# Existence of Solutions for Two-Point Boundary Value Problem of Fractional Differential Equations at Resonance

Authors: Lei Hu; Shuqin Zhang; Ailing Shi
Journal: International Journal of Differential Equations
 (2014)

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## Abstract

We establish the existence results for two-point boundary value problem of fractional differential equations at resonance by means of the coincidence degree theory. Furthermore, a result on the uniqueness of solution is obtained. We give an example to demonstrate our results.


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## Body

## 1. Introduction

Fractional differential equations have been studied extensively. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications such as physics, chemistry, phenomena arising in engineering, economy, and science; see, for example, [1–5].Recently, more and more authors have paid their attentions to the boundary value problems of fractional differential equations; see [6–21]. Moreover, there have been many works related to the existence of solutions for boundary value problems at resonance; see [12–21]. It is considerable that there are many papers that have dealt with the solutions of multipoint boundary value problems of fractional differential equations at resonance (see, e.g., [12, 16]).In [12], Bai and Zhang considered a three-point boundary value problem of fractional differential equations with nonlinear growth given by
(1)D0+αu(t)=f(t,u(t),D0+α-1u(t)),0<t<1,u(0)=0,u(1)=σu(η),
where 1<α≤2, 0<η, σ<1>0, σηα-1=1, D0+α is Riemann-Liouville fractional derivative, and f,g:[0,1]×R2→R are given functions.In [13], Hu et al. have studied a two-point boundary value problem for fractional differential equation at resonance
(2)D0+αx(t)=f(t,x(t),x′(t)),0≤t≤1,x(0)=0,x′(0)=x′(1),
where 1<α≤2, D0+α is Caputo fractional derivative, and f:[0,1]×R2→R satisfies Carathéodory conditions.As far as we know, there are few works on the existence of two-point boundary value problems of the fractional differential equations at resonance. Motivated by the works above, we discuss the existence and uniqueness of solutions for the following nonlinear fractional differential equation:(3)D0+αu(t)=f(t,u(t),D0+α-1u(t),D0+α-2u(t),…,D0+α-(N-1)u(t)),u(0)=D0+α-2u(0)=⋯=D0+α-(N-1)u(0)=0,D0+α-1u(0)=D0+α-1u(1),
where 0<t<1, N-1<α<N, D0+α is Riemann-Liouville fractional derivative, and f:[0,1]×R2→R is continuous function.More precisely, we use the coincidence degree theorem due to Mawhin [22]. The rest of this paper is organized as follows. In Section 2, we give some necessary notations, definitions, and lemmas. In Section 3, we study the existence of solutions of (3) by the coincidence degree theory. Finally, an example is given to illustrate our results in Section 4.The two-point boundary value problem (3) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem
(4)D0+αu(t)=0,u(0)=D0+α-2u(0)=⋯=D0+α-(N-1)u(0)=0,D0+α-1u(0)=D0+α-1u(1),
has u(t)=c1tα-1 as a nontrivial solution.


