Zhang and Liu Advances in Difference Equations (2018) 2018:15 https://doi.org/10.1186/s13662-017-1463-0

RESEARCH

Open Access

Existence of solutions for fractional multi-point boundary value problems at resonance with three-dimensional kernels Wei Zhang and Wenbin Liu* *

Correspondence: [email protected] School of Mathematics, China University of Mining and Technology, Xuzhou, 221116, P.R. China

Abstract In this paper, by using Mawhin’s continuation theorem, we investigate the existence of solutions for a class of fractional differential equations with multi-point boundary value problems at resonance, and the dimension of the kernel for a fractional differential operator is three. An example is given to show our main result. MSC: 34A08; 34B15 Keywords: fractional differential equation; multi-point boundary value problem; resonance; Mawhin’s continuation theorem

1 Introduction Fractional calculus as a generalization of integer-order calculus has been studied for more than 300 years. Concerning the history and basic results on fractional calculus theory, we refer to [1–5]. Fractional differential equations as an application of fractional calculus have recently been viewed as a useful mathematical model applied in numerous fields of science and engineering (see [2–11]). For example, SIS epidemic model can be described by two fractional-order differential equations of the form 

Dα1 S(t) =  – βSI – μS + φI, Dα1 I(t) = βSI – (φ + μ + α)I,

where Dα1 is Caputo fractional derivatives with 0 < α1 ≤ 1, S(t) is the number of individuals in the susceptible class at time t and I(t) is the number of individuals who are infectious at time t (see [7]). In recent years, boundary value problems (BVPs) and initial value problems (IVPs) of fractional differential equations have been discussed widely, and numerous valuable results have been obtained (see [12–30]). And the usual way to investigate the fractional BVPs and IVPs is nonlinear analysis such as variation method (see [12–14]), fixed-point theorems (see [15–18]), upper and lower solutions method (see [19, 20]), coincidence degree theory (see [21–25]). Besides, for the recent advances in other techniques for solving nonlinear problems, see [26–30]. For example, De La Sen et al. [29] studied the following © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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fractional boundary value problems:   Dα1 x1 (t) = f1 t, x1 (t), x2 (t), . . . , xk (t), I β11 x1 (t), I β12 x2 (t), . . . , I β1k xk (t) ,   C α2 D x2 (t) = f1 t, x1 (t), x2 (t), . . . , xk (t), I β21 x1 (t), I β22 x2 (t), . . . , I β2k xk (t) , C

.. . C

  Dαk xk (t) = f1 t, x1 (t), x2 (t), . . . , xk (t), I βk1 x1 (t), I βk2 x2 (t), . . . , I βkk xk (t) ,

which are associated with the boundary conditions

xi (0) + xi (1) = ai ,

k  j=1

I βij xi (ξj ) +

k  j=1



1

I βij xi (ηj ) = bi

xi (s) ds, 0

where k is a natural number, i = 1, 2, . . . , k, 1 ≤ αi < 2, C Dαi denotes the Caputo fractional derivative. By using the shifted Chebyshev and Legendre polynomials approach, the authors obtained the numerical solutions for the above k-dimensional system. A boundary value problem is called resonance if the corresponding homogeneous boundary value problem has a nontrivial solution. Recently, fractional boundary value problems at resonance have attracted many scholars’ attention (see [21–25]). To our knowledge, the most effective method for solving fractional resonance boundary value problems is Mawhin’s continuation theorem (see [31, 32]). For example, in [22], Chen and Tang considered the solvability of the following fractional-order multi-point boundary value problems at resonance by using Mawhin’s continuation theorem. 

(a(t)C Dα0+ u(t)) = f (t, u(t), u (t), C Dα0+ u(t)), t ∈ [0, 1],  C α D0+ u(0) = 0, u(1) = m–1 u(0) = 0, j=1 σj u(ξj ),

where C Dα0+ is the Caputo fractional derivative with 1 < α ≤ 2, 0 < σj ∈ R, ξj ∈ (0, 1), m–1 3 j=1 σj ξj = 1, f : [0, 1] × R → R satisfies the Carathéodory conditions. In [23], Bai and Zhang considered the solvability of the following fractional multipoint boundary value problems at resonance with two-dimensional kernels by employing Mawhin’s continuation theorem. 

α–1 t ∈ (0, 1), Dα0+ u(t) = f (t, u(t), Dα–2 0+ u(t), D0+ u(t)),  3–α α–1 α–1 D0+ u(0) = D0+ u(η), u(1) = m I0+ u(0) = 0, i=1 αi u(ηi ),

where Dα0+ is the standard Riemann-Liouville fractional derivative with 2 < α ≤ 3,   α–1 α–2 0 < η ≤ 1, 0 < η1 < · · · < ηm < 1, m ≥ 2, m = m = 1, f : [0, +∞) × R3 → R i=1 αi ηi i=1 αi ηi satisfies the Carathéodory conditions. In order to make sure that the linear operator Q is well defined, the author assumed that the following condition holds:   m m   ηα–1 ( (α))2 ηα (α) (α – 1) 2α–2 2α–1 – = 0. 1– 1– αi ηi αi ηi R= α (2α – 1) (α – 1) (2α) i=1 i=1 In [24], Jiang studied the existence of solutions for the following fractional multipoint boundary value problems at resonance with two-dimensional kernels by applying

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Mawhin’s continuation theorem. 

a.e. t ∈ [0, 1], Dα0+ u(t) = f (t, u(t), Dα–1 0+ u(t)), m α–1 u(0) = 0, D0+ u(0) = i=1 ai Dα–1 0+ u(ξi ),

Dα–2 0+ u(1) =

n

α–2 j=1 bj D0+ u(ηj ),

where Dα0+ is the standard Riemann-Liouville fractional derivative with 2 < α < 3, 0 <  n n ξ1 < ξ2 < · · · < ξm < 1, 0 < η1 < η2 < · · · < ηn < 1, m i=1 ai = j=1 bj = j=1 bj ηj = 1, f : 2 [0, 1] × R → R satisfies the Carathéodory conditions. In order to make sure that the linear operator Q is well defined, the author assumed that the following condition holds:   m m n n     1 1 3 2 1– 1– bj ηj ai ξi – bj ηj ai ξi2 = 0. 6 4 j=1 i=1 j=1 i=1 Thus, motivated by the results mentioned, in this paper, we discuss the existence of solutions for the following multi-point boundary value problems by using Mawhin’s continuation theorem. ⎧ α–3 α α–2 ⎪ u(t), Dα–1 t ∈ (0, 1), 0+ u(t)), ⎨ D0+ u(t) = f (t, u(t), D0+ u(t), D0+  l 4–α α–1 α–1 I0+ u(t)|t=0 = 0, D0+ u(0) = i=1 αi D0+ u(ξi ), ⎪  n ⎩ α–2 α–3 α–2 β D Dα–3 D0+ u(0) = m 0+ u(1) = j=1 j 0+ u(ηj ), k=1 γk D0+ u(ρk ),

