Spectrum of Linear Difference Operators and the Solvability of Nonlinear Discrete Problems

Abstract

Let 𝑇∈ℕ with 𝑇>5. Let 𝕋∶={1,…,𝑇}. We study the Fučik spectrum Σ of the discrete problem Δ²𝑢(𝑡−1)+𝜆𝑢⁺(𝑡)−𝜇𝑢⁻(𝑡)=0, 𝑡∈𝕋, 𝑢(0)=𝑢(𝑇+1)=0, where 𝑢⁺(𝑡)=max{𝑢(𝑡),0}, 𝑢⁻(𝑡)=max{−𝑢(𝑡),0}. We give an expression of Σ via the matching-extension method. We also use such discrete spectrum theory to study nonlinear boundary value problems of difference equations at resonance.

1. Introduction

Let 𝑇∈ℕ with 𝑇>5. Let 𝕋∶={1,…,𝑇} and 𝕋∶={0,1,…,𝑇,𝑇+1}. For 𝑢∶𝕋→ℝ, we define 𝑢⁺,𝑢⁻∶𝕋→ℝ by 𝑢⁺(𝑡)=max{𝑢(𝑡),0},𝑢⁻(𝑡)=max{−𝑢(𝑡),0}. (1.1) The Fučik spectrum of the problem Δ²𝑢(𝑡−1)+𝜆𝑢⁺(𝑡)−𝜇𝑢⁻(𝑡)=0,𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0 (1.2) is defined as the set Σ of those (𝜆,𝜇) such that (1.2) has nontrivial solutions.

In the past thirty years, the Fučik spectrum of two-point boundary value problems of ordinary differential equations has been extensively studied, see Fučik [1, 2], Ruf [3], and Rynne [4] and references therein. A typical result is the following.

Theorem 1.1. The problem 𝑢^′′(𝑥)+𝜆𝑢⁺(𝑥)−𝜇𝑢⁻(𝑥)=0,𝑥∈(0,1),𝑢(0)=𝑢(1)=0 (1.3) has a nontrivial solution for (𝜆,𝜇)∈ℝ² if and only if (𝜆,𝜇) satisfies(i)for 𝑘≥1 odd, either 𝑘+121√𝜆+𝑘−121√𝜇=1, (1.4) or 𝑘−121√𝜆+𝑘+121√𝜇=1, (1.5)(ii)for 𝑘≥2 even 𝑘21√𝜆+𝑘21√𝜇=1. (1.6)

Of course, the natural question is whether a similar result can be established for the discrete analogue (1.2). The purpose of this paper is to study the structure of Fučik spectrum Σ for (1.2). We give an expression of Σ via the matching-extension method in Sections 2 and 3. In Sections 4 and 5, we also show that Σ contains a curve Σ∩proj_ℝ2𝒮⁺₂ (for the definition of proj_ℝ2𝒮⁺₂, see Definition 5.1 below), which is continuous, strictly decreasing, symmetric with respect to the diagonal, and for each (𝜆,𝜇)∈Σ∩proj_ℝ2𝒮⁺₂, 𝑢 has exactly one simple generalized zero in the interval (1,𝑇). Finally, in Section 6, we apply our Fučik spectrum of (1.2) to study the solvability of nonlinear problem Δ²𝑢(𝑡−1)+𝜆₁𝑢(𝑡)+𝑔(𝑡,𝑢(𝑡))=𝑓(𝑡),𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0, (1.7) where 𝑔∶𝕋×ℝ→ℝ is continuous, 𝑓∶𝕋→ℝ, 𝜆₁ is the first eigenvalue of the linear problem Δ²𝑢(𝑡−1)+𝜆𝑢(𝑡)=0,𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0. (1.8)

2. Matching Continuation

For 𝑟∈(0,∞), we denote the integer part of 𝑟 by [𝑟].

Denote ℕ^∗={0}∪ℕ. (2.1)

Lemma 2.1. For fixed 𝜆,𝜇∈ℝ, the initial value problem Δ²𝑢(𝑡−1)+𝜆𝑢⁺(𝑡)−𝜇𝑢⁻(𝑡)=0,𝑡∈ℕ,𝑢(0)=0,𝑢(1)=1 (2.2)(2.3) has a unique solution 𝑤₊∶ℕ^∗→𝑅.

Proof. Equation (2.2) is equivalent to the recurrence 𝑢(𝑡+1)=2𝑢(𝑡)−𝑢(𝑡−1)−𝜆𝑢⁺(𝑡)+𝜇𝑢⁻(𝑡),𝑡∈ℕ, (2.4) which guarantees the existence and uniqueness of solution 𝑤₊∶ℕ^∗→𝑅.

Lemma 2.2. Let 𝑤₊(𝑡) be a nontrivial solution of the initial value problem (2.2), (2.3) and 𝑤₊(𝑡₀)=0 for some 𝑡₀∈𝕋. Then, 𝑤₊(𝑡₀−1)𝑤₊(𝑡₀+1)<0. (2.5)

Proof. It can be easily deduced from (2.4).

Lemma 2.3. Let 𝑤₊(𝑡,𝜆,𝜇) be the unique solution of (2.2), (2.3) for fixed 𝜆,𝜇∈ℝ. Then, (𝜆,𝜇) is an eigenvalue of (1.2) if and only if 𝑤₊(𝑇+1,𝜆,𝜇)=0. (2.6)

To find (𝜆,𝜇) satisfying (2.6), we consider the linear initial value problem Δ²𝑢(𝑡−1)+𝜆𝑢(𝑡)=0,𝑡∈ℕ,𝑢(0)=0,𝑢(1)=1. (2.7)(2.8) The characteristic equation of (2.7) is 𝑟²+(𝜆−2)𝑟+1=0, (2.9) so 𝑟=2−𝜆±(𝜆−2)²−42. (2.10)

If |𝜆−2|>2, then it can be shown that there are no eigenvalues of (1.2).

In fact, in this case, 𝑟₁∶=2−𝜆+(𝜆−2)²−42,𝑟₂∶=2−𝜆−(𝜆−2)²−42, (2.11)

and the general solution of (2.7) is 𝑢(𝑡)=𝐴𝑟^𝑡₁+𝐵𝑟^𝑡₂. (2.12)

It is easy to check that the function 𝐴𝑟^𝑥₁+𝐵𝑟^𝑥₂,𝑥∈ℝ has at most one zero in ℝ for every 𝐴,𝐵∈ℝ with 𝐴²+𝐵²≠0.

If |𝜆−2|=2, then it can also be shown that there are no eigenvalues of (1.2).

In fact, |𝜆−2|=2 implies that 𝑟₁=𝑟₂∶=2−𝜆2, (2.13)

and the general solution of (2.7) is 𝑢(𝑡)=𝐴𝑟^𝑡₁+𝐵𝑡𝑟^𝑡₁. (2.14)

It is easy to check that the function 𝐴𝑟^𝑥₁+𝐵𝑥𝑟^𝑥₁,𝑥∈ℝ has at most one zero in ℝ for every 𝐴,𝐵∈ℝ with 𝐴²+𝐵²≠0.

