In this section, we apply Lemma 2.8 and Lemma 2.9 to study FBVP (1.1), and we obtain some new results on the existence and uniqueness of positive solutions. The method used here is relatively new to the literature and so are the existence and uniqueness results to the fractional differential equations.

In our considerations, we work in the Banach space

$C[0,1]=\{x:[0,1]\to \mathbf{R}\text{is}\text{continuous}\}$ with the standard norm

$\parallel x\parallel =sup\{|x(t)|:t\in [0,1]\}$. Notice that this space can be equipped with a partial order given by

$x,y\in C[0,1],\phantom{\rule{1em}{0ex}}x\le y\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}x(t)\le y(t)\phantom{\rule{1em}{0ex}}\text{for}t\in [0,1].$

Set $P=\{x\in C[0,1]\mid x(t)\ge 0,t\in [0,1]\}$, the standard cone. It is clear that *P* is a normal cone in $C[0,1]$ and the normality constant is 1.

**Theorem 3.1**
*Assume that*

(H_{1}) $f:[0,1]\times [0,+\mathrm{\infty})\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ *is continuous and* $g:[0,1]\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ *is continuous*;

(H_{2}) $f(t,u,v)$ *is increasing in* $u\in [0,+\mathrm{\infty})$ *for fixed* $t\in [0,1]$ *and* $v\in [0,+\mathrm{\infty})$, *decreasing in* $v\in [0,+\mathrm{\infty})$ *for fixed* $t\in [0,1]$ *and* $u\in [0,+\mathrm{\infty})$, *and* $g(t,u)$ *is increasing in* $u\in [0,+\mathrm{\infty})$ *for fixed* $t\in [0,1]$;

(H_{3}) $g(t,0)\not\equiv 0$ *and* $g(t,\lambda u)\ge \lambda g(t,u)$ *for* $\lambda \in (0,1)$, $t\in [0,1]$, $u\in [0,+\mathrm{\infty})$, *and there exists a constant* $\beta \in (0,1)$ *such that* $f(t,\lambda u,{\lambda}^{-1}v)\ge {\lambda}^{\beta}f(t,u,v)$, $\mathrm{\forall}t\in [0,1]$, $\lambda \in (0,1)$, $u,v\in [0,+\mathrm{\infty})$;

(H_{4}) *there exists a constant* ${\delta}_{0}>0$ *such that* $f(t,u,v)\ge {\delta}_{0}g(t,u)$, $t\in [0,1]$, $u,v\ge 0$.

*Then*:

- (1)
*there exist*
${u}_{0},{v}_{0}\in {P}_{h}$
*and*
$r\in (0,1)$
*such that*
$r{v}_{0}\le {u}_{0}<{v}_{0}$
*and*

*where* $h(t)={t}^{\nu -1}$,

$t\in [0,1]$ *and* $G(t,s)$ *is given as in* (2.2);

- (2)
*FBVP* (1.1) *has a unique positive solution* ${u}^{\ast}$ *in* ${P}_{h}$;

- (3)
*for any* ${x}_{0},{y}_{0}\in {P}_{h}$,

*constructing successively the sequences*

*we have* $\parallel {x}_{n}-{u}^{\ast}\parallel \to 0$ *and* $\parallel {y}_{n}-{u}^{\ast}\parallel \to 0$ *as* $n\to \mathrm{\infty}$.

*Proof* To begin with, from Lemma 2.2, FBVP (1.1) has an integral formulation given by

$u(t)={\int}_{0}^{1}G(t,s)[f(s,u(s),u(s))+g(s,u(s))]\phantom{\rule{0.2em}{0ex}}ds,$

where $G(t,s)$ is given as in (2.2).

Define two operators

$A:P\times P\to E$ and

$B:P\to E$ by

$A(u,v)(t)={\int}_{0}^{1}G(t,s)f(s,u(s),v(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}(Bu)(t)={\int}_{0}^{1}G(t,s)g(s,u(s))\phantom{\rule{0.2em}{0ex}}ds.$

It is easy to prove that *u* is the solution of FBVP (1.1) if and only if $u=A(u,u)+Bu$. From (H_{1}), we know that $A:P\times P\to P$ and $B:P\to P$. In the sequel, we check that *A*, *B* satisfy all the assumptions of Lemma 2.8.