(1.1)

where Dα0+ is the standard Riemann-Liouville fractional derivative with 3 < α ≤ 4, 0 < ξ1 < · · · < ξl < η1 < · · · < ηm < 1, 0 < ρ1 < · · · < ρk < 1, αi , βj , γk ∈ R, f ∈ [0, 1] × R4 → R is a Carathéodory function. Throughout this paper, we assume that the following resonance conditions of (1.1) hold. l m n m n n 2 (H1 ) i=1 αi = 1, j=1 βj = 1, k=1 γk = 1, j=1 βj ηj = 0, k=1 γk ρk = 1, k=1 γk ρk = 1. Compared with previous work in the field, in this paper several new features can be shown as follows. Firstly, to the best of author’s knowledge, there are only few papers that consider the integral-order resonance BVPs with three-dimensional kernels (see [33–36]); and for fractional resonance boundary value problems, most of the discussions are limited to the kernels of operator dimension less than or equal to two. So, our results are a generalization of some previous publications. Secondly, compared with [23, 24, 35, 36], in this paper, based on analysis as proved, we needn’t ensure that the linear operator Q is well defined by assumed conditions. Compared with [33, 34], it is more difficult to define the linear isomorphism operator J and also it is even more difficult to give an example that satisfies all the assumptions in the paper. Furthermore, when we take α = 4, BVPs (1.1) are reduced to the fourth-order differential equation resonance BVPs. We organized the rest of the article as follows. In Section 2, we recall some definitions and lemmas. In Section 3, based on Mawhin’s continuation theorem, we establish an existence theorem for problem (1.1). In Section 4, we present an example to illustrate our main result. In the last section, we give a short conclusion.

2 Preliminaries In this section, we recall some definitions and lemmas which are used throughout this paper.

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Let X and Y be two Banach spaces with the norms · X and · Y , respectively. Define L : dom(L) ⊂ X → Y to be a Fredholm operator with index zero, P : X → X, Q : Y → Y to be two projectors such that Im P = Ker L,

Im L = Ker Q,

X = Ker L ⊕ Ker P,

Y = Im L ⊕ Im Q,

L|dom L∩Ker P : dom L → Im L is invertible, we denote the inverse by Kp . Let be an open ¯ = ∅, the map N : X → Y is called L-compact on , ¯ if bounded subset of X and dom L ∩ ¯ ¯ QN( ) is bounded and KP,Q N = Kp (I – Q)N : → X is compact (see [31, 32]). Lemma 2.1 (see [31, 32]) Let L : dom L ⊂ X → Y be a Fredholm operator of index zero ¯ Assume that the following conditions are satisfied: and N : X → Y be L-compact on . (i) Lu = λNu for any u ∈ (dom L \ Ker L) ∩ ∂ , λ ∈ (0, 1); (ii) Nu ∈/ Im L for any u ∈ Ker L ∩ ∂ ; (iii) deg(QN|Ker L , ∩ Ker L, 0) = 0. ¯ Then the equation Lx = Nx has at least one solution in dom L ∩ . Definition 2.1 (see [2, 3]) The Riemann-Liouville fractional integral of order α > 0 for a function u : (0, +∞) → R is given by α I0+ u(t) =

1 (α)



t

(t – s)α–1 u(s) ds 0

provided that the right-hand side integral is pointwise defined on (0, +∞). Definition 2.2 (see [2, 3]) The Riemann-Liouville fractional derivative of order α > 0 for a function u : (0, +∞) → R is given by Dα0+ u(t) =

dn dn n–α 1 I u(t) = 0+ dt n (n – α) dt n



t

(t – s)n–α–1 u(s) ds,

0

where n = [α] + 1, provided that the right-hand side integral is pointwise defined on (0, +∞). Lemma 2.2 (see [2, 3, 18, 24]) Let α > 0. Assume that u, Dα0+ u ∈ L1 (0, 1), then the following equality holds: α I0+ Dα0+ u(t) = u(t) + c1 t α–1 + c2 t α–2 + · · · + cn t α–n ,

where n = [α] + 1, ci ∈ R, i = 1, 2, . . . , n. Lemma 2.3 (see [2, 3, 18, 24]) Assume that u ∈ L1 (0, 1), α ≥ β ≥ 0, then β

α+β

α I0+ u(t) = I0+ u(t), I0+

β

α–β

α D0+ I0+ u(t) = I0+ u(t).

Lemma 2.4 (see [2, 3, 18, 24]) Assume that α > 0, λ > –1, t > 0, then α λ t = I0+

(λ + 1) α+λ t , (λ + 1 + α)

Dα0+ t λ =

(λ + 1) λ–α t , (λ + 1 – α)

in particular Dα0+ t α–m = 0, m = 1, 2, . . . , n, where n = [α] + 1.

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3 Main result Take   X = u : u, Dα–1 0+ u ∈ C[0, 1] ,

Y = L1 [0, 1].

It is easy to check that X and Y are two Banach spaces with norms    α–2   α–1         u X = max u ∞ , Dα–3 0+ u ∞ , D0+ u ∞ , D0+ u ∞ ,

y Y = y 1 ,

1 respectively, where u ∞ = supt∈[0,1] |u(t)|, y 1 = 0 |y(t)| dt. Define the linear operator L : dom L ⊂ X → Y and the nonlinear operator N : X → Y as follows: Lu(t) = Dα0+ u(t), u(t) ∈ dom L,   α–2 α–1 Nu(t) = f t, u(t), Dα–3 0+ u(t), D0+ u(t), D0+ u(t) ,

u(t) ∈ X,

where   dom L = u ∈ X : Dα0+ u(t) ∈ Y , u satisfies boundary value conditions of (1.1) . Then problem (1.1) is equivalent to the operator equation Lu = Nu, u ∈ dom L. Lemma 3.1 Assume that (H1 ) holds, then the operator L : dom L ⊂ X → Y satisfies   Ker L = u ∈ dom L : u(t) = at α–1 + bt α–2 + ct α–3 , a, b, c ∈ R ,

(3.1)

Im L = {y ∈ Y : Q1 y = Q2 y = Q3 y = 0},

(3.2)

where Q1 y =

l  i=1

 Q3 y =

1



ξi 0

m  j=1

2

(1 – s) y(s) ds – 0

Q2 y =

y(s) ds,

αi

n  k=1

 γk

ρk



ηj

(ηj – s)y(s) ds,

βj 0

(ρk – s)2 y(s) ds.

