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Introduction Consider the singular multi-point boundary-value problem (ϕp(u0(t)))0+λf(t, u(t

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Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 38, pp. 1–13.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

MULTIPLE POSITIVE SOLUTIONS FOR SINGULAR MULTI-POINT BOUNDARY-VALUE PROBLEMS WITH A

POSITIVE PARAMETER

CHAN-GYUN KIM, EUN KYOUNG LEE

Abstract. In this article we study the existence, nonexistence, and multi- plicity of positive solutions for a singular multi-point boundary value problem with positive parameter. We use the fixed point index theory on a cone and a well-known theorem for the existence of a global continuum of solutions to establish our results.

1. Introduction

Consider the singular multi-point boundary-value problem

p(u0(t)))0+λf(t, u(t)) = 0, t∈(0,1), (1.1) u(0) =

m−2

X

i=1

aiu(ξi), u(1) =

m−2

X

i=1

biu(ξi), (1.2)

whereϕp(s) =|s|p−2s, p >1, λa nonnegative real parameter,ξi∈(0,1) with 0<

ξ1< ξ2 <· · ·< ξm−2 <1,ai, bi∈[0,1) with 0≤Pm−2

i=1 ai<1, 0≤Pm−2 i=1 bi<1, andf ∈C((0,1)×[0,∞),(0,∞)). Here,f(t, u) may be singular att= 0 and/or 1 and satisfies the following conditions:

(F1) for all M > 0, there exists hM ∈ A such that f(t, u) ≤ hM(t), for all u∈[0, M] and allt∈(0,1), where

A={h: Z 1/2

0

ϕ−1p Z 1/2 s

h(τ)dτ ds+

Z 1 1/2

ϕ−1p Z s 1/2

h(τ)dτ

ds <∞}: (F2) there exists [α, β] ⊂(0,1) such that limu→∞f(t, u)/up−1 =∞ uniformly

in [α, β].

By a positive solution of problem (1.1)-(1.2), we mean a functionu∈C[0,1]∩ C1(0,1) with ϕp(u0)∈C1(0,1) that satisfies (1.1)-(1.2) and u > 0 in (0,1). Here k · kdenotes the usual maximum norm inC[0,1].

Motivated by the work of Bitsadze [3, 4], the study of multi-point boundary value problem for linear second-order ordinary differential equations was initially done by

2000Mathematics Subject Classification. 34B10, 34B16.

Key words and phrases. Singular problem; multi-point boundary value problems;

positive solution;p-Laplacian; multiplicity.

c

2014 Texas State University - San Marcos.

Submitted July 3, 2013. Published February 5, 2014.

1

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Il’in and Moiseev [13, 14]. Gupta [11] studied three-point boundary value problems for nonlinear ordinary differential equations. Since then, many researchers have studied nonlinear second-order multi-point boundary value problems under various conditions on the nonlinear term. We refer the reader to [2, 8, 9, 15, 19, 22, 23, 24, 25, 27, 28] and references therein.

Problem (1.1)-(1.2) is a singular boundary value problem since f is allowed to have singularity att= 0 and/or 1. Singular problems have been extensively studied in the literature. For the case of two-point boundary value problems, the results were proved in [1, 5, 6, 12, 18, 21, 26, 29, 30] and for multi-point boundary value problems, the results were proved in [8, 9, 19, 22, 24, 28]. However, there are few results for multi-point boundary value problems having nonlinear term which does not satisfyL1-Carath´eodory condition. Recently, in semi-linear case, Sun et al. [24]

studied the following singular three-point boundary-value problem y00+µa(t)g1(t, y) = 0, t∈(0,1)

y(0)−βy0(0) = 0, y(1) =αy(η), (1.3) where µ > 0 is a parameter, β > 0, 0 < η < 1, 0< αη < 1, (1−αη) +β(1− α)>0, a∈ C((0,1),(0,∞)) satisfies 0<R1

0(β+s)(1−s)a(s)ds < ∞, and g1 ∈ C([0,1]×(0,∞),(0,∞)) may be singular at y = 0. Without any monotone or growth conditions imposed on the nonlinearityg1, using fixed point index theorem, they obtained not only the existence results of positive solutions to the problem (1.3), but also the explicit interval about positive parameter µ. Kim [19], in p- Laplacian case, presented some sufficient conditions for one or multiple positive solutions to the problem (1.1)-(1.2), where f(t, u) = h(t)g2(t, u), h ∈ A, g2 ∈ C([0,1]×[0,∞),[0,∞)).

