Existence and Multiplicity of Positive Solutions of a Nonlinear Discrete Fourth-Order Boundary Value Problem

Abstract

we show the existence and multiplicity of positive solutions of the nonlinear discrete fourth-order boundary value problem Δ⁴𝑢(𝑡−2)=𝜆ℎ(𝑡)𝑓(𝑢(𝑡)), 𝑡∈𝕋₂, 𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0, where 𝜆>0, ℎ∶𝕋₂→(0,∞) is continuous, and 𝑓∶ℝ→[0,∞) is continuous, 𝑇>4, 𝕋₂={2,3,…,𝑇}. The main tool is the Dancer's global bifurcation theorem.

1. Introduction

It's well known that the fourth order boundary value problem 𝑢^′′′′(𝑡)((𝑡(),𝑢(0)(1)=𝑓𝑡,𝑢)),𝑡∈0,1=𝑢=𝑢^′′(0)=𝑢^′′(1)=0 (1.1) can describe the stationary states of the deflection of an elastic beam with both ends hinged, (it also models a rotating shaft). The existence and multiplicity of positive solutions of the boundary value problem (1.1) have been considered extensively in the literature, see [1–10]. The existence and multiplicity of positive solutions of the parameterized boundary value problem 𝑢^′′′′(𝑡)(𝑡)𝑓(𝑢(𝑡(),𝑢(0)(1)=𝜆ℎ)),𝑡∈0,1=𝑢=𝑢^′′(0)=𝑢^′′(1)=0 (1.2) have also been studied by several authors, see Bai and Wang [11], Cid et al. [12], and the references therein.

However, relatively little is known about the corresponding discrete fourth-order problems. Let 𝑇>4,𝕋₀={0,1,…,𝑇+2},𝕋₁={1,2,…,𝑇+1},𝕋₂={2,3,…,𝑇}. (1.3)

Zhang et al. [13], and He and Yu [14] used the fixed point index theory in cones to study the following discrete analogueΔ⁴𝑢(𝑡−2)=𝜆ℎ(𝑡)𝑓(𝑢(𝑡)),𝑡∈𝕋₂,𝑢(0)=𝑢(𝑇+2)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0, (1.4)(1.5) where Δ⁴𝑢(𝑡−2) denote the fourth forward difference operator and Δ𝑢(𝑡)=𝑢(𝑡+1)−𝑢(𝑡). It has been pointed out in [13, 14] that (1.4), (1.5) are equivalent to the equation of the form: 𝑢(𝑡)=𝜆^𝑇+1_𝑠=1𝐺(𝑡,𝑠)^𝑇_𝑗=2𝐺₁𝑠,𝑗ℎ𝑗𝑓𝑢𝑗=∶𝐴₀𝑢(𝑡),𝑡∈𝕋₀, (1.6) where 𝐺()=1𝑡,𝑠𝑠()𝑡()𝐺𝑇+2𝑇+2−𝑡,1≤𝑠≤𝑡≤𝑇+2,𝑇+2−𝑠,0≤𝑡≤𝑠≤𝑇+1,₁=1𝑠,𝑗𝑇()()𝑇+1−𝑠𝑗−1,2≤𝑗≤𝑠≤𝑇+1,𝑇+1−𝑗𝑠−1,1≤𝑠≤𝑗≤𝑇. (1.7)

Notice that two distinct Green's functions used in (1.6) make the construction of cones and the verification of strong positivity of 𝐴₀ become more complex and difficult. Therefore, Ma and Xu [15] considered (1.4) with the boundary condition 𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0, (1.8) and introduced the definition of generalized positive solutions:

Definition 1.1. A function 𝑦∶𝕋₀→ℝ⁺ is called a generalized positive solution of (1.4), (1.8), if 𝑦 satisfies (1.4), (1.8), and 𝑦(𝑡)≥0 on 𝕋₁ and 𝑦(𝑡)>0 on 𝕋₂.

Remark 1.2. Notice that the fact 𝑦∶𝕋₀→ℝ⁺ is a generalized positive solution of (1.4), (1.8) does not means that 𝑦(𝑡)≥0 on 𝕋₀. In fact, 𝑦 satisfies
(1) 𝑦(𝑡)≥0 for 𝑡∈𝕋₂;
(2) 𝑦(1)=𝑦(𝑇+1)=0;
(3) 𝑦(0)=−𝑦(2),𝑦(𝑇+2)=−𝑦(𝑇).

Ma and Xu [15] also applied the fixed point theorem in cones to obtain some results on the existence of generalized positive solutions.

It is the purpose of this paper to show some new results on the existence and multiplicity of generalized positive solutions of (1.4), (1.8) by Dancer's global bifurcation theorem. To wit, we get the following.

