# Positive Solutions for Sturm-Liouville Boundary Value Problems in a Banach Space

Authors: Hua Su; Lishan Liu; Yonghong Wu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider the existence of single and multiple positive solutions for a second-order Sturm-Liouville boundary value problem in a Banach space. The sufficient condition for the existence of positive solution is obtained by the fixed point theorem of strict set contraction operators in the frame of the ODE technique. Our results significantly extend and improve many known results including singular and nonsingular cases.


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## Body

## 1. Introduction

Boundary value problems for ordinary differential equations play a very important role in both theoretical study and practical application in many fields. They are used to describe a large number of physical, biological, and chemical phenomena. In this paper, we study the existence of positive solutions for the following second-order nonlinear Sturm-Liouville boundary value problem (BVP) in a Banach SpaceE(1.1)1p(t)(p(t)u′(t))′+f(u(t))=0,0<t<1,αu(0)-βlim⁡t→0+p(t)u′(t)=0,γu(1)+δlimt→1-p(t)u′(t)=0,
where α,β,δ,γ≥0 are constants such that B(t,s)=∫ts‍dτ/p(τ),ω=βγ+αγB(0,1)+αδ>0, and p∈C1((0,1),(0,+∞)). Moreover, p may be singular at t=0 and/or 1.The BVP (1.1) is often referred to as a model for the deformation of an elastic beam under a variety of boundary conditions [1–12]. We notice that previous work is limited to use the completely continuous operators and the function f is required to satisfy some growth condition or assumptions of monotonicity.The aim of this paper is to consider the existence of positive solutions for the more general Sturm-Liouville boundary value problem (1.1) by using the fixed point theorem of strict set contraction operators. Here we allow p to have singularity at t=0,1. The results obtained in this paper improve and generalize many well-known results.The rest of the paper is organized as follows. In Section2, we first present some properties of Green's functions to be used to define a positive operator. Then we approximate the singular second-order boundary value problem by constructing an integral operator. In Section 3, the sufficient condition for the existence of single and multiple positive solutions for the BVP (1.1) is established. In Section 4, we give a example to demonstrate the application of our results.