## 2. Preliminaries

In this section, we present the necessary definitions and lemmas from fractional calculus theory. These definitions and properties can be found in the literature. For more details see [1–3].Definition 1 (see [1]).
The Riemann-Liouville 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 that the right-hand side is pointwise defined on (0,∞).Definition 2 (see [1]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function f:(0,∞)→R is given by
(6)D0+αf(t)=1Γ(n-α)dndtn∫0t‍f(s)(t-s)α-n+1ds,
where n-1<α≤n, provided that the right-hand side is pointwise defined on (0,∞).Lemma 3 (see [1]).
Letn-1<α≤n, u∈C(0,1)⋂L1(0,1); then
(7)I0+αD0+αu(t)=u(t)+C1tα-1+C2tα-2+⋯+Cntα-n,
where Ci∈R, i=1,2,…n.Lemma 4 (see [1]).
Ifα>0, m∈N and D=d/dx. If the fractional derivatives D0+αu(t) and D0+α+mu(t) exist, then
(8)DmD0+αu(t)=D0+α+mu(t).Lemma 5 (see [1]).
The relation(9)Ia+αIa+βf(x)=Ia+α+βf(x)
is valid in following cases β>0, α+β>0, and f(x)∈L1(a,b).Now let us recall some notations about the coincidence degree continuation theorem.LetY, Z be real Banach spaces, let L:dom⁡L⊂Y→Z be a Fredholm map of index zero, and let P:Y→Y, Q:Z→Z be continuous projectors such that ker⁡L=Im⁡P, Im⁡L=ker⁡Q, and Y=ker⁡L⊕ker⁡P, Z=Im⁡L⊕Im⁡Q. It follows that L|dom⁡L∩ker⁡P:dom⁡L∩ker⁡P→Im⁡L is invertible. We denote the inverse of this map by KP. If Ω is an open bounded subset of Y, the map N will be called L-compact on Ω¯ if QN(Ω¯) is bounded and KP,QN=KP(I-Q)N:Ω¯→Y is compact.Theorem 6.
LetL be a Fredholm operator of index zero and N be L-compact on Ω¯. Suppose that the following conditions are satisfied:(1)
Lx≠λNx for each (x,λ)∈[(dom⁡L∖ker⁡L)∩∂Ω]×(0,1);(2)
Nx∉Im⁡L for each x∈ker⁡L∩∂Ω;(3)
deg⁡(JQN|ker⁡L,Ω∩ker⁡L,0)≠0, where Q:Z→Z is a continuous projection as above with Im⁡L=ker⁡Q and J:Im⁡Q→ker⁡L is any isomorphism.Then the equation Lx=Nx has at least one solution in dom⁡L∩Ω¯.


