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ISRN Mathematical Analysis
Volume 2012 (2012), Article ID 830983, 12 pages
doi:10.5402/2012/830983
Research Article

Positive Solutions to Periodic Boundary Value Problems for Four-Order Differential Equations

Huantao Zhu1 and Zhiguo Luo2

1Hunan College of Information, Changsha, Hunan 410200, China
2Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China

Received 26 November 2011; Accepted 9 January 2012

Academic Editor: G. Gripenberg

Copyright © 2012 Huantao Zhu and Zhiguo Luo. 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

We apply fixed point theorem in a cone to obtain sufficient conditions for the existence of single and multiple positive solutions of periodic boundary value problems for a class of four-order differential equations.

1. Introduction

In this paper, we investigate the existence of positive solutions of the following periodic boundary value problem: 𝑦 ( 4 ) ( 𝑡 ) − 𝐴 𝑦 ( 𝑡 ) + 𝐵 𝑦 ( 𝑡 ) = 𝑓 𝑡 , 𝑦 ( 𝑡 ) , 𝑦 𝑦 ( 𝑡 ) , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑦 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , 2 , 3 , ( 1 . 1 ) where 𝐼 = [ 0 , 2 𝜋 ] , 𝐴 and 𝐵 are positive constants with 𝐴 2 > 4 𝐵 , 𝑓 ∈ 𝐶 ( 𝐼 × ( 0 , + ∞ ) × 𝑅 , [ 0 , + ∞ ) ) .

In recent years, the nonlinear periodic boundary value problems have been widely studied by many authors, for example, see [1–7] and the references therein. Many theorems and methods of nonlinear functional analysis, for instance, upper and lower solutions method, fixed point theorems, variational method, and critical point theory, and so on, have been applied to their problems. When positive solutions are discussed, it seems that fixed point theorem in cones is quite effective in dealing with the problems with singularity. In [8], Zhang and Wang proved periodic boundary value problems with singularity 𝑦 ( 𝑡 ) + 𝑘 2 𝑦 𝑦 ( 𝑡 ) = 𝑔 ( 𝑡 , 𝑦 ( 𝑡 ) ) , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑦 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , ( 1 . 2 ) have multiple positive solutions under some conditions, where 𝑔 ( 𝑡 , 𝑦 )   is singular at 𝑦 = 0 , that is, l i m 𝑦 → 0 + 𝑔 ( 𝑡 , 𝑦 ) = + ∞ . ( 1 . 3 ) Relying on a nonlinear alternative of Leray-Schauder type and fixed point theorem, Chu and Zhou [9] discussed the existence of positive solutions for the third-order periodic boundary value problem 𝑦 ( 𝑡 ) + 𝑘 3 𝑦 𝑦 ( 𝑡 ) = 𝑔 ( 𝑡 , 𝑦 ( 𝑡 ) ) , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑦 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , 2 , ( 1 . 4 ) where √ 𝑘 ∈ ( 0 , 1 / 3 ) .  However, relatively few papers have been published on the same problem for four-order differential equations. Recently, by using a maximum principle for operator 𝐿 𝑢 = 𝑢 ( 4 ) − 𝛽 𝑢 + 𝛼 𝑢   in periodic boundary condition and fixed point index theory in cones, Li [10] considered the existence of positive solution for the fourth-order periodic boundary value problem 𝑦 ( 4 ) ( 𝑡 ) − 𝛽 𝑦 [ ] , 𝑦 ( 𝑡 ) + 𝛼 𝑦 ( 𝑡 ) = 𝑔 ( 𝑡 , 𝑦 ( 𝑡 ) ) , 𝑡 ∈ 0 , 1 ( 𝑖 ) ( 0 ) = 𝑦 ( 𝑖 ) ( 1 ) , 𝑖 = 0 , 1 , 2 , 3 , ( 1 . 5 ) where 𝑓 ∶ [ 0 , 1 ] × 𝑅 + → 𝑅 +  is continuous, 𝛼 , 𝛽 ∈ 𝑅 and satisfy 0 < 𝛼 < ( 𝛽 / 2 + 2 𝜋 2 ) 2 , 𝛽 > − 2 𝜋 2 , 𝛼 / 𝜋 4 + 𝛽 / 𝜋 2 > 1 . However, since there appears 𝑦   in nonlinear term 𝑔 , the method in [9] cannot be directly applied to (1.1). The main aim of this paper is to establish sufficient conditions for the existence of positive solutions to the problem (1.1).

To prove our main results, we present an existence theorem.

