Existence and Ulam’s Stability for Conformable Fractional Differential Equations with Constant Coefficients

Abstract

In this article, we develop a standard idea in seeking the solution of linear ODEs to derive the representation of solutions to conformable fractional linear differential equations with constant coefficients by adopting the variation of constants method. In addition, we present the existence of solutions to conformable fractional nonlinear differential equations with constant coefficients under mild conditions on the nonlinear term. Also, we transfer the concepts of Ulam’s stability for ODEs to this type of equation and give the Ulam–Hyers and Ulam–Hyers–Rassias stability results on finite time and infinite time intervals.

1 Introduction

Over the years there were many developments on ODEs and PDEs involving global fractional derivatives of Caputo, Riemann–Liouville or Hadamard type; we refer the reader to the books [1,2,3] and the papers [4,5,6,7,8,9,10,11,12,13,14,15,16,17]. However, with fractional derivatives of Caputo, Riemann–Liouville or Hadamard type one has some complex rules (e.g., the chain rule), and a local fractional derivative involving a limit instead of a singular integral (in global fractional derivatives) called the conformable fractional derivative was proposed in [18,19,20], and basic properties of conformable fractional derivative were given in [19]. (They can also be used to extend Newton mechanics [21].) Recently, a new type of derivative which modifies conformable derivatives was introduced by [22]. In [23,24,25], the authors studied a regular fractional generalization of the Sturm–Liouville eigenvalue problem and discussed the variation of parameters method for conformable fractional differential equations, and also the authors considered Lyapunov-type inequalities for conformable and sequential conformable boundary value problems using higher-order conformable derivatives and integrals. Conformable fractional differential equations were initiated, and the fundamental theory on existence, Sturm’s theorems and stability can also be found in [26,27,28,29,30,31] and the references therein.

In this paper we study the existence of solutions and the Ulam–Hyers stability of the following conformable fractional differential equations with constant coefficients (here $\lambda \in \mathbb {R}{\setminus }\{0\}$) of the type:

$$\begin{aligned} \left\{ \begin{array}{l} \mathfrak {D}_{\beta }^{a}y(x)=\lambda y(x)+g(x,y(x)),~x\in (a,b] \text{ or } (a,\infty ),~0<\beta < 1,\\ y(a)=y_{a}, \end{array} \right. \end{aligned}$$

(1)

where $\mathfrak {D}_{\beta }^{a}y$ is called the conformable fractional derivative (CFD) with lower index a of the function y (which is given in Definition 2.1) and $g\in C([a,b]\times \mathbb {R},\mathbb {R})$. Note the existence of solutions, Sturm’s separation and comparison theorems for (1) with $\lambda =0$ were investigated in [26].

This paper is organized as follows: In Sect. 2, we recall the basic definitions of conformable fractional derivative and integral and some known results. In Sect. 3, we study the following linear Cauchy problem:

$$\begin{aligned} \left\{ \begin{array}{l} \mathfrak {D}_{\beta }^{a}y(x)=\lambda y(x)+f(x),~x\in (a,b],~0<\beta <1,~\lambda \ne 0,\\ y(a)=y_{a}, \end{array} \right. \end{aligned}$$

(2)

when $f\in C([a,b],\mathbb {R})$. With the help of the variation of constants method, we derive the explicit formula for solutions to (2). In Sect. 3, we present several results on the existence of solutions to (1). Section 4 is devoted to Ulam–Hyers and Ulam–Hyers–Rassias stability results for (1) on finite time and infinite time intervals. Examples are then given to illustrate our theory.

2 Preliminaries

Let $I\subseteq [a,\infty )$ and $C(I,\mathbb {R})$ be the Banach space of continuous function from I into $\mathbb {R}$ with the norm $\Vert y\Vert =\sup \limits _{x\in I}|y(x)|$ and $C^1(I,\mathbb {R})=\{y\in C(I,\mathbb {R}): y'\in C(I,\mathbb {R})\}$.

Definition 2.1

(see [18, Definition 2.1]) The CFD with lower index a of a function $y:[a,\infty )\rightarrow \mathbb {R}$ is defined as

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}y(x)= & {} \lim \limits _{\varepsilon \rightarrow 0} \frac{y(x+\varepsilon (x-a)^{1-\beta })-y(x)}{\varepsilon },~x>a,~0<\beta <1,\\ \mathfrak {D}_{\beta }^{a}y(a)= & {} \lim \limits _{x\rightarrow a^{+}}\mathfrak {D}_{\beta }^{a}y(x). \end{aligned}$$

Remark 2.2

If $\mathfrak {D}_{\beta }^{a}y(x_{0})$ exists and is finite, we say that y is $\beta$-differentiable at $x_{0}$.

If $y\in C^1([a,\infty ),\mathbb {R})$, then $\mathfrak {D}_{\beta }^{a}y(x)=(x-a)^{1-\beta }y'(x)$.

Definition 2.3

(see [18, Notation, p. 58]) The conformable fractional integral (CFI) with lower index a of a function $y:[a,\infty )\rightarrow \mathbb {R}$ is written as

$$\begin{aligned} \mathfrak {I}_{\beta }^{a}y(x)=\int _{a}^{x}y(t)d_{\beta }(t,a) =\int _{a}^{x}(t-a)^{\beta -1}y(t)\mathrm{d}t,~t\ge a,~0<\beta <1. \end{aligned}$$

If $a=0$, then we write $d_{\beta }(t,a)$ as $d_{\beta }(t)$.

Remark 2.4

CFI is a special case of the Riemann–Stieltjes integral $\int f(t) dg(t)$ in the simplest form, i.e., set $g(t)= \frac{t^{\beta }}{\beta }$.

Lemma 2.5

(see [18, Lemma 2.8]) For $y\in C^1([a,\infty ),\mathbb {R})$, one has

$$\begin{aligned} \mathfrak {I}_{\beta }^{a}\mathfrak {D}_{\beta }^{a}y(x)=y(x)-y(a),~x>a,~0<\beta <1. \end{aligned}$$

3 Representation of Solutions of the Linear Problem

In this section, we seek the explicit formula of solutions to (2).

Theorem 3.1

Define a fractional exponential function $Y(\cdot )=e^{\lambda \frac{(\cdot -a)^{\beta }}{\beta }}:[a,b]\rightarrow \mathbb {R}$. Then

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}Y(x)=\lambda Y(x). \end{aligned}$$

Proof

Using Remark 2.2, one has

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}Y(x)= & {} (x-a)^{1-\beta }\frac{\hbox {d}Y(x)}{\hbox {d}x}\\= & {} (x-a)^{1-\beta }\lambda (x-a)^{\beta -1}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\\= & {} \lambda e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\\= & {} \lambda Y(x). \end{aligned}$$

The proof is complete. $\square$

Remark 3.2

Obviously, $Y(\cdot )$ for $\mathfrak {D}_{\beta }^{a}y=\lambda y$ is an extension of the classical exponential function $e^{\lambda \cdot }$ for linear ODEs $y'=\lambda y$, which will play an important role in finding a general solution for (2) via Dehamel’s principle.

