On Linear and Nonlinear Fourth-Order Eigenvalue Problems with Nonlocal Boundary Condition

Abstract

We determine the principal eigenvalue of the linear problem 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)=𝜇[𝑢(𝑡)−𝑢^′′(𝑡)],𝑡∈(0,1),  𝑢(0)=𝑢(1)=∫¹₀𝑝(𝑠)𝑢(𝑠)d𝑠, 𝑢^′′(0)=𝑢^′′(1)=∫¹₀𝑞(𝑠)𝑢^′′(𝑠)d𝑠, where 0<𝛽<𝜋² and 𝑝,𝑞∈𝐿[0,1]. Moreover, we investigate the existence of positive solutions for the corresponding nonlinear problem. The proofs of our main results are based upon the Krein-Rutman theorem and fixed point index theory.

1. Introduction

The deformations of an elastic beam can be described by the boundary value problems of the fourth-order ordinary differential equations. For example, an elastic beam in an equilibrium state whose both ends are simply supported can be described by the fourth-order boundary value problem of the form 𝑢⁽⁴⁾(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)(),𝑢(0)(1),𝑡∈0,1=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (1) see Gupta [1, 2]. Owing to its significance in physics, it has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point index theory, and the method of lower and upper solutions; see, for example, [3–16].

Recently, Bai [3] investigated the existence of positive solutions for more general fourth-order nonlocal boundary value problem 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)=𝜆𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)(),𝑢(0)(1),𝑡∈0,1=𝑢=¹₀𝑝(𝑠)𝑢(𝑠)𝑢d𝑠,_′′(0)=𝑢^′′(1)=¹₀𝑞(𝑠)𝑢^′′(𝑠)d𝑠. (2) By using the Krasnoselskii fixed point theorem, the sufficient conditions for the existence of positive solutions of (2) are obtained. We can find that, however, those conditions do not involve the eigenvalues with respect to the relevant linear operator, and those sufficient conditions are not optimal.

The likely reason is that the spectrum structure of the linear eigenvalue problem 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)=𝜇𝑢(𝑡)−𝑢^′′(𝑡(),𝑢(0)(1)=),𝑡∈0,1=𝑢¹₀𝑝(𝑠)𝑢(𝑠)𝑢d𝑠,_′′(0)=𝑢^′′(1)=¹₀𝑞(𝑠)𝑢^′′(𝑠)d𝑠 (3) is not clear.

It is the purpose of this paper to investigate the first eigenvalue of (3) by using the Krein-Rutman theorem [17, 18], and then we use this spectrum result to establish the existence of positive solutions of nonlinear nonlocal problem (2). The existence of positive solution is obtained by means of fixed point index theory under some conditions concerning the first eigenvalue with respect to the relevant linear operator. The obtained sufficient conditions in this paper are optimal. For the concepts and properties of fixed point index theory, we refer the reader to [19].

The rest of the paper is arranged as follows: in Section 2, we study the spectrum of (3). Finally, in Section 3, we give an application of our main result.

2. Spectrum of (3)

Let us assume that(A1)0<𝛽<𝜋², 𝑝,𝑞∈𝐿[0,1], 𝑝(𝑡)≥0, 𝑞(𝑡)≥0, 𝑡∈[0,1],  ∫¹₀𝑝(𝑠)d𝑠<1,  ∫¹₀𝑞(𝑠)sin𝛽𝑠d𝑠+∫¹₀𝑞(𝑠)sin𝛽(1−𝑠)d𝑠<sin𝛽.(A2)𝑓∈𝐶([0,1]×[0,∞)×(−∞,0],[0,∞)).

Definition 1. We say 𝜇 is an eigenvalue of linear problem (see (3)), if (3) has nontrivial solutions.

To study the spectrum of (3), we need several preliminary results.

Let 𝐺₁()=𝑡()𝑠()𝐺𝑡,𝑠1−𝑠,0≤𝑡≤𝑠≤1;1−𝑡,0≤𝑠≤𝑡≤1;₂(𝑡,𝑠)=sin𝛽𝑡sin𝛽(1−𝑠)𝛽sin𝛽,0≤𝑡≤𝑠≤1;sin𝛽𝑠sin𝛽(1−𝑡)𝛽sin𝛽,0≤𝑠≤𝑡≤1. (4)

Lemma 2 (see [7]). Assume (A1) holds. Then one has(i)𝐺_𝑖(𝑡,𝑠)≥0,  for 𝑡,𝑠∈[0,1]; 𝐺_𝑖(𝑡,𝑠)>0,  for 𝑡,𝑠∈(0,1);(ii)𝑏_𝑖𝐺_𝑖(𝑡,𝑡)𝐺_𝑖(𝑠,𝑠)≤𝐺_𝑖(𝑡,𝑠)≤𝐶_𝑖𝐺_𝑖(𝑡,𝑡),  for 𝑡,𝑠∈[0,1],where 𝐶₁=1, 𝑏₁=1; 𝐶₂=1/sin𝛽, 𝑏₂=𝛽sin𝛽.

