JFSA Journal of Function Spaces and Applications 1758-4965 0972-6802 Hindawi Publishing Corporation 567970 10.1155/2013/567970 567970 Research Article Positive Solutions of a Singular Third-Order m-Point Boundary Value Problem Zhou Shaolin Han Xiaoling Ma To College of Mathematics and Information Science Northwest Normal University Lanzhou 730070 China nwnu.edu.cn 2013 19 3 2013 2013 04 12 2012 16 01 2013 2013 Copyright © 2013 Shaolin Zhou and Xiaoling Han. 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.

This paper is concerned with the existence and nonexistence of positive solutions to the singular third-order m-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.

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 . Nonsingular third-order multipoint boundary value problems have been studied by many authors by using different type of techniques, see, for example,  and the references therein. In recent years, singular third-order multipoint boundary value problems have also received much attention, see .

Very recently, motivated by Ma , Sun  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:

aC((0,1),[0,)) and 0<01(1-s)sa(s)ds<;

fC([0,),[0,));

f is superlinear, that is, f0=0,  f=;

f is sublinear, that is, f0=,  f=0,

where f0=limr0+(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 [<xref ref-type="bibr" rid="B15">14</xref>, 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 [<xref ref-type="bibr" rid="B15">14</xref>, 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 [<xref ref-type="bibr" rid="B15">14</xref>, 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-order m-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:

0<i=1m-2αiξi<1;

limu0+sup(f(u)/u)<Λ1 and limu+inf(f(u)/u)>Λ2;

limu0+inf(f(u)/u)>0 and limu+sup(f(u)/u)<L,

where (2)Λ1=2(1+h)-1L,h=12γ,L=[(1+γi=1m-2αi)01(1-s)sa(s)ds]-1,γ=(1-i=1m-2αiξi)-1,Λ2=14[θ2θ1(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 . We find that the nondecreasing condition of f can be removed from Theorem 3.2 in , and the same result is obtained in Theorem 1.

Remark 6.

It is obvious that Theorem 1 is an extension and complement of Theorems 3.1 and 3.2; furthermore, Theorem 2 is also an extension of Theorem 3.3 in .

2. Preliminary Lemmas

In this section, we present some notation and preliminary lemmas.

Let C+[0,1]={uC[0,1]u(t)0,t[0,1]} equipped with the norm u=max0t1|u(t)|.

Lemma 7 (see [<xref ref-type="bibr" rid="B13">15</xref>, Lemma 2.1]).

Suppose that ϕ{φLloc1[0,1]01t(1-t)|φ(t)|dt<}.

Then 0tsϕ(s)ds,  t1(1-s)ϕ(s)dsL1(0,1) and (3)010tsϕ(s)dsdt=01t1(1-s)ϕ(s)dsdt=01s(1-s)ϕ(s)ds.

Let r(0,1). Then (4)limt0+v(t)t1(1-s)ϕ(s)ds=0,

for every vC1[0,r] with v(0)=0, and (5)limt1-w(t)0tsϕ(s)ds=0,

for every wC1[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 solution uC1[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,0st1,(1-s)t2,0ts1,G1(t,s):={(1-t)s,0st1,(1-s)t,0ts1, and δf=i=1m-2αi01G1(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))dsC1(0,r],t1(1-s)a(s)f(u(s))dsC1[r,1).

Thus v(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 Lemma 7, we have from (13) that (16)v(0)=limt0+v(t)=limt0+(1-t)0tsa(s)f(u(s))ds+limt0+tt1(1-s)a(s)f(u(s))ds+limt0+tγδf=0.

Again applying (13), we have (17)v(1)=limt1-v(t)=γδf.

This together with (13) implies that v(1)=i=1m-2αiv(ξi). The claim is proved.

By Lemma 7, we obtain from (11), (12), and (13) that (18)u(t)=0tv(τ)dτ=0t0τ(1-τ)sa(s)f(u(s))dsdτ+0tτ1(1-s)τa(s)f(u(s))dsdτ+12γδft2=120t(2t-t2-s)sa(s)f(u(s))ds+12t1(1-s)t2a(s)f(u(s))ds+12γδft2.

It is easy to see that uC1[0,1]C3(0,1), and moreover, u is a nonnegative solution of the BVP (6).

The proof is complete.