Theorem 3.3

A solution $y\in C([a,b],\mathbb {R})$ of (2) has the following form

$$\begin{aligned} y(x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} +\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}f(t)(t-a)^{\beta -1}\mathrm{d}t. \end{aligned}$$

(3)

Proof

We adopt the variation of constants method to give the representation of solutions to (2). Any solution of (2) should be of the form

$$\begin{aligned} y(x)=e^{\lambda \frac{(x-a)^{\beta }}{\beta }}c(x), \end{aligned}$$

(4)

where $c(\cdot )$ is an unknown continuously differentiable function.

From (4) via Remark 2.2, one has

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}y(x)= & {} \mathfrak {D}_{\beta }^{a} (e^{\lambda \frac{(x-a)^{\beta }}{\beta }}c(x))\\= & {} (x-a)^{1-\beta }e^{\lambda \frac{(x-a)^{\beta }}{\beta }} (x-a)^{\beta -1}c(x)+e^{\lambda \frac{(x-a)^{\beta }}{\beta }}(x-a)^{1-\beta }c'(x)\\= & {} \lambda c(x)e^{\lambda \frac{(x-a)^{\beta }}{\beta }} +e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\mathfrak {D}_{\beta }^{a}c(x)\\= & {} \lambda y(x)+e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\mathfrak {D}_{\beta }^{a}c(x). \end{aligned}$$

This yields that

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}c(x)=e^{-\lambda \frac{(x-a)^{\beta }}{\beta }}f(x). \end{aligned}$$

(5)

From (5) via Lemma 2.5, one has

$$\begin{aligned} c(x)=c(a)+\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}f(t)(t-a)^{\beta -1}\hbox {d}t, \end{aligned}$$

(6)

where $c(a)=y(a)=y_{a}$.

Use (6) in (4), and the proof is complete.$\square$

Remark 3.4

In [18, Example 5.4], the fractional Laplace transform (see [18, Theorem 5.1]) is used to verify the solution of $\mathfrak {D}_{\beta }^{a}y(x)=\lambda y(x)$ with $y(a)=c$ via the fractional exponential function $Y(\cdot )$. One can also apply the fractional Laplace transform to derive the general solution of (2), which is presented in [18, (59)].

4 Existence Results

In this section, we give several existence results for solutions to (1) under different conditions on $\lambda$ and g. First we give the concept of a solution to (1).

Definition 4.1

A function $y\in C^1([a,b],\mathbb {R})$ is called the solution of (1) if y satisfies $\mathfrak {D}_{\beta }^{a}y(x)=\lambda y(x)+g(x,y(x)),~x\in (a,b]$ and $y(a)=y_{a}$.

Following the procedure in Theorem 3.3, it is easy to verify that a function $y\in C([a,a+h],\mathbb {R})$ is the solution of (1) if and only if y is the solution of the integral equation

$$\begin{aligned} y(x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} +\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,y(t))(t-a)^{\beta -1}\mathrm{d}t,~x\in [a,a+h]. \end{aligned}$$

4.1 Case When $\lambda >0$

Let $\mathbb {B}=\{y\in C([a,b],\mathbb {R}):\Vert y-y_{a}e^{\lambda \frac{(\cdot -a)^{\beta }}{\beta }}\Vert \le c\}$ and $\mathbb {D}=\{(x,y):x\in [a,b],y\in \mathbb {B}\}$.

Consider the following assumptions:

$[H_{1}]$: Suppose that $g\in C([a,b],\mathbb {R})$.

$[H_{2}]$: There exists a constant $L>0$ such that $|g(x,y)-g(x,z)|\le L|y-z|$ for $x\in [a,b]$ and $y,z\in \mathbb {R}$.

Now we use the Picárd iterative approach to establish local existence of solutions to (1).

Theorem 4.2

Assume that $[H_{1}]$ and $[H_{2}]$ are satisfied. Then (1) has a unique solution $y\in C([a,a+h],\mathbb {R})$ (here $h=\min \{a,(\frac{\beta \ln \left( \frac{\lambda c}{M}+1\right) }{\lambda })^{^{\frac{1}{\beta }}}\}>0$ and $M=\max \limits _{(x,y)\in \mathbb {D}}|g(x,y(x))|$).

Proof

For any $x\in [a,a+h]$ and $n\in \mathbb {N}$, we let

$$\begin{aligned} \left\{ \begin{array}{l} y_{0}(x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }},\\ y_{n}(x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }} g(t,y_{n-1}(t))(t-a)^{\beta -1}\hbox {d}t. \end{array} \right. \end{aligned}$$

(7)

Note (7) is well defined because of $[H_{1}].$

Step 1. We prove that $y_n\in \mathbb {B},n\in \mathbb {N}$.

We use mathematical induction to establish this.

(i) For $n=1$ and $x\in [a,a+h]$, we have

$$\begin{aligned} \left| y_{1}(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right|= & {} \left| \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,y_{0}(t))(t-a)^{\beta -1}\hbox {d}t\right| \\\le & {} \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,y_{0}(t))|(t-a)^{\beta -1}\hbox {d}t\\\le & {} \frac{M}{\lambda }\left( e^{\lambda \frac{h^{\beta }}{\beta }}-1\right) \le c, \end{aligned}$$

which implies that $\Vert y_{1}-y_{a}e^{\lambda \frac{(\cdot -a)^{\beta }}{\beta }}\Vert \le c.$

(ii) For $n=k$ and $x\in [a,a+h]$, suppose the inequality $\Vert y_{k}-y_{a}e^{\lambda \frac{(\cdot -a)^{\beta }}{\beta }}\Vert \le c$ holds. Now for $n=k+1$ and $x\in [a,a+h]$,

$$\begin{aligned} \left| y_{k+1}(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right|\le & {} \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,y_{k}(t))|(t-a)^{\beta -1}\hbox {d}t\\\le & {} \frac{M}{\lambda }\left( e^{\lambda \frac{h^{\beta }}{\beta }}-1\right) \le c, \end{aligned}$$

which implies that $\Vert y_{k+1}-y_{a}e^{\lambda \frac{(\cdot -a)^{\beta }}{\beta }}\Vert \le c.$

Step 2. We prove $\{y_{n}(x)\}$ defined in (7) is uniformly convergent on $[a,a+h]$.

