Nonlinear boundary value problems for mixed-type fractional equations and Ulam-Hyers stability

Huiwen Wang; Fang Li

doi:10.1515/math-2020-0051

Article Open Access

Nonlinear boundary value problems for mixed-type fractional equations and Ulam-Hyers stability

Huiwen Wang and Fang Li

Published/Copyright: September 10, 2020

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$Open Mathematics$

From the journal Open Mathematics Volume 18 Issue 1

Abstract

In this article, we discuss the nonlinear boundary value problems involving both left Riemann-Liouville and right Caputo-type fractional derivatives. By using some new techniques and properties of the Mittag-Leffler functions, we introduce a formula of the solutions for the aforementioned problems, which can be regarded as a novelty item. Moreover, we obtain the existence result of solutions for the aforementioned problems and present the Ulam-Hyers stability of the fractional differential equation involving two different fractional derivatives. An example is given to illustrate our theoretical result.

Keywords: mixed fractional derivatives; nonlinear boundary value problem; Mittag-Leffler functions; Ulam-Hyers stability

MSC 2010: 34A08; 34A37; 34B10

1 Introduction

The fractional differential equations can model some engineering and scientific disciplines in the fields of physics, chemistry, electrodynamics of complex medium, polymer rheology, etc. In particular, the forward and backward fractional derivatives provide an excellent tool for the description of some physical phenomena such as the fractional oscillator equations and the fractional Euler-Lagrange equations [1,2,3,4,5,6,7,8,9,10,11,12,13,14]. Recently, a linear boundary value problem (BVP) involving both the right Caputo and the left Riemann-Liouville fractional derivatives has been studied by some authors [10,11,12,13,14].

On the other hand, Ulam’s stability problem [15] has been attracted by many famous researchers (see [16,17] and references therein). Recently, studying the stability of Ulam-Hyers for fractional differential equations is gaining much importance and attention [18,19]. However, the Ulam-Hyers stabilities of differential equations involving with the forward and backward fractional derivatives have not yet been investigated.

In this article, we investigate the following BVP of the fractional differential equation with two different fractional derivatives:

(1.1) D 1 − β c ( D 0 + α L + λ ) u ( t ) = f ( t , u ( t ) ) , t ∈ J ≔ ( 0 , 1 ] ,

(1.2) lim t → 0 + t 1 − α u ( t ) = u 0 , ( I 0 + 1 − α u ) ( 0 ) + ( I 0 + q ρ u ) ( 1 ) = 0 ,

where α , β , α + β ∈ ( 0 , 1 ) , λ , ρ , q > 0 , α + ρ > 1 . D 1 − β c is the right Caputo fractional derivative of order β , D 0 + α L is the left Riemann-Liouville fractional derivative of order α , I 0 + 1 − α is the Riemann-Liouville fractional integral and I 0 + q ρ is the Katugampola fractional integral.

The rest of this article is organized as follows. In Section 2, we collect some concepts of fractional calculus. In Section 3, we prove some properties of classical and generalized Mittag-Leffler functions. In Section 4, we present the definition of solution to (1.1) and (1.2). In Section 5, we obtain the existence and uniqueness of solutions to problem (1.1) and (1.2). In Section 6, we present the Ulam-Hyers stability result for Eq. (1.1). An example is given in Section 7 to demonstrate the application of our result.

2 Preliminaries

In this section, we introduce some notations and definitions of fractional calculus. Throughout this article, we denote by C ( J , ℝ ) the Banach space of all continuous functions from J to ℝ , L p ( J , ℝ ) the Banach space of all Lebesgue measurable functions l : J → ℝ with the norm ∥ l ∥ L p = ∫ J | l ( t ) | p d t 1 p < ∞ and A C ( [ a , b ] , ℝ ) the space of all the absolutely continuous functions defined on [ a , b ] .

Definition 2.1

[3,4] The left-sided and the right-sided fractional integrals of order γ for a function x ( t ) ∈ L 1 are defined as

( I a + γ x ) ( t ) = 1 Γ ( γ ) ∫ a t ( t − s ) γ − 1 x ( s ) d s , t > a , γ > 0 , ( I b − γ x ) ( t ) = 1 Γ ( γ ) ∫ t b ( s − t ) γ − 1 x ( s ) d s , t < b , γ > 0 ,

respectively. Here, Γ ( ⋅ ) is the gamma function.

Definition 2.2

[3,4] If x ( t ) ∈ A C ( [ a , b ] , ℝ ) , then the left-sided and the right-sided Riemann-Liouville fractional derivatives D a + γ L x ( t ) and D b − γ L x ( t ) of order γ exist almost everywhere on [ a , b ] and can be written as

( D a + γ L x ) ( t ) = 1 Γ ( 1 − γ ) d d t ∫ a t ( t − s ) − γ x ( s ) d s = d d t ( I a + 1 − γ x ) ( t ) , t > a , 0 < γ < 1 , ( D b − γ L x ) ( t ) = 1 Γ ( 1 − γ ) − d d t ∫ t b ( s − t ) − γ x ( s ) d s = − d d t ( I b − 1 − γ x ) ( t ) , t < b , 0 < γ < 1 ,

respectively.

Definition 2.3

[3,4] If x ( t ) ∈ A C ( [ a , b ] , ℝ ) , then the right-sided Caputo derivative D b − γ c x ( t ) of order γ exists almost everywhere on [ a , b ] and can be written as

( D b − γ c x ) ( t ) = ( D b − γ L [ x ( s ) − x ( b ) ] ) ( t ) , t < b , 0 < γ < 1 .

