Boundary Value Problems
Volume 2010 (2010), Article ID 961496, 12 pages
doi:10.1155/2010/961496
Research Article

Existence of Positive Solutions for Nonlinear Eigenvalue Problems

Sheng-Ping Wang,1 Fu-Hsiang Wong,2 and Fan-Kai Kung2

1Holistic Education Center, Cardinal Tien College of Healthcare and Management, No.171, Zhongxing Rd., Sanxing Township, Yilan County 266, Taiwan
2Department of Mathematics, National Taipei University of Education, 134, Ho-Ping E. Rd, Sec2, Taipei 10659, China

Received 2 June 2009; Accepted 2 February 2010

Academic Editor: Veli Shakhmurov

Copyright © 2010 Sheng-Ping Wang et al. 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 use a fixed point theorem in a cone to obtain the existence of positive solutions of the differential equation, 𝑢 + 𝜆 𝑓 ( 𝑡 , 𝑢 ) = 0 , 0 < 𝑡 < 1 , with some suitable boundary conditions, where 𝜆 is a parameter.

1. Introduction

We consider the existence of positive solutions of the following two-point boundary value problem: 𝐸 𝜆 𝑢 + 𝜆 𝑓 ( 𝑡 , 𝑢 ) = 0 , 0 < 𝑡 < 1 , ( 𝐵 𝐶 ) 𝑢 ( 0 ) = 𝑎 , 𝑢 ( 1 ) = 𝑏 , ( B V P 𝜆 ) where 𝑎 and 𝑏 are nonnegative constants, and 𝑓 ∈ 𝐶 ( [ 0 , 1 ] × [ 0 , ∞ ) , [ 0 , ∞ ) ) .

In the last thirty years, there are many mathematician considered the boundary value problem (BVPλ) with 𝑎 = 𝑏 = 0 , see, for example, Chu et al. [1], Chu et al. [2], Chu and Zhau [3], Chu and Jiang [4], Coffman and Marcus [5], Cohen and Keller [6], Erbe [7], Erbe et al. [8], Erbe and Wang [9], Guo and Lakshmikantham [10], Iffland [11], Njoku and Zanolin [12], Santanilla [13].

In 1993, Wong [14] showed the following excellent result.

Theorem 1 A (see [14]). Assume that [ ] × [ 𝑓 ( 𝑡 , 𝑢 ) ∶ = 𝑝 ( 𝑡 ) ℎ ( 𝑢 ) ∈ 𝐶 ( 0 , 1 0 , ∞ ) ; ( 0 , ∞ ) ) ( 1 . 1 ) is an increasing function with respect to 𝑢 . If there exists a constant 𝐿 such that 𝑐 0 𝑑 𝑢 √ 𝐻 ( 𝑐 ) − 𝐻 ( 𝑢 ) ≤ 𝐿 < ∞ ∀ 𝑐 > 0 , ( 1 . 2 ) where ∫ 𝐻 ( 𝑢 ) ∶ = 𝑢 0 ℎ ( 𝑦 ) 𝑑 𝑦 for 𝑢 ≥ 0 , then, there exists 𝜆 ∗ ∈ ( 0 , 8 𝐿 2 𝑝 0 − 1 ) such that the boundary value problem (BVPλ) with 𝑎 = 𝑏 = 0 has a positive solution in 𝐶 2 ⋂ ( 0 , 1 ) 𝐶 [ 0 , 1 ] for 0 < 𝜆 ≤ 𝜆 ∗ , while there is no such solution for 𝜆 > 𝜆 ∗ in which 𝑝 0 ∶ = m i n { 𝑝 ( 𝑡 ) ∣ 𝑡 ∈ [ 1 / 4 , 3 / 4 ] } .

Seeing such facts, we cannot but ask “whether or not we can obtain a similar conclusion for the boundary value problem (BVPλ).” We give a confirm answer to the question.

First, We observe the following statements.

