# Positive Solutions for (k, n − k) Conjugate Multipoint Boundary Value Problems in Banach Spaces

Authors: Yulin Zhao
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

By means of the fixed point index theory of strict-set contraction operator, we study the existence of positive solutions for the multipoint singular boundary value problem(-1)n-kun(t)=f(t,ut), 0<t<1, n≥2, 1≤k≤n-1, u(0)=∑i=1m-2‍aiu(ξi), u(i)(0)=u(j)(1)=θ, 1≤i≤k−1, 0≤j≤n−k−1	in a real Banach space E, where θ is the zero element of E,0<ξ1<ξ2<⋯<ξm-2<1,ai∈[0,+∞),i=1,2,…,m-2. As an application, we give two examples to demonstrate our results.


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## Body

## 1. Introduction

The theory of ordinary differential equations in Banach spaces has become a new important branch (see, e.g., [1–13] and the references cited therein). In 1988, Guo and Lakshmikantham [4] discussed multiple solutions for two-point boundary value problems of second-order ordinary differential equations in Banach spaces. In [7], Guo obtained the existence of positive solutions for a boundary value problem of nth-order nonlinear impulsive integrodifferential equations in a Banach space by means of fixed point index theory and fixed point theory of completely continuous operators, respectively. Liu et al. in [6] obtained the existence of unbounded nonnegative solutions of a boundary value problem for nth-order impulsive integrodifferential equations on an infinite interval in Banach spaces by means of the Mddotoch fixed point theory in a Banach space. Zhang et al. in [9] dealt with the existence, nonexistence, and multiplicity of positive solutions for a class of nonlinear three-point boundary value problems of nth-order differential equations in Banach spaces. Zhao and Chen in [8, 12] investigated the existence of at least triple positive solutions for nonlinear boundary value problem by upper and low solution methods.In this paper, the author considers the existence of positive solutions of the following higher-order(k,n-k) conjugate multipoint boundary value problems (BVPs):
(1.1)(-1)n-ku(n)(t)=f(t,u(t)),t∈(0,1),n≥2,1≤k≤n-1,u(0)=∑i=1m-2‍aiu(ξi),u(i)(0)=u(j)(1)=θ,1≤i≤k-1,0≤j≤n-k-1
in a real Banach space E, where θ is the zero element of E,0<ξ1<ξ2<⋯<ξm-2<1,ai∈[0,+∞),i=1,2,…,m-2.  f:(0,1)×E→E is continuous and allowed to be singular at t=0 and t=1.In scalar space, because of the widely applied background in mechanics and engineering, the nonlinear higher-order boundary value problems have received much attention (see Chyan and Henderson [Appl. Math. Letters 15 (2002) 767–774]). In [14], Eloe and Ahmad had solved successfully the existence of positive solution to the following nth-order boundary value problems:
(1.2)u(n)(t)+a(t)f(u)=0,t∈(0,1),u(i)(0)=0,i=0,1,…,n-2,u(1)=au(η).
Recently, the existence of solutions and positive solutions of nonlinear (k,n-k) focal boundary value problem
(1.3)(-1)n-ku(n)(t)=f(t,u(t),u′(t),…,u(n-1)(t)),t∈(0,1),u(i)(0)=0,i=0,1,…,k;u(j)(1)=0,j=k+1,k+2,…,n-1,
and its special cases has been studied by many authors (see, e.g., [15–23]).By using the Krasnoselskii fixed point theorem, Eloe and Henderson in [15], Agarwal and O'Regan in [16], and Kong and Wang in [20] have established the existence of solutions for following the (k,n-k) conjugate boundary value problem:
