Volume 2007, Article ID 79090,14pages doi:10.1155/2007/79090
Research Article
Existence of Symmetric Positive Solutions for
an
m
-Point Boundary Value Problem
Yongping Sun and Xiaoping Zhang
Received 23 June 2006; Revised 17 December 2006; Accepted 11 March 2007
Recommended by Colin Rogers
We study the second-order m-point boundary value problem u(t) +a(t)f(t,u(t))=
0, 0< t <1,u(0)=u(1)=m−2
i=1 αiu(ηi), where 0< η1< η2<··· < ηm−2≤1/2,αi>0
fori=1, 2,...,m−2 withim=−12αi<1,m≥3.a: (0, 1)→[0,∞) is continuous, symmetric on the interval (0, 1), and maybe singular att=0 andt=1, f : [0, 1]×[0,∞)→[0,∞) is continuous, and f(·,x) is symmetric on the interval [0, 1] for allx∈[0,∞) and satisfies some appropriate growth conditions. By using Krasnoselskii’s fixed point theorem in a cone, we get some existence results of symmetric positive solutions.
Copyright © 2007 Y. Sun and X. Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Them-point boundary value problems for ordinary differential equations arise in a va-riety of different areas of applied mathematics and physics. In the past few years, the existence of positive solutions for nonlinear second-order multipoint boundary value problems has been studied by many authors by using the Leray-Schauder continuation theorem, nonlinear alternative of Leray Schauder, coincidence degree theory, Krasnosel-skii’s fixed point theorem, Leggett-Wiliams fixed point theorem, or lower- and upper-solutions method (see [1–21] and references therein). On the other hand, there is much current attention focusing on questions of symmetric positive solutions for second-order two-point boundary value problems, for example, Avery and Henderson [22], Henderson and Thompson [23] imposed conditions on f to yield at least three symmetric positive solutions to the problem
y+f(y)=0, 0≤t≤1,
where f :R→[0, +∞) is continuous. Both of the papers [22,23] make an application of an extension of the Leggett-Williams fixed point theorem. Li and Zhang [24] considered the existence of multiple symmetric nonnegative solutions for the second-order bound-ary value problem
−x=f(x,x), 0≤t≤1,
u(0)=u(1)=0, (1.2)
where f :R×R→[0, +∞) is continuous. The main tool is the Leggett-Williams fixed point theorem. Yao [25] gave the existence ofnsymmetric positive solutions and estab-lished a corresponding iterative scheme for the two-point boundary value problem
w(t) +h(t)fw(t)=0, 0< t <1,
αw(0)−βw(0)=0, αw(1) +βw(1)=0, (1.3)
whereα >0, β≥0, and the coefficienth(t) may be singular at both end pointst=0 andt=1. The main tool is the monotone iterative technique. Very recently, by using the Leggett-Wiliams fixed point theorem and a coincidence degree theorem of Mawhin, Kos-matov [26,27] studied the existence of three positive solutions for a multipoint boundary value problem
−u(t)=a(t)ft,u(t),u(t), t∈(0, 1),
u(0)= n
i=1 μiu
ξi
, u(1−t)=u(t), t∈[0, 1], (1.4)
where 0< ξ1< ξ2<···< ξn≤1/2,μi>0 fori=1, 2,...,n, withni=1μi<1,n≥2. In this paper, we are concerned with the existence of symmetric positive solutions for the following second-orderm-point boundary value problem (BVP):
u(t) +a(t)ft,u(t)=0, 0< t <1, (1.5)
u(0)=u(1)=m− 2
i=1 αiu
ηi
, (1.6)
where 0< η1< η2<···< ηm−2≤1/2,αi>0 fori=1, 2,...,m−2, with m−2
i=1 αi<1,m≥ 3.a: (0, 1)→[0,∞) is continuous, symmetric on the interval (0, 1), and may be singular at both end pointst=0 andt=1,f : [0, 1]×[0,∞)→[0,∞) is continuous andf(1−t,x)= f(t,x) for all (t,x)∈[0, 1]×[0,∞). We use Krasnoselskii’s fixed point theorem in cones and combine it with an available transformation to establish some simple criteria for the existence of at least one, at least two, or many symmetric positive solutions to BVP (1.5 )-(1.6).