Theorem 1.3. Let ℎ∶𝕋₂→(0,∞),𝑓∈𝐶(ℝ,[0,∞)), andlim_𝑠→0+𝑓(𝑠)𝑠=𝑓₀∈(0,∞),lim_𝑠→∞𝑓(𝑠)𝑠=𝑓_∞=+∞. (1.9) Assume that there exists 𝐵∈[0,+∞] such that f is nondecreasing on [0,𝐵). Then(i)(1.4), (1.8) have at least one generalized positive solution if 0<𝜆<𝜆₁/𝑓₀;(ii) (1.4), (1.8) have at least two generalized positive solutions if 𝜆₁𝑓₀<𝜆<sup_{()𝑠∈0,𝐵}𝑠𝛾^∗𝑓(𝑠), (1.10) where 𝛾^∗=max_𝑡∈𝕋1∑^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠), 𝐾(𝑡,𝑠) is defined as (2.3) and 𝜆₁ is the first eigenvalue of Δ⁴𝑢(𝑡−2)=𝜆ℎ(𝑡)𝑢(𝑡),𝑡∈𝕋₂,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0. (1.11)

The “dual” of Theorem 1.3 is as follows.

Theorem 1.4. Let ℎ∶𝕋₂→(0,∞),𝑓∈𝐶(ℝ,[0,∞)), andlim_𝑠→0+𝑓(𝑠)𝑠=𝑓₀∈(0,∞),lim_𝑠→∞𝑓(𝑠)𝑠=𝑓_∞=0. (1.12) Assume that there exists 𝐵∈[0,+∞] such that 𝑓 is nondecreasing on [0,𝐵). Then
(i) (1.4), (1.8) have at least a generalized positive solution provided 𝜆>inf_{𝑠∈0,𝑐1𝐵}𝑠𝑐₁𝛾_∗𝑓(𝑠), (1.13)where 𝛾_∗=min_𝑡∈𝕋2∑^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠); (ii)(1.4), (1.8) have at least two generalized positive solutions provided inf_{𝑠∈0,𝑐1𝐵}𝑠𝑐₁𝛾_∗𝑓(𝑠)𝜆<𝜆<¹𝑓₀. (1.14)

The rest of the paper is organized as follows: in Section 2, we present the form of the Green's function of (1.4), (1.8) and its properties, and we enunciate the Dancer's global bifurcation theorem ([16, Corollary  15.2]). In Section 3, we use the Dancer's bifurcation theorem to prove Theorems 1.3 and 1.4 and in Section 4, we finish the paper presenting a couple of illustrative examples.

Remark 1.5. For other results on the existence and multiplicity of positive solutions and nodal solutions for fourth-order boundary value problems based on bifurcation techniques, see [17–21].

2. Preliminaries and Dancer's Global Bifurcation Theorem

Lemma 2.1. Let ℎ∶𝕋₂→ℝ. Then the linear boundary value problemΔ⁴𝑢(𝑡−2)=ℎ(𝑡),𝑡∈𝕋₂,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0 (2.1) has a solution𝑢(𝑡)=^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠),𝑡∈𝕋₁, (2.2) where 𝐾()=(()−()𝑡,𝑠𝑠−1)(𝑇+1−𝑡)2𝑇𝑡−1𝑡−1²−(𝑠−2)𝑠()()−()6𝑇,2≤𝑠≤𝑡≤𝑇+1,𝑡−1)(𝑇+1−𝑡2𝑇𝑠−1𝑠−1²−(𝑡−2)𝑡6𝑇,1≤𝑡≤𝑠≤𝑇. (2.3)

Proof. By a simple summing computation and 𝑢(1)=Δ²𝑢(0)=0, we can obtain𝑢(𝑡)(0)+𝑡()=Δ𝑢)(𝑡−1𝑡−1)(𝑡−26Δ³𝑢(0)+^𝑡−1_𝑠=2(𝑡−𝑠)(𝑡−𝑠−1)(𝑡−𝑠+1)6ℎ(𝑠). (2.4)
This together with 𝑢(𝑇+1)=Δ²𝑢(𝑇)=0, it follows that 𝑢(𝑡)=^𝑇_𝑠=2(𝑇+1−𝑠)(𝑡−1)2𝑇(𝑠−1)−(𝑠−1)²−𝑡(𝑡−2)ℎ(𝑠)+6𝑇_{𝑡−1𝑠=2}(𝑡−𝑠)(𝑡−𝑠−1)(𝑡−𝑠+1)ℎ(𝑠)=6𝑇_{𝑇𝑠=𝑡}(𝑇+1−𝑠)(𝑡−1)2𝑇(𝑠−1)−(𝑠−1)²−𝑡(𝑡−2)ℎ(𝑠)+6𝑇_{𝑡−1𝑠=2}(𝑇+1−𝑡)(𝑠−1)2𝑇(𝑡−1)−(𝑡−1)²−𝑠(𝑠−2)ℎ(𝑠).6𝑇 (2.5) Therefore, (2.2) holds.

