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

Authors: Ruyun Ma; Youji Xu; Chenghua Gao
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

LetT∈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.


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## Body

## 1. Introduction

LetT∈ℕ with T>5. Let 𝕋∶={1,…,T} and 𝕋̂∶={0,1,…,T,T+1}. For u:𝕋→ℝ, we define u+,u-:𝕋→ℝ by (1.1)u+(t)=max{u(t),0},u-(t)=max{-u(t),0}.
The Fučik spectrum of the problem (1.2)Δ2u(t-1)+λu+(t)-μu-(t)=0,t∈𝕋,u(0)=u(T+1)=0
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(1.3)u′′(x)+λu+(x)-μu-(x)=0,x∈(0,1),u(0)=u(1)=0
has a nontrivial solution for (λ,μ)∈ℝ2 if and only if (λ,μ) satisfies(i)
fork≥1 odd, either
(1.4)k+121λ+k-121μ=1,
or
(1.5)k-121λ+k+121μ=1,(ii)
fork≥2 even
(1.6)k21λ+k21μ=1.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∩projℝ2𝒮2+ (for the definition of projℝ2𝒮2+, see Definition 5.1 below), which is continuous, strictly decreasing, symmetric with respect to the diagonal, and for each (λ,μ)∈Σ∩projℝ2𝒮2+, 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 (1.7)Δ2u(t-1)+λ1u(t)+g(t,u(t))=f(t),t∈𝕋,u(0)=u(T+1)=0,
where g:𝕋×ℝ→ℝ is continuous, f:𝕋→ℝ, λ1 is the first eigenvalue of the linear problem (1.8)Δ2u(t-1)+λu(t)=0,t∈𝕋,u(0)=u(T+1)=0.


