# The Existence of Positive Solutions for a Fourth-Order Difference Equation with Sum Form Boundary Conditions

Authors: Yanping Guo; Xuefei Lv; Yude Ji; Yongchun Liang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We consider the fourth-order difference equation:Δ(z(k+1)Δ3u(k-1))=w(k)f(k,u(k)),  k∈{1,2,…,n-1} subject to the boundary conditions: u(0)=u(n+2)=∑i=1n+1g(i)u(i), aΔ2u(0)-bz(2)Δ3u(0)=∑i=3n+1h(i)Δ2u(i-2), aΔ2u(n)-bz(n+1)Δ3u(n-1)=∑i=3n+1h(i)Δ2u(i-2), where a,b>0 and Δu(k)=u(k+1)-u(k) for k∈{0,1,…,n-1},  f:{0,1,…,n}×[0,+∞)→[0,+∞) is continuous. h(i) is nonnegative i∈{2,3,…,n+2}; g(i) is nonnegative for i∈{0,1,…,n}. Using fixed point theorem of cone expansion and compression of norm type and Hölder’s inequality, various existence, multiplicity, and nonexistence results of positive solutions for above problem are derived, which extends and improves some known recent results.


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## Body

## 1. Introduction

Boundary value problems (BVPs) for ordinary differential equations arise in different areas of applied mathematics and so on. The existence of solutions for second order and higher order nonlocal boundary value problems has been studied by several authors; for example, see [1–11] and the references therein. Many authors have also discussed the existence of positive solutions for higher order difference equation BVPs [12, 13], by using fixed point theorem of cone expansion and compression of norm type, sufficient conditions for the existence of positive solutions for fourth-order and third-order difference equation BVPs are established, respectively. Recently, there has been much attention on the existence of positive solutions for the fourth-order differential equations with integral boundary conditions [14–17]. In [17], Zhang and Ge considered the differential equation BVP:
(1)(u(t)x′′′(t))′=w(t)f(t,x(t)),0<t<1,x(0)=x(1)=∫01‍g(s)x(s)ds,ax′′(0)-blim⁡t→0+u(t)x′′′(0)=∫01‍h(s)x′′(s)ds,ax′′(1)+blim⁡t→1-u(t)x′′′(1)=∫01‍h(s)x′′(s)ds,
wherea,b>0, u∈C1([0,1]→[0,+∞)) is symmetric on [0,1], w∈Lp[0,1]for some 1≤p≤+∞, and it is symmetric on the interval[0,1],f:[0,1]×[0,+∞)→[0,+∞)is continuous, andf(1-t,x)=f(t,x)for all(t,x)∈[0,1]×[0,+∞), andg,h∈L1[0,1]are nonnegative, symmetric on[0,1]. The authors made use of fixed point theorem of cone expansion and compression of norm type and Hölder inequality to prove the existence of positive solutions for the above problem.Motivated by the above works, we intend to study the existence and nonexistence of positive solutions of the following fourth-order difference BVP with sum form boundary conditions:(2)Δ(z(k+1)Δ3u(k-1))=w(k)f(k,u(k)),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiik∈{1,2,…,n-1},u(0)=u(n+2)=∑i=1n+1‍g(i)u(i),aΔ2u(0)-bz(2)Δ3u(0)=∑i=3n+1‍h(i)Δ2u(i-2),aΔ2u(n)+bz(n+1)Δ3u(n-1)=∑i=3n+1‍h(i)Δ2u(i-2).Throughout this paper, we make the following assumptions:(A0)
zis symmetric on{2,3,…,n+1}and0<∑k=2n+1‍z(k)<+∞;(A1)
there is am>0such thatw(k)≥m/(n-1)for1≤p≤+∞;(A2)
f:{0,1,…,n}×[0,+∞)→[0,+∞)is continuous;(A3)
h(i)is nonnegative on{2,3,…,n+2}and0≤v<a,g(i)is nonnegative on{0,1,…,n+2}and0≤μ<1, wherev=∑i=3n+1‍h(i),μ=∑i=1n+1‍g(i);(A4)
a,b>0,Q=2ab+a2∑j=2n+1‍(1/z(j))>0.In order to establish the existence of positive solutions of the problem (2), we need the following definitions, theorem, and lemma.Definition 1.
