R E S E A R C H
Open Access
The existence of solutions for nonlinear
fractional multipoint boundary value
problems at resonance
Na Xu, Wenbin Liu
*and Lishun Xiao
*Correspondence: [email protected] College of Sciences, China University of Mining and Technology, Xuzhou, Jiangsu, 221116, P.R. China
Abstract
A class of nonlinear fractional multipoint boundary value problems at resonance is considered in this article. The existence results are obtained by the method of the coincidence degree theory of Mawhin. An example is given to illustrate the results. MSC: 34A08
Keywords: coincidence degree; fractional differential equation; resonance; multipoint boundary conditions
1 Introduction
The subject of fractional calculus has gained considerable popularity during the past decades, due mainly to its frequent appearance in a variety of different areas such as physics, aerodynamics, polymer rheology, etc. (see [–]). Many methods have been intro-duced for solving fractional differential equations (FDEs for short in the remaining), such as the Laplace transform method, the iteration method, the Fourier transform method, etc. (see []).
Recently, there have been many works related to the existence of solutions for multipoint boundary value problems (BVPs for short in the remaining) at nonresonance of FDEs (see [–]). Motivated by the above articles and recent studies on FDEs (see [–]), we con-sider the existence of solutions for a nonlinear fractional multipoint BVPs at resonance in this article.
In [], Zhang and Bai considered the following fractional three-point boundary value problems at resonance:
⎧ ⎨ ⎩
Dα+u(t) =f
t,u(t),Dα+–(n–)u(t), . . . ,Dα+–u(t)
+e(t), <t< ,
In–α
+ u() =D
α–(n–)
+ u() =· · ·=Dα+–u() = , u() =σu(η),
wheren> is a natural number;n– <α≤nis a real number;Dα
+andI+α are the standard
Riemann-Liouville derivative and integral respectively;f : [, ]×Rn→Ris continuous;
e(t)∈L[, ];σ∈(, +∞),η∈(, ) are given constants such thatσ ηα–= . In their article,
they made the operatorLu=Dα+uand gotdim KerL= . In [], Bai discussed fractional m-point boundary value problems at resonance with the case ofdim KerL= .
In , Bai and Jiang studied the fractional differential equation of boundary value problems at resonance with the case ofdim KerL= respectively (see [, ]), and we
can see that they obtained the results by the assumption that a specific algebraic expres-sion is not equal to zero; for example,
R=
αη
α(α)(α– )
(α– )
–
m
i=
αiηiα– –
α– η
α–(α) (α)
–
m
i=
αiηiα– =
is referred to as a condition in []. We will show that the assumption like aboveR= is
not necessary.
In this article, we will use the coincidence degree theory to study the existence of solu-tions for a nonlinear FDEs at resonance which is given by
Dα
+u(t) =f
t,u(t),Dα–
+ u(t),Dα+–u(t)
, <t< , (.)
with boundary conditions
I+–αu(t)|t== , Dα+–u() =
m
i=
aiDα+–u(ξi),
Dα– + u() =
m
i=
biDα+–u(ηi),
(.)
where <α≤; <ξ<· · ·<ξm< ; <η<· · ·<ηm< ;ai,bi∈R;f : [, ]×R→R
with satisfying Carathéodory conditions;Dα
+andIα+are the standard Riemann-Liouville fractional derivative and fractional integral respectively.
BVPs (.)-(.) being at resonance means that the associated linear homogeneous equa-tionDα+u(t) = with boundary conditions (.) hasu(t) =atα–+btα– as a nontrivial
solution, where <t< ,a,b∈R.
We will always suppose that the following conditions hold:
m
i= ai= ,
m
i=
biηi= , m
i=
bi= . (C)
The rest of this article is organized as follows: In Section , we give some definitions, lemmas and notations. In Section , we establish theorems of existence result for BVPs (.)-(.). In Section , we give an example to illustrate our result.
2 Preliminaries
We present here some necessary basic knowledge and definitions of the fractional calculus theory, which can be found in [–].
