# Symmetric Positive Solutions for Nonlinear Singular Fourth-Order Eigenvalue Problems with Nonlocal Boundary Condition

Authors: Fuyi Xu; Jian Liu
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We investigate nonlinear singular fourth-order eigenvalue problems with nonlocal boundary conditionu(4)(t)-λh(t)f(t,u,u′′)=0, 0<t<1, u(0)=u(1)=∫01a(s)u(s)ds, u′′(0)=u′′(1)=∫01b(s)u′′(s)ds, where a,b∈L1[0,1], λ>0, h may be singular at t=0 and/or 1. Moreover f(t,x,y) may also have singularity at x=0 and/or y=0. By using fixed point theory in cones, an explicit interval for λ is derived such that for any λ in this interval, the existence of at least one symmetric positive solution to the boundary value problem is guaranteed. Our results extend and improve many known results including singular and nonsingular cases. The associated Green's function for the above problem is also given.


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## Body

## 1. Introduction

Boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics and so on, and the existence of positive solutions for such problems has become an important area of investigation in recent years. To identify a few, we refer the reader to [1–7] and references therein.At the same time, a class of boundary value problems with nonlocal boundary conditions appeared in heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics. Such problems include two-point, three-point, multipoint boundary value problems as special cases and have attracted the attention of Gallardo [1], Karakostas and Tsamatos [2], and Lomtatidze and Malaguti [3] (and see the references therein). For more information about the general theory of integral equations and their relation to boundary value problems we refer the reader to the book of Corduneanu [8] and Agarwal and O'Regan [9].Motivated by the works mentioned above, in this paper, we study the existence of symmetric positive solutions of the following fourth-order nonlocal boundary value problem (BVP):(1.1)u(4)(t)-λh(t)f(t,u,u′′)=0,0<t<1,u(0)=u(1)=∫01a(s)u(s)ds,u′′(0)=u′′(1)=∫01b(s)u′′(s)ds,
where a,b∈L1[0,1], λ>0, h may be singular at t=0 and/or 1. Moreover f(t,x,y) may also have singularity at x=0 and/or y=0.The main features of this paper are as follows. Firstly, comparing with [4–7], we discuss the boundary value problem with nonlocal boundary conditions, that is, BVP (1.1) including fourth-order two-point, three-point, multipoint boundary value problems as special cases. Secondly, comparing with [4–7], we discuss the boundary value problem when nonlinearity f contains second-derivatives u′′. Thirdly, here we not only allow h have singularity at t=0 and/or 1 but also allow f(t,x,y) have singularity at x=0 and/or y=0. Finally, in [4–7], authors only studied the existence of positive solutions. However, they did not further provide characters of positive solutions, such as symmetry. It is now natural to consider the existence of symmetric positive solutions. To our knowledge, no paper has considered the existence of symmetric positive solutions and nonlinearity f with singularity at x=0 and (or) y=0 for fourth-order equation with nonlocal boundary condition. Hence we improve and generalize the results of [4–7] to some degree, and so it is interesting and important to study the existence of symmetric positive solutions for problem (1.1). The arguments are based upon a specially constructed cone and the fixed point theory for cones.LetK be a cone in a Banach space E and let Kr={x∈K:∥x∥<r},∂Kr={x∈K:∥x∥=r}, and K¯r,R={x∈K:r≤∥x∥≤R}, where 0<r<R<+∞.Our main tool of this paper is the following fixed point theorem.Lemma 1.1 (see [10]).
LetK be a positive cone in real Banach space  E,  0<r<R<+∞, and let  T:K¯r,R→K be a completely continuous operator and such that (i)
∥Tx∥≤∥x∥ for x∈∂KR;(ii)
there existse∈∂K1 such that  x≠Tx+me for any  x∈∂Kr and  m>0.Then T has a fixed point in K¯r,R.Remark 1.2.
If (i) and (ii) are satisfied forx∈∂Kr and x∈∂KR, respectively, then Lemma 1.1 is still true.The following concept will also be utilized.Definition 1.3.
Ifu:[0,1]→R is a continuous function and u(t)=u(1-t) for t∈[0,1], then one says that u is symmetric on [0,1].


