singular fractional boundary value problems

Monotone solutions for

singular fractional boundary value problems

Xinwei Su

and Shuqin Zhang

Department of Mathematics, China University of Mining and Technology, Beijing, 100083, P. R. China Received 10 August 2019, appeared 24 February 2020

Communicated by Jeff R. L. Webb

Abstract. In this paper, we investigate a boundary value problem of fractional differ- ential equation. The nonlinear term includes fractional derivatives and is singular with respect to space variables. By means of Schaefer’s fixed point theorem and Vitali con- vergence theorem, an existence result of monotone solutions is obtained. The proofs are based on regularization and sequential techniques. An example is also given to illustrate our main result.

Keywords: Caputo fractional derivative, monotone solution, boundary value problem, singularity.

2020 Mathematics Subject Classification: 34B16, 26A33.

1 Introduction

In this work, we consider the following boundary value problem (BVP for short) (CD₀^α+u(t) = f(t,u(t)_,u⁰(t)_,^CD^β₀+u(t))_,

u(0) +u(1) =0, u⁰(0) =0, (1.1) where 1 < α < 2, 0 < β < 1, ^CD₀^α+ and ^CD₀^β+ are Caputo fractional derivatives, f(t,x,y,z) is singular at the value 0 of its space variables x,y andz. We establish an existence result of monotone increasing and continuous solutionu(t)of BVP (1.1). Since lim_x→0 f(t,x,y,z) =_∞, it follows from the condition u(0) +u(1) = 0 that there exists ξ ∈ (0, 1)such that u(ξ) = 0 and thusξ is a singular point of f.

Throughout the paper, AC[0, 1] and AC^k[0, 1] are the set of absolutely continuous func- tions on [0, 1] and the set of functions having absolutely continuous kth derivatives on [0, 1] respectively, AC⁰[0, 1] = AC[0, 1] for k = 0. kxk_p = R1

0 |x(t)|^pdt¹_p

is the norm in L^p[0, 1], 1 ≤ p < ∞. The basic space used in this paper is Banach space C¹[0, 1] equipped with the norm kxk_∗ = max{kxk,kx⁰k}, here kxk = max_t∈[0,1]|x(t)|, kx⁰k = max_t∈[0,1]|x⁰(t)|. We say that a monotone increasing function u ∈ C¹[0, 1] is a solution of BVP (1.1) if u satisfies

BCorresponding author. Email: kdxws@163.com

the boundary conditions in (1.1), u(ξ) = 0 for some ξ ∈ (0, 1), ^CD^α₀+u(t) is continuous on (0, 1]\ {ξ}and satisfies the equation in (1.1) fort ∈(0, 1]\ {ξ}.

In recent years, there has been a significant development in ordinary and partial differen- tial equations involving fractional derivatives due to their wide range of applications in varied fields of sciences and engineering. Many research papers have appeared recently concerning the existence of positive solutions for fractional boundary value problems with singularities on time and/or space variables, see, for example, the papers [8,10–12,14,21,23] and the ref- erences therein. In [1–4,6,7,17–20,22], using techniques of nonlinear analysis such as fixed point theorems on cones and nonlinear alternatives combined with the methods of regular- ization and sequential approximation, the authors proved the existence of positive solutions for singular fractional boundary value problems in which the singularities are with respect to space variables.

The singular boundary value problem

(CD₀^α+u(t) + f(t,u(t),u⁰(t),^CD^µ₀+u(t)) =0, u⁰(0) =0, u(1) =0,

is studied in [2], where 1<α<_{2, 0}<µ<_1, f(t,x,y,z)is positive and may be singular at the value 0 of its space variablesx,yandz. f(t,x,y,z)is a L^q-Carathéodory function on[0, 1]× B with q > ¹

α−1, B = (0,∞)×(−_{∞, 0})×(−_{∞, 0}). An existence result of positive solutions in spaceC¹[_{0, 1}]is proved by the combination of regularization and sequential techniques with the Guo–Krasnosel’skii fixed point theorem on cones.

