 3; . . . : ða nÞbn ¼ sn þ l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . n
ð3:20Þ
Let b1 = s – 0, from the second function of (3.20) we have
1
0
C B C X 1 B C B at bl1 bl2 . . . blt C; n ¼ 2; 3; . . . ; bn ¼ n Bsn þ C B a n@ A l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n we can uniquely determine the sequence fbn g1 n¼2 by (3.21), recursively. Note that 0 < jaj – 1 and
1 lim ¼ n!þ1 an n
(
1n ; 0 < jaj < 1; 0;
jaj > 1;
ð3:21Þ
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so there exists a positive number N, such that j an1n j 6 N for n P 2. Then from (2.2) and (3.21)
1
0
C B C B C B X C B jbl1 j:jbl2 j . . . jblt jC jbn j 6 NBjsn j þ C B C B l1 þ l2 þ þ lt ¼ n A @ t ¼ 2; 3; . . . n 1
0
C B C B C B X C B n 6 NBq2 þ jbl1 j:jbl2 j . . . jblt jC; n ¼ 2; 3; . . . ; C B C B l1 þ l2 þ þ lt ¼ n A @ t ¼ 2; 3; . . . n which is a similar form with (3.14). Similarly, we can show that the power series (2.4) converges in a neighborhood of the origin. Finally, if a = n = g, we obtain s1 = u1 = 0. From (2.7) we deduce that u2 = 0 (note that a2 a – 0), and also s2 = 0 according to the definition of sn given in (2.5). Similar, it can be proven that sn = un = 0 for all n P 1. At this point we can reproduce the proof of case a = n, a – g, obtaining again the convergence of series (2.4). h 4. The auxiliary equation in case (H2) and (H3) In this section, we discuss local invertible analytic solutions of auxiliary Eq. (1.4) in cases (H2) and (H3). In order to discuss the existence of analytic solutions of the auxiliary Eq. (1.4) under (H2), we need to introduce Davie’s Lemma. Let h 2 R n Q and ðqn Þn2N be the sequence of partial denominators of the Gauss’ continued fraction for h as above. As in [18], let
Ak ¼
q 1 q ; Ek ¼ max qk ; kþ1 ; gk ¼ k : n P 0jknhk 6 8qk 4 Ek
Let Ak be the set of integers j P 0 such that either j 2 Ak or for some j1 and j2 in Ak, with j2 j1 < Ek, one has j1 < j < j2 and qk divides j j1. For any integer n P 0, define
n lk ðnÞ ¼ max ð1 þ gk Þ 2; qk
ðmn gk þ nÞ
1 1 ; qk
where mn ¼ maxfjj0 6 j 6 n; j 2 Ak g. We then define the function hk : N ! Rþ as follows:
(m
hk ðnÞ ¼
n þgk n
qk
lk ðnÞ;
1; if mn þ qk 2 Ak ; if mn þ qk R Ak :
Let g k ðnÞ :¼ maxðhk ðnÞ; ½qn Þ, and define k(n) by the condition qk(n) 6 n 6 qk(n)+1. Clearly, k(n) is non-decreasing.Moreover,the k function gk is non-negative. Then we are able to state the following result. Lemma 4.1 (Davie’s lemma [19]). Let KðnÞ ¼ n log 2 þ
PkðnÞ j¼0
g j ðnÞ logð2qjþ1 Þ. Then
(a) there is a universal constant c > 0 (independent of n and h) such that
KðnÞ 6 n
! kðnÞ X log qjþ1 þc ; qj j¼0
(b) K(n1) + K(n2) 6 K(n1 + n2) for all n1 and n2, and (c) logjan 1j 6 K(n) K(n 1).
Theorem 4.1. Assume that a = n or a = g but jnj – 1 and (H2) holds, then for any s 2 C, the auxiliary Eq. (1.4) has an analytic solution /(z) in a neighborhood of the origin such that /(0) = 0 and /0 (0) = s. Proof. As in the proof of Theorem 3.1, we seek a solution of (1.4) in power series of the form (2.4). If s = 0, (1.4) has a trivial solution f(z) 0. Assume s – 0 and b1 = s, by using the same arguments as in Theorem 3.1, we can uniquely determine the sequence fbn g1 n¼2 by (3.2) or (3.21), recursively.
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If a = n, then a – g (since a = g forces jnj – 1 and jaj = 1). From (2.5) and (2.7), we obtain sn = 0,un = 0,n = 1, 2, . . ., and
X
ðan aÞbn ¼
at bl1 bl2 . . . blt :
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n
ð4:1Þ
In view of jaj ¼ 1; h 2 R n Q and (2.2), we get
jbn j 6
1 ja 1j
X
n1
jbl1 j:jbl2 j . . . jblt j; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n
ð4:2Þ
t ¼ 2; 3; . . . n To construct a governing series, we consider the following function
VðzÞ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ jsjz s2 z2 6jsjz þ 1 4
ð4:3Þ
which clearly satisfies the equality
ðVðzÞÞ2 : 1 VðzÞ
VðzÞ ¼ jsjz þ
ð4:4Þ
Define the function
Rð1; s; z; VÞ ¼ 0
ð4:5Þ
for (z,V) in a neighborhood of (0,0), where R is defined in (3.6). Similarly to the proof of Theorem 3.1, we can prove that (4.5) 0 s;0;0Þ has a unique analytic solution V(s,z) in a neighborhood of zero, such that V(s,0) = 0, V 0z ðs; 0Þ ¼ RR0z ð1; ¼ jsj and V ð1;s;0;0Þ R(1,s,z,V(s,z)) = 0. So V(s,z) can be expanded into a convergent power series
Vðs; zÞ ¼
1 X
C n zn ; C 1 ¼ jsj:
ð4:6Þ
n¼1
Replacing (4.6) into (4.4) and comparing coefficients as in (3.8) with M = 1, we have
8 C 1 ¼ jsj; > > > P < C l1 C l2 . . . C lt ; n ¼ 2; 3 . . . : Cn ¼ þ l þ þ l ¼ n l > 1 2 t > > : t ¼ 2; 3; . . . n
ð4:7Þ
Note that the series (4.6) converges in a neighborhood of the origin. Hence, there is a constant T > 0 such that
C n 6 T n ; n ¼ 1; 2; . . . :
ð4:8Þ
Now by induction on n we prove
jbn j 6 C n eKðn1Þ ; n ¼ 1; 2; . . . ; where K : N ! R is defined in Lemma 4.1. In fact jb1j = jsj = C1, we assume that jbjj 6 CjeK(j1),j 6 n 1. From Lemma 4.1 and (4.2) we obtain
jbn j 6
1 jan1 1j
X
C l1 :C l2 . . . C lt eKðl1 1ÞþþKðlt 1Þ
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n X 1 C l1 :C l2 . . . C lt 6 n1 eKðn2Þ ja 1j l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n eKðn2Þ ¼ n1 Cn: ja 1j Note that
Kðl1 1Þ þ Kðl2 1Þ þ þ Kðlt 1Þ 6 Kðn 2Þ 6 Kðn 1Þ þ log jan1 1j; then
jbn j 6 C n eKðn1Þ ;
L. Liu / Applied Mathematics and Computation 217 (2011) 7245–7257
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as we claimed. Moreover, according to (4.8) we have jbnj 6 TneK(n1). Note that K(n) 6 n(B(h) + c) for some universal constant c > 0, then
jbn j 6 T n eðn1ÞðBðhÞþcÞ; that is,
1 n1 lim sup jbn jn 6 lim sup Te n ðBðhÞþcÞ ¼ TeBðhÞþc :
n!1
n!1
This implies that the convergence radius of the series (2.4) is at least (TeB(h)+c)1. If a – n, then a = g. It follows that s1 = b1(a n) – 0 and u1 = b1s(a n) – 0. Since jaj = 1, from (3.10) and (2.2), we have
jun j 6
X
1 jan1 1j
jul1 jjul2 j . . . jult j; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n
ð4:9Þ
t ¼ 2; 3; . . . n Using the same majorization as in (4.2), we can also prove the power series (3.11) converges uniformly in the neighborhood P n of the origin. As the proof of convergence of the series (3.18), the power series 1 n¼1 sn z also converges in a neighborhood of n the origin, so there is a 0 < q2 < 1 such that jsn j 6 q2 for all n P 1. Now we prove the series (2.4) is convergent in a neighborhood of the origin. From (3.21), we get
1
0
C B C B X C B at bl1 :bl2 . . . blt C; n ¼ 2; 3; . . . : bn ¼ ðan nÞ1 Bsn þ C B A @ l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n By jnj – 1, a – n and a = g and jan n j P kajn jnk = j1 jnk, it see that
1
0
C B C B X C B jbl1 j:jbl2 j . . . jblt jC; n ¼ 2; 3; . . . : jbn j 6 j1 jnk1 Bqn2 þ C B A @ l1 þ l2 þ þ lt ¼ n
ð4:10Þ
t ¼ 2; 3; . . . n Similar to proof of convergence of the series (3.18), we can show that the power series (2.4) is convergent in a neighborhood of the origin. This completes the proof of Theorem 4.1. h In the case (H3), where a = e2pih with h 2 R n Q a Brjuno number, the constant a is not only on the unit circle in C but also a root of the unity. In this case both the Diophantine condition and Brjuno condition are not satisfied. The difficulty can be overcome by an idea acquired from [20]. Let fDn g1 n¼1 be a sequence defined by
8 D1 ¼ jsj; > > > P < Dl1 :Dl2 . . . Dlt ; n ¼ 2; 3; . . . Dn ¼ C l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð4:11Þ
where C = max{1,jai 1j1,i = 1,2, . . ., p 1}. Theorem 4.2. Suppose that aj – g,j = 1, 2, . . ., p and (H3) holds. Then fbn g1 n¼1 is determined recursively, by b1 = s and
ðan aÞbn ¼ Vðn; aÞ; n ¼ 2; 3; . . . ;
ð4:12Þ
where
Vðn; aÞ ¼
X
at bl1 :bl2 . . . blt :
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3 . . . n If V(mp + 1,a) = 0 for all m = 1, 2, . . ., then Eq. (1.4) has an analytic solution /(z) in a neighborhood of the origin such that / (0) = 0,/0 (0) = s, and /(mp+1)(0) = (mp + 1)!Tmp+1, where all Tmp+1’s are arbitrary constants satisfying the inequality jTmp+1j 6 Dmp+1 and the sequence fDn g1 n¼1 is defined in (4.11). Otherwise, if V(mp + 1,a) – 0 for some m = 1,2, . . ., then Eq. (1.4) has no analytic solutions in any neighborhood of the origin.