## 2. Preliminaries and Lemmas

In this section, we present some lemmas that are important to prove our main results.Lemma 2.1.
Suppose thatd:=∫01m(s)ds≠1,  m∈L1[0,1], y∈C[0,1], then BVP
(2.1)u′′(t)+y(t)=0,0<t<1,(2.2)u(0)=u(1)=∫01m(s)u(s)ds
has a unique solution
(2.3)u(t)=∫01H(t,s)y(s)ds,
where
(2.4)H(t,s)=G(t,s)+11-d∫01G(s,x)m(x)dx,G(t,s)={t(1-s),0≤t≤s≤1,s(1-t),0≤s≤t≤1.Proof.
Integrating both sides of (2.1) on [0,t], we have
(2.5)u′(t)=-∫0ty(s)ds+B.
Again integrating (2.5) from 0 to t, we get
(2.6)u(t)=-∫0t(t-s)y(s)ds+Bt+A.
In particular,
(2.7)u(1)=-∫01(1-s)y(s)ds+B+A,u(0)=A.
By (2.2) we get
(2.8)B=∫01(1-s)y(s)ds.
By G(s,x)=G(x,s) and (2.6), we can get
(2.9)A=u(0)=∫01m(x)u(x)dx=∫01m(x)(-∫0x(x-s)y(s)ds+Bx+A)dx=∫01m(x)(-∫0x(x-s)y(s)ds+x∫01(1-s)y(s)ds)dx+A∫01m(x)dx=∫01m(x)(∫0xs(1-x)y(s)ds+∫x1x(1-s)y(s)ds)dx+Ad=∫01m(x)(∫01G(s,x)y(s)ds)dx+Ad=∫01(∫01G(s,x)m(x)dx)y(s)ds+Ad.
So,
(2.10)A=11-d∫01(∫01G(s,x)m(x)dx)y(s)ds.
By (2.6), (2.8), and (2.10), we have
(2.11)u(t)=-∫0t(t-s)y(s)ds+Bt+A=-∫0t(t-s)y(s)ds+t∫01(1-s)y(s)ds+11-d∫01(∫01G(s,x)m(x)dx)y(s)ds=∫0ts(1-t)y(s)ds+∫t1t(1-s)y(s)ds+11-d∫01(∫01G(s,x)m(x)dx)y(s)ds=∫01G(t,s)y(s)ds+11-d∫01(∫01G(s,x)m(x)dx)y(s)ds=∫01H(t,s)y(s)ds.
This completes the proof of Lemma 2.1.It is easy to verify the following properties ofH(t,s) and G(t,s).Lemma 2.2.
Ifm(t)>0, and  d:=∫01m(s)ds∈(0,1), 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),  G(t,t)≤G(t,s)≤G(s,s),  t,s∈[0,1];(3)
γ1H(s,s)≤H(t,s)≤H(s,s), where γ1 = η1/(1-d+η1)∈(0,1), η1 = ∫01G(x,x)m(x)dx.So we may denote Green's function of boundary value problem(2.12)-u′′(t)=0,0<t<1,u(0)=u(1)=∫01a(s)u(s)ds,-u′′(t)=0,0<t<1,u(0)=u(1)=∫01b(s)u(s)ds
by H1(t,s) and H2(t,s), respectively. By Lemma 2.1, we know that H1(t,s) and H2(t,s) can be written by(2.13)H1(t,s)=G(t,s)+11-∫01a(s)ds∫01G(s,x)a(x)d,H2(t,s)=G(t,s)+11-∫01b(s)ds∫01G(s,x)b(x)dx.
Obviously, H1(t,s) and H2(t,s) have the same properties with H(t,s) in Lemma 2.2.Remark 2.3.
For notational convenience, we introduce the following constants:(2.14)α=∫01a(s)ds,β=∫01b(s)ds,μ=max0≤t≤1∫01H1(t,s)ds,γ=η1-β+η∈(0,1),η=∫01G(x,x)b(x)dx,L=(1+μ)∫01H2(s,s)h(s)ds,l=mint∈[0,1]∫01H2(t,s)h(s)ds.
