# Positive Solutions of a Singular Positone and Semipositone Boundary Value Problems for Fourth-Order Difference Equations

Authors: Chengjun Yuan
Journal: Discrete Dynamics in Nature and Society
 (2010)

---

## Abstract

This paper studies the boundary value problems for the fourth-order nonlinear singular difference equationsΔ4u(i−2)=λα(i)f(i,u(i)), i∈[2,T+2], u(0)=u(1)=0, u(T+3)=u(T+4)=0. We show the existence of positive solutions for positone and semipositone type. The nonlinear term may be singular. Two examples are also given to illustrate the main results. The arguments are based upon fixed point theorems in a cone.


---

## Body

## 1. Introduction

In this paper, we consider the following boundary value problems of difference equations:(1.1)Δ4u(i-2)=λa(i)f(i,u(i)),i∈[2,T+2],u(0)=u(1)=0,u(T+3)=u(T+4)=0.
Here [2,T+2]={2,3,…,T+2} and u:[0,T+4]→R. We will let [a,b] denote the discrete integer set [a,b]={a,a+1,…,b}, and  C([a,b]) denotes the set of continuous function on [a,b] (discrete topology) with norm ∥·∥=maxk∈[a,b]|·|.Due to the wide applications in many fields such as computer science, economics, neural network, ecology, and cybernetics,  the theory of nonlinear difference equations has been widely studied since the 70's of last century. Recently, many literatures on the boundary value of difference equations have appeared. We refer the reader to [1–13] and the references therein, which include work on Agarwal, Elaydi, Eloe, Erber, O'Regan, Henderson, Merdivenci, Yu, and Ma et al., concerning the existence of positive solutions and the corresponding eigenvalue problems. Recently, the existence of positive solutions of fourth-order discrete boundary value problems has been studied by several authors; for example, see [14–16] and the references therein.On the other hand, fourth-order boundary value problems of ordinary value problems have important application in various branches of pure and applied science. They arise in the mathematical modeling of viscoelastic and inelastic flows, deformation of beams and plate deflection theory [17–19]. For example, the deformations of an elastic beam can be described by the boundary value problems of the fourth-order ordinary differential equations. There have been extensive studies on fourth-order boundary value problems with diverse boundary conditions via many methods, for example,  [20–26] and the references therein. We also find that the differential equations on time scales is due to its unification of the theory of differential and difference equations, see [27–30] and the references therein.In this paper, the boundary value problem (1.1) can be viewed as the discrete analogue of the following boundary value problems for ordinary differential equation:(1.2)u(4)(t)=λa(t)f(t,u(t)),t∈(0,1),u(0)=u′(1)=0,u(1)=u′(1)=0.
Equation (1.2) describes an elastic beam in an equilibrium state whose both ends are simply supported. However, very little is known about the existence of solutions of the discrete boundary value problems (1.1). This motivates us to study   (1.1).In this paper, we discuss separately the cases whenf is positone and when f is semipositone; the nonlinear term f is singularity at u=0, and we will prove our two existence results for the problem (1.1) by using Krasnosel'skii fixed point theorem. This paper is organized as follows. In Section 2, starting with some preliminary lemmas, we state the Krasnosel'skii fixed point theorem. In Section 3, we give the sufficient conditions which state the existence of multiple positive solutions to the positone boundary value problem (1.1). In Section 4, we give the sufficient conditions which state the existence of at least one positive solution to the semipositone boundary value problem (1.1).