Lemma 3 (see [3]). Assume (A1) holds. Then for any 𝑔∈𝐶[0,1],  𝑢 solves the problem 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)(𝑡)(),𝑢(0)(1)==𝑔,𝑡∈0,1=𝑢¹₀𝑝(𝑠)𝑢(𝑠)𝑢d𝑠,_′′(0)=𝑢^′′(1)=¹₀𝑞(𝑠)𝑢^′′(𝑠)d𝑠, (5) if and only if 𝑢(𝑡)=¹₀¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑔(𝜏)d𝜏d𝑠, (6) where 𝐾₁(𝑡,𝑠)=𝐺₁()+1𝑡,𝑠1−∫¹₀𝑝(𝑥)d𝑥¹⁰𝐺₁()𝑝(𝑥)𝐾𝑠,𝑥d𝑥,₂(𝑠,𝜏)=𝐺₂(𝑠,𝜏)+𝜌(𝑠)¹₀𝐺₂()𝑞(𝑥)𝜌(𝑠)=𝜏,𝑥d𝑥,sin𝛽𝑠+sin𝛽(1−𝑠)sin𝛽−∫¹₀𝑞(𝑥)sin𝛽𝑥d𝑥−∫¹₀𝑞(𝑥)sin𝛽().1−𝑥d𝑥 (7)

Let 𝑒(𝑡)∶=−𝑡²∫+𝑡+¹⁰𝑝(𝑠)𝑠−𝑠²d𝑠1−∫¹₀𝑝(𝑠)[]d𝑠,𝑡∈0,1. (8) Then, there exist 𝑐,^∼𝑐>0 such that ¹₀𝐾₁(𝑡,𝑠)𝑒(𝑠)d𝑠≤𝑐𝑒(𝑡),𝑡∈[0,1₁₀],𝐾₂(𝑡,𝑠)𝑒(𝑠)d𝑠≤^∼𝑐𝑒(𝑡),𝑡∈[0,1], (9)(10) respectively.

In fact, from (ii) of Lemma 2, we can obtain ¹₀𝐾₁()𝑒(𝑠)=𝑡,𝑠d𝑠₁₀𝐺₁()+1𝑡,𝑠1−∫¹₀𝑝(𝑥)×d𝑥₁₀𝐺₁()𝑝(𝑥)𝑒(𝑠)≤𝑠,𝑥d𝑥d𝑠₁₀𝐺₁()+1𝑡,𝑡1−∫¹₀𝑝(𝑥)d𝑥×¹₀𝐺₁(𝑠,𝑠)𝑝(𝑥)d𝑥𝑒(𝑠)d𝑠=𝐺₁(𝑡,𝑡)¹₀𝑒(𝑠)∫d𝑠+¹⁰𝑝(𝑥)d𝑥1−∫¹₀𝑝(𝑥)×d𝑥₁₀𝐺₁()𝑒(𝑠)=𝑠,𝑠d𝑠−𝑡²+𝑡¹₀𝑒(𝑠)∫d𝑠+¹⁰𝑝(𝑥)d𝑥1−∫¹₀𝑝(𝑥)×d𝑥₁₀𝑠−𝑠²𝑒(𝑠)(𝑡)[],d𝑠≤𝑐𝑒,𝑡∈0,1 (11) where we can take 𝑐=max¹₀𝑒(𝑠)∫d𝑠,¹⁰𝑝(𝑠)d𝑠∫¹₀𝑠−𝑠²𝑝(𝑠)×d𝑠₁₀𝑠−𝑠²𝑒(𝑠)d𝑠, (12) and also we obtain ¹₀𝐾₂()𝑒(𝑠)=𝑡,𝑠d𝑠₁₀𝐺₂(𝑡,𝑠)+𝜌(𝑡)¹₀𝐺₂()𝑞(𝑥)𝑒(𝑠)≤1𝑠,𝑥d𝑥d𝑠sin𝛽¹⁰𝐺₂(𝑡,𝑡)+𝜌(𝑡)¹₀𝐺₂()𝑞(𝑥)(𝑠)𝑥,𝑥d𝑥×𝑒d𝑠≤𝑐₁sin𝛽𝑡sin𝛽(1−𝑡)+𝑐₂≤^∼𝑐𝑒(𝑡), (13) where 𝑐₁=1𝛽sin²𝛽¹⁰𝑒(𝑠)𝑐d𝑠,₂=1sin𝛽¹⁰𝑒(𝑠)d𝑠¹₀𝐺₂(𝑠,𝑠)𝑞(𝑠)d𝑠max_0≤𝑡≤1𝜌(𝑡),_∼𝑐=1−∫¹₀𝑝(𝑠)d𝑠∫¹₀𝑠−𝑠²𝑝(𝑠)𝑐d𝑠¹+𝑐₂. (14)

Let 𝑋=𝑢∈𝐶²[0,1]∶𝑢(0)=𝑢(1)=∫¹₀𝑝(𝑠)𝑢(𝑠)𝑢d𝑠_′′(0)=𝑢^′′(1)=∫¹₀𝑞(𝑠)𝑢^′′(𝑠)(𝑡)d𝑠−𝜀𝑒≤−𝑢^′′(𝑡)≤𝜀𝑒(𝑡), (15) for a certain 𝜀>0, and let 𝑌 be the Banach space 𝐶[0,1] with the norm ‖𝑢‖_∞=max_0≤𝑡≤1|𝑢(𝑡)|. For 𝑢∈𝑋, we have 𝑢(𝑡)=¹₀𝐾₁(𝑡,𝑠)−𝑢^′′(𝑠)d𝑠. (16) Combining (16) and (9) with the fact that −𝜀𝑒(𝑠)≤−𝑢^′′(𝑠)≤𝜀𝑒(𝑠), we conclude that −𝑐𝜀𝑒(𝑡)≤𝑢(𝑡)≤𝑐𝜀𝑒(𝑡),𝑡∈[0,1]. (17) So, we may define the norm of 𝑢∈𝑋 by ‖𝑢‖_𝑋∶=inf𝜀∣𝜀>0,−𝜀𝑒(𝑡)≤−𝑢^′′(𝑡)≤𝜀𝑒(𝑡),𝑡∈[0,1]. (18) This norm is so called 𝑢₀-norm.

It is easy to get the following lemmas.

Lemma 4. (𝑋,||⋅||_𝑋) is a Banach space.