Definition . The Riemann-Liouville fractional integral of orderα> of a functiony: (,∞)→Ris given by
I+α y(t) =
(α) t
(t–s)α–y(s) ds,
Definition . The Riemann-Liouville fractional derivative of orderα> of a function
y: (,∞)→Ris given by
Dα+y(t) =
(n–α)
d dt
n t
y(s) (t–s)α–n+ds,
wheren= [α] + , provided the right side is pointwise defined on (,∞).
Definition .([]) We say that the mapf: [, ]×Rn→Rsatisfies Carathéodory con-ditions with respect toL[, ] if the following conditions are satisfied:
(i) for eachz∈Rn, the mappingt→f(t,z)is Lebesgue measurable;
(ii) for almost everyt∈[, ], the mappingt→f(t,z)is continuous onRn; (iii) for eachE> , there exists aρE∈L[, ]such that, for a.e.t∈[, ]and every
|u| ≤E, we havef(t,u)≤ρE(t).
Lemma . ([]) Assume y(t)∈C[, ]∩L[, ],≤β≤α, then Dβ+Iα+y(t) =Iα+–βy(t).
And, for allα≥,β> –, we have that
Iα
+tβ=
(β+ )
(α+β+ )t
α+β, Dα
+tβ=
(β+ )
(β–α+ )t β–α.
Lemma .([]) Letα> , n= [α] + and assume that y, Dα
+y∈L(, ), then the following
equality holds almost everywhere on[, ],
Iα
+Dα+y
(t) =y(t) –
n
i=
((In–α
+ y)(t))n–i|t= (α–i+ ) t
α–i.
Now, we briefly recall some notations and an abstract existence result, which can be found in []. LetY,Zbe real Banach spaces,L:domL⊂Y→Zbe a Fredholm map of index zero, andP:Y→Y,Q:Z→Zbe continuous projectors such that
ImP=KerL, KerQ=ImL, Y=KerL⊕KerP, Z=ImQ⊕ImL.
It follows thatL|domL∩KerP:domL∩KerP→ImLis invertible. We denote the inverse by
Kp. If is an open bounded subset ofYsuch thatdomL∩ =∅, the mapN:Y→Zwill
be calledL-compact on ifQN(¯) is bounded andKp(I–Q)N: ¯ →Yis compact.
Lemma .([]) Let L be a Fredholm operator of index zero and N be L-compact on ¯. The equation Lx=Nx has at least one solution indomL∩ ¯ if the following conditions are satisfied:
(i) Lx=λNxfor each(x,λ)∈[domL\KerL∩∂ ]×[, ]; (ii) Nx∈/ImLfor eachx∈KerL∩∂ ;
(iii) deg(JQN|KerL,KerL∩ , )= ,
where Q:Z→Z is a projection such thatKerQ=ImL and J:ImQ→KerL is a any iso-morphism.
Lemma .([]) Givenμ> and N= [μ] + , for any x∈C[, ], ci∈R(i= , , . . . ,N– ),
we can define a Banach space
Cμ[, ] =u(t)|u(t) =Iμ+x(t) +ctμ–+ctμ–+· · ·+cN–tμ–(N–),t∈(, )
,
with the norm defined byuCμ=Dμ
+u∞+· · ·+Dμ+–(N–)u∞+u∞.
Lemma .([]) E⊂Cμ[, ]is a sequentially compact set if and only if E is uniformly
bounded and equicontinuous. Here, a uniform bound means that there exists a constant M> with each u∈E, such that
uCμ=Dμ
+u∞+· · ·+D
μ–(N–)
+ u∞+u∞<M,
and equicontinuation means that there exists aδ> with|t–t|<δfor any t,t∈[, ], u∈E andε> , such that
u(t) –u(t)<ε, Dα+–iu(t) –Dα+–iu(t)<ε (i= , , . . . ,N– ).
