# Multiple Solutions of Boundary Value Problems fornth-Order Singular Nonlinear Integrodifferential Equations in Abstract Spaces

Authors: Yanlai Chen; Tingqiu Cao; Baoxia Qin
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

The authors discuss multiple
solutions for thenth-order singular boundary value problems of
nonlinear integrodifferential equations in Banach spaces by means
of the fixed point theorem of cone expansion and compression. An
example for infinite system of scalar third-order singular nonlinear
integrodifferential equations is offered.


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## Body

## 1. Introduction

Singular nonlinear boundary value problems of the ordinary differential equations appeared frequently in applications. With Taliaferro [1] treating the general problem, Callegari and Nachman [2] considered existence questions in boundary layer theory, and Luning and Perry [3] obtained constructive results for generalized Emden-Fowler problems. Results have also been obtained for singular boundary value problems arising in reaction-diffusion theory and in non-Newtonian fluid theory [4]. Singular nonlinear boundary value problems of the ordinary differential equations have made great progress in recent years (please see [5–8]).In the above papers, singular problems are studied in scalar case. In Chen [9], the boundary value problems of a class of nth-order nonlinear integrodifferential equations of mixed type in Banach space are considered, and the existence of three solutions is obtained by using the fixed point index theory. But such equations do not have singular nonlinear terms. As much as we know, there are a few papers ([10–18]) to consider the singular problems in abstract Banach spaces. In Liu [10], the following singular problems in Banach spaces E(1)-x′′t=ft,xt,x′t,∀0<t≤T;x0=x′0=θ,were investigated by constructing a special convex closed set and using Mönch fixed point theorem, where θ denotes the zero element of E. In [10], (1) under certain conditions, there is at least one solution. And, in the methods, under normal circumstances, to investigate the singular problems, at first, one needs to consider the approximation problems which have no singularities. However, in the study of integrodifferential equations in infinite dimensional Banach space, this method is very complicated and difficult.In this paper, not considering approximative problems, informed by the characteristic of nonlinear term, we construct a new cone, and through the cone we create a new special cone. Moreover, through finding the relations fromuc to u(n-2)c (u belongs to the special cone), we triumphantly overcome the singularity and use the fixed point theorem of cone expansion and compression directly to obtain the existence of multiple solutions for singular boundary value problems of nonlinear integrodifferential equations in Banach spaces. Finally, an example of scalar third-order singular nonlinear integrodifferential equations for an infinite system is offered. With the previous methods, one can not get the results in this paper.LetP be a cone in Banach space E which defines a partial ordering in E by x≤y if and only if y-x∈P. Let Pr=u∈P:u<r(r>0). P is said to be normal if there exists a positive constant N such that θ≤x≤y implies x≤Ny, where θ denotes the zero element of E, and the smallest N is called the normal constant of P. For convenience, in the following, we set P as a normal cone and N=1. Let P1={u∈P:u≥u0u}, in which u0∈P and 0<u0<1. Obviously, P1 is a normal cone of E, and the normal constant of P1 also is 1. Cone P1 is the key to overcome the singular nonlinear term (please see the last example).We consider the following singular boundary value problem (SBVP for short) for annth-order nonlinear integrodifferential equations in Banach spaces E: (2)-unt=ft,ut,u′t,…,un-2t,Tut,Sut,0<t<1;ui0=θ,i=0,1,…,n-2,un-11=θ,where J≔[0,1], f∈C[(0,1)×P1∖{θ}×P1∖{θ}×⋯×P1∖{θ}×P1×P1︸,n+1P1], (3)Tut=∫0tkt,susds,Sut=∫01ht,susds,∀t∈J,with k∈CD,R+D=t,s∈J×J:t≥s,h∈CJ×J,R+ (R+ denotes the set of all nonnegative real numbers).f
(
t
,
v
0
,
v
1
,
…
,
v
n
-
1
,
v
n
) is singular at vi=θ(i=0,1,…,n-2), t=0, and/or t=1 if (4)⁡limvi→θ⁡ft,v0,…,vn=+∞,i=0,1,…,n-2,∀t∈(0,1),vk∈P1(k=n-1,n),vj∈P1∖{θ}(j=0,1,…,n-2), (5)limt→0+ft,v0,…,vn=+∞,and/orlimt→1-⁡ft,v0,…,vn=+∞,∀vi∈P1∖{θ}(i=0,1,…,n-2), vj∈P1(j=n-1,n).LetJ′=(0,1). A map u∈Cn-2[J,E]∩Cn[J′,E] is called a solution of SBVP (2) if it satisfies (2).