A functionx(t)is said to be a solution of problem (2) ifx(t)satisfying BVP (2).Definition 2 (see [18]).
LetEbe a real Banach space over  R. A nonempty closed setP⊂Eis said to be a cone provided that(i)
au+bv∈Pfor allu,v∈Pand alla≥0,b≥0and(ii)
u,-u∈Pimpliesu=0.Every coneP⊂Einduces an ordering inEgiven byx≤yif and only ify-x∈P.Theorem 3 (see [18]).
LetPbe a cone in a real Banach spaceE. Assume thatΩ1,Ω2are bounded open sets inEwith0∈Ω1,Ω¯1⊂Ω2. If
(3)A:P∩(Ω¯2∖Ω1)⟶P
is completely continuous such that either(i)
∥Ax∥≤∥x∥,∀x∈P∩∂Ω1and∥Ax∥≥∥x∥,∀x∈P∩∂Ω2,or(ii)
∥Ax∥≥∥x∥,∀x∈P∩∂Ω1and∥Ax∥≤∥x∥,∀x∈P∩∂Ω2,thenAhas at least one fixed point inP∩(Ω¯2∖Ω1).Lemma 4 (Hölder).
Suppose thatu={u1,u2,…,un}is a real-valued column; let
(4)∥u∥p={(∑k=1n‍|uk|p)1/p,0<p<∞,sup⁡k∈{1,2,…,n}|uk|,p=∞,p, q satisfy the condition1/p+1/q=1which are called conjugate exponent, andq=∞forp=1. If1≤p≤∞, then
(5)∥uv∥1≤∥u∥p∥v∥q,
which can be recorded as
(6)∑k=1n‍|ukvk|≤{(∑k=1n‍|uk|p)1/p(∑k=1n‍|vk|q)1/q,1<p<∞,(∑k=1n‍|uk|)(sup⁡k∈{1,2,…,n}|vk|),p=1,(sup⁡k∈{1,2,…,n}|uk|)(∑k=1n‍|vk|),p=∞.


## 2. Preliminaries

LetJ={0,1,…,n+2}; E={u(k):{0,1,…,n+2}→R}is a real Banach space with the norm∥·∥defined by
(7)∥u∥=max⁡k∈J|u(k)|.LetKbe a cone ofE, and
(8)Kr={u∈K:∥u∥≤r},∂Kr={u∈K:∥u∥=r},K¯r,R={u∈K:r≤∥u∥≤R},
where0<r<R.In our main results, we will use the following lemmas and properties.Lemma 5.
Suppose that(A0)and(A1)hold andv≠1; then, for ally∈E, the BVP
(9)-Δ(z(k+1)Δu(k+1))=y(k+2),k∈{1,2,…,n+1},au(2)-bz(2)Δu(2)=∑i=3n+1‍h(i)u(i),au(n+2)+bz(n+1)Δu(n+1)=∑i=3n+1‍h(i)u(i)
has unique solutionugiven by
(10)u(k)=∑i=3n+1‍H(k,i)y(i),
where
(11)H(k,i)=G(k,i)+1a-v∑τ=3n+1‍G(i,τ)h(τ),G(k,i)=1Q{(b+a∑j=kn+1‍1z(j))(b+a∑j=2i-1‍1z(j)),2≤i<k,(b+a∑j=in+1‍1z(j))(b+a∑j=2k-1‍1z(j)),k≤i≤n+2,
whereQ=2ab+a2∑j=2n+1‍(1/z(j)),v=∑i=3n+1‍h(i).Proposition 6.
