# Existence of Nontrivial Solutions for Some Second-Order Multipoint Boundary Value Problems

Authors: Hongyu Li; Junting Zhang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

By using fixed point theorems with lattice structure, the existence of negative solution and sign-changing solution for some second-order multipoint boundary value problems is obtained.


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## Body

## 1. Introduction

In this paper, the following second-order ordinary differential equation will be considered:(1)-x′′t=φt,xt,0≤t≤1,subject to the multipoint boundary condition (2)x0=0,x1=∑i=1m-2βixαi,where βi>0,i=1,2,⋯,m-2;∑i=1m-2βi<1;0<α1<α2<⋯<αm-2<1, and ∑i=1m-2βiαi<1.The multipoint boundary value problems of ordinary differential equations arise in different areas of applied mathematics and physics. In 1992, Gupta studied nonlinear second-order three-point boundary value problems (see [1]). Since then, different types of nonlinear multipoint boundary value problems have been studied. Up to now, many great achievements about multipoint boundary value problems have been made. For example, many authors have investigated the existence of nontrivial solutions for nonlinear multipoint boundary value problems. Most of them have used upper and lower solution method, fixed point index theory, Guo-Krasnosel’skii fixed point theorem, bifurcation theory, fixed point theorems on cones, and so on (see [2–27] and references therein). For instance, in [2], the author considered the second-order multipoint boundary value problem (3)y′′t+fy=0,0≤t≤1,y0=0,y1=∑i=1m-2αiyηi. By using fixed point index and Leray-Schauder degree methods, the author showed existence of multiple sign-changing solutions for the boundary value problem (3). In [14], the authors have considered the following multipoint boundary value problem: (4)-Lφt=λft,φt,0≤t≤1,φ′0=0,φ1=∑i=1m-2βiφηi. The authors have used global bifurcation method to obtain the existence of positive solution of the boundary value problem (4).In recent years, some authors combine the theory of lattice and the theory of topological degree, so they have obtained some fixed point theorems with lattice structure for nonlinear operators which are not assumed to be cone mappings (see [28–34]). At present, a few authors have used those fixed point theorems with lattice structure to study boundary value problems (see [6, 17, 28–37]). For example, in [35], by using fixed point theorems with lattice structure, the authors considered the existence of positive solution and sign-changing solution for integral boundary value problem under sublinear condition. In [37], the authors considered the existence of positive solution for fourth-order differential equation with fixed point theorems with lattice structure. In [6], the author considered the following second-order three-point boundary value problem: (5)-u′′t=gt,ut,0≤t≤1,u0=0,u1=αuβ, where g:[0,1]×(-∞,+∞)→(-∞,+∞) is continuous, 0<α<1,0<β<1. The author used fixed point theorems with lattice structure to study the existence of sign-changing solutions for the boundary value problem (5) under the unilaterally asymptotically linear condition.Motivated by [6, 17, 28–37], we shall study the existence of nontrivial solutions for the boundary value problem (1), (2). In this paper, we assume that the nonlinear term satisfies superlinear conditions concerning the first eigenvalue corresponding to the relevant linear operator. The method we use is fixed point theorems with lattice structure. And we obtain the sufficient condition about the existence of negative solution and sign-changing solution for the boundary value problem (1), (2). The method is different from those of [2, 4]. And the main results are different from those of the work [2, 4]. This paper is arranged as follows. In Section 2, we give some definitions and fixed point theorems with lattice structure. In Section 3, we shall give some lemmas and the main results about the existence of nontrivial solutions (including negative solution and sign-changing solution) for the boundary value problem (1), (2). Finally, in Section 4, some examples are given to illustrate our main results.