## 2. Preliminaries and Several Lemmas

DenoteCn-2[J,E]≔{u:u is a map from J into E and u(n-2)(t) is continuous on J}. The norm of u∈Cn-2[J,E] is defined by (6)un-2=maxi=0,1,…,n-2⁡uic,where(7)uic=maxt∈J⁡uit,i=0,…,n-2.Obviously (Cn-2[J,E],·) is a Banach space.Let(8)Cn-2J,P1≔ui∈CJ,E:uit∈P1,t∈J,i=0,1,2,…,n-2.Obviously, Cn-2[J,P1] is a cone of Cn-2[J,E]. For r>0, we write P1r≔{u∈P1:u<r} and P1r¯≔{u∈P1:u≤r}.Letu:(0,1]→E be continuous. We call the abstract generalized integral ∫01‍u(t)dt convergence if lims→0+⁡∫s1‍u(t)dt exists. Analogously, we can define the convergence of other kinds of abstract generalized integrals.We will useα to denote the Kuratowski measure of noncompactness of set in space E. For details of the Kuratowski measure of noncompactness, please see [19].Lemma 1 (see [19]).
LetH be a bounded set of Cm[J,E]. Suppose that H(m)≔{u(m):u∈H} is equicontinuous. Then,(9)αmH=maxi=0,1,…,m⁡αHiJ=maxi=0,1,…,m⁡maxt∈J⁡αHit⁡,where (10)HiJ≔uit:t∈J,u∈H,i=0,1,2,…,m,Hit≔uit:u∈H,i=0,1,2,…,m.Lemma 2 (see [19]).
IfH⊂C[J,E] is bounded and equicontinuous, then α(H(t)) is continuous on J. Moreover, (11)α∫Jutdt:u∈H≤∫JαHtdt.Lemma 3 (see [19]).
LetH be a bounded set of Cm[J,E]. Then, (12)αmH≥αHJ,αmH≥αH′J,…,αmH≥αHm-1J,αmH≥12αHmJ,where H(i)(J)  (i=0,1,2,…,m) is defined by Lemma 1.Lemma 4 (the fixed point theorem of cone expansion and compression [see [20]]).
P is a cone of real Banach space E. Let (13)PR¯∖Pr≔x∈P:r≤∥x∥≤R,R>r>0.Suppose that A:PR¯∖Pr→P is a strict set contraction such that one of the following two conditions is satisfied: (i)
Ax≱x,∀x∈P, x=r and Ax≰x,∀x∈P, x=R;(ii)
Ax≱x,∀x∈P, x=R and Ax≰x,∀x∈P, x=r.Then, A has at least a fixed point in PR¯∖Pr.


## 3. Main Results and an Example

To continue, let us formulate some conditions.H
1 There exist b∈L[J′,R+], ai∈L[J′,R+](i=0,1,…,n), gi∈C[(0,+∞),(0,+∞)](i=0,1,…,n-2), and hi∈C[0,+∞,[0,+∞)], (i=0,1,…,n), such that(14)ft,v0,v1,…,vn-2,vn-1,vn≤bt+∑i=0n-2aitgivi+hivi+an-1thn-1vn-1+anthnvn,∀t∈0,1,vi∈P1∖θi=0,1,…,n-2,vn-1,vn∈P1,where gi is nonincreasing and hi/gi(i=0,1,…,n-2) and hn-1,hn are nondecreasing.(
H
2
) For any R>r>0, (15)∫01sbs+∑i=0n-2aisgisn-1-in-1-i!r1+hiRgiR+an-1shn-1k∗R+anshnh∗Rds<+∞.And, there exists a R0>0 such that (16)∫01sbs+∑i=0n-2aisgisn-1-in-1-i!R01+hiR0giR0+an-1shn-1k∗R0+anshnh∗R0ds<R0,where b, ai  (i=0,1,…,n), gi  (i=0,1,…,n-2), and hi  (i=0,1,…,n) are defined as in condition (H1), and(17)k∗≔maxt,s∈D⁡kt,s,h∗≔maxt,s∈J×J⁡ht,s.(
H
3
) For any R>r>0, [a,b]⊂(0,1), f is uniformly continuous on (18)a,b×P1R¯∖P1r×P1R¯∖P1r×⋯×P1R¯∖P1r×P1R¯×P1R¯︸n+1,and there exist Li≥0(i=0,1,…,n), such that(19)αft,B0,B1,…,Bn≤∑i=0nLiαBi,∀t∈0,1,Bi⊂P1R¯∖P1ri=0,1,…,n-2,Bn-1,Bn⊂P1R¯.Remark 5.
Obviously, condition(H3) is satisfied automatically when E is finite dimensional.(
H
4
) There exist 0<α<1/2, and φ∈P1∗  (P1∗ denotes the dual cone of P1) such that φ(v)>0 for v>θ. At the same time, one of the n-1 conditions is satisfied (20)φft,v0,v1,…,vnφvj⟶+∞,asφvj>0,vj∈P1,vj⟶0uniformly in t∈[α,1-α], with j∈{0,1,…,n-2}.Remark 6.
