Extremal Solutions for a Caputo-Type Fractional-Order Initial Value Problem

Abstract

In this paper, we study the existence of extremal solutions for a Caputo-type fractional-order initial value problem. By using the monotone iteration technique and the upper–lower solution method, we obtain our existence theorem when the nonlinearity satisfies a reverse-type Lipschitz condition. Note that our nonlinearity depends on the unknown function and its fractional-order derivative.

1. Introduction

Consider the following Caputo-type fractional-order initial value problem:

$$\left\{\begin{array}{c}{}_0{}^{C}D_t^{\alpha }u\left(t\right)=f(t,u\left(t\right),{}_0{}^{C}D_t^{\beta }u\left(t\right)),t\in (0,1),\hfill \\ u\left(0\right)=0,u^{\prime }\left(0\right)=e_0\ge 0,\hfill \end{array}\right.$$

(1)

where ${}_0{}^{C}D_t^{\alpha },{}_0{}^{C}D_t^{\beta }$ are Caputo-type fractional derivatives with $\alpha \in (1,2),\beta \in (0,1)$ and the function f satisfies the following condition:

Hypothesis 1 (H1). 

$f\in C\left(\right[0,1]\times \mathbb{R}\times \mathbb{R},\mathbb{R})$.

It is well known that fractional-order differential equations can accurately describe various nonlinear states or phenomena with memory and genetic characteristics, and they are applied in many fields, such as electrical conduction in biological systems, Brownian motion, turbulence problems, chaos and fractal dynamics, etc. Therefore, the existence and related properties of solutions for nonlinear fractional-order differential equations have very important practical significance and potential application value. In practice, the existence, uniqueness and multiplicity of solutions for fractional-order differential equations are investigated by employing nonlinear operator theory, fixed point theory and the monotone iterative technique. Moreover, there has been a large number of works in this direction (see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23] and the references therein). In [1], the authors used iterative techniques and the upper–lower solution method to study the initial value problem for the following Caputo-type fractional differential equation:

$$\left\{\begin{array}{c}{}_{t_0}{}^{C}D_t^{q}x\left(t\right)=F(t,x\left(t\right)),t\in \left[t_0,t_0+T\right],\hfill \\ x\left(t_0\right)=x_0,\hfill \end{array}\right.$$

where $q\in (0,1),x,x_0\in \mathbb{R}$. In [2], the authors used the Banach contraction principle to study the following initial value problem involving the Caputo–Fabrizio fractional derivative:

$$\left\{\begin{array}{c}{}^{CF}D^{\alpha }x\left(t\right)=g\left(t,x,{}^{CF}D^{\beta }x\right)-g\left(0,x\left(0\right),{}^{CF}D^{\beta }x\left(0\right)\right), 0<t<+\infty ,\hfill \\ x\left(0\right)=x_0\in \mathbb{R},\hfill \end{array}\right.$$

where $0<\beta <\alpha <1,{}^{CF}D^{\alpha }x,{}^{CF}D^{\beta }x$ are Caputo–Fabrizio fractional derivatives, and the nonlinearity g satisfies some Lipschitz condition.

In [3], the authors used the monotone iterative method to study the existence of extremal solutions for the following fractional-order initial value problem involving a Caputo-type generalized fractional derivative and a Steiltjes-type fractional integral:

$$\left\{\begin{array}{c}{}_c{}^{\rho }D_{0+}^{\alpha }y\left(t\right)=f(t,y\left(t\right)),t\in [0,T],\hfill \\ y\left(0\right)=\lambda I_{0+;g}^{\gamma }y\left(T\right)+\kappa :={\displaystyle \frac{\lambda }{\mathsf{\Gamma }\left(\gamma \right)}}{\int }_0^T{\displaystyle \frac{g^{\prime }\left(s\right)y\left(s\right)}{{[g\left(T\right)-g\left(s\right)]}^{1-\gamma }}}ds+\kappa ,{\left.t^{1-\rho }{\displaystyle \frac{dy}{dt}}\right|}_{t=0}=0,\hfill \end{array}\right.$$

where ${}_c{}^{\rho }D_{0+}^{\alpha }$ is the Caputo-type generalized fractional derivative of order $\alpha \in (1,2],\rho >0$.

Motivated by the aforementioned works, we use the monotone iteration technique and the upper–lower solution method to study the existence of extremal solutions for the Caputo-type fractional-order initial value problem (1). When the nonlinearity satisfies a reverse-type Lipschitz condition, we prove that the extremal solutions can be uniformly approximated by some appropriate iterative sequences. Note that our nonlinearity depends on the unknown function and its fractional-order derivative.

2. Preliminaries

In this section, we present some basic results and notations that will be used in this paper.

Definition 1

(see [4,5]). The fractional derivative of a function $f\left(t\right),t\in [0,+\infty )$ in the Caputo sense is defined as

$${}_0{}^{C}D_t^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(n-\alpha )}}{\int }_0^t{(t-s)}^{n-\alpha -1}{f}^{\left(n\right)}\left(s\right)ds,n=\left[\alpha \right]+1,\alpha >0,$$

where $\left[\alpha \right]$ denotes the integer part of the number α.

When α is an integer, the Caputo fractional derivative degenerates into the usual integer-order derivative.