(1)Let 𝑘 ( 𝑡 , 𝑠 ) = 𝑠 ( 1 − 𝑡 ) , f o r 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 𝑡 ( 1 − 𝑠 ) , f o r 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , ( 1 . 3 ) on [ 0 , 1 ] × [ 0 , 1 ] , then 𝑘 ( 𝑡 , 𝑠 ) is the Green's function of the differential equation 𝑢 ( 𝑡 ) = 0 in ( 0 , 1 ) with respect to the boundary value condition 𝑢 ( 0 ) = 𝑢 ( 1 ) = 0 . (2) 𝕂 ∶ = { 𝑢 ∈ 𝐶 [ 0 , 1 ] ∣ 𝑢 ( 𝑡 ) ≥ 0 , m i n 𝑡 ∈ [ 1 / 4 , 3 / 4 ] 𝑢 ( 𝑡 ) ≥ ( 1 / 4 ) ‖ 𝑢 ‖ } , is a cone in the Banach space with ‖ 𝑢 ‖ = s u p 𝑡 ∈ [ 0 , 1 ] | 𝑢 ( 𝑡 ) | .

In order to discuss our main result, we need the follo wing useful lemmas which due to Lian et al. [15] and Guo and Lakshmikantham [10], respectively.

Lemma 1 B (see [10]). Suppose that 𝑘 ( 𝑡 , 𝑠 ) be defined as in ( 1 ) . Then, we have the following results. ( 𝑅 1 ) ( 𝑘 ( 𝑡 , 𝑠 ) / 𝑘 ( 𝑠 , 𝑠 ) ≤ 1 , for 𝑡 ∈ [ 0 , 1 ] and 𝑠 ∈ [ 0 , 1 ] , ) ( 𝑅 2 ) ( 𝑘 ( 𝑡 , 𝑠 ) / 𝑘 ( 𝑠 , 𝑠 ) ≥ 1 / 4 , for 𝑡 ∈ [ 1 / 4 , 3 / 4 ] and 𝑠 ∈ [ 0 , 1 ] . )

Lemma 1 C (see [10, Lemmas 2 . 3 . 3 and 2 . 3 . 1 ]). Let 𝐸 be a real Banach space, and let 𝐶 ⊂ 𝐸 be a cone. Assume that 𝐵 𝜌 ∶ = { 𝑢 ∈ 𝐶 ∣ ‖ 𝑢 ‖ < 𝜌 } and 𝐴 ∶ 𝐵 𝜌 → 𝐶 is completely continuous. Then (1) 𝑖 ( 𝐴 , 𝐵 𝜌 , 𝐶 ) = 0 if 𝐼 𝑛 𝑓 𝑢 ∈ 𝜕 𝐵 𝜌 ‖ 𝐴 𝑢 ‖ > 0 , 𝐴 𝑢 ≠ 𝛼 𝑢 f o r 𝑢 ∈ 𝜕 𝐵 𝜌 ] , , 𝛼 ∈ ( 0 , 1 ( 1 . 4 ) (2) 𝑖 ( 𝐴 , 𝐵 𝜌 , 𝐶 ) = 1 i f 𝐴 𝑢 ≠ 𝛼 𝑢 f o r 𝑢 ∈ 𝜕 𝐵 𝜌 a n d 𝛼 ≥ 1 ,
where 𝑖 ( 𝐴 , 𝐵 𝜌 , 𝐶 ) is the fixed point index of a compact map 𝐴 ∶ 𝐵 𝜌 → 𝐶 , such that 𝐴 𝑢 ≠ 𝑢 for 𝑢 ∈ 𝜕 𝐵 𝜌 , with respect to 𝐵 𝜌 .

2. Main Results

Now, we can state and prove our main result.

Theorem 2.1. Suppose that there exist two distinct positive constants 𝜂 , 𝜃 and a function 𝑔 ∈ 𝐶 ( [ 𝜉 2 , 𝜃 ] ; [ 0 , ∞ ) ) with 𝜃 > m a x { 𝑎 , 𝑏 } ∶ = 𝜉 1 and 𝜉 2 = m i n { 𝑎 , 𝑏 } such that 𝑓 ( 𝑡 , 𝑢 ) ≥ 𝜂 3 / 4 1 / 4 𝑘 1 2 , 𝑠 𝑑 𝑠 − 1 1 o n 4 , 3 4 × 1 4 , [ ] × 𝜉 𝜂 , 𝜂 ( 2 . 1 ) 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑔 ( 𝑢 ) o n 0 , 1 2 , 𝜃 . ( 2 . 2 ) Then (BVPλ) has a positive solution 𝑢 with ‖ 𝑢 ‖ between 𝜂 and 𝜃 if ⎡ ⎢ ⎢ ⎣ 𝜆 ∈ 1 , 2 𝜃 𝜉 1 𝑑 𝑠 √ 𝐺 ( 𝜃 ) − 𝐺 ( 𝑠 ) 2 ⎤ ⎥ ⎥ ⎦ , ( 2 . 3 ) where ⎧ ⎪ ⎨ ⎪ ⎩ 𝐺 ( 𝑢 ) ∶ = 𝑢 𝜉 1 𝜉 𝑔 ( 𝑠 ) 𝑑 𝑠 , i f 𝑢 ∈ 1 , 𝜉 , 𝜃 0 , i f 𝑢 ∈ 2 , 𝜉 1 . ( 2 . 4 )

