# Analytic Solutions of an Iterative Functional Differential Equations Near Regular Points

Authors: Lingxia Liu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

The existence of analytic solutions of an iterative functional differential equation is studied when the given functions are all analytic and when the given functions have regular points. By reducing the equation to another functional equation without iteration of the unknown function an existence theorem is established for analytic solutions of the original equation.


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## Body

## 1. Introduction

Functional differential equations with state-dependent delay have attracted the attentions of many authors in the last few years (see [1–8]). In [6–8], analytic solutions of the state-dependent functional differential equations(1)x′(z)=x(az+bx(z)),αz+βx′(z)=x(az+bx′(z)),fm(x)=G(∑k=0m-1akfk(x))+F(x),m≥2,x∈ℂ
are found. In this paper, we will be concerned with analytic solutions of the functional differential equation(2)αz+βx′(z)=F(∑l=0mclxl(z))+G(z),z∈ℂ,
where α, β, c1, c2,…,cm are complex numbers, β≠0, ∑l=0m|cl|<1, and xl(z)=x(xl-1(z)) that denote the nth iterate of a map x. In general, F, G are given complex-valued functions of a complex variable.In this paper, analytic solutions of nonlinear iterative functional differential equations are investigated. Existence of locally analytic solutions and their construction is given in the case that all given functions exist regular points. As well as in previous work [6–8], we still reduce this problem to find analytic solutions of a differential-difference equation and a functional differential equation with proportional delay. The existence of analytic solutions for such equation is closely related to the position of an indeterminate constant μ depending on the eigenvalue of the linearization of x at its fixed point 0 in the complex plane. For technical reasons, in [6, 7], only the situation of μ off the unit circle in ℂ and the situation of μ on the circle with the Diophantine condition, “|μ|=1, μ is not a root of unity, and log(1/|μn-1|)≤Tlogn, n=2,3,… for some positive constant T”, are discussed. The Diophantine condition requires μ to be far from all roots of unity that the fixed point 0 is irrationally neutral. In this paper, besides the situation that μ is the inside of the unit circle S1, we break the restriction of the Diophantine condition and study the situations that the constant μ in (5) (or μ=b-λ, b is a complex constant, and λ is in (4)) is resonance and a root of unity in the complex plane ℂ near resonance under the Brjuno condition.


## 2. Discussion on Auxiliary Equations

In this section we assume that bothF and G are analytic functions in a neighborhood of the origin, that is, 0 is a regular point, and have power series expansions
(3)F(z)=∑n=0∞anzn,G(z)=∑n=0∞dnzn,a0≠0,d0≠0,z∈ℂ.If there exists a complex constantλ and an invertible function ψ(z) such that ψ(ψ-1(z)+λ) is well defined, then letting x(z)=ψ(ψ-1(z)+λ), we can formally transform (2) into the differential-difference equation
(4)αψ(z)ψ′(z)+βψ′(z+λ)=F(∑l=0mclψ(z+lλ))ψ′(z)+G(ψ(z))ψ′(z).The indeterminate constantλ will be discussed in the following cases:(I1)
ℜλ>0;(I2)
μ=b-λ=e2πiθ, and θ∈ℝ∖ℚ is a Brjuno number [9, 10]; that is, B(θ)=∑k=0∞(log(qk+1)/qk)<∞, where {pk/qk} denotes the sequence of partial fraction of the continued fraction expansion of θ, said to satisfy the Brjuno condition;(I3)
μ=b-λ=e2πiq/p for some integers p∈ℕ with p≥2 and q∈ℤ∖{0}, and α≠e2πil/k for all 1≤k≤p-1 and l∈ℤ∖{0}.Take notationsSp:={z∈ℂ:ℜz>-lnρ/ln|b|,-∞<ℑz<+∞}.A change of variable further transforms (4) into the functional differential equation
(5)αφ(z)φ′(z)+βμφ′(μz)=F(∑l=0mclφ(μlz))φ′(z)+G(φ(z))φ′(z),
where μ is a complex constant. The solution of this equation has properties similar to those of (4). If (5) has an invertible solution φ(z), which satisfies the initial value conditions
(6)φ(0)=0,φ′(0)=τ≠0,
then we can show that x(z)=φ(μφ-1(z)) is an analytic solution of (2).Theorem 1.
