The following results will be used throughout the paper.
Let E=C[0,1], \parallel u\parallel ={max}_{t\in [0,1]}u(t). Then E is a real Banach space with the norm \parallel \cdot \parallel. Let P=\{u\in E:u(t)\ge 0,t\in [0,1]\}, and P is a normal solid cone of E, \{u\in E:u(t)>0,t\in (0,1]\}\subset \stackrel{\circ}{P}=\{x\in Px\text{is an interior point of}P\}.
Let the operators K, F, A be defined by
(Ku)(t)={\int}_{0}^{1}G(t,s)u(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}(Fu)(t)=f(t,u(t)),\phantom{\rule{1em}{0ex}}\text{for}t\in [0,1],u\in E,
(2.1)
and A=KF, respectively, where
G(t,s)=\frac{1}{2}\{\begin{array}{ll}2ts{s}^{2}{t}^{2}s,& 0\le s\le t\le 1,\\ {t}^{2}(1s),& 0\le t\le s\le 1.\end{array}
(2.2)
Remark 1 (1) A,K:E\to E are completely continuous. (2) G(t,s)\ge 0, t,s\in [0,1]. In fact, since it is obvious in the other case, we only need to prove the case 0\le s\le t\le 1. Now we suppose that 0\le s\le t\le 1. Then
G(t,s)=\frac{1}{2}(2ts{s}^{2}{t}^{2}s)=\frac{1}{2}s((t{t}^{2})+(ts))\ge 0.
Definition 2.1 [12]
We call E a lattice under the partial ordering ≤, if sup\{x,y\} and inf\{x,y\} exist for arbitrary x,y\in E.
Remark 2 E=C[0,1] is a lattice under the partial ordering ≤ that is deduced by the cone P=\{u\in E:u(t)\ge 0,t\in [0,1]\} of E.
Definition 2.2 [12]
Let E be a Banach space with a cone P,A:E\u27f6E be a nonlinear operator. We call that A is a unilaterally asymptotically linear operator along {P}_{w}=\{x\in E:x\ge w,w\in E\}, if there exists a bounded linear operator L such that
\underset{x\in {P}_{w},\parallel x\parallel \u27f6\mathrm{\infty}}{lim}\frac{\parallel AxLx\parallel}{\parallel x\parallel}=0.
L is said to be the derived operator of A along {P}_{w} and will be denoted by {A}_{{P}_{w}}^{\prime}. Similarly, we can also define a unilaterally asymptotically linear operator along {P}^{w}=\{x\in E:x\le w,w\in E\}. Specially, if w=\theta, We call that A is a unilaterally asymptotically linear operator along P and −P.
Definition 2.3 [12]
Let D\subseteq E and A:D\u27f6E be a nonlinear operator. A is said to be quasiadditive on lattice, if there exists {v}^{\ast}\in E such that
Ax=A{x}_{+}+A{x}_{}+{v}^{\ast},\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in D,
where {x}_{+}={x}^{+}=sup\{x,\theta \}, {x}_{}={x}^{}=sup\{x,\theta \}.
Remark 3 It is easy to see that the operators F and A=KF defined by (2.1) are both quasiadditive on the lattice E=C[0,1].
Let us list some conditions and preliminary lemmas to be used in this paper.
(H_{1}) f\in C([0,1]\times R,R) is strictly increasing in u, and f(t,u)u>0 for all t\in [0,1], u\in R\mathrm{\setminus}\{0\}.
(H_{2}) {lim}_{u\to +\mathrm{\infty}}\frac{f(t,u)}{u}={\beta}_{1} uniformly on [0,1]. There exists a positive integer {n}_{1} such that
{\lambda}_{2{n}_{1}}^{3}<{\beta}_{1}<{\lambda}_{2{n}_{1}+1}^{3},
where 0<{\lambda}_{1}<{\lambda}_{2}<\cdots <{\lambda}_{n}<\cdots are the positive solutions of the equation
\frac{1}{2}{e}^{\frac{3}{2}\lambda}=cos(\frac{\sqrt{3}}{2}\lambda +\frac{2\pi}{3}).
(2.3)
(H_{3}) {lim}_{u\to 0}\frac{f(t,u)}{u}={\beta}_{0} uniformly on [0,1], and 0<{\beta}_{0}<{\lambda}_{1}^{3}.