Proof. It is clear that (BVPλ) has a solution 𝑢 = 𝑢 ( 𝑡 ) if, and only if, 𝑢 is the solution of the operator equation 𝑢 ( 𝑡 ) = 𝑎 ( 1 − 𝑡 ) + 𝑏 𝑡 + 𝜆 1 0 𝑘 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ∶ = 𝐴 𝑢 ( 𝑡 ) . ( 2 . 5 ) It follows from the definition of 𝕂 in our observation ( 2 ) and Lemma B that m i n [ ] 𝑡 ∈ 1 / 4 , 3 / 4 ( 𝐴 𝑢 ) ( 𝑡 ) = m i n 𝑡 ∈ [ 1 / 4 , 3 / 4 ] 𝑎 ( 1 − 𝑡 ) + 𝑏 𝑡 + 𝜆 1 0 ≥ 1 𝑘 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 4 𝑎 ( 1 − 𝑡 ) + 𝑏 𝑡 + 𝜆 1 0 𝑅 𝑘 ( 𝑠 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 u s i n g 2 ≥ 1 4 𝑎 ( 1 − 𝑡 ) + 𝑏 𝑡 + 𝜆 1 0 𝑅 𝑘 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 u s i n g 1 . ( 2 . 6 ) Hence, m i n 𝑡 ∈ [ 1 / 4 , 3 / 4 ] ( 𝐴 𝑢 ) ( 𝑡 ) ≥ ( 1 / 4 ) ‖ 𝐴 𝑢 ‖ , which implies 𝐴 𝕂 ⊂ 𝕂 . Furthermore, it is easy to check that 𝐴 ∶ 𝕂 → 𝕂 is completely continuous. If there exists a 𝑢 ∈ 𝜕 𝐵 𝜂 ∪ 𝜕 𝐵 𝜃 such that 𝐴 𝑢 = 𝑢 , then we obtain the desired result. Thus, we may assume that 𝐴 𝑢 ≠ 𝑢 f o r 𝑢 ∈ 𝜕 𝐵 𝜂 ∪ 𝜕 𝐵 𝜃 , ( 2 . 7 ) where 𝐵 𝜂 ∶ = { 𝑢 ∈ 𝕂 ∣ ‖ 𝑢 ‖ < 𝜂 } and 𝐵 𝜃 ∶ = { 𝑢 ∈ 𝕂 ∣ ‖ 𝑢 ‖ < 𝜃 } . We now separate the rest proof into the following three steps.Step 1. It follows from the definitions of ‖ 𝑢 ‖ and 𝕂 that, for 𝑢 ∈ 𝜕 𝐵 𝜂 , [ ] , 𝑢 ( 𝑡 ) ≤ ‖ 𝑢 ‖ = 𝜂 f o r 𝑡 ∈ 0 , 1 𝑢 ( 𝑡 ) ≥ m i n 𝑡 ∈ [ 1 / 4 , 3 / 4 ] 1 𝑢 ( 𝑡 ) ≥ 4 1 ‖ 𝑢 ‖ = 4 1 𝜂 f o r 𝑡 ∈ 4 , 3 4 , ( 2 . 8 ) which implies 1 4 1 𝜂 ≤ 𝑢 ( 𝑡 ) ≤ 𝜂 f o r 𝑡 ∈ 4 , 3 4 . ( 2 . 9 ) Hence, by (2.5), 1 ( 𝐴 𝑢 ) 2 = 1 2 ( 𝑎 + 𝑏 ) + 𝜆 1 0 𝑘 1 2 ≥ , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 1 0 𝑘 1 2 𝑓 ≥ , 𝑠 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ( u s i n g 𝜆 ≥ 1 , 𝑎 , 𝑏 ≥ 0 ) 3 / 4 1 / 4 𝑘 1 2 𝑓 , 𝑠 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ≥ 𝜂 3 / 4 1 / 4 𝑘 1 2 , 𝑠 𝑑 𝑠 − 1 3 / 4 1 / 4 𝑘 1 2 , 𝑠 𝑑 𝑠 ‖ 𝑢 ‖ 𝜂 = ‖ 𝑢 ‖ , ( 2 . 1 0 ) which implies ‖ 𝐴 𝑢 ‖ ≥ ‖ 𝑢 ‖ f o r 𝑢 ∈ 𝜕 𝐵 𝜂 . ( 2 . 1 1 ) Hence i n f 𝑢 ∈ 𝜕 𝐵 𝜂 ‖ 𝐴 𝑢 ‖ ≥ i n f 𝑢 ∈ 𝜕 𝐵 𝜂 ‖ 𝑢 ‖ = 𝜂 > 0 . ( 2 . 