Assume that |𝜆−2|<2 and set 2−𝜆=2cos𝜃. (2.15) Then, 𝑟=cos𝜃±𝑖sin𝜃=𝑒^±𝑖𝜃. (2.16) Hence, a general solution of (2.7) is 𝑢(𝑡)=𝐴cos𝜃𝑡+𝐵sin𝜃𝑡. (2.17) From (2.8), it follows that 𝑢(0)=𝐴=0 and 1𝐵=.sin𝜃 (2.18) Since |𝜆−2|<2, it follows from (2.15) that 𝜃≠𝑘𝜋. Thus, 𝐵 is well defined and 1𝑢(𝑡)=sin𝜃sin𝜃𝑡,𝑡∈ℕ^∗. (2.19)

Lemma 2.4. The first eigenvalue of the problem (1.8) is 𝜆₁𝜋=2−2cos,𝑇+1 (2.20) which is simple and the eigenfunction corresponding to 𝜆₁ is 𝜑₁𝜋(𝑡)=sin𝑇+1𝑡,𝑡∈𝕋. (2.21)

Proof. Applying the same method used in [5, Example 7.1], with obvious changes, we can get the desired result.

Lemma 2.5. (i) The line {𝜆₁}×ℝ and the set ℝ×{𝜆₁} are contained in Σ.
(ii) Σ is symmetric with respect to the straight line 𝜆=𝜇.

Proof. (i) From the definition of 𝜆₁ and 𝜑₁ in (2.20) and (2.21), it follows that for any 𝜇∈ℝ, Δ²𝜑₁(𝑡−1)+𝜆₁(𝜑₁)⁺(𝑡)−𝜇(𝜑₁)⁻𝜑(𝑡)=0,𝑡∈𝕋,₁(0)=𝜑₁(𝑇+1)=0. (2.22) This implies that {𝜆₁}×ℝ⊂Σ. Similarly, ℝ×{𝜆₁}⊂Σ.
(ii) Let (𝜆,𝜇)∈Σ and 𝑦 be the corresponding eigenfunction. Then, Δ²𝑦(𝑡−1)+𝜆𝑦⁺(𝑡)−𝜇𝑦⁻(𝑡)=0,𝑡∈𝕋,𝑦(0)=𝑦(𝑇+1)=0. (2.23) Let 𝑧(𝑡)=−𝑦(𝑡) for 𝑡∈𝕋. Then, Δ²𝑧(𝑡−1)+𝜇𝑧⁺(𝑡)−𝜆𝑧⁻(𝑡)=0,𝑡∈𝕋,𝑧(0)=𝑧(𝑇+1)=0, (2.24) which means that (𝜇,𝜆)∈Σ.

In order to construct a nontrivial solution of (2.2), (2.3), we extend 𝑢 to the following function defined on real interval [0,∞): 1𝑢(𝑥)=sin𝜃sin𝜃𝑥,𝑥∈[0,∞). (2.25)

Let 𝛽₁ be the first positive zero of 𝑢, then 𝑢(𝛽₁)=0,𝑢(𝑥)>0,𝑥∈(0,𝛽₁),𝑢(𝑥)=𝐵₁𝜋sin𝛽₁𝑥, (2.26)(2.27) with 𝐵₁∶=(sin(𝜋/𝛽₁))⁻¹. From (2.27) and (2.15), we have that 𝜋𝜆=2−2cos𝛽₁, (2.28) that is, 𝛽₁=𝜋.arccos((2−𝜆)/2) (2.29) Moreover, 𝑢(𝑥) satisfies that Δ²𝑢(𝑡−1)+𝜆𝑢(𝑡)=0,𝑡∈{1,2,…,[𝛽₁]},𝑢(0)=0,𝑢(𝛽₁)=0,𝑢(𝑡)>0,𝑡∈{1,…,[𝛽₁−1]},𝑢([𝛽₁])>0,if𝛽₁>[𝛽₁],𝑢([𝛽₁])=0,if𝛽₁=[𝛽₁],𝑢([𝛽₁]+1)<0, (2.30) where [𝑟] denotes the integer part of 𝑟.

Definition 2.6. Suppose that a function 𝑦∶𝕋→ℝ. If 𝑦(𝑡)=0, then 𝑡 is a zero of 𝑦. If 𝑦(𝑡)=0 and 𝑦(𝑡−1)𝑦(𝑡+1)<0, then 𝑡 is a simple zero of 𝑦. If 𝑦(𝑡−1)𝑦(𝑡)<0, then we say that 𝑦 has a node at the point 𝑠=𝑡. The nodes and simple zeros of 𝑦 are called the simple generalized zeros of 𝑦.

Lemma 2.7. Let Σ=Σ⧵(({𝜆₁}×ℝ)∪(ℝ×{𝜆₁})). (2.31) Then, (𝜆,𝜇)∈Σ implies that 𝜋2−2cos𝑇𝜋<𝜆<4,2−2cos𝑇<𝜇<4. (2.32)

Proof. From (2.20) and (2.28), we see that (𝜆,𝜇)∈Σ if and only if the eigenfunction corresponding to (𝜆,𝜇) has at least simple generalized zero in the open interval (1,𝑇). This means that 1<𝛽₁<𝑇, which implies 𝜋2−2cos𝑇<𝜆<4. (2.33) Similarly, we have that 2−2cos(𝜋/𝑇)<𝜇<4.

Now, let (𝜆,𝜇) satisfy(H)2−2cos(𝜋/𝑇)<𝜆<4,2−2cos(𝜋/𝑇)<𝜇<4.

Let us consider the sequence {𝑢(𝑡,𝜆,𝜇)} induced by (2.4), (2.3), and let 𝛽₁ be the first positive zero of 𝑢 (see (2.27) for the definition of 𝑢). Then, we see from (2.4) that 𝑢([𝛽₁]+1)=2𝑢([𝛽₁])−𝑢([𝛽₁]−1)−𝜆𝑢([𝛽₁]), (2.34) which means that 𝑢([𝛽₁]+1) is uniquely determined by 𝑢([𝛽₁]), 𝑢([𝛽₁]−1) and 𝜆, and is independent of 𝜇.

Now, let 𝑣(𝑡) be the unique solution of the initial value problem Δ²𝑣(𝑡−1)+𝜇𝑣(𝑡)=0,𝑡∈{[𝛽₁]−1,[𝛽₁],[𝛽₁]+1,…},𝑣([𝛽₁])=𝑢([𝛽₁]),𝑣([𝛽₁]+1)=𝑢([𝛽₁]+1). (2.35)(2.36) By a similar method to get 𝑢 and 𝛽₁, we get the following lemma.

Lemma 2.8. Let (H) hold, and let 𝑏 satisfy 𝜋𝜇=2−2cos𝑏. (2.37) Then, 1<𝑏<𝑇.

Proof. From (2.37) and (H), it follows that 𝜋2−2cos𝑇𝜋<2−2cos𝑏<4. (2.38) This is 𝜋cos𝑇𝜋>𝑐𝑜𝑠𝑏𝜋>𝑐𝑜𝑠1, (2.39) which implies 1<𝑏<𝑇.