In [17] the author discussed the existence of positive solutions for the singular fractional boundary value problem

(D₀^α+u(t) + f(t,u(t),u⁰(t),D^µ₀+u(t)) =0, u(₀) =_0, u⁰(₀) =u⁰(₁) =_0,

where 2 < α < 3, 0 < µ < 1, D₀^α+ and D^µ₀+ are the standard Riemann–Liouville fractional derivatives of order α and µ respectively. The function f(t,x,y,z) is positive and may be singular at the value 0 of its arguments x,y and z, moreover, f(t,x,y,z) satisfies the local Carathéodory conditions on[0, 1]×(0,∞)×(0,∞)×(0,∞). By regularization and sequential techniques and by the Guo–Krasnosel’skii fixed point theorem on cones, positive solutions in C¹[0, 1]are obtained.

Although the singular fractional boundary value problems have been investigated widely, the solutions allowing negative values of fractional boundary value problems with singular- ities on space variables are seldom considered. By Schaefer’s fixed point theorem and Vitali convergence theorem, O’Regan and Stanˇek in [13] investigated monotone solutions in space C[0, 1]of the fractional boundary value problem

(CD₀^α+u(t) = f(t,u(t)), u(0) +u(1) =0, u⁰(0) =0,

where 1< α<2, f(t,x)∈C([0, 1]×(_R\{0})). f(t,x)is nonnegative and may be singular at x=_0.

Inspired by above works, we prove the existence of monotone increasing solutions for BVP (1.1). The main tool used in this paper is Schaefer’s fixed point theorem. Our proofs are based on regularization and sequential techniques. Compared with the existing literature,

this paper presents the following new features. Firstly, as far as we know, the existence results of solutions allowing negative values are even less for fractional boundary value problems with singularities on space variables. Our result compensates for this deficiency to some extent. Secondly, the significant difference with the problem discussed in [13] lies in that the nonlinear term f in BVP (1.1) is related to fractional derivatives and permits singularities on all its space variables. That is to say the problem considered in this paper performs a more general form. Moreover, the conditions on f in our paper are more general than those in [13].

2 Preliminaries

In this section, we introduce some notations and preliminary facts which are used throughout this paper.

The Riemann–Liouville fractional integral of order δ > 0 of a function f(t) ∈ L¹(a,b) is defined by (see [9, p. 69])

I_a^δ+f(t) = ¹ Γ(δ)

Z _t

a

(t−s)^δ−1f(s)ds, t> a.

The Riemann–Liouville fractional derivative of orderδ > 0 of a continuous function f on (a,b]is given by (see [9, p. 70])

D_a^δ+f(t) = d

dt n

I_aⁿ+^−δf(t) = ¹ Γ(n−δ)

d dt

nZ _t

a

(t−s)^n−δ−1f(s)ds,

provided that the right-hand side is pointwise defined on(a,b], wherenis the smallest integer greater than or equal toδ. In particular, for δ=n,Dⁿ_a+f(t) = f⁽ⁿ⁾(t).

The Caputo fractional derivative of orderδ > 0 of a function f(t)∈ C(a,b]is defined by (see [9, p. 91])

CD^δ_a+f(t) =D_a^δ+

"

f(t)−

n−1 k

∑

f^(k)(a)

k! (t−a)^k

# ,

provided that the right-hand side is pointwise defined on(a,b], wherenis the smallest integer greater than or equal toδ. In particular, for δ=n,^CDⁿ_a+f(t) = f⁽ⁿ⁾(t).

Remark 2.1. For a function f(t)∈ L¹(a,b), a sufficient condition for the existence of Riemann–

Liouville fractional derivative almost everywhere is that I_aⁿ+^−δf(t) ∈ ACⁿ⁻¹[a,b]. In this case, the function f is said to have a summable fractional derivative of orderδ ([15, Definition 2.4]).

