In the sequel we always take *E* = *C*[0,1] with the norm ||*u*|| = max_{0≤t≤1}|*u*(*t*)| and *P* = {*u* ∈ *C*[0, 1] | *u*(*t*) ≥ 0, 0 ≤ *t* ≤ 1}. Then *P* is a solid cone in *E* and *E* is a lattice under the partial ordering ≤ induced by *P*.

A solution of BVP (*P*) is a fourth differentiable function *u* : [0,1] → *R* such that *u* satisfies (*P*). *u* is said to be a positive solution of BVP (*P*) if *u*(*t*) *>* 0, 0 *< t <* 1. Let *G*(*t, s*) be Green's function of homogeneous linear problem *u*^{(4)}(*t*) = 0, *u*(0) = *u*(1) = *u*'(0) = *u*'(1) = 0. From Yao [11] we have

G\left(t,s\right)=\left\{\begin{array}{cc}\hfill \frac{1}{6}{t}^{2}{\left(1-s\right)}^{2}\left[\left(s-t\right)+2\left(1-t\right)s\right],\hfill & \hfill 0\le t\le s\le 1,\hfill \\ \hfill \frac{1}{6}{s}^{2}{\left(1-t\right)}^{2}\left[\left(t-s\right)+2\left(1-s\right)t\right],\hfill & \hfill 0\le s\le t\le 1,\hfill \end{array}\right.

and

(G_{1}) *G*(*t, s*) ≥ 0, 0 ≤ *t, s* ≤ 1;

(G_{2}) *G*(*t, s*) = *G*(*s, t*);

(G_{3}) *G*(*t, s*) ≥ *p*(*t*)*G*(*τ*; *s*), 0 ≤ *t, s, τ* ≤ 1, where p\left(t\right)=\frac{2}{3}\text{min}\left\{{t}^{2},{\left(1-t\right)}^{2}\right\}.

**Lemma 3.1** Let H\left(t\right)=\frac{1}{2}{t}^{2}{\left(1-t\right)}^{2} and q\left(s\right)=\frac{2}{3}{s}^{2}{\left(1-s\right)}^{2}. Then

q\left(s\right)H\left(t\right)\le G\left(t,s\right)\le H\left(t\right),\phantom{\rule{1em}{0ex}}0\le t,s\le 1.

(3.1)

**Proof**. Since *G*(0, *s*) = *G*(1, *s*) = 0, 0 ≤ *s* ≤ 1, *H*(0) = *H*(1) = 0, then *q*(*s*)*H*(*t*) = *G*(*t, s*) = *H*(*t*) holds for *t* = 0 and *t* = 1. If 0 *< t* ≤ *s* ≤ 1 and *t <* 1, then

\begin{array}{c}G\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right)\phantom{\rule{2.77695pt}{0ex}}=\frac{1}{6}{t}^{2}{\left(1-s\right)}^{2}\left[3s\left(1-t\right)-t\left(1-s\right)\right]\\ \le \frac{1}{2}{t}^{2}{\left(1-s\right)}^{2}s\left(1-t\right)\\ \le \frac{1}{2}{t}^{2}{\left(1-t\right)}^{2}s\left(1-s\right)\\ <\frac{1}{2}{t}^{2}{\left(1-t\right)}^{2}=H\left(t\right),\end{array}

and

\begin{array}{ll}\hfill \frac{G\left(t,s\right)}{H\left(t\right)}\phantom{\rule{2.77695pt}{0ex}}& =\frac{\frac{1}{6}{t}^{2}{\left(1-s\right)}^{2}\left[3s\left(1-t\right)-t\left(1-s\right)\right]}{\frac{1}{2}{t}^{2}{\left(1-t\right)}^{2}}\phantom{\rule{2em}{0ex}}\\ \ge \frac{{\left(1-s\right)}^{2}\left[3s\left(1-t\right)-t\left(1-t\right)\right]}{3{\left(1-t\right)}^{2}}\phantom{\rule{2em}{0ex}}\\ =\frac{{\left(1-s\right)}^{2}\left(3s-t\right)}{3\left(1-t\right)}\ge \frac{2s{\left(1-s\right)}^{2}}{3\left(1-t\right)}\phantom{\rule{2em}{0ex}}\\ \ge \frac{2}{3}{s}^{2}{\left(1-s\right)}^{2}=q\left(s\right).\phantom{\rule{2em}{0ex}}\end{array}

Similarly, (3.1) holds for 0 ≤ *s* ≤ *t <* 1 and *t >* 0. The proof is complete. □

It is well known that the problem (*P*) is equivalent to the integral equation

u\left(t\right)=\mathrm{\lambda}\underset{0}{\overset{1}{\int}}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds.

Let

\left(Au\right)\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds,

(3.2)

\left(Bu\right)\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)u\left(s\right)ds.