## 2. Preliminaries

In this section, we state the preliminary information that we need to prove the main results. From   [28, Definition 2.1], we have the following lemma.Lemma 2.1.
u(i) is a solution of   (1.1)  if only and if
(2.1)u(i)=∑j=2T+2G(i,j)a(j)f(j,u(j)),i∈[0,T+4],
where
(2.2)G(i,j)={(T+4-i)2(j-1)22(i(T+3)2-(T+4+2i)(j-1)3(T+4)3),2≤j≤i+1,i2(T+4-j)26((T+3-i)(T+4+2j)(T+4)3-T+4-j(T+3)2),i+1<j≤T+2.Lemma 2.2.
Green's functionG(t,s) defined by (2.2) has the following properties:
(2.3)C0i2(T+4-i)2(j-1)2(T+4-j)2≤G(i,j)≤(j-1)2(T+4-j)2,G(i,j)≤i2(T+4-i)2,
where C0=1/3(T+4)7.Proof.
For2≤j≤i+1, we have
(2.4)G(i,j)=(T+4-i)2(j-1)22(i(T+3)2-(T+4+2i)(j-1)3(T+4)3)≤(T+4-i)2(j-1)22i(T+3)2≤(T+4-j)2(j-1)2i2(T+3)2≤(j-1)2(T+4-j)2,G(i,j)=(T+4-i)2(j-1)22(i(T+3)2-(T+4+2i)(j-1)3(T+4)3)≤(T+4-i)2(j-1)22i(T+3)2≤i2(T+4-i)2(j-1)22(T+3)2≤i2(T+4-i)2.
On the other  hand,(2.5)G(i,j)=(T+4-i)2(j-1)22(i(T+3)2-(T+4+2i)(j-1)3(T+4)3)≥(T+4-i)2(j-1)22(i(T+4)2-(T+4+2i)(j-1)3(T+4)3)=(T+4-i)2(j-1)26(T+4)3((T+4)((i+1)-j)+2i((T+5)-j))≥2i(T+4-i)2(j-1)2((T+5)-j)6(T+4)3≥i(T+4-i)2(j-1)2((T+4)-j)3(T+4)3≥i2(T+4-i)2(j-1)2((T+4)-j)23(T+4)5≥i2(T+4-i)2(j-1)2((T+4)-j)23(T+4)7.
Then, for2≤j≤i+1, we have
(2.6)C0i2(T+4-i)2(j-1)2(T+4-j)2≤G(i,j)≤(j-1)2(T+4-j)2,G(i,j)≤i2(T+4-i)2.
Fori+1<j≤T+2, we have
(2.7)G(i,j)=i2(T+4-j)26((T+3-i)(T+4+2j)(T+4)3-T+4-j(T+3)2)≤i2(T+4-j)26(T+3-i)(T+4+2j)(T+4)3≤i2(T+4-i)26(T+3-i)(T+4+2j)(T+4)3≤i2(T+4-i)2(T+4+2j)6(T+4)3≤3i2(T+4-i)2(T+4)6(T+4)3≤i2(T+4-i)2,G(i,j)=i2(T+4-j)26((T+3-i)(T+4+2j)(T+4)3-T+4-j(T+3)2)≤i2(T+4-j)26(T+3-i)(T+4+2j)(T+4)3≤(j-1)2(T+4-j)26(T+3-i)(T+4+2j)(T+4)3≤3(j-1)2(T+4-j)2(T+4)26(T+4)3≤(j-1)2(T+4-j)2.
On the other hand,(2.8)G(i,j)=i2(T+4-j)26((T+3-i)(T+4+2j)(T+4)3-T+4-j(T+3)2)≥i2(T+4-j)26((T+3-i)(T+4+2j)(T+4)3-T+3-i(T+3)2)≥i2(T+4-j)2(T+3-i)6((T+8)(T+4)3-1(T+3)2)≥i2(T+4-j)2(T+3-i)62T2+9T+8(T+4)3(T+3)2≥i2(T+4-j)2(T+3-i)31(T+4)3≥i2(T+4-i)2(j-1)2((T+4-j)23(T+4)7.
Then, fori+1<j≤T+2, we have also
(2.9)C0i2(T+4-i)2(j-1)2(T+4-j)2≤G(i,j),G(i,j)≤(j-1)2(T+4-j)2,G(i,j)≤i2(T+4-i)2.We note thatu(t) is a solution of (1.1) if and only if(2.10)u(i)=λ∑j=2T+2G(i,j)a(j)f(j,u(j)),i∈[0,T+4].For our constructions, we will consider the Banach spaceE=C([0,T+4]) equipped with the standard norm ∥u∥=max0≤i≤T+4|u(i)|, u∈E. We define a cone P by (2.11)P={u∈E∣u(i)≥C0i2(T+4-i)2∥u∥,i∈[0,T+4]}.The following theorems will play a major role in our next analysis.Theorem 2.3 (see [1]).
LetX be a Banach space, and let P⊂X be a cone in X. Let Ω1,Ω2 be open subsets of X with 0∈Ω1⊂Ω¯1⊂Ω2, and let S:P→P be a completely continuous operator such that  either  (1)
∥Sw∥≤∥w∥, w∈P∩∂Ω1, ∥Sw∥≥∥w∥, w∈P∩∂Ω2, or(2)
∥Sw∥≥∥w∥, w∈P∩∂Ω1, ∥Sw∥≤∥w∥, w∈P∩∂Ω2. Then,  S has a fixed point in P∩Ω¯2∖Ω1.