Definition 2.4

[5] For ρ , q > 0 , then the Katugampola fractional integral of y ( t ) can be defined as

( I a + q ρ y ) ( t ) = ρ 1 − q Γ ( q ) ∫ a t ( t ρ − τ ρ ) q − 1 τ ρ − 1 y ( τ ) d τ , t > a ,

if the integral exists.

3 Properties of Mittag-Leffler functions

In this section, we prove some properties of the Mittag-Leffler functions.

Definition 3.1

[3,4] For μ , ν > 0 , z ∈ ℝ , the classical Mittag-Leffler functions E μ ( z ) and the generalized Mittag-Leffler functions E μ , ν ( z ) are defined by

E μ ( z ) = ∑ k = 0 ∞ z k Γ ( μ k + 1 ) , E μ , ν ( z ) = ∑ k = 0 ∞ z k Γ ( μ k + ν ) .

Clearly, E μ , 1 ( z ) = E μ ( z ) .

Lemma 3.2

[4,9] Let α ∈ ( 0 , 1 ) , θ > α be arbitrary. The functions E α , E α , α and E α , θ are nonnegative and have the following properties:

For any t ∈ J , E α ( − λ t α ) ≤ 1 , E α , α ( − λ t α ) ≤ 1 Γ ( α ) , E α , θ ( − λ t α ) ≤ 1 Γ ( θ ) .
For any t 1 , t 2 ∈ J ,

| E α ( − λ t 2 α ) − E α ( − λ t 1 α ) | = O ( | t 2 − t 1 | α ) , as t 2 → t 1 , | E α , α ( − λ t 2 α ) − E α , α ( − λ t 1 α ) | = O ( | t 2 − t 1 | α ) , as t 2 → t 1 , | E α , α + 1 ( − λ t 2 α ) − E α , α + 1 ( − λ t 1 α ) | = O ( | t 2 − t 1 | α ) , as t 2 → t 1 .

Lemma 3.3

[4,9] For γ , μ , ν , λ > 0 , t > 0 , 0 < α , β < 1 , the usual derivatives of E μ , ν and the fractional integrals and derivatives of E μ , ν are expressed by

d d t E μ ( − λ t μ ) = − λ t μ − 1 E μ , μ ( − λ t μ ) ;
I 0 + γ ( s ν − 1 E μ , ν ( − λ s μ ) ) ( t ) = 1 Γ ( γ ) ∫ 0 t ( t − s ) γ − 1 s ν − 1 E μ , ν ( − λ s μ ) d s = t γ + ν − 1 E μ , γ + ν ( − λ t μ ) ;
[ D a + ν L ( s − a ) β − 1 E α , β ( − λ ( s − a ) α ) ] ( t ) = ( t − a ) β − ν − 1 E α , β − ν ( − λ ( t − a ) α ) .

Lemma 3.4

For ν , λ > 0 , t > 0 , 0 < α , β < 1 , the fractional integrals and derivatives of E μ , ν are expressed by

[ D a + α L ( s − a ) α − 1 E α , α ( − λ ( s − a ) α ) ] ( t ) = − λ ( t − a ) α − 1 E α , α ( − λ ( t − a ) α ) ;
[ D 1 − β c D 0 + α L s α E α , α + 1 ( − λ s α ) ] ( t ) + λ [ D 1 − β c s α E α , α + 1 ( − λ s α ) ] ( t ) = 0 ;
[ D 1 − β c D 0 + α L s α − 1 E α , α ( − λ s α ) ] ( t ) + λ [ D 1 − β c s α − 1 E α , α ( − λ s α ) ] ( t ) = 0 ;
[ I 0 + q ρ s ν − 1 E α , ν ( − λ s α ) ] ( t ) = t q ρ + ν − 1 ρ q Γ ( q ) ∫ 0 1 s ν − 1 ρ ( 1 − s ) q − 1 E α , ν ( − λ t α s α ρ ) d s , ν ≥ α .

Proof

By using Lemma 3.3, we find
[ D a + α L ( s − a ) α − 1 E α , α ( − λ ( s − a ) α ) ] ( t ) = 1 Γ ( 1 − α ) d d t ∫ a t ( t − s ) − α ( s − a ) α − 1 E α , α ( − λ ( s − a ) α ) d s = d d t [ E α ( − λ ( t − a ) α ) ] = − λ ( t − a ) α − 1 E α , α ( − λ ( t − a ) α ) .
Similarly, we have
(3.1) [ D a + α L ( s − a ) α E α , α + 1 ( − λ ( s − a ) α ) ] ( t ) = E α ( − λ ( t − a ) α ) .
Using (3.1) and ( D 1 − β c 1 ) ( t ) = 0 , we arrive at
[ D 1 − β c D 0 + α L s α E α , α + 1 ( − λ s α ) ] ( t ) + λ [ D 1 − β c s α E α , α + 1 ( − λ s α ) ] ( t ) = [ D 1 − β c ( E α ( − λ s α ) + λ s α E α , α + 1 ( − λ s α ) ) ] ( t ) = ( D 1 − β c 1 ) ( t ) = 0 .
The proof of (3) is similar to that of (2).
Clearly, for ν ≥ α , the integral ∫ 0 t ( 1 − τ ) q − 1 τ ν − 1 ρ E α , ν ( − λ t α τ α ρ ) d τ exists, then we get

[ I 0 + q ρ s ν − 1 E α , ν ( − λ s α ) ] ( t ) = ρ 1 − q Γ ( q ) ∫ 0 t ( t ρ − s ρ ) q − 1 s ρ + ν − 2 E α , ν ( − λ s α ) d s = ρ 1 − q t ρ q + ν − 1 ρ Γ ( q ) ∫ 0 1 ( 1 − τ ) q − 1 τ ν − 1 ρ E α , ν ( − λ t α τ α ρ ) d τ = t ρ q + ν − 1 ρ q Γ ( q ) ∫ 0 1 ( 1 − τ ) q − 1 τ ν − 1 ρ E α , ν ( − λ t α τ α ρ ) d τ .

