# Existence of Three Positive Solutions to Nonlinear Boundary Value Problems

Authors: Li Wang; Jianli Li
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

Criteria are established for existence of three solutions to the
boundary value problemsLx=f(t,x), w1x(0)-w2x′(0)=0=w3x(1)+w4x′(1), where Lx:=-(px′)′+qx. Here, p∈C1[0,1], p>0, q∈C[0,1], q≥0.


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## Body

## 1. Introduction

In this paper, we are concerned with the existence of three positive solutions for the boundary value problem (BVP)(1.1)Lx=f(t,x),0<t<1,(1.2)w1x(0)-w2x'(0)=0,w3x(1)+w4x'(1)=0,
where f∈C([0,1]×[0,+∞)),[0,+∞), wi≥0(i=1,…,4) with ρ:=w2w3+w1w3+w1w4>0 and Lx:=-(p(t)x')'+q(t)x. Here p∈C1([0,1],(0,∞)),q∈C([0,1],[0,∞)). We shall also assume that λ=0 is not an eigenvalue of Lx=λx subject to conditions (1.2). As a consequence, it follows that the the smallest eigenvalue λ1 of the problem Lx=λx subject to (1.2) satisfies λ1>0 and the corresponding eigenfunction φ1(t) does not vanish on (0,1). Without loss of generality, we may assume φ1(t)>0 on (0,1) and ∥φ1∥=max0≤t≤1|φ1(t)|=1.LetG(t,s) denote Green’s function for the problem Lx=0 subject to condition (1.2). It is well known that G(t,s) may be written as (1.3)G(t,s)=1d{ϕ(t)ψ(s),0≤t≤s≤1,ϕ(s)ψ(t),0≤s≤t≤1,
where ϕ(t) and ψ(t) satisfy
(1.4)Lϕ=0,ϕ(0)=w2,ϕ'(0)=w1,Lψ=0,ψ(1)=w4,ψ'(1)=-w3,
and where (1.5)p(t)(ϕ(t)ψ′(t)-ϕ′(t)ψ(t))≡-d.
It may be shown that ϕ(t)≥0 and is increasing on [0,1] while ψ(t)≥0 and is decreasing on [0,1]. As a consequence, it follows that d>0 and, furthermore, we have (1.6)0≤G(t,s)≤G(s,s),0≤t,s≤1.
We define the positive number η,μ by (1.7)η-1:=max0≤t≤1(∫01G(t,s)ds),μ-1:=∫1/43/4G(s,s)ds.For the caseLx=-x′′ (i.e, p(t)≡1,q(t)≡0), the corresponding BVP (1.8)-x′′=f(t,x),0<t<1
subject to (1.2) has attracted considerable attention over the last number of years. Under certain condition, positive solutions of (1.8) and (1.2) are obtained in [1, 2]. In a recent paper, Erbe [3] investigated the existence of multiple positive solutions to (1.1)-(1.2) by applying the fixed point index.The aim of this paper is to establish criteria for the existence of three positive solutions to (1.1) and (1.2), which improve the corresponding result of [3]. Our tool in this paper will be well-known Five Functionals Fixed Point Theorem [4–7].


## 2. Preliminaries

Definition 2.1.
SupposeP is a cone in a Banach. The map α is a nonnegative continuous concave functional on P provided 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 t∈[0,1]. Similarly, the map β is a nonnegative continuous convex functional on P provided 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 t∈[0,1].
Letγ,β,θ be nonnegative, continuous, convex functionals on P and α,ψ be nonnegative, continuous, concave functionals on P. Then, for nonnegative real numbers h,a,b,d and c, we define the convex sets
(2.3)P(γ,c)={x∈P:γ(x)<c},P(γ,α,a,c)={x∈P:a≤α(x),γ(x)≤c},Q(γ,β,d,c)={x∈P:β(x)≤d,γ(x)≤c},P(γ,θ,α,a,b,c)={x∈P:a≤α(x),θ(x)≤b,γ(x)≤c},Q(γ,β,ψ,h,d,c)={x∈P:h≤ψ(x),β(x)≤d,γ(x)≤c}.
To prove our main results, we need the following theorem, which is the Five Functionals Fixed Point Theorem [4].Theorem 2.2.
LetP be a cone in a real Banach space E. Suppose there exist positive numbers c and M, nonnegative, continuous, concave functionals α and ψ on P, and nonnegative, continuous, convex functionals γ,β and θ on P, with
(2.4)α(x)≤β(x),∥x∥≤Mγ(x)
for all x∈P(γ,c)¯. Suppose
(2.5)Φ:P(γ,c)¯→P(γ,c)¯
is completely continuous and there exist nonnegative numbers h,a,k,b, with 0<a<b such that(i)
{x∈P(γ,θ,α,b,k,c):α(x)>b}≠∅ and α(Φx)>b for x∈P(γ,θ,α,b,k,c);(ii)
{x∈Q(γ,β,ψ,h,a,c):β(x)<a}≠∅ and β(Φx)<a for x∈Q(γ,β,ψ,h,a,c);(iii)
α(Φx)>b for x∈P(γ,α,b,c) with θ(Φx)>k;(iv)
β(Φx)<a for x∈Q(γ,β,a,c) with ψ(Φx)<h.Then Φ has at least three fixed points x1,x2,x3∈P(γ,c)¯ such that
(2.6)β(x1)<a,b<α(x2),a<β(x3)withα(x3)<b.