Lemma 2.1 For any f\in C[0,1], u\in {C}^{3}[0,1] is a solution of the following problem:
\{\begin{array}{l}{u}^{\u2034}(t)=f(t),\phantom{\rule{1em}{0ex}}t\in [0,1],\\ u(0)={u}^{\prime}(0)={u}^{\prime}(1)=0\end{array}
if and only if
u(t)
is a solution of the integral equation
u(t)={\int}_{0}^{1}G(t,s)f(s)\phantom{\rule{0.2em}{0ex}}ds,
where G(t,s) is defined by (2.2).
Proof On the one hand, integrating the equation
{u}^{\u2034}(t)=f(t),\phantom{\rule{1em}{0ex}}t\in [0,1]
over [0,t] for three times, we have
u(t)=\frac{1}{2}{\int}_{0}^{t}{(ts)}^{2}f(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{1}{2}{u}^{\u2033}(0){t}^{2}+{u}^{\prime}(0)t+u(0).
Then
{u}^{\prime}(t)=t{\int}_{0}^{t}f(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{t}sf(s)\phantom{\rule{0.2em}{0ex}}ds+{u}^{\u2033}(0)t+{u}^{\prime}(0).
Combining them with boundary condition u(0)={u}^{\prime}(0)={u}^{\prime}(1)=0, we conclude that
{u}^{\u2033}(0)={\int}_{0}^{1}(1s)f(s)\phantom{\rule{0.2em}{0ex}}ds.
Therefore,
\begin{array}{rl}u(t)& =\frac{1}{2}{\int}_{0}^{t}{(ts)}^{2}f(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{1}{2}{t}^{2}{\int}_{0}^{1}(1s)f(s)\phantom{\rule{0.2em}{0ex}}ds\\ =\frac{1}{2}{\int}_{0}^{t}[{(ts)}^{2}{t}^{2}(1s)]f(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{1}{2}{t}^{2}{\int}_{t}^{1}(1s)f(s)\phantom{\rule{0.2em}{0ex}}ds\\ ={\int}_{0}^{1}G(t,s)f(s)\phantom{\rule{0.2em}{0ex}}ds.\end{array}
On the other hand, since
u(t)={\int}_{0}^{1}G(t,s)f(s)\phantom{\rule{0.2em}{0ex}}ds=\frac{1}{2}{\int}_{0}^{t}(2ts{s}^{2}{t}^{2}s)f(s)\phantom{\rule{0.2em}{0ex}}ds+\frac{1}{2}{\int}_{t}^{1}{t}^{2}(1s)f(s)\phantom{\rule{0.2em}{0ex}}ds,
therefore,
\begin{array}{r}{u}^{\prime}(t)={\int}_{0}^{t}(sts)f(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{1}t(1s)f(s)\phantom{\rule{0.2em}{0ex}}ds,\\ {u}^{\u2033}(t)={\int}_{0}^{t}sf(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{t}^{1}(1s)f(s)\phantom{\rule{0.2em}{0ex}}ds,\end{array}
and
{u}^{\u2034}(t)=tf(t)(1t)f(t)=f(t).
Moreover, we get u(0)={u}^{\prime}(0)={u}^{\prime}(1)=0. □
Remark 4 Considering Lemma 2.1, we find that u is a solution of the problem (1.1) if and only if u is a fixed point of the operator A=KF.
From the following lemma, we can obtain the eigenvalues and the algebraic multiplicity of the linear operator K.
Lemma 2.2
The eigenvalues of the linear operator
K
are
\frac{1}{{\lambda}_{1}^{3}}>\frac{1}{{\lambda}_{2}^{3}}>\cdots >\frac{1}{{\lambda}_{n}^{3}}>\cdots
and the algebraic multiplicity of each positive eigenvalue \frac{1}{{\lambda}_{n}^{3}} is equal to 1, where 0<{\lambda}_{1}<{\lambda}_{2}<\cdots <{\lambda}_{n}<\cdots are the positive solutions of (2.3).
Proof Let \overline{\lambda} be a positive eigenvalue of the linear operator K, and y\in E\setminus \{\theta \} be an eigenfunction corresponding to eigenvalue \overline{\lambda}. By Lemma 2.1, we have
\{\begin{array}{l}{y}^{\u2034}(t)=\frac{1}{\overline{\lambda}}y(t),\phantom{\rule{1em}{0ex}}t\in (0,1),\\ y(0)={y}^{\prime}(0)={y}^{\prime}(1)=0.\end{array}
(2.4)
The auxiliary equation of the differential equation (2.4) has roots −μ, \frac{1}{2}\mu (1+\sqrt{3}i), \frac{1}{2}\mu (1\sqrt{3}i), where \mu =\frac{1}{\sqrt[3]{\overline{\lambda}}}. Thus the general solution of (2.4) is
y(t)={C}_{1}{e}^{\mu t}+{C}_{2}{e}^{\frac{1}{2}\mu t}cos\left(\frac{\sqrt{3}}{2}\mu t\right)+{C}_{3}{e}^{\frac{1}{2}\mu t}sin\left(\frac{\sqrt{3}}{2}\mu t\right).
