In this section, we show that problem (1.1) has \(i_{0}\) or \(i_{\infty}\) positive solution(s) for sufficiently large or small λ under \(\alpha(t)\geq t\) on J.
For convenience we introduce the following notation:
$$\beta= \int_{0}^{1}\omega(s)\,ds. $$
Theorem 3.1
Assume (H1)-(H4) hold and
\(\alpha(t)\geq t\)
on
J.
-
(i)
If
\(i_{0}=1\textit{ or }2\), then there exists
\(\lambda_{0}>0\)
such that problem (1.1) has
\(i_{0}\)
positive solution(s) for
\(\lambda>\lambda _{0}\).
-
(ii)
If
\(i_{\infty}=1\textit{ or }2\), then there exists
\(\lambda_{0}>0\)
such that problem (1.1) has
\(i_{\infty}\)
positive solution(s) for
\(0<\lambda <\lambda_{0}\).
-
(iii)
If
\(i_{0}=0\)
or
\(i_{\infty}=0\), then problem (1.1) has no positive solution for sufficiently large or small
λ, respectively.
Proof
Part (i). Noticing that \(f(t,x)>0\) for all t and \(x>0\), we can define
$$m_{r}=\min_{t\in J, \delta r\leq x\leq r} \bigl\{ f(t,x) \bigr\} >0, $$
where \(r>0\).
Since \(0\leq t\leq\alpha(t)\leq1\) on J, for a function \(x\in K\) with \(\|x\|=r\), it follows from \(\delta r\leq x(t)\leq r\) on J that
$$\delta r\leq x \bigl(\alpha(t) \bigr)\leq r\quad \mbox{for }t\in J. $$
Let \(\lambda_{0}=\frac{\Delta r}{m_{r}\beta ab^{2} \gamma}\). Then, for \(x\in K\cap\partial\Omega_{r}\) and \(\lambda>\lambda_{0}\), we have
$$\begin{aligned}[b] (Tx) (t)&=\lambda \int_{0}^{1}H(t,s)\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda m_{r} \int _{0}^{1}\omega(s)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda m_{r}\beta \\ &>\frac{ab^{2}\gamma}{\Delta}\lambda_{0} m_{r}\beta \\ &= r=\|x\|, \end{aligned} $$
which implies that
$$ \|Tx\|>\|x\|,\quad \forall x\in K\cap\partial\Omega_{r}, \lambda> \lambda _{0}. $$
(3.1)
If \(f^{0}=0\), we can choose \(0< r_{1}< r\) such that
$$f(t,x)\leq\frac{\Delta}{a\lambda\beta\gamma D} x, \quad\forall t\in J, 0\leq x\leq r_{1}. $$
Since \(0\leq t\leq\alpha(t)\leq1\) on J, for a function \(x\in K\) with \(\|x\|=r\), it follows from \(0\leq x(t)\leq r_{1}\) on J that
$$0\leq x \bigl(\alpha(t) \bigr)\leq r_{1} \quad\mbox{for }t\in J. $$
Consequently, for any \(t\in J\) and \(x\in K\cap\partial\Omega _{r_{1}}\), (2.8) and (2.11) imply
$$\begin{aligned} (Tx) (t)&=\lambda \int_{0}^{1}H(t,s)\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int _{0}^{1}\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int _{0}^{1}\omega(s)\frac{\Delta}{a\lambda\beta\gamma D}x\bigl( \alpha(s)\bigr)\,ds \\ &\leq \int_{0}^{1}\omega(s)\frac{1}{\beta}\|x\|\,ds \\ &\leq\frac{1}{\beta}\|x\| \int_{0}^{1}\omega(s)\,ds \\ &=\|x\|, \end{aligned}$$
which implies
$$ \|Tx\|\leq\|x\|, \quad\forall x\in K\cap\partial\Omega_{r_{1}}. $$
(3.2)
Thus by (i) of Lemma 2.5, it follows from (3.1) and (3.2) that T has a fixed point x in \(K\cap(\bar{\Omega}_{r}\backslash\Omega _{r_{1}})\) with \(r_{1}\leq\|x\|\leq r\). Lemma 2.3 implies that problem (1.1) has at least one positive solution x with \(r_{1}\leq\|x\|\leq r\).