Consider

$$\begin{aligned} y_{0}(x)+\sum \limits _{k=1}^{\infty }(y_{k}(x)-y_{k-1}(x)),~~x\in [a,a+h], \end{aligned}$$

(8)

and set

$$\begin{aligned} y_{n}(x)=y_{0}(x)+\sum \limits _{k=1}^{n}(y_{k}(x)-y_{k-1}(x)),~~x\in [a,a+h]. \end{aligned}$$

In fact, we only need to prove the uniform convergence of (8) on $x\in [a,a+h]$. To help us note the following estimate via (7),

$$\begin{aligned} |y_{1}(x)-y_{0}(x)|\le & {} \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,y_{0}(t))|(t-a)^{\beta -1}\hbox {d}t\\\le & {} Me^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}(t-a)^{\beta -1}\hbox {d}t\\\le & {} \frac{M}{\beta }e^{\lambda \frac{(x-a)^{\beta }}{\beta }}(x-a)^{\beta }, \end{aligned}$$

and using $[H_{2}]$,

$$\begin{aligned} |y_{2}(x)-y_{1}(x)|\le & {} \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,y_{1}(t))-g(t,y_{0}(t))|(t-a)^{\beta -1}\hbox {d}t\\\le & {} Le^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|y_{1}(t)-y_{0}(t)|(t-a)^{\beta -1}\hbox {d}t\\< & {} \frac{LM}{\beta ^{2}2!}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}(x-a)^{2\beta }. \end{aligned}$$

For a positive integer n and $x\in [a,a+h]$, suppose the following inequality holds:

$$\begin{aligned} |y_{n}(x)-y_{n-1}(x)|\le \frac{ML^{n-1}}{\beta ^{n}n!}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}(x-a)^{n\beta }. \end{aligned}$$

Then, for $x\in [a,a+h]$, using $[H_{2}]$ again,

$$\begin{aligned} |y_{n+1}(x)-y_{n}(x)|\le & {} Le^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|y_{n}(t)-y_{n-1}(t)|(t-a)^{\beta -1}\hbox {d}t\\\le & {} \frac{ML^{n}}{\beta ^{n+1}(n+1)!}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}(x-a)^{(n+1)\beta }. \end{aligned}$$

Therefore, for all $k\in \mathbb {N}$, we have

$$\begin{aligned} |y_{k}(x)-y_{k-1}(x)|\le \frac{ML^{k-1}}{\beta ^{k}k!}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}(x-a)^{k\beta }\le \frac{ML^{k-1}}{\beta ^{k}k!}e^{\lambda \frac{h^{\beta }}{\beta }}h^{k\beta },~x\in [a,a+h]. \end{aligned}$$

Note from (8), we have

$$\begin{aligned} |y_{0}(x)|+\sum \limits _{k=1}^{\infty }|y_{k}(x)-y_{k-1}(x)|\le y_{a}e^{\lambda \frac{h^{\beta }}{\beta }} +\sum \limits _{k=1}^{\infty }\Psi _{k}, \end{aligned}$$

where $\Psi _{k}:=\frac{ML^{k-1}}{\beta ^{k}k!}e^{\lambda \frac{h^{\beta }}{\beta }}h^{k\beta }$. Now $\sum \limits _{k=1}^{\infty }\Psi _{k}$ is uniformly convergent since

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\Psi _{k+1}}{\Psi _{k}}=\lim _{k\rightarrow \infty }\frac{Lh^{\beta }}{\beta (k+1)}=0. \end{aligned}$$

Therefore, $\{y_{n}(x)\}$ is uniformly convergent on $[a,a+h]$. That is, there exists a $y\in C([a,b],\mathbb {R})$ such that $\{y_{n}(x)\}$ uniformly converges to y(x) on $[a,a+h]$.

Step 3. We prove that y(x) is a solution of (1).

Note [$H_2$] and that $\{y_{n}(x)\}$ uniformly converges to y(x) on $[a,a+h]$, so $\{g(x,y_{n}(x)\}$ uniformly converges to a continuous function g(x, y(x)) on $[a,a+h]$.

For $x\in [a,a+h]$, we have

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }y_{n}(x)= & {} y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} +\lim \limits _{n\rightarrow \infty }\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,y_{n-1}(t))(t-a)^{\beta -1}\hbox {d}t\\= & {} y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} +\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(t-a)^{\beta -1}\lim \limits _{n\rightarrow \infty }g(t,y_{n-1}(t))\hbox {d}t\\= & {} y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,y(t))(t-a)^{\beta -1}\hbox {d}t. \end{aligned}$$

Step 4. Uniqueness of solutions.

Suppose $\phi (x)$ is another solution of (1). That is, for $x\in [a,a+h]$,

$$\begin{aligned} \phi (x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,\phi (x))(t-a)^{\beta -1}\hbox {d}t. \end{aligned}$$

Using [$H_2$], one has

$$\begin{aligned} |y(x)-\phi (x)|\le & {} Le^{\lambda \frac{h^{\beta }}{\beta }}\int _{a}^{x}(t-a)^{\beta -1}|y(t)-\phi (x)|\hbox {d}t. \end{aligned}$$

Apply the generalized Gronwall’s inequality in [18, Theorem 2.13], and we have

$$\begin{aligned} |y(x)-\phi (x)|\le 0 \Longrightarrow y(x)=\phi (x),~~~x\in [a,a+h]. \end{aligned}$$

The proof is complete.$\square$

Next, we use Schauder’s fixed point theorem to establish an existence theorem.

$[H_{3}]:$ Suppose $\rho =\frac{L}{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) <1$.

Let $B_{r}:=\{y\in C([a,b],\mathbb {R}):\Vert y\Vert \le r,~r\ge \frac{\omega }{1-\rho }\}$, where $\omega =|y_{a}|e^{\lambda \frac{(b-a)^{\beta }}{\beta }} +\frac{1}{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) \Vert \widetilde{g}\Vert$ and $\widetilde{g}(\cdot )=g(\cdot ,0)$.

Theorem 4.3

Assume that $[H_{1}]$ and $[H_{3}]$ are satisfied. Then (1) has at least one solution $y\in C([a,b],\mathbb {R})$.

Proof

Define an operator T on $B_{r}$ as follows:

$$\begin{aligned} (Ty)(x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,y(t))(t-a)^{\beta -1}\hbox {d}t. \end{aligned}$$

(9)

Step 1. We show that $T(B_{r})\subset B_{r}$.

For any $x\in [a,b]$ and $y\in B_{r}$ we have

$$\begin{aligned} |(Ty)(x)|\le & {} |y_{a}|e^{\lambda \frac{(b-a)^{\beta }}{\beta }} +Le^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|y(t)|(t-a)^{\beta -1}\hbox {d}t\\&+\,e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,0)|(t-a)^{\beta -1}\hbox {d}t\\\le & {} |y_{a}|e^{\lambda \frac{(b-a)^{\beta }}{\beta }}+\frac{L}{\lambda } \left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) \Vert y\Vert +\frac{1}{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) \Vert \widetilde{g}\Vert , \end{aligned}$$

which implies that

$$\begin{aligned} \Vert Ty\Vert\le & {} \frac{L}{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) r+|y_{a}|e^{\lambda \frac{(b-a)^{\beta }}{\beta }} +\frac{1}{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) \Vert \widetilde{g}\Vert \\\le & {} \rho r+\omega \le r. \end{aligned}$$

Step 2. We prove T is continuous.