In this article, letZ=L[, ] with the normy=
|y(s)|dsandY=C
α–[, ] with the
normuY=Dα+–u∞+Dα–+ u∞+u∞. Define the operatorL:domL∩Y→Zby
Lu=Dα+u, (.)
wheredomL={u∈Cα–[, ]|Dα
+u∈Z,usatisfies (.)}. Define the operatorN:Y →Z
by
Nu(t) =ft,u(t),Dα+–u(t),Dα+–u(t), ∀t∈[, ]. (.)
Thus, BVP (.) can be written asLu=Nufor eachu∈domL.
3 Main results
First, let us introduce the following notations for convenience, with settingp∈ {, , . . . ,m– }andq∈Z+withq≥p+ ,
:=
p(p+ )
–
m
i= biηpi+
, :=
q(q– )
–
m
i= biηqi
,
:=
p
–
m
i= aiξip
, :=
q–
–
m
i= aiξiq–
,
:=–, μ:= +
(α)+
(α– ),
ω:= +
(α), ρ:= +
(α)+
(α– ).
(H) There exist functionsh(t),r(t),s(t),w(t),e(t)∈L[, ]and a constantθ∈[, )such that for all(x,y,z)∈R,t∈[, ],
f(t,x,y,z)≤h(t)|x|+r(t)|y|+s(t)|z|+w(t)|z|θ+e(t);
(H) For anyu∈domL,t∈[, ], there exists a constantA> such that ifDα+–u(t) >A,
then either
TNu–TNu< , or TNu–TNu> ;
(H) For anyu∈domL,t∈[, ], there exists a constantB> such that ifDα– + u(t) >B,
then either
TNu–TNu> , or TNu–TNu< .
Theorem . If conditions (C), (H)-(H) hold, then BVPs (.)-(.) have at least one so-lution provided thatρ(h+r+s) < .
In order to obtain our main result, we first present and prove Lemmas .-.. Now, let us define operatorsTj:Z→Z(j= , ) as follows:
Tx(t) =
x(s) ds–
m
i= ai
ξi
x(s) ds, ∀t∈[, ],
Tx(t) =
( –s)x(s) ds–
m
i= bi
ηi
(ηi–s)x(s) ds, ∀t∈[, ].
Lemma . If condition (C) holds and L is defined by (.), then
KerL=atα–+btα–|a,b∈R, ImL={x∈Z|Tjx= ,j= , }.
Proof By (.) and Lemma .,Dα
+u(t) = has a solution
u(t) =D α– + u(t)|t=
(α) t
α–+Dα+–u(t)|t= (α– ) t
α–+I+–αu(t)|t= (α– ) t
α–.
Combining with the condition (.), we getKerL={atα–+btα–|a,b∈R} ∼=R.
Suppose∀x∈ImL, then there existsu∈domLsuch thatx=Lu, i.e.,u∈Y,x=Dα
+u. By
Lemma ., we have
Iα
+x(t) =u(t) – Dα–
+ u(t)|t= (α) t
α––Dα+–u(t)|t= (α– ) t
α––I+–αu(t)|t= (α– ) t
α–.
Then in view of condition (C), (.) and Lemma .,xsatisfies ⎧
⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
x(s) ds–
m
i= ai
ξi
x(s) ds= ,
( –s)x(s) ds–
m
i= bi
ηi
(ηi–s)x(s) ds= .
On the other hand, suppose∀x∈Zand it satisfies (.), letu(t) =Iα
+x(t), thenu∈domL, Dα
+u(t) =x(t), i.e.,x∈ImL. Therefore, we obtain that
ImL={x∈Z|Tjx= ,j= , }.
Lemma . If condition (C) holds, then there exist two constants q ∈ Z+ and p ∈
{, , . . . ,m– }with q≥p+ such that= .
Proof Frommi=ai= , we obtain that for any nonnegative integerl, there existskl– ∈
{lm+ , . . . , (l+ )m}such thatim=aiξikl–= . If else, we obtain that
m
i=aiξikl–= ,kl– =
,lm+ , . . . , (l+ )m. Ifl= , we have
⎛ ⎜ ⎜ ⎜ ⎜ ⎝ · · ·
ξ ξ · · · ξm
..
