, 2018:182

# Monotone iterative method for two-point fractional boundary value problems

• Bo Tang
• Jing Zhao
• Zhenhai Liu
Open Access
Research

## Abstract

In this work, we deal with two-point Riemann–Liouville fractional boundary value problems. Firstly, we establish a new comparison principle. Then, we show the existence of extremal solutions for the two-point Riemann–Liouville fractional boundary value problems, using the method of upper and lower solutions. The performance of the approach is tested through a numerical example.

## Keywords

Riemann–Liouville derivative Boundary value problem Upper and lower solutions Maximal and minimal solutions

## 1 Introduction

The purpose of this paper is to consider the existence of solution for the following nonlinear fractional differential equation with two-point boundary conditions:
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}u(t)=f(t,u(t)),\quad t\in (a,b), 1< \alpha < 2, \\ I^{2-\alpha }_{a^{+}}u(t)\vert _{t=a}=A,\quad u(b)=B, \end{cases}$$
(1.1)
where $$f\in C([a,b]\times R,R)$$; $$A,B\in R$$; $${}^{L}D^{\alpha }_{a^{+}}$$ and $$I^{2-\alpha }_{a^{+}}$$ denote the Riemann–Liouville fractional derivative of order α and the Riemann–Liouville fractional integral of order $$2-\alpha$$, respectively.

The monotone iterative technique, combined with the method of upper and lower solutions, is a powerful tool for proving the existence of solutions for nonlinear ordinary differential equations [1, 2, 3] and for nonlinear Caputo fractional differential equations of order $$0< \alpha <2$$, see [4, 5, 6, 7] and the references therein. Also many people paid attention to the existence result of solution of the initial value problem for fractional differential equations involving Riemann–Liouville fractional derivative of order $$0 < \alpha <1$$, see [8, 9, 10, 11] and the references therein. However, only few papers considered the method of lower and upper solutions for Riemann–Liouville fractional differential equations with order $$1 < \alpha < 2$$.

In this paper, we present a method based on upper and lower solutions to prove the existence of solutions for Riemann–Liouville fractional differential equations (1.1). We establish a new comparison principle and show the existence of extremal solutions for (1.1), applying the monotone iterative technique and the method of lower and upper solutions. Moreover, we consider a numerical example to illustrate the accuracy of the presented method.

## 2 The comparison principle

Let $$C(J)$$ denote the Banach space of all continuous functions from $$J=[a,b]$$ into R with the norm $$\|x\|_{C}=\mbox{sup}_{t \in J}|x(t)|$$. Let $$AC(J)$$ be the space of functions f which are absolutely continuous on J and $$AC^{m}(J)=\{f:J\rightarrow \mathbb{R}\mbox{ and }f^{(m-1)}(x)\in AC(J)\}$$. To define the solutions of (1.1), we also consider $$C_{2-\alpha }(J)=\{x:x\in C(a,b],(t-a)^{2- \alpha }x(t)\in C(J), 1 <\alpha < 2\}$$ with the norm $$\|x\|_{C_{2- \alpha }}=\sup \{(t-a)^{2-\alpha }|x(t)|:t\in J\}$$. Obviously, the space $$C_{2-\alpha }(J)$$ is a Banach space.

We recall the following definitions and basic properties from fractional calculus. For more details, one can see [12].

### Definition 2.1

The integral
$$I^{\alpha }_{a^{+}}f(t)=\frac{1}{\Gamma (\alpha)} \int^{t}_{a}(t-s)^{ \alpha -1}f(s)\,ds,\quad \alpha >0,$$
is called Riemann–Liouville fractional integral of order α, where Γ is the gamma function.

### Definition 2.2

For a function $$f(t)$$ given in the interval $$[0,\infty)$$, the expression
$${}^{L}D^{\alpha }_{a^{+}}f(t)=\frac{1}{\Gamma (n-\alpha)} \biggl( \frac{d}{dt} \biggr)^{n} \int^{t}_{a}(t-s)^{n-\alpha -1}f(s)\,dt,$$
where $$n=[\alpha ]+1$$, $$[\alpha ]$$ denotes the integer part of number $$\alpha >0$$, is called the Riemann–Liouville fractional derivative of order α.