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Proof. If a – g, then we have a = n,s1 = 0 and u1 = 0. Because aj – g,j = 2, . . ., p, then from (2.5) and (2.7), we can determine sn = 0,un = 0 , n = 1, 2 . . .. Thus, we have (4.1) or (4.12). Clearly, (1.4) has a trivial solution /(z) 0 if s = 0. Assume s – 0 and seek a power series solution of (1.4) of the form ( 2.4) as in the proof of Theorem 4.1, where the equality in (4.1) or (4.12) is indispensable. If V(mp + 1,a) – 0 for some natural number m, then the equality in (4.1) or (4.12) does not hold for n = mp + 1 since amp+1 a = 0. In such a circumstance Eq. (1.4) has no formal solutions. When V(mp + 1,a) = 0 for all natural number m, for each m the corresponding bmp+1 in (4.1) or (4.12) has infinitely many choices in C, that is, the formal series solution (2.4) defines a family of solutions with infinitely many parameters. Choose bmp+1 = smp+1 arbitrarily such that
jsmpþ1 j 6 Dmpþ1 ; m ¼ 1; 2 . . . ;
ð4:13Þ
where Dmp+1 is defined by (4.11). In what follows we prove the power series solution (2.4) converges in a neighborhood of the origin. Observe that jan1 1j1 6 C for n – mp + 1. It follows that
X
jbn j 6 C
jbl1 j:jbl2 j . . . jblt j; n – mp þ 1; m ¼ 1; 2; . . . :
l1 þ l2 þ þ lt ¼ n
ð4:14Þ
t ¼ 2; 3; . . . n We consider the implicit functional equation
RðC; s; z; WÞ ¼ 0;
ð4:15Þ
where R is defined in (3.6). Similarly to the proof of Theorem 3.1, we can prove that there exists a unique analytic function W(C,s,z) in a neighborhood of the zero such that W(C,s,0) = 0, W0 z(C,s,0) = jsj. So W(C,s,z) can be expanded into a convergent power series
WðC; s; zÞ ¼
1 X
Dn zn ; D1 ¼ jsj:
ð4:16Þ
n¼1
Moreover, by induction we can claim that
jbn j 6 Dn ; n ¼ 1; 2; . . . : Thus the series (2.4) converges in a neighborhood of the origin.
h
5. Existence of Analytic Solutions In this section, we will show the following theorem. Theorem 5.1. Suppose that conditions of Theorem 3.1, Theorem 4.1 or Theorem 4.2 are satisfied. Then Eq. (1.3) has an analytic solution of the form f(z) = g{/[a/1(z)] p(z)} in a neighborhood of the origin, where g(z) is the inverse function of q(z) and /(z) is a local invertible analytic solutions of Eq. (1.4) in a neighborhood of the origin with /(0) = 0, /0 (0) = s. Proof. By Theorem 3.1, Theorem 4.1 or Theorem 4.2, we can find an analytic solution /(z) of the auxiliary Eq. (1.4) in the form of (2.4), such that /(0) = 0 and /0 (0) = s – 0. Clearly the inverse function /1(z) exists and is analytic in a neighborhood of the origin. Let
uðzÞ ¼ pðzÞ þ qðf ðzÞÞ;
ð5:1Þ
then we have
qðf ðzÞÞ ¼ uðzÞ pðzÞ:
ð5:2Þ 1
Since q(0) = 0 and q0 (z) = f – 0, the inverse function q by g(z), we have
f ðzÞ ¼ g½uðzÞ pðzÞ:
(z) exists and is analytic in a neighborhood of the origin. Denote q1(z)
ð5:3Þ
Eq. (1.3) is changed into
gðuðuðzÞÞ pðuðzÞÞÞ ¼ hðgðuðzÞ pðzÞÞÞ:
ð5:4Þ
Taking
uðzÞ ¼ /ða/1 ðzÞÞ;
ð5:5Þ
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it is easy to see
gðuðuðzÞÞ pðuðzÞÞÞ ¼ gð/ða2 /1 ðzÞÞ pð/ða/1 ðzÞÞÞÞ ¼ hðgð/ða/1 ðzÞÞ pð/ð/1 ðzÞÞÞÞÞ ¼ hðgðuðzÞ pðzÞÞÞ; so u(z) = /(a/1(z)) satisfies Eq. (5.4)consequently f(z) = g(/(a/ 1(z)) p(z)) is an analytic solution of (1.3) in a neighborhood of the origin. This completes the proof of Theorem 5.1. h We now show how to explicitly construct an analytic solution of form (1.3) by means of an example. Consider the following equation:
f ðpðzÞ þ qðf ðzÞÞÞ ¼ hðf ðzÞÞ;
ð4:6Þ
P P1 4zn 3z n 0 z 0 where pðzÞ ¼ 1z ¼ 1 hðzÞ ¼ n¼1 3z ; jzj < 1; pð0Þ ¼ 0; p ð0Þ ¼ n ¼ 3; qðzÞ ¼ 4ðe 1Þ ¼ n¼1 n! ; qð0Þ ¼ 0; q ð0Þ ¼ 1 ¼ 4; P1 2zn z 0 2ð1 e Þ ¼ n¼1 n! ; hð0Þ ¼ 0; h0ð0Þ ¼ g ¼ 2. Because q(0) = 0,q (0) = 1 = 4, then the inverse function g(z) of q(z) exists, then 1 z X zn 1 1 ¼ gðzÞ ¼ ln 1 þ ð1Þn1 n ; jzj < 1; s ¼ ¼ : 4 1 4 n4 n¼1
The characteristic equation
a2 ðn þ gÞa þ ng ¼ a2 a 6 ¼ 0 has two roots a1 =2 = g,a2 = 3 = n and —a1— > 1,—a2— > 1. For a1 =2 = g, by means of Theorem 3.1, the auxiliary equation
gð/ð4zÞ pð/ð2zÞÞÞ ¼ hðgð/ð2zÞ pð/ðzÞÞÞÞ; z 2 C; has an analytic solution /(z) in a neighborhood of the origin such that /(0) = 0 and /0 (0) = s – 0. Let
/ðzÞ ¼
1 X
bn zn ; b1 ¼ s;
n¼1
then s1 ¼ b1 ða nÞ ¼ 5s – 0; u1 ¼ b1 sða nÞ ¼ 54 s – 0, then from (3.10), we have
2 1 5 25 s ¼ s2 ; 6 4 96 ð2Þ2 ð2Þ l1 þl2 ¼2 ! X X 1 1 2 2u1 u2 u31 ¼ 0; c2 ul1 ul2 þ c3 ul1 ul2 ul3 ¼ u3 ¼ 3 6 3! ð2Þ ð2Þ l1 þl2 ¼3 l1 þl2 þl3 ¼3
u2 ¼
X
1
c2 ul1 ul2 ¼
etc. From (3.13) we see
X 1 25 1 25 2 ð5sÞ2 ¼ g 2 sl1 sl2 ¼ s2 s; s2 ¼ u2 4 96 48 2 42 l þl ¼2 1
2
then s2 ¼ 25 s2 . From (3.13) 12
1 s3 ¼ u3 4
X l1 þl2 ¼3
g 2 sl1 sl2 þ
X l1 þl2 þl3 ¼3
! g 3 sl1 sl2 sl3
¼
1 1 2s1 s2 ð5sÞ3 ¼ 0; 32 3 43
so s3 = 0, . . .. From (3.21) we have
/00 ð0Þ 1 25 2 61 2 ¼ s þ 3s2 ¼ s2 ; ðs2 þ a2 b1 Þ ¼ 2! 12 12 ð2Þ2 3 /000 ð0Þ 1 1 61 67 3 ¼ 6 s3 þ 3s3 ¼ s3 ðs3 þ 2a2 b1 b2 þ a3 b1 Þ ¼ b3 ¼ 3 3! 11 12 22 ð2Þ 3
b2 ¼
:
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Because /1(z) is analytic in a neighborhood of the point /(0) = 0, therefore, we can calculate
ð/1 Þ0 ð0Þ ¼
1 /0 ð/1 ð0ÞÞ