## 2. Preliminaries

LetE be an ordered Banach space in which the partial ordering ≤ is induced by a cone P⊂E. P is called normal if there exists a positive constant N>0 such that θ≤u≤v implies u≤Nv. P is called solid if intP≠θ, i.e., P has nonempty interior. P is called total if E=P-P¯. If P is solid, then P is total. For the concepts and the properties about the cones, we refer to [31, 38, 39].We callE a lattice under the partial ordering ≤, if sup{u,v} and inf{u,v} exist for arbitrary u,v∈E.Foru∈E, let (6)u+=supu,θ,u-=sup-u,θ.u+ and u- are called positive part and negative part of u, respectively. Taking u=u++u-, then u∈P. For the definition and the properties of the lattice, we refer to [40].For convenience, we use the following notations:(7)u+=u+,u-=-u-, and clearly (8)u+∈P,u-∈-P,u=u++u-.Definition 1 (see [28–31]).
LetD⊂E and F:D→E be a nonlinear operator. If there exists u∗∈E such that (9)Fu=Fu++Fu-+u∗,∀u∈D, then F is said to be quasi-additive on lattice.
LetB:E→E be a bounded linear operator. If B(P)⊂P, then the operator B is called to be positive.
In this section, we assume thatE is a Banach space, P is a total cone, the partial ordering ≤ in E is induced by P, and E is a lattice in the partial ordering ≤.
LetB:E→E be a positive completely continuous linear operator; B∗ the conjugated operator of B; r(B) a spectral radius of B; and P∗ the conjugated cone of P. Since P⊂E is a total cone, by Krein-Rutman theorem, we can infer that if r(B)≠0, then there exist u¯∈P∖{θ} and f∗∈P∗∖{θ}, such that (10)Bu¯=rBu¯,B∗f∗=rBf∗.
Forδ>0. Let (11)Pf∗,δ=u∈P∣f∗u≥δu. Then P(f∗,δ) is also a cone in E.Definition 2 (see [30, 31, 41]).
If there existu¯∈P∖{θ},f∗∈P∗∖{θ}, and δ>0 such that (10) holds, and B maps P into P(f∗,δ), then the positive linear operator B is said to satisfy H condition.
LetP be a cone of a Banach space E. If u∈(P∖{θ}) is a fixed point of A, then u is said to be a positive fixed point of A. If u∈((-P)∖{θ}) is a fixed point of operator A, then u is said to be a positive fixed point of operator A. If u∈(P∖{θ}) is a fixed point of operator A, then u is said to be a negative fixed point of operator A. If u∉(P∪(-P)) is a fixed point of operator A, then u is said to be a sign-changing fixed point of operator A.
In [30], Sun and Liu considered computation for the topological degree about superlinear operators which are not cone mappings and obtained the following results.Lemma 3.
Let the coneP⊂E be solid, and A:E→E be a completely continuous operator, and A=BF, where B is a positive completely continuous linear operator satisfying H condition and F is quasi-additive on lattice. Assume that
( i ) there exist c1>r-1(B) and u1∈P such that (12)Fu≥c1u-u1,∀u∈P;(ii) there exist 0<c2<r-1(B) and u2∈P such that (13)Fu≥c2u-u2,∀u∈-P;(iii)Aθ=θ, theFréchet derivative Aθ′ of A at θ exists, and 1 is not an eigenvalue ofAθ′.
Then the operatorA has at least one nonzero fixed point.
In [31], Sun further obtained the following result about the existence of sign-changing fixed points for superlinear operators.Lemma 4.
Let the conditions in Lemma3 hold, and β denote the sum of the algebraic multiplicities for all eigenvalues of Aθ′ lying in (1,+∞). In addition, assume that
( i v )   β ≠ 0, β is an even number;
( v )   A ( P ∖ { θ } ) ⊂ i n t P , A ( ( - P ) ∖ { θ } ) ⊂ i n t ( - P ) .Then the operatorA has at least one negative fixed point and one sign-changing fixed point.


