Let

$E=C[0,1]$ be a Banach space with the maximum norm

$\parallel u\parallel ={max}_{0\le t\le 1}|u(t)|$ for

$u\in E$. Define

$P=\{u\in E\mid u(t)\ge 0,t\in [0,1]\}$ and

${B}_{r}=\{u\in E\mid \parallel u\parallel <r\}$. Then

*P* is a total cone in

*E*, that is,

$E=\overline{P-P}$.

${P}^{\ast}$ denotes the dual cone of

*P*, namely,

${P}^{\ast}=\{g\in {E}^{\ast}\mid g(u)\ge 0,\text{for all}u\in P\}$. Let

${E}^{\ast}$ denote the dual space of

*E*, then by Riesz representation theorem,

${E}^{\ast}$ is given by

$\begin{array}{rcl}{E}^{\ast}& =& \{v\mid v\text{is right continuous on}[0,1)\text{and is bounded variation on}[0,1]\\ \text{with}v(0)=0\}.\end{array}$

We assume that the following condition holds throughout this paper.

(H

_{1})

$u(t)\equiv 0$ *is the unique* ${C}^{2}$ *solution of the linear boundary value problem* $\{\begin{array}{l}-(Lu)(t)=0,\phantom{\rule{1em}{0ex}}0<t<1,\\ (cos{\gamma}_{0})u(0)-(sin{\gamma}_{0}){u}^{\prime}(0)=0,\phantom{\rule{2em}{0ex}}(cos{\gamma}_{1})u(1)+(sin{\gamma}_{1}){u}^{\prime}(1)=0.\end{array}$

Let

$\phi ,\psi \in {C}^{2}([0,1],{\mathbf{R}}^{+})$ solve the following inhomogeneous boundary value problems, respectively:

$\{\begin{array}{l}-(L\phi )(t)=0,\phantom{\rule{1em}{0ex}}0<t<1,\\ (cos{\gamma}_{0})\phi (0)-(sin{\gamma}_{0}){\phi}^{\prime}(0)=1,\\ (cos{\gamma}_{1})\phi (1)+(sin{\gamma}_{1}){\phi}^{\prime}(1)=0\end{array}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\{\begin{array}{l}-(L\psi )(t)=0,\phantom{\rule{1em}{0ex}}0<t<1,\\ (cos{\gamma}_{0})\psi (0)-(sin{\gamma}_{0}){\psi}^{\prime}(0)=0,\\ (cos{\gamma}_{1})\psi (1)+(sin{\gamma}_{1}){\psi}^{\prime}(1)=1.\end{array}$

Let ${\kappa}_{1}=1-{\int}_{0}^{1}\phi (\tau )\phantom{\rule{0.2em}{0ex}}d\alpha (\tau )$, ${\kappa}_{2}={\int}_{0}^{1}\psi (\tau )\phantom{\rule{0.2em}{0ex}}d\alpha (\tau )$, ${\kappa}_{3}={\int}_{0}^{1}\phi (\tau )\phantom{\rule{0.2em}{0ex}}d\beta (\tau )$, ${\kappa}_{4}=1-{\int}_{0}^{1}\psi (\tau )\phantom{\rule{0.2em}{0ex}}d\beta (\tau )$.

(H_{2}) ${\kappa}_{1}>0$, ${\kappa}_{4}>0$, $k={\kappa}_{1}{\kappa}_{4}-{\kappa}_{2}{\kappa}_{3}>0$.

**Lemma 2.1** ([7])

*If* (H

_{1})

*and* (H

_{2})

*hold*,

*then BVP* (1.1)

*is equivalent to* $u(t)={\int}_{0}^{1}G(t,s)h(s)f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds,$

*where* $G(t,s)\in C([0,1]\times [0,1],{\mathbf{R}}^{+})$ *is the Green function for* (1.1).

