# Existence Results for a Coupled System of Nonlinear Fractional Boundary Value Problems at Resonance

Authors: Ruijuan Liu; Chunhai Kou; Xiujuan Xie
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

Some new Banach spaces are established. Based on those new Banach spaces and by using the coincidence degree theory, we present the existence results for a coupled system of nonlinear fractional differential equations with multipoint boundary value conditions at resonance case.


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## Body

## 1. Introduction

Fractional differential equations (FDEs) have been of great interest for the last three decades. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, and so forth. Therefore, the theory of FDEs has been developed very quickly. Many qualitative theories of FDEs have been obtained. Many important results can be found in [1–13] and the references cited therein.However, there are few articles which consider the boundary value problems at resonance for a coupled system of nonlinear fractional differential equations. In [14], the authors investigated the existence and uniqueness of solutions for the multipoint boundary value problems for fractional differential equations of the form
(1)Dtαy(t)=f(t,y(t),Dtβy(t)),t∈(0,1),y(0)=0,Dtβy(1)-∑i=1m-2ςiDtβy(ξi)=y0,
where 1<α≤2,  0<β<1,  0<ξi<1(i=1,2,…,m-2),  ςi≥0, with γ=∑i=1m-2ςiξiα-β-1<1 and Dtα represents the standard Riemann-Liouville fractional derivative. And the analysis relies on the Schauder fixed point theorem and the Banach contraction principle.In [15], the following coupled system of nonlinear fractional differential equation boundary value problem:
(2)D0+αu(t)=f(t,v(t),D0+β-1v(t)),0<t<1,D0+βv(t)=g(t,u(t),D0+α-1u(t)),0<t<1,u(0)=v(0)=0,u(1)=σ1u(η1),v(1)=σ2v(η2)
was considered. Where 1<α, β≤2,  0<η1,  η2<1, σ1,σ2>0, σ1η1α-1=σ2η2β-1,  D is the standard Riemann-Liouville fractional derivative and f,g:[0,1]×R2→R are continuous. For more works about coupled system involving fractional differential equations, for instance, see [16, 17].In this paper, we investigate the existence of solutions for the boundary value problem with the following form:(3)D0+αu(t)=f(t,v(t),D0+pu(t)),t∈(0,1),D0+βv(t)=g(t,u(t),D0+qv(t)),t∈(0,1),u(0)=0,D0+pu(1)=∑i=1m-2aiD0+pu(ξi),v(0)=0,D0+qv(1)=∑i=1m-2biD0+qv(ηi),
where 1<α, β≤2, 0<p, q<1, α-p-1, β-q-1≥0, ai,bi≥0, 0<ξi,  and  ηi<1(i=1,2,…,m-2).Moreover, at resonance case,
(4)∑i=1m-2aiξiα-p-1=1,∑i=1m-2biηiβ-q-1=1,
a sufficient condition for the existence of solutions is established by the coincidence degree continuation theorem.


## 2. Background Materials and Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. These definitions and lemmas can be found in [4–6].Definition 1.
The fractional integral of orderα>0 of a function  f:(0,∞)→R is given by
(5)I0+αf(t)=1Γ(α)∫0t(t-s)α-1f(s)ds,
provided the right side is pointwise defined on (0,∞).Definition 2.
The fractional derivative of orderα>0 of a function f:(0,∞)→R is given by
(6)D0+αf(t)=1Γ(n-α)(ddt)n∫0t(t-s)n-α-1f(s)ds,
where n=[α]+1, provided the right side is pointwise defined on (0,∞).
It can be directly verified that the Riemann-Liouville fractional integration and fractional differentiation operators of the power functionstβ yield power functions of the same form. For α≥0,  β>-1, there are
(7)I0+αtβ=Γ(β+1)Γ(β+α+1)tβ+α,D0+αtβ=Γ(β+1)Γ(β-α+1)tβ-α.Lemma 3.
Assume thatα>0, y∈C(0,1)∩L1[0,1], then the differential equation D0+αy(t)=0 has solutions
(8)y(t)=c1tα-1+c2tα-2+⋯+cntα-n,
where ci∈R,  i=1,2,…,n,  n=[α]+1.Lemma 4.