(1.4)(-1)n-ku(n)(t)=f(t,u(t)),t∈(0,1),u(i)(0)=u(i)(1)=0,0≤i≤k-1,0≤j≤n-k-1.Very recently, by using the fixed point theory in a cone for strict-set contraction operators, Jiang and Zhang in [11] have discussed the existence of positive solutions for the above boundary value problem (1.4) in a Banach space, where the nonlinear term f(t,u(t)):I×E→E is continuous but not allowed to have singularity at t=0,1, where I=[0,1].The organization of this paper is as follows. We shall introduce some lemmas and notations in the rest of this section. The preliminary lemmas are in Section2. The main results are given in Section 3. Finally, two examples are presented to demonstrate our main results in Section 4.Let the real Banach spaceE with norm ∥·∥ be partially ordered by a cone P of E, that is, u≤v if and only if v-u∈P, and P* denotes the dual cone of P, that is, P*={φ∣φ∈E*,φ(u)≥0,u∈P}. A cone is called a solid cone if the set of interior points is not empty.The closed balls in spacesE and C[I,E] are denoted by Pr={u∈E:∥u∥≤r}(r>0) and Br={u∈C[I,E]:∥u∥C≤r}, respectively.The basic space used in this paper isC[I,E]. For any u∈C[I,E], evidently, (C[I,E],∥·∥c) is a Banach space with norm ∥u∥C=maxt∈I∥u(t)∥, and Q={u∈C[I,E]:u(t)≥θ for t∈I} is a cone of the Banach space C[I,E]. A function u∈Cn[I,E] is called a positive solution of the boundary value problem (1.1) if it satisfies (1.1) and u∈Q,u(t)≢θ.Letu(t):I→E be continuous, and the abstract generalized integral ∫0a‍u(t)dt(a∈I) can be similarly defined as in the scalar spaces and ∫0a‍u(t)dt∈E. If limε→0+∫0ε‍u(t)dt exist, then we say that the abstract integral is convergent, otherwise the abstract integral is divergent.At the end of this section, we state some definitions and lemmas which will be used in Sections2 and 3 (for details, see [1–3]).Definition 1.1 (Kuratovski noncompactness measure).
LetE be a real Banach space, and S is a bounded set in E. We denote α(S)=inf{δ>0:S=⋃i=1mSi, all the diameters of Si≤δ}.In the following,α(·) denotes the Kuratowski measure of noncompactness in E and C[I,E].Definition 1.2 (strict-set contraction operator).
LetE1,E2 be real Banach spaces, and S⊂E1.   T:S→E2 is a continuous and bounded operator. If there exists a constant k, such that α(T(S))≤kα(S), then T is called a k-set contraction operator. When k<1,T is called a strict-set contraction operator.Lemma 1.3.
IfD⊂C[I,E] is bounded and equicontinuous, then α(D(t)) is continuous on I and α(D)=α(D(I))=supt∈Iα(D(t)), where D(I)={u(t):u∈D,t∈I},D(t)={u(t):u∈D}.Lemma 1.4.
LetP be a cone in a real Banach space E and let  Ω be a nonempty bounded open convex subset of P. Suppose that T:Ω-→P is a strict-set contraction operator and T(Ω-)⊂Ω, where Ω- denotes the closure of in P. Then the fixed-point index i(T,Ω,P)=1.Lemma 1.5.
LetP be a cone in a real Banach space E and let  Ω be a bounded open subset of P, and suppose that T:P∩Ω-→P is a strict-set contraction operator. (i)
Ifθ∈Ω, and Tu≠λu,for allu∈∂Ω∩P,λ≥1. Then i(T,Ω∩P,P)=1.(ii)
If there existsu0∈P∖{θ}, such that u-Tu≠λu,for allu∈∂Ω∩P,λ≥0. Then i(T,Ω∩P,P)=0.


## 2.  The Preliminary Lemmas

To prove the main results, we need the following lemmas.Lemma 2.1 (see [20]).
Letk(t,s) be the Green function for the (k,n-k) conjugate BVP (1.4). Then
(2.1)k(t,s)=1(k-1)!(n-k-1)!{∫0t(1-s)‍vk-1(v+s-t)n-k-1dv,0≤t≤s≤1,∫0s(1-t)‍vn-k-1(v+t-s)k-1dv,0≤s≤t≤1.
Obviously, k(t,s) is continuous on I×I and has the following properties. (G1) There exist nonnegative functions p(t),m(t),q(t)∈C[0,1] such that
(2.2)p(t)q(s)≤k(t,s)≤m(t)q(s)≤q(s),∀t,s∈I,where
(2.3)p(t)=tk(1-t)n-kn-1,m(t)=tk-1(1-t)n-k-1min⁡{k,n-k},q(s)=sn-k(1-s)K(k-1)!(n-k-1)!.(G2) For any τ∈(0,1/2),k(t,s) satisfies
(2.4)k(t,s)≥ρ(τ)k(v,s),t∈Iτ=[τ,1-τ],s,v∈I,where
(2.5)ρ(τ):=min⁡t∈Iτp(t)max⁡s∈[0,1]m(s).Setting(2.6)Φ(t)=(n-1)!(k-1)!(n-k-1)!∫t1‍sk-1(1-s)n-k-1ds.
