Existence of Positive Solutions of a Discrete Elastic Beam Equation

Abstract

Let 𝑇 be an integer with 𝑇≥5 and let 𝕋₂={2,3,…,𝑇}. We consider the existence of positive solutions of the nonlinear boundary value problems of fourth-order difference equations Δ⁴𝑢(𝑡−2)−𝑟𝑎(𝑡)𝑓(𝑢(𝑡))=0, 𝑡∈𝕋₂, 𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0, where 𝑟 is a constant, 𝑎∶𝕋₂→(0,∞),and𝑓∶[0,∞)→[0,∞) is continuous. Our approaches are based on the Krein-Rutman theorem and the global bifurcation theorem.

1. Introduction

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

𝑦^′′′′=𝑓(𝑥,𝑦,𝑦^′′),𝑥∈(0,1),𝑦(0)=𝑦(1)=𝑦^′′(0)=𝑦^′′(1)=0; (1.1) see Gupta [1, 2]. The existence of solutions of (1.3) and (1.4) has been extensively studied; see Gupta [1, 2], Aftabizadeh [3], Yang [4], Del Pino and Manásevich [5], Galewski [6], Yao [7], and the references therein. The existence and multiplicity of positive solutions of the boundary value problem of ordinary differential equations

𝑦^′′′′(𝑥)−𝜆𝑓(𝑥,𝑦(𝑥))=0,𝑥∈(0,1),𝑦(0)=𝑦(1)=𝑦^′′(0)=𝑦^′′(1)=0 (1.2) have also been studied by many authors; see Ma and Wang [8], Ma [9], Bai and Wang [10], Chai [11], Yao and Bai [12] for some references along this line.

Recently, the existence of solutions of boundary value problems (BVPs) of difference equations has received much attention; see Agarwal and Wong [13], Henderson [14], He and Yu [15], Zhang et al. [16], and the references therein. However, relatively little is known about the existence of positive solutions of fourth-order discrete boundary value problems. To our best knowledge, only He and Yu [15] and Zhang et al. [16] dealt with that. In [15], He and Yu studied the existence of positive solutions of the nonlinear fourth-order discrete boundary value problem

Δ⁴𝑢(𝑡−2)−𝑟𝑎(𝑡)𝑓(𝑢(𝑡))=0,𝑡∈𝕋₂,𝑢(0)=𝑢(𝑇+2)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0 (1.3)(1.4) (where 𝑇≥3 is an integer, 𝕋₂∶={2,…,𝑇}, 𝑟 is a parameter, 𝑎∶𝕋₂→[0,∞), 𝑓∈𝐶([0,∞),[0,∞)) satisfies some growth conditions which are not optimal!). The likely reason is that the spectrum structure of the linear eigenvalue problem

Δ⁴𝑢(𝑡−2)=𝜆𝑎(𝑡)𝑢(𝑡),𝑡∈𝕋₂,𝑢(0)=𝑢(𝑇+2)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0 (1.5)(1.6) is not clear. It has been pointed out in [15, 16] that (1.3) and (1.4) are equivalent to the integral equation of the form

𝑢(𝑡)=𝑟^𝑇+1_𝑠=1𝐺(𝑡,𝑠)^𝑇_𝑠=2𝐺₁(𝑠,𝑗)𝑎(𝑗)𝑓(𝑢(𝑗))=∶𝐴₀𝑢(𝑡),𝑗∈𝕋₂, (1.7) where

1𝐺(𝑡,𝑠)=𝐺𝑇+2𝑠(𝑇+2−𝑡),1≤𝑠≤𝑡≤𝑇+2,𝑡(𝑇+2−𝑠),0≤𝑡≤𝑠≤𝑇+1,₁1(𝑡,𝑖)=𝑇(𝑇+1−𝑡)(𝑖−1),2≤𝑖≤𝑡≤𝑇+1,(𝑇+1−𝑖)(𝑡−1),1≤𝑡≤𝑖≤𝑇, (1.8) and other results on the existence of positive solutions of (1.3) and (1.4) can be found in the two papers. Notice that in the integral (1.7), two distinct Green's functions, 𝐺 and 𝐺₁, are used. This makes the construction of cones and the verification of strong positivity of 𝐴₀ more complex and difficult. Therefore, we think that the boundary condition (1.4) is not very suitable for the study of the positive solutions of fourth order difference equations.