ð/1 Þ00 ð0Þ ¼
1 1 ¼ ; /0 ð0Þ s
/00 ð/1 ð0ÞÞð/1 Þ0 ð0Þ 0
n ð/1 Þ000 ð0Þ ¼ þ
2
1
ð/ ð/ ð0ÞÞÞ
¼
/00 ð0Þð/1 Þ0 ð0Þ 0
ð/ ð0ÞÞ
2
¼
61 6
s2 1s 61 ¼ ; 6s s2
o / ð/ ð0ÞÞðð/ Þ ð0ÞÞ þ /00 ð/1 ð0ÞÞð/1 Þ00 ð0Þ ð/0 ð/1 ð0ÞÞÞ2 000
1 0
1
2
ð/0 ð/1 ð0ÞÞÞ4 / ð/ ð0ÞÞð/ Þ ð0Þ 2/0 ð/1 ð0ÞÞ/00 ð/1 ð0ÞÞð/1 Þ0 ð0Þ 00
¼
¼
1
1 0
ð/0 ð/1 ð0ÞÞÞ4
½/000 ð0Þðð/1 Þ0 ð0ÞÞ2 þ /00 ð0Þð/1 Þ00 ð0Þ ð/0 ð0ÞÞ2 ð/0 ð0ÞÞ4 / ð0Þð/1 Þ0 ð0Þ 2/0 ð0Þ/00 ð0Þð/1 Þ0 ð0Þ 00
þ
¼
ð/0 ð0ÞÞ4 201 3 1 61 2 61 2 61 2 1 11 s s2 þ 6 s 6s s þ 6 s s 2s 61 s2 1s 6
s4
43343 ¼ ;: 132s Finally, we determine a solution f(z) of (4.6) at z = 0. Because
f ðzÞ ¼ gðuðzÞ pðzÞÞ ¼ gð/ð2/1 ðzÞÞ pðzÞÞ; then
f ð0Þ ¼ gð/ð0Þ pð0ÞÞ ¼ gð0Þ ¼ 0; f 0 ð0Þ ¼ g 0 ð/ð2/1 ð0Þ pð0ÞÞ½2/0 ð2/1 ð0ÞÞð/1 Þ0 ð0Þ p0 ð0ÞÞ 1 1 5 ; ¼ g 0 ð0Þ½2/0ð0Þð/1 Þ0 ð0Þ p0 ð0Þ ¼ ð2s 3Þ ¼ 32 32 s 2 1 0 1 1 0 00 00 0 f ð0Þ ¼ g ð/ð2/ ð0Þ pð0ÞÞ½2/ ð2/ ð0ÞÞð/ Þ ð0Þ p ð0ÞÞ þ g 0 ð/ð2/1 ð0Þ pð0ÞÞ½4/00 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ2 2/0 ð2/1 ð0ÞÞð/1 Þ00 ð0Þ p00 ð0ÞÞ ¼ g 00 ð0Þ½2/0 ð0Þð/1 Þ0 ð0Þ p0 ð0Þ2 þ g 0 ð0Þ½4/00 ð0Þðð/1 Þ0 ð0ÞÞ2 2/0 ð0Þð/1 Þ00 ð0Þ p00 ð0Þ 1 1 1 61 2 1 61 6 s 2 2s ¼ ð2s 3Þ2 þ ½4 16 4 6 6s s s 195 ; ¼ 16 000 000 f ð0Þ ¼ g ð/ð2/1 ð0Þ pð0ÞÞ½2/0 ð2/1 ð0ÞÞð/1 Þ0 ð0Þ p0 ð0ÞÞ3 þ g 00 ð/ð2/1 ð0Þ pð0ÞÞ 2½2/0 ð2/1 ð0ÞÞð/1 Þ0 ð0Þ p0 ð0ÞÞ ½4/00 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ2 2/0 ð2/1 ð0ÞÞð/1 Þ00 ð0Þ p00 ð0Þ þ g 00 ð/ð2/1 ð0Þ pð0ÞÞ½2/0 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ p0 ð0ÞÞ ½4/00 ð0Þðð/1 Þ0 ð0ÞÞ2 2/0 ð2/1 ð0ÞÞð/1 Þ00 ð0Þ p00 ð0Þ þ g 0 ð/ð2/1 ð0Þ pð0ÞÞ½8/000 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ3 þ 4/00 ð2/1 ð0ÞÞ 2ð/1 Þ0 ð0Þð/1 Þ00 ð0Þ þ 4/00 ð2/1 ð0ÞÞð/1 Þ0 ð0Þð/1 Þ00 ð0Þ 2/0 ð2/1 ð0ÞÞð/1 Þ000 ð0Þ p000 ð0ÞÞ ¼ g 000 ð0Þ½2/0 ð0Þð/1 Þ0 ð0Þ p0 ð0Þ3 þ 3g 00 ð0Þ½2/0 ð0Þð/1 Þ0 ð0Þ p0 ð0Þ ½4/00 ð0Þðð/1 Þ0 ð0ÞÞ2 2/0 ð0Þð/1 Þ00 ð0Þ p00 ð0Þ þ g 0 ð0Þ½8/000 ð0Þðð/1 Þ0 ð0ÞÞ3 þ 12/00 ð0Þð/1 Þ0 ð0Þð/1 Þ00 ð0Þ 2/0 ð0Þð/1 Þ000 ð0Þ p000 ð0Þ
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7257
1 1 61s2 1 61 6 ð2 3Þ3 þ 3 2 ð2 3Þ 4 2 2s 32 6s 6 s 4 2 1 201 3 1 61s 1 61 43343 þ 8 2s s 3 þ 12 18 4 11 132s s 6 s 6s 136135 ;...: ¼ 352 ¼
Thus, the desired solution is
f ðzÞ ¼
5 195 2 136135 3 zþ z þ z þ : 32 32 2112
For a = 3 = n, by means of the same method, we can obtain
f ðzÞ ¼ 0: Remark that the analytic solutions of (1.3) in the cases a = n = g and a = g,a – n,—n— = 1 in (H2), and the cases a = g,a – n and a = n = g in (H3) remain still open. Acknowledgment I thank Professor Jianguo Si and Professor Weinian Zhang for valuable discussions during the preparation of this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
P.M. Diamond, Analytic invariants of mappings of two variables, J. Math. Anal. Appl. 27 (1969) 601–608. M. Kuczma, B. Choczewski, R. Ger, Iterative functional equations, Encycl. Math. Appl., vol. 32, Cambridge University Press, Cambridge, 1990. Fang Qiu, Analytic solutions of an iterative functional equation, Ann. Diff. Equ. 23 (2007) 180–187. Wen Rong Li, Sui Sun Cheng, Analytic solutions of an iterative functional equation, Aequationes Math. 68 (2004) 21–27. V. Mityushev, Functional equations in a class of analytic functions and composite materials, Demonstratio Math. 30 (1997) 63–70. Bing Xu, Weinian Zhang, Analytic solutions of a general nonlinear functional equations near resonance, J. Math. Anal. Appl. 317 (2006) 620–633. A. Járai, On analytic solutions of functional equations, Ann. Univ. Sci. Budapest. Sect. Comput. 14 (1994) 71–77. H.W. Braden, The general analytic solution of a functional equation of addition type, SIAM J. Math. Anal. 28 (1997) 903–923. E.N. Petropoulou, Analytic solutions of a class of linear and nonlinear functional equations, J. Math. Anal. Appl. 261 (2001) 168–176. Lingxia Liu, Local analytic solutions of a functional equation, Appl. Math. Comput. 215 (2009) 644–652. Si Jianguo, On analytic solutions of the equation of invariant curves, C. R. Math. Rep. Acad. Sci. Canada XVII (1995) 49–52. Jianguo Si, Weinian Zhang, Analytic solution of a function for invariant curves, J. Math. Anal. Appl. 259 (2001) 83–93. Sui Sun Cheng, Wenrong Li, Analytic Solutions of Functional Equations, World Scientific, Publishing Co. Pte. Ltd., 2008. A.D. Brjuno, Analytic form of differential equations, Trans. Moscow Math. Soc. 25 (1971) 131–288. S. Marmi, P. Moussa, J.-C. Yoccoz, The Brjuno functions and their regularity properties, Comm. Math. Phys. 186 (1997) 265–293. M.S. Berger, Nonlinearity and Functional Analysis, Academic Press, New York, 1977. C.L. Siegel, Vorlesungenüber Himmelsmechanik, Springer, Verlag, Berlin, 1956. T. Carletti, S. Marmi, Linearization of analytic and non-analytic germs of diffeomorphisms of ðC0 Þ, Bull. Soc. Math. France 128 (2000) 69–85. A.M. Davie, The critical function for the semistandard map, Nonlinearity 7 (1994) 219–229. D. Bessis, S. Marmi, G. Turchetti, On the singularities of divergent majorant series arising from normal form theory, Rend. Mat. Ser. VII 9 (1989) 645– 659.