Suppose that (I1) holds, then (5) has an analytic solution of the form
(7)ψ(z)=∑n=1∞bnb-nz,
in the half plane Sρ for a certain constant ρ>0, which satisfies limℜz→+∞ψ(z)=0.Proof.
SinceF and G are analytic in a neighborhood of the origin and have the power series expansion (3), there exists a positive ϱ such that
(8)|an|≤ϱn-1,|dn|≤ϱn-1,n=2,3,….
Without loss of generality, we can assume thatϱ=1; that is, |an|≤1, |dn|≤1 for n=2,3,…. In fact, let F~(z)=ϱF(ϱ-1z), G~(z)=ϱG(ϱ-1z), and put y=ϱz, β~=ϱβ, u=ϱλ, and ψ~(z)=ϱψ(ϱ-1z). Then (4) can be rewritten as
(9)αψ~(y)ψ~′(y)+β~ψ~′(y+u)=F~(∑l=0mclψ~(y+lu))ψ~′(y)+G~(ψ~(y))ψ~′(y),
in the same form as (4) and |anϱ1-n|≤1, |dnϱ1-n|≤1 for n=2,3,… by (8).
Consider a solutionψ(z) of (4) in the formal Dirichlet series (7); that is, ψ(z)=∑n=1∞bnb-nz, where b is a complex constant and |b|>1. Substituting series (3) and (7) of F, G, and ψ in (4) and comparing coefficients we obtain that
(10)[βb-λ-(a0+d0)]lnbb1=0,(11)[βb-nλ-(a0+d0)]nbn=-α∑i=1n-1ibibn-i+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-ii[dt+at∏k=1t(∑l=0mclblklλ)]bibl1bl2⋯blt,n≥2.
Ifb1=τ=0, then (4) has a trivial solution ψ(z)=0. Assume that b1=τ≠0, because |b|>1; from (10) we have a0+d0=βb-λ. From (11) we obtain that
(12)βb-λ(b-(n-1)λ-1)nbn=-α∑i=1n-1ibibn-i+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-ii[dt+at∏k=1t(∑l=0mclb-lklλ)]bibl1bl2⋯blt,n≥2.
The sequence{bn}n=2∞ is successively determined by (12) in a unique manner.
In what follows we need to prove that the series (7) is convergent in a right-half plane. Since ℜλ>0, so we have
(13)limn→∞|αinβb-λ(b-(n-1)λ-1)|≤limn→∞|α||βb-λ(b-(n-1)λ-1)|=|α||β||b-λ|,ℜλ>0,limn→∞|i[dt+at∏k=1t(∑l=0mclb-lklλ)]nβb-λ(b-(n-1)λ-1)|≤limn→∞|dt|+|at|∏k=1t(∑l=0m|cl|)|βb-λ(b-(n-1)λ-1)|≤2|β||b-λ|,ℜλ>0.
This implies that there exists a constantM>0 such that(14)limn→∞|αinβb-λ(b-(n-1)λ-1)|≤M,limn→∞|i[dt+at∏k=1t(∑l=0mclblklλ)]nβb-λ(b-(n-1)λ-1)|≤M,∀n≥2,ℜλ>0.
Therefore, from (12) we obtain
(15)|bn|≤M(∑i=1n-1|bi||bn-i|+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i|bi||bl1||bl2|⋯|blt|),n≥2.
In order to construct a majorant series of (7), we consider the implicit functional equation
(16)H(z)=|τ|z+M[H2(z)+H2(z)1-H(z)].
Define the function(17)ω(z,H):=ω(z,H,τ,M)=H-|τ|z-M(H2+H21-H)
for (z,H) in a neighborhood of the origin. Then ω(0,0)=0, ωH′(0,0)=1≠0. Thus, there exists a unique function H(z) analytic in a neighborhood of zero; that is, there is a constant δ1>0, as |z|<δ1, the function H(z) is analytic, such that H(0)=0, H′(0)=-ωz′(0,0)/ωH′(0,0)=|τ| and ω(z,H(z))=0. Since H(0)=0, there is a constant δ2>0, such that |H(z)|<1 for |z|<δ2. Therefore, as |z|<δ:=min{δ1,δ2}, the function H(z) satisfies the equation
(18)ω(z,H(z))=H(z)-|τ|z-M[H2(z)+H2(z)1-H(z)]=0.