1 2 ) We now claim that 𝐴 𝑢 ≠ 𝛼 𝑢 , f o r 𝑢 ∈ 𝜕 𝐵 𝜂 , 𝛼 ∈ ( 0 , 1 ) . ( 2 . 1 3 ) In fact, if there exist 𝑢 ∈ 𝜕 𝐵 𝜂 and 𝛼 ∈ ( 0 , 1 ) such that 𝐴 𝑢 = 𝛼 𝑢 , then, by (2.11), ‖ 𝑢 ‖ ≤ ‖ 𝐴 𝑢 ‖ = 𝛼 ‖ 𝑢 ‖ < ‖ 𝑢 ‖ , ( 2 . 1 4 ) which gives a contradiction. This proves that (2.13) holds. Thus, by Lemma C, 𝑖 𝐴 , 𝐵 𝜂 , 𝕂 = 0 . ( 2 . 1 5 ) Step 2. First, we claim that 𝐴 𝑢 ≠ 𝛼 𝑢 f o r 𝑢 ∈ 𝜕 B 𝜃 , 𝛼 > 1 . ( 2 . 1 6 ) Suppose to the contrary that there exist 𝑢 ∈ 𝜕 𝐵 𝜃 and 𝛼 > 1 such that 𝐴 𝑢 = 𝛼 𝑢 . ( 2 . 1 7 ) It is clear that (2.17) is equivalent to 𝑢 𝜆 ( 𝑡 ) + 𝛼 𝑓 ( 𝑡 , 𝑢 ) = 0 . ( 2 . 1 8 ) Since 𝑢 ∈ 𝐶 [ 0 , 1 ] and ‖ 𝑢 ‖ = 𝜃 > 0 , it follows that there exists a 𝑡 ∗ ∈ ( 0 , 1 ) such that 𝑢 𝑡 ∗ = ‖ 𝑢 ‖ = 𝜃 . ( 2 . 1 9 ) Let 𝑡 1 [ ] = m i n { 𝑡 ∈ 0 , 1 ∣ 𝑢 ( 𝑡 ) = 𝜃 } , 𝑡 2 [ ] = m a x { 𝑡 ∈ 0 , 1 ∣ 𝑢 ( 𝑡 ) = 𝜃 } . ( 2 . 2 0 ) Then 0 < 𝑡 1 ≤ 𝑡 ∗ ≤ 𝑡 2 < 1 . From 𝑢 < 0 on ( 0 , 1 ) , we see that 𝑢 ( 𝑡 ) > 0 on ( 0 , 𝑡 1 ) 𝑢 ( 𝑡 ) < 0 on ( 𝑡 2 , 1 ) and 𝑢 ( 𝑡 ) = 0 on [ 𝑡 1 , 𝑡 2 ] . It follows from 𝑢 𝜆 ( 𝑡 ) = − 𝛼 𝜆 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) ≥ − 𝛼 [ ] 𝑔 ( 𝑢 ( 𝑡 ) ) f o r 𝑡 ∈ 0 , 1 ( 2 . 2 1 ) and 𝑢 ( 𝑡 ) = 0 on [ 𝑡 1 , 𝑡 2 ] that 0 < 𝑢 ( 𝑡 ) ≤ 2 𝜆 𝛼 ( 𝐺 ( 𝜃 ) − 𝐺 ( 𝑢 ( 𝑡 ) ) ) f o r 𝑡 ∈ 0 , 𝑡 1 , 0 > 𝑢 ( 𝑡 ) ≥ − 2 𝜆 𝛼 𝑡 ( 𝐺 ( 𝜃 ) − 𝐺 ( 𝑢 ( 𝑡 ) ) ) f o r 𝑡 ∈ 2 . , 1 ( 2 . 2 2 ) Hence, 𝜃 𝑎 𝑑 𝑠 √ ≤ ( 2 𝜆 / 𝛼 ) ( 𝐺 ( 𝜃 ) − 𝐺 ( 𝑠 ) ) 𝑡 1 0 𝑑 𝑡 = 𝑡 1 , 𝜃 𝑏 𝑑 𝑠 √ ≤ ( 2 𝜆 / 𝛼 ) ( 𝐺 ( 𝜃 ) − 𝐺 ( 𝑠 ) ) 1 𝑡 2 𝑑 𝑡 = 1 − 𝑡 2 . ( 2 . 2 3 ) Thus 1 ≥ 1 − 𝑡 2 + 𝑡 1 ≥ 2 √ 2 𝜆 / 𝛼 𝜃 𝜉 1 𝑑 𝑠 √ > 𝐺 ( 𝜃 ) − 𝐺 ( 𝑠 ) 2 𝜆 𝜃 𝜉 1 𝑑 𝑠 √ ⎛ ⎜ ⎜ ⎝ ⎡ ⎢ ⎢ ⎣ 𝐺 ( 𝜃 ) − 𝐺 ( 𝑠 ) ( s i n c e 𝛼 > 1 ) ≥ 1 b e c a u s e 𝜆 ∈ 1 , 2 𝜃 𝜉 1 𝑑 𝑠 √ 𝐺 ( 𝜃 ) − 𝐺 ( 𝑠 ) 2 ⎤ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎠ . ( 2 . 2 4 ) This contradiction implies 𝐴 𝑢 ≠ 𝛼 𝑢 , f o r 𝑢 ∈ 𝜕 𝐵 𝜃 , 𝛼 > 1 . ( 2 . 2 5 ) Therefore, by Lemma C, 𝑖 𝐴 , 𝐵 𝜃 , 𝕂 = 1 . ( 2 . 2 6 ) Step 3. It follows from Steps (1) and (2) and the property of the fixed point index (see, for example, [10, Theorem 2 . 3 . 2 ]) that the proof is complete.