### Lemma 2.1

([12])

Let$$\alpha >0$$, $$m=[\alpha ]+1$$, and let$$x_{m-\alpha }(t)=I^{m-\alpha }_{a^{+}}x(t)$$be the fractional integral of order$$m-\alpha$$. If$$x(t)\in L^{1}(a,b)$$and$$x_{m-\alpha }(t)\in AC^{m}(J)$$, then we have the following equality:
$$\bigl(I^{\alpha }_{a^{+}}\bigr){}^{L}D^{\alpha }_{a^{+}}x(t)=x(t)- \sum_{k=1}^{m}\frac{x ^{(m-k)}_{m-\alpha }(a)}{\Gamma (\alpha -k+1)}(t-a)^{\alpha -k}.$$
For $$1<\alpha <2$$ and $$u\in C_{2-\alpha }(J)$$, we easily get
$$\frac{1}{\Gamma (\alpha -1)}I^{2-\alpha }_{a^{+}}u(t)\vert _{t=a}=\lim _{x\rightarrow a^{+}}\bigl[(t-a)^{2-\alpha }u(t)\bigr].$$
(2.1)
From Lemma 2.1 and simple calculations, we also have the following.

### Lemma 2.2

The linear boundary value problem
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}u(t)+Mu(t)=\sigma (t), \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}u(t)|_{t=a}=A, \quad u(b)=B, \end{cases}$$
(2.2)
whereMis a constant and$$\sigma \in C_{2-\alpha }(J)$$, has the following integral representation of solution:
\begin{aligned} u(t) =&B\frac{(t-a)^{\alpha -1}}{(b-a)^{\alpha -1}}+A\frac{(t-a)^{ \alpha -2}}{\Gamma (\alpha -1)}-A\frac{(t-a)^{\alpha -1}}{\Gamma ( \alpha -1)(b-a)} \\ &{}- \int^{b}_{a}G(t,s) \bigl(\sigma (s)-Mu(s)\bigr)\,ds, \end{aligned}
where
$$G(t,s)=\frac{1}{\Gamma (\alpha)}\textstyle\begin{cases} \frac{(t-a)^{\alpha -1}}{(b-a)^{\alpha -1}}(b-s)^{\alpha -1}-(t-s)^{ \alpha -1}, & a\leq s\leq t\leq b, \\ \frac{(t-a)^{\alpha -1}}{(b-a)^{\alpha -1}}(b-s)^{\alpha -1}, & a\leq t\leq s\leq b. \end{cases}$$
By Lemma 2.2, we may say that $$u\in C_{2-\alpha }(J)$$ is a solution of (1.1) if the following integral equation holds:
\begin{aligned} u(t) =&B\frac{(t-a)^{\alpha -1}}{(b-a)^{\alpha -1}}+A\frac{(t-a)^{ \alpha -2}}{\Gamma (\alpha -1)}-A\frac{(t-a)^{\alpha -1}}{\Gamma ( \alpha -1)(b-a)}- \int^{b}_{a}G(t,s)f\bigl(s,u(s)\bigr)\,ds. \end{aligned}
(2.3)

### Lemma 2.3

([5])

LetGbe the Green function given in Lemma 2.2. Then
1. (1)

$$G(t,s)\geq 0$$for all$$a\leq t,s\leq b$$;

2. (2)

$$\max_{t\in J}G(t,s)=G(s,s)$$, $$s\in J$$;

3. (3)
$$G(s,s)$$has a unique maximum, given by
$$\max_{s\in J}G(s,s)=G\biggl(\frac{a+b}{2},\frac{a+b}{2} \biggr)=\frac{1}{ \Gamma (\alpha)}\biggl(\frac{b-a}{4}\biggr)^{\alpha -1};$$

4. (4)

$${\int^{b}_{a}G(t,s)\,ds\leq \frac{1}{\Gamma (\alpha)}}\frac{(\alpha -1)^{ \alpha -1}}{\alpha^{\alpha +1}}(b-a)^{\alpha }$$.

### Lemma 2.4

Suppose thatMsatisfies the following inequality:
\begin{aligned} 0\leq M< \frac{4^{\alpha -1}(\alpha -1)\Gamma (\alpha)}{(b-a)^{\alpha }}. \end{aligned}
(2.4)
Then problem (2.2) has a unique solution.