Assume thaty∈C(0,1)∩L1[0,1] and D0+αy∈C(0,1)∩L1[0,1], then I0+αD0+αy(t)=y(t)+c1tα-1+c2tα-2+⋯+cntα-n, where ci∈R,  i=1,2,…,n,  n=[α]+1.Lemma 5.
Letα>β>0,t∈[0,1] and y(t)∈L1[0,1], and then
(9)I0+αI0+βy(t)=I0+α+βy(t),D0+αI0+αy(t)=y(t),D0+βI0+αy(t)=I0+α-βy(t).
The following definitions and coincidence degree theory are fundamental in the proofs of our main results.Definition 6.
LetY and Z be Banach spaces. A linear mapping L:dom(L)⊂Y→Z is called a Fredholm mapping if the following two conditions hold:(1)
KerL has a finite dimension;(2)
ImL is closed and has a finite codimension.If L is a Fredholm mapping, its (Fredholm) index is the integer indL=dim KerL-co dim ImL.
And ifL is a Fredholm mapping of index zero, there exist linear continuous projectors P:Y→Y and Q:Z→Z such that ImP=KerL, KerQ=ImL, and Y=KerL⊕KerP,  Z=ImL⊕ImQ. Then it follows that L|domL∩KerP:domL∩KerP→ImL is invertible. We denote the inverse of this map by KP:ImL→domL∩KerP. The generalized inverse of this map is denoted by KP,Q:Y→domL∩KerP, KP,Q=KP(I-Q).Definition 7.
Assume thatL:domL⊂Y→Z is a Fredholm mapping, and  if Ω is an open bounded subset of Y, the map N will be called L-compact on Ω- if QN:Ω→Z is bounded and KP,QN:Ω→Y is compact.
IfImQ is isomorphic to KerL, there exists an isomorphism JNL:ImQ→KerL. We have the coincidence degree continuation theorem which is proved in [18].Theorem 8.
Assume thatL is a Fredholm mapping of index zero and let  N be L-compact on Ω-, where Ω is an open bounded subset of Y. Suppose that the following conditions are satisfied:(1)
Ly≠λNy for each (y,λ)∈[(domL∖KerL)∩∂Q]×(0,1);(2)
Ny∉ImL for each y∈KerL∩∂Q;(3)
deg(JNLQN∣KerL,Ω∩KerL,0)≠0, where Q:Z→Z is a continuous projection as mentioned above with ImL=KerQ and JNL:ImQ→KerL is any isomorphism.Then the equation Ly=Ny has at least one solution in domL∩Ω-.LetZ1=L1[0,1] with the norm ∥x∥1=∫01|x(s)|ds. For 0<p, q<1, we define a linear space
(10)Cp[0,1],Cq[0,1]:Y1=Cp[0,1]={u(t)∣u(t)=I0+px(t),x∈C[0,1],t∈[0,1]},Y2=Cq[0,1]={v(t)∣v(t)=I0+qy(t),y∈C[0,1],t∈[0,1]},
to be accompanied by
(11)∥u∥Y1=maxt∈[0,1]|D0+pu(t)|+maxt∈[0,1]|u(t)|,∥v∥Y2=maxt∈[0,1]|D0+qv(t)|+maxt∈[0,1]|v(t)|.We can prove thatY=Y1×Y2 is a Banach space with the norm∥(u,v)∥Y=max{∥u∥Y1,∥u∥Y2}, and Z=Z1×Z1 is a Banach space, with the norm∥(x,y)∥Z=max{∥x∥1,∥y∥1}.LetL1 to be the linear operator from domL1∩Y1 to Z1 with
(12)domL1={∑i=1m-2aiD0+pu(ξi)u∈Cp[0,1]∣D0+αu∈L1[0,1],u(0)=0,D0+pu(0)=0,D0+py(1)=∑i=1m-2aiD0+pu(ξi)}.
And L1u=D0+αu, u∈domL1.  Let L2 be the linear operator from domL2∩Y2 to Z1 with
(13)domL2={∑j=1m-2bjD0+qv(ηj)v∈Cq[0,1]∣D0+βv∈L1[0,1],v(0)=0,D0+qv(0)=0,D0+qv(1)=∑j=1m-2bjD0+qv(ηj)}.