. . .. ...
ξm ξm · · · ξmm
⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ a a .. . am ⎞ ⎟ ⎟ ⎟ ⎟ ⎠= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ .. . ⎞ ⎟ ⎟ ⎟ ⎟ ⎠.
It is equal to
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
–ξ –ξ · · · –ξm
ξ( –ξ) ξ( –ξ) · · · ξm( –ξm)
..
. . .. ...
ξm–( –ξ) ξm–( –ξ) · · · ξmm–( –ξm)
⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ a a .. . am ⎞ ⎟ ⎟ ⎟ ⎟ ⎠= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ .. . ⎞ ⎟ ⎟ ⎟ ⎟ ⎠.
Since the determinant of coefficients is not equal to zero, we have that ai = (i=
, , . . . ,m), which is a contradiction to condition (C). Ifl∈Z+, we get
⎛ ⎜ ⎜ ⎜ ⎜ ⎝ · · ·
ξlm+ ξlm+ · · · ξmlm+
..
. . .. ...
ξlm+m ξlm+m · · · ξlm+m m ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ a a .. . am ⎞ ⎟ ⎟ ⎟ ⎟ ⎠= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ .. . ⎞ ⎟ ⎟ ⎟ ⎟ ⎠.
Similarly, we can deduce that the determinant of coefficients is not equal to zero, so we have thatai= (i= , , . . . ,m), which is a contradiction to condition (C). Thus, there exists
kl– ∈ {lm+ , . . . , (l+ )m}such that
m
i=aiξikl–= .
Similarly, frommi=bi=
m
i=biηi= , we have that there exists a constantp∈ {, , . . . ,
m– }such that
m
i=
biηpi+= . (.)
Let
S=
(kl– )∈Z+
(p+ )( –mi=biηkil)( –
m i=aiξip)
kl( –
m
i=aiξikl–)
= –
m
i= biηpi+
we shall prove thatSis a finite set. If else, there exists a strict increasing sequence{kln}∞n=
such that
(p+ )( –mi=biηkiln)( –
m i=aiξip)
kln( –
m i=aiξ
kln–
i )
= –
m
i= biηip+.
Sincemi=biηpi+= , we have
m
i=aiξip= . Thus,
–
m
i=
biηpi+= lim kln→+∞
(p+ )( –mi=biηikln)( –
m i=aiξip)
kln( –
m
i=aiξikln–)
= ,
which is a contradiction to (.). Therefore, there exists two constantsp∈ {, , . . . ,m– } andq∈Z+withq≥p+ such that= .
Lemma . If the condition (C) holds and L is defined by (.), then L is a Fredholm
op-erator of index zero. Define the linear opop-erator Kp:ImL→domL∩KerP with KPx=I+αx, then it is the inverse of L. Furthermore, we have
KpxY≤ωx.
Proof For eachp∈ {, , . . . ,m– }andq∈Z+withq≥p+ , define operatorQ:Z→Zby
Qx(t) =Qx(t)
tp–+Qx(t)
tq–, ∀t∈[, ], (.)
where
Qx(t) =
–Tx(t) +Tx(t), Qx(t) =
Tx(t) –Tx(t). (.)
It is clear thatdim ImQ= . It follows from (.), the definition ofTandTthat
Q(Qx)tp–=
–T(Qx)tp–+T(Qx)tp–
=
–(Qx) +(Qx)
=Qx,
(.)
similarly, we can derive that
Q(Qx)tq–= , Q(Qx)tp–= , Q(Qx)tq–=Qx. (.)
Hence, for eachx∈Zandt∈[, ], it follows from the (.)-(.) that
Qx=Q(Qx)tp–+ (Qx)tq–tp–+Q(Qx)tp–+ (Qx)tq–tq–
= (Qx)tp–+ (Qx)tq–=Qx.
For eachx∈ImL, we haveQx= , i.e.,x∈KerQ. On the other hand, for eachx∈KerQ, we have that
⎧ ⎨ ⎩
–Tx+Tx= ,
Tx–Tx= .