### Proof

Define the operator $$T:C_{2-\alpha }(J)\rightarrow C_{2-\alpha }(J)$$ by
\begin{aligned} (Tu) (t) =&B\frac{(t-a)^{\alpha -1}}{(b-a)^{\alpha -1}}+A\frac{(t-a)^{ \alpha -2}}{\Gamma (\alpha -1)}-A\frac{(t-a)^{\alpha -1}}{\Gamma ( \alpha -1)(b-a)} \\ &{}- \int^{b}_{a}G(t,s) \bigl(\sigma (s)-Mu(s)\bigr)\,ds. \end{aligned}
We will show that the operator T has a unique fixed point. Let $$u,v\in C_{2-\alpha }(J)$$. By Lemma 2.3, one has
\begin{aligned} \Vert Tu-Tv\Vert _{C_{2-\alpha }} =&\max_{t\in J}\biggl\{ (t-a)^{2-\alpha }\biggl\vert \int^{b}_{a}G(t,s) \bigl(Mu(s)-Mv(s)\bigr)\,ds\biggr\vert \biggr\} \\ \leq &\max_{t\in J}\biggl\{ (t-a)^{2-\alpha }\frac{M}{\Gamma (\alpha)} \biggl( \frac{b-a}{4}\biggr)^{\alpha -1} \int^{b}_{a}\bigl\vert u(s)-v(s)\bigr\vert \,ds \biggr\} \\ \leq &\max_{t\in J}(t-a)^{2-\alpha }\frac{M}{\Gamma (\alpha)}\biggl( \frac{b-a}{4}\biggr)^{\alpha -1} \int^{b}_{a}(s-a)^{\alpha -2}\,ds\Vert u-v \Vert _{C _{2-\alpha }} \\ \leq &(b-a)^{2-\alpha }\frac{M}{\Gamma (\alpha)}\biggl(\frac{b-a}{4} \biggr)^{ \alpha -1}\frac{(b-a)^{\alpha -1}}{\alpha -1}\Vert u-v\Vert _{C_{2-\alpha }} \\ =&\frac{M(b-a)^{\alpha }}{4^{\alpha -1}(\alpha -1)\Gamma (\alpha)} \Vert u-v\Vert _{C_{2-\alpha }}. \end{aligned}
That is to say, T is a contracting operator on $$C_{2-\alpha }(J)$$. Therefore, the operator T has a unique fixed point, and we get the desired result. □

The key tool to get our main results is the following comparison principle.

### Lemma 2.5

If$$x\in C_{2-\alpha }(J)$$and satisfies the relations
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}x(t)+Mx(t)\geq 0, \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}x(t)|_{t=a}\leq 0, \quad x(b)\leq 0, \end{cases}$$
(2.5)
with
$$0\leq M< \Gamma (\alpha)\frac{\alpha^{\alpha +1}(\alpha -1)^{1- \alpha }}{(b-a)^{\alpha }}.$$
(2.6)
Then, for any$$t\in (a,b)$$, $$x(t)\leq 0$$.

### Proof

Suppose that there exists $$t\in (a,b)$$ such that $$x(t)> 0$$. Let $$x(t^{*})=\max \{x(t):t\in (a,b)\}=\rho$$, $$\rho >0$$. From (2.5), there exist $$q(t)\geq 0$$ and $$A^{*}\leq 0$$, $$B^{*}\leq 0$$ such that
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}x(t)+Mx(t)-q(t)=0, \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}x(t)|_{t=a}=A^{*}, \quad x(b)=B^{*}. \end{cases}$$
By Lemmas 2.2 and 2.3, we obtain that $$\forall t\in (a,b)$$,
\begin{aligned} x(t) =&B^{*}\frac{(t-a)^{\alpha -1}}{(b-a)^{\alpha -1}}+A^{*}\frac{(t-a)^{ \alpha -2}}{\Gamma (\alpha -1)}-A^{*} \frac{(t-a)^{\alpha -1}}{\Gamma (\alpha -1)(b-a)} - \int^{b}_{a}G(t,s) \bigl(q(s)-Mx(s)\bigr)\,ds \\ \leq & \int^{b}_{a}G(t,s) \bigl(Mx(s)-q(s)\bigr)\,ds \leq M \int^{b}_{a}G(t,s)x(s)\,ds \\ \leq &M\frac{1}{\Gamma (\alpha)}\frac{(\alpha -1)^{\alpha -1}}{ \alpha^{\alpha +1}}(b-a)^{\alpha }\rho. \end{aligned}
Let $$t=t^{*}$$, one has
$$\rho \leq M\frac{1}{\Gamma (\alpha)}\frac{(\alpha -1)^{\alpha -1}}{ \alpha^{\alpha +1}}(b-a)^{\alpha }\rho.$$
So
$$M\frac{1}{\Gamma (\alpha)}\frac{(\alpha -1)^{\alpha -1}}{ \alpha^{\alpha +1}}(b-a)^{\alpha }\geq 1,$$
which contradicts $$M\frac{1}{\Gamma (\alpha)}\frac{(\alpha -1)^{ \alpha -1}}{\alpha^{\alpha +1}}(b-a)^{\alpha }<1$$. Hence $$x(t)\leq 0$$, $$\forall t\in (a,b)$$. The proof is complete. □