Definition 2

(see [4,5]). The Riemann–Liouville fractional integral of order α for a function f is defined as

$${}_0{}^{RL}I_t^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-s)}^{\alpha -1}f\left(s\right)ds,\alpha >0,$$

provided that such integral exists.

Lemma 1

(see [6]). Let $\alpha >0$. Then,

$${}_0{}^{RL}I_t^{\alpha }\left({}_0{}^{C}D_t^{\alpha }f\left(t\right)\right)=f\left(t\right)+c_0+c_{1}t+c_2{t}^2+\cdots +c_{n-1}{t}^{n-1},$$

where $c_i\in \mathbb{R},i=0,1,2,\dots ,n,n=\left[\alpha \right]+1$.

Definition 3

(see [4,5]). The Laplace transform of a function $f\left(t\right)$ of a real variable $t\in {\mathbb{R}}^{+}$ is defined by

$$F\left(s\right)=\mathcal{L}\left[f\left(t\right)\right]={\int }_0^{+\infty }f\left(t\right)e^{-st}dt,s=\gamma +i\omega \in \mathbb{C},\gamma >0,$$

and the inverse Laplace transform is defined by

$$f\left(t\right)={\mathcal{L}}^{-1}\left[F\left(s\right)\right]={\displaystyle \frac{1}{2\pi i}}{\int }_{\gamma -i\infty }^{\gamma +i\infty }F\left(s\right)e^{st}ds,i^2=-1.$$

Definition 4

(see [4,5]). The Laplace convolution operator of two functions $h\left(t\right)$ and $\phi \left(t\right)$, given on ${\mathbb{R}}^{+}$, is defined for $x\in {\mathbb{R}}^{+}$ by the integral

$$h\ast \phi =(h\ast \phi )\left(x\right):={\int }_0^{x}h(x-t)\phi \left(t\right)dt.$$

Definition 5

(see [4,5]). The Mittag-Leffler function $E_{\alpha ,\beta }\left(u\right)$ is defined by

$$E_{\alpha ,\beta }\left(u\right):=\sum _{k=0}^{\infty }{\displaystyle \frac{u^k}{\mathsf{\Gamma }(\alpha k+\beta )}},u\in \mathbb{C},\alpha ,\beta >0.$$

Definition 6.

We say that a function w on $t\in [0,1]$ is an upper solution of (1) if it satisfies

$$\left\{\begin{array}{c}{}_0{}^{C}D_t^{\alpha }w\left(t\right)\ge f(t,w\left(t\right),{}_0{}^{C}D_t^{\beta }w\left(t\right)),t\in (0,1),\hfill \\ w\left(0\right)=0,w^{\prime }\left(0\right)\ge e_0.\hfill \end{array}\right.$$

Definition 7.

We say that a function v on $t\in [0,1]$ is a lower solution of (1) if it satisfies

$$\left\{\begin{array}{c}{}_0{}^{C}D_t^{\alpha }v\left(t\right)\le f(t,v\left(t\right),{}_0{}^{C}D_t^{\beta }v\left(t\right)),t\in (0,1),\hfill \\ v\left(0\right)=0,v^{\prime }\left(0\right)\le e_0.\hfill \end{array}\right.$$

In what follows, we will consider some auxiliary linear boundary value problems.

Lemma 2.

Let $\alpha \in (1,2)$ and $y\in C[0,1]$. Then, the initial value problem

$$\left\{\begin{array}{c}{}_0{}^{C}D_t^{\alpha }u\left(t\right)=y\left(t\right),t\in (0,1),\hfill \\ u\left(0\right)=0,u^{\prime }\left(0\right)=e_0\hfill \end{array}\right.$$

(2)

has a solution

$$u\left(t\right)=e_{0}t+{\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-\tau )}^{\alpha -1}y\left(\tau \right)d\tau .$$

Proof. 

From Lemma 1, we have

$$u\left(t\right)=c_1+c_{2}t+{\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-\tau )}^{\alpha -1}y\left(\tau \right)d\tau ,$$

where $c_i\in \mathbb{R},i=1,2$. Note that $u\left(0\right)=0$ implies that $c_1=0$, and

$$u^{\prime }\left(0\right)=c_2=e_0.$$

Therefore, we obtain

$$u\left(t\right)=e_{0}t+{\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-\tau )}^{\alpha -1}y\left(\tau \right)d\tau .$$

This completes the proof. □

Lemma 3.

Let $\alpha \in (1,2),\beta \in (0,1)$, $b_i>0(i=1,2)$ and $y\in C[0,1]$. Then, the initial value problem

$$-{}_0{}^{C}D_t^{\alpha }u\left(t\right)+b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)+b_{2}u\left(t\right)=-y\left(t\right),t\in (0,1),$$

(3)

$$u\left(0\right)=0,u^{\prime }\left(0\right)=e_0$$

(4)

has a solution

$$u\left(t\right)=e_0{g}_3\left(t\right)+{\int }_0^{t}y\left(\tau \right)g_4(t-\tau )d\tau ,$$

where

$$\begin{gathered} g_3\left(t\right)=\sum _{j=0}^{\infty }\sum _{k=0}^{\infty }{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{k\alpha +1+(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+k\beta +2]}}, \\ {g}_4\left(t\right)=\sum _{j=0}^{\infty }\sum _{k=0}^{\infty }{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{(k+1)\alpha -1+(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+k\beta +\alpha ]}}. \end{gathered}$$

(5)

Proof. 