## 3. Singular Positone Problems

Theorem 3.1.
Assume  that the following conditions are satisfied:(H1)
f∈C([2,T+2]×(0,+∞),[0,+∞));(H2)
f(i,u)≤K(i)(g(u)+h(u)) on [2,T+2]×(0,∞) with g>0 continuous and nonincreasing on (0,∞), h≥0 continuous on [0,∞),  and h/g nondecreasing on (0,∞), ∃K0 with g(xy)≤K0g(x)g(y)  for all  x>0,  y>0;(H3)
there exists[α,β]⊂[2,T+2] such that limu→+∞inf(f(i,u)/u)=+∞ for i∈[α,β];(H4)
there exists[α1,β1]⊂[2,T+2] such that limu→0+inf(f(i,u)/u)=+∞ for i∈[α1,β1]. Then, for each r>0, there exists a positive number λ* such that the positone problem (1.1) has at least two positive solutions u1 and u2 with ∥u1∥<r≤∥u2∥ for 0<λ<λ*.Proof.
Now, we define the integral operatorT:P→E by
(3.1)Tu(i)=λ∑j=2T+2G(i,j)a(j)f(j,u(j)),
where P={u∈X∣u(i)≥C0i2(T+4-i)2∥u∥,i∈[0,T+4]}.
It is easy to check thatT(P)⊂P. In fact, for each u∈P, we have by Lemma 2.2 that
(3.2)Tu(i)≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)f(j,u(j)).
This implies ∥Tu∥≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)f(j,u(j)). On the other hand, we have
(3.3)Tu(i)≥C0i2(T+4-i)2λ∑j=2T+2(j-1)2(T+4-j)2a(j)f(j,u(j)).
Thus, we have Tu(i)≥C0i2(T+4-i)2∥Tu∥. In addition, standard argument shows that T is completely continuous.
For  any givenr>0, we fix  it, and take Ωr={u∈E∣∥u∥<r}. Choose
(3.4)λ*=rK02g(C0)∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(j2(T+4-j)2)(g(r)+h(r)).
Foru∈P∩∂Ωr, from (H2) and (3.4), we have
(3.5)Tu(t)=λ∑j=2T+2G(i,j)a(j)f(j,u(j))≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)(g(u)+h(u))≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(u)(1+h(u)g(u))≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(C0j2(T+4-j)2r)(1+h(r)g(r))≤λK02g(C0)∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(j2(T+4-j)2)(g(r)+h(r))<r.
Thus,
(3.6)∥Tu∥<∥u∥,foru∈P∩∂Ωr.
Further, choose a constant M*>0 satisfying that
(3.7)λM*C0σmax0≤t≤1{∑j=αβG(i,j)a(j)}>1,
where σ=minα≤i≤β{i2(T+4-i)2}.
By (H3), there is a constantL>0 such that
(3.8)f(i,u)≥M*u,∀u≥L,i∈[α,β].
Let R=r+L/C0σ and ΩR={u∈E∣∥u∥<R}. For u∈P∩∂ΩR, we have that
(3.9)u(i)≥C0i2(T+4-i)2∥u∥≥C0Ri2(T+4-i)2≥C0Rσ≥L,i∈[α,β].
It follows that
(3.10)f(i,u(i))≥M*u(i)≥M*C0Rσ,i∈[α,β].
Then, foru∈P∩∂ΩR, we have
(3.11)∥Tu∥=λmax0≤t≤1{∑j=2T+2G(i,j)a(j)f(j,u(j))}≥λmax0≤t≤1{∑j=αβG(i,j)a(j)f(j,u(j))}≥λmax0≤t≤1{∑j=αβG(i,j)a(j)M*C0Rσ}≥λM*C0Rσmax0≤t≤1{∑j=αβG(i,j)a(j)}≥R.