Firstly, we prove that

*A* is a mixed monotone operator. In fact, for

${u}_{i},{v}_{i}\in P$,

$i=1,2$ with

${u}_{1}\ge {u}_{2}$,

${v}_{1}\le {v}_{2}$, we know that

${u}_{1}(t)\ge {u}_{2}(t)$,

${v}_{1}(t)\le {v}_{2}(t)$,

$t\in [0,1]$, and by (H

_{2}) and Lemma 2.3,

$A({u}_{1},{v}_{1})(t)={\int}_{0}^{1}G(t,s)f(s,{u}_{1}(s),{v}_{1}(s))\phantom{\rule{0.2em}{0ex}}ds\ge {\int}_{0}^{1}G(t,s)f(s,{u}_{2}(s),{v}_{2}(s))\phantom{\rule{0.2em}{0ex}}ds=A({u}_{2},{v}_{2})(t).$

That is, $A({u}_{1},{v}_{1})\ge A({u}_{2},{v}_{2})$.

Further, it follows from (H

_{2}) and Lemma 2.3 that

*B* is increasing. Next we show that

*A* satisfies the condition (2.5). For any

$\lambda \in (0,1)$ and

$u,v\in P$, by (H

_{3}) we have

$\begin{array}{rcl}A(\lambda u,{\lambda}^{-1}v)(t)& =& {\int}_{0}^{1}G(t,s)f(s,\lambda u(s),{\lambda}^{-1}v(s))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\lambda}^{\beta}{\int}_{0}^{1}G(t,s)f(s,u(s),v(s))\phantom{\rule{0.2em}{0ex}}ds\\ =& {\lambda}^{\beta}A(u,v)(t).\end{array}$

That is,

$A(\lambda u,{\lambda}^{-1}v)\ge {\lambda}^{\beta}A(u,v)$ for

$\lambda \in (0,1)$,

$u,v\in P$. So, the operator

*A* satisfies (2.5). Also, for any

$\lambda \in (0,1)$,

$u\in P$, from (H

_{3}) we know that

$B(\lambda u)(t)={\int}_{0}^{1}G(t,s)g(s,\lambda u(s))\phantom{\rule{0.2em}{0ex}}ds\ge \lambda {\int}_{0}^{1}G(t,s)g(s,u(s))\phantom{\rule{0.2em}{0ex}}ds=\lambda Bu(t),$

that is,

$B(\lambda u)\ge \lambda Bu$ for

$\lambda \in (0,1)$,

$u\in P$. That is, the operator

*B* is sub-homogeneous. Now we show that

$A(h,h)\in {P}_{h}$ and

$Bh\in {P}_{h}$. On the one hand, from (H

_{1}), (H

_{2}) and Lemma 2.4, for any

$t\in [0,1]$, we have

$\begin{array}{rcl}A(h,h)(t)& =& {\int}_{0}^{1}G(t,s)f(s,h(s),h(s))\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int}_{0}^{1}G(t,s)f(s,{s}^{\nu -1},{s}^{\nu -1})\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{1}{\mathrm{\Gamma}(\nu )}h(t){\int}_{0}^{1}{(1-s)}^{\nu -\alpha -1}f(s,1,0)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

On the other hand, also from (H

_{1}), (H

_{2}) and Lemma 2.4, for any

$t\in [0,1]$, we obtain

$\begin{array}{rcl}A(h,h)(t)& =& {\int}_{0}^{1}G(t,s)f(s,h(s),h(s))\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int}_{0}^{1}G(t,s)f(s,{s}^{\nu -1},{s}^{\nu -1})\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \frac{1}{\mathrm{\Gamma}(\nu )}h(t){\int}_{0}^{1}[1-{(1-s)}^{\alpha}]{(1-s)}^{\nu -\alpha -1}f(s,0,1)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

