# Positive Solutions of a Singular Third-Orderm-Point Boundary Value Problem

Authors: Shaolin Zhou; Xiaoling Han
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

This paper is concerned with the existence and nonexistence of positive solutions to the singular third-orderm-point boundary value problem u′′′(t)+a(t)f(u(t))=0,  0<t<1,u(0)=u'(0)=0,u'(1)-∑i=1m-2αiu'(ξi)=λ, where ξi∈[0,1),αi∈[0,∞)  (i=1,2,…,m-2) are constants, λ∈(0,1) is a parameter, f:[0,∞)→[0,∞) is continuous and a(·) is allowed to be singular at t=0 and t=1. The results here essentially extend and improve some known results.


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## Body

## 1. Introduction and the Main Results

Singular boundary value problems for nonlinear ordinary differential equations arise in a variety of areas of applied mathematics, physics, chemistry, and so on. For earlier works, see [1–3]. Nonsingular third-order multipoint boundary value problems have been studied by many authors by using different type of techniques, see, for example, [4–9] and the references therein. In recent years, singular third-order multipoint boundary value problems have also received much attention, see [10–12].Very recently, motivated by Ma [13], Sun [14] considered the third-order three-point boundary value problem
(Pλ)u′′′(t)+a(t)f(u(t))0=0,t∈(0,1),u(0)=u′(0)=0,u′(1)-αu′(η)=λ,
where η∈(0,1),  α∈[0,1/η) are constants and λ∈(0,∞) is a parameter. Under the following assumptions:(H1)
a∈C((0,1),[0,∞)) and 0<∫01‍(1-s)sa(s)ds<∞;(H2)
f∈C([0,∞),[0,∞));(H4*)
f is superlinear, that is, f0=0,  f∞=∞;(H5*)
f is sublinear, that is, f0=∞,  f∞=0,wheref0=limr→0+(f(r)/r),  f∞=limr→+∞(f(r)/r).By using Guo-Krasnosel’skii fixed point theorem, the author established the following results.Theorem A (see [14, Theorem 3.1]).
Suppose that (H1), (H2), and (H4*) hold. Then the problem (Pλ) has at least one positive solution for λ small enough and has no positive solution for λ large enough.Theorem B (see [14, Theorem 3.2]).
Suppose that (H1), (H2), and (H4*) hold. If f nondecreasing, then there exists a positive constant λ* such that the problem (Pλ) has at least one positive solution for λ∈(0,λ*) and has no positive solution for λ∈(λ*,∞).Theorem C (see [14, Theorem 3.3]).
Suppose that (H1), (H2), and (H5*) hold. Then the problem (Pλ) has at least one positive solution for any λ∈(0,∞).Being directly inspired by the previously mentioned works, we will consider the existence and nonexistence of positive solutions to the following third-orderm-point BVP:
(1)u′′′(t)+a(t)f(u(t))=0,0<t<1,(2λ)u(0)=u′(0)=0,u′(1)-∑i=1m-2‍αiu′(ξi)=λ,
where ξi∈[0,1),  αi∈[0,∞)(i=1,2,…,m-2) are constants and λ∈(0,1) is a parameter, a(·) is allowed to be singular at t=0 and t=1. Here, the solution u of BVP of (1), (2λ) is called positive solution if u(t) is positive on (0,1) and satisfies (1) and the boundary conditions (2λ).We assume that (H1), (H2) hold and make the following additional assumptions:(H3)
0<∑i=1m-2αiξi<1;(H4)
limu→0+sup(f(u)/u)<Λ1 and limu→+∞inf(f(u)/u)>Λ2;(H5)
limu→0+inf(f(u)/u)>0 and limu→+∞sup(f(u)/u)<L,where
(2)Λ1=2(1+∥h∥)-1L,∥h∥=12γ,L=[(1+γ∑i=1m-2‍αi)∫01‍(1-s)sa(s)ds]-1,γ=(1-∑i=1m-2‍αiξi)-1,Λ2=14[θ2∫θ1‍(1-s)sa(s)ds]-1.From(H1), we know that there exists t0∈(0,1) such that a(t0)>0. Let θ satisfy 0<θ<t0<1.Our main results are the following.Theorem 1.
Let (H1)–(H4) hold. Then there exists a positive number λ* such that BVP of (1), (2λ) has at least one positive solution for λ∈(0,λ*) and none for λ∈(λ*,∞).Theorem 2.
Let (H1)–(H3) and (H5) hold. Then BVP of (1), (2λ) has at least one positive solution for any λ∈(0,∞).The proof of previous theorems is based on the Schauder fixed-point theorem.Remark 3.
BVP(Pλ) is a special case of (1), (2λ) with α1=α,  ξ1=η, and α2=⋯=αm-2=0,  ξ2=⋯=ξm-2=0.Remark 4.
(
H
4
) allows but do not require the nonlinearity f(u) to be sublinear at zero and infinity; (H5) allows but do not require the nonlinearity f(u) to be sublinear at zero and infinity.Remark 5.
We do not assume any monotonicity condition on the nonlinearity as in [14]. We find that the nondecreasing condition of f can be removed from Theorem 3.2 in [14], and the same result is obtained in Theorem 1.Remark 6.
It is obvious that Theorem1 is an extension and complement of Theorems 3.1 and 3.2; furthermore, Theorem 2 is also an extension of Theorem 3.3 in [14].