□

4 Solutions for BVP

In this section, we present the formulas of solutions to problem (1.1) and (1.2).

Lemma 4.1

[4] For θ > 0 , a general solution of the fractional differential equation D 1 − θ c u ( t ) = 0 is given by

u ( t ) = ∑ i = 0 n − 1 C i ( 1 − t ) i ,

where C i ∈ ℝ , i = 0 , 1 , 2 , … , n − 1 ( n = [ θ ] + 1 ) , and [ θ ] denotes the integer part of the real number θ .

Lemma 4.2

For α , β ∈ ( 0 , 1 ) , D 1 − β c ( D 0 + α L + λ ) u ( t ) = h ( t ) , t ∈ J , then

u ( t ) = C 0 t α E α , α + 1 ( − λ t α ) + C 1 t α − 1 E α , α ( − λ t α ) + ∫ 0 1 K ( t , τ ) h ( τ ) d τ , t ∈ J ,

if the integral exists. Here,

K ( t , τ ) = 1 Γ ( β ) ∫ 0 τ ( t − s ) α − 1 ( τ − s ) β − 1 E α , α ( − λ ( t − s ) α ) d s , 0 < τ < t ≤ 1 , ∫ 0 t ( t − s ) α − 1 ( τ − s ) β − 1 E α , α ( − λ ( t − s ) α ) d s , 0 < t < τ ≤ 1 .

Formally, by Lemma 4.1, for C 0 ∈ ℝ , we obtain ( D 0 + α L + λ ) u ( t ) = C 0 + ( I 1 − β h ) ( t ) . Based on the arguments of [4], we derive

u ( t ) = C 1 t α − 1 E α , α ( − λ t α ) + ∫ 0 t ( t − s ) α − 1 E α , α ( − λ ( t − s ) α ) ( C 0 + ( I 1 − β h ) ( s ) ) d s = C 0 t α E α , α + 1 ( − λ t α ) + C 1 t α − 1 E α , α ( − λ t α ) + 1 Γ ( β ) ∫ 0 t ( t − s ) α − 1 E α , α ( − λ ( t − s ) α ) ∫ s 1 ( τ − s ) β − 1 h ( τ ) d τ d s = C 0 t α E α , α + 1 ( − λ t α ) + C 1 t α − 1 E α , α ( − λ t α ) + 1 Γ ( β ) ∫ 0 t h ( τ ) d τ ∫ 0 τ ( t − s ) α − 1 ( τ − s ) β − 1 E α , α ( − λ ( t − s ) α ) d s + ∫ t 1 h ( τ ) d τ ∫ 0 t ( t − s ) α − 1 ( τ − s ) β − 1 E α , α ( − λ ( t − s ) α ) d s = C 0 t α E α , α + 1 ( − λ t α ) + C 1 t α − 1 E α , α ( − λ t α ) + ∫ 0 1 K ( t , τ ) h ( τ ) d τ .

We define C 1 − α ( [ 0 , 1 ] , ℝ ) = { u ∈ C ( J , ℝ ) : t 1 − α u ( t ) ∈ C ( [ 0 , 1 ] , ℝ ) } with the norm ∥ u ∥ 1 − α = max t ∈ [ 0 , 1 ] t 1 − α | u ( t ) | and we abbreviate C 1 − α ( [ 0 , 1 ] , ℝ ) to C 1 − α .

To prove our results, we make the following assumptions.

Let f : J × ℝ → ℝ be a function such that f ( ⋅ , u ) : J → ℝ is measurable for all u ∈ ℝ and f ( t , ⋅ ) : ℝ → ℝ is continuous for a.e. t ∈ J , and there exists a function φ ∈ L 1 p ( J , ℝ + ) ( 0 < p < min { α , β } ) such that
| f ( t , u ( t ) ) − f ( t , v ( t ) ) | ≤ φ ( t ) ∥ u − v ∥ 1 − α .
sup t ∈ J | f ( t , 0 ) | < ∞ .

For convenience of the following presentation, we set

A ( θ , t ) = t q ρ + θ ρ q Γ ( q ) ∫ 0 1 s θ ρ ( 1 − s ) q − 1 E α , θ + 1 ( − λ t α s α ρ ) d s ; N θ = B θ ρ + 1 , q ρ q Γ ( q ) 1 Γ ( θ + 1 ) ; M 1 = 1 ρ q Γ ( q + 1 ) Γ ( α + 1 ) A ( α , 1 ) + 1 ; M 2 = 1 β 1 − p α − p 1 − p + 1 α 1 − p β − p 1 − p ∥ φ ∥ L 1 p Γ ( α ) Γ ( β ) ; M 3 = 2 Γ ( α + 1 ) Γ ( β + 1 ) ,

where B ( ⋅ , ⋅ ) is the Beta function.