## 3. Main Results

In this section, we will prove the existence results for (3).We use the Banach spaceE=C[0,1] with the norm ∥u∥∞=max⁡0≤t≤1|u(t)|. For α>0, N=[α]+1, we define a linear space
(10)X={u∣u,D0+α-iu∈E,i=1,2,…,N-1}.
By means of the functional analysis theory, we can prove that X is a Banach space with the norm ∥u∥X=∥D0+α-1u∥∞+⋯+∥D0+α-(N-1)u∥∞+∥u∥∞.DefineL to be the linear operator from dom⁡(L)⋂X to E with dom⁡(L)={u∈X∣D0+αu(t)∈E,u(0)=D0+α-2u(0)=⋯=D0+α-(N-1)u(0)=0,D0+α-1u(0)=D0+α-1u(1)} and
(11)Lu=D0+αu,u∈dom⁡(L).
We define N:X→E by
(12)Nu(t)=f(t,u(t),D0+α-1u(t),D0+α-2u(t),…,D0+α-(N-1)u(t)).
Then the problem (3) can be written by Lu=Nu.Lemma 7.
The mappingL:dom⁡(L)⊂E is a Fredholm operator of index zero.Proof.
It is clear that(13)ker⁡(L)={c1tα-1}≅R1.
Let x∈Im⁡L, so there exists a function u∈dom⁡L which satisfies Lu=x. By (11) and Lemma 3, we have
(14)u(t)=I0+αx(t)+c1tα-1+c2tα-2+⋯+cNtα-N.
By u(0)=D0+α-2u(0)=⋯=D0+α-(N-1)u(0)=0, we can obtain c2=⋯=cN=0. Hence
(15)u(t)=I0+αx(t)+c1tα-1.
Then, we have
(16)D0+α-1u(t)=D0+α-1(I0+αx(t)+c1tα-1)=D0+α-1I0+αx(t)+c1Γ(α)Γ(1)=∫0t‍x(s)ds+c1Γ(α).
Taking into account D0+α-1u(0)=D0+α-1u(1), we obtain
(17)∫01‍x(s)ds=0.
On the other hand, suppose x satisfy ∫01‍x(s)ds=0. Let u(t)=I0+αx(t), we can easily prove u(t)∈dom⁡(L).
Thus, we conclude that(18)Im⁡(L)={x:∫01‍x(s)ds=0}.
Consider the linear operators Q:E→E defined by
(19)Qx(t)=∫01‍x(s)ds.
Take x(t)∈E; then
(20)Q(Qx(t))=Q(∫01‍x(s)ds)=∫01‍(∫01‍x(t)dt)ds=∫01‍x(s)ds=Qx(t).
We can see Q2=Q.
Forx(t)∈E in the type x(t)=x(t)-Qx(t)+Qx(t), obviously, x(t)-Qx(t)∈Ker⁡(Q)=Im⁡(L) and Qx(t)∈Im⁡(Q). That is to say, E=Im⁡(L)+Im⁡(Q). If u∈Im⁡(L)⋂Im⁡(Q), we have u=c1; then ∫01‍c1ds=0. As a result c1=0, and we get E=Im⁡(L)⊕Im⁡(Q).
Note thatIndL=dim⁡ker⁡L-codimIm⁡L=0. Then L is a Fredholm mapping of index zero.We can define the operatorsP:X→X, where
(21)Pu=1Γ(α)D0+α-1u(0)tα-1.
For u∈X,
(22)P(Pu)=P(1Γ(α)D0+α-1u(0)tα-1)=1Γ(α)D0+α-1u(0)tα-1=Pu.
So we have P2=P.Note that(23)Ker⁡(P)={u:D0+α-1u(0)=0}.
Since u=u-Pu+Pu, it is easy to say that u-Pu∈Ker⁡(P) and Pu∈Ker⁡(L). So we have X=Ker⁡(P)+Ker⁡(L). If u∈Ker⁡(L)⋂Ker⁡(P), then u=c1tα-1. We can derive c1=0 from D0+α-1c1tα-1|t=0=0. Then
(24)X=Ker⁡(L)⊕Ker⁡(P).
For u∈X,
(25)∥Pu∥X=1Γ(α)|D0+α-1u(0)|·∥tα-1∥X=1Γ(α)|D0+α-1u(0)|·[∥D0+α-(N-1)tα-1∥∞∥tα-1∥∞+∥D0+α-1tα-1∥∞+⋯+∥D0+α-(N-1)tα-1∥∞]=(∑i=1N-1‍1Γ(i)+1Γ(α))|D0+α-1u(0)|=a|D0+α-1u(0)|,
where a=1/Γ(α)+∑i=1N-1‍(1/Γ(i)).We defineKP:Im⁡L→dom⁡L∩ker⁡P by Kpx=I0+αx.Forx∈Im⁡(L), we have