By taking the Laplace transformation in (3), we obtain

$$\begin{array}{c}\hfill \mathcal{L}\left[{}_0{}^{C}D_t^{\alpha }u\left(t\right)\right]-b_1\mathcal{L}\left[{}_0{}^{C}D_t^{\beta }u\left(t\right)\right]-b_2\mathcal{L}\left[u\left(t\right)\right]=\mathcal{L}\left[y\left(t\right)\right].\end{array}$$

(6)

By using the formula of the Laplace transformation, we obtain

$$\begin{array}{c}\hfill s^{\alpha }U\left(s\right)-s^{\alpha -1}u\left(0\right)-s^{\alpha -2}{u}^{\prime }\left(0\right)-b_1[s^{\beta }U\left(s\right)-s^{\beta -1}u\left(0\right)]-b_{2}U\left(s\right)=Y\left(s\right),\end{array}$$

(7)

where $U\left(s\right)=\mathcal{L}\left[u\right(t\left)\right]$, $Y\left(s\right)=\mathcal{L}\left[y\left(t\right)\right]$. Solving (7) gives

$$\begin{array}{c}\hfill U\left(s\right)={\displaystyle \frac{u\left(0\right)s^{\alpha -1}}{s^{\alpha }-b_1{s}^{\beta }-b_2}}-{\displaystyle \frac{b_{1}u\left(0\right)s^{\beta -1}}{s^{\alpha }-b_1{s}^{\beta }-b_2}}+{\displaystyle \frac{u^{\prime }\left(0\right)s^{\alpha -2}}{s^{\alpha }-b_1{s}^{\beta }-b_2}}+{\displaystyle \frac{Y\left(s\right)}{s^{\alpha }-b_1{s}^{\beta }-b_2}}.\end{array}$$

(8)

Let

$$G_1\left(s\right)={\displaystyle \frac{s^{\alpha -1}}{s^{\alpha }-b_1{s}^{\beta }-b_2}},G_2\left(s\right)={\displaystyle \frac{s^{\beta -1}}{s^{\alpha }-b_1{s}^{\beta }-b_2}},$$

$$G_3\left(s\right)={\displaystyle \frac{s^{\alpha -2}}{s^{\alpha }-b_1{s}^{\beta }-b_2}},G_4\left(s\right)={\displaystyle \frac{1}{s^{\alpha }-b_1{s}^{\beta }-b_2}}.$$

Thus, Equation (8) can be rewritten as

$$\begin{array}{c}\hfill U\left(s\right)=u\left(0\right)G_1\left(s\right)-b_{1}u\left(0\right)G_2\left(s\right)+u^{\prime }\left(0\right)G_3\left(s\right)+Y\left(s\right)G_4\left(s\right).\end{array}$$

(9)

By taking the Laplace inverse transformation in (9), we have

$$\begin{array}{c}\hfill {\mathcal{L}}^{-1}\left[U\left(s\right)\right]=u\left(0\right){\mathcal{L}}^{-1}\left[G_1\left(s\right)\right]-b_{1}u\left(0\right){\mathcal{L}}^{-1}\left[G_2\left(s\right)\right]+u^{\prime }\left(0\right){\mathcal{L}}^{-1}\left[G_3\left(s\right)\right]+{\mathcal{L}}^{-1}\left[Y\left(s\right)G_4\left(s\right)\right].\end{array}$$

(10)

Thus, we obtain the solution of (3), which can be expressed as

$$\begin{array}{c}\hfill u\left(t\right)=u\left(0\right)g_1\left(t\right)-b_{1}u\left(0\right)g_2\left(t\right)+u^{\prime }\left(0\right)g_3\left(t\right)+y\left(t\right)\ast g_4\left(t\right),\end{array}$$

(11)

where

$$g_1\left(t\right)={\mathcal{L}}^{-1}\left[G_1\left(s\right)\right]={\mathcal{L}}^{-1}\left[{\displaystyle \frac{s^{\alpha -1}}{s^{\alpha }-b_1{s}^{\beta }-b_2}}\right], g_2\left(t\right)={\mathcal{L}}^{-1}\left[G_2\left(s\right)\right]={\mathcal{L}}^{-1}\left[{\displaystyle \frac{s^{\beta -1}}{s^{\alpha }-b_1{s}^{\beta }-b_2}}\right],$$

$$g_3\left(t\right)={\mathcal{L}}^{-1}\left[G_3\left(s\right)\right]={\mathcal{L}}^{-1}\left[{\displaystyle \frac{s^{\alpha -2}}{s^{\alpha }-b_1{s}^{\beta }-b_2}}\right], g_4\left(t\right)={\mathcal{L}}^{-1}\left[G_4\left(s\right)\right]={\mathcal{L}}^{-1}\left[{\displaystyle \frac{1}{s^{\alpha }-b_1{s}^{\beta }-b_2}}\right]$$

and

$$y\left(t\right)\ast g_4\left(t\right)={\int }_0^t{g}_4(t-\tau )y\left(\tau \right)d\tau .$$

Note that $G_1$ can be expressed by the series

$$G_1\left(s\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{b_2^k{s}^{\alpha -(k+1)\beta -1}}{{(s^{\alpha -\beta }-b_1)}^{k+1}}},s\in (0,1),$$

and, using the formula (5.25) in ([5], P155), we have

$$g_1\left(t\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{b_2^k{t}^{k\alpha }}{k!}}{E}_{\alpha -\beta ,k\beta +1}^{\left(k\right)}\left(b_1{t}^{\alpha -\beta }\right).$$

