Analytic Solutions of an Iterative Functional Differential Equations Near Regular Points

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.

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 equations𝑥^′(𝑧)=𝑥(𝑎𝑧+𝑏𝑥(𝑧)),𝛼𝑧+𝛽𝑥^′(𝑧)=𝑥𝑎𝑧+𝑏𝑥^′(𝑧),𝑓^𝑚(𝑥)=𝐺^𝑚−1_𝑘=0𝑎_𝑘𝑓^𝑘(𝑥)+𝐹(𝑥),𝑚≥2,𝑥∈ℂ (1) are found. In this paper, we will be concerned with analytic solutions of the functional differential equation𝛼𝑧+𝛽𝑥^′(𝑧)=𝐹^𝑚_𝑙=0𝑐_𝑙𝑥^𝑙(𝑧)+𝐺(𝑧),𝑧∈ℂ, (2) where 𝛼, 𝛽, 𝑐₁, 𝑐₂,…,𝑐_𝑚 are complex numbers, 𝛽≠0, ∑^𝑚_𝑙=0|𝑐_𝑙|<1, and 𝑥^𝑙(𝑧)=𝑥(𝑥^𝑙−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/|𝜇^𝑛−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 both 𝐹 and 𝐺 are analytic functions in a neighborhood of the origin, that is, 0 is a regular point, and have power series expansions 𝐹(𝑧)=^∞_𝑛=0𝑎_𝑛𝑧^𝑛,𝐺(𝑧)=^∞_𝑛=0𝑑_𝑛𝑧^𝑛,𝑎₀≠0,𝑑₀≠0,𝑧∈ℂ. (3)

If there exists a complex constant 𝜆 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 𝛼𝜓(𝑧)𝜓^′(𝑧)+𝛽𝜓^′(𝑧+𝜆)=𝐹^𝑚_𝑙=0𝑐_𝑙𝜓(𝑧+𝑙𝜆)𝜓^′(𝑧)+𝐺𝜓(𝑧)𝜓^′(𝑧). (4)

The indeterminate constant 𝜆 will be discussed in the following cases:(I1)ℜ𝜆>0; (I2)𝜇=𝑏^−𝜆=𝑒^{2𝜋𝑖𝜃}, and 𝜃∈ℝ⧵ℚ is a Brjuno number [9, 10]; that is, 𝐵(𝜃)=∑^∞_𝑘=0(log(𝑞_𝑘+1)/𝑞_𝑘)<∞, where {𝑝_𝑘/𝑞_𝑘} denotes the sequence of partial fraction of the continued fraction expansion of 𝜃, said to satisfy the Brjuno condition;(I3)𝜇=𝑏^−𝜆=𝑒^{2𝜋𝑖𝑞/𝑝} for some integers 𝑝∈ℕ with 𝑝≥2 and 𝑞∈ℤ⧵{0}, and 𝛼≠𝑒^{2𝜋𝑖𝑙/𝑘} for all 1≤𝑘≤𝑝−1 and 𝑙∈ℤ⧵{0}.

Take notations 𝑆_𝑝∶={𝑧∈ℂ∶ℜ𝑧>−ln𝜌/ln|𝑏|,−∞<ℑ𝑧<+∞}.

A change of variable further transforms (4) into the functional differential equation 𝛼𝜑(𝑧)𝜑^′(𝑧)+𝛽𝜇𝜑^′𝜇𝑧=𝐹^𝑚_𝑙=0𝑐_𝑙𝜑𝜇^𝑙𝑧𝜑^′(𝑧)+𝐺𝜑(𝑧)𝜑^′(𝑧), (5) 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 𝜑(0)=0,𝜑^′(0)=𝜏≠0, (6) 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 𝜓(𝑧)=^∞_𝑛=1𝑏_𝑛𝑏^−𝑛𝑧, (7) in the half plane 𝑆_𝜌 for a certain constant 𝜌>0, which satisfies lim_{ℜ𝑧→+∞}𝜓(𝑧)=0.