(26)LKpx=LI0+αx=D0+αI0+αx=x.
For u∈dom⁡(L)⋂Ker⁡(P), we have D0+α-1u(0)=0. And for u∈dom⁡(L), the coefficients c1,…,cN in the expressions
(27)u=I0+αD0+αu(t)+c1tα-1+c2tα-2+⋯+cNtα-N
are all equal to zero. Thus, we obtain
(28)KpLu=I0+αD0+αu=u.
This shows that Kp=(Ldom⁡(L)⋂Ker⁡(P))-1. Again for each x∈Im⁡(L),
(29)∥Kpx∥X=∥I0+αx∥X=∥I0+αx∥∞+∥D0+α-1I0+αx∥∞+⋯+∥D0+α-(N-1)I0+αx∥∞;≤(∑i=1N-1‍1Γ(i+1)+1Γ(α+1))∥x∥∞=b∥x∥∞,
where b=1/Γ(α+1)+∑i=1N-1‍(1/Γ(i+1)).Lemma 8.
AssumeΩ⊂Y is an open bounded subset such that dom⁡L⋂Y≠∅; then map N is L-compact on Ω¯Proof.
By the continuity off, we can get that QN(Ω¯) and KP(I-Q)N(Ω¯) are bounded. So, in view of the Arzela-Ascoli theorem, we need only to prove that KP(I-Q)N(Ω¯) is equicontinuous. From the continuity of f, there exists a constant r>0, such that |(I-Q)N(u(t))|≤r, for all u∈Ω¯, t∈[0,1].
For0≤t1≤t2≤1, u∈Ω, we have
(30)|KP,QNu(t2)-KP,QNu(t1)|=1Γ(α)|∫0t2‍(t2-s)α-1(I-Q)N(u(s))ds-∫0t1‍(t1-s)α-1(I-Q)N(u(s))ds∫0t2‍(t2-s)α-1(I-Q)N(u(s))ds|≤rΓ(α)∫0t1‍[(t1-s)α-1-(t2-s)α-1]ds+rΓ(α)∫t1t2‍(t2-s)α-1ds=rΓ(α+1)(t2α-t1α).
Furthermore, we have
(31)|D0+α-iKP,QNu(t2)-D0+α-iKP,QNu(t1)|=1Γ(i)|∫0t2‍(t2-s)i-1(I-Q)N(u(s))dsk-∫0t1‍(t1-s)i-1(I-Q)N(u(s))ds∫0t2‍(t2-s)i-1(I-Q)N(u(s))ds|≤rΓ(i)∫0t1‍[(t1-s)i-1-(t2-s)i-1]ds+rΓ(i)∫t1t2‍(t2-s)i-1ds=rΓ(i+1)(t2i-t1i),
where i=1,2,…,N-1. Since tα and ti are uniformly continuous on [0,1], we can get that KP(I-Q)N:Ω¯→Y is compact. The proof is completed.To obtain our main results, we need the following conditions.(H1)
There exist functionsφ,ψi∈L1[0,1], i=1,N, such that for all u∈R2, t∈[0,1],
(32)|f(t,x1,x2,…,xN)|≤φ+ψ1|x1|+ψ2|x2|+⋯+ψN|xN|.(H2)
There exists a constantA>0 such that for every y∈R, if |x2|>A for all t∈[0,1], then
(33)f(t,x1,x2,…,xN)≠0.(H3)
There exists a constantD>0 such that, for each ci, i=1,2 satisfying min⁡{|c1|,|c2|}>D. We have either at least one of the following:
(34)c1N(c1tα-1)>0or(35)c1N(c1tα-1)<0.(H4)
∑i=2Nρi<1, where ρi+1=(a+b)∥ψi∥1, i=1,2,…,N.Lemma 9.
Ω
1
=
{
u
∈
dom
⁡
(
L
)
∖
Ker
⁡
(
L
)
∣
L
u
=
λ
N
u
,
λ
∈
[
0,1
]
} is bounded.Proof.
Foru∈Ω1, λ≠0 and Lu=λNu. By (12), Lu=λNu∈Im⁡(L)=Ker⁡(Q); that is,
(36)λ∫01‍f(t,u(t),D0+α-1u(t),D0+α-2u(t),…,D0+α-(N-1)u(t))dt=0.
By the integral mean value theorem, there exits a constant t0∈[0,1] such that
(37)f(t0,u(t0),D0+α-1u(t0),D0+α-2u(t0),…,D0+α-(N-1)u(t0))=0.
Form (H2), we can get |D0+α-1u(t0)|≤A.
Again foru∈Ω1, (I-P)u∈dom⁡(L)∖Ker⁡(L) and LPu=0. From (29), we have
(38)∥(I-P)u∥X=∥KpL(I-P)u∥X=∥KpLu∥X≤b∥Nu∥∞.
