Fractional-order boundary value problems with Katugampola ...

Mahmudov and Emin Advances in Difference Equations (2018) 2018:81 https://doi.org/10.1186/s13662-018-1538-6

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Fractional-order boundary value problems with Katugampola fractional integral conditions Nazim I. Mahmudov*

and Sedef Emin

*

Correspondence: [email protected] Department of Mathematics, Eastern Mediterranean University, Gazimagusa, Turkish Republic of Northern Cyprus

Abstract In this paper, we study existence (uniqueness) of solutions for nonlinear fractional differential equations with Katugampola fractional integral conditions. Several fixed point theorems are used for sufficient conditions of existence (uniqueness) solutions of nonlinear differential equations such as Banach’s contraction principle, the Leray–Schauder nonlinear alternative, and Krasnoselskii’s fixed point theorem. Applications of the main results are also presented. Keywords: Fractional differential equations; Katugampola derivative; Integral boundary conditions

1 Introduction In recent years, boundary value problems for nonlinear fractional differential equations have been studied by several researchers. In fact, fractional differential equations have played an important role in physics, chemical technology, biology, economics, control theory, signal and image processing, see [1–25] and the references cited therein. Boundary value problems of fractional differential equations and inclusions involve different kinds of boundary conditions such as nonlocal, integral, and multipoint boundary conditions. The fractional integral boundary conditions were introduced lately in [26] and nonlocal conditions were presented by Bitsadze, see [3]. In [4] authors gave sufficient criteria for existence of solutions for the following Caputo fractional differential equation: Dq x(t) = f t, x(t) ,

0 < t < T,

subject to nonlocal generalized Riemann–Liouville fractional integral boundary conditions of the form x(0) = γ

ρ 1–α (α) 1–β

x(T) = δ

ρ (β)

ξ

sρ–1 x(s) ds := γ ρ I α x(ξ ), (t ρ – sρ )1–α

ε

sρ–1 x(s) ds := δ ρ I β x(ε), (ξ ρ – sρ )1–β

0

0

0 < ξ , ε < T,

© 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.

Mahmudov and Emin Advances in Difference Equations (2018) 2018:81

Page 2 of 17

where Dq denotes the Caputo fractional derivative of order q, ρ I z , z ∈ {α, β}, is the generalized Riemann–Liouville fractional integral of order z > 0, ρ > 0, ξ , ε arbitrary, with ξ , ε ∈ (0, T), γ , δ ∈ R and f : [0, T] × R → R is a continuous function. In [5] authors established the existence of solutions for the following nonlinear Riemann– Liouville fractional differential equation subject to nonlocal Erdelyi–Kober fractional integral conditions: Dq x(t) = f t, x(t) , x(0) = 0,

t ∈ (0, T),

αx(T) =

m

βi Iηγii ,δi x(ξi ),

i=1 γ ,δ

where 1 < q ≤ 2, Dq is the standard Riemann–Liouville fractional derivative of order q, Iηii i is the Erdelyi–Kober fractional integral of order δi > 0 with ηi > 0 and γi ∈ R, i = 1, 2, . . . , m, f : [0, T] × R → R is a continuous function and α, βi ∈ R, ξi ∈ (0, T), i = 1, 2, . . . , m, are given constants. Motivated by the above papers, in this paper, we study the sufficient conditions of existence (uniqueness) solutions of nonlocal boundary conditions for the following nonlinear fractional differential equation of order α ∈ (2, 3]: ⎧ c α ⎪ D x(t) = f (t, x(t)), t ∈ [0, T], ⎪ ⎪ ⎪ ⎪ ⎨x(T) = β ρ I q x(), 0 < < T, ⎪ ⎪x (T) = γ ρ I q x (η), 0 < η < T, ⎪ ⎪ ⎪ ⎩ x (T) = δ ρ I q x (ζ ), 0 < ζ < T,

(1)

where Dα is the Caputo fractional derivative. ρ I q is the Katugampola integral of q > 0, ρ > 0, f : [0, T] × R → R is a continuous function. The rest of the paper is organized as follows. In Sect. 2, we recall some definitions and lemmas that we need in the sequel. In Sect. 3, several fixed point theorems are used to give sufficient conditions for existence (uniqueness) of solutions of (1) such as Banach’s contraction principle, Krasnoselskii’s fixed point theorem, and the Leray–Schauder nonlinear alternative. In Sect. 4, some illustrating examples are given.