Lemma 2.9. Let (H) hold. Then, (2.35), (2.36) has a solution of the form 𝑣(𝑡)=𝐵₂sin𝜋(𝑡−𝛼₂)𝑏, (2.40) where 𝛼₂ is the unique solution of 𝐵₁sin(𝜋[𝛽₁]𝛽₁)=𝐵₂sin𝜋([𝛽₁]−𝛼₂)𝑏,𝐵₁sin(𝜋([𝛽₁]+1)𝛽₁)=𝐵₂sin𝜋([𝛽₁]+1−𝛼₂)𝑏, (2.41) in ([𝛽₁],[𝛽₁+1]) if 𝛽₁>[𝛽₁], and 𝛼₂=𝛽₁ if 𝛽₁=[𝛽₁]; 𝐵₂ is a negative constant satisfying 𝐵₂sin𝜋([𝛽₁]+1−𝛼₂)𝑏=2𝐵₁sin𝜋([𝛽₁])𝛽₁−𝐵₁sin𝜋([𝛽₁]−1)𝛽₁−𝜆𝐵₁sin𝜋([𝛽₁])𝛽₁. (2.42)

Proof. The relation (2.40) can be deduced by the similar method to get (2.27).
If 𝛽₁>[𝛽₁], then we define a function 𝐹∶[[𝛽₁],[𝛽₁+1]]→ℝ by 𝜋𝐹(𝛼)=sin𝛽₁([𝛽₁])sin𝜋([𝛽₁]+1−𝛼)𝑏−sin𝜋([𝛽₁]−𝛼)𝑏sin𝜋([𝛽₁]+1)𝛽₁. (2.43) It is easy to check that 𝐹([𝛽₁+1])=−sin𝜋([𝛽₁]−[𝛽₁+1])𝑏𝜋sin𝛽₁([𝛽₁+1])<0,𝐹([𝛽₁𝜋])=sin𝛽₁([𝛽₁])sin𝜋([𝛽₁+1]−[𝛽₁])𝑏>0. (2.44) Moreover, for 𝛼∈([𝛽₁],[𝛽₁+1]), 𝜕𝐹𝜋𝜕𝛼(𝛼)=−𝑏𝜋sin𝛽₁([𝛽₁])cos𝜋([𝛽₁+1]−𝛼)𝑏+𝜋𝑏cos𝜋([𝛽₁]−𝛼)𝑏𝜋sin𝛽₁([𝛽₁+1])<0, (2.45) (here, we use the fact that 𝑏>1, see Lemma 2.8). So, there is a unique 𝛼₂∈([𝛽₁],[𝛽₁+1]), such that 𝐹(𝛼₂)=0.
If 𝛽₁=[𝛽₁], then 𝑣(𝛽₁)=0. This together with (2.4) imply that 𝑣(𝛽₁+1)=−𝑣(𝛽₁−1), (2.46) that is, 𝐵₂sin𝜋(𝛽₁+1−𝛼₂)𝑏=−𝐵₂sin𝜋(𝛽₁−1−𝛼₂)𝑏. (2.47) So, we may take that 𝛼₂=𝛽₁. (2.48)
Finally, it follows from (2.4) and the fact 𝑢([𝛽₁]+1)=𝑣([𝛽₁]+1) (see (2.36)) that 𝐵₂ satisfies (2.42).

Now, for 𝑥∈ℝ, let 𝑣(𝑥)∶=𝐵₂sin𝜋(𝑥−𝛼₂)𝑏, (2.49) which is an extension of the function 𝑣∶(ℕ^∗∩[[𝛽₁−1],∞))→ℝ defined in (2.40).

Denote 𝛽₂∶=𝛼₂+𝑏, (2.50)

Then 𝑣(𝑥) satisfies Δ²𝑣(𝑡−1)+𝜇𝑣(𝑡)=0,𝑡∈{[𝛽₁+1],…,[𝛽₂]},𝑣([𝛽₁])=𝑢([𝛽₁]),𝑣([𝛽₁+1])=𝑢([𝛽₁+1]),𝑣(𝛽₂)=0,𝑣(𝑡)<0,𝑡∈{[𝛽₁+1],…,[𝛽₂−1]},𝑣([𝛽₂])<0,if𝛽₂>[𝛽₂],𝑣([𝛽₂])=0,if𝛽₂=[𝛽₂],𝑣([𝛽₂]+1)>0. (2.51)

Definition 2.10. We say that 𝛼₂ defined in Lemma 2.8 is the feasible initial phase of the family of functions sin(𝜋(𝑥−𝛼)/𝑏). If 𝛼₂ is the feasible initial phase of the family of functions sin(𝜋(𝑥−𝛼)/𝑏), then we say that 𝑣(𝑥)𝑣(𝑡)∶=𝐵₂sin𝜋(𝑡−𝛼₂)𝑏,𝑡∈[[𝛽₁],[𝛽₂+1]] (2.52) is the matching extension of 𝑢(𝑥),𝑥∈[0,[𝛽₁+1]].

Remark 2.11. Notice that 𝑣(𝑥)≢𝑢(𝑥) for 𝑥∈[[𝛽₁],[𝛽₁+1]] in general. However, 𝑣(𝑡)≡𝑢(𝑡) for 𝑡∈{[𝛽₁],[𝛽₁+1]}.

Obviously, we may repeat the above process to “extend” 𝑢(𝑡) to a sequence defined on ℕ in a unique manner. Moreover, this sequence coincides with the sequence 𝑤₊(𝑡) in Lemma 2.1.

Let 𝛼₁∶=0. Recall that 𝛽₁ is defined by (2.29)𝛽₁=𝜋,arccos((2−𝜆)/2) (2.53) and 𝛼₂=𝛽₁ if 𝛽₁=[𝛽₁], and 𝛼₂ is the unique solution of 𝜋sin𝛽₁([𝛽₁])sin𝜋([𝛽₁]+1−𝛼)𝑏−sin𝜋([𝛽₁]−𝛼)𝑏sin𝜋([𝛽₁]+1)𝛽₁=0. (2.54) Recall 𝛽₂∶=𝛼₂+𝑏, (2.55) with 𝜋𝑏=,arccos((2−𝜇)/2) (2.56) Repeating the above process, we may take 𝛼_𝑘=𝛽_𝑘−1 if 𝛽_𝑘−1=[𝛽_𝑘−1], and 𝛼_𝑘 is the unique solution of 𝐹_𝑘(𝛼)=0, (2.57) where 𝐹_𝑘(𝛼)∶=sin𝜋([𝛽_𝑘−1]−𝛼_𝑘−1)𝛽_𝑘−1−𝛼_𝑘−1sin𝜋([𝛽_𝑘−1+1]−𝛼)𝛽_𝑘−𝛼_𝑘−sin𝜋([𝛽_𝑘−1]−𝛼)𝛽_𝑘−𝛼_𝑘sin𝜋([𝛽_𝑘−1+1]−𝛼_𝑘−1)𝛽_𝑘−1−𝛼_𝑘−1. (2.58) Put 𝛽_𝑘=𝛼_𝑘+𝛽₁𝛼if𝑘isodd,_𝑘+𝑏if𝑘iseven. (2.59)

Recall 𝐵₁𝜋∶=(sin𝛽₁)⁻¹, (2.60) and define 𝐵_𝑘sin𝜋([𝛽_𝑘−1]+1−𝛼_𝑘)𝛽_𝑘−𝛼_𝑘=2𝐵_𝑘−1sin𝜋([𝛽_𝑘−1])𝛽_𝑘−1−𝛼_𝑘−1−𝐵_𝑘−1sin𝜋([𝛽_𝑘−1]−1)𝛽_𝑘−1−𝛼_𝑘−1−𝜈𝐵_𝑘−1sin𝜋([𝛽_𝑘−1])𝛽_𝑘−1−𝛼_𝑘−1. (2.61) with 𝜈=𝜇if𝑘isodd,𝜆if𝑘iseven. (2.62)

Finally, it is easy to check that 𝐹_𝑘([𝛽_𝑘−1])>0,𝐹_𝑘([𝛽_𝑘−1+1])<0,𝜕𝐹_𝑘𝜋𝜕𝛼(𝛼)=−𝛽_𝑘−𝛼_𝑘sin𝜋([𝛽_𝑘−1]−𝛼_𝑘−1)𝛽_𝑘−1−𝛼_𝑘−1cos𝜋([𝛽_𝑘−1+1]−𝛼)𝛽_𝑘−𝛼_𝑘+𝜋𝛽_𝑘−𝛼_𝑘cos𝜋([𝛽_𝑘−1]−𝛼)𝛽_𝑘−𝛼_𝑘sin𝜋([𝛽_𝑘−1+1]−𝛼_𝑘−1)𝛽_𝑘−1−𝛼_𝑘−1<0. (2.63) So, there exists a unique 𝛼_𝑘∈([𝛽_𝑘−1],[𝛽_𝑘−1+1]) which satisfies (2.57).

