# Global Structure of Positive Solutions of a Discrete Problem with Sign-Changing Weight

Authors: Ruyun Ma; Chenghua Gao; Xiaoling Han; Xiaoqiang Chen
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

LetT>5 be an integer, 𝕋={1,2,…,T}. We are concerned with the global structure
of positive solutions set of the discrete second-order boundary value problems
Δ2u(t-1)+rm(t)f(u(t))=0,t∈𝕋,u(0)=u(T+1)=0,
where r∈ℝ is a parameter, m:𝕋→ℝ changes its sign and m(t)≠0fort∈𝕋.


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## Body

## 1. Introduction

LetT>5 be an integer, 𝕋={1,2,…,T}. In this paper, we are concerned with the global structure of positive solutions set of the discrete second-order boundary value problems (1.1)Δ2u(t-1)+rm(t)f(u(t))=0,t∈T,(1.2)u(0)=u(T+1)=0,
where r∈ℝ is a parameter, m:𝕋→ℝ changes its sign, and m(t)≠0 for t∈𝕋.The boundary value problems with sign-changing weight arise from a selection-migration model in population genetics, see Fleming [1]. m changes sign corresponds to the fact that an allele A1 holds an advantage over a rival allele A2 at the same points and is at a disadvantage at others. The parameter r corresponds to the reciprocal of the diffusion. So, the existence and multiplicity of positive solutions with sign-changing weight in continuous case has been studied by many authors, see, for example, [2–8] and the references therein.For the discrete case, there are many literature dealing with difference equations similar to (1.1) subject to various boundary value conditions. We refer to [9–17] and the references therein. However, there are few papers to discuss the existence of positive solutions to (1.1) and (1.2) if m(t) changes sign on 𝕋. Maybe the main reason is the spectrum of the following linear eigenvalue problems (1.3)Δ2u(t-1)+λm(t)u(t)=0,t∈T,u(0)=u(T+1)=0
is not clear when m changes its sign on 𝕋.In 2008, Shi and Yan [18] investigated the spectrum of the second-order boundary value problems of difference equations (1.4)-Δ[p(t-1)Δu(t-1)]+q(t)u(t)=λm(t)u(t),t∈T,(p(0)Δu(0)-p(T)Δu(T))=K(u(0)u(T)),
where K=(k11k12k21k22) is a symmetric and positive definite 2×2 matrix. p:{0,1,…,T}→ℝ, q:𝕋→ℝ, and m:𝕋→ℝ satisfy(A1)
m:𝕋→ℝ changes its sign, and m(t)≠0 for t∈𝕋.(A2)
p(t-1)≥0,q(t)≥0,p(t-1)+q(t)>0,fort∈𝕋, and  p(0)+k11≠0,  p(T)>0.They proved that (1.4) has T eigenvalues λ1≤λ2≤⋯≤λT which have T corresponding linearly independent and orthogonal eigenfunctions.However, it's easy to see that (1.3) is not included in (1.4) under the condition (A2). Also, Shi and Yan [18] provided no information about the sign of the eigenvalues and no information about the corresponding eigenfunctions.In this paper, we will show that (1.3) has two principal eigenvalues λm,-<0<λm,+, and the corresponding eigenfunctions we denote by ψm,- and ψm,+ don't change their sign on 𝕋̂:={0,1,…,T,T+1}. Based on this result, using Rabinowitz's global bifurcation theorem [19], we will discuss the global structure of positive solutions set of (1.1) and (1.2).The assumptions we are interested in this paper are as follows:(H1)
f∈C(ℝ,ℝ) with sf(s)>0 for s≠0;(H2)
f0=lim|s|→0(f(s)/s)∈(0,∞);(H3)
f∞=lim|s|→+∞(f(s)/s)=0;(H3′)
f∞=∞.Our main result is the following.Theorem 1.1.
Assume that (H1), (H2) hold.(i)
If (H3) holds, then there exist0<λ*≤λm,+/f0 and λm,-/f0≤λ*<0 such that (1.1), (1.2) has at least one positive solution for r∈(-∞,λ*)∪(λ*,∞).(ii)
If (H3′) holds, then there exist ρ*≤λm,-/f0<0 and ρ*≥λm,+/f0>0 such that (1.1), (1.2) has at least one positive solution for r∈(ρ*,0)∪(0,ρ*).The rest of the paper is arranged as follows. In Section2, the existence of two principal eigenvalues of (1.3), and some properties of these two eigenvalues will be discussed. In Section 3, we will prove our main result.