It is the purpose of this paper to assume the fourth-order difference equation (1.3) subject to a new boundary condition of the form

𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0. (1.9) This will make our approaches much more simple and natural, and only one Green's function is needed. However, the classical definitions of positive solutions are useless for (1.3) and (1.9) any more. We have to adopt the following new definition of positive solutions.

Definition 1.1. Denote 𝕋₁∶={1,2,…,𝑇+1},𝕋₀∶={0,1,…,𝑇+1,𝑇+2}. (1.10) A function 𝑦∶𝕋₀→ℝ⁺ is called a positive solution of (1.3) and (1.9) if 𝑦 satisfies (1.3), (1.9), and 𝑦(𝑡)≥0 on 𝕋₂ and 𝑦(𝑡)≢0 on 𝕋₂.

Remark 1.2. Notice that the fact 𝑦∶𝕋₀→ℝ⁺ is a positive solution of (1.3) and (1.9) does not mean 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).

Remark 1.3. In [17], Eleo and Henderson defined a kind of positive solutions which actually are sign-change solutions. In Definition 1.1, a positive solution may allow to take nonpositive value at 𝑡=0 and 𝑡=𝑇+2. We think it is useful in this case that 𝑇 is large enough.

In the rest of the paper, we will use global bifurcation technique; see Dancer [18, Theorem 2] or Ma and Xu [19, Lemma 2.1], to deal with (1.3) and (1.9). To do this, we have to study the spectrum properties of (1.5) and (1.9). This will be done in Section 2. Finally, in Section 3, we will state and prove our main result.

2. Eigenvalues

Recall

𝕋₂∶={2,3,…,𝑇},𝕋₁∶={1,2,…,𝑇+1},𝕋₀∶={0,1,…,𝑇+1,𝑇+2}. (2.1) Let

𝐸∶={𝑢∣𝑢∶𝕋₀→ℝ}. (2.2) Then, dim𝐸=𝑇+3, and 𝐸 is a Banach space with the norm

‖𝑢‖_𝐸=max{|𝑢(𝑗)|∣𝑗∈𝕋₀}. (2.3) Let

𝑌∶={𝑢∣𝑢∶𝕋₂→ℝ}. (2.4) Then 𝑌 is a Banach space with the norm

‖𝑢‖_𝑌=max{|𝑢(𝑗)|∣𝑗∈𝕋₂}. (2.5) Let

𝐸₀∶={𝑢∈𝐸∣𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0}. (2.6) Then the operator 𝜒∶𝐸₀→𝑌

𝜒(−𝑢(2),0,𝑢(2),𝑢(3),…,𝑢(𝑇),0,−𝑢(𝑇))∶=(𝑢(2),𝑢(3),…,𝑢(𝑇)), (2.7) is a homomorphism.

In this paper, we assume that

(𝐻1)𝑎∶𝕋₂→(0,∞).

Definition 2.1. We say that 𝜆 is an eigenvalue of linear problem Δ⁴𝑢(𝑡−2)=𝜆𝑎(𝑡)𝑢(𝑡),𝑡∈𝕋₂,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0 (2.8) if (2.8) has a nontrivial solution.

In the rest of this section, we will prove the existence of the first eigenvalue of (2.8).

Theorem 2.2. Equation (2.8) has an algebraically simple eigenvalue 𝜆₁, with an eigenfunction 𝜑 satisfying(i)𝜑(𝑡)>0 on 𝕋₂;(ii)𝜑(0)=−𝜑(2); 𝜑(𝑇+2)=−𝜑(𝑇).Moreover, there is no other eigenvalue whose eigenfunction is nonnegative on 𝕋₂.

To prove Theorem 2.2, we need several preliminary results.