The organization of this paper is as follows. InSection 2, we present some neces-sary definitions and preliminary results that will be used to prove our main results. In
(1.5)-(1.6). Then we will prove the existence of two or many positive solutions in
Section 4, wherenis an arbitrary natural number.
2. Preliminaries and lemmas
In this section, we introduce some necessary definitions and preliminary results that will be used to prove our main results. A functionwis said to be concave on [0, 1] if
wrt1+ (1−r)t2≥rwt1+ (1−r)wt2, r,t1,t2∈[0, 1]. (2.1)
A functionwis said to be symmetric on [0, 1] if
w(t)=w(1−t), t∈[0, 1]. (2.2)
A functionu∗is called a symmetric positive solution of BVP (1.5)-(1.6) ifu∗(t)>0,
u∗(1−t)=u∗(t),t∈[0, 1], and (1.5) and (1.6) are satisfied.
We will consider the Banach spaceC[0, 1] equipped with norm u =max0≤t≤1|u(t)|.
Set
C+[0, 1]=w∈C[0, 1] :w(t)≥0,t∈[0, 1]. (2.3)
We consider first them-point BVP:
u+h(t)=0, 0< t <1, (2.4)
u(0)=u(1)=m
−2
i=1
αiuηi, (2.5)
where 0< η1< η2<···< ηm−2<1.
Lemma2.1. Letmi=−12αi=1,h∈C[0, 1]. Then them-point BVP (2.4)-(2.5) has a unique
solution
u(t)= 1
0H(t,s)h(s)ds, (2.6)
where
H(t,s)=G(t,s) +E(s), (2.7)
G(x,y)= ⎧ ⎨ ⎩
x(1−y), 0≤x≤y≤1,
y(1−x), 0≤y≤x≤1, E(s)= 1 1−m−2
i=1 αi m−2
i=1 αiG
ηi,s
. (2.8)
Proof. From (2.4), we have
u(t)= − t
In particular,
u(0)=A,
u(1)= − 1
0(1−s)h(s)ds+B+A,
uηi
= − ηi 0
ηi−s
h(s)ds+Bηi+A.
(2.10)
Combining with (2.5), we conclude that
B= 1
0(1−s)h(s)ds,
A= 1
1−m−2
i=1 αi m−2
i=1 αi
1
0G
ηi,sh(s)ds.
(2.11)
Therefore, them-point BVP (2.4)-(2.5) has a unique solution
u(t)= − t
0(t−s)h(s)ds+t 1
0(1−s)h(s)ds+
1 1−m−2
i=1 αi m−2
i=1 αi
1
0G
ηi,s
h(s)ds
= 1
0G(t,s)h(s)ds+ 1
0E(s)h(s)ds= 1
0H(t,s)h(s)ds.
(2.12)
This completes the proof.
Lemma 2.2. Suppose0< η1< η2<···< ηm−2≤1/2,αi>0 for i=1, 2,...,m−2, with m−2
i=1 αi<1. Then
(1)H(t,s)≥0,t,s∈[0, 1],H(t,s)>0,t,s∈(0, 1);
(2)G(1−t, 1−s)=G(t,s),t,s∈[0, 1];
(3)γH(s,s)≤H(t,s)≤H(s,s),t,s∈[0, 1], where
γ=
m−2
i=1 αiηi 1−m−2
i=1 αi+im=−12αiηi
. (2.13)
Proof. The conclusions (1), (2), and the second inequality of (3) are evident. Now we prove that the first inequality of (3) holds. In fact, from 0< η1< η2<···< ηm−2≤1/2,
we know 1−ηi≥ηi, thus fors∈[0, 1], we have
Gηi,s= ⎧ ⎨ ⎩
1−ηi
s, 0≤s≤ηi
ηi(1−s), ηi≤s≤1
≥ηis(1−s)=ηiG(s,s), (2.14)
which means that
αiG
ηi,s
and summing both sides from 1 tom−2, we get
m−2
i=1 αiG
ηi,s
≥
m−2
i=1 αiηi
G(s,s). (2.16)
So
m−2
i=1 αiG
ηi,s
+ m−2
i=1
αiηiE(s)≥ m−2
i=1 αiηi
G(s,s) +E(s). (2.17)
Thus
1−m− 2
i=1 αi+
m−2
i=1 αiηi
E(s)≥
m−2
i=1 αiηi
G(s,s) +E(s)=
m−2
i=1 αiηi
H(s,s).