0

Proof If Lu = Dα0+ u = 0, by Lemma 2.2, we have u(t) = at α–1 + bt α–2 + ct α–3 + dt α–4 ,

a, b, c, d ∈ R.

4–α u(t)|t=0 = 0 that d = 0, then It follows from the boundary condition I0+

u(t) = at α–1 + bt α–2 + ct α–3 . So,   Ker L ⊂ u ∈ dom L : u(t) = at α–1 + bt α–2 + ct α–3 , a, b, c ∈ R .

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Conversely, take u(t) = t α–1 + t α–2 + t α–3 , it is easy to check that Dα0+ u = 0 and u(t) satisfies boundary value conditions of (1.1). Thus,   u ∈ dom L : u(t) = at α–1 + bt α–2 + ct α–3 , a, b, c ∈ R ⊂ Ker L. For y ∈ Im L, there exists u ∈ dom L such that Dα0+ u(t) = y(t). By Lemmas 2.2 and 2.4, com4–α bined with the boundary condition I0+ u(t)|t=0 = 0, one has α y(t) + c1 t α–1 + c2 t α–2 + c3 t α–3 . u(t) = I0+

Considering the boundary conditions Dα–1 0+ u(0) =

l 

αi Dα–1 0+ u(ξi ),

Dα–2 0+ u(0) =

i=1

Dα–3 0+ u(1) =

m 

βj Dα–2 0+ u(ηj ),

j=1

n 

γk Dα–3 0+ u(ρk ),

k=1

by Lemmas 2.3 and 2.4, we obtain Dα–1 0+ u(0) = c1 (α) =

l 

αi Dα–1 0+ u(ξi )

i=1

=

l 

 αi

 y(s) ds + c1 (α)

0

i=1

=

ξi

l 



ξi

y(s) ds + c1 (α),

αi 0

i=1

Dα–2 0+ u(0) = c2 (α – 1) =

m 

βj Dα–2 0+ u(ηj )

j=1

=

m 



(ηj – s)y(s) ds + c1 (α)ηj + c2 (α – 1)

βj 0

j=1

=



ηj

m 



ηj

(ηj – s)y(s) ds + c2 (α – 1)

βj 0

j=1

and Dα–3 0+ u(1) = =

1 2



1

0

n 

1 (1 – s)2 y(s) ds+ c1 (α) + c2 (α – 1) + c3 (α – 2) 2

γk Dα–3 0+ u(ρk )

k=1

=

n 

γk

k=1

1 γk 2 n

=

   ρk 1 1 (ρk – s)2 y(s) ds + c1 (α)ρk2 + c2 (α – 1)ρk + c3 (α – 2) 2 0 2

k=1



ρk 0

1 (ρk – s)2 y(s) ds+ c1 (α) + c2 (α – 1) + c3 (α – 2). 2

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Thus, Q1 y = Q2 y = Q3 y = 0,

(3.3)

that is, Im L ⊂ {y ∈ Y : Q1 y = Q2 y = Q3 y = 0}. α Conversely, let y ∈ Y satisfy (3.3), taking u(t) = I0+ y(t), we can easily check that u ∈ dom L  and Lu(t) = y(t). Then we have {y ∈ Y : Q1 y = Q2 y = Q3 y = 0} ⊂ Im L.

Let R1 , R2 , R3 : Y → Y be three linear operators defined as R1 y =

μ(μ + 1)(μ + 2) [A11 Q1 y + A12 Q2 y + A13 Q3 y], (μ, ν, ω)

R2 y =

ν(ν + 1)(ν + 2) [A21 Q1 y + A22 Q2 y + A23 Q3 y], (μ, ν, ω)

R3 y =

ω(ω + 1)(ω + 2) [A31 Q1 y + A32 Q2 y + A33 Q3 y], (μ, ν, ω)

where   (μ + 1)(μ + 2) l αi ξ μ   i=1 i  (μ, ν, ω) =  (ν + 1)(ν + 2) li=1 αi ξiν    (ω + 1)(ω + 2) l αi ξ ω i i=1

 μ+1 (μ + 2) m j=1 βj ηj m (ν + 2) j=1 βj ηjν+1  ω+1 (ω + 2) m j=1 βj ηj

  2(1 – nk=1 γk ρkμ+2 )    2(1 – nk=1 γk ρkν+2 )  .  n 2(1 – k=1 γk ρkω+2 ) 

We note   a11   (μ, ν, ω) :=  a21   a31

a12 a22 a32

 a13   a23  ,  a33 

here Aij (i, j = 1, 2, 3) are the algebraic complements of aij . Lemma 3.2 Assume that (H1 ) holds, then there exist constants μ ∈ {1, 2, . . . , m – 1}, ν ∈ Z+ , ν ≥ μ + 1 and ω ∈ Z+ large enough numbers such that (μ, ν, ω) = 0.  Proof By li=1 αi = 1, we have that, for each s ∈ Z+ , there exists ks ∈ {sl + 1, . . . , (s + 1)l}   such that li=1 αi ξiks = 0. If not, we get li=1 αi ξiks = 0, ks ∈ {sl + 1, . . . , (s + 1)l}, that is, ⎛

ξ1sl+1 ⎜ ξ sl+2 ⎜ 1 ⎜ . ⎜ . ⎝ . ξ1(s+1)l

ξ2sl+1 ξ2sl+2 .. . (s+1)l ξ2

··· ··· .. . ···

⎞⎛ ⎞ ξlsl+1 α1 ⎜ ⎟ ξlsl+2 ⎟ ⎟ ⎜ α2 ⎟ ⎜ ⎟ .. ⎟ ⎟ ⎜ . ⎟ = 0. . ⎠ ⎝ .. ⎠ αl ξl(s+1)l

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Since  sl+1 ξ  1  ξ sl+2  1  .  .  .  (s+1)l ξ 1

 ξlsl+1  ξlsl+2   sl+1 (ξj – ξi ) = 0. ..  = (ξ1 ξ2 · · · ξl ) .  1≤i
··· ··· .. .