Theorem 1.1 (see [11]). Let 𝐸 be a Banach space and 𝑃 a cone in 𝐸 . Suppose Λ 1 and Λ 2 are open subsets of 𝐸 such that 0 ∈ Λ 1 ⊂ Λ 1 ⊂ Λ 2   and suppose that 𝑇 ∶ 𝑃 ∩ Λ 2 Λ 1 ⟶ 𝑃 ( 1 . 6 ) is a completely continuous operator. If one of the following conditions is satisfied:(i) ‖ 𝑇 𝑥 ‖ ≤ ‖ 𝑥 ‖ for 𝑥 ∈ 𝑃 ∩ 𝜕 Λ 1 , ‖ 𝑇 𝑥 ‖ ≥ ‖ 𝑥 ‖   for 𝑥 ∈ 𝑃 ∩ 𝜕 Λ 2 ,(ii) ‖ 𝑇 𝑥 ‖ ≥ ‖ 𝑥 ‖   for 𝑥 ∈ 𝑃 ∩ 𝜕 Λ 1 , ‖ 𝑇 𝑥 ‖ ≤ ‖ 𝑥 ‖   for 𝑥 ∈ 𝑃 ∩ 𝜕 Λ 2 . Then 𝑇 has a fixed point in 𝑃 ∩ ( Λ 2 / Λ 1 ) .

2. Preliminaries

In this section, we present some preliminary results which will be needed in Section 3.

Let 𝛼 > 0 , and for any function 𝑢 ∈ 𝐶 ( 𝐼 ) ,  we defined the operator 𝐽 𝛼 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝛼 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 , ( 2 . 1 ) where 𝐺 𝛼 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝑒 ( 𝑡 , 𝑠 ) = 𝛼 ( 𝑡 − 𝑠 ) + 𝑒 𝛼 ( 2 𝜋 − 𝑡 + 𝑠 ) 𝑒 2 𝛼 2 𝜋 𝛼 𝑒 − 1 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 2 𝜋 , 𝛼 ( 𝑠 − 𝑡 ) + 𝑒 𝛼 ( 2 𝜋 − 𝑠 + 𝑡 ) 𝑒 2 𝛼 2 𝜋 𝛼 − 1 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 2 𝜋 . ( 2 . 2 ) By a direct calculation, we easily obtain 0 2 𝜋 𝐺 𝛼 ( 𝑡 , 𝑠 ) 𝑑 𝑠 = 𝛼 − 2 , 𝑚 𝛼 ∶ = m i n 𝑠 , 𝑡 ∈ 𝐼 𝐺 𝛼 ( 𝑡 , 𝑠 ) = 2 𝑒 𝛼 𝜋 𝑒 2 𝛼 2 𝜋 𝛼 − 1 , 𝑀 𝛼 ∶ = m a x 𝑠 , 𝑡 ∈ 𝐼 𝐺 𝛼 ( 𝑡 , 𝑠 ) = 1 + 𝑒 2 𝛼 𝜋 𝑒 2 𝛼 2 𝜋 𝛼 . − 1 ( 2 . 3 ) Set 𝜌 = √ 𝐴 + 𝐴 2 − 4 𝐵 2 , 𝜆 = √ 𝐴 − 𝐴 2 − 4 𝐵 2 , o r 𝜌 = √ 𝐴 − 𝐴 2 − 4 𝐵 2 , 𝜆 = √ 𝐴 + 𝐴 2 − 4 𝐵 2 , ( 2 . 4 ) then 𝜌 , 𝜆 ∈ 𝑅 . Now, we consider the problem − 𝑢 ′ ( 𝑡 ) + 𝜆 2 𝑢 ( 𝑡 ) = 𝑓 𝑡 , 𝐽 𝜌 𝑢 , 𝜌 2 𝐽 𝜌 𝑢 𝑢 − 𝑢 , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑢 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 . ( 2 . 5 )

Lemma 2.1. If 𝑢 ∈ 𝐶 2 ( 𝐼 )   is a (positive) solution of problem (2.5), then 𝑦 = 𝐽 𝜌 𝑢 ∈ 𝐶 4 ( 𝐼 )   is a (positive) solution of problem (1.1). Moreover, the problem (2.5) is equivalent to integral equation 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜆 𝐽 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 . ( 2 . 6 )