(2.18)
Subsequently,
E(s)≥
m−2
i=1 αiηi 1−m−2
i=1 αi+im=−12αiηi
H(s,s)=γH(s,s). (2.19)
Therefore,
H(t,s)=G(t,s) +E(s)≥E(s)≥γH(s,s), t,s∈[0, 1]. (2.20)
This completes the proof.
Lemma2.3. Letim=−12αi=1,0< η1< η2<···< ηm−2<1,h(t)be symmetric on[0, 1]. Then the unique solutionu(t)of BVP (2.4)-(2.5) is symmetric on[0, 1].
Proof. For anyt,s∈[0, 1], from (2.7) andLemma 2.2, we have
u(1−t)= 1
0H(1−t,s)h(s)ds= 1
0G(1−t,s)h(s)ds+ 1
0E(s)h(s)ds
= 0
1G(1−t, 1−s)h(1−s)d(1−s) + 1
0E(s)h(s)ds
= 1
0G(t,s)h(s)ds+ 1
0E(s)h(s)ds
= 1
0H(t,s)h(s)ds =u(t).
(2.21)
Therefore,
u(1−t)=u(t), t∈[0, 1], (2.22)
Without loss of generality, all constantsηiin the boundary value condition (1.6) are placed in the interval (0, 1/2] because of the symmetry of the solution.
Lemma2.4. Letαi>0fori=1, 2,...,m−2with m−2
i=1 αi<1,0< η1< η2<···< ηm−2<1, h∈C+[0, 1]. Then the unique solutionu(t)of BVP (2.4)-(2.5) is nonnegative on[0, 1], and ifh(t)≡0, thenu(t)is positive on[0, 1].
Proof. Leth∈C+[0, 1]. From the fact thatu(t)= −h(t)≤0, t∈[0, 1], we know that
u(t) is concave on [0, 1]. From (2.5) and (2.6), we have
u(1)=u(0)= 1
0H(0,s)h(s)ds= 1
0E(s)h(s)ds≥0. (2.23)
It follows thatu(t)≥0,t∈[0, 1], and ifh(t)≡0, thenu(t)>0,t∈[0, 1].
From the proof ofLemma 2.4, we know that ifmi=−12αi>1,h∈C+[0, 1], then the BVP (2.4)-(2.5) has no positive solution. So in order to obtain positive solution of the BVP (2.4)-(2.5), in the rest of the paper we assume thatmi=−12αi∈(0, 1).
Lemma2.5. Letim=−12αi∈(0, 1),0< η1< η2<···< ηm−2≤1/2,h∈C+[0, 1]. Then the unique solutionu(t)of BVP (2.4)-(2.5) satisfies
min
t∈[0,1]u(t)≥γ u , (2.24) whereγis as inLemma 2.2.
Proof. Applying (2.6) andLemma 2.2, we find that fort∈[0, 1],
u(t)= 1
0H(t,s)h(s)ds≤ 1
0H(s,s)h(s)ds. (2.25)
Therefore,
u ≤ 1
0H(s,s)h(s)ds. (2.26)
On the other hand, for anyt∈[0, 1], by (2.7) andLemma 2.2, we have
u(t)=
1
0H(t,s)h(s)ds≥ 1
0γH(s,s)h(s)ds=γ 1
0H(s,s)h(s)ds. (2.27)
From (2.26) and (2.27) we know that (2.24) holds.
