In this section, we use Lemmas 2.4, 2.5 to study problem (1.1) and present two new results on the existence and uniqueness of monotone positive solutions. The main results obtained here are relatively new in the literature.
In our considerations we shall consider the Banach space \(E= C^{1}[0,1]\) equipped with the norm \(\u\=\max\{\max_{0\leq t\leq1} u(t),\max_{0\leq t\leq1} u'(t)\}\). In order to find monotone positive solutions, we consider the closed convex cone of nonnegative increasing functions \(P=\{u\in Eu(t)\geq0,u'(t)\geq0,\forall t\in [0,1]\}\). Note that this induces an order relation \(\dot{\leq}\) in E by defining \(u \dot{\leq} v\) if and only if \(vu\in P\). Clearly, this cone is normal. That is, if \(u \dot{\leq} v\), then \(u(t)\leq v(t)\), \(u'(t)\leq v'(t)\), \(t\in[0,1]\). Therefore, \(\u\\leq\v\\) and the normality constant is 1.
Let \(G(t,s)\) be the Green function of the linear problem \(u^{(4)}(t)=0\) with the boundary conditions in (1.1); from [3] we know that
$$ G(t,s)=\frac{1}{6} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} s^{2}(3ts),& 0\leq s\leq t\leq1,\\ \ t^{2}(3st),& 0\leq t\leq s\leq1. \end{array}\displaystyle \right . $$
(3.1)
Then problem (1.1) is equivalent to the integral equation
$$u(t)=\int_{0}^{1}G(t,s)f \bigl(s,u(s),u'(s) \bigr)\,dsg \bigl(u(1) \bigr)\phi(t), \quad\mbox{where } \phi (t)= \frac{1}{2} t^{2}\frac{1}{6} t^{3}, t\in[0,1]. $$
From [17], we give the following properties of the Green function \(G(t,s)\) and \(\phi(t)\).
Lemma 3.1
For any
\(t,s\in[0,1]\), we have
$$\begin{aligned}& \frac{1}{3}s^{2}t^{2}\leq G(t,s)\leq \frac{1}{2}st^{2}, \qquad\frac{1}{3}t^{2}\leq \phi(t) \leq\frac{1}{2}t^{2},\\& \frac{1}{2}s^{2}t \leq \frac{\partial G(t,s)}{\partial t}\leq st, \qquad\frac{1}{2}t\leq\phi'(t) \leq2t. \end{aligned}$$
Theorem 3.2
Assume that
 (H_{1}):

\(f(t,x,y):[0,1]\times[0,+\infty)\times[0,+\infty)\rightarrow [0,+\infty)\)
and
\(g:[0,+\infty)\rightarrow(\infty,0]\);
 (H_{2}):

\(f(t,x,y)\)
is increasing in
\(x \in[0,+\infty)\)
for fixed
\(t \in [0,1]\)
and
\(y \in[0,+\infty)\), decreasing in
\(y \in[0,+\infty)\)
for fixed
\(t \in[0,1]\)
and
\(x\in[0,+\infty)\), and
\(g(x)\)
is decreasing in
\(x \in[0,+\infty)\);
 (H_{3}):

\(g(\lambda x) \leq\lambda g(x)\)
for
\(\lambda\in(0,1)\), \(x\in [0,+\infty)\), and there exists a constant
\(\alpha\in(0,1)\)
such that
\(f(t,\lambda x,\lambda^{1} y) \geq\lambda^{\alpha} f(t,x,y)\), \(\forall t\in[0,1]\), \(\lambda\in(0,1)\), \(x,y\in[0,+\infty)\);
 (H_{4}):

there exists a constant
\(\sigma> 0\)
such that
\(f(t,x,y) \geq \sigma\geqg(x)>0\), \(t\in[0,1]\), \(x,y\geq0\).