To the authors’ knowledge, in the case of p-Laplacian, there is no result about the global structure of positive solutions for parameterλ ∈(0,∞) to multi-point boundary-value problems with the nonlinear term admitting stronger singularity thanL1(0,1) att= 0 and/or 1. The following is the main result in this paper.

Theorem 1.1. Assume that(F1)and(F2)hold. Assume in addition thatf(t, u) = h(t)g(t, u), whereh∈ A andg∈C((0,1)×[0,∞),(0,∞))satisfies

(A1) for all N > 0 and all > 0, there exists δ = δ(N, ) > 0 such that if u, v∈[0, N] and|u−v|< δ, then|g(t, u)−g(t, v)|< , for all t∈(0,1), (A2) inf{g(t, u)|t∈(0,1), u∈[0,∞)}>0.

Then there exists λ > 0 such that problem (1.1)-(1.2) has at least two positive solutions for λ∈(0, λ), at least one positive solution for λ=λ and no positive solution forλ > λ.

The above result is an extension of previous works for two-point boundary-value problems by Choi [5], Wong [26], Dalmasso [6], Ha and Lee [12], Lee [21], Xu and Ma [29], and Kim [18].

The rest of this article is organized as follows. In Section 2, the operator for problem (1.1)-(1.2) is introduced, and well-known facts such as Picone-type identity and Global continuation theorem are presented. In Section 3, the proofs of our results (Theorem 3.4 and Theorem 1.1) and examples for nonlinear term to illustrate our results are given.

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2. Preliminaries

First we introduce the operator corresponding to problem (1.1)-(1.2). Through- out this section we assume that (F1) holds. Set

K={u∈C[0,1] :ui s a nonnegative concave function on [0,1],usatisfies (1.2)}.

Then K is an ordered cone in C[0,1]. For (λ, u) ∈ [0,∞)× K, we define xλ,u : [0,1]→Rasxλ,u(t) =x1λ,u(t)−x2λ,u(t), where

x1λ,u(t) =A−1

m−2

X

i=1

ai Z ξi

0

ϕ−1p hZ t s

λf(τ, u(τ))dτi ds+

Z t 0

ϕ−1p hZ t s

λf(τ, u(τ))dτi ds and

x2λ,u(t) =B−1

m−2

X

i=1

bi

Z 1 ξi

ϕ−1p hZ s t

λf(τ, u(τ))dτi ds+

Z 1 t

ϕ−1p hZ s t

λf(τ, u(τ))dτi ds.

Here

A= 1−

m−2

X

i=1

ai, B= 1−

m−2

X

i=1

bi.

For λ > 0, limt→0+xλ,u(t) < 0 and limt→1xλ,u(t) > 0. Indeed we can rewrite x1λ,u(t) as

x1λ,u(t)

=A−1

m−2

X

i=1

ai

Z ξi t

ϕ−1p hZ s t

λf(τ, u(τ))dτi ds+

Z t 0

ϕ−1p hZ t s

λf(τ, u(τ))dτi ds

. By (F1), there existsh2∈ Asuch that

0≤ Z t

0

ϕ−1p hZ t s

λf(τ, u(τ))dτi ds≤

Z t 0

ϕ−1p hZ t s

h2(τ)dτi ds, and

lim

t→0+

Z t 0

ϕ−1p hZ t s

λf(τ, u(τ))dτi ds= 0.

Clearly limt→0+x2λ,u(t)>0, and thus limt→0+xλ,u(t)<0. In a similar manner we can show limt→1xλ,u(t)>0. Since xλ,u is continuous and strictly increasing in (0,1), there exists a unique zeroAλ,u∈(0,1) such thatxλ,u(Aλ,u) = 0. Forλ= 0, we may takeA0,u= 0 sincex0,u≡0. Then, for (λ, u)∈[0,∞)× K,

A−1

m−2

X

i=1

ai Z ξi

0

ϕ−1p hZ Aλ,u s

λf(τ, u(τ))dτi ds+

Z Aλ,u 0

ϕ−1p hZ Aλ,u s

λf(τ, u(τ))dτi ds

=B−1

m−2

X

i=1

bi

Z 1 ξi

ϕ−1p hZ s Aλ,u

λf(τ, u(τ))dτi ds+

Z 1 Aλ,u

ϕ−1p hZ s Aλ,u

λf(τ, u(τ))dτi ds.

