# Multiple Positive Solutions of a Second Order Nonlinear Semipositonem-Point Boundary Value Problem on Time Scales

Authors: Chengjun Yuan; Yongming Liu
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

In this paper, we study a general second-orderm-point boundary value problem for nonlinear singular dynamic equation on time scales uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)+λq(t)f(t,u(t))=0, t∈(0,1)𝕋, u(ρ(0))=0, u(σ(1))=∑i=1m-2αiu(ηi). This paper shows the existence of multiple positive solutions if f is semipositone and superlinear. The arguments are based upon fixed-point theorems in a cone.


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## Body

## 1. Introduction

In this paper, we consider the following dynamic equation on time scales:(1.1)uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)+λq(t)f(t,u(t))=0,t∈(0,1)𝕋,u(ρ(0))=0,u(σ(1))=∑i=1m-2αiu(ηi),
where αi≥0,0<ηi<ηi+1<1; for all i=1,2,…,m-2; f,q,a and b satisfy(C1)
q∈L is continuously and nonnegative function and there exists t0∈(ρ(0),σ(1)) s.t. q(t0)>0, q(t) may be singular at t=ρ(0),σ(1);(C2)
a∈C([0,1],[0,+∞)), b∈C([0,1],(-∞,0]).In the past few years, the boundary value problems of dynamic equations on time scales have been studied by many authors (see [1–15] and references therein). Recently, multiple-point boundary value problems on time scale have been studied, for instance, see [1–9].In 2008, Lin and Du [2] studied the m-point boundary value problem for second-order dynamic equations on time scales:(1.2)uΔ∇(t)+f(t,u)=0,t∈(0,T)∈𝕋,u(0)=0,u(T)=∑i=1m-2kiu(ξi),
where 𝕋 is a time scale. This paper deals with the existence of multiple positive solutions for second-order dynamic equations on time scales. By using Green's function and the Leggett-Williams fixed point theorem in an appropriate cone, the existence of at least three positive solutions of the problem is obtained.In 2009, Topal and Yantir [1] studied the general second-order nonlinear m-point boundary value problems (1.1) with no singularities and the case. The authors deal with the determining the value of λ; the existences of multiple positive solutions of (1.1) are obtained by using the Krasnosel'skii and Legget-William fixed point theorems.Motivated by the abovementioned results, we continue to study the general second-order nonlinearm-point boundary value problem (1.1), but the nonlinear term may be singularity and semipositone.In this paper, the nonlinear termf of (1.1) is suit to and semipositone and the superlinear case, we will prove our two existence results for 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 main result which state the sufficient conditions for the m-point boundary value problem (1.1) to have existence of positive solutions.


## 2. Preliminaries

In this section, we state the preliminary information that we need to prove the main results. From Lemmas2.1 and 2.3 in [1], we have the following lemma.Lemma 2.1 (see [1]).
Assuming that (C2) holds. Then the equations(2.1)ϕ1Δ∇(t)+a(t)ϕ1Δ(t)+b(t)ϕ1(t)=0,t∈(0,1)𝕋,ϕ1(ρ(0))=0,ϕ1(σ(1))=1,(2.2)ϕ2Δ∇(t)+a(t)ϕ2Δ(t)+b(t)ϕ2(t)=0,t∈(0,1)𝕋,ϕ2(ρ(0))=1,ϕ2(σ(1))=0
have unique solutions ϕ1 and ϕ2, respectively, and(a)
ϕ1 is strictly increasing on [ρ(0),σ(1)],(b)
ϕ2 is strictly decreasing on [ρ(0),σ(1)].For the rest of the paper we need the following assumption:(C3)
0<∑i=1m-2αiϕ1(ηi)<1.Lemma 2.2 (see [1]).
Assuming that (C2) and (C3) hold. Lety∈C[ρ(0),σ(1)]. Then boundary value problem
(2.3)xΔ∇(t)+a(t)xΔ(t)+b(t)x(t)+y(t)=0,t∈(0,1)𝕋,x(ρ(0))=0,x(σ(1))=∑i=1m-2αix(ηi)
is equivalent to integral equation
(2.4)x(t)=∫ρ(0)σ(1)H(t,s)p(s)y(s)∇s+Aϕ1(t),
where
(2.5)p(t)=ea(ρ(t),ρ(0)),A=11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αi∫ρ(0)σ(1)H(ηi,s)p(s)y(s)∇s,(2.6)H(t,s)=1ϕ1Δ(ρ(0)){ϕ1(s)ϕ2(t),s≤t,ϕ1(t)ϕ2(s),t≤s.Proof.