ChoosingB1=|τ| and putting
(19)H(z)=∑n=1∞Bnb-nz
in (18), we can determine all coefficients recursively by B1=|τ| and
(20)Bn=M(∑i=1n-1BiBn-i+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-iBiBl1Bl2⋯Blt),n≥2.
Moreover, it is easy to see from (15) (21)|bn|≤Bn,n=1,2,….
It follows that the power series(22)ϕ(z)=∑n=1∞bnzn
is also convergent as |z|<δ. So there exists ρ≤δ such that Dirichlet series (7) is convergent in Sρ.
Furthermore, one has(23)limℜz→+∞b-z=limℜz→+∞b-ℜz[(cos(ℑzlnb))-isin(ℑzlnb)]=0.
Thuslimℜz→+∞ψ(z)=limℜz→+∞∑n=1∞bnb-nz=0. The proof is complete.We observe thatμ=b-λ is inside the unit circle of (I1) but on the unit circle in the rest cases. Next we devote attention to the existence of analytic solutions of (4) under the Brjuno condition. To do this, we first recall briefly the definition of Brjuno numbers and some basic facts. As stated in [11], for a real number θ, we let [θ] denote its integer part, and {θ}=θ-[θ] its fractional part. Then every irrational number θ has a unique expression of the Gauss' continued fraction
(24)θ=a0+θ0=a0+1a1+θ1=⋯,
denoted simply by θ=[a0,a1,…,an,…], where aj’s and θj’s are calculated by the algorithm: (a) a0=[θ], θ0={θ} and (b) an=[1/θn-1], θn={1/θn-1} for all n≥1. Define the sequences (pn)n∈ℕ and (qn)n∈ℕ as follows:
(25)q-2=1,q-1=0,qn=anqn-1+qn-2,p-2=0,p-1=1,pn=anpn-1+pn-2.It is easy to show thatpn/qn=[a0,a1,…,an]. Thus, for every θ∈ℝ∖ℚ we associate, using its convergence, an arithmetical function B(θ)=∑n≥0(log(qn+1)/qn). We say that θ is a Brjuno number or that it satisfies Brjuno condition if B(θ)<+∞. The Brjuno condition is weaker than the Diophantine condition. For example, if an+1≤cean for all n≥0, where c>0 is a constant, then θ=[a0,a1,…,an,…] is a Brjuno number but is not a Diophantine number. So, the case (I2) contains both Diophantine condition and a part of μ=b-λ “near” resonance. let
(26)Ak={n≥0∣∥nθ∥≤18qk},Ek=max(qk,qk+14),ηk=qkEk.LetAk* be the set of integers j≥0 such that either j∈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≥0, define
(27)lk(n)=max((1+ηk)nqk-2,(mnηk+n)1qk-1),
where mn=max{j∣0≤j≤n,j∈Ak*}. We then define the function hk:ℕ→ℝ+ as follows:
(28)hk(n)={mn+ηknqk-1,ifmn+qk∈Ak*,lk(n),ifmn+qk∉Ak*.Letgk(n):=max(hk(n),[n/qk]), and define k(n) by the condition qk(n)≤n≤qk(n)+1. Clearly, k(n) is nondecreasing. Moreover, the function gk is nonnegative. Then we are able to state the following result.Lemma 2 (Davie’s lemma [12]).
LetK(n)=nlog2+∑j=0k(n)gj(n)log(2qj+1). Then(a)
there is a universal constantγ>0 (independent of n and θ) such that
(29)K(n)≤n(∑j=0k(n)logqj+1qj+γ),(b)
K(n1)+K(n2)≤K(n1+n2) for all n1 and n2, and(c)
-log|b-λn-1|≤K(n)-K(n-1).Now we state and prove the following theorem under Brjuno condition.Theorem 3.