We can also calculate $g_i(i=2,3,4)$ as follows:

$$g_2\left(t\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{b_2^k{t}^{(k+1)\alpha -\beta }}{k!}}{E}_{\alpha -\beta ,\alpha -\beta +1+k\beta }^{\left(k\right)}\left(b_1{t}^{\alpha -\beta }\right),$$

and

$$g_3\left(t\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{b_2^k{t}^{k\alpha +1}}{k!}}{E}_{\alpha -\beta ,k\beta +2}^{\left(k\right)}\left(b_1{t}^{\alpha -\beta }\right),g_4\left(t\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{b_2^k{t}^{(k+1)\alpha -1}}{k!}}{E}_{\alpha -\beta ,\alpha +k\beta }^{\left(k\right)}\left(b_1{t}^{\alpha -\beta }\right),$$

where $E_{\alpha ,\beta }^{\left(k\right)}\left(y\right)={\displaystyle \frac{d^k}{d{y}^k}}{E}_{\alpha ,\beta }\left(y\right)$.

Note that

$$E_{\alpha ,\beta }^{\left(k\right)}\left(y\right)={\displaystyle \frac{d^k}{d{y}^k}}{E}_{\alpha ,\beta }\left(y\right)=\sum _{j=0}^{\infty }{\displaystyle \frac{(j+k)!y^j}{j!\mathsf{\Gamma }\left[\right(j+k)\alpha +\beta ]}},(k=0,1,2,···).$$

Thus, we have

$$\begin{array}{ll}{g}_1\left(t\right)=& {\displaystyle \sum _{j=0}^{\infty }\sum _{k=0}^{\infty }}{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{k\alpha +(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+k\beta +1]}},\\ g_2\left(t\right)=& {\displaystyle \sum _{j=0}^{\infty }\sum _{k=0}^{\infty }}{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{(k+1)\alpha -\beta +(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+\alpha -\beta +1+k\beta ]}},\end{array}$$

and $g_3,g_4$ are found in (5). Consequently, we obtain

$$\begin{array}{ll}u\left(t\right)=& u\left(0\right){\displaystyle \sum _{j=0}^{\infty }\sum _{k=0}^{\infty }}{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{k\alpha +(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+k\beta +1]}}\\ & -b_{1}u\left(0\right){\displaystyle \sum _{j=0}^{\infty }\sum _{k=0}^{\infty }}{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{(k+1)\alpha -\beta +(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+\alpha -\beta +1+k\beta ]}}\\ & +u^{\prime }\left(0\right){\displaystyle \sum _{j=0}^{\infty }\sum _{k=0}^{\infty }}{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{t}^{k\alpha +1+(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+k\beta +2]}}\\ & +{\int }_0^{t}y\left(\tau \right){\displaystyle \sum _{j=0}^{\infty }\sum _{k=0}^{\infty }}{\displaystyle \frac{(j+k)!b_2^k{b}_1^j{(t-\tau )}^{(k+1)\alpha -1+(\alpha -\beta )j}}{k!j!\mathsf{\Gamma }\left[\right(j+k\left)\right(\alpha -\beta )+k\beta +\alpha ]}}d\tau .\end{array}$$

Note that the initial condition in (4) implies that

$$u\left(t\right)=e_0{g}_3\left(t\right)+{\int }_0^{t}y\left(\tau \right)g_4(t-\tau )d\tau .$$

This completes the proof. □

Lemma 4

(Comparison principle). Let $\alpha \in (1,2),\beta \in (0,1)$, $b_i>0(i=1,2)$. If there exists a function u on $[0,1]$ satisfying

$$\left\{\begin{array}{c}{}_0{}^{C}D_t^{\alpha }u\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)-b_{2}u\left(t\right)\ge 0,t\in (0,1),\hfill \\ u\left(0\right)=0,u^{\prime }\left(0\right)\ge e_0,\hfill \end{array}\right.$$

(12)

then $u\left(t\right)\ge 0,t\in [0,1]$.

Proof. 

Let ${}_0{}^{C}D_t^{\alpha }u\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)-b_{2}u\left(t\right)=y\left(t\right)\ge 0,t\in (0,1)$ and $u^{\prime }\left(0\right)-e_0=d_0\ge 0$. Then, we have

$$-{}_0{}^{C}D_t^{\alpha }u\left(t\right)+b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)+b_{2}u\left(t\right)=-y\left(t\right),t\in (0,1),$$

(13)

$$u\left(0\right)=0,u^{\prime }\left(0\right)=e_0+d_0.$$

(14)

From (13) and Lemma 3, we have

$$u\left(t\right)=u^{\prime }\left(0\right)g_3\left(t\right)+{\int }_0^{t}y\left(\tau \right)g_4(t-\tau )d\tau ,$$

and (14) implies that

$$u\left(t\right)=(e_0+d_0)g_3\left(t\right)+{\int }_0^{t}y\left(\tau \right)g_4(t-\tau )d\tau .$$