Remark 2.2. It has been pointed out in [15] that (2.1) is equivalent to the summation equation of the form𝑢(𝑡)=^𝑇_𝑠=2𝐺₁(𝑡,𝑠)^𝑇_𝑗=2𝐺₁𝑠,𝑗ℎ𝑗,𝑡∈𝕋₁. (2.6) It is easy to verify that (2.2) and (2.6) are equivalent.
By a similar method in [9], it follows that 𝐾(𝑡,𝑠) satisfies𝐾(𝑡,𝑠)≤Φ(𝑠)for𝑠∈𝕋₁,𝑡∈𝕋₁,𝐾(𝑡,𝑠)≥𝑐(𝑡)Φ(𝑠)for𝑠∈𝕋₁,𝑡∈𝕋₁, (2.7) where Φ(𝑠)=√3()𝑇27𝑇𝑠−1²−(𝑠−2)𝑠^3/2𝑇,1≤𝑠≤2√+1,3(()−()𝑠)27𝑇𝑇+1−𝑠)(2𝑇𝑠−1𝑠−2^3/2,𝑇2+1<𝑠≤𝑇+1, (2.8)𝑐(𝑡)=3√3𝑇²−𝑡(𝑡−2)(𝑡−1)2𝑇²+1^3/2𝑇,1≤𝑡≤23√+1,3(𝑇+1−𝑡)[2𝑇(𝑡−1)−𝑡(𝑡−2)]2𝑇²+1^3/2,𝑇2+1<𝑡≤𝑇+1. (2.9) Moreover, we have that𝐾(𝑡,𝑠)≥𝑐₁Φ(𝑠),for𝑠∈𝕋₁,𝑡∈𝕋₂. (2.10) here 𝑐₁=3√3𝑇²/2(𝑇²+1)^3/2.
Let 𝑋 be a real Banach space with a cone 𝐾 such that 𝑋=𝐾−𝐾. Let us consider the equation:𝑥=𝜇(𝐿𝑥+𝑁𝑥),𝜇∈ℝ,𝑥∈𝑋 (2.11) under the assumptions:
(A1) The operators 𝐿,𝑁∶𝑋→𝑋 are compact. Furthermore, 𝐿 is linear, ‖𝑁𝑥‖_𝑋/‖𝑥‖_𝑋→0 as ‖𝑥‖_𝑋→0, and (𝐿+𝑁)(𝐾)⊆𝐾. (A2) The spectral radius 𝑟(𝐿) of 𝐿 is positive. Denote 𝜇₀=𝑟(𝐿)⁻¹. (A3)𝐿 is strongly positive.Dancer's global bifurcation theorem is the following.

Theorem 2.3 (see [16, Corollary  15.2]). Let 𝑆₊∶=𝜇,𝑥∈ℝ×𝑋∣𝜇,𝑥isasolutionof(2.11)with𝑥>0and𝜇>0. (2.12) If (A1) and (A2) are satisfied, then (𝜇₀,0) is a bifurcation point of (2.11) and 𝑆₊ has an unbounded solution component 𝒞₊ which passes through (𝜇₀,0). Additionally, if (A3) is satisfied, then (𝜇,𝑥)∈𝒞₊ and 𝜇≠𝜇₀ always implies𝑥>0 and 𝜇>0.

3. Proof of the Main Results

Before proving Theorem 1.3, we state some preliminary results and notations. Let 𝜌∶=4sin²𝜋(𝑡)𝜋()2𝑇,𝑒∶=sin𝑡−1𝑇,𝑡∈𝕋₁,𝑋∶=𝑢∣𝑢∶𝕋₀⟶ℝ,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0. (3.1)(3.2) Then 𝑋 is a Banach space under the normal:‖𝑢‖_𝑋𝛾∶=inf𝜌∣−𝛾𝑒(𝑡)≤−Δ²𝑢(𝑡−1)≤𝛾𝑒(𝑡),𝑡∈𝕋₁. (3.3) See [22] for the detail.

Let 𝐾∶=𝑢∈𝑋∣Δ²𝑢(𝑡−1)≤0,𝑢(𝑡)≥0,𝑡∈𝕋₁. (3.4) Then 𝐾 is normal and has a nonempty interior and 𝑋=𝐾−𝐾.

Let 𝑌={𝑢∣𝑢∶𝕋₂→ℝ}. Then 𝑌 is a Banach space under the norm: ‖𝑢‖_∞=max_𝑡∈𝕋2|𝑢(𝑡)|. (3.5) Define ℒ∶𝑋→𝑌 by setting ℒ𝑢∶=Δ⁴𝑢(𝑡−2),𝑢∈𝑋. (3.6)

It is easy to check that ℒ⁻¹∶𝑌→𝑋 is compact.

Lemma 3.1. Let ℎ∈𝑌 with ℎ≥0 and ℎ(𝑡₀)>0 for some 𝑡₀∈𝕋₂, and ℒ𝑢−ℎ=0. (3.7) Then 𝑢∈int𝐾.

