# Existence of Three Positive Solutions form-Point Discrete Boundary Value Problems with p-Laplacian

Authors: Yanping Guo; Wenying Wei; Yuerong Chen
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We consider the multi-point discrete boundary value problem with one-dimensionalp-Laplacian operator
Δ(ϕp(Δu(t−1))+q(t)f(t,u(t),Δu(t))=0, t∈{1,…,n−1} subject to the boundary conditions: u(0)=0, u(n)=∑i=1m−2aiu(ξi), where ϕp(s)=|s|p−2s,p>1,ξi∈{2,…,n−2} with 1<ξ1<⋯<ξm−2<n−1 and ai∈(0,1),0<∑i=1m−2ai<1. Using a new fixed point theorem due to Avery and Peterson, we study the existence of at least three positive solutions to the above boundary value problem.


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## Body

## 1. Introduction

The second-order differential and difference boundary value problems arise in many branches of both applied and basic mathematics and have been extensively studied in literature. We refer the reader to some recent results for second-order nonlinear two-point [1–6] and multipoint [7–9] boundary value problems. The main tools used in the above works are fixed point theorems.Recently, Feng and Ge in [9] considered the following multipoint BVPs:(1.1)(ϕp(u(t-1))′)′+q(t)f(t,u(t),u′(t))=0,t∈(0,1),u(0)=0,u(n)=∑i=1m-2aiu(ξi).
The authors obtained sufficient conditions that guarantee the existence of at least three positive solutions by using fixed point theorems due to Avery-Peterson.In this work, we study the existence of multiple positive solutions to the discrete boundary value problem for the one-dimensionalp-Laplacian:(1.2)Δ(ϕp(Δu(t-1)))+q(t)f(t,u(t),Δu(t))=0,t∈{1,…,n-1}u(0)=0,u(n)=∑i=1m-2aiu(ξi),
where Δu(t)=u(t+1)-u(t) for t∈{0,1,…,n-1}, Δ2u(t)=u(t+2)-2u(t+1)+u(t), for t∈{0,1,…,n-2}, and ϕp(s)=|s|p-2s, p>1, ξi∈{2,…,n-2} with 1<ξ1<⋯<ξm-2<n-1.In order to study the existence of at least three positive solutions to the above boundary value problem, we assume thatai, f,q satisfy the following.(H1)
ai∈(0,1) satisfy  0<∑i=1m-2ai<1.(H2)
f:{1,…,n-1}×[0,+∞)×R→(0,+∞) is continuous.(H3)
q(t)>0 for t∈{1,2,…,n-1}.We will depend on an application of a fixed point theorems due to Avery and Peterson, which deals with fixed points of a cone-preserving operator defined on an ordered Banach space to obtain our main results.


## 2. Preliminaries

For the convenience of readers, we provide some background material from the theory of cones in Banach spaces. In this section, we also state Avery-Peterson's fixed point theorem.Definition 2.1.
LetE be a real Banach space over R. A nonempty convex closed set P⊂E is said to be a cone of E if it satisfies the following conditions:(i)
au+bv∈P  for all  u,v∈p  and all  a≥0, b≥0;(ii)
u,-u∈P implies u=0.Every cone P⊂E induces an ordering in E given by x≤y if and only if y-x∈P.Definition 2.2.
An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.Definition 2.3.
