AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 951824 10.1155/2014/951824 951824 Research Article Nodal Solutions for Some Second-Order Semipositone Integral Boundary Value Problems Lu Huiqin Wang Yang http://orcid.org/0000-0002-4911-9016 Liu Yansheng Infante Gennaro 1 School of Mathematical Sciences Shandong Normal University Jinan, Shandong 250014 China sdnu.edu.cn 2014 2742014 2014 22 12 2013 31 03 2014 27 4 2014 2014 Copyright © 2014 Huiqin Lu et al. 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.

Using bifurcation techniques, we first prove a global bifurcation theorem for nonlinear second-order semipositone integral boundary value problems. Then the existence and multiplicity of nodal solutions of the above problems are obtained. Finally, an example is worked out to illustrate our main results.

1. Introduction

In this paper, we consider the existence and multiplicity of nodal solutions for the following nonlinear second-order semipositone integral boundary value problems (BVP for short): (1)x′′(t)+λf(x(t))=0,0<t<1,x(0)=0,x(1)=01a(s)x(s)ds, where λ>0 is a parameter, fC(R,R), and aL[0,1] is nonnegative with 0<01a2(s)ds<1.

Boundary value problems with integral boundary conditions for ordinary differential equations arise in different areas of applied mathematics and physics. Moreover, they include two, three, multipoint, and nonlocal boundary value problems as special cases. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to  and the references therein.

In , utilizing the fixed point index and Leray-Schauder degree theory, Zhang and Sun obtained some existence results for multiple solutions including sign-changing solutions under some technical hypotheses for the following integral boundary value problem: (2)x′′(t)+f(x(t))=0,0<t<1x(0)=0,x(1)=01a(s)x(s)ds, where fC(R,R), f(0)=0, aL[0,1] is nonnegative with 01a2(s)ds<1.

The purpose of this paper is to investigate the existence and multiplicity of sign-changing solutions of BVP (1), having a given number of zeros (so-called “nodal solution”). The existence of such solutions has been investigated for many types of nonlinear Sturm-Liouville problems with separated boundary conditions and multipoint boundary conditions in many recent papers; see . Recently, Sun et al.  studied for the following m-point boundary value problems: (3)u′′(t)+f(u(t))=0,0<t<1u(0)=0,u(1)=i=1m-2αiu(ηi), where λ>0 is a parameter, fC(R,R), f(0)=0, αi(0,1), i=1,2,,m-2, 0<i=1m-2αi<1, 0<η1<η2<<ηm-2<1. By using Rabinowitz’s global bifurcation theorem, they obtained the existence and multiplicity of nodal solutions when f0(0,+), where (4)f0=limt0f(t)t. To the authors’ knowledge, there are few papers that have considered the existence of nodal solutions for integral boundary value problems. In , Zhang and Sun have obtained sign-changing solutions of BVP (2), but no information is obtained regarding the number of zeros of the solution.

Motivated by [1, 15, 16], in this paper we investigate the existence and multiplicity of nodal solutions for BVP (1). The main features of this paper are as follows. First, the nonlinear term f is semipositone, and f0(0,+), where f0 is defined as in (4). Next, the methods used here are Rabinowitz’s global bifurcation theorem and some of the techniques used in , which are entirely different from [1, 7, 8]. Finally, the results we obtained are the existence of at least any given even number of nodal solutions.

Now we give some notations and a global bifurcation theorem which will be used in Section 3. Let E be a real Banach space; Rabinowitz studied a nonlinear eigenvalue problem of the form (5)u=λLu+H(λ,u), where λ>0 is a parameter, L:EE is a compact linear map, H:εR×EE is completely continuous, and H(λ,u)=o(u) for u near 0 uniformly on bounded λ intervals. A solution of (5) is a pair (λ,u)ε which satisfies (5). The closure of the set of nontrivial solutions of (5) is denoted by . If there exist μR+=[0,+) and 0vE such that v=μLv, μ is said to be a positive eigenvalue of L and v is said to be an eigenfunction corresponding to μ. The set of positive eigenvalues of L will be denoted by r(L). The algebraic multiplicity of μr(L) is dimj=1N((I-μL)j), where N(A) denotes the null space of A. The following was shown in Theorem 1.3 and Theorem 1.25 of Rabinowitz  and Theorem 2 of Dancer .