First we show that the unique solution of (2.3) can be represented by (2.4). From Lemma 2.1, we know that the homogenous part of (2.3) has two linearly independent solution ϕ1 and ϕ2 since
(2.7)|ϕ1(ρ(0))ϕ1Δ(ρ(0))ϕ2(ρ(0))ϕ2Δ(ρ(0))|=-ϕ1Δ(ρ(0))≠0.
Now by the method of variations of constants, we can obtain the unique solution of (2.3) which can be represented by (2.4) where A and H are as in (2.5) and (2.6), respectively. Next we check the function defined in (2.4) is the solution of the boundary value problem (2.3). For this purpose we first show that (2.4) satisfies (2.3). From the definition of Green's function (2.6), we get
(2.8)x(t)=1ϕ1Δ(ρ(0))(∫ρ(0)tϕ1(s)ϕ2(t)p(s)y(s)∇s+∫tσ(1)ϕ1(t)ϕ2(s)p(s)y(s)∇s)+Aϕ1(t).
Hence, the derivatives xΔ and xΔ∇ are as follows:
(2.9)xΔ(t)=1ϕ1Δ(ρ(0))(ϕ2Δ(t)∫ρ(0)tϕ1(s)p(s)y(s)∇s+ϕ1Δ(t)∫tσ(1)ϕ2(s)p(s)y(s)∇s)+Aϕ1Δ(t),xΔ∇(t)=1ϕ1Δ(ρ(0))(ϕ2Δ∇(t)∫ρ(0)ρ(t)ϕ1(s)p(s)y(s)∇s+ϕ2Δ(t)ϕ1(t)p(t)y(t)+ϕ1Δ∇(t)∫ρ(t)σ(1)ϕ2(s)p(s)y(s)∇s+ϕ1Δ(t)ϕ2(t)p(t)y(t))+Aϕ1Δ∇(t).
Replacing the derivatives in (2.3), we deduce that
(2.10)xΔ∇(t)+a(t)xΔ(t)+b(t)x(t)=A(ϕ1Δ∇(t)+a(t)ϕ1Δ(t)+b(t)ϕ1(t))+(1ϕ1Δ(ρ(0))∫ρ(0)tϕ1(s)p(s)y(s)∇s)(ϕ2Δ∇(t)+a(t)ϕ2Δ(t)+b(t)ϕ2(t))+(1ϕ1Δ(ρ(0))∫tσ(1)ϕ2(s)p(s)y(s)∇s)(ϕ1Δ∇(t)+a(t)ϕ1Δ(t)+b(t)ϕ1(t))+1ϕ1Δ(ρ(0))(ϕ2Δ∇(t)∫tρ(t)ϕ1(s)p(s)y(s)∇s+ϕ1Δ∇(t)∫ρ(t)tϕ2(s)p(s)y(s)∇s)+1ϕ1Δ(ρ(0))(ϕ2Δ(t)ϕ1(t)-ϕ1Δ(t)ϕ2(t))p(t)y(t)=1ϕ1Δ(ρ(0))(ϕ2Δ∇(t)(ρ(t)-t)ϕ1(t)p(t)y(t)-ϕ1Δ∇(t)ϕ2(t)p(t)y(t)+ϕ2Δ(t)ϕ1(t)p(t)y(t)-ϕ1Δ(t)ϕ2(t)p(t)y(t))=1ϕ1Δ(ρ(0))p(t)y(t)(ϕ2Δ(t)ϕ1(t)-ϕ1Δ(t)ϕ2(t))+1ϕ1Δ(ρ(0))p(t)y(t)(ρ(t)-t)(ϕ2Δ∇(t)ϕ1(t)-ϕ1Δ∇(t)ϕ2(t))=1ϕ1Δ(ρ(0))p(t)y(t){(ϕ2Δ(t)ϕ1(t)-ϕ1Δ(t)ϕ2(t))+(ρ(t)-t)(ϕ2Δ(t)ϕ1(t)-ϕ1Δ(t)ϕ2(t))∇}=1ϕ1Δ(ρ(0))p(t)y(t)(ϕ2Δ(ρ(t))ϕ1(ρ(t))-ϕ1Δ(ρ(t))ϕ2(ρ(t)))=1ϕ1Δ(ρ(0))p(t)y(t)eΘa(ρ(t),ρ(0))(-ϕ1Δ(ρ(0)))=-y(t).
