Existence of solutions for fractional differential equations with three-point boundary conditions at

Existence of solutions for fractional differential equations with three-point boundary conditions at

resonance in R ⁿ

Fu-Dong Ge

and Hua-Cheng Zhou

1College of Information Science and Technology, Donghua University, Shanghai 201620, P. R. China

2Academy of Mathematics and Systems Science, Academia Sinica, Beijing 100190, P. R. China

Received 7 June 2014, appeared 1 January 2015 Communicated by Nickolai Kosmatov

Abstract.In this paper, by applying the coincidence degree theory which was first intro- duced by Mawhin, we obtain an existence result for the fractional three-point boundary value problems inRⁿ, where the dimension of the kernel of fractional differential op- erator with the boundary conditions can take any value in {1, 2, . . . ,n}. This is our novelty. Several examples are presented to illustrate the result.

Keywords: fractional differential equations, resonance, coincidence degree theory.

2010 Mathematics Subject Classification: 34A08, 34B10, 34B40.

1 Introduction

In this paper, we are concerned with the existence of solutions for the following fractional three-point boundary value problems (BVPs) at resonance inRⁿ:







D₀^α+x(t) = f(t,x(t),D₀^α+⁻¹x(t)), 1<α≤2, t ∈(0, 1), x(₀) =_θ, D^α₀+⁻¹x(₁) = AD^α₀+⁻¹x(ξ)_,

(1.1)

where D₀^α+ and I₀^α+ are the Riemann–Liouville differentiation and integration; θ is the zero vector inRⁿ; Ais a square matrix of order nsatisfying rank(I−A)< n; ξ ∈ (_{0, 1})_{is a fixed} constant; f:[0, 1]×_Rⁿ×_Rⁿ→_Rⁿis a Carathéodory function, that is,

(i) for each (u,v)∈_Rⁿ×_Rⁿ, t7→ f(t,u,v)is measurable on[0, 1]; (ii) for a.e.t∈ [_{0, 1}]_, (u,v)7→ f(t,u,v)is continuous onRⁿ×_Rⁿ;

(iii) for every compact setΩ ⊆ _Rⁿ×_Rⁿ, the function ϕ_Ω(t) =sup{|f(t,u,v)| : (u,v) ∈ _Ω}

∈ L¹[0, 1], where|x|=max{|x_i|, i=1, 2, . . . ,n}, the norm of x= (x₁,x₂, . . . ,x_n)^>inRⁿ.

BCorresponding author. Email: hczhou@amss.ac.cn

The system (1.1) is said to be at resonance inRⁿif det(I−A) =0, i.e., dim ker(I−A)≥1, otherwise, it is said to be non-resonant. In the past three decades, many authors investigated the existence of solutions for the fractional differential equations with the boundary value conditions. The attempts on det(I−A) 6= 0, non-resonance case, for fractional differential equations are available in [1–3,10,11,17,21–23], and the attempts on det(I− A) = 0 and n ≤ 2, resonance case, can be found in [4–6,8,9,13,14,18–20], and the references therein.

However, to the best of our knowledge, almost all results derived in these papers are for the casen = 1 with dim kerL = 0 or 1 and for the casen = 2 with dim kerL = 2. It is still open for the casen≥3. So we study this issue in this paper.

For instance, whenn=1, consider the following problems







D₀^α+u(t) = f(t,u(t),D^α₀+⁻¹u(t)), 0<t <1, I₀²+^−αu(t)|_t=0 =0, D^α₀+⁻¹u(1) =^m_∑⁻²

i=₁

β_iD₀^α+⁻¹u(ξ_i) ^(1.2) where 1 < α ≤ 2, ξ_i ∈ (0, 1), β_i ∈ _R i = 1, 2, . . . ,m−2, 0 < ξ₁ < ξ2 < · · · < ξ_i < 1, f: [0, 1]×_R×_R→_Ris a continuous function. It follows from the argument above that(1.2) is resonant when∑^m_i=⁻1²β_i =1 and it is non-resonant when∑^m_i=⁻1²β_i 6=1.

Further, in order to apply the coincidence degree theory of Mawhin [15], we suppose additionally that Asatisfies rank(I−A)<nand one of the following conditions

(a₁) Ais idempotent, that is, A² = A, or;

(a₂) A²= I, where I stands for the identity matrix of size n.

It is also obvious that dim ker(I−A)can take any value in{1, 2 . . . ,n}for suitableA, which is surely generalize the previous efforts. However, we point out that without the above assump- tions, it will be difficult to construct the projector Q as (3.1) below. This is the reason why we only choose the two special cases of A. Removing such an assumption, for the general A satisfying rank(I−A)<n, the problem (1.1) may be a challenging problem, which is also an issue of our further research.

In particular, when A = I, it is clear that A satisfies(a₁)and (a₂). It is also obvious that det(I−A) = 0, so under this boundary condition, the system (1.1) is at resonance. Besides, kerL= {(c₁,c₂, . . . ,c_n)^>t^α−1 :c_i ∈_R,i=1, 2, . . . ,n}and dim kerL=n, whereLis defined by (2.2) below. For A=0, it is clear that det(I−A) =1, kerL={0}, so, in this case, the system (1.1) is non-resonant.