Let $\{y_{n}\}$ be a sequence with $y_{n}\rightarrow y$ in $B_{r}$. For each $x\in [a,b]$,

$$\begin{aligned} |(Ty_{n})(x)-(Ty)(x)|\le & {} Le^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}(t-a)^{\beta -1}\mathrm{d}t\Vert y_{n}-y\Vert \\\le & {} \frac{L}{\beta }e^{\lambda \frac{(b-a)^{\beta }}{\beta }}(b-a)^{\beta }\Vert y_{n}-y\Vert , \end{aligned}$$

so

$$\begin{aligned} \Vert Ty_{n}-Ty\Vert \le \frac{L}{\beta }e^{\lambda \frac{(b-a)^{\beta }}{\beta }}(b-a)^{\beta }\Vert y_{n}-y\Vert . \end{aligned}$$

This implies that T is continuous.

Step 3. We show $T(B_{r})$ is equicontinuous.

Let $x_{1},x_{2}\in [a,b]$. Without loss of generality, let $x_{1}<x_{2}$ and $x_{2}-x_{1}<\delta$. For a given $\varepsilon >0$ and $y\in B_{r}$, we have

$$\begin{aligned}&|Ty(x_{2})-Ty(x_{1})|\\&\quad \le \left( |y_{a}|+\int _{a}^{x_{1}} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,y(t))-g(t,0)|(t-a)^{\beta -1}\hbox {d}t\right) \\&\qquad \left( e^{\lambda \frac{(x_{2}-a)^{\beta }}{\beta }}-e^{\lambda \frac{(x_{1}-a)^{\beta }}{\beta }}\right) \\&\qquad +\int _{a}^{x_{1}}\left( e^{\lambda \frac{(x_{2}-a)^{\beta }}{\beta }}-e^{\lambda \frac{(x_{1}-a)^{\beta }}{\beta }}\right) e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,0)|(t-a)^{\beta -1}\hbox {d}t\\&\qquad +\int _{x_{1}}^{x_{2}}e^{\lambda \frac{(x_{2}-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,y(t))|(t-a)^{\beta -1}\hbox {d}t\\&\quad \le \lambda \left( |y_{a}|+\frac{rL}{\beta }(x_{1}-a)^{\beta }+\frac{\Vert \widetilde{g}\Vert }{\beta }(x_{1}-a)^{\beta }\right) (\xi -a)^{\beta -1}e^{\lambda \frac{(\xi -a)^{\beta }}{\beta }}(x_{2}-x_{1})\\&\qquad +\frac{1}{\beta } (rL+\Vert \widetilde{g}\Vert )e^{\lambda \frac{(b-a)^{\beta }}{\beta }}((x_{2}-a)^{\beta }-(x_{1}-a)^{\beta })\\&\quad \le \lambda \left( |y_{a}|+\frac{rL}{\beta }(x_{1}-a)^{\beta }+\frac{\Vert \widetilde{g}\Vert }{\beta }(x_{1}-a)^{\beta }\right) (\xi -a)^{\beta -1}e^{\lambda \frac{(\xi -a)^{\beta }}{\beta }}(x_{2}-x_{1})\\&\qquad +\frac{1}{\beta }(rL+\Vert \widetilde{g}\Vert )e^{\lambda \frac{(x_{2}-a)^{\beta }}{\beta }}(x_{2}-x_{1})^{\beta },~\xi \in (x_{1},x_{2}), \end{aligned}$$

which tends to zero as $\delta \rightarrow 0$. Thus, $T(B_{r})$ is equicontinuous.

From the Arzela–Ascoli theorem, $T(B_{r})$ is relatively compact. Schauder’s fixed point theorem guarantees that T has a fixed point in $B_{r}$, and we are finished.$\square$

4.2 Case When $\lambda <0$

Now we use the contraction mapping principle to establish an existence and uniqueness result.

Theorem 4.4

Assume that $[H_{1}]$ and $[H_{2}]$ are satisfied. Then (1) has a unique solution $y\in C([a,a+h_1],\mathbb {R})$ (here $h_1>0$ is a sufficiently small number).

Proof

Consider the operator T defined in (9) on $C([a,a+h_1],\mathbb {R})$. For $y,z\in C([a,a+h_1],\mathbb {R})$ and each $x\in [a,a+h_1]$, one has

$$\begin{aligned} |(T y)(x)-(T z)(x)| \le \widetilde{\rho }\Vert y-z\Vert , \end{aligned}$$

where $\widetilde{\rho }=\frac{L}{-\lambda }\left( 1-e^{\lambda \frac{h_1^{\beta }}{\beta }}\right)$. This implies that $\Vert T y-T z\Vert \le \widetilde{\rho }\Vert y-z\Vert$. Now choose a sufficiently small $h_1>0$ so $\widetilde{\rho }<1$. Now one can apply the contraction mapping principle to obtain the result.$\square$

Next, we use Krasnoselskii fixed point theorem to establish an existence result. Consider the following assumptions:

$[H_{4}]:$ There exist positive constants $\widetilde{L}$ and G such that $|g(x,y)|\le \widetilde{L}|y|+G$ for any $x\in [a,b]$ and $y\in \mathbb {R}$.

$[H_{5}]:$ Set $\overline{\rho }=\frac{\widetilde{L}}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right) <1$.

Let $B_{\widetilde{r}}:=\{y\in C([a,b],\mathbb {R}):\Vert y\Vert \le \widetilde{r}~\text{ and }~\widetilde{r}\ge \frac{\widetilde{\omega }}{1-\overline{\rho }}\}$, $\widetilde{\omega }=|y_{a}| +\frac{G}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right)$.

Theorem 4.5

Assume that $[H_{1}]$, $[H_{4}]$ and $[H_{5}]$ are satisfied. Then (1) has at least one solution $y\in C([a,b],\mathbb {R})$.

Proof

Consider the operator T defined in (9) on $B_{\widetilde{r}}$, and divide it into two operators A and P on $B_{\widetilde{r}}$ as follows:

$$\begin{aligned} (A y)(x)= & {} y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }},\\ (P y)(x)= & {} \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,y(t))(t-a)^{\beta -1}\mathrm{d}t. \end{aligned}$$

Step 1. We show that $(A+P)(B_{\widetilde{r}})\subset B_{\widetilde{r}}$.

For any $x\in [a,b]$ and $y\in B_{\widetilde{r}}$ we have

$$\begin{aligned}&|(Ay)(x)+(Py)(x)|\\&\quad \le |y_{a}|+e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(\widetilde{L}|y(t)|+G)(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le |y_{a}|+\widetilde{L}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} \int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|y(t)|(t-a)^{\beta -1}\mathrm{d}t\\&\qquad +Ge^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le |y_{a}|+\frac{\widetilde{L}}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right) \Vert y\Vert +\frac{G}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right) , \end{aligned}$$

which implies that

$$\begin{aligned} \Vert Ay+Py\Vert\le & {} \frac{\widetilde{L}}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right) \widetilde{r}+|y_{a}| +\frac{G}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right) \\\le & {} \overline{\rho } r+\widetilde{\omega }\le \widetilde{r}. \end{aligned}$$

Step 2. Note $A: B_{\widetilde{r}} \rightarrow B_{\widetilde{r}}$ is a contraction mapping.

Step 3. The operator P is continuous.