### Remark 2.1

Note that the following inequality
$$\frac{1}{\Gamma (\alpha)}\frac{(\alpha -1)^{\alpha -1}}{ \alpha^{\alpha +1}}(b-a)^{\alpha }\leq \frac{1}{\Gamma (\alpha)} \frac{(b-a)^{ \alpha }}{(\alpha -1)4^{\alpha -1}}$$
holds, i.e., (2.4) implies (2.6).

For problem (1.1), we list the definitions of upper and lower solutions below.

### Definition 2.3

A function $$\varphi \in C_{2-\alpha }(J)$$ is called a lower solution of problem (1.1) if it satisfies
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}\varphi (t)\geq f(t,\varphi (t)), \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}\varphi (t)|_{t=a}\leq A, \quad \varphi (b)\leq B. \end{cases}$$
(2.7)
Analogously, the function $$\phi \in C_{2-\alpha }(J)$$ is called an upper solution of problem (1.1) if it satisfies
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}\phi (t)\leq f(t,\phi (t)), \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}\phi (t)|_{t=a}\geq A, \quad \phi (b)\geq B. \end{cases}$$
(2.8)
The following assumptions will be used in the sequel:
1. (H)
Let ϕ and φ be a couple of upper and lower solutions of (1.1), and let (2.4) hold. $$f:J\times R\rightarrow R$$ satisfies
$$f(t,u)-f(t,v)\leq -M(u-v)\quad \mbox{for } \varphi (t)\leq v\leq u\leq \phi (t).$$
(2.9)

## 3 The main result

In this section, we prove the existence of extremal solutions of problem (1.1) by the monotone iterative technique.

### Theorem 3.1

Let (H) hold. Suppose that$$\eta,\theta \in C_{2-\alpha }(J)$$such that
\begin{aligned}& \textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}\eta (t)+M\eta (t)=f(t,\varphi (t))+M\varphi (t), \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}\eta (t)|_{t=a}= A, \quad \eta (b)= B, \end{cases}\displaystyle \end{aligned}
(3.1)
\begin{aligned}& \textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}\theta (t)+M\theta (t)=f(t,\phi (t))+M\phi (t), \quad t\in (a,b), \\ I^{2-\alpha }_{a^{+}}\theta (t)|_{t=a}= A, \quad \theta (b)= B. \end{cases}\displaystyle \end{aligned}
(3.2)
Then$$\varphi (t)\leq \eta (t)\leq \theta (t)\leq \phi (t)$$, and$$\theta (t)$$, $$\eta (t)$$are an upper and a lower solution of (1.1), respectively.