## 2. Preliminary Lemmas

In this section, we present some notation and preliminary lemmas.LetC+[0,1]={u∈C[0,1]∣u(t)≥0,t∈[0,1]} equipped with the norm ∥u∥=max0≤t≤1|u(t)|.Lemma 7 (see [15, Lemma 2.1]).
Suppose thatϕ∈{φ∈Lloc1[0,1]∣∫01‍t(1-t)|φ(t)|dt<∞}.(i)
Then∫0t‍sϕ(s)ds,  ∫t1‍(1-s)ϕ(s)ds∈L1(0,1) and
(3)∫01‍∫0t‍sϕ(s)dsdt=∫01‍∫t1‍(1-s)ϕ(s)dsdt=∫01‍s(1-s)ϕ(s)ds.(ii)
Letr∈(0,1). Then
(4)limt→0+v(t)∫t1‍(1-s)ϕ(s)ds=0,
for everyv∈C1[0,r] with v(0)=0, and
(5)limt→1-w(t)∫0t‍sϕ(s)ds=0,
for everyw∈C1[r,1] with w(1)=0.Lemma 8.
Suppose that (H1)–(H3) hold, then BVP
(6)u′′′(t)+a(t)f(u(t))=0,0<t<1,u(0)=u′(0)=0,u′(1)-∑i=1m-2‍αiu′(ξi)=0,
has a unique nonnegative solutionu∈C1[0,1]∩C3(0,1) which can be represented as
(7)u(t)=∫01‍G(t,s)a(s)f(u(s))ds+12γδft2,
where
(8)G(t,s):=12{(2t-t2-s)s,0≤s≤t≤1,(1-s)t2,0≤t≤s≤1,G1(t,s):={(1-t)s,0≤s≤t≤1,(1-s)t,0≤t≤s≤1,
and δf=∑i=1m-2αi∫01‍G1(ξi,s)a(s)f(u(s))ds.Proof.
The proof of the uniqueness is standard and hence is omitted here. Now we prove the existence of the solution.
From (H1)–(H3) and Lemma 7, we conclude that the integration in (7) is well defined. Let v=u′; then BVP (6) may be reduced to boundary value problems
(9)v′′(t)+a(t)f(u(t))=0,0<t<1,(10)v(0)=0,v(1)-∑i=1m-2‍αiv(ξi)=0,(11)u′(t)=v(t),0<t<1,(12)u(0)=0.
We claim that (9), (10) have a nonnegative solution v which can be represented as
(13)v(t)=∫01‍G1(t,s)a(s)f(u(s))ds+γδft.
In fact, from (H1)–(H3) and Lemma 7, for each r∈(0,1), sa(s)f(u(s))∈L1[0,r] and (1-s)a(s)f(u(s))∈L1[r,1]. Combining the continuity of sa(s)f(u(s)) and (1-s)a(s)f(u(s)), we have
(14)∫0t‍sa(s)f(u(s))ds∈C1(0,r],∫t1‍(1-s)a(s)f(u(s))ds∈C1[r,1).
Thusv(t)∈C1(0,1). Moreover
(15)v′(t)=-∫0t‍sa(s)f(u(s))ds+∫t1‍(1-s)a(s)f(u(s))ds+γδf.
Similarity,v′(t)∈C1(0,1). From (15), we get v′′(t)=-a(t)f(u(t)),  t∈(0,1).
By Lemma7, we have from (13) that
(16)v(0)=limt→0+v(t)=limt→0+(1-t)∫0t‍sa(s)f(u(s))ds+limt→0+t∫t1‍(1-s)a(s)f(u(s))ds+limt→0+tγδf=0.
Again applying (13), we have
(17)v(1)=limt→1-v(t)=γδf.
This together with (13) implies that v(1)=∑i=1m-2αiv(ξi). The claim is proved.
By Lemma7, we obtain from (11), (12), and (13) that
(18)u(t)=∫0t‍v(τ)dτ=∫0t‍∫0τ‍(1-τ)sa(s)f(u(s))dsdτ+∫0t‍∫τ1‍(1-s)τa(s)f(u(s))dsdτ+12γδft2=12∫0t‍(2t-t2-s)sa(s)f(u(s))ds+12∫t1‍(1-s)t2a(s)f(u(s))ds+12γδft2.
It is easy to see thatu∈C1[0,1]∩C3(0,1), and moreover, u is a nonnegative solution of the BVP (6).
The proof is complete.Lemma 9 (see [14, Lemmas 2.2 and 2.3]).
For any(t,s)∈[0,1]×[0,1], one has(i)
q(t)G(1,s)≤G(t,s)≤G(1,s)=(1/2)(1-s)s, where q(t)=t2,(ii)
(∂/∂t)G(t,s)=G1(t,s), and 0≤G1(t,s)≤G1(s,s)=(1-s)s.Lemma 10.
Suppose that (H1)–(H3) hold; then the unique nonnegative solution u of (6) satisfies
(19)mint∈[θ,1]u(t)≥θ2∥u∥.The proof is similar to Lemma 2.4 in [14].Lemma 11.
Suppose that (H1)–(H3) hold. Let j∈{1,2,…,m-2} and r∈(-∞,0). Then BVP
(20)w′′(t)=0,0<τ<t<1,(21)w(τ)=0,w(1)-∑i=jm-2‍αiw(ξi)=r
has a unique solutionw satisfying w(t)≤0 on [τ,1].Proof.
From (20) and (21), we obtain w(t)=(t-τ)w′(τ). Again applying (21), we have
(22)(1-τ)w′(τ)-∑i=jm-2‍αi(ξi-τ)w′(τ)=r,w′(τ)(1-τ-∑i=jm-2‍αi(ξi-τ))=r.
Now setd:=1-τ-∑i=jm-2αi(ξi-τ); then we have that
(23)d>1-τ-∑i=jm-2‍αi(ξi-τξi)=(1-τ)(1-∑i=jm-2‍αiξi)>0.
This together with the fact thatr∈(-∞,0) implies that w′(τ)<0. Thus w(t)≤0 on [τ,1]. The proof is complete.