Therefore the function defined in (2.4) satisfies (2.3). Further we obtain that the boundary value conditions are satisfied by (2.4). The first condition follows from (2.5) and (2.6) and Lemma 2.1. Now we verify the second boundary condition. Since
(2.11)H(σ(1),s)=1ϕ1Δ(ρ(0))ϕ1(s)ϕ2(σ(1))=0,
we obtain that
(2.12)x(σ(1))=∫ρ(0)σ(1)H(σ(1),s)p(s)y(s)∇s+Aϕ1(σ(1))=A.
On the other hand, by using (2.5), we find that
(2.13)∑i=1m-2αix(ηi)=∑i=1m-2αi(∫ρ(0)σ(1)H(ηi,s)p(s)y(s)∇s+Aϕ1(ηi))=∑i=1m-2αi(∫ρ(0)σ(1)H(ηi,s)p(s)y(s)∇s+11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiϕ1(ηi)∫ρ(0)σ(1)H(ηi,s)p(s)y(s)∇s)=11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αi∫ρ(0)σ(1)H(ηi,s)p(s)y(s)∇s=A.
Combining the two equations above finishes the proof.Lemma 2.3.
Green's functionH(t,s) has the following properties:
(2.14)H(t,s)≤H(t,t),ϕ1Δ(ρ(0))∥ϕ1∥∥ϕ2∥H(t,t)H(s,s)≤H(t,s)≤H(s,s),H(t,t)≤ϕ1(t)∥ϕ2∥ϕ1Δ(ρ(0)).Lemma 2.4.
Assume that (C2) and (C3) hold. Letu be a solution of boundary value problem (1.1) if and only if u is a solution of the following integral equation:
(2.15)u(t)=∫ρ(0)σ(1)G(t,s)p(s)q(s)f(s,u(s))∇s,
where
(2.16)G(t,s)=H(t,s)+11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiH(ηi,s)ϕ1(t).The proofs of the Lemmas2.3 and 2.4 can be obtained easily by Lemmas 2.1 and 2.2.Lemma 2.5.
Green's functionG(t,s) defined by (2.16) has the following properties:
(2.17)C2ϕ1(t)H(s,s)≤G(t,s)≤C1H(s,s),G(t,s)≤C3ϕ1(t),
where
(2.18)C1=1+∥ϕ1∥1-∑i=1m-2αiϕ1(ηi)∑i=1m-2αi,C2=ϕ1Δ(ρ(0))∥ϕ1∥∥ϕ2∥11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiH(ηi,ηi),C3=∥ϕ2∥ϕ1Δ(ρ(0))+11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiH(ηi,ηi).Proof.
From Lemma2.3, we have
(2.19)G(t,s)≤H(s,s)+11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiH(s,s)∥ϕ1∥≤C1H(s,s),G(t,s)≤∥ϕ2∥ϕ1Δ(ρ(0))ϕ1(t)+11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiH(ηi,ηi)ϕ1(t)≤C3ϕ1(t),G(t,s)≥11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiH(ηi,s)ϕ1(t)≥11-∑i=1m-2αiϕ1(ηi)∑i=1m-2αiϕ1Δ(ρ(0))∥ϕ1∥∥ϕ2∥H(ηi,ηi)H(s,s)ϕ1(t)≥C2ϕ1(t)H(s,s).
The proof is complete.The following theorems will play major role in our next analysis.Theorem 2.6 (see [16]).