Then:

(1)
there exist
\(u_{0},v_{0} \in P_{h}\)
and
\(r \in(0,1)\)
such that
\(rv_{0} \dot{\leq} u_{0} \dot{<} v_{0}\)
and
$$\begin{aligned}& u_{0}(t) \leq\int^{1}_{0} G(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& u_{0}'(t) \leq\int^{1}_{0} G_{t}(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi'(t),\quad t \in[0,1], \\& v_{0}(t) \geq\int^{1}_{0} G(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& v_{0}'(t) \geq\int^{1}_{0} G_{t}(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi'(t),\quad t \in[0,1], \end{aligned}$$
where
\(h(t)=t^{2}\), \(t\in[0,1]\)
and
\(G(t,s)\)
is given as in (3.1);

(2)
problem (1.1) has a unique monotone positive solution
\(u^{*}\)
in
\(P_{h}\);

(3)
for any
\(x_{0},y_{0}\in P_{h}\), constructing successively the sequences
$$\begin{aligned}& x_{n}(t) = \int^{1}_{0} G(t,s)f \bigl(s,x_{n1}(s),y_{n1}'(s) \bigr)\,dsg \bigl(x_{n1}(1) \bigr)\phi(t),\quad n=1,2,\ldots, \\& y_{n}(t) = \int^{1}_{0} G(t,s)f \bigl(s,y_{n1}(s),x_{n1}'(s) \bigr)\,dsg \bigl(y_{n1}(1) \bigr)\phi'(t),\quad n=1,2,\ldots, \end{aligned}$$
we have
\(\ x_{n}u^{*} \\rightarrow0\)
and
\(\ y_{n}u^{*} \\rightarrow0\)
as
\(n\rightarrow\infty\).
Proof
Define two operators \(A:P\times P\rightarrow E\) and \(B:P\rightarrow E\) by
$$A(u,v) (t)=\int^{1}_{0} G(t,s)f \bigl(s,u(s),v'(s) \bigr)\,ds,\qquad Bu(t)=g \bigl(u(1) \bigr)\phi (t),\quad t \in[0,1]. $$
Then
$$\bigl(A(u,v) \bigr)'(t)=\int_{0}^{1}G_{t}(t,s)f \bigl(s,u(s),v'(s) \bigr)\,ds,\qquad (Bu)'(t)=g \bigl(u(1) \bigr) \phi '(t),\quad t\in[0,1]. $$
Evidently, u is the solution of problem (1.1) if and only if \(u=A(u,u)+Bu\). For \(u,v\in P\), we know that \(u(t),v(t)\geq0\), \(u'(t),v'(t)\geq0\), \(t\in [0,1]\). From (H_{1}) and Lemma 3.1, we have \(A(u,v)(t)\geq0\), \(Bu(t)\geq0\), \((A(u,v))'(t)\geq0\), \((Bu)'(t)\geq 0\), \(t\in[0,1]\). Therefore, \(A(u,v)\in P\), \(Bu\in P\). That is, \(A:P\times P\rightarrow P\) and \(B:P\rightarrow P\). In the sequel we check that A, B satisfy all assumptions of Lemma 2.4.
Firstly, we prove that A is a mixed monotone operator. In fact, for \(u_{i},v_{i}\in P\), \(i=1,2\) with \(u_{1}\dot{\geq} u_{2}\), \(v_{1}\dot{\leq} v_{2}\), we know that \(u_{1}(t)\geq u_{2}(t)\), \(v_{1}(t)\leq v_{2}(t)\), \(u_{1}'(t)\geq u_{2}'(t)\), \(v_{1}'(t)\leq v_{2}'(t)\), \(t\in[0,1]\) and by (H_{2}) and Lemma 3.1, we have
$$\begin{aligned}& A (u_{1},v_{1}) (t)=\int_{0}^{1}G(t,s)f \bigl(s,u_{1}(s),v_{1}'(s) \bigr)\,ds\\& \hphantom{A (u_{1},v_{1}) (t)}\geq\int _{0}^{1}G(t,s)f \bigl(s,u_{2}(s),v_{2}'(s) \bigr)\,ds\\& \hphantom{A (u_{1},v_{1}) (t)}=A (u_{2},v_{2}) (t),\\& \bigl(A(u_{1},v_{1}) \bigr)'(t)=\int _{0}^{1}G_{t}(t,s)f \bigl(s,u_{1}(s),v_{1}'(s) \bigr)\,ds\\& \hphantom{\bigl(A(u_{1},v_{1}) \bigr)'(t)}\geq\int _{0}^{1}G_{t}(t,s)f \bigl(s,u_{2}(s),v_{2}'(s) \bigr)\,ds\\& \hphantom{\bigl(A(u_{1},v_{1}) \bigr)'(t)}=\bigl(A(u_{2},v_{2}) \bigr)'(t). \end{aligned}$$
That is, \(A(u_{1},v_{1})\dot{\geq} A (u_{2},v_{2})\).