## 3. Main Results

For convenience, we list the following conditions.( C 1 )   φ : [ 0,1 ] × R 1 → R 1 is continuous, φ(t,0)=0,∀t∈[0,1].( C 2 ) The sequence of positive solutions of the equation (14)siny=∑i=1m-2βisin⁡αiyis(15)0<λ1<λ2<⋯<λn<λn+1<⋯.( C 3 )   l i m x → 0 ( φ t , x / x ) = η uniformly on t∈[0,1].LetX=C[0,1] with supremum norm x=sup0≤t≤1|x(t)|. Set P={x∈X∣x(t)≥0,t∈[0,1]}, the P is a solid cone in X. And under the partial order ≤ which is induced by P, X is a lattice.In the following, we define some operatorsA,B, and Φ: (16)Axt=∫01Kt,sφs,xsds,t∈0,1,(17)Bxt=∫01Kt,sxsds,t∈0,1,(18)Φxt=φt,xt,t∈0,1,where (19)Kt,s=gt,s+t∑i=1m-2βigαi,s1-∑i=1m-2βiαi,(20)gt,s=t1-s,0≤t≤s≤1,s1-t,0≤s≤t≤1.Obviously, A=BΦ, and the nontrivial fixed points of the operator A are nontrivial solutions of the boundary value problem (1), (2) (see [3]).Lemma 5 (see [2]).
Letμ be a positive number, and the linear operator B be defined by (17). Eigenvalues of the linear operator μB are (21)μλ1,μλ2,⋯,μλn,⋯,and algebraic multiplicity of μ/λn is equal to 1, where λn is defined by (C2).Lemma 6.
The linear operatorB satisfies H condition.Proof.
By(C2), Lemma 5, and the definition of the spectral radius, we know that (22)rB=supλ∈1/λn,n=1,2,⋯λ=1λ1>0.
By (20), we have (23)gαi,s≥αi1-αis1-s,∀s∈0,1.(24)gt,s≤s1-s,∀t,s∈0,1.
By (19) and (23), we have (25)Kt,s≥t∑i=1m-2βigαi,s1-∑i=1m-2βiαi≥t∑i=1m-2βiαi1-αis1-s1+∑i=1m-2βi1-αi,∀t,s∈0,1.
From (24) and (25), we have (26)Kt,s≥t∑i=1m-2βiαi1-αi1+∑i=1m-2βi1-αigτ,s,∀τ,t,s∈0,1.
By (19), we have (27)Kt,s≥t∑i=1m-2βigαi,s1-∑i=1m-2βiαi≥t∑i=1m-2βiτgαi,s1-∑i=1m-2βiαi≥t∑i=1m-2βiαi1-αi1+∑i=1m-2βi1-αi∑i=1m-2βiτgαi,s1-∑i=1m-2βiαi,∀τ,t,s∈0,1.Hence, by adding (26) to (27), we have (28)2Kt,s≥t∑i=1m-2βiαi1-αi1+∑i=1m-2βi1-αigτ,s+∑i=1m-2βiτgαi,s1-∑i=1m-2βiαi, i.e., (29)Kt,s≥MtKτ,s,∀τ,t,s∈0,1,where M=∑i=1m-2βiαi(1-αi)/2(1+∑i=1m-2βi(1-αi)).
Let(30)B∗xt=∫01K∗t,sxsds,∀t∈0,1,where K∗(t,s)=K(s,t). Obviously, r(B∗)=r(B)=1/λ1>0. By Krein-Rutman theorem, there exist xt∈P∖{θ} and x∗(t)∈P∖{θ} such that (31)Bxt=rBxt,(32)B∗x∗t=rBx∗t.
By (29) and (32), we obtain (33)x∗s=r-1BB∗x∗s=r-1B∫01K∗s,tx∗tdt=r-1B∫01Kt,sx∗tdt≥Mr-1B∫01tKτ,sx∗tdt=Mr-1B∫01tx∗tdtKτ,s,∀τ,s∈0,1.Set (34)f∗u=∫01x∗tutdt,∀u∈X,t∈0,1.Obviously, f∗∈P∗∖{θ}, and by (34), for u∈X, we have (35)f∗Bu=∫01x∗tButdt=∫01x∗tdt∫01Kt,susds=∫01∫01Kt,sx∗tusdtds=∫01∫01K∗s,tx∗tdtusds=∫01rBx∗susds=rBf∗u.That is, (36)B∗f∗=rBf∗.From (33) and (35), we have (37)f∗Bu=rB∫01x∗susds≥M∫01x∗ttdt∫01Kτ,susds=M∫01tx∗tdtBuτ,∀τ∈0,1. By (37), we have (38)f∗Bu≥δBu,where δ=M∫01tx∗(t)dt>0.
Therefore, from (31), (36), and (38), it is easy to know that the linear operator B satisfies H condition.Theorem 7.
Suppose thatC1, (C2), and (C3) hold. In addition, assume that there exists γ>0 such that(39)liminfx→+∞φt,xx≥λ1+γ,uniformlyont∈0,1;(40)limsupx→-∞φt,xx≤λ1-γ,uniformlyont∈0,1.If η≠λ1,λ2,⋯,λn,⋯, where λi is defined by (C2), then the boundary value problem (1), (2) has at least one nontrivial solution.Proof.