If \(f^{\infty}=0\), we can choose \(0<\varepsilon<\frac{\Delta}{a\gamma D \lambda\beta\varepsilon}\) and \(l>0\) such that
$$f(t,x)\leq\varepsilon x\quad \mbox{for }t\in J\mbox{ and }x\geq l. $$
Letting \(\zeta=\max_{t\in J, x\in[0,l]}f(t,x)\), then
$$0\leq f(t,x)\leq\varepsilon x+\zeta\quad \mbox{for }t\in J\mbox{ and }x\in[0, \infty). $$
Since \(0\leq t\leq\alpha(t)\leq1\) on J, for a function \(x\in K\) with \(\|x\|=r\), it follows from \(x(t)\geq l\) or \(0\leq x(t)\leq l\) on J that
$$x \bigl(\alpha(t) \bigr)\geq l\quad \mbox{or}\quad 0\leq x \bigl(\alpha(t) \bigr)\leq l\quad \mbox{for }t\in J. $$
Let \(r_{2}>\max \{2r,\frac{a\gamma D \lambda\zeta\beta}{\Delta -a\gamma D \lambda\beta\varepsilon} \}\). Then for \(t\in J\) and \(x\in K\cap\partial\Omega_{r_{2}}\), (2.8) and (2.11) imply
$$\begin{aligned} (Tx) (t)&=\lambda \int_{0}^{1}H(t,s)\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int _{0}^{1}\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int _{0}^{1}\omega(s) \bigl(\varepsilon x\bigl( \alpha(s)\bigr)+\zeta\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int _{0}^{1}\omega(s) \bigl(\varepsilon\|x\|+\zeta\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda\beta (\varepsilon r_{2}+\zeta) \\ &< r_{2}, \end{aligned}$$
which implies
$$ \|Tx\|\leq\|x\|,\quad \forall x\in K\cap\partial\Omega_{r_{2}}. $$
(3.3)
Thus by (ii) of Lemma 2.5, it follows from (3.1) and (3.3) that T has a fixed point x in \(K\cap(\bar{\Omega}_{r_{2}}\backslash\Omega _{r})\) with \(r\leq\|x\|\leq r_{2}\). Lemma 2.3 implies that problem (1.1) has at least one positive solution x with \(r\leq\|x\|\leq r_{2}\).
Turning to \(f^{0}=f^{\infty}=0\). Choose two numbers \(r_{3}\) and \(r_{4}\) satisfying
$$ 0< r_{1}< r_{3}< \delta r_{4}< r_{4}< \delta r_{2}< r_{2}< +\infty. $$
(3.4)
Similar to the proof of (3.1), there exists \(\lambda_{0}>0\) such that for \(\lambda>\lambda_{0}\)
$$ \|Tx\|>\|x\|,\quad \forall x\in K\cap\partial\Omega_{r_{i}}, i=3,4, $$
(3.5)
which together with (3.2) and (3.3) shows that T has a fixed point \(x_{1}\) in \(K\cap(\Omega_{r_{3}}\backslash\Omega_{r_{1}})\) and a fixed point \(x_{2}\) in \(K\cap(\bar{\Omega}_{r_{2}}\backslash\Omega_{r_{4}})\) with
$$r_{1}\leq\|x_{1}\|\leq r_{3}< r_{4}\leq \|x_{2}\|\leq r_{2}. $$
Consequently, it follows from Lemma 2.3 that problem (1.1) has two positive solutions for \(\lambda>\lambda_{0}\) if \(f^{0}=f^{\infty}=0\).
Part (ii). Noticing that \(f(t,x)>0\) for all t and \(x>0\), we can define
$$M_{r}=\max_{t\in J, 0\leq x\leq r} \bigl\{ f(t,x) \bigr\} >0, $$
where \(r>0\).