Obviously, μ>0, 0<l<L<∞.Now we define an integral operatorS:C[0,1]→C[0,1] by(2.15)(Sv)(t)=∫01H1(t,τ)v(τ)dτ.
Then, we have(2.16)(Sv)′′(t)=-v(t),0<t<1,Sv(0)=Sv(1)=∫01a(s)Sv(s)ds.Lemma 2.4.
The fourth-order nonlocal boundary value problem (1.1) has a positive solution if and only if the following integral-differential boundary value problem
(2.17)v′′(t)+λh(t)f(t,Sv(t),-v(t))=0,0<t<1,v(0)=v(1)=∫01b(s)v(s)ds
has a positive solution.Proof.
In fact, ifu is a positive solution of (BVP) (1.1), let u(t)=Sv(t), then v=-u′′. This implies that u′′=-v is a solution of (2.17). Conversely, if v is a positive solution of (2.17). Let u(t)=Sv(t), by (2.16), u′′(t)=(Sv)′′(t)=-v. Thus u(t)=Sv(t) is a positive solution of (BVP) (1.1). This completes the proof of Lemma 2.4.So we will concentrate our study on (2.17). Let C+[0,1]={x∈C[0,1]:x≥0} and(2.18)K={x∈C+[0,1]:x(t)is  symmetric  and  concave  function  on[0,1],mint∈[0,1]x(t)≥γ∥x∥};∥·∥ is the supremum norm on C+[0,1]. It is easy to see that K is a cone in C[0,1] and K¯r,R⊂K⊂C+[0,1]. Now we define an operator T:K∖{0}→C+[0,1] by(2.19)(Tv)(t)=λ∫01H2(t,s)h(s)f(s,Sv(s),-v(s))ds,t∈[0,1].
Clearly v is a solution of the BVP (2.17) if and only if v is a fixed point of the operator T.In the rest of the paper, we make the following assumptions:(H1)
a,b∈L1[0,1], a(t)≥0, b(t)≥0, a(1-t)=a(t),b(1-t)=b(t), α,β∈(0,1);(H2)
h∈C((0,1),[0,+∞)), h(1-t)=h(t), 0<∫01H2(s,s)h(s)ds<+∞;(H3)
f(t,u,v)∈C((0,1)×(0,+∞)×(-∞,0),[0,+∞)), f((1-t),u,v)=f(t,u,v), and for any 0<r1<R1<+∞, 0<r2<R2<+∞(2.20)limn→+∞supu∈K¯r1,R1,-v∈K¯r2,R2∫D(n)H2(s,s)h(s)f(s,u(s),v(s))ds=0,
where D(n)=[0,1/n]∪[(n-1)/n,1].Remark 2.5.
If (H1) holds, then for all t,s∈[0,1], we have
(2.21)H1(1-t,1-s)=H1(t,s),H2(1-t,1-s)=H2(t,s).Lemma 2.6.
Assume that conditions(H1), (H2), and (H3) hold. Then T:K¯r,R→C+[0,1] is a completely continuous operator.Proof.
Firstly, for anyr>0, we will show
(2.22)supv∈∂Krλ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds<+∞.
At the same time, this implies that T:K∖{0}→C+[0,1] is well defined.
In fact, by(H3), for any r>0, there exists a natural number m such that
(2.23)supv∈∂Krλ∫D(m)H2(s,s)h(s)f(s,Sv(s),-v(s))ds<1.
For any v∈∂Kr, let v(t0)=maxt∈[0,1]|v(t)| = r. It follows from the concavity of v(t) on [0,1] that
(2.24)v(t)≥{rtt0,0≤t≤t0,r1-t0(1-t),t0≤t≤1.
So we obtain
(2.25)v(t)≥{rt,0≤t≤t0,r(1-t),t0≤t≤1.
Consequently, from (2.25) for any t∈[1/m,(m-1)/m], we have r/m≤v(t)≤r and
(2.26)lmrm=rmmint∈[1/m,(m-1)/m]∫01H1(t,s)ds≤Sv(t)≤rmaxt∈[1/m,(m-1)/m]∫01H1(t,s)ds≤μr,
where lm=mint∈[1/m,(m-1)/m]∫01H1(t,s)ds and μ is defined in Remark 2.3. Let
(2.27)M1=max{f(t,x,y):(t,x,y)∈[1m,m-1m]×[lmrm,μr]×[-r,-rm]}.