And L1v=D0+βv,  v∈domL2. Define L to be the linear operator from domL∩Y to Z with
(14)domL={(u,v)∈Y∣u∈domL1,v∈domL2},
and L(u,v)=(L1u,L2v). We define N:Y→Z by setting
(15)N(u,v)=(N1(u(t),v(t)),N2(u(t),v(t))),
where N1(u(t),v(t))=f(t,v(t),D0+pu(t)),   N2(u(t),v(t))=g(t,u(t),D0+qv(t)).Then the coupled system of BVPs (3) can be written as L(u,v)=N(u,v).Lemma 9 (see [19]).
F
⊂
C
μ
[
0,1
],  0<μ<1, is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here uniformly bounded means that there exists M>0 such that for every u∈F,
(16)∥y∥Cμ=maxt∈J|D0+μu(t)|+maxt∈J|u(t)|<M,
and equicontinuous means that for all ε>0,  ∃δ>0, for all t1,t2∈[0,1],  |t1-t2|<δ,  u∈F, there hold |u(t1)-u(t2)|<ε,  |D0+βu(t1)-D0+βu(t2)|<ε.Definition 10 (see [19]).
One  says thatf:[0,1]×R2→R satisfies the Carathéodory  condition if (i) for each (x,y)∈R2, the function t∈[0,1]→f(t,x,y)∈R is measurable on [0,1]; (ii) for a.e. t∈[0,1], the function (x,y)∈R2→f(t,x,y)∈R is continuous on R2; (iii) for each r>0, there exists φr(t)∈L1[0,1] such that |f(t,x,y)|≤φr(t) for a.e. t∈[0,1] and all (x,y)∈R2 with x2+y2≤r.


## 3. Main Results

In this section, we will give and prove our main results.Lemma 11.
Letf,g∈Z1,  J=[0,1],  ∑i=1m-2aiξiα-p-1=1,  ∑i=1m-2biηiβ-q-1=1. Then u,v∈Y are, respectively, the solutions of the following boundary value problems:
(17)D0+αu(t)=f(t),t∈J,u(0)=0,D0+pu(1)=∑i=1m-2aiD0+pu(ξi),(18)D0+βv(t)=g(t),t∈J,v(0)=0,D0+qv(1)=∑i=1m-2biD0+qv(ηi)
if and only if u,v satisfied
(19)u(t)=c1tα-1+I0+αf(t),∑i=1m-2aiξiα-p-1∫01(1-s)α-p-1f(s)ds=∫0ξi(ξi-s)α-p-1f(s)ds,v(t)=c2tβ-1+I0+βg(t),∑i=1m-2biηiβ-q-1∫01(1-s)β-q-1g(s)ds=∫0ηi(ηi-s)β-q-1g(s)ds.Proof.
(
u
,
v
) in Y1×Y2 are the solutions of the following boundary value problem:
(20)I0+αD0+αu(t)=I0+αf(t)=1Γ(α)∫0t(t-s)α-1f(s)ds,I0+βD0+βu(t)=I0+βg(t)=1Γ(β)∫0t(t-s)β-1g(s)ds.
According to Lemma 4, we have
(21)u(t)=c1tα-1+c2tα-2+I0+αf(t),v(t)=c3tβ-1+c4tβ-2+I0+βg(t).
By  u(0)=v(0)=0, we get c2=c4=0, and  then u(t)=c1tα-1+I0+αf(t), v(t)=c3tβ-1+I0+βg(t).
Using  (7) and Lemma 5, we get that
(22)D0+pu(t)=D0+pI0+αf(t)+c1Γ(α)Γ(α-p)tα-p-1=I0+α-pf(t)+c1Γ(α)Γ(α-p)tα-p-1,D0+qv(t)=D0+qI0+βg(t)+c3Γ(β)Γ(β-q)tβ-q-1=I0+β-qg(t)+c3Γ(α)Γ(β-q)tβ-q-1.