Assume that0≤v<a; then
(12)H(k,i)>0,G(k,i)>0,k,i∈{2,3,…,n+2}.Proposition 7.
Assume thatk,i∈{2,3,…,n+2}; then
(13)1Qb2≤G(k,i)≤G(i,i)≤1QD,G(n+4-k,n+4-i)=G(k,i),
where
(14)D=(b+a∑j=2n+1‍1z(j))2,k,i∈{2,3,…,n+2}.Proposition 8.
Suppose that0≤v<a; then
(15)1Qvb2γ<H(k,i)≤H(i,i)≤1QγDa,iiiiiiiiiiiiiiiiiiiiiik,i∈{2,3,…,n+2},
where
(16)D=(b+a∑j=2n+1‍1z(j))2,γ=1a-v,iiiiiiiiiiiiiiiiiiiiiiiiiik,i∈{2,3,…,n+2}.Proof.
From Lemma5 and Proposition 7, we have
(17)H(k,i)=G(k,i)+1a-v∑τ=3n+1‍G(i,τ)h(τ)>1a-v∑τ=3n+1‍G(i,τ)h(τ)≥1a-v1Qb2∑τ=3n+1‍h(τ)=1Qγb2v.
On the other hand, usingG(k,i)≤G(i,i)≤(1/Q)D, we get
(18)H(k,i)=G(k,i)+1a-v∑τ=3n+1‍G(i,τ)h(τ)≤G(i,i)+1a-v∑τ=3n+1‍G(i,τ)h(τ)≤1QD+1QD1a-v∑τ=3n+1‍h(τ)≤1QD(1+va-v)=1QDaγ.Lemma 9.
Suppose thatμ≠1, for ally∈E; the BVP
(19)-Δ2u(k-1)=y(k),k∈{1,2,…,n+1},u(0)=u(n+2)=∑i=1n+1‍g(i)u(i)
has unique solutionugiven by
(20)u(k)=∑i=1n+1‍H1(k,i)y(i),
where
(21)H1(k,i)=G1(k,i)+11-μ∑τ=1n+1‍G1(τ,i)g(τ),G1(k,i)={in+2(n+2-k),0≤i<k,kn+2(n+2-i),k≤i<n+2.Proof.
From the properties of difference operator, we can get(22)-Δu(k)+Δu(k-1)=y(k);
then we have
(23)-Δu(1)+Δu(0)=y(1),-Δu(2)+Δu(1)=y(2),-Δu(3)+Δu(2)=y(3),⋮-Δu(k-1)+Δu(k-2)=y(k-1).
It can imply that
(24)-Δu(k-1)+Δu(0)=∑i=1k-1‍y(i).
LetΔu(0)=A; we have
(25)-Δu(k-1)=A-∑i=1k-1‍y(i).
That is
(26)u(k)-u(k-1)=A-∑i=1k-1‍y(i);
thus,
(27)u(1)-u(0)=A,u(2)-u(1)=A-∑i=11‍y(i),u(3)-u(2)=A-∑i=12‍y(i),⋮u(k)-u(k-1)=A-∑i=1k-1‍y(i).
We can get(28)u(k)=u(0)+kA-∑j=1k-1‍∑i=1j‍y(i),u(n+2)=u(0)+(n+2)A-∑j=1n+1‍∑i=1j‍y(i).
From the boundary conditions, we have(29)A=1n+2∑j=1n+1‍∑i=1j‍y(i);
thus,
(30)u(k)=∑i=1n+1‍g(i)u(i)+kn+2∑j=1n+1‍∑i=1j‍y(i)-∑j=1k-1‍∑i=1j‍y(i).
