# Nodal Solutions for Some Second-Order Semipositone Integral Boundary Value Problems

Authors: Huiqin Lu; Yang Wang; Yansheng Liu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

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.


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## Body

## 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, f∈C(R,R), and a∈L[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 [1–12] and the references therein.In [1], 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 f∈C(R,R), f(0)=0, a∈L[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 [13–18]. Recently, Sun et al. [16] 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, f∈C(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=lim⁡t→0f(t)t.
To the authors’ knowledge, there are few papers that have considered the existence of nodal solutions for integral boundary value problems. In [1], 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 [16], 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 Section3. 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:E→E is a compact linear map, H:ε≡R×E→E 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 μ∈R+=[0,+∞) and 0≠v∈E 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 μ∈r(L) is dim⁡∪j=1∞N((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 [19] and Theorem 2 of Dancer [20].Theorem A.
Ifμ∈r(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

LetX=C[0,1] with the norm ∥x∥=max⁡t∈[0,1]|x(t)|, Y={x∈C1[0,1], x(0)=0, x(1)=∫01‍a(s)x(s)ds} with the norm ∥x∥1=max⁡{∥x∥,∥x′∥}, Z={x∈C2[0,1], x(0)=0, x(1)=∫01‍a(s)x(s)ds} with the norm ∥x∥2=max⁡{∥x∥,∥x′∥,∥x′′∥}. Then X, Y, Z are Banach spaces.For anyC1 function x, if x(t0)=0, then t0 is said to be a simple zero of x if x′(t0)≠0. For any integer k≥1 and any ν∈{±}, as in [15], we define sets Tkν⊂Z consisting of the set of functions x∈Z satisfying the following conditions:(i)
x(0)=0, νx′(0)>0, and x′(1)≠0;(ii)
x′ has only simple zeros in (0,1) and has exactly k such zeros;(iii)
x has a zero strictly between each two consecutive zeros of x′.Note thatTk-=-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 operatorsK on Y by
(7)(Kx)(t)=∫01‍k(t,s)x(s)ds,
where
(8)k(t,s)=G(t,s)+t∫01‍G(τ,s)a(τ)dτ1-∫01‍sa(s)ds,G(t,s)={t(1-s),0≤t≤s≤1s(1-t),0≤s≤t≤1.It is easy to prove the following lemma.Lemma 1.
The linear operatorK:Y→Y 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 functionG:(0,+∞)→R by
(9)G(η)=sinη-∫01‍a(s)sin(ηs)ds,η∈(0,+∞).Lemma 2.
All the zeros ofG(η) are simple.Proof.
Suppose thatη is a double zero of G(η); that is,
(10)sinη=∫01‍a(s)sin(ηs)ds,cos⁡η=∫01‍sa(s)cos⁡(ηs)ds.
Hence,
(11)1=(∫01‍a(s)sin(ηs)ds)2+(∫01‍sa(s)cos⁡(ηs)ds)2≤∫01‍a2(s)ds∫01‍sin2(ηs)ds+∫01‍s2a2(s)ds∫01‍cos⁡2(ηs)ds≤∫01‍a2(s)ds<1,
which shows that (10) cannot hold, and so G(η) has only simple zeros.Lemma 3.
Suppose thata(s) is symmetrical in [0,1]. Then G(η) has no zero on [0,π/2] and, for each i≥1, G(η) has exactly one zero ηi=2iπ on I2i:=((2i-(1/2))π,(2i+(1/2))π).Proof.
Since0<∫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,i≥1.
For each integeri≥1, 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, 2∫01‍a(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 integeri≥1, 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.(i)
Considerη¯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).(ii)
Considerη¯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)2≤∫01‍a2(s)ds∫01‍sin2(η~is)ds+∫01‍s2a2(s)ds∫01‍cos⁡2(η~is)ds≤∫01‍a2(s)ds<1,
which is a contradiction.(iii)
Considerηi<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 i≥1.As the proof of Lemma 4 in [1], it is easy to obtain the following lemma.Lemma 4.
(1) For eachk≥1, ηk∈Ik≔((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 eigenvalueλn(n=1,2,…) of K is 1.Lemma 5.
Suppose thata(s) is symmetrical in [0,1]. Then(1)
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);(2)
φni(t)∈T2i+ for i=1,2,….Proof.
From Lemmas3 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 operatorsF 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 thatA:Y→Y 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.Fory∈Y, 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)t∫01‍a(s)ds/(1-ξ∫01‍a(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={x∈X∣x(t)+ω(t)≥∥x(t)+ω(t)∥e(t),t∈[0,1]}.
Obviously W∈X is a closed convex set, and, for each τ>0,
(21)τW={y=τx∣x∈W}={y∈X∣1τy(t)+ω(t)≥∥1τy(t)+ω(t)∥e(t),t∈[0,1]}={y∈X∣y(t)+τω(t)≥∥y(t)+τω(t)∥e(t),t∈[0,1]}.Lemma 7.
LetM0 be a positive number such that f(x)≥-M0, for each x∈R. Then(1)
A:Y→W;(2)
for each0<τ1<τ2<∞, τ1W⊂τ2W.Proof.
(1) For eachx∈Y, 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))+M0≥0, 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]dτds≥0.
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∥=max⁡t∈[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)ds≤y(1)∫01‍a(s)(1-ξ)ds+(1-∫01‍a(s)ds)∫01‍a(s)(1-ξ)ds=y(1)1-ξ∫01‍a(s)ds∫01‍a(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)ξt≤y(ξ)ξ≤y(1)1-ξ∫01‍a(s)ds∫01‍a(s)ξ(1-ξ)ds.
It follows from (25) and (26) that
(27)y(1)≥ξ(1-ξ)∫01‍a(s)ds1-ξ∫01‍a(s)ds∥y∥.
Then we have from the concavity of y that
(28)y(t)≥(y(1)-y(0))t=y(1)t≥ξ(1-ξ)∫01‍a(s)ds1-ξ∫01‍a(s)ds∥y∥t=∥y∥e(t);
that is,
(29)Ax(t)+ω(t)≥∥Ax(t)+ω(t)∥e(t),t∈[0,1].
This implies that A:Y→W, and, therefore, conclusion (1) holds.
(2) SinceAθ=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 x∈τ1W, 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 x∈τ2W. Thus, τ1W⊂τ2W.