## 2. Existence of Two Principal Eigenvalues to(1.3)

Recall that𝕋={1,2,…,T} and 𝕋̂={0,1,…,T+1}. Let X={u:𝕋̂→ℝ∣u(0)=u(T+1)=0}. Then X is a Banach space under the norm ∥u∥X=maxt∈𝕋̂|u(t)|. Let Y={u∣u:𝕋→ℝ}. Then Y is a Banach space under the norm ∥u∥Y=maxt∈𝕋|u(t)|.It's well known that the operatorχ:X→Y, (2.1)χ(0,u(1),u(2),…,u(T),0)=(u(1),u(2),…,u(T))
is a homomorphism.Define the operatorL:X→Y by (2.2)Lu(t)=-Δ2u(t-1),t∈T.In this section, we will discuss the existence of principal eigenvalues for (1.3) with m:𝕋→ℝ that changes its sign. The main idea we will use aries from [20, 21].Theorem 2.1.
Equation (1.3) has two principal eigenvalues λm,+ and λm,- such that λm,-<0<λm,+, and the corresponding eigenfunctions, we denoted by ψm,+ and ψm,- do not change sign on 𝕋̂.Proof.
Consider, for fixedλ, the eigenvalue problems
(2.3)Lu-λm(t)u(t)=μu(t),t∈T,u(0)=u(T+1)=0.
By Kelley and Peterson [22, Theorem  7.6], for fixed λ, (2.3) has T-simple eigenvalues
(2.4)μm,1(λ)<μm,2(λ)<⋯<μm,T(λ),
and the corresponding eigenfunction ψm,k(λ,t) has exactly k-1 simple generalized zeros.
Thus,λ is a principal eigenvalue of (1.3) if and only if μm,1(λ)=0.
On the other hand, let(2.5)Sm,λ={∑t=0T|Δϕ(t)|2-λ∑t=1Tm(t)ϕ(t)2:ϕ∈X,∑t=1Tϕ(t)2=1}.
Clearly, Sm,λ is bounded below. Then μm,1(λ)=infSm,λ.
For fixedϕ∈X, λ→∑t=0T|Δϕ(t)|2-λ∑t=1Tm(t)ϕ2(t) is an affine and so concave function. As the infimum of any collection of concave functions is concave, it follows that λ→μm,1(λ) is a concave function. Also, by considering test functions ϕ1,ϕ2∈X such that ∑t=1Tm(t)ϕ12(t)<0 and ∑t=1Tm(t)ϕ22(t)>0, it is easy to see that μm,1(λ)→-∞ as λ→±∞. Thus, λ→μm,1(λ) is an increasing function until it attains its maximum and is a decreasing function thereafter.
Sinceμ1(0)>0, λ→μm,1(λ) must have exactly two zeros. Thus, (1.3) has exactly two principal eigenvalues, λm,+>0 and λm,-<0, and the corresponding eigenfunctions we denoted by ψm,+ and ψm,- don't change sign on 𝕋̂.Remark 2.2.
From the proof of Theorem2.1, it is not difficult to see that the following results hold.(i)
Ifm(t)≥0 on 𝕋 and there exist at least one point t0∈𝕋 such that m(t0)>0, then (1.3) has only one principal eigenvalue λm,+>0.(ii)
Ifm(t)≤0 on 𝕋 and there exist at least one point t0∈𝕋 such that m(t0)<0, then (1.3) has only one principal eigenvalue λm,-<0.Now, we give some properties for the above principal eigenvalue(s).Theorem 2.3.