In paper[16], the authors investigated the following second differential system (u⁰⁰(t) = f(t,u(t),u⁰(t)), 0<t<1,

u⁰(0) =θ, u(1) = Au(η), (1.3)

where f: [_{0, 1}]×_Rⁿ×_Rⁿ→_Rⁿ is a Carathéodory function and the square matrixAsatisfies the condition(a₁)or (a₂). Therefore, it is more natural to ask whether there exists a solution when the order of the derivative is fractional. In this paper, we offer an answer by considering the system (1.1).

The goal of this paper is to study the existence of solutions for the fractional differential equations with boundary value conditions when n ≥ 3. The layout of this paper will be as follows: in Section 2, we give some necessary background and some preparations for our consideration. The statement and the proof of our main result will be given in Section 3 by the coincidence degree theory of Mawhin [15]. In Section 4, we present several examples to illustrate the main result.

2 Background materials and preliminaries

In this section, we introduce some necessary definitions and lemmas which will be used later.

For more details, we refer the reader to [7,12,15], and the references therein.

Definition 2.1 ([12]). The fractional integral of order α > 0 of a function x: (0,∞) → _R is given by

I₀^α+x(t) = ¹ Γ(α)

Z _t

0

(t−s)^α−1x(s)ds, provided the right-hand side is pointwise defined on(0,∞).

Remark 2.2. The notation I₀^α+x(t)|_t=₀ means that the limit is taken at almost all points of the right-sided neighborhood(0,ε) (ε>0)of 0 as follows:

I₀^α+x(t)|_t=0 = lim

t→0⁺I₀^α+x(t).

Generally, I₀^α+x(t)|_t=0is not necessarily zero. For instance, letα∈(0, 1), x(t) =t^−α. Then I₀^α+t^−α|_t=0 = lim

t→0⁺

1 Γ(α)

Z _t

0

(t−s)^α−1s^−αds= lim

t→0⁺Γ(1−α) =_Γ(1−α).

Definition 2.3 ([12]). The fractional derivative of orderα > 0 of a function x: (0,∞) → _R is given by

D₀^α+x(t) = ¹ Γ(n−α)

d dt

nZ _t

0

x(s)

(t−s)^{α−n+1 ds,}

wheren = [α] +1, provided the right-hand side is pointwise defined on(0,∞).

Lemma 2.4 ([12]). Assume that x ∈ C(0,+_∞)∩L_loc(0,+_∞)with a fractional derivative of order α>0that belongs to C(0,+_∞)∩L_loc(0,+_∞). Then

I₀^α+D₀^α+x(t) =x(t) +c₁t^α−1+c₂t^α−2+· · ·+c_nt^α−N,

for some c_i ∈ _R, i=1, . . . ,N, where N is the smallest integer greater than or equal toα.

For anyx(t) = (x₁(t),x2(t), . . . ,xn(t))^>∈ _Rⁿ, the fractional derivative of orderα>0 of x is defined by

D₀^α+x(t) = (D₀^α+x₁(t),D^α₀+x₂(t), . . . ,D^α₀+x_n(t))^> ∈_Rⁿ.

The following definitions and coincidence degree theory are fundamental in the proof of our main result. One can refer to [7,15].

Definition 2.5. LetXandYbe normed spaces. A linear operatorL: dom(L)⊂X→Yis said to be a Fredholm operator of index zero provided that

(i) imLis a closed subset ofY, and (ii) dim kerL=_{codim imL}<+_∞.

It follows from Definition2.5that there exist continuous projectorsP: X →XandQ:Y→ Ysuch that

imP=_kerL, kerQ=_imL, X =_kerL⊕_kerP, Y=_imL⊕_imQ

and the mapping L|_domL∩kerP: domL∩kerP → imL is invertible. We denote the inverse of L|_domL∩kerP by KP: imL → domL∩kerP. The generalized inverse of L denoted by K_P,Q: Y → domL∩kerP is defined by K_P,Q = K_P(I−Q). Furthermore, for every isomor- phism J: imQ → kerL, we can obtain that the mapping K_P,Q+JQ: Y → domL is also an isomorphism and for allx∈domL, we know that

(K_P,Q+JQ)⁻¹x = (L+J⁻¹P)x. (2.1) Definition 2.6. Let L be a Fredholm operator of index zero, let Ω ⊆ X be a bounded subset and domL∩_Ω6= _∅. Then the operator N: Ω→Yis called to be L-compact inΩif

(i) the mappingQN: Ω→Yis continuous andQN(_Ω)⊆Yis bounded, and (ii) the mappingK_P,QN: Ω→Xis completely continuous.

Assume thatLis defined in Definition2.6andN: Ω→YisL-compact. For anyx ∈_{Ω, by} (2.1), we shall see that

Lx = (K_P,Q+JQ)⁻¹x−J⁻¹Px= (K_P,Q+JQ)^−1hIx−K_P,QJ⁻¹Px−JQJ⁻¹Pxi

= (K_P,Q+JQ)⁻¹(Ix−Px).

Then we can equivalently transform the existence problem of the equation Lx = Nx, x ∈ _Ω into a fixed point problem of the operatorP+ (K_P,Q+JQ)Nin Ω.

