Research Article | Open Access

Zhengxin Zhou, "On the Equivalence of Differential Equations in the Sense of Coincidence Reflecting Functions", *Abstract and Applied Analysis*, vol. 2015, Article ID 456364, 8 pages, 2015. https://doi.org/10.1155/2015/456364

# On the Equivalence of Differential Equations in the Sense of Coincidence Reflecting Functions

**Academic Editor:**Jaume Giné

#### Abstract

We use a new method for constructing some differential equations which are equivalent to a given equation in the sense of having the same reflecting function. We completely solve the problem: when is a polynomial differential equation equivalent to a given polynomial differential equation? Many sufficient conditions have been established for one differential equation to be equivalent to a given differential equation. We apply the obtained results to study the boundary value problem of two equivalent differential equations.

#### 1. Introduction

By [1, 2] we know that some polynomial differential systemwhere and are polynomial functions, can be transformed to an equation of the formwhere are polynomial in , . The fact that systems with a homogeneous nonlinearity can be transformed to (2) with has been exploited in a number of previous papers [1–4]. The limit cycles of (1) correspond to -periodic solutions of (2). This fact has been used to facilitate certain computations and to provide some information about the global phase portrait of the system. In this paper we will use the method of Mironenko [5] to construct some differential equations which are equivalent to (2) in the sense of having the same reflecting function.

Now, we simply introduce the concept of the reflecting function.

Consider differential systemwhich has a continuously differentiable right-hand side and general solution . For each such system, the* reflecting function* is defined as [5].

If system (3) is -periodic with respect to , then is the Poincaré mapping of (3) over the period . Thus, the solution of (3) defined on is -periodic if and only if is a fixed point of .

A differentiable function is a reflecting function of system (3) if and only if it is a solution of the Cauchy problem

If the reflecting functions of two differential systems coincide in their common domain, then these systems are said to be* equivalent* [5].

If is the reflecting function of (3), then it is also the reflecting function of any systemwhere is an arbitrary vector function such that the solutions of every system above are uniquely determined by its initial conditions. The systems (3) and (5) have the same operators of translation along solutions on the symmetric time interval , and therefore the initial data of solutions of the boundary value problems of the form , where is an arbitrary function, coincide for such systems.

In general, it is very difficult to find out the reflecting function of (3), so to writ out system (5) is difficult, too. How to judge if two systems are equivalent when we do not know their reflecting function? This is a very interesting problem! Mironenko in [5–7] has studied it and obtained some good results.

Lemma 1 (see [6]). *If continuously differentiable vector functions are solutions of differential system**then all perturbed systems of the form**where are arbitrary continuous scalar odd functions, are equivalent to each other and to system (3).*

The theory of reflecting function has been used for studying the qualitative behavior of solutions of differential systems by many authors [5–13].

Now, for a given equation (2), we present a method for constructing other first-order differential equations which are equivalent to (2), that is, to find out the solution of (6) with . However, it is impossible to find all the solutions in most cases. Therefore, we take only polynomial solutions of (6), that is, its solutions of the formwhere the coefficients are assumed to be differentiable functions on and is not to be zero identically.

Bel’skii in papers [8–10] has discussed, respectively, when a first-order polynomial differential equation is equivalent to a linear equation and Riccati equation () and Abel equation (). In this paper, we will in detail discuss when is a first-order equation equivalent to (2) with . We have obtained the sufficient conditions for (6) has solution in form of (8) and we give the explicit expressions of . At the same time, in addition, we also find out some nonpolynomial solutions of (6). The results of this paper generalize the results of Bel’skii [8–10], and the method of proof in this paper is more simple and efficient than in the previous papers.

#### 2. Main Results

Now, we consider (2) and assume that is not to be zero identically in and . By Bel’skii [8], we know that if , then the degree of polynomial is equal to . In the following, we suppose that and .

Theorem 2. *For (2), suppose that and . Then the function of the form (8) is a solution of (6), if and only if**Therefore, can be represented in the form*

*Proof. *Let us substitute the function of the form (8) into (6); we getEquating the coefficients of the like power of , we obtainwhere and , when .

From (16) when we get . Taking in (16), we have . Similarly for , , we getthat is, relation (11) is true.

Substituting (11) into (16) and simplifying we get (10).

Using relation (11) and taking in (16), we getit implies relation (12).

Using relation (12) and taking in (16), we havesubstituting into the above, we get that relation (13) is held.

Substituting (11)–(13) into (8), we getwhich completes the proof.

We rewrite as follows:in which

Lemma 3. *If , , , then*

*Proof. *As , , so (21) goes toequating the coefficients of and ; from here we get relations (23) and (24).