## 2. Matching Continuation

Forr∈(0,∞), we denote the integer part of r by [r].Denote(2.1)ℕ*={0}∪ℕ.Lemma 2.1.
For fixedλ,μ∈ℝ, the initial value problem
(2.2)Δ2u(t-1)+λu+(t)-μu-(t)=0,t∈ℕ,(2.3)u(0)=0,u(1)=1
has a unique solution w+:ℕ*→R.Proof.
Equation (2.2) is equivalent to the recurrence
(2.4)u(t+1)=2u(t)-u(t-1)-λu+(t)+μu-(t),t∈ℕ,
which guarantees the existence and uniqueness of solution w+:ℕ*→R.Lemma 2.2.
Letw+(t) be a nontrivial solution of the initial value problem (2.2), (2.3) and w+(t0)=0 for some t0∈𝕋. Then,
(2.5)w+(t0-1)w+(t0+1)<0.Proof.
It can be easily deduced from (2.4).Lemma 2.3.
Letw+(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
(2.6)w+(T+1,λ,μ)=0.To find(λ,μ) satisfying (2.6), we consider the linear initial value problem
(2.7)Δ2u(t-1)+λu(t)=0,t∈ℕ,(2.8)u(0)=0,u(1)=1.
The characteristic equation of (2.7) is
(2.9)r2+(λ-2)r+1=0,
so
(2.10)r=2-λ±(λ-2)2-42.If|λ-2|>2, then it can be shown that there are no eigenvalues of (1.2).In fact, in this case,(2.11)r1∶=2-λ+(λ-2)2-42,r2∶=2-λ-(λ-2)2-42,and the general solution of (2.7) is (2.12)u(t)=Ar1t+Br2t.It is easy to check that the functionAr1x+Br2x,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 (2.13)r1=r2∶=2-λ2,and the general solution of (2.7) is (2.14)u(t)=Ar1t+Btr1t.It is easy to check that the functionAr1x+Bxr1x,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
(2.15)2-λ=2cosθ.
Then,
(2.16)r=cosθ±isinθ=e±iθ.
Hence, a general solution of (2.7) is
(2.17)u(t)=Acosθt+Bsinθt.
From (2.8), it follows that u(0)=A=0 and
(2.18)B=1sinθ.
Since |λ-2|<2, it follows from (2.15) that θ≠kπ. Thus, B is well defined and
(2.19)u(t)=1sinθsinθt,t∈ℕ*.Lemma 2.4.
The first eigenvalue of the problem (1.8) is
(2.20)λ1=2-2cosπT+1,
which is simple and the eigenfunction corresponding to λ1 is
(2.21)φ1(t)=sinπT+1t,t∈𝕋.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 ofλ1 and φ1 in (2.20) and (2.21), it follows that for any μ∈ℝ,
(2.22)Δ2φ1(t-1)+λ1(φ1)+(t)-μ(φ1)-(t)=0,t∈𝕋,φ1(0)=φ1(T+1)=0.
This implies that {λ1}×ℝ⊂Σ. Similarly, ℝ×{λ1}⊂Σ.
(ii) Let(λ¯,μ¯)∈Σ and y be the corresponding eigenfunction. Then,
(2.23)Δ2y(t-1)+λ¯y+(t)-μ¯y-(t)=0,t∈𝕋,y(0)=y(T+1)=0.
Let z(t)=-y(t) for t∈𝕋̂. Then,
(2.24)Δ2z(t-1)+μ¯z+(t)-λ¯z-(t)=0,t∈𝕋,z(0)=z(T+1)=0,
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,∞): (2.25)ũ(x)=1sinθsinθx,x∈[0,∞).Letβ1 be the first positive zero of ũ, then (2.26)ũ(β1)=0,ũ(x)>0,x∈(0,β1),(2.27)ũ(x)=B1sinπβ1x,