2 Preliminaries In this section, we recall some basic definitions of fractional calculus [1, 2, 27] and some auxiliary lemmas which we need later. Definition 1 ([1]) The Riemann–Liouville fractional integral of order p > 0 of a continuous function f : (0, ∞) → R is defined by J p f (t) =

1 (p)

t

(t – s)p–1 f (s) ds, 0

provided the right-hand side is point-wise defined on (0, ∞), where is the gamma func ∞ tion defined by (p) = 0 e–s sp–1 ds.


Page 3 of 17

Definition 2 ([1]) The Riemann–Liouville fractional derivative of order p > 0 of a continuous function f : (0, ∞) → R is defined by p

D0+ f (t) =

n t 1 d (t – s)n–p–1 f (s) ds, (n – p) dt 0

n – 1 < p < n,

where n = [p] + 1, [p] denotes the integer part of a real number p. Definition 3 The Caputo derivative of order p for a function f : [0, ∞) → R can be written as

p c p D f (t) = D0+

f (t) –

n–1 k t k=0

k!

f (k) (0) ,

t > 0, n – 1 < p < n.

Remark 4 If f (t) ∈ C n [0, ∞), then c

Dp f (t) =

1 (n – p)

t 0

f (n) (s) ds = I n–p f (n) (t), (t – s)p+1–n

t > 0, n – 1 < p < n.

Definition 5 [28] Katugampola integral of order q > 0 and ρ > 0, of a function f (t), for all 0 < t < ∞, is defined as ρ q

I f (t) =

ρ 1–q (q)

0

t

sρ–1 f (s) ds, (t ρ – sρ )1–q

provided the right-hand side is point-wise defined on (0, ∞). Remark 6 ([28]) The above definition corresponds to the one for Riemann–Liouville fractional integral of order q > 0 when ρ = 1, while the famous Hadamard fractional integral follows for ρ → 0, that is, 1 lim I f (t) = ρ→0 (q) ρ q

t t q–1 f (s) ds. log s s 0

Lemma 7 ([4]) Let ρ, q > 0 and p > 0 be the given constants. Then the following formula holds: ) t p+ρq ( p+ρ ρ I t = p+ρq+ρ . ( ρ ) ρ q

ρ q p

Lemma 8 ([1]) For q > 0 and x ∈ C(0, T)∩L(0, T). Then the fractional differential equation c q D x(t) = 0 has a unique solution x(t) = c0 + c1 t + · · · + cn–1 t n–1 , and the following formula holds: I q Dq x(t) = x(t) + c0 + c1 t + · · · + cn–1 t n–1 , where ci ∈ R, i = 0, 1, . . . , n – 1, and n – 1 ≤ q < n.


Page 4 of 17

Lemma 9 Let 2 < α ≤ 3 and β, γ , δ ∈ R. Then, for any y ∈ C([0, T], R), x is a solution of the following nonlinear fractional differential equation with Katugampola fractional integral conditions: ⎧ c α ⎪ D x(t) = y(t), t ∈ [0, T], ⎪ ⎪ ⎪ ⎪ ⎨x(T) = β ρ I q x(), 0 < < T, ⎪ ⎪x (T) = γ ρ I q x (η), 0 < η < T, ⎪ ⎪ ⎪ ⎩ x (T) = δ ρ I q x (ζ ), 0 < ζ < T,

(2)

if and only if ρ q α 1 β I J y() – J α y(T)

1 (β, ) 1

2 (β, ) – – t γ ρ I q J α–1 y(η) – J α–1 y(T)

1 (γ , η) 1 (β, )

3 (β, ) 2 (β, ) 2 (γ , η) 1 – +

1 (δ, ζ ) 2 1 (β, ) 1 (β, ) 1 (γ , η)

2 (γ , η)t t 2 α–2 – J y(T) – δ ρ I q J α–2 y(ζ ) , +

1 (γ , η) 2

x(t) = J α y(t) +

(3)

where 1 ξ ρq

1 (α, ξ ) = 1 – α q = 0, ρ (q + 1) 1+ρ ξ ρq+1 ( ρ ) ,

2 (α, ξ ) = T – α q ρ ( 1+ρq+ρ ) ρ

2+ρ ξ ρq+2 ( ρ ) .

3 (α, ξ ) = T 2 – α q ρ ( 2+ρq+ρ ) ρ

(4)

(5)

(6)

Proof Using Lemma 8, the general solution of the nonlinear fractional differential equation in (2) can be represented as x(t) = c0 + c1 t + c2 t 2 + J α y(t),

c0 , c1 , c2 ∈ R.

