JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi Publishing Corporation 10.1155/2015/928105 928105 Research Article Existence of Positive Solutions for a Fourth-Order m-Point Boundary Value Problem Yang Liu Shen Chunfang Falset Jesús G. Department of Mathematics Hefei Normal UniversityHefei, Anhui 230061 China hftc.edu.cn 2015 1852015 2015 03 08 2014 05 02 2015 1852015 2015 Copyright © 2015 Liu Yang and Chunfang Shen. 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.

By using Krasnosel’skii’s fixed point theorem and the fixed point index theorem in the special function space, we obtain some sufficient conditions for the existence of positive solutions of fourth-order boundary value problem with multipoint boundary conditions. Applications of our results to some special problems are also discussed.

1. Introduction

In this paper, we study the existence of positive solutions of the boundary value problem consisting of the nonlinear fourth-order differential equation(1)u(4)(t)+a(t)f(u(t))=0,t[0,1]and the multipoint boundary condition (2)u′′′0=0,u′′0=0,u(0)=0,u(1)=i=1m-2βiu(ξi)or (3)u′′′(1)=0,u′′(1)=0,u(1)=0,u(0)=i=1m-2βiu(ξi),where 0<ξ1<ξ2<<ξm-2<1, 0<βi<1, i=1,2,,m-2, i=1m-2βi<1 and fC([0,+),[0,+)), a(t)C([0,1],[0,+)), and there exist t0[0,1] such that a(t0)>0.

It is well known that boundary value problems of fourth-order ordinary differential equations are used to describe a large number of physical, biological, and chemical phenomena. For an example, (1) is often used to describe the deformation of an elastic beam under a variety of boundary conditions . For this reason, there is a wide literature that deals with the existence and the multiplicity of solutions for fourth-order boundary value problems; see .

Krasnoselskiis theorem in a cone and fixed point index theorem have often been used to study the existence and multiplicity of positive solutions of nonlocal boundary value problems over the last several years. As to fourth-order Bvps, by using Krasnoselskiis fixed point theorem, Graef et al.  established the existence results of positive solutions for the nonlinear fourth-order ordinary differential equation(4)u(4)(t)+λg(t)f(u(t))=0,t[0,1]with the boundary conditions(5)u(0)=u(1)=u′′(0)=u′′(p)-u′′(1)=0.

However, the aforementioned papers mainly considered the two or three point boundary conditions. There are very few works on the multipoint boundary value problem for fourth-order ordinary differential equations. As to m-point boundary value problem ((1),  (2)) or ((1),  (3)), there is no existence results of solution or positive solution. The goal of this paper is to fill this gap. In this paper, by constructing an available integral operator and combining fixed point theorem and fixed point index theorem, we establish some sufficient conditions of the existence of positive solutions for problems ((1),  (2)) and ((1),  (3)).

The rest of this paper is organized as follows. In Section 2, we given some preliminaries and lemmas for use later. Section 3 is devoted to the existence and multiplicity of positive solutions of problem ((1),  (2)). In Section 4, we establish the existence results of positive solutions for problem ((1),  (3)). In Section 5, we give some examples to demonstrate the main results of this paper.

2. Preliminaries Definition 1.

Let E be a real Banach space over the real numbers. A nonempty convex closed set PE is said to be a cone provided that

auP, for all uP, a0,

u,-uP implies u=0.

Definition 2.

An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Let E be the Banach space C[0,1] endowed with the norm u=max0t1ut. We denote (6)f0=limu0+f(u)u,f=limuf(u)u.

Lemma 3 (see [<xref ref-type="bibr" rid="B20">20</xref>]).

Let E be a Banach space and KE be a cone. Assume Ω1 and Ω2 are open bounded subsets of E with 0Ω1Ω¯1Ω2, and let (7)A:K(Ω¯2Ω1)Kbe a completely continuous operator such that (8)Auu,uKΩ1,Auu,uKΩ2or (9)Auu,uKΩ1,Auu,uKΩ2;then, A has a fixed point in K(Ω¯2Ω1).

Lemma 4 (see [<xref ref-type="bibr" rid="B20">20</xref>]).