## 2. Preliminaries and Lemmas

In this paper, we denote by(E,∥·∥1) a real Banach space. A nonempty closed convex subset P in E is said to be a cone if λP∈P for λ≥0 and P⋂{-P}={θ}, where θ denotes the zero element of E. The cone P defines a partial ordering in E by x≤y if and only if y-x∈P. Recall that the cone P is said to be normal if there exists a positive constant λ such that 0≤x≤y implies ∥x∥1≤λ∥y∥1.In this paper, we assumeP⊆E is normal, and without loss of generality, we may assume that the normal constant of P is 1. Let J=[0,1], and
(2.1)C(J,E)={u:J→E∣u(t)continuous},Ci(J,E)={u:J→E∣u(t)isi-ordercontinuouslydifferentiable},i=1,2,….
For u=u(t)∈C(J,E), let ∥u∥=maxt∈J∥u(t)∥1, then C(J,E) is a Banach space with the norm ∥·∥.Definition 2.1.
A functionu(t) is said to be a positive solution of the boundary value problem (1.1) if u∈C([0,1],E)⋂C1((0,1),E) satisfies u(t)>0,t∈(0,1], pu′∈C1((0,1),E) and the BVP (1.1).We notice that ifu(t) is a positive solution of the BVP (1.1) and p∈C1(0,1), then u∈C2(0,1).Now we denote byG(t,s) the Green's functions for the following boundary value problem:
(2.2)1p(t)(p(t)u′(t))′=0,0<t<1,u(0)-lim⁡t→0+βp(t)u′(t)=0,γu(1)+limt→1-δp(t)u′(t)=0.
It is well known that G(t,s) can be written by
(2.3)G(t,s)=1ω{(β+αB(0,s))(δ+γB(t,1)),0≤s≤t≤1,(β+αB(0,t))(δ+γB(s,1)),0≤t≤s≤1,
where B(t,s)=∫ts‍dτ/p(τ),ω=αδ+αγB(0,1)+βγ>0.It is easy to verify the following properties ofG(t,s): (I)
G(t,s)≤G(s,s)≤(1/ω)(β+αB(0,1))(δ+γB(0,1))<+∞;(II)
G(t,s)≥ρG(s,s), for any t∈[a,b]⊂(0,1),s∈[0,1], where
(2.4)ρ=min⁡{δ+γB(b,1)δ+γB(0,1),β+αB(0,a)β+αB(0,1)}.Throughout this paper, we adopt the following assumptions:(H1)
p∈C1((0,1),(0,+∞)) and satisfies
(2.5)0<∫01‍dsp(s)<+∞,0<e=∫01‍G(s,s)p(s)ds<+∞.(H2)
f:P→P is a uniformly continuous function and there exists M>0 such that for any bounded set B⊂C(J,E), we have
(2.6)α(f(B(t)))≤Mα(B(t)),t∈J,2Meρ<1,where α(·) denotes the Kuratowski measure of noncompactness in E.The following Lemmas play an important role in this paper (see [13]).Lemma 2.2.
LetB⊂C(J,E) be bounded and equicontinuous on J, then αc(B)=supt∈Jα(B(t)).Lemma 2.3.
LetB⊂C(J,E) be bounded and equicontinuous on J, then α(B(t)) is continuous on J and
(2.7)α({∫J‍u(t)dt:u∈B})≤∫J‍α(B(t))dt.Lemma 2.4.
LetB⊂C(J,E) be a bounded set on J. Then α(B(t))≤2αc(B).Now, for the given[a,b]⊂(0,1) and the ρ as in (II), we introduce
(2.8)K={u∈C(J,P):u(t)≥ρu(s),t∈[a,b],s∈[0,1]}.
It is easy to check that K is a cone in C(J,E) and for u(t)∈K,t∈[a,b], we have ∥u(t)∥1≥ρ∥u∥.Next, we define an operatorT:K→C(J,P) given by
(2.9)Tu(t)=∫01‍G(t,s)p(s)f(u(s))ds,∀u∈K,t∈[0,1].
Clearly, u is a solution of the BVP (1.1) if and only if u is a fixed point of the operator T.Through direct calculation, by (II) and forv∈K,t∈[a,b],s∈J, we have
(2.10)Tu(t)=∫01‍G(t,s)p(s)f(u(s))ds≥ρ∫01‍G(s,s)p(s)f(u(s))ds=ρTu(s).
So, this implies that TK⊂K.Lemma 2.5.
Assume that,  hold. ThenT:K→K is a strict set contraction operator.Proof.
Firstly, The continuity ofT is easily obtained. In fact, if vn,v∈K and vn→v in the norm in C(J,E), then for any t∈J, we get
(2.11)‖Tvn(t)-Tv(t)‖1≤∫01‍G(s,s)p(s)‖f(vn(s))-f(v(s))‖1ds,
so, by the uniformly continuity of f, we have
(2.12)‖Tvn-Tv‖=sup⁡t∈J‖Tvn(t)-Tv(t)‖1⟶0.
This implies that Tvn→Tv in C(J,E), that is, T is continuous.