This can be guaranteed by the following lemma, which is also the main tool in this paper.

Lemma 2.7([15]). LetΩ⊂X be bounded, L be a Fredholm mapping of index zero and N be L-compact onΩ. Suppose that the following conditions are satisfied:

(i) Lx 6=λNx for every(x,λ)∈((domL\kerL)∩_∂Ω)×(0, 1); (ii) Nx∈/imL for every x∈kerL∩_∂Ω;

(iii) deg(JQN|_{kerL∩∂Ω},Ω∩kerL, 0)6=0, with Q: Y→Y a continuous projector such thatkerQ= imL and J: imQ→kerL is an isomorphism.

Then the equation Lx= Nx has at least one solution indomL∩_Ω.

In this paper, we utilize spacesX,Yintroduced as

X=ⁿx(t)∈_Rⁿ:x(t) =I₀^α₊⁻¹u(t), u∈C([0, 1],Rⁿ), t∈ [0, 1]^o

with the norm kxk = max{kxk_∞,kD₀^α₊⁻¹xk_∞}and Y = L¹([0, 1],Rⁿ) with the norm kyk₁ = R1

0 |y(s)|ds, respectively, wherek · k_∞ represents the sup-norm.

We have the following compactness criterion on subsetF ofX(see, e.g., [19]).

Lemma 2.8. F ⊂ X is a sequentially compact set if and only if F is uniformly bounded and equicon- tinuous which are understood in the following sense:

(1) there exists an M>0such that for every x∈ F,kxk ≤M;

(2) for any givenε>0, there exists aδ>0such that

|x(t₁)−x(t₂)|<_ε, |D₀^α+⁻¹x(t₁)−D^α₀+⁻¹x(t₂)|<_ε, for t₁,t₂∈ [_{0, 1}]_,|t₁−t₂|<_δ, ∀x∈ F.

Now we define the linear operatorL: domL⊆ X→Yby

Lx:=D₀^α+x, (2.2)

where domL = ⁿx∈ X:D^α₀+x ∈Y, x(0) =θ, D₀^α+⁻¹x(1) =AD₀^α+⁻¹x(ξ)^o. Define N: X → Y by

Nx(t):= f(t,x(t),D₀^α+⁻¹x(t)), t ∈[0, 1]. (2.3) Then the problem can be equivalently rewritten as Lx= Nx.

Lemma 2.9. The operator L defined above is a Fredholm operator of index zero.

Proof. For anyx∈domL, by Lemma2.4 andx(0) =θ, we obtain

x(t) =I₀^α₊Lx(t) +ct^α−1, c∈ _Rⁿ_, t ∈[_{0, 1}]_{, (2.4)} which, together with D₀^α+⁻¹x(1) = AD₀^α+⁻¹x(ξ), yields

kerL={x ∈X: x(t) =ct^α−1, t∈ [0, 1], c∈ker(I−A)}w^ker(I−A)t^α−1. (2.5) Now we claim that

imL={y∈Y: g(y)∈im(I−A)}, (2.6) where g: Y→_Rⁿ is a continuous linear operator defined by

g(y):= ^A Γ(α)

Z _ξ

0 y(s)ds− ^I Γ(α)

Z ₁

0 y(s)ds. (2.7)

Actually, for anyy ∈imL, there exists a functionx∈domLsuch thaty= Lx. It follows from (2.4)thatx(t) =I₀^α+y(t) +ct^α−1. Together withD^α₀+⁻¹x(1) = AD^α₀+⁻¹x(ξ), we obtain

A Γ(α)

Z _ξ

0 y(s)ds− ^I Γ(α)

Z ₁

0 y(s)ds= (I−A)c, c∈_Rⁿ, which means thatg(y)∈im(I−A).

On the other hand, for any y ∈ Y satisfying g(y) ∈ im(I −A), there exists a constant c^∗ such that g(y) = (I−A)c^∗. Let x^∗(t) = I₀^α+y(t) +c^∗t^α−1. A straightforward computation shows thatx^∗(0) =θ andD₀^α+⁻¹x^∗(1) =AD₀^α+⁻¹x^∗(ξ). Hence,x^∗ ∈domLandy(t) =D₀^α+x^∗(t), which implies thaty∈imL.

Next, we setρ_A=k(I−A), where k=

(1, if the hypothesis(a₁)holds, i.e., A² = A;

1

2, if the hypothesis(a₂)holds, i.e.,A² = I. (2.8) For A²= A, we have

ρ²_A = (I−A)² = I−2A+A² = I−A=ρ_A,

(I−ρ_A)(ξ^αA−I) = A(ξ^αA−I) =ξ^αA²−A= (ξ^α−1)A= (ξ^α−1)(I−ρ_A).

(2.9) For A²= I, we have

ρ²_A= ¹

4(I−A)² = ¹

4(I−2A+A²) = ¹

2(I−A) =ρ_A, (I−ρ_A)(ξ^αA−I) = ¹

2(I+A)(ξ^αA−I)

= ¹

2[ξ^α−I+ξ^αA²−A] = ¹

2(ξ^α−1)(I+A) = (ξ^α−1)(I−ρ_A).