Let E be a Banach space and let KE be a cone. For ρ>0, define Kρ={uK:uρ} and assume that A:KρK is a completely continuous operator such that Auu for xKρ={uK:u=ρ}. Then, (10)Auu,uKρ,theni(A,Kρ,K)=1,Auu,uKρ,theni(A,Kρ,K)=0.

3. Existence Results of Problems (<xref ref-type="disp-formula" rid="EEq1.1">1</xref>) and (<xref ref-type="disp-formula" rid="EEq1.2">2</xref>)

We begin with the fourth-order m-point boundary value problem (11)u4t+yt=0,t0,1,(12)u′′′(0)=0,u′′(0)=0,u(0)=0,u(1)=i=1m-2βiu(ξi).

Lemma 5.

Denoting ξ0=0, ξm-1=1, β0=βm-1=0, and y(t)C[0,1], problem (11),  (12) has the unique solution(13)u(t)=01G(t,s)y(s)ds,where (14)Gt,s=-16s3+16-12s+12s2-16k=0i-1βks3+k=im-2βkξk-16ξk2+12ξks-12s2·1-k=0m-1βk-1,ts,-16t3+12st2-12s2t+16-12s+12s2-16k=0i-1βks3+k=im-2βkξk-16ξk2+12ξks-12s2·1-k=0m-1βk-1,ts,for ξi-1sξi, i=1,2,,m-1.

Proof.

Let G(t,s) be Green’s function of problem -u(4)(t)=0 with boundary condition (12). We can suppose (15)G(t,s)=a3t3+a2t2+a1t+a0,ts,ξi-1sξi,i=1,2,,m-1,b3t3+b2t2+b1t+b0,ts,ξi-1sξi,i=1,2,,m-1,where ai, bi, i=0,1,2,3 are unknown coefficients. Considering the properties of Green’s function and boundary condition (12), we have(16)a3s3+a2s2+a1s+a0=b3s3+b2s2+b1s+b0,3a3s2+2a2s+a1=3b3s2+2b2s+b1,6a3s+2a2=6b3s+2b2,6a3-6b3=1,a3=0,a2=0,a1=0,b3+b2+b1+b0=k=0i=1βka3ξk3+a2ξk2+a1ξk+a0+k=im-2βkb3ξk3+b2ξk2+b1ξk+b0.A straightforward calculation shows that(17)a3=a2=a1=0,a0=-16s3+16-12s+12s2-16k=0i-1βks3+k=im-2βkξk-16ξk2+12ξks-12s2·1-k=0m-1βk-1,b3=-16,b2=s2,b1=-s22,b0=16-12s+12s2-16k=0i-1βks3+k=im-2βkξk-16ξk2+12ξks-12s2·1-k=0m-1βk-1.These give the explicit expression of Green’s function. Then, we have (18)ut=01Gt,sysds.

Lemma 6.

One can see that G(t,s)0, t,s[0,1].

Proof.

For ξi-1sξi, i=1,2,,m-1,(19)G(t,s)t=0,ts,ξi-1sξi,-12t-s2,ts,ξi-1sξi.Then, G(t,s)/t0, 0t, s1, which induces that G(t,s) is decreasing on t. By a simple computation, we see(20)G1,s=-16+12s-12s2+16-12s+12s2-16k=0i-1βks3+k=im-2βkξk-16ξk2+12ξks-12s2·1-k=0m-1βk-1=16k=0m-1βk1-s3+16k=im-212βks-ξk3·1-k=0m-1βk-10.This ensures that G(t,s)0, t,s[0,1].

Lemma 7.

If y(t)0, t[0,1] and u(t) is the solution of problem (11),  (12), then (21)min0t1utγmax0t1ut,where γ=i=1m-2βi(1-ξi)/(1-i=1m-2βiξi).

Proof.