It is obvious that 0≤Φ(t)≤1,t∈[0,1], and by the properties of the Euler integral, we have
(2.7)Φ(0)=1,Φ(1)=0,‖Φ‖=1.In order to abbreviate our discussion, we give the following assumptions.(C0)
∑i=1m-2‍aiΦ(ξi)<1.(C1)
f∈C[(0,1)×P,P] and ∥f(t,u)∥≤g(t)∥h(u)∥,t∈(0,1),u∈P, where h:P→P is continuous and bounded and g:(0,1)→(0,+∞) is continuous and satisfies ∫01‍g(s)ds<+∞.(C2)
For anyr>0 and [a,b]⊂(0,1),  f(t,u) is uniformly continuous on [a,b]×Pr.(C3)
There exists constantL≥0 such that for any t∈(0,1) and the bounded set D⊂P(2.8)α(f(t,D))≤Lα(D),
where
(2.9)2L{1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai}⋅max⁡s∈Iq(s)<1.Lemma 2.2.
Let∑i=1m-2‍aiΦ(ξi)≠1 and (C1) be satisfied, then the problem
(2.10)(-1)n-ku(n)(t)=f(t,u(t)),t∈I=[0,1],n≥2,1≤k≤n-1,(2.11)u(0)=∑i=1m-2aiu(ξi),(2.12)u(i)(0)=u(j)(1)=θ,1≤i≤k-1,0≤j≤n-k-1
has a unique solution
(2.13)u(t)=∫01‍K(t,s)f(s,u(s))ds,t∈I,
where
(2.14)K(t,s)=k(t,s)+(1-∑i=1m-2aiΦ(ξi))-1Φ(t)∑i=1m-2aik(ξi,s),0≤t,s≤1.Proof.
According to the definitions of generalized integral in abstract space, the proof of this lemma is similar to the proof in scalar spaces, so we omit it.Foru∈C[I,E], we define an operator T by
(2.15)(Tu)(t)=∫01‍K(t,s)f(s,u(s))ds,t∈I.Lemma 2.3.
Suppose that(C0)–(C3)  hold. Then T:C[I,E]∩Br→C[I,E] is a strict-set contraction operator.Proof.
Foru∈u∈C[I,P], it follows from (C1) that T is well defined and bounded operator. If B∈C[I,P] is a bounded subset of C[I,P], then TB is bounded.
Next we prove thatT is continuous on C[I,E]∩Br. Let {uj},{u}⊂C[I,E]∩Br, and ∥uj-u∥C→0(j→∞). Hence {uj} is a bounded subset of C[I,E]∩Br. Thus, there exists r>0 such that r=supj∥uj∥C<∞ and ∥u∥C≤r.
According to continuity off, for all ε>0, there exists J>0 such that(2.16)‖f(t,xj(t))-f(t,x(t))‖<ε[1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai]∫01‍q(s)ds,
for j≥J,for allt∈I.
Then,(2.17)‖(Tuj)(t)-(Tu)(t)‖≤∫01‍K(t,s)‖f(s,xj(s))-f(s,(s))‖ds≤[1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai]∫01‍q(s)‖f(s,xj(s))===============================-f(s,x(s))‖ds<ε.
Therefore, for all ε>0, for any t∈I and j≥J, we get
(2.18)‖(Tuj)(t)-(Tu)(t)‖⟶0.
This implies T is continuous on C[I,E]∩Br. By the properties of continuity of G(t,s), it is easy to see that T is equicontinuous on I.
For anyS⊂C[I,E]∩Br, it is easy to get that functions T(S)={Tu∣u∈S} are uniformly bounded. By Lemma 1.3, we get
(2.19)α(TS)=sup⁡t∈Iα((TS)(t)),
where (TS)(t)={(Tu)(t):u∈S,t∈I is fixed}.
Write(2.20)D={∫01‍K(t,s)f(s,u(s))ds:u∈S},(2.21)Dε={∫ε1-ε‍K(t,s)f(s,u(s))ds:u∈S},(0<ε<12).