By the boundary conditions, ∑i=1m-2aiξiα-p-1=1,  ∑i=1m-2biηiβ-q-1=1, we have
(23)∑i=1m-2aiξiα-p-1∫01(1-s)α-p-1f(s)ds=∫0ξi(ξi-s)α-p-1f(s)ds,(24)∑i=1m-2biηiβ-q-1∫01(1-s)β-q-1g(s)ds=∫0ηi(ηi-s)β-q-1g(s)ds.
Conversely, suppose that (23) and (24) hold. In view of Lemma 5, we can easily verify that u,v are the solutions of (17) and (18), respectively. Then, we complete the proof.Lemma 12.
The mappingL:dom(L)⊂Y→Z is a Fredholm operator of index zero.Proof.
SinceKerL1={u∣D0+αu(t)=0}, KerL2={v∣D0+βv(t)=0}, it can be directly verified that KerL1={u∣u=c1tα-1,c1∈R}, KerL2={v∣v=c2tβ-1,c2∈R}. It is clear that KerL={c1tα-1,c2tβ-1∣c1,c2∈R}.
Let(x,y)∈ImL; then there exists (u,v)∈domL such that (x,y)=L(u,v), that is, u∈Y1, x=D0+αu and v∈Y2, y=D0+βv. From these results, we can easily deduce the results
(25)ImL={(x,y)∈Z∣xsatisfies(23),ysatisfies(24)}.
Lett∈(0,1), ∑i=0m-2aiξiα-p-1=1,  ∑i=0m-2biηiβ-q-1=1. Assume that
(26)Λ1=∑i=1m-2ai(ξiα-p-1∫01(1-s)α-p-1skdsΛ1=Λ1=---∫0ξi(ξi-s)α-p-1skds)≠0,Λ2=∑i=1m-2bi(ηiβ-q-1∫01(1-s)β-q-1skdsΛ2=Λ2=---∫0ηi(ηi-s)β-q-1skds)≠0.
Consider the auxiliary mapping,
(27)Q1:Z⟶R:Q1x=∑i=1m-2ai(∫0ξi(ξi-s)α-p-1x(s)dsξiα-p-1∫01(1-s)α-p-1x(s)dsQ1x====-∫0ξi(ξi-s)α-p-1x(s)ds),Q2:Z⟶R:Q2y=∑i=1m-2bi(ηiβ-q-1∫01(1-s)β-q-1y(s)dsQ1x===-∫0ηi(ηi-s)β-q-1y(s)ds).
It is obvious that  Q1,Q2 are  continuous linear mappings. Also we have that Q1(tk)=Λ1, Q2(tk)=Λ2. Take the mapping Q:Z→Z defined by Q(x,y)=(Q1′x,Q2′y). Q1′x=(Q1x/Λ1)tk, Q2′x=(Q2x/Λ2)tk, where x,y∈Z. Note that  (x,y)∈ImL implies Q(x,y)=(Q1x,Q2y)=(0,0). Evidently,
(28)ImQ={(x,y)∣x=c1tk,y=c2tk,c1,c2∈R},
and Q:Z→Z is a continuous linear projector. In fact, for any (x,y)∈Z, we have
(29)Q1(Q1′x)=Q1(Q1xΛ1tk)=Q1xΛ1Q1(tk)=Q1xΛ1Λ1=Q1x,Q2(Q2′y)=Q2(Q2yΛ2tk)=Q2yΛ2Q2(tk)=Q2yΛ2Λ2=Q2y,Q1′2x=Q1′(Q1′x)=Q1(Q1′x)Λ1tk=Q1xΛ1tk=Q1′x,Q2′2y=Q2′(Q2′y)=Q2(Q2′y)Λ2tk=Q2yΛ2tk=Q2′y,
that is to say, Q:Z→Z is idempotent.
Observe that(x,y)∈ImL leads to Q(x,y)=(0,0) and (x,y)∈KerQ, where (0,0) is the element in Z. Conversely, if (x,y)∈KerQ, we can have that Q(x,y)=(0,0), that is to say, (x,y)∈ImL. So, KerQ=ImL.