To sum up, we may define Γ_𝑘(𝑥)∶=𝐵_𝑘sin𝜋(𝑥−𝛼_𝑘)𝛽_𝑘−𝛼_𝑘,𝑥∈[[𝛽_𝑘−1],[𝛽_𝑘+1]], (2.64) which can be thought as the “matching extension” of Γ_𝑘−1. Notice that 𝛼_𝑘 is the left zero of Γ_𝑘 on [[𝛽_𝑘−1],[𝛽_𝑘+1]], and 𝛽_𝑘 is the right zero of Γ_𝑘 on [[𝛽_𝑘−1],[𝛽_𝑘+1]].

Clearly, we may start the discuss from the IVP Δ²𝑢(𝑡−1)+𝜆𝑢(𝑡)=0,𝑡∈𝕋,𝑢(0)=0,𝑢(1)=−1 (2.65) which has unique solution 𝑤₋(𝑡,𝜆,𝜇). Then, we may uniquely determine the 𝑘th segmental arc Γ_𝑘 on [[𝛿_𝑘−1],[𝛿_𝑘+1]] and the feasible initial phase 𝛾_𝑘 via matching extension method, and accordingly, we get the left zero 𝛾_𝑘 and the right zero 𝛿_𝑘 of Γ_𝑘 on [[𝛿_𝑘−1],[𝛿_𝑘+1]].

3. The Main Result

The main result of this paper is the following discrete analogue of Theorem 1.1.

Theorem 3.1. (𝜆,𝜇)∈ℝ² is an eigenvalue of (1.2) if and only if (𝜆,𝜇) satisfies(i) for 𝑘≥2 even 𝑘2𝜋+𝑘arccos((2−𝜆)/2)2𝜋+arccos((2−𝜇)/2)^𝑘−1_𝑗=1(𝛼_𝑗+1−𝛽_𝑗)=𝑇+1 (3.1)(ii)for 𝑘≥1 odd, either 𝑘+12𝜋+arccos((2−𝜆)/2)𝑘−12𝜋+arccos((2−𝜇)/2)^𝑘−1_𝑗=1(𝛼_𝑗+1−𝛽_𝑗)=𝑇+1 (3.2) or 𝑘−12𝜋+arccos((2−𝜆)/2)𝑘+12𝜋+arccos((2−𝜇)/2)^𝑘−1_𝑗=1(𝛾_𝑗+1−𝛿_𝑗)=𝑇+1. (3.3)

Proof. (i) Let 𝑘≥1 be even. Let Γ_𝑚∶={(𝑥,𝐵₁𝜋sin𝛽₁(𝑥−𝛼_𝑚))∣𝑥=[𝛼_𝑚],…,[𝛽_𝑚𝑘+1]},𝑚=2𝑗−1,𝑗∈{1,…,2},{(𝑥,𝐵₂𝜋sin𝑏(𝑥−𝛼_𝑚))∣𝑥=[𝛼_𝑚],…,[𝛽_𝑚+1]},𝑚=2𝑗,𝑗∈{1,…,𝑘−22},{(𝑥,𝐵₂𝜋sin𝑏(𝑥−𝛼_𝑚))∣𝑥=[𝛼_𝑚],…,[𝛽_𝑚]},𝑚=𝑘. (3.4) Here, 𝛽₁=𝜋𝜋arccos((2−𝜆)/2),𝑏=.arccos((2−𝜇)/2) (3.5) Then, a nontrivial solution of (1.2) can be constructed via matching extension method to Γ₁ and Γ₂, …, Γ_𝑘−1 and Γ_𝑘, respectively. Moreover, 𝑘𝑇+1=2𝜋+𝑘arccos((2−𝜆)/2)2𝜋+arccos((2−𝜇)/2)^𝑘−1_𝑗=1(𝛼_𝑗+1−𝛽_𝑗). (3.6)
(ii) The case that 𝑘 is odd can be treated by the similar way.

Example 3.2. We may use Mathematica 5.0 to give a numerical example for Theorem 3.1.
Take 𝑘=2,𝑇=3.
Take 𝛽₁=2.2, then 𝜆≐1.7153703202231987.
Take 𝑏=1.8198598, then 𝜇≐2.3097214213167683.
Using the relation 𝜋sin𝛽₁([𝛽₁])sin𝜋([𝛽₁+1]−𝛼)𝑏=sin𝜋([𝛽₁]−𝛼)𝑏𝜋sin𝛽₁([𝛽₁+1]), (3.7) we may find 𝛼₂≐2.1801402269418726. (3.8) Since 2.2+1.8198598+(2.1801402269418726−2.2)≐4, (3.9) we may think that (1.7153703202231987,2.3097214213167683) is an eigenvalue of the problem Δ²𝑢(𝑡−1)+𝜆𝑢⁺(𝑡)−𝜇𝑢⁻(𝑡)=0,𝑡∈{1,2,3},𝑢(0)=𝑢(4)=0. (3.10) Now, by the recurrence relation 𝑢(𝑡+1)=2𝑢(𝑡)−𝑢(𝑡−1)−1.7153703202231987𝑢⁺(𝑡)+2.309739789282126𝑢⁻(𝑡),𝑢(0)=0,𝑢(1)=1, (3.11) it is easily to compute that 𝑢(4)≐−4.6600427960896695×10⁻⁸≈0. (3.12) This is a desired numerical result.

4. Some Properties of 𝛼₂

Proposition 4.1. If 𝜆=𝜇 and 𝛽_𝑖=[𝛽_𝑖], then 𝛽_𝑗=𝛼_𝑗+1 for 𝑗=1,2,….

Proof. In fact, if 𝜆=𝜇 and 𝛽_𝑖=[𝛽_𝑖], then 𝛽₁𝜋=𝑏=,arccos((2−𝜆)/2) (4.1) which implies that 𝛼_𝑚,𝛽_𝑚∈ℕ^∗. (4.2)

Proposition 4.2. Let (H) hold. Assume that 𝜆 is such that 𝛽₁=[𝛽₁1]+2. (4.3) Then, for every 𝜇>1, we have that 𝛼₂=[𝛽₁1]+2. (4.4)

Proof. By 𝛽₁=[𝛽₁]+1/2, it follows that sin𝜋[𝛽₁]𝛽₁=−sin𝜋([𝛽₁]+1)𝛽₁. (4.5) This together with 𝐹(𝛼₂)=0 imply that sin𝜋([𝛽₁]−𝛼₂)𝛽₂−𝛼₂=−sin𝜋([𝛽₁]+1−𝛼₂)𝛽₂−𝛼₂, (4.6) and accordingly, sin𝜋([𝛽₁]−𝛼₂)𝛽₂−𝛼₂+sin𝜋([𝛽₁]+1−𝛼₂)𝛽₂−𝛼₂=2sin𝜋([𝛽₁]+1/2−𝛼₂)𝛽₂−𝛼₂cos𝜋/2𝛽₂−𝛼₂=0. (4.7) By (H), we have that 𝛽₂−𝛼₂>1. Therefore, [𝛽₁1]+2−𝛼₂=0. (4.8)

Proposition 4.3. Let (H) hold. Then,(i)𝛽₁>[𝛽₁]+1/2 implies [𝛽₁]+1/2<𝛼₂<𝛽₁,(ii)𝛽₁<[𝛽₁]+1/2 implies [𝛽₁]+1/2>𝛼₂>𝛽₁.

