# On a Singular Second-Order Multipoint Boundary Value Problem at Resonance

Authors: S. A. Iyase; O. F. Imaga
Journal: International Journal of Differential Equations
 (2017)

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## Abstract

The aim of this paper is to derive existence results for a second-order singular multipoint boundary value problem at resonance using coincidence degree arguments.


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## Body

## 1. Introduction

In this paper we derive existence results for the second-order singular multipoint boundary value problem of the form(1)x′′t=ft,xt,x′t+gt,0<t<1,x′0=0,x1=∑i=1m-2aixξi,where f:[0,1]×R2→R is Caratheodory’s function (i.e., for each (x,y)∈R2 the function f(·,x,y) is measurable on [0,1]; for a.e. t∈[0,1], the function f(t,·,·) is continuous on R2). Let ξi∈(0,1), i=1,2,…,m-2, 0<ξ1<ξ2<⋯<ξm-2<1, ai∈(0,1)i=1,2,…,m-2, and ∑i=1m-2‍ai=1, where f and g have singularity at t=1.In [1] Gupta et al. studied the above equation when f and g have no singularity and ∑i=1m-2‍ai≠1. They obtained existence of a C1[0,1] solution by utilising the Leray-Schauder continuation principle. In [2] Ma and O’Regan derived existence results for the same equation when f and g have a singularity at t=1 and ∑i=1m-2‍ai≠1. They also utilised the Leray-Schauder continuation method. These results correspond to the nonresonance case. The purpose of this article is therefore to derive existence results for (1) when ∑i=1m-2‍ai=1 (the resonance case) and when f and g have a singularity at t=1. We shall employ coincidence degree arguments in obtaining our results. In this case, the methods used in [1, 2] are not valid.Research on singular differential equations is important because singular differential equations are useful in the modeling of many problems in the physical and engineering sciences; see [3].In general singular boundary value problems can be difficult to solve because they may blow up near the singularity. The existence and multiplicity of solutions for second-order nonsingular boundary value problems have been extensively studied by many researchers. However to the best of our knowledge the corresponding problem for second-order differential equations at resonance and with a singularity had not received much attention in the literature. For recent results in these directions see [1, 2, 4–9] and references therein.The rest of this paper is organised as follows. In Section2, we present some definitions, lemmas, and theorems necessary for obtaining our main results. In Section 3, we derive some lemmas and the main theorem. In what follows we shall utilise the following assumptions:(A0)
Forξi∈(0,1), i=1,2,…,m-2, 0<ξ1<ξ2<⋯<ξm-2<1 and ∑i=1m-2‍ai=1.(A1)
There exista(t),c(t)∈L1[0,1] with (1-t)a(t),(1-t)c(t),b(t)∈L1[0,1] and ft,x,y≤atx+bty+c(t), a.e., t∈[0,1], (x,y)∈R2.(A2)
g:[0,1]→R is such that ∫01‍1-tgt<∞.