Proof. If 𝑢 ∈ 𝐶 2 ( 𝐼 ) ,  then 𝐽 𝜌 𝑢 ∈ 𝐶 4 ( 𝐼 )   and 𝐽 𝜌 𝑢 ( 𝑡 ) = − 𝑢 ( 𝑡 ) + 𝜌 2 𝐽 𝜌 𝑢 𝐽 ( 𝑡 ) , 𝑡 ∈ 𝐼 , 𝜌 𝑢 ( 𝑖 ) 𝐽 ( 0 ) = 𝜌 𝑢 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 . ( 2 . 7 ) Thus, 𝑦 ( 𝑡 ) = − 𝑢 ( 𝑡 ) + 𝜌 2 𝐽 𝜌 𝑢 ( 𝑡 ) , 𝑦 ( 𝑡 ) = − 𝑢 ( 𝑡 ) + 𝜌 2 𝐽 𝜌 𝑢 𝑦 ( 𝑡 ) , ( 4 ) ( 𝑡 ) = − 𝑢 ( 𝑡 ) + 𝜌 2 𝐽 𝜌 𝑢 ( 𝑡 ) = − 𝑢 ( 𝑡 ) − 𝜌 2 𝑢 ( 𝑡 ) + 𝜌 4 𝐽 𝜌 𝑢 ( 𝑡 ) . ( 2 . 8 ) Then, 𝑦 ( 4 ) ( 𝑡 ) − 𝐴 𝑦 ( 𝑡 ) + 𝐵 𝑦 ( 𝑡 ) = − 𝑢 ( 𝑡 ) + 𝐴 − 𝜌 2 𝜌 𝑢 ( 𝑡 ) + 4 − 𝐴 𝜌 2 𝐽 + 𝐵 𝜌 𝑢 ( 𝑡 ) = − 𝑢 ( 𝑡 ) + 𝜆 2 𝑢 ( 𝑡 ) = 𝑓 𝑡 , 𝐽 𝜌 𝑢 , 𝜌 2 𝐽 𝜌 𝑢 − 𝑢 = 𝑓 𝑡 , 𝑦 ( 𝑡 ) , 𝑦 . ( 𝑡 ) ( 2 . 9 ) On the other hand, 𝑦 ( 0 ) = 𝑦 ( 2 𝜋 ) , 𝑦 ( 0 ) = 𝑦 ( 2 𝜋 ) , 𝑦 ( 0 ) = − 𝑢 ( 0 ) + 𝜌 2 𝐽 𝜌 𝑢 ( 0 ) = − 𝑢 ( 2 𝜋 ) + 𝜌 2 𝐽 𝜌 𝑢 ( 2 𝜋 ) = 𝑦 𝑦 ( 2 𝜋 ) , ( 0 ) = − 𝑢 ( 0 ) + 𝜌 2 𝐽 𝜌 𝑢 ( 0 ) = − 𝑢 ( 2 𝜋 ) + 𝜌 2 𝐽 𝜌 𝑢 ( 2 𝜋 ) = 𝑦 ( 2 𝜋 ) . ( 2 . 1 0 ) Hence, if 𝑢 ∈ 𝐶 2 ( 𝐼 ) is a solution of problem (2.5), then 𝑦 = 𝐽 𝜌 𝑢 ∈ 𝐶 4 ( 𝐼 ) is a solution of problem (1.1). And, if 𝑢 ∈ 𝐶 2 ( 𝐼 ) is a positive function, noting that 𝐺 𝜌 ( 𝑡 , 𝑠 ) > 0 for any 𝑠 , 𝑡 ∈ 𝐼 , we have 𝑦 = 0 2 𝜋 𝐺 𝜌 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 > 0 . ( 2 . 1 1 ) Noting that, for any function ℎ ∈ 𝐶 ( 𝐼 ) , linear problem − 𝑢 ( 𝑡 ) + 𝜆 2 𝑢 𝑢 ( 𝑡 ) = ℎ ( 𝑡 ) , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑢 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , ( 2 . 1 2 ) has a unique solution 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜆 ( 𝑡 , 𝑠 ) ℎ ( 𝑠 ) 𝑑 𝑠 , ( 2 . 1 3 ) one can easily obtain that (2.6) holds. The proof is complete.

In the following application, we take 𝐸 = 𝐶 2 ( 𝐼 ) with the supremum norm ‖ ⋅ ‖ and define 𝑃 𝜆 = 𝑢 ∈ 𝐸 ∶ 𝑢 ( 𝑡 ) ≥ 0 f o r a l l 𝑡 ∈ 𝐼 , m i n 𝑡 ∈ 𝐽 𝑢 ( 𝑡 ) ≥ 𝛿 𝜆 ‖ 𝑢 ‖ , ( 2 . 1 4 ) where 𝛿 𝜆 = 𝑚 𝜆 / 𝑀 𝜆 ∈ ( 0 , 1 ) .

One easily checks and verifies that 𝑃 𝜆   is a cone in 𝐸 . For any 𝑟 > 0 , let 𝑃 𝜆 𝑟 = { 𝑢 ∈ 𝑃 𝜆 ∶ ‖ 𝑢 ‖ < 𝑟 } ,  then 𝜕 𝑃 𝜆 𝑟 = { 𝑢 ∈ 𝑃 𝜆 ∶ ‖ 𝑢 ‖ = 𝑟 } .  For any 𝑢 ∈ 𝐸 , define mapping 𝑇 𝜆 ∶ 𝐸 → 𝐸 by 𝑇 𝜆 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜆 𝐽 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 , ( 2 . 1 5 ) then the fixed point of 𝑇 𝜆 in 𝐸 is a positive solution of (2.5).

Lemma 2.2. For any 𝜍 > 𝜏 > 0 , 𝑇 𝜆 ∶ 𝑃 𝜆 𝜍 ⧵ 𝑃 𝜆 𝜏 → 𝑃 𝜆   is completely continuous.