## 3. Main Results

Theorem 8.
Suppose thata(s) is symmetrical in [0,1], f0∈(0,∞), and there exists M0>0 such that f(x)≥-M0 for each x∈R. 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 (1)
Cni+∩({λ}×Y)≠∅ for each i≥1,λ≥(2iπ)2/f0;(2)
Cni-∩({λ}×Y)≠∅ for each i≥1,λ≥(2iπ)2/f0.Proof.
Sincef0∈(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(∥u∥1) 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)∣λ∈R+} 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.
IfH(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 k≠ni.
Consider (33) as a bifurcation problem from the trivial solution. From Theorem A and f0∈(0,∞), it follows that, for each integer i≥1, ∑‍ possess a maximal subcontinuum Cni⊆E 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), ∥w∥1=o(|α|) for α near 0.
By (18) and the continuity of the operator A:Y→Z, the set Cniν lies in R×Z and the injection Cniν→R×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.
SinceT2i is open in Z and φni(t)∈T2i+, we know that
(36)xα=φni+wα∈T2i+
for 0≠α sufficiently small. Then there exists ϵ0>0 such that, for ϵ∈(0,ϵ0), we have
(37)(λ,x)∈Φ2i,(Cni∖{((2iπ)2f0,0)})∩Bϵ⊂Φ2i,
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 [15], 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 thatCniν 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.
Letλ≥(2iπ)2/f0 be fixed. For each 0<τ<λ and x∈Z, (τ,x) is a solution of (18), and by Lemma 7, x=τAx∈τW⊂λW. Thus,
(39)x(t)+λω(t)≥∥x(t)+λω(t)∥e(t),t∈[0,1],
since
(40)ω(t)=M0ω0(t)≤M0t1-ξ∫01‍a(s)ds∫01‍(1-s)ds<(M0+1)t2(1-ξ∫01‍a(s)ds)≤c1e(t),t∈[0,1],
where c1≥(M0+1)/2ξ∫01‍a(s)(1-ξ)ds, and so
(41)x(t)≥(∥x∥-λ∥ω∥-λc1)e(t),t∈[0,1].
LetR(λ)=2(λ∥ω∥+λc1). Then for each (τ,x)∈Σ, τ≠0, ∥x∥≥R(λ), we have x(t)≥0 for t∈[0,1]. This implies that
(42)Cni+∩([0,λ]×{x∈Y∣∥x∥=R(λ)})=∅(i≥1),Cni-∩([0,λ]×{x∈Y∣∥x∥=R(λ)})=∅(i≥1).
Thus, the conclusion holds and the proof is complete.Immediately, from Theorem8, we have the following result.Theorem 9.
Suppose that all the conditions of Theorem8 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 thatf0=lim⁡x→0(104π2sin⁡(x)/sin⁡(x))=104π2, so, (2iπ)2/f0=4i2π2/104π2=i2/26.Next, we check that all the conditions of Theorem9 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].



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