Note that from the non-negativity of $y,g_3,g_4$ and $e_0+d_0\ge 0$, we have

$$u\left(t\right)\ge 0,t\in [0,1].$$

This completes the proof. □

3. Main Results

Let $E=\left\{u\in C[0,1]:{}_0{}^{C}D_t^{\alpha }u,{}_0{}^{C}D_t^{\beta }u\in C[0,1]\right\}$. Then (see Lemma 2), we define an operator $\mathsf{\Theta }:E\to E$ as

$$\left(\mathsf{\Theta }u\right)\left(t\right)=e_{0}t+{\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-\tau )}^{\alpha -1}f(\tau ,u\left(\tau \right),{}_0{}^{C}D_t^{\beta }u\left(\tau \right))d\tau ,t\in [0,1].$$

(15)

If there exists $u^{\ast }\in E\setminus \left\{0\right\}$ such that $\mathsf{\Theta }{u}^{*}=u^{*}$, then $u^{*}$ is a nontrivial solution for (1). Moreover, (H1) implies that $\mathsf{\Theta }:E\to E$ is a completely continuous operator.

Now, we list our assumptions as follows:

Hypothesis 2 (H2). 

There exist $w_0,v_0$ on $t\in [0,1]$ such that they are the upper and lower solutions for (1), respectively, and $v_0\left(t\right)\le w_0\left(t\right),t\in [0,1]$;

Hypothesis 3 (H3). 

There exist positive constants $b_1,b_2$ such that

$$f(t,x_1,y_1)-f(t,x_2,y_2)\ge b_2(x_1-x_2)+b_1(y_1-y_2),t\in [0,1],x_i,y_i\in \mathbb{R},i=1,2.$$

Theorem 1.

Suppose that (H1)–(H3) hold. Then, there exist monotone iterative sequences $\left\{v_n\right\},\left\{w_n\right\}\subset \left[v_0,w_0\right]$ such that $v_n⟶v^{*},w_n⟶w^{*}$ as $n⟶\infty$ uniformly in $\left[v_0,w_0\right]$, and $v^{*},w^{*}$ are the minimal and maximal solutions of (1) in $\left[v_0,w_0\right]$, respectively.

Proof. 

We define sequences ${\left\{w_n\right\}}_{n=1}^{\infty }$ and ${\left\{v_n\right\}}_{n=1}^{\infty }$ as follows:

$$\left\{\begin{array}{cc}\hfill & {}_0{}^{C}D_t^{\alpha }{w}_n\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_n\left(t\right)-b_2{w}_n\left(t\right)\hfill \\ & =f(t,w_{n-1}\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_{n-1}\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{w}_{n-1}\left(t\right)-b_2{w}_{n-1}\left(t\right),t\in (0,1),\hfill \\ & w_n\left(0\right)=0,w_n^{\prime }\left(0\right)=e_0,\hfill \end{array}\right.$$

(16)

and

$$\left\{\begin{array}{cc}\hfill & {}_0{}^{C}D_t^{\alpha }{v}_n\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{v}_n\left(t\right)-b_2{v}_n\left(t\right)\hfill \\ & =f(t,v_{n-1}\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_{n-1}\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{v}_{n-1}\left(t\right)-b_2{v}_{n-1}\left(t\right),t\in (0,1),\hfill \\ & v_n\left(0\right)=0,v_n^{\prime }\left(0\right)=e_0,\hfill \end{array}\right.$$

(17)

In what follows, we will prove the following claims:

Claim 1: For any $t\in [0,1]$, $v_0\left(t\right)\le v_1\left(t\right)\le w_1\left(t\right)\le w_0\left(t\right)$.

Let $z_v\left(t\right)=v_1\left(t\right)-v_0\left(t\right),t\in [0,1]$. Then, note that $v_0$ is a lower solution. Thus, we have

$$\begin{array}{l}{}_0{}^{C}D_t^{\alpha }{z}_v\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{z}_v\left(t\right)-b_2{z}_v\left(t\right)\\ =[{}_0{}^{C}D_t^{\alpha }{v}_1\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)-b_2{v}_1\left(t\right)]-[{}_0{}^{C}D_t^{\alpha }{v}_0\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right)-b_2{v}_0\left(t\right)]\\ \ge f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right)-b_2{v}_0\left(t\right)+b_1{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right)+b_2{v}_0\left(t\right)-f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))\\ =0,\end{array}$$

and

$$z_v\left(0\right)=v_1\left(0\right)-v_0\left(0\right)=0,z_v^{\prime }\left(0\right)=v_1^{\prime }\left(0\right)-v_0^{\prime }\left(0\right)\ge 0.$$

Now, Lemma 4 implies that $z_v\left(t\right)\ge 0$, i.e., $v_1\left(t\right)\ge v_0\left(t\right),t\in [0,1]$.

Let $z_w\left(t\right)=w_0\left(t\right)-w_1\left(t\right),t\in [0,1]$. Then, note that $w_0$ is an upper solution. Thus, we obtain

$$\begin{array}{l}{}_0{}^{C}D_t^{\alpha }{z}_w\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{z}_w\left(t\right)-b_2{z}_w\left(t\right)\\ =[{}_0{}^{C}D_t^{\alpha }{w}_0\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-b_2{w}_0\left(t\right)]-[{}_0{}^{C}D_t^{\alpha }{w}_1\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right)-b_2{w}_1\left(t\right)]\\ \ge f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-b_2{w}_0\left(t\right)\\ -[f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-b_2{w}_0\left(t\right)]\\ =0,\end{array}$$

and

$$z_w\left(0\right)=w_0\left(0\right)-w_1\left(0\right)=0,z_w^{\prime }\left(0\right)=w_0^{\prime }\left(0\right)-w_1^{\prime }\left(0\right)\ge 0.$$

Now, Lemma 4 implies that $z_w\left(t\right)\ge 0$, i.e., $w_0\left(t\right)\ge w_1\left(t\right),t\in [0,1]$.