(3.3)

**Lemma 3.2** Let *B* be defined by (3.3). Then B is a *u*_{0}*-* bounded linear operator.

**Proof**. Let {u}_{0}\left(t\right)=H\left(t\right)=\frac{1}{2}{t}^{2}{\left(1-t\right)}^{2}, *t* ∈ [0,1]. For any *u* ∈ *P*\{*θ*}, by Lemma 3.1

we have

\underset{0}{\overset{1}{\int}}q\left(s\right)H\left(t\right)u\left(s\right)ds\le Bu\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)u\left(s\right)ds\le \underset{0}{\overset{1}{\int}}H\left(t\right)u\left(s\right)ds.

Take arbitrarily 0<{\epsilon}_{0}<\frac{1}{2}, then

{u}_{0}\left(t\right)\underset{{\epsilon}_{0}}{\overset{1-{\epsilon}_{0}}{\int}}q\left(s\right)u\left(s\right)ds\le Bu\left(t\right)\le \u2225u\u2225\cdot {u}_{0}\left(t\right).

Let \zeta ={\int}_{{\epsilon}_{0}}^{1-{\epsilon}_{0}}q\left(s\right)u\left(s\right)ds>0, *η* = ||*u*|| *>* 0. Then

\zeta {u}_{0}\le Bu\le \eta {u}_{0}.

This indicates that *B* : *E* → *E* is a *u*_{0}*-* bounded linear operator. □

From Lemma 2.1 we have *r*(*B*) ≠ 0 and *r*^{-}^{1}(*B*) is the only eigenvalue of *B*. Denote λ_{1} = *r*^{-}^{1}(*B*).

Now let us list the following conditions which will be used in this article:

(H_{1}) there exist constants *α* and *β* with *α > β* ≥ 0 satisfying

\underset{u\to +\infty}{\text{lim}\text{inf}}\frac{f\left(t,u\right)}{u}\ge \alpha ,\underset{u\to -\infty}{\text{lim}\text{sup}}\frac{f\left(t,u\right)}{u}\le \beta ,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right].\phantom{\rule{0.3em}{0ex}}

(3.4)

(H_{2}) there exists a constant *γ* ≥ 0 satisfying

\underset{u\to 0}{\text{lim}\text{sup}}\left|\frac{f\left(t,u\right)}{u}\right|\le \gamma ,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right].

(3.5)

(H_{3}) \underset{u\to +\infty}{\text{lim}}\frac{f\left(t,u\right)}{u}=+\infty.

**Theorem 3.1** Suppose that (H_{1}) and (H_{2}) hold. Then for any \mathrm{\lambda}\in \left(\frac{{\mathrm{\lambda}}_{1}}{\alpha},\frac{{\mathrm{\lambda}}_{1}}{\iota}\right), BVP (*P*) has at least one nontrivial solution, where λ_{1} = *r*^{-}^{1}(*B*) is the only eigenvalue of *B, B* is denoted by (3.3), *ι* = max{*β, γ*}.

**Proof**. Let (*Fu*)(*t*) = *f*(*t, u*(*t*)). Then *A* = *BF*, where *A* is denoted by (3.2). By Remark 2.2, *F* is quasi-additive on lattice. Applying the Arzela-Ascoli theorem and a standard argument, we can prove that *A* : *E* → *E* is a completely continuous operator.

Now we show that λ*A* = λ*BF* has at least one nontrivial fixed point, which is the nontrivial solution of BVP (*P*).

On account of (G_{3}) we have that p\left(t\right)=\frac{2}{3}\text{min}\left\{{t}^{2},{\left(1-t\right)}^{2}\right\}\in P\backslash \left\{\theta \right\} such that

G\left(t,s\right)\ge p\left(t\right)G\left(\tau ,s\right),\phantom{\rule{1em}{0ex}}0\le t,\phantom{\rule{2.77695pt}{0ex}}s,\phantom{\rule{2.77695pt}{0ex}}\tau \le 1.

Notice that Bu\left(t\right)={\int}_{0}^{1}G\left(t,s\right)u\left(s\right)ds, where *G*(*t, s*) ≥ 0, *G*(*t, s*) ∈ *C*([0,1] × [0,1]). From Lemma 3.2 *B* is a *u*_{0}*-* bounded linear operator. By Lemma 2.1 we have *r*(*B*) ≠ 0 and λ_{1} = *r*^{-}^{1}(*B*) is the only eigenvalue of *B*. Then there exist \stackrel{\u0304}{\phi}\in P\backslash \left\{\theta \right\} and *g** ∈ *P**\{*θ*} such that (2.3) holds. Notice that λ *>* 0, from Remark 2.3, λ*B* satisfies *H* condition.