Proof. It is enough to show that there exist two constants 𝑟₁,𝑟₂∈(0,∞) such that 𝑟₁𝑒(𝑡)≤−Δ²𝑢(𝑡−1)≤𝑟₂𝑒(𝑡),𝑡∈𝕋₁. (3.8) In fact, we have from (3.7) that−Δ²𝑢(𝑡−1)=^𝑇_𝑠=2𝐺₁(𝑡,𝑠)ℎ(𝑠),𝑡∈𝕋₁. (3.9) This together with the relation ((𝑡−1)(𝑇+1−𝑡)/𝑇)𝐺₁(𝑠,𝑠)≤𝐺₁(𝑡,𝑠)≤(𝑡−1)(𝑇+1−𝑡)/𝑇 implies that^𝑇_𝑠=2𝐺₁()ℎ(𝑠()𝑠,𝑠)𝑡−1)(𝑇+1−𝑡𝑇≤^𝑇_𝑠=2𝐺₁(𝑡,𝑠)ℎ(𝑠)≤‖ℎ‖_∞(𝑡−1)(𝑇+1−𝑡)𝑇. (3.10) Combining (3.9) with (3.10) and the fact that 𝑐₁𝜋()sin𝑡−1𝑇≤(𝑡−1)(𝑇+1−𝑡)𝑇≤𝑐₂𝜋()sin𝑡−1𝑇,𝑡∈𝕋₁ (3.11) for some constants 𝑐₁,𝑐₂∈(0,∞), we conclude that (3.8) is true.

Let 𝜁∈𝐶(ℝ,ℝ) be such that 𝑓(𝑢)=𝑓₀𝑢+𝜁(𝑢), (3.12) clearly lim_|𝑢|→0𝜁(𝑢)𝑢=0. (3.13) Let us consider ℒ𝑢=𝜆ℎ(⋅)𝑓₀𝑢+ℎ(⋅)𝜁(𝑢) (3.14) as a bifurcation problem from the trivial solution 𝑢≡0.

By (1.4), (1.8), it follows that if 𝑢(𝑡)∈𝑋 is one solution of (1.4), (1.8), then 𝑢(𝑡) satisfies 𝑢(0)=−𝑢(2),𝑢(𝑇+2)=−𝑢(𝑇). So, (𝑢(0),0,𝑢(2),…,𝑢(𝑇),0,𝑢(𝑇+2)) is a solution of (1.4), (1.8), if and only if, (0,𝑢(2),…,𝑢(𝑇),0) solves the operator equation 𝑢(𝑡)=𝜆^𝑇_𝑠=2𝐺(𝑡,𝑠)ℎ(𝑠)𝑓(𝑢(𝑠)),𝑡∈𝕋₁. (3.15)

Now, let 𝐽∶𝑌→𝑋 be the linear operator: 𝐽(𝑢(2),𝑢(3),…,𝑢(𝑇))=(−𝑢(2),0,𝑢(2),𝑢(3),…,𝑢(𝑇),0,−𝑢(𝑇)),𝑢∈𝑌. (3.16)

Let 𝐿,𝑁∶𝑋→𝑋 be the operators: 𝐿𝑢∶=(𝐽∘ℒ)⁻¹ℎ(⋅)𝑓₀𝑢,𝑁𝑢∶=(𝐽∘ℒ)⁻¹(ℎ(⋅)𝜁(𝑢)), (3.17)(3.18) respectively. Then Lemma 3.1 yields that 𝐿∶𝑋→𝑋 is strongly positive. Moreover, [16, Theorem  7.c] implies 𝑟(𝐿)>0.

Now, it follows from Theorem 2.3 that there exists a continuum 𝒞₊⊆𝜇,𝑥∈ℝ×𝑋∣𝜇,𝑥isasolutionof(1.4),(1.8)with𝑥>0and𝜇>0, (3.19) which joins (𝑟(𝐿)⁻¹,0) with infinity in (0,∞)×𝐾 and 𝜇,𝑥∈𝒞₊,𝜇≠𝑟(𝐿)⁻¹⟹𝑥>0,𝜇>0. (3.20) It is easy to check that 𝑟(𝐿)⁻¹=𝜆₁𝑓₀. (3.21)

Lemma 3.2. Let ℎ₁,ℎ₂∈𝑌 with ℎ₁≥ℎ₂>0. Then the eigenvalue problems ℒ𝑢(𝑡)=𝜆ℎ_𝑖(𝑡)𝑢(𝑡),𝑡∈𝕋₂,𝑖=1,2 (3.22) have the principal eigenvalue 𝜆_𝑖, 𝑖=1,2 such that 𝜆₁≤𝜆₂. Moreover, the corresponding eigenfunctions 𝜓_𝑖 are positive in 𝕋₂.