with B1∶=(sin(π/β1))-1. From (2.27) and (2.15), we have that (2.28)λ=2-2cosπβ1,
that is, (2.29)β1=πarccos((2-λ)/2).
Moreover, ũ(x) satisfies that (2.30)Δ2ũ(t-1)+λũ(t)=0,t∈{1,2,…,[β1]},ũ(0)=0,ũ(β1)=0,ũ(t)>0,t∈{1,…,[β1-1]},ũ([β1])>0,ifβ1>[β1],ũ([β1])=0,ifβ1=[β1],ũ([β1]+1)<0,
where [r] denotes the integer part of r.Definition 2.6.
Suppose that a functiony:𝕋̂→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(2.31)Σ̃=Σ∖(({λ1}×ℝ)∪(ℝ×{λ1})).
Then, (λ,μ)∈Σ̃ implies that
(2.32)2-2cosπT<λ<4,2-2cosπT<μ<4.Proof.
From (2.20) and (2.28), we see that (λ,μ)∈Σ̃ 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
(2.33)2-2cosπT<λ<4.
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 ũ (see (2.27) for the definition of ũ). Then, we see from (2.4) that (2.34)u([β1]+1)=2u([β1])-u([β1]-1)-λu([β1]),
which means that u([β1]+1) is uniquely determined by u([β1]), u([β1]-1) and λ, and is independent of μ.Now, letv(t) be the unique solution of the initial value problem
(2.35)Δ2v(t-1)+μv(t)=0,t∈{[β1]-1,[β1],[β1]+1,…},(2.36)v([β1])=u([β1]),v([β1]+1)=u([β1]+1).
By a similar method to get ũ and β1, we get the following lemma.Lemma 2.8.
Let (H) hold, and letb satisfy
(2.37)μ=2-2cosπb.
Then, 1<b<T.Proof.
From (2.37) and (H), it follows that
(2.38)2-2cosπT<2-2cosπb<4.
This is
(2.39)cosπT>cosπb>cosπ1,
which implies 1<b<T.Lemma 2.9.
Let (H) hold. Then, (2.35), (2.36) has a solution of the form
(2.40)v(t)=B2sinπ(t-α2)b,
where α2 is the unique solution of
(2.41)B1sin(π[β1]β1)=B2sinπ([β1]-α2)b,B1sin(π([β1]+1)β1)=B2sinπ([β1]+1-α2)b,
in ([β1],[β1+1]) if β1>[β1], and α2=β1 if β1=[β1]; B2 is a negative constant satisfying
(2.42)B2sinπ([β1]+1-α2)b=2B1sinπ([β1])β1-B1sinπ([β1]-1)β1-λB1sinπ([β1])β1.Proof.
The relation (2.40) can be deduced by the similar method to get (2.27).
Ifβ1>[β1], then we define a function F:[[β1],[β1+1]]→ℝ by
(2.43)F(α)=sinπβ1([β1])sinπ([β1]+1-α)b-sinπ([β1]-α)bsinπ([β1]+1)β1.
It is easy to check that
(2.44)F([β1+1])=-sinπ([β1]-[β1+1])bsinπβ1([β1+1])<0,F([β1])=sinπβ1([β1])sinπ([β1+1]-[β1])b>0.
Moreover, for α∈([β1],[β1+1]),
(2.45)∂F∂α(α)=-πbsinπβ1([β1])cosπ([β1+1]-α)b+πbcosπ([β1]-α)bsinπβ1([β1+1])<0,
(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.
Ifβ1=[β1], then v(β1)=0. This together with (2.4) imply that
(2.46)v(β1+1)=-v(β1-1),
that is,
(2.47)B2sinπ(β1+1-α2)b=-B2sinπ(β1-1-α2)b.
So, we may take that
(2.48)α2=β1.