Suppose that (I2) holds. Then (4) has an analytic solution ψ of the form (7) in the half plane Sρ={z∈ℂ:ℜz>-lnρ/ln|b|,-∞<ℑz<+∞} for a certain constant ρ>0, which satisfies limℜz→+∞ψ(z)=0.Proof.
As in the proof of Theorem1, we find a solution in the form of the Dirichlet series (7). Using the same method as above mentioned, for chosen b1=τ we can uniquely determine the sequence {bn}n=2∞ recursively by (12). In fact, in view of (I2) we see that μ=b-λ satisfies the conditions of Lemma 2, and from (12) we have
(30)|bn|≤M1|b-(n-1)λ-1|({∑l1+l2+⋯+lt=n-it=1,2,…,n-i}∑i=1n-1|bi||bn-i|≤M1|b-(n-1)λ-1|+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i|bi||bl1||bl2|⋯|blt|),n≥2,
where M1=max{|α|/|β|,2/|β|}>0.
To construct a governing series of (7), we consider the implicit functional equation
(31)ω(z,U,τ,M1)=0,
where ω is defined in (17). Similarly to the proof of Theorem 1, using the implicit function theorem we can prove that (31) has a unique analytic solution U(z,τ,M1) in a neighborhood of the origin; that is, there is a constant δ3>0, as |z|<δ3, the function U(z,τ,M1) is analytic such that U(0,τ,M1)=0 and Uz′(0,τ,M1)=|τ|. Thus U(z,τ,M1) in (31) can be expanded into a convergent series
(32)U(z,τ,M1)=∑n=1∞Cnzn,
in a neighborhood of the origin. Replacing (32) into (31) and comparing coefficients, we obtain that C1=|τ| and
(33)Cn=M1(∑i=1n-1CiCn-i+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-iCiCl1Cl2⋯Clt),n≥2.
Note that the series (32) converges in a neighborhood of the origin. So, there is a constant T>0 such that
(34)Cn<Tn,n≥1.
Now, we can deduce, by induction, that|bn|≤CneK(n-1) for n≥1, where K:ℕ→ℝ is defined in Lemma 2. In fact |b1|=|τ|=C1, for inductive proof, and we assume that |bj|≤CjeK(j-1), j≤n. From (30) and Lemma 2 we obtain
(35)|bn+1|≤M1|b-nλ-1|×({∑l1+l2+⋯+lt=n-i+1t=1,2,…,n-i+1}∑i=1n|bi||bn-i+1|+∑i=1n∑l1+l2+⋯+lt=n-i+1t=1,2,…,n-i+1|bi||bl1||bl2|⋯|blt|)≤M1|b-nλ-1|[∑i=1nCiCn-i+1eK(i-1)+K(n-i)≤M1|b-nλ-1|+∑i=1n∑l1+l2+⋯+lt=n-i+1t=1,2,…,n-i+1CiCl1Cl2⋯Clt≤M1|b-nλ-1|×eK(i-1)+K(l1-1)+⋯+K(lt-1){∑i=1n}].
Note that(36)K(i-1)+K(n-i)≤K(n-1)≤K(n)+log|b-λn-1|,K(l1-1)+K(l2-1)+⋯+K(lt-1)≤K(n-i+1-t)≤K(n-i),K(i-1)+K(n-i)≤K(n-1)≤K(n)+log|b-λn-1|,
then
(37)|bn+1|≤eK(n-1)M1|b-nλ-1|×(∑i=1nCiCn-i+1+∑i=1n∑l1+l2+⋯+lt=n-i+1t=1,2,…,n-i+1CiCl1Cl2⋯Clt)≤Cn+1eK(n)
as desired. Moreover, from Lemma 2, we know that K(n)≤n(B(θ)+γ) for some universal constant γ>0. Then from (34) we have |bn|≤CneK(n-1)≤Tne(n-1)(B(θ)+γ); that is, limn→∞sup(|bn|1/n)≤limn→∞sup(Te((n-1)/n)(B(θ)+γ))=TeB(θ)+γ, which shows that the series (7) converges for |z|<ρ=min{δ3,(TeB(θ)+γ)-1}. So does series (22) in Sρ. Similarly limℜz→+∞ψ(z)=0, as proved in Theorem 1.