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Existence of analytic solutions of an iterative functional equation q Lingxia Liu Department of Mathematics, Weifang University, Weifang, Shandong 261061, PR China
a r t i c l e
i n f o
Keywords: Iterative functional equation Analytic solution Auxiliary equation Convergent power series
a b s t r a c t This paper is concern analytic solutions of an iterative functional equation of the form
f ðpðzÞ þ qðf ðzÞÞÞ ¼ hðf ðzÞÞ; z 2 C: Byconstructing a convergent power series solution of an auxiliary equation
gð/ða2 zÞ pð/ðazÞÞÞ ¼ hðgð/ðazÞ pð/ðzÞÞÞÞ; z 2 C; analytic solutions of the original equation are obtained. We discuss not only these a appeared in the auxiliary equation at the hyperbolic case 0 < jaj – 1 and resonance, i.e., at a root of the unity, but also those a near resonance (i.e., near a root of the unity) under Brjuno condition. Ó 2011 Elsevier Inc. All rights reserved.
1. Introduction Invariant curves of the area preserving maps play an important role in the theory of periodic stability of discrete dynamical systems. A common and useful method to understand behaviors of a discrete dynamical system generated by the iteration of a self-mapping is to find a simple structures in its phase space and to describe the dynamics in terms of the effect caused by the presence of these structures. In particular, invariant curve, in dimension two, is one of such structures. Diamond [1] researched the existence of analytic invariant curves for two-dimensional maps of the form
Tðx; yÞ ¼ ðx þ y; yð1 þ bxk Þ þ Fðx; yÞÞ: For an investigation of other related problems the interested reader is referred to [2–12] and the monograph [13]. In particular, Wen Rong Li and Sui Sun Cheng [4] discussed the analytic solutions of functional equation
f ðpðzÞ þ bf ðzÞÞ ¼ hðzÞ
ð1:1Þ
by using the method of majorant series. The purpose of this paper is to find invertible and linearizable invariant analytic curves of the 2-D complex map T : C2 ! C2 ; ðz; wÞ # ðz1 ; w1 Þ, defined by
z1 ¼ pðzÞ þ qðwÞ; w1 ¼ hðwÞ:
ð1:2Þ
Throughout this paper, we assume that p(z), q(z) and h(z) are analytic in a neighborhood of the origin, and p(0) = 0,h(0) = 0,q(0) = 0,p0 (0) = n – 0,h0 (0) = g – 0 and q0 (0) = f – 0. Clearly, map T has a fixed point O = (0,0) with the Jacobian matrix
q
Supported by the Natural Science Foundation of Shandong Province (2006ZRB01066). E-mail address: [email protected]
0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.02.015
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L. Liu / Applied Mathematics and Computation 217 (2011) 7245–7257
A ¼ DTð0Þ ¼
n
f
0 g
at O. Its characteristic polynomial is
PA ðkÞ ¼ a2 ðn þ gÞa þ ng: As is well known that a curve w = f(z) is said to be an invariant curve of T if w1 = f(z1). Thus, we see that map T has an invariant curve w = f(z) if and only if f satisfies the functional equation
f ðpðzÞ þ qðf ðzÞÞÞ ¼ hðf ðzÞÞ; z 2 C:
ð1:3Þ
The results of this paper can be regarded as a generalization of the results obtained in [4]. From the above assumptions, it is easy to see that the inverse function of q(z) exists and is analytic in a neighborhood of the q(0) = 0. We denote the inverse function of q(z) by g(z) and consider the following equation
gð/ða2 zÞ pð/ðazÞÞÞ ¼ hðgð/ðazÞ pð/ðzÞÞÞÞ; z 2 C;
ð1:4Þ
which is called the auxiliary equation of (1.3). We first construct analytic solutions of (1.4) in the cases: (H1) 0 < jaj – 1. P log qkþ1 (H2) a = e2pih, where h 2 R n Q is a Brjuno number ([14] and [15]), i.e., BðhÞ ¼ 1 < 1, where {pk/qk} denotes the k¼0 qk sequence of partial fraction of the continued fraction expansion of h, said to satisfy the Brjuno condition. (H3) a = e2pi q/p for some integers p 2 N with p P 2 and q 2 Z n f0g, and a – e2pil/k for all 1 6 k 6 p 1 and l 2 Z n f0g. Observe that a is the inside or outside of the unit circle S1 in the case of (H1) but on S1 in the rest cases. More difficulties are encountered for a on S1, as mentioned in the so-called ‘‘small-divisor problem’’ (seen in [16] p. 22 and p. 146 and [17]). Under Diophantine condition: a = e2pih, where h 2 R n Q and there exist constants f > 0 and r > 0 such that jan 1j P f1nr for all n P 1, the number a 2 S1 is ‘‘far’’ from all roots of the unity and was considered in different settings [10–12]. In [10] the case of (H3), where a is a root of the unity, was also discussed for a general class of iterative equations. Since then, we have been striving to give a result of analytic solutions for those a ‘‘near’’ a root of the unity, i.e., neither being roots of the unity nor satisfying the Diophantine condition. The Brjuno condition in (H2) provides such a chance for us. As stated in [18], for a real number h, we let [h] denote its integer part and {h} = h [h] its fractional part. Then every irrational number h has a unique expression of the Gauss’ continued fraction
h ¼ a0 þ h0 ¼ a0 þ
1 ¼ ; a1 þ h1
denoted simply by h = [a0,a1, . . ., an, . . .], where aj’s and hj’s are calculated by the algorithm: (a) a0 = [h], h0 = {h}, and (b) 1 1 an ¼ ½hn1 ; hn ¼ fhn1 g for all n P 1. Define the sequences ðpn Þn2N and ðqn Þn2N as follows:
q2 ¼ 1; q1 ¼ 0; qn ¼ an qn1 þ qn2 ; p2 ¼ 0; p1 ¼ 1; pn ¼ an pn1 þ pn2 : It is easy to show that pn/qn = [a0,a1, . . ., an]. Thus, For every h 2 R n Q we associate, using its convergence, an arithmetical P log q function BðhÞ ¼ nP0 qnnþ1 . We say that h is a Brjuno number or that it satisfies Brjuno condition if B(h) <+1. The Brjuno condition is weaker than the Diophantine condition. For example, if anþ1 6 cean for all n P 0, where c > 0 is a constant, then h = [a0,a1, . . ., an, . . .] is a Brjuno number but is not a Diophantine number. So, the case (H2) contains both Diophantine condition and a part of a ‘‘near’’ resonance. In this paper, considering the Brjuno condition instead of the Diophantine one, we discuss not only the cases (H1) and (H3) but also (H2) for analytic invariant curves of the mapping T defined in (1.2).