Proof. Since 𝐹 and 𝐺 are analytic in a neighborhood of the origin and have the power series expansion (3), there exists a positive 𝜚 such that 𝑎_𝑛≤𝜚^𝑛−1,𝑑_𝑛≤𝜚^𝑛−1,𝑛=2,3,…. (8)
Without loss of generality, we can assume that 𝜚=1; that is, |𝑎_𝑛|≤1, |𝑑_𝑛|≤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 𝛼𝜓𝑦𝜓^′𝑦+𝛽𝜓^′𝑦+𝑢=𝐹^𝑚_𝑙=0𝑐_𝑙𝜓𝑦+𝑙𝑢𝜓^′𝑦+𝐺𝜓𝑦𝜓^′𝑦, (9) in the same form as (4) and |𝑎_𝑛𝜚^1−𝑛|≤1, |𝑑_𝑛𝜚^1−𝑛|≤1 for 𝑛=2,3,… by (8).
Consider a solution 𝜓(𝑧) of (4) in the formal Dirichlet series (7); that is, 𝜓(𝑧)=∑^∞_𝑛=1𝑏_𝑛𝑏^−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 𝛽𝑏^−𝜆−𝑎₀+𝑑₀ln𝑏𝑏₁=0,𝛽𝑏^−𝑛𝜆−𝑎₀+𝑑₀𝑛𝑏_𝑛=−𝛼^𝑛−1_𝑖=1𝑖𝑏_𝑖𝑏_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝑖𝑑_𝑡+𝑎_𝑡^𝑡_𝑘=1^𝑚_𝑙=0𝑐_𝑙𝑏^{𝑙𝑘𝑙𝜆}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥2. (10)(11)
If 𝑏₁=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 𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}−1𝑛𝑏_𝑛=−𝛼^𝑛−1_𝑖=1𝑖𝑏_𝑖𝑏_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝑖𝑑_𝑡+𝑎_𝑡^𝑡_𝑘=1^𝑚_𝑙=0𝑐_𝑙𝑏^{−𝑙𝑘𝑙𝜆}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥2. (12)
The sequence {𝑏_𝑛}^∞_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 lim_𝑛→∞𝛼𝑖𝑛𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}−1≤lim_𝑛→∞|𝛼|𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}=|𝛼|−1𝛽𝑏^−𝜆,ℜ𝜆>0,lim_𝑛→∞𝑖𝑑_𝑡+𝑎_𝑡∏^𝑡_𝑘=1∑^𝑚_𝑙=0𝑐_𝑙𝑏^{−𝑙𝑘𝑙𝜆}𝑛𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}−1≤lim_𝑛→∞𝑑_𝑡+𝑎_𝑡∏^𝑡_𝑘=1∑^𝑚_𝑙=0𝑐_𝑙𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}≤2−1𝛽𝑏^−𝜆,ℜ𝜆>0. (13)
This implies that there exists a constant 𝑀>0 such thatlim_𝑛→∞𝛼𝑖𝑛𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}−1≤𝑀,lim_𝑛→∞𝑖𝑑_𝑡+𝑎_𝑡∏^𝑡_𝑘=1∑^𝑚_𝑙=0𝑐_𝑙𝑏^{𝑙𝑘𝑙𝜆}𝑛𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}−1≤𝑀,∀𝑛≥2,ℜ𝜆>0. (14)
Therefore, from (12) we obtain 𝑏_𝑛≤𝑀^𝑛−1_𝑖=1𝑏_𝑖𝑏_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥2. (15)
In order to construct a majorant series of (7), we consider the implicit functional equation 𝐻(𝑧)=|𝜏|𝑧+𝑀𝐻²(𝑧)+𝐻²(𝑧)(𝑧).1−𝐻 (16)
Define the function 𝜔(𝑧,𝐻)∶=𝜔(𝑧,𝐻,𝜏,𝑀)=𝐻−|𝜏|𝑧−𝑀𝐻²+𝐻²1−𝐻 (17) for (𝑧,𝐻) in a neighborhood of the origin. Then 𝜔(0,0)=0, 𝜔^′_𝐻(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)=−𝜔^′_𝑧(0,0)/𝜔^′_𝐻(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 𝜔(𝑧,𝐻(𝑧))=𝐻(𝑧)−|𝜏|𝑧−𝑀𝐻²(𝑧)+𝐻²(𝑧)1−𝐻(𝑧)=0. (18)
Choosing 𝐵₁=|𝜏| and putting 𝐻(𝑧)=^∞_𝑛=1𝐵_𝑛𝑏^−𝑛𝑧 (19) in (18), we can determine all coefficients recursively by 𝐵₁=|𝜏| and 𝐵_𝑛=𝑀^𝑛−1_𝑖=1𝐵_𝑖𝐵_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝐵_𝑖𝐵_𝑙1𝐵_𝑙2⋯𝐵_𝑙𝑡,𝑛≥2. (20)
Moreover, it is easy to see from (15) 𝑏_𝑛≤𝐵_𝑛,𝑛=1,2,…. (21)
It follows that the power series 𝜙(𝑧)=^∞_𝑛=1𝑏_𝑛𝑧^𝑛 (22) is also convergent as |𝑧|<𝛿. So there exists 𝜌≤𝛿 such that Dirichlet series (7) is convergent in 𝑆_𝜌.
Furthermore, one has lim_{ℜ𝑧→+∞}𝑏^−𝑧=lim_{ℜ𝑧→+∞}𝑏^−ℜ𝑧[(cos(ℑ𝑧ln𝑏))−𝑖sin(ℑ𝑧ln𝑏)]=0. (23)
Thus lim_{ℜ𝑧→+∞}𝜓(𝑧)=lim_{ℜ𝑧→+∞}∑^∞_𝑛=1𝑏_𝑛𝑏^−𝑛𝑧=0. The proof is complete.