In view of the definition of Caputo fractional derivative, ^CD_a^δ+f(t) = D_a^δ+f(t)_for δ ∈ _Nand

CD^δ_a+f(t) = D^δ_a+f(t)−_∑ⁿ_k=⁻₀¹_Γ(^f_k⁽₋^k)(a)

δ+1)(t−a)^k−δ _forδ ∈/ _N(see (2.4.6) in [9]), thus this is also a sufficient condition for the existence of Caputo fractional derivative. It is worth mentioning that the solutionu(t)in our main result not only has summable fractional derivative^CD₀^α+u(t) on[_{0, 1}]but also has continuous fractional derivative^CD₀^α+u(_t)_on(_{0, 1}]\ {ξ}. For more details, see Step 3 in the proof of Theorem4.1and Remark4.2in Section 4.

Remark 2.2. The following properties are useful for our discussion.

(i) ([9, Lemma 2.8]) I_a^δ+ :C[a,b]→C[a,b]forδ >0.

(ii) ([9, Lemma 2.3]) If δ > 0, γ > 0, δ+γ > 1 and f ∈ L^p(a,b) (1 ≤ p ≤ _∞), then I_a^δ+I_a^γ+f(t) =I_a^δ+^+γf(t)_, t∈[a,b]_.

(iii) ([9, Theorem 2.2]) Ifn−1< δ ≤n and f(t)∈Cⁿ[a,b], then ^CD_a^δ+f(t) = I_aⁿ+^−δf⁽ⁿ⁾(t), t∈ [a,b].

(iv) ([9, Lemma 2.21]) Ifδ >0 and f ∈C[a,b], then^CD^δ_a+I_a^δ+f(t) = f(t), t∈[a,b].

(v) ([17, Lemma 2.1]) I_a^δ+ : L¹[a,b] → L¹[a,b] _forδ ∈ (_{0, 1})_and I_a^δ+ : L¹[a,b] → AC^[δ]−1[a,b] forδ ≥1, where[δ]means the integral part ofδ.

For convenience, in the following discussion we use I^α, ^CD^α and D^α to denote I₀^α+, ^CD₀^α+

andD₀^α+, respectively.

A sequence {φn} ⊂ L¹[0, 1] is said to have uniformly absolutely continuous integrals on [0, 1] if for any e > 0, there exists δ > 0 such that if E ⊂ [0, 1] and meas(E) < δ, then R

E|φ_n(t)|dt<efor alln∈_N(see [5, p. 178]). To prove the main result, we need the following Vitali convergence theorem and nonlinear alternative.

Lemma 2.3([5, pp. 178–179] Vitali convergence theorem). Let{φ_n} ⊂L¹[0, 1],lim_n→+_∞φ_n(t) = φ(t)for a.e. t∈[_{0, 1}]and|φ(t)|<_∞for a.e. t∈ [_{0, 1}]. Then the following statements are equivalent.

(1) φ∈L¹[0, 1]andlimn→+_∞kφn−φk₁ =0.

(2) The sequence{φ_n}has uniformly absolutely continuous integrals on[0, 1].

Lemma 2.4([16, p. 29] Schaefer’s fixed point theorem). Let X be a Banach space and T :X→ X be completely continuous. Then the following alternative holds. Either the equation x = λT(x)has a solution for everyλ∈[_{0, 1}]or the set A={x ∈X: x=λTx for someλ∈(_{0, 1})}is unbounded.

Denote R0 = _R\ {0}, R⁺ = [0,+_∞) and R₀⁺ = (0,+_∞). We work with the following conditions on the function f in (1.1).

(H1) f∈C([0, 1]×_R0×_R⁺₀×_R⁺₀), limx→0f(t,x,y,z) =lim_y→0⁺f(t,x,y,z) =lim_z→0⁺f(t,x,y,z) = +_∞and f(t,x,y,z)≥ mt^2−α for(t,x,y,z)∈[0, 1]×_R0×_R0⁺×_R0⁺.

(H2) f(t,x,y,z) ≤ ρ(t)g(|x|,y,z) + p(|x|) +q(y,z) _for (t,x,y,z) ∈ [_{0, 1}]×_R0×_R0⁺×_R⁺₀, where ρ(t) is nonnegative on [0, 1], g(x,y,z) ∈ C(_R⁺×_R⁺×_R⁺)is nonnegative and nondecreasing in all its arguments, p(x) ∈ C(_R⁺₀) is nonnegative and nonincreasing, q(_y,z)∈ C(_R⁺₀ ×_R⁺₀)is nonnegative and nonincreasing in all its arguments.