Since \(0\leq t\leq\alpha(t)\leq1\) on J, for a function \(x\in K\) with \(\|x\|=r\), it follows from \(0\leq x(t)\leq r\) on J that
$$0\leq x \bigl(\alpha(t) \bigr)\leq r\quad \mbox{for }t\in J. $$
Let \(\lambda_{0}\leq\frac{\Delta r}{M_{r}\beta a D\gamma}\). Then, for \(x\in K\cap\partial\Omega_{r}\) and \(0<\lambda<\lambda_{0}\), we have
$$\begin{aligned} (Tx) (t)&=\lambda \int_{0}^{1}H(t,s)\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int _{0}^{1}\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}M_{r}\lambda \int _{0}^{1}\omega(s)\,ds \\ &=\frac{a\gamma D}{\Delta}M_{r}\lambda\beta \\ &< \frac{a\gamma D}{\Delta}M_{r}\lambda_{0}\beta \\ &\leq r, \end{aligned}$$
which implies that
$$ \|Tx\|< \|x\|, \quad\forall x\in K\cap\partial\Omega_{r}, 0< \lambda < \lambda_{0}. $$
(3.6)
If \(f_{0}=\infty\), we can choose \(0< r_{1}< r\) such that
$$f(t,x)\geq\frac{\Delta}{ab^{2}\delta\lambda\beta\gamma} x, \quad \forall t\in J, 0\leq x\leq r_{1}. $$
Since \(0\leq t\leq\alpha(t)\leq1\) on J, for a function \(x\in K\) with \(\|x\|=r\), it follows from \(0\leq x(t)\leq r_{1}\) on J that
$$0\leq x \bigl(\alpha(t) \bigr)\leq r_{1}\quad \mbox{for }t\in J. $$
Consequently, for any \(t\in J\) and \(x\in K\cap\partial\Omega _{r_{1}}\), (2.8) and (2.11) imply
$$\begin{aligned} (Tx) (t)&=\lambda \int_{0}^{1}H(t,s)\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)\frac{\Delta}{ab^{2}\delta\lambda\beta\gamma} x\bigl( \alpha (s)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)\frac{\Delta}{ab^{2}\delta\lambda\beta\gamma}\delta\| x\| \,ds \\ &\geq\|x\|, \end{aligned}$$
which implies
$$ \|Tx\|\geq\|x\|, \quad \forall x\in K\cap\partial\Omega_{r_{1}}. $$
(3.7)
Thus by (ii) of Lemma 2.5, it follows from (3.6) and (3.7) that T has a fixed point x in \(K\cap(\bar{\Omega}_{r}\backslash\Omega _{r_{1}})\) with \(r_{1}\leq\|x\|\leq r\). Lemma 2.3 shows that problem (1.1) has at least one positive solution x with \(r_{1}\leq\|x\|\leq r\).
If \(f_{\infty}=\infty\), we can choose sufficiently large \(\varepsilon >0\) and \(l>0\) such that
$$f(t,x)\geq\varepsilon x\quad \mbox{for }t\in J\mbox{ and }x\geq l, $$
where ε satisfies
$$\frac{ab^{2}\gamma}{\Delta}\lambda\varepsilon\delta\beta\geq1. $$
Since \(0\leq t\leq\alpha(t)\leq1\) on J, it follows from \(x(t)\geq l\) on J that
$$x \bigl(\alpha(t) \bigr)\geq l\quad \mbox{for }t\in J. $$
Let \(r_{2}>\max \{2r,\frac{l}{\delta} \}\). Then for \(t\in J\) and \(x\in K\cap\partial\Omega_{r_{2}}\) we have
$$x(t)\geq\delta\|x\|> l. $$
Hence, for \(t\in J\) and \(x\in K\cap\partial\Omega_{r_{2}}\), it follows from (2.8) and (2.11) that
$$\begin{aligned} (Tx) (t)&=\lambda \int_{0}^{1}H(t,s)\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)f\bigl(s,x\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)\varepsilon x\bigl(\alpha(s) \bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int _{0}^{1}\omega(s)\varepsilon\delta\|x\|\,ds \\ &=\frac{ab^{2}\gamma}{\Delta}\lambda\varepsilon \delta\|x\|\beta \\ &\geq\|x\|, \end{aligned}$$
which implies
$$ \|Tx\|\geq\|x\|,\quad \forall x\in K\cap\partial\Omega_{r_{2}}. $$
(3.8)
Thus by (i) of Lemma 2.5, it follows from (3.6) and (3.8) that T has a fixed point x in \(K\cap(\bar{\Omega}_{r_{2}}\backslash\Omega _{r})\) with \(r\leq\|x\|\leq r_{2}\). Lemma 2.3 shows that problem (1.1) has at least one positive solution x with \(r\leq\|x\|\leq r_{2}\).