Lemma 2.3. For each ℎ=(ℎ(2),…,ℎ(𝑇)), the linear problem Δ⁴𝑢(𝑡−2)=ℎ(𝑡),𝑡∈𝕋₂,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0 (2.9) has a unique solution 𝑢(𝑡)=^𝑇_𝑠=2𝐻(𝑡,𝑠)^𝑇_𝑗=2𝐻(𝑠,𝑗)ℎ(𝑗),𝑡∈𝕋₁, (2.10) where 1𝐻(𝑡,𝑠)=𝑇(𝑡−1)(𝑇+1−𝑠),1≤𝑡≤𝑠≤𝑇,(𝑠−1)(𝑇+1−𝑡),2≤𝑠≤𝑡≤𝑇+1. (2.11)

Proof. Let Δ²𝑢(𝑡−2)=𝑤(𝑡−1) for 𝑡∈𝕋₂. Then (2.9) is equivalent to the system Δ²𝑤(𝑡−1)=ℎ(𝑡),𝑡∈𝕋₂,Δ²𝑢(𝑡−1)=𝑤(𝑡),𝑡∈𝕋₂,𝑤(1)=𝑤(𝑇+1)=0,𝑢(1)=𝑢(𝑇+1)=0. (2.12) From Kelly and Peterson [20, Theorem 6.8 and Example 6.12], it follows that 𝑤(𝑡)=−^𝑇_𝑠=2𝐻(𝑡,𝑠)ℎ(𝑠),𝑡∈𝕋₁𝑢(𝑡)=−^𝑇_𝑠=2𝐻(𝑡,𝑠)𝑤(𝑠),𝑡∈𝕋₁. (2.13) Therefore, (2.10) holds.

Denote

𝜌∶=4sin²𝜋2𝑇,𝑒(𝑡)∶=sin𝜋(𝑡−1)𝑇,𝑡∈𝕋₁. (2.14) Then

Δ²𝑒(𝑡−1)+𝜌𝑒(𝑡)=0,𝑡∈𝕋₂,𝑒(1)=𝑒(𝑇+1)=0. (2.15) Notice that

{(𝑡−1𝑇,𝑒(𝑡))∣𝑡∈𝕋₁}⊂{(𝑥,sin𝜋𝑥)∣𝑥∈[0,1]}. (2.16) From the assumption 𝑇≥5, we have

𝜌≤4sin²𝜋2⋅5<4sin²𝜋6=1. (2.17)

Lemma 2.4. Let 𝑣∈𝐸 satisfy 𝑣(1)=𝑣(𝑇+1)=Δ²𝑣(0)=Δ²𝑣(𝑇)=0 and −𝛾𝑒(𝑡)≤−Δ²𝑣(𝑡−1)≤𝛾𝑒(𝑡),𝑡∈𝕋₂, (2.18) where 𝛾∈(0,∞). Then −𝛾𝜌𝛾𝑒(𝑡)≤𝑣(𝑡)≤𝜌𝑒(𝑡),𝑡∈𝕋₁. (2.19)

Proof. From (2.18), we get −𝛾^𝑇_𝑠=2𝐻(𝑡,𝑠)𝑒(𝑠)≤−^𝑇_𝑠=2𝐻(𝑡,𝑠)Δ²𝑣(𝑡−1)≤𝛾^𝑇_𝑠=2𝐻(𝑡,𝑠)𝑒(𝑠),𝑡∈𝕋₂. (2.20) This is −𝛾𝜌𝛾𝑒(𝑡)≤𝑣(𝑡)≤𝜌𝑒(𝑡),𝑡∈𝕋₂. (2.21) Combining this with the boundary conditions 𝑣(1)=𝑣(𝑇+1)=Δ²𝑣(0)=Δ²𝑣(𝑇)=0, it concludes that −𝛾𝜌𝛾𝑒(𝑡)≤𝑣(𝑡)≤𝜌𝑒(𝑡),𝑡∈𝕋₁. (2.22)

Let

𝑋∶={𝑢∈𝐸₀∣−𝛾𝑒(𝑡)≤−Δ²𝑢(𝑡−1)≤𝛾𝑒(𝑡),𝑡∈𝕋₁} (2.23) for some 𝛾∈(0,∞). Since 𝛾<𝛾/𝜌, from Lemma 2.4 and (2.17), we may define

‖𝑢‖_𝑋𝛾∶=inf{𝜌∣−𝛾𝑒(𝑡)≤−Δ²𝑢(𝑡−1)≤𝛾𝑒(𝑡),𝑡∈𝕋₁}. (2.24) For any 𝑥,𝑦∈𝑋, we have from the definition of ‖⋅‖_𝑋 that