The next theorem is devoted to the case of (I3), whereb-λ is not only on the unit circle in ℂ but also a root of unity. In this case the Diophantine condition and Brjuno condition are not satisfied. The idea of our proof is acquired from [13]. Let {An}n=1∞ be a sequence defined by A1=τ and
(38)An=ξM1(∑i=1n-1AiAn-i+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-iAiAl1Al2⋯Alt),n≥2,
where ξ=max{1,|αi-1|-1,i=1,2…,p-1} and M1 is defined in Theorem 3.Theorem 4.
Suppose that (I3) holds andp is given as above mentioned. Let {bn}n=1∞ be determined recursively by b1=τ and
(39)βb-λ(b-(n-1)λ-1)nbn=Ω(n,λ),n=2,3,…,
where
(40)Ω(n,λ)=-α∑i=1n-1ibibn-i+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-ii[dt+at∏k=1t(∑l=0mclb-lklλ)]bibl1bl2⋯blt.
IfΩ(νp+1,λ)=0 for all ν=1,2,…, then (4) has an analytic solution ψ(z)=ϕ(b-z) in the half plane Sρ:={z∈ℂ:ℜz>-lnρ/ln|b|,-∞<ℑz<+∞} for a certain ρ>0, where ϕ is an analytic function of the form (22) in Uρ(0)={z∣|z|<ρ} such that ϕ(0)=0, and ϕ(νp+1)(0)=(νp+1)!τνp+1, for all ν=0,1,2,…, where τνp+1′s are arbitrary constants satisfying the inequality |τνp+1|≤Aνp+1 and the sequence {An}n=1∞ is defined in (38). The other derivatives at 0 satisfy that ϕ(i)(0)=i!bi for i≠νp+1. Otherwise, if Ω(νp+1,λ)≠0 for some ν=1,2,…, then (4) has no analytic solutions in the half plane Sρ:={z∈ℂ:ℜz>-lnρ/ln|b|,-∞<ℑz<+∞} for any ρ>0.Proof.
Analogously to the proof of Theorem1, we seek for a solution of (4) in Dirichlet (7). Without loss of generality, as in the proof of Theorem 1 we still assume that ρ=1 in (8). Taking (3) and (7) in (4) and defining b1=τ≠0, we obtain (12) or (39). If Ω(νp+1,λ)=0 for all natural numbers ν, then for each ν, b-νpλ-1=0, the corresponding bνp+1 has infinitely many choices in ℂ; that is, the formal series solution (7) or (22) defines a family of solutions with infinitely many parameters. Choose bνp+1=τνp+1 arbitrarily such that
(41)|τνp+1|≤Aνp+1,ν=0,1,2,…,
where Aνp+1 is defined by (38). In what follows, we prove that the formal series solution (7) converges in a neighborhood of the origin. Observe that |b-nλ-1|-1≤ξ for n≠νp. It follows from (12) or (39) that
(42)|bn|≤ξM1({∑l1+l2+⋯+lt=n-it=1,2,…,n-i}∑i=1n-1|bi||bn-i|≤ξM1+∑i=1n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i|bi||bl1||bl2|⋯|blt|)
for all n≠νp+1, ν=0,1,2,…, where M1 is defined in Theorem 3. Let
(43)V(z,τ,ξM1)=∑n=1∞Anzn,A1=|τ|.
It is easy to check that there exists a constantρ>0, for |z|<ρ, and (43) satisfies the implicit functional equation
(44)ω(z,V,τ,ξM1)=0,
where ω is defined in (17). Moreover, similarly to the proof of Theorem 1, we can prove that (44) has a unique analytic solution V(Z,τ,ξM1) in a neighborhood of the origin such that V(0,τ,ξM1)=0 and Vz′(0,τ,ξM1)=|τ|≠0. Thus (43) converges in a neighborhood of the origin. Moreover, it is easy to show that, by induction,
(45)|bn|≤An,n=1,2,….