Now by Lemma 4(39)|D0+α-1u(0)|≤|D0+α-1u(t0)|+|∫0t0‍D0+αu(s)ds|≤|D0+α-1u(t0)|+|t0|max⁡0≤t≤t0|D0+αu(t)|≤|D0+α-1u(t0)|+∥D0+αu(t)∥∞≤A+∥Lu∥∞=A+∥Nu∥∞.
That is,
(40)|D0+α-1u(0)|≤A+∥Nu∥∞.
From (25) and (38), we have
(41)∥u∥X=∥Pu+(I-P)u∥X≤∥Pu∥X+∥(I-P)u∥X≤a|D0+α-1u(0)|+b∥Nu∥∞.
Furthermore, it follows from (40) and (H1) that
(42)∥u∥X≤(a|D0+α-1u(0)|+b∥Nu∥∞)≤a(A+∥Nu∥∞)+b∥Nu∥∞=aA+(a+b)∥Nu∥∞≤aA+(a+b)×∥f(t,u(t),D0+α-1u(t),D0+α-2u(t),…,D0+α-(N-1)u(t))∥∞≤aA+(a+b)(+⋯+∥ψN∥1∥D0+α-(N-1)u∥∞∥φ∥1+∥ψ∥1∥u∥∞+∥ψ2∥1∥D0+α-1u∥∞+⋯+∥ψN∥1∥D0+α-(N-1)u∥∞)=aA+(a+b)∥φ∥1+ρ2∥u∥∞+ρ3∥D0+α-1u∥∞+ρ4∥D0+α-2u∥∞+⋯+ρN+1∥D0+α-(N-1)u∥∞.
By the definition ∥u∥X and (H4), it is easy to see that ∥D0+α-1u∥∞,…,∥D0+α-(N-1)u∥∞ and ∥u∥∞ are bounded. So, Ω1 is bounded.Lemma 10.
Ω
2
=
{
u
∈
Ker
⁡
(
L
)
:
N
u
∈
Im
⁡
(
L
)
} is bounded.Proof.
Letu∈Ker⁡(L), so we have u=c1tα-1, c1∈R. For Nu∈Im⁡(L)=Ker⁡(Q),
(43)∫01‍f(t,c1tα-1,c1Γ(α),…,Γ(α)Γ(N-1)c1tN-2)dt=0.
By the integral mean value theorem, there exits a constant t1∈[0,1] such that
(44)f(t1,c1t1α-1,c1Γ(α),…,Γ(α)Γ(N-1)c1t1N-2)=0.
From (H2), it follows that |c1|≤A/Γ(α). Hence, Ω2 is bounded.Lemma 11.
Ω
3
=
{
u
∈
Ker
⁡
(
L
)
:
λ
u
+
(
1
-
λ
)
Q
N
u
=
0
,
λ
∈
[
0,1
]
} is bounded.Proof.
Letu∈Ker⁡(L), so we have u=c1tα-1, c1∈R. If λ=0, then |c1|≤D. If λ=1, we have c1=0.
Ifλ≠0 and λ≠1, then
(45)λc1tα-1+(1-λ)QN(u)=0.
It follows that
(46)λc1tβ-1+(1-λ)×∫01‍f(t,c1tα-1,c1Γ(α),…,Γ(α)Γ(N-1)c1tN-2)dt=0.
Then we get
(47)λc12tα-1+(1-λ)∫01‍c1f(t,c1tα-1,…,c2Γ(α))dt=0,
which, together with (H3), implies |c1|≤D. Here, Ω3 is bounded.Remark 12.
If the other parts of (H3) hold, then the set Ω3′={u∈Ker⁡(L):-λu+(1-λ)QNu=(0,0),λ∈[0,1]} is bounded.Theorem 13.
Suppose (H1)–(H4) hold; then the problem (3) has at least one solution in Y.Proof.
LetΩ be a bounded open set of Y, such that ⋃i=13Ω¯i⊂Ω. It follows from Lemma 8, N is L-compact on Ω. By Lemmas 9, 10, and 11, we get the following:(1)
Lu≠λNu, for every u∈[(dom⁡L∖Ker⁡L)⋂∂Ω]×(0,1);(2)
Nu∉Im⁡L for every u∈Ker⁡L⋂∂Ω;(3)
letH(u,λ)=±λIu+(1-λ)JQNu, where I is the identical operator. Via the homotopy property of degree, we obtain that
(48)deg⁡(JQN|ker⁡L,Ω∩ker⁡L,0)=deg⁡(H(·,0),Ω∩ker⁡L,0)=deg⁡(H(·,1),Ω∩ker⁡L,0)=deg⁡(I,Ω∩ker⁡L,0)=1≠0.Applying Theorem 6, we conclude that Lu=Nu has at least one solution in dom⁡L∩Ω¯.Under the stronger conditions imposed onf, we can prove the uniqueness of solutions to the (3) studied above.Theorem 14.
Suppose the conditions (H1) in the theorem are replaced by the following conditions.(H1)′