Becausef(t,v0,v1,…,vn) is singular at vi=θ(i=0,1,…,n-2), condition (H4) is easy to be satisfied. And only one of the n-1 conditions is satisfied.(
H
5
) There exist 0<α<1/2, and φ∈P1∗  (P1∗ denotes the dual cone of P1) such that φ(v)>0 for v>θ. At the same time, one of the following n-1 conditions is satisfied: (21)φft,v0,v1,…,vnφvj⟶+∞,asφvj>0,vj∈P1,vj⟶+∞uniformly in t∈[α,1-α], with j∈{0,1,…,n-2}.Remark 7.
In condition(H5), only one of the n-1 conditions is satisfied.To avoid singularity, let(22)Q≕u∈Cn-2J,P1:ui0=θ,i=0,1,…,n-2,uit≥tn-1-in-1-i!uisi=0,1,…,n-2,t,s∈J.Obviously, Q is a normal cone in Cn-2[J,E], and the normal constant of Q is 1.Lemma 8.
Supposeu∈Q. Then, (23)un-2=un-2c,(24)ujt≥tn-1-jn-1-j!un-2c,j=0,1,…,n-2,∀t∈J.Proof.
For anyu∈Q, that is, u(i)(0)=θ(i=0,1,…,n-2), (25)uit=∫0tui+1sds≤tui+1c≤ui+1c,i=0,1,…,n-3,∀t∈J.Therefore, (26)uic≤ui+1c,i=0,1,…,n-3,which implies that (27)un-2=un-2c.Because of u(n-2)(t)≥tu(n-2)(s),∀t,s∈J, and the normal characters of P, it is easy to get (28)un-2t≥tun-2c,∀t∈J.Hence,(29)ujt=1n-3-j!∫0tt-sn-3-jun-2sds≥1n-3-j!∫0tt-sn-3-jsun-2τds=tn-1-jn-1-j!un-2τ,j=0,1,…,n-3,∀t,τ∈J.It follows from (28) and (29) and the normal characters of P that (30)ujt≥tn-1-jn-1-j!un-2c,j=0,1,…,n-2,∀t∈J.By (27) and (30), the conclusion holds.Remark 9.
Formula (23) implies that the norm of u∈Q is decided by (n-2)th-order derivative u(n-2).Remark 10.
Inequality (24) implies that u(n-2) controls distance between θ and u(l)(l=0,1,…,n-2). This is one of the keys to apart from the singularities of the nonlinear term f.Lemma 11.
Forj=1,2,…,n-1, the following conclusion holds: (31)tj1-1-sjj!≤Gj+1t,s≤1-1-sjj!,where (32)
G
j
t
,
s
≔
t
j
-
1
-
t
-
s
j
-
1
j
-
1
!
,
0
≤
s
≤
t
≤
1
;
t
j
-
1
j
-
1
!
,
0
≤
t
≤
s
≤
1
.Proof.
In fact, forj=1,2,…,n-1,  ∀s0∈J, we get (33)supt∈J⁡Gj+1t,s0=max⁡sup0≤s0≤t≤1⁡tj-t-s0jj!,sup0≤t≤s0≤1⁡tjj!=1j!max⁡1-1-s0j,s0j.Consider(34)1-1-s0j≥s0j,∀s0∈J.It follows from (33) and (34) that (35)supt∈J⁡Gj+1t,s0=1j!1-1-s0j,∀s0∈J;that is,(36)supt∈J⁡Gj+1t,s=1j!1-1-sj,∀s∈J.On the other hand, for j=1,2,…,n-1, it is easy to get (37)inf0<t≤1⁡Gj+1t,stj=1j!min⁡inf0≤s<t≤1⁡tj-t-sjtj,inf0<t≤s≤1⁡tjtj=1j!min⁡inf0≤s<t≤1⁡1-1-stj,1=1j!1-1-sj.It follows from (36) and (37) that (38)tj1-1-sjj!≤Gj+1t,s≤1-1-sjj!,∀t∈0,1.Since for t=0 (38) holds, we get (31).Lemma 12.
Suppose conditions(H1) and (H2) are satisfied. Then, for any R>r>0, A:QR¯∖Qr→Q, in which the operator A is defined by(39)Aut≔∫01Gnt,sfs,us,u′s,…,un-2s,Tus,Susds,∀t∈J,where (40)
G
n
t
,
s
≕
t
n
-
1
-
t
-
s
n
-
1
n
-
1
!
,
0
≤
s
<
t
≤
1
;
t
n
-
1
n
-
1
!
,
0
≤
t
≤
s
≤
1
.Proof.