−𝜌𝑒‖𝑥‖_𝑋≤−Δ²𝑥(𝑡−1)≤𝜌𝑒‖𝑥‖_𝑋,𝑡∈𝕋₁,−𝜌𝑒‖𝑦‖_𝑋≤−Δ²𝑦(𝑡−1)≤𝜌𝑒‖𝑦‖_𝑋,𝑡∈𝕋₁. (2.25) It follows that

−𝜌𝑒(‖𝑥‖_𝑋+‖𝑦‖_𝑋)≤−Δ²(𝑥(𝑡−1)+𝑦(𝑡−1))≤𝜌𝑒(‖𝑥‖_𝑋+‖𝑦‖_𝑋),𝑡∈𝕋₁. (2.26) Thus, 𝑥+𝑦∈𝑋, and moreover,

‖𝑥+𝑦‖_𝑋≤‖𝑥‖_𝑋+‖𝑦‖_𝑋. (2.27) Therefore, ‖⋅‖_𝑋 is a norm of 𝑋, and (𝑋,‖⋅‖_𝑋) is a normed linear space. Since dim𝑋=𝑇−1, (𝑋,‖⋅‖_𝑋) is actually a Banach space. Let

𝑃∶={𝑢∈𝑋∣Δ²𝑢(𝑡−1)≤0for𝑡∈𝕋₂;𝑢(𝑡)≥0for𝑡∈𝕋₁}. (2.28) Then the cone 𝑃 is normal and has nonempty interior int𝑃.

Lemma 2.5. For 𝑢∈𝑋, ‖𝑢‖_∞,1≤(𝑇+1)²‖Δ²𝑢‖_∞,2,‖Δ²𝑢‖_∞,2≤𝜌‖𝑢‖_𝑋, (2.29) where ‖𝑢‖_∞,1∶=max{|𝑢(𝑗)|∣𝑢∈𝕋₁},‖Δ²𝑢‖_∞,2∶=max{|Δ²𝑢(𝑗−1)|∣𝑗∈𝕋₂}. (2.30)

Proof. (i) From the relation
𝑢(𝑡)=^𝑇_𝑠=2𝐻(𝑡,𝑠)(−Δ²𝑢(𝑠−1)),𝑡∈𝕋₁, (2.31) it follows that ‖𝑢‖_∞,1≤(𝑇+1)²‖Δ²𝑢‖_∞,2. (2.32)
(ii) By (2.14) and the fact that 𝑒(𝑡)>0 on 𝕋₂, it follows that there exists 𝛾>0, such that
|−Δ²𝑢(𝑡−1)|≤𝛾𝑒(𝑡),𝑡∈𝕋₂. (2.33) Let 𝛾₀∶=inf{𝛾∣|−Δ²𝑢(𝑡−1)|≤𝛾𝑒(𝑡),𝑡∈𝕋₂}. (2.34) Then |−Δ²𝑢(𝑡−1)|≤𝛾₀𝑒(𝑡),𝑡∈𝕋₂. (2.35) This implies ‖Δ²𝑢‖_∞,2≤𝜌⋅(𝛾₀/𝜌), and accordingly ‖Δ²𝑢‖_∞,2≤𝜌‖𝑢‖_𝑋.

