**Lemma 2.1** *For a given function* y\in C(J,\mathbb{R}), *the following problem*

\{\begin{array}{l}{D}^{\alpha}u(t)=y(t),\\ {D}^{\beta}u(0)=u(0)=0,\end{array}

(2.1)

*has a unique solution* u(t)={I}^{\alpha}y(t), *where* *I* *is the fractional integral and* {I}^{\alpha}y(t)={\int}_{0}^{t}\frac{{(t-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )}y(s)\phantom{\rule{0.2em}{0ex}}ds, 1<\alpha \le 2, 0<\beta \le 1 *and* 0<\alpha -\beta \le 1.

*Proof* One can reduce equation {D}^{\alpha}u(t)=y(t) to an equivalent integral equation

u(t)={I}^{\alpha}y(t)+{c}_{1}{t}^{\alpha -1}+{c}_{2}{t}^{\alpha -2}

(2.2)

for some {c}_{1},{c}_{2}\in \mathbb{R}.

By u(0)=0, it follows {c}_{2}=0. Consequently, the general solution of (2.2) is

u(t)={I}^{\alpha}y(t)+{c}_{1}{t}^{\alpha -1}.

(2.3)

Thus, we have

\begin{array}{rl}{D}^{\beta}u(t)& ={I}^{\alpha -\beta}y(t)+{c}_{1}\frac{\mathrm{\Gamma}(\alpha )}{\mathrm{\Gamma}(\alpha -\beta )}{t}^{\alpha -\beta -1}\\ ={\int}_{0}^{t}\frac{{(t-s)}^{\alpha -\beta -1}}{\mathrm{\Gamma}(\alpha -\beta )}y(s)\phantom{\rule{0.2em}{0ex}}ds+{c}_{1}\frac{\mathrm{\Gamma}(\alpha )}{\mathrm{\Gamma}(\alpha -\beta )}{t}^{\alpha -\beta -1}.\end{array}

(2.4)

By the condition {D}^{\beta}u(0)=0, it follows that {c}_{1}=0. Therefore, we have u(t)={I}^{\alpha}y(t).

Conversely, by a direct computation, we can get {D}^{\alpha}u(t)=y(t) and {D}^{\beta}u(t)={I}^{\alpha -\beta}y(t). It is easy to verify u(t)={I}^{\alpha}y(t) satisfies (2.1).

This completes the proof. □

Combined with Lemma 2.1, we see that (1.1) can be translated into the following system

y(t)=f(t,y(t),{I}^{\alpha -\beta}y(t),{I}^{\alpha}y(t)),

(2.5)

where y(t)={D}^{\alpha}u(t), \mathrm{\forall}t\in J and {I}^{\alpha}, {I}^{\alpha -\beta} are the standard fractional integrals.

Now, we list for convenience the following condition:

(H_{1}) There exist {y}_{0},{z}_{0}\in C(J,\mathbb{R}) satisfying {y}_{0}\le {z}_{0} such that

\{\begin{array}{l}{y}_{0}(t)\le f(t,{y}_{0}(t),{I}^{\alpha -\beta}{y}_{0}(t),{I}^{\alpha}{y}_{0}(t)),\\ {z}_{0}(t)\ge f(t,{z}_{0}(t),{I}^{\alpha -\beta}{z}_{0}(t),{I}^{\alpha}{z}_{0}(t)).\end{array}

(H_{2}) There exists a function M\in C(J,(-1,+\mathrm{\infty})) such that

f(t,u(t),{I}^{\alpha -\beta}u(t),{I}^{\alpha}u(t))-f(t,v(t),{I}^{\alpha -\beta}v(t),{I}^{\alpha}v(t))\ge -M(t)(u-v)(t),

where {y}_{0}\le v\le u\le {z}_{0}, \mathrm{\forall}t\in J.

(H_{3}) There exist functions N,K,L\in C(J,[0,+\mathrm{\infty})) such that

\begin{array}{r}f(t,u(t),{I}^{\alpha -\beta}u(t),{I}^{\alpha}u(t))-f(t,v(t),{I}^{\alpha -\beta}v(t),{I}^{\alpha}v(t))\\ \phantom{\rule{1em}{0ex}}\le N(t)(u-v)(t)+K(t){I}^{\alpha -\beta}(u-v)(t)+L(t){I}^{\alpha}(u-v)(t),\end{array}

where {y}_{0}\le v\le u\le {z}_{0}, \mathrm{\forall}t\in J.