We observe that 𝜇=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 𝜃=𝑎₀+𝜃₀=𝑎₀+1𝑎₁+𝜃₁=⋯, (24) denoted simply by 𝜃=[𝑎₀,𝑎₁,…,𝑎_𝑛,…], where 𝑎_𝑗’s and 𝜃_𝑗’s are calculated by the algorithm: (a) 𝑎₀=[𝜃], 𝜃₀={𝜃} and (b) 𝑎_𝑛=[1/𝜃_𝑛−1], 𝜃_𝑛={1/𝜃_𝑛−1} for all 𝑛≥1. Define the sequences (𝑝_𝑛)_𝑛∈ℕ and (𝑞_𝑛)_𝑛∈ℕ as follows: 𝑞₋₂=1,𝑞₋₁=0,𝑞_𝑛=𝑎_𝑛𝑞_𝑛−1+𝑞_𝑛−2,𝑝₋₂=0,𝑝₋₁=1,𝑝_𝑛=𝑎_𝑛𝑝_𝑛−1+𝑝_𝑛−2. (25)

It is easy to show that 𝑝_𝑛/𝑞_𝑛=[𝑎₀,𝑎₁,…,𝑎_𝑛]. Thus, for every 𝜃∈ℝ⧵ℚ we associate, using its convergence, an arithmetical function 𝐵(𝜃)=∑_𝑛≥0(log(𝑞_𝑛+1)/𝑞_𝑛). 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 𝑎_𝑛+1≤𝑐𝑒^𝑎𝑛 for all 𝑛≥0, where 𝑐>0 is a constant, then 𝜃=[𝑎₀,𝑎₁,…,𝑎_𝑛,…] 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 𝐴_𝑘=‖‖≤1𝑛≥0∣𝑛𝜃8𝑞_𝑘,𝐸_𝑘=max𝑞_𝑘,𝑞_𝑘+14,𝜂_𝑘=𝑞_𝑘𝐸_𝑘. (26)

Let 𝐴^∗_𝑘 be the set of integers 𝑗≥0 such that either 𝑗∈𝐴_𝑘 or for some 𝑗₁ and 𝑗₂ in 𝐴_𝑘, with 𝑗₂−𝑗₁<𝐸_𝑘, one has 𝑗₁<𝑗<𝑗₂ and 𝑞_𝑘 divides 𝑗−𝑗₁. For any integer 𝑛≥0, define 𝑙_𝑘(𝑛)=max1+𝜂_𝑘𝑛𝑞_𝑘−2,𝑚_𝑛𝜂_𝑘1+𝑛𝑞_𝑘−1, (27) where 𝑚_𝑛=max{𝑗∣0≤𝑗≤𝑛,𝑗∈𝐴^∗_𝑘}. We then define the function ℎ_𝑘∶ℕ→ℝ₊ as follows: ℎ_𝑘(𝑛)=𝑚_𝑛+𝜂_𝑘𝑛𝑞_𝑘−1,if𝑚_𝑛+𝑞_𝑘∈𝐴^∗_𝑘,𝑙_𝑘(𝑛),if𝑚_𝑛+𝑞_𝑘∉𝐴^∗_𝑘. (28)

Let 𝑔_𝑘(𝑛)∶=max(ℎ_𝑘(𝑛),[𝑛/𝑞_𝑘]), and define 𝑘(𝑛) by the condition 𝑞_𝑘(𝑛)≤𝑛≤𝑞_𝑘(𝑛)+1. Clearly, 𝑘(𝑛) is nondecreasing. Moreover, the function 𝑔_𝑘 is nonnegative. Then we are able to state the following result.

Lemma 2 (Davie’s lemma [12]). Let 𝐾(𝑛)=𝑛log2+∑^𝑘(𝑛)_𝑗=0𝑔_𝑗(𝑛)log(2𝑞_𝑗+1). Then(a) there is a universal constant 𝛾>0 (independent of 𝑛 and 𝜃) such that 𝐾(𝑛)≤𝑛^𝑘(𝑛)_𝑗=0log𝑞_𝑗+1𝑞_𝑗+𝛾, (29)(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 𝑆_𝜌={𝑧∈ℂ∶ℜ𝑧>−ln𝜌/ln|𝑏|,−∞<ℑ𝑧<+∞} for a certain constant 𝜌>0, which satisfies lim_{ℜ𝑧→+∞}𝜓(𝑧)=0.