Proof of Theorem 2.2. For 𝑢∈𝐸, define a linear operator 𝐾∶𝑋→𝑌 and 𝐽∶𝑌→𝑋 by 𝐾𝑢(𝑡)∶=^𝑇_𝑠=2𝐻(𝑡,𝑠)^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑎(𝑗)𝑢(𝑗),𝑡∈𝕋₁,𝐽(𝑢(1),…,𝑢(𝑇+1))=(−𝑢(2),𝑢(1),…,𝑢(𝑇+1),−𝑢(𝑇)). (2.36)(2.37) Then (2.8) can be written as 𝑢=𝜆𝐽∘𝐾𝑢,𝑢∈𝑋. (2.38) Since 𝑋 is finite dimensional, we have that 𝐾∶𝑋→𝑌 is compact. Obviously, 𝐽∘𝐾(𝑃)⊆𝑃.
Next, we show that 𝐽∘𝐾∶𝑃→𝑃 is strongly positive.
Since 𝑎(𝑡) is positive on 𝕋₂, there exists a constant 𝑘>0 such that 𝑎(𝑡)>𝑘 on 𝕋₂.
For 𝑢∈𝑃⧵{0}, we have that
^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑎(𝑗)𝑢(𝑗)≥𝑘^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑢(𝑗)>0,𝑠∈𝕋₁. (2.39) It follows that there exists 𝑘₁>0 such that ^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑎(𝑗)𝑢(𝑗)≥𝑘₁𝑒(𝑠). (2.40) Also, for 𝑢∈𝑃⧵{0}, we have from the fact 𝐽∘𝐾𝑢∈𝑃 and 𝐾𝑢≠0 in 𝕋₂ that ^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑎(𝑗)𝑢(𝑗)≤max_𝑡∈𝕋2𝑎(𝑡)⋅^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑢(𝑗)≤𝑘₂𝑒(𝑠),𝑠∈𝕋₁, (2.41) for some constant 𝑘₂>0. By (2.39) and (2.41), we get 𝑘₁𝑒(𝑠)≤^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑎(𝑗)𝑢(𝑗)≤𝑘₂𝑒(𝑠),𝑠∈𝕋₁. (2.42) Thus 𝑘₁^𝑇_𝑠=2𝐻(𝑡,𝑠)𝑒(𝑠)≤^𝑇_𝑠=2𝐻(𝑡,𝑠)^𝑇_𝑗=2𝐻(𝑠,𝑗)𝑎(𝑗)𝑢(𝑗)≤𝑘₂^𝑇_𝑠=2𝐻(𝑡,𝑠)𝑒(𝑠),𝑡∈𝕋₁. (2.43) Since ^𝑇_𝑠=21𝐻(𝑡,𝑠)𝑒(𝑠)=𝜌𝑒(𝑡),𝑡∈𝕋₁. (2.44) Using (2.43) and (2.44), it follows that 𝑘₁𝜌𝑘𝑒(𝑡)≤(𝐾𝑢)(𝑡)≤²𝜌𝑒(𝑡),𝑡∈𝕋₁. (2.45) Therefore, 𝐽∘𝐾𝑢∈int𝑃.
Now, by the Krein-Rutman theorem [21, Theorem 7.𝐶; 20, Theorem 19.3], 𝐾 has an algebraically simple eigenvalue 𝜆>0 with an eigenvector 𝜑(⋅)∈int𝑃. Moreover, there is no other eigenvalue with an eigenfunction in 𝑃.

3. The Main Result

In this section, we will make the following assumptions:

(H2) 𝑓∶[0,∞)→[0,∞) is continuous and 𝑓(𝑠)>0 for 𝑠>0;

(H3) 𝑓₀,𝑓_∞∈(0,∞), where

𝑓₀=lim_𝑢→0+𝑓(𝑢)𝑢,𝑓_∞=lim_𝑢→+∞𝑓(𝑢)𝑢. (3.1)

Remark 3.1. It is not difficult to see that (H2) and (H3) imply that there exists a constant 𝑎₀∈(0,∞) such that 𝑓(𝑢)≥𝑎₀𝑢,𝑢∈[0,∞). (3.2)

Theorem 3.2. Let (H1), (H2), and (H3) hold. Assume that either 𝜆₁𝑟𝑓₀𝜆<1<¹𝑟𝑓_∞ (3.3) or 𝜆₁𝑟𝑓_∞𝜆<1<¹𝑟𝑓₀. (3.4) Then (1.3) and (1.9) have at least one positive solution.

Remark 3.3. Recently, Ma and Xu [19] considered the nonlinear fourth-order problem 𝑢⁽⁴⁾(𝑡)=𝑓(𝑡,𝑢(𝑡),𝑢^′′(𝑡)),𝑡∈(0,1),𝑢(0)=𝑢(1)=𝑢^′′(0)=𝑢^′′(1)=0 (3.5) under some conditions involved the generalized eigenvalues of the linear problem 𝑢^′′′′=𝜆[𝐴(𝑡)𝑢−𝐵(𝑡)𝑢^′′],0<𝑡<1,𝑢(0)=𝑢(1)=𝑢^′′(0)=𝑢^′′(1)=0. (3.6) Our main result, Theorem 3.2, needs 𝑓₀,𝑓_∞∈(0,∞); see (H3). However, in [19, Theorem 3.1], some weaker conditions of the form 𝑎₀(𝑡)𝑢−𝑏₀𝑝−𝜉₁(𝑡,𝑢,𝑝)≤𝑓(𝑡,𝑢,𝑝)≤𝑎⁰(𝑡)𝑢−𝑏⁰𝑝+𝜉₂(𝑡,𝑢,𝑝) (3.7) are used.