## 3. Proof of the Main Results

In this section, we will prove our main results.Proof of Theorem1.
We divide the proof into three steps.
Step
1.We first prove the existence of positive solutions to (1), (2λ) for sufficiently small λ:λ>0.
Leth be the unique solution of
(24)u′′′(t)=0,0<t<1,u(0)=u′(0)=0,u′(1)-∑i=1m-2‍αiu′(ξi)=1.
Thenh(t)=(1/2)γt2. Let v=u-λh; then u is a positive solution of BVP (1), (2λ) if and only if v=u-λh is a nonnegative solution of BVP
(25)v′′′(t)+a(t)f(v(t)+λh(t))=0,0<t<1,v(0)=v′(0)=0,v′(1)-∑i=1m-2‍αiv′(ξi)=0.
Letf~(x)=sup0≤s≤xf(s). Since limu→0+sup(f~(u)/u)<Λ1; then there exists a positive number λ1 such that
(26)f~(λ1+λ1∥h∥)≤Λ1(λ1+λ1∥h∥)=2λ1L.
Define a closed convex subset inC+[0,1] by
(27)D={v∈C+[0,1]∣v(t)≤λ1,t∈[0,1]}
and an operatorT:D→C+[0,1] by
(28)Tv(t):=∫01‍G(t,s)a(s)f(v(s)+λh(s))ds+12γδft2,
where δf=∑i=1m-2αi∫01‍G1(ξi,s)a(s)f(v(s)+λh(s))ds. Modeling the proof of Lemma 2.3 in [10], we can show that T is a completely continuous operator. From Lemma 8, we know that  v is a nonnegative solution of (25) if and only if v is a fixed point of T.
Suppose thatλ<λ1; we claim that T:D→D.
In fact, from Lemma9 and (28), we have
(29)0≤Tv(t)≤f~(λ1+λ1∥h∥)[∑i=1m-2‍αi∫01‍(1-s)12∫01‍(1-s)sa(s)ds(λ1+λ1∥h∥p)+12γ∑i=1m-2‍αi∫01‍(1-s)sa(s)ds]=12f~(λ1+λ1∥h∥)L-1≤λ1.
The claim is proved. Using the Schauder fixed point theorem, we conclude thatT has a fixed point v in D, and then u=v+λh is a positive solution of (1), (2λ).
Step
2.We verify that BVP of (1), (2λ) has no positive solutions for λ large enough.
Suppose to the contrary that BVP of (1), (2λ) has at least one positive solution for any λ>0. Then there exist 0<λ1<λ2<⋯<λn<⋯, with limn→∞λn=∞, such that for any positive integer n, BVP of (1), (2λ) has a positive solution un. Thus vn=un-λnh is a nonnegative solution to (25). On the one hand, we have
(30)∥vn+λnh∥≥λn∥h∥=12γλn→∞,(n→∞).
On the other hand, sincelimu→+∞inf(f(u)/u)>Λ2, there exists r0>0 such that f(u)≥Λ2u, for any u∈[r0,∞). Let n be large enough that θ2∥vn+λnh∥≥r0. By Lemma 10, we have
(31)inft∈[θ,1]vn(t)≥θ2∥vn∥,inft∈[θ,1]h(t)=12γθ2=θ2∥h∥.
This implies that(32)inft∈[θ,1](vn(t)+λnh(t))≥θ2(∥vn∥+λn∥h∥)≥θ2∥vn+λnh∥.
Thus(33)∥vn+λnh∥≥∥vn∥=∫01‍G(1,s)a(s)f(vn(s)+λnh(s))ds+12γ∑i=1m-2αi∫01‍G1(ξi,s)a(s)f(vn(s)+λnh(s))ds≥∫θ1‍G(1,s)a(s)f(vn(s)+λnh(s))ds≥12Λ2(vn(s)+λnh(s))∫θ1‍(1-s)sa(s)ds=2∥vn+λnh∥,
which is a contradiction.
Step
3. Let B={λ∣  BVP of (1), (2λ) has at least one positive solution} and λ*=supB; then 0<λ*<∞. We show that (1), (2λ) have positive solution for any λ∈(0,λ*). From the definition of λ*, we know that, for any λ∈(0,λ*), there exists λ~>λ such that (1), (2λ) have positive solution uλ~.
Now we consider the following third-orderm-point boundary value problem:
(34)u′′′(t)+a(t)(F~u)(t)=0,0<t<1,u(0)=u′(0)=0,u′(1)-∑i=1m-2‍αiu′(ξi)=λ,
where
(35)(F~u)(t)={f(uλ~(t)),ifu(t)>uλ~(t),f(u(t)),if0≤u(t)≤uλ~(t),f(0),ifu(t)<0.
SinceF~ is bounded, by Schauder fixed point theorem, the problem (34) has a solution uλ.