Therefore, by the first part of the Fixed Point Theorem2.3, T has a fixed point y with r≤∥u2∥≤R.
Finally, choose a constantM*>0 satisfying that
(3.12)λM*C0max0≤t≤1{∑j=α1β1G(i,j)a(j)j2(T+4-j)2}>1.
By(H4), there is a constant δ>0 and δ<r such that
(3.13)f(i,u)≥M*u,∀u≤δ,i∈[α1,β1].
Let r*=δ/2 and Ωr*={u∈E∣∥u∥<r*}. For u∈P∩∂Ωr*, we have
(3.14)u(i)≥C0i2(T+4-i)2∥u∥≥C0r*i2(T+4-i)2.
It follows that
(3.15)f(i,u(i))≥M*u(i)≥M*C0r*i2(T+4-i)2,i∈[α1,β1].
Then, foru∈P∩∂Ωr*, we have
(3.16)∥Tu∥=λmax0≤t≤1{∑j=2T+2G(i,j)a(j)f(j,u(j))}≥λmax0≤t≤1{∑j=α1β1G(i,j)a(j)f(j,u(j))}≥λmax0≤t≤1{∑j=α1β1G(i,j)a(j)M*C0r*j2(T+4-j)2}≥λM*C0r*max0≤t≤1{∑j=α1β1G(i,j)a(j)j2(T+4-j)2}≥r*.
Therefore, by the first part of the Fixed Point Theorem2.3, T has a fixed point u1 with r*≤∥u1∥≤r. It follows from (3.6) that ∥u1∥≠r.
Then, for eachr>0, there exists a positive number λ* such that the positone problem (1.1) has at least two positive solutions ui(i=1,2) with r*≤∥u1∥<r≤∥u2∥≤R for 0<λ<λ*.From the proof of Theorem3.1, we have the following result.Theorem 3.2.
Assume that (H1)–(H3) are satisfied. Then, for eachr>0, there exists a positive number λ* such that the positone problem (1.1) has at least one positive solution u2 with r≤∥u2∥ for 0<λ<λ*.Theorem 3.3.
Assume that (H1), (H2), and (H4) are satisfied. Then,  for eachr>0, there exists a positive number λ* such that the positone problem (1.1) has at least one positive solution u1 with ∥u1∥<r for 0<λ<λ*.Example 3.4.
Consider the  boundary value problem(3.17)Δ4u(i-2)=λa(i)(u-α+uβ(sin2u+1)),i∈[2,T+2],u(0)=u(1)=0,u(T+3)=u(T+4)=0,
where 0<α<1<β are constants. Then,  for each  r>0, there exists a positive number λ* such that the problem (3.17) has at least two positive solutions for 0<λ<λ*.
In fact, it is clear that(3.18)f(i,u)=u-α+uβ(sin2u+1),limu→0+f(i,u)u=+∞,limu→+∞f(i,u)u=+∞.
LettingK(i)=1,  g(u)=u-α,  and h(u)=2uβ, we have
(3.19)f(i,u)≤K(i)(g(u)+h(u)),K0=1
with  g>0 continuous and nonincreasing on (0,∞), h≥0 continuous on [0,∞), and  h/g=2uα+β nondecreasing on (0,∞);  K0=1 with g(xy)=g(x)g(y)≤K0g(x)g(y) for ∀x>0, y>0.
Then, by Theorem3.1, for each given r>0, we choose
(3.20)λ*=C0αr1+α(1+rα+β)∑j=2T+2j-2α(j-1)2(T+4-j)2(1-α)a(j),
such that the problem (3.17) has at least two positive solutions for 0<λ<λ*.