**Theorem 2.1** *Assume that* (H_{1}) *and* (H_{2}) *hold*. *Then problem* (2.5) *has the minimal and maximal solution* {y}^{\ast}, {z}^{\ast} *in the ordered interval* [{y}_{0},{z}_{0}]. *Moreover*, *there exist explicit monotone iterative sequences* \{{y}_{n}\},\{{z}_{n}\}\subset [{y}_{0},{z}_{0}] *such that* {lim}_{n\to \mathrm{\infty}}{y}_{n}(t)={y}^{\ast}(t) *and* {lim}_{n\to \mathrm{\infty}}{z}_{n}(t)={z}^{\ast}(t), *where* {y}_{n}(t), {z}_{n}(t) *are defined as*

\begin{array}{r}{y}_{n}(t)=\frac{1}{1+M(t)}[f(t,{y}_{n-1}(t),{I}^{\alpha -\beta}{y}_{n-1}(t),{I}^{\alpha}{y}_{n-1}(t))+M(t){y}_{n-1}(t)],\\ \phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,n=1,2,\dots ,\\ {z}_{n}(t)=\frac{1}{1+M(t)}[f(t,{z}_{n-1}(t),{I}^{\alpha -\beta}{z}_{n-1}(t),{I}^{\alpha}{z}_{n-1}(t))+M(t){z}_{n-1}(t)],\\ \phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J,n=1,2,\dots ,\end{array}

(2.6)

*and*

{y}_{0}\le {y}_{1}\le \cdots \le {y}_{n}\le \cdots \le {y}^{\ast}\le {z}^{\ast}\le \cdots \le {z}_{n}\le \cdots \le {z}_{1}\le {z}_{0}.

(2.7)

*Proof* Define an operator Q:[{y}_{0},{z}_{0}]\to C(J,\mathbb{R}) by x=Q\eta, where *x* is the unique solution of the corresponding linear problem corresponding to \eta \in [{y}_{0},{z}_{0}] and

Q\eta =\frac{1}{1+M(t)}[f(t,\eta (t),{I}^{\alpha -\beta}\eta (t),{I}^{\alpha}\eta (t))+M(t)\eta (t)].

(2.8)

Then, the operator *Q* has the following properties:

\begin{array}{r}\text{(a)}\phantom{\rule{1em}{0ex}}{y}_{0}\le Q{y}_{0},\phantom{\rule{2em}{0ex}}Q{z}_{0}\le {z}_{0};\\ \text{(b)}\phantom{\rule{1em}{0ex}}Q{h}_{1}\le Q{h}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}{h}_{1},{h}_{2}\in [{y}_{0},{z}_{0}],{h}_{1}\le {h}_{2}.\end{array}

(2.9)

Firstly, we show that (a) holds. Let {y}_{1}=Q{y}_{0}, p={y}_{1}-{y}_{0}. By (H_{1}) and the definition of *Q*, we know that

\begin{array}{rl}p(t)& =\frac{1}{1+M(t)}[f(t,{y}_{0}(t),{I}^{\alpha -\beta}{y}_{0}(t),{I}^{\alpha}{y}_{0}(t))+M(t){y}_{0}(t)]-{y}_{0}(t)\\ \ge \frac{1}{1+M(t)}[{y}_{0}(t)+M(t){y}_{0}(t)]-{y}_{0}(t)\\ =0.\end{array}

Thus, we can obtain p(t)\ge 0, \mathrm{\forall}t\in J. That is, {y}_{0}\le Q{y}_{0}. Similarly, we can prove that Q{z}_{0}\le {z}_{0}. Then, (a) holds.

Secondly, let q=Q{h}_{2}-Q{h}_{1}, by (2.8) and (H_{2}), we have

\begin{array}{rl}q(t)=& \frac{1}{1+M(t)}[f(t,{h}_{2}(t),{I}^{\alpha -\beta}{h}_{2}(t),{I}^{\alpha}{h}_{2}(t))+M(t){h}_{2}(t)]\\ -\frac{1}{1+M(t)}[f(t,{h}_{1}(t),{I}^{\alpha -\beta}{h}_{1}(t),{I}^{\alpha}{h}_{1}(t))+M(t){h}_{1}(t)]\\ \ge & \frac{1}{1+M(t)}[-M(t)({h}_{2}-{h}_{1})(t)+M(t)({h}_{2}-{h}_{1})(t)]\\ =& 0.\end{array}

Hence, we have q(t)\ge 0, \mathrm{\forall}t\in J. That is, Q{h}_{2}\ge Q{h}_{1}. Then, (b) holds.

Now, put

{y}_{n}=Q{y}_{n-1},\phantom{\rule{2em}{0ex}}{z}_{n}=Q{z}_{n-1},\phantom{\rule{1em}{0ex}}n=1,2,\dots .

(2.10)

By (2.9), we can get

{y}_{0}\le {y}_{1}\le \cdots \le {y}_{n}\le \cdots \le {z}_{n}\le \cdots \le {z}_{1}\le {z}_{0}.