Theorem A.

If μr(L) is simple, then possess a maximal subcontinuum ζμ which can be decomposed into two subcontinua Cμ+ and Cμ- such that, for some neighborhood B of (μ,0), (λ,u)Cμ+(Cμ-)B, and (λ,u)(μ,0) imply (λ,u)=(λ,αv+w), where α>0(α<0) and |λ-μ|=o(1), w=o(|α|) for α near 0. Moreover, either Cμ+ and Cμ- are both unbounded or Cμ+Cμ-{(μ,0)}.

This paper is arranged as the follows: some preliminaries and some lemmas are given including the study of the eigenvalues and eigenfunctions of the linearization of BVP (1) in Section 2. The main results are proved by using Theorem A in Section 3. A concrete example is given to illustrate the application of the main results in Section 4.

2. Some Preliminaries and Lemmas

Let X=C[0,1] with the norm x=maxt[0,1]|x(t)|, Y={xC1[0,1], x(0)=0, x(1)=01a(s)x(s)ds} with the norm x1=max{x,x}, Z={xC2[0,1], x(0)=0, x(1)=01a(s)x(s)ds} with the norm x2=max{x,x,x}. Then X, Y, Z are Banach spaces.

For any C1 function x, if x(t0)=0, then t0 is said to be a simple zero of x if x(t0)0. For any integer k1 and any ν{±}, as in , we define sets TkνZ consisting of the set of functions xZ satisfying the following conditions:

x(0)=0, νx(0)>0, and x(1)0;

x has only simple zeros in (0,1) and has exactly k such zeros;

x has a zero strictly between each two consecutive zeros of x.

Note that Tk-=-Tk+ and let Tk=Tk+Tk-. It is easy to see that the sets Tk+ and Tk- are disjoint and open in Z. Let E=R×Y under the product topology, Φk+=R×Tk+, Φk-=R×Tk-, and Φk=R×Tk.

In the following, we give some information on the spectrum structure of the linear integral boundary value problem corresponding to BVP (1): (6)x′′(t)+λx(t)=0,0<t<1x(0)=0,x(1)=01a(s)x(s)ds.

Define the operators K on Y by (7)(Kx)(t)=01k(t,s)x(s)ds, where (8)k(t,s)=G(t,s)+t01G(τ,s)a(τ)dτ1-01sa(s)ds,G(t,s)={t(1-s),0ts1s(1-t),0st1.

It is easy to prove the following lemma.

Lemma 1.

The linear operator K:YY is completely continuous. Moreover, (λ,x)(0,)×C2[0,1] is a solution of (6) if and only if (λ,x)E is a solution of the operator equation x=λKx.

We now define a function G:(0,+)R by (9)G(η)=sinη-01a(s)sin(ηs)ds,η(0,+).

Lemma 2.

All the zeros of G(η) are simple.

Proof.

Suppose that η is a double zero of G(η); that is, (10)sinη=01a(s)sin(ηs)ds,cosη=01sa(s)cos(ηs)ds. Hence, (11)1=(01a(s)sin(ηs)ds)2+(01sa(s)cos(ηs)ds)201a2(s)ds01sin2(ηs)ds+01s2a2(s)ds01cos2(ηs)ds01a2(s)ds<1, which shows that (10) cannot hold, and so G(η) has only simple zeros.

Lemma 3.

Suppose that a(s) is symmetrical in [0,1]. Then G(η) has no zero on [0,π/2] and, for each i1, G(η) has exactly one zero ηi=2iπ on I2i:=((2i-(1/2))π,(2i+(1/2))π).

Proof.

Since 0<01a2(s)ds<1 and 01a(s)ds(01a2(s)ds)1/2, we have (12)0<01a(s)ds<1.

Now, (12) implies that G(η)>0 on (0,π/2]; that is, G(η) has no zero in this interval, and also (13)G((2i-12)π)<0,G((2i+12)π)>0,i1.

For each integer i1, by the symmetry of a(s) in [0,1], we have (14)01a(s)sin(2iπs)ds=01a(1-t)sin[2iπ(1-t)]dt=-01a(1-t)sin2iπtdt=-01a(t)sin2iπtdt. So, 201a(s)sin(2iπs)ds=0; that is, 01a(s)sin(2iπs)ds=0.

