Existence of periodic solutions for p-Laplacian neutral Rayleigh equation
© Min He and Shen; licensee Springer. 2014
Received: 3 May 2013
Accepted: 30 January 2014
Published: 14 February 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.
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; , .
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.
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 , .
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.
is continuous and takes a bounded set into bounded set.
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 
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 .
(A4) There exists a constant such that , .
where a is defined in (3.8).
In the following, we prove the above claim in two cases.
Case 1. If , , then (3.7) holds.
Therefore, (3.7) is also satisfied for case 2.
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.
where , , .
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.
therefore, (3.19) is satisfied.
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.
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.
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. □
where , , , , , . Obviously we get , .
By (4.2), we obtain , , .
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.
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).
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