Proof. As in the proof of Theorem 1, 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 {𝑏_𝑛}^∞_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 𝑏_𝑛≤𝑀₁𝑏^{−(𝑛−1)𝜆}−1^𝑛−1_𝑖=1𝑏_𝑖𝑏_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥2, (30) where 𝑀₁=max{|𝛼|/|𝛽|,2/|𝛽|}>0.
To construct a governing series of (7), we consider the implicit functional equation 𝜔𝑧,𝑈,𝜏,𝑀₁=0, (31) 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 𝑈^′_𝑧(0,𝜏,𝑀₁)=|𝜏|. Thus 𝑈(𝑧,𝜏,𝑀₁) in (31) can be expanded into a convergent series 𝑈𝑧,𝜏,𝑀₁=^∞_𝑛=1𝐶_𝑛𝑧^𝑛, (32) in a neighborhood of the origin. Replacing (32) into (31) and comparing coefficients, we obtain that 𝐶₁=|𝜏| and 𝐶_𝑛=𝑀₁^𝑛−1_𝑖=1𝐶_𝑖𝐶_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝐶_𝑖𝐶_𝑙1𝐶_𝑙2⋯𝐶_𝑙𝑡,𝑛≥2. (33)
Note that the series (32) converges in a neighborhood of the origin. So, there is a constant 𝑇>0 such that 𝐶_𝑛<𝑇^𝑛,𝑛≥1. (34)
Now, we can deduce, by induction, that |𝑏_𝑛|≤𝐶_𝑛𝑒^{𝐾(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 |𝑏_𝑗|≤𝐶_𝑗𝑒^{𝐾(𝑗−1)}, 𝑗≤𝑛. From (30) and Lemma 2 we obtain 𝑏_𝑛+1≤𝑀₁𝑏^−𝑛𝜆×−1_𝑛𝑖=1𝑏_𝑖𝑏_{𝑛−𝑖+1}+^𝑛_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖+1𝑡=1,2,…,𝑛−𝑖+1}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡≤𝑀₁𝑏^−𝑛𝜆−1^𝑛_𝑖=1𝐶_𝑖𝐶_{𝑛−𝑖+1}𝑒^{𝐾(𝑖−1)+𝐾(𝑛−𝑖)}+^𝑛_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖+1𝑡=1,2,…,𝑛−𝑖+1}𝐶_𝑖𝐶_𝑙1𝐶_𝑙2⋯𝐶_𝑙𝑡×𝑒^{𝐾(𝑖−1)+𝐾(𝑙1−1)+⋯+𝐾(𝑙𝑡−1)}. (35)
Note that𝐾()()()(𝑛)𝑏𝑖−1+𝐾𝑛−𝑖≤𝐾𝑛−1≤𝐾+log^−𝜆𝑛,𝐾𝑙−1₁−1+𝐾𝑙₂−1+⋯+𝐾𝑙_𝑡()(),𝐾()()()(𝑛)𝑏−1≤𝐾𝑛−𝑖+1−𝑡≤𝐾𝑛−𝑖𝑖−1+𝐾𝑛−𝑖≤𝐾𝑛−1≤𝐾+log^−𝜆𝑛−1, (36) then 𝑏_𝑛+1≤𝑒^{𝐾(𝑛−1)}𝑀₁𝑏^−𝑛𝜆×−1_𝑛𝑖=1𝐶_𝑖𝐶_{𝑛−𝑖+1}+^𝑛_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖+1𝑡=1,2,…,𝑛−𝑖+1}𝐶_𝑖𝐶_𝑙1𝐶_𝑙2⋯𝐶_𝑙𝑡≤𝐶_𝑛+1𝑒^𝐾(𝑛) (37) 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 |𝑏_𝑛|≤𝐶_𝑛𝑒^{𝐾(𝑛−1)}≤𝑇^𝑛𝑒^{(𝑛−1)(𝐵(𝜃)+𝛾)}; that is, lim_𝑛→∞sup(|𝑏_𝑛|^1/𝑛)≤lim_𝑛→∞sup(𝑇𝑒^{((𝑛−1)/𝑛)(𝐵(𝜃)+𝛾)})=𝑇𝑒^{𝐵(𝜃)+𝛾}, which shows that the series (7) converges for |𝑧|<𝜌=min{𝛿₃,(𝑇𝑒^{𝐵(𝜃)+𝛾})⁻¹}. So does series (22) in 𝑆_𝜌. Similarly lim_{ℜ𝑧→+∞}𝜓(𝑧)=0, as proved in Theorem 1.
The next theorem is devoted to the case of (I3), where 𝑏^−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 {𝐴_𝑛}^∞_𝑛=1 be a sequence defined by 𝐴₁=𝜏 and 𝐴_𝑛=𝜉𝑀₁^𝑛−1_𝑖=1𝐴_𝑖𝐴_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝐴_𝑖𝐴_𝑙1𝐴_𝑙2⋯𝐴_𝑙𝑡,𝑛≥2, (38) where 𝜉=max{1,|𝛼^𝑖−1|⁻¹,𝑖=1,2…,𝑝−1} and 𝑀₁ is defined in Theorem 3.