Then
{y}^{\prime}(t)={C}_{1}(\mu {e}^{\mu t})+{C}_{2}\mu {e}^{\frac{1}{2}\mu t}cos(\frac{\sqrt{3}}{2}\mu t+\frac{\pi}{3})+{C}_{3}\mu {e}^{\frac{1}{2}\mu t}sin(\frac{\sqrt{3}}{2}\mu t+\frac{\pi}{3}).
Applying the condition y(0)={y}^{\prime}(0)=0, we obtain {C}_{2}={C}_{1}, {C}_{3}=\sqrt{3}{C}_{1}, where {C}_{1}\ne 0.
Applying the second condition {y}^{\prime}(1)=0, that is,
{e}^{\mu}+2{e}^{\frac{1}{2}\mu}cos(\frac{\sqrt{3}}{2}\mu +\frac{2\pi}{3})=0.
(2.5)
Considering (2.3), we see that μ is one of {\lambda}_{1},{\lambda}_{2},\dots ,{\lambda}_{n},\dots , that is,
\frac{1}{{\lambda}_{1}^{3}}>\frac{1}{{\lambda}_{2}^{3}}>\cdots >\frac{1}{{\lambda}_{n}^{3}}>\cdots
are eigenvalues of the linear operator K and the eigenfunction corresponding to the eigenvalue \frac{1}{{\lambda}_{n}^{3}} is
{y}_{n}(t)=C({e}^{{\lambda}_{n}t}{e}^{\frac{1}{2}{\lambda}_{n}t}cos\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)+\sqrt{3}{e}^{\frac{1}{2}{\lambda}_{n}t}sin\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)),
(2.6)
where C is a nonzero constant.
Next we prove that the algebraic multiplicity of the eigenvalue \frac{1}{{\lambda}_{n}^{3}} is 1. From (2.6), any two eigenfunctions corresponding to the same eigenvalue \frac{1}{{\lambda}_{n}^{3}} are merely nonzero constant multiples of each other, that is,
dimker(\frac{1}{{\lambda}_{n}^{3}}IK)=dimker(I{\lambda}_{n}^{3}K)=1.
Now we show that
ker(I{\lambda}_{n}^{3}K)=ker{(I{\lambda}_{n}^{3}K)}^{2}.
Obviously, we only need to show that
ker{(I{\lambda}_{n}^{3}K)}^{2}\subset ker(I{\lambda}_{n}^{3}K).
In fact, for any y\in ker{(I{\lambda}_{n}^{3}K)}^{2}, (I{\lambda}_{n}^{3}K)y is an eigenfunction of linear operator K corresponding to the eigenvalue \frac{1}{{\lambda}_{n}^{3}} if (I{\lambda}_{n}^{3}K)y\ne \theta. Considering (2.6), there exists a nonzero constant σ such that
(I{\lambda}_{n}^{3}K)y(t)=\sigma ({e}^{{\lambda}_{n}t}{e}^{\frac{1}{2}{\lambda}_{n}t}cos\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)+\sqrt{3}{e}^{\frac{1}{2}{\lambda}_{n}t}sin\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)).
By direct computation, we have
\{\begin{array}{l}{y}^{\u2034}(t)+{\lambda}_{n}^{3}y(t)\\ \phantom{\rule{1em}{0ex}}=\sigma {\lambda}_{n}^{3}({e}^{{\lambda}_{n}t}{e}^{\frac{1}{2}{\lambda}_{n}t}cos(\frac{\sqrt{3}}{2}{\lambda}_{n}t)+\sqrt{3}{e}^{\frac{1}{2}{\lambda}_{n}t}sin(\frac{\sqrt{3}}{2}{\lambda}_{n}t)),\phantom{\rule{1em}{0ex}}t\in [0,1],\\ y(0)={y}^{\prime}(0)={y}^{\prime}(1)=0.\end{array}
(2.7)
It is easy to see that the solution for the corresponding homogeneous equation of (2.7) is of the form
{y}_{0}(t)={C}_{1}{e}^{{\lambda}_{n}t}+{C}_{2}{e}^{\frac{1}{2}{\lambda}_{n}t}cos\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)+{C}_{3}{e}^{\frac{1}{2}{\lambda}_{n}t}sin\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right).