Now, letB⊂K be a bounded set. It follows from  that there exists a positive number L such that ∥f(v)∥≤L for any v∈B. Then, we can get
(2.13)‖Tv(t)‖1≤Le<∞,∀t∈J,v∈B,
where e is as defined in. So, T(B)⊂K is a bounded set in K.
For anyε>0, by (H1), there exists a δ′>0 such that
(2.14)∫0δ′‍G(s,s)p(s)≤ε6L,∫1-δ′1‍G(s,s)p(s)≤ε6L.
Let P=maxt∈[δ′,1-δ′]p(t). It follows from the uniform continuity of G(t,s) on [0,1]×[0,1] that there exists δ>0 such that
(2.15)|G(t,s)-G(t′,s)|≤ε3PL,|t-t′|<δ,t,t′∈[0,1],s∈[0,1].
Consequently, when |t-t′|<δ,t,t′∈[0,1],v∈B, we have
(2.16)‖Tv(t)-Tv(t′)‖1=‖∫01‍(G(t,s)-G(t′,s))p(s)f(v(s))ds‖1≤∫0δ′‍|G(t,s)-G(t′,s)|p(s)‖f(v(s))‖1ds+∫δ′1-δ′‍|G(t,s)-G(t′,s)|p(s)‖f(v(s))‖1ds+∫1-δ′1‍|G(t,s)-G(t′,s)|p(s)‖f(v(s))‖1ds≤2L∫0δ′‍G(s,s)p(s)ds+2L∫1-δ′1‍G(s,s)p(s)ds+PL∫01‍|G(t,s)-G(t′,s)|ds≤ε.
This implies that T(B) is an equicontinuous set on J. Therefore, by Lemma 2.2, we have
(2.17)α(T(B))=sup⁡t∈Jα(T(B)(t)).
Without loss of generality, by condition, we may assume thatp(t) is singular at t=0,1. So, there exist {ani},{bni}⊂(0,1),{ni}⊂N with {ni} being a strictly increasing sequence and limi→+∞ni=+∞ such that
(2.18)0<⋯<ani<⋯<an1<bn1<⋯<bni<⋯<1,p(t)≥ni,t∈(0,ani]∪[bni,1),p(ani)=p(bni)=ni,(2.19)limi→+∞ani=0,limi→+∞bni=1.
Next, we let
(2.20)pni(t)={ni,t∈(0,ani]∪[bni,1),p(t),t∈[ani,bni].
Then, from the above discussion we know that pni is continuous on J for every i∈N and
(2.21)pni(t)≥p(t),pni(t)⟶p(t),∀t∈(0,1),asi⟶+∞.
For anyε>0, by (2.19) and, there exists an i0 such that for any i>i0, we have
(2.22)2L∫0ani‍G(s,s)p(s)ds<ε2,2L∫bni1‍G(s,s)p(s)ds<ε2.
Therefore, for any bounded set B⊂C(J,E), by Lemmas 2.3 and 2.4,, the above discussion and noting that pni(t)≤p(t),t∈(0,1), as t∈J,i>i0, we have that
(2.23)α(T(B)(t))=α({∫01‍G(t,s)p(s)f(v(s))ds∣v∈B})≤α({∫01‍G(t,s)[p(s)-pni(s)]f(v(s))ds∣v∈B})+α({∫01‍G(t,s)pni(s)f(v(s))ds∣v∈B})≤2L∫0ani‍G(s,s)p(s)ds+2L∫bni1‍G(s,s)p(s)ds+∫01‍α(G(t,s)pni(s)f(v(s))∣v∈B)ds≤ε+ρ∫01‍G(s,s)p(s)α(f(v(s))∣v∈B)ds≤ε+2Meρα(B).
As ε is arbitrarily, we get
(2.24)α(T(B)(t))≤2Meραc(B),t∈J.
So, it follows from (2.17) and (2.24) that for any bounded set B⊂C(J,E), we have
(2.25)αc(T(B))≤2Meραc(B).
And note that 2Meρ<1, we have T:K→K is a strict set contraction operator. The proof is completed.Remark 2.6.
WhenE=R, (2.6) naturally holds. In this case, we may take M as 0, consequently, T:K→K is a completely continuous operator. Furthermore, if p(t)≡1,t∈J, Dalmasso [1] used the following condition:
(2.26)0<e=∫01‍sα(1-s)βp(s)ds<+∞,α,β∈[0,1).
Clearly, our condition  is weaker than (2.26).Our main tool used in this paper is the following fixed point index theorem of cone.Theorem 2.7 (see [13]).
Suppose thatE is a Banach space, K⊂E is a cone, and let the Ω1,Ω2 be two bounded open sets of E such that θ∈Ω1,Ω¯1⊂Ω2. Let operator T:K∩(Ω¯2∖Ω1)→K be strict set contraction. Suppose that one of the following two conditions holds:(i)
∥Tx∥≤∥x∥, for all x∈K∩∂Ω1,∥Tx∥≥∥x∥, for all x∈K∩∂Ω2;(ii)
∥Tx∥≥∥x∥, for all x∈K∩∂Ω1,∥Tx∥≤∥x∥, for all x∈K∩∂Ω2.Then T has at least one fixed point in K∩(Ω¯2∖Ω1).Theorem 2.8 (see [13]).
SupposeE is a real Banach space, K⊂E is a cone. Let Ωr={u∈K:∥u∥≤r}, and let the operator T:Ωr→K be completely continuous and satisfy Tx≠x, for all x∈∂Ωr. Then(i)
If∥Tx∥≤∥x∥, for all x∈∂Ωr, then i(T,Ωr,K)=1;(ii)
If∥Tx∥≥∥x∥, for all x∈∂Ωr, then i(T,Ωr,K)=0.