Finally, it follows from (2.4) and the fact u([β1]+1)=v([β1]+1) (see (2.36)) that B2 satisfies (2.42).Now, forx∈ℝ, let (2.49)ṽ(x)∶=B2sinπ(x-α2)b,
which is an extension of the function v:(ℕ*∩[[β1-1],∞))→ℝ defined in (2.40).Denote(2.50)β2∶=α2+b,Thenṽ(x) satisfies (2.51)Δ2ṽ(t-1)+μṽ(t)=0,t∈{[β1+1],…,[β2]},ṽ([β1])=ũ([β1]),ṽ([β1+1])=ũ([β1+1]),ṽ(β2)=0,ṽ(t)<0,t∈{[β1+1],…,[β2-1]},ṽ([β2])<0,ifβ2>[β2],ṽ([β2])=0,ifβ2=[β2],ṽ([β2]+1)>0.Definition 2.10.
We say thatα2 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 ṽ(x)(2.52)ṽ(t)∶=B2sinπ(t-α2)b,t∈[[β1],[β2+1]]
is the matching extension of ũ(x),x∈[0,[β1+1]].Remark 2.11.
Notice thatṽ(x)≢ũ(x) for x∈[[β1],[β1+1]] in general. However, ṽ(t)≡ũ(t) for t∈{[β1],[β1+1]}.Obviously, we may repeat the above process to “extend” ũ(t) to a sequence defined on ℕ in a unique manner. Moreover, this sequence coincides with the sequence w+(t) in Lemma 2.1.Letα1∶=0. Recall that β1 is defined by (2.29)(2.53)β1=πarccos((2-λ)/2),
and α2=β1 if β1=[β1], and α2 is the unique solution of (2.54)sinπβ1([β1])sinπ([β1]+1-α)b-sinπ([β1]-α)bsinπ([β1]+1)β1=0.
Recall (2.55)β2∶=α2+b,
with (2.56)b=πarccos((2-μ)/2),
Repeating the above process, we may take αk=βk-1 if βk-1=[βk-1], and αk is the unique solution of (2.57)Fk(α)=0,
where (2.58)Fk(α)∶=sinπ([βk-1]-αk-1)βk-1-αk-1sinπ([βk-1+1]-α)βk-αk-sinπ([βk-1]-α)βk-αksinπ([βk-1+1]-αk-1)βk-1-αk-1.
Put (2.59)βk={αk+β1ifkisodd,αk+bifkiseven.Recall(2.60)B1∶=(sinπβ1)-1,
and define (2.61)Bksinπ([βk-1]+1-αk)βk-αk=2Bk-1sinπ([βk-1])βk-1-αk-1-Bk-1sinπ([βk-1]-1)βk-1-αk-1-νBk-1sinπ([βk-1])βk-1-αk-1.
with (2.62)ν={μifkisodd,λifkiseven.Finally, it is easy to check that(2.63)Fk([βk-1])>0,Fk([βk-1+1])<0,∂Fk∂α(α)=-πβk-αksinπ([βk-1]-αk-1)βk-1-αk-1cosπ([βk-1+1]-α)βk-αk+πβk-αkcosπ([βk-1]-α)βk-αksinπ([βk-1+1]-αk-1)βk-1-αk-1<0.
So, there exists a unique αk∈([βk-1],[βk-1+1]) which satisfies (2.57).To sum up, we may define(2.64)Γk(x)∶=Bksinπ(x-αk)βk-αk,x∈[[βk-1],[βk+1]],
which can be thought as the “matching extension” of Γk-1. Notice that αk is the left zero of Γk on [[βk-1],[βk+1]], and βk is the right zero of Γk on [[βk-1],[βk+1]].Clearly, we may start the discuss from the IVP(2.65)Δ2u(t-1)+λu(t)=0,t∈𝕋,u(0)=0,u(1)=-1
which has unique solution w-(t,λ,μ). Then, we may uniquely determine the kth segmental arc Γ̂k on [[δk-1],[δk+1]] and the feasible initial phase γk via matching extension method, and accordingly, we get the left zero γk and the right zero δk of Γ̂k on [[δk-1],[δk+1]].