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. Main Results

We make the following assumptions:(H1)
f(t,u)∈C([ρ(0),σ(1)]×[0,+∞),(-∞,+∞)), moreover there exists a function g(t)∈L1([ρ(0),σ(1)],(0,+∞)) such that f(t,u)≥-g(t), for any t∈(ρ(0),σ(1)), u∈[0,+∞).(H1*)
f(t,u)∈C((ρ(0),σ(1))×[0,+∞),(-∞,+∞)) may be singular at t=ρ(0),σ(1), moreover there exists a function g(t)∈L1((ρ(0),σ(1)),(0,+∞)) such that f(t,u)≥-g(t), for any t∈(ρ(0),σ(1)), u∈[0,+∞).(H2)
f(t,0)>0, for t∈[ρ(0),σ(1)].(H3)
There exists[θ1,θ2]∈(ρ(0),σ(1)) such that limu↑+∞mint∈[θ1,θ2](f(t,u)/u)=+∞.(H4)
∫ρ(0)σ(1)H(s,s)p(s)q(s)f(s,z)∇s<+∞ for any z∈[0,m], m>0 is any constant.In fact, we only consider the boundary value problem(3.1)xΔ∇(t)+a(t)xΔ(t)+b(t)x(t)+λq(t)[f(t,[x(t)-v(t)]*)+g(t)]=0,λ>0,x(ρ(0))=0,x(σ(1))=∑i=1m-2αix(ηi),
where (3.2)y(t)*={y(t),y(t)≥0,0,y(t)<0,
and v(t)=λ∫ρ(0)σ(1)G(t,s)p(s)q(s)g(s)∇s, which is the solution of the boundary value problem (3.3)vΔ∇(t)+a(t)vΔ(t)+b(t)v(t)+λq(t)g(t)=0,v(ρ(0))=0,v(σ(1))=∑i=1m-2αiv(ηi).
From Lemma 2.1, it is easy to verify that v(t)≤λC0ϕ1(t) and C0=C3∫ρ(0)σ(1)q(s)g(s)∇s.We will show that there exists a solutionx for boundary value problem (3.1) with x(t)≥v(t),t∈[ρ(0),σ(1)]. If this is true, then u(t)=x(t)-v(t) is a nonnegative solution (positive on (ρ(0),σ(1))) of boundary value problem (3.1). Since for any t∈(ρ(0),σ(1)), from (3.4)(u(t)+v(t))Δ∇+a(t)(u(t)+v(t))Δ+b(t)(u(t)+v(t))=-λq(t)[f(t,u)+g(t)],
we have (3.5)uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)=-λq(t)f(t,u).
As a result, we will concentrate our study on boundary value problem (3.1).We note thatx(t) is a solution of (3.1) if and only if(3.6)x(t)=λ∫ρ(0)σ(1)G(t,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s.For our constructions, we will consider the Banach spaceX=C[ρ(0),σ(1)] equipped with standard norm ∥x∥=max0≤t≤1|x(t)|,x∈X. We define a cone P by(3.7)P={x∈X∣x(t)≥C2C1ϕ1(t)∥x∥,t∈[ρ(0),σ(1)]},
where ϕ1 is defined by Lemma 2.1 (namely, ϕ1 is solution (2.1)). Define an integral operator T:P→X by(3.8)Tx(t)=λ∫ρ(0)σ(1)G(t,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s.
Notice, from (3.8) and Lemma 2.5, we have Tx(t)≥0 on [0,1] for x∈P and (3.9)Tx(t)=λ∫ρ(0)σ(1)G(t,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s≤C1λ∫ρ(0)σ(1)H(s,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s,
then ∥Tx∥≤C1λ∫ρ(0)σ(1)H(s,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s.On the other hand, we have(3.10)Tx(t)=λ∫ρ(0)σ(1)G(t,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s≥λ∫ρ(0)σ(1)C2ϕ1(t)H(s,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s≥C2C1ϕ1(t)λ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)(f(t,[x(t)-v(t)]*)+g(t))∇s≥C2C1ϕ1(t)∥Tx∥.
Thus, T(P)⊂P. In addition, standard arguments show that T(P)⊂P and T is a compact, and completely continuous.Theorem 3.1.
Suppose that (H1)-(H2) hold. Then there exists a constant λ¯>0 such that, for any 0<λ≤λ¯, boundary value problem (1.1) has at least one positive solution.Proof.
Fixδ∈(0,1). From (H2), let 0<ɛ<1 be such that
(3.11)f(t,z)≥δf(t,0),forρ(0)≤t≤σ(1),0≤z≤ɛ.
Suppose that
(3.12)0<λ<ɛ2cf¯(ɛ):=λ¯
where f¯(ɛ)=maxρ(0)≤t≤σ(1),0≤z≤ɛ{f(t,z)+g(t)} and c=C1∫ρ(0)σ(1)H(s,s)p(s)q(s)∇s. Since
(3.13)limz↓0f¯(z)z=+∞,f¯(ɛ)ɛ<12cλ,
there exists a R0∈(0,ɛ) such that
(3.14)f¯(R0)R0=12cλ.