Thus, G(2iπ)=0. That is, G(η) has one zero ηi=2iπ on each interval I2i((2i-(1/2))π,(2i+(1/2))π).

For any fixed integer i1, suppose that G(η) has another zero η¯i on I2i. In view of the continuity of G(η) and (13), then G(η) has the third zero η~i on I2i. Without loss of generality, we may assume that η¯i<η~i. We have the following three cases to consider.

Consider η¯i<ηi<η~i. By (9) and (12), we have (15)G(ηi)=G(2iπ)=cos(2iπ)-01sa(s)cos(2iπs)ds=1-01sa(s)cos(2iπs)ds>0.

From (13) and Lemma 2, it is easy to see that G(ηi)<0, which contradicts to (15).

Consider η¯i<η~i<ηi. From (13) and Lemma 2, it is easy to see that G(η~i)<0. So, we have (16)sin(η~i)=01a(s)sin(η~is)ds,0<cos(η~i)<01sa(s)cos(η~is)ds. Hence, (17)1<(01a(s)sin(η~is)ds)2+(01sa(s)cos(η~is)ds)201a2(s)ds01sin2(η~is)ds+01s2a2(s)ds01cos2(η~is)ds01a2(s)ds<1, which is a contradiction.

Consider ηi<η¯i<η~i. Similar to the proof of Case (ii), we can also lead to a contradiction.

Therefore, G(η) has exactly one zero ηi=2iπ on I2i((2i-(1/2))π,(2i+(1/2))π) for each i1.

As the proof of Lemma 4 in , it is easy to obtain the following lemma.

Lemma 4.

(1) For each k1, ηkIk((k-(1/2))π,(k+(1/2))π) is one zero of G(η) if and only if λk=ηk2 is an eigenvalue of K. In addition, φk(t)=sin(ηkt) is an eigenfunction corresponding to λk and φk(t)Tk+.

(2) The algebraic multiplicity of each positive eigenvalue λn  (n=1,2,) of K is 1.

Lemma 5.

Suppose that a(s) is symmetrical in [0,1]. Then

there exists a subsequence {λni} of the eigenvalue sequence {λn} of K such that λni=(2iπ)2  (i=1,2,), and the eigenfunction φni corresponding to λni is φni(t)=sin(2iπt);

φni(t)T2i+ for i=1,2,.

Proof.

From Lemmas 3 and 4, conclusion (1) can be obtained immediately. Noticing that φni(t)=sin(2iπt), i=1,2,, it is easy to check that φni(t)T2i+ for i=1,2,.

Define the operators F and A on Y by

( F x ) ( t ) = f ( x ( t ) ) and Ax(t)=(KFu)(t) for t[0,1], respectively, where the operator K is defined as in (7).

It is easy to see that A:YY is completely continuous. By direct computation, we can easily get the following lemma.

Lemma 6.

( λ , x ) ( 0 , ) × C 2 [ 0,1 ] is a solution of BVP (1) if and only if (λ,x)E is a solution of equation (18)x=λAx.

For yY, by the mean-value theorem for the integral, there exists a point ξ(0,1) such that (19)y(1)=01a(s)y(s)ds=y(ξ)01a(s)ds. Let x0(t)=1, ω0(t)=K(x0(t)) for each t[0,1] and e(t)=ξ(1-ξ)t01a(s)ds/(1-ξ01a(s)ds). Let ω(t)=M0ω0(t) for each t[0,1], where M0>0 is a constant to be defined later. The set W is defined by (20)W={xXx(t)+ω(t)x(t)+ω(t)e(t),t[0,1]}. Obviously WX is a closed convex set, and, for each τ>0, (21)τW={y=τxxW}={yX1τy(t)+ω(t)1τy(t)+ω(t)e(t),t[0,1]}={yXy(t)+τω(t)y(t)+τω(t)e(t),t[0,1]}.

Lemma 7.

Let M0 be a positive number such that f(x)-M0, for each xR. Then

A:YW;

for each 0<τ1<τ2<, τ1Wτ2W.

Proof.