## 3.   The Main Results

Denote(3.1)f0=lim⁡∥u∥1→0+‖f(u)‖1‖u‖1,f∞=lim⁡∥u∥1→∞‖f(u)‖1‖u‖1.
We now present our main results by Theorems 3.1 and 3.2.Theorem 3.1.
Suppose that conditions,  hold. Assume thatf also satisfies(A1)
f(u)≥ru*,    ρr≤∥u∥1≤r;(A2)
f(u)≤Ru*,    0≤∥u∥1≤R,where u* and u* satisfy
(3.2)ρ‖∫ab‍G(s,s)p(s)u*(s)ds‖1≥1,‖u*‖∫01‍G(s,s)p(s)ds≤1.
Then, the boundary value problem (1.1) has a positive solution u such that ∥u∥ is between r and R.Proof of Theorem3.1.
Without loss of generality, we suppose thatr<R. For any u∈K, we have
(3.3)‖u(t)‖1≥ρ‖u‖,t∈[a,b].
We now define two open subset Ω1 and Ω2 of C(J,E)(3.4)Ω1={u∈C(J,E):‖u‖<r},Ω2={u∈C(J,E):‖u‖<R}.
For u∈K⋂∂Ω1, by (3.3), we have
(3.5)r=‖u‖≥‖u(t)‖1≥ρ‖u‖=ρr,t∈[a,b].
Foru∈K∩∂Ω1, if  holds, we have
(3.6)‖Tu(t)‖1=‖∫01‍G(t,s)p(s)f(u(s))ds‖1≥‖∫ab‍G(t,s)p(s)u*(s)rds‖1≥ρr‖∫ab‍G(s,s)p(s)u*(s)ds‖1≥r=‖u‖,t∈J.
Therefore, we have
(3.7)‖Tu‖=max⁡t∈[0,1]‖Tu(t)‖1≥‖u‖,∀u∈K∩∂Ω1.
On the other hand, as u∈K∩∂Ω2, we have u(t)≤∥u∥=R and by, we know
(3.8)‖Tu(t)‖1=‖∫01‍G(t,s)p(s)f(u(s))ds‖1≤R‖∫01‍G(t,s)p(s)u*(s)ds‖1≤R‖u*‖∫01‍G(s,s)p(s)ds≤R=‖u‖.
Thus
(3.9)‖T(u)‖=max⁡t∈[0,1]‖Tu(t)‖1≤‖u‖,∀u∈K∩∂Ω2.
Therefore, by (3.7), (3.9), Theorem 2.7 and r<R, we have that T has a fixed point u∈K∩(Ω2∖Ω¯1). Obviously, u is a positive solution of the problem (1.1) and r<∥u∥<R. The proof of Theorem 3.1 is complete.Theorem 3.2.
Suppose that conditions,, and  in Theorem3.1 hold. Assume that f also satisfies(A3)
f0=0;(A4)
f∞=0.Then, the boundary value problem (1.1) has at least two solutions.Proof of Theorem3.2.
Firstly, by condition, we can havelim∥u∥1→0+(∥f(u)∥1/∥u∥1)=0. Then, there exists an adequately small positive number m>0 such that m∫ab‍G(s,s)p(s)ds≤1 and there exists a constant ρ*∈(0,r) such that
(3.10)‖f(u)‖1≤m‖u‖1,0<∥u∥1≤ρ*,u≠0.
Set Ωρ*={u∈K:∥u∥<ρ*}, for any u∈∂Ωρ*, by (3.10), we have
(3.11)‖f(u)‖1≤m‖u‖1≤mρ*.
For u∈∂Ωρ*, we have
(3.12)‖Tu(t)‖1=‖∫01‍G(t,s)p(s)f(u(s))ds‖1≤∫01‍G(t,s)p(s)‖f(u(s))‖1≤∫ab‍G(t,s)p(s)mρ*ds≤ρ*m∫ab‍G(s,s)p(s)ds≤ρ*=‖u‖.
Therefore, we have
(3.13)‖Tu‖≤‖u‖,∀u∈∂Ωρ*.
Then by Theorem 2.8, we have
(3.14)i(T,Ωρ*,K)=1.
Next, by condition (A4), we have lim∥u∥1→∞(∥f(u)∥1/∥u∥1)=0. Then, there exists an adequately small positive number m¯>0 such that m¯∫ab‍G(s,s)p(s)ds≤1, and there exists a constant ρ0>0 such that
(3.15)‖f(u)‖1≤m¯‖u‖1,∥u∥1>ρ0.
We choose a constant ρ*>r, obviously, ρ*<r<ρ*. Set Ωρ*={u∈K:∥u∥<ρ*}, then for any u∈∂Ωρ*, by (3.15), we have
(3.16)‖f(u)‖1≤m¯‖u‖1≤m¯ρ*.
For u∈∂Ωρ*, we have
(3.17)‖Tu(t)‖1=‖∫01‍G(t,s)p(s)f(u(s))ds‖1≤∫01‍G(t,s)p(s)‖f(u(s))‖1≤∫ab‍G(t,s)p(s)m¯ρ*ds≤ρ*m¯∫ab‍G(s,s)p(s)ds≤ρ*=‖u‖.
Therefore, we have
(3.18)‖Tu‖≤∥u∥,∀u∈∂Ωρ*.
Then by Theorem 2.8, we have
(3.19)i(T,Ωρ*,K)=1.
Finally, setΩr={u∈K:∥u∥<r}, For any u∈∂Ωr, by, Lemma 2.3 and proceeding as for the proof of Theorem 3.1, we have
(3.20)‖Tu‖≥‖u‖,∀u∈∂Ωr.
Then by Theorem 2.8, we have
(3.21)i(T,Ωr,K)=0.
Therefore, by (3.14), (3.19), (3.21) and ρ*<R<ρ*, we have
(3.22)i(T,Ωr∖Ω¯ρ*,k)=-1,i(T,Ωρ*∖Ω¯r,k)=1.
Then T have fixed points u1∈Ωr∖Ω¯ρ* and u2∈Ωρ*∖Ω¯r. Obviously, u1 and u2 are all positive solutions of the BVP (1.1). The proof of Theorem 3.2 is complete.