Lemma 9 (see [<xref ref-type="bibr" rid="B15">14</xref>, Lemmas 2.2 and 2.3]).

For any (t,s)[0,1]×[0,1], one has

q(t)G(1,s)G(t,s)G(1,s)=(1/2)(1-s)s, where q(t)=t2,

(/t)G(t,s)=G1(t,s), and 0G1(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)θ2u.

The proof is similar to Lemma 2.4 in .

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 solution w 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 set d:=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 that r(-,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 Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

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.

Let h be the unique solution of (24)u′′′(t)=0,0<t<1,u(0)=u(0)=0,u(1)-i=1m-2αiu(ξi)=1.

Then h(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.

Let f~(x)=sup0sxf(s). Since limu0+sup(f~(u)/u)<Λ1; then there exists a positive number λ1 such that (26)f~(λ1+λ1h)Λ1(λ1+λ1h)=2λ1L.

Define a closed convex subset in C+[0,1] by (27)D={vC+[0,1]v(t)λ1,t[0,1]}

and an operator T:DC+[0,1] by (28)Tv(t):=01G(t,s)a(s)f(v(s)+λh(s))ds+12γδft2, where δf=i=1m-2αi01G1(ξi,s)a(s)f(v(s)+λh(s))ds. Modeling the proof of Lemma 2.3 in , 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 that λ<λ1; we claim that T:DD.

In fact, from Lemma 9 and (28), we have (29)0Tv(t)f~(λ1+λ1h)[i=1m-2αi01(1-s)1201(1-s)sa(s)ds(λ1+λ1hp)+12γi=1m-2αi01(1-s)sa(s)ds]=12f~(λ1+λ1h)L-1λ1.

The claim is proved. Using the Schauder fixed point theorem, we conclude that T 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 limnλn=, 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+λnhλnh=12γλn,(n).

On the other hand, since limu+inf(f(u)/u)>Λ2, there exists r0>0 such that f(u)Λ2u, for any u[r0,). Let n be large enough that θ2vn+λnhr0. By Lemma 10, we have (31)inft[θ,1]vn(t)θ2vn,inft[θ,1]h(t)=12γθ2=θ2h.

This implies that (32)inft[θ,1](vn(t)+λnh(t))θ2(vn+λnh)θ2vn+λnh.

Thus (33)vn+λnhvn=01G(1,s)a(s)f(vn(s)+λnh(s))ds+12γi=1m-2αi01G1(ξi,s)a(s)f(vn(s)+λnh(s))dsθ1G(1,s)a(s)f(vn(s)+λnh(s))ds12Λ2(vn(s)+λnh(s))θ1(1-s)sa(s)ds=2vn+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-order m-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)),if  u(t)>uλ~(t),f(u(t)),if  0u(t)uλ~(t),f(0),if  u(t)<0.

Since F~ is bounded, by Schauder fixed point theorem, the problem (34) has a solution uλ.

By Lemma 8, uλ satisfies (36)uλ(t)=01G(t,s)a(s)(F~u)(s)ds+12γt2i=1m-2αi01G1(ξi,s)a(s)×(F~u)(s)ds+12γλt2, thus uλ0.

Let I={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 that Ω.

If uλ(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).

If uλ(1)>uλ~(1), we claim that there exists j{1,2,,m-2} such that w(ξj)>0.

In fact, from the fact that w(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.

Let j0=min{jj{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 that Ω.

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 obtain w(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.

If j0=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 Lemma 11, 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 that Ω=; thus I=. Up to now, the problem (1), (2λ) has a solution uλ.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

Since limu0+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)Rλmax{2r1,r2,λh} such that (43)f(u)f(2Rλ),for  u[0,2Rλ]. Defining a closed convex subset in C+[0,1] by (44)D1={vC+[0,1]v(t)Rλ,t[0,1]}. For each vD1, we have 0v+λh2Rλ. By (28) and Lemma 9, we obtain that (45)Tv(t)12f(2Rλ)(i=1m-201(1-s)sa(s)ds12f(2Rλo)+γi=1m-2αi01(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 vD1, 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)limu0+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.

Acknowledgments

The authors are grateful to the anonymous referee for his or her constructive comments and suggestions which led to improvement of the original paper. X. Han is supported by the NNSF of China (no.11101335).

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