However, the determinant of coefficients is as follows
–
–
= –=
then we haveTjx= (j= , ), i.e.,x∈ImL. Thus,KerQ=ImL.
Take anyx∈Zin the typex= (x–Qx) +Qx, obviously,x–Qx∈KerQ=ImLandQx∈ ImQ, soZ=ImL+ImQ. For anyx∈ImL∩ImQwithx=atp–+btq–, by Lemma ., we
have
asp–+bsq–ds–
m
i= ai
ξi
asp–+bsq–ds= ,
( –s)asp–+bsq–ds–
m
i= bi
ηi
(ηi–s)
asp–+bsq–ds= .
That is, ⎧ ⎨ ⎩
a+b= ,
a+b= ,
but the determinant of coefficients is as follows
= –= ,
we can deduce thata=b= . Hence,ImL∩ImQ= . Furthermore, we getZ=ImL⊕ ImQ. Therefore,dim KerL=dim ImQ=codim ImL= , which means thatLis a Fredholm operator of index zero.
Let operatorP:Y→Yand
Pu(t) =D α– + u(t)|t=
(α) t α–+D
α– + u(t)|t= (α– ) t
α–, ∀t∈[, ]. (.)
It is easy to calculate thatPu(t) =Pu(t); furthermore,Pis a continuous linear projector. Obviously
KerP=u∈Y|D+α–u() =Dα+–u() = .
It is clear thatY=KerL⊕KerP.
For anyx∈ImL, in view of the definition of operatorsKpandL, we haveLKPx=LI+αx= Dα
u∈domL. Therefore, by Lemma . and definitions of operatorsKpandL, we know that
(KpL)u=u, which implies thatKp= [L|domL∩KerP]–. By the definition ofKp, we have
Kpx(t) =I+αx(t) =
(α) t
(t–s)α–x(s) ds, ∀x∈ImL.
It follows from Lemma . that
Dα–
+(Kpx)(t) =
t
(t–s)x(s) ds, Dα–
+ (Kpx)(t) =
t
x(s) ds.
Then, we have
Kpx∞≤
(α)x, D α–
+ (Kpx)∞≤ x, D+α–(Kpx)∞≤ x.
By the definition of the norm in spaceY, we getKpxY≤ωx.
Lemma . Assume ⊂Y is an open bounded subset such thatdomL∩ ¯ =∅, and N is defined by (.), then N is L-compact on ¯.
Proof In order to proveNisL-compact, we only need to prove thatQN(¯) is bounded and
Kp(I–Q)N(u) : ¯ →Yis compact. Since the functionf satisfies Carathéodory conditions
andu∈ ¯, for eachE> , there exists aρE(t)∈L[, ] such that, for a.e.t∈[, ] and every
|u| ≤E, we havef ≤ρE. By the definition of operatorsQandKp on the interval [, ], it
is easy to get thatQN(¯) andKp(I–Q)N(¯) are bounded. Thus, there exists a constant
r> with eacht∈[, ], such that|QNu(t)| ≤r. For all ≤t<t≤, <α≤,u∈ ¯, we have
Kp(I–Q)Nu(t) –Kp(I–Q)Nu(t)
=
(α) t
(t–s)α–– (t–s)α–
(I–Q)Nu(s) ds
+ t
t
(t–s)α–(I–Q)Nu(s) ds
≤
(α)
t
(t–s)α–– (t–s)α–r+ρE(s)ds
+ t
t
(t–s)α–
r+ρE(s)ds
!
≤
(α)
t
ρE(s)(t–s)α–ds–
t
ρE(s)(t–s)α–ds
! + r
(α+ )
tα–tα,
and
Dα–K
p(I–Q)Nu(t) –Dα–Kp(I–Q)Nu(t)
= t
(I–Q)Nu(s) ds– t
(I–Q)Nu(s) ds
≤r(t–t) +
t
t
Since tα is uniformly continuous on [, ] and ρ
E(t)∈L[, ], so Kp(I –Q)N(¯) and
Dα–K
p(I–Q)N(¯) are equicontinuous. By Lemma ., we get thatKp(I–Q)N:Y→Yis
completely continuous.