By(C1), we easily know that A:X→X is a completely continuous operator, and B:X→X is a bounded positive linear completely continuous operator (see [3]). By Lemma 6, we know that the linear operator B satisfiesH condition.
Forx∈X, let (41)x+t=xt,xt≥0,0,xt<0,x-t=xt,xt≤0,0,xt>0,and then x(t)=x+(t)+x-(t).
Byφ(t,0)=0, we have (42)Φx=φt,xt=φt,x+t+x-t=φt,x+t+φt,x-t=Φx++Φx-.From (42), we know that Φ is quasi-additive on lattice.
From (39) and (40), there exists C>0 such that (43)φt,xx≥λ1+γ4,∀x≥C,t∈0,1,(44)φt,xx≤λ1-γ4,∀x≤-C,t∈0,1.
By (43) and (44), we have (45)φt,x≥λ1+γ4x,∀x≥C,t∈0,1,(46)φt,x≥λ1-γ4x,∀x≤-C,t∈0,1.
LetC~=max0≤t≤1,x≤Cφt,x. Then by (45) and (46), we have (47)φt,x≥λ1+γ4x-C~,∀x≥0,t∈0,1,φt,x≥λ1-γ4x-C~,∀x≤0,t∈0,1,i.e., (48)Φx≥h1x-C~,∀x∈P,Φx≥h2x-C~,∀x∈-P, where h1=λ1+γ/4,h2=λ1-γ/4. Obviously, we have (49)h1>r-1B,h2<r-1B.
In the following, we prove thatAθ′=ηB.
In fact, byφ(t,0)=0,∀t∈[0,1], we have Aθ=θ. From (C3), ∀ϵ>0,∃δ>0, when 0<x<δ, we have (50)φt,xx-η<ϵ,i.e.,(51)φt,x-ηx<ϵx,∀t∈0,1,0<x<δ.So (52)Φx-ηx≤ϵx,∀x<δ.
Therefore, by (52), we have (53)Ax-Aθ-ηBx=BΦx-ηx≤B·Φx-ηx≤ϵB·x,∀x<δ.So(54)limx→0Ax-Aθ-ηBxx=0,i.e.,(55)Aθ′=ηB.
Sincer(Aθ′)=ηr(B), we know that 1 is not the eigenvalue of Aθ′ by Lemma 6 and (C2).
By the above proof, we know that the conditions of Lemma3 hold. So by Lemma 3, the boundary value problem (1), (2) has at least one nontrivial solution.Theorem 8.
Assume that(C1)-(C3), (39), and (40) are satisfied. In addition, suppose that φ(t,x)x>0,∀t∈[0,1],x≠0, and λ2n0<η<λ2n0+1, where n0 is a natural number. Then the boundary value problem (1), (2) has at least one negative solution and one sign-changing solution.Proof.
By (17), for ∀x∈P∖{θ}, we have (56)Bxt=∫01Kt,sxsds=∫01gt,sxsds+t∑i=1m-2βi∫01gαi,sxsds1-∑i=1m-2βiαi≤t1-t∫01xsds+t∑i=1m-2βi1-∑i=1m-2βiαi∫01xsds=t1-t+t∑i=1m-2βi1-∑i=1m-2βiαi∫01xsds.(57)Bxt=∫01gt,sxsds+t∑i=1m-2βi∫01gαi,sxsds1-∑i=1m-2βiαi≥t∑i=1m-2βi∫01gαi,sxsds1-∑i=1m-2βiαi.
From (56) and (57), we obtain that (58)BP∖θ⊂intP.Similarly, we know that (59)B-P∖θ⊂int-P.Since φ(t,x)x>0,∀t∈[0,1],x≠0, we have φ(t,x)>0,∀x>0,t∈[0,1], and φ(t,x)<0,∀x<0,t∈[0,1]. So we have (60)Φx∈P∖θ,∀x∈P∖θ.(61)Φx∈-P∖θ,∀x∈-P∖θ.By (58)-(61), we have (62)AP∖θ⊂intP,A-P∖θ⊂int-P.
Letβ be the sum of algebraic multiplicities for all the eigenvalues ofAθ′, lying in the interval (1,∞). By (55), Lemma 5, and λ2n0<η<λ2n0+1, we know that (63)β=2n0.
By (62) and (63), we know that the conditions (iv) and (v) in Lemma 4 hold. By the proof of Theorem 7, the conditions (i), (ii), and (iii) in Lemma 4 are satisfied. Therefore, by Lemma 4, the boundary value problem (1), (2) has at least one negative solution and one sign-changing solution.