For t ∈ J , y > p , applying the Hölder inequality, there hold the following estimates:

(4.1) ∫ a t ( t − s ) y − 1 φ ( s ) d s ≤ ∫ a t ( t − s ) y − 1 1 − p d s 1 − p ∥ φ ∥ L 1 p = 1 − p y − p 1 − p ( t − a ) y − p ∥ φ ∥ L 1 p , t > a ,

(4.2) ∫ t b ( s − t ) y − 1 φ ( s ) d s ≤ ∫ t b ( s − t ) y − 1 1 − p d s 1 − p ∥ φ ∥ L 1 p = 1 − p y − p 1 − p ( b − t ) y − p ∥ φ ∥ L 1 p , t < b .

Lemma 4.3

(4.3) ∫ 0 τ [ ( t 1 − s ) α − 1 E α , α ( − λ ( t 1 − s ) α ) − ( t 2 − s ) α − 1 E α , α ( − λ ( t 2 − s ) α ) ] ( τ − s ) β − 1 d s ≤ O ( ( t 2 − t 1 ) α ) , 0 < τ < t 1 < t 2 ≤ 1 ,

(4.4) ∫ 0 t 1 ( t 1 − s ) α − 1 − ( t 2 − s ) α − 1 ( τ − s ) β − 1 d s ≤ ( τ − t 2 ) β − 1 O ( ( t 2 − t 1 ) α ) , 0 < t 1 < t 2 < τ ≤ 1 .

Proof

For 0 < t 1 < t 2 ≤ 1 , it follows from Lemma 3.2 and the mean value theorem that

which yields

∫ 0 τ [ ( t 1 − s ) α − 1 E α , α ( − λ ( t 1 − s ) α ) − ( t 2 − s ) α − 1 E α , α ( − λ ( t 2 − s ) α ) ] ( τ − s ) β − 1 d s ≤ ∫ 0 τ ( τ − s ) β − 1 d s ⋅ O ( ( t 2 − t 1 ) α ) ≔ O ( ( t 2 − t 1 ) α ) , for 0 < τ < t 1 < t 2 ≤ 1 .

For 0 < t 1 < t 2 < τ ≤ 1 , by the mean value theorem, one can see that

∫ 0 t 1 [ ( t 1 − s ) α − 1 − ( t 2 − s ) α − 1 ] ( τ − s ) β − 1 d s ≤ ∫ 0 t 1 [ ( t 1 − s ) α − 1 − ( t 2 − s ) α − 1 ] d s ⋅ ( τ − t 1 ) β − 1 ≤ 1 α [ t 2 α − t 1 α + ( t 2 − t 1 ) α ] ⋅ ( τ − t 1 ) β − 1 ≤ ( τ − t 2 ) β − 1 ⋅ O ( ( t 2 − t 1 ) α ) .

□

Let us define

k 1 ( t , τ ) = 1 Γ ( β ) ∫ 0 τ ( t − s ) α − 1 ( τ − s ) β − 1 E α , α ( − λ ( t − s ) α ) d s , 0 < τ < t ≤ 1 , k 2 ( t , τ ) = 1 Γ ( β ) ∫ 0 t ( t − s ) α − 1 ( τ − s ) β − 1 E α , α ( − λ ( t − s ) α ) d s , 0 < t < τ ≤ 1 ,

clearly,

∫ 0 1 K ( t , τ ) d τ = ∫ 0 t k 1 ( t , τ ) d τ + ∫ t 1 k 2 ( t , τ ) d τ ,

one can obtain the following estimates.

Lemma 4.4

(4.5) | k 1 ( t , τ ) | ≤ ( t − τ ) α − 1 Γ ( β + 1 ) Γ ( α ) , 0 < τ < t ≤ 1 ,

(4.6) | k 2 ( t , τ ) | ≤ ( τ − t ) β − 1 Γ ( α + 1 ) Γ ( β ) , 0 < t < τ ≤ 1 ,

(4.7) ∫ 0 1 | K ( t 2 , τ ) − K ( t 1 , τ ) | φ ( τ ) d τ = O ( ( t 2 − t 1 ) α − p ) + O ( ( t 2 − t 1 ) β − p ) ,

(4.8) ∫ 0 1 | K ( t 2 , τ ) − K ( t 1 , τ ) | d τ = O ( ( t 2 − t 1 ) α ) + O ( ( t 2 − t 1 ) β ) ,

(4.9) ∫ 0 1 K ( t , τ ) φ ( τ ) d τ ≤ M 2 ,

(4.10) ∫ 0 1 K ( t , τ ) d τ ≤ M 3 .

Proof

Since

∫ 0 τ ( t − s ) α − 1 ( τ − s ) β − 1 d s ≤ ∫ 0 τ ( t − τ ) α − 1 ( τ − s ) β − 1 d s = τ β ( t − τ ) α − 1 β , 0 < τ < t ≤ 1 , ∫ 0 t ( t − s ) α − 1 ( τ − s ) β − 1 d s ≤ ∫ 0 t ( t − s ) α − 1 ( τ − t ) β − 1 d s = t α ( τ − t ) β − 1 α , 0 < t < τ ≤ 1 ,

(4.5) and (4.6) hold. Moreover, (4.1) and (4.2) imply

(4.11) ∫ t 1 t 2 k 1 ( t 2 , s ) φ ( s ) d s ≤ ∫ t 1 t 2 ( t 2 − s ) α − 1 φ ( s ) Γ ( β + 1 ) Γ ( α ) d s ≤ 1 − p α − p 1 − p ( t 2 − t 1 ) α − p Γ ( β + 1 ) Γ ( α ) ∥ φ ∥ L 1 p ,

(4.12) ∫ t 1 t 2 k 2 ( t 1 , s ) φ ( s ) d s ≤ ∫ t 1 t 2 ( s − t 1 ) β − 1 φ ( s ) Γ ( α + 1 ) Γ ( β ) d s ≤ 1 − p β − p 1 − p ( t 2 − t 1 ) β − p Γ ( α + 1 ) Γ ( β ) ∥ φ ∥ L 1 p .