We will use the following assumptions.
(A1) 0< η1< η2<···< ηm−2≤1/2,αi>0 fori=1, 2,...,m−2, withim=−12αi<1; (A2)a: (0, 1)→[0,∞) is continuous, symmetric on (0, 1), and
0< 1
0H(s,s)a(s)ds <+∞; (2.28)
Define
K=w∈C+[0, 1] :w(t) is symmetric, concave on [0, 1], min
0≤t≤1w(t)≥γ w
.
(2.29)
It is easy to see thatK is a cone of nonnegative functions inC[0, 1]. Define an integral operatorT:E→Eby
Tu(t)= 1
0H(t,s)a(s)f
s,u(s)ds, t∈[0, 1]. (2.30)
It is easy to see that BVP (1.5)-(1.6) has a solutionu=u(t) if and only ifuis a fixed point of the operatorTdefined by (2.30).
Lemma2.6. Suppose that(A1),(A2), and(A3)hold, thenT is completely continuous and T(K)⊂K.
Proof. (Tu)(t)= −a(t)f(t,u(t))≤0 implies thatTuis concave, thus from Lemmas2.3,
2.4, and2.5, we know thatT(K)⊂K. Now we will prove that the operatorTis completely continuous. Forn≥2, defineanby
an(t)= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
inf0<s≤1/na(s), 0< t≤1n,
a(t), 1
n< t <1−d
1
n,
inf1−1/n≤s<1a(s), 1−1
n≤t <1,
(2.31)
and defineTn:K→Kby
Tnu(t)=
1
0H(t,s)an(s)f
s,u(s)ds. (2.32)
Obviously,Tnis compact onKfor anyn≥2 by an application of Ascoli-Arzela theorem [28]. DenoteBR= {u∈K: u ≤R}. We claim thatTn converges onBR uniformly to
T asn→ ∞. In fact, letMR=max{f(s,x) : (s,x)∈[0, 1]×[0,R]}, thenMR<∞. Since 0<01H(s,s)a(s)ds <+∞, by the absolute continuity of integral, we have
lim
n→∞ e(1/n)H(s,s)a(s)ds=0, (2.33)
wheree(1/n)=[0, 1/n]∪[1−1/n, 1]. So, for anyt∈[0, 1], fixedR >0, andu∈BR,
T
nu(t)−Tu(t)=
1
0
a(s)−an(s)
H(t,s)fs,u(s)ds
≤MR
1
0
a(s)−an(s)H(t,s)ds
≤MR
e(1/n)a(s)H(s,s)ds−→0(n−→ ∞),
where we have used assumptions (A1), (A2), and (A3) and the fact thatH(t,s)≤H(s,s) fort,s∈[0, 1]. Hence the completely continuous operatorTnconverges uniformly toT asn→ ∞on any bounded subset ofK, and thereforeTis completely continuous.
We will use the following notations:
f0=lim inf x→+0 tmin∈[0,1]
f(t,x)
x , f∞=lim infx→+∞ tmin∈[0,1] f(t,x)
x ,
f0=lim sup
x→+0 tmax∈[0,1] f(t,x)
x , f
∞=lim sup x→+∞ tmax∈[0,1]
f(t,x)
x ,
Λ= 1
0H(s,s)a(s)ds
−1 .
(2.35)
Now we formulate a fixed point theorem which will be used in the sequel (cf. [29,30]).
Theorem2.7. LetEbe a Banach space and letK⊂Ebe a cone inE. AssumeΩ1andΩ2 are open subsets ofEwith0∈Ω1andΩ1⊂Ω2, letT:K∩(Ω2\Ω1)→Kbe a completely continuous operator such that
(A) Tu ≤ u , for allu∈K∩∂Ω1and Tu ≥ u , for allu∈K∩∂Ω2; or
(B) Tu ≥ u , for allu∈K∩∂Ω1and Tu ≤ u , for allu∈K∩∂Ω2. ThenThas a fixed point inK∩(Ω2\Ω1).