At first, we show that the operatorA defined by (39) is reasonable for u∈QR¯∖Qr with any R>r>0. In fact, for u∈QR¯∖Qr with R≥uit≥(tn-1-i/n-1-i!)r, by (H1) and (H2), (41)∫01sbs+∑i=0n-2aisgisn-1-in-1-i!r1+hiRgiR+an-1shn-1k∗R+anshnh∗Rds<+∞,which implies that (Au)(t) defined by (39) is reasonable for t∈J.
Next, we show that(Au)(t)∈Cn-2[J,P1]. For u(i)(t)∈P1(i=0,1,…,n-2),(42)Tut=∫0tkt,susds≥∫0tkt,su0usds≥u0∫0tkt,susds=u0Tut,Sut=∫01ht,susds≥∫01ht,su0usds≥u0∫01ht,susds=u0Sut,which implies (Tu)(t)∈P1 and (Su)(t)∈P1. This together with(43)f∈C0,1×P1∖θ×P1∖θ×⋯×P1∖θ×P1×P1︸n+1,P1,gives (44)Ault=∫01Gn-lt,sfus,u′s,…,un-2s,Tus,Susds≥∫01Gn-lt,su0fus,u′s,…,un-2s,Tus,Susds≥u0∫01Gn-lt,sfus,u′s,…,un-2s,Tus,Susds=u0Ault,∀t∈J,l=0,1,…,n-2.Therefore, (Au)(t)∈Cn-2[J,P1] holds.
Finally, by Lemma11 and (39) and (40), one can see (45)Aul0=θl=0,1,…,n-2,Ault=∫01Gn-lt,sfs,us,u′s,…,un-2s,Tus,Susds≥tn-1-ln-1-l!∫01Gn-lτ,sfs,us,u′s,…,un-2s,Tus,Susds=tn-1-ln-1-l!Aulτ,∀t,τ∈J,l=0,1,…,n-2.It follows from (44) and (45) that A:QR¯∖Qr→Q.Lemma 13.
Let coneP be normal and let conditions (H1) and (H2) be satisfied. Then, u∈QR¯∖Qr is a fixed point of operator A if and only if u∈Cn[J′,E]∩QR¯∖Qr is a solution for SBVP (2).Proof.
By Lemma12, A:QR¯∖Qr→Q. For u∈Cn[J,E], Taylor’s formula with the integral remainder term gives (46)ut=∑i=0n-1tii!ui0+1n-1!∫0tt-sn-1unsds,∀t∈J.Substituting (47)un-10=un-11-∫01unsdsinto (46), we get (48)ut=∑i=0n-2tii!ui0+tn-1n-1!un-11-1n-1!∫01tn-1unsds-∫0tt-sn-1unsds,∀t∈J.Let u∈Cn[(0,1),E]∩Cn-2[J,E] be the solution of SBVP (2). Then, (48) implies (49)ut=1n-1!∫01tn-1fs,us,u′s,…,un-2s,Tus,Susds-∫0tt-sn-1fs,us,u′s,…,un-2s,Tus,Susds.Comparing this with (39) and (40), we have u(t)=(Au)(t), which means u(t) is the fixed point of the operator A in QR¯∖Qr.
On the other hand, letu(t)∈QR¯∖Qr be the fixed point of the operator A. By (39) and (40), (50)ujt=Aujt=1n-1-j!∫01tn-1-jfs,us,u′s,…,un-2s,Tus,Susds-∫0tt-sn-1-jfs,us,u′s,…,un-2s,Tus,Susds,where j=1,2,…,n-1. It follows by taking t=0 and t=1 in (50) that (51)uj0=θ,i=0,1,…,n-2,un-11=θ,un-1t=∫t1fs,us,u′s,…,un-1s,Tus,Susds,t∈J′;that is,(52)unt=-ft,ut,u′t,…,un-1t,Tut,Sut,t∈J′.Then, (51)-(52) imply that u is the solution for SBVP (2) in Cn[(0,1),E]∩Cn-2[J,E].Lemma 14.
Suppose conditions(H1)–(H3) are satisfied. Let(53)Dlt≔∫01Gn-lt,sfs,us,…,un-2s,Tus,Susds:u∈B,l=0,1,…,n-2,with t∈J, B⊂QR¯∖Qr. Then,(54)αDlt≤∑i=0n-2LiαBiJ+Ln-1k∗αBJ+Lnh∗αBJ,l=0,1,…,n-2,with BiJ≔uis:s∈J,u∈Bi=0,1,…,n-2.Proof.