Turning to \(f_{0}=f_{\infty}=\infty\). Choose two numbers \(r_{3}\) and \(r_{4}\) satisfying (3.4). Similar to the proof of (3.6), there exists \(\lambda_{0}>0\) such that for \(0<\lambda<\lambda_{0}\)
$$ \|Tx\|< \|x\|,\quad \forall x\in K\cap\partial\Omega_{r_{i}}, i=3,4, $$
(3.9)
which together with (3.7) and (3.8) shows that T has a fixed point \(x_{1}\) in \(K\cap(\Omega_{r_{3}}\backslash\Omega_{r_{1}})\) and a fixed point \(x_{2}\) in \(K\cap(\bar{\Omega}_{r_{2}}\backslash\Omega_{r_{4}})\) with
$$r_{1}\leq\|x_{1}\|\leq r_{3}< r_{4}\leq \|x_{2}\|\leq r_{2}. $$
Consequently, it follows from Lemma 2.3 that problem (1.1) has two positive solutions for \(0<\lambda<\lambda_{0}\) if \(f_{0}=f_{\infty }=\infty\).
Part (iii). If \(i_{0}=0\), then \(f_{0}>0\) and \(f_{\infty}>0\). It follows that there exist positive numbers \(\eta_{1}>0\), \(\eta_{2}>0\), \(h_{1}>0\), and \(h_{2}>0\) such that \(h_{1}< h_{2}\) and, for \(t \in J\), \(0< x\leq h_{1}\), we have
$$ f(t,x)\geq\eta_{1}x, $$
(3.10)
and, for \(t \in J\), \(x\geq h_{2}\), we have
$$ f(t,x)\geq\eta_{2}x. $$
(3.11)
Let
$$\eta=\min \biggl\{ \eta_{1},\eta_{2}, \min \biggl\{ \frac{f(t,x)}{x}:t \in J, \delta h_{1}\leq x \leq h_{2} \biggr\} \biggr\} >0. $$
Thus, for \(t \in J\), \(x \geq\delta h_{1}\), we have
$$ f(t,x)\geq\eta x, $$
(3.12)
and, for \(t \in J\), \(x \leq h_{1}\), we have
$$ f(t,x)\geq\eta x. $$
(3.13)
Assume \(y\in K\) is a positive solution of problem (1.1). We will show that this leads to a contradiction for \(\lambda> \lambda_{0}=[ab^{2}\gamma\eta\delta\beta]^{-1}\Delta\).