(7)

By using the first integral condition of problem (2) and applying the Katugampola integral on (7), we obtain c0 + c1 T + c2 T 2 + J α y(T) = βc0

1+ρ ρq 1 ρq+1 ( ρ ) + βc 1 ρ q (q + 1) ρ q ( 1+ρq+ρ ) ρ

2+ρ ρq+2 ( ρ ) + β ρ I q J α y(). + βc2 q ρ ( 2+ρq+ρ ) ρ


Page 5 of 17

After collecting the similar terms in one part, we have the following equation: 1+ρ ρq 1 ρq+1 ( ρ ) + c1 T – βc1 q c0 1 – β q ρ (q + 1) ρ ( 1+ρq+ρ ) ρ + c2

2+ρ ρq+2 ( ρ ) T – βc2 q ρ ( 2+ρq+ρ ) ρ

2

= β ρ I q J α y() – J α y(T).

(8)

Rewriting equation (8) by using (4), (5), and (6), we obtain c0 1 (β, ) + c1 2 (β, ) + c2 3 (β, ) = β ρ I q J α y() – J α y(T).

(9)

Then, taking the derivative of (7) and using the second integral condition of (2), we get x (T) = c1 + 2c2 T + J α–1 y(T).

(10)

Now, applying the Katugampola integral on (10), we have c1 + 2c2 T + J α–1 y(T) = γ c1

ηρq 1 q ρ (q + 1)

+ 2c2 γ

1+ρ ηρq+1 ( ρ ) + γ ρ I q J α–1 y(η). ρ q ( 1+ρq+ρ ) ρ

(11)

The above equation (11) implies that 1+ρ ηρq ηρq+1 ( ρ ) 1 + 2c2 T – γ q c1 1 – γ q ρ (q + 1) ρ ( 1+ρq+ρ ) ρ = γ ρ I q J α–1 y(η) – J α–1 y(T).

(12)

Also, by using (4) and (5), equation (12) can be written as c1 1 (γ , η) + 2c2 2 (γ , η) = γ ρ I q J α–1 y(η) – J α–1 y(T).

(13)

By using the last integral condition of (2) and applying Katugampola integral operator on the second derivative of (10), we have 2c2 + J α–2 y(T) = 2δc2

ζ ρq 1 + δ ρ I q J α–2 y(ζ ). q ρ (q + 1)

Hence, we obtain the following equation: ζ ρq 1 = δ ρ I q J α–2 y(ζ ) – J α–2 y(T). 2c2 1 – δ q ρ (q + 1)

(14)


Page 6 of 17

By using (4), equation (14) can be written as 2c2 1 (δ, ζ ) = δ ρ I q J α–2 y(ζ ) – J α–2 y(T).

(15)

Moreover, equation (15) implies that c2 =

ρ q α–2 1 δ I J y(ζ ) – J α–2 y(T) . 2 1 (δ, ζ )

(16)

Substituting the values of (16) in (13), we get c1 =

ρ q α–1 1 γ I J y(η) – J α–1 y(T)

1 (γ , η) –

ρ q α–2

2 (γ , η) δ I J y(ζ ) – J α–2 y(T) .

1 (γ , η) 1 (δ, ζ )

(17)

Now, substituting the values of (16) and (17) in (9), we obtain c0 =

ρ q α 1 β I J y() – J α y(T)

1 (β, ) –

ρ q α–1

2 (β, ) γ I J y(η) – J α–1 y(T)

1 (β, ) 1 (γ , η)

–

ρ q α–2

3 (β, ) δ I J y(ζ ) – J α–2 y(T) 2 1 (β, ) 1 (δ, ζ )

+

ρ q α–2

2 (β, ) 2 (γ , η) δ I J y(ζ ) – J α–2 y(T) .

1 (β, ) 1 (γ , η) 1 (δ, ζ )

(18)

Finally, substituting the values of (18), (17), and (16) in equation (7), we obtain the general solution of problem (2) which is (3). Converse is also true by using the direct computation.