Proof. Let {𝑢_𝑛} be Cauchy sequence of (𝑋,||⋅||_𝑋). Then, we have 𝑢_𝑛−𝑢_𝑚𝑋→0,(𝑛,𝑚→∞). (19) From (18) and (19), for all 𝜀>0, there exists 𝑁>0 such that −𝜀𝑒(𝑡)≤−𝑢_𝑛^′′(𝑡)−𝑢_𝑚^′′(𝑡)(𝑡),[]≤𝜀𝑒𝑡∈0,1,∀𝑛,𝑚>𝑁. (20) Hence, 𝑢_𝑛^′′−𝑢_𝑚^′′_∞≤𝜀‖𝑒‖_∞→0,(𝑛,𝑚→∞), (21) which implies that {𝑢_𝑛^′′} is a Cauchy sequence of (𝐶[0,1],||⋅||_∞). According to the completeness of (𝐶[0,1],||⋅||_∞), there exists 𝑢₀∈𝐶²[0,1] such that 𝑢_𝑛^′′−𝑢₀^′′_∞→0,(𝑛→∞). (22) Let 𝑚→∞ in (20). Then, we have −𝜀𝑒(𝑡)≤−𝑢_𝑛^′′(𝑡)−𝑢₀^′′(𝑡)(𝑡),[],(),≤𝜀𝑒𝑡∈0,1𝑛>𝑁 (23) which implies that 𝑢_𝑛−𝑢_0𝑋≤𝜀,(𝑛>𝑁). (24) Hence, for Cauchy sequence {𝑢_𝑛}⊂(𝑋,||⋅||_𝑋), there exists 𝑢₀∈𝑋 satisfying 𝑢_𝑛−𝑢_0𝑋→0,(𝑛→∞). (25) Therefore, (𝑋,||⋅||_𝑋) is a Banach space.

Lemma 5. 𝑋∩𝐶⁴[0,1] is compactly embedded in 𝑋.

Proof. Let {𝑢_𝑛}⊂𝑋∩𝐶⁴[0,1] be bounded. Then, according to the fact that 𝐶⁴[0,1] is compactly embedded in 𝐶²[0,1], there exist {𝑢_𝑛𝑗}⊂{𝑢_𝑛} and 𝑢₀∈𝐶²[0,1] such that 𝑢_𝑛^′′_𝑗−𝑢₀^′′_∞→0,𝑗→∞. (26) From (26), for all 𝜀>0, there exists 𝑁>0 such that −𝜀≤−𝑢_𝑛^′′_𝑗(𝑡)−𝑢₀^′′(𝑡)≤𝜀,𝑡∈[0,1],𝑛_𝑗>𝑁. (27) Since 𝑒(𝑡)≥∫¹₀𝑠−𝑠²𝑝(𝑠)d𝑠1−∫¹₀𝑝(𝑠)[]d𝑠,𝑡∈0,1, (28) combining this fact with (27) we have −1−∫¹₀𝑝(𝑠)d𝑠∫¹₀𝑠−𝑠²𝑝(𝑠)(𝑡)𝑢d𝑠𝜀𝑒≤−_𝑛^′′_𝑗(𝑡)−𝑢₀^′′(𝑡)≤1−∫¹₀𝑝(𝑠)d𝑠∫¹₀𝑠−𝑠²𝑝(𝑠)(𝑡)[]d𝑠𝜀𝑒,𝑡∈0,1,𝑛_𝑗>𝑁, (29) which implies 𝑢_𝑛𝑗−𝑢_0𝑋≤1−∫¹₀𝑝(𝑠)d𝑠∫¹₀𝑠−𝑠²𝑝(𝑠)d𝑠𝜀→0,𝑛_𝑗→∞. (30) Therefore, 𝑋∩𝐶⁴[0,1] is compactly embedded in 𝑋.

Let 𝑃∶=𝑢∈𝑋∣𝑢(𝑡)≥0,𝑢^′′(𝑡)≤0,𝑡∈[0,1]. (31) Then, the cone 𝑃 is normal and has nonempty interior int𝑃.

In the rest of this section, we will prove the existence of the first eigenvalue of (3). To wit, we get the following.

Theorem 6. Assume (A1) holds. Then (3) has an algebraically simple eigenvalue 𝜇₁, with an eigenfunction 𝜑₁∈int𝑃, and there is no other eigenvalue with a positive eigenfunction.

Remark 7. If 𝑝=𝑞≡0, then 𝜇₁ can be explicitly given by 𝜇₁=𝜋⁴−𝛽𝜋²1+𝜋² (32) and the corresponding eigenfunction 𝜑₁(𝑡)=sin𝜋𝑡, 𝑡∈[0,1].