Apart from the singularities, let(55)Dl,δt≕∫δ1-δGn-lt,sfs,us,…,un-1s,Tus,Susds:u∈B,l=0,1,…,n-2,0<δ<12,t∈J.By conditions (H1) and (H2), for any t∈J, u∈B, one can see that(56)∫01Gn-lt,sfs,us,…,un-2s,Tus,Susds-∫δ1-δGn-lt,sfs,us,…,un-2s,Tus,Susds≤∫0δsbs+∑i=0n-2aisgisn-1-in-1-i!r1+hiRgiR+an-1shn-1k∗R+anshnh∗Rds+∫1-δ1sbs+∑i=0n-2aisgisn-1-in-1-i!r1+hiRgiR+an-1shn-1k∗R+anshnh∗Rds,l=0,1,…,n-2.By virtue of absolute continuity of the Lebesgue integrable function and (56), it is easy to see that (57)dHDl,δt,Dlt⟶0,asδ⟶0,∀t∈J,l=0,1,…,n-2,in which dH(Dl,δ(t),Dl(t)) denotes the Hausdorff distance between Dl,δ(t) and Dl(t). Therefore, (58)αDlt=limδ→0⁡αDl,δt,∀t∈J,l=0,1,…,n-2.
Now, we check that(59)αDn-2t≤∑i=0n-2LiαBiJ+Ln-1k∗αBJ+Lnh∗αBJ.For B⊂QR¯∖Qr, it is easy to see that A(B)⊂Cn-1[J,P1]⊂Cn-2[J,E] is bounded, which implies that Dl,δ(l=0,1,…,n-2) are bounded. Since(60)∫δ1-δGn-lt,sfs,us,…,un-2s,Tus,Susds∈1-2δco¯Gn-lt,sfs,us,…,un-2s,Tus,Sus:s∈δ,1-δ,we have(61)αDn-2,δt=α∫δ1-δG2t,sfs,us,…,un-2s,Tus,Susds:u∈B≤α1-2δco¯G2t,sfs,us,…,un-2s,Tus,Sus:s∈δ,1-δ,u∈B<αG2t,sfs,us,…,un-2s,Tus,Sus:s∈δ,1-δ,u∈B≤αfIδ,BIδ,…,Bn-2Iδ,TBIδ,SuIδ,where Iδ=δ,1-δ, BlIδ=uls:s∈Iδ,u∈B  l=0,1,…,n-2, TBIδ=Tus:s∈Iδ,u∈B, and (SB)(Iδ)={(Su)(s):s∈Iδ,u∈B}. Take r1=(δn-1/n-1!)r, R1=Rmax⁡{1,k∗,h∗}, such that (62)BlIδ⊂P1R1¯∖P1r1,l=0,1,…,n-2,TBIδ,SBIδ⊂P1R1¯.Obviously, (TB)⊂C[J,E] and B⊂C[J,E]; moreover, both of them are equacontinuous on Iδ. It follows from (62), (H3), and Lemmas 1 and 2 that (63)αfIδ,BIδ,…,Bn-2Iδ,TBIδ,SuIδ=maxs∈Iδ⁡αfs,BIδ,…,Bn-2Iδ,TBIδ,SuIδ≤∑i=0n-2LiαBiIδ+Ln-1αTBIδ+LnαSBIδ,αTBIδ=maxs∈Iδ⁡αTBs=maxs∈Iδ⁡α∫0sks,τuτdτ:u∈B≤∫0sαks,τBτdτ≤k∗αBJ.Analogously, it is easy to get (64)αSBIδ≤h∗αBJ.It follows from (H3), (61), (63), and (64) that (65)αDn-2,δt≤∑i=0n-2LiαBiJ+Ln-1k∗αBJ+Lnh∗αBJ.Hence, by (58), we know (59) is true, and the conclusion holds.Lemma 15.
Let conditions(H1), (H2), and (H3) be satisfied. Suppose that 0≤γ<1, in which γ=∑i=0n-3Li+Ln-1k∗+Lnh∗+2Ln-2. Then, A is a strict set contraction from Q¯R∖Qr into Q.Proof.
By Lemma12, (H1), and (H2), it is easy to see that A(QR¯∖Qr)⊂Q and A is a bounded operator. We check that A(QR¯∖Qr)→Q is continuous. In fact, let um,u∈QR¯∖Qr, um-un-2→0(m→∞). For i=0,1,…,n-2, it is easy to get (66)R≥umit≥tn-1-in-1-i!r,R≥∥uit∥≥tn-1-in-1-i!r,∀t∈J.For ∀t∈J, by (31) and (50), (67)Aumn-2t-Aun-2t=∫01G2t,sfs,ums,um′s,…,umn-2s,Tums,Sumsds-∫01G2t,sfs,us,u′s,…,un-2s,Tus,Susds≤∫01sfs,ums,um′s,…,umn-2s,Tums,Sums-fs,us,u′s,…,un-2s,Tus,Susds.By (66) and conditions (H1) and (H2), it is easy to get (68)fs,ums,um′s,…,umn-2s,Tums,Sums-fs,us,u′s,…,un-2s,Tus,Sus≤2sbs+∑i=0n-2aisgisn-1-in-1-i!r1+hiRgiR+an-1shn-1k∗R+anshnh∗R.From (67), by Lebesgue dominated convergence theorem, combined with the equicontinuity of {Aum(n-2)}m=1+∞ and the continuity of (Au)(n-2)(t), we have (69)Aumn-2t-Aun-2t⟶0,asm⟶∞,uniformly for t∈J. Therefore, (70)limm→∞⁡Aumn-2-Aun-2c=limm→∞⁡maxt∈J⁡Aumn-2t-Aun-2t=maxt∈J⁡limm→∞⁡Aumn-2t-Aun-2t=0.Combining this with (23), we get (71)Aum-Aun-2⟶0,asm⟶∞.Hence, A:QR¯∖Qr→Q is continuous.