Theorem 4. Suppose that (I3) holds and 𝑝 is given as above mentioned. Let {𝑏_𝑛}^∞_𝑛=1 be determined recursively by 𝑏₁=𝜏 and 𝛽𝑏^−𝜆𝑏^{−(𝑛−1)𝜆}−1𝑛𝑏_𝑛=Ω(𝑛,𝜆),𝑛=2,3,…, (39) where Ω(𝑛,𝜆)=−𝛼^𝑛−1_𝑖=1𝑖𝑏_𝑖𝑏_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝑖𝑑_𝑡+𝑎_𝑡^𝑡_𝑘=1^𝑚_𝑙=0𝑐_𝑙𝑏^{−𝑙𝑘𝑙𝜆}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡. (40)
If Ω(𝜈𝑝+1,𝜆)=0 for all 𝜈=1,2,…, then (4) has an analytic solution 𝜓(𝑧)=𝜙(𝑏^−𝑧) in the half plane 𝑆_𝜌∶={𝑧∈ℂ∶ℜ𝑧>−ln𝜌/ln|𝑏|,−∞<ℑ𝑧<+∞} for a certain 𝜌>0, where 𝜙 is an analytic function of the form (22) in 𝑈_𝜌(0)={𝑧∣|𝑧|<𝜌} such that 𝜙(0)=0, and 𝜙^(𝜈𝑝+1)(0)=(𝜈𝑝+1)!𝜏_𝜈𝑝+1, for all 𝜈=0,1,2,…, where 𝜏^′_𝜈𝑝+1𝑠 are arbitrary constants satisfying the inequality |𝜏_𝜈𝑝+1|≤𝐴_𝜈𝑝+1 and the sequence {𝐴_𝑛}^∞_𝑛=1 is defined in (38). The other derivatives at 0 satisfy that 𝜙^(𝑖)(0)=𝑖!𝑏_𝑖 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 𝑆_𝜌∶={𝑧∈ℂ∶ℜ𝑧>−ln𝜌/ln|𝑏|,−∞<ℑ𝑧<+∞} for any 𝜌>0.

Proof. Analogously to the proof of Theorem 1, 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 𝑏_𝜈𝑝+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 𝑏_𝜈𝑝+1=𝜏_𝜈𝑝+1 arbitrarily such that 𝜏_𝜈𝑝+1≤𝐴_𝜈𝑝+1,𝜈=0,1,2,…, (41) where 𝐴_𝜈𝑝+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 𝑏_𝑛≤𝜉𝑀₁^𝑛−1_𝑖=1𝑏_𝑖𝑏_𝑛−𝑖+^𝑛−1_{𝑖=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}𝑏_𝑖𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡 (42) for all 𝑛≠𝜈𝑝+1, 𝜈=0,1,2,…, where 𝑀₁ is defined in Theorem 3. Let 𝑉𝑧,𝜏,𝜉𝑀₁=^∞_𝑛=1𝐴_𝑛𝑧^𝑛,𝐴₁=|𝜏|. (43)
It is easy to check that there exists a constant 𝜌>0, for |𝑧|<𝜌, and (43) satisfies the implicit functional equation 𝜔𝑧,𝑉,𝜏,𝜉𝑀₁=0, (44) 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 𝑉^′_𝑧(0,𝜏,𝜉𝑀₁)=|𝜏|≠0. Thus (43) converges in a neighborhood of the origin. Moreover, it is easy to show that, by induction, 𝑏_𝑛≤𝐴_𝑛,𝑛=1,2,…. (45)
By inequality (45) we see that the series (22) converges in 𝑈_𝜌(0)={𝑧∣|𝑧|<𝜌}. Thus series (7) converges in 𝑆_𝜌. 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=𝑒^{2𝜋𝑖𝜃}, where 𝜃∈ℝ⧵ℚ is a Brjuno number, 𝐵(𝜃)=∑^∞_𝑘=0(log(𝑞_𝑘+1)/𝑞_𝑘)<∞, where {𝑝_𝑘/𝑞_𝑘} denotes the sequence of partial fraction of the continued fraction expansion of 𝜃, said to satisfy the Brjuno condition;(H3)𝜇=𝑒^{2𝜋𝑖𝑞/𝑝} for some integers 𝑝∈ℕ with 𝑝≥2 and 𝑞∈ℤ⧵{0}, and 𝜇≠𝑒^{2𝜏𝑖𝑙/𝑘} for all 1≤𝑘≤𝑝−1 and 𝑙∈ℤ⧵{0}.

