In this section, we apply Lemma 2.6 to establish the existence of positive solutions for problem (1.5). We begin by introducing the following conditions on \(F(t,u,v)\):
- (H3):
-
There exist two positive constants r, R with \((\rho _{2}+1)\phi_{q}(r)<(\rho_{1}+1)\phi_{q}(\delta R)\) such that
$$\begin{aligned}& F(t,u,v)\leq\frac{1}{\eta}r,\quad \forall t\in J, |u|+|v|\leq( \rho_{2}+1)\phi_{q}(r), \end{aligned}$$
(3.1)
$$\begin{aligned}& F(t,u,v)\geq\frac{1}{\rho\beta\delta} R,\quad \forall t\in J, |u|+|v|\geq( \rho_{1}+1)\phi_{q}(\delta R), \end{aligned}$$
(3.2)
where
$$\eta=\frac{1}{4}\gamma_{2}, \qquad \sigma=\rho_{1} \delta^{q-1} , \qquad \beta = \int_{0}^{1}s(1-s)\,ds=\frac{1}{6}. $$
Theorem 3.1
Assume that (H1)-(H3) hold. Then we have the following conclusions:
-
(i)
Problem (2.5) has (at least) one positive solution
\(x\in K\)
such that
$$ \delta r\leq x(t)\leq\frac{1}{\delta} R, \quad t\in J. $$
(3.3)
-
(ii)
Problem (1.5) has (at least) one positive solution
y
such that
$$ \left \{ \textstyle\begin{array}{@{}l} y(t)=\int_{0}^{1}H_{1}(t,s)\phi_{q}(x(s))\,ds,\quad t\in J;\\ \|y\|\leq\rho_{2}\phi_{q}(\|x\|);\\ y(t)\geq\sigma\phi_{q}(\|x\|), \quad t\in J. \end{array}\displaystyle \right . $$
(3.4)
We further have
$$ \sigma\phi_{q}(r)\leq y(t) \leq\rho_{2}\phi_{q} \biggl(\frac{1}{\delta}R \biggr), \quad t\in J. $$
(3.5)
Proof
Let T be the cone preserving, completely continuous operator that was defined by (2.16).
Let \(x\in K\) with \(\|x\|=r\). Then \(0\leq\phi_{q}(x(t))\leq\phi _{q}(r)\), \(t\in J\), and
$$0\leq \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\leq\rho_{2} \phi_{q}(r). $$
And hence for \(x\in K\) with \(\|x\|=r\), we have
$$\biggl| \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\biggr|+\bigl|{-}\phi_{q} \bigl(x(t) \bigr)\bigr|\leq (\rho_{2}+1)\phi_{q}(r). $$
Then it follows from (3.1) that
$$\begin{aligned} \|Tx\|&=\max_{t\in J} \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int _{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ &\leq\eta\frac{1}{\eta}r=r. \end{aligned}$$
(3.6)
Now if we let \(\Omega_{1}=\{x\in K:\|x\|< r\}\), then (3.6) shows that
$$ \|Tx\|\leq\|x\|, \quad x\in\partial\Omega_{1}. $$
(3.7)
Further, let
$$ R_{1}=\frac{1}{\delta}R, $$
(3.8)
and
$$\Omega_{2}=\bigl\{ x\in K:\|x\|< R_{1}\bigr\} . $$
Then \(x\in K\) and \(\|x\|=R_{1}\) implies
$$x(t)\geq\delta x(s),\quad t,s\in J, $$
that is,
$$x(t)\geq\delta R_{1}=R, \quad t\in J. $$
Hence, \(\phi_{q}(x(t))\geq\phi_{q}(R)\) for all \(t\in J\), and
$$\begin{aligned} \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\geq\rho_{1} \int _{0}^{1}\phi_{q} \bigl(x(\tau) \bigr)\,d\tau \geq\rho _{1}\phi_{q}\bigl(\delta\|x\|\bigr). \end{aligned}$$
So
$$\biggl|\int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\biggr|+\bigl|{-}\phi_{q} \bigl(x(t) \bigr)\bigr|\geq (\rho_{1}+1)\phi_{q}(\delta R). $$
Using condition (3.2), it follows from \(x\in K\) and \(\|x\|=R_{1}\) that
$$\begin{aligned} \|Tx\|&=\max_{t\in J} \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int _{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ &\geq\rho\beta\frac{1}{\rho\beta\delta}R=\frac{R}{\delta}=R_{1}, \end{aligned}$$
that is, \(x\in\partial\Omega_{2}\) implies
$$ \|Tx\|\geq \|x\|. $$
(3.9)
It follows from Lemma 2.6 that problem (2.5) has (at least) one positive solution \(x\in\bar{\Omega}_{2}\setminus\Omega_{1}\) satisfying (3.3).