Remark 3.4. The first eigenvalue 𝜇₁ of the linear problem Δ²𝑢(𝑡−1)+𝜇𝑢(𝑡)=0,𝑡∈{2,…,𝑇},𝑢(1)=𝑢(𝑇+1)=0, (3.8) is 𝜇₁=4sin²(𝜋/2𝑇), and the first eigenvalue 𝜆₁ of the linear problem Δ⁴𝑢(𝑡−1)+𝜆𝑢(𝑡)=0,𝑡∈{2,…,𝑇},𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0, (3.9) is 𝜆₁=16sin⁴𝜋.2𝑇 (3.10) It is easy to check that the function 𝑓(𝑠)∶=2𝜆₁𝜆𝑠,𝑠∈(0,1),₁𝑠2+32𝜆₁,𝑠∈[1,∞). (3.11) Then, for each 𝑟∈(1/2,2), the condition (3.3) holds.

Remark 3.5. The condition (3.3) or (3.4) is optimal since for any 𝜖>0, the linear problem Δ⁴𝑢(𝑡−2)−(𝜆₁−𝜖)𝑎(𝑡)𝑢(𝑡)=0,𝑡∈𝕋₂,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0, (3.12) has the unique solution 𝑢(𝑡)≡0. In fact, 𝜆₁ is the least eigenvalue of the linear problem Δ⁴𝑢(𝑡−2)−𝜇𝑎(𝑡)𝑢(𝑡)=0,𝑡∈𝕋₂,𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0. (3.13)

To prove Theorem 3.2, we define 𝐿∶𝐷(𝐿)→𝑌 by

𝐿𝑢(𝑡)∶=−Δ⁴𝑢(𝑡−2),𝑢∈𝐷(𝐿),𝑡∈𝕋₂ (3.14) where

𝐷(𝐿)={𝑢∈𝑋∣𝑢(1)=𝑢(𝑇+1)=Δ²𝑢(0)=Δ²𝑢(𝑇)=0}. (3.15) It is easy to check that 𝐿⁻¹∶𝑌→𝑋 is compact.

Let 𝜁,𝜉∶[0,∞)→ℝ be such that

𝑓(𝑢)=𝑓₀𝑢+𝜁(𝑢),𝑓(𝑢)=𝑓_∞𝑢+𝜉(𝑢). (3.16) Obviously, (H3) implies

lim_𝑢→0+𝜁(𝑢)𝑢=0,lim_𝑢→+∞𝜉(𝑢)𝑢=0. (3.17) Let

𝜉̃(𝜏)=max_{0≤|𝑠|≤𝜏}|𝜉(𝑠)|. (3.18) Then 𝜉̃ is nondecreasing and

lim_𝜏→∞𝜉̃(𝜏)𝜏=0. (3.19)

Let us consider

𝐿𝑢+𝜆𝑟𝑎(𝑡)𝑓₀𝑢+𝜆𝑟𝑎(𝑡)𝜁(𝑢)=0,𝜆>0, (3.20) as a bifurcation problem for the trivial solution 𝑢≡0. It is easy to check that (3.20) can be converted to the equivalent equation

𝑢(𝑡)=𝜆[^𝑇𝑇_𝑠=2𝑗=2𝐻(𝑡,𝑠)𝐻(𝑠,𝑗)(𝑟𝑎(𝑗)𝑓₀𝑢(𝑗)+𝑟𝑎(𝑠)𝜁(𝑢(𝑠)))]∶=(𝜆𝐿⁻¹[𝑟𝑎(⋅)𝑓₀𝑢(⋅)](𝑡)+𝜆𝐿⁻¹[𝑟𝑎(⋅)𝜁(𝑢(⋅))](𝑡)). (3.21)