(2.10)

It follows from (2.9) and (2.10) thatρ_A satisfies the following properties

ρ²_A= ρ_A, (I−ρ_A)(ξ^αA−I) = (ξ^α−1)(I−ρ_A). (2.11) Furthermore, we note that ify=ct^α−1,c∈_Rⁿ, then

g(y) = ^A Γ(α)

Z _ξ

0 y(s)ds− ^I Γ(α)

Z ₁

0 y(s)ds= (ξ^αA−I)c

Γ(α+1) ^{. (2.12)} Define the continuous linear mappingQ: Y→Yby

Qy(t):= ^Γ(α+1)

ξ^α−₁ (I−ρ_A)g(y)t^α−1, t ∈[0, 1], y∈Y. (2.13) By (2.11), it is easy to verify Q²y = Qy, that is, Q is a projection operator. The equality kerQ=imLfollows from the trivial fact that

y∈kerQ⇔ g(y)∈ker(I−ρ_A)⇔g(y)∈imρ_A⇔g(y)∈im(I−A)⇔y∈imL.

Therefore, we getY=_kerQ⊕imQ=_imL⊕imQ.

Finally, we shall prove that imQ= kerL. Indeed, for anyz ∈ imQ, letz = Qy, y ∈Y. By (2.11), we have

k(I−A)z(t) =ρ_Az(t) =ρ_AQy(t) = ^Γ(α+1)

ξ^α−1 ρ_A(I−ρ_A)g(y)t^α−1 =θ,

which impliesz ∈kerL. Conversely, for eachz ∈kerL, there exists a constantc^∗ ∈ker(I−A) such thatz =c^∗t^α−1 fort∈[0, 1]. By (2.11) and (2.12), we derive

Qz(t) = ^Γ(α+1)

ξ^α−1 (I−ρ_A)g(c^∗t^α−1)t^α−1= c^∗t^α−1 =z(t)_, t∈[_{0, 1}]_,

which implies thatz ∈ imQ. Hence we know that imQ = kerL. Then the operator L is a Fredholm operator of index zero. The proof is complete.

Define the operatorP: X→Xas follows:

Px(t) = ¹

Γ(α)(I−ρ_A)D^α₀₊⁻¹x(₀)t^α−1. (2.14) Lemma 2.10. The mapping P: X→X defined as above is a continuous projector such that

imP=kerL, X =kerL⊕kerP and the linear operator KP: imL→domL∩kerP can be written as

K_Py(t) = I₀^α₊y(t), also

K_P = (L|_domL∩kerP)⁻¹ and kK_Pyk ≤_1/Γ(α)kyk₁.

Proof. The continuity of P is obvious. By the first identity of (2.11), we have (I−ρ_A)² = (I−ρ_A), which implies that the mapping Pis a projector. Moreover, if v ∈imP, there exists anx ∈Xsuch thatv= Px. By the first identity of (2.11) again, we see that

1

Γ(α)(I−A)(I−ρ_A)D^α₀₊⁻¹x(0) = ¹

kΓ(α)^ρA(I−ρ_A)D^α₀₊⁻¹x(0) =0,

which gives usv∈_kerL. Conversely, ifv∈_kerL, thenv(t) =c∗t^α−1for somec∗∈ _ker(I−A)_, and we deduce that

Pv(t) = ¹

Γ(α)(I−ρ_A)D^α₀₊⁻¹v(₀)t^α−1= (I−ρ_A)c∗t^α−1 =c∗t^α−1 =v(t)_, t∈ [_{0, 1}]_, which gives us v∈imP. Thus, we get that kerL=imPand consequentlyX=kerL⊕kerP.

Moreover, lety∈imL. There exists x∈ domLsuch thaty= Lx, and we obtain K_Py(t) =x(t) +ct^α−1,

where c∈ _Rⁿ satisfiesc= Ac. It is easy to see thatK_Py ∈domLandK_Py ∈kerP. Therefore, K_P is well defined. Further, fory∈imL, we have

L(K_Py(t)) =D₀^α₊(K_Py(t)) =y(t) and for x∈_domL∩_kerP, we obtain that

K_P(Lx(t)) =x(t) +c₁t^α−1+c₂t^α−1,

for some c₁,c₂ ∈ _Rⁿ. In view of x ∈ domL∩kerP, we know that c₁ = c₂ = 0. Therefore, (K_PL)x(t) = x(t). This shows thatK_P = (L|_domL∩kerP)⁻¹. Finally, by the definition of K_P, we derive

kD^α₀₊⁻¹K_Pyk_∞ =

Z _t

0 y(s)ds ∞

≤ kyk₁ (2.15)

and

kK_Pyk_∞ =

1 Γ(α)

Z _t

0

(t−s)^α−1y(s)ds ∞

≤ ¹

Γ(α)kyk₁. (2.16) It follows from(2.15)and(2.16)that

kK_Pyk=max{kD^α₀₊⁻¹K_Pyk_∞,kK_Pyk_∞} ≤max

kyk₁, 1 Γ(α)kyk₁

= ¹

Γ(α)kyk₁. (2.17) Then Lemma2.10is proved.