It is observed from (2.2) that problem (1.5) has (at least) one positive solution y such that
$$y= \int_{0}^{1}H_{1}(t,s)\phi_{q} \bigl(x(s) \bigr)\,ds, \quad t\in J \mbox{ and } \|y\| \leq\rho_{2} \phi_{q}\bigl(\|x\|\bigr). $$
Moreover, since \(x\in K\), we get for \(t\in J\),
$$\begin{aligned} y(t)&= \int_{0}^{1}H_{1}(t,s)\phi_{q} \bigl(x(s) \bigr)\,ds \\ & \geq\rho_{1} \int_{0}^{1}\phi_{q} \bigl(x(s) \bigr)\,ds \\ & \geq\rho_{1}\phi_{q}\bigl(\delta\|x\|\bigr) \\ & =\sigma\phi_{q}\bigl(\|x\|\bigr). \end{aligned}$$
Then we get (3.4).
Further, it follows from (3.3) and (3.4) that (3.5) holds. □
In Theorem 3.2 we assume the following condition on \(F(t,u,v)\):
- (H4):
-
There exist two positive constants r, R with \((\rho _{2}+1)\phi_{q}(r)< R\) such that:
$$\begin{aligned}& F(t,u,v)\geq\frac{1}{\rho_{1}\phi_{p}(\rho_{1}+1)\beta\delta}\phi_{p} \bigl(|u|+|v|\bigr),\quad \forall t\in J, |u|+|v|\leq(\rho_{2}+1)\phi_{q}(r), \end{aligned}$$
(3.10)
$$\begin{aligned}& F(t,u,v)\leq\frac{1}{2\eta\phi_{p}(\rho_{2}+1)} \phi_{p}\bigl(|u|+|v|\bigr),\quad \forall t\in J, |u|+|v|\geq R, \end{aligned}$$
(3.11)
and we write
$$ M=\max_{\forall t\in J, |u|+|v|\leq R}F(t,u,v). $$
(3.12)
Theorem 3.2
Assume that (H1)-(H2) and (H4) hold. Then we have the following conclusions:
-
(i)
Problem (2.5) has (at least) one positive solution
\(x\in K\)
such that
$$ \delta r\leq x(t)\leq\max \biggl\{ 2\eta M, \phi_{p} \biggl( \frac{R}{\phi _{q}(\delta)(\rho_{1}+1)} \biggr) \biggr\} , \quad t\in J. $$
(3.13)
-
(ii)
Problem (1.5) has (at least) one positive solution
y
such that (3.4) holds. We further have
$$ \sigma\delta\phi_{q}(r)\leq y(t) \leq\max \biggl\{ \rho_{2} \phi _{q}(2\eta M), \frac{\rho_{2}R}{\phi_{q}(\delta)(\rho_{1}+1)} \biggr\} ,\quad t\in J. $$
(3.14)
Proof
Let \(x\in K\) with \(\|x\|=r\). Then \(0\leq\phi _{q}(x(t))\leq\phi_{q}(r)\), \(t\in J\), and
$$0\leq \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\leq\rho_{2} \phi_{q}(r). $$
And hence for \(x\in K\) with \(\|x\|=r\), we have
$$0\leq\biggl| \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\biggr|+\bigl|{-}\phi _{q} \bigl(x(t) \bigr)\bigr|\leq(\rho_{2}+1)\phi_{q}(\delta r), $$
and it follows from condition (3.10) that
