Abstract

We investigate an analytic solution of the second-order differential equation with a state derivative dependent delay of the form . Considering a convergent power series of an auxiliary equation with the relation , we obtain an analytic solution . Furthermore, we characterize a polynomial solution when is a polynomial.

1. Introduction

The functional differential equation,where all , , provides a mathematical model for a physical or biological system in which the rate of change of system is determined not only by its present state, but also by its history (see [1, 2]). In recent years, many authors studied the existence and the uniqueness of an analytic solution of a variety of these equations. In 1984, Eder [3] classified solutions of the functional differential equation by using the Banach fixed point theorem. Let and be nonzero complex constants and a complex function. The first-order functional differential equation , , and has been studied by Si and Cheng [4], Qiu and Liu [5], and Zhang [6], respectively.

In 2001 [7], Si and Wang investigated the existence of analytic solution of the second-order functional differential equation:In 2009, Liu and Li [8] studied the equationObserve that (3) can be reduced to (2) by setting .

Next, the equationhas been studied by Si and Wang [9].

In order to obtain analytic solutions of (4), they constructed a corresponding auxiliary equation with parameter . The existence of solutions of an auxiliary equation depends on the condition of a parameter , such as is in the unit circle and is a root of unity which satisfies the Diophantine condition.

In this paper, we study the existence of analytic solutions of the second-order differential equation with a state derivative dependent delay of the formIf , then (5) reduces to (4).

Note that in this paper, we will study three cases of parameter in a corresponding auxiliary equation. One of them is the case that is a root of unity satisfying the Brjuno condition.

To construct an auxiliary equation, we setThenwhere is a complex constant. In particular, we haveApplying relations (6) and (8) to (5), we obtainWe construct the corresponding equation by differentiating both sides of (9) with respect to . This yields

2. Analytic Solutions of (10)

Consider the auxiliary equationwith initial value conditions and , where are complex numbers. Observe that if is an analytic solution of (11), then (10) has an analytic solution of the from . Equation (11) can be reduced equivalently to the integro-differential equationwhere and To construct analytic solutions of (12), we separate our study on the conditions of the parameter as follows:(H1);(H2), where is a Brjuno number; that is, , where denotes the sequence of partial fraction of the continued fraction expansion of ;(H3) for some integers with and , and for all and .

From now on, we let be an analytic function in a neighborhood of the origin. Then we represent by a power series .

Theorem 1. Let satisfy condition (H1). Then (11) has an analytic solutionin a neighborhood of the origin such that , , where is a nonzero complex number.

Proof. Since is analytic in a neighborhood of the origin, there exists a constant such that for . Substituting (13) into (12) and comparing coefficients of , we getand in general for The first expression allows us to choose and the second expression implies Consequently, the sequence is successively determined by the last expression in a unique manner. This implies that (11) has a formal power series solution.
Next, we show that the power series converges in a neighborhood of the origin. Since and for , there exists a positive constant such thatLet us define a power series , where a positive sequence is determined by , and for It follows that for That is, is a majorant series of We show that is analytic in a neighborhood of the origin. Note that if we let , thenConsider the equationSince is continuous in a neighborhood of the origin, , and , the implicit function theorem implies that there exists a unique function which is analytic in a neighborhood of the origin with a positive radius. Because is a majorant series of , is also analytic in a neighborhood of the origin with a positive radius. This completes the proof.

Now, we consider an analytic solution of the auxiliary equation (11) in the case of satisfies condition In order to study the existence of under the Brjuno condition, we will recall the definition of Brjuno number and some basic facts. As stated in [10], for a real number , we let be an integer part of and let be a fractional part of . Observe that if is an irrational number, then it has a unique expression of Gauss’s continued fractiondenoted simply by , where ’s and ’s are calculated by the following algorithm:(a), ;(b), for all .Define the sequences and by the following recursive relation:Note that . For each , we consider an arithmetical function When satisfies condition , we call it a Brjuno number. Consider in which for each , , where is a positive constant. We can show that is a Brjuno number but is not a Diophantine number. Therefore, Brjuno condition is weaker than the Diophantine condition. So, condition contains both Diophantine condition and a part of near resonance.

Let and be the sequence of partial denominators of Gauss’s continued fraction for As in [10], letLet be the set of integers such that either or for some and in , with , one has and divides . For any nonnegative integer , we definewhere . Let be a function defined byLet , and let be defined by condition . Note that is nondecreasing.

Lemma 2 (Davie’s lemma [11]). LetThen(a)there is a universal constant (independent of and ) such that(b) for all and ,(c).

Theorem 3. Assume that satisfies condition (H2). Then there exists an analytic solutionof (11) in a neighborhood of the origin such that , , where is a nonzero complex number.