Further, we show B is increasing. For any \(u,v\in P\) with \(u\dot{\leq } v\), we know that \(u(t)\leq v(t)\), \(u'(t)\leq v'(t)\), \(t\in [0,1]\). It follows from (H_{1}), (H_{2}) and Lemma 3.1 that \(Bu(t)\leq Bv(t)\), \((Bu)'(t)\leq(Bv)'(t)\), \(t\in[0,1]\). That is, \(Bu\dot{\leq} Bv\). Next we show that operator A satisfies the condition (2.1). For any \(\lambda\in(0,1)\) and \(u,v\in P\), by (H_{3}) we have
$$\begin{aligned}& A\bigl(\lambda u,\lambda^{1} v\bigr) (t)=\int ^{1}_{0}G(t,s)f\bigl(s,\lambda u(s), \lambda^{1} v'(s)\bigr)\,ds \\& \hphantom{ A\bigl(\lambda u,\lambda^{1} v\bigr) (t)} \geq \lambda^{\alpha} \int^{1}_{0}G(t,s)f \bigl(s,u(s),v'(s)\bigr)\,ds \\& \hphantom{ A\bigl(\lambda u,\lambda^{1} v\bigr) (t)} =\lambda^{\alpha} A(u,v) (t), \\& \begin{aligned}[b] \bigl(A\bigl(\lambda u,\lambda^{1} v\bigr) \bigr)'(t)&=\int^{1}_{0}G_{t}(t,s)f \bigl(s,\lambda u(s),\lambda^{1} v'(s)\bigr)\,ds \\ &\geq \lambda^{\alpha} \int^{1}_{0}G_{t}(t,s)f \bigl(s,u(s),v'(s)\bigr)\,ds \\ &=\bigl(\lambda^{\alpha} A(u,v)\bigr)'(t). \end{aligned} \end{aligned}$$
That is, \(A(\lambda u,\lambda^{1} v) \dot{\geq} \lambda^{\alpha} A(u,v)\) for \(\lambda\in(0,1)\), \(u,v\in P\). So operator A satisfies (2.1). Also, for any \(\lambda\in(0,1)\), \(u\in P\), from (H_{3}) we know that
$$\begin{aligned}& B(\lambda u) (t)=g \bigl(\lambda u(1) \bigr)\phi(t)\geq\lambda\bigl(g \bigl(u(1) \bigr) \phi (t)\bigr)=\lambda Bu(t), \\& \bigl(B(\lambda u) \bigr)'(t)=g \bigl(\lambda u(1) \bigr) \phi'(t)\geq\lambda\bigl(g \bigl(u(1) \bigr)\phi '(t)\bigr)=( \lambda Bu)'(t), \end{aligned}$$
that is, \(B(\lambda u)\dot{\geq} \lambda Bu\) for \(\lambda\in (0,1)\), \(u\in P\). That is, operator B is subhomogeneous. Now we show that \(A(h,h)\in P_{h}\) and \(Bh\in P_{h}\). On the one hand, from (H_{1}), (H_{2}) and Lemma 3.1, for any \(t\in[0,1]\), we have