Proof. (i) By the definition of 𝐹, (see (2.43)), we have that 𝐹(𝛽₁𝜋)=sin𝛽₁([𝛽₁])sin𝜋([𝛽₁+1]−𝛽₁)𝑏−sin𝜋([𝛽₁]−𝛽₁)𝑏sin𝜋([𝛽₁+1])𝛽₁. (4.9) Since 𝛽₁>[𝛽₁]+1/2, we have that 𝜋sin(𝛽₁[𝛽₁𝜋])>−sin(𝛽₁[𝛽₁+1]), (4.10) which together with (4.9) implies that 𝐹(𝛽₁𝜋)≤sin𝛽₁([𝛽₁])[sin𝜋([𝛽₁+1]−𝛽₁)𝑏+sin𝜋([𝛽₁]−𝛽₁)𝑏]<0. (4.11) Since [𝛽₁+1]−𝛽₁<𝛽₁−[𝛽₁]. Thus, 𝐹(𝛽₁)<𝐹(𝛼₂)=0. (4.12) From the fact that 𝐹 is decreasing on ([𝛽₁],[𝛽₁]+1), it follows that 𝛽₁>𝛼₂.
On the other hand, we have from (2.43) that 𝐹([𝛽₁1]+2𝜋)=sin𝛽₁([𝛽₁𝜋])sin𝜋2𝑏+sin𝜋2𝑏sin𝛽₁([𝛽₁+1]). (4.13) Combining this with the fact that [𝛽₁+1]−𝛽₁<𝛽₁−[𝛽₁], it concludes that 𝐹([𝛽₁]+1/2)>0. So, [𝛽₁]+1/2<𝛼₂.
(ii)𝛽₁<[𝛽₁]+1/2 yields that sin(𝜋[𝛽₁]𝛽₁)<−sin(𝜋[𝛽₁+1]𝛽₁), (4.14) which together with (4.9) and the fact that [𝛽₁+1]−𝛽₁>𝛽₁−[𝛽₁] imply that 𝐹(𝛽₁𝜋)≥sin𝛽₁([𝛽₁])[sin𝜋([𝛽₁+1]−𝛽₁)𝑏+sin𝜋([𝛽₁]−𝛽₁)𝑏]>0. (4.15) Since 𝐹 is decreasing on ([𝛽₁],[𝛽₁]+1), it follow that 𝛽₁<𝛼₂.
Finally, we may use (4.13) and the fact [𝛽₁+1]−𝛽₁>𝛽₁−[𝛽₁] to deduce that 𝐹([𝛽₁1]+2)<𝐹(𝛼₂)=0, (4.16) which implies that [𝛽₁]+1/2>𝛼₂.

5. Some Properties of Σ∩proj_ℝ2𝒮⁺₂

Definition 5.1. For 𝑘∈ℕ and 𝜈∈{+,−}, let 𝑆^𝜈_𝑘 denote the set of functions 𝑢∶𝕋→ℝ satisfying(1)𝑢 has only simple generalized zeros in [1,𝑇];(2)𝑢 has exactly 𝑘−1 simple generalized zeros in [1,𝑇];(3)𝜈𝑢(1)>0.
Let 𝒮⁺₂∶=ℝ²×𝑆⁺₂. (5.1)

An immediate consequence of Lemma 2.2 is the following.

Lemma 5.2. If 𝑢 of a nontrivial solution of (1.2), then 𝑢∈𝑆^𝜈_𝑘 for some 𝑘∈ℕ and 𝜈∈{+,−}.

In the rest of this section, one will pay one’s attention to the study of Σ∩𝑆⁺₂.

Let (H) hold and define a function 𝐺∶(1,𝑇)×([𝛽₁],[𝛽₁]+1)→ℝ by 𝜋𝐺(𝑏,𝛼)=sin𝛽₁([𝛽₁])sin𝜋([𝛽₁+1]−𝛼)𝑏−sin𝜋([𝛽₁]−𝛼)𝑏𝜋sin𝛽₁([𝛽₁+1]). (5.2) Then, from Lemma 2.8, one has that for each 𝑏∈(𝜋/arccos((2−𝜇)/2),𝑇), there exists a unique 𝛼₂(𝑏)∈([𝛽₁],[𝛽₁+1]) such that 𝐺(𝑏,𝛼₂(𝑏))=0. (5.3)

Lemma 5.3. Let (H) hold. Then,(i)𝛼₂(𝑏) is continuous on (𝜋/arccos((2−𝜇)/2),𝑇),(ii)|𝑑𝛼₂/𝑑𝑏|<1/𝑏.

Proof. (i) From (5.2), we see that 𝐺(𝑏,𝛼) is continuous and 𝜕𝐺𝜋𝜕𝛼=−𝑏𝜋sin(𝛽₁[𝛽₁])cos𝜋([𝛽₁+1]−𝛼)𝑏+𝜋𝑏cos𝜋([𝛽₁]−𝛼)𝑏𝜋sin(𝛽₁[𝛽₁+1]). (5.4) This together with the facts that 𝛼∈([𝛽₁],[𝛽₁+1]) and 𝑏>1 imply that 𝜕𝐺/𝜕𝛼₂<0. So, the implicit function theorem yields that 𝛼₂(𝑏) is continuous on (𝜋/arccos((2−𝜇)/2),𝑇).
(ii) From (5.2), 𝜕𝐺𝜕𝑏=−𝜋([𝛽₁+1]−𝛼)𝑏²𝜋sin(𝛽₁[𝛽₁])cos𝜋([𝛽₁+1]−𝛼)𝑏+𝜋([𝛽₁]−𝛼)𝑏²cos𝜋([𝛽₁]−𝛼)𝑏𝜋sin(𝛽₁[𝛽₁+1]). (5.5) Denote 𝐴(𝑏,𝛼₂𝜋)=sin(𝛽₁[𝛽₁])cos𝜋([𝛽₁+1]−𝛼₂)𝑏,𝐵(𝑏,𝛼₂)=cos𝜋([𝛽₁]−𝛼₂)𝑏𝜋sin(𝛽₁[𝛽₁+1]). (5.6) Then, 𝐴(𝑏,𝛼₂)>0,𝐵(𝑏,𝛼₂)<0, (5.7) and subsequently, |𝑑𝛼₂𝑑𝑏(𝑏)|=|−𝜕𝐺(𝑏,𝛼₂)/𝜕𝑏𝜕𝐺(𝑏,𝛼₂)/𝜕𝛼₂|=|−−(𝜋([𝛽₁+1]−𝛼₂)/𝑏²)𝐴(𝑏,𝛼₂)+(𝜋([𝛽₁]−𝛼₂)/𝑏²)𝐵(𝑏,𝛼₂)−(𝜋/𝑏)𝐴(𝑏,𝛼₂)+(𝜋/𝑏)𝐵(𝑏,𝛼₂)|=|(𝜋([𝛽₁+1]−𝛼₂)/𝑏²)𝐴(𝑏,𝛼₂)+(𝜋([𝛽₁]−𝛼₂)/𝑏²)(−𝐵(𝑏,𝛼₂))−(𝜋/𝑏)𝐴(𝑏,𝛼₂)−(𝜋/𝑏)(−𝐵(𝑏,𝛼₂|))<|(𝜋([𝛽₁+1]−𝛼₂)/𝑏²)𝐴(𝑏,𝛼₂)−(𝜋([𝛽₁]−𝛼₂)/𝑏²)(−𝐵(𝑏,𝛼₂))−(𝜋/𝑏)𝐴(𝑏,𝛼₂)−(𝜋/𝑏)(−𝐵(𝑏,𝛼₂|))=|(𝜋([𝛽₁+1]−𝛼₂)/𝑏²)𝐴(𝑏,𝛼₂)+(𝜋(−[𝛽₁]+𝛼₂)/𝑏²)(−𝐵(𝑏,𝛼₂))−(𝜋/𝑏)𝐴(𝑏,𝛼₂)−(𝜋/𝑏)(−𝐵(𝑏,𝛼₂|≤1))𝑏, (5.8) since max{𝜋([𝛽₁+1]−𝛼₂)/𝑏²,𝜋(−[𝛽₁]+𝛼₂)/𝑏²}≤𝜋/𝑏².