3. The existence of single positive solution
In this section, we will impose growth conditions on f which allow us to applyTheorem 2.7with regard to obtaining the existence of at least one symmetric positive solution for BVP (1.5)-(1.6). We obtain the following existence results.
Theorem3.1. Assume that (A1), (A2), and (A3) hold. If there exist two constantsR1,R2 with0< R1≤γR2such that
(D1) f(t,x)≤ΛR1, for all (t,x)∈[0, 1]×[γR1,R1], and f(t,x)≥(1/γ)ΛR2, for all
(t,x)∈[0, 1]×[γR2,R2]; or
(D2) f(t,x)≥(1/γ)ΛR1, for all (t,x)∈[0, 1]×[γR1,R1], and f(t,x)≤ΛR2, for all
(t,x)∈[0, 1]×[γR2,R2],
then BVP (1.5)-(1.6) has at least one symmetric positive solutionu∗satisfying
R1≤u∗≤R2. (3.1)
Proof. We only prove the case (D1). Let
Ω1=
u:u∈E, u < R1, Ω2=
u:u∈E, u < R2. (3.2)
t∈[0, 1], we have
Tu(t)= 1
0H(t,s)a(s)f
s,u(s)ds≤ 1
0H(s,s)a(s)f
s,u(s)ds
≤ΛR1 1
0H(s,s)a(s)ds=R1= u .
(3.3)
Therefore,
Tu ≤ u , u∈K∩∂Ω1. (3.4)
On the other hand, foru∈K∩∂Ω2, we haveu(s)∈[γR2,R2],s∈[0, 1], which imply that f(s,u(s))≥(1/γ)ΛR2. Thus fort∈[0, 1], we have
Tu(t)= 1
0H(t,s)a(s)f
s,u(s)ds≥1 γΛR2
1
0H(t,s)a(s)ds
≥1 γΛR2
1
0γH(s,s)a(s)ds=R2= u ,
(3.5)
which implies that
Tu ≥ u , u∈K∩∂Ω2. (3.6)
Therefore, from (3.4), (3.6), andTheorem 2.7, it follows thatT has a fixed point
u∗∈K∩(Ω2\Ω1). So, u∗ is a symmetric positive solution of BVP (1.5)-(1.6) with
R1≤ u∗ ≤R2.
Theorem3.2. Assume that (A1), (A2), and (A3) hold. If one of the following conditions is satisfied:
(D3) f0>(1/γ2)Λand f∞<Λ(particularly, f0= ∞and f∞=0),
(D4) f0<Λandf∞>(1/γ2)Λ(particularly, f0=0and f∞= ∞),
then BVP (1.5)-(1.6) has at least one symmetric positive solution.
Proof. We only prove the case (D3). From f0>(1/γ2)Λ, we know that there existsR1>0
such that f(s,x)≥(1/γ2)Λxfor (s,x)∈[0, 1]×[0,R1]. LetΩ
1= {u:u∈E, u < R1},
then foru∈K∩∂Ω1andt∈[0, 1], we have
Tu(t)=
1
0H(t,s)a(s)f
s,u(s)ds≥ 1 γ2Λ
1
0H(t,s)a(s)u(s)ds
≥ 1 γ2Λ
1
0γG(s,s)a(s)γ u ds= u .
(3.7)
Therefore,
Tu ≥ u , u∈K∩∂Ω1. (3.8)
On the other hand, from f∞<Λwe know that there existsR >0 such that f(s,x)≤Λx
|u < R2}. Then, foru∈K∩∂Ω2, we have u(s)≥γ u =γR2> R, which implies that f(u(s))≤Λu(s) fors∈[0, 1]. Thus,
Tu(t)= 1
0H(t,s)a(s)f
s,u(s)ds≤ 1
0H(t,s)a(s)Λu(s)ds ≤Λ 1
0H(s,s)a(s) u ds= u .