Let $h\left(t\right)=w_1\left(t\right)-v_1\left(t\right),t\in [0,1]$. Then, from (H3), we have

$$\begin{array}{l}{}_0{}^{C}D_t^{\alpha }h\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }h\left(t\right)-b_{2}h\left(t\right)\\ =[{}_0{}^{C}D_t^{\alpha }{w}_1\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right)-b_2{w}_1\left(t\right)]-[{}_0{}^{C}D_t^{\alpha }{v}_1\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)-b_2{v}_1\left(t\right)]\\ =[f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-b_2{w}_0\left(t\right)]\\ -[f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right)-b_2{v}_0\left(t\right)]\\ \ge b_2(w_0\left(t\right)-v_0\left(t\right))+b_1({}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))\\ -b_2(w_0\left(t\right)-v_0\left(t\right))-b_1({}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))\\ =0,\end{array}$$

and

$$h\left(0\right)=w_1\left(0\right)-v_1\left(0\right)=0,h^{\prime }\left(0\right)=w_1^{\prime }\left(0\right)-v_1^{\prime }\left(0\right)=0.$$

Now, Lemma 4 implies that $h\left(t\right)\ge 0$, i.e., $w_1\left(t\right)\ge v_1\left(t\right),t\in [0,1]$.

As a result, Claim 1 holds.

Claim 2: $w_1,v_1$ are upper and lower solutions of (1), respectively.

From (16) and (H3), we have

$$\begin{array}{l}{}_0{}^{C}D_t^{\alpha }{w}_1\left(t\right)=f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))+b_1{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right)+b_2{w}_1\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-b_2{w}_0\left(t\right)\\ =f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))-f(t,w_1\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right))+f(t,w_1\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right))\\ +b_1{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right)+b_2{w}_1\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-b_2{w}_0\left(t\right)\\ \ge f(t,w_1\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right))+b_2(w_0\left(t\right)-w_1\left(t\right))+b_1({}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right))\\ -b_2(w_0\left(t\right)-w_1\left(t\right))-b_1({}_0{}^{C}D_t^{\beta }{w}_0\left(t\right)-{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right))\\ =f(t,w_1\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_1\left(t\right)),\end{array}$$

and

$$w_1\left(0\right)=0,w_1^{\prime }\left(0\right)=e_0.$$

Using Definition 6, $w_1$ is an upper solution of (1). Furthermore, from (17) and (H3), we have

$$\begin{array}{l}-{}_0{}^{C}D_t^{\alpha }{v}_1\left(t\right)=-f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)-b_2{v}_1\left(t\right)+b_1{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right)+b_2{v}_0\left(t\right)\\ =-f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))+f(t,v_1\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right))-f(t,v_1\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right))\\ -b_1{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)-b_2{v}_1\left(t\right)+b_1{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right)+b_2{v}_0\left(t\right)\\ \ge -f(t,v_1\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right))+b_2(v_1\left(t\right)-v_0\left(t\right))+b_1({}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)-{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))\\ -b_2(v_1\left(t\right)-v_0\left(t\right))-b_1({}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)-{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))\\ =-f(t,v_1\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_1\left(t\right)),\end{array}$$

and

$$v_1\left(0\right)=0,v_1^{\prime }\left(0\right)=e_0.$$

Using Definition 7, $v_1$ is a lower solution of (1).

Therefore, Claim 2 holds.

If $w_1,v_1$ are regarded as basic functions, we can repeat the processes of Claims 1–2 and obtain the following conclusion:

$$v_1\le v_2\le w_2\le w_1,$$

and $w_2,v_2$ are upper and lower solutions of (1), respectively.

Consequently, by using mathematical induction, we can obtain the non-decreasing lower solutions’ sequence ${\left\{v_n\right\}}_{n=1}^{\infty }$ and the non-increasing upper solutions’ sequence ${\left\{w_n\right\}}_{n=1}^{\infty }$ satisfying

$$v_0\le v_1\le \cdots v_n\le \cdots \le w_n\le w_{n-1}\le \cdots \le w_1\le w_0.$$

We easily find that ${\left\{v_n\right\}}_{n=1}^{\infty }$ and ${\left\{w_n\right\}}_{n=1}^{\infty }$ are uniformly bounded in E, and the monotone bounded theorem guarantees that these sequences converge. Thus, we write

$$\underset{n\to \infty }{lim}{v}_n\left(t\right)=v^{*}\left(t\right),\underset{n\to \infty }{lim}{w}_n\left(t\right)=w^{*}\left(t\right),t\in [0,1].$$

Let $n\to \infty$ in (16) and (17). Thus, we have

$$\left\{\begin{array}{cc}\hfill & {}_0{}^{C}D_t^{\alpha }{w}^{*}\left(t\right)-\cancel{b_1{}_0{}^{C}D_t^{\beta }{w}^{*}\left(t\right)}-\cancel{b_2{w}^{*}\left(t\right)}\hfill \\ & =f(t,w^{*}\left(t\right),{}_0{}^{C}D_t^{\beta }{w}^{*}\left(t\right))-\cancel{b_1{}_0{}^{C}D_t^{\beta }{w}^{*}\left(t\right)}-\cancel{b_2{w}^{*}\left(t\right)},t\in (0,1),\hfill \\ & w^{*}\left(0\right)=0,{\left(w^{*}\right)}^{\prime }\left(0\right)=e_0,\hfill \end{array}\right.$$