$$\begin{aligned}& \begin{aligned}[b] A(h,h) (t)&=\int^{1}_{0} G(t,s)f\bigl(s,h(s),h'(s)\bigr)\,ds \\ &=\int^{1}_{0} G(t,s)f\bigl(s,s^{2},2s \bigr)\,ds \\ & \leq \int^{1}_{0}\frac{1}{2}t^{2}sf \bigl(s,s^{2},2s\bigr)\,ds\leq\frac{1}{2}\int^{1}_{0}s f(s,1,0)\,ds\cdot h(t), \end{aligned} \\& A(h,h) (t)\geq \int^{1}_{0}\frac{1}{3}t^{2}s^{2}f \bigl(s,s^{2},2s\bigr)\,ds \geq\frac{1}{3}\int ^{1}_{0}s^{2} f(s,0,2)\,ds\cdot h(t). \end{aligned}$$
On the other hand, also from (H_{1}), (H_{2}) and Lemma 3.1, for any \(t\in[0,1]\), we obtain
$$\begin{aligned} \bigl(A(h,h)\bigr)'(t)&=\int^{1}_{0} G_{t}(t,s)f\bigl(s,s^{2},2s\bigr)\,ds \\ &\leq\int^{1}_{0} stf(s,1,0)\,ds= \frac{1}{2} \int^{1}_{0}s f(s,1,0)\,ds\cdot h'(t), \\ \bigl(A(h,h)\bigr)'(t)&\geq \int^{1}_{0} \frac{1}{2}s^{2}t f(s,0,2)\,ds=\frac{1}{4} \int ^{1}_{0}s^{2} f(s,0,2)\,ds\cdot h'(t). \end{aligned}$$
Let
$$c_{1}=\frac{1}{4}\int^{1}_{0}s^{2}f(s,0,2) \,ds,\qquad c_{2}=\frac{1}{2}\int^{1}_{0}sf(s,1,0) \,ds. $$
From (H_{2}), (H_{4}), we have
$$c_{2}\geq c_{1}\geq\frac{1}{4}\int ^{1}_{0}s^{2}\sigma \,ds= \frac{1}{12} \sigma>0, $$
and in consequence,
$$\begin{aligned}& c_{1}h(t)\leq A(h,h) (t)\leq c_{2}h(t),\\& (c_{1}h)'(t)=c_{1}h'(t) \leq \bigl(A(h,h) \bigr)'(t)\leq c_{2}h'(t)=(c_{2}h)'(t), \quad t\in[0,1]. \end{aligned}$$
Thus, \(c_{1}h\dot{\leq}A(h,h)\dot{\leq}c_{2}h\). That is, \(A(h,h)\in P_{h}\). Similarly, from (H_{1}), (H_{2}) and Lemma 3.1, for any \(t\in[0,1]\), we have
$$\begin{aligned}& \frac{1}{3}g(1)h(t)=g(1)\frac{1}{3}t^{2} \leq Bh(t)=g \bigl(h(1) \bigr)\phi(t)\leq g(1)\frac{1}{2}t^{2}= \frac{1}{2}g(1) h(t), \\& \frac{1}{4}g(1)h'(t)=g(1)\frac{1}{2}t \leq(Bh)'(t)=g \bigl(h(1) \bigr)\phi'(t)\leq g(1)2t=g(1) h'(t). \end{aligned}$$
Let \(c_{3}=\frac{1}{4}g(1)\), \(c_{4}=g(1)\). Then, from (H_{1}), (H_{4}), we have \(c_{4}\geq c_{3}>0\) and thus
$$\begin{aligned}& c_{3}h(t)\leq Bh(t)\leq c_{4}h(t),\\& (c_{3}h)'(t)=c_{3}h'(t) \leq(Bh)'(t)\leq c_{4}h'(t)=(c_{4}h)'(t), \quad t\in[0,1]. \end{aligned}$$
Therefore, \(c_{3}h\dot{\leq}Bh\dot{\leq}c_{4}h\). That is, \(Bh\in P_{h}\). Hence the condition (i) of Lemma 2.4 is satisfied.