### Proof

By Lemma 2.4, η and θ are well defined. Let $$m(t)=\varphi (t)-\eta (t)$$. Then
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}m(t)+Mm(t)={}^{L}D^{\alpha }_{a^{+}}\varphi (t)-{}^{L}D ^{\alpha }_{a^{+}}\eta (t)+M\varphi (t)-M\eta (t)\geq 0, \\ I^{2-\alpha }_{a^{+}}m(t)\vert _{t=a}\leq 0, \quad m(b)\leq 0. \end{cases}$$
(3.3)
By Lemma 2.5, we have $$m(t)\leq 0$$, that is, $$\varphi (t) \leq \eta (t)$$, $$\forall t\in (a,b)$$. A similar argument using the property of upper solution of problem (1.1) gives $$\theta (t) \leq \phi (t)$$, $$\forall t\in (a,b)$$.
Again, let $$\omega (t)=\eta (t)-\theta (t)$$. By (2.9) , we have
$$\textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}\omega (t)+M\omega (t)={}^{L}D^{\alpha }_{a^{+}} \eta (t)-{}^{L}D^{\alpha }_{a^{+}}\theta (t)+M\eta (t)-M\theta (t) \\ \hphantom{{}^{L}D^{\alpha }_{a^{+}}\omega (t)+M\omega (t)} =f(t,\varphi (t))-f(t,\phi (t))-M(\phi (t)-\varphi (t))\geq 0, \\ I^{2-\alpha }_{a^{+}}\omega (t)|_{t=a}\leq 0, \quad \omega (b)\leq 0. \end{cases}$$
(3.4)
By Lemma 2.5 again, we also have $$\omega (t)\leq 0$$, that is, $$\eta (t)\leq \theta (t)$$, $$\forall t\in (a,b)$$. Then
$$\varphi (t)\leq \eta (t)\leq \theta (t)\leq \phi (t),\quad t\in (a,b).$$
Next, we prove that $$\eta (t)$$ is the lower solution of (1.1). Note that
\begin{aligned} {}^{L}D^{\alpha }_{a^{+}}\eta (t) =&f\bigl(t,\varphi (t) \bigr)+M\varphi (t)-M \eta (t) \\ =&f\bigl(t,\varphi (t)\bigr)+M\varphi (t)-M\eta (t)-f\bigl(t,\eta (t)\bigr)+f \bigl(t,\eta (t)\bigr) \geq f\bigl(t,\eta (t)\bigr),\quad t\in (a,b). \end{aligned}
Furthermore, by $$I^{2-\alpha }_{a^{+}}\eta (t)|_{t=a}=A$$ and $$\eta (b)=B$$ and the definition of lower solution, we easily get that $$\eta (t)$$ is a lower solution of (1.1). Similarly, $$\theta (t)$$ is an upper solution of (1.1). The proof is complete. □

### Theorem 3.2

Suppose (H) holds, then there exist monotone iterative sequences$$\{u_{n}\}, \{v_{n}\}\subset [\varphi,\phi ]$$such that$$u_{n}\rightarrow u^{*}$$, $$v_{n}\rightarrow v^{*}$$ ($$n\rightarrow \infty$$) uniformly in$$[\varphi,\phi ]$$, and$$u^{*}$$, $$v^{*}$$are a minimal and a maximal generalized solution of (1.1) in$$[\varphi,\phi ]$$, respectively.

### Proof

For any $$u_{n-1}, v_{n-1}\in C_{2-\alpha }(J)$$, $$n\geq 1$$, we may define two sequences $$\{u_{n}\}, \{v_{n}\}\subset [\varphi,\phi ]$$ satisfying the following equation:
\begin{aligned}& \textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}u_{n}(t)+Mu_{n}(t)=f(t,u_{n-1}(t))+Mu_{n-1}(t),\quad t \in (a,b), \\ I^{2-\alpha }_{a^{+}}u_{n}(t)\vert _{t=a}= A, \quad u_{n}(b)= B. \end{cases}\displaystyle \end{aligned}
(3.5)
\begin{aligned}& \textstyle\begin{cases} {}^{L}D^{\alpha }_{a^{+}}v_{n}(t)+Mv_{n}(t)=f(t,v_{n-1}(t))+Mv_{n-1}(t),\quad t \in (a,b), \\ I^{2-\alpha }_{a^{+}}v_{n}(t)\vert _{t=a}= A, \quad v_{n}(b)= B. \end{cases}\displaystyle \end{aligned}
(3.6)
By Lemma 2.4, $$\{u_{n}\}$$ and $$\{v_{n}\}$$ are well defined. Now, using Theorem 3.1 and induction, we immediately conclude that
$$\varphi =u_{0}\leq u_{1}\leq \cdots \leq u_{n}\leq \cdots \leq v_{n}\leq \cdots \leq v_{1}\leq v_{0}= \phi.$$

Using the standard arguments, moreover, it is easy to show that $$\{(t-a)^{2-\alpha }u_{n}\}$$ and $$\{(t-a)^{2-\alpha }v_{n}\}$$ are uniformly bounded and equicontinuous in $$C(J)$$. By the Arzela–Ascoli theorem, we obtain that $$(t-a)^{2-\alpha }u_{n}\rightarrow (t-a)^{2- \alpha }u^{*}$$, $$(t-a)^{2-\alpha }v_{n}\rightarrow (t-a)^{2-\alpha }v ^{*}$$ ($$n\rightarrow \infty$$) uniformly in J, i.e., $$u_{n}\rightarrow u^{*}$$, $$v_{n}\rightarrow v^{*}$$ (as $$n\rightarrow \infty$$) in $$C_{2-\alpha }(J)$$ and that $$u^{*},v^{*}\in [\varphi,\phi ]$$ are solutions of problem (1.1).