2. Substituting power series in Eq. (1.4) Let
pðzÞ ¼
1 X n¼1
an zn ;
hðzÞ ¼
1 X
c n zn ;
gðzÞ ¼
n¼1
1 X n¼1
g n zn ;
a1 ¼ n; c1 ¼ g; g 1 ¼ s ¼
1 : f
ð2:1Þ
Without loss of generality , we can assume that
jan j 6 1; jcn j 6 1; jg n j 6 1; n ¼ 2; 3; . . . :
ð2:2Þ
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In fact, since the series 1 X
nþ
an zn1 ;
gþ
n¼2
1 X
cn zn1 ;
sþ
1 X
n¼2
g n zn1
n¼2
are uniformly convergent in a neighborhood of O, there exists a constant q > 0 such that
jan j 6 qn1 ; jcn j 6 qn1 ; jg n j 6 qn1 ; n ¼ 2; 3; . . . :
ð2:3Þ
Introducing new functions
~ ~ðzÞ ¼ qpðq1 zÞ; /ðzÞ ¼ q/ðq1 zÞ; p ~ hðzÞ ¼ qhðq1 zÞ; g~ðzÞ ¼ qgðq1 zÞ; ~ we see that /ðzÞ satisfies
~ g~ð/ð ~ azÞÞÞ ¼ hð ~ azÞ p ~ ~ a2 zÞ p ~ð/ð ~ð/ðzÞÞÞÞ; g~ð/ð which is the same form of (1.4). From (2.3) and the following expansions 1 X
~ðzÞ ¼ qpðq1 zÞ ¼ nz þ p ~ hðzÞ ¼ qhðq1 zÞ ¼ gz þ
n¼2 1 X
an q1n zn ; cn q1n zn ;
n¼2
and
g~ðzÞ ¼ qgðq1 zÞ ¼ sz þ
1 X
g n q1n zn ;
n¼2
we get janq1nj 6 1, jcnq1nj 6 1, jgnq1nj 6 1, n = 2, 3, . . .. Let 1 X
/ðzÞ ¼
bn zn
ð2:4Þ
n¼1
be the expansion of a formal solution of (1.4). Substituting to /, p, h, g, their power series (2.4) and (2.1) respectively in (1.4), we have
1 3n 9n > > > > 6 > C 7 B > 7 6 = 1 1 C B X X X 7 6 C B n n7 6 cn gn6 at bl1 bl2 . . . blt Cz 7 Bbn a > C 7 > > > 6 n¼1 B n¼1 n¼1 > A 5 > @ l1 þ l2 þ þ lt ¼ n > > 4 > > > > ; : t ¼ 1; 2; . . . ; n 3n 2 1 0 8 > > > > > >
2
0
7 6 C B 7 6X C X 7 61 B C B n n n7 ¼ gn6 b a b b . . . b a z a B n t l1 l2 lt C 7 : 6 C B 7 6 n¼1 @ n¼1 A l1 þ l2 þ þ lt ¼ n 5 4 t ¼ 1; 2; . . . ; n 1 X
Let
X
sn ¼ bn an
at bl1 bl2 . . . blt ;
un ¼
X
l1 þ l2 þ þ lt ¼ n
l1 þ l2 þ þ lt ¼ n
t ¼ 1; 2 . . . ; n
t ¼ 1; 2; . . . ; n
g t sl1 sl2 . . . slt ; ð2:5Þ
then we obtain 1 X n¼1
un an zn ¼
1 X
X
n¼1
l1 þ l2 þ þ lt ¼ n t ¼ 1; 2; . . . ; n
ct ul1 ul2 . . . ult zn :
ð2:6Þ
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Comparing coefficients we obtain
8 ða gÞu1 ¼ 0; > > > P < ðan gÞu ¼ ct ul1 ul2 . . . ult ; n ¼ 2; 3; . . . ; n l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð2:7Þ
From (2.5), it follows that s1 = b1(a n) and u1 = b1s(a n). And then from the first equation of (2.7) we have b1s(a n)(a g) = 0. In view of s – 0 and (a n)(a g) = 0, we can choose b1 = s – 0. 3. Auxiliary equation in case (H1) In this section, we discuss Eq. (1.4) in the case 0 < jaj – 1. Theorem 3.1. Assume that a 2 (H1), then for any s 2 C, the auxiliary Eq. (1.4) has an analytic solution /(z) in a neighborhood of the origin such that /(0) = 0 and /0 (0) = s. Proof. If a – g, then a = n. Thus, we have s1 = 0 and u1 = 0. From (2.5) and (2.7) it is easily seen that sn = 0 and un = 0 for all n P 1. Consequently,
X
bn an ¼ nbn þ
at bl1 bl2 . . . blt ;
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n that is,
8 ða nÞb1 ¼ 0; > > > n P < at bl1 bl2 . . . blt ; n ¼ 2; 3; . . . : ða nÞbn ¼ l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð3:1Þ
In view of a = n and the condition (H1), the second equality of (3.1) can be written as
bn ¼
X
1
an a
at bl1 bl2 . . . blt ; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n
ð3:2Þ
Then for b1 = s – 0, we can uniquely determine the sequence fbn g1 n¼2 by (3.2), recursively. In what follows we prove the convergence of series (2.4) in a neighborhood of the origin. Since
lim
n!1
1 ¼ an a
(
a1 ; 0 < jaj < 1; 0;
jaj > 1;
there exists a positive number M such that
1 an a 6 M;
n P 2:
ð3:3Þ
From (2.2), (3.2) and (3.3), we see
X
jbn j 6 M
jbl1 j:jbl2 j jblt j; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n
ð3:4Þ
t ¼ 2; 3; . . . ; n To construct a governing series, we consider the implicit functional equation
ðM þ 1ÞðGðzÞÞ2 ðjsjz þ 1ÞGðzÞ þ jsjz ¼ 0; i.e.,
GðzÞ ¼ jsjz þ M
ðGðzÞÞ2 : 1 GðzÞ
ð3:5Þ
Define the function
RðM; s; z; GÞ ¼ ðM þ 1ÞG2 ðjsjz þ 1ÞG þ jsjz
ð3:6Þ
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for (z,G) in a neighborhood of (0,0). Then R(M,s,0,0) = 0, R0G ðM; s; 0; 0Þ ¼ 1 – 0. Thus, there exists a unique function G(M,s,z), 0 s;0;0Þ analytic in a neighborhood of zero, such that G(M,s,0) = 0, G0 ðM; s; 0Þ ¼ RR0z ðM; ¼ jsj and R(M,s,z,G(M,s,z)) = 0. G(M,s,z) can G ðM;s;0;0Þ be expanded into a convergent series
GðM; s; zÞ ¼
1 X
An z n :
ð3:7Þ
n¼1
Replacing (3.7) into (3.5) and comparing coefficients, we obtain that
8 A1 ¼ jsj; > > > P < Al1 Al2 . . . ; Alt ; n ¼ 2; 3; . . . : An ¼ M þ l þ þ l ¼ n l > 1 2 t > > : t ¼ 2; 3; . . . ; n
ð3:8Þ
Furthermore,
jbn j 6 An ; n ¼ 1; 2; . . . :
ð3:9Þ
In fact, jb1j = jsj = A1. For inductive proof we assume that jbjj 6 Aj, j 6 n 1. Observe that in (3.4), jblj j 6 Alj ; j ¼ 1; 2; . . . ; k, because 1 6 l1, . . ., lk 6 n 1. From (3.8) we know jbnj 6 An and (3.9) is proved. By the convergence of (3.7) and the inequality (3.9), we see that the series (2.4) converges uniformly in a neighborhood of the origin. If a – n, then a = g. It is easy to see that s1 = b1(a n) – 0 and u1 = b1s(a n) – 0. The second equality of (2.7) can be written as
un ¼
X
1
an a
ct ul1 ul2 . . . ult ; n ¼ 2; 3; . . . : ð3:10Þ
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n
As the proof of convergence of the power series (2.4), we can show that power series
UðzÞ ¼
1 X
un zn
ð3:11Þ
n¼1
converges in a neighborhood of the origin. Thus, there is a 0 < q1 < 1 such that
jun j 6 qn1 :
ð3:12Þ
Now we show the convergence of the power series From (2.5), we get
P1
n¼1 sn z
n
.