Proof . Let 𝐿_𝑖∶𝑋→𝑋 be the operator 𝐿_𝑖𝑢∶=𝜆(𝐽∘ℒ)⁻¹ℎ_𝑖(⋅)𝑢,𝑖=1,2. (3.23)
Then Lemma 3.1 yields that 𝐿_𝑖∶𝑋→𝑋 is strongly positive. By Krein-Rutman theorem [16, Theorem  7.c] the spectral radius 𝑟(𝐿_𝑖)>0 and there exist 𝜓_𝑖∈𝑋 with 𝜓_𝑖>0 on 𝕋₂ such that 𝐿_𝑖𝜓_𝑖(𝑡)=𝑟𝐿_𝑖𝜓_𝑖(𝑡),𝑖=1,2. (3.24) That is, the eigenvalue problems (3.22) have the principal eigenvalues 𝜆_𝑖=1/𝑟(𝐿_𝑖), and 𝜓_𝑖(𝑡) is the corresponding eigenfunctions of 𝜆_𝑖, 𝑖=1,2.
Next, we prove 𝜆₁≤𝜆₂. Since ∑^𝑇_𝑡=2Δ⁴𝜓₁(𝑡−2)𝜓₂(𝑡)=∑^𝑇_𝑡=2𝜓₁(𝑡)Δ⁴𝜓₂(𝑡−2), it follows that^𝑇_𝑡=2ℎ₁(𝑡)𝜓₁(𝑡)𝜓₂(𝑡)≥^𝑇_𝑡=2𝜆₂𝜆₂ℎ₂(𝑡)𝜓₂(𝑡)𝜓₁(𝑡)=^𝑇_𝑡=21𝜆₂Δ⁴𝜓₂(𝑡−2)𝜓₁(𝑡)=^𝑇_𝑡=21𝜆₂𝜓₂(𝑡)Δ⁴𝜓₁(𝑡−2)=^𝑇_𝑡=2𝜓₂(𝑡)𝜆₂𝜆₁ℎ₁(𝑡)𝜓₁(𝑡)=𝜆₁𝜆₂^𝑇_𝑡=2ℎ₁(𝑡)𝜓₁(𝑡)𝜓₂(𝑡). (3.25) Therefore, 𝜆₁≤𝜆₂.

Suppose that 𝕋_𝑎={𝑎+1,𝑎+2,…,𝑏−1} is a strict subset of 𝕋₂ and ℎ_𝑎 denote the restriction of ℎ on 𝕋_𝑎. Consider the linear eigenvalue problems: Δ⁴𝑢(𝑡−2)=𝜆ℎ_𝑎(𝑡)𝑓₀𝑢(𝑡),𝑡∈𝕋_𝑎,𝑢(𝑎)=𝑢(𝑏)=Δ²𝑢(𝑎−1)=Δ²𝑢(𝑏−1)=0. (3.26) Then we get the following result.

Lemma 3.3. Let 𝜆₁ is the principal eigenvalue of (3.17), then (3.26) has only one principal eigenvalue 𝜆_𝑎 such that 0<𝜆₁<𝜆_𝑎.

Proof. It is not difficult to prove that (3.26) has only one principal eigenvalue 𝜆_𝑎>0 by Lemma 3.1, and the corresponding eigenfunction 𝜓_𝑎>0 on 𝕋_𝑎. So we only to verify that 0<𝜆₁<𝜆_𝑎.
Let 𝜓₁ be the corresponding eigenfunction of 𝜆₁, we have that ^𝑏−1_𝑡=𝑎+1Δ⁴𝜓_𝑎(𝑡−2)𝜓₁(𝑡)=^𝑏−1_𝑡=𝑎+1Δ⁴𝜓₁(𝑡−2)𝜓_𝑎(𝑡)−𝜓₁(𝑏)Δ²𝜓_𝑎(𝑏−2)−𝜓_𝑎(𝑏−1)Δ²𝜓₁(𝑏−1)−Δ²𝜓_𝑎(𝑎)𝜓₁(𝑎)−𝜓_𝑎(𝑎+1)Δ²𝜓₁()>𝑎−1_{𝑏−1𝑡=𝑎+1}Δ⁴𝜓₁(𝑡−2)𝜓_𝑎(𝑡). (3.27) So ^𝑏−1_𝑡=𝑎+1ℎ(𝑡)𝜓_𝑎(𝑡)𝜓₁(𝑡)=^𝑏−1_𝑡=𝑎+11𝜆_𝑎Δ⁴𝜓_𝑎(𝑡−2)𝜓₁(𝑡)>^𝑏−1_𝑡=𝑎+11𝜆_𝑎Δ⁴𝜓₁(𝑡−2)𝜓_𝑎(𝑡)=^𝑏−1_𝑡=𝑎+1𝜓_𝑎(𝑡)𝜆_𝑎𝜆₁ℎ(𝑡)𝜓₁(𝑡)=𝜆₁𝜆_𝑎^𝑏−1_𝑡=𝑎+1ℎ(𝑡)𝜓_𝑎(𝑡)𝜓₁(𝑡). (3.28) Thus 0<𝜆₁<𝜆_𝑎.