Because
(31)kn+2∑j=1n+1‍∑i=1j‍y(i)=kn+2×[y(1)+(y(1)+y(2))+(y(1)+y(2)+y(3))+⋯+(y(1)+y(2)+⋯+y(n+1))]=kn+2[(n+1)y(1)+ny(2)+(n-1)y(3)+⋯+y(n+1)]=kn+2(n+1)y(1)+kn+2ny(2)+⋯+kn+2y(n+1),∑j=1k-1‍∑i=1j‍y(i)=y(1)+(y(1)+y(2))+⋯+(y(1)+y(2)+⋯+y(k-1))=(k-1)y(1)+(k-2)y(2)+⋯+y(k-1),
we have
(32)kn+2∑j=1n+1‍∑i=1j‍y(i)-∑j=1k-1‍∑i=1j‍y(i)=[k(n+1)n+2-(k-1)]y(1)+[knn+2-(k-2)]y(2)+⋯+[kn+2(n-k+3)-1]y(k-1)+[kn+2(n-k+2)]y(k)+⋯+kn+2y(n+1).
Thus, we get
(33)u(k)=∑i=1n+1‍g(i)u(i)+∑i=1n+1‍G1(k,i)y(i),
where
(34)G1(k,i)={in+2(n+2-k),0≤i<k,kn+2(n+2-i),k≤i<n+2.
Multiplying the above equation withg(k), summing them from1ton+1, we have
(35)∑k=1n+1‍g(k)u(k)=∑k=1n+1‍g(k)∑i=1n+1‍g(i)u(i)+∑k=1n+1‍g(k)∑i=1n+1‍G1(k,i)y(i),∑i=1n+1‍g(i)u(i)=11-∑k=1n+1g(k)∑k=1n+1‍g(k)∑i=1n+1‍G1(k,i)y(i).
So, we can get
(36)u(k)=∑i=1n+1‍G1(k,i)y(i)+11-∑k=1n+1g(k)∑k=1n+1‍g(k)∑i=1n+1‍G1(k,i)y(i)=∑i=1n+1‍H1(k,i)y(i).Proposition 10.
Assume that0≤μ<1; then
(37)H1(k,i)>0,G1(k,i)>0,k,i∈{1,2,…,n+1},H1(k,i)≥0,G1(k,i)≥0,k,i∈J.Proposition 11.
Fork,i∈J, one has
(38)1n+2e(i)e(k)≤G1(k,i)≤G1(i,i)=i(1-in+2)=e(i)≤n+24,G1(n+2-k,n+2-i)=G1(k,i),
wheree(i)=(i/(n+2))(n+2-i).Proposition 12.
If0≤μ<1, then, for allk,i∈J, one has
(39)ρe(i)≤H1(k,i)≤H1(i,i)≤γ*in+2(n+2-i)≤n+24γ*,
whereγ*=1/(1-μ), ρ=∑τ=1n+1e(τ)g(τ)/(n+2)(1-μ).We construct a cone onEby
(40)K={{0,1,…,n}andmin⁡k∈Ju(k)≥δ*∥u∥u∈E:u≥0,Δ2u(k)≤0on⁡{0,1,…,n},andmin⁡k∈Ju(k)≥δ*∥u∥},
where
(41)δ*=2(n+3)vb2ρ3(n+2)γ*a(b+a∑j=2n+1(1/z(j)))2.
Obviously,Kis a closed convex cone ofE.Define an operatorT:E→Eas
(42)(Tu)(k)=∑τ=1n+1‍H1(k,τ)∑i=3n+1‍H(τ+1,i)w(i-2)f(i-2,u(i-2)).
Let
(43)Ψ(k,i)=∑τ=1n+1‍H1(k,τ)H(τ+1,i).
Then we can obtain the following properties.Proposition 13.
If(A3)and(A4)hold, then
(44)0<Ψ(k,i)≤(n+2)(n+1)4(1-μ)H(i,i),iiiiiiiiiiiik∈J,i∈{2,3,…,n+2}.Proposition 14.
If(A3)and(A4)hold, then, for all k∈J,i∈{2,3,…,n+2}, one has
(45)(n+1)(n+3)6Qvb2γρ<Ψ(k,i)≤(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2a.Lemma 15.