Remark 2.11. The constant _Γ(¹_α) in (2.16) is sharp, and its value can not be improved. Actually, one can prove the following proposition.

Proposition 2.12. Letα∈(1, 2]and define the linear mapping T: L¹[0, 1]→C[0, 1]by (Ty)(t) = ¹

Γ(α)

Z _t

0

(t−s)^α−1y(s)ds.

ThenkTk= _Γ(¹

α).

Proof. Indeed, fromk(Ty)(t)k_∞≤ _Γ(¹

α)kyk₁, we have kTk ≤ ¹

Γ(α)^{. (2.18)}

On the other hand, forε∈(0, 1), let y(t) =





 1

ε, 0≤t ≤ε, 0, ε<t ≤1.

A direct computation shows thatkyk₁=1, and

|Ty(1)|= ¹

Γ(α)[1−δ(ε)],

whereδ(ε) =1− ¹⁻⁽¹_αε^−ε)α >0. It is easy to verify that δis an increasing function with respect to ε and lim_ε→0δ(ε) = 0. Thus, kTk ≥ _Γ(¹

α)[1−2δ(ε)]. As ε was chosen arbitrarily, we have kTk ≥ _Γ(¹

α). This together with (2.18) leads to conclusion.

Lemma 2.13. Let f be a Carathéodory function. Then N defined by(2.3)is L-compact.

Proof. LetΩbe a bounded subset in X. By the hypothesis(iii)on the function f, there exists a functionϕ_Ω(t)∈ L¹[0, 1]such that for allx∈_Ω,

|f(t,x(t),D₀^α+⁻¹x(t))| ≤ ϕ_Ω(t), a.e.t ∈[0, 1], (2.19) which, along with (2.7) and (2.13), implies

kQyk₁ =

Γ(α+₁)

ξ^α−1 (I−ρ_A)g(y)

Z ₁

0 s^α−1ds

≤ (kAk+1)kI−ρ_Ak

|1−ξ^α| kϕ_Ωk₁ <_∞.

(2.20)

This shows thatQN(_Ω)⊆ Y is bounded. The continuity ofQN follows from the hypothesis on f and the Lebesgue dominated convergence theorem.

Next, we shall show thatK_P,QN is completely continuous. First, for anyx∈ _{Ω, we have} K_P,QNx(t) =K_P(I−Q)Nx(t) =K_PNx(t)−K_PQNx(t)

= I₀^α₊Nx(t)− ^Γ(α+1)

ξ^α−₁ (I−ρ_A)g(Nx(t))I₀^α₊t^α−1. By Lemma2.8, it is easy to know thatK_P,QNis continuous.

From (2.19) and (2.7), we derive that

|g(Nx(t))|=

A Γ(α)

Z _ξ

0 f(s,x(s),D₀^α+⁻¹x(s))ds− ^I Γ(α)

Z ₁

0 f(s,x(s),D₀^α+⁻¹x(s))ds

≤ kAk+1

Γ(α) kϕ_Ωk₁.

(2.21)

From (2.21), we obtain

kK_P,QNxk= I₀^α₊Nx(t)−^Γ(α+1)

ξ^α−1 (I−ρ_A)g(Nx(t))I₀^α₊t^α−1

≤ kϕ_Ωk₁+ ^Γ(α+1)kI−ρ_Ak

Γ(α)|ξ^α−1| |g(Nx(t))|

Z _t

0

(t−s)^α−1s^α−1ds

≤ kϕ_Ωk₁+ ^Γ(α+₁)kI−ρ_Ak(kAk+₁)

Γ(2α)|ξ^α−1| kϕ_Ωk₁< _∞, which shows thatK_P,QNΩis uniformly bounded inX. Noting that

b^p−a^p ≤(b−a)^p for anyb≥ a>0, 0< p≤1, (2.22) for any t₁,t2∈ [0, 1]witht₁ <t2, we shall see that

|K_P,QNx(t₂)−K_P,QNx(t₁)|

= ¹ Γ(α)

Z _t1

0

[(t₂−s)^α−1−(t₁−s)^α−1]Nx(s)ds+

Z _t2

t₁

(t₂−s)^α−1Nx(s)ds

−^Γ(α+1)

ξ^α−1 (I−ρ_A)g(Nx(t))[I₀^α₊t^α₂⁻¹−I₀^α₊t^α₁⁻¹]

≤ ¹ Γ(α)

Z _t1

0

(t₂−t₁)^α−1ϕ_Ω(s)ds+ ¹ Γ(α)

Z _t2

t1

ϕ_Ω(s)ds + ^Γ(α+1)kI−ρ_Ak(kAk+1)

Γ(2α)|ξ^α−1| kϕ_Ωk₁|t^2α₂ ⁻¹−t^2α₁ ⁻¹|

→0 ast2→t₁ and

D^α₀+⁻¹K_P,QNx(t₂)−D₀^α+⁻¹K_P,QNx(t₁) =

Z _t2

t1

Nx(s)ds

≤

Z _t2

t1

ϕ_Ω(s)ds→0 ast₂ →t₁. Then we get that K_P,QNΩis equicontinuous inX. By Lemma (2.8),K_P,QNΩ⊆ Xis relatively compact. Thus we can conclude that the operator NisL-compact continuous inΩ. The proof is complete.