The mapα is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E provided that  α:P→[0,∞) is continuous and
(2.1)α(tx+(1-t)y)≥tα(x)+(1-t)α(y)
for all x,y∈P and 0≤t≤1. Similarly, we say that the map β is a nonnegative continuous convex functional on a cone P of a real Banach space E provided that β:P→[0,∞) is continuous and
(2.2)β(tx+(1-t)y)≤tβ(x)+(1-t)β(y)
for all x,y∈P and 0≤t≤1.Letγ and θ be nonnegative continuous convex functionals on a cone P, let α be a nonnegative continuous concave functional on a cone P, and let ψ be a nonnegative continuous functional on a cone P. Then for positive real numbers a,b,c, and d, we define the following convex sets:(2.3)P(γ,d)={u∈P∣γ(u)<d},P(γ,α,b,d)={u∈P∣b≤α(u),γ(u)≤d},P(γ,θ,α,b,c,d)={u∈P∣b≤α(u),θ(u)≤c,γ(u)≤d}
and a closed set(2.4)R(γ,ψ,a,d)={u∈P∣a≤ψ(u),γ(u)≤d}.To prove our results, we need the following fixed point theorem due to Avery and Peterson in [1].Theorem 2.4.
LetP be a cone in a real Banach space E. Let γ and θ be nonnegative continuous convex functionals on P, let α be a nonnegative continuous concave functional on P, and let  ψ be a nonnegative continuous functional on P satisfying ψ(λu)≤λψ(u) for 0≤λ≤1, such that for some positive numbers M and d,
(2.5)α(u)≤ψ(u),∥u∥≤Mγ(u)
for all u∈P(γ,d)¯. Suppose that
(2.6)T:P(γ,d)¯→P(γ,d)¯
is completely continuous and there exist positive numbers a,b, and c with a<b such that(S1)
{u∈P(γ,θ,α,b,c,d)∣α(u)>b}≠∅  and α(Tu)>b for u∈P(γ,θ,α,b,c,d);(S2)
α(Tu)>b  for  u∈P(γ,α,b,d)  with  θ(Tu)>c;(S3)
0∈̅R(γ,ψ,a,d)  and  ψ(Tu)<a for  u∈R(γ,ψ,a,d)  with ψ(u)=a.Then T has at least three fixed points u1,u2,u3∈P(γ,d)¯, such that
γ(ui)≤d   for i=1,2,3,b<α(u1)a<ψ(u2), with α(u2)<b,ψ(u3)<a.


## 3. Related Lemmas

Let the Banach spaceE={u:{0,1,…,n}→R} be endowed with the ordering x≤y if x(t)≤y(t) for all t∈{0,1,…,n}, and the maximum norm(3.1)∥u∥=max{maxt∈{0,1,…,n}|u(t)|,maxt∈{0,1,…,n-1}|Δu(t)|}.
Then, we define the cone P in E by(3.2)P={u∈E∣u(t)≥0,t∈{0,1,…,n};u(0)=0,Δ2u(t)≤0,t∈{0,1,…,n-2}}.
Let k be a natural number, such that k<min{ξ1,n-ξm-2}.Let the nonnegative continuous concave functionalα, the nonnegative continuous convex functionals θ, γ, and the nonnegative continuous functional ψ be defined on the cone P by(3.3)γ(u)=maxt∈{0,1,…,n-1}|Δu(t)|,ψ(u)=θ(u)=maxt∈{0,1,…,n}|u(t)|,α(u)=mint∈{k+1,…,n-k-1}|u(t)|
for u∈P.In order to prove our main results, we need the following lemma.Lemma 3.1.
Ifu∈P, then
(3.4)maxt∈{0,1,…,n}|u(t)|≤nmaxt∈{0,1,…,n-1}|Δu(t)|,that  is,θ(u)≤nγ(u).Proof.
Suppose that the maximum ofu occurs at t0∈{0,1,…,n}; by the definition of the cone P, we know Δu(t+1)≤Δu(t), and then,
(3.5)u(t0)=u(t0)-u(0)=Δu(0)+Δu(1)+⋯+Δu(t0-1)≤t0Δu(0)≤nΔu(0)≤nmaxt∈{0,1,…,n-1}|Δu(t)|.
So, we have
(3.6)maxt∈{0,1,…,n}|u(t)|≤nmaxt∈{0,1,…,n-1}|Δu(t)|.