(H3) lim_x→+_∞ ^g(x,x,x)

x = 0. p(λx) ≤ λ^−σp(x) for some σ ∈ (0,^α−₂¹) and for any λ ∈ (0, 1], x∈_R⁺₀. ρ(t), p(t²)andq(_mΓ(3−α)t,^mΓ_Γ(⁽₃³₋⁻_β^α₎⁾t^2−β)∈ L^ν[0, 1]for someν∈ ( ¹

α−1,_2σ¹). Remark 2.5. In [13], the nonlinear term satisfies f(t,x) ≤ g(|x|) + _|x^A_|ν, where A > 0 is a constant and ν > 0 is a suitable small number. It is easy to verify that the simple function p(x) = _x¹ω for 0 < ω < ^α−₂¹ fulfils the conditions (H2) and (H3) with ω ≤ σ < ^α−₂¹ and ν∈( ¹

α−1,_2σ¹ ).

Remark 2.6. By Lemma 2.1 and 2.2 in [2], for any f(t) ∈ L^ν[0, 1] with ν > ¹

α−1, I^α−1f(t) ∈ C[_{0, 1}]_and

Rt

0(t−s)^α−2f(s)ds

≤ ^t_d^dµ1kfk_{ν, where}d = (α−₂)µ+_{1 and}µ = ^ν

ν−1. Thus we can know easily lim_t→0⁺ I^α−1f(t) = 0. Similarly, I^αf(t)∈ C[0, 1]and lim_t→0⁺I^αf(t) = 0. The continuity of I^αf(t) on [0, 1]can also be derived from the continuity of I^α−1f(t), Remark2.2 (i) and (ii).

3 Auxiliary regular problem

This section deals with an auxiliary regular problem. We prove its solvability and give the properties of its solutions. We also state a necessary lemma and its useful corollary.

Consider the integral equation defined by u(t) = ¹

Γ(α)

Z _t

0

(t−s)^α−1f_n(s,u(s),u⁰(s),^CD^βu(s))ds

− ¹ 2Γ(α)

Z ₁

0

(1−s)^α−1f_n(s,u(s),u⁰(s),^CD^βu(s))ds,

(3.1)

where

f_n(t,x,y,z) =







f(t,x,χn(y),χn(z)), x≥ ¹_n,

n 2

f(t,¹_n,χn(y),χn(z))(_n¹+x) + f(t,−¹_n,χn(y),χn(z))(¹_n−x), |x| ≤ ¹_n,

f(t,x,χ_n(y)_,χ_n(z))_, x≤ −¹_n,

and

χ_n(τ) =

(τ, τ≥ ¹_n,

1

n, τ≤ ¹_n. Then the conditions (H1) and (H2) give

(K1) f_n∈ C([0, 1]×_R×_R×_R)and f_n(t,x,y,z)≥mt^2−α for(t,x,y,z)∈[0, 1]×_R×_R×_R.

(K2) f_n(t,x,y,z) ≤ ρ(t)g(|x|+1,y+1,z+1) +p(¹_n) +q(¹_n,¹_n) for (t,x,y,z) ∈ [0, 1]×_R× R⁺×_R⁺, f_n(t,x,y,z) ≤ ρ(t)g(|x|+_1,y+_1,z+₁) + p(|x|) +q(y,z) _for (t,x,y,z) ∈ [0, 1]×_R0×_R⁺₀ ×_R⁺₀.

Define an operatorT_nby the formula T_nu(t) = ¹

Γ(α)

Z _t

0

(t−s)^α−1f_n(s,u(s)_,u⁰(s)_,^CD^βu(s))ds

− ¹ 2Γ(α)

Z ₁

0

(1−s)^α−1f_n(s,u(s),u⁰(s),^CD^βu(s))ds.

(3.2)

Obviously, the fixed points ofTnare exactly the solutions of integral equation (3.1).

Lemma 3.1. Suppose that (H1) holds. Then T_n:C¹[0, 1]→C¹[0, 1]is completely continuous.