3 Main results Let us denote by C = C([0, T], R) the Banach space of all continuous functions from [0, T] → R endowed with a topology of uniform convergence with the norm defined by x = sup{|x(t)| : t ∈ [0, T]}. We define an operator H : C → C on problem (1) as ρ q α 1 β I J f s, x(s) ()

1 (β, ) 1

2 (β, ) α – J f s, x(s) (T) – –t

1 (γ , η) 1 (β, ) × γ ρ I q J α–1 f s, x(s) (η) – J α–1 f s, x(s) (T)

3 (β, ) 2 (β, ) 2 (γ , η) 1 – +

1 (δ, ζ ) 2 1 (β, ) 1 (β, ) 1 (γ , η)

2 (γ , η)t t 2 α–2 – J f s, x(s) (T) +

1 (γ , η) 2 – δ ρ I q J α–2 f s, x(s) (ζ ) .

(Hx)(t) = J α f s, x(s) (t) +

(19)


Page 7 of 17

Also, we define the notations =

1 Tα + (α + 1) | 1 (β, )|(α + 1) ) α+ρq ( α+ρ ρ α + T × |β| α+ρq+ρ ( ρ ) ρ q | 2 (β, )| 1 +T + | 1 (γ , η)|(α) | 1 (β, )| ) ηα–1+ρq ( α–1+ρ ρ α–1 + T × |γ | α–1+ρq+ρ ( ) ρq ρ | 3 (β, )| 1 + | 1 (δ, ζ )|(α – 1) 2| 1 (β, )| | 2 (β, )|| 2 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (β, )|| 1 (γ , η)| | 1 (γ , η)| 2 ) ζ α–2+ρq ( α–2+ρ ρ + T α–2 × |δ| α–2+ρq+ρ q ρ ( ) ρ

+

(20)

and 1 =

) α+ρq ( α+ρ |β| ρ | 1 (β, )|(α + 1) ( α+ρq+ρ ) ρq ρ ) ηα–1+ρq ( α–1+ρ | 2 (β, )| |γ | ρ +T α–1+ρq+ρ | 1 (γ , η)|(α) | 1 (β, )| ( ) ρq ρ |δ| | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )|(α – 1) 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| ) ζ α–2+ρq ( α–2+ρ | 2 (γ , η)|T T 2 ρ + + . | 1 (γ , η)| 2 ( α–2+ρq+ρ ) ρq ρ +

(21)

In the following subsections, we prove existence (uniqueness) results for the boundary value problem (1) by using Banach’s fixed point theorem, the Leray–Schauder nonlinear alternative, and Krasnoselskii’s fixed point theorem.

3.1 Existence and uniqueness result Theorem 10 Let f : [0, T] × R → R be a continuous function. Assume that: (S1 ) |f (t, x) – f (t, y)| ≤ L x – y for all t ∈ [0, T], L > 0, x, y ∈ R; (S2 ) L < 1, where is defined by (20). Then the boundary value problem (1) has a unique solution on [0, T]. Proof By using the operator H, which is defined by (19), we obtain (Hx)(t) – (Hy)(t) ≤ J α f s, x(s) – f s, y(s) (T) +

|β| ρ q α I J f s, x(s) – f s, y(s) () | 1 (β, )|


1 J α f s, x(s) – f s, y(s) (T) | 1 (β, )| | 2 (β, )| |γ | +T + | 1 (γ , η)| | 1 (β, )| × ρ I q J α–1 f s, x(s) – f s, y(s) (η) 1 | 2 (β, )| + +T | 1 (γ , η)| | 1 (β, )| × J α–1 f s, x(s) – f s, y(s) (T) 1 | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 × J α–2 f s, x(s) – f s, y(s) (T) | 3 (β, )| | 2 (β, )|| 2 (γ , η)| |δ| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 × ρ I q J α–2 f s, x(s) – f s, y(s) (ζ ) ρ q α 1 ≤ L x – y J α (1)(T) + |β| I J (1)() | 1 (β, )| + J α (1)(T) 1 | 2 (β, )| + + T |γ | ρ I q J α–1 (1)(η) | 1 (γ , η)| | 1 (β, )| + J α–1 (1)(T) 1 | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 ρ q α–2 + |δ| I J (1)(ζ ) + | 1 (γ , η)| 2 + J α–2 (1)(T) +

Tα 1 + (α + 1) | 1 (β, )|(α + 1) ) α+ρq ( α+ρ ρ α + T × |β| α+ρq+ρ ( ρ ) ρ q | 2 (β, )| 1 +T + | 1 (γ , η)|(α) | 1 (β, )| ( α–1+ρ ) ηα–1+ρq ρ α–1 × |γ | α–1+ρq+ρ +T ( ) ρq ρ 1 | 3 (β, )| + | 1 (δ, ζ )|(α – 1) 2| 1 (β, )|