(1) For each xY, let y=Ax+ω=L(Fx+M0x0). By direct computation we have (22)y′′(t)+(f(x(t))+M0)=0,0<t<1y(0)=0,y(1)=01a(s)y(s)ds. Since f(x(t))+M00, then y′′(t)0, and so y is a concave function on [0,1]. From (7), it is easy to see that (23)y(1)=11-01sa(s)ds×01G(τ,s)a(τ)[f(x(s))+M0]dτds0. Using the concavity of y and the boundary condition y(0)=0, y(1)0, we can see that y(t)0 for each t[0,1] and y=maxt[0,1]y(t); we have from the concavity of y that (24)y(t)y(1)-y(ξ)1-ξ(t-1)+y(1),t[0,ξ]. By (19), we have y(ξ)=y(1)/01a(s)ds. Hence, (25)y(t)y(1)01a(s)(1-ξ)ds+(1-t)(1-01a(s)ds)01a(s)(1-ξ)dsy(1)01a(s)(1-ξ)ds+(1-01a(s)ds)01a(s)(1-ξ)ds=y(1)1-ξ01a(s)ds01a(s)(1-ξ)ds,t[0,ξ]. From the concavity of y, we have for each t[ξ,1] that (26)y(t)y(ξ)ξty(ξ)ξy(1)1-ξ01a(s)ds01a(s)ξ(1-ξ)ds. It follows from (25) and (26) that (27)y(1)ξ(1-ξ)01a(s)ds1-ξ01a(s)dsy. Then we have from the concavity of y that (28)y(t)(y(1)-y(0))t=y(1)tξ(1-ξ)01a(s)ds1-ξ01a(s)dsyt=ye(t); that is, (29)Ax(t)+ω(t)Ax(t)+ω(t)e(t),t[0,1]. This implies that A:YW, and, therefore, conclusion (1) holds.

(2) Since Aθ=θ, from (1), we see that (30)ω(t)=Aθ(t)+ω(t)Aθ(t)+ω(t)e(t)=ω(t)e(t),t[0,1]. For each xτ1W, we have (31)x(t)+τ1ω(t)x(t)+τ1ω(t)e(t),t[0,1]. Then by (30) and (31), we have (32)x(t)+τ2ω(t)=x(t)+τ1ω(t)+(τ2-τ1)ω(t)[x(t)+τ1ω(t)+(τ2-τ1)ω(t)]e(t)x(t)+τ2ω(t)e(t),t[0,1]. This implies that xτ2W. Thus, τ1Wτ2W.

3. Main Results Theorem 8.

Suppose that a(s) is symmetrical in [0,1], f0(0,), and there exists M0>0 such that f(x)-M0 for each xR. Then for each integer i>0 and each ν=+, or −, there exists an unbounded maximal subcontinuum Cniν of solutions of BVP (1) in Φ2iν{((2iπ)2/f0,0)}, which meets {((2iπ)2/f0,0)} in Σ and satisfies

Cni+({λ}×Y) for each i1,  λ(2iπ)2/f0;

Cni-({λ}×Y) for each i1,  λ(2iπ)2/f0.

Proof.

Since f0(0,), the operator equation (18) can be rewritten as (33)x=λf0Kx+H(λ,x). Here H(λ,x)=λAx-λf0Kx and K is defined as in (7). Obviously, it is easy to see that H(λ,u)=o(u1) for u near 0 uniformly on bounded λ intervals. Notice that K is a compact linear map on Y. A solution of BVP (1) is a pair (λ,x)E. By f0(0,), the known curve of solutions {(λ,0)λR+} will henceforth be referred to as the trivial solutions. The closure of the set on nontrivial solutions of BVP (1) will be denoted by Σ as in Theorem A.

If H(λ,x)0, then (33) becomes a linear system (34)x=λf0Kx. By Lemmas 3, 4, and 5, (34) possesses an increasing subsequence {λni/f0}={(2iπ)2/f0} of simple eigenvalues sequence {λn/f0} and (2iπ)2/f0+ as i+. Any eigenfunction φk(t)=sin(ηkt) corresponding to λk/f0 is in Tk+. Moreover, φni(t)T2i+ for i=1,2, and φk(t)T2i+ for kni.