By (H1)–(H3), we have
(2.28)supv∈∂Krλ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds≤supv∈∂Krλ∫D(m)H2(s,s)h(s)f(s,Sv(s),-v(s))ds+supv∈∂Krλ∫1/m(m-1)/mH2(s,s)h(s)f(s,Sv(s),-v(s))ds≤1+M1λ∫01H2(s,s)h(s)ds<+∞,
that is, (2.22) holds. This also implies that T(B) is uniformly bounded for any bounded set B⊂K¯r,R from (2.28).
Next we prove thatT is equicontinuous on K¯r,R. In fact, by (H3) for any ɛ>0, there exists a natural number k such that
(2.29)supv∈K¯r,Rλ∫D(k)H2(s,s)h(s)f(s,Sv(s),-v(s))ds<ɛ4.
Let
(2.30)M2=max{f(t,x,y):(t,x,y)∈[1k,k-1k]×[lkrk,μr]×[-r,-rk]},
where lk=mint∈[1/k,(k-1)/k]∫01H1(t,s)ds. Since H2(t,s) is uniformly continuous on [0,1]×[0,1], for the above ɛ>0 and fixed s∈[1/k,(k-1)/k], there exists δ>0 such that
(2.31)|H2(t,s)-H2(t′,s)|≤(2λLM2)-1(μ+1)H2(s,s)ɛ
for |t-t′|<δ and t,t′∈[0,1]. Consequently, when |t-t′|<δ and t,t′∈[0,1], we have
(2.32)|Tv(t)-Tv(t′)|≤2supv∈K¯r,Rλ∫D(k)H2(s,s)h(s)f(s,Sv(s),-v(s))ds+λsupv∈K¯r,R∫1/k(k-1)/k|H2(t,s)-H2(t′,s)|h(s)f(s,Sv(s),-v(s))ds<ɛ.
This implies that T(K¯r,R) is equicontinuous. Then by the Arzela-Ascoli theorem T:K¯r,R→C+[0,1] is compact.
Finally, we show thatT:K¯r,R→C+[0,1] is continuous. Assume vn,v0∈K¯r,R and ∥vn-v0∥→0(n→∞). Then r≤∥vn∥≤R and r≤∥v0∥≤R. For any ɛ>0, by (H3), there exists a natural number m>0 such that
(2.33)supv∈K¯r,Rλ∫D(m)H2(s,s)h(s)f(s,Sv(s),-v(s))ds<ɛ4.
On the other hand, by (2.25), for any t∈[1/m,(m-1)/m], we have
(2.34)rm≤vn(t)≤R,lmrm≤Svn(t)≤μR,n=0,1,2,…,
where lm=mint∈[1/m,(m-1)/m]∫01H1(t,s)ds and μ is defined by Remark 2.3.
Sincef(t,x,y) is uniformly continuous in [1/m,(m-1)/m]×[lmr/m,μR]×[-R,-r/m], we have that
(2.35)limn→+∞|f(s,Svn(s),-vn(s))-f(s,Sv0(s),-v0(s))|=0
holds uniformly on s∈[1/m,(m-1)/m]. Then the Lebesgue dominated convergence theorem yields that
(2.36)∫1/m(m-1)/mλH2(s,s)h(s)|f(s,Svn(s),-vn(s))-f(s,Sv0(s),-v0(s))|ds→0,asn→∞.
Thus for above ɛ>0, there exists a natural number; N, for n>N, we have
(2.37)∫1/m(m-1)/mλH2(s,s)h(s)|f(s,Svn(s),-vn(s))-f(s,Sv0(s),-v0(s))|ds<ɛ2.
It follows from (2.33) and (2.37) that when n>N,
(2.38)∥Tvn-Tv0∥≤∫1/m(m-1)/mλH2(s,s)h(s)|f(s,Svn(s),-vn(s))-f(s,Sv0(s),-v0(s))|ds+2supv∈K¯r,R∫D(m)λH2(s,s)h(s)f(s,Sv(s),-v(s))ds<ɛ2+2×ɛ4=ɛ.