## 3. Main Result

In this section, we shall obtain existence results for BVP (1.1) and (1.2) by using the Five Functional Fixed Point Theorem.By [3], it is well known that BVP associated with (1.1), (1.2) is equivalent to the operator equation (3.1)x=Ax,x∈C[0,1],
where (3.2)(Ax)(t)=∫01G(t,s)f(s,x(s))ds.
Now with X=C[0,1],∥x∥=max0≤t≤1|x(t)|, it is easy to see that A:X→X is completely continuous. We define a cone P⊂X by (3.3)P:={x∈X:x(t)≥0,min1/4≤t≤3/4x(t)≥σ∥x∥},
where σ is defined by (3.4)σ:=min{G(t,s)G(s,s):14≤t≤34,0≤s≤1}.
By (1.3) and the properties of φ(t),ψ(t), we have (3.5)σ:=min{ϕ(1/4)ϕ(1),ψ(3/4)ψ(0)}.
Clearly, 0<σ<1 and G(t,s)≥σG(s,s) for 1/4≤t≤3/4,0≤s≤1.Lemma 3.1.
The operatorA maps P into P.Proof.
Letx∈P. From (1.6) and the condition of f, we see that Ax≥0. Next, for x∈P, we have
(3.6)|(Ax)(t)|=(Ax)(t)=∫01G(t,s)f(s,x(s))ds≤∫01G(s,s)f(s,x(s))ds.
Hence,
(3.7)∥Ax∥≤∫01G(s,s)f(s,x(s))ds.
Now, from G(t,s)≥σG(s,s) for 1/4≤t≤3/4,0≤s≤1, we have
(3.8)mint∈[1/4,3/4](Ax)(t)=mint∈[1/4,3/4]∫01G(t,s)f(s,x(s))ds≥σ∫01G(s,s)f(s,x(s))ds≥σ∥Ax∥.
This show that Ax∈P, which completes this proof.Theorem 3.2.
Let0<a<b and μb<ησc,  and suppose f(t,x) satisfies the following conditions:(H1)
f(t,x)<ηa for 0≤t≤1 and 0≤x≤a,(H2)
f(t,x)≥(μb)/σ for 1/4≤t≤3/4 and b≤x≤b/σ,(H3)
f(t,x)≤ηc for 0≤t≤1 and 0≤x≤c.Then the BVP (1.1)-(1.2) has at least three positive solutions.Proof.
Theorem2.2 will be applied. We begin by defining the nonnegative continuous concave functional α,ψ and the nonnegative continuous convex functional β,θ,γ on P(3.9)ψ(x)=mint∈[0,1]x(t),β(x)=θ(x)=maxt∈[0,1]x(t),α(x)=mint∈[1/4,3/4]x(t),γ(x)=∥x∥.
It is clear that α(x)≤β(x) for all x∈P.
First, we shall show that the operatorA maps P(γ,c)¯ into P(γ,c)¯. Let x∈P(γ,c)¯. Thus we have 0≤x(t)≤c for 0≤t≤1. Using (H3), we have
(3.10)|(Ax)(t)|=(Ax)(t)=∫01G(t,s)f(s,x(s))ds≤ηc∫01G(t,s)ds≤c.
Hence
(3.11)γ(Ax)=∥Ax∥≤c.
Therefore, we have shown that A:P(γ,c)¯→P(γ,c)¯.
We next prove that Condition (i) of Theorem2.2 holds. Let x≡(1/2)(b+k),k=b/σ. Then
(3.12)α(x)=12(b+k)>b,θ(x)=12(b+k)<k,γ(x)=12(b+k)<c,
which shows that {x∈P(γ,θ,α,b,k,c),α(x)>b}≠∅. Let x∈P(γ,θ,α,b,k,c). Then α(x)>b,θ(x)<k=b/σ imply that
(3.13)b<x(t)<bσ,fort∈[14,34].
By (H2) we can obtain
(3.14)α(Ax)=mint∈[1/4,3/4](Ax)(t)=mint∈[1/4,3/4]∫01G(t,s)f(s,x(s))ds≥σ∫01G(s,s)f(s,x(s))ds>σ∫1/43/4G(s,s)f(s,x(s))ds≥μb∫1/43/4G(s,s)ds=b.
