**Definition 2.1** Let

*E* be a Banach space, a nonempty convex closed set

$P\subset E$ is said to be a cone provided the following hypotheses are satisfied:

- (i)
if $u\in P$, $\lambda \u2a7e0$, then $\lambda u\in P$;

- (ii)
if $u\in P$ and $-u\in P$, then $u=0$.

Every cone

$P\subset E$ induces a partial ordering ‘⩽’ on E defined by

$u\u2a7dv\phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}v-u\in P.$

**Definition 2.2** Let $(E,\u2a7d)$ be an ordered Banach space. An operator $\phi :E\to E$ is said to be nondecreasing (nonincreasing) provided that $\phi (u)\u2a7d\phi (v)$ ($\phi (u)\u2a7e\phi (v)$) for all $u,v\in E$ with $u\u2a7dv$. If the inequality is strict, then *φ* is said to be strictly nondecreasing (nonincreasing).

**Definition 2.3** Let

$E={C}^{1}[0,1]$,

$u\in E$ is said to be concave on

$[0,1]$ if

$u(\lambda {x}_{1}+(1-\lambda ){x}_{2})\u2a7e\lambda u({x}_{1})+(1-\lambda )u({x}_{2})$

for any ${x}_{1},{x}_{2}\in [0,1]$ and $\lambda \in [0,1]$.

We consider the Banach space $E={C}^{1}[0,1]$ equipped with the norm $\parallel u\parallel =max\{{\parallel u\parallel}_{\mathrm{\infty}},{\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}}\}$, where ${\parallel u\parallel}_{\mathrm{\infty}}={max}_{x\in [0,1]}|u(x)|$. In this paper, a symmetric positive solution ${u}^{\ast}$ of (1.1) means a function which is symmetric and positive on $(0,1)$ and satisfies equation (1.1) as well as the boundary conditions (1.2).

In this paper, we always suppose that the following assumptions hold:

(H_{1}) $f\in C([0,1]\times [0,+\mathrm{\infty})\times R,[0,+\mathrm{\infty}))$, $f(x,u,v)=f(1-x,u,-v)$ for $x\in [\frac{1}{2},1]$, and $f(x,u,v)>0$ for all $(x,u,v)\in [0,1]\times [0,+\mathrm{\infty})\times R$;

(H_{2}) $f(x,\cdot ,v)$ is nondecreasing for each $(x,v)\in [0,\frac{1}{2}]\times R$, $f(x,u,\cdot )$ is nondecreasing for $(x,u)\in [0,\frac{1}{2}]\times [0,+\mathrm{\infty})$;

(H_{3}) $p\in {L}^{1}[0,1]$ is nonnegative and $0\le \mu <1$, where $\mu ={\int}_{0}^{1}p(s)\phantom{\rule{0.2em}{0ex}}ds$.

Denote

$\begin{array}{c}{C}^{+}[0,1]=\{u\in E:u(x)\u2a7e0,x\in [0,1]\},\hfill \\ P=\{u\in E:u(x)\u2a7e0\text{is concave and}u(x)=u(1-x),x\in [0,1]\}.\hfill \end{array}$

It is easy to see that *P* is a cone in *E*.

For any

$y\in {C}^{+}[0,1]$, suppose that

*u* is a solution of the following BVP:

$\{\begin{array}{c}{u}^{\u2033}(x)+f(x,y(x),{y}^{\prime}(x))=0,\phantom{\rule{1em}{0ex}}0<x<1,\hfill \\ u(0)=u(1)={\int}_{0}^{1}p(s)u(s)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

Then we can easily get the solution:

$u(x)={\int}_{0}^{1}H(x,t)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt,$

(2.1)

where

$H(x,t)=G(x,t)+\frac{1}{1-\mu}{\int}_{0}^{1}G(t,s)p(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}G(t,s)=\{\begin{array}{cc}t(1-s),\hfill & 0\u2a7dt\u2a7ds\u2a7d1,\hfill \\ s(1-t),\hfill & 0\u2a7ds\u2a7dt\u2a7d1\hfill \end{array}$

and

$\mu ={\int}_{0}^{1}p(s)\phantom{\rule{0.2em}{0ex}}ds.$

During the process of getting the above solution, we can also know

${u}^{\prime}(x)={\int}_{0}^{1}(1-t)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt-{\int}_{0}^{x}f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt$

(2.2)

for $x\in [0,1]$.

**Lemma 2.1** *If* (H

_{3})

*is satisfied*,

*the following results are true*:

- 1.
$H(x,t)\u2a7e0$, *for* $x,t\in [0,1]$; $H(x,t)>0$ *for* $x,t\in (0,1)$.