Proof. Letu∈C¹[0, 1]. Using Remark2.2(i) and (iii) we have^CD^βu(t) =_Γ(¹

1−β)

Rt

0(t−s)^−βu⁰(s)ds and ^CD^βu(t) ∈ C[0, 1]. Thus, in view of (3.2), Remark 2.2 (i) and (K1) ensure Tnu(t) ∈ C[0, 1]. Moreover, according to (K1), Remark 2.2 (i), (ii) and (iv), we know (T_nu)⁰(t) =

Γ(α1−1)

Rt

0(t−s)^α−2f_n(s,u(s)_,u⁰(s)_,^CD^βu(s))dsand(T_nu)⁰(t)∈C[_{0, 1}]. So we haveT_n: C¹[_{0, 1}]→ C¹[0, 1].

T_nis a continuous operator. In fact, let{u_k} ⊂C¹[0, 1]be such that lim_k→+_∞ku_k−uk_∗ =0, thenu(t)∈ C¹[0, 1]. Since

|^CD^βu_k(t)−^CD^βu(t)| ≤ ¹ Γ(₁−β)

Z _t

0

(t−s)^−β|u⁰_k(s)−u⁰(s)|ds

≤ ku⁰_k−u⁰k Γ(1−β)

Z _t

0

(t−s)^−βds≤ ku⁰_k−u⁰k Γ(2−β)^,

we getk^CD^βu_k−^CD^βuk → 0 and thuskf_n(t,u_k,u⁰_k,^CD^βu_k)− f_n(t,u,u⁰,^CD^βu)k → 0 as k → +∞. Note that

|Tnu_k(t)−Tnu(t)|

=

1 Γ(α)

Z _t

0

(t−s)^α−1hfn(s,u_k(s),u⁰_k(s),^CD^βu_k(s))− fn(s,u(s),u⁰(s),^CD^βu(s))ⁱds

− ¹ 2Γ(α)

Z ₁

0

(1−s)^α−1hfn(s,u_k(s),u⁰_k(s),^CD^βu_k(s))− fn(s,u(s),u⁰(s),^CD^βu(s))ⁱds

≤ ¹ Γ(α)

Z _t

0

(t−s)^α−1

fn(s,u_k(s),u⁰_k(s),^CD^βu_k(s))− fn(s,u(s),u⁰(s),^CD^βu(s)) ds

+ ¹

2Γ(α)

Z ₁

0

(1−s)^α−1f_n(s,u_k(s),u⁰_k(s),^CD^βu_k(s))− f_n(s,u(s),u⁰(s),^CD^βu(s))ds

≤ kf_n(t,u_k,u⁰_k,^CD^βu_k)− f_n(t,u,u⁰,^CD^βu)k Γ(α)

Z _t

0

(t−s)^α−1ds

+ kf_n(t,u_k,u⁰_k,^CD^βu_k)− f_n(t,u,u⁰,^CD^βu)k 2Γ(α)

Z ₁

0

(1−s)^α−1ds

= kfn(t,u_k,u⁰_k,^CD^βu_k)− fn(t,u,u⁰,^CD^βu)k Γ(α+1)

t^α+ ¹

2

≤ ³kf_n(t,u_k,u⁰_k,^CD^βu_k)− f_n(t,u,u⁰,^CD^βu)k

2Γ(α+1) ^,

and

|(T_nu_k)⁰(t)−(T_nu)⁰(t)|

=

1 Γ(α−1)

Z _t

0

(t−s)^α−2hf_n(s,u_k(s),u⁰_k(s),^CD^βu_k(s))− f_n(s,u(s),u⁰(s),^CD^βu(s))ⁱds

≤ ¹

Γ(α−₁)

Z _t

0

(t−s)^α−2f_n(s,u_k(s),u⁰_k(s),^CD^βu_k(s))− f_n(s,u(s),u⁰(s),^CD^βu(s))ds

≤ kf_n(t,u_k,u⁰_k,^CD^βu_k)− f_n(t,u,u⁰,^CD^βu)k Γ(α−1)

Z _t

0

(t−s)^α−2ds

= kf_n(t,u_k,u⁰_k,^CD^βu_k)− f_n(t,u,u⁰,^CD^βu)k

Γ(α) ^t

α−1

≤ kfn(t,u_k,u⁰_k,^CD^βu_k)− fn(t,u,u⁰,^CD^βu)k

Γ(α) ^.