Proof of Theorem 6. For 𝑢∈𝑋, define a linear operator 𝑇∶𝑋→𝑌 by (𝑇𝑢)(𝑡)∶=¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏d𝑠. (33) Then, by the definition of 𝐾₁, we have (𝑡)=𝑇𝑢)(₁₀𝐺₁()+1𝑡,𝑠1−∫¹₀𝑝(𝑥)d𝑥¹⁰𝐺₁()𝑝(𝑥)×𝑠,𝑥d𝑥₁₀𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏=)d𝜏d𝑠_𝑡0𝑠()+11−𝑡1−∫¹₀𝑝(𝑥)d𝑥¹⁰𝐺₁()𝑝(𝑥)×𝑠,𝑥d𝑥₁₀𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏+)d𝜏d𝑠_1𝑡𝑡()+11−𝑠1−∫¹₀𝑝(𝑥)d𝑥¹⁰𝐺₁()𝑝(𝑥)×𝑠,𝑥d𝑥₁₀𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏d𝑠. (34) Therefore, we can obtain (𝑇𝑢)^′(𝑡)=^𝑡₀−𝑠¹₀𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏+)d𝜏d𝑠_1𝑡(1−𝑠)¹₀𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏())d𝜏d𝑠,𝑇𝑢^′′(𝑡)=−𝑡¹₀𝐾₂(𝑡,𝜏)𝑢(𝜏)−𝑢^′′(𝜏+())d𝜏𝑡−1¹₀𝐾₂(𝑡,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏=−¹₀𝐾₂(𝑡,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏. (35)(36)
We claim that 𝑇∶𝑃→𝑃. In fact, for 𝑢∈𝑋, let ||𝑢||_𝑋=𝛼. Then, from (18) and (17) we have, for 𝑡∈[0,1], −𝛼𝑒(𝑡)≤−𝑢^′′(𝑡)(𝑡),(𝑡)(𝑡)(𝑡).≤𝛼𝑒−𝑐𝛼𝑒≤𝑢≤𝑐𝛼𝑒 (37) On the one hand, from (36)-(37) and (10) we have −(𝑇𝑢)^′′(𝑡)=¹₀𝐾₂(𝑡,𝑠)𝑢(𝑠)−𝑢^′′(𝑠≤)d𝑠₁₀𝐾₂(]𝑒(𝑠)≤𝑡,𝑠)[𝑐𝛼+𝛼d𝑠_∼𝑐(𝑐𝛼+𝛼)𝑒(𝑡). (38) On the other hand, from (36)-(37) and (10) we have −(𝑇𝑢)^′′(𝑡)=¹₀𝐾₂(𝑡,𝑠)𝑢(𝑠)−𝑢^′′(𝑠)d𝑠≥−¹₀𝐾₂(𝑡,𝑠)[𝑐𝛼+𝛼]𝑒(𝑠)d𝑠≥−^∼𝑐(𝑐𝛼+𝛼)𝑒(𝑡). (39) Therefore, 𝑇(𝑋)⊆𝑋. If 𝑢∈𝑃, then 𝑢(𝑠)≥0, 𝑢^′′(𝑠)≤0 on [0,1]. According to (36) and the fact that 𝐾₂(𝑡,𝑠)≥0,0≤𝑡,𝑠≤1, (40) we have (𝑇𝑢)^′′(𝑡)=−¹₀𝐾₂(𝑡,𝑠)𝑢(𝑠)−𝑢^′′(𝑠)d𝑠≤0. (41) Furthermore, according to (41), the definition of 𝑇, and the fact that 𝐾₁(𝑡,𝑠)≥0,0≤𝑡,𝑠≤1, (42) we have (𝑇𝑢)(𝑡)=¹₀𝐾₁(𝑡,𝑠)¹₀𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏=)d𝜏d𝑠₁₀𝐾₁(𝑡,𝑠)−(𝑇𝑢)^′′(𝑠)d𝑠≥0. (43) Then 𝑢∈𝑃, and accordingly, 𝑇(𝑃)⊆𝑃.
Now, since 𝑇(𝑋)⊂𝑋∩𝐶⁴[0,1] and 𝑋∩𝐶⁴[0,1] is compactly embedded in 𝑋, we have that 𝑇∶𝑋→𝑋 is compact.
Next, we show that 𝑇∶𝑃→𝑃 is strongly positive.
For 𝑢∈𝑃⧵{𝜃}, it is easy to check that there exist 𝜌_1𝑢,𝜌_2𝑢>0, such that for 𝑡∈[0,1](𝑇𝑢)(𝑡)≥𝜌_1𝑢𝑒(𝑡),−(𝑇𝑢)^′′(𝑡)≥𝜌_2𝑢𝑒(𝑡). (44) In fact, for every 𝑢∈𝑃⧵{𝜃}, it follows from Lemma 2 that (𝑡)=𝑇𝑢)(₁₀𝐺₁()+1𝑡,𝑠1−∫¹₀𝑝(𝑥)d𝑥¹⁰𝐺₁()𝑝(𝑥)×𝐺𝑠,𝑥d𝑥₂(𝑠,𝜏)+𝜌(𝑠)¹₀𝐺₂()𝑞(𝑥)×𝑢(𝜏)𝜏,𝑥d𝑥−𝑢^′′(𝜏≥)d𝜏d𝑠₁₀𝐺₁(𝑡,𝑡)𝐺₁()+𝐺𝑠,𝑠₁(𝑠,𝑠)1−∫¹₀𝑝(𝑥)d𝑥¹⁰𝐺₁()𝑝(𝑥)×𝐺𝑥,𝑥d𝑥₂(𝑠,𝜏)+𝜌(𝑠)¹₀𝐺₂()𝑞(𝑥)×𝑢(𝜏)𝜏,𝑥d𝑥−𝑢^′′(𝜏)d𝜏d𝑠=𝜌_1𝑢𝑒(𝑡), (45) where 𝜌_1𝑢=¹₀𝐺₁()×𝐺𝑠,𝑠₂(𝑠,𝜏)+𝜌(𝑠)¹₀𝐺₂()𝑞(𝑥)×𝑢(𝜏)𝜏,𝑥d𝑥−𝑢^′′(𝜏−())d𝜏d𝑠>0,𝑇𝑢^′′(𝑡)=¹₀𝐺₂(𝑡,𝑠)+𝜌(𝑡)¹₀𝐺₂()𝑞(𝑥)×𝑢(𝑠)𝑠,𝑥d𝑥−𝑢^′′(𝑠≥)d𝑠𝛽sin𝛽¹₀𝐺₂(𝑡,𝑡)𝐺₂()(𝑡)𝐺𝑠,𝑠+𝜌₂(𝑠,𝑠)¹₀𝐺₂()𝑞(𝑥)×𝑢(𝑠)𝑥,𝑥d𝑥−𝑢^′′(𝑠=)d𝑠𝛽sin𝛽𝐺₂(𝑡,𝑡)+𝜌(𝑡)¹₀𝐺₂()𝑞(𝑥)×𝑥,𝑥d𝑥₁₀𝐺₂(𝑠,𝑠)𝑢(𝑠)−𝑢^′′(𝑠(𝑡)d𝑠≥𝜌)¹₀sin𝛽𝑥sin𝛽()𝑞(𝑥)×1−𝑥d𝑥₁₀𝐺₂(𝑠,𝑠)𝑢(𝑠)−𝑢^′′(𝑠)d𝑠≥𝜌_2𝑢𝑒(𝑡), (46)(47) where 𝜌_2𝑢=min_[]𝑡∈0,1𝜌(𝑡)¹₀sin𝛽𝑥sin𝛽()𝑞(𝑥)×1−𝑥d𝑥₁₀𝐺₂(𝑠,𝑠)𝑢(𝑠)−𝑢^′′(𝑠×1)d𝑠4+∫¹₀𝑝(𝑠)𝑠−𝑠²d𝑠1−∫¹₀𝑝(𝑠)d𝑠⁻¹>0. (48)
Therefore, it follows from (44) that 𝑇𝑢∈int𝑃.
Now, on the one hand, by the Krein-Rutman theorem ([17, Theorem 7.C] and [18, Theorem 19.3]), 𝑇 has an algebraically simple eigenvalue 𝑟(𝑇)>0 with an eigenfunction 𝜑₁∈int𝑃. Moreover, there is no other eigenvalue with a positive eigenfunction. On the other hand, we have from the definition of 𝑇 and Lemma 3 that (3) is equivalent to the integral equation 𝑢(𝑡)=𝜇(𝑇𝑢)(𝑡). (49) Therefore, 𝜇₁=(𝑟(𝑇))⁻¹ with a positive eigenfunction 𝜑₁ is a simple eigenvalue of (3). Moreover, for (3), there is no other eigenvalue with a positive eigenfunction.