## 4. Application

In order to illustrate the application of our results, we give an example in this section.Example 4.1.
Consider the following singular boundary value problem (SBVP):(4.1)3t3(13t3u′(t))+160[u1/2+u1/3]=θ,0<t<1,u(0)-lim⁡t→0+13t3u′(t)=θ,u(1)+lim⁡t→1-13t3u′(t)=θ,
where
(4.2)β=γ=δ=1,α=3,p(t)=13t3,f(u)=160(u1/2+u1/3).
Then obviously,
(4.3)∫01‍1p(t)dt=32,f∞=0,f0=0.
By computing, we know that the Green's function is
(4.4)G(t,s)=17{(1+3s)(2-t),0≤s≤t≤1,(1+3t)(2-s),0≤t≤s≤1.
It is easy to note that 0≤G(s,s)≤1 and conditions,,,  hold.
Next, by computing, we know thatρ=0.44. We choose r=3,u*=104, as 1.32=ρr≤∥u∥=max{u(t),t∈J}≤3 and ρ∥∫ab‍G(s,s)p(s)u*(s)ds∥=1.3>1, because of the monotone increasing of f(u) on [0,∞), then
(4.5)f(u)≥f(1.32)=359.3≥312=ru*,1.32≤‖u‖≤3.
Thus condition (A1) holds. Hence by Theorem 3.2, SBVP (4.1) has at least two positive solutions u1,u2 and 0<∥u1∥<3<∥u2∥.



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