This implies that T:K¯r,R→C+[0,1] is continuous. Thus T:K¯r,R→C+[0,1] is completely continuous. This completes the proof of Lemma 2.6.Lemma 2.7.
One hasT(K¯r,R)⊂K.Proof.
Forv∈K¯r,R,t∈[0,1], we have
(2.39)(Tv)(t)=λ∫01H2(t,s)h(s)f(s,Sv(s),-v(s))d≤λ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds.
Thus
(2.40)∥Tv∥≤λ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds.
On the other hand, by Lemma 2.2 we have
(2.41)(Tv)(t)=λ∫01H2(t,s)h(s)f(s,Sv(s),-v(s))ds≥γλ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds
which implies mint∈[0,1](Tv)(t)≥γ∥Tv∥.
In addition, forv∈K,t∈[0,1], by Remark 2.5, (H2), and (H3) we have
(2.42)(Sv)(1-t)=∫01H1(1-t,τ)v(τ)dτ=∫10H1(1-t,1-τ)v(1-τ)d(1-τ)=∫01H1(t,τ)v(τ)dτ=(Sv)(t),(2.43)(Tv)(1-t)=λ∫01H2(1-t,s)h(s)f(s,Sv(s),-v(s))ds,=λ∫10H2(1-t,1-s)h(1-s)f(1-s,Sv(1-s),-v(1-s))d(1-s)=λ∫01H2(t,s)h(s)f(s,Sv(s),-v(s))ds=(Tv)(t),
that is, (Tv)(1-t)=(Tv)(t), t∈[0,1]. Therefore, (Tv)(t) is symmetric on [0,1]. Obviously, (Tv)(t) is concave on [0,1]. Consequently, T(K¯r,R)⊂K. This completes the proof of Lemma 2.7.


## 3. The Main Results

Our main results of this paper are as follows. For notational convenience, we let(3.1)f0=lim sup|x|+|y|→0x>0,y<0maxt∈[0,1]f(t,x,y)|x|+|y|,f∞=lim inf|x|+|y|→+∞x>0,y<0mint∈[0,1]f(t,x,y)|x|+|y|,f∞=lim sup|x|+|y|→+∞x>0,y<0maxt∈[0,1]f(t,x,y)|x|+|y|,f0=lim inf|x|+|y|→0x>0,y<0mint∈[0,1]f(t,x,y)|x|+|y|.Theorem 3.1.
Suppose that conditions(H1), (H2), and (H3) are satisfied. Further assume that the following condition (H4) holds:(H4)
Lf0<lf∞.Then the BVP (1.1) has at least one symmetric positive solution for any
(3.2)λ∈(1lf∞,1Lf0),
where L and l are defined by Remark 2.3.Proof.
Letλ satisfy (3.2) and at ɛ>0 be chosen such that
(3.3)f∞-ɛ>0,1(f∞-ɛ)l≤λ≤1(f0+ɛ)L.
Next, by (H4) there exists r0>0 such that
(3.4)f(t,x,y)≤(f0+ɛ)(|x|+|y|),∀t∈[0,1],0<|x|+|y|<r0,x>0,y<0.
Take r=r0/(μ+1). Notice that
(3.5)0<|Sv|+|v|≤(μ+1)∥v∥=(μ+1)r=r0,0≤t≤1.
It follows from (3.4) and (3.5) that, for any v∈∂Kr,
(3.6)∥Tv∥=maxt∈[0,1]λ∫01H2(t,s)h(s)f(s,Sv(s),-v(s))ds≤λ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds≤λ(f0+ɛ)∫01H2(s,s)h(s)(|Sv(s)|+|v(s)|)ds=λ(f0+ɛ)(μ+1)r∫01H2(s,s)h(s)ds=λL(f0+ɛ)r≤r=∥v∥.
Thus, ∥Tv∥≤∥v∥, for all v∈∂Kr.