Since u(4)(t)0, t[0,1], then u′′′(t) is decreasing on [0,1]. Considering u′′′(0)=0, we have u′′′(t)0, t[0,1]. Thus, u′′(t) is decreasing on [0,1]. Considering this together with the boundary condition u′′(0)=0, we conclude that u′′(t)0. Then, u(t) is concave on [0,1]. Taking into account that u(0)=0, we get that (22)max0t1u(t)=u(0),min0t1u(t)=u(1).From the concavity of u(t), we have (23)ξiu1-u0u(ξi)-u(0).Multiplying both sides with βi and considering the boundary condition, we have (24)min0t1ut=u1i=1m-2βi1-ξi1-i=1m-2βiξiu0=γmax0t1ut.Problem (1),  (2) has a solution u=u(t) if and only if u solves the operator equation (25)ut=01Gt,sasfusds:=Au(t),0t1.Denote the cone (26)K={uE:u0,min0t1u(t)γu}.Since K is a cone, by Lemma 7 we have that A(K)K. For the convenience, we denote (27)M=max0t101G(t,s)a(s)ds,m=min0t101G(t,s)a(s)ds.

Theorem 8.

Problem (1),  (2) has at least one positive solution if

f0=0, f= or

f0=, f=0.

Proof.

Firstly, we consider case (1). By f0=0, we choose ε1>0 satisfying ε1M1. Then, there exists H1>0 such that(28)fuε1u,uH1.Define Ω1={uE:u<H1}; then, for uKΩ1, we have (29)Au=max0t101Gt,sasfusds01G(t,s)a(s)ds×ε1×uu.On the other hand, by f=, choose ε2 satisfying ε2mγ1. Then, there exists H¯2>0 such that (30)f(u)ε2u,foruH¯2.Let H2=max{2H1,1/γH¯2} and let Ω2={uE:u<H2}. Then, for uKΩ2, we see uγ1/γH¯2=H¯2. Then, (31)Au=max0t101Gt,sasfusdsε2muε2mγuu.Then, by Lemma 3, A has a fixed point u, which implies problem (1),  (2) has at least one positive solution K(Ω2Ω¯1).

Next, the case f0=, f=0 is considered. For f0=, there exists H3>0 such that (32)fuε3u,for0<uH3,where ε3>0 satisfies ε3mγ1. Denote Ω3={uE:u<H3}; then, for uKΩ3, we have (33)Au=max0t101Gt,sasfusdsε3muε3mγuu.Considering f=0, there exists H¯4>0 such that (34)f(u)ε4u,foruH¯4,where ε4>0 satisfies ε4M1. We distinguish two cases to discuss.

Case 1. Suppose that f is bounded; then, there exists N>0 satisfying f(u)<N. Taking H4=max{2H3,NM1} and Ω4={uE:u<H4}, then for uKΩ4, we have (35)Au=max0t1Aut=max0t101G(t,s)a(s)f(u(s))dsNMH4u.

Case 2. Suppose that f is unbounded. Since fC([0,+),[0,+)), then there certainly exists H4>max{2H3,H4¯} such that(36)f(u)f(H4),0uH4.For uKΩ4, where Ω4={uE:u<H4}, we have (37)Au=max0t101Gt,sasfusdsMf(H4)ε4MH4u.So, in either case, we can choose H4>0 such that Auu, for uKΩ4. Then, Lemma 3 implies that A has a fixed point. Consequently, problem (1),  (2) has at least one positive solution.

Theorem 9.

Problem (1),  (2) has at least one positive solution in one of the following cases:

f0=a(0,+), f=b(0,+), and aM<1, bmγ>1;

f0=a(0,+), f=b(0,+), and bM<1, amγ>1;

f0=a(0,+), f=, and aM<1;

f0=a, f=0, and amγ>1;

f0=0, f=b(0,+), and bmγ>1;

f0=, f=b, and bM<1.

Proof.

We consider the case f0=a(0,+), f=b(0,+) firstly. Since f0=a(0,+), for ε5>0 satisfying (a+ε5)M1, there exists H5>0 such that (38)fua+ε5u,foruH5.Define Ω5={uE:u<H5}; then, for xKΩ5, we see (39)Au=max0t101Gt,sasfusds(a+ε5)Muu.Considering f=b, there exists H¯60 and ε6 satisfies (b-ε6)mγ1 such that (40)fub-ε6u,foruH¯6.Let H6=max{2H5,1/γH¯6} and let Ω6={uE:u<H6}; then, for uKΩ6, we get uH¯6; then, (41)Au=max0t101Gt,sasfusds(b-ε6)mu(b-ε6)mγuu.Then, by Lemma 3, problem (1),  (2) has at least one positive solution. As for other cases (4),(5),(6),(7),(8), the proof is considerably analogous with the case (1),(2),(3) and is omitted here.