From the proof process of Theorem 2.2, we have that for each fixed 𝜆>0, the operator 𝐾∶𝑋→𝑋,

𝐾𝑢(𝑡)=^𝑇𝑇_𝑠=2𝑗=2𝐻(𝑡,𝑠)𝐻(𝑠,𝑗)𝑟𝑎(𝑗)𝑓₀𝑢(𝑗) (3.22) is compact and strongly positive. Define 𝐹∶[0,∞)×𝑋→𝑋 by

𝐹(𝜆,𝑢)∶=𝜆[^𝑇𝑇_𝑠=2𝑗=2𝐻(𝑡,𝑠)𝐻(𝑠,𝑗)𝑟𝑎(𝑗)𝜁(𝑢(𝑗))]. (3.23) Then we have from (3.17) and Lemma 2.5 that

‖𝐹(𝜆,𝑢)‖_𝑋=𝑜(‖𝑢‖_𝑋),as‖𝑢‖_𝑋→0, (3.24) locally uniformly in 𝜆. Now, we have from a version of Dancer [18, Theorem 2], see Ma [9, Lemma 2.1] for details, to conclude that there exists an unbounded connected subset 𝒞 in the set

𝜆{(𝜆,𝑢)∈(0,∞)×𝑃∶𝑢=𝜆𝐾𝑢+𝐹(𝜆,𝑢),𝑢∈int𝑃}∪{(¹𝑟𝑓₀,0)} (3.25) such that (𝜆₁/(𝑟𝑓₀),0)∈𝒞.