Theorem 4. *Suppose that , , , and**where**Then the function in (8) is a solution of (6); that is, (2) is equivalent to equation**where is an arbitrary continuous odd function.*

*Proof. *For , by Lemma 3, substituting (12) and (13) into (10), we obtainSubstituting (24) into this relation, we obtainBy (11) and (23), we getAs , (37) is an identity when . For , solving (37), we haveSubstituting (38) into (11)–(13), we obtain (29)–(32).

On the other hand, by relation (10), for , we haveSubstituting (29)–(32) and (37) into the above, we obtainwhich deduces that relation (26) is true. By Theorem 2 and in (10) taking we get that relations (27) and (28) are held.

By Lemma 1 and Theorem 2, the proof is completed.

Theorem 5. *Supposing that ,**where**where and are arbitrary constants. Then the function of the form (8) is a solution of (6); that is, (2) is equivalent to (34).*

*Proof. *As , and , by Lemma 3, we haveUsing , it is not difficult to check that (37) is an identity when . For , in relation (10) taking and computing, it follows thatwhich implies thatCombining relation (12), it yieldssolving this equation we get the expression (45).

Substituting (49) into (13) implies (42). Substituting (42)–(45) into equation and simplifying we get . By Theorem 2, the proof is finished.

Similarly, we get the following.

Theorem 6. *Supposing that , and ,**where** and are arbitrary constants. Then (2) is equivalent to (34).*

Theorem 7. *Supposing that , ,**where ** are arbitrary constants. Then (2) is equivalent to (34).*

By the literature [5], we know the following.

Corollary 8. *If all the conditions of one of the above four theorems (Theorems 4–7) are satisfied and (2) is -periodic in , then the initial data of solutions of the boundary value problems of the form , where is an arbitrary function, coincide for systems (2) and (34).*

*Example 9. *The equationhas the only one -periodic solution . For this equation, we haveBy Theorem 6, we getwhere and are constants:Asthenthat is, the first condition of Theorem 6 is satisfied. ThusTaking and , we getTaking and , we haveTherefore, (54) is equivalent to equationwhere are arbitrary continuously odd functions. By Corollary 8, (63) has only one solution such that .

Now, taking in the above theorems, we get the following corollaries.

Corollary 10. *Suppose that ; then**where** and are arbitrary constants; . Then the Abel equation**is equivalent to equation**where is an arbitrary continuously odd function.*

Corollary 11. *Suppose that , , and ;**where** and are arbitrary constants. Then Abel equation (66) is equivalent to (67).*

Corollary 12. *Suppose that , , and ;**where** are arbitrary constants. Then Abel equation (66) is equivalent to (67).*

*Example 13. *Abel equationhas three constant solutions , , and . For this equation, we have , , , , and .

Thusand .

Solving , we getwhere , are arbitrary constants. ThereforeTaking and , we obtainTaking and , we getConsequently, by Lemma 1 and Corollary 10, Abel equation (72) is equivalent to equationwhere are arbitrary continuously odd functions. By Corollary 8, (78) has only three solutions such that ( is a nonzero constant).

*Remark 14. *The conclusions of the above three corollaries have been proven by Bel’skii and Mironenko in [9].

Because (6) is a linear equation, if are solutions of (6), then also is solution of (6), where are arbitrary constants. In the following, we will find out some solutions of (6) in the other form.

Theorem 15. *For the first-order differential equation , if are solutions of (6), then , , is a solution of (6), too.*

*Proof. *Asand ,Thus, the proof is completed.

By Example 9 and Theorem 15 we get that equationis also equivalent to equationwhere are arbitrary continuously odd functions, , are arbitrary constants, and are the same as in Example 9. This equation has only one solution such that .

Using Theorem 15 and Corollary 10 we get the following.

Corollary 16. *Supposing that ,**Then Abel equation**is equivalent to equation**where are arbitrary continuously odd functions and is a constant:**in which**, are arbitrary constants, and . Consider*

Theorem 17. *For equation**is a solution of (6). Thus, (89) is equivalent to equation**where is an arbitrary continuously odd function, is an arbitrary differentiable function, and is a continuous function.**This result is easy to be proven.*

Obviously, from Theorem 17, we see that if is a polynomial of , then the corresponding equation (6) has at least one polynomial solution . This is implied by taking in (90).

By Theorem 17, (72) is also equivalent toIf in (92) we put , then we obtainwhere is an arbitrary continuously odd function. This equation has only three solutions such that ( is a nonzero constant).

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The work was supported by the NSF of Jiangsu of China under Grant no. BK2012682 and the NSF of China under Grant nos. 11271026 and 61374010.

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#### Copyright

Copyright © 2015 Zhengxin Zhou. 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.