Then, by an ordinary differential equation method, we see that the general solution of (2.7) is of the form
y(t)={y}_{0}(t)+{y}_{1}^{\ast}(t)+{y}_{2}^{\ast}(t),\phantom{\rule{1em}{0ex}}t\in [0,1],
where
{y}_{1}^{\ast}(t)=\frac{1}{3}\sigma {\lambda}_{n}t{e}^{{\lambda}_{n}t}
is the special solution of the equation
{y}^{\u2034}(t)+{\lambda}_{n}^{3}y(t)=\sigma {\lambda}_{n}^{3}{e}^{{\lambda}_{n}t},
and
\begin{array}{rl}{y}_{2}^{\ast}(t)& =\frac{1}{3}\sigma {\lambda}_{n}t{e}^{\frac{1}{2}{\lambda}_{n}t}(cos\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)+\sqrt{3}sin\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right))\\ =\frac{2}{3}\sigma {\lambda}_{n}t{e}^{\frac{1}{2}{\lambda}_{n}t}cos(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n}t)\end{array}
is the special solution of the equation
{y}^{\u2034}(t)+{\lambda}_{n}^{3}y(t)=\sigma {\lambda}_{n}^{3}{e}^{\frac{1}{2}{\lambda}_{n}t}(cos\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)\sqrt{3}sin\left(\frac{\sqrt{3}}{2}{\lambda}_{n}t\right)).
Then
\begin{array}{c}{\left({y}_{1}^{\ast}\right)}^{\prime}(t)=\frac{1}{3}\sigma {\lambda}_{n}{e}^{{\lambda}_{n}t}+\frac{1}{3}\sigma {\lambda}_{n}^{2}t{e}^{{\lambda}_{n}t},\hfill \\ \begin{array}{rl}{\left({y}_{2}^{\ast}\right)}^{\prime}(t)=& \frac{2}{3}\sigma {\lambda}_{n}{e}^{\frac{1}{2}{\lambda}_{n}t}cos(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n}t)+\frac{1}{3}\sigma {\lambda}_{n}^{2}t{e}^{\frac{1}{2}{\lambda}_{n}t}cos(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n}t)\\ +\frac{\sqrt{3}}{3}\sigma {\lambda}_{n}^{2}t{e}^{\frac{1}{2}{\lambda}_{n}t}sin(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n}t).\end{array}\hfill \end{array}
Applying the condition y(0)={y}^{\prime}(0)=0, we obtain {C}_{2}={C}_{1}, {C}_{3}=\sqrt{3}{C}_{1}. From the condition {y}^{\prime}(1)=0, we obtain
{({y}_{0})}^{\prime}(1)+{\left({y}_{1}^{\ast}\right)}^{\prime}(1)+{\left({y}_{2}^{\ast}\right)}^{\prime}(1)=0.
(2.8)
From (2.5), we have {({y}_{0})}^{\prime}(1)=0. Thus it follows from (2.8) that
\begin{array}{r}{e}^{{\lambda}_{n}}+{\lambda}_{n}{e}^{{\lambda}_{n}}+2{e}^{\frac{1}{2}{\lambda}_{n}}cos(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n})+{\lambda}_{n}{e}^{\frac{1}{2}{\lambda}_{n}}cos(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n})\\ \phantom{\rule{1em}{0ex}}+\sqrt{3}{\lambda}_{n}{e}^{\frac{1}{2}{\lambda}_{n}}sin(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n})=0,\end{array}
which implies that
{e}^{{\lambda}_{n}}+{e}^{\frac{1}{2}{\lambda}_{n}}cos(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n})+\sqrt{3}{e}^{\frac{1}{2}{\lambda}_{n}}sin(\frac{\pi}{3}\frac{\sqrt{3}}{2}{\lambda}_{n})=0.
That is
\frac{1}{2}{e}^{\frac{3}{2}{\lambda}_{n}}=cos\frac{\sqrt{3}}{2}{\lambda}_{n},
which is a contradiction of
\frac{1}{2}{e}^{\frac{3}{2}{\lambda}_{n}}=cos(\frac{\sqrt{3}}{2}{\lambda}_{n}+\frac{2\pi}{3}).