Define an operator

$A:E\to E$ as follows:

$(Au)(t)={\int}_{0}^{1}G(t,s)h(s)f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}u\in E.$

(2.1)

It is easy to show that $A:E\to E$ is a completely continuous nonlinear operator, and if $u\in E$ is a fixed point of *A*, then *u* is a solution of BVP (1.1) by Lemma 2.1.

For any

$u\in E$, define a linear operator

$K:E\to E$ as follows:

$(Ku)(t)={\int}_{0}^{1}G(t,s)h(s)u(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}u\in E.$

(2.2)

It is easy to show that

$K:E\to E$ is a completely continuous nonlinear operator and

$K(P)\subset P$ holds. By [

7], the spectral radius

$r(K)$ of

*K* is positive. The Krein-Rutman theorem [

21] asserts that there are

$\varphi \in P\setminus \{0\}$ and

$\omega \in {P}^{\ast}\setminus \{0\}$ corresponding to the first eigenvalue

${\lambda}_{1}=1/r(K)$ of

*K* such that

${\lambda}_{1}K\varphi =\varphi $

(2.3)

and

${\lambda}_{1}{K}^{\ast}\omega =\omega ,\phantom{\rule{2em}{0ex}}\omega (1)=1.$

(2.4)

Here

${K}^{\ast}:{E}^{\ast}\to {E}^{\ast}$ is the dual operator of

*K* given by:

$\left({K}^{\ast}v\right)(s)={\int}_{0}^{s}{\int}_{0}^{1}G(t,\tau )h(\tau )\phantom{\rule{0.2em}{0ex}}dv(t)\phantom{\rule{0.2em}{0ex}}d\tau ,\phantom{\rule{1em}{0ex}}v\in {E}^{\ast}.$

The representation of

${K}^{\ast}$, the continuity of

*G* and the integrability of

*h* imply that

$\omega \in {C}^{1}[0,1]$. Let

$e(t):={\omega}^{\prime}(t)$. Then

$e\in P\setminus \{0\}$, and (2.4) can be rewritten equivalently as

$r(K)e(s)={\int}_{0}^{1}G(t,s)h(s)e(t)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{2em}{0ex}}{\int}_{0}^{1}e(t)\phantom{\rule{0.2em}{0ex}}dt=1.$

(2.5)

**Lemma 2.2** ([7])

*If* (H_{1}) *holds*, *then there is* $\delta >0$ *such that* ${P}_{0}=\{u\in P\mid {\int}_{0}^{1}u(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\ge \delta \parallel u\parallel \}$ *is a subcone of* *P* *and* $K(P)\subset {P}_{0}$.

**Lemma 2.3** ([22])

*Let* *E* *be a real Banach space and* $\mathrm{\Omega}\subset E$ *be a bounded open set with* $0\in \mathrm{\Omega}$. *Suppose that* $A:\overline{\mathrm{\Omega}}\to E$ *is a completely continuous operator*. (1) *If there is* ${y}_{0}\in E$ *with* ${y}_{0}\ne 0$ *such that* $u\ne Au+\mu {y}_{0}$ *for all* $u\in \partial \mathrm{\Omega}$ *and* $\mu \ge 0$, *then* $deg(I-A,\mathrm{\Omega},0)=0$. (2) *If* $Au\ne \mu u$ *for all* $u\in \partial \mathrm{\Omega}$ *and* $\mu \ge 1$, *then* $deg(I-A,\mathrm{\Omega},0)=1$. *Here* deg *stands for the Leray*-*Schauder topological degree in* *E*.

**Lemma 2.4** *Assume that* (H_{1}), (H_{2}) *and the following assumptions are satisfied*:

(C_{1}) *There exist* $\varphi \in P\setminus \{0\}$, $\omega \in {P}^{\ast}\setminus \{0\}$ *and* $\delta >0$ *such that* (2.3), (2.4) *hold and* *K* *maps* *P* *into* ${P}_{0}$.