Letx∈P and ν∈(0,1) be such that x=νT(x), we claim that ∥x∥≠R0. In fact
(3.15)∥Tx(t)∥≤νλ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)[f(s,[x(s)-v(s)]*)+g(s)]∇s≤λ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)[f(s,[x(s)-v(s)]*)+g(s)]∇s≤λ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)max0≤s≤1;0≤z≤R0[f(s,z)+g(s)]∇s≤λ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)f¯(R0)∇s≤λcf¯(R0),
that is,
(3.16)f¯(R0)R0≥1cλ>12cλ=f¯(R0)R0
which implies that ∥x∥≠R0. Let U={x∈P:∥x∥<R0}. By nonlinear alternative of Leray-Schauder type theorem, T has a fixed point x∈U¯. Moreover, combing (3.8), (3.28), and R0<ɛ, we obtain that
(3.17)x(t)=λ∫ρ(0)σ(1)G(t,s)p(s)q(s)[f(s,[x(s)-v(s)]*)+g(s)]∇s≥λ∫ρ(0)σ(1)G(t,s)p(s)q(s)[δf(s,0)+g(s)]∇s≥λ[δ∫ρ(0)σ(1)G(t,s)p(s)q(s)f(s,0)∇s+∫ρ(0)σ(1)G(t,s)p(s)q(s)g(s)∇s]>λ∫ρ(0)σ(1)G(t,s)p(s)q(s)g(s)∇s=v(t)fort∈(ρ(0),σ(1)).
Let u(t)=x(t)-v(t)>0. Then (1.1) has a positive solution u and ∥u∥≤∥x∥≤R0<1. This completes the proof of Theorem 3.1.Theorem 3.2.
Suppose that (H1*) and (H3)-(H4) hold. Then there exists a constant λ*>0 such that, for any 0<λ≤λ*, boundary value problem (1.1) has at least one positive solution.Proof.
LetΩ1={x∈P:∥x∥<R1}, where R1=max{1,r} and r=(C1C3/C2)∫ρ(0)σ(1)q(s)g(s)∇s. Choose
(3.18)λ*=min{1,R1[C1∫ρ(0)σ(1)H(s,s)p(s)q(s)[max0≤z≤R1f(s,z)+g(s)]∇s]-1}.
Then for any x∈P∩∂Ω1, then ∥x∥=R1 and x(s)-v(s)≤x(s)≤∥x∥, we have
(3.19)∥Tx(t)∥≤λ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)[f(s,[x(s)-v(s)]*)+g(s)]∇s≤λ∫ρ(0)σ(1)C1H(s,s)p(s)q(s)[f(s,[x(s)-v(s)]*)+g(s)]∇s≤λC1∫ρ(0)σ(1)H(s,s)p(s)q(s)[max0≤z≤R1f(s,z)+g(s)]∇s≤R1=∥x∥.
This implies that
(3.20)∥Tx∥≤∥x∥,x∈P∩∂Ω1.
On the other hand, choose a constantN>0 such that
(3.21)λC22N2C1∫θ1θ2H(s,s)ϕ1(s)p(s)q(s)∇sminθ1≤t≤θ2ϕ1(t)≥1.
By assumption (H3), for any t∈[θ1,θ2], there exists a constant B>0 such that
(3.22)f(t,z)z>N,namely,f(t,z)>Nz,forz>B.
Choose R2=max{R1+1,2λr,2C1(B+1)/C2minθ1≤t≤θ2ϕ1(t)}, and let Ω2={x∈P:∥x∥<R2}, then for any x∈P∩∂Ω2, we have
(3.23)x(t)-v(t)=x(t)-λ∫ρ(0)σ(1)G(t,s)p(s)q(s)g(s)∇s≥x(t)-λC3ϕ1(t)∫ρ(0)σ(1)p(s)q(s)g(s)∇s≥x(t)-C1x(t)C2∥x∥λC3∫ρ(0)σ(1)p(s)q(s)g(s)∇s≥x(t)-x(t)R2λr≥(1-λrR2)x(t)≥12x(t)≥0,t∈[0,1].
Then,
(3.24)minθ1≤t≤θ2{x(t)-v(t)}≥minθ1≤t≤θ2{12x(t)}≥minθ1≤t≤θ2{C22C1ϕ1(t)∥x∥}=R2C2minθ1≤t≤θ2ϕ1(t)2C1≥B+1>B.