- 2.
$G(1-x,1-t)=G(x,t)$, $G(x,t)\u2a7dG(x,x)$ *for* $x,t\in [0,1]$.

For any

$y\in {C}^{+}[0,1]$,

$T:P\to E$ is defined

$(Ty)(x)={\int}_{0}^{1}H(x,t)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\text{for}x\in [0,1].$

(2.3)

**Lemma 2.2** *If* (H_{3}) *is satisfied*, $T:P\to P$ *is completely continuous*, *i*.*e*., *T is continuous and compact*. *Moreover*, *T* *is nondecreasing provided that* (H_{2}) *holds*.

*Proof* For any

$y\in P$, from the definition of

*Ty*, we know

$\{\begin{array}{c}{(Ty)}^{\u2033}(x)+f(x,y(x),{y}^{\prime}(x))=0,\phantom{\rule{1em}{0ex}}0<x<1,\hfill \\ (Ty)(0)=(Ty)(1)={\int}_{0}^{1}p(s)(Ty)(s)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

Obviously, *Ty* is concave. From the expression of *Ty*, combining with Lemma 2.1, we know that *Ty* is nonnegative on $[0,1]$. We now prove that *Ty* is symmetric about $\frac{1}{2}$.

For

$x\in [0,\frac{1}{2}]$, then

$(1-x)\in [\frac{1}{2},1]$, and

$\begin{array}{rcl}(Ty)(1-x)& =& {\int}_{0}^{1}(G(1-x,t)+\frac{1}{1-\mu}{\int}_{0}^{1}G(t,s)p(s)\phantom{\rule{0.2em}{0ex}}ds)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int}_{0}^{1}G(1-x,t)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{1-\mu}{\int}_{0}^{1}{\int}_{0}^{1}G(t,s)p(s)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int}_{1}^{0}G(1-x,1-t)f(1-t,y(1-t),{y}^{\prime}(1-t))\phantom{\rule{0.2em}{0ex}}d(1-t)\\ +\frac{1}{1-\mu}{\int}_{0}^{1}{\int}_{0}^{1}G(t,s)p(s)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int}_{0}^{1}G(x,t)f(1-t,y(t),-{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{1-\mu}{\int}_{0}^{1}{\int}_{0}^{1}G(t,s)p(s)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int}_{0}^{1}G(x,t)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt+\frac{1}{1-\mu}{\int}_{0}^{1}{\int}_{0}^{1}G(t,s)p(s)f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt\\ =& (Ty)(x).\end{array}$

Similarly, we have

$(Ty)(1-x)=(Ty)(x)\phantom{\rule{1em}{0ex}}\text{for}x\in [\frac{1}{2},1].$

So,

$TP\subset P$. The continuity of

*T* is obvious. We now prove that

*T* is compact. Let

$\mathrm{\Omega}\subset P$ be a bounded set. Then there exists

*R* such that

$\mathrm{\Omega}=\{y\in P\mid \parallel y\parallel \u2a7dR\}.$

For any

$y\in \mathrm{\Omega}$, we have

$0\u2a7df(s,y(s),{y}^{\prime}(s))\u2a7dmax\{f(s,y,{y}^{\prime})\mid s\in [0,1],y\in [0,R],{y}^{\prime}\in [-R,R]\}=:M.$

Therefore, from (2.3), we have

${\parallel (Ty)\parallel}_{\mathrm{\infty}}\u2a7dM+\frac{\mu}{1-\mu}M=\frac{M}{1-\mu},\phantom{\rule{2em}{0ex}}{\parallel {(Ty)}^{\prime}\parallel}_{\mathrm{\infty}}\u2a7dM.$

So,

$\parallel (Ty)\parallel $ is uniformly bounded. Now we prove

*Ty* is equi-continuous. For

$0\u2a7d{x}_{1}\u2a7d{x}_{2}\u2a7d\frac{1}{2}$, we have

$\begin{array}{r}|(Ty)({x}_{2})-(Ty)({x}_{1})|\\ \phantom{\rule{1em}{0ex}}=\left|{\int}_{0}^{1}(G({x}_{2},t)-G({x}_{1},t))f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\right|\\ \phantom{\rule{1em}{0ex}}\u2a7d\{\begin{array}{cc}{\int}_{0}^{1}|({x}_{2}-{x}_{1})(1-t)|f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt,\hfill & 0\u2a7d{x}_{1}\u2a7d{x}_{2}\u2a7dt\u2a7d1,\hfill \\ {\int}_{0}^{1}|t({x}_{1}-{x}_{2})|f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt,\hfill & 0\u2a7dt\u2a7d{x}_{1}\u2a7d{x}_{2}\u2a7d1,\hfill \\ {\int}_{0}^{1}|t(1-{x}_{2})-{x}_{1}(1-t)|f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt,\hfill & 0\u2a7d{x}_{1}\u2a7dt\u2a7d{x}_{2}\u2a7d1.\hfill \end{array}\\ \phantom{\rule{1em}{0ex}}\u2a7dM|{x}_{2}-{x}_{1}|.\end{array}$