3 Main result

In this section, we shall present and prove our main result.

Theorem 3.1. Let f be a Carathéodory function and the following conditions hold:

(H₁) There exist three nonnegative functions a,b,c∈L¹[0, 1]such that

|f(t,u,v)| ≤a(t)|u|+b(t)|v|+c(t), for all t∈ [0, 1], u,v∈_Rⁿ

(H2) There exists a constant A₁ > 0such that for x ∈ domL, if|D₀^α+⁻¹x(t)|> A₁ for all t∈ [0, 1], then

Z ₁

ξ

f(s,x(s),D₀^α+⁻¹x(s))ds∈/im(I−A).

(H₃) There exists a constant A₂ > 0 such that for any e ∈ _Rⁿ satisfying e = Ae and min₁≤i≤n{|e_i|}> A2, either

he,QNei ≤0, min

1≤i≤n{|e_i|}> A₂, or else

he,QNei ≥0, min

1≤i≤n{|e_i|}> A₂, whereh·,·iis the scalar product inRⁿ.

Then the BVPs(1.1)have at least one solution in space X provided that

(kI−ρ_Ak+1)(kak₁+kbk₁)<_Γ(α). (3.1) Proof. We shall construct an open bounded subsetΩinXsatisfying all assumption of Lemma 2.7. Let

Ω1 ={x ∈domL\kerL: Lx=λNxfor someλ∈[0, 1]}. (3.2) For anyx ∈ _Ω1, x ∈/ kerL, we get that λ 6= 0. Since Nx ∈ imL = kerQ, by the definition of imL, we haveg(Nx)∈im(I−A)_{, where}

g(Nx) = ^A Γ(α)

Z _ξ

0 f(s,x(s),D^α₀+⁻¹x(s))ds− ^I Γ(α)

Z ₁

0 f(s,x(s),D^α₀+⁻¹x(s))ds.

Hence

Z ₁

ξ

f(s,x(s)_,D₀^α+⁻¹x(s))ds

=−_Γ(α)g(Nx)−(A−I)

Z _ξ

0 f(s,x(s),D₀^α+⁻¹x(s))ds∈ im(I−A).

(3.3)

It follows from hypothesis(H₂) and (3.3) that there exists t₀ ∈ [0, 1]such that |D₀^α+⁻¹x(t₀)| ≤ A₁. Then by D^α₀₊⁻¹x(0) =D₀^α₊⁻¹x(t₀)−Rt

0 D^α₀₊x(s)ds, we deduce that

|D^α₀₊⁻¹x(0)| ≤A₁+kD₀^α₊xk₁= A₁+kLxk₁ ≤ A₁+kNxk₁, which implies

kPxk=

1

Γ(α)(I−ρ_A)D^α₀₊⁻¹x(0)t^α−1

≤ kI−ρ_Ak

Γ(α) (A₁+kNxk₁). (3.4) Further, again forx ∈_Ω1, since imP=kerL,X= kerL⊕kerP, we have(I−P)x∈ domL∩ kerPandLPx=θ. Then

k(I−P)xk=kK_PL(I−P)xk ≤ kK_PLxk ≤ ¹

Γ(α)kLxk₁≤ ¹

Γ(α)kNxk_{1. (3.5)} From(3.4)and(3.5), we can conclude that

kxk=kPx+ (I−P)xk ≤ kPxk+k(I−P)xk ≤ kI−ρ_Ak

Γ(α) ^A1+kI−ρ_Ak+1

Γ(α) kNxk₁. (3.6) Moreover, by the definition ofN and the hypothesis(H₁), we see that

kNxk₁=

Z ₁

0

|f(s,x(s),D^α₀+⁻¹x(s))|dt≤ kak₁kxk_∞+kbk₁kD₀^α+⁻¹xk_∞+kck₁. (3.7)

Then

kxk ≤ kI−ρ_Ak

Γ(α) ^A1+ kI−ρ_Ak+1

Γ(α) (kak₁kxk_∞+kbk₁kD₀^α+⁻¹xk_∞+kck₁). (3.8) From(3.8)andkxk_∞≤ kxk, we derive

kxk_∞ ≤

kI−ρAk

Γ(α) A₁+ ^kI−_Γ^ρ₍^Ak+1

α) (kbk₁kD₀^α+⁻¹xk_∞+kck₁) 1− ^kI−_Γ^ρ₍^Ak+1

α) kak₁ ^{, (3.9)}

which, together withkD₀^α+⁻¹xk_∞ ≤ kxk,(3.8)and(3.9), gives us kD^α₀+⁻¹xk_∞ ≤

kI−ρ_Ak

Γ(α) A₁+^kI−_Γ^ρ₍^Ak+1

α) kck₁ 1−^kI−_Γ^ρ₍^Ak+1

α) (kak₁+kbk₁))

= kI−ρ_AkA₁+ (kI−ρ_Ak+1)kck₁

Γ(α)−(kI−ρ_Ak+₁)(kak₁+kbk₁)^{. (3.10)} It follows from(3.9)and(3.10)thatΩ1is bounded.