By inequality (45) we see that the series (22) converges in Uρ(0)={z∣|z|<ρ}. Thus series (7) converges in Sρ. This completes the proof.The following theorem shows that each analytic solution of (4) leads to an analytic solution of (5). We shall discuss (5) in the following cases:(H1)
0<|μ|≠1;(H2)
μ=e2πiθ, where θ∈ℝ∖ℚ is a Brjuno number, B(θ)=∑k=0∞(log(qk+1)/qk)<∞, where {pk/qk} denotes the sequence of partial fraction of the continued fraction expansion of θ, said to satisfy the Brjuno condition;(H3)
μ=e2πiq/p for some integers p∈ℕ with p≥2 and q∈ℤ∖{0}, and μ≠e2τil/k for all 1≤k≤p-1 and l∈ℤ∖{0}.Theorem 5.
Suppose that (H1) holds and thata0≠0, d0≠0. Then in a neighborhood of the origin (5) has an analytic solution φ satisfying φ(0)=0, φ′(0)=η.Proof.
Let(46)φ(z)=∑n=1∞bnzn,b1=η
be the formal series of the solution φ for (5). We are going to determine {bn}n=1∞. Substituting (3) and (46) into (5) and comparing coefficients, we obtain
(47)[βμ-(a0+d0)]b1=0,(48)(n+1)[βμn+1-(a0+d0)]bn+1=-α∑i=0n-1(i+1)bi+1bn-i+∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i(i+1)[dt+at∏k=1t(∑l=0mclμllk)]×bi+1bl1bl2⋯blt,n≥1.
Ifb1=η=0, then (5) has a trivial solution φ(z)=0. Assume that b1=η≠0; from (47) we can choose βμ=a0+d0, then (48) can be changed into
(49)(n+1)βμ(μn-1)bn+1=-α∑i=0n-1(i+1)bi+1bn-i+∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i(i+1)[dt+at∏k=1t(∑l=0mclμllk)]×bi+1bl1bl2⋯blt,n≥1.
From (49) the sequence {bn}n=2∞ is determined uniquely in the recursive way.
Now we show the convergence of series (46) near zero. Since the power series in (3) are both convergent for |z|<σ, for any fixed r∈(0,σ) there exists a constant M2>0 such that
(50)|an|≤M2rn,|dn|≤M2rn.
Note that since1≤i≤n-1, then there exists some positive number M3 as follows:
(51)|(i+1)∏k=1t(∑l=0mclμllk)(n+1)βμ(μn-1)|≤1|β||μ||μn-1|≤M3,|i+1(n+1)βμ(μn-1)|≤1|β||μ||μn-1|≤M3,|α(i+1)(n+1)βμ(μn-1)|≤|α||β||μ||μn-1|≤M3,
then we have
(52)|bn+1|≤L({∑l1+l2+⋯+lt=n-it=1,2,…,n-i}∑i=0n-1|bi+1||bn-i|+∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i1rt|bi+1||bl1||bl2|⋯|blt|),n≥1,
where L:=max{2M2M3,M3}>0. Let
(53)Θ(z,W)=W-|η|z-L(W2/r1-(W/r)+W2)
for (z,W) from a neighborhood of the origin. Since Θ(0,0)=0,ΘW′(0,0)=1≠0, there exists a unique function W(z), analytic in a neighborhood of the origin, such that W(0)=0, W′(0)=|η|≠0, and Θ(z,W(z))=0. Then define a sequence {Dn}n=1∞ by D1=|η|≠0 and
(54)Dn+1=L({∑l1+l2+⋯+lt=n-it=1,2,…,n-i}∑i=0n-1Di+1Dn-i+∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i1rtDi+1Dl1Dl2⋯Dlt),n≥1.
By (52) we see that
(55)|bn|≤Dn,n≥1.