From (H

_{2}), (H

_{4}), we have

$f(s,1,0)\ge f(s,0,1)\ge {\delta}_{0}g(s,0)\ge 0.$

Since

$g(t,0)\not\equiv 0$, we get

${\int}_{0}^{1}f(s,1,0)\phantom{\rule{0.2em}{0ex}}ds\ge {\int}_{0}^{1}f(s,0,1)\phantom{\rule{0.2em}{0ex}}ds\ge {\delta}_{0}{\int}_{0}^{1}g(s,0)\phantom{\rule{0.2em}{0ex}}ds>0,$

So,

${l}_{2}h(t)\le A(h,h)(t)\le {l}_{1}h(t)$,

$t\in [0,1]$; and hence we have

$A(h,h)\in {P}_{h}$. Similarly,

$\frac{1}{\mathrm{\Gamma}(\nu )}h(t){\int}_{0}^{1}[1-{(1-s)}^{\alpha}]{(1-s)}^{\nu -\alpha -1}g(s,0)\phantom{\rule{0.2em}{0ex}}ds\le Bh(t)\le \frac{1}{\mathrm{\Gamma}(\nu )}h(t){\int}_{0}^{1}{(1-s)}^{\nu -\alpha -1}g(s,1)\phantom{\rule{0.2em}{0ex}}ds,$

from $g(t,0)\not\equiv 0$, we easily prove $Bh\in {P}_{h}$. Hence the condition (i) of Lemma 2.8 is satisfied.

In the following, we show the condition (ii) of Lemma 2.8 is satisfied. For

$u,v\in P$, and any

$t\in [0,1]$, from (H

_{4}),

$A(u,v)(t)={\int}_{0}^{1}G(t,s)f(s,u(s),v(s))\phantom{\rule{0.2em}{0ex}}ds\ge {\delta}_{0}{\int}_{0}^{1}G(t,s)g(s,u(s))\phantom{\rule{0.2em}{0ex}}ds={\delta}_{0}Bu(t).$

Then we get

$A(u,v)\ge {\delta}_{0}Bu$, for

$u,v\in P$. Finally, an application of Lemma 2.8 implies: there exist

${u}_{0},{v}_{0}\in {P}_{h}$ and

$r\in (0,1)$ such that

$r{v}_{0}\le {u}_{0}<{v}_{0}$,

${u}_{0}\le A({u}_{0},{v}_{0})+B{u}_{0}\le A({v}_{0},{u}_{0})+B{v}_{0}\le {v}_{0}$; the operator equation

$A(u,u)+Bu=u$ has a unique solution

${u}^{\ast}$ in

${P}_{h}$; for any initial values

${x}_{0},{y}_{0}\in {P}_{h}$, constructing successively the sequences

${x}_{n}=A({x}_{n-1},{y}_{n-1})+B{x}_{n-1},\phantom{\rule{2em}{0ex}}{y}_{n}=A({y}_{n-1},{x}_{n-1})+B{y}_{n-1},\phantom{\rule{1em}{0ex}}n=1,2,\dots ,$

we have

${x}_{n}\to {u}^{\ast}$ and

${y}_{n}\to {u}^{\ast}$ as

$n\to \mathrm{\infty}$. That is,

FBVP (1.1) has a unique positive solution

${u}^{\ast}$ in

${P}_{h}$; for

${x}_{0},{y}_{0}\in {P}_{h}$, the sequences

satisfy $\parallel {x}_{n}-{u}^{\ast}\parallel \to 0$ and $\parallel {y}_{n}-{u}^{\ast}\parallel \to 0$ as $n\to \mathrm{\infty}$. □

**Theorem 3.2** *Assume* (H_{1}), (H_{2}) *and*

(H_{5}) *there exists a constant* $\beta \in (0,1)$ *such that* $g(t,\lambda u)\ge {\lambda}^{\beta}g(t,u)$, $\mathrm{\forall}t\in [0,1]$, $\lambda \in (0,1)$, $u\in [0,+\mathrm{\infty})$, *and* $f(t,\lambda u,{\lambda}^{-1}v)\ge \lambda f(t,u,v)$ *for* $\lambda \in (0,1)$, $t\in [0,1]$, $u,v\in [0,+\mathrm{\infty})$;

(H_{6}) $f(t,0,1)\not\equiv 0$ *for* $t\in [0,1]$ *and there exists a constant* ${\delta}_{0}>0$ *such that* $f(t,u,v)\le {\delta}_{0}g(t,u)$, $t\in [0,1]$, $u,v\ge 0$.