Theorem 5. Suppose that (H1) holds and that 𝑎₀≠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 𝜑(𝑧)=^∞_𝑛=1𝑏_𝑛𝑧^𝑛,𝑏₁=𝜂 (46) be the formal series of the solution 𝜑 for (5). We are going to determine {𝑏_𝑛}^∞_𝑛=1. Substituting (3) and (46) into (5) and comparing coefficients, we obtain 𝛽𝜇−𝑎₀+𝑑₀𝑏₁(=0,𝑛+1)𝛽𝜇^𝑛+1−𝑎₀+𝑑₀𝑏_𝑛+1=−𝛼^𝑛−1_𝑖=0(𝑖+1)𝑏_𝑖+1𝑏_𝑛−𝑖+^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}(𝑖+1)𝑑_𝑡+𝑎_𝑡^𝑡_𝑘=1^𝑚_𝑙=0𝑐_𝑙𝜇^𝑙𝑙𝑘×𝑏_𝑖+1𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥1. (47)(48)
If 𝑏₁=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 (𝑛+1)𝛽𝜇𝜇^𝑛−1𝑏_𝑛+1=−𝛼^𝑛−1_𝑖=0(𝑖+1)𝑏_𝑖+1𝑏_𝑛−𝑖+^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}(𝑖+1)𝑑_𝑡+𝑎_𝑡^𝑡_𝑘=1^𝑚_𝑙=0𝑐_𝑙𝜇^𝑙𝑙𝑘×𝑏_𝑖+1𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥1. (49)
From (49) the sequence {𝑏_𝑛}^∞_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 𝑎_𝑛≤𝑀₂𝑟^𝑛,𝑑_𝑛≤𝑀₂𝑟^𝑛. (50)
Note that since 1≤𝑖≤𝑛−1, then there exists some positive number 𝑀₃ as follows: (𝑖+1)∏^𝑡_𝑘=1∑^𝑚_𝑙=0𝑐_𝑙𝜇^𝑙𝑙𝑘(𝑛+1)𝛽𝜇𝜇^𝑛≤1−1𝛽𝜇𝜇^𝑛−1≤𝑀₃,𝑖+1(𝑛+1)𝛽𝜇𝜇^𝑛≤1−1𝛽𝜇𝜇^𝑛−1≤𝑀₃,𝛼(𝑖+1)(𝑛+1)𝛽𝜇𝜇^𝑛≤|𝛼|−1𝛽𝜇𝜇^𝑛−1≤𝑀₃, (51) then we have 𝑏_𝑛+1≤𝐿^𝑛−1_𝑖=0𝑏_𝑖+1𝑏_𝑛−𝑖+^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}1𝑟^𝑡𝑏_𝑖+1𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥1, (52) where 𝐿∶=max{2𝑀₂𝑀₃,𝑀₃}>0. Let Θ()𝜂𝑊𝑧,𝑊=𝑊−𝑧−𝐿²/𝑟1−(𝑊/𝑟)+𝑊² (53) for (𝑧,𝑊) from a neighborhood of the origin. Since Θ(0,0)=0,Θ^′_𝑊(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 {𝐷_𝑛}^∞_𝑛=1 by 𝐷₁=|𝜂|≠0 and 𝐷_𝑛+1=𝐿^𝑛−1_𝑖=0𝐷_𝑖+1𝐷_𝑛−𝑖+^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}1𝑟^𝑡𝐷_𝑖+1𝐷_𝑙1𝐷_𝑙2⋯𝐷_𝑙𝑡,𝑛≥1. (54)
By (52) we see that 𝑏_𝑛≤𝐷_𝑛,𝑛≥1. (55)
Let 𝑃(𝑧)=^∞_𝑛=1𝐷_𝑛𝑧^𝑛,𝐷₁=𝜂, (56) with the recursive law of {𝐷_𝑛}^∞_𝑛=1. Then 𝑃²(𝑧)=^∞_𝑛=0𝐷_𝑛+1𝑧^𝑛+1∞_𝑛=1𝐷_𝑛𝑧^𝑛=^∞_𝑛=1^𝑛−1_𝑖=0𝐷_𝑖+1𝐷_𝑛−𝑖𝑧^𝑛+1,𝑃²(𝑧)/𝑟(𝑃(𝑧))(𝑧)𝑃(𝑧)1−/𝑟=𝑃/𝑟(𝑃(𝑧))=1−/𝑟_∞𝑛=0𝐷_𝑛+1𝑧^𝑛+1∞_𝑛=1𝑃(𝑧)𝑟^𝑛=^∞_𝑛=0𝐷_𝑛+1𝑧^𝑛+1∞_𝑛=11𝑟^𝑛∞_𝑛=1𝐷_𝑛𝑧^𝑛𝑛=^∞_𝑛=0𝐷_𝑛+1𝑧^𝑛+1∞_{𝑛=1𝑙1+𝑙2+⋯+𝑙𝑡=𝑛𝑡=1,2,…,𝑛}1𝑟^𝑡𝐷_𝑙1𝐷_𝑙2⋯𝐷_𝑙𝑡𝑧^𝑛=^∞_𝑛=1^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}1𝑟^𝑡𝐷_𝑖+1𝐷_𝑙1𝐷_𝑙2⋯𝐷_𝑙𝑡𝑧^𝑛+1. (57)
Then we have 𝑃²(𝑧)/𝑟1−(𝑃(𝑧)/𝑟)+𝑝²(𝑧)=1𝐿^∞_𝑛=1𝐷_𝑛+1𝑧^𝑛+1=1𝐿𝑃(𝑧)−𝜂𝑧, (58) that is 𝑃(𝑧)−𝜂𝑃𝑧−𝐿²(𝑧)/𝑟1−(𝑃(𝑧)/𝑟)+𝑝²(𝑧)=0. (59)
This shows that 𝑃(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 Theorem 3.