## 3. The Main Result

The main result of this paper is the following discrete analogue of Theorem1.1.Theorem 3.1.
(λ,μ)∈ℝ2 is an eigenvalue of (1.2) if and only if (λ,μ) satisfies(i)
fork≥2 even
(3.1)k2πarccos((2-λ)/2)+k2πarccos((2-μ)/2)+∑j=1k-1(αj+1-βj)=T+1(ii)
fork≥1 odd, either
(3.2)k+12πarccos((2-λ)/2)+k-12πarccos((2-μ)/2)+∑j=1k-1(αj+1-βj)=T+1 or
(3.3)k-12πarccos((2-λ)/2)+k+12πarccos((2-μ)/2)+∑j=1k-1(γj+1-δj)=T+1.Proof.
(i) Letk≥1 be even. Let
(3.4)Γm∶={{(x,B1sinπβ1(x-αm))∣x=[αm],…,[βm+1]},m=2j-1,j∈{1,…,k2},{(x,B2sinπb(x-αm))∣x=[αm],…,[βm+1]},m=2j,j∈{1,…,k-22},{(x,B2sinπb(x-αm))∣x=[αm],…,[βm]},m=k.
Here,
(3.5)β1=πarccos((2-λ)/2),b=πarccos((2-μ)/2).
Then, a nontrivial solution of (1.2) can be constructed via matching extension method to Γ1 and Γ2, …, Γk-1 and Γk, respectively. Moreover,
(3.6)T+1=k2πarccos((2-λ)/2)+k2πarccos((2-μ)/2)+∑j=1k-1(αj+1-βj).
(ii) The case thatk is odd can be treated by the similar way.Example 3.2.
We may useMathematica 5.0 to give a numerical example for Theorem 3.1.
Takek=2,T=3.
Takeβ1=2.2, then λ≐1.7153703202231987.
Takeb=1.8198598, then μ≐2.3097214213167683.
Using the relation(3.7)sinπβ1([β1])sinπ([β1+1]-α)b=sinπ([β1]-α)bsinπβ1([β1+1]),
we may find
(3.8)α2≐2.1801402269418726.
Since
(3.9)2.2+1.8198598+(2.1801402269418726-2.2)≐4,
we may think that (1.7153703202231987,2.3097214213167683) is an eigenvalue of the problem
(3.10)Δ2u(t-1)+λu+(t)-μu-(t)=0,t∈{1,2,3},u(0)=u(4)=0.
Now, by the recurrence relation
(3.11)u(t+1)=2u(t)-u(t-1)-1.7153703202231987u+(t)+2.309739789282126u-(t),u(0)=0,u(1)=1,
it is easily to compute that
(3.12)u(4)≐-4.6600427960896695×10-8≈0.
This is a desired numerical result.