In the following we show the condition (ii) of Lemma 2.4 is satisfied. For \(u,v \in P\) and any \(t\in[0,1]\), from (H_{4}) and Lemma 3.1,
$$\begin{aligned}& \begin{aligned}[b] A(u,v) (t)&=\int^{1}_{0} G(t,s)f\bigl(s,u(s),v'(s)\bigr)\,ds \\ &\geq \int^{1}_{0}\frac{1}{3}t^{2}s^{2}f \bigl(s,u(s),v'(s)\bigr)\,ds \geq\frac{1}{3}t^{2}\int ^{1}_{0} s^{2}\sigma \,ds \\ &=\frac{1}{9}\sigma t^{2}\geq\frac{1}{9}t^{2} \bigl[g\bigl(u(1)\bigr)\bigr]=\frac{2}{9}\bigl[g\bigl(u(1)\bigr)\bigr] \frac{1}{2}t^{2} \\ &\geq\frac{2}{9}\bigl[g\bigl(u(1)\bigr)\bigr]\phi(t)= \frac{2}{9}Bu(t), \end{aligned} \\& \begin{aligned}[b] \bigl(A(u,v)\bigr)'(t)&=\int ^{1}_{0} G_{t}(t,s)f \bigl(s,u(s),v'(s)\bigr)\,ds \\ &\geq \int^{1}_{0}\frac{1}{2}ts^{2}f \bigl(s,u(s),v'(s)\bigr)\,ds \geq\frac{1}{2}t\int ^{1}_{0} s^{2}\sigma \,ds \\ &=\frac{1}{6}\sigma t\geq\frac{1}{6}t\bigl[g\bigl(u(1)\bigr)\bigr]= \frac{1}{12}\bigl[g\bigl(u(1)\bigr)\bigr]2t \\ &\geq \frac{1}{12}\bigl[g\bigl(u(1)\bigr)\bigr]\phi'(t)= \frac{1}{12}(Bu)'(t). \end{aligned} \end{aligned}$$
Let \(\delta_{0}=\frac{1}{12}\). Then
$$A(u,v) (t)\geq\delta_{0}Bu(t),\qquad \bigl(A(u,v) \bigr)'(t) \geq\delta_{0}(Bu)'(t),\quad t\in[0,1]. $$
Therefore, we get \(A(u,v) \dot{\geq} \delta_{0}Bu\) for \(u,v\in P\). Finally, an application of Lemma 2.4 implies: there exist \(u_{0},v_{0} \in P_{h}\) and \(r \in(0,1)\) such that \(rv_{0}\dot{ \leq} u_{0}\dot{ <} v_{0}\), \(u_{0} \dot{\leq} A(u_{0},v_{0})+Bu_{0} \dot{\leq} A(v_{0},u_{0})+Bv_{0} \dot{\leq} v_{0}\); the operator equation \(A(u,u)+Bu=u\) has a unique solution \(u^{*}\) in \(P_{h}\); for any initial values \(x_{0},y_{0} \in P_{h}\), constructing successively the sequences
$$x_{n}=A(x_{n1},y_{n1})+Bx_{n1},\qquad y_{n}=A(y_{n1},x_{n1})+By_{n1},\quad n=1,2, \ldots, $$
we have \(x_{n} \rightarrow u^{*}\) and \(y_{n} \rightarrow u^{*}\) as \(n \rightarrow\infty\). That is,
$$\begin{aligned}& u_{0}(t) \leq\int^{1}_{0} G(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& u_{0}'(t) \leq\int^{1}_{0} G_{t}(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi'(t),\quad t \in[0,1], \\& v_{0}(t) \geq\int^{1}_{0} G(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& v_{0}'(t) \geq\int^{1}_{0} G_{t}(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi'(t),\quad t \in[0,1]; \end{aligned}$$
problem (1.1) has a unique positive solution \(u^{*}\) in \(P_{h}\) and \(u^{*}(t)\) is monotone increasing; for any \(x_{0},y_{0}\in P_{h}\), constructing successively the sequences
$$\begin{aligned}& x_{n}(t) = \int^{1}_{0}G(t,s)f \bigl(s,x_{n1}(s),y_{n1}'(s) \bigr)\,dsg \bigl(x_{n1}(1) \bigr)\phi(t),\quad n=1,2,\ldots, \\& y_{n}(t) = \int^{1}_{0} G(t,s)f \bigl(s,y_{n1}(s),x_{n1}'(s) \bigr)\,dsg \bigl(y_{n1}(1) \bigr)\phi'(t),\quad n=1,2,\ldots, \end{aligned}$$
we have \(\ x_{n}u^{*} \\rightarrow0\) and \(\ y_{n}u^{*} \\rightarrow0\) as \(n\rightarrow\infty\). □
Theorem 3.3
Assume (H_{1}), (H_{2}) and
 (H_{5}):

there exists a constant
\(\alpha\in(0,1)\)
such that
\(g(\lambda x) \leq\lambda^{\alpha} g(x)\), \(\forall \lambda\in(0,1)\), \(x\in[0,+\infty)\), and
\(f(t,\lambda x,\lambda^{1} y) \geq\lambda f(t,x,y)\)
for
\(\lambda\in(0,1)\), \(t \in[0,1]\), \(x,y\in [0,+\infty)\);
 (H_{6}):

\(f(t,0,2)\not\equiv0\)
for
\(t\in[0,1]\)
and there exists a constant
\(\sigma> 0\)
such that
\(f(t,x,y) \leq\sigma\leqg(x)\), \(t\in [0,1]\), \(x,y\geq0\).