$$\begin{aligned} \|Tx\|&=\max_{t\in J} \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int _{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ &\geq\rho_{1}\beta\frac{1}{\rho_{1}\beta\delta\phi _{p}(\rho_{1}+1)}\phi_{p} \bigl[( \rho_{1}+1)\phi_{q}(\delta r) \bigr]=r, \end{aligned}$$
(3.15)
that is, \(x\in\partial\Omega_{1}\) implies that
$$ \|Tx\|\geq\|x\|. $$
(3.16)
From (3.11) and (3.12), we have
$$ F(t,u,v)\leq M+\frac{1}{2\eta\phi_{p}(\rho_{2}+1)} \phi_{p}\bigl(|u|+|v|\bigr),\quad (t,u,v)\in J\times [0,\infty)\times(-\infty, 0 ]. $$
(3.17)
Further, let
$$ R_{2}>\max \biggl\{ 2\eta M, \phi_{p} \biggl( \frac{R}{\phi_{q}(\delta)(\rho _{1}+1)} \biggr) \biggr\} $$
(3.18)
and
$$\Omega_{3}=\bigl\{ x\in K:\|x\|< R_{2}\bigr\} . $$
Noticing that for \(x\in\partial\Omega_{3}\) we have
$$R< (\rho_{1}+1)\phi_{q}(\delta R_{2})\leq \biggl|\int_{0}^{1}H_{1}(s,\tau)\phi _{q} \bigl(x(\tau) \bigr)\,d\tau\biggr|+\bigl| {-}\phi_{q} \bigl(x(t) \bigr)\bigr|. $$
Thus, for \(x\in\partial\Omega_{3}\), it follows from (3.17) that
$$\begin{aligned} \|Tx\|&=\max_{t\in J} \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int _{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ &\leq\eta \int_{0}^{1} \biggl(M+\frac{1}{2\eta\phi_{p}(\rho _{2}+1)} \phi_{p} \bigl((\rho_{2}+1)\phi_{q}(R_{2}) \bigr) \biggr)\,ds \\ & =\eta M+\frac{1}{2}R_{2} \\ & < R_{2}, \end{aligned}$$
that is, \(x\in\partial\Omega_{3}\) implies
$$ \|Tx\|< \|x\|. $$
(3.19)
It now follows from Lemma 2.6 that problem (2.5) has (at least) one positive solution \(x\in\bar{\Omega}_{3}\setminus\Omega_{1}\) satisfying (3.13).
It follows from (2.2) that problem (1.5) has (at least) one positive solution y. Similar to the proof of (3.5), one can show that y satisfies (3.14). □
Theorem 3.3
Assume that (H1)-(H2), (3.2) of (H3) and (3.10) of (H4) hold. In addition, let
f
satisfies the following condition:
- (H5):
-
Let
l, \(\zeta_{i} \) (\(i=1,2\)) and
L
satisfy
$$0< l< \delta\zeta_{1}< \zeta_{1}< \delta\zeta_{2}< \zeta_{2}< L< \infty. $$
If
$$\max_{\forall t\in J, \phi_{q}(\delta\zeta_{i})\leq|u|+|v|\leq (\rho_{2}+1)\phi_{q}(\zeta_{i})}F(t,u,v)< \frac{1}{\eta}\zeta_{i}, \quad i=1,2, $$
then we have the following conclusions:
-
(i)
Problem (2.5) has (at least) two positive solutions
\(x_{1},x_{2}\in K\)