## 4. Some Properties ofα2

Proposition 4.1.
Ifλ=μ and βi=[βi], then βj=αj+1 for j=1,2,….Proof.
In fact, ifλ=μ and βi=[βi], then
(4.1)β1=b=πarccos((2-λ)/2),
which implies that
(4.2)αm,βm∈ℕ*.Proposition 4.2.
Let (H) hold. Assume thatλ is such that
(4.3)β1=[β1]+12.
Then, for every μ>1, we have that
(4.4)α2=[β1]+12.Proof.
Byβ1=[β1]+1/2, it follows that
(4.5)sinπ[β1]β1=-sinπ([β1]+1)β1.
This together with F(α2)=0 imply that
(4.6)sinπ([β1]-α2)β2-α2=-sinπ([β1]+1-α2)β2-α2,
and accordingly,
(4.7)sinπ([β1]-α2)β2-α2+sinπ([β1]+1-α2)β2-α2=2sinπ([β1]+1/2-α2)β2-α2cosπ/2β2-α2=0.
By (H), we have that β2-α2>1. Therefore,
(4.8)[β1]+12-α2=0.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 ofF, (see (2.43)), we have that
(4.9)F(β1)=sinπβ1([β1])sinπ([β1+1]-β1)b-sinπ([β1]-β1)bsinπ([β1+1])β1.
Since β1>[β1]+1/2, we have that
(4.10)sin(πβ1[β1])>-sin(πβ1[β1+1]),
which together with (4.9) implies that
(4.11)F(β1)≤sinπβ1([β1])[sinπ([β1+1]-β1)b+sinπ([β1]-β1)b]<0.
Since [β1+1]-β1<β1-[β1]. Thus,
(4.12)F(β1)<F(α2)=0.
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
(4.13)F([β1]+12)=sinπβ1([β1])sinπ2b+sinπ2bsinπβ1([β1+1]).
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
(4.14)sin(π[β1]β1)<-sin(π[β1+1]β1),
which together with (4.9) and the fact that [β1+1]-β1>β1-[β1] imply that
(4.15)F(β1)≥sinπβ1([β1])[sinπ([β1+1]-β1)b+sinπ([β1]-β1)b]>0.
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
(4.16)F([β1]+12)<F(α2)=0,
which implies that [β1]+1/2>α2.


## 5. Some Properties ofΣ∩projℝ2𝒮2+

Definition 5.1.
Fork∈ℕ and ν∈{+,-}, let Skν denote the set of functions u:𝕋̂→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(5.1)𝒮2+∶=ℝ2×S2+.An immediate consequence of Lemma2.2 is the following.Lemma 5.2.
Ifu of a nontrivial solution of (1.2), then u∈Skν for some k∈ℕ and ν∈{+,-}.In the rest of this section, one will pay one’s attention to the study ofΣ∩S2+.Let (H) hold and define a functionG:(1,T)×([β1],[β1]+1)→ℝ by (5.2)G(b,α)=sinπβ1([β1])sinπ([β1+1]-α)b-sinπ([β1]-α)bsinπβ1([β1+1]).
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 (5.3)G(b,α2(b))=0.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
(5.4)∂G∂α=-πbsin(πβ1[β1])cosπ([β1+1]-α)b+πbcosπ([β1]-α)bsin(πβ1[β1+1]).
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),
(5.5)∂G∂b=-π([β1+1]-α)b2sin(πβ1[β1])cosπ([β1+1]-α)b+π([β1]-α)b2cosπ([β1]-α)bsin(πβ1[β1+1]).
Denote
(5.6)A(b,α2)=sin(πβ1[β1])cosπ([β1+1]-α2)b,B(b,α2)=cosπ([β1]-α2)bsin(πβ1[β1+1]).
Then,
(5.7)A(b,α2)>0,B(b,α2)<0,
and subsequently,
(5.8)|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))|≤1b,
since max{π([β1+1]-α2)/b2,π(-[β1]+α2)/b2}≤π/b2.Theorem 5.4.
Let (H) hold. Then, for eachμ∈(2-2cos(π/T),4), there exists at most one λ∈(2-2cos(π/T),4) such that
(5.9)πarccos((2-λ)/2)+πarccos((2-μ)/2)+(α2-β1)=T+1.Proof.
Set(5.10)g(b)∶=b+(α2(b)-β1).
Then, by Lemma 5.3,
(5.11)g′(b)=1+α2′(b)>1-1b>0,
which implies that g is strictly increasing in b, and accordingly, the function
(5.12)ĝ(μ)∶=πarccos((2-μ)/2)+(α2(πarccos((2-μ)/2))-β1)
is strictly decreasing and continuous in μ. Therefore, the function
(5.13)λ=2-2cos(πT+1-ĝ(μ))
is strictly decreasing and continuous in μ.Remark 5.5.
Letλ2 be the second eigenvalue of (2.18), (2.19). Then, (λ2,λ2)∈Σ∩projℝ2𝒮2+.Since(5.14){πarccos((2-μ)/2)∣μ∈(2-2cosπT,4)}=(1,T),there exists an open intervalI⊂(2-2cos(π/T),4) such that (5.15){πarccos((2-μ)/2)∣μ∈I}=(2,T-1).Theorem 5.6.
For everyμ∈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,
(5.16)1<πarccos((2-μ)/2)+(α2(πarccos((2-μ)/2))-β1)<T,μ∈I,
which implies that
(5.17)T>(T+1)-{πarccos((2-μ)/2)+(α2(πarccos((2-μ)/2))-β1)}>1,μ∈I.
Combining this with the fact that
(5.18){πarccos((2-λ)/2)∣λ∈(2-2cosπT,4)}=(1,T),
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