3. An Application of Theorem 6

For convenience, we introduce the following notations: 𝑓^∞=limsup_{|𝑢|+|𝑣|→+∞}max_𝑡∈[0,1]𝑓(𝑡,𝑢,𝑣)|𝑢|+|𝑣|,𝑓₀=liminf_{|𝑢|+|𝑣|→0+}min_𝑡∈[0,1]𝑓(𝑡,𝑢,𝑣)|𝑢|+|𝑣|,𝑓⁰=limsup_{|𝑢|+|𝑣|→0+}max_𝑡∈[0,1]𝑓(𝑡,𝑢,𝑣)|𝑢|+|𝑣|,𝑓_∞=liminf_{|𝑢|+|𝑣|→+∞}min_𝑡∈[0,1]𝑓(𝑡,𝑢,𝑣)|𝑢|+|𝑣|,𝐵_𝑟=𝑢∈𝑋∣‖𝑢‖_𝑋<𝑟,𝜕𝐵_𝑟=𝑢∈𝑋∣‖𝑢‖_𝑋=𝑟(𝑟>0). (50)

Theorem 8. Assume that (A1) and (A2) hold, and 0≤𝑓^∞<𝑓₀≤+∞. Then, for each 𝜆 satisfying 𝜇₁𝑓₀𝜇<𝜆<¹𝑓^∞, (51) there exists at least one positive solution of (2).

Theorem 9. Assume that (A1) and (A2) hold, and 0≤𝑓⁰<𝑓_∞≤+∞. Then, for each 𝜆 satisfying 𝜇₁𝑓_∞𝜇<𝜆<¹𝑓⁰, (52) there exists at least one positive solution of (2).

Remark 10. Bai [3] proved existence of positive solutions via Guo-Krasnoselskii fixed point theorem under some conditions which do not involve the eigenvalue of (3). While our Theorems 8 and 9 are established under (51) or (52) which are related to the eigenvalue of (3). Our Theorems 8 and 9 cover an undefined case in [3]. Consider the following boundary value problem: 𝑢⁽⁴⁾(𝑡)+𝜋²4𝑢^′′(𝑡)=3𝜋⁴+141+𝜋²𝑢(𝑡)−𝑢^′′(𝑡−2)41+𝜋²sin𝑢(𝑡)−𝑢^′′(𝑡(),𝑢(0)(1)),𝑡∈0,1=𝑢=𝑢^′′(0)=𝑢^′′(1)=0. (53) In this case, 𝜇₁=3𝜋⁴/4(1+𝜋²), 𝑓(𝑡,𝑢,𝑣)=((3𝜋⁴+1)/4(1+𝜋²))(𝑢−𝑣)−(2/4(1+𝜋²))sin(𝑢−𝑣), 𝜆=1, and 𝜇₁𝑓_∞=3𝜋⁴3𝜋⁴+1<1=𝜆<3𝜋⁴3𝜋⁴=𝜇−1𝑓¹⁰. (54) According to Theorem 9, the above boundary value problem has at least one positive solution. For the above boundary value problem, however, we cannot obtain the above conclusion by [3, Theorem 3.1] since 𝑓_∞=3𝜋⁴+141+𝜋²𝜋<20<36<³4+2√2+22√2=𝜂₁𝜆. (55) Moreover, (51) and (52) are optimal. In order to illustrate this point, consider the problem 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)=𝜋⁴−𝛽𝜋²1+𝜋²𝑢(𝑡)−𝑢^′′(𝑡𝑢(𝑡))+𝜌,𝑢^′′(𝑡)(),𝑢(0)(1),𝑡∈0,1=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (56) where 𝜌𝑥,𝑦=𝑥−𝑦²√,0≤𝑥−𝑦≤1,𝑥−𝑦,1≤𝑥−𝑦<∞. (57) In this case, 𝜇₁=(𝜋⁴−𝛽𝜋²)/(1+𝜋²), 𝑓(𝑡,𝑢,𝑣)=((𝜋⁴−𝛽𝜋²)/(1+𝜋²))[𝑢−𝑣]+𝜌(𝑢,𝑣), 𝜆=1, and 𝜇₁𝑓₀𝜇=𝜆=¹𝑓^∞,𝜇₁𝑓_∞𝜇=𝜆=¹𝑓⁰. (58) However, (56) has no positive solution. In fact, suppose on the contrary that (56) has a positive solution 𝑢. Multiplying the first equation of (56) with sin𝜋𝑡 and integrating from 0 to 1, we get 0=¹₀𝜌𝑢(𝑡),𝑢^′′(𝑡)sin𝜋𝑡d𝑡>0, (59) which is a contradiction.