The proof is complete.By Lemma3.1 and the definitions, the functionals defined above satisfy(3.7)1[n/k]+1θ(u)≤α(u)≤θ(u)=ψ(u),∥u∥=max{θ(u),γ(u)}≤nγ(u)
for all u∈P(γ,d)¯⊂P. Therefore, condition (2.5) is satisfied.Now, we show that(1/([n/k]+1))θ(u)≤α(u). Here, we also suppose θ(u)=u(t0), and by the definitions of α and the cone P, we can distinguish two cases.(i)α(u)=u(k+1), then we certainly have t0≥k+1, and
(3.8)u(t0)=Δu(0)+⋯+Δu(k)+Δu(k+1)+⋯+Δu(2k)+Δu(2k+1)+⋯+Δu(3k)+⋯+Δu([t0k]k)+Δu([t0k]k+1)+⋯+Δu(t0-1)≤([t0k]+1)u(k+1)≤([nk]+1)u(k+1),
that is, 1/([n/k]+1)u(t0)≤u(k+1).(ii)α(u)=u(n-k-1), then t0≤n-k-1, u(n)≤u(n-k-1) and
(3.9)u(t0)=u(n)+(n-t0k-Δu(n-1)-⋯-Δu(n-k-1)-Δu(n-k-2)-⋯-Δu(n-2k-1)-Δu(n-2k-2)-⋯-Δu(n-[n-t0k]k-1)-Δu(n-[n-t0k]k-2)-⋯-Δu(t0))≤u(n)+([n-t0k]+1)[u(n-k-1)-u(n)]≤([n-t0k]+1)u(n-k-1)≤([nk]+1)u(n-k-1),
that is, (1/([n/k]+1))u(t0)≤u(n-k-1). So, we have (1/([n/k]+1))θ(u)≤α(u).Lemma 3.2.
Assume  that(H1)–(H3) hold. Then, for any x∈E,x(t)≥0,
(3.10)Δ(ϕp(Δu(t-1))+q(t)f(t,x(t),Δx(t))=0,t∈{1,…,n-1},(3.11)u(0)=0,u(n)=∑i=1m-2aiu(ξi)
has a unique solution u(t) given by
(3.12)u(t)=∑j=0t-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i)))
or
(3.13)u(t)=∑i=1m-2ai∑j=0ξi-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i)))-∑j=tn-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i))),
where Au satisfies
(3.14)∑i=1m-2ai∑j=0ξi-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i)))=∑j=0n-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i))).Proof.
For anyx∈E suppose that u is a solution of the BVPs (3.7) and (3.11). According to the property of the difference operator, it follows that
(3.15)ϕp(Δu(1))-ϕp(Δu(0))=-q(1)f(1,x(1),Δx(1)),ϕp(Δu(2))-ϕp(Δu(1))=-q(2)f(2,x(2),Δx(2)),⋯ϕp(Δu(t))-ϕp(Δu(t-1))=-q(t)f(t,x(t),Δx(t)),
then
(3.16)ϕp(Δu(t))-ϕp(Δu(0))=-∑i=1tq(i)f(i,x(i),Δx(i)),ϕp(Δu(t))=ϕp(Δu(0))-∑i=1tq(i)f(i,x(i),Δx(i)).
Let ϕp(Δu(0))=Au,
(3.17)Δu(t)=ϕp-1(Au-∑i=1tq(i)f(i,x(i),Δx(i))),
and by
(3.18)Δu(0)=u(1)-u(0)=ϕp-1(Au),Δu(1)=u(2)-u(1)=ϕp-1(Au-∑i=11q(i)f(i,x(i),Δx(i))),⋯Δu(t-1)=u(t)-u(t-1)=ϕp-1(Au-∑i=1t-1q(i)f(i,x(i),Δx(i))),
then
(3.19)u(t)-u(0)=∑j=0t-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i))).
