Monotone iterative method for two-point fractional boundary value problems

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.

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 [1–3] and for nonlinear Caputo fractional differential equations of order $0< \alpha <2$, see [4–7] 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 [8–11] 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)

where M is 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])

Let G be the Green function given in Lemma 2.2. Then

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

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

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

Lemma 2.4

Suppose that M satisfies 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:

• (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.

Figure 1

The graphs of $u_{n}$ and $v_{n}$, $n=0,1,2,3$, for example

Table 1 The performance of the error value $E(n)$, for example