(18)

and

$$\left\{\begin{array}{cc}\hfill & {}_0{}^{C}D_t^{\alpha }{v}^{*}\left(t\right)-\cancel{b_1{}_0{}^{C}D_t^{\beta }{v}^{*}\left(t\right)}-\cancel{b_2{v}^{*}\left(t\right)}\hfill \\ & =f(t,v^{*}\left(t\right),{}_0{}^{C}D_t^{\beta }{v}^{*}\left(t\right))-\cancel{b_1{}_0{}^{C}D_t^{\beta }{v}^{*}\left(t\right)}-\cancel{b_2{v}^{*}\left(t\right)},t\in (0,1),\hfill \\ & v^{*}\left(0\right)=0,{\left(v^{*}\right)}^{\prime }\left(0\right)=e_0,\hfill \end{array}\right.$$

(19)

Note that from (18) and (19) (see also (1) and (15)), we have

$$v^{*}\left(t\right)=\left(\mathsf{\Theta }{v}^{*}\right)\left(t\right),w^{*}\left(t\right)=\left(\mathsf{\Theta }{w}^{*}\right)\left(t\right),t\in [0,1],$$

i.e., $v^{*},w^{*}$ are solutions for (1).

Finally, we prove that $v^{*}$ and $w^{*}$ are extremal solutions for (1) in $\left[v_0,w_0\right]$. Let $u\in \left[v_0,w_0\right]$ be any solution for (1). We assume that $v_m\left(t\right)\le u\left(t\right)\le w_m\left(t\right),t\in [0,1]$, for some m. Let $p\left(t\right)=u\left(t\right)-v_{m+1}\left(t\right),q\left(t\right)=w_{m+1}\left(t\right)-u\left(t\right),t\in [0,1]$. Then, from (1), (16) and (17), as well as (H3), we have

$$\begin{array}{l}{}_0{}^{C}D_t^{\alpha }p\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }p\left(t\right)-b_{2}p\left(t\right)\\ =[{}_0{}^{C}D_t^{\alpha }u\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)-b_{2}u\left(t\right)]-[{}_0{}^{C}D_t^{\alpha }{v}_{m+1}\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{v}_{m+1}\left(t\right)-b_2{v}_{m+1}\left(t\right)]\\ \ge f(t,u\left(t\right),{}_0{}^{C}D_t^{\beta }u\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)-b_{2}u\left(t\right)\\ +b_1{}_0{}^{C}D_t^{\beta }{v}_{m+1}\left(t\right)+b_2{v}_{m+1}\left(t\right)-f(t,v_{m+1}\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_{m+1}\left(t\right))\\ \ge b_2(u\left(t\right)-v_{m+1}\left(t\right))+b_1({}_0{}^{C}D_t^{\beta }u\left(t\right)-{}_0{}^{C}D_t^{\beta }{v}_{m+1}\left(t\right))\\ -b_2(u\left(t\right)-v_{m+1}\left(t\right))-b_1({}_0{}^{C}D_t^{\beta }u\left(t\right)-{}_0{}^{C}D_t^{\beta }{v}_{m+1}\left(t\right))\\ =0,t\in (0,1),\\ p\left(0\right)=u\left(0\right)-v_{m+1}\left(0\right)=0,p^{\prime }\left(0\right)=u^{\prime }\left(0\right)-v_{m+1}^{\prime }\left(0\right)=0,\end{array}$$

and

$$\begin{array}{l}{}_0{}^{C}D_t^{\alpha }q\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }q\left(t\right)-b_{2}q\left(t\right)\\ =[{}_0{}^{C}D_t^{\alpha }{w}_{m+1}\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }{w}_{m+1}\left(t\right)-b_2{w}_{m+1}\left(t\right)]-[{}_0{}^{C}D_t^{\alpha }u\left(t\right)-b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)-b_{2}u\left(t\right)]\\ \ge f(t,w_{m+1}\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_{m+1}\left(t\right))-b_1{}_0{}^{C}D_t^{\beta }{w}_{m+1}\left(t\right)-b_2{w}_{m+1}\left(t\right)\\ +b_1{}_0{}^{C}D_t^{\beta }u\left(t\right)+b_{2}u\left(t\right)-f(t,u\left(t\right),{}_0{}^{C}D_t^{\beta }u\left(t\right))\\ \ge b_2(w_{m+1}\left(t\right)-u\left(t\right))+b_1({}_0{}^{C}D_t^{\beta }{w}_{m+1}\left(t\right)-{}_0{}^{C}D_t^{\beta }u\left(t\right))\\ -b_2(w_{m+1}\left(t\right)-u\left(t\right))-b_1({}_0{}^{C}D_t^{\beta }{w}_{m+1}\left(t\right)-{}_0{}^{C}D_t^{\beta }u\left(t\right))\\ =0,t\in (0,1),\\ q\left(0\right)=w_{m+1}\left(0\right)-u\left(0\right)=0,q^{\prime }\left(0\right)=w_{m+1}^{\prime }\left(0\right)-u^{\prime }\left(0\right)=0.\end{array}$$

Now, Lemma 4 implies that $p,q\left(t\right)\ge 0$, i.e., $w_{m+1}\left(t\right)\ge u\left(t\right)\ge v_{m+1}\left(t\right),t\in [0,1]$.