Proof. For any 𝑢 ∈ 𝑃 𝜆 𝜍 ⧵ 𝑃 𝜆 𝜏 , 𝜏 ≤ ‖ 𝑢 ‖ ≤ 𝜍    and 𝛿 𝜆 𝜏 ≤ 𝑢 ( 𝑡 ) ≤ 𝜍 for all 𝑡 ∈ 𝐼 . Thus, if 𝑢 ∈ 𝑃 𝜆 𝜍 ⧵ 𝑃 𝜆 𝜏 , 𝐽 𝜌 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜌 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 𝜏 0 2 𝜋 𝐺 𝜌 𝛿 ( 𝑡 , 𝑠 ) 𝑑 𝑠 ≥ 𝜆 𝜏 𝜌 2 . ( 2 . 1 6 ) It is easy to see that 𝑇 𝜆 is continuous and completely continuous since 𝑓 ∶ 𝐼 × ( 0 , + ∞ ) × 𝑅 → [ 0 , + ∞ )   is continuous. Next, we show that 𝑇 𝜆 ∶ 𝑃 𝜆 𝜍 ⧵ 𝑃 𝜆 𝜏 → 𝑃 𝜆 .  Since 𝑓 ( 𝑠 , ( 𝐽 𝜌 𝑢 ) ( 𝑠 ) , 𝜌 2 ( 𝐽 𝜌 𝑢 ) ( 𝑠 ) − 𝑢 ( 𝑠 ) ) ≥ 0   for 𝑢 ∈ 𝑃 𝜆 𝜍 ⧵ 𝑃 𝜆 𝜏 , 𝑇 𝜆 𝑢 ≥ 0 . On the other hand, ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ = m a x 𝑡 ∈ 𝐼 0 2 𝜋 𝐺 𝜆 𝐽 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝑀 𝜆 0 2 𝜋 𝑓 𝐽 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 𝑇 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 , 𝜆 𝑢 ( 𝑡 ) ≥ 𝑚 𝜆 0 2 𝜋 𝑓 𝐽 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ . ( 2 . 1 7 ) The proof is complete.

3. Positive Solutions of (1.1)

In this section, we make the following hypotheses. ( 𝐻 𝜆 1 ) There exist nonnegative functions ℎ ( 𝑢 ) , 𝑔 ( 𝑢 ) ∈ 𝐶 ( ( 0 , + ∞ ) ( 0 , + ∞ ) )   and 𝑝 ( 𝑡 , 𝑣 ) , 𝑞 ( 𝑡 , 𝑣 ) ∈ 𝐶 ( 𝐽 × 𝑅 )  such that 𝑓 ( 𝑡 , 𝑢 , 𝑣 ) ≤ 𝑝 ( 𝑡 , 𝑣 ) ℎ ( 𝑢 ) + 𝑞 ( 𝑡 , 𝑣 ) 𝑔 ( 𝑢 ) ( 3 . 1 ) for all ( 𝑡 , 𝑢 , 𝑣 ) ∈ 𝐼 × ( 0 , ∞ ) × 𝑅   and s u p 𝑢 > 0 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝜌 𝑢 𝑔 − 2 𝑢 m a x 𝛿 𝜆 − 1 𝑢 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝑢 ∫ 0 2 𝜋 𝜌 𝑝 ( 𝑡 , 𝑣 ) 𝑑 𝑡 ℎ − 2 𝑢 + ∫ 0 2 𝜋 𝜌 𝑞 ( 𝑡 , 𝑣 ) 𝑑 𝑡 𝑔 − 2 𝑢 𝑔 𝜌 − 2 𝛿 𝜆 𝑢 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ > 𝑀 𝜆 , ( 3 . 2 ) where 𝑔 is nonincreasing and ℎ / 𝑔 is nondecreasing on ( 0 , ∞ ) . ( 𝐻 𝜆 2 ) One has l i m i n f 𝑢 → 0 + ∫ m i n 0 2 𝜋 𝑓 ( 𝑡 , 𝑤 , 𝑣 ) 𝑑 𝑡 , 𝜌 − 2 𝛿 𝜆 𝑢 ≤ 𝑤 ≤ 𝜌 − 2 𝛿 𝑢 , 𝜆 − 1 𝑢 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝑢 𝑢 > 𝑚 𝜆 − 1 . ( 3 . 3 ) ( 𝐻 𝜆 3 ) One has l i m i n f 𝑢 → + ∞ ∫ m i n 0 2 𝜋 𝑓 ( 𝑡 , 𝑤 , 𝑣 ) 𝑑 𝑡 , 𝜌 − 2 𝛿 𝜆 𝑢 ≤ 𝑤 ≤ 𝜌 − 2 𝛿 𝑢 , 𝜆 − 1 𝑢 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝑢 𝑢 > 𝑚 𝜆 − 1 . ( 3 . 4 )

Under the above hypotheses, we can obtain the following result.