By (C1), for any u∈S, we have
(2.22)Hε=‖∫ε1-ε‍K(t,s)f(s,u(s))ds-∫01‍K(t,s)f(s,u(s))ds‖≤c0∫0ε‍K(t,s)g(s)ds+c0∫1-ε1‍K(t,s)g(s)ds,
where c0:=maxu∈Br∥h(u)∥.
It follows fromHε and (C1) that the Hausdorff metric dH(Dε,D)→0,(ε→0+). Thus
(2.23)lim⁡ε→0+α(Dε)=α(D).
We next shall estimateα(Dε). For any u∈C[I,E], by ∫ε1-ε‍u(t)dt∈(1-2ε)co-{u(t):t∈I}, then(2.24)α(Dε)=α({∫ε1-ε‍K(t,s)f(s,u(s))ds:u∈S})≤(1-2ε)α(co-({K(t,s)f(s,u(s)):s∈[ε,1-ε],u∈S}))≤α({k(t,s)f(s,u(s)):s∈[ε,1-ε],u∈S})+(1-∑i=1m-2aiΦ(ξi))-1α({∑i=1m-2aik(ξi,s)f(s,u(s)):s∈[ε,1-ε],u∈S})≤(1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2)⋅max⁡s∈Iq(s)⋅α(f(Iε×S(Iε))),
where Iε=[ε,1-ε],S(Iε)={u(t):t∈Iε}.
On the other hand, using a similar method as in the proof of Lemma 2 in [11] we can get that
(2.25)α(S(I))≤2α(S).
Therefore, it follows from (2.19), (2.25) that
(2.26)α(f(Iε×S(Iε)))=sup⁡t∈Iεα(f(t,S(Iε)))≤L⋅α(S(Iε))≤L⋅α(S(I))≤2L⋅α(S).
Thus, we have
(2.27)α(Dε)≤2L(1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai)⋅max⁡s∈Iq(s)⋅α(S).
Combining with (2.24), we get
(2.28)α(D)≤2L(1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai)⋅max⁡s∈Iq(s)⋅α(S).
Therefore, we have
(2.29)α(TS)≤2L(1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai)⋅max⁡s∈Iq(s)⋅α(S),∀S∈P∩Br.
Notice that by (2.9) we claim that T:C[I,E]∩Br→C[I,E] is a strict-set contraction. The proof is complete.Further, we construct a coneΩ by(2.30)Ω={u∈Q:u(t)≥ρ*(τ)u(s),∀t∈Iτ,∀s∈I},
where Iτ=[τ,1-τ], and
(2.31)ρ*(τ)=min⁡{ρ(τ),min⁡t∈IτΦ(t)},
where ρ(τ) is defined in (G2). It is easy to see that Ω is a closed convex cone of C[I,E] and Ω⊂Q.Lemma 2.4.
Suppose that (C0)–(C3)  hold. Then, T(Ω)⊂Ω.Proof.
From(G2), (2.6), (2.15), and (2.30), for any u∈Ω,t∈Tτ,t′∈I, we obtain
(2.32)(Tu)(t)=∫01‍K(t,s)f(s,u(s))ds≥ρ(τ)∫01‍k(t′,s)f(s,u(s))ds+(1-∑i=1m-2aiΦ(ξi))-1Φ(t)∫01‍∑i=1m-2aik(ξi,s)f(s,u(s))ds≥ρ*(τ)(∫01‍k(t′,s)f(s,u(s)))ds+(1-∑i=1m-2aiΦ(ξi))-1∫01‍∑i=1m-2aik(ξi,s)f(s,u(s))ds≥ρ*(τ)[∫01‍K(t′,s)f(s,u(s))ds]=ρ*(τ)(Tu)(t′),t′∈I.
Therefore, T(u)∈Ω, that is, T(Ω)⊂Ω.


## 3. The Main Results

Let(3.1)M*:=max⁡s∈Iq(s)⋅(1+(1-∑i=1m-2aiΦ(ξi))-1∑i=1m-2ai)∫01‍g(s)ds.Theorem 3.1.