Proof of Theorem 1.3. We divide the proof into three steps.
Let {(𝜇_𝑛,𝑦_𝑛)}⊂𝒞₊ be such that 𝜇_𝑛+𝑦_𝑛𝑋→∞,𝑛→∞. (3.29) Then Δ⁴𝑦_𝑛(𝑡−2)=𝜇_𝑛ℎ(𝑡)𝑓𝑦_𝑛(𝑡),𝑡∈𝕋₂,𝑦_𝑛(1)=𝑦_𝑛(𝑇+1)=Δ²𝑦_𝑛(0)=Δ²𝑦_𝑛(𝑇)=0. (3.30)
Step 1. We show that there exists a constant 𝑀 such that 𝜇_𝑛∈(0,𝑀] for all 𝑛.
Suppose on the contrary that lim_𝑛→∞𝜇_𝑛=∞. (3.31) Let 𝑣_𝑛=𝑦_𝑛/‖𝑦_𝑛‖_𝑋. Then it follows from (3.30) that Δ⁴𝑣_𝑛(𝑡−2)=𝜇_𝑛ℎ(𝑡)𝑓𝑦_𝑛(𝑡)𝑦_𝑛(𝑡)𝑣_𝑛(𝑡),𝑡∈𝕋₂,𝑣_𝑛(1)=𝑣_𝑛(𝑇+1)=Δ²𝑣_𝑛(0)=Δ²𝑣_𝑛(𝑇)=0. (3.32) Since𝑓(𝑠)inf𝑠∣𝑠>0∶=𝑀₀>0, (3.33) there exists a constant 𝑀₀>0, such that 𝑓𝑦_𝑛(𝑡)𝑦_𝑛(𝑡)>𝑀₀>0. (3.34) Let 𝜆^∗ be the principal eigenvalue of the linear eigenvalue problems: Δ⁴𝑣(𝑡−2)=𝜆ℎ(𝑡)𝑀₀𝑣(𝑡),𝑡∈𝕋₂,𝑣(1)=𝑣(𝑇+1)=Δ²𝑣(0)=Δ²𝑣(𝑇)=0. (3.35) Combining (3.31) and (3.34) with the relation (3.32), using Lemma 3.2, we get 0<𝜇_𝑛≤𝜆^∗. (3.36) This contradicts (3.31). So 𝜇_𝑛∈(0,𝑀] for all 𝑛.
Step 2. We show that 𝒞₊ joins (𝜆₁/𝑓₀,0) with (0,∞).
Assume that there exist 𝛿>0 and {(𝜇_𝑛,𝑦_𝑛)}⊂𝒞₊ such that 0<𝛿≤𝜇_𝑛≤𝑀;𝑦_𝑛𝑋⟶∞,𝑛⟶∞. (3.37)
First, we show that 𝑦_𝑛𝑋⟶∞⟹𝑦_𝑛∞⟶∞. (3.38)
Suppose on the contrary that𝑦_𝑛∞≤𝑀₁ (3.39) for some 𝑀₁>0 (independent on 𝑛). Then it follows from (3.30) and 0<𝛿≤|𝜇_𝑛|≤𝑀 that Δ⁴𝑦_𝑛∞≤𝑀‖ℎ‖_∞sup𝑓(𝑠)∣0<𝑠≤𝑀₁, (3.40) and subsequently, {‖𝑦_𝑛‖_𝑋} is bounded. This is a contradiction. So, (3.38) holds.
Next, we show that 𝑦_𝑛∞⟶∞⇒min𝑦_𝑛(𝑡)∣𝑡∈𝕋₂⟶∞. (3.41)
In fact, 𝑦_𝑛(𝑡)=𝜇_𝑛^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠)𝑓𝑦_𝑛(𝑠),𝑡∈𝕋₁. (3.42) This together with (2.7) imply that (3.41) is valid.
Finally, we have from the facts that min{𝑦_𝑛(𝑡)∣𝑡∈𝕋₂}→∞ and 0<𝛿≤|𝜇_𝑛|≤𝑀 that 𝜇_𝑛𝑓𝑦_𝑛(𝑡)𝑦_𝑛(𝑡)⟶∞,𝑛⟶∞forany𝑡∈𝕋_𝑎. (3.43) Consider the following linear eigenvalue problems: Δ⁴𝑣(𝑡−2)=𝜆ℎ_𝑎(𝑡)𝑣(𝑡),𝑡∈𝕋_𝑎,𝑣(𝑎)=𝑣(𝑏)=Δ²𝑣(𝑎−1)=Δ²𝑣(𝑏−1)=0. (3.44) By Lemma 3.3 and (3.32), (3.44) has a positive principal eigenvalue 𝜆_𝑎, and𝜇_𝑛𝑓𝑦_𝑛(𝑡)𝑦_𝑛(𝑡)≤𝜆_𝑎, (3.45) which contradicts (3.43). Thus lim_𝑛→∞𝜇_𝑛=0.
Step 3. Fixed 𝜆 such that 0<𝜆<sup_{()𝑠∈0,𝐵}𝑠𝛾^∗𝑓(𝑠). (3.46) Then there exists 𝑏∈(0,𝐵] such that 𝑏0<𝜆<𝛾^∗𝑓(𝑏). (3.47) We show that there is no (𝜇,𝑢)∈𝒞₊ such that ‖𝑢‖_∞𝑏=𝑏,0<𝜇<𝛾^∗𝑓(𝑏). (3.48) In fact, if there exists (𝜂,𝑦)∈𝒞₊ satisfying (3.48), then 𝑦(𝑡)=𝜂^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠)𝑓𝑦(𝑠)≤𝜂𝛾^∗𝑓(𝑏)=𝜂𝛾^∗𝑓(𝑏)𝑏⋅𝑏 (3.49) for 𝑡∈𝕋₁, and subsequently, 𝜂≥𝑏/𝛾^∗𝑓(𝑏). Therefore, no (𝜇,𝑢)∈𝒞₊ satisfies (3.48).Now, combining the conclusions in Steps 2 and 3, using the fact that no (𝜇,𝑢)∈𝒞₊ satisfies (3.48), it concludes that for every 𝜆∈(𝜆₁/𝑓₀,𝑏/𝛾^∗𝑓(𝑏)), (1.4), (1.8) has at least two generalized positive solutions in 𝒞₊. For arbitrary 𝜆∈(0,sup_{𝑠∈(0,𝐵)}(𝑠/𝛾^∗𝑓(𝑠))), we may find 𝑏=𝑏(𝜆) satisfying (3.47). So, for every 𝜆∈(𝜆₁/𝑓₀,sup_{𝑠∈(0,𝐵)}(𝑠/𝛾^∗𝑓(𝑠))), (1.4), (1.8) has at least two generalized positive solutions in 𝒞₊.