Suppose that(A0)–(A4)hold; ifu∈Eis a solution of the equation
(46)u(k)=(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2)),
thenuis a solution of the BVP (2).Lemma 16.
Assume that(A0)–(A4)hold; thenT(K)⊂KandT:K→Kis completely continuous.Proof.
From above works, for allu∈K, we have
(47)Δ2(Tu)(k-1)iii=-∑i=3n+1‍H(k+1,i)w(i-2)f(i-2,u(i-1))≤0,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiik∈{1,2,…,n+1}.
Because
(48)(Tu)(0)=(Tu)(n+2)=∑τ=1n+1‍H1(0,τ)×∑i=3n+1‍H(τ+1,i)w(i-2)f(i-2,u(i-2))=∑τ=1n+1‍H1(n+2,τ)×∑i=3n+1‍H(τ+1,i)w(i-2)f(i-2,u(i-2))=∑τ=1n+1‍11-μ∑j=1n+1‍G1(j,τ)G(j)×∑i=3n+1‍H(τ+1,i)w(i-2)f(i-2,u(i-2))≥0.
That is to say(Tu)(k)≥0fork∈J.
On the other hand, fork∈J, we have
(49)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≤(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2×a∑i=3n+1‍w(i-2)f(i-2,u(i-2)).
Similarly, we can get
(50)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≥(n+2)(n+1)6Qvb2γρ∑i=3n+1‍w(i-2)f(i-2,u(i-2))=δ*(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2×a∑i=3n+1‍w(i-2)f(i-2,u(i-2))≥δ*∥Tu∥,k∈J.
So,Tu∈KandT(K)⊂K. It is easy to see thatT:K→Kis completely continuous.


## 3. The Existence of One Positive Solution

In this part, we apply Theorem3 and Lemma 4 to prove the existence of one positive solution for BVP (2). We need consider the following cases for:p>1,p=1, andp=∞.Let(51)fβ=lim⁡n→βsup⁡max⁡k∈Jf(k,u)u,fβ=lim⁡u→βinf⁡min⁡k∈Jf(k,u)u,
whereβdenotes0or∞, and
(52)B=max⁡{(n+2)(n+1)4(1-μ)(∑i=3n+1‍|H|p)1/p(∑i=3n+1‍|w|q)1/q,(n+2)(n+1)4Qγγ*a×(b+a∑j=2n+1‍1z(j))(∑i=3n+1‍|w|)},η=[(n+2)(n+1)γγ*(b+a∑j=2n+1‍1z(j))2a]-1.Remark 17.
If we only consider the casep=∞, then we can take
(53)B=(n+2)(n+1)4Qγ(b+a∑j=2n+1‍1z(j))2(∑i=3n+1‍|w|).Firstly, the following theorem deals with the casep>1.Theorem 18.
Suppose that(A0)–(A4)hold. If there exist two constantsr, Rwith0<r<δ*Rsuch that(C1)
f(k,u)≤B-1rfor all(k,u)∈J×[δ*r,r]andf(k,u)≥(4η/mδ*)QRfor all(k,u)∈J×[δ*R,R];or(C2)
f(k,u)≥(4η/mδ*)QRfor all(k,u)∈J×[δ*r,r]andf(k,u)≤B-1Rfor all(k,u)∈J×[δ*R,R], then BVP (2) has at least one positive solution.Proof.
We only consider condition(C1). Foru∈K, from the definition ofKwe obtain that
(54)min⁡k∈Ju(k)≥δ*∥u∥.
So, foru∈∂Kr, we haveu(k)∈[δ*r,r], which implies thatf(k,u(k))≤B-1γ. Thus, fork∈J, from(C1)we get
(55)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≤(n+2)(n+1)4(1-μ)B-1γ∑i=3n+1‍H(i,i)w(i-2)≤(n+2)(n+1)4(1-μ)(∑i=3n+1‍|H|p)1/p(∑i=3n+1‍|w|q)1/qB-1γ=r=∥u∥;
that is,u∈∂Krimplies that
(56)∥Tu∥≤∥u∥.