Then:

(1)
there exist
\(u_{0},v_{0} \in P_{h}\)
and
\(r \in(0,1)\)
such that
\(rv_{0}\dot{ \leq} u_{0} \dot{<} v_{0}\)
and
$$\begin{aligned}& u_{0}(t) \leq\int^{1}_{0} G(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& u_{0}'(t) \leq\int^{1}_{0} G_{t}(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi'(t),\quad t \in[0,1], \\& v_{0}(t) \geq\int^{1}_{0} G(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& v_{0}'(t) \geq\int^{1}_{0} G_{t}(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi'(t), \quad t \in[0,1], \end{aligned}$$
where
\(h(t)=t^{2}\), \(t\in[0,1]\)
and
\(G(t,s)\)
is given as in (3.1);

(2)
problem (1.1) has a unique monotone positive solution
\(u^{*}\)
in
\(P_{h}\);

(3)
for any
\(x_{0},y_{0}\in P_{h}\), constructing successively the sequences
$$\begin{aligned}& x_{n}(t) = \int^{1}_{0} G(t,s)f \bigl(s,x_{n1}(s),y_{n1}'(s) \bigr)\,dsg \bigl(x_{n1}(1) \bigr)\phi(t),\quad n=1,2,\ldots, \\& y_{n}(t) = \int^{1}_{0} G(t,s)f \bigl(s,y_{n1}(s),x_{n1}'(s) \bigr)\,dsg \bigl(y_{n1}(1) \bigr)\phi'(t), \quad n=1,2,\ldots, \end{aligned}$$
we have
\(\ x_{n}u^{*} \\rightarrow0\)
and
\(\ y_{n}u^{*} \\rightarrow0\)
as
\(n\rightarrow\infty\).
Sketch of the proof
Consider two operators A, B defined in the proof of Theorem 3.2. Similarly, from (H_{1}), (H_{2}), we obtain that \(A:P\times P\rightarrow P\) is a mixed monotone operator and \(B:P\rightarrow P\) is increasing. From (H_{5}), we have
$$A \bigl(\lambda u,\lambda^{1}v \bigr)\dot{\geq} \lambda A(u,v);\qquad B( \lambda u)\dot {\geq} \lambda^{\alpha}Bu, \quad\mbox{for } \lambda\in(0,1), u,v\in P. $$
Since \(f(t,0,2)\not\equiv0\), we get \(\int^{1}_{0}s^{2}f(s,0,2)\,ds>0\), and in sequence, \(c_{2}\geq c_{1}>0\), here \(c_{1}\), \(c_{2}\) are defined in the proof of Theorem 3.2. So we can easily prove that \(A(h,h)\in P_{h}\). From (H_{6}), we know that \(g(1)>0\), and from the proof of Theorem 3.2 we get \(Bh\in P_{h}\). For \(u,v \in P\) and any \(t\in[0,1]\), from (H_{6}),
$$\begin{aligned}& \begin{aligned}[b] A(u,v) (t)&=\int^{1}_{0} G(t,s)f\bigl(s,u(s),v'(s)\bigr)\,ds \\ &\leq \int^{1}_{0}\frac{1}{2}t^{2}sf \bigl(s,u(s),v'(s)\bigr)\,ds \leq\frac{1}{2}t^{2}\int ^{1}_{0} s\sigma \,ds \\ &=\frac{1}{4}\sigma t^{2}\leq\frac{1}{4}t^{2} \bigl[g\bigl(u(1)\bigr)\bigr]=\frac{3}{4}\bigl[g\bigl(u(1)\bigr)\bigr] \frac{1}{3}t^{2} \\ &\leq \frac{3}{4}\bigl[g\bigl(u(1)\bigr)\bigr]\phi(t)= \frac{3}{4}Bu(t), \end{aligned} \\& \begin{aligned}[b] \bigl(A(u,v)\bigr)'(t)&=\int ^{1}_{0} G_{t}(t,s)f \bigl(s,u(s),v'(s)\bigr)\,ds \\ &\leq \int^{1}_{0} tsf\bigl(s,u(s),v'(s) \bigr)\,ds\leq t\int^{1}_{0} s\sigma \,ds \\ &=\frac{1}{2}\sigma t\leq\frac{1}{2}t\bigl[g\bigl(u(1)\bigr)\bigr] \leq \bigl[g\bigl(u(1)\bigr)\bigr]\phi'(t)=(Bu)'(t). \end{aligned} \end{aligned}$$