Let $\{y_{n}\}$ be a sequence with $y_{n}\rightarrow y$ in $B_{\widetilde{r}}$. Denote $g_{n}(\cdot )=g(\cdot ,y_{n})$ and $g(\cdot )=g(\cdot ,y)$. Note $e^{-\lambda \frac{(\cdot -a)^{\beta }}{\beta }}(\cdot -a)^{\beta -1}g_{n}\rightarrow e^{-\lambda \frac{(\cdot -a)^{\beta }}{\beta }}(\cdot -a)^{\beta -1}g$ as $n\rightarrow \infty$ and $|g_{n}-g|\le 2(L\widetilde{r}+G)$ and $e^{-\lambda \frac{(\cdot -a)^{\beta }}{\beta }}(\cdot -a)^{\beta -1}2(L\widetilde{r}+G):[a,b]\rightarrow \mathbb {R}$ is integrable.

Use the Lebesgue dominated convergence theorem, and we have that

$$\begin{aligned} |(Py_{n})(x)-(Py)(x)|\le & {} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g_n(t)-g(t)|(t-a)^{\beta -1}\mathrm{d}t,~x\in [a,b] \end{aligned}$$

tends to zero as $n\rightarrow \infty$. Thus, P is continuous.

Step 4. We show $P(B_{\widetilde{r}})$ is equicontinuous.

Let $x_{1},x_{2}\in [a,b]$. Without loss of generality, let $x_{1}<x_{2}$ and $x_{2}-x_{1}<\delta$ for some $\delta >0$. For a given $\varepsilon >0$ and $y\in B_{\widetilde{r}}$, we have

$$\begin{aligned}&|Py(x_{2})-Py(x_{1})|\\&\quad \le |\lambda |(\xi -a)^{\beta -1}(x_{2}-x_{1})\int _{a}^{x_{1}} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(\widetilde{L}|y(t)|+G)(t-a)^{\beta -1}\mathrm{d}t\\&\qquad +\int _{x_{1}}^{x_{2}}e^{\lambda \frac{(x_{2}-a)^{\beta }}{\beta }} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(\widetilde{L}|y(t)|+G)(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le \frac{|\lambda |}{\beta }(\xi -a)^{\beta -1}(\widetilde{L}\widetilde{r}+G) e^{-\lambda \frac{(x_{1}-a)^{\beta }}{\beta }}(x_{1}-a)^{\beta }(x_{2}-x_{1})\\&\qquad +\frac{1}{\beta }(\widetilde{L}\widetilde{r}+G)e^{-\lambda \frac{(x_{2}-a)^{\beta }}{\beta }}(x_{2}-x_{1})^{\beta },~\xi \in (x_{1},x_{2}), \end{aligned}$$

which tends to zero as $\delta \rightarrow 0$. Thus, $P(B_{\widetilde{r}})$ is equicontinuous.

From the Arzela–Ascoli theorem, $P(B_{\widetilde{r}})$ is relatively compact. Finally, the Krasnoselskii fixed point theorem completes the proof.$\square$

Next we consider the existence of a solution on $[a,\infty )$. Consider the following assumptions:

$[H'_{2}]:$ Suppose that $g\in C([a,\infty ),\mathbb {R})$. There exists a constant $\widetilde{L}>0$ such that $|g(x,y)-g(x,z)|\le \widetilde{L}|y-z|$ for $x\in [a,\infty )$ and $y,z\in \mathbb {R}$.

$[H'_{4}]:$ There exist positive constants $\overline{L}$ and $\overline{G}$ such that $|g(x,y)|\le \overline{L}|y|+\overline{G}$ for any $x\in [a,\infty )$ and $y\in \mathbb {R}$.

Theorem 4.6

Assume that $[H'_{2}]$ and $[H'_{4}]$ and $\lambda >0$ are satisfied. Then (1) has a unique solution $y\in C([a,\infty ),\mathbb {R})$.

Proof

From Theorem 4.2, (1) has a unique solution in $C([a,a+h],\mathbb {R})$. Assume that the solution y(x) admits a maximal existence interval, denoted by $[a,b)\subset [a,\infty )$ and $\lim \limits _{x\rightarrow b}|y(x)|=\infty$. In fact, for all $x\in [a,b)$, we have

$$\begin{aligned} |y(x)|\le & {} |y_{a}|e^{\lambda }\frac{(b-a)^{\beta }}{\beta }+\overline{L} \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|y(t)|(t-a)^{\beta -1}\mathrm{d}t\\&+\,\overline{G}\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(t-a)^{\beta -1}\mathrm{d}t\\\le & {} |y_{a}|e^{\lambda }\frac{(b-a)^{\beta }}{\beta }+\frac{\overline{G}}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right) \\&+\,\overline{L}e^{\lambda }\frac{(b-a)^{\beta }}{\beta } \int _{a}^{x}(t-a)^{\beta -1}|y(t)|\mathrm{d}t. \end{aligned}$$

Let $l_{1}=|y_{a}|e^{\lambda }\frac{(b-a)^{\beta }}{\beta }+\frac{\overline{G}}{-\lambda }\left( 1-e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\right)$ and $l_{2}=\overline{L}e^{\lambda }\frac{(b-a)^{\beta }}{\beta }$. Therefore,

$$\begin{aligned} |y(x)|\le l_{1}+l_{2}\int _{a}^{x}(t-a)^{\beta -1}|y(t)|\mathrm{d}t. \end{aligned}$$

Apply the generalized Gronwall’s inequality in [18, Theorem 2.13], and we have

$$\begin{aligned} |y(x)| \le l_{1}e^{l_{2}\frac{(b-a)^{\beta }}{\beta }}:=l_{3}<\infty . \end{aligned}$$

This implies $\Vert y\Vert <l$ on [a, b) when $l=l_{1}+l_{3}$. This contradicts the assumption that [a, b) is the maximal existence interval. The proof is complete.$\square$

Remark 4.7

Assume that $[H'_{2}]$ and $[H'_{4}]$ and $\lambda <0$ are satisfied. Then (1) has a unique solution $y\in C([a,\infty ),\mathbb {R})$.

5 Ulam–Hyers and Ulam–Hyers–Rassias Stability Results

In this section, we discuss Ulam–Hyers stability and Ulam–Hyers–Rassias stability of (1) on finite time and infinite time intervals.

Let $J:=[a,b]$ or $[a,\infty )$$\varepsilon >0$ and $\varphi \in C(J,\mathbb {R})$. Consider (1) and

$$\begin{aligned} |\mathfrak {D}_{\beta }^{a}z(x)-\lambda z(x)-g(x,z(x))|\le \varepsilon ,~0<\beta <1,~x\in J, \end{aligned}$$

(10)

and

$$\begin{aligned} |\mathfrak {D}_{\beta }^{a}z(x)-\lambda z(x)-g(x,z(x))|\le \varepsilon \varphi (x),~0<\beta <1,~x\in J. \end{aligned}$$

(11)

Definition 5.1

(1) is called Ulam–Hyers stable if there exists a constant $N>0$ such that for each $\varepsilon >0$ and for each solution $z\in C(J,\mathbb {R})$ of (10) there exists a solution $y\in C(J,\mathbb {R})$ of (1) with

$$\begin{aligned} |z(x)-y(x)|\le N\varepsilon ,~x\in J. \end{aligned}$$

Remark 5.2

A function $z\in C(J,\mathbb {R})$ is a solution of (10) if and only if there exists a function $h\in C(J,\mathbb {R})$ such that $(i)~|h(x)|\le \varepsilon ,x\in J$, (ii) $\mathfrak {D}_{\beta }^{a}z(x)=\lambda z(x)+g(x,z(x))+h(x),x\in J$.