For 0 < τ < t 1 < t 2 ≤ 1 , by (4.3), we get

| k 1 ( t 2 , τ ) − k 1 ( t 1 , τ ) | ≤ 1 Γ ( β ) ∫ 0 τ [ ( t 2 − s ) α − 1 E α , α ( − λ ( t 2 − s ) α ) − ( t 1 − s ) α − 1 E α , α ( − λ ( t 1 − s ) α ) ] ( τ − s ) β − 1 d s = O ( ( t 2 − t 1 ) α ) ,

furthermore,

(4.13) ∫ 0 t 1 | k 1 ( t 2 , τ ) − k 1 ( t 1 , τ ) | φ ( τ ) d τ ≤ ∥ φ ∥ L 1 ⋅ O ( ( t 2 − t 1 ) α ) ≔ O ( ( t 2 − t 1 ) α ) ,

(4.14) ∫ 0 t 1 | k 1 ( t 2 , τ ) − k 1 ( t 1 , τ ) | d τ ≔ O ( ( t 2 − t 1 ) α ) .

For 0 < t 1 < t 2 < τ ≤ 1 , by (4.4) and Lemma 3.2, we find

| k 2 ( t 2 , τ ) − k 2 ( t 1 , τ ) | ≤ 1 Γ ( β ) 1 Γ ( α ) ∫ 0 t 1 [ ( t 1 − s ) α − 1 − ( t 2 − s ) α − 1 ] ( τ − s ) β − 1 d s + ∫ 0 t 1 ( t 2 − s ) α − 1 ( τ − s ) β − 1 | E α , α ( − λ ( t 1 − s ) α ) − E α , α ( − λ ( t 2 − s ) α ) | d s + 1 Γ ( α ) ∫ t 1 t 2 ( t 2 − s ) α − 1 ( τ − s ) β − 1 d s ≤ ( τ − t 2 ) β − 1 ⋅ O ( ( t 2 − t 1 ) α ) + ( τ − t 2 ) β − 1 Γ ( α ) Γ ( β ) ∫ t 1 t 2 ( t 2 − s ) α − 1 d s ≔ ( τ − t 2 ) β − 1 ⋅ O ( ( t 2 − t 1 ) α ) ,

then by (4.2), we arrive at

(4.15) ∫ t 2 1 | k 2 ( t 2 , τ ) − k 2 ( t 1 , τ ) | φ ( τ ) d τ ≤ ∫ t 2 1 ( τ − t 2 ) β − 1 φ ( τ ) d τ ⋅ O ( ( t 2 − t 1 ) α ) ≔ O ( ( t 2 − t 1 ) α ) ,

(4.16) ∫ t 2 1 | k 2 ( t 2 , τ ) − k 2 ( t 1 , τ ) | d τ ≔ O ( ( t 2 − t 1 ) α ) .

From (4.13), (4.15), (4.11) and (4.12), it follows that

∫ 0 1 | K ( t 2 , τ ) − K ( t 1 , τ ) | φ ( τ ) d τ ≤ ∫ 0 t 1 | k 1 ( t 2 , s ) − k 1 ( t 1 , s ) | φ ( s ) d s + ∫ t 1 t 2 k 1 ( t 2 , s ) φ ( s ) d s + ∫ t 2 1 | k 2 ( t 2 , s ) − k 2 ( t 1 , s ) | φ ( s ) d s + ∫ t 1 t 2 k 2 ( t 1 , s ) φ ( s ) d s = O ( ( t 2 − t 1 ) α − p ) + O ( ( t 2 − t 1 ) β − p ) .

Similarly, one can conclude from (4.14), (4.16), (4.5) and (4.6) that

∫ 0 1 | K ( t 2 , τ ) − K ( t 1 , τ ) | d τ ≔ O ( ( t 2 − t 1 ) α ) + O ( ( t 2 − t 1 ) β ) .

By (4.5), (4.6), (4.1) and (4.2), we observe that

∫ 0 1 K ( t , τ ) φ ( τ ) d τ = ∫ 0 t k 1 ( t , τ ) φ ( τ ) d τ + ∫ t 1 k 2 ( t , τ ) φ ( τ ) d τ ≤ M 2 , ∫ 0 1 K ( t , τ ) d τ ≤ M 3 , ∫ 0 1 K ( t , τ ) | f ( τ , 0 ) | d τ ≤ ∫ 0 1 K ( t , τ ) d τ ⋅ sup t ∈ J | f ( t , 0 ) | ≤ M 3 sup t ∈ J | f ( t , 0 ) | . □

For the sake of convenience, we adopt the following notation:

( F u ) ( t ) = ∫ 0 1 K ( t , τ ) f ( τ , u ( τ ) ) d τ = ∫ 0 t k 1 ( t , τ ) f ( τ , u ( τ ) ) d τ + ∫ t 1 k 2 ( t , τ ) f ( τ , u ( τ ) ) d τ .