Consider (33) as a bifurcation problem from the trivial solution. From Theorem A and f0(0,), it follows that, for each integer i1, possess a maximal subcontinuum CniE which can be decomposed into two subcontinua Cni+,  Cni- such that, for some neighborhood B of ((2iπ)2/f0,0), (35)(λ,x)Cni+(Cni-)B,(λ,x)((2iπ)2f0,0), implying (λ,x)=(λ,αφni+w), where α>0  (α<0) and |λ-((2iπ)2/f0)|=o(1), w1=o(|α|) for α near 0.

By (18) and the continuity of the operator A:YZ, the set Cniν lies in R×Z and the injection CniνR×Z is continuous. Moreover, note that Cniν({0}×Z)=. So, Cniν is also a continuum in R×Z, and the above properties hold in R×Z.

Since T2i is open in Z and φni(t)T2i+, we know that (36)xα=φni+wαT2i+ for 0α sufficiently small. Then there exists ϵ0>0 such that, for ϵ(0,ϵ0), we have (37)(λ,x)Φ2i,(Cni{((2iπ)2f0,0)})  BϵΦ2i, where Bϵ is an open ball in R×Z of radius ϵ centered at ((2iπ)2/f0,0). Since T2iν is open in Z, it can follow, similar to the proof of Proposition 4.1 in , that (38)(λ,x)Cni(R×T2i)x=0, which means Cni{((2iπ)2/f0,0)}Φ2i=. Consequently, Cni lies in Φ2i{((2iπ)2/f0,0)}.

Similarly we can obtain that Cniν lies in Φ2iν{((2iπ)2/f0,0)}  (ν=+ or -). Noticing that Tk+Tk-=, it can be obtained that Cni+Cni-=((2iπ)2/f0,0). From Theorem A, we know that Cni+ and Cni- are unbounded in R×Z.

Let λ(2iπ)2/f0 be fixed. For each 0<τ<λ and xZ, (τ,x) is a solution of (18), and by Lemma 7, x=τAxτWλW. Thus, (39)x(t)+λω(t)x(t)+λω(t)e(t),t[0,1], since (40)ω(t)=M0ω0(t)M0t1-ξ01a(s)ds01(1-s)ds<(M0+1)t2(1-ξ01a(s)ds)c1e(t),t[0,1], where c1(M0+1)/2ξ01a(s)(1-ξ)ds, and so (41)x(t)(x-λω-λc1)e(t),t[0,1].

Let R(λ)=2(λω+λc1). Then for each (τ,x)Σ, τ0, xR(λ), we have x(t)0 for t[0,1]. This implies that (42)Cni+([0,λ]×{xYx=R(λ)})=(i1),Cni-([0,λ]×{xYx=R(λ)})=(i1). Thus, the conclusion holds and the proof is complete.

Immediately, from Theorem 8, we have the following result.

Theorem 9.

Suppose that all the conditions of Theorem 8 hold. Then, for each λ>(2iπ)2/f0, BVP (1) has at least 2i nodal solutions un1+,un1-,un2+,un2-,,uni+,uni- in Y such that uni+ has (2i-1) zeros in (0,1) and is positive near t=0 and uni- has (2i-1) zeros in (0,1) and is negative near t=0.

4. An Example

Consider the following nonlinear second-order integral boundary value problem: (43)x′′(t)+f(x(t))=0,0<t<1x(0)=0,x(1)=01(s-s2)x(s)ds, where f(x)=104π2sin(x).

By direct computation, it is easy to see that f0=limx0(104π2sin(x)/sin(x))=104π2, so, (2iπ)2/f0=4i2π2/104  π2=i2/26.

Next, we check that all the conditions of Theorem 9 hold. Take a(s)=s-s2. It is clear that 0<01a2(s)ds=1/30<1, and a(s) is symmetrical in [0,1], and a(s)L[0,1] is nonnegative. Since f(x)=104π2sin(x)104π2, f0=104π2(0,), and λ=1. It follows from Theorem 9, when i2/26<1, we have i=1,2,3,4,5, so the boundary value problem (43) has at least 10 nodal solutions in C1[0,1].

Conflict of Interests

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

Acknowledgment

This research is supported by the Reward Fund for Excellent Young and Middle-Aged Scientists of Shandong Province (BS2011SF022), China.

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