Letm:𝕋→ℝ change its sign. Assume that there exists m′:𝕋→ℝ such that m(t)≤m′(t) for t∈𝕋. Then the followings hold.(i)
Ifm′ changes sign on 𝕋, then λm′,-≤λm,-, λm′,+≤λm,+;(ii)
Ifm′≥0, then 0<λm′,+≤λm,+.Proof.
For convenience, we only prove the case (i). It can be seen that forλ>0, Sm,λ≥Sm′,λ, which implies μm,1(λ)≥μm′,1(λ) and consequently, λm,+≥λm′,+.
On the other hand, forλ<0, Sm,λ≤Sm′,λ, which implies μm,1(λ)≤μm′,1(λ) and consequently, λm,-≥λm′,-.Suppose that𝕋0={a+1,a+2,…,b-1} is a strict subset of 𝕋, and m𝕋0 denote the restriction of m on 𝕋0. Consider the linear eigenvalue problems (2.6)Δ2u(t-1)+λm(t)u(t)=0,t∈T0,u(a)=u(b)=0.Then we get the following result.Theorem 2.4.
(i) Ifm(t)>0 for t∈𝕋0, then (2.6) has only one positive principal eigenvalue λm𝕋0,+ such that 0<λm,+<λm𝕋0,+.
(ii) Ifm(t)<0 for t∈𝕋0, then (2.6) has only one principal eigenvalue λm𝕋0,- such that λm𝕋0,-<λm,-.
(iii) Ifm changes its sign on 𝕋0, then (2.6) has two principal eigenvalue λm𝕋0,-<0 and λm𝕋0,+>0 such that
(2.7)λm,+<λmT0,+,λmT0,-<λm,-.Proof.
Consider the following problems:(2.8)-Δ2u(t-1)-λm(t)u(t)=μu(t),t∈T0,u(a)=u(b)=0.
Let μm𝕋0,k denote the kth eigenvalue of (2.8), and ψm𝕋0,k the corresponding eigenfunction which has exactly k-1 generalized zeros in (a,b). Let E={u:{a,…,b}→ℝ∣u(a)=u(b)=0} be a Banach space under the norm ∥u∥E=max{|u(t)|:t∈{a,a+1,…,b-1,b}}. Let
(2.9)SmT0,λ={∑t=ab-1|Δϕ(t)|2-λ∑t=a+1b-1m(t)ϕ(t)2:ϕ∈E,∑t=a+1b-1ϕ(t)2=1}.
We get μm𝕋0,1=infSm𝕋0,λ. Similar to the proof of Theorem 2.1, we get the following assertions.(i)
Ifm(t)>0 for t∈𝕋0, then (2.6) has only one principal eigenvalue λm𝕋0,+>0.(ii)
Ifm(t)<0 for t∈𝕋0, then (2.6) has only one negative principal eigenvalue λm𝕋0,-<0.(iii)
Ifm changes its sign on 𝕋0, then (2.6) has two principal eigenvalue λm𝕋0,-<0 and λm𝕋0,+>0.
Now, we prove that the inequalities in (i), (ii), and (iii) hold. For convenience, suppose that∑t=a+1b-1ψm𝕋0,12(λ,t)=1.
Letψ̃m𝕋0,1 denote the extension of ψm𝕋0,1 by zero on 𝕋, that is,
(2.10)ψ̃mT0,1(λ,t)={ψmT0,1(λ,t),t∈T0,0,t∈T∖T0.
Then,
(2.11)μmT0,1(λ)=∑t=ab-1|ΔψmT0,1(λ,t)|2-λ∑t=a+1b-1m(t)ψmT0,12(λ,t)=∑t=1T|Δψ̃mT0,1(λ,t)|2-λ∑t=1Tm(t)ψ̃mT0,12(λ,t)>inf⁡v∈X⁡{∑t=1T|Δv(λ,t)|2-λ∑t=1Tm(t)v2(λ,t),∑t=1Tv2(λ,t)=1}=μm,1(λ),
which implies the desired results.