Let(56)P(z)=∑n=1∞Dnzn,D1=|η|,
with the recursive law of {Dn}n=1∞. Then
(57)P2(z)=(∑n=0∞Dn+1zn+1)(∑n=1∞Dnzn)=∑n=1∞∑i=0n-1Di+1Dn-izn+1,P2(z)/r1-(P(z)/r)=P(z)P(z)/r1-(P(z)/r)=(∑n=0∞Dn+1zn+1)(∑n=1∞(P(z)r)n)=(∑n=0∞Dn+1zn+1)(∑n=1∞1rn(∑n=1∞Dnzn)n)=(∑n=0∞Dn+1zn+1)(∑n=1∞∑l1+l2+⋯+lt=nt=1,2,…,n1rtDl1Dl2⋯Dltzn)=∑n=1∞∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i1rtDi+1Dl1Dl2⋯Dltzn+1.
Then we have(58)P2(z)/r1-(P(z)/r)+p2(z)=1L∑n=1∞Dn+1zn+1=1L(P(z)-|η|z),
that is
(59)P(z)-|η|z-L[P2(z)/r1-(P(z)/r)+p2(z)]=0.
This shows thatP(0)=0,P′(0)=|η|, and Θ(z,P(z))=0. So we have P(z)=W(z). It follows that the power series (56) converges in a neighborhood of the origin. Therefore, from (55) we see that (46) converges in a neighborhood of the origin. The proof is complete.In the case (H2) we obtain similarly an analogue to Theorem3.Theorem 6.
Suppose that (H2) holds and thata0≠0, d0≠0. Then in a neighborhood of the origin (5) has an analytic solution φ(z) satisfying φ(0)=0,φ′(0)=η≠0.In the case (H3) we also obtain similarly an analogue to Theorem4.Theorem 7.
Suppose that (H3) holds,a0≠0, d0≠0, and p is given as above mentioned. Let {bn}n=1∞ be determined recursively by b1=η and
(60)(n+1)βμ(μn-1)bn+1=Ξ(n+1,μ),
where
(61)Ξ(n+1,μ)=-α∑i=0n-1(i+1)bi+1bn-i+∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i(i+1)[dt+at∏k=1t(∑l=0mclμllk)]×bi+1bl1bl2⋯blt,n≥1.
IfΞ(lp+1,μ)=0 for all l=1,2,…, then (5) has an analytic solution in a neighborhood of the origin such that φ(0)=0, φ′(0)=η≠0, and φ(lp+1)(0)=(lp+1)!ηlp+1, where ηlp+1′s are arbitrary constants satisfying the inequality |ηlp+1|≤dlp+1, l=1,2,… and the sequence {dn}n=1∞ is defined as follows: d1=|η| and
(62)dn+1=ΓL1(∑i=0n-1di+1dn-i+∑i=0n-1∑l1+l2+⋯+lt=n-it=1,2,…,n-i1rtdi+1dl1dl2⋯dlt),n≥1,
where Γ=max{1,|μi-1|-1,i=1,2…,p-1}, and L1=max{2/|β|,|α|/|β|}. Otherwise, if Ξ(lp+1,μ)≠0 for some l=1,2,…, then (5) has no analytic solutions in any neighborhood of the origin.


## 3. Analytic Solutions of the Original (2)

Having known analytic solutions of the auxiliary equation (4) and (5), we can give results to the original (2).Theorem 8.
Suppose that the conditions of Theorems1, 3, or 4 are satisfied. Then (2) has an analytic solution x(z)=ψ(ψ-1(z)+λ) in a neighborhood of the origin, where ψ(z) is an analytic solution of (4) in the half plane Sρ.Proof.
In view of Theorems1, 3, or 4, we may find a sequence {bn}n=1∞ such that the function ψ(z) of the form (7) is an analytic solution of (4) in the half plane Sρ. As in Theorem 1, ψ(z)=ϕ(b-z). Since ϕ′(0)≠0, the function ϕ-1 is analytic in a neighborhood of the point ϕ(0)=0. Thus ψ-1(z)=-lnϕ-1(z)/lnb is analytic in a neighborhood of the origin. Define x(z)=ψ(ψ-1(z)+λ) which is analytic clearly. Note that
(63)x′(z)=ψ′(ψ-1(z)+λ)ψ′(ψ-1(z)),xk(z)=ψ(ψ-1(z)+kλ),
then from (4)
(64)αz+βx′(z)=αz+βψ′(ψ-1(z)+λ)ψ′(ψ-1(z))=F(∑l=0mclψ(ψ-1(z)+lλ))+G(ψ(ψ-1(z)))=F(∑l=0mclxl(z))+G(z).