In fact, if \(\|y\|\leq h_{1}\), (3.13) shows that
$$f(t,y)\geq\eta y,\quad \mbox{for }t\in J. $$
On the other hand, if \(\|y\|>h_{1}\), then
$$\min_{t\in J}y(t) \geq\delta\|y\|>\delta h_{1}. $$
Since \(0\leq t\leq\alpha(t)\leq1\), it follows from \(y(t)>\delta h_{1}\) on \(t\in J\) that \(y(\alpha(t))>\delta h_{1}\) on \(t\in J\), which, together with (3.12), shows that
$$f(t,y \bigl(\alpha(t) \bigr)\geq\eta y \bigl(\alpha(t) \bigr),\quad t\in J. $$
Since \((Ty)(t)=y(t)\), for \(\lambda> \lambda_{0}\), it follows from (2.8) and (2.11) that
$$\begin{aligned} \|y\|&=\|Ty\| \\ &= \max_{ t\in J}\lambda \int _{0}^{1}H(t,s)w(s)f\bigl(s,y\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int_{0}^{1}\omega (s)f\bigl(s,y\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int_{0}^{1}\omega (s)\eta y\bigl(\alpha(s)\bigr) \,ds \\ &\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int_{0}^{1}\omega (s)\eta\delta\|y\|\,ds \\ &=\frac{ab^{2}\gamma}{\Delta}\lambda\eta\delta\|y\|\beta \\ &>\frac{ab^{2}\gamma}{\Delta}\lambda_{0} \eta\delta\|y\|\beta \\ &=\|y\|, \end{aligned}$$
which is a contradiction.
If \(i_{\infty}=0\), then \(f^{0}<\infty\) and \(f^{\infty}<\infty\). It follows that there exist positive numbers \(\eta_{3}>0\), \(\eta_{4}>0\), \(h_{3}>0\), and \(h_{4}>0\) such that \(h_{3}< h_{4}\) and, for \(t \in J\), \(0< x\leq h_{3}\), we have
$$ f(t,x)\leq\eta_{3}x, $$
(3.14)
and, for \(t \in J\), \(x\geq h_{4}\), we have
$$ f(t,x)\leq\eta_{4}x. $$
(3.15)
Let
$$\eta^{*}=\max \biggl\{ \eta_{3},\eta_{4}, \max \biggl\{ \frac{f(t,x)}{x}:t \in J, h_{3}\leq x \leq h_{4} \biggr\} \biggr\} >0. $$
Thus, we have
$$ f(t,x)\leq\eta^{*} x,\quad t \in J, x \in[0,\infty). $$
(3.16)
Since \(0\leq t\leq\alpha(t)\leq1\) on J, it follows from \(0\leq x(t)\leq h_{3}\), \(x(t)\geq h_{4}\), and \(h_{3}\leq x(t)\leq h_{4}\) on J that \(0\leq x(\alpha(t))\leq h_{3}\), \(x(\alpha(t))\geq h_{4}\), and \(h_{3}\leq x(\alpha(t))\leq h_{4}\) on J, respectively.
Assume \(y\in K\) is a positive solution of problem (1.1). We will show that this leads to a contradiction for \(0<\lambda< \lambda_{0}=[a\gamma D \eta^{*}\beta]^{-1}\Delta\).
Since \((Ty)(t)=y(t)\), for \(0<\lambda< \lambda_{0}\), it follows from (2.8) and (2.11) that
$$\begin{aligned} \|y\|&=\|Ty\| \\ &= \max_{ t\in J}\lambda \int _{0}^{1}H(t,s)w(s)f\bigl(s,y\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int_{0}^{1}\omega (s)f\bigl(s,y\bigl(\alpha(s) \bigr)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda \int_{0}^{1}\omega (s)\eta^{*}y\bigl( \alpha(s)\bigr)\,ds \\ &\leq\frac{a\gamma D}{\Delta}\lambda\eta^{*}\|y\| \int _{0}^{1}\omega(s)\,ds \\ &=\frac{a\gamma D}{\Delta}\lambda\eta^{*}\|y\|\beta \\ &< \frac{a\gamma D}{\Delta}\lambda_{0} \eta^{*}\|y\|\beta \\ &=\|y\|, \end{aligned}$$
which is a contradiction. □
Theorem 3.2 is a direct consequence of the proof of Theorem 3.1(iii). Under the conditions of Theorem 3.2 we are able to give explicit intervals of λ such that (1.1) has no positive solution.