Finally, we prove that $$u^{*}$$ and $$v^{*}$$ are a minimal and a maximal solution of (1.1) in $$[\varphi,\phi ]$$, respectively. Let $$u(t)\in C_{2-\alpha }(J)$$ be any solution of (1.1). Suppose that there exists a positive integer n such that $$u_{n}(t)\leq u(t) \leq v_{n}(t)$$, $$t\in J$$. Let $$\lambda (t)=u_{n+1}(t)-u(t)$$. By (2.9), we have
\begin{aligned} {}^{L}D^{\alpha }_{a^{+}}\lambda (t)+M\lambda (t) =&{}^{L}D^{\alpha }_{a ^{+}}\bigl(u_{n+1}(t)-u(t) \bigr)+M\bigl(u_{n+1}(t)-u(t)\bigr) \\ =&f\bigl(t,u_{n}(t)\bigr)+M\bigl(u_{n}(t)-u_{n+1}(t) \bigr)-f\bigl(t,u(t)\bigr)+M\bigl(u_{n+1}(t)-u(t)\bigr) \\ \geq &0,\quad t\in (a,b). \end{aligned}
Besides, $$I^{2-\alpha }_{a^{+}}\lambda (t)|_{t=a}=0$$ and $$\lambda (b)=0$$. By Lemma 2.5, we get $$\lambda \leq 0$$, that is, $$u_{n+1}(t)\leq u(t)$$. Similar to the proof of above, we get $$u(t)\leq v_{n+1}(t)$$. Since $$u_{0}\leq u(t)\leq v_{0}$$, then $$u_{n}\leq u(t)\leq v_{n}$$, by induction, taking the limit $$n\rightarrow \infty$$, we obtain $$u^{*}\leq u(t)\leq v^{*}$$. This completes the proof. □

## 4 Numerical example

We apply the previous analysis and general numerical scheme to an example to verify the performance of the proposed approach.