## 3. The Proof of the Main Result

First, we deal with the caser>0.Recall thatL:X→Y, (3.1)Lu(t)=-Δ2u(t-1).
Let ζ,ξ∈C(ℝ,ℝ) be such that (3.2)f(u)=f0u+ζ(u),f(u)=f∞u+ξ(u).
Clearly, (3.3)lim⁡|u|→0⁡ζ(u)u=0,lim⁡|u|→∞⁡ξ(u)u=0.
Let (3.4)ξ̃(u)=max⁡0≤s≤u|ξ(s)|.
Then, ξ̃ is nondecreasing and (3.5)lim⁡|u|→∞⁡ξ̃(u)u=0.
Let us consider (3.6)Lu-λm(t)rf0u-λm(t)rζ(u)=0,
as a bifurcation problem from the trivial solution u≡0.Equation (3.6) can be converted to the equivalent equation (3.7)u(t)=λL-1[m(⋅)rf0u(⋅)+m(⋅)rζ(u(⋅))](t).
Further, we note that ∥L-1[m(·)ζ(u(·))]∥=o(∥u∥) for u near 0 in X, since (3.8)‖L-1[m(⋅)ζ(u(⋅))]‖=max⁡t∈T|∑s=1TG(t,s)m(s)ζ(u(s))|≤C⋅max⁡s∈T|m(s)|‖ζ(u(⋅))‖,
where (3.9)G(t,s)=1T+1{(T+1-t)s,0≤s≤t≤T+1,(T+1-s)t,0≤t≤s≤T+1.The results of Rabinowitz [19] for (3.6) can be stated as follows: from (λm,+/rf0,0), there emanates an unbounded continua 𝒞+ of positive solutions in ℝ×X.It is clear that any solution of (3.6) of the form (1,u) yields a solution u of (1.1) and (1.2). So, we focus on the shape of 𝒞+ under the conditions (H1)–(H3) or (H1)–(H3′), and we will show that 𝒞+ crosses the hyperplane {1}×X in ℝ×X.Lemma 3.1.
Suppose that (H1)–(H3) hold. LetJ=[a,b] be a given compact interval in (0,∞). Then for all r∈J, there exists MJ>0 such that all possible positive solution u of (1.1) and (1.2) satisfy ∥u∥X≤MJ.Proof.
Suppose on the contrary that there exists a sequence{yn} of positive solutions for (3.6) with {μn}⊂J and ∥yn∥X→∞. Let α∈(0,1/bQ), where Q=∑s=1TG(s,s)|m(s)|. Then, by (H3), there exists uα>0 such that u>uα implies f(u)<αu.
LetKα=maxu∈[0,uα]f(u) and let An={t∈𝕋̂∣yn(t)≤uα} and Bn={t∈𝕋̂∣yn(t)>uα}. Then we have
(3.10)yn(t)=μn∑s=1TG(t,s)m(s)f(yn(s))=μn∑AnG(t,s)m(s)f(yn(s))+μn∑BnG(t,s)m(s)f(yn(s))≤μnKαQ+μn∑BnG(t,s)m(s)f(yn(s))
for t∈𝕋̂. Thus,
(3.11)1μn≤KαQ‖yn‖X+∑BnG(t,s)m(s)f(yn(s))‖yn‖X.
On Bn,yn(s)>uα implies f(yn(s))/∥yn∥X<f(yn(s))/yn(s)<α. Thus,
(3.12)1μn≤KαQ‖yn‖X+α∑BnG(s,s)|m(s)|≤KαQ‖yn‖X+αQ.
Since 0<a≤μn≤b for all n, we have 1/μn≥1/b for all n, and, thus,
(3.13)1b≤KαQ‖yn‖X+αQ.
By the fact ∥yn∥X→∞ as n→∞, we get
(3.14)1b≤αQ<1b.
This contradiction completes the proof.Lemma 3.2.
Suppose that (H1)–(H3) hold. ThenProjℝ𝒞+⊃[λm,+/rf0,+∞).Proof.
Assume on the contrary thatsup{λ∣(λ,y)∈𝒞+}<∞, then, there exists a sequence {(μn,yn)}∈𝒞+ such that
(3.15)lim⁡n→∞‖yn‖X=∞,μn<C0
for some positive constant C0 independent of n, since 𝒞+ is unbounded. On the other hand, μn>0 for all n∈ℕ, since (0,0) is the only solution of (3.6) for λ=0 and 𝒞+∩({0}×X)=∅. Meanwhile, {(μn,yn)} satisfy
(3.16)Δ2yn(t-1)+μnm(t)f(yn(t))yn(t)yn(t)=0,t∈T,yn(0)=yn(T+1)=0.
By (H3), there exist a positive constant Lf>0 such that f(u)≤Lfu. Define a function χ:𝕋→[0,∞) by
(3.17)χ(t)={Lf,m(t)>0,0,m(t)<0.
Then, m(t)(f(yn(t))/yn(t))≤χ(t)m(t). Let γ+ be the principal eigenvalue of linear eigenvalue problem
(3.18)Δ2v(t-1)+γχ(t)m(t)v(t)=0,t∈T,v(0)=v(T+1)=0.
Then, by Remark 2.2, γ+>0. Subsequently, by Theorem 2.3, we know that
(3.19)γ+≤μn.
This combine with Lemma 3.1, limn→∞μn=∞, which contradicts (3.15). Thus,
(3.20)ProjRC+⊃[λm,+rf0,∞).Now, by Lemma3.2, 𝒞+ crosses the hyperplane {1}×X in ℝ×X, and, then, Theorem 1.1(i) holds. To obtain Theorem 1.1(ii), we need to prove the following Lemma.Lemma 3.3.
