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}}vu\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(1x,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(1x),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(1s),\hfill & 0\u2a7dt\u2a7ds\u2a7d1,\hfill \\ s(1t),\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}(1t)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(1x,1t)=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(1x)\in [\frac{1}{2},1], and
\begin{array}{rcl}(Ty)(1x)& =& {\int}_{0}^{1}(G(1x,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(1x,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(1x,1t)f(1t,y(1t),{y}^{\prime}(1t))\phantom{\rule{0.2em}{0ex}}d(1t)\\ +\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(1t,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)(1x)=(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 equicontinuous. 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})(1t)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}(1t)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 ArzelàAscoli 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}(1s)f(s,{y}_{2}(s),{y}_{2}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds{\int}_{0}^{1}(1s)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. □