Therefore, the algebraic multiplicity of the eigenvalue \frac{1}{{\lambda}_{n}^{3}} is 1. □
Lemma 2.3 Suppose that (H_{1}) holds and y\in P\mathrm{\setminus}\{\theta \} is a solution of the (1.1), then y\in \stackrel{\circ}{P}. Similarly, if y\in (P)\mathrm{\setminus}\{\theta \} is a solution of the (1.1), then y\in \stackrel{\circ}{P}.
Proof The proof is obvious. □
Lemma 2.4 Suppose that (H_{1})(H_{3}) hold. Then the operator A is Fréchet differentiable at θ and ∞, and {A}_{\theta}^{\prime}={\beta}_{0}K, {A}_{\mathrm{\infty}}^{\prime}={\beta}_{1}K.
Proof Since (H_{3}): {lim}_{u\to 0}\frac{f(t,u)}{u}={\beta}_{0} uniformly on [0,1]. That is, for any \epsilon >0, there exists \delta >0 such that
f(t,u){\beta}_{0}u\le \epsilon u,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],0<u<\delta .
From (H_{1}), it is easy to see that \mathrm{\forall}t\in [0,1], f(t,0)=0. Then, for any u\in E with \parallel u\parallel <\delta, we have
\begin{array}{rl}(AuA\theta {\beta}_{0}Ku)(t)& =K(Fu{\beta}_{0}u)(t)\\ \le \parallel K\parallel \underset{s\in [0,1]}{max}f(s,u(s)){\beta}_{0}u(s)\le \parallel K\parallel \parallel u\parallel \epsilon .\end{array}
Then
\parallel AuA\theta {\beta}_{0}Ku\parallel \le \parallel K\parallel \parallel u\parallel \epsilon ,\phantom{\rule{1em}{0ex}}\parallel u\parallel <\delta .
Thus,
\underset{\parallel u\parallel \to 0}{lim}\frac{\parallel AuA\theta {\beta}_{0}Ku\parallel}{\parallel u\parallel}=0,
which means {A}_{\theta}^{\prime}={\beta}_{0}K.
Since (H_{2}): {lim}_{u\to +\mathrm{\infty}}\frac{f(t,u)}{u}={\beta}_{1} uniformly on [0,1]. That is, for any \epsilon >0, there exists R>0 such that
f(t,u){\beta}_{1}u\le \epsilon u,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],u>R.
Let {M}_{R}={max}_{0\le u\le R}f(t,u). Then
(Fu)(t){\beta}_{1}u(t)=f(t,u(t)){\beta}_{1}u(t)\le {M}_{R}+{\beta}_{1}R+\epsilon \parallel u\parallel ,\phantom{\rule{1em}{0ex}}t\in [0,1],u\in E.
Thus,
(Au{\beta}_{1}Ku)(t)=K(Fu{\beta}_{1}u)(t)\le \parallel K\parallel ({M}_{R}+{\beta}_{1}R+\epsilon \parallel u\parallel ).
Then
\underset{\parallel u\parallel \to +\mathrm{\infty}}{lim}\frac{\parallel Au{\beta}_{1}Ku\parallel}{\parallel u\parallel}\le \epsilon \parallel K\parallel .
Therefore, {A}_{\mathrm{\infty}}^{\prime}={\beta}_{1}K. □
Remark 5 Suppose (H_{2}) holds. Similar to Lemma 2.4, we have
{A}_{P}^{\prime}={A}_{P}^{\prime}={A}_{\mathrm{\infty}}^{\prime}={\beta}_{1}K.
Lemma 2.5 [13]
Suppose that E is an ordered Banach space with a lattice structure, P is a normal solid cone in E, and the nonlinear operator A is quasiadditive on the lattice. Assume that

(i)
A is strongly increasing on P and −P;

(ii)
both {A}_{P}^{\prime} and {A}_{P}^{\prime} exist with r({A}_{P}^{\prime})>1 and r({A}_{P}^{\prime})>1; 1 is not an eigenvalue of {A}_{P}^{\prime} or {A}_{P}^{\prime} corresponding a positive eigenvector;

(iii)
A\theta =\theta; the Fréchet derivative {A}_{\theta}^{\prime} of A at θ is strongly positive, and r({A}_{\theta}^{\prime})<1;

(iv)
the Fréchet derivative {A}_{\mathrm{\infty}}^{\prime} of A at ∞ exists; 1 is not an eigenvalue of {A}_{\mathrm{\infty}}^{\prime}; the sum β of the algebraic multiplicities for all eigenvalues of {A}_{\mathrm{\infty}}^{\prime} lying in the interval (1,\mathrm{\infty}) is an even number.
Then A has at least three nontrivial fixed points containing one signchanging fixed point.