Lemma . Suppose (H)-(H) hold, then the set ={u∈domL\KerL:Lu=λNu,λ∈
[, ]}is bounded.
Proof Taking anyu∈ , then we haveLu=λNu, which yieldsλ= and Nu∈ImL= KerQ, i.e.,QNu= for allt∈[, ]. It follows from (H) and (H) that there existst∈[, ] such that|Dα–
+u(t)|+|D+α–u(t)| ≤A+B. Then we can get that
Dα–
+ u(t) =Dα+–u(t) +
t t
Dα
+u(s) ds,
Dα+–u(t) =Dα+–u(t) + t
t
Dα+–u(s) ds.
Furthermore, we have that, with settingM=A+B,
Dα+–u()≤Dα+–u(t)∞≤Dα+–u(t)+Dα+u
≤M+Lu≤M+Nu,
(.)
Dα–
+ u()≤D
α–
+ u(t)∞≤D+α–u(t)+D
α– + u∞
≤Dα+–u(t)+D+α–u(t)+Dα+u
≤M+Lu≤M+Nu.
(.)
By (.)-(.) and Lemma ., we have that
PuY≤μ
M+Nu
.
As before, for any u ∈ , we have (I –P)(u)∈domL∩KerP and LP(u) = . From
Lemma . and for eachλ∈(, ], we can get
(I–P)(u)Y=KpL(I–P)(u)Y=Kp(Lu)Y≤ωNu.
Furthermore, we have
uY≤(I–P)(u)Y+P(u)Y≤ρNu+μM.
By (H) and the definition ofN, we have
uY≤ρ
hu∞+rDα+–u∞sDα+–u∞+wDα+–u
θ
∞+D
,
where D=e+μM/ρ. Sincemax{u∞,D+α–u∞,Dα+–u∞} ≤ uY andρ(h+
r+s) < hold true, we can get that
u∞≤ ρ
–ρh
rDα+–u∞+sDα+–u∞+wDα+–u
θ
∞+D
which yield that
Dα+–u∞≤ ρ
–ρh–ρr
sDα+–u∞+wDα+–u
θ
∞+D
.
Furthermore, from the previous inequalities, we know that
Dα+–u∞≤ ρ
–ρh–ρr–ρs
wDα+–u
θ
∞+D
.
Sinceθ∈[, ), there exist constantsm,m,m> such that
Dα–
+ u∞≤m, Dα+–u∞≤m, u∞≤m.
Therefore, is bounded.
Lemma . Suppose (H) and (H) hold, then the set ={u∈KerL:Nu∈ImL} is bounded.
Proof For any u∈ and a,b ∈R, then u(t) =atα–+btα– andQNu= . By (H),
we get that|Dα–
+u(t)|=|a(α)| ≤A, then we have|a| ≤A/(α). By (H), we have that
|Dα–
+ u(t)|=|at(α) +b(α – )| ≤ B, thus |b| ≤ (B+A)/(α – ). Therefore, is
bounded.
Lemma . If the first parts of (H) and (H) hold, then the set ={u∈KerL:λJ–u+
( –λ)QNu= ,λ∈[, ]}is bounded.
Proof Taking anyu∈ anda,b∈R, we haveu(t) =atα–+btα–. For allt∈[, ], we
define the isomorphismJ–:KerL→ImQby
J–atα–+btα–=a
t
p–+b t
q–.
By the definition of the set , we can get that
⎧ ⎨ ⎩
λa+ ( –λ)–TNatα–+btα–+TNatα–+btα–= , λb+ ( –λ)
TN
atα–+btα–– TN
atα–+btα–= . (.)
Ifλ= , we have
TNatα–+btα–= , TNatα–+btα–= .