To prove Theorems 8 and 9, we need the following preliminary results.

Lemma 11. For every 𝑢∈𝑋, there exist 𝐿₀,𝐿₁>0 such that 𝐿₀‖𝑢‖_∞≤𝑢^′_∞≤𝑢^′′_∞≤𝐿₁‖𝑢‖_𝑋. (60)

Proof. (1) By 𝑢(0)=∫¹₀𝑝(𝑠)𝑢(𝑠)d𝑠, there is 𝑢(𝑡)=∫^𝑡₀𝑢^′(𝑠)d𝑠+∫¹₀𝑝(𝑠)𝑢(𝑠)d𝑠, so |𝑢(𝑡)|≤¹₀𝑢^′(𝑠)d𝑠+¹₀𝑝(𝑠𝑢(𝑠≤𝑢)|)|d𝑠_′∞+‖𝑢‖_∞¹₀𝑝(𝑠)d𝑠. (61) Hence 𝐿₀‖𝑢‖_∞≤𝑢^′_∞, (62) where 𝐿₀=1−∫¹₀𝑝(𝑠)d𝑠.
(2) By 𝑢(0)=𝑢(1), there is a 𝜉∈(0,1) such that 𝑢^′(𝜉)=0, and so, for 𝑡∈[0,1]−𝑢^′(𝑡)=^𝜉_𝑡𝑢^′′(𝑠)d𝑠. (63) Hence |𝑢^′(𝑡)|≤|∫^𝜉_𝑡|𝑢^′′(𝑠)|d𝑠|≤∫¹₀|𝑢^′′(𝑠)|d𝑠≤||𝑢^′′||_∞, and accordingly, 𝑢^′_∞≤𝑢^′′_∞. (64)
(3) We have from (18) that |𝑢^′′(𝑡)|≤||𝑢||_𝑋𝑒(𝑡), which implies that 𝑢^′′(𝑡≤1)4+∫¹₀𝑝(𝑠)𝑠−𝑠²d𝑠1−∫¹₀𝑝(𝑠)‖𝑢‖d𝑠^𝑋, (65) and consequently 𝑢^′′_∞≤𝐿₁‖𝑢‖_𝑋, (66) where 𝐿₁=(1/4)+(∫¹₀𝑝(𝑠)(𝑠−𝑠²)d𝑠/(1−∫¹₀𝑝(𝑠)d𝑠)).

Let (𝐴𝑢)(𝑡)∶=𝜆¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′′(𝜏[])d𝜏d𝑠,𝑡∈0,1,𝜆>0. (67) It is easy to show that 𝐴∶𝑃→𝑃 is a completely continuous operator. In addition, we can verify that the nonzero fixed points of the operator 𝐴 are positive solutions of the problem (2).

Lemma 12 (see [19]). Let 𝐸 be Banach space, 𝑃 a cone in 𝐸, and Ω(𝑃) a bounded open set in 𝑃. Suppose that 𝐴∶Ω(𝑃)→𝑃 is a completely continuous operator. If there exists 𝑢₀∈𝑃⧵{𝜃} such that 𝑢−𝐴𝑢≠𝜇𝑢₀,∀𝑢∈𝜕Ω(𝑃),𝜇≥0, (68) then the fixed point index 𝑖(𝐴,Ω(𝑃),𝑃)=0.

Lemma 13 (see [19]). Let 𝐸 be Banach space, 𝑃 a cone in 𝐸, and Ω(𝑃) a bounded open set in 𝑃 with 𝜃∈Ω(𝑃). Suppose that 𝐴∶Ω(𝑃)→𝑃 is a completely continuous operator. If 𝐴𝑢≠𝜇𝑢,∀𝑢∈𝜕Ω(𝑃),𝜇≥1, (69) then the fixed point index 𝑖(𝐴,Ω(𝑃),𝑃)=1.