ξ2sl+1 ξ2sl+2 .. . ξ2(s+1)l

···

  So, αi = 0 (i = 1, 2, . . . , l), which is a contradiction with li=1 αi = 1. Similarly, by m j=1 βj = 1  m μ+1 and m β η = 0, there exists μ ∈ {1, 2, . . . , m – 1} such that β η =  0. Otherwise, j=1 j j j=1 j j  μ+1 β η = 0, μ = 0, 1, 2, . . . , m – 1, that is, we have m j=1 j j ⎛

⎞⎛ ⎞ ηm β1 ⎟ 2 ⎟⎜ ηm ⎟ ⎜ β2 ⎟ ⎟ ⎜ .. ⎟ ⎜ .. ⎟ ⎟ = 0. . ⎠⎝ . ⎠ m βm ηm

··· ··· .. .

η2 η22 .. . η2m

η1 ⎜ η2 ⎜ 1 ⎜ . ⎜ . ⎝ . η1m

···

Because   η1   η2  1  .  .  .   ηm 1

η2 η22 .. . η2m

··· ··· .. . ···

 ηm  2  ηm   (ηj – ηi ) = 0. ..  = η1 η2 · · · ηm .  1≤i
So, βj = 0 (j = 1, 2, . . . , m), which has conflicts with  S = ks ∈ Z : +

l

μ m ks +1 i=1 αi ξi j=1 βj ηj l (ks + 1) i=1 αi ξiks

(μ + 1)

m

=

j=1 βj

m 

= 1. Set 

βj ηjμ+1

.

j=1

We show that S is a finite set. If else, there exists a strictly monotonic sequence {kst }∞ t=1 such that l

kst +1 μ m i=1 αi ξi j=1 βj ηj  ks (kst + 1) li=1 αi ξi t

(μ + 1)

=

m 

μ+1

βj ηj

.

j=1

It follows from 0 < ξ1 < · · · < ξl < η1 < · · · < ηm < 1, m  j=1

μ+1 βj ηj

= lim

kst →∞

= lim

kst →∞

m

μ+1 j=1 βj ηj

= 0 that

l

kst +1 μ m i=1 αi ξi j=1 βj ηj  ks (kst + 1) li=1 αi ξi t

(μ + 1)

l

μ m kst +1 i=1 αi ξi j=1 βj (ηj /ξl ) l (kst + 1) i=1 αi (ξi /ξl )kst

ξl (μ + 1)

= ∞.

l

μ i=1 αi ξi

= 0 and

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It is a contradiction. Thus, there exist μ ∈ {1, 2, . . . , m – 1}, ν ∈ Z+ , ν ≥ μ + 1 such that A33 = 0. Therefore, we have   a11   lim (μ, ν, ω) =  a21 ω→∞   0

a12 a22 0

 a13   a23  = 2A33 = 0.  2 

So, if we make sure ω ∈ Z+ (the set of positive integers) is large enough, it can be such that (μ, ν, ω) = 0.  Lemma 3.3 Assume that (H1 ) holds, then L : dom L ⊂ X → Y is a Fredholm operator of index zero. The linear projector operator P : X → X and Q : Y → Y can be defined as follows: 1 1 1 α–1 α–2 Dα–1 Dα–2 Dα–3 u(0)t α–3 , + + 0+ u(0)t 0+ u(0)t (α) (α – 1) (α – 2) 0+       (Qy)(t) = R1 y(t) t μ–1 + R2 y(t) t ν–1 + R3 y(t) t ω–1 , (Pu)(t) =

where taking μ, ν, ω satisfies Lemma 3.2. Proof By the definition of P, we can easily check that P is a continuous linear projector operator and satisfies Im P = Ker L, X = Ker P ⊕ Ker L. According to Lemma 3.2, there exist constants μ ∈ {1, 2, . . . , m – 1}, ν ∈ Z+ , ν ≥ μ + 1 and ω ∈ Z+ large enough, such that (μ, ν, ω) = 0. So, Q is a well-defined operator. It is clear that Q is a continuous linear operator and dim Im Q = 3. By the definitions of R1 , R2 , R3 , we can calculate the following equations:   R1 (R1 y)t μ–1 = R1 y,   R2 (R1 y)t μ–1 = 0,   R3 (R1 y)t μ–1 = 0,

  R1 (R2 y)t ν–1 = 0,   R2 (R2 y)t ν–1 = R2 y,   R3 (R2 y)t ν–1 = 0,

  R1 (R3 y)t ω–1 = 0,   R2 (R3 y)t ω–1 = 0,   R3 (R3 y)t ω–1 = R3 y.

Thus, 

        Q2 y (t) = Q R1 y(t) t μ–1 + R2 y(t) t ν–1 + R3 y(t) t ω–1         = R1 R1 y(t) t μ–1 + R2 y(t) t ν–1 + R3 y(t) t ω–1 t μ–1         + R2 R1 y(t) t μ–1 + R2 y(t) t ν–1 + R3 y(t) t ω–1 t ν–1         + R3 R1 y(t) t μ–1 + R2 y(t) t ν–1 + R3 y(t) t ω–1 t ω–1       = R1 y(t) t μ–1 + R2 y(t) t ν–1 + R3 y(t) t ω–1 = Qy(t).

So, Q is a projector operator. From Lemma 3.1, we have Im L ⊂ Ker Q. Now, we show the fact that Ker Q ⊂ Im L. In fact, for y ∈ Ker Q, thus Qy = 0, then we get a system of linear equations with respect to Q1 y, Q2 y, Q3 y as follows: ⎧ ⎪ ⎨ A11 Q1 y + A12 Q2 y + A13 Q3 y = 0, A21 Q1 y + A22 Q2 y + A23 Q3 y = 0, ⎪ ⎩ A31 Q1 y + A32 Q2 y + A33 Q3 y = 0.