Theorem 3.1. Assume that ( 𝐻 𝜆 1 ) and ( 𝐻 𝜆 2 )   are satisfied, then there exist two positive constants 𝛼 , 𝛽    such that (1.1) has at least positive solution 𝑦 with 𝜌 − 2 𝛼 < ‖ 𝑦 ‖ < 𝜌 − 2 𝛽 . ( 3 . 5 )
Assume ( 𝐻 𝜆 1 ) and ( 𝐻 𝜆 3 ) are satisfied, then there exist two positive constants 𝛽 , 𝛾 such that (1.1) has at least positive solution 𝑦 with 𝜌 − 2 𝛽 < ‖ 𝑦 ‖ < 𝜌 − 2 𝛾 . ( 3 . 6 )
Assume ( 𝐻 𝜆 1 ) , ( 𝐻 𝜆 2 ) , and ( 𝐻 𝜆 3 ) are satisfied, then there exist positive constants 𝛼 , 𝛽 , 𝛾    such that (1.1) has at least two positive solutions 𝑦 1 , 𝑦 2 with 𝜌 − 2 ‖ ‖ 𝑦 𝛼 < 1 ‖ ‖ < 𝜌 − 2 ‖ ‖ 𝑦 𝛽 < 2 ‖ ‖ < 𝜌 − 2 𝛾 . ( 3 . 7 )