≤L

Page 8 of 17


Page 9 of 17

| 2 (β, )|| 2 (γ , η)| | 2 (γ , η)|T T 2 + + + | 1 (β, )|| 1 (γ , η)| | 1 (γ , η)| 2 α–2+ρ ( ρ ) ζ α–2+ρq × |δ| α–2+ρq+ρ + T α–2 x – y q ρ ( ) ρ = L x – y for any x, y ∈ C and for each t ∈ [0, T]. This implies that Hx – Hy ≤ L x – y . As L < 1, the operator H : C → C is a contraction mapping. As a result, the boundary value problem (1) has a unique solution on [0, T].

3.2 Existence results Lemma 11 Let E be a Banach space, C be a closed, convex subset of E, U be an open subset of C and 0 ∈ U. Suppose that A : U → C is a continuous, compact map. Then either (i) A has a fixed point in U, or (ii) there are x ∈ ∂U (the boundary of U in C) and μ ∈ (0, 1) with x = μA(x). Theorem 12 Let f : [0, T] × R → R be a continuous function. Assume that: (S3 ) There exist a nonnegative function ∈ C([0, T], R) and a nondecreasing function : [0, ∞) → (0, ∞) such that f (t, u) ≤ (t) |u|

for any (t, u) ∈ [0, T] × R;

(S4 ) There exists a constant M > 0 such that M > 1, (M) where in (20). Then problem (1) has at least one solution on [0, T]. Proof Let Bd = {x ∈ C : x ≤ d} be a closed bounded subset in C([0, T], R). Notice that problem (1) is equivalent to the problem of finding a fixed point of H. As a first step, we show that the operator H, which is defined by (19), maps bounded sets into bounded sets in C([0, T], R). Then, for t ∈ [0, T], we have H(x)(t) ≤ J α f s, x(s) (T) ρ q α 1 |β| I J f s, x(s) () | 1 (β, )| + J α f s, x(s) (T) 1 | 2 (β, )| + +T | 1 (γ , η)| | 1 (β, )| × |γ |J α–1 f s, x(s) (η) + J α–1 f s, x(s) (T) 1 | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 +


Page 10 of 17

× J α–2 f s, x(s) (T) + |δ| ρ I q J α–2 f s, x(s) (ζ ) ≤ x J α (s)(T) ( x ) ρ q α |β| I J (s)() | 1 (β, )| + J α (s)(T) ( x ) | 2 (β, )| + +T | 1 (γ , η)| | 1 (β, )| × |γ |ρ I q J α–1 (s)(η) + J α–1 (s)(T) ( x ) | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 α–2 × J (s)(T) + |δ| ρ I q J α–2 (s)(ζ ) 1 Tα + ≤ (d) (α + 1) | 1 (β, )|(α + 1) ( α+ρ ) α+ρq ρ α × |β| α+ρq+ρ + T ( ρ ) ρ q | 2 (β, )| 1 +T + | 1 (γ , η)|(α) | 1 (β, )| ) ηα–1+ρq ( α–1+ρ ρ α–1 × |γ | α–1+ρq+ρ + T ( ) ρq ρ | 3 (β, )| 1 + | 1 (δ, ζ )|(α – 1) 2| 1 (β, )| | 2 (β, )|| 2 (γ , η)| | 2 (γ , η)|T T 2 + + + | 1 (β, )|| 1 (γ , η)| | 1 (γ , η)| 2 α–2+ρ ( ρ ) ζ α–2+ρq + T α–2 × |δ| α–2+ρq+ρ q ρ ( ) ρ +

= (d), which leads to H(x) ≤ (d). By (S4 ) there exists d > 0 such that (d) < d. Next, we show that the map H : C([0, T], R) → C([0, T], R) is completely continuous. Therefore, to prove that the map H is completely continuous, we show that H is a map from bounded sets into equicontinuous sets of C([0, T], R). Let us choose t1 , t2 from the interval [0, T] and also t1 < t2 . Then we have (Hx)(t2 ) – (Hx)(t1 ) ≤ J α f s, x(s) (t2 ) – J α f s, x(s) (t1 ) |t2 – t1 | q α–1 f s, x(s) (η) + J α–1 f s, x(s) (T) |γ | J | 1 (γ , η)| 1 | 2 (γ , η)| |t 2 – t 2 | + |t2 – t1 | + 2 1 | 1 (δ, ζ )| | 1 (γ , η)| 2