(3.4)

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Since the determinant of coefficiency for (3.4) is 2 (μ, ν, ω) = 0, we get that Q1 y = Q2 y = Q3 y = 0, thus Ker Q ⊂ Im L. Therefore, Ker Q = Im L. For y ∈ Y , set y = (y – Qy) + Qy, then (y – Qy) ∈ Ker Q = Im L, Qy ∈ Im Q. So, y = Im L + Im Q. Furthermore, for any y ∈ Im L ∩ Im Q, there exist constants a, b, c ∈ R such that y(t) = at μ–1 + bt ν–1 + ct ω–1 and Q1 y = Q2 y = Q3 y = 0. Then we also get a system of linear equations with respect to a, b, c as follows: ⎧ l a μ b ν c ω ⎪ ⎪ i=1 αi ( μ ξi + ν ξi + ω ξi ) = 0, ⎪  ⎪ ⎨ m β [ a ημ+1 + b ην+1 + c ηω+1 ] = 0, j j=1 j μ(μ+1) j ω(ω+1) j n ν(ν+1)  μ+2 2a 2b ⎪ (1 – γ ρ ) + (1 – nk=1 γk ρkν+2 ) ⎪ k=1 k k ν(ν+1)(ν+2) ⎪ μ(μ+1)(μ+2)  ⎪ ⎩ + 2c (1 – nk=1 γk ρkω+2 ) = 0. ω(ω+1)(ω+2)

(3.5)

By simple calculation, we obtain that the determinant of coefficiency for (3.5) is (μ, ν, ω) = 0. μ(μ + 1)(μ + 2)ν(ν + 1)(ν + 2)ω(ω + 1)(ω + 2) Thus, (3.5) only have zero solutions, that is, a = b = c = 0. It means that Im Q ∩ Im L = {0}. Therefore, Y = Im Q⊕Im L. From the above, we get dim Ker L = dim Im Q = co dim Im L = 3. So, L is a Fredholm operator of index zero.  Lemma 3.4 Assume that (H1 ) holds, define the linear operator Kp : Im L → dom L ∩ Ker P by (Kp y)(t) =

1 (α)



t

(t – s)α–1 y(s) ds,

y ∈ Im L,

0

then Kp is the inverse of L|dom L∩Ker P and Kp y X ≤ y 1 , ∀y ∈ Im L. Proof For y ∈ Im L, then Q1 y = Q2 y = Q3 y = 0, which is combined with the definition of Kp and Lemmas 2.2-2.4, we can check that Kp y ∈ dom L ∩ Ker P. So, Kp is well defined on Im L. Obviously, (LKp )y(t) = y(t), ∀y ∈ Im L. For u(t) ∈ dom L, by Lemma 2.2, we have α (Kp L)u(t) = I0+ Dα0+ u(t) = u(t) + c1 t α–1 + c2 t α–2 + c3 t α–3 + c4 t α–4 ,

c1 , c2 , c3 , c4 ∈ R.

4–α Then 0 = I0+ [(Kp L)u(t)]|t=0 = c4 . It follows from P[(Kp L)u(t)] = 0 and c1 t α–1 + c2 t α–2 + c3 t α–3 ∈ Ker L = Im P that c1 t α–1 + c2 t α–2 + c3 t α–3 = –Pu(t). Then (Kp L)u(t) = u(t) – Pu(t). Therefore, if u(t) ∈ dom L ∩ Ker P, we have (Kp L)u(t) = u(t). So, Kp is the inverse of L|dom L∩Ker P . Moreover, by Lemma 2.3 we have the following inequations:

 1   1 y(s) ds ≤ y 1 , (α) 0 0  t  1   α–3     2  ds ≤ 1 D Kp y ≤ 1 y(s) ds ≤ y 1 , y(s) (t – s) 0+ 2 0 2 0  t  1  α–2      D Kp y ≤ y(s) ds = y 1 , y(s) ds ≤ (t – s) 0+ |Kp y| ≤

1 (α)



t

  (t – s)α–1 y(s) ds ≤

0

 α–1  D Kp y ≤ 0+



t

  y(s) ds ≤

0

So, Kp y X ≤ y 1 , ∀y ∈ Im L.

0



1

 y(s) ds = y 1 .

0



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Lemma 3.5 Assume that (H1 ) holds and ⊂ X is an open bounded subset with dom L ∩ ¯ = ∅, then N is L-compact on . ¯ Proof From f : [0, 1] × R4 → R satisfies the Carathéodory conditions, we can get that ¯ and (I – Q)N( ) ¯ are bounded, that is, there exist constants  ,  > 0 such that QN( ) ¯ a.e. t ∈ [0, 1]. So, we only need to show that Kp (I – Q)N : |QNu| ≤  , |(I – Q)Nu| ≤ , u ∈ , ¯ is bounded. It follows from the Lebesgue ¯ → X is compact. By Lemma 3.4, Kp (I – Q)N( ) ¯ → X is continuous. For 0 ≤ t1 < dominated convergence theorem that Kp (I – Q)N : ¯ we have t2 ≤ 1, u ∈ ,   Kp (I – Q)Nu(t1 ) – Kp (I – Q)Nu(t2 )    t2  1  t1 α–1 α–1 (t1 – s) (I – Q)Nu(s) ds – (t2 – s) (I – Q)Nu(s) ds =  (α) 0 0   t1     1  α–1 α–1  (t (I – Q)Nu(s) ds – s) – (t – s) ≤ 1 2   (α) 0  t2   1  + (t2 – s)α–1 (I – Q)Nu(s) ds  (α) t1  t1  t2     (t2 – s)α–1 – (t1 – s)α–1 ds + (t2 – s)α–1 ds ≤ (α) 0 (α) t1 =

 α α  t –t . (α + 1) 2 1

¯ is equicontinuous. In Since t α is uniformly continuous on [0, 1], we get Kp (I – Q)N( ) addition, because the following equations hold:  α–2 Dα–2 0+ Kp (I – Q)Nu(t2 ) – D0+ Kp (I – Q)Nu(t1 ) =

t1

 α–3 Dα–3 0+ Kp (I – Q)Nu(t2 ) – D0+ Kp (I – Q)Nu(t1 ) =

t2

t2

t1

Dα–1 0+ Kp (I – Q)Nu(s) ds, Dα–2 0+ Kp (I – Q)Nu(s) ds,

¯ we simply indicate the equicontinuity of Dα–1 0+ Kp (I – Q)N( ). In fact,   α–1 D Kp (I – Q)Nu(t1 ) – Dα–1 Kp (I – Q)Nu(t2 ) 0+ 0+   t2    (I – Q)Nu(s) ds ≤ (t2 – t1 ). = t1