Definition 5.3

(1) is called Ulam–Hyers–Rassias stable if there exists a constant $\widetilde{N}>0$ such that for each $\varepsilon >0$ and for each solution $z\in C(J,\mathbb {R})$ of (11) there exists a solution $y\in C(J,\mathbb {R})$ of (1) with

$$\begin{aligned} |z(x)-y(x)|\le \widetilde{N}\varepsilon \varphi (x),~x\in J. \end{aligned}$$

Remark 5.4

A function $z\in C(J,\mathbb {R})$ is a solution of inequality (11) if and only if there exists a function $\tilde{h}\in C(J,\mathbb {R})$ such that $(i)~|\tilde{h}(x)|\le \varepsilon \varphi (x),x\in J$, $(ii)~\mathfrak {D}_{\beta }^{a}z(x)=\lambda z(x)+g(x,z(x))+\tilde{h}(x),x\in J$.

5.1 Case When $\lambda >0$

In this section, we discuss Ulam–Hyers stability and Ulam–Hyers–Rassias stability of (1) when $\lambda > 0$ on the finite time interval [a, b].

Lemma 5.5

Let $\lambda >0$ and let $z\in C([a,b],\mathbb {R})$ be a solution of (10). Then z is a solution of the inequality

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad \le \frac{\varepsilon }{\lambda }\left( e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-1\right) , \end{aligned}$$

where $y_{a}=y(a)$ and $x\in [a,b]$.

Proof

From Remark 5.2, the solution of the equation

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}z(x)=\lambda z(x)+g(x,z(x))+h(x),~x\in [a,b] \end{aligned}$$

can be written as

$$\begin{aligned} z(x)= & {} y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\\&+\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}h(t)(t-a)^{\beta -1}\mathrm{d}t. \end{aligned}$$

For all $x\in [a,b]$, one has

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad = \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|h(t)|(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le \frac{\varepsilon }{\lambda }\left( e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-1\right) , \end{aligned}$$

and we are finished.$\square$

Lemma 5.6

Let $\lambda >0$, and let $z\in C([a,b],\mathbb {R})$ be a solution of (11). Then z is a solution of the inequality:

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\\le & {} \varepsilon e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t, \end{aligned}$$

where $y_{a}=y(a)$ and $x\in [a,b]$.

Proof

From Remark 5.4, the solution of the equation

$$\begin{aligned} \mathfrak {D}_{\beta }^{a}z(x)=\lambda z(x)+g(x,z(x))+\tilde{h}(x),~x\in [a,b] \end{aligned}$$

can be written as

$$\begin{aligned} z(x)= & {} y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }} g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\\&+\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\tilde{h}(t)(t-a)^{\beta -1}\mathrm{d}t. \end{aligned}$$

For all $x\in [a,b]$, one has

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad = \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|\tilde{h}(t)|(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le \varepsilon e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t, \end{aligned}$$

and we are finished.$\square$

Theorem 5.7

Assume that $[H_{1}]$, $[H_{2}]$ and $[H_{3}]$ are satisfied. Then (1) with $\lambda >0$ is Ulam–Hyers stable on [a, b].

Proof

Let $z\in C([a,b],\mathbb {R})$ be a solution of (10). Let (see Theorem 4.3 and step 4 in the proof of Theorem 4.2) y be the unique solution of (1) with $\lambda >0$, that is,

$$\begin{aligned} y(x)=y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}+\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }} g(t,y(t))(t-a)^{\beta -1}\mathrm{d}t. \end{aligned}$$

From Lemma 5.5, we have

$$\begin{aligned} |z(x)-y(x)|\le & {} \left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} -\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&+\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|g(t,z(t))-g(t,y(t))|(t-a)^{\beta -1}\mathrm{d}t\\\le & {} \frac{\varepsilon }{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}{-}1\right) {+}L\int _{a}^{x}e^{\lambda \frac{(x{-}a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|z(t)-y(t)|(t-a)^{\beta -1}\mathrm{d}t\\\le & {} \frac{\varepsilon }{\lambda }\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) +\rho \Vert y-z\Vert , \end{aligned}$$

so

$$\begin{aligned} \Vert z-y\Vert \le \frac{\varepsilon }{\lambda (1-\rho )}\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) . \end{aligned}$$

Thus, we have

$$\begin{aligned} |z(x)-y(x)|\le N\varepsilon ,~N=\frac{1}{\lambda (1-\rho )}\left( e^{\lambda \frac{(b-a)^{\beta }}{\beta }}-1\right) . \end{aligned}$$

The proof is complete.$\square$

Next, we consider the following assumption:

$[H_{6}]:$ There exists a function $\varphi (\cdot )\in C([a,b],\mathbb {R})$ such that

$$\begin{aligned} \int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t\le \widetilde{C}\varphi (x),~\widetilde{C}>0. \end{aligned}$$

Theorem 5.8

Assume that $[H_{1}]$, $[H_{3}]$ and $[H_{6}]$ are satisfied. Then (1) with $\lambda >0$ is Ulam–Hyers–Rassias stable on [a, b].

Proof

Let $z\in C([a,b],\mathbb {R})$ be a solution of inequality (11), and let y be a solution (see Theorem 4.3) of (1). From Lemma 5.6, we have

$$\begin{aligned} |z(x)-y(x)|\le & {} \varepsilon e^{\lambda \frac{(b-a)^{\beta }}{\beta }} \widetilde{C}\varphi (x)+L\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|z(t)-y(t)|(t-a)^{\beta -1}\mathrm{d}t\\\le & {} \varepsilon e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\widetilde{C}\varphi (x)+\rho \Vert y-z\Vert , \end{aligned}$$

so

$$\begin{aligned} \Vert z-y\Vert \le \frac{\varepsilon \widetilde{C}}{1-\rho }e^{\lambda \frac{(b-a)^{\beta }}{\beta }}\varphi (x). \end{aligned}$$

Thus, we have

$$\begin{aligned} |z(x)-y(x)|\le \widetilde{N}\varepsilon \varphi (x),~\widetilde{N}=\frac{\widetilde{C}}{1-\rho }e^{\lambda \frac{(b-a)^{\beta }}{\beta }}. \end{aligned}$$

The proof is complete.$\square$

5.2 Case When $\lambda <0$

In this part, we discuss Ulam–Hyers stability and Ulam–Hyers–Rassias stability of (1) when $\lambda <0$ on the interval $[a,\infty )$.