Let

Ω2={x ∈kerL:Nx ∈imL}. (3.11)

For anyx∈ _Ω2, it follows fromx∈kerLthatx =et^α−1for somee∈ker(I−A), and it follows fromNx ∈imLthatg(Nx)∈im(I−A). By a similar argument as above, by hypothesis(H₂), we arrive at|D^α₀+⁻¹x(t0)|=|e|_Γ(α)≤ A₁. Thus we get that

kxk ≤ |e|_Γ(α)≤ A₁.

That is,Ω2is bounded inX. If the first part of(H₃)holds, denote

Ω3={x∈kerL:−λx+ (₁−λ)QNx=_θ, t ∈[_{0, 1}]}, then for any x∈ _Ω3, we know that

x=et^α−1 withe∈ker(I−A)andλx = (1−λ)QNx.

If λ = 0, we have Nx ∈ kerQ = imL, then x ∈ _Ω2, by the argument above, we get that kxk ≤ A₁. Moreover, ifλ∈ (0, 1]and if|e|> A₂, by the hypothesis (H₃), we deduce that

0<λ|e|² =λhe,ei= (1−λ)he,QNei ≤0,

which is a contradiction. Then kxk = ket^α−1k ≤ max{|e|,Γ(α)|e|}. That is to say, Ω3 is bounded. If other part of(H₃)holds, we take

Ω3= {x∈kerL:λx+ (1−λ)QNx=θ, t∈ [0, 1]}.

By using the same arguments as above, we can conclude that Ω3is also bounded.

In the sequel, we will show that all conditions of Lemma2.7are satisfied.

Assume that Ω is a bounded open subset of X such that ∪³_i=1Ωi ⊆ Ω. It follows from Lemmas 2.9and 2.13that L is a Fredholm operator of index zero and Nis L-compact onΩ.

By the definition of Ωand the argument above, in order to complete the theorem, we only need to prove that the condition (iii)of Lemma2.7is also satisfied. For this purpose, let

H(x,λ) =±λx+ (₁−λ)QNx, (3.12)

where we let the isomorphism J: imQ → kerL be the identical operator. Since Ω3 ⊆ _Ω, H(x,λ) 6= 0 for (x,λ) ∈ kerL∩∂Ω×[0, 1], then by the homotopy property of degree, we obtain

deg(JQN|_{kerL∩∂Ω},Ω∩kerL, 0)

=deg(H(·, 0),Ω∩kerL, 0)

=_deg(H(·, 1),Ω∩kerL, 0)

=deg(±Id,Ω∩kerL, 0) =±16=0.

(3.13)

Thus(H₃)of Lemma2.7 is fulfilled and Theorem3.1 is proved. The proof is complete.

4 Examples

In this section, we shall present three examples to illustrate our main result in R³ with dim kerL = 1, dim kerL = 2, dim kerL = 3, respectively, which surely generalize the pre- vious results [4–6,8,9,13,14,18–20], where the dimension of dim kerLis only 1 or 2.

Example 4.1. Let us consider the following system with dim kerL=1 inR³.



























































D₀³²+x₁(t) = ¹ 148

4t¹²x₃(t) +√¹

πD₀¹²+x₁(t)−4

, t∈ (0, 1), D₀³²+x₂(t) = ¹

148

t⁻¹²x₂(t) + √²

πD₀¹²+x₁(t)

, t∈(0, 1), D₀³²+x₃(t) = ^4t+1

148 , t∈ (0, 1), x₁(0) =x₂(0) =x₃(0) =0, D₀¹²+x₁(1) =−3D₀¹²+x₁

1 2

+3D₀¹²+x₂ 1

2

−3D₀¹²+x₃ 1

2

, D₀¹²+x₂(1) =−5D₀¹²+x₁

1 2

+5D₀¹²+x₂ 1

2

−5D₀¹²+x₃ 1

2

, D₀¹²+x3(1) =−D₀¹²+x₁

1 2

+D₀¹²+x2

1 2

−D₀¹²+x3

1 2

.

(4.1)

Letα=_3/2,ξ =_{1/2 and}

A=





−3 3 −3

−5 5 −5

−1 1 −1



. (4.2)

It is clear that A² = Aand dim ker(I−A) = 1. Define the function f: [0, 1]×_R³×_R³ →_R³ by

f(t,u,v) = ¹ 148















4t¹²x₃+√¹

πy₁−4 t⁻¹²x₂+√²

πy₁ 4t+1















(4.3)

for all t∈ [0, 1]andu= (x₁,x₂,x₃),v = (y₁,y₂,y₃)∈ _R³. Then problem (4.1) has one solution if and only if problem (1.1), with Aand f defined by (4.2), (4.3), has one solution. Hence we need only to verify that the conditions of Theorem 3.1are satisfied.