¯ Since t is uniformly continuous on [0, 1], thus Dα–1 0+ Kp (I – Q)N( ) is equicontinuous. By ¯  the Arzelà-Ascoli theorem, we obtain that Kp (I – Q)N : → X is compact. In order to obtain our main results, we suppose that the following conditions are satisfied: (H2 ) There exist nonnegative functions a(t), b(t), c(t), d(t), e(t) ∈ Y such that, for any (u1 , u2 , u3 , u4 ) ∈ R4 , t ∈ (0, 1),   f (t, u1 , u2 , u3 , u4 ) ≤ a(t)|u1 | + b(t)|u2 | + c(t)|u3 | + d(t)|u4 | + e(t)

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and a 1 + b 1 + c 1 + d 1 <

1 , (q + 1)

1 1 1 where q = max{ (α) + (α–1) + (α–2) , 52 }. (H3 ) There exists a constant L > 0 for any u(t) ∈ dom L, if inft∈[0,1] |Dα–1 0+ u(t)| > L, then

Dα–1 0+ u(t)Q1 Nu(t) > 0,

∀t ∈ [0, 1]

(3.6)

Dα–1 0+ u(t)Q1 Nu(t) < 0,

∀t ∈ [0, 1];

(3.7)

or

(H4 ) There exists a constant M > 0 for any u(t) ∈ dom L, if inft∈[0,1] |Dα–2 0+ u(t)| > M, then Dα–2 0+ u(t)Q2 Nu(t) > 0,

∀t ∈ [0, 1]

(3.8)

Dα–2 0+ u(t)Q2 Nu(t) < 0,

∀t ∈ [0, 1];

(3.9)

or

(H5 ) There exists a constant G > 0 for any u(t) ∈ dom L, if inft∈[0,1] |Dα–3 0+ u(t)| > G, then Dα–3 0+ u(t)Q3 Nu(t) > 0,

∀t ∈ [0, 1]

(3.10)

Dα–3 0+ u(t)Q3 Nu(t) < 0,

∀t ∈ [0, 1].

(3.11)

or

Lemma 3.6 Suppose that (H2 )-(H5 ) hold, set   1 = u ∈ dom L \ Ker L : Lu = λNu, λ ∈ (0, 1) . Then 1 is bounded. Proof For u ∈ 1 , we have Nu ∈ Im L = Ker Q, that is, Q1 (Nu(t)) = Q2 (Nu(t)) = Q3 (Nu(t)) = 0. Thus, from (H3 )-(H5 ), we obtain that there exist constants t1 , t2 , t3 ∈ [0, 1] α–2 α–1 such that |Dα–3 0+ u(t3 )| ≤ G, |D0+ u(t2 )| ≤ M and |D0+ u(t1 )| ≤ L. Since, for all t ∈ [0, 1], the following equations hold:  α–1 Dα–1 0+ u(t) = D0+ u(t1 ) +

 α–2 Dα–2 0+ u(t) = D0+ u(t2 ) +

 α–3 Dα–3 0+ u(t) = D0+ u(t3 ) +

t t1

Dα0+ u(s) ds,

t t2

Dα–1 0+ u(s) ds,

t t3

Dα–2 0+ u(s) ds.

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Then we have  α–1        D u(0) ≤ Dα–1 u(t) ≤ Dα–1 u(t1 ) + Dα u ≤ L + Lu 1 ≤ L + Nu 1 , 0+ 0+ 0+ 0+ ∞ 1         α–2 D u(0) ≤ Dα–2 u(t) ≤ Dα–2 u(t2 ) + Dα–1 u ≤ L + M + Nu 1 , 0+ 0+ 0+ 0+ ∞ ∞ and  α–3        D u(0) ≤ Dα–3 u(t) ≤ Dα–3 u(t3 ) + Dα–2 u ≤ L + M + G + Nu 1 . 0+ 0+ 0+ 0+ ∞ ∞ By the definition of P and Lemma 2.4, we get 1 α–1 2 α–2 α–3 Dα–3 0+ Pu(t) = D0+ u(0)t + D0+ u(0)t + D0+ u(0), 2 α–1 α–2 Dα–2 0+ Pu(t) = D0+ u(0)t + D0+ u(0),

α–1 Dα–1 0+ Pu(t) = D0+ u(0).

Thus,       α–3  1  α–1 D Pu ≤ D u(0) + Dα–2 u(0) + Dα–3 u(0) ≤ 5 L + 2M + G + 5 Nu 1 , 0+ 0+ 0+ 2 0+ 2 2   α–2   α–2   α–1 D Pu ≤ D u(0) + D u(0) ≤ 2L + M + 2 Nu 1 , 0+ 0+ 0+   α–1   α–1 D Pu = D u(0) ≤ L + Nu 1 . 0+ 0+ Because of     α–2  α–3 1  α–1 1 1 D u(0) + D u(0) D u(0) + (α) 0+ (α – 1) 0+ (α – 2) 0+  L L+M L+M+G 1 1 1 ≤ + + + + + Nu 1 . (α) (α – 1) (α – 2) (α) (α – 1) (α – 2)

|Pu| ≤

Therefore,    α–2   α–1         Pu X = max Pu ∞ , Dα–3 0+ Pu ∞ , D0+ Pu ∞ , D0+ Pu ∞ ≤ qL + pM + rG + q Nu 1 ,

(3.12)

1 1 1 + (α–2) , 2}, r = (α–2) . Also, for u ∈ 1 , u ∈ dom L \ Ker L, then (I – where p = max{ (α–1) P)u ∈ dom L ∩ Ker P, LPu = 0, from Lemma 3.4, we have

    (I – P)u = Kp L(I – P)u = Kp Lu X ≤ Lu 1 ≤ Nu 1 . X X

(3.13)

It follows from (3.12) and (3.13) that     u X = Pu + (I – P)uX ≤ Pu X + (I – P)uX ≤ pM + qL + rG + (q + 1) Nu 1 .

(3.14)

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By (H2 ), we have   α–2      Nu 1 ≤ a 1 u ∞ + b 1 Dα–3 0+ u ∞ + c 1 D0+ u ∞  α–1  + d 1 D u + e 1 ∞

0+

  ≤ a 1 + b 1 + c 1 + d 1 pM + qL + rG + (q + 1) Nu 1 + e 1 . 

(3.15)

Substituting (3.15) into (3.14), one gets u X ≤ (q + 1)

( a 1 + b 1 + c 1 + d 1 )(pM + qL + rG) + e 1 + pM + qL + rG. 1 – (q + 1)( a 1 + b 1 + c 1 + d 1 ) 

So, 1 is bounded. Lemma 3.7 Suppose that (H3 )-(H5 ) hold, set 2 = {u ∈ Ker L : Nu ∈ Im L}. Then 2 is bounded.