Theorem 10.

Assume following conditions are satisfied:

f0=f=0;

there exist r>0 such that f(u)r/m for u[r,r/γ];

then, problem (1),  (2) has at least two positive solutions u1,u2 satisfying (42)0u1ru2.

Proof.

Considering f0=0, similarly with Theorem 8, we claim there exists r1(0,r) such that(43)iA,Kr1,K=1,whereKr1=uE:u<r1.Since f=0, similarly with above, there exists r2(r,+) such that(44)iA,Kr2,K=1,whereKr2=uE:u<r2.For uKKr, Kr={uE:u<r}, (45)Au=max0t101Gt,sasfusdsm×rm=ru.Hence, i(A,Kr,K)=0.

Therefore, (46)i(A,KrKr1,K)=-1,i(A,Kr2Kr,K)=1.Thus, there exists at least two positive solutions u1,u2 such that (47)0<u1ru2.

Theorem 11.

Assume the following conditions are satisfied:

f0=f=;

there exists R>0 such that f(u)R/M for u[R,R/γ];

then, problem (1),  (2) has at least two positive solutions u1,u2 satisfying (48)0u1Ru2.

Proof.

Considering f0=, f=, we claim there exists R1(0,R), R2(R,+) such that(49)iA,KR1,K=0,whereKR1=uE:u<R1,iA,KR2,K=0,whereKR2=uE:u<R2.For uKKR, KR={uE:u<R}, (50)Au=max0t101Gt,sasfusdsM×RM=Ru.Hence, i(A,KR,K)=1. Therefore, (51)i(A,KRKR1,K)=1,i(A,KR2KR,K)=-1.Thus, there exists at least two positive solutions u1,u2 such that (52)0<u1Ru2.

4. Existence Results of Problem (<xref ref-type="disp-formula" rid="EEq1.1">1</xref>),  (<xref ref-type="disp-formula" rid="EEq1.3">3</xref>) Lemma 12.

Denoting ξ0=0, ξm-1=1, and β0=βm-1=0, Green’s function of problem (53)-u(4)(t)=0,(54)u′′′(1)=0,u′′(1)=0,u(1)=0,u(0)=i=1m-2βiu(ξi)is(55)H(t,s)=16t3-12st2+12s2t+k=0i-1βk16ξk3-12ξk2s+12ξks2+16k=im-2βks3·1-k=0m-1βk-1,ts,ξi-1sξi16s3+k=0i-1βk(16ξk3-12ξk2s+12ξks2)+16k=im-2βks3·1-k=0m-1βk-1,ts,ξi-1sξi,for i=1,2,,m-1.

Proof.

For ξi-1sξi, i=1,2,,m-1, we suppose(56)H(t,s)=a3t3+a2t2+a1t+a0,ts,b3t3+b2t2+b1t+b0,ts.Considering the properties of Green’s function together with the boundary condition (54), we have(57)a3s3+a2s2+a1s+a0=b3s3+b2s2+b1s+b0,3a3s2+2a2s+a1=3b3s2+2b2s+b1,6a3s+2a2=6b3s+2b2,6a3-6b3=1,b3=0,6b3+2b2=0,3b3+2b2+b1=0,a0=k=0i=1βka3ξk3+a2ξk2+a1ξk+a0+k=im-2βkb3ξk3+b2ξk2+b1ξk+b0.Hence, (58)a3=16,a2=-12s,a1=12s2,a0=k=0i-1βk16ξk3-12ξk2s+12ξks2+16k=im-2βks3·1-k=0m-1βk-1,b3=0,b2=0,b1=0,b0=16s3+k=0i-1βk16ξk3-12ξk2s+12ξks2+16k=im-2βks3·1-k=0m-1βk-1.The proof of Lemma 12 is completed.

Lemma 13.

One can see that H(t,s)0, t,s[0,1].

Proof.