So we obtain lim_k→+∞kT_nu_k−T_nuk_∗ =0. Therefore, T_nis a continuous operator.

Furthermore,Tn is completely continuous. Suppose thatΩ⊂ C¹[0, 1]is bounded and let M_n = _sup{kf_n(t,u,u⁰,^CD^βu)k_,u ∈ _Ω}_{, here} M_n is well defined because ^CD^βu(t) ≤ _Γ(^ku0k

2−β). Then we have

|Tnu(t)| ≤ ¹ Γ(α)

Z _t

0

(t−s)^α−1fn(s,u(s),u⁰(s),^CD^βu(s))ds

+ ¹

2Γ(α)

Z ₁

0

(1−s)^α−1f_n(s,u(s),u⁰(s),^CD^βu(s))ds

≤ ^Mn Γ(α)

Z _t

0

(t−s)^α−1ds+ ^Mn 2Γ(α)

Z ₁

0

(1−s)^α−1ds≤ ^3Mn 2Γ(α+1)^,

and

|(T_nu)⁰(t)| ≤ ¹ Γ(α−1)

Z _t

0

(t−s)^α−2f_n(s,u(s),u⁰(s),^CD^βu(s))ds

≤ ^Mn Γ(α−1)

Z _t

0

(t−s)^α−2ds≤ ^Mn Γ(α)^.

Therefore, Tn(_Ω) is bounded. Now we are in the position to prove Tn(_Ω) ⊂ C¹[0, 1] is an equicontinuous set. Let t₁,t₂ ∈ [0, 1] and t₁ < t₂, then |T_nu(t₂)−T_nu(t₁)| ≤ _Γ^M₍ⁿ

α)(t₂−t₁) by the mean value theorem and|(T_nu)⁰(t)| ≤ _Γ^M₍ⁿ

α). Moreover,

|(Tnu)⁰(t₂)−(Tnu)⁰(t₁)|

= ¹

Γ(α−1)

Z _t2

0

(t₂−s)^α−2fn(s,u(s),u⁰(s),^CD^βu(s))ds

−

Z _t1

0

(t₁−s)^α−2fn(s,u(s),u⁰(s),^CD^βu(s))ds

≤ ¹

Γ(α−1)

Z _t1

0

(t₁−s)^α−2−(t2−s)^α−2 fn(s,u(s),u⁰(s),^CD^βu(s))ds

+ ¹

Γ(α−1)

Z _t2

t1

(t2−s)^α−2fn(s,u(s),u⁰(s),^CD^βu(s))ds

≤ ^Mn Γ(α−1)

Z _t1

0

(t₁−s)^α−2−(t₂−s)^α−2ds+ ^Mn Γ(α−1)

Z _t2

t1

(t₂−s)^α−2ds

= ^Mn Γ(α) ^t

α−1

1 −t^α₂⁻¹+2(t2−t₁)^α−1.

Keeping in mind that the function t^α−1 is uniformly continuous on [0, 1], we have T_n(_Ω) is equicontinuous. Consequently, the Arzelà–Ascoli theorem guarantees that T_n is a completely continuous operator. The proof of Lemma3.1is finished.

The next lemma presents the existence of fixed points for the operatorTn.

Lemma 3.2. Assume that the conditions (H1), (H2) and (H3) are satisfied. Then T_nhas a fixed point in C¹[0, 1]for any n∈_N.