DefineH : [0,∞)× K →C[0,1] as

H(λ, u)(t) =









A−1Pm−2 i=1 aiRξi

0 ϕ−1p RAλ,u

s λf(τ, u(τ))dτ ds +Rt

0ϕ−1p RAλ,u

s λf(τ, u(τ))dτ

ds, 0≤t≤Aλ,u,

B−1Pm−2 i=1 bi

R1

ξiϕ−1p Rs

Aλ,uλf(τ, u(τ))dτ ds +R1

Aλ,uϕ−1p Rs

Aλ,uλf(τ, u(τ))dτ

ds, Aλ,u≤t≤1.

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In view of the definition ofAλ,u,H(λ, u) is well-defined,kH(λ, u)k=H(λ, u)(Aλ,u), andH(λ, u)∈ Kfor all (λ, u)∈[0,∞)× K(see, e.g., [8, Lemma 2.2]).

Lemma 2.1. Problem (1.1)-(1.2) has a positive solution u if and only if H(λ,·) has a fixed pointuinK forλ >0.

Proof. We assume that u is a positive solution of problem (1.1)-(1.2). If λ = 0, u ≡0 by the facts that 0 ≤ Pm−2

i=1 ai < 1 and 0 ≤ Pm−2

i=1 bi < 1. Thus λ > 0.

Since u0 is strictly decreasing in (0,1), u ∈ K. From the fact that u satisfies (BC), max{u(0), u(1)}< u(ξj) for some 1≤j≤m−2, and there exists a unique Au∈(0,1) such that u0(Au) = 0. Integrating (Pλ) fromstoAu, we have

u0(s) =ϕ−1p h λ

Z Au s

f(τ, u(τ))dτi

. (2.1)

Again integrating (2.1) from 0 tot, we have u(t) =u(0) +

Z t 0

ϕ−1p hZ Au s

λf(τ, u(τ))dτi

ds, t∈[0,1).

Thenu(ξi) =u(0) +Rξi

0 ϕ−1p RAu

s λf(τ, u(τ))dτ dsand u(0) =

m−2

X

i=1

aiu(ξi)

=

m−2

X

i=1

aiu(0) +

m−2

X

i=1

ai Z ξi

0

ϕ−1p hZ Au

s

λf(τ, u(τ))dτi ds.

Thus

u(0) =A−1

m−2

X

i=1

ai

Z ξi 0

ϕ−1p hZ Au s

λf(τ, u(τ))dτi ds.

Similarly, integrating (2.1) fromtto 1, u(t) =u(1) +

Z 1 t

ϕ−1p hZ s Au

λf(τ, u(τ))dτi

ds, t∈(0,1]

and

u(1) =B−1

m−2

X

i=1

bi Z 1

ξi

ϕ−1p hZ s Au

λf(τ, u(τ))dτi ds.

Then, by the definition ofAλ,u,Au=Aλ,uand consequentlyH(λ, u)≡u.

Conversely, if we assume that there existsu∈ Ksuch thatH(λ, u) =uforλ >0, then one can easily see thatuis a positive solution of problem (1.1)-(1.2).

Lemma 2.2. Let M >0 be given and let{(λn, un)} be a sequence in[0,∞)× K with |λn|+kunk ≤ M. If Aλn,un → 0 (or 1) as n → ∞, then λn → 0 and kH(λn, un)k →0as n→ ∞.

Proof. We only prove the caseAλn,un →0 as n→ ∞since the other case can be showed in a similar manner. By the definition ofAλ,u, we can easily knowλn→0 as n → ∞. By (F1), there exists hM ∈ A such that f(t, u) ≤hM(t), t ∈ (0,1), u∈[0, M]. For sufficiently largen, we haveAλn,un< ξ1,

0≤H(λn, un)(0) =A−1

m−2

X

i=1

ai Z ξi

0

ϕ−1p hZ Aλn,un s

λnf(τ, un(τ))dτi ds

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≤λnA−1 Z ξm−2

0

ϕ−1p hZ ξm−2

s

hM(τ)dτi ds, and

kH(λn, un)k=H(λn, un)(0) +λn

Z Aλn,un 0

ϕ−1p hZ Aλn,un s

hM(τ)dτi ds.