Suppose that (H1), (H2), and (H3′) hold. Then Projℝ𝒞+⊃(0,λm,+/rf0).Proof.
Let{(μn,yn)}⊂𝒞+ be such that |μn|+∥yn∥X→∞ as n→∞. Then,
(3.21)Δ2yn(t-1)+μnm(t)f(yn(t))=0,t∈T,yn(0)=yn(T+1)=0.
If {∥yn∥} is bounded, say, ∥yn∥≤M1, for some M1 independent of n, then we may assume that
(3.22)lim⁡n→∞μn=∞.
Note that
(3.23)f(yn(t))yn(t)≥inf⁡{f(s)s∣0<s≤M1}>0.
Then, there exist two constants M3>0,M2>0 such that
(3.24)0<M2<f(yn(t))yn(t)<M3.
Define two functions χ1,χ2:𝕋→(0,∞) by
(3.25)χ1={M2,m(t)>0,M3,m(t)<0.χ2={M3,m(t)>0,M2,m(t)<0.
Let η*,η* be the positive principal eigenvalue of the linear eigenvalue problems
(3.26)Δ2v(t-1)+ηχ2(t)m(t)v(t)=0,t∈T,(3.27)v(0)=v(T+1)=0,(3.28)Δ2v(t-1)+ηχ1(t)m(t)v(t)=0,t∈T,(3.29)v(0)=v(T+1)=0,
respectively.
Combining (3.22) and (3.24) with the relation
(3.30)Δ2yn(t-1)+rμnm(t)f(yn(t))yn(t)yn(t)=0,t∈T,yn(0)=yn(T+1)=0,
using Theorem 2.3, we get
(3.31)η*r≤μn≤η*r.
This contradicts (3.22). So, {∥yn∥X} is bounded uniformly for all n∈ℕ.
Now, taking{(μn,yn)}⊂𝒞+ be such that
(3.32)‖yn‖X⟶+∞,asn⟶+∞.
We show that limn→∞μn=0.
Suppose on the contrary that, choosing a subsequence and relabeling if necessary,μn≥b0 for some constant b0>0. By (3.32), there exists t0∈𝕋 such that yn(t0)=∥yn∥X and yn(t0)→+∞ as n→+∞. Thus,
(3.33)μnf(yn(t0))yn(t0)⟶+∞,asn⟶+∞.
Now, the proof can be divided into two cases.
Case 1 (m(t0)>0).
Consider the following linear eigenvalue problems(3.34)Δ2v(t-1)+αm(t)v(t)=0,t=t0,v(t0-1)=v(t0+1)=0.
By Theorem 2.4, (3.34) has a positive principal eigenvalue α+, and
(3.35)μnf(yn(t0))yn(t0)≤α+,
which contradicts (3.33).Case 2 (m(t0)<0).
Since(μn,yn) is a solution of (3.7), we get
(3.36)0≥Δ2yn(t0-1)=-μnrm(t0)f(yn(t0))>0.
This is a contradiction.
Thus,limn→∞μn=0.
At last, we deal with the caser<0.
Let us consider(3.37)Lu-λrm(t)f0u-λrm(t)ζ(u)=0
as a bifurcation problem from the trivial solution u≡0. The results of Rabinowitz [19] for (3.37) can be stated as follows: from (λm,-/-rf0,0), there emanates an unbounded continua 𝒞- of positive solutions in ℝ×X.
It is clear that any solution of (3.37) of the form (-1,u) yields a solution u of (1.1) and (1.2). Now, our proofs focus on the shape of 𝒞-. It will be proved that when (H1)–(H3) hold, then Projℝ𝒞-⊃(-∞,λm,-/-rf0), and when (H1)–(H3′) hold, Projℝ𝒞-⊃(λm,-/-rf0,0), that is, 𝒞- crosses the hyperplane {-1}×X in ℝ×X. Since the proof is similar to the case r>0, we omit it.Remark 3.4.
As an application of Theorem1.1, let us consider nonlinear discrete problem with indefinite weight
(3.38)Δ2u(t-1)+rm̂(t)f̂(u(t))=0,t∈T,u(0)=0,u(3)=0,
where 𝕋={1,2}, m̂:𝕋→ℝ with m̂(1)=1 and m̂(2)=-1, and
(3.39)f̂(s)={s,s∈[-1,1],2s1+s4,s∈(-∞,-1)∪(1,∞).
Then,
(3.40)f̂0=1,f̂∞=0.
To find the principal eigenvalues of linear eigenvalue problem(3.41)Δ2u(t-1)+λm̂(t)u(t)=0,t∈T,(3.42)u(0)=0,u(3)=0,
we rewrite (3.41) to the recursive sequence
(3.43)u(t+1)=[2-λm̂(t)]u(t)-u(t-1).
This together with the initial value condition
(3.44)u(0)=0,u(1)=1
imply that
(3.45)u(2)=[2-λm̂(1)]u(1)-u(0)=2-λ,u(3)=[2-λm̂(2)]u(2)-u(1)=3-λ2.
The last equation together with the boundary value condition u(3)=0 imply that
(3.46)λm̂,-=-3,λm̂,+=3.
Thus by Theorem 1.1(i), (3.38), has a positive solution if r∈(-∞,-3)∪(3,∞).



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