Remark 2.2. It follows from the conclusion of Theorem 2.1 that the positive constant 𝜃 and nonnegative function g ( 𝑢 ) satisfy 𝜃 𝜉 1 𝑑 𝑠 √ 𝐺 ≥ 1 ( 𝜃 ) − 𝐺 ( 𝑠 ) √ 2 . ( 2 . 2 7 )

There are many functions 𝑔 ( 𝑢 ) and positive constants 𝜃 satisfying (2.27). For example, Suppose that 𝑀 ∈ ( 0 , 8 ] and 𝜃 ∈ ( 𝜉 1 , ∞ ) . Let 𝑔 ( 𝑢 ) ∶ = 𝑀 ( 𝜃 − 𝜉 1 ) on [ 𝜉 2 , 𝜃 ] , then 𝐺 ( 𝑢 ) = 𝑀 ( 𝜃 − 𝜉 1 ) ( 𝑢 − 𝜉 1 ) on [ 𝜉 1 , 𝜃 ] and

𝜃 𝜉 1 1 √ 𝐺 1 ( 𝜃 ) − 𝐺 ( 𝑢 ) 𝑑 𝑢 = 𝑀 𝜃 − 𝜉 1 𝜃 𝜉 1 1 √ = 1 𝜃 − 𝑢 𝑑 𝑢 𝑀 𝜃 − 𝜉 1 2 √ 𝜃 − 𝜉 1 = 2 √ 𝑀 ≥ 1 √ 2 . ( 2 . 2 8 )

Remark 2.3. We now define m a x 𝑓 0 ∶ = l i m 𝑢 → 0 + m a x [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 , m i n 𝑓 0 ∶ = l i m 𝑢 → 0 + m i n [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 , m a x 𝑓 ∞ ∶ = l i m 𝑢 → ∞ m a x [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 , m i n 𝑓 ∞ ∶ = l i m 𝑢 → ∞ m i n [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 . ( 2 . 2 9 ) A simple calculation shows that 3 / 4 1 / 4 𝑘 1 2 3 , 𝑠 𝑑 𝑠 = . 3 2 ( 2 . 3 0 )

Then, we have the following results.