This shows thatx(z)=ψ(ψ-1(z)+λ) satisfies (2). The proof is complete.Under the hypothesis of Theorem1 the origin 0 is a hyperbolic fixed point of x, but under hypotheses of Theorems 3 and 4 it is not. Actually, when the constant μ=b-λ satisfies the Brjuno condition, the norm of the eigenvalue of the linearized of x at 0 equals 1, but the eigenvalue is not a root of unity. Under (I3), the fixed point 0 of x is a resonance.When0<|b|<1, the same method is applicable and a similarly result can be obtained; that is, there exists a constant ρ>0 such that (4) has an analytic solution in the left-half plane {z∈ℂ∣ℜz<-lnρ/ln|b|,-∞<ℑz<+∞}.Theorem 9.
Under one of the conditions in Theorems5, 6, or 7, (2) has an analytic solution of the form x(z)=φ(μφ-1(z)) in a neighborhood of the origin, where φ(z) is an analytic solution of (5) in a neighborhood of the origin.Proof.
In Theorems5–7 we have found a solution φ(z) of (5) in the form (46), which is analytic near 0. Since φ(0)=0, φ′(0)=η≠0, the function φ-1 is also analytic near 0. Thus x(z)=φ(μφ-1(z)) is analytic. Moreover,
(65)x′(z)=μφ′(μφ-1(z))φ′(φ-1(z)),xl(z)=φ(μlφ-1(z)),
so from (5) we have
(66)αz+βx′(z)=αz+βμφ′(μφ-1(z))φ′(φ-1(z))=F(∑l=0mclφ(μlφ-1(z)))+G(φ(φ-1(z)))=F(∑l=0mclxl(z))+G(z).
Therefore,x(z)=φ(μφ-1(z)) satisfies (2). The proof is complete.


## 4. Example

Example 1.
Consider the equation(67)z+x′(z)=4e(1/2)x(x(z))-ez-176.
It is in the form of (2), where α=β=1,  c0=c1=0,  c2=1/2,  m=2,  F(z)=4ez-(13/3)=∑n=1∞(4zn/n!)-(1/3), and G(z)=(3/2)-ez=-∑n=1∞(zn/n!)+(1/2). Clearly both G and F are analytic near 0, a0=F(0)=-1/3, d0=G(0)=1/2, a1=F′(0)=4, and d1=G′(0)=-1. For arbitrary |b|>1, let λ=ln6/ln|b|, the 0<|b-λ|<1. By Theorem 1, there is a constant ρ>0 such that the corresponding auxiliary equation
(68)ψ(z)ψ′(z)+ψ′(z+λ)=F(12ψ(z+2λ))ψ′(z)+G(ψ(z))ψ′(z),
and (67) itself have an analytic solution in the half plane Sρ={z∣ℜz>-(lnρ/lnb),-∞<ℑz<+∞}. In the routine in the proofs of our theorems we can calculate the solutions
(69)x(z)=16z-3536z2-5511944z3+⋯.Example 2.
Consider the equation(70)z-x′(z)=12x(z)-14x(x(z))+ez-14.
It is in the form of (2), where α=1, β=-1, c0=0, c1=-1/2, c2=1/4, m=2, a0=1/2, a1=-1, an=0(n≥2), d0=1/4, d1=1, F(z)=-z+(1/2), and G(z)=ez-(3/4)=∑n=1∞(zn/n!)+(1/4). Clearly, ∑l=02|cl|<1 and μ=(1/β)(a0+d0)=-3/4 such that 0<|μ|<1. By Theorem 5 the corresponding auxiliary equation
(71)φ(z)φ′(z)-μφ′(μz)=(12φ(μz)-14φ(μ2z))φ′(z)+(eφ(z)-14)φ′(z)
and (70) itself have an analytic solution each in a neighborhood of the origin. Proofs of our theorems provide a method to calculate the solution
(72)x(z)=μz+18μ(μ-2)z2+13![116μ(μ-1)(μ-2)(μ+2)-1]z3+⋯.



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