There exist positive constantsai, i=0,1,…,N-1, such that, for all (x1,x2,…,xN), (y1,y2,…,yN)∈RN, one has
(49)|f(t,x1,x2,…,xN)-f(t,y1,y2,…,yN)|≤a0|x1-y1|+⋯+aN-1|xN-yN|.(H1)′′
There exist constantsli, i=1,2,…,N-1, such that for all (x1,x2,…,xN), (y1,y2,…,yN)∈RN, one has
(50)|f(t,x1,x2,…,xN)-f(t,y1,y2,…,yN)|≥-l0|x1-y1|+l1|x2-y2|-l2|x3-y3|-⋯-lN-1|xN-yN|.Then, the BVP (3) has a unique solution, provided that
(51)al0l1+aa0+a0c+∑i=2N-1‍alil1+(a+c)∑i=1N-1‍ai<1.Proof.
Letyi=0, i=1,2,…,N, and φ1=|f(t,0,…,0)|; then the condition (H1) is satisfied. According to Theorem 13, BVP (3) has at least one solution. Suppose ui∈Y, i=1,2 are two solutions of (3); then
(52)D0+αui(t)=f(t,ui(t),D0+α-1ui(t),D0+α-2ui(t),…,D0+α-(N-1)ui(t)),i=1,2.
Note that u=u1-u2, so u satisfy the equation
(53)D0+αu=f(t,u1,D0+α-1u1,…,D0+α-(N-1)u1)-f(t,u2,D0+α-1u2,…,D0+α-(N-1)u2).
According to Im⁡(L)=Ker⁡(Q), we have
(54)∫01‍f(t,u1,D0+α-1u1,…,D0+α-(N-1)u1)-f(t,u2,D0+α-1u2,…,D0+α-(N-1)u2)dt=0.
By the integral mean value theorem, there exists η∈[0,1], such that
(55)f(η,u1(η),D0+α-1u1(η),…,D0+α-(N-1)u1(η))-f(η,u2(η),D0+α-1u2(η),…,D0+α-(N-1)u2(η))=0.
By (H1)′′, we have
(56)0=|-f(η,u2(η),D0+α-1u2(η),…,D0+α-(N-1)u2(η))f(η,u1(η),D0+α-1u1(η),…,D0+α-(N-1)u1(η))kk-f(η,u2(η),D0+α-1u2(η),…,D0+α-(N-1)u2(η))|≥-l0|u(η)|+l1|D0+α-1u(η)|-l2|D0+α-2u(η)|-⋯-lN-1|D0+α-(N-1)u(η)|.
We can have
(57)|D0+α-1u(η)|≤l0l1|u(η)|+l2l1|D0+α-2u(η)|+⋯+lN-1l1|D0+α-(N-1)u(η)|≤l0l1∥u∥∞+∑i=2N-1‍lil1∥D0+α-iu∥∞.
Thus, we can obtain
(58)|D0+α-1u(0)|≤|D0+α-1u(η)|+|∫0η‍D0+αu(s)ds|≤|D0+α-1u(η)|+|η|max⁡0≤t≤η|D0+αu(t)|≤l0l1∥u∥∞+∑i=2N-1‍lil1∥D0+α-iu∥∞+∥D0+αu(t)∥∞=l0l1∥u∥∞+∑i=2N-1‍lil1∥D0+α-iu∥∞+∥Lu∥∞.
According to (25), (38), and (58), we have
(59)∥u∥X=∥Pu+(I-P)u∥X≤∥Pu∥X+∥(I-P)u∥X=al0l1∥u∥∞+∑i=2N-1‍alil1∥D0+α-iu∥∞+a∥Lu∥∞+c∥Lu∥∞;≤al0l1∥u∥∞+∑i=2N-1‍alil1∥D0+α-iu∥∞+(a+c)(a0∥u∥∞+∑i=1N-1‍ai∥D0+β-iu∥∞).
From the definition of ∥u∥X and the assumption (51), we have ∥u∥=0, so that u1=u2.


## 4. Example

Let us consider the following boundary value problems:(60)D0+2.5u(t)=t5+19D0+1.5u(t)+sin2(D0+0.5u(t))+arctanu(t),0<t<1,u(0)=D0+0.5u(0)=0,D0+1.5u(0)=D0+1.5u(1).
Corresponding to the problem (3), we have that α=2.5 and
(61)f(t,x,y,z)=t5+arctanx+19y+sin2(z).
Moreover,
(62)|f(t,x,y,z)|≤15+π2+19|y|+1.
We can get that the condition (H1) holds; that is, φ=(12+5π)/10, ψ1=ψ3=0, and ψ2=1/9. Taking A=25, D=19, we can calculate that (H2)–(H4) hold.Hence, by Theorem13, we obtain that (60) has at least one solution.



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*Source: 632434-2014-08-05.xml*