By Lemma8, uλ satisfies
(36)uλ(t)=∫01‍G(t,s)a(s)(F~u)(s)ds+12γt2∑i=1m-2‍αi∫01‍G1(ξi,s)a(s)×(F~u)(s)ds+12γλt2,
thus uλ≥0.
LetI={t∈(0,1]∣uλ(t)>uλ~(t)} and Ω={t∈(0,1]∣w(t)>0}, where w(t)=uλ′(t)-uλ~′(t),  t∈[0,1]. We will show that I=∅. Noticing that, if uλ′(t)>uλ~′(t) holds for any t∈(0,1), combining with uλ′(0)=uλ~′(0), we get uλ(t)>uλ~(t). Thus we prove that Ω=∅; then we have I=∅.
Suppose to the contrary thatΩ≠∅.
Ifuλ′(1)<uλ~′(1), then, from Ω≠∅, and the continuity of w(t), there exists (a,b)⊂Ω such that w′(a)=w(b)=0. Moreover, w′′(t)=0 in (a,b). Thus w(t)≡0 in (a,b). This contradicts with the fact that w(t)>0 in (a,b).
Ifuλ′(1)>uλ~′(1), we claim that there exists j∈{1,2,…,m-2} such that w(ξj)>0.
In fact, from the fact thatw(1)-∑i=1m-2αiw(ξi)=λ-λ~<0 and w(1)>0, we have that ∑i=1m-2αiw(ξi)>0. Thus, there exists j∈{1,2,…,m-2} such that w(ξj)>0.
Letj0=min{j∣j∈{1,2,…,m-2} such that w(ξj)>0}; then, we only need to deal with the following four cases.
Case
1. w(t)>0 in (0,1). In this case, we have
(37)w′′(t)=0,0<t<1,w(0)=0,w(1)-∑i=1m-2‍αiw(ξi)=λ-λ~<0.
We easily verify that(38)w(t)=(λ-λ~)γt<0,in(0,1).
This contradicts with the fact thatΩ≠∅.
Case
2. There exists τ∈(0,1) such that w(τ)=0 and w(t)>0  in (0,τ). In this case, we have
(39)w′′(t)=0,0<t<τ,w(0)=0,w(τ)=0.
We easily obtainw(t)≡0 in [0,τ], a contradiction again.
Case
3. There exists τ∈(0,1) such that w(τ)=0 and w(t)>0 in (τ,1]. In this case, if j0>1, then, for any i∈{1,2,…,j0-1}, we have w(ξi)<0. Thus
(40)r:=w(1)-∑i=j0m-2‍αiw(ξi)=w(1)-∑i=1m-2‍αiw(ξi)+∑i=1j0-1‍αiw(ξi)=λ-λ~+∑i=1j0-1‍αiw(ξi)≤λ-λ~<0.
Ifj0=1, then r:=w(1)-∑i=j0m-2αiw(ξi)=λ-λ~<0, and w(t) satisfies
(41)w′′(t)=0,τ<t<1,w(τ)=0,w(1)-∑i=j0m-2‍αiw(ξi)=r<0.
By Lemma11, we also have w(t)≤0 in [τ,1], a contradiction again.
Case
4. There exists [a,b]⊂(0,1) such that w(a)=w(b)=0 and w(t)>0 in (a,b). The same as Case 2, we can lead to a contradiction.
Summarizing the previous discussion, we assert thatΩ=∅; thus I=∅. Up to now, the problem (1), (2λ) has a solution uλ.Proof of Theorem2.
Sincelimu→0+inf(f(u)/u)>0, there exists μ>0, and r1>0 such that f(u)≥μu,  u∈[0,r1]. Next we consider two cases: f is bounded or f is unbounded.
Case
1. Suppose that f is bounded, that is, f(u)≤M, for all u∈[0,∞). By Schauder fixed point theorem the problem of (1), (2λ) has a positive solution.
Case
2. If f is unbounded. Since limu→+∞sup(f(u)/u)<L, there exists a positive number r2 such that f(u)≤Lu, for u∈[r2,∞). Since f is unbounded, for any λ∈(0,∞), we are able to choose
(42)Rλ≥max{2r1,r2,λ∥h∥}
such that
(43)f(u)≤f(2Rλ),foru∈[0,2Rλ].
Defining a closed convex subset in C+[0,1] by
(44)D1={v∈C+[0,1]∣v(t)≤Rλ,t∈[0,1]}.
For each v∈D1, we have 0≤v+λh≤2Rλ. By (28) and Lemma 9, we obtain that
(45)Tv(t)≤12f(2Rλ)(∑i=1m-2‍∫01‍(1-s)sa(s)ds12f(2Rλo)+γ∑i=1m-2‍αi∫01‍(1-s)sa(s)ds)≤12L·2RλL-1=Rλ.
That is,Tv(t)∈D1. By using the Schauder fixed point theorem, we assert that T has a fixed point v∈D1, and then u=v+λh is a positive solution of BVP of (1), (2λ).