Proof of Theorem 8. It follows from the first inequality of (51) that there exists 𝑟₀>0, such that 𝜆𝑓(𝑡,𝑢,𝑣)≥𝜇₁(|𝑢|+|𝑣|),∀𝑡∈[0,1],0≤|𝑢|+|𝑣|≤𝑟₀. (70) Let 𝜑₁ be the positive eigenfunction of 𝑇 (see (33)) corresponding to 𝜇₁. Thus 𝜑₁=𝜇₁𝑇𝜑₁.
Let 𝑟₁=𝐿₀𝑟₀/(1+𝐿₀)𝐿₁. Then for every 𝑢∈𝜕𝐵_𝑟1∩𝑃, we have from Lemma 11 that 0≤𝑢(𝑡)−𝑢^′′(𝑡)≤‖𝑢‖_∞+𝑢^′′_∞≤1+𝐿₀𝐿₀‖𝑢‖_𝑋≤𝑟₀. (71) It follows from (70) that (𝐴𝑢)(𝑡)≥𝜇₁¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏d𝑠=𝜇₁(𝑇𝑢)(𝑡),𝑡∈[0,1]. (72)
We may suppose that 𝐴 has no fixed points on 𝜕𝐵_𝑟1∩𝑃 (otherwise, the proof is finished). Now we show that 𝑢−𝐴𝑢≠𝜇𝜑₁,∀𝑢∈𝜕𝐵_𝑟1∩𝑃,𝜇≥0. (73)
Suppose the contrary, that there exists 𝑢₁∈𝜕𝐵_𝑟1∩𝑃 and 𝜏₁≥0 such that 𝑢₁−𝐴𝑢₁=𝜏₁𝜑₁. Hence 𝜏₁>0 and 𝑢₁=𝐴𝑢₁+𝜏₁𝜑₁≥𝜏₁𝜑₁. (74) Put 𝜏^∗∶=sup𝜏∣𝑢₁≥𝜏𝜑₁. (75) It is easy to see that 𝜏^∗≥𝜏₁>0 and 𝑢₁≥𝜏^∗𝜑₁. We find from 𝑇(𝑃)⊂𝑃 that 𝜇₁𝑇𝑢₁≥𝜏^∗𝜇₁𝑇𝜑₁=𝜏^∗𝜑₁. (76) Therefore, by (72), we have 𝑢₁=𝐴𝑢₁+𝜏₁𝜑₁≥𝜇₁𝑇𝑢₁+𝜏₁𝜑₁≥𝜏^∗𝜑₁+𝜏₁𝜑₁, (77) which contradicts the definition of 𝜏^∗. Hence (73) is true, and we have from Lemma 12 that 𝑖𝐴,𝐵_𝑟1∩𝑃,𝑃=0. (78)
It follows from the second inequality of (51) that there exists 0<𝜎<1 and 𝑟₂>𝑟₁ such that 𝜆𝑓(𝑡,𝑢,𝑣)≤𝜎𝜇₁(|𝑢|+|𝑣|),∀𝑡∈[0,1],|𝑢|+|𝑣|≥𝑟₂. (79)
Let 𝑇₁𝑢=𝜎𝜇₁𝑇𝑢, 𝑢∈𝑋. Then 𝑇₁∶𝑋→𝑋 is a bounded linear operator and 𝑇₁(𝑃)⊂𝑃.
Let 𝑀=max_{0≤𝑡,𝑠,𝜏≤1}𝐾₁(𝑡,𝑠)𝐾₂(×𝑠,𝜏)sup_{𝑢∈𝐵𝑟3∩𝑃}𝜆¹₀𝑓𝜏,𝑢(𝜏),𝑢^′′(𝜏)d𝜏, (80) where 𝑟₃=𝐿₀𝑟₂/(1+𝐿₀)𝐿₁. It is clear that 𝑀<+∞.
Let 𝑊=𝑢∈𝑃∣𝑢=𝜇𝐴𝑢,0≤𝜇≤1. (81) In the following, we prove that 𝑊 is bounded.
For any 𝑢∈𝑊, set 𝐸(𝜏)={𝜏∈[0,1]∣𝑢(𝜏)−𝑢^′′(𝜏)≥𝑟₂}. Then, 𝑢(𝑡)=𝜇(𝐴𝑢)(𝑡)≤𝜆¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′′(𝜏)d𝜏d𝑠=𝜆¹_0𝐸(𝜏)𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′′(𝜏)d𝜏d𝑠+𝜆¹_{0[0,1]⧵𝐸(𝜏)}𝐾₁(𝑡,𝑠)𝐾₂()𝑓×(𝜏)𝑠,𝜏𝜏,𝑢,𝑢^′′(𝜏)d𝜏d𝑠≤𝜎𝜇₁¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏d𝑠+𝜆¹_{0[0,1]⧵𝐸(𝜏)}𝐾₁(𝑡,𝑠)𝐾₂()𝑓×(𝜏)𝑠,𝜏𝜏,𝑢,𝑢^′′(𝜏)≤𝑇d𝜏d𝑠₁𝑢(𝑡)+𝑀,𝑡∈[0,1]. (82) Thus, ((𝐼−𝑇₁)𝑢)(𝑡)≤𝑀, for every 𝑡∈[0,1]. Since 𝜇₁ is the first eigenvalue of 𝑇 and 0<𝜎<1, the first eigenvalue of 𝑇₁, (𝑟(𝑇₁))⁻¹>1. Therefore, the inverse operator (𝐼−𝑇₁)⁻¹ exists and 𝐼−𝑇₁⁻¹=𝐼+𝑇₁+𝑇²₁+⋯+𝑇^𝑛₁+⋯. (83) It follows from 𝑇₁(𝑃)⊂𝑃 that (𝐼−𝑇₁)⁻¹(𝑃)⊂𝑃. So, we have 𝑢(𝑡)≤(𝐼−𝑇₁)⁻¹𝑀, 𝑡∈[0,1], and we conclude that 𝑊 is bounded.
Select 𝑟₄>max{𝑟₁,sup𝑊}. Then, from the invariance property of the fixed point index, we have 𝑖𝐴,𝐵_𝑟4∩𝑃,𝑃=𝑖𝜃,𝐵_𝑟4∩𝑃,𝑃=1. (84) By (78) and (84), we have that 𝑖𝐴,𝐵_𝑟4∩𝑃⧵𝐵_𝑟1∩𝑃,𝑃=𝑖𝐴,𝐵_𝑟4∩𝑃,𝑃−𝑖𝐴,𝐵_𝑟1∩𝑃,𝑃=1. (85) Then 𝐴 has at least one fixed point on (𝐵_𝑟4∩𝑃)⧵(𝐵_𝑟1∩𝑃). This means that the boundary value problem (2) has at least one positive solution.