Consider the following problem:
$$\textstyle\begin{cases} {}^{L}D^{\frac{3}{2}}_{0^{+}}u(t)+\frac{1}{30}u^{2}(t)e^{t}+ \frac{1}{30}u(t)e^{t}+\frac{1}{30}e^{t}=0, \quad t\in (0,1), \\ I^{1/2}_{0^{+}}u(t)|_{t=0}=0, \quad u(1)=0. \end{cases}$$
(4.1)
Taking $$u_{0}(t)\equiv 0$$, $$v_{0}(t)\equiv 1$$, we have
\begin{aligned}& \textstyle\begin{cases} {}^{L}D^{\frac{3}{2}}_{0^{+}}0+\frac{1}{30}e^{t}\geq 0, \quad t\in (0,1), \\ I^{1/2}_{0^{+}}u_{0}(t)|_{t=0}=0, \quad u_{0}(1)=0, \end{cases}\displaystyle \\& \textstyle\begin{cases} {}^{L}D^{\frac{3}{2}}_{0^{+}}1+\frac{e^{t}}{10}=\frac{e^{t}}{10}-\frac{1}{2\sqrt{ \pi }}t^{-\frac{3}{2}}\leq 0, \quad t\in (0,1), \\ I^{1/2}_{0^{+}}v_{0}(t)|_{t=0}=0, \quad v_{0}(1)=1\geq 0, \end{cases}\displaystyle \end{aligned}
which shows that $$u_{0}(t)$$ and $$v_{0}(t)$$ are a lower and an upper solution of (4.1), respectively. On the other hand, note that $$f(t,u(t))=-\frac{1}{30}u^{2}(t)e^{t}-\frac{1}{30}u(t)e^{t}- \frac{1}{30}e^{t}$$ is valid for
$$f(t,x)-f(t,y)\leq -\frac{1}{30}(x-y), \quad \forall t\in [0,1],$$
where $$u_{0}(t)\leq y\leq x\leq v_{0}(t)$$. Hence the constant M used in the algorithm is $$M=\frac{1}{30}<\frac{\sqrt{\pi }}{2}$$. To sum up, condition (H) of Theorem 3.2 is satisfied. Then (4.1) has two extremal generalized solutions $$u^{*},v^{*}\in [u _{0},v_{0}]$$ which are obtained by taking limits from its iterative sequences.
Applying Lemma 2.2 and (4.1) to (3.5) and (3.6), we have
\begin{aligned} u_{n}(t)=\frac{1}{30} \int^{1}_{0}G(t,s)\bigl[e^{s}u^{2}_{n-1}(s)+ \bigl(e^{s}-1\bigr)u _{n-1}(s)+u_{n}(s)+e^{s} \bigr]\,ds, \end{aligned}
(4.2)
where
$$G(t,s)=\frac{2}{\sqrt{\pi }}\textstyle\begin{cases} t^{\frac{1}{2}}(1-s)^{\frac{1}{2}}-(t-s)^{\frac{1}{2}}, \quad 0\leq s\leq t\leq 1, \\ t^{\frac{1}{2}}(1-s)^{\frac{1}{2}}, \quad 0\leq t\leq s\leq 1, \end{cases}$$
and a similar formula for $$v_{n}(t)$$. Let $$u_{ni}\approx u_{n}(t_{i})$$, $$f_{j}^{n}=e^{s_{j}}u^{2}_{n}(s_{j})+(e^{s_{j}}-1)u_{n}(s_{j})+e^{s _{j}}$$, and $$G_{ij}=G(t_{i},s_{j})$$. Then, using the composite trapezoidal quadrature formula to approximate the integral on the right-hand side of (4.2), we can obtain the following linear system:
\begin{aligned} u_{ni}=\frac{h}{30}\sum _{j=1}^{N}G_{ij}u_{nj}+ \frac{h}{30} \sum_{j=1}^{N}G_{ij}f_{j}^{n-1}, \end{aligned}
(4.3)
where $$t_{i}=ih$$, $$h=\frac{1}{N}$$, $$0< i\leq N$$, $$N\in N^{+}$$. So, (4.3) can be written as a matrix-vector system:
$$\biggl(I-\frac{h}{30}(G_{ij}) \biggr)U_{n}= \biggl( \frac{h}{30}(G_{ij}) \biggr)F_{n-1},$$
where $$U_{n}=(u_{n1},u_{n2},\ldots,u_{nN})^{T}$$, $$F_{n-1}=(f _{1}^{n-1},f_{2}^{n-1},\ldots,f_{N}^{n-1})^{T}$$ and I is an identity matrix.
Here, for a given accuracy ϵ, we take $$u_{n}$$ and $$v_{n}$$ as ϵ-accurate approximations of $$u^{*}$$ and $$v^{*}$$, respectively, according to the stopping criteria $$E(n)< \epsilon$$, $$E(n)$$ is defined by $$E(n)=\max \{|v_{n}(t)-u_{n}(t)|:t \in (0,1]\}$$. We found that for $$\epsilon =10^{-10}$$, at $$n=N=4$$, error values $$E(4)<\epsilon$$. Table 1 displays $$E(n)$$ versus n for selected values of n, and the graphs of $$u_{n}$$, $$v_{n}$$, for selected values of n, are plotted in Fig. 1.
Table 1

The performance of the error value $$E(n)$$, for example

n

0

1

2

3

4

E(n)

1.0000

0.0247

8.3537×10−7

4.4207×10−9

2.3160×10−11

## Notes

### Acknowledgements

The authors are very grateful to Professor Wang Dongling of Northwestern University for his help and suggestions in the process of numerical calculation.

### Authors’ contributions

The authors have made the same contribution. All authors read and approved the final manuscript.

### Funding

The authors were supported by the National Natural Science Foundation of China (Nos. 11671101, 11661012) and Special Funds of Guangxi Distinguished Experts Construction Engineering, NSF of Guangxi (No. 2018GXNSFDA118167), Qinzhou University project (No. 2018KYQD03).

### Competing interests

The authors declare that they have no competing interests.

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