Theorem 6. Suppose that (H2) holds and that 𝑎₀≠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 Theorem 4.

Theorem 7. Suppose that (H3) holds, 𝑎₀≠0, 𝑑₀≠0, and 𝑝 is given as above mentioned. Let {𝑏_𝑛}^∞_𝑛=1 be determined recursively by 𝑏₁=𝜂 and (𝑛+1)𝛽𝜇𝜇^𝑛−1𝑏_𝑛+1=Ξ𝑛+1,𝜇, (60) where Ξ𝑛+1,𝜇=−𝛼^𝑛−1_𝑖=0(𝑖+1)𝑏_𝑖+1𝑏_𝑛−𝑖+^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}(𝑖+1)𝑑_𝑡+𝑎_𝑡^𝑡_𝑘=1^𝑚_𝑙=0𝑐_𝑙𝜇^𝑙𝑙𝑘×𝑏_𝑖+1𝑏_𝑙1𝑏_𝑙2⋯𝑏_𝑙𝑡,𝑛≥1. (61)
If  Ξ(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)!𝜂_𝑙𝑝+1, where 𝜂^′_𝑙𝑝+1𝑠 are arbitrary constants satisfying the inequality |𝜂_𝑙𝑝+1|≤𝑑_𝑙𝑝+1, 𝑙=1,2,… and the sequence {𝑑_𝑛}^∞_𝑛=1 is defined as follows: 𝑑₁=|𝜂| and 𝑑_𝑛+1=Γ𝐿₁^𝑛−1_𝑖=0𝑑_𝑖+1𝑑_𝑛−𝑖+^𝑛−1_{𝑖=0𝑙1+𝑙2+⋯+𝑙𝑡=𝑛−𝑖𝑡=1,2,…,𝑛−𝑖}1𝑟^𝑡𝑑_𝑖+1𝑑_𝑙1𝑑_𝑙2⋯𝑑_𝑙𝑡,𝑛≥1, (62) where Γ=max{1,|𝜇^𝑖−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 Theorems 1, 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 𝑆_𝜌.