Lemma 5.9

Let $\lambda <0$, and let $z\in C([a,\infty ),\mathbb {R})$ be a solution of (10). Then z is a solution of the inequality:

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad \le \frac{\varepsilon }{-\lambda }\left( 1-e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right) , \end{aligned}$$

where $y_{a}=y(a)$ and $x\in [a,\infty )$.

Proof

The proof is similar to that in Lemma 5.5, and from Remark 5.2, for all $x\in [a,\infty )$, one has

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }} g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad = \varepsilon e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x} e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le \frac{\varepsilon }{-\lambda }\left( 1-e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right) . \end{aligned}$$

The proof is complete.$\square$

Lemma 5.10

Let $\lambda <0$, and let $z\in C([a,\infty ),\mathbb {R})$ be a solution of (11). Then z is a solution of the inequality:

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad \le \varepsilon e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t, \end{aligned}$$

where $y_{a}=y(a)$ and $x\in [a,\infty )$.

Proof

The proof is similar to the proof of Lemma 5.6 and from Remark 5.4, for all $x\in [a,\infty )$, one has

$$\begin{aligned}&\left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}-\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&\quad = \int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|\widehat{h}(t)|(t-a)^{\beta -1}\mathrm{d}t\\&\quad \le \varepsilon e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t. \end{aligned}$$

The proof is complete.$\square$

Consider the following assumptions:

$[H_{7}]:$ Let $\mu =\frac{L}{-\lambda }<1$.

$[H_{8}]:$ There exists a function $\varphi (\cdot )\in C([a,\infty ),\mathbb {R})$ such that

$$\begin{aligned} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t\le \overline{C} \varphi (x),~\overline{C}>0. \end{aligned}$$

Theorem 5.11

Assume that $[H'_{2}]$, $[H'_{4}]$ and $[H_{7}]$ are satisfied. Then (1) with $\lambda <0$ is Ulam–Hyers stable on $[a,\infty )$.

Proof

Let $z\in C([a,\infty ),\mathbb {R})$ be a solution of (10), and let $y\in C([a,\infty ),\mathbb {R})$ be the unique solution (see Remark 4.7) of (1) on $[a,\infty )$. From Lemma 5.9, we have

$$\begin{aligned} |z(x)-y(x)|\le & {} \left| z(x)-y_{a}e^{\lambda \frac{(x-a)^{\beta }}{\beta }} -\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}g(t,z(t))(t-a)^{\beta -1}\mathrm{d}t\right| \\&+\,L\int _{a}^{x}e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|z(t)-y(t)|(t-a)^{\beta -1}\mathrm{d}t\\\le & {} \frac{\varepsilon }{-\lambda }\left( 1-e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right) +\frac{L}{-\lambda }\left( 1-e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right) \Vert y-z\Vert \\\le & {} \frac{\varepsilon }{-\lambda }+\frac{L}{-\lambda }\Vert y-z\Vert , \end{aligned}$$

so

$$\begin{aligned} \Vert z-y\Vert \le \frac{\varepsilon }{-\lambda (1-\mu )} =\frac{\varepsilon }{-\lambda -L}:=\overline{N}. \end{aligned}$$

Thus, we have

$$\begin{aligned} |z(x)-y(x)|\le \overline{N}\varepsilon . \end{aligned}$$

The proof is complete.$\square$

Theorem 5.12

Assume that $[H'_{2}]$,$[H'_{4}]$, $[H_{7}]$ and $[H_{8}]$ are satisfied. Then (1) with $\lambda <0$ is Ulam–Hyers–Rassias stable on $[a,\infty )$.

Proof

Let $z\in C([a,\infty ),\mathbb {R})$ be a solution of (11), and let $y\in C([a,\infty ),\mathbb {R})$ be the unique solution of (1) on $[a,\infty )$. From Lemma 5.10, we have

$$\begin{aligned} |z(x)-y(x)|\le & {} \varepsilon \overline{C}\varphi (x)+L\int _{a}^{x} e^{\lambda \frac{(x-a)^{\beta }}{\beta }}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}|z(t)-y(t)|(t-a)^{\beta -1}\mathrm{d}t\\\le & {} \varepsilon \overline{C}\varphi (x)+\frac{L}{-\lambda }\left( 1-e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\right) \Vert y-z\Vert \le \varepsilon \overline{C}\varphi (x)+\frac{L}{-\lambda }\Vert y-z\Vert , \end{aligned}$$

so

$$\begin{aligned} \Vert z-y\Vert \le \frac{\varepsilon \overline{C}}{1-\mu }\varphi (x). \end{aligned}$$

Thus, we have

$$\begin{aligned} |z(x)-y(x)|\le \widehat{N}\varepsilon \varphi (x),~\widehat{N}=\frac{\overline{C}}{1-\mu }. \end{aligned}$$

The proof is complete.$\square$

6 Examples

In this section, examples are given to illustrate our theoretical results.

Example 6.1

Let $\beta =\frac{1}{2}$, $\lambda =1$ and $y_{1}=1$. Consider

$$\begin{aligned} \left\{ \begin{array}{l} \mathfrak {D}_{\frac{1}{2}}^{1}y(x)=y(x)+\frac{x}{4+4x^{2}}\sin (y(x)),~x\in (1,2],\\ y(1)=1, \end{array} \right. \end{aligned}$$

(12)

and

$$\begin{aligned} \left| \mathfrak {D}_{\frac{1}{2}}^{1}y(x)-y(x)-\frac{x}{4+4x^{2}}\sin (y(x)) \right| \le \varepsilon ,~x\in (1,2], \end{aligned}$$

(13)

and

$$\begin{aligned} \left| \mathfrak {D}_{\frac{1}{2}}^{1}y(x)-y(x)-\frac{x}{4+4x^{2}} \sin (y(x))\right| \le \varepsilon \varphi (x),~x\in (1,2]. \end{aligned}$$

(14)

Let $g(x,y(x))=\frac{x}{4+4x^{2}}\sin (y(x))$ for arbitrary $x\in [a,b]$. Then

$$\begin{aligned} |g(x,y(x))-g(x,z(x))|\le \frac{1}{8}|y(x)-z(x)|. \end{aligned}$$

Obviously, $[H_{1}]$ and $[H_{2}]$ are satisfied. Note $M=L=\frac{1}{8}$, $\rho =\frac{1}{8}(e^{2}-1)=0.7986$, $N=\frac{1}{1-0.7986}(e^{2}-1)=31.7232$, $\widetilde{C}=2$ and $\widetilde{N}=\frac{2}{1-0.7986}e^{2}$$=73.3769$. Choose $c=\frac{1}{8}(e^{2}-1)=0.7986$ and $h=\min \{1,(\frac{1}{2}\ln (8c+1))^{2}\}=1$.