By
(3.20)Δu(n-1)=u(n)-u(n-1)=ϕp-1(Au-∑i=1n-1q(i)f(i,x(i),Δx(i))),Δu(n-2)=u(n-1)-u(n-2)=ϕp-1(Au-∑i=1n-2q(i)f(i,x(i),Δx(i))),⋯Δu(t)=u(t+1)-u(t)=ϕp-1(Au-∑i=1tq(i)f(i,x(i),Δx(i))),
so that
(3.21)u(n)-u(t)=∑j=tn-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i))),u(t)=u(n)-∑j=tn-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i))),
Using the boundary condition (3.11), we can easily obtain
(3.22)u(t)=∑j=0t-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i)))
or
(3.23)u(t)=∑i=1m-2ai∑j=0ξi-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i)))-∑j=tn-1ϕp-1(Au-∑i=1jq(i)f(i,x(i),Δx(i))),
where Au satisfies (3.14).
On the other hand, it is easy to verify that ifu is the solution of (3.12) or (3.13), then u is a solution of (3.10) and (3.11).Lemma 3.3.
For anyu∈E, u(t)≥0, there exists a unique Au∈(-∞,+∞) satisfying (3.14). Moreover, there is a unique n0∈{1,…,n-1}, such that
(3.24)∑i=1n0-1q(i)f(i,u(i),Δu(i))<Au≤∑i=1n0q(i)f(i,u(i),Δu(i)).Proof.
Let(3.25)φ(t)=∑i=1m-2ai∑j=0ξi-1ϕp-1(t-∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=0n-1ϕp-1(t-∑i=1jq(i)f(i,u(i),Δu(i))),t∈(-∞,+∞),
so that
(3.26)φ(0)=∑i=1m-2ai∑j=0ξi-1ϕp-1(-∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=0n-1ϕp-1(-∑i=1jq(i)f(i,u(i),Δu(i)))=∑j=0n-1ϕp-1(∑i=1jq(i)f(i,u(i),Δu(i)))-∑i=1m-2ai∑j=0ξi-1ϕp-1(∑i=1jq(i)f(i,u(i),Δu(i)))>∑i=1m-2ai[∑j=0n-1ϕp-1(∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=0ξi-1ϕp-1(∑i=1jq(i)f(i,u(i),Δu(i)))]≥0,φ(∑i=1n-1q(i)f(i,u(i),Δu(i)))=∑i=1m-2ai∑j=0ξi-1ϕp-1(∑i=j+1n-1q(i)f(i,u(i),Δu(i)))-∑j=0n-1ϕp-1(∑i=j+1n-1q(i)f(i,u(i),Δu(i)))<∑i=1m-2ai[∑j=0ξi-1ϕp-1(∑i=j+1n-1q(i)f(i,u(i),Δu(i)))-∑j=0n-1ϕp-1(∑i=j+1n-1q(i)f(i,u(i),Δu(i)))]≤0.
By the continuity of φ(t), we know that there exists at least one
(3.27)Au∈(0,∑i=1n-1q(i)f(i,u(i),Δu(i)))⊂(-∞,+∞)
satisfying (3.14).
On the other hand,(3.28)φ′(t)=∑i=1m-2ai∑j=0ξi-1(ϕp-1)′(t-∑i=1jq(t)f(i,u(i),Δu(i)))-∑j=0n-1(ϕp-1)′(t-∑i=1jq(i)f(i,u(i),Δu(i)))<∑i=1m-2ai[∑j=0ξi-1(ϕp-1)′(t-∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=0n-1(ϕp-1)′(t-∑i=1jq(i)f(i,u(i),Δu(i)))]≤0.
Then φ(t) is strictly increasing on t∈(-∞,+∞).
So, there exists a uniqueAu∈(-∞,+∞) satisfying (3.14).
Moreover, there exists an0∈{1,…,n-1}, such that
(3.29)∑i=1n0-1q(i)f(i,u(i),Δu(i))<Au≤∑i=1n0q(i)f(i,u(i),Δu(i)).Lemma 3.4.
