Journal of Applied Mathematics

Volume 2013 (2013), Article ID 726076, 9 pages

http://dx.doi.org/10.1155/2013/726076

Research Article

Analytic Solutions of an Iterative Functional Differential Equations Near Regular Points

Lingxia Liu

Department of Mathematics, Weifang University, Weifang, Shandong 261061, China

Received 13 August 2012; Accepted 5 March 2013

Academic Editor: Juan Torregrosa

Copyright © 2013 Lingxia Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 equations

are found. In this paper, we will be concerned with analytic solutions of the functional differential equation

where , , , are complex numbers, , , and that denote the th iterate of a map . In general, , 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 depending on the eigenvalue of the linearization of at its fixed point 0 in the complex plane. For technical reasons, in [6, 7], only the situation of off the unit circle in and the situation of on the circle with the Diophantine condition, “, is not a root of unity, and , for some positive constant ”, are discussed. The Diophantine condition requires to be far from all roots of unity that the fixed point 0 is irrationally neutral. In this paper, besides the situation that is the inside of the unit circle , we break the restriction of the Diophantine condition and study the situations that the constant in (5) (or , is a complex constant, and is in (4)) is resonance and a root of unity in the complex plane near resonance under the Brjuno condition.

2. Discussion on Auxiliary Equations

In this section we assume that both and are analytic functions in a neighborhood of the origin, that is, 0 is a regular point, and have power series expansions

If there exists a complex constant and an invertible function such that is well defined, then letting , we can formally transform (2) into the differential-difference equation

The indeterminate constant will be discussed in the following cases:

(I1); (I2), and is a Brjuno number [9, 10]; that is, , where denotes the sequence of partial fraction of the continued fraction expansion of , said to satisfy the Brjuno condition;(I3) for some integers with and , and for all and .

Take notations .

A change of variable further transforms (4) into the functional differential equation

where is a complex constant. The solution of this equation has properties similar to those of (4). If (5) has an invertible solution , which satisfies the initial value conditions

then we can show that is an analytic solution of (2).

Theorem 1. Suppose that (I1) holds, then (5) has an analytic solution of the form

in the half plane for a certain constant , which satisfies .

Proof. Since and are analytic in a neighborhood of the origin and have the power series expansion (3), there exists a positive such that

Without loss of generality, we can assume that ; that is, , for . In fact, let , , and put , , , and . Then (4) can be rewritten as

in the same form as (4) and , for by (8).

Consider a solution of (4) in the formal Dirichlet series (7); that is, , where is a complex constant and . Substituting series (3) and (7) of , , and in (4) and comparing coefficients we obtain that

If , then (4) has a trivial solution . Assume that , because ; from (10) we have . From (11) we obtain that

The sequence 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 , so we have

This implies that there exists a constant such that

Therefore, from (12) we obtain

In order to construct a majorant series of (7), we consider the implicit functional equation

Define the function

for in a neighborhood of the origin. Then , . Thus, there exists a unique function analytic in a neighborhood of zero; that is, there is a constant , as , the function is analytic, such that , and . Since , there is a constant , such that for . Therefore, as , the function satisfies the equation

Choosing and putting

in (18), we can determine all coefficients recursively by and

Moreover, it is easy to see from (15)

It follows that the power series

is also convergent as . So there exists such that Dirichlet series (7) is convergent in .

Furthermore, one has

Thus . The proof is complete.

We observe that 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 , we let denote its integer part, and its fractional part. Then every irrational number has a unique expression of the Gauss' continued fraction

denoted simply by , where ’s and ’s are calculated by the algorithm: (a) , and (b) , for all . Define the sequences and as follows:

It is easy to show that . Thus, for every we associate, using its convergence, an arithmetical function . We say that is a Brjuno number or that it satisfies Brjuno condition if . The Brjuno condition is weaker than the Diophantine condition. For example, if for all , where is a constant, then is a Brjuno number but is not a Diophantine number. So, the case (I2) contains both Diophantine condition and a part of “near” resonance. let

Let be the set of integers such that either or for some and in , with , one has and divides . For any integer , define

where . We then define the function as follows:

Let , and define by the condition . Clearly, is nondecreasing. Moreover, the function is nonnegative. Then we are able to state the following result.

Lemma 2 (Davie’s lemma [12]). Let . Then

(a) there is a universal constant (independent of and ) such that

(b) for all and , and(c).

Now we state and prove the following theorem under Brjuno condition.

Theorem 3. Suppose that (I2) holds. Then (4) has an analytic solution of the form (7) in the half plane for a certain constant , which satisfies .

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 we can uniquely determine the sequence recursively by (12). In fact, in view of (I2) we see that satisfies the conditions of Lemma 2, and from (12) we have

where .

To construct a governing series of (7), we consider the implicit functional equation

where 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 in a neighborhood of the origin; that is, there is a constant , as , the function is analytic such that and . Thus in (31) can be expanded into a convergent series

in a neighborhood of the origin. Replacing (32) into (31) and comparing coefficients, we obtain that and

Note that the series (32) converges in a neighborhood of the origin. So, there is a constant such that

Now, we can deduce, by induction, that for , where is defined in Lemma 2. In fact , for inductive proof, and we assume that , . From (30) and Lemma 2 we obtain

Note that

then

as desired. Moreover, from Lemma 2, we know that for some universal constant . Then from (34) we have ; that is, , which shows that the series (7) converges for . So does series (22) in . Similarly , as proved in Theorem 1.