Proof. We now imitate the proof of Theorem 1 with approximate new bound. The sequence is defined similar to the proof of Theorem 1. Note that and . Since and is analytic near the origin, there exists a positive constant so that for To construct a governing series of , we let be a nonnegative sequence determined by , and for all From this construction, we can demonstrate that a power series satisfies the implicit functional equationwith and . This yields the power series converges in a neighborhood of the origin. Hence, there exists a positive constant such that for .
Let be a function defined in Lemma 2. By mathematical induction, we can show that for Lemma 2 yields . This implies that has a convergence radius at least . The proof is completed.

Finally, we consider the case that satisfies condition . In this case, is not only on the unit circle, but also a root of unity. Let be a sequence defined by , with , , andwhere is a positive constant defined as in the proof of Theorem 3.

Theorem 4. Assume that satisfies condition (H3). Let be a power series determined by , , andwhereIf for , then (11) has an analytic solution in a neighborhood of the origin such that , , and , where is an arbitrary constant satisfying , where is defined as in (32).
Otherwise, if for some , then (11) has no analytic solution in a neighborhood of the origin.

Proof. Observe that if then is a trivial analytic solution of (11). So, we consider only the case .
If for some positive number , then . But implies , which is a contradiction. This concludes that (11) has no analytic solution in a neighborhood of the origin.
Assume that for . Then . So, there are infinitely many choices of . Choose so that , where is defined in (32).
Note that for , where We can see thatfor .
Likewise, the remaining proof is similar to one of Theorem 1. Consider the implicit functional equationSince , , the implicit function theorem implies that there exists a unique function which is analytic in a neighborhood of the origin with a positive radius. We can show that the power series in which is determined by (32) satisfies (36). Moreover, for . That is, is a majorant series of . Then converges in a neighborhood of the origin. This completes the proof.

Theorem 5. Let be an analytic solution in a neighborhood of the origin of (11), with , , which is obtained from Theorem 1, Theorem 3, or Theorem 4. Then (10) has an analytic solution of the form in a neighborhood of the origin.

Proof. Since , is analytic in a neighborhood of .
Let . ThenThat is, is an analytic solution of (10). The proof is completed.

3. Analytic Solutions of (5) via (10)

In this section, we construct an analytic solution of (5) from an analytic solution of (10). Assume that is an analytic solution of the functional differential equation (5) in a neighborhood of the origin. Since is analytic in a neighborhood of the origin, can be represented by Taylor’s seriesLet , where and and for Since , we have Since and , it follows that This implies that  = .

Since , we have .

By using mathematical induction, we can show thatfor , where is a polynomial with nonnegative coefficients.

Therefore, the explicit form of an analytic solution of our equation iswhere denotes .

4. Polynomial Solutions of (5)

In this section, we let be a polynomial. Then, we investigate the polynomial solution of (5).

Theorem 6. For a polynomial , the equationhas a nontrivial polynomial solution if and only if with or with .

Proof.
Necessity. Assume that is a nontrivial polynomial solution of (41). Let with .
Observe that when . This implies . From now on, we let .
We consider cases.
Case 1 (). That is, . Equation (41) becomeswhere .
If , then (42) changes to .
Next, we consider .
Comparing coefficient of in (42), we have .
Equation (42) is reduced towhere .
Comparing coefficient of in (43), we have Then, repeating the above method, we obtain and also have .
By choosing an arbitrary nonzero , say , both situations yield a nontrivial solution .
Case 2 (). That is, , where .
Equation (41) becomeswhere .
By comparing coefficient of constant term and in (44), we obtain for .
Next, we consider .
Comparing coefficient of in (44), we have , which implies . Therefore, (44) is reduced towhere .
Comparing coefficient of in (45), we have . Continuing this process, we obtain , where . By comparing coefficient of together with , we obtain a nontrivial solution  +, where is an arbitrary constant. Note that if then .
Case 3 (). We consider 2 subcases.
Subcase 3.1 (). Equation (41) becomeswhere . By using method of undetermined coefficient, we obtain . Consequently, (46) is reduced towhere .
Comparing coefficient of in (47), we have . If , then . That is, which is a contradiction. Therefore, . Substituting this relation in (46) and repeating this process, we get for . Using this fact in (47) and then comparing the coefficient of , we obtain . This yields , which is a contradiction.
Subcase 3.2 (). Equation (41) becomeswhere . Comparing coefficient of in (48) together with , we have . Then, repeating the above method, we obtain which is a contradiction.
Thus, (41) has a nontrivial polynomial solution when with or with .
Sufficiency. It follows from the proof of Cases and in necessity part.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The first author is supported in part by Development and Promotion of Science and Talents Project (DPST). The second author is supported by National Research Council of Thailand and Khon Kaen University, Thailand (Grant no. kku fmis(580010)).