Let \(\delta_{0}=1\). Then
$$A(u,v) (t)\leq\delta_{0}Bu(t),\qquad \bigl(A(u,v) \bigr)'(t) \leq\delta_{0}(Bu)'(t),\quad t\in[0,1]. $$
Therefore, we get \(A(u,v) \dot{\leq} \delta_{0}Bu\) for \(u,v\in P\). Finally, an application of Lemma 2.5 implies: there exist \(u_{0},v_{0} \in P_{h}\) and \(r \in(0,1)\) such that \(rv_{0}\dot{ \leq} u_{0}\dot{ <} v_{0}\), \(u_{0} \dot{\leq} A(u_{0},v_{0})+Bu_{0} \dot{\leq} A(v_{0},u_{0})+Bv_{0} \dot{\leq} v_{0}\); the operator equation \(A(u,u)+Bu=u\) has a unique solution \(u^{*}\) in \(P_{h}\); for any initial values \(x_{0},y_{0} \in P_{h}\), constructing successively the sequences
$$x_{n}=A(x_{n1},y_{n1})+Bx_{n1},\qquad y_{n}=A(y_{n1},x_{n1})+By_{n1},\quad n=1,2, \ldots, $$
we have \(x_{n} \rightarrow u^{*}\) and \(y_{n} \rightarrow u^{*}\) as \(n \rightarrow\infty\). That is,
$$\begin{aligned}& u_{0}(t) \leq\int^{1}_{0} G(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& u_{0}'(t) \leq\int^{1}_{0} G_{t}(t,s)f \bigl(s,u_{0}(s),v_{0}'(s) \bigr)\,dsg \bigl(u_{0}(1) \bigr)\phi'(t), \quad t \in[0,1], \\& v_{0}(t) \geq\int^{1}_{0} G(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi (t),\quad t\in[0,1], \\& v_{0}'(t) \geq\int^{1}_{0} G_{t}(t,s)f \bigl(s,v_{0}(s),u_{0}'(s) \bigr)\,dsg \bigl(v_{0}(1) \bigr)\phi'(t),\quad t \in[0,1]; \end{aligned}$$
problem (1.1) has a unique positive solution \(u^{*}\) in \(P_{h}\) and \(u^{*}(t)\) is monotone increasing; for any \(x_{0},y_{0}\in P_{h}\), constructing successively the sequences
$$\begin{aligned}& x_{n}(t) = \int^{1}_{0}G(t,s)f \bigl(s,x_{n1}(s),y_{n1}'(s) \bigr)\,dsg \bigl(x_{n1}(1) \bigr)\phi(t),\quad n=1,2,\ldots, \\& y_{n}(t) = \int^{1}_{0} G(t,s)f \bigl(s,y_{n1}(s),x_{n1}'(s) \bigr)\,dsg \bigl(y_{n1}(1) \bigr)\phi'(t),\quad n=1,2,\ldots, \end{aligned}$$
we have \(\ x_{n}u^{*} \\rightarrow0\) and \(\ y_{n}u^{*} \\rightarrow0\) as \(n\rightarrow\infty\).
Remark 3.4
Comparing Theorems 3.23.3 with the main results in [17], we provide some alternative approaches to study the same type of problems under different conditions. Our results can guarantee the existence of a unique monotone positive solution and the existence of upperlower solutions, which are seldom seen in the literature.