Letλ1 be the first eigenvalue of (1.8). Then, (6.1)λ1=2-2cosπT+1.
The eigenfunction corresponding to λ1 is   sin(πt/(T+1)).Let us consider the nonlinear problem(6.2)Δ2u(t-1)+λ1u(t)+g(t,u(t))=f(t),t∈𝕋,u(0)=u(T+1)=0,
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
(6.3)|g(t,s)|≤p(t)+q|s|,s∈ℝ,t∈𝕋,(A3)
there exist a function functionsa,A:𝕋→ℝ and constants r<0<R, such that
(6.4)g(t,s)≥A(t),s≥R,t∈𝕋,g(t,s)≤a(t),s≤r,t∈𝕋.
Obviously, (A3) implies
(6.5)liminfs→±∞g(t,s)s≥0.Denote(6.6)g+∞(t)∶=liminfs→+∞g(t,s),g-∞(t)∶=lim sups→-∞g(t,s).Lemma 6.1 (Sturm separation theorem; [7, Theorem 6.5]).
Two linearly independent solutions of(6.7)Δ2y(t-1)+q(t)y(t)=0,t∈𝕋
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]).
Letq:𝕋→ℝ. We say that (6.7) is disconjugate on 𝕋̂ provided that no nontrivial solution of (6.7) has two or more generalized zeros on 𝕋̂.Lemma 6.3 (Sturm comparison theorem; [7, Theorem 8.12]).
Assume thatq1(t)≥q2(t) on 𝕋. If Δ2y(t-1)+q1(t)y(t)=0 is disconjugate on 𝕋̂, then Δ2y(t-1)+q2(t)y(t)=0 is disconjugate on 𝕋̂.Let us denote byD the Hilbert space (6.8)D∶={u:𝕋→ℝ},with the inner product(6.9)〈u,v〉D∶=∑t=1Tu(t)v(t),u,v∈D,and the norm(6.10)∥u∥D∶=〈u,v〉D1/2,u,v∈D.Let us denote byD* the Hilbert space (6.11)D*∶={u:𝕋̂→ℝ∣u(0)=u(T+1)=0},with the inner product(6.12)〈u,v〉D*∶=∑t=1Tu(t)v(t),u,v∈D*,and the norm(6.13)∥u∥D∶=〈u,v〉D*1/2,u,v∈D.Foru∈D, define j:D→D* by (6.14)j(u(1),u(2),…,u(T))=(0,u(1),u(2),…,u(T),0),thenj is a natural isomorphism.Obviously,(6.15)〈u,v〉D=〈ju,jv〉D*,(u,v)∈D.
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(6.16)u(t)=∑j=1TajsinjπtT+1be the fourier series ofu∈D*. Then, one will write (6.17)u(t)=u0(t)+ũ(t),where(6.18)u0(t)=a1sinπtT+1,ũ(t)=∑j=2TajsinjπtT+1.Using the same method to prove [7, Lemma 1.4] with obvious changes, we obtain the following.Lemma 6.4.
Assume that for eachn∈ℕ, one has 0≤χn(t) for t∈𝕋, and for each t∈𝕋, χn(t)→0 as n→∞. Then, there exists a constant ρ>0, such that for all u∈D*,
(6.19)∑t=1T[Δ2u(t-1)+λ1u2(t)+χn(t)u(t)][u0(t)-ũ(t)]≥ρ∥ũ∥2.Letψ1 be such that
(6.20)Δ2ψ1(t-1)+λ⋄ψ1+(t)-μ⋄ψ1-(t)=0,t∈𝕋,(6.21)ψ1(0)=ψ1(T+1)=0,
then ψ1>0 on 𝕋.Assume that(C1)
Letχ+,χ-∈D be such that there exist two point (λ⋄,μ⋄)∈Σ∩S1± and (λ⋄,μ⋄)∈Σ∩S2+ with
(6.22)λ⋄≤χ+(t),μ⋄≤χ-(t),∀t∈𝕋,
and the strict inequalities λ⋄<χ+(t),μ⋄<χ-(t) hold for some t∈𝕋(6.23)λ⋄≥χ+(t),μ⋄≥χ-(t),∀t∈𝕋,
and the strict inequalities λ⋄>χ+(t),μ⋄>χ-(t) hold for some t∈𝕋∖{τ̂}, with τ̂ is the unique generalized zero of ψ2 on (0,T+1), and
(6.24)Δ2ψ2(t-1)+λ⋄ψ2+(t)-μ⋄ψ1-(t)=0,t∈𝕋,ψ2(0)=ψ2(T+1)=0,
with ψ2(1)>0.Theorem 6.5.
Let (A1)–(A3) hold. Assume that there existence(λ*,μ*)∈Σ∩S2+, such that
(6.25)lim sups→+∞g(t,s)s≤λ*-λ1,lim sups→-∞g(t,s)s≤μ*-λ1.
Suppose that the strict inequalities in (6.25) hold on some t0∈𝕋∖{τ̂}. Then, BVP (6.2) has at least one solutions provided
(6.26)∑t=1Tg-∞(t)sinπtT+1<∑t=1Tf(t)sinπtT+1<∑t=1Tg+∞(t)sinπtT+1.Remark 6.6.
Let us consider the nonlinear boundary value problem(6.27)Δ2u(t-1)+λ1u(t)+g(u(t))=f(t),t∈{1,2,3},u(0)=u(4)=0,
where f:{1,2,3}→ℝ is a fixed function,
(6.28)g(s)={1.1s,s≥01.7s,s<0.
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 Theorem6.5, we need the following.Lemma 6.7.
Let (C1) holds. Then, the Dirichlet problem(6.29)Δ2u(t-1)+χ+(t)u+(t)-χ-(t)u-(t)=0,t∈𝕋,(6.30)u(0)=u(T+1)=0,
has only the trivial solution.Proof.
Suppose on the contrary thatu is a nontrivial solution of (6.29), (6.30) with
(6.31)u(1)>0.
We claim that the number of generalized zeros ofu in (0,T+1) is 0 or 1.