Proof For u ∈ 2 , then u ∈ Ker L and Nu ∈ Im L = Ker Q, that is, there exist constants a, b, c ∈ R such that u(t) = at α–1 + bt α–2 + ct α–3 , QNu(t) = 0. Thus, Q1 Nu(t) = Q2 Nu(t) = Q3 Nu(t) = 0. By (H3 )-(H5 ), there exist constants t4 , t5 , t6 ∈ [0, 1] such that |Dα–3 0+ u(t6 )| ≤ G, α–2 α–1 |D0+ u(t5 )| ≤ M and |D0+ u(t4 )| ≤ L, that is,  α–1     α–2    D u(t4 ) = a (α) ≤ L, D u(t5 ) = a (α)t5 + b (α – 1) ≤ M, 0+ 0+     α–3 D u(t6 ) = a (α)t 2 /2 + b (α – 1)t6 + c (α – 2) ≤ G. 0+ 6 Then |a| ≤

L , (α)

|b| ≤

M+L , (α – 1)

|c| ≤

G + M + (3L/2) . (α – 2)

Therefore, u ∞ ≤ |a| + |b| + |c| ≤

M+L G + M + (3L/2) L + + , (α) (α – 1) (α – 2)

 α–1  D u ≤ (α)|a| ≤ L, 0+ ∞  α–2  D u ≤ (α)|a| + (α – 1)|b| ≤ 2L + M, 0+ ∞

 α–3  D u ≤ 1 (α)|a| + (α – 1)|b| + (α – 2)|c| 0+ ∞ 2 ≤ 3L + 2M + G. So, 2 is bounded. Lemma 3.8 Suppose that (H3 )-(H5 ) hold, set   3 = u ∈ Ker L : ϑλJu + (1 – λ)QNu = 0, λ ∈ [0, 1] .



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Then 3 is bounded, where ϑ = 1, if (3.6), (3.8), (3.10) hold and ϑ = –1, if (3.7), (3.9), (3.11) hold, J : Ker L → Im Q is the linear isomorphism defined by   J at α–1 + bt α–2 + ct α–3 =

  μ–1 1 a1 t + b1 t ν–1 + c1 t ω–1 , (μ, ν, ω)

∀a, b, c ∈ R,

where a1 = μ(μ + 1)(μ + 2)(A11 a + A12 b + A13 c), b1 = ν(ν + 1)(ν + 2)(A21 a + A22 b + A23 c), c1 = ω(ω + 1)(ω + 2)(A31 a + A32 b + A33 c). Proof Without loss of generality, we suppose that (3.6), (3.8), (3.10) hold, then for any u ∈ 3 , there exist constants a, b, c ∈ R and λ ∈ [0, 1] such that u(t) = at α–1 + bt α–2 + ct α–3 and λJu + (1 – λ)QNu = 0. Therefore, we get a system of linear equations with respect to x1 , x2 , x3 as follows: ⎧ ⎪ ⎨ A11 x1 + A12 x2 + A13 x3 = 0, A21 x1 + A22 x2 + A23 x3 = 0, ⎪ ⎩ A31 x1 + A32 x2 + A33 x3 = 0,

(3.16)

where x1 = λa + (1 – λ)Q1 Nu(t), x2 = λb + (1 – λ)Q2 Nu(t), x3 = λc + (1 – λ)Q3 Nu(t). Since the determinant of coefficiency for (3.16) is 2 (μ, ν, ω) = 0, we get x1 = x2 = x3 = 0, that is, λa + (1 – λ)Q1 Nu(t) = 0,

(3.17)

λb + (1 – λ)Q2 Nu(t) = 0,

(3.18)

λc + (1 – λ)Q3 Nu(t) = 0.

(3.19)

If λ = 1, one has a = b = c = 0. Obviously, 3 is bounded. If λ ∈ [0, 1), it follows from (3.6) L and (3.17) that we can get |a| ≤ (α) ; otherwise, by (3.6) and (3.17), a contradiction will be obtained: 0 < λa2 (α) = –(1 – λ)a (α)Q1 Nu(t) < 0. Similarly, from (3.8) and (3.18), we have |b| ≤ contradiction

M+L ; (α–1)

if not, by (3.8) and (3.18), we get a

0 < λb2 (α – 1) = –(1 – λ)b (α – 1)Q2 Nu(t) < 0. From (3.10) and (3.19), we can derive |c| ≤ a contradiction

M+G+(3L/2) ; (α–2)

else, by (3.10) and (3.19), we obtain

0 < λc2 (α – 2) = –(1 – λ)c (α – 2)Q3 Nu(t) < 0. Similarly to the discussion of Lemma 3.7, 3 is bounded.



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Theorem 3.1 Suppose that (H1 )-(H5 ) hold. Then problem (1.1) has at least one solution in X. ! ¯ i ⊂ . By Lemma 3.5, N is LProof Set be a bounded open set of X such that 3i=1 ¯ From Lemmas 3.6 and 3.7, we get compact on . (i) Lu = λNu for any (u, λ) ∈ [(dom L \ Ker L) ∩ ∂ ] × (0, 1), (ii) Nu ∈ Im L for any u ∈ Ker L ∩ ∂ . In the following, we only need to check that (iii) of Lemma 2.1 is satisfied. Take H(u, λ) = ϑλJu + (1 – λ)QNu, where ϑ is defined as before. According to Lemma 3.8, we derive H(u, λ) = 0 for all u ∈ Ker L ∩ ∂ . Thus, it follows from the homotopy of degree that   deg{QN|Ker L , ∩ Ker L, 0} = deg H(·, 0), ∩ Ker L, 0   = deg H(·, 1), ∩ Ker L, 0 = deg{ϑJ, ∩ Ker L, 0} = 0. Then, by Lemma 2.1, we can get that the operator function Lu = Nu has at least one so¯ which is equivalent to problem (1.1) that has at least one solution lution in dom L ∩ , in X. 