## 2. Preliminaries

In this section we state some definitions, theorems, and lemmas that will be used in the subsequent section.Definition 1.
LetX and Z be real Banach spaces. One says that the linear operator L:dom⁡L⊂X→Z is a Fredholm mapping of index zero if Ker⁡L and Z/Im⁡L are of finite dimension, where Im⁡L denotes the image of L.As a result of Definition1, we will require the continuous projections P:X→X, Q:Z→Z such that Im⁡P=Ker⁡L, Ker⁡Q=Im⁡L, X=Ker⁡L⊕Ker⁡P, Z=Im⁡L⊕Im⁡Q, and Ldom⁡L∩Ker⁡P:dom⁡L∩Ker⁡P→Im⁡L is an isomorphism.Definition 2.
LetL be a Fredholm mapping of index zero and Ω a bounded open subset of X such that dom⁡L∩Ω≠ϕ. The map N:X→Z is called L-compact on Ω¯, if the map QN(Ω¯) is bounded and Kp(I-Q) is compact, where one denotes by Kp:Im⁡L→dom⁡L∩Ker⁡P the generalised inverse of L. In addition N is L-completely continuous if it is L-compact on every bounded Ω⊂X.Theorem 3 (see [10]).
LetL be a Fredholm operator of index zero and let N be L-compact on Ω¯. Assume that the following conditions are satisfied: (i)
Lx≠λNx for every (x,λ)∈[(dom⁡L∖Ker⁡L)∩∂Ω]×(0,1).(ii)
Nx∉Im⁡L, for every x∈Ker⁡L∩∂Ω.(iii)
deg⁡(QNKer⁡L∩∂Ω,Ω∩Ker⁡L,0)≠0, with Q:Z→Z being a continuous projection such that Ker⁡Q=Im⁡L. Then the equation Lx=Nx has at least one solution in dom⁡L∩Ω¯.In what follows, we shall make use of the following classical spaces,C[0,1], C1[0,1], L1[0,1], and L∞[0,1]. Let AC[0,1] denote the space of all absolute continuous functions on [0,1], AC1[0,1] = {x∈C1[0,1]:x′(t)∈AC[0,1]}, Lloc⁡1[0,1] = {x:x[0,d]∈L1[0,d]for  every  compact  interval[0,d]⊆[0,1)}.A
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}.LetZ be the Banach space defined by(2)Z=y∈Lloc⁡10,1:1-tyt∈L10,1, with the norm(3)yZ=∫011-tytdt. Let X be the Banach space (4)X=x∈C10,1:x∈C0,1,limt→1-⁡1-tx′texists,with the norm(5)xX=max⁡x∞,1-tx′t∞wherex∞=supt∈0,1⁡xt.We denote the norm in L1[0,1] by ·1. We define the linear operator L:dom⁡L⊂X→Z by(6)Lx=x′′t,where dom⁡L={x∈X:x′(0)=0,x(1)=∑i=1m-2‍aix(ξi)} and N:X→Z is defined by(7)Nx=ft,xt,x′t+gt.Then boundary value problem (1) can be written as(8)Lx=Nx.Lemma 4 (see [2]).
Lety∈Z. Then (i)
∫0t‍y(s)ds∈L1[0,1].(ii)
limt→1-⁡(1-t)∫0t‍y(s)ds=0.Lemma 5.
If∑i=1m-2‍ai=1 then (i)
Ker⁡L={x∈dom⁡L:x(t)=c,c∈R,t∈[0,1]};(ii)
Im⁡L=y∈Z:∑i=1m-2‍ai∫ξi1‍∫0s‍y(τ)dτds=0;(iii)
L:dom⁡L⊂X→Z is a Fredholm operator of index zero and the continuous operator Q:Z→Z can be defined by (9)Qy=eth∑i=1m-2ai∫ξi1∫0syτdτds,where h=∑i=1m-2‍ai[e+ξi-eξi-1]≠0.(iv)
The linear operatorKp:Im⁡L:→dom⁡L∩Ker⁡P can be defined as (10)Kpy=∫0t∫0syτdτds.(v)
KpyX≤yZ for all y∈Z.Proof.
(i) It is obvious that(11)Ker⁡L=x∈dom⁡L:xt=c,c∈R.
(ii) We show that(12)Im⁡L=y∈Z:∑i=1m-2ai∫ξi1∫0syτdτds=0. To do this, we consider the problem(13)x′′t=yt,and we show that (13) has a solution x(t) satisfying x′(0)=0, x(1)=∑i=1m-2‍aix(ξi) if and only if(14)∑i=1m-2ai∫ξi1∫0syτdτds=0.Suppose (13) has a solution x(t) satisfying x′(0)=0, x(1)=∑i=1m-2‍aix(ξi); then we obtain from (13) that(15)xt=x0+∫0t∫0syτdτds, and applying the boundary conditions we get(16)∑i=1m-2ai∫0ξi∫0syτdτds=∫01∫0syτdτds, since ∑i=1m-2‍ai=1, and using (i) of Lemma 4 we get(17)∑i=1m-2ai∫ξi1∫0syτdτds=0. On the other hand if (14) holds, let x0∈R; then x(t)=x0+∫0t‍∫0s‍y(τ)dτds, where y∈Z and x′(t)∈ACloc⁡[0,1). Then from Lemma 4  ∫0t‍y(τ)dτ∈L1[0,1] and limt→1-(1-t)x′(t) = limt→1-(1-t)∫0t‍y(τ)dτ=0. Hence(18)x′′t=yt.