Theorem 5.4. Let (H) hold. Then, for each 𝜇∈(2−2cos(𝜋/𝑇),4), there exists at most one 𝜆∈(2−2cos(𝜋/𝑇),4) such that 𝜋+𝜋arccos((2−𝜆)/2)arccos((2−𝜇)/2)+(𝛼₂−𝛽₁)=𝑇+1. (5.9)

Proof. Set 𝑔(𝑏)∶=𝑏+(𝛼₂(𝑏)−𝛽₁). (5.10) Then, by Lemma 5.3, 𝑔^′(𝑏)=1+𝛼^′₂1(𝑏)>1−𝑏>0, (5.11) which implies that 𝑔 is strictly increasing in 𝑏, and accordingly, the function 𝜋𝑔(𝜇)∶=arccos((2−𝜇)/2)+(𝛼₂(𝜋arccos((2−𝜇)/2))−𝛽₁) (5.12) is strictly decreasing and continuous in 𝜇. Therefore, the function 𝜋𝜆=2−2cos()𝑇+1−𝑔(𝜇) (5.13) is strictly decreasing and continuous in 𝜇.

Remark 5.5. Let 𝜆₂ be the second eigenvalue of (2.18), (2.19). Then, (𝜆₂,𝜆₂)∈Σ∩proj_ℝ2𝒮⁺₂.

Since {𝜋𝜋arccos((2−𝜇)/2)∣𝜇∈(2−2cos𝑇,4)}=(1,𝑇), (5.14)

there exists an open interval 𝐼⊂(2−2cos(𝜋/𝑇),4) such that {𝜋arccos((2−𝜇)/2)∣𝜇∈𝐼}=(2,𝑇−1). (5.15)

Theorem 5.6. For every 𝜇∈𝐼, there exists a unique 𝜆∈(2−2cos(𝜋/𝑇),4), such that (5.9) holds.

Proof. Using the fact that −1<𝛼₂−𝛽₁<1, it concludes that for 𝜇∈𝐼, 𝜋1<arccos((2−𝜇)/2)+(𝛼₂(𝜋arccos((2−𝜇)/2))−𝛽₁)<𝑇,𝜇∈𝐼, (5.16) which implies that 𝜋𝑇>(𝑇+1)−{arccos((2−𝜇)/2)+(𝛼₂(𝜋arccos((2−𝜇)/2))−𝛽₁)}>1,𝜇∈𝐼. (5.17) Combining this with the fact that {𝜋𝜋arccos((2−𝜆)/2)∣𝜆∈(2−2cos𝑇,4)}=(1,𝑇), (5.18) it follows that (5.9) has at least one solution for every 𝜇∈𝐼. The uniqueness can be deduced from Theorem 5.4.

Remark 5.7. Espinoza [6] proved that the discrete analogue of Fučik spectrum of the Laplacian with Dirichlet boundary condition contains a curve which is continuous, strictly decreasing, and symmetric with respect to the diagonal. However, Espinoza [6] gave no information about the construction of the Fučik spectrum. Our main results, Theorems 5.4 and 3.1, are established for ordinary difference equation only. However, much more information about the construction of the Fučik spectrum is contained in these theorems.

6. Applications to Nonlinear Problems

Let 𝜆₁ be the first eigenvalue of (1.8). Then, 𝜆₁𝜋=2−2cos.𝑇+1 (6.1) The eigenfunction corresponding to 𝜆₁ is sin(𝜋𝑡/(𝑇+1)).

Let us consider the nonlinear problem Δ²𝑢(𝑡−1)+𝜆₁𝑢(𝑡)+𝑔(𝑡,𝑢(𝑡))=𝑓(𝑡),𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0, (6.2) where 𝑔 and 𝑓 satisfy(A1)𝑓∶𝕋→ℝ;(A2)𝑔∶𝕋×ℝ→ℝ is continuous, and there exist a function 𝑝∶𝕋→ℝ and a constant 𝑞>0, such that |𝑔(𝑡,𝑠)|≤𝑝(𝑡)+𝑞|𝑠|,𝑠∈ℝ,𝑡∈𝕋, (6.3)(A3) there exist a function functions 𝑎,𝐴∶𝕋→ℝ and constants 𝑟<0<𝑅, such that 𝑔(𝑡,𝑠)≥𝐴(𝑡),𝑠≥𝑅,𝑡∈𝕋,𝑔(𝑡,𝑠)≤𝑎(𝑡),𝑠≤𝑟,𝑡∈𝕋. (6.4) Obviously, (A3) implies liminf_𝑠→±∞𝑔(𝑡,𝑠)𝑠≥0. (6.5)

Denote 𝑔_+∞(𝑡)∶=liminf_𝑠→+∞𝑔(𝑡,𝑠),𝑔^−∞(𝑡)∶=limsup_{𝑠→−∞}𝑔(𝑡,𝑠). (6.6)

Lemma 6.1 (Sturm separation theorem; [7, Theorem 6.5]). Two linearly independent solutions of Δ²𝑦(𝑡−1)+𝑞(𝑡)𝑦(𝑡)=0,𝑡∈𝕋 (6.7) cannot have a common zero. If a nontrivial solution of (6.7) has a zero at 𝑡₁ and a generalized zeros at 𝑡₂>𝑡₁, then any second linear independent solution has a generalized zero in (𝑡₁,𝑡₂]. If a nontrivial solution of (6.7) has a generalized zero at 𝑡₁ and a generalized zero at 𝑡₂>𝑡₁, then any second linearly independent solution has a generalized zero in [𝑡₁,𝑡₂].

Definition 6.2 (see [7, Definition 6.3]). Let 𝑞∶𝕋→ℝ. We say that (6.7) is disconjugate on 𝕋 provided that no nontrivial solution of (6.7) has two or more generalized zeros on 𝕋.

Lemma 6.3 (Sturm comparison theorem; [7, Theorem 8.12]). Assume that 𝑞₁(𝑡)≥𝑞₂(𝑡) on 𝕋. If Δ²𝑦(𝑡−1)+𝑞₁(𝑡)𝑦(𝑡)=0 is disconjugate on 𝕋, then Δ²𝑦(𝑡−1)+𝑞₂(𝑡)𝑦(𝑡)=0 is disconjugate on 𝕋.

Let us denote by 𝐷 the Hilbert space 𝐷∶={𝑢∶𝕋→ℝ}, (6.8)

with the inner product ⟨𝑢,𝑣⟩_𝐷∶=^𝑇_𝑡=1𝑢(𝑡)𝑣(𝑡),𝑢,𝑣∈𝐷, (6.9)

and the norm ‖𝑢‖_𝐷∶=⟨𝑢,𝑣⟩_𝐷^1/2,𝑢,𝑣∈𝐷. (6.10)

Let us denote by 𝐷^∗ the Hilbert space 𝐷^∗∶={𝑢∶𝕋→ℝ∣𝑢(0)=𝑢(𝑇+1)=0}, (6.11)

with the inner product ⟨𝑢,𝑣⟩_𝐷^∗∶=^𝑇_𝑡=1𝑢(𝑡)𝑣(𝑡),𝑢,𝑣∈𝐷^∗, (6.12)

and the norm ‖𝑢‖_𝐷∶=⟨𝑢,𝑣⟩_𝐷^1/2_∗,𝑢,𝑣∈𝐷. (6.13)

For 𝑢∈𝐷, define 𝑗∶𝐷→𝐷^∗ by 𝑗(𝑢(1),𝑢(2),…,𝑢(𝑇))=(0,𝑢(1),𝑢(2),…,𝑢(𝑇),0), (6.14)

then 𝑗 is a natural isomorphism.