On the other hand, for the aboveɛ, by (H4), there exists R0>0 such that
(3.7)f(t,x,y)>(f∞-ɛ)(|x|+|y|),t∈[0,1],|x|+|y|≥R0,x>0,y<0.
Let R=max{2r,γ-1R0} and φ(t)≡1, t∈[0,1]. Then R>r and φ(t)∈∂K1.
In the following we showv≠Tv+mφ(m>0). Otherwise, there exists v0∈∂KR and m0>0 such that v0=Tv0+m0φ.   Let ξ=min{v0(t):t∈[0,1]} and notice that for any s∈[0,1],
(3.8)|Sv0(s)|+|v0(s)|≥mins∈[0,1][|Sv0(s)|+|v0(s)|]≥mins∈[0,1]|v0(s)|≥γ∥v0∥≥R0.
Consequently for any t∈[0,1], we have
(3.9)v0(t)=λ∫01H2(t,s)h(s)f(s,Sv0(s),-v0(s))ds+m0φ(t)=λ∫01H2(t,s)h(s)f(s,Sv0(s),-v0(s))ds+m0≥λ(f∞-ɛ)∫01H2(t,s)h(s)(|Sv0(s)|+|v0(s)|)ds+m0≥λ(f∞-ɛ)∫01H2(t,s)h(s)v0(s)ds+m0≥λ(f∞-ɛ)ξmint∈[0,1]∫01H2(t,s)h(s)ds+m0≥ξ+m0>ξ.
This implies that ξ>ξ, which is a contradiction. It follows from Lemma 1.1 that T has a fixed point v* with r<|v*|<R. Thus v* is a symmetric positive solution of the BVP (2.17). Consequently, by Lemma 2.4, one can obtain that BVP (1.1) has a symmetric positive solution. This completes the proof of Theorem 3.1.Remark 3.2.
Since0<l<L<+∞, we easily obtain 0<1/lf∞<1, 1/Lf0>1. Thus 1∈(1/lf∞,1/Lf0); so when λ=1, Theorem 3.1 always holds.Remark 3.3.
From Theorem3.1, we can see that f(t,x,y) need not be superlinear or sublinear. In fact, Theorem 3.1 still holds, if one of the following conditions is satisfied.(i)
Iff∞=∞, f0>0, then for each λ∈(0,1/Lf0).(ii)
Iff∞=∞, f0=0, then for each λ∈(0,+∞).(iii)
Iff∞>l-1>0, f0=0, then for each λ∈(1/lf∞,+∞).Theorem 3.4.
Suppose that conditions(H1), (H2), and (H3) are satisfied. Further assume that the following condition (H5) holds.(H5)
Lf∞<lf0.Then the BVP (1.1) has at least one symmetric positive solution for any
(3.10)λ∈(1lf0,1Lf∞),
where L and l are defined by Remark 2.3.Proof.
Letλ satisfy (3.10) and let ɛ1>0 be chosen such that L-1-ɛ1>0 and λf∞<L-1-ɛ1. By (H5), there exists μR0′ such that
(3.11)f(t,x,y)≤1λ(L-1-ɛ1)(|x|+|y|),|x|+|y|≥μR0′,x>0,y<0,t∈[0,1].
Let
(3.12)M0=supv∈∂KR0′λ∫01H2(s,s)h(s)f(s,Sv(s),-v(s))ds.
Then M0<+∞ by (2.22). Take R1>max{R0′,M0/Lɛ1}, then M0<LR1ɛ1.
Notice thatv∈∂KR0′ implies that
(3.13)v(t)≤∥v∥=R0′,Sv(t)≤maxt∈[0,1]∫01H1(t,s)ds∥v∥=μR0′.
So for any v∈∂KR1, let
(3.14)D(Sv,-v)={t∈[0,1]:(Sv,-v)∈[μR0′,+∞)×(-∞,-R0′]},
and then for any t∈D(Sv,-v), clearly μR0′≤|Sv|+|v|≤(μ+1)∥v∥=(μ+1)R1.