Proof. In view of Lemma 2.4 and Lemma 3.1, it is sufficient to prove the set An = {u ∈ C¹[0, 1]:u=λT_nufor someλ∈(0, 1)}is bounded. For anyu∈ A_n, we have

u(t) = ^λ Γ(α)

Z _t

0

(t−s)^α−1fn(s,u(s),u⁰(s),^CD^βu(s))ds

− ^λ 2Γ(α)

Z ₁

0

(1−s)^α−1f_n(s,u(s),u⁰(s),^CD^βu(s))ds,

(3.3)

u⁰(t) = ^λ Γ(α−1)

Z _t

0

(t−s)^α−2f_n(s,u(s)_,u⁰(s)_,^CD^βu(s))ds

≥ ^mλ Γ(α−1)

Z _t

0

(t−s)^α−2s^2−αds

= ^mλt

Γ(α−₁)

Z ₁

0

(1−s)^α−2s^2−αds=_mλΓ(3−α)t ≥0

(3.4)

by (K1). According to (3.3) and (3.4), one hasu(0) +u(1) = 0,u⁰(0)> 0 on(0, 1]. Thus there exists ξ ∈ (0, 1) such thatu(ξ) = 0. It follows that |u(t)| = |u(t)−u(ξ)| ≤ ku⁰k|t−ξ| and hencekuk ≤ ku⁰k. Since ^CD^βu(t)≥ 0 by (3.4) and^CD^βu(t)≤ _Γ(^k₂^u₋^0k

β), applying the conditions (H2), (H3) and (K2) we can derive

u⁰(t)≤ ¹ Γ(α−1)

Z _t

0

(t−s)^α−2

ρ(s)g(|u(s)|+1,u⁰(s) +1,^CD^βu(s) +1) +p 1

n

+q 1

n,1 n

ds

≤ ¹

Γ(α−1)

Z _t

0

(t−s)^α−2

ρ(s)g(ku⁰k+1,ku⁰k+1, ku⁰k

Γ(2−β)+1) +p 1

n

+q 1

n,1 n

ds

≤ ^g(ku⁰k+1,ku⁰k+1,_Γ(^k₂^u₋^0k_β) +1) Γ(α−1)

Z _t

0

(t−s)^α−2ρ(s)ds+ ^p(¹_n) +q(¹_n,¹_n) Γ(α)

≤Cg

ku⁰k+1,ku⁰k+1, ku⁰k Γ(2−β)+1

+ ^p(_n¹) +q(¹_n,_n¹) Γ(α) ^, here C = max_{t∈[0,1] Γ(α}¹₋₁₎Rt

0(t−s)^α−2ρ(s)ds, C is well defined by Remark 2.6 and (H3). In particular, the inequality

1≤ ^Cg(ku⁰k+1,ku⁰k+1,_Γ(^k₂^u₋^0k

β)+1)

ku⁰k + ^p(¹_n) +q(_n¹,¹_n) ku⁰k_Γ(α) is fulfilled. The condition lim_x→+_∞ ^g(x,x,x)

x =0 in (H3) guarantees that there exists L>0 such that

Cg(ku⁰k+1,ku⁰k+1,_Γ(^k₂^u₋^0k_β)+1)

ku⁰k + ^p(¹_n) +q(_n¹,¹_n) ku⁰k_Γ(α) <1

forku⁰k> L. Consequently, we obtainkuk ≤ ku⁰k ≤ Lforu∈ An. Therefore, Anis bounded and we complete the proof.

Lemma 3.2 shows that the integral equation (3.1) admits a solution u_n in C¹[0, 1] for any n∈_N. The properties of solutions to (3.1) are collected in the following lemma.

Lemma 3.3. Let the conditions (H1), (H2) and (H3) be valid and unbe solution of (3.1). Then (1) u_n(0) +u_n(1) = 0, u⁰_n(0) = 0, u⁰_n(t) ≥ mΓ(3−α)t and there exists ξ_n ∈ (0, 1) such that

un(ξn) =0.

(2) |un(t)| ≥ ^mΓ(3₂^−α)|t²−ξ²_n|.

(3) {un(t),n∈_N}is a compact subset of C¹[0, 1].

(4) There exists a constant l∈ (0, 1)such that l ≤ξn<1for any n∈_N.

Proof. Proof of (1). Similar to (3.4), the condition (K1) ensures u⁰_n(t) ≥ mΓ(3−α)t. Other assertions in (1) are obvious so we omit their proofs.