(C

_{2})

*There exists a continuous operator* $H:E\to P$ *such that* $\underset{\parallel u\parallel \to +\mathrm{\infty}}{lim}\frac{\parallel Hu\parallel}{\parallel u\parallel}=0.$

(C_{3}) *There exist a bounded continuous operator* $F:E\to E$ *and* ${u}_{0}\in E$ *such that* $Fu+{u}_{0}+Hu\in P$ *for all* $u\in E$.

(C_{4}) *There exist* ${v}_{0}\in E$ *and* $\zeta >0$ *such that* $KFu\ge {\lambda}_{1}(1+\zeta )Ku-KHu-{v}_{0}$ *for all* $u\in E$.

*Let* $A=KF$,

*then there exists* $R>0$ *such that* *where* ${B}_{R}=\{u\in E\mid \parallel u\parallel <R\}$.

*Proof* Choose a constant

${L}_{0}={(\delta {\lambda}_{1})}^{-1}(1+{\zeta}^{-1})+\parallel K\parallel >0$. From (C

_{2}), for

$0<{\epsilon}_{0}<{L}_{0}^{-1}$, there exists

${R}_{1}>0$ such that

$\parallel u\parallel >{R}_{1}$ implies

$\parallel Hu\parallel <{\epsilon}_{0}\parallel u\parallel .$

(2.6)

Now we shall show

$u\ne KFu+\mu \varphi \phantom{\rule{1em}{0ex}}\text{for any}u\in \partial {B}_{R}\text{and}\mu \ge 0,$

(2.7)

provided that *R* is sufficiently large.

In fact, if (2.7) is not true, then there exist

${u}_{1}\in \partial {B}_{R}$ and

${\mu}_{1}\ge 0$ satisfying

${u}_{1}=KF{u}_{1}+{\mu}_{1}\varphi .$

(2.8)

Since

$\varphi \in P\setminus \{0\}$,

$e(t)\in P\setminus \{0\}$,

${\int}_{0}^{1}\varphi (t)e(t)\phantom{\rule{0.2em}{0ex}}dt>0$. Multiply (2.8) by

$e(t)$ on both sides and integrate on

$[0,1]$. Then, by (C

_{4}), (2.5), we get

$\begin{array}{c}{\int}_{0}^{1}{u}_{1}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}={\int}_{0}^{1}(KF{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\mu}_{1}{\int}_{0}^{1}\varphi (t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\ge {\lambda}_{1}(1+\zeta ){\int}_{0}^{1}{\int}_{0}^{1}G(t,s)h(s){u}_{1}(s)\phantom{\rule{0.2em}{0ex}}dse(t)\phantom{\rule{0.2em}{0ex}}dt-{\int}_{0}^{1}(KH{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt-{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}={\lambda}_{1}(1+\zeta ){\int}_{0}^{1}{\int}_{0}^{1}G(t,s)h(s){u}_{1}(s)e(t)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{2em}{0ex}}-{\int}_{0}^{1}{\int}_{0}^{1}G(t,s)h(s)(H{u}_{1})(s)e(t)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt-{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}={\lambda}_{1}(1+\zeta ){\int}_{0}^{1}[{\int}_{0}^{1}G(t,s)h(s)e(t)\phantom{\rule{0.2em}{0ex}}dt]{u}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\hfill \\ \phantom{\rule{2em}{0ex}}-{\int}_{0}^{1}[{\int}_{0}^{1}G(t,s)h(s)e(t)\phantom{\rule{0.2em}{0ex}}dt](H{u}_{1})(s)\phantom{\rule{0.2em}{0ex}}ds-{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}={\lambda}_{1}(1+\zeta )r(K){\int}_{0}^{1}e(s){u}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds-r(K){\int}_{0}^{1}(H{u}_{1})(s)e(s)\phantom{\rule{0.2em}{0ex}}ds-{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}=(1+\zeta ){\int}_{0}^{1}{u}_{1}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt-r(K){\int}_{0}^{1}(H{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt-{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