LetB⊂QR¯∖Qr be bounded, so A(B)⊂Cn-1[J,P1]⊂Cn-2[J,E] is bounded. It is easy to prove that (A(B))(n-1) is bounded, so (A(B))(n-2) is equicontinuous. By Lemma 1, (72)αn-2AB=maxl=0,1,…,n-2⁡maxt∈J⁡αABlt,where α((A(B))(l)(t))=α({(Au)(l)(t):u∈S})(t is fixed, l=0,1,…,n-2). On account of Lemmas 14 and 3, it is easy to see that (73)αABt=αD0t≤∑i=0n-2LiαBiJ+Ln-1k∗αBJ+Lnh∗αBJ≤∑i=0n-3Li+Ln-1k∗+Lnh∗+2Ln-2αn-2B=γαn-2B.Similarly,(74)αABlt≤γαn-2B,l=1,2,…,n-2.Thus, we get αn-2(A(B))≤γαn-2(B) by (72), (73), and (74). Since A is bounded and continuous and 0≤γ<1, the conclusion holds.Theorem 16.
Suppose that the conditions(H1), (H2), (H3), (H4), and (H5) are satisfied. Then, SBVP (2) has at least two solutions u∗ and u∗ in Cn[(0,1),E]∩Cn-2[J,E], satisfying (75)0<u∗n-2<R0<u∗n-2.Proof.
Suppose that the conditions(H1), (H2), and (H4) are satisfied. For a l∈{0,1,2,…,n-2}, 0<δ<1/2, let M>(n-1-l)!/δn-1-l(1-(1-δ)n-1-l)(1-2δ). There is a 0<β<R0, such that (76)φft,ut,u′t,…,un-2t,Tut,Sut≥Mφult,t∈δ,1-δ,ui∈P1∖θ,i=0,1,…,n-2,un-2≤β.
Now, for any0<r1<β, we show that (77)Au≰1u,u∈Q,un-2=r1,in which ≤1 is partial ordering defined by Q. In fact, suppose that there is a u1∈Q, u1n-2=r1, such that Au1≤1u1. Then, for l∈{0,1,2,…,n-2}, t∈[δ,1-δ], we have (78)u1lt≥Au1lt=∫01Gn-lt,sfs,u1s,u1′s,…,u1n-2s,Tu1s,Su1sds≥δn-1-l1-1-δn-1-ln-1-l!∫δ1-δfs,u1s,u1′s,…,u1n-2s,Tu1s,Su1sds.By (76) and (78), one can see that (79)φu1lt≥φδn-1-l1-1-δn-1-ln-1-l!∫δ1-δfs,u1s,u1′s,…,u1n-2s,Tu1s,Su1sds=δn-1-l1-1-δn-1-ln-1-l!∫δ1-δφfs,u1s,u1′s,…,u1n-2s,Tu1s,Su1sds≥δn-1-l1-1-δn-1-ln-1-l!∫δ1-δMφu1lsds.Hence, (80)∫δ1-δφu1ltdt≥δn-1-l1-1-δn-1-l1-2δn-1-l!M∫δ1-δφu1lsds.Then, it follows from M(δn-1-l(1-1-δn-1-l)(1-2δ)/(n-1-l)!)>1, φ(u1(l)(t))≥0 and (80) that (81)φu1lt=0,∀t∈δ,1-δ.Since u1∈Q, by (24), we get (82)u1lt≥tn-1-ln-1-l!u1n-2c≥δn-1-ln-1-l!u1n-2c,∀t∈δ,1-δ,which implies (83)u1n-2c=0.It contradicts u1n-2=r1>0. Thus, (77) holds.
On the other hand, by conditions(H1), (H2), and (H5), for l∈{0,1,2,…,n-2}, 0<δ<1/2, let (84)M>n-1-l!δn-1-l1-1-δn-1-l1-2δ.There exists τ>0 such that (85)φft,ut,u′t,…,un-2t,Tut,Sut≥Mφun-2t,t∈δ,1-δ,ui∈P1∖θ,i=0,1,…,n-2,ul≥τ.Set (86)R1>maxn-1-l!τδn-1-l,R0.We show that (87)Au≰1u,u∈Q,un-2=R1.In fact, by (24), if there is u1∈Q, u1n-2=R1 such that Au1≤1u1, then (88)u1lt≥tn-1-ln-1-l!u1n-2c,∀t∈δ,1-δ.It is easy to see by (86) and (88) that (89)mint∈δ,1-δ⁡u1lt≥tn-1-ln-1-l!u1n-2=δR1>τ.In the same way, similar to the proof of (77), (87) holds.