Obviously, {y}_{n}, {z}_{n} satisfy

\begin{array}{r}{y}_{n}(t)=f(t,{y}_{n-1}(t),{I}^{\alpha -\beta}{y}_{n-1}(t),{I}^{\alpha}{y}_{n-1}(t))-M(t)({u}_{n}-{y}_{n-1})(t),\\ {z}_{n}(t)=f(t,{z}_{n-1}(t),{I}^{\alpha -\beta}{z}_{n-1}(t),{I}^{\alpha}{z}_{n-1}(t))-M(t)({z}_{n}-{z}_{n-1})(t).\end{array}

(2.11)

Employing the same arguments used in Ref. [17], we see that \{{y}_{n}\}, \{{z}_{n}\} converge to their limit functions {y}^{\ast}, {z}^{\ast}, respectively. That is, {lim}_{n\to \mathrm{\infty}}{y}_{n}(t)={y}^{\ast}(t) and {lim}_{n\to \mathrm{\infty}}{z}_{n}(t)={z}^{\ast}(t). Moreover, {y}^{\ast}(t), {z}^{\ast}(t) are solutions of (2.5) in [{y}_{0},{z}_{0}]. (2.7) is true.

Finally, we prove that {y}^{\ast}(t), {z}^{\ast}(t) are the minimal and the maximal solution of (2.5) in [{y}_{0},{z}_{0}]. Let w\in [{y}_{0},{z}_{0}] be any solution of (2.5), then Qw=w. By {y}_{0}\le w\le {z}_{0}, (2.9) and (2.10), we can obtain

{y}_{n}\le w\le {z}_{n},\phantom{\rule{1em}{0ex}}n=1,2,\dots .

(2.12)

Thus, taking limit in (2.12) as n\to +\mathrm{\infty}, we have {y}^{\ast}\le w\le {z}^{\ast}. That is, {y}^{\ast}, {z}^{\ast} are the minimal and maximal solution of (2.5) in the ordered interval [{y}_{0},{z}_{0}], respectively.

This completes the proof. □

**Theorem 2.2** *Let* N(t)\ge -M(t). *Assume conditions* (H_{1})-(H_{3}) *hold*. *If*

\lambda (t)=N(t)+\frac{K(t){t}^{\alpha -\beta}}{\mathrm{\Gamma}(\alpha -\beta +1)}+\frac{L(t){t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}<1,

*then problem* (2.5) *has a unique solution* x(t)\in [{y}_{0},{z}_{0}].

*Proof* By Theorem 2.1, we have proved that {y}^{\ast}, {z}^{\ast} are the minimal and maximal solution of (2.5) and

{y}_{0}(t)\le {y}^{\ast}(t)\le {z}^{\ast}(t)\le {z}_{0}(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in J.

Now, we are going to show that problem (2.5) has a unique solution *x*, *i.e.*, {y}^{\ast}(t)={z}^{\ast}(t)=x(t).

Let p(t)={z}^{\ast}(t)-{y}^{\ast}(t), by (H_{3}), we have

\begin{array}{rl}0& \le p(t)\le f(t,{z}^{\ast}(t),{I}^{\alpha -\beta}{z}^{\ast}(t),{I}^{\alpha}{z}^{\ast}(t))-f(t,{y}^{\ast}(t),{I}^{\alpha -\beta}{y}^{\ast}(t),{I}^{\alpha}{y}^{\ast}(t))\\ \le N(t)({z}^{\ast}-{y}^{\ast})(t)+K(t){I}^{\alpha -\beta}({z}^{\ast}-{y}^{\ast})(t)+L(t){I}^{\alpha}({z}^{\ast}-{y}^{\ast})(t)\\ =N(t)p(t)+K(t){\int}_{0}^{t}\frac{{(t-s)}^{\alpha -\beta -1}}{\mathrm{\Gamma}(\alpha -\beta )}p(s)\phantom{\rule{0.2em}{0ex}}ds+L(t){\int}_{0}^{t}\frac{{(t-s)}^{\alpha -1}}{\mathrm{\Gamma}(\alpha )}p(s)\phantom{\rule{0.2em}{0ex}}ds\\ \le [N(t)+\frac{K(t){t}^{\alpha -\beta}}{\mathrm{\Gamma}(\alpha -\beta +1)}+\frac{L(t){t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}]\underset{t\in J}{max}p(t)\\ \triangleq \lambda (t)\underset{t\in J}{max}p(t),\end{array}

which implies that {max}_{t\in J}p(t)\le 0. Since p(t)\ge 0, then it holds p(t)=0. That is, {y}^{\ast}(t)={z}^{\ast}(t). Therefore, problem (2.5) has a unique solution x\in [{y}_{0},{z}_{0}]. □

Let x(t) be the unique solution of (2.5). Noting that x\in [{y}_{0},{z}_{0}] and u(t)={I}^{\alpha}x(t), we can easily obtain the following theorem.

**Theorem 2.3** *Let all conditions of Theorem * 2.2 *hold*. *Then problem* (1.1) *has a unique solution* u\in [{I}^{\alpha}{y}_{0},{I}^{\alpha}{z}_{0}], \mathrm{\forall}t\in J.