Now, the conditions in Theorems 4.2 and 4.3 are fulfilled. Thus, (12) admits a solution $y\in C([1,2],\mathbb {R})$. The solution of (12) is given by

$$\begin{aligned} y(x) =e^{2(x-1)^{1/2}}+\int _{1}^{x}e^{2(x-1)^{1/2}}e^{-2(t-1)^{1/2}} (t-1)^{-1/2}\frac{t}{4+4t^{2}}\sin (y(t))\mathrm{d}t,~x\in [1,2]. \end{aligned}$$

$\bullet$ Let $z\in C([1,2],\mathbb {R})$ be a solution of inequality (13). Then for every $x\in [1,2]$, there exists a function $h(x)=\varepsilon e^{-x}\in C([1,2],\mathbb {R})$ such that $|h(x)|\le \varepsilon$ and $\mathfrak {D}_{\frac{1}{2}}^{1}y(x)=y(x)+\frac{x}{4+4x^{2}}\sin (y(x))+h(x)$. Now $\rho =0.7986<1$, so $[H_{3}]$ holds. Choose $N=31.7232$. From Theorem 5.7, we have

$$\begin{aligned} |z(x)-y(x)|\le N\varepsilon . \end{aligned}$$

Thus, Eq. (12) is Ulam–Hyers stable on [1, 2] with $N=31.7232$.

$\bullet$ Let $z\in C([1,2],\mathbb {R})$ be a solution of inequality (14). Then for every $x\in [1,2]$, there exists a function $\widetilde{h}(x)=\frac{\varepsilon }{x} e^{2(x-1)^{1/2}}\in C([1,2],\mathbb {R})$ such that $|\widetilde{h}(x)|\le \varepsilon e^{2(x-1)^{1/2}}:=\varepsilon \varphi (x)$. Moreover,

$$\begin{aligned} \int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t= & {} \int _{1}^{x}e^{-2(t-1)^{1/2}}e^{2(t-1)^{1/2}}(t-1)^{-1/2}\mathrm{d}t\\\le & {} 2e^{2(x-1)^{1/2}}, \end{aligned}$$

so $[H_{6}]$ holds. Choose $\widetilde{N}=73.3769$. From Theorem 5.8, we have

$$\begin{aligned} |z(x)-y(x)|\le \widetilde{N}\varepsilon \varphi (x). \end{aligned}$$

Thus, Eq. (12) is Ulam–Hyers–Rassias stable on [1, 2] with $\widetilde{N}=73.3769$.

Example 6.2

Let $\beta =\frac{1}{2}$, $\lambda =-1$ and $y_{1}=1$. Consider

$$\begin{aligned} \left\{ \begin{array}{l} \mathfrak {D}_{\frac{1}{2}}^{1}y(x)=-y(x)+\frac{x}{1+x^{2}}\sin (y(x)),~x\in (1,\infty ),\\ y(1)=1. \end{array} \right. \end{aligned}$$

(15)

and

$$\begin{aligned} \left| \mathfrak {D}_{\frac{1}{2}}^{1}y(x)+y(x)-\frac{x}{1+x^{2}}\sin (y(x)) \right| \le \varepsilon ,~x\in (1,\infty ), \end{aligned}$$

(16)

and

$$\begin{aligned} \left| \mathfrak {D}_{\frac{1}{2}}^{1}y(x)+y(x)-\frac{x}{1+x^{2}}\sin (y(x)) \right| \le \varepsilon \varphi (x),~x\in (1,\infty ). \end{aligned}$$

(17)

Let $g(x,y(x))=\frac{x}{1+x^{2}}\sin (y(x))$ for arbitrary $x\in [1,\infty )$. Note

$$\begin{aligned} |g(x,y(x))-g(x,z(x))|\le & {} \frac{1}{2}|y(x)-z(x)|,\\ |g(x,y(x))|\le & {} \frac{1}{2}|y(x)|. \end{aligned}$$

Let $L=\widetilde{L}=\frac{1}{2}$ and $G=0$. Note $\mu =\frac{1}{2}$, $\overline{C}=2$, $\overline{N}=2$, $\widehat{N}=4$, $l_{1}=1$, $l_{2}=\frac{1}{2}$ and $l=l_{1}+l_{2}=1.5$. From Remark 4.7, (15) admits a solution $y\in C([1,\infty ),\mathbb {R})$. Then, the solution of (15) satisfies the following integral equations:

$$\begin{aligned} y(x)=e^{-2(x-1)^{1/2}}+\int _{1}^{x}e^{-2(x-1)^{1/2}}e^{2(t-1)^{1/2}} (t-1)^{-1/2} \frac{t}{1+t^{2}}\sin (y(t))\mathrm{d}t,~x\in [1,\infty ). \end{aligned}$$

$\bullet$ Let $z\in C([1,\infty ),\mathbb {R})$ be a solution of inequality (16). Then for every $x\in [1,\infty )$, there exists a function $\overline{h}(x)=\varepsilon e^{-x}\in C([1,\infty ),\mathbb {R})$ such that $|\overline{h}(x)|\le \varepsilon$ and $\mathfrak {D}_{\frac{1}{2}}^{1}y(x)=-y(x)+\frac{x}{1+x^{2}}\sin (y(x))+\varepsilon e^{-x}$. Choose $\overline{N}=2$. From Theorem 5.11, we have

$$\begin{aligned} |z(x)-y(x)|\le \overline{N}\varepsilon . \end{aligned}$$

Thus, Eq. (15) is Ulam–Hyers stable on $[1,\infty )$ with $\overline{N}=2$.

$\bullet$ Let $z\in C([1,\infty ),\mathbb {R})$ be a solution of inequality (17). Then for every $x\in [1,\infty )$, there exists a function $\widehat{h}(x)=\frac{\varepsilon }{x} e^{-(x-1)^{1/2}}\in C([1,\infty ),\mathbb {R})$ such that $|\widehat{h}(x)|\le \varepsilon e^{-(x-1)^{1/2}}:=\varepsilon \varphi (x)$. Note

$$\begin{aligned}&e^{\lambda \frac{(x-a)^{\beta }}{\beta }}\int _{a}^{x}e^{-\lambda \frac{(t-a)^{\beta }}{\beta }}\varphi (t)(t-a)^{\beta -1}\mathrm{d}t\\&\quad =e^{-2(x-1)^{1/2}}\int _{1}^{x}e^{2(t-1)^{1/2}}e^{-(t-1)^{1/2}}(t-1)^{-1/2}\mathrm{d}t\\&\quad \le e^{-2(x-1)^{1/2}}\int _{1}^{x}e^{(t-1)^{1/2}}(t-1)^{-1/2}\mathrm{d}t\\&\quad \le 2e^{-(x-1)^{1/2}}\left( 1-e^{-(x-1)^{1/2}}\right) \le 2e^{-(x-1)^{1/2}}, \end{aligned}$$

so $[H_{8}]$ holds. Choose $\widehat{N}=4$. From Theorem 5.12, we have

$$\begin{aligned} |z(x)-y(x)|\le \widehat{N}\varepsilon \varphi (x). \end{aligned}$$

Thus, Eq. (15) is Ulam–Hyers–Rassias stable on $[1,\infty )$ with $\widehat{N}=4$.