## 4. Singular Semipositone Problems

Before we prove our next main result,  we first state a result.Lemma 4.1.
The boundary value problem(4.1)Δ4w(i-2)=λa(i)e(i),i∈[2,T+2],w(0)=w(1)=0,w(T+3)=w(T+4)=0
has a solution w with w(t)≤c0i2(T+4-i)2, where c0=∑j=2T+2a(j)e(j).In fact,  from Lemma2.1,   (4.1) has solution (4.2)w(t)=∑j=2T+2G(i,j)a(j)e(j).
According to Lemma 2.2, we have (4.3)w(t)≤i2(T+4-i)2∑j=2T+2a(j)e(j)=c0i2(T+4-i)2.Theorem 4.2.
Assume that the following conditions are satisfied:(B1)
f:[2,T+2]×(0,∞)→R is continuous and there exists a function e∈C([2,T+2],(0,+∞)) with f(i,u)+e(i)≥0 for (i,u)∈[2,T+2]×(0,∞);(B2)
f*(i,u)=f(i,u)+e(i)≤K(i)(g(u)+h(u)) on [2,T+2]×(0,∞) with g>0 continuous and nonincreasing on (0,∞), h≥0 continuous on [0,∞),  and h/g nondecreasing on (0,∞);(B3)
∃K0 with g(xy)≤K0g(x)g(y) for all  x>0,y>0;(B4)
there exists[α,β]⊂[2,T+2] such that limu→+∞inf(f(i,u)/u)=+∞ for i∈[α,β]. Then, for each r>0, there exists a positive number λ* such that the semipositone problem (1.1) has at least one positive solution for 0<λ<λ*.Proof.
To show that  (1.1) has a nonnegative solution, we will look at the boundary value problem
(4.4)Δ4y(i-2)=λa(i)f*(i,y(i)-φ(i)),i∈[2,T+2],y(0)=y(1)=0,y(T+3)=y(T+4)=0,
where φ(i)=λw(i)  and  w is as in Lemma 4.1.
We will show, using Theorem2.3, that there exists a solution y to (4.4) with y(i)>φ(i) for i∈[2,T+2]. If this is true, then u(i)=y(i)-φ(i)(0≤i≤T+4) is a nonnegative solution (positive on [2,T+2]) of (1.1), since
(4.5)Δ4u(i-2)=Δ4(y(i-2)-φ(i-2))=λa(i)f*(i,y(i)-φ(i))-λa(i)e(i)=λa(i)[f(i,y(i)-φ(i))+e(i)]-λa(i)e(i)=λa(i)f(i,y(i)-φ(i))=λa(i)f(i,u(i)),i∈[0,T+4].
Next, letT:K→E be defined by
(4.6)(Ty)(i)=λ∑j=2T+2G(i,j)a(j)f*(j,y(j)-φ(j)),0≤i≤T+4.
In addition, standard argument shows  that T(P)⊂P and T is completely continuous.
For any  givenr>0, fix it. We choose
(4.7)λ*=min{C0r2c0,rK02a0(g(r)+h(r))},
where a0=g(C0/2)∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(j2(T+4-j)2).
Now, let(4.8)Ωr={y∈E∣∥y∥<r}.
We show that
(4.9)∥Ty∥≤∥y∥fory∈P∩∂Ωr.
To see this, let y∈P∩∂Ωr. Then,  ∥y∥=r and y(t)≥C0i2(T+4-i)2r for i∈[0,T+4]. Now, for i∈[0,T+4], the Lemma 4.1 implies
(4.10)y(i)-φ(i)≥C0ri2(T+4-i)2-λc0i2(T+4-i)2≥(C0r-λc0)i2(T+4-i)2≥C0r2i2(T+4-i)2>0,
so for i∈[0,T+4],  we have