Let(x,y)=(x,y)-Q(x,y)+Q(x,y)=(I-Q)(x,y)+Q(x,y), where (x,y)∈Z is an arbitrary element. Since Q(x,y)∈ImQ and (I-Q)(x,y)∈KerQ, we obtain that Z=ImQ+KerQ. Take (x0,y0)∈ImQ∩KerQ, (x0,y0) can be written as (x0,y0)=(c1tk,c2tk), c1,c2∈R. For (x0,y0)∈ImQ then since (x0,y0)∈KerQ=ImL, by (27), we get that
(30)Q(x0,y0)=Q(c1tk,c2tk)=(c1Q1(tk),c2Q2(tk))=(c1Λ1,c2Λ2)=(0,0),
which implies that (x0,y0)=(0,0). Therefore, ImQ∩KerQ={0,0}, and  thus Z=ImQ⊕KerQ=ImQ⊕ImL.
Now,dim KerL=1=dim ImQ=co dim KerQ=co dim ImL<+∞, and observing that ImL is closed in Z, so L is a Fredholm mapping of index zero.
LetP:Y→Y be defined by
(31)(P1u)(t)=D0+pu(t)|t=0tα-1,t∈J,u∈Y1,(P2v)(t)=D0+qv(t)|t=0tβ-1,t∈J,v∈Y2,
and P(u,v)=(P1u,P2v)  it is clear that P1, P2, P are linear continuous projectors and ImP=KerL={c1tα-1,c2tβ-1∣c1,c2∈R}. Also, proceeding as the proof of Lemma 12, we can show that Y=ImP⊕KerP=KerL⊕KerP.Lemma 13.
Assume thatf,g:[0,1]×R2→R satisfies the Carathéodory conditions
(32)∑i=1m-2aiξiα-p-1=1,∑i=1m-2biηiβ-q-1=1,α-p-1≥0,β-q-1≥0,Λ1≠0,Λ2≠0.
Then KP(I-Q)N:Y→Y is completely continuous.Here we omit reasoning process, and  it is easy to prove.Consider the mappingKP:ImL→domL∩KerP,KP(x,y)=(I0+αx,I0+βy),
(33)(KPx)(t)=I0+αx(t)=1Γ(α)∫0t(t-s)α-1x(s)ds,x∈ImL1,(KPy)(t)=I0+βy(t)=1Γ(β)∫0t(t-s)β-1y(s)ds,y∈ImL2.For(x,y)∈ImL, we have (KPL)(x,y)=KP(L(x,y))=(x,y).(
L
K
P
)
(
x
,
y
)
=
L
(
K
P
(
x
,
y
)
)
=
(
x
,
y
),  for all (x,y)∈ImL. Thus, KP=(LP)-1, where LP=L|domL∩KerP:domL∩KerP→ImL; hence, for each (x,y)∈ImL and t∈J, we have
(34)∥KP(x,y)∥Y=∥(I0+αx,I0+βy)∥Y=maxt∈J{∥I0+αx∥Y1,∥I0+βy∥Y2}=maxt∈J{∥D0+qI0+βy∥∥D0+pI0+αx∥+∥I0+αx∥,=maxt∈J=∥D0+qI0+βy∥+∥I0+βx∥}=maxt∈J{|1Γ(α-p)∫0t(t-s)α-p-1x(s)ds|+=maxt∈J+|1Γ(α)∫0t(t-s)α-1x(s)ds|,+=maxt∈J|1Γ(β-q)∫0t(t-s)β-q-1y(s)ds|+=maxt∈J+|1Γ(β)∫0t(t-s)β-1y(s)ds|}≤maxt∈J{(1Γ(α-p)+1Γ(α))∥x∥1,≤≤≤maxt∈J(1Γ(β-q)+1Γ(β))∥y∥1}=maxt∈J{η1∥x∥1,η2∥y∥1},
where
(35)η1=1Γ(α-p)+1Γ(α),η2=1Γ(β-q)+1Γ(β).