ThuskH(λn, un)k →0 asn→ ∞sincehM ∈ Aandλn→0 asn→ ∞.

Lemma 2.3. H : [0,∞)× K → Kis completely continuous.

Proof. By Lemma 2.2, Ascoli-Arzel`a theorem, and Lebesgue dominated conver- gence theorem, one can easily show the completely continuity ofH(e.g., see [1, 19]).

Thus we omit the proof here.

Next we introduce the generalized Picone identity due to Jaros and Kusano ([16]). Let us consider the following operators:

lp[y]≡(ϕp(y0))0+q(t)ϕp(y), Lp[z]≡(ϕp(z0))0+Q(t)ϕp(z).

Theorem 2.4([20, p 382]). Letq(t)andQ(t)be measurable functions on an inter- val I. If y and z are any functions such that y, z, ϕp(y0),ϕp(z0)are differentiable a.e. onI andz(t)6= 0fort∈I, then the following holds

d dt

n|y|pϕp(z0)

ϕp(z) −yϕp(y0)o

= (q−Q)|y|p

|y0|p+ (p−1)|yz0

z |p−pϕp(y)y0ϕp

z0 z

−ylp[y] + |y|p ϕp(z)Lp[z].

(2.2) Remark 2.5. By Young’s inequality, we have

|y0|p+ (p−1)|yz0

z |p−pϕp(y)y0ϕp z0 z

≥0, and the equality holds if and only ify0=yz0/z in (a, b).

Finally we recall a well-known theorem for the existence of a global continuum of solutions by Leray and Schauder [17].

Theorem 2.6 ([31, Corollary 14.12]). LetX be a Banach space withX 6={0}and letK be an ordered cone in X. Consider

x=H(µ, x), (2.3)

whereµ∈[0,∞)and x∈ K. IfH : [0,∞)× K → K is completely continuous and H(0, x) = 0 for all x∈ K. Then the solution component C of (2.2) in [0,∞)× K which contains(0,0) is unbounded.

3. Main results

Since H(0, u) = 0 and H(λ,0) 6= 0 if λ 6= 0, by Lemma 2.3, Theorem 2.6, we obtain the following proposition.

Proposition 3.1. Assume that (F1) holds. Then there exists an unbounded con- tinuum C emanating from (0,0) in the closure of the set of positive solutions of problem (1.1)-(1.2)in [0,∞)× K.

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To see the shape ofC, we need lemmas regarding λ-direction block anda priori estimate. Using the generalized Picone identity (Theorem 2.4) and the properties of thep-sine function [7, 32], we obtain the following two lemmas.

Lemma 3.2. Assume that (F1) and(F2)hold. Then there existsλ >¯ 0 such that if problem (1.1)-(1.2)has a positive solutionuλ, thenλ≤λ.¯

Proof. Letuλbe a positive solution of problem (1.1)-(1.2). Sincef(t, u)>0 for all (t, u)∈(0,1)×[0,∞), by (F2), there existsC1>0 such that

f(t, u)> C1ϕp(u) foru∈[0,∞), t∈[α, β]. (3.1) It is easy to check thatw(t) =Sqq(t−α)/(β−α)), whereSq is theq-sine func- tion and 1p+1q = 1, is a solution of

p(w0(t)))0+ πq β−α

p

ϕp(w(t)) = 0, t∈(α, β), w(α) =w(β) = 0.

Takingy =w, z=uλ,q(t) = (πq/(β−α))p andQ(t) =λf(t, uλ)/ϕp(uλ) in (2.2) and integrating (2.2) fromαtoβ, by Remark 2.5,

Z β α

πq

β−α p

−λf(t, uλ) ϕp(uλ)

|w|pdt≥0.

It follows from (3.1) that πq

β−α p

−λC1

Z β α

|w|pdt≥0,

and thus the proof is complete.