On the other hand, foru∈∂KR, we haveu(k)∈[δ*R,R], which implies thatf(k,u(k))≥(4η/mδ*)QR; therefore fork∈J, from(C1)we have
(57)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≥(n+1)(n+3)6Qvb2γρ4ηmδ*QR∑i=3n+1‍w(i-2)≥δ*(n+1)(n+2)4Qγ*γ×(b+a∑j=2n+1‍1z(j))2a4ηmδ*QRm=R=∥u∥;
that is,u∈∂KRimplies that
(58)∥Tu∥≥∥u∥.
From the above works, we apply (i) of Theorem3 to yield thatThas a fixed pointu*∈K¯r,R, r≤∥u*∥≤Randu*(k)≥δ*∥u*∥>0,k∈J. Thus, it follows that BVP (2) has at least one positive solutionu*.The following theorem deals with the casep=1.Theorem 19.
Suppose that(A0)–(A4)hold and (C1) or (C2) holds. Then BVP (2) has at least one positive solution.Proof.
Let(∑i=3n+1|H(i,i)|)(sup⁡i∈{3,4,…,n+1}⁡|w(i-2)|)replace (∑i=3n+1|H(i,i)|p)1/p(∑i=3n+1|w(i-2)|q)1/q and repeat the argument of Theorem 18.Finally we consider the casep=∞.Theorem 20.
Suppose that(A0)–(A4)hold and (C1) or (C2) holds. Then BVP (2) has at least one positive solution.Proof.
Similar to the proof of Theorem18, forx∈∂Kr, we have
(59)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≤(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2aB-1r∑i=3n+1‍w(i-2)≤(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2aB-1r∑i=3n+1‍|w|≤r=∥u∥.
So, forx∈∂Kr, we have∥Tu∥≤∥u∥. And from the proof of Theorem 18,∥Tu∥≥∥u∥,k∈Jforu∈∂KR. Thus Theorem 20 is proved.Theorem 21.
Assume that(A0)–(A4)hold. If one of the following conditions is satisfied(C3)
f0>4ηQ/mδ*andf∞<1/B(particularly,f0=∞andf∞=0);(C4)
f0<1/Bandf∞>4ηQ/mδ*2 (particularly,f0=0andf∞=∞),then BVP (2) has at least one positive solution.Proof.
We only consider the case(C3). The proof of case(C4)is similar to case(C3). Consideringf0>4ηQ/mδ*2, there existsr1>0such thatf(k,u)≥(f0-ε1)ufork∈J,u∈[0,r1], whereε1>0satisfiesf0-ε1≥4ηQ/mδ*2. Then, fork∈J,u∈∂Kr1, from(C3)we have
(60)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≥(n+1)(n+3)6Qvb2γρ∑i=3n+1‍w(i-2)f(i-2,u(i-2))≥δ*2(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2a(f0-ε1)×∑i=3n+1‍w(i-2)u(i-2)≥δ*2(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2a(f0-ε1)m∥u∥≥δ*2(n+2)(n+1)4Qγ*γ(b+a∑j=2n+1‍1z(j))2a4ηQmδ*2m∥u∥≥∥u∥;
that is,x∈∂Kr1implies that∥Tu∥≥∥u∥.
Next, consideringf∞<1/B, then there existsR¯1>0such that
(61)f(k,u)≤(f∞+ε2)u,k∈J,u∈(R¯1,∞),
whereε2>0satisfiesB(f∞+ε2)<1; assume that
(62)M=max⁡0≤u≤R¯1,k∈Jf(k,u);
then
(63)f(k,u)≤M+(f∞+ε2)uɛ.
ChoosingR1>max⁡{r1,R¯1,MB(1-B(f∞+ε2))-1}.