Let (C0)–(C3)  hold. In addition, assume that the following conditions are satisfied: (C4) di:=M*ci<1,(i=1,2), where
(3.2)c1:=lim⁡-‖u‖→∞‖h(u)‖‖u‖,c2:=lim⁡-‖u‖→∞‖h(u)‖‖u‖,(C5) P is a solid cone, and there exist u0∈Int(P) and τ∈(0,1/2) such that f(t,u)≥l(t)u0,l(t)∈[Iτ,R+], and
(3.3)l0=min⁡t∈Iτp(t)∫τ1-τ‍q(s)l(s)ds≥1,for all t∈Iτ=[τ,1-τ],u≥u0.
Then problem (1.1) has at least two positive solutions u1,u2, and u1(t)≥l0u0, for t∈Iτ,u≥u0.Proof.
We first show that there existsβ>0, such that
(3.4)‖v‖≥β,for anyv≥u0.
If (3.4) is not true, then there exist the sequences {vn}n=1∞ satisfying vn≥u0, and ∥vn∥<1/n,(n=1,2,…). Thus u0≤θ, which is contradiction with u0∈int(P).
Let(3.5)ci′=1+ciM*2M*,(i=1,2).
Then
(3.6)ci′>ci,di′=ci′M*<1.
From (C4) and (3.6), there exist two positive constants r1,r2 with
(3.7)0<r1<β,r2>max⁡{β,2‖u0‖}
such that
(3.8)‖h(u)‖≤ci′‖u‖,u∈Br1∩P,‖h(u)‖≤ci′‖u‖,u∈P∖Br2.
Therefore, for any u∈P, we have
(3.9)‖h(u)‖≤ci′‖u‖+M,
where M=sup{∥h(u)∥:u∈Br2}.
Choose(3.10)r3=r2+(1-d2′)-1M*M.
In the following, let
(3.11)Ω1={u∈Q:∣‖u‖<r1},(3.12)Ω2={∈Q:∣∥u∥<r3},(3.13)Ω3={u∈Q:∣∥u∥<r3,ift∈Iτ,u(t)≥u0}.
It is to see from (3.7) that Ω3 is nonempty for 2u0∈Ω3, which implies Ω3≠∅. Obviously, Ωi⊂Q  (i=1,2,3) are nonempty, convex, open sets and Ω-1=Q∩Br1,Ω-2=Q∩Br2,Ω-3={u∈Ω-2:u(t)≥u0,t∈Iτ}. So
(3.14)Ω1⊂Ω2,Ω3⊂Ω2,Ω1∩Ω3=∅.
For anyu∈Ω-2, form (3.8) and (3.9), we have
(3.15)‖(Tu)(t)‖≤(∫01‍K(t,s)g(s)ds)(c2′‖u‖+M)≤∫01‍g(s)ds⋅(c2′‖u‖+M)⋅(1+(1-∑i=1m-2‍aiΦ(ξi))-1∑i=1m-2‍ai)max⁡s∈Iq(s)=d2′‖u‖+M*M<d2′r3+M*M<r3,
which implies
(3.16)T(Ω-2)⊂Ω2.
From (2.15) and (3.7), we get
(3.17)T(Ω-1)⊂Ω1.
For anyu∈Ω-3, it follows from (3.16) that ∥Tu∥<r3. According to (C5) and (G1), we can obtain
(3.18)(Tu)(t)≥∫τ1-τ‍K(t,s)f(s,u(s))ds≥∫τ1-τ‍K(t,s)l(s)u0ds≥min⁡t∈Iτp(t)∫τ1-τ‍q(s)l(s)ds⋅u0=l0u0≥u0,∀t∈Iτ,
which implies
(3.19)T(Ω-3)⊂Ω3.
Combining (3.16)–(3.19) with Lemma 1.4, we have i(T,Ωi,P)=1,i=1,2,3. Furthermore, using the fixed point index theory, we obtain successively
(3.20)i(T,Ω2∖(Ω-1∪Ω-3),P)=i(T,Ω2,P)-i(T,Ω1,P)-i(T,Ω3,P)=-1.
ThenT has at least two fixed points u1 and u2 which satisfy u1∈Ω3 and u2∈Ω2∖(Ω-1∪Ω-3). Then Theorem 3.1 is proved.Theorem 3.2.
Assume that (C0)–(C3), d2<1 hold. Suppose further that (C5′)  P is a cone of the real Banach space E, and there exist u0∈P∖{θ} and τ∈(0,1/2) such that f(t,u)≥l(t)u0,l(t)∈[Iτ,R+], for for all t∈Iτ,u≥u0.