Moreover,

$|{(Ty)}^{\prime}({x}_{2})-{(Ty)}^{\prime}({x}_{1})|=\left|{\int}_{{x}_{1}}^{{x}_{2}}f(t,y(t),{y}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\right|\u2a7dM|{x}_{2}-{x}_{1}|.$

And the similar results can be obtained for $\frac{1}{2}\u2a7d{x}_{1}\u2a7d{x}_{2}\u2a7d1$ and $0\u2a7d{x}_{1}\u2a7d\frac{1}{2}\u2a7d{x}_{2}\u2a7d1$.

The Arzelà-Ascoli theorem guarantees that *T* Ω is relatively compact, which means *T* is compact.

Finally, we show that *Ty* is nondecreasing about *y*.

For any

${y}_{i}(x)\in P$ (

$i=1,2$) with

${y}_{1}(x)\u2a7d{y}_{2}(x)$. By the properties of a cone, we have

${y}_{2}(x)-{y}_{1}(x)\in P$ for

$x\in [0,1]$. Then

${y}_{2}(x)-{y}_{1}(x)\u2a7e0$ is concave and symmetric about

$\frac{1}{2}$. Therefore,

$\{\begin{array}{c}{y}_{2}^{\prime}(x)\u2a7e{y}_{1}^{\prime}(x)\phantom{\rule{1em}{0ex}}\text{for}x\in [0,\frac{1}{2}],\hfill \\ {y}_{2}^{\prime}(x)\u2a7d{y}_{1}^{\prime}(x)\phantom{\rule{1em}{0ex}}\text{for}x\in [\frac{1}{2},1].\hfill \end{array}$

Hence, for

$x\in [0,\frac{1}{2}]$, by (H

_{2}) and the definition of

*Ty*, we have

$\begin{array}{rl}(T{y}_{1})(x)-(T{y}_{2})(x)=& {\int}_{0}^{1}H(x,t)f(t,{y}_{1}(t),{y}_{1}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\\ -{\int}_{0}^{1}H(x,t)f(t,{y}_{2}(t),{y}_{2}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\u2a7d0.\end{array}$

Furthermore, we have

$\begin{array}{rl}{(T{y}_{2})}^{\prime}(x)-{(T{y}_{1})}^{\prime}(x)=& {\int}_{0}^{1}(1-s)f(s,{y}_{2}(s),{y}_{2}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds-{\int}_{0}^{1}(1-s)f(s,{y}_{1}(s),{y}_{1}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{0}^{x}f(s,{y}_{1}(s),{y}_{1}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds-{\int}_{0}^{x}f(s,{y}_{2}(s),{y}_{2}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

In order to prove

$(T{y}_{2})(x)-(T{y}_{1})(x)$ is concave, we need to prove

${(T{y}_{2})}^{\prime}(x)-{(T{y}_{1})}^{\prime}(x)$ is nonincreasing. Let

$0\u2a7d{x}_{1}\u2a7d{x}_{2}\u2a7d\frac{1}{2}$, then

$\begin{array}{c}{(T{y}_{2})}^{\prime}({x}_{2})-{(T{y}_{1})}^{\prime}({x}_{2})-{(T{y}_{2})}^{\prime}({x}_{1})+{(T{y}_{1})}^{\prime}({x}_{1})\hfill \\ \phantom{\rule{1em}{0ex}}={\int}_{{x}_{1}}^{{x}_{2}}f(s,{y}_{1}(s),{y}_{1}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds-{\int}_{{x}_{1}}^{{x}_{2}}f(s,{y}_{2}(s),{y}_{2}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\u2a7d0.\hfill \end{array}$

A similar result can be obtained for $x\in [\frac{1}{2},1]$. And it is easy to see that $(T{y}_{2})(x)-(T{y}_{1})(x)$ is symmetric about $\frac{1}{2}$. So, $(T{y}_{2}-T{y}_{1})\in P$ and thus *T* is nondecreasing. □