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
$G\left(t,s\right)=\left\{\begin{array}{cc}\hfill \frac{1}{6}{t}^{2}{\left(1s\right)}^{2}\left[\left(st\right)+2\left(1t\right)s\right],\hfill & \hfill 0\le t\le s\le 1,\hfill \\ \hfill \frac{1}{6}{s}^{2}{\left(1t\right)}^{2}\left[\left(ts\right)+2\left(1s\right)t\right],\hfill & \hfill 0\le s\le t\le 1,\hfill \end{array}\right.$
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 $p\left(t\right)=\frac{2}{3}\text{min}\left\{{t}^{2},{\left(1t\right)}^{2}\right\}$.
Lemma 3.1 Let
$H\left(t\right)=\frac{1}{2}{t}^{2}{\left(1t\right)}^{2}$ and
$q\left(s\right)=\frac{2}{3}{s}^{2}{\left(1s\right)}^{2}$. 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
$\begin{array}{c}G\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right)\phantom{\rule{2.77695pt}{0ex}}=\frac{1}{6}{t}^{2}{\left(1s\right)}^{2}\left[3s\left(1t\right)t\left(1s\right)\right]\\ \le \frac{1}{2}{t}^{2}{\left(1s\right)}^{2}s\left(1t\right)\\ \le \frac{1}{2}{t}^{2}{\left(1t\right)}^{2}s\left(1s\right)\\ <\frac{1}{2}{t}^{2}{\left(1t\right)}^{2}=H\left(t\right),\end{array}$
and
$\begin{array}{ll}\hfill \frac{G\left(t,s\right)}{H\left(t\right)}\phantom{\rule{2.77695pt}{0ex}}& =\frac{\frac{1}{6}{t}^{2}{\left(1s\right)}^{2}\left[3s\left(1t\right)t\left(1s\right)\right]}{\frac{1}{2}{t}^{2}{\left(1t\right)}^{2}}\phantom{\rule{2em}{0ex}}\\ \ge \frac{{\left(1s\right)}^{2}\left[3s\left(1t\right)t\left(1t\right)\right]}{3{\left(1t\right)}^{2}}\phantom{\rule{2em}{0ex}}\\ =\frac{{\left(1s\right)}^{2}\left(3st\right)}{3\left(1t\right)}\ge \frac{2s{\left(1s\right)}^{2}}{3\left(1t\right)}\phantom{\rule{2em}{0ex}}\\ \ge \frac{2}{3}{s}^{2}{\left(1s\right)}^{2}=q\left(s\right).\phantom{\rule{2em}{0ex}}\end{array}$
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 ${u}_{0}\left(t\right)=H\left(t\right)=\frac{1}{2}{t}^{2}{\left(1t\right)}^{2}$, 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 quasiadditive on lattice. Applying the ArzelaAscoli theorem and a standard argument, we can prove that A : E → E is a completely continuous operator.
Now we show that λA = λBF has at least one nontrivial fixed point, which is the nontrivial solution of BVP (P).
On account of (G
_{3}) we have that
$p\left(t\right)=\frac{2}{3}\text{min}\left\{{t}^{2},{\left(1t\right)}^{2}\right\}\in P\backslash \left\{\theta \right\}$ 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, λB 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
$f\left(t,u\right)\ge \left(\alpha \epsilon \right)uM,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}u\ge 0,$
(3.6)
$f\left(t,u\right)\ge \left(\beta +\epsilon \right)uM,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}u\le 0,$
(3.7)
$\leftf\left(t,u\right)\right\phantom{\rule{2.77695pt}{0ex}}\le \left(\gamma +\epsilon \right)\leftu\right,\phantom{\rule{1em}{0ex}}\forall t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}\leftu\right\phantom{\rule{2.77695pt}{0ex}}\le r.$
(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 LeraySchauder 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. □