By the first parts of (H) and (H), similar to the proof of Lemma ., then
|a| ≤ A
(α), |b| ≤
B+A (α– ).
Therefore, is bounded.
If <λ< , we get that |Dα–u(t)| ≤A and|Dα–u(t)| ≤B, similar to the proof of
Lemma ., is bounded. If else, we have thatTNu–TNu< andTNu– TNu> . It contradicts (.), thus is bounded.
Remark . If the other parts of (H) and (H) hold, then the set ={u∈KerL:
–λJ–u+ ( –λ)QNu= ,λ∈[, ]}is bounded.
Now with Lemmas .-. in hands, we can begin to prove our main result - Theo-rem ..
Proof of Theorem . Assume that is a bounded open set ofY with "i= ¯ ⊂ . By Lemma .,NisL-compact on ¯. Then by Lemmas . and ., we have
(i) Lu=λNufor every(u,λ)∈[domL\KerL∩∂ ]×[, ]; (ii) Nu∈/ImLfor everyu∈KerL∩∂ .
Finally, we will prove that (iii) of Lemma . is satisfied. We letIas the identity operator in the Banach spaceYandH(u,λ) =±λJ–(u) + ( –λ)QN(u), according to Lemma . (or
Remark .) we know that for allu∈∂ ∩KerL,H(u,λ)= . By the homotopic property of degree, we have
deg(JQN|KerL,KerL∩ , ) =deg
H(·, ),KerL∩ ,
=degH(·, ),KerL∩ ,
=deg(±I,KerL∩ , )= ,
so (iii) of Lemma . is satisfied.
Consequently, by Lemma ., the equationLu=Nuhas at least one solution indomL∩
¯. Namely, BVPs (.)-(.) have at least one solution in the spaceY.
According to Theorem ., we have the following corollary.
Corollary . Suppose that (H) is replaced by the following condition,
(H) there exist functionsh(t),r(t),s(t),w(t),e(t)∈L[, ]and a constantθ∈[, )such that for all(x,y,z)∈R,t∈[, ],
f(t,x,y,z)≤h(t)|x|+r(t)|y|+s(t)|z|+w(t)|y|θ+e(t),
or
f(t,x,y,z)≤h(t)|x|+r(t)|y|+s(t)|z|+w(t)|x|θ+e(t),
4 An example
Example Consider the following boundary value problem for allt∈(, ):
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Du(t) = u(t) +
D
+u(t)
+
D
+u(t)
+
sinu(t) +cos
t+ ,
I
+u() = , D
+u() =
D +u + D +u , D
+u() = –D
+u + D
+u . (.)
Letα= /, and a= /, ξ= /, a= /, ξ= /,b= –,η= /,b= ,η=
/. We can get that the condition (C) holds, i.e.,i=ai= ,
i=biηi= ,
i=bi= .
Moreover,
f(t,x,y,z) = x+
y+
z+
sin
/x+cost+ .
Thus, we have
f(t,x,y,z)≤ |x|+
|y|+
|z|+
|x|
/+ .
Taking h= /, r= /, s= /, w= /, e= , = ( –
i=biηi)/, =
( –i=biηi)/, = –
i=aiξi, = –
i=aiξi, μ= +–(/) +–(/),
ω= +–(/), ρ= + –(/) +–(/), we can calculate that (H)-(H) hold. Furthermore, we can get
ρh+r+s
= ..
By Corollary ., the BVP (.) has at least one solution inC/[, ].
Competing interests
The authors declare that they have no competing interests.
Author’s contributions
NX designed all the steps of proof in this research and also wrote the article. WBL suggested many good ideas in this article. LSX helped to draft the first manuscript and gave an example to illustrate our result. All authors read and approved the final manuscript.
Acknowledgement
The authors would like to acknowledge the anonymous referee for many helpful comments and valuable suggestions on this article. This work is sponsored by Fundamental Research Funds for the Central Universities (2012LWB44).
Received: 17 January 2012 Accepted: 18 May 2012 Published: 28 June 2012 References
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doi:10.1186/1687-2770-2012-65