Therefore, applying mathematical induction, we obtain $v_n\left(t\right)\le u\left(t\right)\le w_n\left(t\right)$ for any $n\in \mathbb{N},t\in [0,1]$. Note that u is any solution for (1) in $\left[v_0,w_0\right]$, and let $n\to \infty$. Thus, we have $v^{*}\left(t\right)\le u\left(t\right)\le w^{*}\left(t\right),t\in [0,1]$. This completes the proof. □

In what follows, we will provide some examples to verify the conditions of our theorem.

Example 1.

Let $\alpha =1.1,\beta =0.1$, $w_0\left(t\right)=t^{1.5}$, $v_0\left(t\right)=-t^{1.5}$, $t\in [0,1]$, $f(t,x,y)=0.7(x+y),x,y\in \mathbb{R}$. Then, we have

$$f(t,x_1,y_1)-f(t,x_2,y_2)={\displaystyle \frac{7}{10}}(x_1-x_2)+{\displaystyle \frac{7}{10}}(y_1-y_2),$$

and

$${}_0{}^{C}D_t^{1.1}{w}_0=1.47t^{0.4},{}_0{}^{C}D_t^{0.1}{w}_0=1.07t^{1.4}.$$

Note that $0.7(t^{1.5}+1.07t^{1.4})\le 1.47t^{0.4},t\in [0,1]$, and we obtain

$$\left\{\begin{array}{c}1.47t^{0.4}={}_0{}^{C}D_t^{\alpha }{w}_0\left(t\right)\ge f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))=0.7(t^{1.5}+1.07t^{1.4}),t\in (0,1),\hfill \\ w_0\left(0\right)=0,w_0^{\prime }\left(0\right)=0,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}-1.47t^{0.4}={}_0{}^{C}D_t^{\alpha }{v}_0\left(t\right)\le f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))=-0.7(t^{1.5}+1.07t^{1.4}),t\in (0,1),\hfill \\ v_0\left(0\right)=0,v_0^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

Therefore, all the conditions of Theorem 1 are satisfied.

Example 2.

Let $\alpha =1.5,\beta =0.5$, $w_0\left(t\right)=t^{1.5}$, $v_0\left(t\right)=-t^{1.5}$, $t\in [0,1]$, $f(t,x,y)={\displaystyle \frac{2\sqrt{\pi }}{3\sqrt{\pi }+1}}(x+y),x,y\in \mathbb{R}$. Then, we have

$$f(t,x_1,y_1)-f(t,x_2,y_2)={\displaystyle \frac{2\sqrt{\pi }}{3\sqrt{\pi }+1}}(x_1-x_2)+{\displaystyle \frac{2\sqrt{\pi }}{3\sqrt{\pi }+1}}(y_1-y_2),$$

and

$${}_0{}^{C}D_t^{1.5}{w}_0={\displaystyle \frac{3}{4}}\sqrt{\pi },{}_0{}^{C}D_t^{0.5}{w}_0={\displaystyle \frac{3}{4}}\sqrt{\pi }t.$$

Note that ${\displaystyle \frac{2\sqrt{\pi }}{3\sqrt{\pi }+1}}(t^{1.5}+{\displaystyle \frac{3}{4}}\sqrt{\pi }t)\le {\displaystyle \frac{3}{4}}\sqrt{\pi },t\in [0,1]$, and we obtain

$$\left\{\begin{array}{c}{\displaystyle \frac{3}{4}}\sqrt{\pi }={}_0{}^{C}D_t^{\alpha }{w}_0\left(t\right)\ge f(t,w_0\left(t\right),{}_0{}^{C}D_t^{\beta }{w}_0\left(t\right))={\displaystyle \frac{2\sqrt{\pi }}{3\sqrt{\pi }+1}}(t^{1.5}+{\displaystyle \frac{3}{4}}\sqrt{\pi }t),t\in (0,1),\hfill \\ w_0\left(0\right)=0,w_0^{\prime }\left(0\right)=0,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}-{\displaystyle \frac{3}{4}}\sqrt{\pi }={}_0{}^{C}D_t^{\alpha }{v}_0\left(t\right)\le f(t,v_0\left(t\right),{}_0{}^{C}D_t^{\beta }{v}_0\left(t\right))=-{\displaystyle \frac{2\sqrt{\pi }}{3\sqrt{\pi }+1}}(t^{1.5}+{\displaystyle \frac{3}{4}}\sqrt{\pi }t),t\in (0,1),\hfill \\ v_0\left(0\right)=0,v_0^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

Therefore, all conditions of Theorem 1 are satisfied.

4. Conclusions

In this paper, we use a reverse-type Lipschitz condition to study the existence of extremal solutions for the Caputo-type fractional-order initial value problem (1). We obtain our existence theorem via the monotone iteration technique and the upper–lower solution method. As is noted in [1], it is of interest to construct algorithms for iterative techniques that avoid the application of Mittag-Leffer functions and obtain directly successive approximations. This is achieved in this paper; however, it is important to note that this function plays a crucial and indispensable role in the study of fractional-order equations.