(4.11)(Ty)(i)=λ∑j=2T+2G(i,j)a(j)f*(j,y(j)-φ(j))≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)[g(y(j)-φ(j))+h(y(j)-φ(j))]=λ∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(y(j)-φ(j)){1+h(y(j)-φ(j))g(y(j)-φ(j))}≤λ∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(C0r2j2(T+4-j)2){1+h(r)g(r)}ds≤λK02g(C02)(g(r)+h(r))∑j=2T+2(j-1)2(T+4-j)2a(j)K(j)g(j2(T+4-j)2)=λK02a0(g(r)+h(r))<r.
This yields ∥Ty∥≤r=∥y∥, so (4.9) is satisfied.
Further, choose a constantM*>0 satisfying that
(4.12)λM*C02σmax0≤t≤1{∑j=αβG(i,j)a(j)}>1,
where σ=minα≤i≤β{i2(T+4-i)2}.
By(B4), there is a constant L>0 such that
(4.13)f*(i,x)≥M*x,∀x≥L,i∈[α,β].
Let R=r+max{2λc0C1/C2,2C1L/C2σ} and ΩR={y∈E∣∥y∥<R}.
Next, we show that(4.14)∥Ty∥≥∥y∥,fory∈P∩∂ΩR.
To see this,  let y∈P∩∂ΩR. We have
(4.15)y(t)-φ(t)≥C0i2(T+4-i)2∥y∥-λc0i2(T+4-i)2≥C02Ri2(T+4-i)2≥C02Rσ≥L,i∈[α,β].
It follows that, for y∈P∩∂ΩR, we have
(4.16)f*(i,y(i)-φ(i))≥M*(y(i)-φ(i))≥M*C02Rσ,i∈[α,β].
Then, we have(4.17)∥Ty∥=λmax0≤i≤T+4{∑j=2T+2G(i,j)a(j)f*(j,y(j)-φ(j))}≥λmax0≤i≤T+4{∑j=αβG(i,j)a(j)f*(j,y(j)-φ(j))}≥λmax0≤i≤T+4{∑j=αβG(i,j)a(j)M*C02Rσ}≥λM*C02Rσmax0≤i≤T+4{∑j=αβG(i,j)a(j)}≥R.
This yields ∥Ty∥≥∥y∥, so (4.14) holds.
Therefore, by the first part of the Fixed Point Theorem2.3, T has a fixed point y with r≤∥y∥≤R, since
(4.18)y(i)-φ(i)≥C0i2(T+4-i)2r-λc0i2(T+4-i)2≥(C0r-λc0)i2(T+4-i)2>0,i∈[0,T+4].
Namely, u=y-φ is a positive solution of the semipositone problem (1.1).
Then,  for eachr>0, there exists a positive number λ* such that the semipositone problem (1.1) has at least one positive solution for 0<λ<λ*.Example 4.3.
Consider the boundary value problem(4.19)Δ4y(i-2)=λa(i)(u-α+uβ-sin(iu+i1/2))=0,i∈[2,T+2],y(0)=y(1)=0,y(T+3)=y(T+4)=0,
where 0<α<1<β are constants. Then,  for each r>0, there exists a positive number λ* such that the problem (4.19) has at least one positive solution for 0<λ<λ*.
To see this, we will apply Theorem4.2 (here λ*>0 will be chosen later). From
(4.20)f(t,u)=u-α+uβ-sin(iu+i1/2),
we let
(4.21)g(u)=u-α,h(u)=uβ+2,K(i)=1,e(t)=1,K0=1.
It is clear that0≤f(i,u)+e(i)≤K(i)(g(u)+h(u)), g(xy)≤K0g(x)g(y),  and limu→+∞inf(f(i,u)/u)=+∞,  i∈[α,β]⊂[2,T+2] hold.
Then, the (B1)–(B4) of Theorem4.2 hold. Now, we have
(4.22)c0=∑j=2T+2a(j),a0=2αC0-α∑j=2T+2j-2α(j-1)2(T+4-j)2(1-α))a(j).
For eachr>0, we can  choose
(4.23)λ*=min{C2r2c0C1,r1+αK0a0(1+(r+1)α+β)}.
Thus, all the conditions of Theorem 4.2 are satisfied, so the  existence of positive solution is guaranteed for 0<λ<λ*.



---

*Source: 312864-2010-11-02.xml*