Hence, α(Ax)>b for all x∈P(γ,θ,α,b,k,c) and so Condition (i) of Theorem 2.2 holds.
Next, we verify that Condition (ii) of Theorem2.2 is satisfied. Take x≡σa,h=σa, then
(3.15)γ(x)=σa<c,ψ(x)=σa=h,β(x)=σa<a.
From this we know that {x∈Q(γ,β,ψ,h,a,c),β(x)<a}≠∅. Let x∈Q(γ,β,ψ,h,a,c). Then we have β(x)≤a, which lead to 0≤x(t)≤a, for t∈[0,1]. In view of (H1), we have
(3.16)β(Ax)=maxt∈[0,1](Ax)(t)=maxt∈[0,1]∫01G(t,s)f(s,x(s))ds≤ηa·maxt∈[0,1]∫01G(t,s)ds=a.
Hence, β(Ax)<a for all x∈Q(γ,β,ψ,h,a,c). Thus, Condition (ii) of Theorem 2.2 is fulfilled.
We shall next show that Condition (iii) of Theorem2.2 is met. Observe that for x∈P(3.17)θ(Ax)=maxt∈[0,1](Ax)(t)=maxt∈[0,1]∫01G(t,s)f(s,x(s))ds≤∫01G(s,s)f(s,x(s))ds.
On the other hand,
(3.18)α(Ax)=mint∈[1/4,3/4](Ax)(t)=mint∈[1/4,3/4]∫01G(t,s)f(s,x(s))ds≥σ∫01G(s,s)f(s,x(s))ds.(3.17) together (3.18) implies that
(3.19)α(Ax)≥σθ(Ax),x∈P.
Letx∈P(γ,α,b,c) with θ(Ax)>k=b/σ. Then, it follows from (3.19) that
(3.20)α(Ax)≥σθ(Ax)>b.
Thus, α(Ax)>b for all x∈P(γ,α,b,c) with θ(Ax)>b/σ. Hence, Condition (iii) of Theorem 2.2 holds.
Finally, we shall prove that Condition (iv) of Theorem2.2 is fulfilled. Let x∈Q(γ,β,a,c) and ψ(Ax)<h=σa. Then 0≤x(t)≤a,t∈[0,1]. By (H1), we have
(3.21)β(Ax)=maxt∈[0,1](Ax)(t)=maxt∈[0,1]∫01G(t,s)f(s,x(s))ds<ηa·maxt∈[0,1]∫01G(t,s)ds=a.
Thus, Condition (iv) of Theorem 2.2 is satisfied.
Now, an application of Theorem2.2 ensures that the BVP (1.1) and (1.2) has at least three positive solutions x1,x2,x3 such that
(3.22)β(x1)<a,b<α(x2),a<β(x3)withα(x3)<b.
This proof is complete.Remark 3.3.
This Theorem improves the Corollary2.5 in [3].Example 3.4.
For simplicity, we consider the boundary value problem(3.23)-x′′=f(t,x),0<t<1,x(0)=x'(1)=0,
where
(3.24)f(t,x)={110|sint|+20x5,x≤1,110|sint|+20,x≥1.
By direct calculation we can obtain that η=2,μ=4,σ=1/4. Set a=1/2,b=1,c=12, so the nonlinear term f satisfies
(3.25)f(t,x)≤0.1+20×(12)5<1=ηa,(t,x)∈[0,1]×[0,12],f(t,x)>20>16=(μb)σ,(t,x)∈[14,34]×[1,4],f(t,x)<21<24=ηc,(t,x)∈[0,1]×[0,12].
Then the conditions in Theorem 3.2 are all satisfied, so the boundary value problem (3.23) has at least three positive solutions x1,x2,x3 such that
(3.26)max0≤t≤1x1(t)<12,1<min1/4≤t≤3/4x2(t),1<max0≤t≤1x3(t)withmin1/4≤t≤3/4x3(t)<1.



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