Theorem 3.2
Assume (H1)-(H4) hold and
\(\alpha(t)\geq t\)
on
J.
-
(i)
If there exists
\(l>0\)
such that
\(f(t,x)\geq lx\)
for
\(t\in J\)
and
\(x\in[0,\infty)\), then there exists
\(\lambda_{0}>0\)
such that problem (1.1) has no positive solution for
\(\lambda>\lambda_{0}\).
-
(ii)
If there exists
\(L>0\)
such that
\(f(t,x)\leq Lx\)
for
\(t\in J\)
and
\(x\in[0,\infty)\), then there exists
\(\lambda_{0}>0\)
such that problem (1.1) has no positive solution for
\(0<\lambda<\lambda_{0}\).
Theorem 3.3
Assume (H1)-(H4) hold, \(\alpha(t)\geq t\)
on
J
and
\(i_{0}=i_{\infty}=0\). Then problem (1.1) has at least one positive solution in
K
provided
$$ \frac{\Delta}{ab^{2}\gamma\beta\delta\max\{f_{\infty},f^{0}\}} < \lambda < \frac{\Delta}{aD\gamma\beta\min\{f_{\infty},f^{0}\}}. $$
(3.17)
Proof
We give the proof under two cases of \(f_{\infty}>f^{0}\) and \(f_{\infty}< f^{0}\).
If \(f_{\infty}>f^{0}\), then (3.17) implies that
$$\frac{\Delta}{ab^{2}\gamma\beta\delta f_{\infty}}< \lambda< \frac{\Delta }{aD\gamma\beta f^{0}}. $$
It is easy to see that there exists \(\varepsilon>0\) such that
$$\frac{\Delta}{ab^{2}\gamma\beta\delta(f_{\infty}-\varepsilon)} \leq\lambda \leq\frac{\Delta}{aD\gamma\beta(f^{0}+\varepsilon)}. $$
Now, considering \(f^{0}\) and \(f_{\infty}\), there exists \(r_{1}>0\) such that \(f(t,x)\leq(f^{0}+\varepsilon)x\) for \(t\in J\) and \(0\leq x\leq r_{1}\).
Since \(0\leq t\leq\alpha(t)\leq1\), it follows from \(0\leq x(t)\leq r_{1}\) on J that \(0\leq x(\alpha(t))\leq r_{1}\). Hence, similar to the proof of (3.2), for \(x\in K\cap\partial\Omega_{r_{1}}\) we have
$$\|Tx\|\leq\frac{a\gamma D}{\Delta}\lambda \int_{0}^{1}\omega (s)f \bigl(s,x \bigl(\alpha(s) \bigr) \bigr)\,ds\leq\frac{a\gamma D}{\Delta}\lambda \bigl(f^{0}+\varepsilon \bigr)\beta\|x\|\leq\|x\|. $$
On the other hand, there exists \(L>0\) with \(L>r_{1}\) such that \(f(t,x)\geq(f_{\infty}-\varepsilon)x\) for \(t\in J\) and \(x\geq L\).
Since \(0\leq t\leq\alpha(t)\leq1\), it follows from \(0\leq x(t)\leq r_{1}\) on J that \(x(\alpha(t))\geq L\).
Let \(r_{2}=\max \{2r_{1},\frac{L}{\delta} \}\) and it follows that \(x(t)\geq\delta\|x\|\geq L\) for \(t\in J\) and \(x\in K\cap\partial\Omega _{r_{2}}\). Similar to the proof of (3.8), for \(t\in J\) and \(x\in K\cap \partial\Omega_{r_{2}}\) we have
$$\|Tx\|\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int_{0}^{1}\omega (s)f \bigl(s,x \bigl(\alpha(s) \bigr) \bigr)\,ds\geq\frac{ab^{2}\gamma}{\Delta}\lambda(f_{\infty }-\varepsilon)\beta \delta\|x\|\geq\|x\|. $$
It follows from Lemma 2.5 that T has a fixed point in \(K\cap(\bar{\Omega}_{r_{2}}\backslash\Omega_{r_{1}})\). Consequently, problem (1.1) has a positive solution.