Check(H₁)of Theorem3.1: for somer∈_{R, where}|u|=|(u₁,u₂,u₃)|=max{|u₁|,|u₂|,|u₃|}

letΩ= {(u,v)∈_R³×_R³: |u| ≤r,|v| ≤r}and let ϕ_Ω(t) = ₁₄₈¹ 4rt¹² +rt⁻¹² +^√3

πr+4t+4

∈ L¹[0, 1]. SincekAk=max{_∑ⁿ_j=1|a_ij|,j=1, 2, . . . ,n}=15, let

a(t) = ¹

148(4t¹² +t⁻¹²), b(t) = ³ 148√

π, c(t) = ^4t+₄

148 . (4.4)

It is easy to see that(H₁)of Theorem3.1and the condition(kI−ρ_Ak+1)(kak₁+kbk₁)<_Γ(α) hold.

Check (H₂)of Theorem 3.1: noting that for any x = (x₁,x₂,x₃),y = (y₁,y₂,y₃) ∈ _R³, we have

f3(t,x₁,x2,x3,y₁,y2,y3) = ^4t+₁ 148 ≥ ¹

148 >0.

This together with im(I−A) ={τ(1, 0, 0) +ζ(0, 1, 0):τ,ζ ∈_R}yields Z ₁

1 2

f(s,x(s)_,D₀¹²+x(s))ds∈/im(I−A)_.

Check(H₃)of Theorem3.1: for anyy ∈L¹([0, 1],R³), by (2.13), we haveρ_A = I−Aand Qy(t) = ^Γ(α+1)

ξ^α−1 (I−ρ_A)g(y)t^α−1 = ³

√

√ π

2−4Ag(y)t^α−1, (4.5) where

g(y) = √^2A π

Z _1/2

0 y(s)ds−√^2I π

Z ₁

0 y(s)ds.

For anye ∈_R³satisfying e= Ae,ecan be written as

e =σ(_{3, 5, 1})^>_{, for}σ∈ _R.

By (2.3), we have

N et¹²

(t) = ¹

148(4σt+1.5σ−4, 8σ, 4t+1)^> (4.6) and

g(N(et^α−1)(t)) = ^A 148√

π

(2.5σ−4, 8σ, 2)^>− ^I 148√

π

(7σ−8, 16σ, 6)^>. (4.7) It follows from (4.5), (4.6) and (4.7) that

Q Net¹²

= ³

√

√ π

2−4Ag Net¹²

t¹² = ^3t

12

148(√

2−4)(−10.5σ,−17.5σ,−3.5σ)^>

and

D

e,QNet¹²E

=− ^367.5 148(√

2−4)^σ

2t¹² ≥0.

Therefore, (4.1) admits at least one solution.

Example 4.2. Consider the following system with dim kerL=2 inR³.



































































D₀³²+x₁(t) =









 1

36, |D₀^1/2+ x₁(t)|<1;

D₀¹²+x₁(t) +^hD^1/2₀+ x₁(t)ⁱ⁻¹−1

36 , |D₀^1/2+ x₁(t)| ≥1, D

3 2

0⁺x2(t) = |x₂(t)|+|x₃(t)|

36 ,

D₀³²+x₃(t) =−^x3(t) 36 ,

x₁(0) =x₂(0) =x₃(0) =0, D₀¹²+x₁(1) =D₀¹²+x₁

1 2

, D

1 2

0⁺x2(1) =−D

1 2

0⁺x2

1 2

+2D

1 2

0⁺x3

1 2

, D₀¹²+x₃(1) =D₀¹²+x₃

1 2

.

(4.8)

Letα=3/2,ξ =1/2, for allt∈ [0, 1],u= (x₁,x₂,x₃),v= (y₁,y₂,y₃)∈_R³,

f(t,u,v) = ¹ 36















(1, |y₁|<1;

y₁+1/y₁−1, |y₁| ≥1

|x₂|+|x₃|

−x₃















(4.9)

and

A=





1 0 0

0 −1 2

0 0 1



. (4.10)

It is not difficult to see that A² = I and dim ker(I−A) = 2. Then problem (4.8), with Aand f defined by (4.10) and (4.9), has one solution if and only if problem (1.1) has one solution.

Check (H₁)of Theorem3.1: for somer ∈ _R,Ω= {(u,v)∈ _R³×_R³ : |u| ≤ r,|v| ≤ r}, let ϕ_Ω(t) = ₁₂¹r+_36r¹ + ₃₆¹ ∈ L¹[0, 1]. SincekAk=3, let

a(t) = ¹

18, b(t) = ¹

36, c(t) = ¹

36. (4.11)

One can see that the condition(H₁)of Theorem3.1is satisfied.

Check (H2) of Theorem 3.1: it follows from the definition of f that |f₁| > ₃₆¹ > 0. This together with im(I−A) ={σ₀(0, 0, 1):σ₀ ∈_R}implies that the condition(H₂)of Theorem3.1 is satisfied.

Check (H3)of Theorem3.1: since dim ker(I−A) =2, for anye ∈ _R³ satisfyinge = Ae,e can be written as

e=σ₁(1, 0, 0)^T+σ₂(0, 1, 1)^T, forσ_i ∈_R, i=1, 2.

For anyy∈ L¹([0, 1],R³), by (2.13) andρ_A = ¹₂(I−A), we have Qy(t) = ^Γ(α+1)

ξ^α−1 (I−ρ_A)g(y)t^α−1= ³

√

√ π 2−₄

I+A

2 g(y)t^α−1, (4.12)