Thus, foru∈KR1, we obtain
(64)∥(Tu)(k)∥=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≤(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)f(i-2,u(i-2))≤(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)(M+(f∞+ε2)∥u∥)≤(n+2)(n+1)4(1-μ)(∑i=3n+1‍|H|p)1/p(∑i=3n+1‍|w|q)1/q×(M+(f∞+ε2)∥u∥)≤B(M+(f∞+ε2)∥u∥)≤BM+(f∞+ε2)∥u∥B<R1-R1(f∞+ε2)B+(f∞+ε2)∥u∥B≤R1-(f∞+ε2)∥u∥B+(f∞+ε2)∥u∥B=R1;
that is,u∈∂KR1, we have∥Tu∥<∥u∥.
From above works and (ii) of Theorem3 we know thatThas a fixed pointu*∈K¯r1,R1, r1≤u*<R1, andu*(k)≥δ*∥u*∥>0, k∈J. Thus, it implies that BVP (2) has a positive solutionu*.Theorem 22.
Suppose that(A0)–(A4)hold. If there exist two constantsr2,R2with0<r2<R2such that(C5)
f(k,·)is nondecreasing on[0,R2]for allk∈J;(C6)
f(k,δ*r2)≥(4ηQ/mδ*)r2andf(k,R2)≤B-1R2for allk∈J,then the BVP (2) has at least one positive solution.Proof.
Foru∈K, from the definition ofKwe have min⁡k∈Ju(t)≥δ*∥u∥. So, foru∈∂Kr2, we haveu(k)≥δ*∥u∥=δ*r2,k∈J, from conditions (C5) and (C6), we obtain
(65)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≥∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,δ*r2)≥(n+1)(n+3)6Qvb2γρ4ηQmδ*r2∑i=3n+1‍w(i-2)=δ*(n+2)(n+1)4Qγ*γ×(b+a∑j=2n+1‍1z(j))2a4ηQmδ*r2∑i=3n+1‍w(i-2)=r2=∥u∥,
that is, foru∈∂Kr2, we can imply that
(66)∥Tu∥≥∥u∥.
On the other hand, foru∈∂KR2, we haveu(k)≤R2, k∈J, this together with(C5)and(C6), we have
(67)(Tu)(k)=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≤∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,R2)≤(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)f(i-2,R2)≤(n+2)(n+1)4(1-μ)(∑i=3n+1‍|H|p)1/p(∑i=3n+1‍|w|q)1/qB-1R2=R2=∥u∥,
that is, foru∈∂KR2, we can imply that
(68)∥Tu∥≤∥u∥.
From above works and Theorem3, we prove thatThas a fixed pointu*∈K¯r2,R2,r2≤∥u*∥≤R2, andu*(k)≥δ*∥u*∥>0,k∈J. So the BVP (2) has at least one positive solution.


## 4. The Existence of Multiple Positive Solutions

Theorem 23.
Assume that(A0)–(A4)hold, and the following two conditions hold:(C7)
f0>4ηQ/mδ*2 andf∞>4ηQ/mδ*2 (particularly,f0=f∞=∞);(C8)
there existsl>0such thatmax⁡k∈J,u∈∂Kl⁡f(k,u)<B-1l.
Then BVP (2) has at least two positive solutionsu*(k), u**(k), which satisfy
(69)0<∥u**∥<l<∥u*∥.Proof.
We can take two constantsr, Rand suppose that0<r<l<R. Iff0>4ηQ/mδ*2; from the proof of Theorem 21 we have
(70)∥Tu∥≥∥u∥,u∈∂Kr.
Iff∞>4ηQ/mδ*2, similarly, we have
(71)∥Tu∥≥∥u∥,u∈∂KR.
On the other hand, from(C8), foru∈∂Kl, we have
(72)∥Tu∥=∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))<(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)B-1l≤(n+2)(n+1)4(1-μ)(∑i=3n+1‍|H|p)1/p(∑i=3n+1‍|w|q)1/qB-1l=l;
that is,
(73)∥Tu∥<l=∥u∥,u∈∂Kl.