Assume that(H1)–(H3) hold. If x∈E,x(t)≥0, then the unique solution u(t) of the BVPs  (3.10) and (3.11) has the following properties:(i)
Δ2u(t)≤0, and u(t)≥0;(ii)
there exists a uniquen0∈{0,1,…,n}, such that u(n0)=maxt∈{0,1,…,n}u(t), which n0 is given in Lemma 3.3.Proof.
Suppose thatu(t) is the solution of (3.10) and (3.11). Then we have the following.(i)
By Lemma3.2, it is easy to see that Δ2u(t)≤0. Without loss of generality, we assume that u(n)=min{u(0),u(n)}. By Δ2u(t)≤0, we know that u(t)≥u(n),t∈{0,1,…,n}. So we get u(n)=∑i=1m-2aiu(ξi)≥∑i=1m-2aiu(n), that is, (1-∑i=1m-2ai)u(n)≥0. Hence u(n)≥0. So, from the concavity of u(t), we know that u(t)≥0,t∈{0,1,…,n}.(ii)
From(3.30)u(n0)=∑j=0n0-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i))),u(n0+1)=∑j=0n0-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))+ϕp-1(Au-∑i=1n0q(i)f(i,u(i),Δu(i))),
and by Lemma 3.3, we have Au-∑i=1n0q(i)f(i,u(i),Δu(i))<0, so that u(n0)>u(n0+1).
Also(3.31)u(n0-1)=∑j=0n0-2ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))=u(n0)-ϕp-1(Au-∑i=1n0-1q(i)f(i,u(i),Δu(i))),
and one arrives at u(n0)≥u(n0-1). So, u(n0)=maxt∈{0,1,…,n}u(t).
If there existn1,n2∈{0,1,…,n},n1<n2 such that Δu(n1)=Δu(n2), then
(3.32)0=ϕp(Δu(n1))-ϕp(Δu(n2))=∑n1+1n2q(i)f(i,x(i),Δx(i))>0,
which is a contradiction.Lemma 3.5.
For anyu∈P, define the operator
(3.33)(Tu)(t)={∑j=0t-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i))),0≤t≤n0,∑i=1m-2ai∑j=0ξi-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=tn-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i))),n0+1≤t≤n.
Then T:P→P is completely continuous.Proof.
Using the continuity off and the definition of T, it is easy to show that T:P→P is continuous. Next, we prove that  T is completely continuous.
Suppose that the sequence{ui}⊆P is bounded, then there exists M>0, such that ui(j)≤M, for any i=1,2,…,j=0,1,…,n. By the continuity of f, ϕp-1 and Au are bounded, and we know that there exists M′>0, such that |Tui(t)|≤M′, for t∈{0,1,…,n} and i=1,2,…,n,…. In view of the bounded sequence {Tui(0)}, there exists {ui0}⊆{ui}, such that limi→∞Tui0(0)=a0. For the bounded sequence {Tui0(1)}, there exists {ui1}⊆{ui0}, such that limi→∞Tui1(1)=a1. By repetition in this way, we have that there exists {uij}⊆{uij-1} for j=2,3,…,n, such that limi→∞Tuij(j)=aj. Let y={a0,a1,…,an}; by the definition of the norm on E, there exists {uin}⊆{ui}, such that limi→∞Tuin(j)=y.
Hence,T:P→P is completely continuous.


## 4. Existence of Triple Positive Solutions to (1.2)

We are now ready to apply Avery-Peterson's fixed point theorem to the operatorT to give sufficient conditions for the existence of at least three positive solutions to the BVPs (1.2).Now for convenience we introduce the following notations. Let(4.1)L=ϕp-1(∑i=1n-1q(i)),M=min{∑j=k+1[n/2]ϕp-1(∑i=j+1[n/2]q(i)),∑j=[n/2]n-k-1ϕp-1(∑i=[n/2]+1jq(i))},N=max{∑j=0[n/2]ϕp-1(∑i=j+1[n/2]+1q(i)),∑i=1m-2ai∑j=0ξi-1ϕp-1(∑i=j+1n-1q(i))+∑j=[n/2]n-1ϕp-1(∑i=[n/2]+1jq(i))}.Theorem 4.1.