+


× J α–2 f s, x(s) (T) + |δ| ρ I q J α–2 f s, x(s) (ζ ) t1 t2 ( x )(s) (t2 – s)α–1 – (t1 – s)α–1 ds + ≤ (t2 – s)α–1 ds (α) 0 t1 ( x ) |γ |J α–1 (s)(η) + J α–1 (s)(T) + |t2 – t1 | | 1 (γ , η)| | 2 (γ , η)| |t2 + t1 | | 1 (γ , η)| α–2 ρ q α–2 + J (s)(T) + |δ| I J (s)(ζ ) + | 1 (δ, ζ )| 2 | 1 (δ, ζ )| t1 t2 (d) (t2 – s)α–1 – (t1 – s)α–1 ds + (t2 – s)α–1 ds ≤ (α) 0 t1 ) ηα–1+ρq ( α–1+ρ (d) 1 ρ α–1 + |t2 – t1 | + |γ | α–1+ρq+ρ T | 1 (γ , η)| (α) ( ) ρq ρ | 2 (γ , η)| |t2 + t1 | | 1 (γ , η)| 1 + + (α – 1) | 1 (δ, ζ )| 2 | 1 (δ, ζ )| ) ζ α–2+ρq ( α–2+ρ ρ α–2 × T + |δ| α–2+ρq+ρ . ( ) ρq ρ

Page 11 of 17

(22)

It is clear that the right-hand side of (22) is independent of x. Therefore, as t2 – t1 → 0, inequality (22) tends to zero. That means H is equicontinuous, and by the Arzelà–Ascoli theorem, the operator H : C([0, T], R) → C([0, T], R) is completely continuous. In the last step we show that the operator H has a fixed point. Let H(x) = x be a solution. Then, for t ∈ [0, T], Hx = x ≤ x , which implies that x ≤ 1. ( x ) In view of (S4 ), there exists positive M such that x = M. Let us set U = x ∈ C [0, T], R : x < M . Then the operator H : U → C([0, T], R) is continuous and completely continuous. From the choice of U, there is no x ∈ ∂U such that x = μHx for some μ ∈ (0, 1). It can be proved by using contraction. Assume that there exists x ∈ ∂U such that x = μHx for some μ ∈ (0, 1). Then x = μHx ≤ Hx ≤ x , x ≤ 1. ( x ) This contradicts x > 1. ( x )


Page 12 of 17

Consequently, by the nonlinear alternative of Leray–Schauder type, we conclude that H has a fixed point x ∈ U, which is a solution of problem (1). This completes the proof. Theorem 13 ([29]) Let M be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A, B be the operators such that (a) Ax + By ∈ M whenever x, y ∈ M; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists z ∈ M such that z = Az + Bz. Theorem 14 Let f : [0, T] × R → R be a continuous function, and let condition (S1 ) hold. In addition, the function f satisfies the assumptions: (S5 ) There exists a nonnegative function ∈ C([0, T], R) such that f (t, u) ≤ (t) for any (t, u) ∈ [0, T] × R. (S6 ) L1 < 1, where 1 is defined by (21). Then the boundary value problem (1) has at least one solution on [0, T]. Proof We first define the new operators H1 and H2 as (H1 x)(t) = J α f s, x(s) (t) –

1 J α f s, x(s) (T)

1 (β, )

2 (β, ) 1 – t J α–1 f s, x(s) (T) +

1 (γ , η) 1 (β, ) 1

3 (β, ) 2 (β, ) 2 (γ , η) + –

1 (δ, ζ ) 2 1 (β, ) 1 (β, ) 1 (γ , η)

2 (γ , η)t t 2 α–2 – J f s, x(s) (T) +

1 (γ , η) 2

(23)

and (H2 x)(t) =

β ρ q α I J f s, x(s) ()

1 (β, ) γ

2 (β, ) – t ρ I q J α–1 f s, x(s) (η) –

1 (γ , η) 1 (β, )

3 (β, ) 2 (β, ) 2 (γ , η) δ – –

1 (δ, ζ ) 2 1 (β, ) 1 (β, ) 1 (γ , η)

2 (γ , η)t t 2 ρ q α–2 – I J f s, x(s) (ζ ). +

1 (γ , η) 2

(24)