Proof. In view of Theorems 1, 3, or 4, we may find a sequence {𝑏_𝑛}^∞_𝑛=1 such that the function 𝜓(𝑧) of the form (7) is an analytic solution of (4) in the half plane 𝑆_𝜌. 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 𝑥^′(𝑧)=𝜓^′𝜓⁻¹(𝑧)+𝜆𝜓^′𝜓⁻¹(𝑧),𝑥^𝑘(𝑧)=𝜓𝜓⁻¹(𝑧)+𝑘𝜆, (63) then from (4) 𝛼𝑧+𝛽𝑥^′(𝑧)=𝛼𝑧+𝛽𝜓^′𝜓⁻¹(𝑧)+𝜆𝜓^′𝜓⁻¹(𝑧)=𝐹^𝑚_𝑙=0𝑐_𝑙𝜓𝜓⁻¹(𝑧)+𝑙𝜆+𝐺𝜓𝜓⁻¹(𝑧)=𝐹^𝑚_𝑙=0𝑐_𝑙𝑥^𝑙(𝑧)+𝐺(𝑧). (64)
This shows that 𝑥(𝑧)=𝜓(𝜓⁻¹(𝑧)+𝜆) satisfies (2). The proof is complete.

Under the hypothesis of Theorem 1 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.

When 0<|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 Theorems 5, 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 Theorems 5–7 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, 𝑥^′(𝑧)=𝜇𝜑^′𝜇𝜑⁻¹(𝑧)𝜑^′𝜑⁻¹(𝑧),𝑥^𝑙(𝑧)=𝜑𝜇^𝑙𝜑⁻¹(𝑧), (65) so from (5) we have 𝛼𝑧+𝛽𝑥^′(𝑧)=𝛼𝑧+𝛽𝜇𝜑^′𝜇𝜑⁻¹(𝑧)𝜑^′𝜑⁻¹(𝑧)=𝐹^𝑚_𝑙=0𝑐_𝑙𝜑𝜇^𝑙𝜑⁻¹(𝑧)+𝐺𝜑𝜑⁻¹(𝑧)=𝐹^𝑚_𝑙=0𝑐_𝑙𝑥^𝑙(𝑧)+𝐺(𝑧). (66)
Therefore, 𝑥(𝑧)=𝜑(𝜇𝜑⁻¹(𝑧)) satisfies (2). The proof is complete.

4. Example

Example 1. Consider the equation 𝑧+𝑥^′(𝑧)=4𝑒^{(1/2)𝑥(𝑥(𝑧))}−𝑒^𝑧−176. (67)
It is in the form of (2), where 𝛼=𝛽=1,  𝑐₀=𝑐₁=0,  𝑐₂=1/2,  𝑚=2,  𝐹(𝑧)=4𝑒^𝑧−(13/3)=∑^∞_𝑛=1(4𝑧^𝑛/𝑛!)−(1/3), and 𝐺(𝑧)=(3/2)−𝑒^𝑧=−∑^∞_𝑛=1(𝑧^𝑛/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 𝜓(𝑧)𝜓^′(𝑧)+𝜓^′()1𝑧+𝜆=𝐹2𝜓(𝑧+2𝜆)𝜓^′(𝑧)+𝐺𝜓(𝑧)𝜓^′(𝑧), (68) and (67) itself have an analytic solution in the half plane 𝑆_𝜌={𝑧∣ℜ𝑧>−(ln𝜌/ln𝑏),−∞<ℑ𝑧<+∞}. In the routine in the proofs of our theorems we can calculate the solutions 𝑥(𝑧)=16𝑧−35𝑧36²−551𝑧1944³+⋯. (69)

Example 2. Consider the equation 𝑧−𝑥^′(𝑧)=12𝑥(𝑧)−14𝑥(𝑥(𝑧))+𝑒^𝑧−14. (70)
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, 𝑎_𝑛=0(𝑛≥2), 𝑑₀=1/4, 𝑑₁=1, 𝐹(𝑧)=−𝑧+(1/2), and 𝐺(𝑧)=𝑒^𝑧−(3/4)=∑^∞_𝑛=1(𝑧^𝑛/𝑛!)+(1/4). Clearly, ∑²_𝑙=0|𝑐_𝑙|<1 and 𝜇=(1/𝛽)(𝑎₀+𝑑₀)=−3/4 such that 0<|𝜇|<1. By Theorem 5 the corresponding auxiliary equation 𝜑(𝑧)𝜑^′(𝑧)−𝜇𝜑^′=1𝜇𝑧2𝜑−1𝜇𝑧4𝜑𝜇²𝑧𝜑^′(𝑧)+𝑒^𝜑(𝑧)−14𝜑^′(𝑧) (71) 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 𝑥(𝑧)1=𝜇𝑧+8𝜇𝜇−2𝑧²+113!𝜇𝑧16𝜇−1𝜇−2𝜇+2−1³+⋯. (72)

Acknowledgments

This work was supported by the Natural Science Foundation of Shandong Province (ZR2012AM017) and the Natural Science Foundation of Shandong Province (2011ZRA07006).