Since

| f ( t , u ( t ) ) | ≤ | f ( t , u ( t ) ) − f ( t , 0 ) | + | f ( t , 0 ) | ≤ φ ( t ) ∥ u ∥ 1 − α + | f ( t , 0 ) | ,

then by (4.9) and (4.10),

(4.17) | ( F u ) ( t ) ) | ≤ ∫ 0 1 K ( t , τ ) φ ( τ ) d τ ⋅ ∥ u ∥ 1 − α + ∫ 0 1 K ( t , τ ) | f ( τ , 0 ) | d τ ≤ M 2 ∥ u ∥ 1 − α + M 3 sup t ∈ J | f ( t , 0 ) | .

Lemma 4.5

Assume that (H1) and (H2) hold. For u ∈ C 1 − α , t ∈ J , ( F u ) ( t ) satisfies the following relations:

( F u ) ( t ) ∈ A C ( J , ℝ ) ;
[ D 0 + α L ( F u ) ] ( t ) = − λ ( F u ) ( t ) + ( I 1 − β f ) ( t ) ;
[ D 1 − β c D 0 + α L ( F u ) ] ( t ) + λ [ D 1 − β c ( F u ) ] ( t ) = f ( t , u ( t ) ) ;
[ I 0 + q ρ ( F u ) ] ( t ) = t q ρ ρ q Γ ( q ) ∫ 0 1 ( 1 − s ) q − 1 ( F u ) ( t s 1 ρ ) d s .

Proof

For every finite collection { ( a j , b j ) } 1 ≤ j ≤ n on J with ∑ j = 1 n ( b j − a j ) → 0 , noting that (4.7) and (4.8), we derive

Hence, ( F u ) ( t ) is absolutely continuous on J. Furthermore, for almost all t ∈ J , [ L D 0 + α ( F u ) ( s ) ] ( t ) exists and from the fact ( F u ) ( s ) = ∫ 0 s ( s − τ ) α − 1 E α , α ( − λ ( s − τ ) α ) ( I 1 − β f ) ( τ , u ( τ ) ) d τ and Lemma 3.3, we get

[ D 0 + α L ( F u ) ( s ) ] ( t ) = 1 Γ ( 1 − α ) d d t ∫ 0 t ∫ 0 s ( t − s ) − α ( s − τ ) α − 1 E α , α ( − λ ( s − τ ) α ) ( I 1 − β f ) ( τ ) d τ d s = 1 Γ ( 1 − α ) d d t ∫ 0 t ( I 1 − β f ) ( τ ) ∫ τ t ( t − s ) − α ( s − τ ) α − 1 E α , α ( − λ ( s − τ ) α ) d s d τ = d d t ∫ 0 t ( I 1 − β f ) ( τ ) E α ( − λ ( t − τ ) α ) d τ = − λ ( F u ) ( t ) + ( I 1 − β f ) ( t ) .

Thus,

[ D 1 − β c D 0 + α L ( F u ) ( s ) ] ( t ) + λ [ D 1 − β c ( F u ) ( s ) ] ( t ) = [ D 1 − β c ( − λ ( F u ) ( s ) + ( I 1 − β f ) ( s ) ) ] ( t ) + λ [ D 1 − β c ( F u ) ( s ) ] ( t ) = f ( t , u ( t ) ) .

It follows from (4.17) that ∫ 0 1 ( 1 − s ) q − 1 ( F u ) ( t s 1 ρ ) d s exists and

[ I 0 + q ρ ( F u ) ] ( t ) = ρ 1 − q Γ ( q ) ∫ 0 t ( t ρ − s ρ ) q − 1 s ρ − 1 ( F u ) ( s ) d s = t q ρ ρ q Γ ( q ) ∫ 0 1 ( 1 − s ) q − 1 ( F u ) ( t s 1 ρ ) d s . □

Lemma 4.6

Assume that (H1) and (H2) hold. A function u is a solution of the following fractional integral equation:

(4.18) u ( t ) = ( P u ) ( t ) + ( Q u ) ( t )

if and only if u is a solution of problem ( 1.1 ) and ( 1.2 ), where

( P u ) ( t ) = − u 0 Γ ( α ) ( 1 + A ( α − 1 , 1 ) ) A ( α , 1 ) t α E α , α + 1 ( − λ t α ) + u 0 Γ ( α ) t α − 1 E α , α ( − λ t α ) , ( Q u ) ( t ) = − 1 ρ q Γ ( q ) ∫ 0 1 ( 1 − τ ) q − 1 ( F u ) ( τ 1 ρ ) d τ A ( α , 1 ) t α E α , α + 1 ( − λ t α ) + ( F u ) ( t ) .

Proof

(Sufficiency) Let u be the solution of (1.1) and (1.2), Lemmas 3.3, 3.4, 4.2 and 4.5 imply

u ( t ) = a t α E α , α + 1 ( − λ t α ) + b t α − 1 E α , α ( − λ t α ) + ( F u ) ( t ) , t ∈ J , ( I 0 + 1 − α u ) ( t ) = a t E α , 2 ( − λ t α ) + b E α ( − λ t α ) + [ I 0 + 1 − α ( F u ) ] ( t ) , [ I 0 + q ρ u ] ( t ) = a A ( α , t ) + b A ( α − 1 , t ) + t q ρ ρ q Γ ( q ) ∫ 0 1 ( 1 − s ) q − 1 ( F u ) ( t s 1 ρ ) d s ,

where a , b are constants. Using the boundary value condition (1.2), we derive that b = u 0 Γ ( α ) and

a A ( α , 1 ) + u 0 Γ ( α ) ( 1 + A ( α − 1 , 1 ) ) + 1 ρ q Γ ( q ) ∫ 0 1 ( 1 − τ ) q − 1 ( F u ) ( τ 1 ρ ) d τ = 0 ,

which means

a = − u 0 Γ ( α ) ( 1 + A ( α − 1 , 1 ) ) + 1 ρ q Γ ( q ) ∫ 0 1 ( 1 − τ ) q − 1 ( F u ) ( τ 1 ρ ) d τ A ( α , 1 ) .