Then we consider a closed, bounded, convex, and nonempty subset of the Banach space X as Bd = x ∈ C : x < d

with ≤ d,

where is defined by (20). Now, we show that H1 x + H2 y ∈ Bd for any x, y ∈ Bd , where H1


and H2 are denoted by (23) and (24), respectively. H1 x + H2 y ≤ J α f s, x(s) (T) α 1 J f s, x(s) (T) + |β| ρ I q J α f s, x(s) () | 1 (β, )| 1 | 2 (β, )| + +T | 1 (γ , η)| | 1 (β, )| × J α–1 f s, x(s) (T) + |γ | ρ I q J α–1 f s, x(s) (η) | 3 (β, )| | 2 (β, )|| 2 (γ , η)| 1 + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 α–2 × J f s, x(s) (T) + |δ| ρ I q J α–2 f s, x(s) (ζ ) +

≤ J α (s)(T) α 1 J (s)(T) + |β| ρ I q J α (s)() | 1 (β, )| 1 | 2 (β, )| + +T | 1 (γ , η)| | 1 (β, )| × J α–1 (s)(T) + |γ | ρ I q J α–1 (s)(η) | 3 (β, )| | 2 (β, )|| 2 (γ , η)| 1 + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 α–2 × J (s)(T) + |δ| ρ I q J α–2 (s)(ζ ) Tα 1 ≤ + (α + 1) | 1 (β, )|(α + 1) ) α+ρq ( α+ρ ρ × T α + |β| α+ρq+ρ + ( ρ ) ρ q | 2 (β, )| 1 +T + | 1 (γ , η)|(α) | 1 (β, )| ( α–1+ρ ) ηα–1+ρq ρ α–1 + |γ | α–1+ρq+ρ × T ( ) ρq ρ 1 | 3 (β, )| + | 1 (δ, ζ )|(α – 1) 2| 1 (β, )| | 2 (β, )|| 2 (γ , η)| | 2 (γ , η)|T T 2 + + + | 1 (β, )|| 1 (γ , η)| | 1 (γ , η)| 2 α–2+ρ ( ρ ) ζ α–2+ρq × T α–2 + |δ| α–2+ρq+ρ ( ) ρq ρ +

= ≤ d. Therefore, it is clear that H1 x + H2 y ≤ d. Hence, H1 x + H2 y ∈ Bd .

Page 13 of 17


Page 14 of 17

The next step is related to the compactness and continuity of the operator H1 . The proof is similar to that of Theorem 12. Finally, we show that the operator H2 is a contraction. By using assumption (S1 ), |β| ρ q α I J f s, x(s) – f s, y(s) () | 1 (β, )| |γ | | 2 (β, )| + +T | 1 (γ , η)| | 1 (β, )| × ρ I q J α–1 f s, x(s) – f s, y(s) (η) |δ| | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 × ρ I q J α–2 f s, x(s) – f s, y(s) (ζ ) |β| ρ q α I J (1)() ≤ L x – y | 1 (β, )| |γ | | 2 (β, )| + + T ρ I q J α–1 (1)(η) | 1 (γ , η)| | 1 (β, )| |δ| | 3 (β, )| | 2 (β, )|| 2 (γ , η)| + + | 1 (δ, ζ )| 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 ρ q α–2 + I J (1)(ζ ) + | 1 (γ , η)| 2 ( α+ρ ) α+ρq |β| ρ ≤L α+ρq+ρ | 1 (β, )|(α + 1) ( ρ ) ρ q | 2 (β, )| |γ | +T + | 1 (γ , η)|(α) | 1 (β, )|

H2 x – H2 y ≤

) ηα–1+ρq ( α–1+ρ ρ α–1+ρq+ρ ( ) ρq ρ | 3 (β, )| | 2 (β, )|| 2 (γ , η)| |δ| + + | 1 (δ, ζ )|(α – 1) 2| 1 (β, )| | 1 (β, )|| 1 (γ , η)| | 2 (γ , η)|T T 2 + + | 1 (γ , η)| 2 ) ζ α–2+ρq ( α–2+ρ ρ x – y × α–2+ρq+ρ ( ) ρq ρ ×

= L1 x – y , which means H2 x – H2 y ≤ L1 x – y . As L1 < 1, the operator H2 is a contraction. For this reason, problem (1) has at least one solution on [0, T].