Obviously, ⟨𝑢,𝑣⟩_𝐷=⟨𝑗𝑢,𝑗𝑣⟩_𝐷∗,(𝑢,𝑣)∈𝐷. (6.15) So, in the rest of the paper, one will use ⟨𝑢,𝑣⟩, ‖𝑢‖ to denote the inner product and the norm in 𝐷 (or 𝐷^∗), respectively.

Let 𝑢(𝑡)=^𝑇_𝑗=1𝑎_𝑗sin𝑗𝜋𝑡𝑇+1 (6.16)

be the fourier series of 𝑢∈𝐷^∗. Then, one will write 𝑢(𝑡)=𝑢⁰(𝑡)+𝑢(𝑡), (6.17)

where 𝑢⁰(𝑡)=𝑎₁sin𝜋𝑡,𝑇+1𝑢(𝑡)=^𝑇_𝑗=2𝑎_𝑗sin𝑗𝜋𝑡.𝑇+1 (6.18)

Using the same method to prove [7, Lemma 1.4] with obvious changes, we obtain the following.

Lemma 6.4. Assume that for each 𝑛∈ℕ, one has 0≤𝜒_𝑛(𝑡) for 𝑡∈𝕋, and for each 𝑡∈𝕋, 𝜒_𝑛(𝑡)→0 as 𝑛→∞. Then, there exists a constant 𝜌>0, such that for all 𝑢∈𝐷^∗, ^𝑇_𝑡=1[Δ²𝑢(𝑡−1)+𝜆₁𝑢²(𝑡)+𝜒_𝑛(𝑡)𝑢(𝑡)][𝑢⁰(𝑡)−𝑢(𝑡)]≥𝜌‖𝑢‖². (6.19)

Let 𝜓₁ be such that Δ²𝜓₁(𝑡−1)+𝜆_⋄𝜓⁺₁(𝑡)−𝜇_⋄𝜓⁻₁𝜓(𝑡)=0,𝑡∈𝕋,₁(0)=𝜓₁(𝑇+1)=0, (6.20)(6.21) then 𝜓₁>0 on 𝕋.

Assume that(C1) Let 𝜒₊,𝜒₋∈𝐷 be such that there exist two point (𝜆_⋄,𝜇_⋄)∈Σ∩𝑆^±₁ and (𝜆^⋄,𝜇^⋄)∈Σ∩𝑆⁺₂ with 𝜆_⋄≤𝜒₊(𝑡),𝜇_⋄≤𝜒₋(𝑡),∀𝑡∈𝕋, (6.22) and the strict inequalities 𝜆_⋄<𝜒₊(𝑡),𝜇_⋄<𝜒₋(𝑡) hold for some 𝑡∈𝕋𝜆^⋄≥𝜒₊(𝑡),𝜇^⋄≥𝜒₋(𝑡),∀𝑡∈𝕋, (6.23) and the strict inequalities 𝜆^⋄>𝜒₊(𝑡),𝜇^⋄>𝜒₋(𝑡) hold for some 𝑡∈𝕋⧵{𝜏}, with 𝜏 is the unique generalized zero of 𝜓₂ on (0,𝑇+1), and Δ²𝜓₂(𝑡−1)+𝜆^⋄𝜓⁺₂(𝑡)−𝜇^⋄𝜓⁻₁𝜓(𝑡)=0,𝑡∈𝕋,₂(0)=𝜓₂(𝑇+1)=0, (6.24) with 𝜓₂(1)>0.

Theorem 6.5. Let (A1)–(A3) hold. Assume that there existence (𝜆^∗,𝜇^∗)∈Σ∩𝑆⁺₂, such that limsup_𝑠→+∞𝑔(𝑡,𝑠)𝑠≤𝜆^∗−𝜆₁,limsup_{𝑠→−∞}𝑔(𝑡,𝑠)𝑠≤𝜇^∗−𝜆₁. (6.25) Suppose that the strict inequalities in (6.25) hold on some 𝑡₀∈𝕋⧵{𝜏}. Then, BVP (6.2) has at least one solutions provided ^𝑇_𝑡=1𝑔^−∞(𝑡)sin𝜋𝑡<𝑇+1^𝑇_𝑡=1𝑓(𝑡)sin𝜋𝑡<𝑇+1^𝑇_𝑡=1𝑔_+∞(𝑡)sin𝜋𝑡.𝑇+1 (6.26)

Remark 6.6. Let us consider the nonlinear boundary value problem Δ²𝑢(𝑡−1)+𝜆₁𝑢(𝑡)+𝑔(𝑢(𝑡))=𝑓(𝑡),𝑡∈{1,2,3},𝑢(0)=𝑢(4)=0, (6.27) where 𝑓∶{1,2,3}→ℝ is a fixed function, 𝑔(𝑠)=1.1𝑠,𝑠≥01.7𝑠,𝑠<0. (6.28) Obviously, 𝑓 and 𝑔 satisfy (A1)–(A3). From Example 3.2, it follows that (6.25) hold. Since 𝑔_−∞=−∞ and 𝑔_+∞=+∞, it follows that (6.26) holds.

To prove Theorem 6.5, we need the following.

Lemma 6.7. Let (C1) holds. Then, the Dirichlet problem Δ²𝑢(𝑡−1)+𝜒₊(𝑡)𝑢⁺(𝑡)−𝜒₋(𝑡)𝑢⁻(𝑡)=0,𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0, (6.29)(6.30) has only the trivial solution.