By (3.4) and (3.5), there exist *r >* 0, *M >* 0 and 0<\epsilon <\text{min}\left\{\frac{\alpha \mathrm{\lambda}-{\mathrm{\lambda}}_{1}}{\mathrm{\lambda}},\frac{{\mathrm{\lambda}}_{1}-\beta \mathrm{\lambda}}{\mathrm{\lambda}},\frac{{\mathrm{\lambda}}_{1}-\gamma \mathrm{\lambda}}{\mathrm{\lambda}}\right\} such that

f\left(t,u\right)\ge \left(\alpha -\epsilon \right)u-M,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}u\ge 0,

(3.6)

f\left(t,u\right)\ge \left(\beta +\epsilon \right)u-M,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}u\le 0,

(3.7)

\left|f\left(t,u\right)\right|\phantom{\rule{2.77695pt}{0ex}}\le \left(\gamma +\epsilon \right)\left|u\right|,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}\left|u\right|\phantom{\rule{2.77695pt}{0ex}}\le r.

(3.8)

By (3.6) and (3.7), we have (2.4) and (2.5) hold, where *a*_{1} = *α - ε, a*_{2} = *β* + *ε*.

Let *B*_{1} = λ*B*. Then {r}^{-1}\left({B}_{1}\right)=\frac{{\mathrm{\lambda}}_{1}}{\mathrm{\lambda}}. Obviously, for any \mathrm{\lambda}\in \left(\frac{{\mathrm{\lambda}}_{1}}{\alpha},\frac{{\mathrm{\lambda}}_{1}}{\iota}\right), *a*_{1}*> r*^{-}^{1}(*B*_{1}), *a*_{2}*< r*^{-}^{1}(*B*_{1}). From Lemma 2.2 there exists *R*_{0}*>* 0 such that for any *R >* max{*R*_{0}, *r*},

\text{deg}\left(I-\mathrm{\lambda}A,{T}_{R},\phantom{\rule{2.77695pt}{0ex}}\theta \right)=0.

(3.9)

Let *B*_{2} = λ(*γ* + *ε*)*B*. From (3.8) we have |*λAu*| ≤ *B*_{2}|*u*|, also r\left({B}_{2}\right)=\frac{\mathrm{\lambda}\left(\gamma +\epsilon \right)}{{\mathrm{\lambda}}_{1}}\le 1. Without loss of generality we assume that *λA* has no fixed point on ∂*T*_{
r
} , where *T*_{
r
} = {*u* ∈ *C*[0,1] | ||*u*|| *< r*}. By Lemma 2.3 we have

\text{deg}\left(I-\mathrm{\lambda}A,{T}_{r},\phantom{\rule{2.77695pt}{0ex}}\theta \right)=1,

(3.10)

It is easy to see from (3.9) and (3.10) that *λA* has at least one nontrivial fixed point. Thus problem (*P*) has at least one nontrivial solution. □

**Remark 3.1** If *α* = +∞, *β* = *γ* = 0, then for any λ *>* 0 problem (*P*) has at least one nontrivial solution.

**Theorem 3.2** Suppose that (H_{1}) holds. Assume *f*(*t*, 0) ≡ 0, ∀*t* ∈ [0,1] and

\underset{u\to 0}{\text{lim}}\frac{f\left(t,u\right)}{u}=\rho .

(3.11)

Then for any \mathrm{\lambda}\in \left(\frac{{\mathrm{\lambda}}_{1}}{\alpha},\frac{{\mathrm{\lambda}}_{1}}{\beta}\right) and \mathrm{\lambda}\ne \frac{{\mathrm{\lambda}}_{1}}{\rho}, BVP (*P*) has at least one nontrivial solution.

**Proof**. Since *f*(*t*, 0) ≡ 0, ∀*t* ∈ [0,1], then *Aθ* = *θ*. By (3.11) we have that the Frechet derivative {A}_{\theta}^{\prime} of *A* at *θ* exists and

\left({A}_{\theta}^{\prime}u\right)\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)\rho u\left(s\right)ds.

Notice that \mathrm{\lambda}\ne \frac{{\mathrm{\lambda}}_{1}}{\rho}, then 1 is not an eigenvalue of \mathrm{\lambda}{A}_{\theta}^{\prime}. By the famous Leray-Schauder theorem there exists *r >* 0 such that

\text{deg}\left(I-\mathrm{\lambda}A,{T}_{r},\theta \right)=\text{deg}\left(I-\mathrm{\lambda}{A}_{\theta}^{\prime},{T}_{r},\theta \right)={\left(-1\right)}^{\kappa}\ne 0,

(3.12)

where *κ* is the sum of algebraic multiplicities for all eigenvalues of \mathrm{\lambda}{A}_{\theta}^{\prime} lying in the interval (0, 1). From the proof of Theorem 3.1 we have that (3.9) holds for any \mathrm{\lambda}\in \left(\frac{{\mathrm{\lambda}}_{1}}{\alpha},\frac{{\mathrm{\lambda}}_{1}}{\beta}\right). By (3.9) and (3.12), *λA* has at least one nontrivial fixed point. Thus problem (*P*) has at least one nontrivial solution. □