such that
$$ \delta l\leq x_{1}(t)\leq\zeta_{1}< \delta \zeta_{2}\leq x_{2}(t)\leq L,\quad t\in J. $$
(3.20)
-
(ii)
Problem (1.5) has (at least) two positive solutions
\(y_{1}\), \(y_{2}\)
such that, for
\(i=1,2\),
$$ \left \{ \textstyle\begin{array}{@{}l} y_{i}(t)=\int_{0}^{1}H_{1}(t,s)\phi_{q}(x_{i}(s))\,ds, \quad t\in J;\\ \|y_{i}\|\leq\rho_{2}\phi_{q}(\|x_{i}\|);\\ y_{i}(t)\geq\sigma\phi_{q}(\|x_{i}\|), \quad t\in J. \end{array}\displaystyle \right . $$
(3.21)
We further have
$$ \left \{ \textstyle\begin{array}{@{}l} y_{1}(t)> \sigma\phi_{q}(l), \quad t\in J;\\ \|y_{2}\|\leq\rho_{2}\phi_{q}(L). \end{array}\displaystyle \right . $$
(3.22)
Proof
If (3.10) of (H4) holds, similar to the proof of (3.16), we can prove that
$$ \|Tx\|\geq\|x\|,\quad x\in K, \|x\|=l. $$
(3.23)
If (3.2) of (H3) holds, similar to the proof of (3.9), we have
$$ \|Tx\|\geq\|x\|,\quad x\in K, \|x\|=L. $$
(3.24)
Finally, we show that
$$ \|Tx\|< \|x\|,\quad x\in K, \|x\|=\zeta_{i}, i=1,2. $$
(3.25)
In fact, for \(x\in K\) with \(\|x\|=\zeta_{i}\) (\(i=1,2\)) then we have
$$x(t)\leq\|x\|= \zeta_{i},\quad i=1,2, $$
and
$$\rho_{1}\phi_{q}(\delta\zeta_{i})\leq \int_{0}^{1}H_{1}(s,\tau)\phi _{q} \bigl(x(\tau) \bigr)\,d\tau\leq\rho_{2} \phi_{q}(\zeta_{i}),\quad i=1,2. $$
Therefore,
$$\begin{aligned} &\phi_{q}(\delta\zeta_{i})< (\rho_{1}+1) \phi_{q}(\delta\zeta_{i})\leq \biggl| \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\biggr|+\bigl|{-}\phi_{q} \bigl(x(s) \bigr)\bigr|\leq (\rho_{2}+1)\phi_{q}(\zeta_{i}),\\ &\quad i=1,2, \end{aligned}$$
and hence it follows from (H5) that
$$\begin{aligned} \|Tx\|&=\max_{t\in J} \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int _{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ &< \eta\frac{1}{\eta}\zeta_{i}=\zeta_{i},\quad i=1,2, \end{aligned}$$
which shows that (3.25) holds.
Applying Lemma 2.6 to (3.23), (3.24) and (3.25) shows that problem (2.5) has (at least) two positive solutions \(x_{1}\), \(x_{2}\) with \(x_{1}\in K_{l,\zeta_{1}}=\{x \in K, l \leq\|x\| \leq\zeta_{1} \}\), \(x_{2}\in K_{\zeta_{2},L}=\{x \in K, \zeta_{2}\leq\|x\| \leq L \}\). Hence, since for \(x_{i}\in K\) we have \(x_{i}(t)\geq \delta\|x_{i}\|\), \(t \in J\), \(i=1,2\), it follows that (3.20) holds.