Proof of Theorem 1.4. We divide the proof into three steps.
Step 1. We show that there exists a positive constant 𝛽>0 such that inf𝜇∣𝜇,𝑢∈𝒞₊=∶𝛽>0. (3.50) Suppose on the contrary that there exists {(𝜇_𝑛,𝑦_𝑛)}⊂𝒞₊ such that 𝜇_𝑛→0⁺,as𝑛→∞. (3.51) Then we have from (3.32), (3.51), 𝑓₀∈(0,∞) and 𝑓_∞=0 that 𝑣_𝑛𝑋→0,as𝑛→∞. (3.52) However, this contradicts with the fact that ‖𝑣_𝑛‖_𝑋=1 for all 𝑛∈ℕ. Therefore, (3.50) holds.
Step 2. We show that for any closed interval 𝐼⊂[𝛽,∞), there exists 𝑀_𝐼>0 such that sup‖𝑢‖∣𝜇,𝑢∈𝒞₊≤𝑀_𝐼. (3.53) Suppose on the contrary that there exists {(𝜇_𝑛,𝑦_𝑛)}⊂𝒞₊ with 𝜇_𝑛⊂𝐼,𝑦_𝑛𝑋⟶∞as𝑛⟶∞. (3.54) Then by (3.38), 𝑦_𝑛∞⟶∞,𝑛⟶∞. (3.55) and subsequently min_𝑡∈𝕋2𝑦_𝑛(𝑡)≥𝑐₁𝑦_𝑛∞⟶∞. (3.56) This together with (3.32) and 𝑓₀∈(0,∞) and 𝑓_∞=0 that 𝑣_𝑛|_𝕋2∞⟶0,𝑛⟶∞. (3.57) However, this contradicts with the fact that min_𝑡∈𝕋2𝑣_𝑛(𝑡)≥𝑐₁,𝑛∈ℕ. (3.58) Therefore, (3.53) holds.
Step 3. Fixed 𝜆 such that 𝜆>inf_{𝑠∈0,𝑐1𝐵}𝑠𝑐₁𝛾_∗𝑓(𝑠). (3.59) Then there exists 𝑙∈(0,𝑐₁𝐵) such that 𝑙𝜆>𝛾_∗𝑐₁𝑓(𝑙). (3.60) We show that there is no (𝜂,𝑦)∈𝒞₊ such that 𝑦_∞=𝑙𝑐₁𝑙𝜂>𝛾_∗𝑐₁𝑓(𝑙). (3.61) Suppose on the contrary that there exists (𝜂,𝑦)∈𝐶₊ satisfying (3.61). Then for 𝑡∈𝕋₂, 𝑦(𝑡)=𝜂^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠)𝑓𝑦(𝑠)≥𝜂^𝑇_𝑠=2𝐾(t,𝑠)ℎ(𝑠)𝑓𝑐₁𝑦_∞=𝜂^𝑇_𝑠=2𝐾(𝑡,𝑠)ℎ(𝑠)𝑓(𝑙)≥𝜂𝛾_∗𝑓(𝑙)=𝜂𝛾_∗𝑓(𝑙)𝑙⋅𝑙, (3.62) and subsequently, 𝜂≤𝑙/𝑐₁𝛾_∗𝑓(𝑙). Therefore, there is no (𝜂,𝑦)∈𝒞₊ such that (3.61) holds.
Now, combining the conclusions in Steps 2 and 3, using the fact that no (𝜇,𝑢)∈𝒞₊ satisfying (3.61), it concludes that for every 𝜆∈(𝑙/𝑐₁𝛾_∗𝑓(𝑙),𝜆₁/𝑓₀), (1.4), (1.8) has at least two generalized positive solutions in 𝒞₊. For arbitrary 𝜆∈(inf_{𝑠∈(0,𝑐1𝐵)}(𝑠/𝑐₁𝛾_∗𝑓(𝑠)),∞), we may find 𝑙=𝑙(𝜆) satisfying (3.60). So, for every 𝜆∈(inf_{𝑠∈(0,𝑐1𝐵)}(𝑠/𝛾_∗𝑓(𝑠)),𝜆₁/𝑓₀), (1.4), (1.8) has at least two generalized positive solutions in 𝒞₊.