Then problem (1.1) has at least one positive solutions u with u(t)≥l0u0,t∈Iτ,u≥u0, where l0 is defined in (3.3).Proof .
As in the proof of Theorem3.1, we need only to show that T has one positive fixed point u with u(t)≥l0u0,t∈Iτ,l0≥1.
Chooser0 satisfying r0>max{β,2∥u0∥} and let
(3.21)Ω0={u∈C[I,E]:‖u‖<r0,u(t)≥u0,∀t∈Iτ}.
Obviously, Ω0 is a bounded closed convex set in C[I,E].  Ω0≠∅, for u*(t)≡u0∈Ω0. Let u∈Ω0. As the proof of (3.19), we have T(Ω0)⊂Ω0, where T is given by (2.15). Thus, it follows from Lemmas 2.3 and 1.4 that T has a fixed point u∈Ω0. The proof is complete.Theorem 3.3.
Assume that (C0)–(C3) and the following conditions hold.(C6) There exists R0>0 such that supPR0∥h(u)∥<R0(M*)-1, where PR0={u∈P:∥u∥<R0}.(C7) There exists φ∈P* such that φ(u)>0 for any u>θ, and
(3.22)lim⁡‖u‖→+∞min⁡t∈Iτφ(f(t,u))φ(u)>M0,where τ∈(0,1/2), and
(3.23)M0:=(min⁡t∈Iτp(t)⋅ρ*(τ)∫τ1-τ‍q(s)ds)-1.(C8) There exists φ∈P* such that φ(u)>0 for any u>θ, and
(3.24)lim⁡‖u‖→0min⁡t∈Iτφ(f(t,u))φ(u)>M0,
where τ∈(0,1/2), and M0 is defined in (3.23). Then problem (1.1) has at least two positive solutions.Proof.
If(C6) holds, then there exists ε1 satisfying 0<ε1<R0(M*)-1 such that ∥h(u)∥≤(R0(M*)-1-ε1). For any u∈Ω∩∂PR0,t∈Iτ, it follows from (C1) and (2.15) that
(3.25)‖Tu‖≤∫01‍q(s)g(s)‖h(u)‖ds+(1-∑i=1m-2‍aiΦ(ξi))-1∑i=1m-2‍ai∫01‍q(s)g(s)‖h(u)‖ds≤max⁡s∈Iτq(s)(1+(1-∑i=1m-2‍aiΦ(ξi))-1∑i=1m-2‍ai)∫01‍g(s)ds⋅(R0(M*)-1-ε1)=(R0(M*)-1-ε1)M*<R0,
which implies Tu≠λu,for allu∈Ω∩∂PR0, and λ≥1. From Lemma 1.5 (i), we have
(3.26)i(T,Ω∩∂PR0,Ω)=1.
According to(C7), there exist ε2>0 and R2>R0>0, for any t∈Iτ, such that
(3.27)φ(f(t,u))≥(M0+ε2)φ(u),‖u‖≥R2.
Let U2={u∈C[I,E]:∥u∥<R2}. We need only to show that u-Tu≠λe, for any u∈Ω∩∂U2, and λ≥0,e∈P with ∥e∥=1. If it is false, then there exists u*∈Ω∩∂U2 and λ0≥0, such that
(3.28)u*-Tu*=λ0e,
which implies
(3.29)u*(t)≥Tu*(t)=∫01‍K(t,s)f(s,u*(s))ds≥p(t)∫τ1-τ‍q(s)f(s,u*(s))ds.
Hence, for any φ∈P*, we have
(3.30)φ(u*(t))≥φ(Tu*(t))=∫01‍K(t,s)φ(f(s,u*(s)))ds≥∫τ1-τ‍p(t)q(s)(M0+ε2)φ(u*(s))ds≥min⁡t∈Iτp(t)∫τ1-τ‍q(s)(M0+ε2)φ(u*(s))ds≥(M0+ε2)min⁡t∈Iτp(t)⋅ρ*(τ)∫τ1-τ‍q(s)ds⋅φ(u*(t)).