Proof. First, we assume that ( 𝐻 𝜆 1 ) and ( 𝐻 𝜆 2 ) are satisfied. From the condition ( 𝐻 𝜆 1 ) ,  one can obtain that there exist a 𝛽 > 0 such that 𝜌 𝛽 𝑔 − 2 𝛽 m a x ( 𝛿 𝜆 − 1 ) 𝛽 ≤ 𝑣 ≤ ( 1 − 𝛿 𝜆 ) 𝛽 ∫ 0 2 𝜋 𝜌 𝑝 ( 𝑡 , 𝑣 ) 𝑑 𝑡 ℎ − 2 𝛽 + ∫ 0 2 𝜋 𝜌 𝑞 ( 𝑡 , 𝑣 ) 𝑑 𝑡 𝑔 − 2 𝛽 𝑔 𝜌 − 2 𝛿 𝜆 𝛽 > 𝑀 𝜆 . ( 3 . 8 ) For any 𝑢 ∈ 𝜕 𝑃 𝜆 𝛽 , 𝛿 𝜆 𝛽 ≤ 𝑢 ( 𝑡 ) ≤ 𝛽   for all 𝑡 ∈ 𝐼 and 𝛽 𝛿 𝜆 𝜌 2 ≤ 𝐽 𝜌 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜌 𝛽 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝜌 2 , 𝛿 𝜆 𝛽 − 𝛽 ≤ 𝜌 2 𝐽 𝜌 𝑢 ( 𝑡 ) − 𝑢 ( 𝑡 ) ≤ 𝛽 − 𝛿 𝜆 𝛽 . ( 3 . 9 ) Thus, for 𝑢 ∈ 𝜕 𝑃 𝜆 𝛽 , from ( 𝐻 𝜆 1 ) ,  we have 𝑇 𝜆 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜆 𝐽 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝑀 𝜆 0 2 𝜋 𝑝 𝑠 , 𝜌 2 𝐽 𝜌 𝑢 ℎ 𝐽 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝜌 𝑢 ( 𝑠 ) + 𝑞 𝑠 , 𝜌 2 𝐽 𝜌 𝑢 𝑔 𝐽 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝜌 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝑀 𝜆 0 2 𝜋 𝑝 𝑠 , 𝜌 2 𝐽 𝜌 𝑢 ℎ 𝐽 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝜌 𝑢 ( 𝑠 ) 𝑔 𝐽 𝜌 𝑢 ( 𝑠 ) + 𝑞 𝑠 , 𝜌 2 𝐽 𝜌 𝑢 𝑔 𝐽 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝜌 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝑀 𝜆 0 2 𝜋 𝑝 𝑠 , 𝜌 2 𝐽 𝜌 𝑢 ℎ 𝜌 ( 𝑠 ) − 𝑢 ( 𝑠 ) − 2 𝛽 𝑔 𝜌 − 2 𝛽 + 𝑞 𝑠 , 𝜌 2 𝐽 𝜌 𝑢 𝑔 𝜌 ( 𝑠 ) − 𝑢 ( 𝑠 ) − 2 𝛿 𝜆 𝛽 𝑑 𝑠 ≤ 𝑀 𝜆 𝑔 𝜌 − 2 𝛿 𝜆 𝛽 𝑔 𝜌 − 2 𝛽 m a x 𝛿 𝜆 − 1 𝛽 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝛽 0 2 𝜋 𝜌 𝑝 ( 𝑡 , 𝑣 ) 𝑑 𝑡 ℎ − 2 𝛽 + 0 2 𝜋 𝜌 𝑞 ( 𝑡 , 𝑣 ) 𝑑 𝑡 𝑔 − 2 𝛽 < 𝛽 = ‖ 𝑢 ‖ , ( 3 . 1 0 ) which implies that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ < ‖ 𝑢 ‖ , ∀ 𝑢 ∈ 𝜕 𝑃 𝜆 𝛽 . ( 3 . 1 1 )
From ( 𝐻 𝜆 2 )   is satisfied, there exists a positive constant 𝛼 < 𝛽   such that m i n 0 2 𝜋 𝑓 ( 𝑡 , 𝑤 , 𝑣 ) 𝑑 𝑡 , 𝜌 − 2 𝛿 𝜆 𝛼 ≤ 𝑤 ≤ 𝜌 − 2 𝛿 𝛼 , 𝜆 − 1 𝛼 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝛼 > 𝑚 𝜆 − 1 𝛼 . ( 3 . 1 2 ) For any 𝑢 ∈ 𝜕 𝑃 𝜆 𝛼 , 𝛿 𝜆 𝛼 ≤ 𝑢 ( t ) ≤ 𝛼 for all 𝑡 ∈ 𝐼 and 𝛿 𝜆 𝛼 𝜌 2 ≤ 𝐽 𝜌 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜌 𝛼 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝜌 2 , 𝛿 𝜆 𝛼 − 𝛼 ≤ 𝜌 2 𝐽 𝜌 𝑢 ( 𝑡 ) − 𝑢 ( 𝑡 ) ≤ 𝛼 − 𝛿 𝜆 𝛼 , ( 𝑇 𝑢 ) ( 𝑡 ) = 0 2 𝜋 𝐺 𝜆 𝐽 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝑚 𝜆 0 2 𝜋 𝑓 𝐽 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝑚 𝜆 m i n 0 2 𝜋 𝑓 ( 𝑡 , 𝑤 , 𝑣 ) 𝑑 𝑡 , 𝜌 − 2 𝛿 𝜆 𝛼 ≤ 𝑤 ≤ 𝜌 − 2 𝛿 𝛼 , 𝜆 − 1 𝛼 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝛼 > 𝛼 = ‖ 𝑢 ‖ , ( 3 . 1 3 ) which implies that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ > ‖ 𝑢 ‖ , ∀ 𝑢 ∈ 𝜕 𝑃 𝜆 𝛼 . ( 3 . 1 4 ) From (3.11) and (3.14) and Theorem 1.1, one can obtain that 𝑇 𝜆 has a fixed point 𝑢 in   𝑃 𝜆 𝛽 ⧵ 𝑃 𝜆 𝛼 with 𝛼 < ‖ 𝑢 ‖ < 𝛽 .  Hence, 𝑦 = 𝐽 𝜌 𝑢   is a positive solution of (1.1) with 𝜌 − 2 𝛼 < ‖ 𝑦 ‖ < 𝜌 − 2 𝛽 .
Next, we assume that ( 𝐻 𝜆 1 ) and ( 𝐻 𝜆 3 ) are satisfied. In this case, we have (3.11).
Suppose that ( 𝐻 𝜆 3 ) is satisfied, there exists a positive constant 𝛾 > 𝛽 such that m i n 0 2 𝜋 𝑓 ( 𝑡 , 𝑤 , 𝑣 ) 𝑑 𝑡 , 𝜌 − 2 𝛿 𝜆 𝛾 ≤ 𝑤 ≤ 𝜌 − 2 𝛿 𝛾 , 𝜆 − 1 𝛾 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝛾 > 𝑚 𝜆 − 1 𝛾 . ( 3 . 1 5 ) For any 𝑢 ∈ 𝜕 𝑃 𝜆 𝛾 , 𝛿 𝜆 𝛾 ≤ 𝑢 ( 𝑡 ) ≤ 𝛾   for all 𝑡 ∈ 𝐼 and 𝛿 𝜆 𝛾 𝜌 2 ≤ 𝐽 𝜌 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝜌 𝛾 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝜌 2 , 𝛿 𝜆 𝛾 − 𝛾 ≤ 𝜌 2 𝐽 𝜌 𝑢 ( 𝑡 ) − 𝑢 ( 𝑡 ) ≤ 𝛾 − 𝛿 𝜆 𝛾 , ( 𝑇 𝑢 ) ( 𝑡 ) = 0 2 𝜋 𝐺 𝜆 𝐽 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝑚 𝜆 0 2 𝜋 𝑓 𝐽 𝑠 , 𝜌 𝑢 ( 𝑠 ) , 𝜌 2 𝐽 𝜌 𝑢 ( 𝑠 ) − 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝑚 𝜆 m i n 0 2 𝜋 𝑓 ( 𝑡 , 𝑤 , 𝑣 ) 𝑑 𝑡 , 𝜌 − 2 𝛿 𝜆 𝛾 ≤ 𝑤 ≤ 𝜌 − 2 𝛿 𝛾 , 𝜆 − 1 𝛾 ≤ 𝑣 ≤ 1 − 𝛿 𝜆 𝛾 > 𝛾 = ‖ 𝑢 ‖ , ( 3 . 1 6 ) which implies that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ > ‖ 𝑢 ‖ , ∀ 𝑢 ∈ 𝜕 𝑃 𝜆 𝛾 . ( 3 . 1 7 ) From (3.11) and (3.17) and Theorem 1.1, one can obtain that 𝑇 𝜆 has a fixed point 𝑢 in   𝑃 𝜆 𝛾 ⧵ 𝑃 𝜆 𝛽 with 𝛽 < ‖ 𝑢 ‖ < 𝛾 .  Thus, 𝑦 = 𝐽 𝜌 𝑢   is a positive solution of (1.1) with 𝜌 − 2 𝛽 < ‖ 𝑦 ‖ < 𝜌 − 2 𝛾 .
Assume that ( 𝐻 𝜆 1 ) , ( 𝐻 𝜆 2 ) , and ( 𝐻 𝜆 3 ) are satisfied. Repeating the above argument, one can obtain that 𝑇 𝜆 has a fixed point 𝑢 1 in   𝑃 𝜆 𝛽 ⧵ P 𝜆 𝛼   and a fixed point 𝑢 2 in   𝑃 𝜆 𝛾 ⧵ 𝑃 𝜆 𝛽 with ‖ ‖ 𝑢 𝛼 < 1 ‖ ‖ ‖ ‖ 𝑢 < 𝛽 < 2 ‖ ‖ < 𝛾 . ( 3 . 1 8 ) Hence, 𝑦 1 = 𝐽 𝜌 𝑢 1 , 𝑦 2 = 𝐽 𝜌 𝑢 2   are two positive solutions of (1.1) with 𝜌 − 2 ‖ ‖ 𝑦 𝛼 < 1 ‖ ‖ < 𝜌 − 2 ‖ ‖ 𝑦 𝛽 < 2 ‖ ‖ < 𝜌 − 2 𝛾 . ( 3 . 1 9 ) The proof is complete.