## 4. Examples

We consider second-order four-point boundary value problem(64)-x′′t=φt,xt,0≤t≤1,x0=0,x1=13x13+12x12.By simple calculations,λ1≈5.602, λ2≈42.32, λ3≈99.97, and λ4≈148.87 are solutions of the equation (65)sin⁡x=13sin⁡x3+12sin⁡x2.Example 1.
Choose(66)φt,x=8x+t-1x,t∈0,1,x∈4,+∞,30+5t3x-1-3t,t∈0,1,x∈1,4,3x-31+tx2,t∈0,1,x∈-1,1,8-t7x+1-6+3t,t∈0,1,x∈-8,-1,2x+t-1x3,t∈0,1,x∈-∞,-8.
By (66), it is easy to know that φ:[0,1]×(-∞,+∞)→(-∞,+∞) is continuous, and φ(t,0)=0,∀t∈[0,1]. By calculation, η=3<λ1. We can choose γ=2. Then we have (67)liminfx→+∞φt,xx=8≥λ1+γ,limsupx→-∞φt,xx=2≤λ1-γ.So by Theorem 7, the boundary value problem (64) has at least one nontrivial solution.Example 2.
Choose(68)φt,x=10x+1-tx3,t∈0,1,x∈8,+∞,31-3t7x-1+51+t,t∈0,1,x∈1,8,50x+1+tx5/3,t∈0,1,x∈-1,1,-21-4t26x+1-51+t,t∈0,1,x∈-27,-1,x+1-tx3,t∈0,1,x∈-∞,-27.
By (68), we know that φ:[0,1]×(-∞,+∞)→(-∞,+∞) is continuous, φ(t,0)=0,∀t∈[0,1], and φ(t,x)x>0,∀t∈[0,1],x≠0. By calculation, λ2≤η=50<λ3. We can choose γ=4. Then we have (69)liminfx→+∞φt,xx=10≥λ1+γ,limsupx→-∞φt,xx=1≤λ1-γ.So by Theorem 8, the boundary value problem (64) has at least one negative solution and one sign-changing solution.Example 3.
Choose(70)φt,x=4532x2+1-tx3,t∈0,1,x∈8,+∞,31-3t7x-1+61+t,t∈0,1,x∈1,8,60x+1+tx5/3,t∈0,1,x∈-1,1,-31-4t26x+1-61+t,t∈0,1,x∈-27,-1,x+1-tx3,t∈0,1,x∈-∞,-27.
By (70), we know that φ:[0,1]×(-∞,+∞)→(-∞,+∞) is continuous, φ(t,0)=0,∀t∈[0,1], and φ(t,x)x>0,∀t∈[0,1],x≠0. By calculation, λ2≤η=60<λ3. We can choose γ=3. Then we have (71)liminfx→+∞φt,xx=+∞≥λ1+γ,limsupx→-∞φt,xx=1≤λ1-γ.So by Theorem 8, the boundary value problem (64) has at least one negative solution and one sign-changing solution.



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