It is easy to see thatφ(u*(t))>0. In fact, if φ(u*(t))=0, since u*∈Ω∩∂U2, then we have φ(u*(t))≥ρ*(τ)φ(u*(s))≥0 and consequently ∥u*∥=0, which contradicts with ∥u*∥=R2. By (3.29) and (3.30), we get 1<(M0+ε2)mint∈Iτp(t)ρ*(τ)∫τ1-τ‍q(s)ds≤1, which is a contradiction. It follows from Lemma 1.5 (ii) that
(3.31)i(T,Ω∩∂U2,Ω)=0.
If(C8) holds, then there exist ε3>0 and R1>0 with R1<R0, such that
(3.32)φ(f(t,u))≥(M0+ε3)φ(u),‖u‖≥R1.
Let U1={u∈C[I,E]:∥u∥<R1}. As in the proof of (3.31), for any u∈Ω∩∂U1,t∈Iτ, we get
(3.33)i(T,Ω∩∂U1,Ω)=0.
Notice that U-1⊂PR0,P-R0⊂U2. Thus, it follows from (3.26), (3.31), and (3.33) that
(3.34)(i(T,Ω∩(PR0∖U-1)),Ω)=i(T,Ω∩PR0,Ω)-i(T,Ω∩U1,Ω)=1,(i(T,Ω∩(U2∖P-R0)),Ω)=i(T,Ω∩U2,Ω)-i(T,Ω∩PR0,Ω)=-1.
ThenT has at least two fixed points u1 and u2 which satisfy u1∈Ω∩(PR0∖U-1) and u2∈Ω∩(U2∖P-R0). Then Theorem 3.3 is proved.
Similarly to the proofs of Theorem3.3, we can easily get the following corollaries.Corollary 3.4.
Assume that (C0)–(C3), d1<1 and (C7) hold. Then problem (1.1) has at least one positive solution.Corollary 3.5.
Assume that (C0)–(C3), d2<1 and (C8) hold. Then problem (1.1) has at least one positive solution.


## 4. Examples

Now we present two examples to illustrate our main results.Example 4.1.
Consider the boundary value problems inE=RN (N-dimensional Euclidean space and ∥u∥=max1≤p≤N|up|<+∞)
(4.1)(-1)6up(10)(t)=1+tt(1-t)(sin2up+1+ln(1+up2)+t(1-t)H(up)),0<t<1,up(0)=19up(14)+12up(12),p=1,2,3,…u(i)(0)=u(j)(1)=0,1≤i≤3,0≤j≤5,
where n=10,k=4, and
(4.2)H(up)={23×105up+N‖u0‖×1012up2,up∈[0,∥u0∥N],(23×105+1×1012)up,up∈[∥u0∥N,1+r3(∥u0∥)+∥u0∥N],23×105up+1+r3(∥u0∥)+∥u0∥N×1012up,up∈[1+r3(∥u0∥)+∥u0∥N,+∞),
where r3(·) is as defined in (3.10). Then the problem (4.1) can be regarded as a BVP of the form (1.1) in E. In this situation, u=(u1,…,uP,…,uN)∈RN,f=(f1,…,fP,…,fN),θ=(0,0,…,0)∈RN and a1=1/9,ξ1=1/4,a2=ξ2=1/2,g=(g1,…,gP,…,gN),h=(h1,…,hP,…,hN), in which
(4.3)fp(t,u1,u2,…,uN)=1+tt(1-t)(sin2up+1+ln(1+up2)+t(1-t)H(up)),p(t)=t4(1-t)69,m(t)=t3(1-t)54,q(s)=s6(1-s)43!⋅5!.
Obviously, f:(0,1)×E→E is continuous. Taking τ=1/4. By direct account, we have
(4.4)∑i=1m-2‍aiΦ(ξi)=19Φ(14)+12Φ(12)≈0.219<1.
Now we show that (C1)–(C5) hold.First, it is easy to see that∥f(t,u)∥≤g(t)∥h(u)∥,t∈(0,1),u∈E, with
(4.5)gp(t)=2t(1-t),hp(u)=sin⁡2up+1+ln⁡(1+up2)+12H(up),
and ∫01‍gp(s)ds=2π<+∞.For sufficiently large numberR>0, take P={u=(u1,…,up,…,uN)∈RN:up≥0,∥u∥≤R,p=1,2,…,N}.Thenh:P→P is continuous and bounded. Therefore, (C1) is satisfied. In addition, (C2) and (C3) are automatically satisfied in N-dimensional Euclidean space RN.As far as(C4) is concerned, we get
(4.6)M*=2πmax⁡s∈[0,1]s6(1-s)43!5!⋅(1+(1-0.219)-1×(19+12))≈1.86×10-5,c1=lim⁡-‖u‖→∞‖h(u)‖‖u‖=13×105,c2=lim⁡-‖u‖→∞‖h(u)‖‖u‖=13×105,di=M*ci≈0.62<1.