Assume that conditions(H1)–(H3) hold. Let 0<a<b<([n/k]+1)b<d, and suppose that f satisfies the following conditions:(A1)
f(t,u,v)≤ϕp(d/L), for (t,u,v)∈{0,1,…,n}×[0,nd]×[-d,d];(A2)
f(t,u,v)>ϕp(([n/k]+1)b/M), for (t,u,v)∈{k+1,…,n-k-1}×[b,([n/k]+1)b]×[-d,d];(A3)
f(t,u,v)<ϕp(a/N), for (t,u,v)∈{0,1,…,n}×[0,a]×[-d,d].Then BVPs (1.2) have at least three positive solutions u1, u2, and u3 such that
(4.2)maxt∈{0,1,…,n-1}|Δui(t)|≤d,i=1,2,3,b<mint∈{k+1,…,n-k-1}|u1(t)|,maxt∈{0,1,…,n}|u1(t)|≤b,a<maxt∈{0,1,…,n}|u2(t)|<([nk]+1)b,withmint∈{k+1,…,n-k-1}|u2(t)|<b,maxt∈{0,1,…,n}|u3(t)|<a.Proof.
The BVPs (1.2) have a solution u=u(t) if and only if u solves the operator equation u=Tu. Thus we set out to verify that the operator T satisfies Avery-Peterson's fixed point theorem which will prove the existence of three fixed points of T which satisfy the conclusion of the theorem. Now the proof is divided into some steps.
(1) We will show that (A1) implies that T:P(r,d)¯→P(r,d)¯.
In fact, foru∈P(r,d)¯, there is γ(u)=maxt∈{0,1,…,n-1}|Δu(t)|≤d. With Lemma 3.1, there is maxt∈{0,1,…,n}|u(t)|≤nd, and then condition (A1) implies f(t,u,v)≤ϕp(d/L). On the other hand, for u∈P, there is Tu∈P, then Tu is concave on t∈{0,1,…,n}, and max|ΔTu(t)|=max{|ΔTu(0)|,|ΔTu(n-1)|}, and so
(4.3)γ(Tu)=maxt∈{0,1,…,n-1}|ΔTu(t)|=max{|ΔTu(0)|,|ΔTu(n-1)|}=max{ϕp-1(Au),|ϕp-1(Au-∑i=1n-1q(i)f(i,u(i),Δu(i)))|}≤ϕp-1(∑i=1n-1q(i)ϕp(dL))=dLL=d.
Thus, T:P(r,d)¯→P(r,d)¯ holds.
(2)We show that condition (S1) in Theorem 2.4 holds.
We takeu(t)=([n/k]+1)b, for t∈{1,2,…,n}, and u(0)=0. It is easy to see that u(t)∈P(γ,θ,α,b,([n/k]+1)b,d) and α(u)=([n/k]+1)b>b. Hence {u∈P(γ,θ,α,b,([n/k]+1)b,d)∣α(u)>b}≠∅. Thus for u∈P(γ,θ,α,b,([n/k]+1)b,d), there is b≤u(t)≤([n/k]+1)b, |Δu(t)|≤d. Hence by condition (A2) of this theorem, one has f(t,u,v)>ϕp(([n/k]+1)b/M) for t∈[k+1,n-k-1]. By Lemma 3.4 and combining the conditions on α and P, we have
(4.4)α(Tu)=mink+1≤t≤n-k-1|(Tu)(t)|≥1([n/k]+1)maxt∈{0,…,n}|Tu(t)|=1([n/k]+1)(Tu)(n0)=1([n/k]+1)min{∑j=0n0-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i))),∑i=1m-2ai∑j=0ξi-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=n0n-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))}≥1([n/k]+1)min{∑j=k+1n0-1ϕp-1(∑i=j+1n0-1q(i)f(i,u(i),Δu(i))),∑j=n0n-k-1ϕp-1(∑i=n0+1jq(i)f(i,u(i),Δu(i)))}≥1([n/k]+1)min{∑j=k+1[n/2]ϕp-1(∑i=j+1[n/2]q(i)f(i,u(i),Δu(i))),∑j=[n/2]n-k-1ϕp-1(∑i=[n/2]+1jq(i)f(i,u(i),Δu(i)))}>1([n/k]+1)min{∑j=k+1[n/2]ϕp-1(∑i=j+1[n/2]q(i)ϕp(([n/k]+1)bM)),∑j=[n/2]n-k-1ϕp-1(∑i=[n/2]+1jq(i)ϕp(([n/k]+1)bM))}=1([n/k]+1)([n/k]+1)bMmin{∑j=k+1[n/2]ϕp-1(∑i=j+1[n/2]q(i)),∑j=[n/2]n-k-1ϕp-1(∑i=[n/2]+1jq(i))}=bMM=b.
