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Existence of periodic solutions for p-Laplacian neutral Rayleigh equation
Advances in Difference Equations volume 2014, Article number: 67 (2014)
In this paper, we use topological degree theory to establish new results on the existence of periodic solutions for a p-Laplacian neutral Rayleigh equation.
In this paper, we consider the following second-order p-Laplacian neutral functional differential equation:
where for and ; σ and c are given constants with ; , . The conjugate exponent of p is denoted by q, i.e. . f, g, β, e, and τ are real continuous functions on ℝ; τ, β, and e are periodic with periodic T, is a constant; , .
As we know, the p-Laplace Rayleigh equation with a deviating argument is applied in many fields such as physics, mechanics, engineering technique fields, and so on. The existence of a periodic solution for the second-order p-Laplacian Rayleigh equations with a deviating argument as follows:
has been extensively studied in [1–4]. In recent years, Lu et al. [5–7] used Mawhin’s continuation theory to do research to the existence of a periodic solution for p-Laplacian neutral Rayleigh equation. They obtained some existence results of periodic solutions to p-Laplacian neutral Rayleigh equations.
In the research mentioned above, the corresponding nonlinear terms of the second-order p-Laplacian Rayleigh equation did not include a variable coefficient. Only little literature discussed this kind of p-Laplacian Rayleigh equation. For more details refer to [8–11]. Here, we focus on  by Liang Feng et al. They discussed the existence of the solution to the following equation:
They established sufficient conditions for the existence of a T-periodic solution of (1.4). But their conclusions are founded on the prerequisite , which does not satisfy (1.1). Another significance of the paper is that the result is related to the deviating argument , while the conclusions in those existing papers mentioned above have no relation with .
2 Preliminary results
For convenience, throughout this paper, we will adopt the following assumptions:
(H1) , , ;
(H2) , ;
It is obvious that with norm and with norm are two Banach spaces.
Now we define a linear operator , .
If , then A has continuous bounded inverse on and
Lemma 2.2 If and , then
Proof We know that is a periodic function. So for . When , from Lemma 2.1, we have
which implies . That is to say (2.1) holds. If , we can also prove that (2.1) is true in the same way. Thus Lemma 2.2 is proved.
Now we consider the following equation in :
is continuous and takes a bounded set into bounded set.
Let us define , , , and
It is clear that if is the solution to (2.2), then u satisfies the abstract equation
where the operator is given by
α is a continuous function which sends bounded sets of into bounded sets of ℝ, and it is a completely continuous mapping. For more details as regards the meaning of α, please refer to . □
Lemma 2.3 
Let Ω be an open bounded set in . Suppose that the following conditions hold:
For each , the equation has no solution on ∂ Ω.
The equation has no solution on .
The Brouwer degree of Ϝ, .
Then (2.2) has at least one T-periodic solution in .
Lemma 2.4 
is open bounded and symmetric with respect to . If and , , , then is an odd number.
3 Main results
Theorem 3.1 Suppose that the following conditions hold:
(A1) , or , .
(A2) The sign of is unchanged in the interval .
(A3) There exist constants , and such that
(A4) There exists a constant such that , .
Then (1.1) has at least one solution with periodic T if there exists a constant such that the following condition holds:
where a is defined in (3.8).
Proof Consider the homotopic equation of (1.1) as follows:
Let be a possible T-periodic solution to (3.2). By integrating both sides of (3.2) over [0, T], we have
Let , by the condition (A1), we know that has a unique inverse denoted by ; noting that , we get
Based on the condition of (A2) and the integral mean value theorem, there exists such that
By the condition (A3)(2) for a given , when such that
Now we can claim that there are two constants a and b such that
In the following, we prove the above claim in two cases.
Case 1. If , , then (3.7) holds.
Case 2. If , , by (3.5), (3.6), and the condition (A3)(1), we have
When , according to the Minkowski inequality, we have
When , from , , , we have
Therefore, (3.7) is also satisfied for case 2.
For any , there exists , such that , where k is an integer. Then
By (3.7), we have
At first, we prove that there is a constant such that
By (3.10), we only need to prove that is bounded in order to prove (3.11).
If , then is obviously bounded.
If , then , that is, is bounded as well.
If , we prove that is bounded in the following.
By multiplying both sides of (3.2) by and integrating them over , we have
Let , , then
By (3.12) and (3.13), we get
where , , .
By elementary analysis, we know that there is a constant which satisfies such that
By (3.16), one has
By Lemma 2.2, together with (3.15) and (3.17), we can derive
From (3.1) and (3.18), we know that also is bounded in this case. Based on the above, we can derive the result that has a bound; therefore, (3.11) holds.
Secondly, we prove that there is a constant such that
Based on (1.1), together with (3.13) and the condition (A2)(1), we get
Because , there exists such that . Noting that , we have
From Lemma 2.1, we derive
therefore, (3.19) is satisfied.
Let , then (3.2) is equivalent to the following equation:
Then is a T-periodic solution of (3.2) if y is a T-periodic solution of (3.20). Let
since f, g are continuous and A has a continuous inverse, the mapping in (3.21) is continuous and takes bounded sets into bounded sets.
In addition, (3.20) can be represented as
Let ; is a constant, , then (3.22) has no solution on ∂ Ω for . In fact, if is a solution to (3.22) on ∂ Ω, then or . If , then . That is to say, . This is a contradiction with (3.11). Similarly, . Then (3.22) satisfies the condition (i) of Lemma 2.3.
If , y is a constant and , then and , . By (A4), we obtain . Therefore
on . This indicates that (3.22) satisfies the condition (ii) of Lemma 2.3.
We know , then for , we have or . By (3.23) and the condition (A4), we conclude that , , . Based on Lemma 2.4, we get .
Based on the above, (3.22) satisfies all the conditions of Lemma 2.3. So does (3.20). By Lemma 2.3, (3.20) has at least one T-periodic solution, then (1.1) has also at least a periodic solution. □
Consider the following equation:
where , , , , , . Obviously we get , .
If we take , , then the condition (A4) of Theorem 3.1 is satisfied and
By (4.2), we obtain , , .
If we choose , , then when , we have
We calculate easily that
Based on the above, we know that (4.1) satisfies all conditions included in Theorem 3.1; therefore, (4.1) has at least one T-periodic solution.
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The authors wish to thank the anonymous referee for his/her valuable suggestions to this paper. This work is supported by the NSFC of China (11171085).
The authors declare that they have no competing interests.
ZMH performed all the steps of proof in this research and also wrote the paper. JHS suggested many good ideas that made this paper possible and helped to improve the manuscript. All authors read and approved the final manuscript.