If \(f_{\infty}< f^{0}\), then (3.17) shows that
$$\frac{\Delta}{ab^{2}\gamma\beta\delta f^{0}}< \lambda< \frac{\Delta }{aD\gamma\beta f_{\infty}}. $$
It is easy to see that there exists \(\varepsilon>0\) such that
$$\frac{\Delta}{ab^{2}\gamma\beta\delta(f^{0}-\varepsilon)} \leq\lambda\leq \frac{\Delta}{aD\gamma\beta(f_{\infty}+\varepsilon)}. $$
Now, turning to \(f^{0}\) and \(f_{\infty}\), there exists \(r_{1}>0\) such that \(f(t,x)\geq(f^{0}-\varepsilon)x\) for \(t\in J\) and \(0\leq x\leq r_{1}\).
Since \(0\leq t\leq\alpha(t)\leq1\), it follows from \(0\leq x(t)\leq r_{1}\) that \(0\leq x(\alpha(t))\leq r_{1}\) on J. Hence, similar to the proof of (3.8), for \(x\in K\cap\partial\Omega_{r_{1}}\) we have
$$\|Tx\|\geq\frac{ab^{2}\gamma}{\Delta}\lambda \int_{0}^{1}\omega (s)f \bigl(s,x \bigl(\alpha(s) \bigr) \bigr)\,ds\geq\frac{ab^{2}\gamma}{\Delta}\lambda \bigl(f^{0}-\varepsilon \bigr)\beta\delta\|x\|\geq\|x\|. $$
On the other hand, there exists \(L>0\) with \(L>r_{1}\) such that \(f(t,x)\leq(f_{\infty}+\varepsilon)x\) for \(t\in J\) and \(x\geq L\).
Letting \(\zeta=\max_{t\in J, x\in[0,L]}f(t,x)\), then
$$0\leq f(t,x)\leq(f_{\infty}+\varepsilon)x+\zeta \quad \mbox{for }t\in J \mbox{ and }x\in[0, \infty). $$
Since \(0\leq t\leq\alpha(t)\leq1\) on J, it follows from \(x(t)\geq L\) or \(0\leq x(t)\leq L\) on J that
$$x \bigl(\alpha(t) \bigr)\geq L \quad \mbox{or}\quad 0\leq x \bigl(\alpha(t) \bigr)\leq L\quad \mbox{for }t\in J. $$
Let \(r_{2}>\max \{2r,\frac{a\gamma D \lambda\zeta\beta}{\Delta -a\gamma D \lambda\beta(f_{\infty}+\varepsilon)} \}\). Then, for \(t\in J\) and \(x\in K\cap\partial\Omega_{r_{2}}\), similar to the proof of (3.3) we get
$$\|Tx\|\leq\frac{a\gamma D}{\Delta}\lambda \int_{0}^{1}\omega (s)f \bigl(s,x \bigl(\alpha(s) \bigr) \bigr)\,ds\leq\frac{a\gamma D}{\Delta}\lambda\beta \bigl((f_{\infty}+ \varepsilon)r_{2}+\zeta \bigr)< r_{2}=\|x\|. $$
It follows from Lemma 2.5 that T has a fixed point in \(K\cap(\bar{\Omega}_{r_{2}}\backslash\Omega_{r_{1}})\). Consequently, problem (1.1) has a positive solution. □
Corollary 3.1
Assume (H1)-(H4) hold, \(\alpha(t)\geq t\)
on
J
and
\(i_{0}=i_{\infty}=0\). Then problem (1.1) has at least one positive solution in
K
provided
$$\frac{\Delta}{ab^{2}\gamma\beta\delta\max\{f^{\infty},f_{0}\}} < \lambda < \frac{\Delta}{aD\gamma\beta\min\{f^{\infty},f_{0}\}}. $$
Proof
The proof is similar to that of Theorem 3.3. □