Therefore we have
(4.5)α(Tu)>b∀u∈P(γ,θ,α,b,([nk]+1)b,d).
Consequently, condition (S1) in Theorem 2.4 is satisfied.
(3)We now prove that (S2) in Theorem 2.4 holds.
With (3.7), we have
(4.6)α(Tu)≥1([n/k]+1)θ(Tu)>1([n/k]+1)([nk]+1)b=b
for u∈P(γ,α,b,d) with θ(Tu)>([n/k]+1)b. Hence, condition (S2) in Theorem 2.4 is satisfied.
(4)Finally, we prove that (S3) in Theorem 2.4 is also satisfied.
Sinceψ(0)=0<a, so 0∈̅R(γ,ψ,a,d). Suppose that u∈R(γ,ψ,a,d) with ψ(u)=a. Then, by the condition (S3) of this theorem,
(4.7)ψ(Tu)=maxt∈{0,1,…,n}|Tu(t)|=Tu(n0)=max{∑j=0n0-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i))),∑i=1m-2ai∑j=0ξi-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))-∑j=n0n-1ϕp-1(Au-∑i=1jq(i)f(i,u(i),Δu(i)))}≤max{∑j=0[n/2]ϕp-1(∑i=j+1[n/2]+1q(i)f(i,u(i),Δu(i))),∑i=1m-2ai∑j=0ξi-1ϕp-1(∑i=j+1n-1q(i)f(i,u(i),Δu(i)))+∑j=[n/2]n-1ϕp-1(∑i=[n/2]+1jq(i)f(i,u(i),Δu(i)))}≤max{∑j=0[n/2]ϕp-1(∑i=j+1[n/2]+1q(i)ϕp(aN)),∑i=1m-2ai∑j=0ξi-1ϕp-1(∑i=j+1n-1q(i)ϕp(aN))+∑j=[n/2]n-1ϕp-1(∑i=[n/2]+1jq(i)ϕp(aN))}=aNmax{∑j=0[n/2]ϕp-1(∑i=j+1[n/2]+1q(i)),∑i=1m-2ai∑j=0ξi-1ϕp-1(∑i=j+1n-1q(i))+∑j=[n/2]n-1ϕp-1(∑i=[n/2]+1jq(i))}=aNN=a.
Thus condition (S3) in Theorem 2.4 holds.Therefore an application of Theorem2.4 implies that BVPs (1.2) have at least three positive solutions u1, u2, and u3 such that(4.8)maxt∈{0,1,…,n-1}|Δui(t)|≤d,i=1,2,3,b<mint∈{k+1,…,n-k-1}|u1(t)|,maxt∈{0,1,…,n}|u1(t)|≤b,a<maxt∈{0,1,…,n}|u2(t)|<([nk]+1)b,withmint∈{k+1,…,n-k-1}|u2(t)|<b,maxt∈{0,1,…,n}|u3(t)|<a.



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*Source: 538431-2009-11-11.xml*