Lemma 3.3. Assume that (F1) and (F2) hold, and let J = [D, E] be a compact subset of (0,∞). Then there exists MJ >0 such that if uis a positive solution of problem (1.1)-(1.2)with λ∈J, then kuk ≤MJ.

Proof. Suppose on the contrary that there exists a sequence {un} of positive so- lutions of problem (1.1)-(1.2) with λn instead ofλ, and {λn} ⊂ J = [D, E] and kunk → ∞asn→ ∞. It follows from the concavity ofun for allnthat

un(t)≥min{α,1−β}kunk, t∈(α, β). (3.2) TakeC= 2D−1q/(β−α))p>0. By (F2), there existsK >0 such thatf(t, u)>

p(u), fort ∈(α, β),u > K. From the assumption, we get kuNk >(min{α,1− β})−1K, for sufficiently large N. Therefore, by (3.2), we have

f(t, uN(t))> Cϕp(uN(t)), t∈(α, β).

As in the proof of Lemma 3.2, if we takey(t) =Sqq(t−α)/(β−α)) andz=uN, by Theorem 2.4 and Remark 2.5,

C≤D−1 πq β−α

p .

This contradicts the choice ofC, and thus the proof is complete.

Setting λ = sup{µ > 0: for all λ ∈ (0, µ), there exists at least two positive solutions of problem (1.1)-(1.2), thenλ>0 is well-defined. Indeed by Proposition 3.1,Cemanates from (0,0), and problem (1.1)-(1.2) has a small solution near (0,0) for λ ∈ (0, s) with small s > 0. On the other hand, for any M > 0, define

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CM = {(λ, u) ∈ C : kuk ≥ M} and the projection of CM to the λ-axis as ΛM. Then, by Lemma 3.2 and Lemma 3.3, for large M, ΛM = (0, aM], whereaM >0 and it is decreasing in M. This implies that, for any interval (0, s) with small s >0, problem (1.1)-(1.2) also has a large solution forλ∈(0, s). Thus λ >0 is well-defined. Moreover it follows from an easy compactness argument that problem (1.1)-(1.2) has at least two positive solution forλ∈(0, λ) and at least one positive solution forλ=λ.

The following is the first result in this work.

Theorem 3.4. Assume that (F1) and (F2)hold. Then there exists λ ≥λ >0 such that problem (1.1)-(1.2)has at least two positive solutions for λ∈(0, λ), at least one positive solution forλ∈[λ, λ], and no positive solution forλ > λ. Proof. Defineλ= sup{λ: problem (1.1)-(1.2) has at least one positive solution}.

Then by Lemma 3.2, λ ≤λ <∞. We only consider the case λ < λ, since the proof is done for the case λ. For λ∈ [λ, λ), there exists ˆλ∈[λ, λ) such that (1.1)-(1.2) with ˆλ instead ofλ, has a positive solution, say ˆu. Consider the modified problem

p(u0(t)))0+λf¯(t, u(t)) = 0, t∈(0,1), u(0) =

m−2

X

i=1

aiu(ξi), u(1) =

m−2

X

i=1

biu(ξi), (3.3) where ¯f(t, u) =f(t, γ(t, u)) andγ: (0,1)×R→Ris defined as

γ(t, u) =



 ˆ

u(t), ifu >u(t),ˆ u, if 0≤u≤u(t),ˆ 0, ifu <0.

Then all solutions u of (3.3) are concave and non-trivial. Define Tλ : C[0,1] → C[0,1] as

Tλ(u)(t) =













A−1Pm−2 i=1 ai

Rξi

0 ϕ−1p RAˆ

s λf¯(τ, u(τ))dτ ds +Rt

0ϕ−1p RAˆ

s λf¯(τ, u(τ))dτ

ds, 0≤t≤Aˆ

B−1Pm−2 i=1 biR1

ξiϕ−1p Rs

Aˆλf¯(τ, u(τ))dτ ds +R1

Aˆϕ−1p Rs

Aˆλf¯(τ, u(τ))dτ

ds, Aˆ≤t≤1.

where ˆAsatisfies A−1

m−2

X

i=1

ai

Z ξi 0

ϕ−1p hZ Aˆ s

λf¯(τ, u(τ))dτi ds+

Z Aˆ 0

ϕ−1p hZ Aˆ s

λf¯(τ, u(τ))dτi ds

=B−1

m−2

X

i=1

bi

Z 1 ξi

ϕ−1p hZ s Aˆ

λf¯(τ, u(τ))dτi ds+

Z 1 Aˆ

ϕ−1p hZ s Aˆ

λf¯(τ, u(τ))dτi ds.