The operator KP(I-Q)N:Y→Y is continuous in view of the continuity of the function f and g. Let Ω⊂Y be bounded; that is to say, there exists a positive constant M>0 such that ||(x,y)||Y≤M for all (x,y)∈Ω.We can derive that(36)max(x,y)∈Ω|KP(I-Q)N(x,y)|=max(x,y)∈Ω{|I0+α{Nx-QNx}|,|I0+β{Ny-QNy}|}=max(x,y)∈Ω{I0+β(g(t,y(t),D0+qy))|I0+α{f(t,x(t),D0+px)-Q(f(t,x(t),D0+px))}|,=max(x,y)∈Ω==|I0+β{g(t,y(t),D0+qx)-Q(g(t,y(t),D0+qy))}|}≤max(x,y)∈Ω{1Γ(α)∥(g(t,y(t),D0+qy))g(t,y(t),D0+qy)≤max(x,y)∈Ω--Q--Q-Q(g(t,y(t),D0+qy))∥1,≤max(x,y)∈Ω--1Γ(β)∥(g(t,y(t),D0+qy))g(t,y(t),D0+qy)≤max(x,y)∈Ω------Q(g(t,y(t),D0+qy))∥11Γ(α)}≤max(x,y)∈Ω{1Γ(α)M1,1Γ(β)M2}.
Here,
(37)M1=max∥x∥Cp≤M∥f(t,x(t),D0+px)-Q(f(t,x(t),D0+px))∥1,M2=max∥y∥Cq≤M∥f(t,y(t),D0+qy)-Q(f(t,y(t),D0+qy))∥1.
By Lemma 5,
(38)maxx∈Ω|D0+pKP(I-Q)Nx|=maxx∈Ω|D0+pI0+α{Nx-QNx}|=maxx∈Ω|I0+α{f(t,x(t),D0+px)-Q(f(t,x(t),D0+px))}|≤1Γ(α-p)∥f(t,x(t),D0+px)-Q(f(t,x(t),D0+px))∥1≤1Γ(α-p)M1.
Similarly, we can prove that maxy∈Ω|D0+qKP(I-Q)Ny|≤(1/Γ(β-q))M2.Hence,(39)∥KP(I-Q)N(x,y)∥Y≤max(x,y)∈Ω{(1Γ(α-p)+1Γ(α))M1,m≤max(x,y)∈Ω(1Γ(β-q)+1Γ(β))M2},
that is to say, KP(I-Q)N(Ω)⊂Y is bounded.  On the other hand, let
(40)M3=max∥x∥Y1≤Mmaxt∈J|f(t,x(t),D0+px)-Q(f(t,x(t),D0+px))|<+∞,M4=max∥y∥Y2≤Mmaxt∈J|f(t,y(t),D0+qy)-Q(f(t,y(t),D0+qy))|<+∞.
For every (x,y)∈Ω, 0≤t1≤t2≤1,   for all ε>0, there is δ=min{(Γ(α-p)/(M1+1))ε,(Γ(β-q)/(M2+1))ε}, and t2-t1<δ such that
(41)|D0+pKP(I-Q)Nx(t2)-D0+pKP(I-Q)Nx(t1)|=|I0+α-p[f(t2,x(t2),D0+px(t2))======--Q(f(t2,x(t2),D0+px(t2)))(t2,x(t2),D0+px(t2))]-I0+α-p-I0+α-p[(f(t1,x(t1),D0+px(t1)))f(t1,x(t1),D0+px(t1))-I0+α-p-I0+α-p--Q(f(t1,x(t1),D0+px(t1)))]|=1Γ(α-p)|∫t1t2(t-s)α-p-1=1Γ(α-p)====×[f(s,x(s),D0+px(s))=1Γ(α-p)--===--Q(f(s,x(s),D0+px(s)))]ds∫t1t2(t-s)α-p-1f(s,x(s),D0+px(s))|≤1Γ(α-p)M1δ<ε,(42)|D0+qKP(I-Q)Ny(t2)-D0+qKP(I-Q)Ny(t1)|<ε.
This ensures that KP(I-Q)N(Ω) is equicontinuous too. Then an application of Ascoli-Arezela theorem ensures that KP(I-Q)N:Y→Y is completely continuous.Theorem 14.