4 Examples In this section, some examples are illustrated to show our results.


Page 15 of 17

Example 1 Consider the following nonlinear fractional differential equation with Katugampola fractional integral conditions: ⎧ c 5/2 ⎪ ⎪ D x(t) = ⎨ ⎪ ⎪ ⎩

sin2 (π t) |x(t)| ( (et +10) |x(t)|+1

1 5 13 I x(3/8), 2 5 1 x ( 12 ) = 12 I 3 x(2/5).

x( 21 ) =

+ 1),

t ∈ [0, 12 ],

x ( 12 ) =

1 5 13 I x(1/3), 2

(25)

Here, α = 5/2, T = 12 , β = 1/2, γ = 1/2, δ = 1/2, = 3/8, η = 1/3, ζ = 2/5, ρ = 5, q = 13 , and f (t, x) =

sin2 (πt) |x| + 1 . (et + 10) |x| + 1

1 1 x – y . Then, assumption (S1 ) is satisfied with L = 10 . Hence, we have |f (t, x) – f (t, y)| ≤ 10 1 3 1 1 1 2 By using the Matlab program, ω1 ( 2 , 8 ) = 0.9361, ω1 ( 2 , 3 ) = 0.9475, ω1 ( 2 , 5 ) = 0.9289, ω2 ( 12 , 38 ) = 0.4779, ω2 ( 12 , 13 ) = 0.4838, ω3 ( 12 , 38 ) = 0.4922, and = 1.2261 are found. Therefore, L = 0.1226 < 1, which implies that assumption (S2 ) holds true. By using Theorem 10, the boundary value problem (25) has a unique solution on [0, 12 ].

Example 2 Consider the following nonlinear fractional differential equation with Katugampola fractional integral conditions: √ ⎧ |x(t)| x2 (t) t 2 +1 5/2 ⎪ √ ⎪ x(t) = ( )( + + 12 ), t ∈ [0, 12 ], D ⎪ 10 |x(t)|+1 ⎨ 2(1+ |x(t)|) 5 1 5 1 x ( 12 ) = 12 I 3 x(1/3), x( 21 ) = 12 I 3 x(3/8), ⎪ ⎪ ⎪ 5 1 ⎩ 1 x ( 2 ) = 12 I 3 x(2/5),

(26)

where α = 5/2, T = 12 , β = 1/2, γ = 1/2, δ = 1/2, = 3/8, η = 1/3, ζ = 2/5, ρ = 5, q = 13 . Moreover, √ 2 u2 1 (t 2 + 1)(|u| + 1) |u| f (t, u) = t + 1 + ≤ . + √ 10 |u| + 1 2(1 + |u|) 2 10 2

By using assumption (S3 ), it is easy to see that (t) = t 10+1 and (|u|) = |u| + 1. Moreover, = 18 and = 1.2261 which was found in the previous example. Now, we need to show that there exists M > 0 such that M > 1, (M) and such M > 0 exists if 1 – > 0. By using direct computation = 0.1533 < 1, assumption (S4 ) is satisfied. Hence, by using Theorem 12, the boundary value problem (26) has at least one solution on [0, 12 ].


Page 16 of 17

Example 3 Consider the following nonlinear fractional differential equation with Katugampola fractional integral conditions: ⎧ c 5/2 ⎪ ⎪ ⎨ D x(t) =

9 sin2 (π t) |x(t)| ( + 1), t ∈ [0, 12 ], (et +10) |x(t)|+1 5 1 5 1 x( 21 ) = 12 I 3 x(3/8), x ( 12 ) = 12 I 3 x(1/3), ⎪ ⎪ ⎩ 1 5 1 x ( 2 ) = 12 I 3 x(2/5).

(27)

Here, α = 5/2, T = 12 , β = 1/2, γ = 1/2, δ = 1/2, = 3/8, η = 1/3, ζ = 2/5, ρ = 5, q = 13 , and f (t, x) =

9 sin2 (πt) |x| + 1 . (et + 10) |x| + 1

9 9 Since |f (t, x) – f (t, y)| ≤ 10 |x – y|, then it implies that L = 10 means (S1 ) is satisfied but (S2 ), which is L < 1, is not satisfied. [L = 1.10358 > 1.] Therefore, we consider (S5 ) which is

f (t, x) ≤

9 |x| 18 +1 ≤ t = (t). (et + 10) |x| + 1 (e + 10)

By using (21), 1 = 0.0561 is found. It is obvious that L1 = 0.05049 < 1. So, (S6 ) is satisfied. Hence, by using Theorem 14, the boundary value problem (27) has at least one solution on [0, 12 ]. Acknowledgements The authors would like to thank the anonymous referees and the editor for their constructive suggestions on improving the presentation of the paper. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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