It is easy to check thatTλis completely continuous onC[0,1], anduis a solution of (3.3) if and only ifu=Tλu. It follows from the definition ofγ and the continuity of f that there exists R1 >0 such thatkTλuk < R1 for all u∈C[0,1]. Then by Schauder fixed point theorem, there exists uλ ∈C[0,1] such thatTλuλ=uλ, and uλis a positive solution of (Mλ).

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We first claim thatuλ(0) ≤u(0). If the claim is not true,ˆ uλ(0) >u(0). Putˆ x(t) =uλ(t)−u(t). Thenˆ

0< x(0) =uλ(0)−u(0) =ˆ

m−2

X

i=1

aix(ξi)≤

m−2

X

i=1

aix(ξj)< x(ξj),

wherex(ξj) = max{x(ξi)|1≤i≤m−2}. Similarly,x(1)< x(ξj). Thus, there exists σ∈(0,1) anda∈[0, σ) such that x(σ) = maxt∈[0,1]x(t)>0,x0(σ) = 0, x(a) = 0, and x(t)>0 fort ∈(a, σ]. Sinceλ <λ, forˆ t ∈(a, σ], (ϕp(u0λ(t)))0 >(ϕp(ˆu0(t)))0 and integrating this fromt toσ, u0λ(t)<uˆ0(t). Again integrating from atoσ, we havex(σ) =uλ(σ)−u(σ)ˆ < uλ(a)−u(a) =ˆ x(a). This is a contradiction. Thus the claim is proved. Similarly, we haveuλ(1)≤u(1). Next we show thatˆ uλ(t)≤u(t)ˆ fort∈(0,1). If it is not true, it follows fromuλ(0)≤u(0) andˆ uλ(1)≤u(1) thatˆ there exists an interval [t1, t2]⊂[0,1] such thatuλ(t1) = ˆu(t1),uλ(t2) = ˆu(t2) and uλ(t)>u(t) for allˆ t∈(t1, t2). Then

p(u0λ(t)))0>(ϕp(ˆu0(t)))0, t∈(t1, t2) (3.4) and we can choose an interval [b, c] ⊂[t1, t2] such that u0λ(b)>uˆ0(b) and u0λ(c)<

ˆ

u0(c). Using (3.4), we can get the contradiction

0>[ϕp(u0λ(c))−ϕp(u0λ(b))]−[ϕp(ˆu0(c))−ϕp(ˆu0(b))]

= Z c

b

p(u0λ(t))]0−[ϕp(ˆu0(t))]0 dt >0.

Therefore, by the definition ofγ,uλturns out a positive solution of problem (1.1)- (1.2). Furthermore, by Lemma 3.3 and the complete continuity ofH, we can show that problem (1.1)-(1.2), withλ instead ofλ, has a positive solutionu, and thus

the proof is complete.

Now we considerf(t, u) =h(t)g(t, u) and letube a positive solution of problem (1.1)-(1.2), withλ instead ofλ.

Lemma 3.5. Assume that(F1)and(F2)hold. Assume in addition that gsatisfies the conditions (A1) and (A2). Then, for all λ∈ (0, λ), there exists δλ >0 such that αλ(t) =u(t) +δλ satisfies

p0λ(t)))0+λh(t)g(t, αλ(t))<0, t∈(0,1). (3.5) Proof. Letλbe fixed in (0, λ). Put

= 1

2[λ/λ−1] inf

t∈(0,1)g(t, u(t))>0.

By (A1), there existsδλ>0 such that ifu, v∈[0,kuk+ 1] and|u−v|< δλ, then

|g(t, u)−g(t, v)|< , t∈(0,1). Putαλ(t) =u(t) +δλ. Then (ϕp0λ(t)))0+λf(t, αλ(t)) = (ϕp(u0(t)))0+λf(t, u(t) +δλ)

=h(t)[−λg(t, u(t)) +λg(t, u(t) +δλ)].