8 s1 ¼ ða nÞs; > > > P < g t sl1 sl2 . . . slt ; n ¼ 2; 3; . . . : ssn ¼ un l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð3:13Þ
For given un and s1 = (a n)s – 0, we can uniquely determine the sequence fsn g1 n¼2 by (3.13), recursively, and
1
0
C B C X 1B C B g t sl1 sl2 . . . slt C; n ¼ 2; 3; . . . : sn ¼ Bun C sB A @ l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n In view of (3.12) and (2.2) and f ¼ 1s – 0,
0
1
C B C B X C B n jsl1 j:jsl2 j . . . jslt jC; jsn j 6 jfjBq1 þ C B A @ l1 þ l2 þ þ lt ¼ n
n ¼ 2; 3; . . . :
t ¼ 2; 3; . . . n To construct a governing series we consider the implicit functional equation
fq21 z2 fq21 z2 ð1 þ fÞðWðzÞÞ2 1 þ þ jða nÞsjz WðzÞ þ jða nÞsjz þ ¼ 0; 1 q1 z 1 q1 z
ð3:14Þ
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i.e.,
WðzÞ ¼ jða nÞsjz þ
fq21 z2 ðWðzÞÞ2 þf : 1 q1 z 1 WðzÞ
ð3:15Þ
Now let us define the function
2 2 2 2 e WÞ ¼ ð1 þ fÞW 2 1 þ fq1 z þ jða nÞsjz W þ jða nÞsjz þ fq1 z Rðz; 1 q1 z 1 q1 z e e 0 ð0; 0Þ ¼ 1 – 0, there exists a unique function W(z), anafor (z, W) in a neighborhood of the zero. Since Rð0; 0Þ ¼ 0 and R W eR 0z ð0; 0Þ 0 ¼j ða nÞs j – 0. So W(z) can be expanded into a conlytic in a neighborhood of zero, such that Wð0Þ ¼ 0; W ð0Þ ¼ 0 eR W ð0; 0Þ vergent power series
WðzÞ ¼
1 X
Bn zn
ð3:16Þ
n¼1
uniformly in a neighborhood of zero. Replacing (3.16) into (3.15) and comparing coefficients, we obtain
8 B1 ¼ jða nÞsj; > > 1 0 > > > > > < C B C B P C B n > ; n ¼ 2; 3; . . . B ¼ f q þ B . . . B B B n l l l > tC 1 1 2 > C B > > l1 þ l2 þ þ lt ¼ n A @ > > : t ¼ 2; 3; . . . n then by induction and using (3.14) it is not difficult to show that
jsn j 6 Bn ; n ¼ 1; 2; . . . :
ð3:17Þ
By the convergence of (3.16) and the inequality (3.17) we see that the series
SðxÞ ¼
1 X
sn zn
ð3:18Þ
n¼1
converges uniformly in a neighborhood of the origin. Thus, there is a 0 < q2 < 1 such that
jsn j 6 qn2 :
ð3:19Þ
Now we prove the convergence of the power series (2.4) in the case a – n,a = g. From (2.5) we see
8 ða nÞb1 ¼ s1 ; > > > P < n at bl1 bl2 . . . blt ; n ¼ 2; 3; . . . : ða nÞbn ¼ sn þ l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . n
ð3:20Þ
Let b1 = s – 0, from the second function of (3.20) we have
1
0
C B C X 1 B C B at bl1 bl2 . . . blt C; n ¼ 2; 3; . . . ; bn ¼ n Bsn þ C B a n@ A l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n we can uniquely determine the sequence fbn g1 n¼2 by (3.21), recursively. Note that 0 < jaj – 1 and
1 lim ¼ n!þ1 an n
(
1n ; 0 < jaj < 1; 0;
jaj > 1;
ð3:21Þ
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so there exists a positive number N, such that j an1n j 6 N for n P 2. Then from (2.2) and (3.21)
1
0
C B C B C B X C B jbl1 j:jbl2 j . . . jblt jC jbn j 6 NBjsn j þ C B C B l1 þ l2 þ þ lt ¼ n A @ t ¼ 2; 3; . . . n 1
0
C B C B C B X C B n 6 NBq2 þ jbl1 j:jbl2 j . . . jblt jC; n ¼ 2; 3; . . . ; C B C B l1 þ l2 þ þ lt ¼ n A @ t ¼ 2; 3; . . . n which is a similar form with (3.14). Similarly, we can show that the power series (2.4) converges in a neighborhood of the origin. Finally, if a = n = g, we obtain s1 = u1 = 0. From (2.7) we deduce that u2 = 0 (note that a2 a – 0), and also s2 = 0 according to the definition of sn given in (2.5). Similar, it can be proven that sn = un = 0 for all n P 1. At this point we can reproduce the proof of case a = n, a – g, obtaining again the convergence of series (2.4). h 4. The auxiliary equation in case (H2) and (H3) In this section, we discuss local invertible analytic solutions of auxiliary Eq. (1.4) in cases (H2) and (H3). In order to discuss the existence of analytic solutions of the auxiliary Eq. (1.4) under (H2), we need to introduce Davie’s Lemma. Let h 2 R n Q and ðqn Þn2N be the sequence of partial denominators of the Gauss’ continued fraction for h as above. As in [18], let
Ak ¼
q 1 q ; Ek ¼ max qk ; kþ1 ; gk ¼ k : n P 0jknhk 6 8qk 4 Ek
Let Ak be the set of integers j P 0 such that either j 2 Ak or for some j1 and j2 in Ak, with j2 j1 < Ek, one has j1 < j < j2 and qk divides j j1. For any integer n P 0, define
n lk ðnÞ ¼ max ð1 þ gk Þ 2; qk
ðmn gk þ nÞ
1 1 ; qk
where mn ¼ maxfjj0 6 j 6 n; j 2 Ak g. We then define the function hk : N ! Rþ as follows:
(m
hk ðnÞ ¼
n þgk n
qk
lk ðnÞ;
1; if mn þ qk 2 Ak ; if mn þ qk R Ak :
Let g k ðnÞ :¼ maxðhk ðnÞ; ½qn Þ, and define k(n) by the condition qk(n) 6 n 6 qk(n)+1. Clearly, k(n) is non-decreasing.Moreover,the k function gk is non-negative. Then we are able to state the following result. Lemma 4.1 (Davie’s lemma [19]). Let KðnÞ ¼ n log 2 þ
PkðnÞ j¼0
g j ðnÞ logð2qjþ1 Þ. Then
(a) there is a universal constant c > 0 (independent of n and h) such that
KðnÞ 6 n
! kðnÞ X log qjþ1 þc ; qj j¼0
(b) K(n1) + K(n2) 6 K(n1 + n2) for all n1 and n2, and (c) logjan 1j 6 K(n) K(n 1).
Theorem 4.1. Assume that a = n or a = g but jnj – 1 and (H2) holds, then for any s 2 C, the auxiliary Eq. (1.4) has an analytic solution /(z) in a neighborhood of the origin such that /(0) = 0 and /0 (0) = s. Proof. As in the proof of Theorem 3.1, we seek a solution of (1.4) in power series of the form (2.4). If s = 0, (1.4) has a trivial solution f(z) 0. Assume s – 0 and b1 = s, by using the same arguments as in Theorem 3.1, we can uniquely determine the sequence fbn g1 n¼2 by (3.2) or (3.21), recursively.
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If a = n, then a – g (since a = g forces jnj – 1 and jaj = 1). From (2.5) and (2.7), we obtain sn = 0,un = 0,n = 1, 2, . . ., and
X
ðan aÞbn ¼
at bl1 bl2 . . . blt :
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n
ð4:1Þ
In view of jaj ¼ 1; h 2 R n Q and (2.2), we get
jbn j 6
1 ja 1j
X
n1
jbl1 j:jbl2 j . . . jblt j; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n
ð4:2Þ
t ¼ 2; 3; . . . n To construct a governing series, we consider the following function
VðzÞ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ jsjz s2 z2 6jsjz þ 1 4
ð4:3Þ
which clearly satisfies the equality