Then (C4) is satisfied.On the other hand,P is a solid cone, for all u0∈Int(P). For all t∈Iτ=[1/4,3/4], we have
(4.7)fp(t,u)≥1×1012(1+t)up≥l(t)u0(p),foru≥u0,
where u0=(u0(1),…,u0(p),…,u0(N)),l(t)=1×1012(1+t), and
(4.8)l0=min⁡t∈[1/4,3/4]p(t)∫1/43/4‍q(s)l(s)ds≈7.055>1.
So (C5) is also satisfied. For example, if u0={1,1,…,1}, we let β=1, then r3(∥u0∥)+(∥u0∥)/N+1≈9.745×107(2N+1). By Theorem 3.1, the problem (4.1) has at least two positive solutions in E.Example 4.2.
Consider the following boundary value problems inl∞:
(4.9)(-1)3up(9)(t)=πt(ln(1+up2p)+t(1-t)upp|sinup+1|+H(up)),0<t<1,up(0)=14up(34),p=1,2,3,…u(i)(0)=u(j)(1)=0,1≤i≤5,0≤j≤2,
where n=9,k=6, and
(4.10)H(up)={8×108up,up∈[0,116×10-8],100up4,up∈[116×10-8,1×104],8×108up+(10-8×1010)up,up∈[1×104,+∞).
Let P={u=(u1,…,up,…)∈l∞:up≥0,p=1,2,…},∥u∥=supp≥1|up|<+∞. Then l∞ is a Banach space. Thus the problem (4.9) can be regarded as a BVP of the form (1.1) in l∞. In this situation, u=(u1,…,uP,…)∈l∞,f=(f1,…,fP,…),θ=(0,0,…)∈l∞ and a1=1/4,ξ=3/4,g=(g1,…,gP,…,),h=(h1,…,hP,…,), in which
(4.11)fp(t,u1,u2,…,)=πt(ln(1+up2p)+t(1-t)upp|sinup+1|+H(up)),p(t)=t6(1-t)38,m(t)=t5(1-t)23,q(s)=s3(1-s)65!⋅2!.
Obviously, f:(0,1)×E→E is continuous taking τ=1/8. By direct account, we have
(4.12)∑i=1m-2‍aiΦ(ξi)=14Φ(34)≈0.0804<1,M*=1.099×10-4,ρ*(τ)=min⁡{ρ(τ),min⁡[1/8,7/8]Φ(t)}≈min⁡{0.6471,0.0674}=0.0674.
Let
(4.13)gp(t)=πt,hp(u)=ln⁡((1+up2p)+up2p|sin⁡up+1|+H(up)),
and ∫01‍gp(s)ds=2π<+∞,
(4.14)M0=(min⁡t∈[1/8,7/8]p(t)⋅ρ*(τ)∫1/87/8‍q(s)ds)-1≈7.0811×108.
Similar to the proofs of Example 4.1, we can show that (C1)–(C3) hold. Taking R0=108/3, then supu∈PR0{∥h(u)∥}<M*R0, which implies (C6), holds.Foru∈P, choose φ∈P* with φ(u)=∑p=1+∞‍up/p2. Obviously, φ(u)>0 for u>θ, and
(4.15)lim⁡‖u‖→+∞min⁡t∈[1/8,7/8]φ(f(t,u))φ(u)=lim⁡‖u‖→+∞min⁡t∈[1/8,7/8]πt⋅∑p=1+∞1/p2{ln⁡(1+up2/p)+up/2p|sin⁡up+1|+H(up)}∑p=1+∞up/p2>8×108>M0,
which implies (C7) is satisfied. Similarly, we can show that (C8) holds. By Theorem 3.3, the problem (4.9) has at least two positive solutions in E.



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*Source: 727468-2012-08-08.xml*