4. Some Examples

In this section, we will apply our results to two examples.

For convenience, set 𝑇=12, then 𝕋₁={1,2,…,13}, 𝕋₂={2,3,…,12}.

Example 4.1. Let us consider the boundary value problem Δ⁴𝑢(𝑡−2)=𝜆𝑓(𝑢(𝑡)),𝑡∈𝕋₂,𝑢(1)=𝑢(13)=Δ²𝑢(0)=Δ²𝑢(12)=0, (4.1) where 𝑓(𝑢)=(],()arctan𝑢,𝑢∈0,1000𝑢−1000²+arctan1000,𝑢∈(1000,∞). (4.2) Clearly, 𝑓(𝑢) is nondecreasing, 𝑓₀=1, 𝑓_∞=∞. Take 𝐵=1000. By a simple computation, it follows that inf_{𝑠∈(0,1000)}(𝑓(𝑠)/𝑠)=arctan1000/1000≈0.00157, 𝜆₁=16sin⁴(𝜋/24)≈0.0048 and 𝛾^∗=max_𝑡∈𝕋1∑¹²_𝑠=2𝐾(𝑡,𝑠)=1629/6, then 0.0048≈16sin⁴𝜋=𝜆24𝑓¹₀<sup_{()𝑠∈0,1000}𝑠𝑓(𝑠)𝛾^∗=61629inf_{𝑠∈(0,1000)}𝑓(𝑠)/𝑠≈2.34631. (4.3) So, Theorem 1.3(i) implies that (4.1) has at least one generalized positive solution for 𝜆0<𝜆<¹𝑓₀≈0.0048; (4.4) Theorem 1.3(ii) implies that (4.1) has at least two generalized positive solutions for 𝜆₁𝑓₀1<𝜆<𝛾^∗16sin⁴(𝜋/24)−1≈2.34631. (4.5)

Example 4.2. Let us consider the boundary value problem: Δ⁴𝑢(𝑡−2)=𝜆𝑓(𝑢(𝑡)),𝑡∈𝕋₂,𝑢(1)=𝑢(13)=Δ²𝑢(0)=Δ²𝑢(12)=0, (4.6) where 𝑓(𝑢)=𝑒^𝑢−1𝑢(],√,𝑢∈0,30𝑒𝑢−30+³⁰−1().30,𝑢∈30,∞ (4.7) Obviously, 𝑓(𝑢) is nondecreasing in [0,∞), so 𝑓₀=lim_𝑢→0(𝑓(𝑢)/𝑢)=1, 𝑓_∞=lim_𝑢→0(𝑓(𝑢)/𝑢)=0. By a simple computation, it follows that 𝜆₁=16sin⁴(𝜋/24) and 𝛾_∗=min_𝑡∈𝕋2∑¹²_s=2𝐾(𝑡,𝑠)=143/2, 𝑐₁=216√3/145√145≈0.214. Take 𝐵=50. Since sup_{𝑠∈(0,10.715)}(𝑓(𝑠)/𝑠)=(𝑒^10.715−1)/10.715≈4202.0726, it follows that 0.0000155≈inf_{𝑠∈0,𝑐1𝐵}𝑠𝑐₁𝛾_∗𝑓(𝑠)=1sup_{𝑠∈(0,10.715)}𝑓(𝑠)/𝑠𝑐₁𝛾_∗<𝜆₁𝑓₀=16sin⁴𝜋24≈0.0048. (4.8) Therefore, (i) of Theorem 1.4 implies that(4.6) has at least one generalized positive solution for 𝜆>inf_{𝑠∈0,𝑐1𝐵}𝑠𝑐₁𝛾_∗𝑓(𝑠)≈0.0000155; (4.9)(ii) of Theorem 1.4 implies that (4.6) has at least two generalized positive solutions for 0.0000155<𝜆<16sin⁴𝜋.24 (4.10)

Acknowledgments

This paper was written when the first author visited Tabuk University, Tabuk, during May 16-June 13, 2011 and he is very thankful to the administration of Tabuk University for providing him the hospitalities during the stay. The second author gratefully acknowledges the partial financial support from the Deanship of Scientific Research (DSR) at King Abdulaziz University, Jeddah. The authors thank the referees for their valuable comments.