(3.9)
Hence we have
Tu ≤ u , u∈K∩∂Ω2. (3.10)
Therefore, from (3.8), (3.10), and Theorem 2.7, it follows that T has a fixed point
u∗∈K∩(Ω2\Ω1), and thusu∗is a symmetric positive solution of BVP (1.5)-(1.6).
Theorem3.3. Assume that (A1), (A2), and (A3) hold. If there exists two constantsR1,R2 with0< R1≤R2such that
(D5) f(t,·)is nondecreasing on[0,R2]for allt∈[0, 1],
(D6) f(s,γR1)≥(1/γ)ΛR1, and f(t,R2)≤ΛR2for allt∈[0, 1],
then BVP (1.5)-(1.6) has at least one symmetric positive solutionu∗satisfying
R1≤u∗≤R2. (3.11)
Proof. Let
Ω1=
u:u∈E, u < R1, Ω2=
u:u∈E, u < R2. (3.12)
Foru∈K, fromLemma 2.5, we know that min0≤t≤1u(t)≥γ u . Therefore, foru∈K∩ ∂Ω1, we haveu(s)≥γ u =γR1fors∈[0, 1], thus by (D5) and (D6), we have
Tu(t)= 1
0H(t,s)a(s)f
s,u(s)ds≥ 1
0H(t,s)a(s)f
s,γR1ds
≥ 1
0γH(s,s)a(s)
1
γΛR1ds=R1= u .
(3.13)
Therefore,
Tu ≥ u , u∈K∩∂Ω1. (3.14)
On the other hand, foru∈K∩∂Ω2, we haveu(s)≤R2fors∈[0, 1], thus by (D5) and
(D6), we have
Tu(t)= 1
0H(t,s)a(s)f
s,u(s)ds≤ 1
0H(t,s)a(s)f
s,R2ds
≤ 1
0H(s,s)a(s)ΛR2ds=R2= u .
Hence we have
Tu ≤ u , u∈K∩∂Ω2. (3.16)
Therefore, from (3.14), (3.16), andTheorem 2.7, it follows thatThas a fixed pointu∗∈ K∩(Ω2\Ω1) satisfying R1≤ u∗ ≤R2,u∗is a symmetric positive solution of BVP
(1.5)-(1.6).
4. The existence of many positive solutions
Now we discuss the multiplicity of positive solutions for BVP (1.5)-(1.6). We obtain the following existence results.
Theorem4.1. Assume that (A1), (A2) and (A3) hold. In addition, suppose that
(D7) f0>(1/γ2)Λand f∞>(1/γ2)Λ(particularly, f0= f∞= ∞); (D8)there exists a constantρ1such that
f(s,x)≤Λρ1, (s,x)∈[0, 1]×γρ1,ρ1. (4.1)
Then BVP (1.5)-(1.6) has at least two symmetric positive solutionsu1andu2 satisfy-ing0< u1 ≤ρ1≤ u2 .
Proof. At first, in view of f0>(1/γ2)Λ, there existsr∈(0,ρ1) such that
f(s,x)≥ 1
γ2Λx, (s,x)∈[0, 1]×[0,r]. (4.2)
SetΩr= {u:u∈E, u < r}. Then foru∈K∩∂Ωr, we have
Tu(t)= 1
0G(t,s)a(s)f
s,u(s)ds≥ 1
0G(t,s)a(s)
1
γ2Λu(s)ds
≥ 1 γ2Λ
1
0γG(s,s)a(s)γ u ds= u ,
(4.3)
which implies that
Tu ≥ u , u∈K∩∂Ωr. (4.4)
Next, since f∞>(1/γ2)Λ, there existsR∈(ρ1,∞) such that
f(s,x)≥ 1
γ2Λx, (s,x)∈[0, 1]×[0,R]. (4.5)
that f(s,u(s))≥(1/γ2)Λu(s)≥(1/γ)Λ u . Thus,
Tu(t)= 1
0G(t,s)a(s)f
s,u(s)ds≥ 1
0G(t,s)a(s)
1
γΛ u ds
≥1 γΛ
1
0γG(s,s)a(s) u ds= u ,
(4.6)
which implies that
Tu ≥ u , u∈K∩∂ΩR. (4.7)
Finally, setΩρ1= {u:u∈E, u < ρ1}. For anyu∈K∩∂Ωρ1, we haveu(s)∈[γρ1,ρ1], s∈[0, 1]. Thus, from (2.30) and (D8), we obtain
Tu(t)= 1
0G(t,s)a(s)f
s,u(s)ds≤ 1
0G(t,s)a(s)Λu(s)ds ≤Λ 1
0G(s,s)a(s) u ds= u ,
(4.8)
which yields
Tu ≤ u , u∈K∩∂Ωρ1. (4.9)
Hence, sincer < ρ1< R, from (4.4), (4.7), and (4.9), it follows fromTheorem 2.7thatT
has a fixed pointu1∈K∩(Ωρ1\Ωr), and a fixed pointu2∈K∩(ΩR\Ωρ1). Both are
symmetric positive solutions of BVP (1.5)-(1.6).