Proof of Theorem 9. It follows from the second inequality of (52) that there exists 𝑟₀>0, such that 𝜆𝑓(𝑡,𝑢,𝑣)≤𝜇₁(|𝑢|+|𝑣|),∀𝑡∈[0,1],0≤|𝑢|+|𝑣|≤𝑟₀. (86) Let 𝑟₁=𝐿₀𝑟₀/(1+𝐿₀)𝐿₁. Then, for every 𝑢∈𝜕𝐵_𝑟1∩𝑃, we have from Lemma 11 that 0≤𝑢(𝑡)−𝑢^′′(𝑡)≤‖𝑢‖_∞+𝑢^′′_∞≤1+𝐿₀𝐿₁𝐿₀‖𝑢‖_𝑋≤𝑟₀. (87) It follows from (86) that (𝐴𝑢)(𝑡)≤𝜇₁¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑢(𝜏)−𝑢^′′(𝜏)d𝜏d𝑠=𝜇₁(𝑇𝑢)(𝑡),𝑡∈[0,1]. (88) Suppose there exists 𝑢₂∈𝜕𝐵_𝑟1∩𝑃 and 𝜏₂≥1 satisfying 𝐴𝑢₂=𝜏₂𝑢₂. We may suppose that 𝐴 has no fixed points on 𝜕𝐵_𝑟1∩𝑃 (otherwise, the proof is finished). So 𝜏₂>1. By (88), we have 𝜏₂𝑢₂=𝐴𝑢₂≤𝜇₁𝑇𝑢₂. By induction, we have 𝜏^𝑛₂𝑢₂≤𝜇^𝑛₁𝑇^𝑛𝑢₂, for all 𝑛∈𝑁. Thus 𝑇^𝑛≥𝑇^𝑛𝑢_2𝑋𝑢_2𝑋≥𝜏^𝑛₂𝑢_2𝑋𝜇^𝑛₁𝑢_2𝑋=𝜏^𝑛₂𝜇^𝑛₁. (89) By Gelfand’s formula, we have 𝑟(𝑇)=lim_𝑛→∞^𝑛𝑇^𝑛≥𝜏₂𝜇₁>1𝜇₁, (90) which is a contradiction with 𝑟(𝑇)=1/𝜇₁. Hence 𝐴𝑢≠𝜏𝑢,∀𝑢∈𝜕𝐵_𝑟1∩𝑃,𝜏≥1, (91) and we have from Lemma 13 that 𝑖𝐴,𝐵_𝑟1∩𝑃,𝑃=1. (92)
It follows from the first inequality of (52) that there exist 𝜀>0 and 𝑅₀>0 such that 𝜆𝑓(𝑡,𝑢,𝑣)≥𝜇₁+𝜀(|𝑢|+|𝑣|),|𝑢|+|𝑣|≥𝑅₀. (93) From (A2), we know that there exists 𝑑≥0 such that 𝜆𝑓(𝑡,𝑢,𝑣)≥𝜇₁𝑢|+|𝑣[]|𝑢|+|𝑣|+𝜀(||)−𝑑,𝑡∈0,1,0≤<+∞. (94)
We can take 𝑅>𝑟₁ large enough, such that 𝜌_1𝑢≥𝑑∬¹₀𝐾₁(𝑠,𝑠)𝐾₂(𝑠,𝜏)d𝜏d𝑠𝜀∫¹₀𝑠−𝑠²𝑝(𝑠)d𝑠/1−∫¹₀𝑝(𝑠),d𝑠 (95) where 𝜌_1𝑢 is defined as in (46). Then by (94) and (33), for all 𝑢∈𝑃,  ‖𝑢‖_𝑋≥𝑅, 𝑡∈[0,1], we have (𝐴𝑢)(𝑡)=𝜆¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′′(𝜏)≥𝜇d𝜏d𝑠₁+𝜀¹₀𝐾₁(𝑡,𝑠)𝐾₂()×𝑢(𝜏)𝑠,𝜏−𝑢^′′(𝜏)d𝜏d𝑠−𝑑¹₀𝐾₁(𝑡,𝑠)𝐾₂(𝑠,𝜏)d𝜏d𝑠≥𝜇₁(𝑇𝑢)(𝑡)+𝜀𝜌_1𝑢𝑒(𝑡)−𝑑¹₀𝐾₁(𝑠,𝑠)𝐾₂(𝑠,𝜏)d𝜏d𝑠≥𝜇₁(𝑇𝑢)(𝑡)+𝜀𝜌_1𝑢∫¹₀𝑠−𝑠²𝑝(𝑠)d𝑠1−∫¹₀𝑝(𝑠)d𝑠−𝑑¹₀𝐾₁(𝑠,𝑠)𝐾₂(𝑠,𝜏)d𝜏d𝑠≥𝜇₁(𝑇𝑢)(𝑡). (96)
Hence, for every 𝑢∈𝑃,  ‖𝑢‖_𝑋≥𝑅, we have (𝐴𝑢)(𝑡)≥𝜇₁(𝑇𝑢)(𝑡),𝑡∈[0,1]. (97) Let 𝜑₁ be the positive eigenfunction of 𝑇 corresponding to 𝜇₁. Thus 𝜑₁=𝜇₁𝑇𝜑₁.
We may suppose that 𝐴 has no fixed points on 𝜕𝐵_𝑅∩𝑃 (otherwise, the proof is finished). Now, we show that 𝑢−𝐴𝑢≠𝜏𝜑₁,∀𝑢∈𝜕𝐵_𝑅∩𝑃,𝜏≥0. (98)
Suppose the contrary, that there exists 𝑢₃∈𝜕𝐵_𝑅∩𝑃 and 𝜏₃≥0 such that 𝑢₃−𝐴𝑢₃=𝜏₃𝜑₁. (99) Hence 𝜏₃>0 and 𝑢