4 Example Example 4.1 Consider the boundary value problems ⎧ 3.5 0.5 1.5 ⎪ u(t), D2.5 t ∈ (0, 1), 0+ u(t)), ⎨ D0+ u(t) = f (t, u(t), D0+ u(t), D0+ 2 0.5 2.5 2.5 D0+ u(0) = i=1 αi D0+ u(ξi ), I0+ u(t)|t=0 = 0, ⎪  3 ⎩ 1.5 0.5 D0.5 D0+ u(0) = 2j=1 βj D1.5 0+ u(ηj ), 0+ u(1) = k=1 γk D0+ u(ρk ), where   1.5 2.5 f t, u(t), D0.5 0+ u(t), D0+ u(t), D0+ u(t)  1.5  2.5   1 = g1 (t)D0.5 0+ u(t) – 180g2 (t)h1 D0+ u(t) + 3h2 D0+ u(t) + sin u(t) + cos t, 8   0, 0 ≤ t ≤ 3/4, 0, 0 ≤ t ≤ 1/2, g1 (t) = g2 (t) = 1, 3/4 < t ≤ 1, 1, 1/2 < t ≤ 1, ⎧ ⎪ ⎨ –1, κ < –1, hi (κ) = κ, –1 ≤ κ ≤ 1, i = 1, 2. ⎪ ⎩ 1, κ > 1. Take α1 = –1, β1 = 9,

α2 = 2, β2 = –8,

1 ξ1 = , 4 2 η1 = , 3

1 ξ2 = , 2 3 η2 = , 4

(4.1)

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γ1 = 1,

γ2 = –3,

γ2 = 3,

1 ρ1 = , 4

1 ρ2 = , 2

3 ρ3 = , 4

Page 17 of 19

μ = 1,

ν = 2,

ω = 3,

then

(1, 2, 3) = –

     f t, u(t), D0.5 u(t), D1.5 u(t), D2.5 u(t)  ≤ 1 D0.5 u(t) + 185. 0+ 0+ 0+ 0+ 8

575 , 2048

Let a(t) = c(t) = d(t) = 0, b(t) = 18 , e(t) = 185, then a 1 + b 1 + c 1 + d 1 =

1 1 2 < = . 8 7 q+1

2.5 Take L = 1, note v(t) = 3h2 (D2.5 0+ u(t)) + sin u(t) + cos t, if |D0+ u(t)| > 1, then

 2.5 D2.5 0+ u(t)Q1 Nu(t) = D0+ u(t)



1/2



1/2

v(t) dt +

v(t) dt > 0.

1/4

0

Take M = 1, since    9 



2/3

3/4

(2/3 – t)v(t) dt – 8 0

0

  85 (3/4 – t)v(t) dt  ≤ , 4

if |D1.5 0+ u(t)| > 1, then D1.5 0+ u(t)Q2 Nu(t)   = D1.5 u(t) 9 0+ 



2/3

1/2

(3/4 – t)v(t) dt

0

0



2/3

– 1620

3/4

(2/3 – t)v(t) dt – 8 (2/3 – t)h1 D1.5 0+ u(t)





3/4

dt + 1440 1/2



(3/4 – t)h1 D1.5 0+ u(t)

> 0. Take G = 23,040, because of  1   2  (1 – t) v(t) dt –  0

1/4

 (1/4 – t) v(t) dt + 3

0



3/4

2



0



1 1/2

3/4

+ 540 1/2

1415 , ≤ 96

(1/2 – t)2 v(t) dt

0

(3/4 – t) v(t) dt – 180

–3

1/2

2

  (1 – t)2 h1 D1.5 0+ u(t) dt

     (3/4 – t)2 h1 D1.5 u(t) dt 0+ 



 dt

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if |D0.5 0+ u(t)| > G, then D0.5 0+ u(t)Q3 Nu(t)  1  2 = D0.5 u(t) (1 – t) v(t) dt – 0+ 0



3/4

–3 0

1/4 0

(3/4 – t)2 v(t) dt – 180





3/4

1

  (1 – t)2 h1 D1.5 0+ u(t) dt

  1 (3/4 – t) h1 D1.5 0+ u(t) dt + 8



1/2



1

2

+ 540

(1/2 – t)2 v(t) dt

0

1/2



1/2

(1/4 – t)2 v(t) dt + 3

(1 – t) 3/4

2

D0.5 0+ u(t) dt

> 0. By Theorem 3.1, boundary value problems (4.1) have at least one solution.

5 Conclusion In this paper, some sufficient conditions are established for the existence of solutions for a class of fractional multi-point boundary value problems at resonance with threedimensional kernel. The main result of this paper is obtained by using Mawhin’s continuation theorem. Compared with previous work, the main difficulties in this paper are as follows. First, as we know, by the way of Mawhin’s continuation theorem, the higher the dimension of kernel is, the more difficult it is to construct the projections P and Q. Second, the main difficulty lies in evaluating prior bounds. Third, it is difficult to construct an example. Acknowledgements This research is supported by the National Natural Science Foundation of China (No. 11271364). Competing interests The authors declare that they have no competing interests. Authors’ contributions The authors have made equal contributions to each part of this paper. All the authors read and approved the final manuscript.

Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 22 September 2017 Accepted: 28 December 2017 References 1. Podlubny, I: Geometric and physical interpretation of fractional integration and fractional differentiation. Fract. Calc. Appl. Anal. 5(4), 367-386 (2002). Dedicated to the 60th anniversary of Prof. Francesco Mainardi 2. Miller, KS, Ross, B: An Introduction to the Fractional Calculus and Fractional Differential Equations. Wiley, New York (1993) 3. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam (2006) 4. Zhou, Y, Wang, J, Zhang, L: Basic Theory of Fractional Differential Equations. World Scientific, Singapore (2016) 5. Podlubny, I: Fractional Differential Equations: An Introduction to Fractional Derivatives, Fractional Differential Equations, to Methods of Their Solution and Some of Their Applications. Academic Press, San Diego (1998) 6. Magin, RL: Fractional calculus models of complex dynamics in biological tissues. Comput. Math. Appl. 59(5), 1586-1593 (2010) 7. El-Saka, HA: The fractional-order SIS epidemic model with variable population size. J. Egypt. Math. Soc. 22(1), 50-54 (2014) 8. Shabibi, M, Postolache, M, Rezapour, S, Vaezpour, SM: Investigation of a multi-singular pointwise defined fractional integro-differential equation. J. Math. Anal. 7(5), 61-77 (2016) 9. Shabibi, M, Rezapour, Sh, Vaezpour, S: A singular fractional integro-differential equation. UPB Sci. Bull., Ser. A, Appl. Math. Phys. 79(1), 109-118 (2017)

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