Proof of (2). Using (1), one has easily |u_n(t)| = Rt

ξnu⁰_n(s)ds

≥ _mΓ(3−α) Rt

ξnsds =

mΓ(3−α)

2 |t²−ξ²_n|.

Proof of (3). In order to apply the Arzelà–Ascoli theorem, we need to prove {u_n(t)} is bounded in C¹[0, 1] and {u⁰_n(t)} is equicontinuous. Firstly, we prove {u_n(t)} is bounded.

In view of (1), we get

ku_nk ≤ ku⁰_nk, ^CD^βu_n(t)≥mΓ(3−α) Γ(1−β)

Z _t

0

(t−s)^−βsds= ^mΓ(3−α) Γ(3−β) ^t

2−β.

We also know ^CD^βu_n(t) ≤ _Γ^k₍^u0nk

2−β). Thus, for t ∈ (_{0, 1}]\ {ξ_n}, by (H2), (K2), (1) and (2) we derive

fn(t,un(t),u⁰_n(t),^CD^βun(t))≤ρ(t)g(|un(t)|+1,u⁰_n(t) +1,^CD^βun(t) +1) +p(|u_n(t)|) +q(u⁰_n(t)_,^CD^βu_n(t))

≤ρ(t)g

ku⁰_nk+1,ku⁰_nk+1, ku⁰_nk Γ(2−β)+1

+p

mΓ(3−α)

2 |t²−ξ²_n|

+q

mΓ(3−α)t,mΓ(3−α) Γ(3−β) ^t

2−β

. Hence,

u⁰_n(t)≤ ^g(ku⁰_nk+_1,ku⁰_nk+_1,Γ^k₍^u0nk

2−β)+₁) Γ(α−1)

Z _t

0

(t−s)^α−2ρ(s)ds

+ ¹

Γ(α−1)

Z _t

0

(t−s)^α−2p

mΓ(3−α)

2 |s²−ξ²_n|

ds

+ ¹

Γ(α−1)

Z _t

0

(t−s)^α−2q

mΓ(3−α)s,mΓ(3−α) Γ(3−β) ^s

2−β

ds.

(3.5)

Furthermore, by (H3) and Remark2.6, we can let C1 = _max

t∈[0,1]

1 Γ(α−1)

Z _t

0

(_t−s)^α−2ρ(_s)_{ds, (3.6)} C₂= max

t∈[0,1]

1 Γ(α−1)

Z _t

0

(t−s)^α−2q

mΓ(3−α)s,mΓ(3−α) Γ(3−β) ^s

2−β

ds (3.7)

and by the Hölder inequality one has Z _t

0

(t−s)^α−2p

mΓ(3−α)

2 |s²−ξ²_n|

ds

≤ [(α−₂)µ+₁]^−1µ _{Z t}

0

p^ν

mΓ(3−α)

2 |s²−ξ_n²|

ds ¹_ν

,

(3.8)

here (α−2)µ+1 > 0 and ¹_µ+ ¹

ν = 1, µis well defined by the choice of ν in condition (H3).

Next we estimate the integral on the right side of (3.8).

Z _t

0 p^ν

mΓ(3−α)

2 |s²−ξ²_n|

ds

≤

Z ₁

0 p^ν

mΓ(3−α)

2 |s²−ξ²_n|

ds

=

Z _ξn

0 p^ν

mΓ(3−α)

2 |s²−ξ²_n|

ds+

Z ₁

ξn

p^ν

mΓ(3−α)

2 |s²−ξ²_n|

ds

= I₁+I₂.

(3.9)

In view of the monotone property of pand (H3), we get I₁ ≤

Z _ξn

0 p^ν

mΓ(3−α)

2 ξ_n(ξ_n−s)

ds=

Z ₁

0 ξ_np^ν

mΓ(3−α)

2 ξ²_n(1−s)

ds

≤ Aξ¹_n^−2σν Z ₁

0

(p(1−s))^νds≤ A Z ₁

0

(p(s))^νds≤ A Z ₁

0

(p(s²))^νds=C3<+_∞,

(3.10)