(iii) Fory∈Z, we define the projection Qy as(19)Qy=eth∑i=1m-2ai∫ξi1∫0syτdτds,t∈0,1, where h=∑i=1m-2‍ai[e+ξi-eξi-1]≠0.
We show thatQ:Z→Z is well defined and bounded.(20)Qyt≤eth∑i=1m-2ai∫011-sysds=1h∑i=1m-2aiyZet,QyZ≤∫011-tQytdt≤1h∑i=1m-2aiyZ∫011-tetdt=1h∑i=1m-2aiyZetZ.In addition it is easily verified that(21)Q2y=Qy,y∈Z. We therefore conclude that Q:Z→Z is a projection. If y∈Im⁡L, then from (14) Qy(t)=0. Hence Im⁡L⊆Ker⁡Q. Let y1=y-Qy; that is, y1∈Ker⁡Q. Then(22)∑i=1m-2ai∫ξi1∫0sy1τdτds=∑i=1m-2ai∫ξi1∫0syτdτds-1h∑i=1m-2∫ξi1∫01yτdτds·h=0. Thus, y1∈Im⁡L and therefore Ker⁡Q⊆Im⁡L and hence Z=Im⁡L+Im⁡Q=Im⁡L+R. It follows that since Im⁡L∩R={0}, then Z=Im⁡L⊕Im⁡Q. Therefore(23)dim⁡Ker⁡L=dim⁡Im⁡Q=dim⁡R=codim⁡Im⁡L=1.This implies that L is Fredholm mapping of index zero.
(iv) We defineP:X→X by(24)Px=x0, and clearly P is continuous and linear and P2x=P(Px)=Px(0)=x(0)=Px and Ker⁡P=x∈X:x0=0. We now show that the generalised inverse Kp:Im⁡L→dom⁡L∩Ker⁡P of L is given by(25)Kpy=∫0t∫0syτdτds. For y∈Im⁡L we have(26)LKpyt=Kpyt′′=yt and for x∈dom⁡L∩Ker⁡P we know that (27)KpLxt=∫0t∫0sx′′τdτds=∫0tt-sx′′sds=xt-x′0t-x0=xtsince x∈dom⁡L∩Ker⁡P, x(0)=0, and Px=0.
This shows thatKp=(Ldom⁡L∩Ker⁡P)-1.
(v)(28)Kpy∞≤maxt∈0,1⁡∫0tt-sysds≤∫011-sysds≤yZ,1-tKpy′∞≤maxt∈0,1⁡∫0t1-sysds≤yZ.We conclude that(29)KpyX≤yZ.Lemma 6.
The operatorN:X→Z defined by (30)Nxt=ft,xt,x′t+gt,t∈0,1is L-completely continuous.Proof.
SupposeΩ is an open bounded subset of X. Let R1=sup⁡xX:x∈Ω. From condition (A1) and each xn∈Ω we have(31)Nxnt≤ft,xnt,xn′t+gt≤atR1+btR11-t+rt+gt=φt. We can deduce from (A1) and (A2) that φ(t)∈Z: (32)QNxnt=eth∑i=1m-2ai∫ξi1∫0sNxnτdτds≤1h∑i=1m-2ai∫011-sNxnxdset≤1h∑i=1m-2ai∫011-sφsdset≤1h∑i=1m-2aiφZet,t∈0,1,QNxnZ≤1h∑i=1m-2aiφZ∫011-tetdt=1h∑i=1m-2aiφZetZ.This shows that QN(Ω¯) is bounded in Z and QN is continuous by using the Lebesgue Dominated Convergence Theorem. Next we show that KP,QN(Ω¯)=KP(I-Q)N(Ω¯) is compact.
By using (31) we derive (33)KPNnt≤∫0tt-sNxnsds≤∫011-sφsds=φZ,Nxnt-QNxnt≤Nxnt+QNxnt≤φt+1h∑i=1m-2aiφZet=αrt,KP,QNxnt=KpNxn-QNxnt≤KPαrt≤αrZ.This indicates that the sequence {KP,QNxn(Ω¯)} is uniformly bounded in C[0,1]. Also for t∈[0,1)(34)1-tKP,QNxn′t=1-t∫0tNxns-QNxnsds≤∫011-sαrsds≤αrZ. Hence the sequence KP,QNxn(t) is bounded in C[0,1] and limt→1-(1-t)(KP,QNxn)′(t)=0. Thus KP,QNxn(t) is bounded in X.
Next we show that the sequenceKP,QNxnt is equicontinuous. Let t1,t2∈[0,1], t1<t2; then (35)KP,Qxnt2-KP,Qxnt1=∫t1t2KP,QNxn′sds≤∫t1t2∫0sNxnτ-QNxnτdτds≤∫t1t2∫0sαrτdτds, for every t1,t2∈[0,1]. By (i) of Lemma 4  ∫0s‍αr(τ)dτ∈L1[0,1]. Thus the sequence KP,QNxnt is equicontinuous on [0,1] and by Arzela-Ascoli Theorem is convergent. Next we prove that the sequence {1-tKP,QNxn)′ is also equicontinuous on [0,1]. We have for t∈[0,1](36)1-tKP,QNxn′t′=-KP,QNxn′t+1-tKP,QNxn′′t≤∫0tNxns-QNxnsds+1-tNxnt≤∫0tαrsds+1-tφt=ψt. Using (i) of Lemma 4 and the fact that αr(t) and φ(t) are in Z we conclude that ψ(t)∈L1[0,1]. Therefore(37)1-t2KP,QNxn′t2-1-t1KP,QNxn′t1=∫t1t21-sKP,QNxn′s′ds≤∫t1t21-sKP,QNxn′s′ds≤∫t1t2ψsds. The sequence {(1-t)(KP,QNxn)′(t)} is therefore equicontinuous on [0,1) and therefore converges to some (1-t)(KP,QNx0)′(t)∈C[0,1] with limt→1-(1-t)[(1-t)(KP,QNxn)′(t)]=0, t∈[0,1).
We then conclude thatKP,Q is relatively compact and since QN(Ω¯) is bounded we conclude from Definition 2 that N is L-compact on every bounded subset Ω of X and hence N is L-completely continuous.