ðVðzÞÞ2 : 1 VðzÞ
VðzÞ ¼ jsjz þ
ð4:4Þ
Define the function
Rð1; s; z; VÞ ¼ 0
ð4:5Þ
for (z,V) in a neighborhood of (0,0), where R is defined in (3.6). Similarly to the proof of Theorem 3.1, we can prove that (4.5) 0 s;0;0Þ has a unique analytic solution V(s,z) in a neighborhood of zero, such that V(s,0) = 0, V 0z ðs; 0Þ ¼ RR0z ð1; ¼ jsj and V ð1;s;0;0Þ R(1,s,z,V(s,z)) = 0. So V(s,z) can be expanded into a convergent power series
Vðs; zÞ ¼
1 X
C n zn ; C 1 ¼ jsj:
ð4:6Þ
n¼1
Replacing (4.6) into (4.4) and comparing coefficients as in (3.8) with M = 1, we have
8 C 1 ¼ jsj; > > > P < C l1 C l2 . . . C lt ; n ¼ 2; 3 . . . : Cn ¼ þ l þ þ l ¼ n l > 1 2 t > > : t ¼ 2; 3; . . . n
ð4:7Þ
Note that the series (4.6) converges in a neighborhood of the origin. Hence, there is a constant T > 0 such that
C n 6 T n ; n ¼ 1; 2; . . . :
ð4:8Þ
Now by induction on n we prove
jbn j 6 C n eKðn1Þ ; n ¼ 1; 2; . . . ; where K : N ! R is defined in Lemma 4.1. In fact jb1j = jsj = C1, we assume that jbjj 6 CjeK(j1),j 6 n 1. From Lemma 4.1 and (4.2) we obtain
jbn j 6
1 jan1 1j
X
C l1 :C l2 . . . C lt eKðl1 1ÞþþKðlt 1Þ
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n X 1 C l1 :C l2 . . . C lt 6 n1 eKðn2Þ ja 1j l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . n eKðn2Þ ¼ n1 Cn: ja 1j Note that
Kðl1 1Þ þ Kðl2 1Þ þ þ Kðlt 1Þ 6 Kðn 2Þ 6 Kðn 1Þ þ log jan1 1j; then
jbn j 6 C n eKðn1Þ ;
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as we claimed. Moreover, according to (4.8) we have jbnj 6 TneK(n1). Note that K(n) 6 n(B(h) + c) for some universal constant c > 0, then
jbn j 6 T n eðn1ÞðBðhÞþcÞ; that is,
1 n1 lim sup jbn jn 6 lim sup Te n ðBðhÞþcÞ ¼ TeBðhÞþc :
n!1
n!1
This implies that the convergence radius of the series (2.4) is at least (TeB(h)+c)1. If a – n, then a = g. It follows that s1 = b1(a n) – 0 and u1 = b1s(a n) – 0. Since jaj = 1, from (3.10) and (2.2), we have
jun j 6
X
1 jan1 1j
jul1 jjul2 j . . . jult j; n ¼ 2; 3; . . . :
l1 þ l2 þ þ lt ¼ n
ð4:9Þ
t ¼ 2; 3; . . . n Using the same majorization as in (4.2), we can also prove the power series (3.11) converges uniformly in the neighborhood P n of the origin. As the proof of convergence of the series (3.18), the power series 1 n¼1 sn z also converges in a neighborhood of n the origin, so there is a 0 < q2 < 1 such that jsn j 6 q2 for all n P 1. Now we prove the series (2.4) is convergent in a neighborhood of the origin. From (3.21), we get
1
0
C B C B X C B at bl1 :bl2 . . . blt C; n ¼ 2; 3; . . . : bn ¼ ðan nÞ1 Bsn þ C B A @ l1 þ l2 þ þ lt ¼ n t ¼ 2; 3; . . . ; n By jnj – 1, a – n and a = g and jan n j P kajn jnk = j1 jnk, it see that
1
0
C B C B X C B jbl1 j:jbl2 j . . . jblt jC; n ¼ 2; 3; . . . : jbn j 6 j1 jnk1 Bqn2 þ C B A @ l1 þ l2 þ þ lt ¼ n
ð4:10Þ
t ¼ 2; 3; . . . n Similar to proof of convergence of the series (3.18), we can show that the power series (2.4) is convergent in a neighborhood of the origin. This completes the proof of Theorem 4.1. h In the case (H3), where a = e2pih with h 2 R n Q a Brjuno number, the constant a is not only on the unit circle in C but also a root of the unity. In this case both the Diophantine condition and Brjuno condition are not satisfied. The difficulty can be overcome by an idea acquired from [20]. Let fDn g1 n¼1 be a sequence defined by
8 D1 ¼ jsj; > > > P < Dl1 :Dl2 . . . Dlt ; n ¼ 2; 3; . . . Dn ¼ C l1 þ l2 þ þ lt ¼ n > > > : t ¼ 2; 3; . . . ; n
ð4:11Þ
where C = max{1,jai 1j1,i = 1,2, . . ., p 1}. Theorem 4.2. Suppose that aj – g,j = 1, 2, . . ., p and (H3) holds. Then fbn g1 n¼1 is determined recursively, by b1 = s and
ðan aÞbn ¼ Vðn; aÞ; n ¼ 2; 3; . . . ;
ð4:12Þ
where
Vðn; aÞ ¼
X
at bl1 :bl2 . . . blt :
l1 þ l2 þ þ lt ¼ n t ¼ 2; 3 . . . n If V(mp + 1,a) = 0 for all m = 1, 2, . . ., then Eq. (1.4) has an analytic solution /(z) in a neighborhood of the origin such that / (0) = 0,/0 (0) = s, and /(mp+1)(0) = (mp + 1)!Tmp+1, where all Tmp+1’s are arbitrary constants satisfying the inequality jTmp+1j 6 Dmp+1 and the sequence fDn g1 n¼1 is defined in (4.11). Otherwise, if V(mp + 1,a) – 0 for some m = 1,2, . . ., then Eq. (1.4) has no analytic solutions in any neighborhood of the origin.
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Proof. If a – g, then we have a = n,s1 = 0 and u1 = 0. Because aj – g,j = 2, . . ., p, then from (2.5) and (2.7), we can determine sn = 0,un = 0 , n = 1, 2 . . .. Thus, we have (4.1) or (4.12). Clearly, (1.4) has a trivial solution /(z) 0 if s = 0. Assume s – 0 and seek a power series solution of (1.4) of the form ( 2.4) as in the proof of Theorem 4.1, where the equality in (4.1) or (4.12) is indispensable. If V(mp + 1,a) – 0 for some natural number m, then the equality in (4.1) or (4.12) does not hold for n = mp + 1 since amp+1 a = 0. In such a circumstance Eq. (1.4) has no formal solutions. When V(mp + 1,a) = 0 for all natural number m, for each m the corresponding bmp+1 in (4.1) or (4.12) has infinitely many choices in C, that is, the formal series solution (2.4) defines a family of solutions with infinitely many parameters. Choose bmp+1 = smp+1 arbitrarily such that
jsmpþ1 j 6 Dmpþ1 ; m ¼ 1; 2 . . . ;
ð4:13Þ
where Dmp+1 is defined by (4.11). In what follows we prove the power series solution (2.4) converges in a neighborhood of the origin. Observe that jan1 1j1 6 C for n – mp + 1. It follows that
X
jbn j 6 C
jbl1 j:jbl2 j . . . jblt j; n – mp þ 1; m ¼ 1; 2; . . . :
l1 þ l2 þ þ lt ¼ n
ð4:14Þ
t ¼ 2; 3; . . . n We consider the implicit functional equation
RðC; s; z; WÞ ¼ 0;
ð4:15Þ
where R is defined in (3.6). Similarly to the proof of Theorem 3.1, we can prove that there exists a unique analytic function W(C,s,z) in a neighborhood of the zero such that W(C,s,0) = 0, W0 z(C,s,0) = jsj. So W(C,s,z) can be expanded into a convergent power series
WðC; s; zÞ ¼
1 X
Dn zn ; D1 ¼ jsj:
ð4:16Þ
n¼1
Moreover, by induction we can claim that