The next theorem is devoted to the case of (I3), where is not only on the unit circle in 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 be a sequence defined by and

where and is defined in Theorem 3.

Theorem 4. Suppose that (I3) holds and is given as above mentioned. Let be determined recursively by and

where

If for all , then (4) has an analytic solution in the half plane for a certain , where is an analytic function of the form (22) in such that , and , for all , where are arbitrary constants satisfying the inequality and the sequence is defined in (38). The other derivatives at satisfy that for . Otherwise, if for some , then (4) has no analytic solutions in the half plane for any .

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 in (8). Taking (3) and (7) in (4) and defining , we obtain (12) or (39). If for all natural numbers , then for each , , the corresponding has infinitely many choices in ; that is, the formal series solution (7) or (22) defines a family of solutions with infinitely many parameters. Choose arbitrarily such that

where is defined by (38). In what follows, we prove that the formal series solution (7) converges in a neighborhood of the origin. Observe that for . It follows from (12) or (39) that

for all , , where is defined in Theorem 3. Let

It is easy to check that there exists a constant , for , and (43) satisfies the implicit functional equation

where is defined in (17). Moreover, similarly to the proof of Theorem 1, we can prove that (44) has a unique analytic solution in a neighborhood of the origin such that and . Thus (43) converges in a neighborhood of the origin. Moreover, it is easy to show that, by induction,

By inequality (45) we see that the series (22) converges in . Thus series (7) converges in . 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);(H2), where is a Brjuno number, , where denotes the sequence of partial fraction of the continued fraction expansion of , said to satisfy the Brjuno condition;(H3) for some integers with and , and for all and .

Theorem 5. Suppose that (H1) holds and that , . Then in a neighborhood of the origin (5) has an analytic solution satisfying , .

Proof. Let

be the formal series of the solution for (5). We are going to determine . Substituting (3) and (46) into (5) and comparing coefficients, we obtain

If , then (5) has a trivial solution . Assume that ; from (47) we can choose , then (48) can be changed into

From (49) the sequence 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 , for any fixed there exists a constant such that

Note that since , then there exists some positive number as follows:

then we have

where . Let

for from a neighborhood of the origin. Since , there exists a unique function , analytic in a neighborhood of the origin, such that , , and . Then define a sequence by and

By (52) we see that

Let

with the recursive law of . Then

Then we have

that is

This shows that , and . So we have . 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 , . Then in a neighborhood of the origin (5) has an analytic solution satisfying .

In the case (H3) we also obtain similarly an analogue to Theorem 4.

Theorem 7. Suppose that (H3) holds, , , and is given as above mentioned. Let be determined recursively by and

where

If   for all , then (5) has an analytic solution in a neighborhood of the origin such that , , and , where are arbitrary constants satisfying the inequality , and the sequence is defined as follows: and

where , and . Otherwise, if for some , 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 in a neighborhood of the origin, where is an analytic solution of (4) in the half plane .

Proof. In view of Theorems 1, 3, or 4, we may find a sequence such that the function of the form (7) is an analytic solution of (4) in the half plane . As in Theorem 1, . Since , the function is analytic in a neighborhood of the point . Thus is analytic in a neighborhood of the origin. Define which is analytic clearly. Note that

then from (4)

This shows that satisfies (2). The proof is complete.

Under the hypothesis of Theorem 1 the origin is a hyperbolic fixed point of , but under hypotheses of Theorems 3 and 4 it is not. Actually, when the constant satisfies the Brjuno condition, the norm of the eigenvalue of the linearized of at equals 1, but the eigenvalue is not a root of unity. Under (I3), the fixed point of is a resonance.

When , the same method is applicable and a similarly result can be obtained; that is, there exists a constant such that (4) has an analytic solution in the left-half plane .

Theorem 9. Under one of the conditions in Theorems 5, 6, or 7, (2) has an analytic solution of the form in a neighborhood of the origin, where is an analytic solution of (5) in a neighborhood of the origin.

Proof. In Theorems 5–7 we have found a solution of (5) in the form (46), which is analytic near . Since , , the function is also analytic near . Thus is analytic. Moreover,

so from (5) we have

Therefore, satisfies (2). The proof is complete.

4. Example

Example 1. Consider the equation

It is in the form of (2), where ,  ,  ,  ,  , and . Clearly both and are analytic near , , , , and . For arbitrary , let , the . By Theorem 1, there is a constant such that the corresponding auxiliary equation

and (67) itself have an analytic solution in the half plane . In the routine in the proofs of our theorems we can calculate the solutions

Example 2. Consider the equation

It is in the form of (2), where , , , , , , , , , , , , and . Clearly, and such that . By Theorem 5 the corresponding auxiliary equation

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

Acknowledgments

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

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