Suppose on the contrary that the number of generalized zeros ofu 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<τ̂).
In this case, we claim that(6.32)Δ2ψ(t-1)+λ⋄ψ(t)=0
is disconjugate on [0,t1].
Suppose on the contrary that there exists a solutionψ* 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 Lemma6.3,
(6.33)Δ2u(t-1)+χ+(t)u(t)=0
is disconjugate on [0,t1]. However, this contradicts the fact that t1 is a generalized zero of u.Case 2 (t1>τ̂).
The case can be treated by the same method as in Case1 with obvious changes.Case 3 (t1=τ̂).
Subcase 3.1 (t1=τ̂ and ψ2(τ̂)=0).
It is easy to check that(6.34)Δ2ψ(t-1)-μ⋄ψ(t)=0
is disconjugate on [τ̂,t2]. The fact and Lemma 6.3 yield
(6.35)Δ2v(t-1)-χ-(t)v(t)=0,t∈{τ̂,…,t2-1}
is disconjugate on [τ̂,t2], and subsequently, u cannot have two generalized zero in [τ̂,t2]. This is a contradiction.Subcase 3.2 (t1=τ̂ and ψ2(τ̂)<0).
We note that the general solution of (6.35) has the form
(6.36)v(t)=c1sin(θt+c2),t∈{τ̂-1,…,t2},
with
(6.37)θ∶=arccos2-μ⋄2,μ⋄∈(0,4).
Denote by d(μ⋄) the distance between any two consecutive zeros of sin(θt+c2). Then,
(6.38)d(μ⋄)=πθ,(6.39)(T+1)-τ̂<d(μ⋄)<(T+1)-(τ̂-1).
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 [τ̂-1,t2], which implies that (6.35) is disconjugate on [τ̂-1,t2].
This fact and Lemma6.3 yield
(6.40)Δ2v(t-1)-χ-(t)v(t)=0,t∈{τ̂,…,t2-1}
is disconjugate on [τ̂-1,t2], and subsequently, u cannot have two generalized zero in [τ̂-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
(6.41)u(t)>0,t∈𝕋.
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
(6.42)0=∑t=1T[χ+(t)u+(t)ψ1(t)-λ⋄u(t)ψ1(t)]=∑t=1T(χ+(t)-λ⋄)u(t)ψ1(t)>0,
a contradiction.
Ifu has a unique generalized zeros τu in (0,T+1), then from Lemma 6.3 and the similar method to deal with Case 1, it follows that
(6.43)τu=τ̂.
Thus,
(6.44)Δ2u(t-1)+χ+(t)u+(t)=0,t∈{1,…,τ̂-1},Δ2u(t-1)-χ-(t)u-(t)=0,t∈{τ̂,…,T+1}.
This together with the facts
(6.45)Δ2ψ2(t-1)+λ⋄ψ2+(t)=0,t∈{1,…,τ̂-1},Δ2ψ2(t-1)-μ⋄ψ2-(t)=0,t∈{τ̂,…,T+1}
imply that
(6.46)0=∑t=1τ̂-1[χ+(t)u+(t)ψ2+(t)-λ⋄u+(t)ψ2+(t)]+∑t=τ̂T[χ-(t)u-(t)ψ2-(t)-μ⋄u-(t)ψ2-(t)]<0,
a contradiction.
Therefore,u(t)=0 on 𝕋.Proof of Theorem6.5.
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
(6.47)Δ2u(t-1)+λ1u(t)+(1-r)δu(t)+rg(t,u(t))=rh(t),r∈[0,1],u(0)=u(T+1)=0.
We will show that there exists R>0 such that (6.47), (6.29) has no solution u with
(6.48)∥u∥=R.
Similar to the proof of [7, Remark 1.2], we may prove that there exist two positive constants q1 and q2, such that
(6.49)g(t,s)=γ(t,s)s+h(t,s),t∈𝕋,s∈ℝ,
where 0≤γ(t,s)≤q1, |h(t,s)|≤q2 for all s∈ℝ and t∈𝕋.
Let us define(6.50)ĝ1(t,s)={min{g(t,s),1},s≥1min{g(t,s),-1},s≤-1,g*(t,s)=g(t,s)-ĝ1(t,s),γ(t,s)={g*(t,s)s,|s|≥1,g*(t,s|s|)s,0<|s|<1,0,s=0.
Assume to the contrary that there exists a sequence{(rn,un)}⊆[0,1]×𝒟* with ∥un∥→∞, such that
(6.51)Δ2un(t-1)+λ1un(t)+(1-rn)δun(t)+rng(t,un(t))=rnh(t),t∈𝕋,un(0)=un(T+1)=0.
Set(6.52)χn(t)=(1-rn)δ+rnγ(t,un(t)).
Then we may assume that χn→χ in 𝒟*. Moreover, it follows from (6.5), (6.25), and (6.49) that χ(t)≥0 on t∈𝕋, and
(6.53)χ(t)≤λ⋄-λ1on{t∈𝕋∣ψ2(t)<0},χ(t)≤μ⋄-λ1on{t∈𝕋∣ψ2(t)<0},
with the strict inequalities on some t0∈𝕋∖{τ̂}. 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.
Theorem6.5 improves the main results of Rodriguez [8] and R. Ma and H. Ma [9, 10], where the nonlinearities are not jumping at infinity.



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