*Then*:

- (1)
*there exist*
${u}_{0},{v}_{0}\in {P}_{h}$
*and*
$r\in (0,1)$
*such that*
$r{v}_{0}\le {u}_{0}<{v}_{0}$
*and*

*where* $h(t)={t}^{\nu -1}$,

$t\in [0,1]$ *and* $G(t,s)$ *is given as in* (2.2);

- (2)
*FBVP* (1.1) *has a unique positive solution* ${u}^{\ast}$ *in* ${P}_{h}$;

- (3)
*for any* ${x}_{0},{y}_{0}\in {P}_{h}$,

*constructing successively the sequences*

*we have* $\parallel {x}_{n}-{u}^{\ast}\parallel \to 0$ *and* $\parallel {y}_{n}-{u}^{\ast}\parallel \to 0$ *as* $n\to \mathrm{\infty}$.

*Sketch of the proof* Consider two operators

*A*,

*B* defined in the proof of Theorem 3.1. Similarly, from (H

_{1}), (H

_{2}), we obtain that

$A:P\times P\to P$ is a mixed monotone operator and

$B:P\to P$ is increasing. From (H

_{5}), we have

$A(\lambda u,{\lambda}^{-1}v)\ge \lambda A(u,v);\phantom{\rule{2em}{0ex}}B(\lambda u)\ge {\lambda}^{\beta}Bu,\phantom{\rule{1em}{0ex}}\text{for}\lambda \in (0,1),u,v\in P.$

From (H

_{2}), (H

_{6}), we have

$g(s,0)\ge \frac{1}{{\delta}_{0}}f(s,0,1),\phantom{\rule{2em}{0ex}}f(s,1,0)\ge f(s,0,1),\phantom{\rule{1em}{0ex}}s\in [0,1].$

Since

$f(t,0,1)\not\equiv 0$, we get

So, we can easily prove that

$A(h,h)\in {P}_{h}$,

$Bh\in {P}_{h}$. For

$u,v\in P$, and any

$t\in [0,1]$, from (H

_{6}),

$A(u,v)(t)={\int}_{0}^{1}G(t,s)f(s,u(s),v(s))\phantom{\rule{0.2em}{0ex}}ds\le {\delta}_{0}{\int}_{0}^{1}G(t,s)g(s,u(s))\phantom{\rule{0.2em}{0ex}}ds={\delta}_{0}Bu(t).$

Then we get

$A(u,v)\le {\delta}_{0}Bu$, for

$u,v\in P$. Finally, an application of Lemma 2.9 implies: there exist

${u}_{0},{v}_{0}\in {P}_{h}$ and

$r\in (0,1)$ such that

$r{v}_{0}\le {u}_{0}<{v}_{0}$,

${u}_{0}\le A({u}_{0},{v}_{0})+B{u}_{0}\le A({v}_{0},{u}_{0})+B{v}_{0}\le {v}_{0}$; the operator equation

$A(u,u)+Bu=u$ has a unique solution

${u}^{\ast}$ in

${P}_{h}$; for any initial values

${x}_{0},{y}_{0}\in {P}_{h}$, constructing successively the sequences

${x}_{n}=A({x}_{n-1},{y}_{n-1})+B{x}_{n-1},\phantom{\rule{2em}{0ex}}{y}_{n}=A({y}_{n-1},{x}_{n-1})+B{y}_{n-1},\phantom{\rule{1em}{0ex}}n=1,2,\dots ,$

we have

${x}_{n}\to {u}^{\ast}$ and

${y}_{n}\to {u}^{\ast}$ as

$n\to \mathrm{\infty}$. That is,

FBVP (1.1) has a unique positive solution

${u}^{\ast}$ in

${P}_{h}$; for

${x}_{0},{y}_{0}\in {P}_{h}$, the sequences

satisfy $\parallel {x}_{n}-{u}^{\ast}\parallel \to 0$ and $\parallel {y}_{n}-{u}^{\ast}\parallel \to 0$ as $n\to \mathrm{\infty}$. □

From Remark 2.10 and similar to the proofs of Theorems 3.1-3.2, we can prove the following conclusions.