Remark 4.2. From the proof, we know that if (D8) holds and f0>(1/γ2)Λ(or f∞> (1/γ2)Λ), then BVP (1.5)- (1.6) has a symmetric positive solutionusatisfying 0< u ≤ ρ1(or u ≥ρ1).
In a similar way, we can get the following results.
Theorem4.3. Assume that (A1), (A2), and (A3) hold. If the following conditions are satis-fied.
(D9) f0<Λandf∞<Λ(particularly, f0=f∞=0).
(D10)There exists a constantρ2such that
f(s,x)≥ 1
γ2Λρ2, (s,x)∈[0, 1]×
γρ2,ρ2. (4.10)
Then BVP (1.5)-(1.6) has at least two symmetric positive solutionsu1andu2 satisfy-ing0< u1 < ρ2< u2 .
Remark 4.4. If (D10) holds and f0<Λ(or f∞<Λ), then BVP (1.5)-(1.6) has a symmetric
positive solutionusatisfying 0< u ≤ρ2(or u ≥ρ2).
Theorem4.5. Assume that (A1), (A2), and (A3) hold. If there exist2npositive numbersrk,
(D11) f(s,x)≤Λrk for (s,x)∈[0, 1]×[γrk,rk], and f(s,x)≥(1/γ)ΛRk for (s,x)∈ [0, 1]×[γRk,Rk],k=1, 2,...,n; or
(D12) f(s,x)≥(1/γ)Λrk for (s,x)∈[0, 1]×[γrk,rk], and f(s,x)≤ΛRk for (s,x)∈ [0, 1]×[γRk,Rk],k=1, 2,...,n,
then BVP (1.5)-(1.6) hasnsymmetric positive solutionsuk satisfyingrk≤ uk ≤Rk for
k=1, 2,...,n.
Theorem4.6. Assume that (A1), (A2), and (A3) hold. If there exist2npositive numbers r1< R1< r2< R2<···< rn< Rnsuch that
(D13) f(t,·)is nondecreasing on[0,Rn]for allt∈[0, 1];
(D14) f(s,γrk)≥(1/γ)Λrk, and f(s,Rk)≤ΛRk,k=1, 2,...,nfor alls∈[0, 1],
then BVP (1.5)-(1.6) hasnsymmetric positive solutionsuk satisfyingrk≤ uk ≤Rk,k= 1, 2,...,n.
Acknowledgments
The author thanks the referees for valuable suggestions and comments. This project was supported by the NSF of Zhejiang Province of China (Y605144), the Education Depart-ment of Zhejiang Province of China (20051897), and Zhejiang University of Media and Communications (2006016).
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Yongping Sun: Department of Electron and Information, Zhejiang University of Media and Communications, Hangzhou 310018, Zhejiang, China
Email address:mathysun@yahoo.com.cn
Xiaoping Zhang: Department of Electron and Information, Zhejiang University of Media and Communications, Hangzhou 310018, Zhejiang, China