(i)Suppose that m a x 𝑓 0 ∶ = 𝐶 1 ∈ [ 0 , 𝑀 ) ⊆ [ 0 , 8 ) . Taking 𝜖 = 𝑀 − 𝐶 1 > 0 , there exists 1 > 𝜃 1 > 0 ( 𝜃 1 can be chosen small arbitrarily) such that m a x [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 ≤ 𝜖 + 𝐶 1 = 𝑀 o n 0 , 𝜃 1 . ( 2 . 3 1 ) Hence, 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑀 𝑢 ≤ 𝑀 𝜃 1 [ ] × 𝜉 o n 0 , 1 2 , 𝜃 1 ⊂ [ ] × 0 , 1 0 , 𝜃 1 . ( 2 . 3 2 ) It follows from Remark 2.2 that the hypothesis (2.2) of Theorem 2.1 is satisfied if 𝜆 ∈ [ 1 , 8 / 𝑀 ] .(ii)Suppose that m i n 𝑓 ∞ ∶ = 𝐶 2 ∈ ( 1 2 8 / 3 , ∞ ] . Taking 𝜖 = 𝐶 2 − 1 2 8 / 3 > 0 , there exists 𝜂 1 > 0 ( 𝜂 1 can be chosen large arbitrarily) such that m i n [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 ≥ − 𝜖 + 𝐶 2 = 1 2 8 3 1 o n 4 𝜂 1 . , ∞ ( 2 . 3 3 ) Hence, 𝑓 ( 𝑡 , 𝑢 ) ≥ 1 2 8 3 𝑢 ≥ 1 2 8 3 1 4 𝜂 1 ≥ 3 2 3 𝜂 1 1 o n 4 , 3 4 × 1 4 𝜂 1 , 𝜂 1 ⊂ [ ] × 1 0 , 1 4 𝜂 1 , , ∞ ( 2 . 3 4 ) which satisfies the hypothesis (2.1) of Theorem 2.1.(iii)Suppose that m i n 𝑓 0 ∶ = 𝐶 3 ∈ ( 1 2 8 / 3 , ∞ ] . Taking 𝜖 = 𝐶 3 − 1 2 8 / 3 > 0 , there exists 1 > 𝜂 2 > 0 ( 𝜂 2 can be chosen small arbitrarily) such that m i n [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 ≥ − 𝜖 + 𝐶 3 = 1 2 8 3 o n 0 , 𝜂 2 . ( 2 . 3 5 ) Hence, 𝑓 ( 𝑡 , 𝑢 ) ≥ 1 2 8 3 𝑢 ≥ 1 2 8 3 1 4 𝜂 2 = 3 2 3 𝜂 2 1 o n 4 , 3 4 × 1 4 𝜂 2 , 𝜂 2 ⊂ [ ] × 0 , 1 0 , 𝜂 2 , ( 2 . 3 6 ) which satisfies the hypothesis (2.1) of Theorem 2.1.(iv)Suppose that m a x 𝑓 ∞ ∶ = 𝐶 4 ∈ [ 0 , 𝑀 ) ⊆ [ 0 , 8 ) . Taking 𝜖 = 𝑀 − 𝐶 4 > 0 , there exists a 𝛿 > 0 ( 𝛿 can be chosen large arbitrarily) such that m a x [ ] 𝑡 ∈ 0 , 1 𝑓 ( 𝑡 , 𝑢 ) 𝑢 ≤ 𝜖 + 𝐶 4 [ = 𝑀 o n 𝛿 , ∞ ) . ( 2 . 3 7 ) Hence, we have the following two cases.

Case i. Assume that m a x 𝑡 ∈ [ 0 , 1 ] 𝑓 ( 𝑡 , 𝑢 ) is bounded, say [ ] × [ 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝐿 o n 0 , 1 0 , ∞ ) , ( 2 . 3 8 ) for some constant 𝐿 . Taking 𝜃 2 = 𝐿 / 𝑀 > 1 (since 𝐿 can be chosen large arbitrarily, 𝜃 2 can be chosen large arbitrarily, too), 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝐿 = 𝑀 𝜃 2 [ ] × o n 0 , 1 0 , 𝜃 2 ⊂ [ ] × [ 0 , 1 0 , ∞ ) . ( 2 . 3 9 )