4. A Similar Problem

In this section, we use the idea in Sections 2 and 3 to consider the following problem: − 𝑦 ( 4 ) ( 𝑡 ) + 𝑘 4 𝑦 ( 𝑡 ) = 𝑓 𝑡 , 𝑦 ( 𝑡 ) , 𝑦 ′ 1 ( 𝑡 ) , 𝑡 ∈ 𝐼 , 0 < 𝑘 < 2 , 𝑦 ( 𝑖 ) ( 0 ) = 𝑦 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , 2 , 3 , ( 4 . 1 ) where 𝐼 = [ 0 , 2 𝜋 ] , 𝑓 ∈ 𝐶 ( 𝐼 × ( 0 , + ∞ ) × 𝑅 , [ 0 , + ∞ ) ) .

Let 𝑢 ∈ 𝐶 2 ( 𝐼 )   and ∫ 𝑦 = 𝐽 𝑢 = 0 2 𝜋 𝐾 ( 𝑡 , 𝑠 ) 𝑢 ( 𝑠 ) 𝑑 𝑠 ,  where ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝐾 ( 𝑡 , 𝑠 ) = c o s 𝑘 ( 𝜋 − 𝑡 + 𝑠 ) 2 𝑘 s i n 𝑘 𝜋 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 2 𝜋 , c o s 𝑘 ( 𝜋 + 𝑡 − 𝑠 ) 2 𝑘 s i n 𝑘 𝜋 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 2 𝜋 . ( 4 . 2 ) Then, 0 2 𝜋 𝐾 ( 𝑡 , 𝑠 ) 𝑑 𝑠 = 𝑘 − 2 , 𝐽 𝑢 ( 𝑡 ) = 𝑢 ( 𝑡 ) − 𝑘 2 𝐽 𝑢 ( 𝑡 ) , 𝑡 ∈ 𝐼 , 𝐽 𝑢 ( 𝑖 ) ( 0 ) = 𝐽 𝑢 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , ( 4 . 3 ) and one easily check that (4.1) is equivalent to the problem − 𝑢 ( 𝑡 ) + 𝑘 2 𝑢 ( 𝑡 ) = 𝑓 𝑡 , 𝐽 𝑢 , 𝑢 − 𝑘 2 𝑢 𝐽 𝑢 , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑢 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 . ( 4 . 4 ) If 𝑢 ∈ 𝐶 2 ( 𝐼 ) is a (positive) solution of problem (4.4), then 𝑦 = 𝐽 𝑢 ∈ 𝐶 4 ( 𝐼 )   is a (positive) solution of problem (4.1). Moreover, the problem (4.4) is equivalent to integral equation 𝑢 ( 𝑡 ) = 0 2 𝜋 𝐺 𝑘 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝐽 𝑢 ( 𝑠 ) , 𝑢 ( 𝑠 ) − 𝑘 2 𝐽 𝑢 ( 𝑠 ) 𝑑 𝑠 . ( 4 . 5 ) For any 𝑢 ∈ 𝐸 , define mapping 𝑇 ∶ 𝐸 → 𝐸 by ( 𝑇 𝑢 ) ( 𝑡 ) = 0 2 𝜋 𝐺 𝑘 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝐽 𝑢 ( 𝑠 ) , 𝑢 ( 𝑠 ) − 𝑘 2 𝐽 𝑢 ( 𝑠 ) 𝑑 𝑠 . ( 4 . 6 ) For any 𝜍 > 𝜏 > 0 , one can obtain that 𝑇 ∶ 𝑃 𝑘 𝜍 ⧵ 𝑃 𝑘 𝜏 → 𝑃 𝑘   is completely continuous.

Similar to the proof of Theorem 3.1, we can obtain the following result.