Similar to the proof of (3.4) and (3.5), one can show that (3.21) and (3.22) hold. □
In Theorem 3.4 we assume the following condition on \(F(t,u,v)\):
- (H6):
-
There exist two constants r, R with \(0< r< \delta R\) such that
$$\begin{aligned}& F(t,u,v)\leq\frac{1}{\eta}r, \quad \forall(t,u,v)\in J\times \bigl[\rho _{1}\phi_{q}(\delta r),\rho_{2} \phi_{q}(r) \bigr]\times \bigl[-\phi_{q}(r),-\phi _{q}(\delta r) \bigr]; \\& F(t,u,v)\geq\frac{1}{\rho\beta}R, \quad \forall(t,u,v)\in J\times \bigl[ \rho_{1}\phi_{q}(\delta R),\rho_{2} \phi_{q}(R) \bigr]\times \bigl[-\phi _{q}(R),- \phi_{q}(\delta R) \bigr]. \end{aligned}$$
Theorem 3.4
Assume that (H1)-(H2) and (H6) hold. Then we have the following conclusions:
-
(i)
Problem (2.5) has (at least) one positive solution
\(x\in K\)
such that
$$ \delta r\leq x(t)\leq R,\quad t\in J. $$
(3.26)
-
(ii)
Problem (1.5) has (at least) one positive solution
y
such that (3.4) holds. We further have
$$ \sigma\phi_{q}(r)\leq y(t) \leq\rho_{2} \phi_{q}(R), \quad t\in J. $$
(3.27)
Proof
For \(x\in K\) with \(\|x\|=r\), we have \(x(s)\in[\delta r,r]\), and
$$\begin{aligned}& \int_{0}^{1}H_{1}(s,\tau) \phi_{q} \bigl(x(\tau) \bigr)\,d\tau\in \bigl[\rho_{1}\phi _{q}(\delta r),\rho_{2}\phi_{q}(r) \bigr],\\& - \phi_{q} \bigl(x(s) \bigr)\in \bigl[-\phi _{q}(r),- \phi_{q}(\delta r) \bigr], \quad\forall s\in J. \end{aligned}$$
Then, for \(t\in J\), we have
$$\begin{aligned} (Tx) (t)&= \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int_{0}^{1}H_{1}(s,\tau)\phi _{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ & \leq \int_{0}^{1}\gamma_{2}G_{2}(s,s)\,ds \cdot\frac {1}{\eta}r \\ & =r, \end{aligned}$$
i.e.
\(x\in\partial\Omega_{1}\) implies
$$ \|Tx\|\leq\|x\|. $$
(3.28)
On the other hand, for \(x\in K\) with \(\|x\|=R\), we have \(x\in[\delta R,R]\), and \(\phi_{q}(x)\in[-\phi_{q}(R), -\phi_{q}(\delta R)]\), \(\int _{0}^{1}H_{1}(s,\tau)\phi_{q}(x(\tau))\,d\tau\in[\rho_{1}\phi_{q}(\delta R),\rho_{2}\phi_{q}(R)]\), \(\forall s\in J\). Then, for \(t\in J\), we have
$$\begin{aligned} (Tx) (t)&= \int_{0}^{1}H_{2}(t,s)F \biggl(s, \int_{0}^{1}H_{1}(s,\tau)\phi _{q} \bigl(x(\tau) \bigr)\,d\tau,-\phi_{q} \bigl(x(s) \bigr) \biggr)\,ds \\ & \geq \int_{0}^{1}\rho G_{2}(s,s)\,ds\cdot \frac {6}{\rho}R \\ & =R, \end{aligned}$$
i.e.
\(x\in\partial\Omega_{2}\) implies
$$ \|Tx\|\geq\|x\|. $$
(3.29)
Applying Lemma 2.6 to (3.28) and (3.29) shows that T has a fixed point \(x\in K\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\) with (3.26). It is observed from (2.2) that problem (1.5) has one solution y such that
$$y(t)= \int_{0}^{1}H_{1}(t,s)\phi_{q} \bigl(x(s) \bigr)\,ds, \quad \mbox{and}\quad y(t)\in \bigl[\sigma \phi_{q}( r),\rho_{2}\phi_{q}(R) \bigr],\quad t \in J. $$
□
We remark that the condition (H6) in Theorem 3.4 can be replaced by the following condition:
- (H7):
-
There exist two constants r, R with \(0< r<\delta R\) such that
$$\begin{aligned}& F(t,u,v)\geq\frac{1}{\rho\beta}r, \quad \forall(t,u,v)\in J\times \bigl[ \rho_{1}\phi_{q}(\delta r),\rho_{2} \phi_{q}(r) \bigr]\times \bigl[-\phi_{q}(r),\phi _{q}(\delta r) \bigr]; \\& F(t,u,v)\leq\frac{1}{\eta}R, \quad \forall(t,u,v)\in J\times \bigl[\rho _{1}\phi_{q}(\delta R),\rho_{2} \phi_{q}(R) \bigr]\times \bigl[-\phi_{q}(R),\phi _{q}(\delta R) \bigr]. \end{aligned}$$
Corollary 3.1
If the condition (H6) in Theorem
3.4
is replaced by (H7), respectively, then the conclusions of Theorem
3.4
also hold.