Now we can see that (4.18) holds.

(Necessity) Let u satisfy (4.18). According to Lemma 3.4(2), (3) and Lemma 4.5(3), [ D 1 − β c D 0 + α L u ] ( t ) exists and D 1 − β c ( D 0 + α L + λ ) u ( t ) = f ( t , u ( t ) ) for t ∈ J . Clearly, the boundary value condition (1.2) holds and hence the necessity is proved.□

5 Existence and uniqueness result

In this section, we deal with the existence and uniqueness of solutions to problem (1.1) and (1.2).

Theorem 5.1

Assume that (H1) and (H2) are satisfied, then problem ( 1.1 ) and ( 1.2 ) has a unique solution u ∈ C 1 − α if M 1 M 2 < 1 .

Proof

We consider an operator ℱ : C 1 − α → C 1 − α defined by

( ℱ u ) ( t ) = ( P u ) ( t ) + ( Q u ) ( t ) .

Clearly, ℱ is well defined and the fixed point of ℱ is the solution of problem (1.1) and (1.2).

Let B r = { u ∈ C 1 − α : ∥ u ∥ 1 − α ≤ r } be a bounded set in C 1 − α , where

r ≥ | u 0 | α 1 + N α − 1 A ( α , 1 ) + α + M 1 M 3 sup t ∈ J | f ( t , 0 ) | 1 − M 1 M 2 .

For u ∈ B r , it follows from Lemmas 3.2(i) and (4.17) that

∥ P u ∥ 1 − α ≤ | u 0 | α 1 + N α − 1 A ( α , 1 ) + α , ∥ Q u ∥ 1 − α ≤ M 1 ⋅ [ M 2 r + M 3 sup t ∈ J | f ( t , 0 ) | ]

and hence

∥ ( ℱ u ) ∥ 1 − α ≤ ∥ P u ∥ 1 − α + ∥ Q u ∥ 1 − α ≤ | u 0 | α 1 + N α − 1 A ( α , 1 ) + α + M 1 ⋅ [ M 2 r + M 3 sup t ∈ J | f ( t , 0 ) | ] ≤ r .

Then for t ∈ J , u ∈ B r , ℱ u ∈ B r .

For any u , v ∈ B r , t ∈ J , by (4.9),

| ( F u ) ( t ) − ( F v ) ( t ) | ≤ ∫ 0 1 K ( t , τ ) | f ( τ , u ( τ ) ) − f ( τ , v ( τ ) ) | d τ ≤ ∫ 0 1 K ( t , τ ) φ ( τ ) d τ ⋅ ∥ u − v ∥ 1 − α ≤ M 2 ∥ u − v ∥ 1 − α ,

and thus

| ( ℱ u ) ( t ) − ( ℱ v ) ( t ) | ≤ ∫ 0 1 ( 1 − τ ) q − 1 ( F u ) ( τ 1 ρ ) − ( F v ) ( τ 1 ρ ) d τ A ( α , 1 ) Γ ( α + 1 ) ρ q Γ ( q ) + | ( F u ) ( t ) − ( F v ) ( t ) | ≤ M 1 M 2 ∥ u − v ∥ 1 − α .

Furthermore, ∥ ℱ u − ℱ v ∥ 1 − α < ∥ u − v ∥ 1 − α . The proof now can be finished by using the Banach contraction mapping principle.□

6 Ulam-Hyers stability

Let ϵ ˜ be a positive real number. We consider Eq. (1.1) with inequality

(6.1) | D 1 − β c ( D 0 + α L + λ ) v ( t ) − f ( t , v ( t ) ) | ≤ ϵ ˜ , t ∈ J .

Definition 6.1

Eq. (1.1) is Ulam-Hyers stable if there exists a real number c > 0 such that for each solution v ( t ) of inequality (6.1) there exists a solution u of Eq. (1.1) with

| v ( t ) − u ( t ) | ≤ c ϵ ˜ , t ∈ J .

Remark 6.2

A function v ∈ C 1 − α is a solution of inequality (6.1) if and only if there exists a function g ∈ C 1 − α such that (i) | g ( t ) | ≤ ϵ ˜ ; (ii) D 1 − β c ( D 0 + α L + λ ) v ( t ) = f ( t , v ( t ) ) + g ( t ) .

Let

v ˜ ( t ) = − u 0 Γ ( α )(1 + A ( α − 1,1)) A ( α ,1) t α E α , α + 1 ( − λ t α ) + u 0 Γ ( α ) t α − 1 E α , α ( − λ t α ) − 1 ρ q Γ ( q ) ∫ 0 1 (1 − τ ) q − 1 ( F v ) ( τ 1 ρ ) d τ A ( α ,1) t α E α , α + 1 ( − λ t α ) + ( F v ) ( t ) ,

where

( F v ) ( t ) = ∫ 0 1 K ( t , τ ) f ( τ , v ( τ ) ) d τ .