Case ii. Assume that m a x 𝑡 ∈ [ 0 , 1 ] 𝑓 ( 𝑡 , 𝑢 ) is unbounded, then there exist a 𝜃 2 ≥ m a x { 𝛿 , 𝜉 2 } ( 𝜃 2 can be chosen large arbitrarily) and 𝑡 0 ∈ [ 0 , 1 ] such that 𝑓 𝑡 ( 𝑡 , 𝑢 ) ≤ 𝑓 0 , 𝜃 2 [ ] × o n 0 , 1 0 , 𝜃 2 . ( 2 . 4 0 ) It follows from 𝜃 2 ≥ 𝛿 and (2.37) that 𝑓 𝑡 ( 𝑡 , 𝑢 ) ≤ 𝑓 0 , 𝜃 2 ≤ 𝑀 𝜃 2 [ ] × 𝜉 o n 0 , 1 2 , 𝜃 2 ⊂ [ ] × 0 , 1 0 , 𝜃 2 . ( 2 . 4 1 )

By Cases (i), (ii) and Remark 2.2, we see that the hypothesis (2.2) of Theorem 2.1 is satisfied if 𝜆 ∈ [ 1 , 8 / 𝑀 ] .

We immediately conclude the following corollaries.

Corollary 2.4. (BVPλ) has at least one positive solution for 𝜆 ∈ [ 1 , 8 𝑀 ] if one of the following conditions holds: ( 𝐻 1 ) m a x 𝑓 0 = 𝐶 1 ∈ [ 0 , 𝑀 ) ⊆ [ 0 , 8 ) , m i n 𝑓 ∞ = 𝐶 2 ∈ ( 1 2 8 / 3 , ∞ ] , ( 𝐻 2 ) m i n 𝑓 0 = 𝐶 3 ∈ ( 1 2 8 / 3 , ∞ ] , m a x 𝑓 ∞ = 𝐶 4 ∈ [ 0 , 𝑀 ) ⊆ [ 0 , 8 ) .

Proof. It follows from Remark 2.3 and Theorem 2.1 that the desired result holds, immediately.

Corollary 2.5. Let ( 𝐻 3 ) m i n 𝑓 ∞ = 𝐶 2 , m i n 𝑓 0 = 𝐶 3 ∈ ( 1 2 8 / 3 , ∞ ] , ( 𝐻 4 ) 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑀 𝜃 ∗ on [ 0 , 1 ] × [ 𝜉 2 , 𝜃 ∗ ] for some 𝑀 ∈ ( 0 , 8 ] and 𝜃 ∗ > 0 .Then, for 𝜆 ∈ [ 1 , 8 / 𝑀 ] , (BVPλ) has at least two positive solutions 𝑢 1 and 𝑢 2 such that ‖ ‖ 𝑢 0 < 1 ‖ ‖ < 𝜃 ∗ < ‖ ‖ 𝑢 2 ‖ ‖ . ( 2 . 4 2 )

Proof. It follows from Remark 2.3 that there exist two real numbers 𝜂 2 < 𝜃 ∗ < 𝜂 1 satisfying 𝑓 ( 𝑡 , 𝑢 ) ≥ 3 2 3 𝜂 1 1 o n 4 , 3 4 × 1 4 𝜂 1 , 𝜂 1 , 𝑓 ( 𝑡 , 𝑢 ) ≥ 3 2 3 𝜂 2 1 o n 4 , 3 4 × 1 4 𝜂 2 , 𝜂 2 . ( 2 . 4 3 ) Hence, by Theorem 2.1 and Remark 2.2, we see that for each 𝜆 ∈ [ 1 , 8 / 𝑀 ] , there exist two positive solutions 𝑢 1 and 𝑢 2 of (BVPλ) such that 𝜂 2 < ‖ ‖ 𝑢 1 ‖ ‖ < 𝜃 ∗ < ‖ ‖ 𝑢 2 ‖ ‖ < 𝜂 1 . ( 2 . 4 4 ) Thus, we complete the proof.