Proof of Theorem 3.2. It is clear that any solution of (3.20) of the form (1,𝑢) yields a solutions 𝑢 of (1.3) and (1.9). We will show that 𝒞 crosses the hyperplane {1}×𝑋 in ℝ×𝑋. To do this, it is enough to show that 𝒞 joins (𝜆₁/𝑟𝑓₀,0) to (𝜆₁/𝑟𝑓_∞,∞). Let (𝜇_𝑛,𝑦_𝑛)∈𝒞 satisfy 𝜇_𝑛+‖𝑦_𝑛‖_𝑋→∞,𝑛→∞. (3.26) We note that 𝜇_𝑛>0 for all 𝑛∈ℕ, since (0,0) is the only solution of (3.20) for 𝜆=0 and 𝒞∩({0}×𝑋)=∅.Case 1 (𝜆₁/𝑟𝑓_∞<1<𝜆₁/𝑟𝑓₀). In this case, we show that
(𝜆₁𝑟𝑓_∞,𝜆₁𝑟𝑓₀)⊆{𝜆∈ℝ∣∃(𝜆,𝑢)∈𝒞}. (3.27) We divide the proof into two steps.Step 1. We show that if there exists a constant number 𝑀>0 such that 𝜇_𝑛⊂(0,𝑀], (3.28) then 𝐶^𝜈_𝑘 joins (𝜆₁/𝑟𝑓₀,0) to (𝜆₁/𝑟𝑓_∞,∞).
From (3.28), we have that ‖𝑦‖_𝑋→∞. We divide the equation
𝐿𝑦_𝑛+𝜇_𝑛𝑟𝑎(𝑡)𝑓_∞𝑦_𝑛+𝜇_𝑛𝑟𝑎(𝑡)𝜉(𝑦_𝑛)=0 (3.29) by ‖𝑦_𝑛‖_𝑋 and set 𝑦_𝑛=𝑦_𝑛/‖𝑦‖_𝑋. Since 𝑦_𝑛 is bounded in 𝑋, choosing a subsequence and relabelling if necessary, we see that 𝑦_𝑛→𝑦 for some 𝑦∈𝑋 with ‖𝑦‖=1. Moreover, from (3.19) and the fact that 𝜉̃ is nondecreasing, we have that lim_𝑛→∞|𝜉(𝑦_𝑛(𝑡))|‖𝑦‖_𝑋=0, (3.30) since𝜉(𝑦_𝑛(𝑡))‖𝑦_𝑛‖_𝑋≤𝜉̃(|𝑦_𝑛(𝑡)|)‖𝑦_𝑛‖_𝑋≤𝜉̃(‖𝑦_𝑛‖_∞,1)‖𝑦_𝑛‖_𝑋≤𝜉̃(𝜌(𝑇+1)²‖𝑦_𝑛‖_𝑋)‖𝑦_𝑛‖_𝑋. (3.31) Thus, 𝑦(𝑡)∶=^𝑇𝑇_𝑠=2𝑗=2𝐺(𝑡,𝑠)𝐺(𝑠,𝑗)𝜇𝑟𝑎(𝑗)𝑓_∞𝑦(𝑗), (3.32) where 𝜇∶=lim_𝑛→∞𝜇_𝑛, choosing a subsequence and relabelling if necessary. Thus, 𝐿𝑦+𝜇𝑟𝑎(𝑡)𝑓_∞𝑦=0. (3.33) By Theorem 2.2, we have 𝜆𝜇=¹𝑟𝑓_∞. (3.34) Thus, 𝒞 joins (𝜆₁/𝑟𝑓₀,0) to (𝜆₁/𝑟𝑓_∞,∞).Step 2. We show that there exists a constant 𝑀 such that 𝜇_𝑛∈(0,𝑀] for all 𝑛.
By [9, Lemma 2.1], we only need to show that 𝐴 has a linear minorant 𝑉 and there exists a (𝜇,𝑦)∈(0,∞)×𝑃 such that ‖𝑦‖_𝑋=1 and 𝜇𝑉𝑦≥𝑦.
By Remark 3.1, there exist constants 𝑎₀∈(0,∞) such that
𝑓(𝑢)≥𝑎₀𝑢,𝑢∈[0,∞). (3.35) For 𝑢∈𝑋, let 𝑉𝑢(𝑡)∶=^𝑇𝑇_𝑠=2𝑗=2𝐺(𝑡,𝑠)𝐺(𝑠,𝑗)𝑎₀𝑢(𝑠). (3.36) Then 𝑉 is a linear minorant of 𝑅. Moreover, 𝑉(𝑒(𝑡)𝜌)=^𝑇𝑇_𝑠=2𝑗=2𝐺(𝑡,𝑠)𝐺(𝑠,𝑗)𝑎₀𝑒(𝑗)𝜌=1𝜌²𝑎₀^𝑇_𝑠=2≥𝑐𝐺(𝑡,𝑠)𝑒(𝑠)𝜌𝑒(𝑡) (3.37) for some constant 𝑐>0, independent of 𝑡∈𝕋₀. So, 𝑐⁻¹𝑒𝑉(𝜌𝑒)≥𝜌. (3.38) Therefore, it follows [9, Lemma 2.1] that |𝜂_𝑛|≤𝑐⁻¹. (3.39)Case 2 (𝜆₁/𝑟𝑓₀<1<𝜆₁/𝑟𝑓_∞). In this case, if (𝜂_𝑛,𝑦_𝑛)∈𝒞 is such that lim_𝑛→∞(𝜂_𝑛+‖𝑦_𝑛‖_𝑋)=∞,lim_𝑛→∞𝜂_𝑛=∞, (3.40) then (𝜆₁𝑟𝑓₀,𝜆₁𝑟𝑓_∞)⊂{𝜆∈(0,∞)∣(𝜆,𝑢)∈𝒞} (3.41) and, moreover, ({1}×𝑋)∩𝒞≠∅. (3.42) Assume that there exists 𝑀>0, such that for all 𝑛∈ℕ, 𝜂_𝑛∈(0,𝑀]. (3.43) Applying a similar argument to that used in Step 1 of Case 1, after taking a subsequence and relabelling if necessary, if follows that (𝜂_𝑛,𝑦_𝑛𝜆)→(¹𝑟𝑓_∞,∞),𝑛→∞. (3.44) Again 𝒞 joins (𝜆₁/𝑟𝑓₀,0) to (𝜆₁/𝑟𝑓_∞,∞) and the result follows.

Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. This paper is supported by the NSFC (no. 10671158), the NSF of Gansu Province (no. 3ZS051-A25-016), NWNUKJCXGC- 03-17, the Spring-sun program (no. Z2004-1-62033), SRFDP (no. 20060736001), and the SRF for ROCS, SEM (2006[311]).