Assume thatf,g:[0,1]×R2→R satisfy the Carathéodory conditions, t∈[0,1],  Λ1,Λ2≠0, 1<α,  β≤2,  0<p,  q<1,  α-p-1≥0,  β-q-1≥0, and ∑i=1m-2aiξiα-p-1=1,  ∑i=1m-2biηiβ-q-1=1. Assume that all of the following hypotheses hold.(H1)
There exist nonnegative functionsai,bi,ci∈Z, i=1,2,t∈J, such that for all (x,y)∈R2,
(43)|f(t,x,y)|≤a1(t)|x|+b1(t)|y|+c1(t),|g(t,x,y)|≤a2(t)|x|+b2(t)|y|+c2(t).(H2)
0<η1′η1′′<1, 0<η2′η2′′<1, where η1,η2 is defined by (35) and η1′′=∥a1∥1+∥b1∥1, η2′′=∥a2∥1+∥b2∥1, η1′=(Γ(p+1)+1)/Γ(α-p)+η1,η2′=(Γ(q+1)+1)/Γ(β-q)+η2.(H3)
There exist constantsT1,T2>0 such that Q1(Nu)≠0, Q2(Nv)≠0, for each u∈domL1,  v∈domL2 satisfying |D0+pu(t)|>T1, |D0+qv(t)|>T2,  for all t∈[0,1].(H4)
There exist constantsS1,S2>0 such that for any c1,c2∈R, if  |c1|>S1, |c2|>S2, either
(44)c1Λ1Q1(N(c1tα-1))<0,c2Λ2Q2(N(c2tβ-1))<0,or else
(45)c1Λ1Q1(N(c1tα-1))>0,c2Λ2Q2(N(c2tβ-1))>0.
Then the problem (3) has at least one solution in domL.Proof.
The proof consists of four main steps.
Step   1. Set Ω1={(u,v)∈domL∖KerL∣L(u,v)=λN(u,v),λ∈[0,1]}, and prove Ω1 is bounded. In fact, for (u,v)∈Ω1, then (u,v)∈domL∖KerL and L(u,v)=λN(u,v), so λ≠0 and N(u,v)∈ImL=KerQ⊂Z. Hence, Q(N(u,v))=0, that is, Q1(N1u(t))=0,Q2(N2v(t))=0. From (H3), we have that t0∈[0,1] such that |D0+pu(t0)|≤T1,  |D0+qv(t0)|≤T2.
Again, for(u,v)∈Ω1, we get
(46)∥Pu∥Y1=maxu∈Ω1|D0+pPu|+maxu∈Ω1|Pu|=maxu∈Ω1|D0+p((D0+pu(t0))tp)|+maxu∈Ω1|(D0+pu(t0))tp|≤|D0+pu(t0)|Γ(p+1)+|D0+pu(t0)|,∥Pv∥Y2≤|D0+qv(t0)|Γ(q+1)+|D0+qv(t0)|.
In view of Lemma 5, we have
(47)|D0+pu(t0)|=|D0+pu(t0)-I0+α-p(D0+pu(t0))|≤|D0+pu(t0)|+1Γ(α-p)∥D0+pu(t)∥1≤T1+1Γ(α-p)∥L1u∥1≤T1+1Γ(α-p)∥Nu∥1.
Similarly, we can also obtain |D0+qv(t0)|≤T2+(1/Γ(β-q))∥L2v∥1.
Thus,(48)∥P(u,v)∥Y≤max{[Γ(p+1)+1][T1+1Γ(α-p)∥Nu∥1],≤max≤[Γ(q+1)+1][T2+1Γ(β-q)∥Nv∥1]}.
According to Lemma 13, we have
(49)∥(I-P)(u,v)∥Y=∥KPL(I-P)(u,v)∥Y≤max{η1∥L(I-P)u∥1,η2∥L(I-P)v∥1}=max{η1∥Lu∥1,η2∥Lv∥1}≤max{η1∥Nu∥1,η2∥Nv∥1}.
Combining (48) and (49), we can obtain
(50)∥(u,v)∥Y=∥(I-P)u+Pu∥Y≤∥P(u,v)∥Y+∥(I-P)(u,v)∥Y=max{[Γ(p+1)+1]T1+η1∥Nu∥1,m=max[Γ(q+1)+1]T2+η2∥Nv∥1}.