For ξi-1sξi, i=1,2,,m-1, (59)H(t,s)t=12s-t2,ts,ξi-1sξi,0,ts,ξi-1sξi.Then, H(t,s)/t0, 0t, s1, which implies that H(t,s) is increasing on t. The fact that(60)H0,s=k=0i-1βk16ξk3-12ξk2s+12ξks2+16k=im-2βks3·1-k=0m-1βk-10ensures that H(t,s)0, t,s[0,1].

Lemma 14.

If y(t)0, t[0,1], and u(t) is the solution of problem (53),  (54), then (61)min0t1utγ1max0t1ut,where γ1=i=1m-2βiξi/(1-i=1m-2βi(1-ξi)).

Proof.

It follows from the same methods as Lemma 7 that u(t) is concave on [0,1]. Taking into account that u(1)=0, one see that u(t) is increasing on [0,1] and(62)max0t1u(t)=u(1),min0t1u(t)=u(0).From the concavity of u(t), we have (63)ξi(u(1)-u(0))u(ξi)-u(0).Multiplying both sides with βi and considering the boundary condition, we have (64)i=1m-2βiξiu(1)1-i=1m-2βi1-ξiu(0).Thus, (65)min0t1utγ1max0t1ut.Problem (1),  (3) has a solution u=u(t) if and only if u solves the operator equation (66)u(t)=01H(t,s)a(s)f(u)ds:=A1u(t),0t1.Denote the cone (67)K1={uE:u0,min0t1u(t)γ1u}.It is obvious that A1(K1)K1. We denote (68)M1=max0t101H(t,s)a(s)ds,m1=min0t101H(t,s)a(s)ds.

Theorem 15.

Problem (1),  (3) has at least one positive solution if

f0=0, f=,

f0=, f=0.

Theorem 16.

Problem (1),  (3) has at least one positive solution in one of the following cases:

f0=a(0,+), f=b(0,+), and aM1<1, bm1γ1>1;

f0=a(0,+), f=b(0,+), and bM1<1, am1γ1>1;

f0=a(0,+), f=, and aM1<1;

f0=a, f=0, and am1γ1>1;

f0=0, f=b(0,+), and bm1γ1>1;

f0=, f=b, and bM1<1.

Theorem 17.

Assume following conditions are satisfied:

f0=f=0;

there exist r>0 such that f(u)r/m1 for u[r,r/γ1];

then, problem (1),  (3) has at least two positive solutions u1, u2 satisfying (69)0u1ru2.

Theorem 18.

Assume following conditions are satisfied:

f0=f=;

there exists R>0 such that f(u)R/M1 for u[R,R/γ1];

then, problem (1),  (3) has at least two positive solutions u1,u2 satisfying (70)0u1Ru2.

5. Examples

In this section, we given some examples to illustrate the main results established in this paper.

Example 1.

Consider the fourth-order boundary value problem (71)u(4)(t)+π1-t2uα(t)=0,0t1,u′′′(0)=u′′(0)=u(0)=0,u(1)=12u23,where α(0,1)(1,). Then, problems (71) have at least one positive solution. In fact, when α(0,1), f0= and f=0; when α(1,), f0=0 and f0=. From Theorem 8, there exists at least one positive solution of problem (71).

Example 2.

Consider the fourth-order boundary value problem (72)u(4)+1-t1/2u1/2+u5/3=0,0t1,u′′′(1)=u′′(1)=u(1)=0,u(0)=23u12.We see that (73)H(t,s)=16t3-12st2+12s2t+13s3,ts,0s23,12s3,ts,0s23,16t3-12st2+12s2t+12s2-14s+124,ts,23s1,16s3+12s2-14s+124,ts,23s1,γ1=12,M=max0t101H(t,s)a(s)ds=371920,m=min0t101H(t,s)a(s)ds=313840.It is easy to check that f0=f=. Taking R=8, we see (74)f(u)1536037,u[8,16].From Theorem 18, there exist at least two positive solutions of problem (72) such that (75)u18u2.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work is sponsored by the NSFC (11201109), Anhui Provincial Natural Science Foundation (1408085QA07), and the Higher School Natural Science Project of Anhui Province (KJ2014A200).

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