Finally, by condition(H2), one can see (90)Au≱1u,u∈Q,∥u∥n-2=R0.In fact, if there is u1∈Q, u1n-2=R0 such that Au1≥1u1, then (91)θ≤u1n-2t≤Au1n-2t=∫01G2t,sfs,u1s,u1′s,…,u1n-2s,Tu1s,Su1sds,∀t∈J.Hence, (92)R0=u1n-2≤∫01sbs+∑i=0n-2aisgisn-1-in-1-i!R01+hiR0giR0+an-1shn-1k∗R0+anshnh∗R0ds<R0.This is a contradiction. Therefore, (90) is true.
Above all, we setR=R1, r=r1. By (H1), (H2), (H3), and Lemma 15, we know that A:Q¯R1∖Qr1→Q is a strict set contraction. By virtue of (77), (87), and (90), applying Lemma 4 twice, we obtain that operator A has at least one fixed point in QR1∖Q¯R0 and QR0∖Q¯r1, respectively. By Lemma 13, SBVP (2) has at least two solutions u∗∈QR1∖Q¯R0 and u∗∈QR0∖Q¯r1 satisfying 0<u∗n-2<R0<u∗n-2.An application of Theorem16 is as follows.Example 17.
Consider SBVP of infinite system for scalar nonlinear third-order singular integrodifferential equations:(93)-un′′′t=14nt1-t+t46n3/2u2nt1/2+t4unt4n+2+1-t1/25nun+1′t1/3+1-t1/23n+12supn∈1,2,…⁡un′t2+t24∫0t11+t-sunsds+t56∫01t+sunsds,0<t<1;un0=un′0=un′′1=0,n=1,2,….Conclusion 18.
Infinite system (93) has at least two solutions: (94)unt,n=1,2,3,…,vnt,n=1,2,3,…such that (95)unt⟶0,vnt⟶0,asn⟶∞,0<supn⁡maxt∈0,1⁡un′′t<1<supn⁡maxt∈0,1⁡vn′′t.Proof.
LetJ=[0,1], E=:C0={u=(u1,u2,…,un,…):un→0} with norm u=supn⁡un, and P={u=(u1,…,un,…)∈C0:un≥0,n=1,2,…}. Obviously P is a normal cone in E and the normal constant N=1. Let u0=(1/4,1/9,…,1/(n+1)2,…). Then, it is easy to see u0∈P, 0<u0=1/4<1, and P1={u∈P:un≥u0nu,n=1,2,….}. Infinite system (93) can be regarded as SBVP of the form (2) in E. In this situation, (96)kt,s=11+t-s∈CD,R+,D=t,s∈J×J:s≤t,ht,s=t+s∈CJ×J,R+,u=u1,u2,…,un,…,v=v1,v2,…,vn,…,w=w1,w2,…,wn,…,x=x1,x2,…,xn,…,f=(f1,f2,…,fn,…), with (97)fnt,u,v,w,x=14nt1-t+t46n3/2u2n1/2+t4un4n+2+1-t1/25nvn+11/3+1-t1/23n+12v2+t24wn+t56xn.Obviously, for (t,u,v,w,x)∈(0,1)×P1∖{θ}×P1∖{θ}×P1×P1, it is easy to see that (98)6n3/2u2n1/2≥6n3/22n+1u1/2>0,u1/24n+2≥un1/24n+2≥14n+2n+1u1/2,5nvn+11/3≥5n1n+22/3v1/3>0,112v2≥13n+12v2,wn≥1n+12w,xn≥1n+12x,n=1,2,…,which implies that (99)fn≤14nt1-t+2n+1t46n3/2u1/2+u1/2t44n+2+1-t1/25n1/n+22/3v1/3+1-t1/23n+12v2+t24wn+t56xn,n=1,2,….Since un→0, vn→0, wn→0, xn→0, as n→+∞, one can see that |fn|→0, as n→+∞. That is, f∈E. Obviously, f∈P. By (99), we can see (100)f≤14t1-t+t42u1/2+u1/2t412+32/31-t1/25v1/3+1-t1/212v2+t24w+t56x.On the other hand, it follows from (97) and (98) that (101)fnt,u,v,w,x≥14nt1-t+t46n3/2u1/2+t44n+2n+1u1/2+1-t1/25n1/n+22/3v1/3+1-t1/23n+12v2+t24wn+t56xn≥1n+12n+124nt1-t+n+12t46n3/2u1/2+n+1t44n+2u1/2+n+22/3n+121-t1/25nv1/3+1-t1/23v2+t24w+t56x,n=1,2,….It is easy to get (102)n+12n≥4,n+12n3/2≥3,n+14n+2≥16,n+22/3n+125n≥432/35,n=1,2,….It follows from (100), (101), and (102) that (103)fnt,u,v,w,x≥1n+1214t1-t+t42u1/2+u1/2t46+432/31-t1/25v1/3+1-t1/23v2+t24w+t56x≥1n+12f=u0nf,n=1,2,….So, f∈C[(0,1)×P1∖{θ}×P1∖{θ}×P1×P1,P1] and f(t,v0,v1,v2,v3) is singular at vi=θ(i=0,1), t=0, and/or t=1. Thus, by (100), condition (H1) holds for (104)bt=14t1-t,a0t=12t4,a1t=121-t1/2,a2t=t24,a3t=t56,g0y=1y1/2,g1y=1y1/3,h2y=h3y=y,h0y=y1/2,h1y=y2.