From (H1), for each u∈Ω1, we have
(51)∥Nu∥1=∫01|Nu(s)|ds=∫01|f(s,u(s),D0+pu(s))|ds≤∫01|a1(s)||u(s)|ds+∫01|b1(s)||D0+pu(s)|ds+∫01|c1(s)|ds≤(∫01|a1(s)|ds+∫01|b1(s)|ds)∥u∥Y1+∥c1∥1=η1′′∥u∥Y1+η3.
Also, we can obtain that
(52)∥Nv∥1=∫01|Nv(s)|ds≤η2′′∥v∥Y2+η4,
where η3=∥c1∥1,η4=∥c2∥1. Thus, by  (H2) and (50)–(52) we can derive that
(53)∥(u,v)∥Y≤max{[Γ(p+1)+1]T1+η1η31-η1η1′′,≤mmax[Γ(q+1)+1]T2+η2η41-η2η2′′},
which clearly states that Ω1 is bounded.
Step   2. Set Ω2={(u,v)∈KerL∣N(u,v)∈ImL} and prove that  Ω2 is bounded. Let (u,v)∈Ω2; then
(54)(u,v)∈KerL={(u,v)∈domL∣u=c1tα-1,v=v=c2tβ-1,∀c1,c2∈R,t∈(0,1)}.
Since N(u,v)∈ImL=KerQ, we have Q(N(u,v))=(0,0), that is, Q1(N(c1tα-1))=0, Q2(N(c2tβ-1))=0. Taking account of (H4), |c1|≤S1, |c2|≤S2, which implies that Ω2 is bounded.
Step   3. Set Ω3={(u,v)∈KerL∣-λ(u,v)+(1-λ)JNLQN(u,v)=(0,0), λ∈[0,1]}; if (35) holds, we will set
(55)Ω3={(u,v)∈KerL∣λ(u,v)+(1-λ)JNLQN(u,v)=(0,0)=(0,0),λ∈[0,1]},
if (44) holds, where JNL:ImQ→KerL is a linear isomorphism defined as
(56)JNL(c1tk)=c1tα-1,JNL(c2tk)=c2tβ-1,∀c1,c2∈R,t∈(0,1).
And then we will prove that  Ω3 is bounded. Without loss of generality, in the following part of the proof, we assume that (35) holds in (H4), for (u,v)∈Ω3; we have u=c1tα-1, v=c2tβ-1, c1,c2∈R,  t∈[0,1], and λ(u,v)=(1-λ)JNLQN(u,v) Thus,
(57)λtα-1=(1-λ)Q1(N(c1tα-1))c1Λ1tα-1,λtβ-1=(1-λ)Q2(N(c2tβ-1))c2Λ2tβ-1.
Therefore, via (H4) and (35), we have |c1|≤S1, |c2|≤S2, which show that Ω3 is bounded.
Step  4. Let  Ω be an bounded open set such that Ω⊃⋃i=13Ω-i and prove that deg(JNLQN|KerL,Ω∩KerL,0)≠0. The operator N is L-compact on Ω- due to the fact that QN(Ω-) is bounded and KP,QN=KP(I-P)N:Ω-→Y is compact by Lemma 13; then by Step 1 and Step 2, we have(i)
L(u,v)≠λN(u,v) for each ((u,v),λ)∈[(domL∖KerL)∩∂Ω]×(0,1);(ii)
N(u,v)∉ImL for each (u,v)∈KerL∩∂Ω.
DefineH((u,v),λ)=-λI(u,v)+(1-λ)JNLQN(u,v), where I is the identity operator in Y. According to the arguments in Step 3, we have H((u,v),λ)≠0, for all (u,v)∈KerL∩∂Ω, and therefore, via the homotopy property of degree, we obtain that
(58)deg(JNLQN|KerL,Ω∩KerL,0)=deg(H(·,0),Ω∩KerL,0)=deg(H(·,1),Ω∩KerL,0)=deg(-I,Ω∩KerL,0)=-1≠0,
which certifies the condition (3) of Theorem 8. Then, applying Theorem 8, we conclude that the problem (3) has at least one solution in domL∩Ω-. The proof is complete.


## 4. Conclusion

In this paper, we obtain the existence results of muti-point boundary value problem at resonance for a coupled system of nonlinear fractional differential equations by means of the coincidence degree theory.



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*Source: 267386-2013-08-21.xml*