Proof. Suppose on the contrary that 𝑢 is a nontrivial solution of (6.29), (6.30) with 𝑢(1)>0. (6.31)
We claim that the number of generalized zeros of 𝑢 in (0,𝑇+1) is 0 or 1.
Suppose on the contrary that the number of generalized zeros of 𝑢 in (0,𝑇+1) is large than 1. Let 𝑡₁,𝑡₂∈𝕋 be the first two positive generalized zeros of 𝑢 with 𝑡₁<𝑡₂.
We divide the proof into three cases.
Case 1 (𝑡₁<𝜏). In this case, we claim that Δ²𝜓(𝑡−1)+𝜆^⋄𝜓(𝑡)=0 (6.32) is disconjugate on [0,𝑡₁].
Suppose on the contrary that there exists a solution 𝜓^∗ of (6.32) which is linearly independent of 𝜓₂ and has two consecutive generalized zeros in [0,𝑡₁]. Then, from Lemma 6.1, 𝜓₂ has a generalized zero in (0,𝑡₁]. This is impossible. Therefore, the claim is true.
Now, from the above claim and Lemma 6.3, Δ²𝑢(𝑡−1)+𝜒₊(𝑡)𝑢(𝑡)=0 (6.33) is disconjugate on [0,𝑡₁]. However, this contradicts the fact that 𝑡₁ is a generalized zero of 𝑢.Case 2 (𝑡₁>𝜏). The case can be treated by the same method as in Case 1 with obvious changes.Case 3 (𝑡₁=𝜏). Subcase 3.1 (𝑡₁=𝜏 and 𝜓₂(𝜏)=0). It is easy to check that Δ²𝜓(𝑡−1)−𝜇^⋄𝜓(𝑡)=0 (6.34) is disconjugate on [𝜏,𝑡₂]. The fact and Lemma 6.3 yield Δ²𝑣(𝑡−1)−𝜒₋(𝑡)𝑣(𝑡)=0,𝑡∈{𝜏,…,𝑡₂−1} (6.35) is disconjugate on [𝜏,𝑡₂], and subsequently, 𝑢 cannot have two generalized zero in [𝜏,𝑡₂]. This is a contradiction.Subcase 3.2 (𝑡₁=𝜏 and 𝜓₂(𝜏)<0). We note that the general solution of (6.35) has the form 𝑣(𝑡)=𝑐₁sin(𝜃𝑡+𝑐₂),𝑡∈{𝜏−1,…,𝑡₂}, (6.36) with 𝜃∶=arccos2−𝜇^⋄2,𝜇^⋄∈(0,4). (6.37) Denote by 𝑑(𝜇^⋄) the distance between any two consecutive zeros of sin(𝜃𝑡+𝑐₂). Then, 𝑑(𝜇^⋄𝜋)=𝜃,(𝑇+1)−𝜏<𝑑(𝜇^⋄)<(𝑇+1)−(𝜏−1). (6.38)(6.39)
Combining (6.36) with (6.39) and using the definition of generalized zero, it follows that 𝑣(𝑡) has at most one generalized zero in [𝜏−1,𝑡₂], which implies that (6.35) is disconjugate on [𝜏−1,𝑡₂].
This fact and Lemma 6.3 yield Δ²𝑣(𝑡−1)−𝜒₋(𝑡)𝑣(𝑡)=0,𝑡∈{𝜏,…,𝑡₂−1} (6.40) is disconjugate on [𝜏−1,𝑡₂], and subsequently, 𝑢 cannot have two generalized zero in [𝜏−1,𝑡₂]. This is a contradiction.
Therefore, 𝑢 has at most one generalized zero in (0,𝑇+1).If the number of generalized zeros is in (0,𝑇+1) is 0, then 𝑢(𝑡)>0,𝑡∈𝕋. (6.41) Multiplying (6.29) by 𝜓₁(𝑡) and (6.20) by 𝑢(𝑡) and subtracting, and then taking the summation from 1 to 𝑇, it follows that 0=^𝑇_𝑡=1[𝜒₊(𝑡)𝑢⁺(𝑡)𝜓₁(𝑡)−𝜆_⋄𝑢(𝑡)𝜓₁(𝑡)]=^𝑇_𝑡=1(𝜒₊(𝑡)−𝜆_⋄)𝑢(𝑡)𝜓₁(𝑡)>0, (6.42) a contradiction.
If 𝑢 has a unique generalized zeros 𝜏_𝑢 in (0,𝑇+1), then from Lemma 6.3 and the similar method to deal with Case 1, it follows that 𝜏_𝑢=𝜏. (6.43) Thus, Δ²𝑢(𝑡−1)+𝜒₊(𝑡)𝑢⁺Δ(𝑡)=0,𝑡∈{1,…,𝜏−1},₂𝑢(𝑡−1)−𝜒₋(𝑡)𝑢⁻(𝑡)=0,𝑡∈{𝜏,…,𝑇+1}. (6.44) This together with the facts Δ²𝜓₂(𝑡−1)+𝜆^⋄𝜓⁺₂Δ(𝑡)=0,𝑡∈{1,…,𝜏−1},₂𝜓₂(𝑡−1)−𝜇^⋄𝜓⁻₂(𝑡)=0,𝑡∈{𝜏,…,𝑇+1} (6.45) imply that 𝜏0=⁻¹_𝑡=1[𝜒₊(𝑡)𝑢⁺(𝑡)𝜓⁺₂(𝑡)−𝜆^⋄𝑢⁺(𝑡)𝜓⁺₂+(𝑡)]_𝑇𝑡=𝜏[𝜒₋(𝑡)𝑢⁻(𝑡)𝜓⁻₂(𝑡)−𝜇^⋄𝑢⁻(𝑡)𝜓⁻₂(𝑡)]<0, (6.46) a contradiction.
Therefore, 𝑢(𝑡)=0 on 𝕋.

Proof of Theorem 6.5. The idea is the same as in the proof of [7, Theorem 1.9]. Let 𝛿<min{𝜆^⋄−𝜆₁,𝜇^⋄−𝜆₁} and define the homotopy family Δ²𝑢(𝑡−1)+𝜆₁𝑢(𝑡)+(1−𝑟)𝛿𝑢(𝑡)+𝑟𝑔(𝑡,𝑢(𝑡))=𝑟ℎ(𝑡),𝑟∈[0,1],𝑢(0)=𝑢(𝑇+1)=0. (6.47) We will show that there exists 𝑅>0 such that (6.47), (6.29) has no solution 𝑢 with ‖𝑢‖=𝑅. (6.48)
Similar to the proof of [7, Remark 1.2], we may prove that there exist two positive constants 𝑞₁ and 𝑞₂, such that 𝑔(𝑡,𝑠)=𝛾(𝑡,𝑠)𝑠+ℎ(𝑡,𝑠),𝑡∈𝕋,𝑠∈ℝ, (6.49) where 0≤𝛾(𝑡,𝑠)≤𝑞₁, |ℎ(𝑡,𝑠)|≤𝑞₂ for all 𝑠∈ℝ and 𝑡∈𝕋.
Let us define 𝑔₁𝑔(𝑡,𝑠)=min{𝑔(𝑡,𝑠),1},𝑠≥1min{𝑔(𝑡,𝑠),−1},𝑠≤−1,_∗(𝑡,𝑠)=𝑔(𝑡,𝑠)−𝑔₁𝑔(𝑡,𝑠),𝛾(𝑡,𝑠)=^∗(𝑡,𝑠)𝑠𝑔,|𝑠|≥1,_∗𝑠(𝑡,|𝑠|)𝑠,0<|𝑠|<1,0,𝑠=0. (6.50)
Assume to the contrary that there exists a sequence {(𝑟_𝑛,𝑢_𝑛)}⊆[0,1]×𝒟^∗ with ‖𝑢_𝑛‖→∞, such that Δ²𝑢_𝑛(𝑡−1)+𝜆₁𝑢_𝑛(𝑡)+(1−𝑟_𝑛)𝛿𝑢_𝑛(𝑡)+𝑟_𝑛𝑔(𝑡,𝑢_𝑛(𝑡))=𝑟_𝑛𝑢ℎ(𝑡),𝑡∈𝕋,_𝑛(0)=𝑢_𝑛(𝑇+1)=0. (6.51)
Set 𝜒_𝑛(𝑡)=(1−𝑟_𝑛)𝛿+𝑟_𝑛𝛾(𝑡,𝑢_𝑛(𝑡)). (6.52) Then we may assume that 𝜒_𝑛→𝜒 in 𝒟^∗. Moreover, it follows from (6.5), (6.25), and (6.49) that 𝜒(𝑡)≥0 on 𝑡∈𝕋, and 𝜒(𝑡)≤𝜆^⋄−𝜆₁on{𝑡∈𝕋∣𝜓₂(𝑡)<0},𝜒(𝑡)≤𝜇^⋄−𝜆₁on{𝑡∈𝕋∣𝜓₂(𝑡)<0}, (6.53) with the strict inequalities on some 𝑡₀∈𝕋⧵{𝜏}. Now, by the standard arguments, see the proof of [7, Theorems 1.5 and 1.9], we may get the desired contradiction.

Remark 6.8. Theorem 6.5 improves the main results of Rodriguez [8] and R. Ma and H. Ma [9, 10], where the nonlinearities are not jumping at infinity.

Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. This work Supported by the NSFC (no. 11061030), the Fundamental Research Funds for the Gansu Universities.