Now,
(3.25)∥Tx(t)∥≥max0≤t≤1λ∫ρ(0)σ(1)C2ϕ1(t)H(s,s)p(s)q(s)[f(s,[x(s)-v(s)]*)+g(s)]∇s≥max0≤t≤1λC2ϕ1(t)∫ρ(0)σ(1)H(s,s)p(s)q(s)f(s,[x(s)-v(s)]*)∇s≥λC2minθ1≤t≤θ2ϕ1(t)∫θ1θ2H(s,s)p(s)q(s)f(s,x(s)-v(s))∇s≥λC2minθ1≤t≤θ2ϕ1(t)∫θ1θ2H(s,s)p(s)q(s)N(x(s)-v(s))∇s≥λC2minθ1≤t≤θ2ϕ1(t)∫θ1θ2H(s,s)p(s)q(s)N2x(s)∇s≥λC22N2C1minθ1≤t≤θ2ϕ1(t)∫θ1θ2H(s,s)p(s)q(s)ϕ1(s)∥x∥∇s≥λC22N2C1minθ1≤t≤θ2ϕ1(t)∫θ1θ2H(s,s)ϕ1(s)p(s)q(s)∇s∥x∥≥∥x∥.⇒∥Tx∥≥∥x∥,x∈P∩∂Ω2.
Condition (2.1) of Krasnosel'skii’s fixed-point theorem is satisfied. So T has a fixed point x with R1≤∥x∥<R2 such that
(3.26)xΔ∇(t)+a(t)xΔ(t)+b(t)x(t)=-λq(s)(f(s,[x(s)-v(s)]*)+g(s)),0<t<1,x(ρ(0))=0,x(σ(1))=∑i=1m-2αix(ηi).
Since r<∥x∥,
(3.27)x(t)-v(t)≥C2C1ϕ1(t)∥x∥-λ∫ρ(0)σ(1)G(t,s)p(s)q(s)g(s)∇s≥C2C1ϕ1(t)∥x∥-ϕ1(t)λC3∫ρ(0)σ(1)p(s)q(s)g(s)∇s≥C2C1ϕ1(t)∥x∥-C2C1ϕ1(t)λr≥C2C1ϕ1(t)r-C2C1ϕ1(t)λr≥(1-λ)C2C1rϕ1(t)>0.
Let u(t)=x(t)-v(t), then u(t) is a positive solution of boundary value problem (1.1). This completes the proof of Theorem 3.2.Since condition (H1) implies conditions (H1*) and (H4), and from proof of Theorems 3.1 and 3.2, we immediately have the following theorem.Theorem 3.3.
Suppose that (H1)–(H3) hold. Then boundary value problem (1.1) has at least two positive solutions for λ>0 sufficiently small.Proof.
On the hand, fixδ∈(0,1). From (H2), let 0<ɛ<min{1,r/2} be such that
(3.28)f(t,z)≥δf(t,0),forρ(0)≤t≤σ(1),0≤z≤ɛ.
Choose
(3.29)ɛ2cf¯(ɛ):=λ¯,
where f¯(ɛ)=maxρ(0)≤t≤σ(1),0≤z≤ɛ{f(t,z)+g(t)}, c=C1∫ρ(0)σ(1)H(s,s)p(s)q(s)∇s, and r=(C1C3/C2)∫ρ(0)σ(1)q(s)g(s)∇s.
On the other hand, setΩ1={x∈P:∥x∥<R1}, where R1=1+r. Choose
(3.30)λ*=min{1,R1[C1∫ρ(0)σ(1)H(s,s)p(s)q(s)[max0≤z≤R1f(s,z)+g(s)]∇s]-1}.
So, let(3.31)0<λ<min{λ¯,λ*}.
From0<λ<min{λ¯,λ*}, we have 0<λ<λ¯, from proof of Theorem 3.1, we know that (1.1) has a positive solution u1 and ∥u1∥≤∥u1∥≤R0<r/2. Further, also from 0<λ<min{λ¯,λ*}, we have 0<λ<λ*, from proof of Theorem 3.2, we know that (1.1) has a positive solution u2 and ∥u2∥≥R1/2>r/2. Then (1.1) has at least two positive solutions u1 and u2. This completes the proof of Theorem 3.3.


## 4. Examples

To illustrate the usefulness of the results, we give some examples.Example 4.1.