For anyR>r>0, by (100), we have, with k∗=1 and h∗=2, (105)∫01s4s1-s+2s42r1+R+s1-s1/22sr31+R4/3+s34R+s662Rds<+∞.On the other hand, taking R0=1, by (105), we have(106)∫01s4s1-s+2s4+s1-s1/22s31+1+s34+s662ds=π8+25+965+116+121<1.By (105) and (106), condition (H2) is satisfied.
For anyR>r>0, [a,b]⊂(0,1), it is clear that f is uniformly continuous on (107)a,b×P1R¯∖P1r×P1R¯∖P1r×⋯×P1R¯∖P1r×P1R¯×P1R¯︸n+1.Let f=f1+f2+f3, with (108)f1=f11,f21,…,fn1,…,f2=f12,f22,…,fn2,…,f3=f13,f23,…,fn3,…,where (109)fn1t,u,v,w,x=14nt1-t+t46n3/2u2n1/2+t4un4n+2+1-t1/25nvn+11/3+1-t1/23n+12v2,fn2t,u,v,w,x=t44wn,fn3t,u,v,w,x=t56xn,n=1,2,….For any (110)z=z1,z2,…,zn,…∈f1t,B0,B1,B2,B3∀t∈a,b,B0,B1⊂P1R¯∖P1r,B2,B3⊂P1R¯,by (99) and (109), we get (111)zn≤14na1-b+2n+1b46n3/2r1/2+R1/2b44n+2+1-a1/25n1/n+22/3r1/3+1-a1/23n+12R2,n=1,2,….So, the relative compactness of f1(t,B0,B1,B2,B3) in C0 follows directly from a known result (see [21]): a bounded set X of C0 is relatively compact if and only if (112)limn→∞⁡supz∈X⁡max⁡Zk:k≥n=0.Hence, (113)αf1t,B0,B1,B2,B3=0,∀t∈a,b,B0,B1⊂P1R¯∖P1r,B2,B3⊂P1R¯.By (109), it is easy to get (114)αf2t,B0,B1,B2,B3≤14αB2,∀t∈0,1,B0,B1⊂P1R¯∖P1r,B2,B3⊂P1R¯,(115)αf3t,B0,B1,B2,B3≤16αB3,∀t∈0,1,B0,B1⊂P1R¯∖P1r,B2,B3⊂P1R¯.Combining this with (109), (113), and (114), one can see (116)αft,B0,B1,B2,B3≤αf1t,B0,B1,B2,B3+αf2t,B0,B1,B2,B3+αf3t,B0,B1,B2,B3≤14αB2+16αB3,∀t∈a,b,B0,B1⊂P1R¯∖P1r,B2,B3⊂P1R¯.Therefore, condition (H3) holds.
For anyu∈P1, define φ by φ(u)=u1. It is easy to see φ∈P1∗, φ=1. Let [a,b]=[1/4,3/4], for (117)t,u,v,w,x∈14,34×P1∖θ×P1∖θ×P1×P1,and it is easy to see 0<(1/4)u≤u1≤u, 0<(1/4)v≤v1≤v. Thus, (97), (101), and (102) imply that (118)lim∥u∥→0⁡φft,u,v,w,xφu=lim∥u∥→0⁡f1t,u,v,w,xu1≥lim∥u∥→0⁡1/211u1/2u1≥lim∥u∥→0⁡1211u3/2=+∞,(119)lim∥v∥→+∞⁡φft,u,v,w,xφv=lim∥v∥→+∞⁡f1t,u,v,w,xv1≥lim∥v∥→+∞⁡v224v1≥lim∥v∥→+∞⁡v24=+∞.The conditions (H4) and (H5) follow from (118) and (119). It is easy to see that (120)γ=14+13=712<1for(121)L0=0,L1=0,L2=14,L3=16.Therefore, by Theorem 16, our conclusion holds.



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