Remark 6.3

Let v ∈ C 1 − α be a solution of inequality (6.1) with lim t → 0 + t 1 − α v ( t ) = u 0 , ( I 0 + 1 − α v ) ( 0 ) + ( I 0 + q ρ v ) ( 1 ) = 0 . Then, v is a solution of the inequality | v ( t ) − v ˜ ( t ) | ≤ M 1 M 3 ϵ ˜ .

Indeed, by Remark 6.2, we have

D 1 − β c ( D 0 + α L + λ ) v ( t ) = f ( t , v ( t ) ) + g ( t ) ,

lim t → 0 + t 1 − α v ( t ) = u 0 , ( I 0 + 1 − α v ) ( 0 ) + ( I 0 + q ρ v ) ( 1 ) = 0 .

Applying the same arguments as in the proof of Lemma 4.6, we obtain

v ( t ) = − u 0 Γ ( α ) ( 1 + A ( α − 1 , 1 ) ) A ( α , 1 ) t α E α , α + 1 ( − λ t α ) + u 0 Γ ( α ) t α − 1 E α , α ( − λ t α ) − 1 ρ q Γ ( q ) ∫ 0 1 ( 1 − τ ) q − 1 ( G v ) ( τ 1 ρ ) d τ A ( α , 1 ) t α E α , α + 1 ( − λ t α ) + ( G v ) ( t ) ,

where

( G v ) ( t ) = ∫ 0 1 K ( t , τ ) [ f ( τ , v ( τ ) ) + g ( τ ) ] d τ .

Therefore, by (4.10), we conclude that | v ( t ) − v ˜ ( t ) | ≤ M 1 M 3 ϵ ˜ .

Theorem 6.4

Assume that (H1) and (H2) are satisfied, then Eq. ( 1.1 ) is Ulam-Hyers stable if M 1 M 2 < 1 .

Proof

Let v ∈ C 1 − α be a solution of inequality (6.1) with lim t → 0 + t 1 − α v ( t ) = u 0 , ( I 0 + 1 − α v ) ( 0 ) + ( I 0 + q ρ v ) ( 1 ) = 0 . u denotes the unique solution of the following problem:

D 1 − β c ( D 0 + α L + λ ) u ( t ) = f ( t , u ( t ) ) , t ∈ J , lim t → 0 + t 1 − α u ( t ) = u 0 , ( I 0 + 1 − α u ) ( 0 ) + ( I 0 + q ρ u ) ( 1 ) = 0 .

It is easy to check that

then

∥ v − u ∥ 1 − α ≤ M 1 M 3 ϵ ˜ 1 − M 1 M 2 ,

that is, Eq. (1.1) is Ulam-Hyers stable.□

7 Example

As an example, we consider here the following BVP of the fractional differential equation with two different fractional derivatives:

(7.1) D 1 − 1 5 c ( D 0 + 3 5 L + 3 ) u ( t ) = 1 30 t 10 sin ( t 2 5 u ( t ) + t 3 4 ) , t ∈ J ≔ ( 0 , 1 ] ,

(7.2) lim t → 0 + t 2 5 u ( t ) = u 0 , ( I 0 + 2 5 u ) ( 0 ) + ( I 0 + 3 2 u ) ( 1 ) = 0 .

Corresponding to (1.1) and (1.2), we recognize that α = 3 5 , β = 1 5 , λ = 3 , ρ = 2 , q = 3 and

f ( t , u ( t ) ) = 1 30 t 10 sin ( t 2 5 u ( t ) + t 3 4 ) .

The space C 1 − α = { u ∈ C ( J , ℝ ) : t 2 5 u ( t ) ∈ C ( [ 0 , 1 ] , ℝ ) } with the norm ∥ u ∥ 2 5 = max t ∈ [ 0 , 1 ] t 2 5 | u ( t ) | .

It is not difficult to obtain

| f ( t , u ( t ) ) − f ( t , v ( t ) ) | ≤ φ ( t ) ∥ u − v ∥ 2 5 ,

where φ ( t ) = 1 30 t 10 ∈ L 1 p [ 0 , 1 ] ( p = 3 20 ) and ∥ φ ∥ L 1 p = 3 3 20 30 . Moreover,

| f ( t , 0 ) | = 1 30 t 10 | sin t 3 4 | ≤ 1 30 ,

and thus sup t ∈ J | f ( t , 0 ) | ≤ 1 30 .

By direct computation, we find

A 3 5 , 1 = 1 8 Γ ( 3 ) ∫ 0 1 s 3 10 ( 1 − s ) 2 E 3 5 , 8 5 ( − 3 s 3 10 ) d s ≈ 0.005 , M 1 = 1 ρ q Γ ( q + 1 ) Γ ( α + 1 ) A ( α , 1 ) + 1 = 1 8 Γ ( 4 ) Γ 8 5 A 3 5 , 1 + 1 ≈ 5.6 ; M 2 = 1 β 1 − p α − p 1 − p + 1 α 1 − p β − p 1 − p ∥ φ ∥ L 1 p Γ ( α ) Γ ( β ) = 5 17 9 17 20 + 5 3 ( 17 ) 17 20 3 3 20 30 Γ 3 5 Γ 1 5 ≈ 0.16 ,

consequently, M 1 M 2 ≈ 0.9 < 1 , by Theorems 5.1 and 6.4, problem (7.1) and (7.2) has a unique solution and Eq. (7.1) is Ulam-Hyers stable.

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Received: 2019-08-13

Revised: 2020-03-04

Accepted: 2020-04-28

Published Online: 2020-09-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2020-0051

Keywords for this article

mixed fractional derivatives; nonlinear boundary value problem; Mittag-Leffler functions; Ulam-Hyers stability

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