Theorem 4.1. Assume that ( 𝐻 𝑘 1 ) and ( 𝐻 𝑘 2 ) are satisfied, then there exist two positive constants 𝛼 , 𝛽   such that (4.1) has at least positive solution 𝑦 with 𝑘 − 2 𝛼 < ‖ y ‖ < 𝑘 − 2 𝛽 . ( 4 . 7 )
Assume ( 𝐻 𝑘 1 ) and ( 𝐻 𝑘 3 ) are satisfied, then there exist two positive constants 𝛽 , 𝛾 such that (4.1) has at least positive solution 𝑦 with 𝑘 − 2 𝛽 < ‖ 𝑦 ‖ < 𝑘 − 2 𝛾 . ( 4 . 8 )
Assume ( 𝐻 𝑘 1 ) , ( 𝐻 𝑘 2 ) , and ( 𝐻 𝑘 3 ) are satisfied, then there exist positive constants 𝛼 , 𝛽 , 𝛾 such that (4.1) has at least two positive solutions 𝑦 1 , 𝑦 2 with 𝑘 − 2 ‖ ‖ 𝑦 𝛼 < 1 ‖ ‖ < 𝑘 − 2 ‖ ‖ 𝑦 𝛽 < 2 ‖ ‖ < 𝑘 − 2 𝛾 , ( 4 . 9 ) where ( 𝐻 𝑘 𝑗 ) ( 𝑗 = 1 , 2 , 3 )   is condition obtained by replacing 𝜆 and 𝜌 by 𝑘 in the condition ( 𝐻 𝜆 𝑗 ) defined in Section 3.

Example 4.2. Consider the differential equation 𝑦 ( 4 ) ( 𝑡 ) − 𝜇 𝑦 ′ 1 ( 𝑡 ) + 𝑦 ( 𝑡 ) = 𝑒 2 0 𝜋 𝜇 𝑦 ′ ′ ( 𝑡 ) + 1 + 𝑡 𝑦 𝜇 𝑦 ( 𝑡 ) , 𝑡 ∈ 𝐼 , ( 𝑖 ) ( 0 ) = 𝑦 ( 𝑖 ) ( 2 𝜋 ) , 𝑖 = 0 , 1 , 2 , 3 , ( 4 . 1 0 ) where 𝜇 > 4 is a constant.
Let 𝜆 = 𝜇 − √ 𝜇 2 − 4 2 , 𝜌 = 𝜇 + √ 𝜇 2 − 4 2 𝑒 , 𝑓 ( 𝑡 , 𝑢 , 𝑣 ) = 𝑣 + ( 1 + 𝑡 ) ( 𝜇 𝑢 ) − 1 , 1 2 0 𝜋 𝜇 ℎ ( 𝑢 ) ≡ 1 , 𝑔 ( 𝑢 ) = 𝑢 𝑒 , 𝑝 ( 𝑡 , 𝑣 ) = 𝑣 2 0 𝜋 𝜇 , 𝑞 ( 𝑡 , 𝑣 ) = 1 + 𝑡 2 0 𝜋 𝜇 2 . ( 4 . 1 1 ) Then, for all ( 𝑡 , 𝑢 , 𝑣 ) ∈ 𝐼 × ( 0 , ∞ ) × 𝑅 , 0 < 𝑓 ( 𝑡 , 𝑢 , 𝑣 ) ≤ 𝑝 ( 𝑡 , 𝑣 ) ℎ ( 𝑢 ) + 𝑞 ( 𝑡 , 𝑣 ) 𝑔 ( 𝑢 ) . ( 4 . 1 2 ) Noting that s u p 𝑢 > 0 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑢 m a x ( 𝛿 𝜆 − 1 ) 𝑢 ≤ 𝑣 ≤ ( 1 − 𝛿 𝜆 ) 𝑢 ∫ 0 2 𝜋 𝜌 𝑝 ( 𝑡 , 𝑣 ) 𝑑 𝑡 ℎ − 2 𝑢 𝜌 / 𝑔 − 2 𝑢 + ∫ 0 2 𝜋 𝑔 𝜌 𝑞 ( 𝑡 , 𝑣 ) 𝑑 𝑡 − 2 𝛿 𝜆 𝑢 ⎫ ⎪ ⎬ ⎪ ⎭ > 1 0 𝜇 𝛿 𝜆 𝑒 + 3 𝜋 𝜇 − 1 𝜌 2 , l i m 𝜇 → + ∞ 𝑀 𝜆 𝜇 𝛿 𝜆 = 1 4 𝜋 , l i m 𝜇 → + ∞ 𝜌 2 𝜇 = 1 , ( 4 . 1 3 ) we obtain that ( 𝐻 𝜆 1 ) holds when 𝜇 is sufficiently large. On the other hand, it is easy to check that ( 𝐻 𝜆 2 ) is satisfied since 𝑓 ( 𝑡 , 𝑢 , 𝑣 ) → + ∞   as 𝑢 → 0 + for any 𝑡 ∈ 𝐼 and 𝑣 ∈ 𝑅 . Hence, (4.10) has at least a positive solution when 𝜇 is sufficiently large.

Acknowledgment

A Project Supported by the NNSF of China (10871063) and ScientificResearch Fund of human Provincial Education Department (10C0258).

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