Consider the boundary value problem(4.1)uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)+λ(ua(t)+1(t-t2)1/2cos(2πu(t)))=0,λ>0,u(ρ(0))=0,u(σ(1))=∑i=1m-2αiu(ηi),
where a>1. Then, if λ>0 is sufficiently small, (4.1) has a positive solution u with u(t)>0 for t∈(0,1).
To see this, we will apply Theorem3.2 with
(4.2)q(t)=1,f(t,u)=ua(t)+1(t-t2)1/2cos(2πu(t)),g(t)=1(t-t2)1/2.
Clearly
(4.3)f(t,0)=1(t-t2)1/2>0,f(t,u)+g(t)≥ua(t)>0,limu↑+∞f(t,u)u=+∞,
for t∈(ρ(0),σ(1)),u>0. Namely, (H1*) and (H2)–(H4) hold. From ∫ρ(0)σ(1)(1/((σ(1)-ρ(0))s-(s-ρ(0))2)1/2)∇s=π, set R1=C1π and m=maxρ(0)≤t≤σ(1){p(t)}+1, we have
(4.4)∫ρ(0)σ(1)C1H(s,s)p(s)[max0≤z≤R1f(s,z)+g(s)]∇s≤∫ρ(0)σ(1)mC1∥ϕ1∥∥ϕ2∥ϕ1Δ(ρ(0))[(C1π)a+1((σ(1)-ρ(0))s-(s-ρ(0))2)1/2]∇s≤mC1∥ϕ1∥∥ϕ2∥ϕ1Δ(ρ(0))((C1π)a+π)
and λ*=min{1,ϕ1Δ(ρ(0))/m∥ϕ1∥∥ϕ2∥((C1aπa-1+1)}. Now, if λ<λ*, Theorem 3.2 guarantees that (4.1) has a positive solution u with ∥u∥≥2.Example 4.2.
Consider the boundary value problem(4.5)uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)+λ(u2(t)-92u(t)+2)=0,0<t<1,λ>0,u(ρ(0))=0,u(σ(1))=∑i=1m-2αiu(ηi).
Then, if λ>0 is sufficiently small, (4.5) has two solutions ui with ui(t)>0 for t∈(0,1),i=1,2.
To see this, we will apply Theorem3.3 with
(4.6)f(t,u)=u2(t)-92u(t)+2,g(t)=4.
Clearly
(4.7)q(t)=0,f(t,0)=2>0,f(t,u)+g(t)≥1516>0,limu↑+∞f(t,u)u=+∞.
Namely, (H1)–(H3) hold. Let δ=1/4, ɛ=1/4 and c=∫ρ(0)σ(1)C1H(s,s)p(s)∇s, then we may have
(4.8)λ¯=18c(max0≤z≤ɛf(t,z)+4)=148c.
Now, if λ<λ¯, Theorem 3.2 guarantees that (4.5) has a positive solution u1 with ∥u1∥≤1/4.
Next, letR1=p, where p=4cC3/C2+1 is a constant, then we have
(4.9)∫ρ(0)σ(1)C1H(s,s)p(s)[max0≤z≤R1f(s,z)+g(s)]∇s=∫ρ(0)σ(1)C1H(s,s)p(s)[92+4]∇s=17c2
and λ*=min{1,2p/17c}. Now, if 0<λ<λ*, Theorem 3.1 guarantees that (4.5) has a positive solution u2 with ∥u2∥≥p.
So, since all the conditions of Theorem3.3 are satisfied, if λ<min{λ¯,λ*}, Theorem 3.3 guarantees that (4.5) has two solutions ui with ui(t)>0  (i=1,2).Example 4.3.
Consider the boundary value problem(4.10)uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)+λ(ua(t)+cos(2πu(t)))=0,0<t<1,λ>0,u(ρ(0))=0,u(σ(1))=∑i=1m-2αiu(ηi),
where a>1. Then, if λ>0 is sufficiently small, (4.10) has two solutions ui with ui(t)>0 for t∈(0,1),i=1,2.
To see this we will apply Theorem3.3 with
(4.11)f(t,u)=ua(t)+cos(2πu(t)),g(t)=2.
Clearly
(4.12)f(t,0)=1>0,f(t,u)+g(t)≥ua(t)+1>0,limu↑+∞f(t,u)u=+∞,fort∈(ρ(0),σ(1)).
Namely, (H1)–(H3) hold. Let δ=1/2, ɛ=1/8 and c=∫ρ(0)σ(1)C1H(s,s)p(s)∇s, then we may have
(4.13)ɛ2c(max0≤x≤ɛf(t,x)+2)≥116c((1/8)a+3):=λ¯.