Applying Theorem3, we can prove thatThas a fixed pointu**∈K¯r,land a fixed pointu*∈K¯l,R. Thus, we prove that BVP (2) has two positive solutions u*, u**. From above formula we obtain∥u*∥≠land∥u**∥≠l. So0<∥u**∥<l<∥u*∥.Remark 24.
From the proof of Theorem23 we obtain that if(C8)holds andf0>4ηQ/mδ*2(orf∞>4ηQ/mδ*2), then BVP (2) has at least one positive solutionu. It satisfies0<∥u∥<l(orl<∥u∥).Using a similar method we can obtain the following results.Theorem 25.
Suppose that(A0)–(A4)hold, and the following two conditions hold(C9)
f0<1/Bandf∞<1/B;(C10)
there existsL>0such thatmin⁡k∈J,u∈∂K2⁡f(k,u)>(4ηQ/mδ*)L,then BVP (2) has at least two positive solutionsu*(k),u**(k), which satisfy
(74)0<∥u**∥<L<∥u*∥.Remark 26.
If(C10)holds andf0<1/B(orf∞<1/B), then BVP (2) has at least one positive solutionusatisfying0<∥u∥<L(orL<∥u∥).Theorem 27.
Assume that(A0)–(A4)hold. If there exist2npositive numbersdt,Dt,t=1,2,…,n, withd1<δ*D1<D1<d2<δ*D2<D2<⋯<dn<δ*Dn<Dnsuch that(C11)
f(k,u)≤B-1dt,for(k,u)∈J×(δ*dt,dt)andf(k,u)≥(4ηQ/mδ*)Dtfor(k,u)∈J×[δ*Dt,Dt],k=1,2,…,n;or(C12)
f(k,u)≥(4ηQ/mδ*)dt,for(k,u)∈J×(δ*dt,dt)andf(k,u)≤B-1Dtfor(k,u)∈J×[δ*Dt,Dt],t=1,2,…,n,then BVP (2) has at leastnpositive solutionsuksatisfyingdt≤∥ut∥≤Dt, t=1,2,…,n.Theorem 28.
Assume that(A0)–(A4)hold. If there exist2npositive numbersdt,Dt,t=1,2,…,n, withd1<D1<d2<D2<⋯<dn<Dnsuch that(C13)
f(k,·)is nondecreasing on[0,Dn]for allk∈J;(C14)
f(k,δ*dk)≥(4ηQ/mδ*)dt, andf(k,Dt)≤B-1Dt,t=1,2,…,n,then BVP (2) has at leastnpositive solutionsutsatisfyingdt≤∥ut∥≤Dt,t=1,2,…,n.


## 5. The Nonexistence of Positive Solution

Theorem 29.
Suppose that(A0)–(A4)hold, andBf(k,u)<u, for allk∈J, u>0; then BVP (2) has no positive solution.Proof.
Assume thatu(k)is a positive solution of BVP (2). Thenu∈K, u(k)>0, k∈{0,1,…,n+2}and
(75)∥Tu∥=∥u∥=max⁡k∈J∥u(k)∥=max⁡k∈J∑i=3n+1‍Ψ(k,i)w(i-2)f(i-2,u(i-2))≤(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)f(i-2,u(i-2))<(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)B-1u(i-2)≤B-1∥u∥(n+2)(n+1)4(1-μ)∑i=3n+1‍H(i,i)w(i-2)≤B-1∥u∥(n+2)(n+1)4(1-μ)(∑i=3n+1‍|H|p)1/p(∑i=3n+1‍|w|q)1/q=∥u∥,
which is a contradiction.Similarly, we can get the following result.Theorem 30.
Suppose that(A0)–(A4)hold and (mδ*2/4ηQ)f(k,u)>u, for all k∈J, u>0; then BVP (2) has no positive solution.



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*Source: 578672-2014-07-17.xml*