In addition, for anyv∈∂KR1, let v1(t)=min{v(t),R0′}, then v1∈∂KR0′. Thus, for any v∈∂KR1, we have
(3.15)∥Tv∥=maxt∈[0,1]λ∫01H2(t,s)h(s)f(s,Sv(s),-v(s))ds≤maxt∈[0,1]λ∫D(Sv,-v)H2(t,s)h(s)f(s,Sv(s),-v(s))ds+λ∫[0,1]/D(Sv,-v)H2(s,s)h(s)f(s,Sv(s),-v(s))ds≤1λ(L-1-ɛ1)λ∫01H2(s,s)h(s)(|Sv(s)|+|v(s)|)ds+λ∫01H2(s,s)h(s)f(s,Sv1(s),-v1(s))ds≤(L-1-ɛ1)(μ+1)R1∫01H2(s,s)h(s)ds+M0≤(L-1-ɛ1)LR1+M0<R1=∥v∥.
Therefore ∥Tv∥≤∥v∥ for any v∈∂KR1.
Next, letλ satisfy (3.10). Choose ɛ2>0 such that l-1+ɛ2<λf0. Then from (H5), there exists 0<δ<(μ+1)R1 such that
(3.16)f(t,x,y)≥1λ(l-1+ɛ2)(|x|+|y|),0<|x|+|y|≤δ,x>0,y<0,t∈[0,1].
Let r1=δ/(μ+1) and ϕ(t)≡1, t∈[0,1], then r1<R1 and ϕ∈∂K1.
Now we provev≠Tv+mϕ(m>0). Otherwise, there exists v0∈∂Kr1 and m0>0 such that v0=Tv0+m0ϕ. Let ζ=min{v0(t):t∈[0,1]} and apply that |v0(s)|+|Sv0(s)|<(μ+1)r1=δ; then for any t∈[0,1], we have
(3.17)v0(t)=λ∫01H2(t,s)h(s)f(s,Sv0(s),-v0(s))ds+m0ϕ(t)=λ∫01H2(t,s)h(s)f(s,Sv0(s),-v0(s))ds+m0≥1λ(l-1+ɛ2)λ∫01H2(t,s)h(s)(|Sv0(s)|+|v0(s)|)ds+m0≥(l-1+ɛ2)∫01H2(t,s)h(s)v0(s)ds+m0≥(l-1+ɛ2)ζmint∈[0,1]∫01H2(t,s)h(s)ds+m0≥(1+lɛ2)ζ+m0>ζ.
This implies that ζ>ζ, which is a contradiction. It follows from Lemma 1.1 that T has a fixed point v** with r1<|v**|<R1. Thus v** is a symmetric positive solution of the BVP (2.17). By Lemma 2.4, BVP (1.1) has a symmetric positive solution.Remark 3.5.
From Theorem3.4 we can see that the conclusions still hold, if one of the following conditions is satisfied.(i)
Iff∞<L-1, f0=∞, then for each λ∈(0,1/Lf∞).(ii)
Iff∞=0, f0=+∞, then for each λ∈(0,+∞).(iii)
Iff∞=0, f0>l-1>0, then for each λ∈(1/lf0,+∞).Remark 3.6.
Because of singularity ofh, f, it seems to be difficult to prove our results by using the norm-type expansion and compression theorem. In addition, we need to point out that we not only obtain the existence of symmetric positive solutions of BVP (1.1) but also get the explicit interval about λ, which is different from the previous papers (see [4, 5]).Remark 3.7.
If conditions(H1), (H2) and the following condition (H3′) are satisfied,(H3′)
f(t,u)∈C((0,1)×(0,+∞),[0,+∞)), f((1-t),u)=f(t,u), and for any 0<r<R<+∞,(3.18)limn→+∞supu∈K¯r,R∫D(n)H(s,s)h(s)f(s,u(s))ds=0,
where D(n)=[0,1/n]∪[(n-1)/n,1]. Then our results can be applied to the second-order singular boundary value problem with nonlocal boundary condition:
(3.19)u′′(t)+λh(t)f(t,u(t))=0,0<t<1,u(0)=u(1)=∫01a(s)u(s)ds,
under the corresponding to conditions of Theorems 3.1 and 3.4.



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*Source: 187827-2010-03-22.xml*