Now, if 0<λ<λ¯ then 0<λ<ɛ/2c(max0≤x≤ɛf(t,x)+2), Theorem 3.2 guarantees that (4.10) has a positive solution u1 with ∥u1∥≤1/8.
Next, letR1=p, where p=4cC3/C2+1 is a constant, then we have
(4.14)∫ρ(0)σ(1)C1H(s,s)p(s)[max0≤z≤R1f(s,z)+g(s)]∇s≤∫ρ(0)σ(1)C1H(s,s)p(s)[pa+2]∇s=(pa+2)c
and λ*=min{1,p/(pa+2)c}. Now, if 0<λ<λ*, then 0<λ<p(∫ρ(0)σ(1)C1H(s,s)p(s)[max0≤z≤R1f(s,z)+g(s)]∇s)-1, Theorem 3.1 guarantees that (4.10) has a positive solution u2 with ∥u2∥≥1.
So, ifλ<min{λ¯,λ*}, Theorem 3.3 guarantees that (4.10) has two solutions ui with ui(t)>0  (i=1,2).Example 4.4.
Let𝕋={0,1/4,2/4,3/4,1,5/4,…}. We consider the following four point boundary value problem:
(4.15)uΔ∇(t)+125uΔ(t)-165u(t)+λ(ua(t)+cos(2πu(t)))=0,λ>0,u(0)=0,u(54)=12u(14)+14u(12),
where a(t)=12/5, b(t)=-16/5, and q(t)=1. Then, if λ>0 is sufficiently small, (4.15) has two solutions ui with ui>0 (i=1,2).
Letϕ1 and ϕ2 be the solutions of the following linear boundary value problems, respectively,
(4.16)uΔ∇(t)+125uΔ(t)-165u(t)=0,u(0)=0,u(54)=1,uΔ∇(t)+125uΔ(t)-165u(t)=0,u(0)=1,u(54)=0.
It is evident (form the Corollaries 4.24 and 4.25 and Theorem 4.28 of [17]) that
(4.17)ϕ1(t)=(5/4)4t-(1/2)4t(5/4)5-(1/2)5,ϕ2(t)=(5/4)5(1/2)4t-(1/2)5(5/4)4t(5/4)5-(1/2)5.
Also ϕ1 satisfies (C3). Green's function is
(4.18)H(t,s)=10249279{(54)4s-(12)4s(54)5(12)4t-(12)5(54)4t,s≤t,(54)4t-(12)4t(54)5(12)4s-(12)5(54)4s,t≤s,
and p(t)=(2/5)4t-1 follows from eα(t,t0)=(1+αh)(t-t0)/h on 𝕋=hℕ.
To see this, we will apply Theorem3.3 with
(4.19)f(t,u)=ua(t)+cos(2πu(t)),g(t)=2.
Clearly
(4.20)f(t,0)=1>0,f(t,u)+g(t)≥ua(t)+1>0,limu↑+∞f(t,u)u=+∞,fort∈(ρ(0),σ(1)).
Namely, (H1)–(H3) hold. Let δ=1/2, ɛ=1/8 and c=∫ρ(0)σ(1)C1H(s,s)p(s)∇s, then we may have
(4.21)ɛ2c(max0≤x≤ɛf(t,x)+2)≥116c((1/8)a+3):=λ¯.
Now, if 0<λ<λ¯ then 0<λ<ɛ/2c(max0≤x≤ɛf(t,x)+2), Theorem 3.2 guarantees that (4.15) has a positive solution u1 with ∥u1∥≤1/8.
Next, letR1=4cC3/C2+1 is a constant, then we have
(4.22)∫ρ(0)σ(1)[max0≤z≤R1f(s,z)+g(s)]∇s≤∫ρ(0)σ(1)C1H(s,s)p(s)[R1a+2]∇s=(pa+2)c
and λ*=min{1,R1/(R1a+2)c}. Now, if 0<λ<λ*, then 0<λ<R1(∫ρ(0)σ(1)C1H(s,s)[max0≤z≤R1f(s,z)+g(s)]∇s)-1, Theorem 3.1 guarantees that (4.15) has a positive solution u2 with ∥u2∥≥1.
So, ifλ<min{λ¯,λ*}, Theorem 3.3 guarantees that (4.15) has two solutions ui with ui(t)>0  (i=1,2).



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