(2.9)

Thus,

${\int}_{0}^{1}{u}_{1}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\le {\zeta}^{-1}(r(K){\int}_{0}^{1}(H{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt).$

(2.10)

By (2.9),

${\int}_{0}^{1}(KH{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt=r(K){\int}_{0}^{1}(H{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt$ holds. Then (2.3), (2.6) and (2.10) imply

$\begin{array}{c}{\int}_{0}^{1}({u}_{1}(t)+(KH{u}_{1})(t)+(K{u}_{0})(t))e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le {\zeta}^{-1}(r(K){\int}_{0}^{1}(H{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt)\hfill \\ \phantom{\rule{2em}{0ex}}+r(K){\int}_{0}^{1}(H{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{1}(K{u}_{0})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le {\zeta}^{-1}(1+\zeta )r(K){\int}_{0}^{1}(H{u}_{1})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\zeta}^{-1}{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{1}(K{u}_{0})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le {\zeta}^{-1}(1+\zeta )r(K){\epsilon}_{0}\parallel {u}_{1}\parallel +{L}_{1},\hfill \end{array}$

(2.11)

where ${L}_{1}={\zeta}^{-1}{\int}_{0}^{1}{v}_{0}(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{1}(K{u}_{0})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt$ is a constant.

(C

_{3}) shows

$F{u}_{1}+{u}_{0}+H{u}_{1}\in P$ and (C

_{1}) implies

${\mu}_{1}\varphi ={\mu}_{1}{\lambda}_{1}K{\phi}_{1}\in {P}_{0}$. Then (C

_{1}), (2.8) and Lemma 2.2 tell us that

${u}_{1}+KH{u}_{1}+K{u}_{0}=KF{u}_{1}+{\mu}_{1}\varphi +KH{u}_{1}+K{u}_{0}=K(F{u}_{1}+H{u}_{1}+{u}_{0})+{\mu}_{1}\varphi \in {P}_{0}.$

The definition of

${P}_{0}$ yields

$\begin{array}{rcl}{\int}_{0}^{1}({u}_{1}+KH{u}_{1}+K{u}_{0})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt& \ge & \delta \parallel {u}_{1}+KH{u}_{1}+K{u}_{0}\parallel \\ \ge & \delta \parallel {u}_{1}\parallel -\delta \parallel KH{u}_{1}\parallel -\delta \parallel K{u}_{0}\parallel .\end{array}$

(2.12)

It follows from (2.6), (2.11) and (2.12) that

$\begin{array}{rcl}\parallel {u}_{1}\parallel & =& {\delta}^{-1}{\int}_{0}^{1}({u}_{1}+KH{u}_{1}+K{u}_{0})(t)e(t)\phantom{\rule{0.2em}{0ex}}dt+\parallel KH{u}_{1}\parallel +\parallel K{u}_{0}\parallel \\ \le & {\epsilon}_{0}{(\delta {\lambda}_{1})}^{-1}(1+{\zeta}^{-1})\parallel {u}_{1}\parallel +{L}_{1}{\delta}^{-1}+{\epsilon}_{0}\parallel K\parallel \cdot \parallel {u}_{1}\parallel +\parallel K{u}_{0}\parallel \\ =& {\epsilon}_{0}{L}_{0}\parallel {u}_{1}\parallel +{L}_{2},\end{array}$

(2.13)

where ${L}_{2}=\parallel K{u}_{0}\parallel +{L}_{1}{\delta}^{-1}$ is a constant.

Since

$0<{\epsilon}_{0}{L}_{0}<1$, then (2.13) deduces that (2.7) holds provided that

*R* is sufficiently large such that

$R>max\{{L}_{2}/(1-{\epsilon}_{0}{L}_{0}),{R}_{1}\}$. By (2.13) and Lemma 2.3, we have

□