**Corollary 3.3** *Let* $g\equiv 0$.

*Assume that* *f* *satisfies the conditions of Theorem * 3.1

*and* $f(t,0,1)\not\equiv 0$.

*Then*: (i)

*there exist* ${u}_{0},{v}_{0}\in {P}_{h}$ *and* $r\in (0,1)$ *such that* $r{v}_{0}\le {u}_{0}<{v}_{0}$ *and* *where* $h(t)={t}^{\nu -1}$,

$t\in [0,1]$ *and* $G(t,s)$ *is given as in* (2.2); (ii)

*the FBVP* $\{\begin{array}{c}-{D}_{{0}^{+}}^{\nu}y(t)=f(t,y(t),y(t)),\phantom{\rule{1em}{0ex}}0<t<1,n-1<\nu \le n,\hfill \\ {y}^{(i)}(0)=0,\phantom{\rule{1em}{0ex}}0\le i\le n-2,\hfill \\ {[{D}_{{0}^{+}}^{\alpha}y(t)]}_{t=1}=0,\phantom{\rule{1em}{0ex}}1\le \alpha \le n-2,\hfill \end{array}$

*has a unique positive solution* ${u}^{\ast}$ *in* ${P}_{h}$; (iii)

*for any* ${x}_{0},{y}_{0}\in {P}_{h}$,

*constructing successively the sequences* *we have* $\parallel {x}_{n}-{u}^{\ast}\parallel \to 0$ *and* $\parallel {y}_{n}-{u}^{\ast}\parallel \to 0$ *as* $n\to \mathrm{\infty}$.

**Corollary 3.4** *Let* $f\equiv 0$.

*Assume that* *g* *satisfies the conditions of Theorem * 3.2

*and* $g(t,0)\not\equiv 0$ *for* $t\in [0,1]$.

*Then*: (i)

*there exist* ${u}_{0},{v}_{0}\in {P}_{h}$ *and* $r\in (0,1)$ *such that* $r{v}_{0}\le {u}_{0}<{v}_{0}$ *and* ${u}_{0}(t)\le {\int}_{0}^{1}G(t,s)g(s,{u}_{0}(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}{v}_{0}(t)\ge {\int}_{0}^{1}G(t,s)g(s,{v}_{0}(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,1],$

*where* $h(t)={t}^{\nu -1}$,

$t\in [0,1]$ *and* $G(t,s)$ *is given as in* (2.2); (ii)

*the FBVP* $\{\begin{array}{c}-{D}_{{0}^{+}}^{\nu}y(t)=g(t,y(t)),\phantom{\rule{1em}{0ex}}0<t<1,n-1<\nu \le n,\hfill \\ {y}^{(i)}(0)=0,\phantom{\rule{1em}{0ex}}0\le i\le n-2,\hfill \\ {[{D}_{{0}^{+}}^{\alpha}y(t)]}_{t=1}=0,\phantom{\rule{1em}{0ex}}1\le \alpha \le n-2,\hfill \end{array}$

*has a unique positive solution* ${u}^{\ast}$ *in* ${P}_{h}$; (iii)

*for any* ${x}_{0},{y}_{0}\in {P}_{h}$,

*constructing successively the sequences* ${x}_{n+1}(t)={\int}_{0}^{1}G(t,s)g(s,{x}_{n}(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}{y}_{n+1}(t)={\int}_{0}^{1}G(t,s)g(s,{y}_{n}(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}n=0,1,2,\dots ,$

*we have* $\parallel {x}_{n}-{u}^{\ast}\parallel \to 0$ *and* $\parallel {y}_{n}-{u}^{\ast}\parallel \to 0$ *as* $n\to \mathrm{\infty}$.