From this, ifαλ does not satisfy (3.5), there existst0∈(0,1) such that

−λg(t0, u(t0)) +λg(t0, u(t0) +δλ)≥0, and then

g(t0, u(t0) +δλ)≥λ

λg(t0, u(t0)).

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By the choice ofδλ,

≥ λ

λ −1

g(t0, u(t0)),

which contradicts the choice of. This completes the proof.

Proof of Theorem 1.1. Suppose on the contrary thatλ< λ. Letλbe fixed with λ≤λ < λ. Then by showing that (1.1)-(1.2) has at least two positive solutions forλ∈[λ, λ), we get a contradiction to the definition ofλ, which completes the proof. By Lemma 3.5, there exists δλ > 0 such that αλ(t) = u(t) +δλ satisfies (3.5). Consider the modified problem

p(u0(t)))0+λh(t)g(t, γ1(t, u(t))) = 0, u(0) =

m−2

X

i=1

aiu(ξi), u(1) =

m−2

X

i=1

biu(ξi), (3.6) whereγ1: (0,1)×R→[0,∞) is defined as

γ1(t, u) =





αλ(t), ifu > αλ(t), u, if 0≤u≤αλ(t), 0, ifu <0.

Letube a positive solution of (3.6). Set

Ω ={u∈C[0,1]| −1< u(t)< αλ(t), t∈[0,1]}.

Then Ω is bounded and open inC[0,1]. We claim that ifuis a positive solution of (3.6), thenu∈Ω∩ K. Indeed, by the similar argument as in the proof of Theorem 3.4, 0≤u(t)≤αλ(t),t∈[0,1] and

u(0) =

m−2

X

i=1

aiu(ξi)≤

m−2

X

i=1

aiαλi)

=

m−2

X

i=1

ai(ui) +δλ)<

m−2

X

i=1

aiui) +δλ

=u(0) +δλλ(0).

Similarly, αλ(1)> u(1). If the claim is not true, then there exists [t0, t1]⊂(0,1) with t0 ≤ t1 such that 0 < u(t) = αλ(t), t ∈ [t0, t1] and 0 < u(t) < αλ(t), t∈(t0−δ1, t11)\[t0, t1] for some δ1>0. Sinceαλ satisfies (3.5),

max

t∈[t0−δ1,t11]{(ϕp0λ(t)))0+λh(t)g(t, αλ(t))}=−1<0.

By condition (A1), there existsδ2>0 such that if|u−v|< δ2andu, v∈[0,kαλk], then

|g(t, u)−g(t, v)|< 2,

where 2 = 1[2λmaxt∈[t0−δ1,t11]h(t)]−1 > 0, and then there exists an interval [a, b]⊂(t0−δ1, t11) such that

(u−αλ)0(a)>0, (u−αλ)0(b)<0 and

−δ2< γ(t, u(t))−αλ(t) =u(t)−αλ(t)≤0, t∈[a, b].

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Consequently,

ϕp(u0(a))−ϕp0λ(a))>0, ϕp(u0(b))−ϕp0λ(b))<0, g(t, γ(t, u(t)))< g(t, αλ(t)) +2, t∈[a, b].

Then, by the choice of2,

0> ϕp(u0(b))−ϕp0λ(b))−ϕp(u0(a)) +ϕpλ0(a)),

= [ϕp(u0(b))−ϕp(u0(a))]−[ϕp0λ(b))−ϕp0λ(a))]

= Z b

a

{(ϕp(u0(t)))0−(ϕp0λ(t)))0}dt

= Z b

a

{−λh(t)g(t, γ(t, u(t)))−(ϕp0λ(t)))0}dt

>

Z b a

{−λh(t)[g(t, αλ(t)) +2]−(ϕp0λ(t)))0}dt

>

Z b a

(−λh(t)2−[(ϕp0λ(t)))0+λh(t)g(t, αλ(t))])dt

≥ Z b

a

(−λ2h(t) +1)dt≥0.