Corollary 2.6. Let ( 𝐻 5 ) m a x 𝑓 0 = 𝐶 1 , m a x 𝑓 ∞ = 𝐶 4 ∈ [ 0 , 𝑀 ) ⊆ [ 0 , 8 ) , ( 𝐻 6 ) 𝑓 ( 𝑡 , 𝑢 ) ≥ ( 3 2 / 3 ) 𝜂 ∗ on [ 1 / 4 , 3 / 4 ] × [ ( 1 / 4 ) 𝜂 ∗ , 𝜂 ∗ ] , for some 𝜂 ∗ > 0 . Then, for 𝜆 ∈ [ 1 , 8 / 𝑀 ] , (BVPλ) has at least two positive solutions 𝑢 1 and 𝑢 2 such that ‖ ‖ 𝑢 0 < 1 ‖ ‖ < 𝜂 ∗ < ‖ ‖ 𝑢 2 ‖ ‖ . ( 2 . 4 5 )

Proof. It follows from Remark 2.3 that there exist two real numbers 𝜃 1 < 𝜂 ∗ < 𝜃 2 satisfying 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑀 𝜃 1 [ ] × 𝜉 o n 0 , 1 2 , 𝜃 1 , 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑀 𝜃 2 [ ] × 𝜉 o n 0 , 1 2 , 𝜃 2 . ( 2 . 4 6 ) Hence, by Theorem 2.1 and Remark 2.2, we see that, for each 𝜆 ∈ [ 1 , 8 / 𝑀 ] , (BVPλ) has two positive solutions 𝑢 1 and 𝑢 2 such that 𝜃 1 < ‖ ‖ 𝑢 1 ‖ ‖ < 𝜂 ∗ < ‖ ‖ 𝑢 2 ‖ ‖ < 𝜃 2 . ( 2 . 4 7 ) Thus, we completed the proof.

3. Examples

To illustrate the usage of our results, we present the following examples.

Example 3.1. Consider the following boundary value problem: 𝑢 ( 𝑡 ) + 𝜆 𝑢 𝑒 𝑢 1 + 𝑡 2 = 0 i n ( 0 , 1 ) , 𝐵 𝐶 1 𝑢 ( 0 ) = 𝑎 = 1 , 𝑢 ( 1 ) = 𝑏 = 1 . ( B V P . 1 ) Clearly, m a x 𝑓 0 [ [ = 1 ∈ 0 , 𝑀 ) ⊆ 0 , 8 ) , m i n 𝑓 ∞ = ∞ ∈ 1 2 8 3 . , ∞ ( 3 . 1 ) If we take 𝑀 = 2 , then it follows from ( 𝐻 1 ) of Corollary 2.4 that (BVP.1) has a solution if 𝜆 ∈ [ 1 , 4 ] .

Example 3.2. Consider the following boundary value problem: 𝑢 ( 𝑡 ) + 𝜆 𝑢 ( 1 − 𝑡 ) + 𝐾 ( 1 − 𝑒 − 𝑢 ) 1 = 0 i n ( 0 , 1 ) , 𝐾 + 4 > 1 2 8 3 , 𝐵 𝐶 2 𝑢 ( 0 ) = 𝑎 = 1 , 𝑢 ( 1 ) = 𝑏 = 2 . ( B V P . 2 ) Clearly, m i n 𝑓 0 1 = 𝐾 + 4 ∈ 1 2 8 3 , , ∞ m a x 𝑓 ∞ [ [ = 1 ∈ 0 , 𝑀 ) ⊆ 0 , 8 ) . ( 3 . 2 ) If we take 𝑀 = 2 , then it follows from ( 𝐻 2 ) of Corollary 2.4 that (BVP.2) has a solution if 𝜆 ∈ [ 1 , 4 ] .

Example 3.3. Consider the following boundary value problem: 𝑢 ( 𝑡 ) + 𝜆 𝑢 3 / 2 + 𝑢 1 / 2 / ( 1 + 𝑡 ) = 0 i n ( 0 , 1 ) , 𝐵 𝐶 3 𝑢 ( 0 ) = 𝑎 = 0 , 𝑢 ( 1 ) = 𝑏 = 1 . ( B V P . 3 ) Clearly, if we take 𝑀 = 2 and 𝜃 ∗ = 1 , m i n 𝑓 ∞ ] , = ∞ ∈ ( 1 2 8 / 3 , ∞ m i n 𝑓 0 ] , [ ] × [ ] . = ∞ ∈ ( 1 2 8 / 3 , ∞ 𝑓 ( 𝑡 , 𝑢 ) ≤ 2 o n 0 , 1 0 , 1 ( 3 . 3 ) Hence, it follows from Corollary 2.5 that (BVP.3) has two solutions if 𝜆 ∈ [ 1 , 4 ] .

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