jbn j 6 Dn ; n ¼ 1; 2; . . . : Thus the series (2.4) converges in a neighborhood of the origin.
h
5. Existence of Analytic Solutions In this section, we will show the following theorem. Theorem 5.1. Suppose that conditions of Theorem 3.1, Theorem 4.1 or Theorem 4.2 are satisfied. Then Eq. (1.3) has an analytic solution of the form f(z) = g{/[a/1(z)] p(z)} in a neighborhood of the origin, where g(z) is the inverse function of q(z) and /(z) is a local invertible analytic solutions of Eq. (1.4) in a neighborhood of the origin with /(0) = 0, /0 (0) = s. Proof. By Theorem 3.1, Theorem 4.1 or Theorem 4.2, we can find an analytic solution /(z) of the auxiliary Eq. (1.4) in the form of (2.4), such that /(0) = 0 and /0 (0) = s – 0. Clearly the inverse function /1(z) exists and is analytic in a neighborhood of the origin. Let
uðzÞ ¼ pðzÞ þ qðf ðzÞÞ;
ð5:1Þ
then we have
qðf ðzÞÞ ¼ uðzÞ pðzÞ:
ð5:2Þ 1
Since q(0) = 0 and q0 (z) = f – 0, the inverse function q by g(z), we have
f ðzÞ ¼ g½uðzÞ pðzÞ:
(z) exists and is analytic in a neighborhood of the origin. Denote q1(z)
ð5:3Þ
Eq. (1.3) is changed into
gðuðuðzÞÞ pðuðzÞÞÞ ¼ hðgðuðzÞ pðzÞÞÞ:
ð5:4Þ
Taking
uðzÞ ¼ /ða/1 ðzÞÞ;
ð5:5Þ
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it is easy to see
gðuðuðzÞÞ pðuðzÞÞÞ ¼ gð/ða2 /1 ðzÞÞ pð/ða/1 ðzÞÞÞÞ ¼ hðgð/ða/1 ðzÞÞ pð/ð/1 ðzÞÞÞÞÞ ¼ hðgðuðzÞ pðzÞÞÞ; so u(z) = /(a/1(z)) satisfies Eq. (5.4)consequently f(z) = g(/(a/ 1(z)) p(z)) is an analytic solution of (1.3) in a neighborhood of the origin. This completes the proof of Theorem 5.1. h We now show how to explicitly construct an analytic solution of form (1.3) by means of an example. Consider the following equation:
f ðpðzÞ þ qðf ðzÞÞÞ ¼ hðf ðzÞÞ;
ð4:6Þ
P P1 4zn 3z n 0 z 0 where pðzÞ ¼ 1z ¼ 1 hðzÞ ¼ n¼1 3z ; jzj < 1; pð0Þ ¼ 0; p ð0Þ ¼ n ¼ 3; qðzÞ ¼ 4ðe 1Þ ¼ n¼1 n! ; qð0Þ ¼ 0; q ð0Þ ¼ 1 ¼ 4; P1 2zn z 0 2ð1 e Þ ¼ n¼1 n! ; hð0Þ ¼ 0; h0ð0Þ ¼ g ¼ 2. Because q(0) = 0,q (0) = 1 = 4, then the inverse function g(z) of q(z) exists, then 1 z X zn 1 1 ¼ gðzÞ ¼ ln 1 þ ð1Þn1 n ; jzj < 1; s ¼ ¼ : 4 1 4 n4 n¼1
The characteristic equation
a2 ðn þ gÞa þ ng ¼ a2 a 6 ¼ 0 has two roots a1 =2 = g,a2 = 3 = n and —a1— > 1,—a2— > 1. For a1 =2 = g, by means of Theorem 3.1, the auxiliary equation
gð/ð4zÞ pð/ð2zÞÞÞ ¼ hðgð/ð2zÞ pð/ðzÞÞÞÞ; z 2 C; has an analytic solution /(z) in a neighborhood of the origin such that /(0) = 0 and /0 (0) = s – 0. Let
/ðzÞ ¼
1 X
bn zn ; b1 ¼ s;
n¼1
then s1 ¼ b1 ða nÞ ¼ 5s – 0; u1 ¼ b1 sða nÞ ¼ 54 s – 0, then from (3.10), we have
2 1 5 25 s ¼ s2 ; 6 4 96 ð2Þ2 ð2Þ l1 þl2 ¼2 ! X X 1 1 2 2u1 u2 u31 ¼ 0; c2 ul1 ul2 þ c3 ul1 ul2 ul3 ¼ u3 ¼ 3 6 3! ð2Þ ð2Þ l1 þl2 ¼3 l1 þl2 þl3 ¼3
u2 ¼
X
1
c2 ul1 ul2 ¼
etc. From (3.13) we see
X 1 25 1 25 2 ð5sÞ2 ¼ g 2 sl1 sl2 ¼ s2 s; s2 ¼ u2 4 96 48 2 42 l þl ¼2 1
2
then s2 ¼ 25 s2 . From (3.13) 12
1 s3 ¼ u3 4
X l1 þl2 ¼3
g 2 sl1 sl2 þ
X l1 þl2 þl3 ¼3
! g 3 sl1 sl2 sl3
¼
1 1 2s1 s2 ð5sÞ3 ¼ 0; 32 3 43
so s3 = 0, . . .. From (3.21) we have
/00 ð0Þ 1 25 2 61 2 ¼ s þ 3s2 ¼ s2 ; ðs2 þ a2 b1 Þ ¼ 2! 12 12 ð2Þ2 3 /000 ð0Þ 1 1 61 67 3 ¼ 6 s3 þ 3s3 ¼ s3 ðs3 þ 2a2 b1 b2 þ a3 b1 Þ ¼ b3 ¼ 3 3! 11 12 22 ð2Þ 3
b2 ¼
:
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Because /1(z) is analytic in a neighborhood of the point /(0) = 0, therefore, we can calculate
ð/1 Þ0 ð0Þ ¼
1 /0 ð/1 ð0ÞÞ
ð/1 Þ00 ð0Þ ¼
1 1 ¼ ; /0 ð0Þ s
/00 ð/1 ð0ÞÞð/1 Þ0 ð0Þ 0
n ð/1 Þ000 ð0Þ ¼ þ
2
1
ð/ ð/ ð0ÞÞÞ
¼
/00 ð0Þð/1 Þ0 ð0Þ 0
ð/ ð0ÞÞ
2
¼
61 6
s2 1s 61 ¼ ; 6s s2
o / ð/ ð0ÞÞðð/ Þ ð0ÞÞ þ /00 ð/1 ð0ÞÞð/1 Þ00 ð0Þ ð/0 ð/1 ð0ÞÞÞ2 000
1 0
1
2
ð/0 ð/1 ð0ÞÞÞ4 / ð/ ð0ÞÞð/ Þ ð0Þ 2/0 ð/1 ð0ÞÞ/00 ð/1 ð0ÞÞð/1 Þ0 ð0Þ 00
¼
¼
1
1 0
ð/0 ð/1 ð0ÞÞÞ4
½/000 ð0Þðð/1 Þ0 ð0ÞÞ2 þ /00 ð0Þð/1 Þ00 ð0Þ ð/0 ð0ÞÞ2 ð/0 ð0ÞÞ4 / ð0Þð/1 Þ0 ð0Þ 2/0 ð0Þ/00 ð0Þð/1 Þ0 ð0Þ 00
þ
¼
ð/0 ð0ÞÞ4 201 3 1 61 2 61 2 61 2 1 11 s s2 þ 6 s 6s s þ 6 s s 2s 61 s2 1s 6
s4
43343 ¼ ;: 132s Finally, we determine a solution f(z) of (4.6) at z = 0. Because
f ðzÞ ¼ gðuðzÞ pðzÞÞ ¼ gð/ð2/1 ðzÞÞ pðzÞÞ; then
f ð0Þ ¼ gð/ð0Þ pð0ÞÞ ¼ gð0Þ ¼ 0; f 0 ð0Þ ¼ g 0 ð/ð2/1 ð0Þ pð0ÞÞ½2/0 ð2/1 ð0ÞÞð/1 Þ0 ð0Þ p0 ð0ÞÞ 1 1 5 ; ¼ g 0 ð0Þ½2/0ð0Þð/1 Þ0 ð0Þ p0 ð0Þ ¼ ð2s 3Þ ¼ 32 32 s 2 1 0 1 1 0 00 00 0 f ð0Þ ¼ g ð/ð2/ ð0Þ pð0ÞÞ½2/ ð2/ ð0ÞÞð/ Þ ð0Þ p ð0ÞÞ þ g 0 ð/ð2/1 ð0Þ pð0ÞÞ½4/00 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ2 2/0 ð2/1 ð0ÞÞð/1 Þ00 ð0Þ p00 ð0ÞÞ ¼ g 00 ð0Þ½2/0 ð0Þð/1 Þ0 ð0Þ p0 ð0Þ2 þ g 0 ð0Þ½4/00 ð0Þðð/1 Þ0 ð0ÞÞ2 2/0 ð0Þð/1 Þ00 ð0Þ p00 ð0Þ 1 1 1 61 2 1 61 6 s 2 2s ¼ ð2s 3Þ2 þ ½4 16 4 6 6s s s 195 ; ¼ 16 000 000 f ð0Þ ¼ g ð/ð2/1 ð0Þ pð0ÞÞ½2/0 ð2/1 ð0ÞÞð/1 Þ0 ð0Þ p0 ð0ÞÞ3 þ g 00 ð/ð2/1 ð0Þ pð0ÞÞ 2½2/0 ð2/1 ð0ÞÞð/1 Þ0 ð0Þ p0 ð0ÞÞ ½4/00 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ2 2/0 ð2/1 ð0ÞÞð/1 Þ00 ð0Þ p00 ð0Þ þ g 00 ð/ð2/1 ð0Þ pð0ÞÞ½2/0 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ p0 ð0ÞÞ ½4/00 ð0Þðð/1 Þ0 ð0ÞÞ2 2/0 ð2/1 ð0ÞÞð/1 Þ00 ð0Þ p00 ð0Þ þ g 0 ð/ð2/1 ð0Þ pð0ÞÞ½8/000 ð2/1 ð0ÞÞðð/1 Þ0 ð0ÞÞ3 þ 4/00 ð2/1 ð0ÞÞ 2ð/1 Þ0 ð0Þð/1 Þ00 ð0Þ þ 4/00 ð2/1 ð0ÞÞð/1 Þ0 ð0Þð/1 Þ00 ð0Þ 2/0 ð2/1 ð0ÞÞð/1 Þ000 ð0Þ p000 ð0ÞÞ ¼ g 000 ð0Þ½2/0 ð0Þð/1 Þ0 ð0Þ p0 ð0Þ3 þ 3g 00 ð0Þ½2/0 ð0Þð/1 Þ0 ð0Þ p0 ð0Þ ½4/00 ð0Þðð/1 Þ0 ð0ÞÞ2 2/0 ð0Þð/1 Þ00 ð0Þ p00 ð0Þ þ g 0 ð0Þ½8/000 ð0Þðð/1 Þ0 ð0ÞÞ3 þ 12/00 ð0Þð/1 Þ0 ð0Þð/1 Þ00 ð0Þ 2/0 ð0Þð/1 Þ000 ð0Þ p000 ð0Þ
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1 1 61s2 1 61 6 ð2 3Þ3 þ 3 2 ð2 3Þ 4 2 2s 32 6s 6 s 4 2 1 201 3 1 61s 1 61 43343 þ 8 2s s 3 þ 12 18 4 11 132s s 6 s 6s 136135 ;...: ¼ 352 ¼
Thus, the desired solution is
f ðzÞ ¼
5 195 2 136135 3 zþ z þ z þ : 32 32 2112
For a = 3 = n, by means of the same method, we can obtain
f ðzÞ ¼ 0: Remark that the analytic solutions of (1.3) in the cases a = n = g and a = g,a – n,—n— = 1 in (H2), and the cases a = g,a – n and a = n = g in (H3) remain still open. Acknowledgment I thank Professor Jianguo Si and Professor Weinian Zhang for valuable discussions during the preparation of this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
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