This is a contradiction. Thus the claim is proved. Define

M u(t) =









A−1Pm−2 i=1 aiRξi

0 ϕ−1p RAu

s λf(τ, γ1(τ, u(τ)))dτ ds +Rt

0ϕ−1p RAu

s λf(τ, γ1(τ, u(τ)))dτ

ds, 0≤t≤Au,

B−1Pm−2 i=1 biR1

ξiϕ−1p Rs

Auλf(τ, γ1(τ, u(τ)))dτ ds +R1

Auϕ−1p Rs

Auλf(τ, γ1(τ, u(τ)))dτ

ds, Au≤t≤1,

whereAu is defined as A−1

m−2

X

i=1

ai

Z ξi 0

ϕ−1p hZ Au s

λf(τ, γ1(τ, u(τ)))dτi ds

+ Z Au

0

ϕ−1p hZ Au s

λf(τ, γ1(τ, u(τ)))dτi ds

=B−1

m−2

X

i=1

bi Z 1

ξi

ϕ−1p hZ s Au

λf(τ, γ1(τ, u(τ)))dτi ds

+ Z 1

Au

ϕ−1p hZ s Au

λf(τ, γ1(τ, u(τ)))dτi ds.

ThenM :K → Kis completely continuous, and uis a positive solution of (3.6) if and only if u =M u onK. By simple calculation, there existsR1 > 0 such that kM uk< R1for allu∈ Kand Ω⊂BR1. Applying [10, Lemma 2.3.1] withO=BR1,

i(M, BR1∩ K,K) = 1.

By the above claim and excision property,

i(M,Ω∩ K,K) =i(M, BR1∩ K,K) = 1.

Since problem (1.1)-(1.2) is equivalent to problem (3.6) on Ω∩ K, we conclude (1.1)-(1.2) has a positive solution in Ω∩ K. AssumeH(λ,·) has no fixed point in

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∂Ω∩ K, since otherwise the proof is done. Then,i(H(λ,·),Ω∩ K,K) is well-defined, and

i(H(λ,·),Ω∩ K,K) =i(M,Ω∩ K,K) = 1 (3.7) sinceM u=H(λ, u) foru∈Ω∩ K. By Lemma 3.2, (1.1)-(1.2) withλN0 instead of λhas no solution inK forλN0> λ. Thus, for any open subsetOin X,

i(H(λN0,·),O ∩ K,K) = 0.

Bya prioriestimate (Lemma 3.3) withI= [λ, λN0], there existsR2(> R1) such that all possible positive solutionsuof (1.1)-(1.2) withµ instead ofλfor µ∈[λ, λN0], satisfykuk< R2.

Defineh: [0,1]×(BR2∩ K)→ K as

h(τ, u) =H(τ λN0+ (1−τ)λ, u).

Thenhis completely continuous on [0,1]× K, and it satisfies thath(τ, u)6=ufor all (τ, u)∈[0,1]×(∂BR2∩ K). By the property of homotopy invariance,

i(H(λ,·), BR2∩ K,K) =i(H(λN0,·), BR2∩ K,K) = 0.

By (3.7) and the additivity property,

i(H(λ,·),(BR2\Ω)∩ K,K) =−1.

Thus problem (1.1)-(1.2) has another positive solution in (BR2\Ω)∩ K. This com-

pletes the proof.

Finally, we give the examples for the nonlinear term to illustrate our results.

Example 3.6. (1) Put f1(t, u) = [t(1−t)]−p+1/(u+1)exp(u). Then, it is easily verified thatf1 satisfies the assumptions of Theorem 3.4.

(2) Putf2(t, u) = (1−t)−α1g(t, u), whereg(t, u) =c1t−β1+c2(uq+ 1). Thenf2 satisfies the assumptions of Theorem 1.1 ifα1, β1< p,c1≥0,c2>0, andq > p−1.

Acknowledgments. Chan-Gyun Kim was supported by National Research Foun- dation of Korea Grant funded by the Korean Government (Ministry of Education, Science and Technology, NRF-2011-357-C00006) Eun Kyoung Lee was supported by a 2-Year Research Grant of Pusan National University.

This work was done when the first author visited College of William and Mary.

He would like to thank Department of Mathematics, College of William and Mary for warm hospitality, and thank Professor Junping Shi for constant encouragement and helpful discussions.

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Chan-Gyun Kim

Department of Mathematics, College of William and Mary, Williamsburg, Virginia 23187-8795, USA

E-mail address:cgkim75@gmail.com

Eun Kyoung Lee (corresponding author)

Department of Mathematics and Education, Pusan National University, Busan 609-735, Korea

E-mail address:eunkyoung165@gmail.com

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