## 3. Main Result

In this section we will state and prove the main existence results for problem (1).Theorem 7.
Assume that the following conditions are satisfied:(H1)
There exists a positive constantB1 such that, for each x∈dom⁡L, if xt>B1 for all t∈[0,1] then (38)QNxt=eth∑i=1m=2ai∫ξi1∫0sfτ,xτ,x′τ+gτdτds≠0.(H2)
There exists a positive constantB2 such that for c∈R and c>B2 either (i)QN(c)≥0 or (ii)QN(c)≤0. Then (1) has at least one solution in X provided(39)aZ+b1<12.To prove Theorem7, we first establish some lemmas.Lemma 8.
LetΩ1={x∈dom⁡L∖Ker⁡L:Lx=λNx,λ∈(0,1)} then Ω1 is bounded in X.Proof.
Letx∈Ω1. We let Lx=λNx, 0<λ<1. Since λ≠0 it is clear that Nx∈Im⁡L=Ker⁡Q; hence QNx=0 for all t∈[0,1]. Therefore by assumption (H1) there exist t0∈[0,1] such that xt0<B1. Now (40)∫0t∫0sx′′τdτds=∫0t0t0-sx′′sds=xt0-x0(41)PxX=x0=xt0+∫0t0t0-sx′′sds≤xt0+∫011-sx′′sds≤B1+LxZ≤B1+NxZ.We note that (I-P)x∈dom⁡L∩Ker⁡P:(42)I-PxX=KPLI-PxX≤KPLxX≤LxZ<NxZ.From (41) and (42) we get(43)xX=Px+I-PxX≤PxX+I-PxX<B1+2NxZ. From the definition of N we obtain(44)NxZ=1-tNxt1=∫011-tft,xt,x′t+gtdt≤∫011-tatxt+bt1-tx′t+1-trt+1-tgtdt≤aZx∞+b11-tx′∞+rZ+gZ. From (43) and (44) we get(45)xX<B1+2aZxX+b1xX+rZ+gZ. Since 1-2[aZ+b1]>0 we obtain that(46)xX<B1+2rZ1-2aZ+b1+2gZ1-2aZ+b1.Therefore Ω1 is bounded in X.Lemma 9.
The setΩ2={x∈Ker⁡L:Nx∈Im⁡L} is a bounded subset of X.Proof.
Letx∈Ω2 with x(t)=c, c∈R. Then QN(c)=0 implies N(c)∈Im⁡L=Ker⁡Q. We therefore derive from (H2) that(47)xX=c=max⁡c,1-t0=c<B2.Lemma 10.
The setsΩ3+={x∈Ker⁡L:λx+(1-λ)QNx=0,λ∈[0,1]} and Ω3-={x∈Ker⁡L:-λx+(1-λ)QNx=0,λ∈[0,1]} are bounded in X provided (H2)(i) and (H2)(ii) are satisfied simultaneously.Proof.
IfQN(c)≥0 then, for x∈Ω3+ with x(t)=c, c∈R, we have(48)λc=-1-λQNc. If λ=0, it follows from (48) that N(c)∈Ker⁡Q=Im⁡L; that is, N(c)∈Ω2, and therefore by Lemma 9. we have xX≤B2. However if λ∈(0,1) and c>B2 then using assumption (H2)(i) we obtain the contradiction(49)λc2=-1-λcQNc≤0. Thus xX=c<B2. Hence Ω3+ is bounded in X. We can use the same argument to prove that Ω3- is also bounded in X.Proof of Theorem7.
We show that the conditions of Theorem3 are satisfied where Ω is an open and bounded set such that ⋃i=13Ωi⊂Ω. It is easily seen that conditions (i) and (ii) of Theorem 3 are satisfied by using Lemmas 8 and 9. To verify the third condition we set H(x,λ)=±λx+(1-λ)QNx. We choose the isomorphism J:Im⁡Q→Ker⁡L defined by J(c)=c, c∈R. By Lemma 10, we derive that H(x,λ)≠0 for all (x,λ)∈(Ker⁡L∩∂Ω)×[0,1]. Hence(50)deg⁡QNKer⁡L,Ω∩Ker⁡L,0=deg⁡±J,Ω∩Ker⁡L,0≠0. Therefore problem (1) has at least one solution in X.



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