## 4. Example

Consider the boundary value problem(46)u′′′(t)+1t(1-t)·u2(t)+u(t)u(t)+176(7+cosu(t))=0,hhhhhu′′′(t)+1t(1-t)·u2(t)+u(t)u(t)+1760<t<1,(47λ)u(0)=u′(0)=0,u′(1)-∑i=14‍iu′(18i)=λ,
where αi=i,  ξi=(1/8i)(i=1,2,3,4),  a(t)=1/t(1-t),  t∈(0,1), and f(u)=((u2+u)/(u+176))(7+cosu),  u∈[0,+∞). Obviously (H1), and (H2), and (H3) hold. By calculating, we have γ=2,  ∥h∥=1,  L=1/21,  Λ1=1/21. Let θ=1/4, then Λ2=16/3. We easily verify that
(47)limu→0+supf(u)u=122<Λ1,limu→+∞inff(u)u=6>Λ2,
that is, (H4) is satisfied. Therefore, Theorem 1 now guarantees that there exists a positive number λ* such that BVP of (46), (47λ) has at least one positive solution for λ∈(0,λ*) and none for λ∈(λ*,∞).But we cannot apply Theorem B [14, Theorem 3.2]. In fact, f does not satisfy monotonicity condition. Moreover, condition (H4*) of Theorem B does not hold.



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*Source: 567970-2013-03-19.xml*