We recall that \(J =[j_{*} , j^{*}]\subset (0,1)\) is a closed interval of positive measure such that \(a_{i}(t) >0\) for all \(t \in J\) and \(i=1,2\). Let \(\gamma =\min \{ j_{*},1-j^{*}\}>0\). Define the cone
$$ \mathcal{K} = \Bigl\{ (w_{1},w_{2}) \in \mathcal{P}: \min _{J} w_{i}(t) \geq \gamma \Vert w_{i} \Vert _{\infty } \text{ for } i=1,2 \Bigr\} . $$
Then we have the following:
Lemma 3.1
For a given cone
\(\mathcal{P} \)
in
X, we have
$$ T_{\lambda, \mu,\delta }(\mathcal{P}) \subset \mathcal{K}. $$
(8)
Proof
For any \((w_{1},w_{2}) \in \mathcal{P}\), we first find
$$\begin{aligned} A_{\lambda, \mu,\delta }(w_{1},w_{2}) (t) \leq& \delta (1+\rho \nu _{1}+\rho_{1} \nu_{2}) \\ &{} + \int_{0}^{1} G(s,s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds \\ &{} +\rho_{1} \int_{0}^{1} \biggl( \int_{0}^{1}G(\tau, s) g_{2}(\tau)\,d\tau \biggr) a_{1}(s) f_{1}(w_{1}+p_{\lambda, \mu }, w_{2}+q_{\lambda, \mu })\,ds \\ &{} +\rho \int_{0}^{1} \biggl( \int_{0}^{1}G(\tau, s) g_{1}(\tau)\,d\tau \biggr) a _{2}(s) f_{2}(w_{1}+p_{\lambda, \mu }, w_{2}+q_{\lambda, \mu })\,ds \end{aligned}$$
for all \(t \in [0,1]\). Thus we obtain
$$\begin{aligned} \bigl\Vert A_{\lambda, \mu,\delta }(w_{1},w_{2}) \bigr\Vert _{\infty } \leq& \delta (1+ \rho \nu_{1}+ \rho_{1} \nu_{2}) \\ &{} + \int_{0}^{1} G(s,s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds+\rho_{1} M_{1} +\rho M_{2}, \end{aligned}$$
(9)
where
$$ M_{1}:= \int_{0}^{1} \biggl( \int_{0}^{1}G(\tau, s) g_{2}(\tau)\,d\tau \biggr) a_{1}(s) f_{1}(w_{1}+p_{\lambda, \mu }, w_{2}+q_{\lambda, \mu })\,ds $$
and
$$ M_{2}:= \int_{0}^{1} \biggl( \int_{0}^{1}G(\tau, s) g_{1}(\tau)\,d\tau \biggr) a _{2}(s) f_{2}(w_{1}+p_{\lambda, \mu }, w_{2}+q_{\lambda, \mu })\,ds. $$
Similarly, we have
$$\begin{aligned}& \bigl\Vert B_{\lambda, \mu,\delta }(w_{1},w_{2}) \bigr\Vert _{\infty } \\& \quad \leq \delta (1+ \rho \nu_{2}+ \rho_{2} \nu_{1}) \\& \qquad {} + \int_{0}^{1} G(s,s) a_{2}(s)f_{2}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds+\rho_{2} M_{2} +\rho M_{1}. \end{aligned}$$
(10)
Then by (9) we find that, for all \(t \in J\),
$$\begin{aligned} A_{\lambda, \mu,\delta }(w_{1},w_{2}) (t) =&\delta \bigl[1+(\rho \nu_{1}+ \rho_{1} \nu_{2})t\bigr] \\ &{} + \int_{0}^{1} G(t,s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds+ t\rho_{1} M_{1} +t \rho M_{2} \\ =&\delta \bigl[1+(\rho \nu_{1}+\rho_{1} \nu_{2})t\bigr]+ \int_{0}^{t} s(1-t)a_{1}(s)f _{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q_{\lambda,\mu }) \,ds \\ &{} + \int_{t}^{1} t(1-s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds + t\rho_{1} M_{1} +t \rho M_{2} \\ \geq& \delta \bigl[1+(\rho \nu_{1}+\rho_{1} \nu_{2})j_{*}\bigr]+\bigl(1-j^{*}\bigr) \int _{0}^{t} s a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q_{\lambda, \mu })\,ds \\ &{} + j_{*} \int_{t}^{1} (1-s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w _{2}+q_{\lambda,\mu })\,ds + j_{*}( \rho_{1} M_{1} + \rho M_{2} ) \\ \geq& \min \bigl\{ j_{*}, 1-j^{*} \bigr\} \biggl( \delta (1+\rho \nu_{1}+\rho_{1} \nu _{2}) \\ &{} + \int_{0}^{t} s(1-s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds \\ &{} + \int_{t}^{1} s(1-s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds + \rho_{1} M_{1} + \rho M_{2} \biggr) \\ =& \min \bigl\{ j_{*}, 1-j^{*} \bigr\} \biggl( \delta (1+\rho \nu_{1}+\rho_{1} \nu _{2}) \\ &{} + \int_{0}^{1} G(s,s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q _{\lambda,\mu })\,ds+ \rho_{1} M_{1} + \rho M_{2} \biggr) \\ \geq& \gamma \bigl\Vert A_{\lambda, \mu }(w_{1},w_{2}) \bigr\Vert _{\infty }. \end{aligned}$$
By the same argument, using (10), we also have that \(B_{\lambda, \mu,\delta }(w_{1},w_{2})(t)\geq \gamma \|B_{\lambda, \mu }(w_{1},w_{2})\|_{\infty }\) for all \(t \in J\). □
As a consequence of the lemma, note that if \((w_{1},w_{2})\) is a solution of (5), then by (7) we have
$$\begin{aligned} \inf_{t \in J} w_{i} (t) \geq \gamma \Vert w_{i} \Vert _{\infty } \quad \text{for } i=1,2. \end{aligned}$$
(11)
Lemma 3.2
Assume (H). If (5) has a positive solution at
\(( \bar{\lambda },\bar{\mu })\), then (5) also has a positive solution at
\((\lambda, \mu)\)
for all
\((\lambda, \mu) \leq ( \bar{ \lambda },\bar{\mu })\).
Proof
Let \((\bar{w}_{1},\bar{w}_{2})\) be a positive solution of (5) at \((\bar{\lambda },\bar{\mu })\), and let \((\lambda, \mu) \in [0, \infty)^{2} \setminus \{(0,0)\}\) with \((\lambda, \mu)\leq (\bar{\lambda },\bar{\mu })\). Then \((\bar{w}_{1},\bar{w} _{2})\) is a supersolution of (5) at \((\lambda, \mu)\) since \(p_{\lambda,\mu } (t) \leq p_{\bar{\lambda },\bar{\mu }}(t)\) and \(q_{\lambda,\mu } (t) \leq q_{\bar{\lambda },\bar{\mu }}(t)\) for \(t \in (0,1)\) and \(f_{i}\) is increasing for each \(i=1,2\). Clearly, \((0,0)\) is a subsolution of (5). Notice that \((0,0)\) is not a solution of (5) since \(f_{i}(p_{\lambda,\mu }(t),q_{\lambda,\mu }(t)) >0 \) for \(t \in (0,1)\) and \(i=1,2\). Since \((\bar{w}_{1}, \bar{w}_{2})\neq(0,0)\) and \((\bar{w}_{1},\bar{w}_{2})\geq (0,0)\), by Theorem 2.1, (5) has a positive solution at \((\lambda, \mu)\). □
Lemma 3.3
Assume (H) and (H1). Then there exists
\(( \tilde{\lambda },\tilde{\mu })>(0,0)\)
such that (5) has a positive solution for all
\((\lambda, \mu) \leq (\tilde{\lambda }, \tilde{\mu })\).
Proof
Let \((\psi_{1},\psi_{2})\) be the unique solution of the following system:
$$\begin{aligned} \textstyle\begin{cases} \psi_{1}''(t)+a_{1}(t)=0 & t \in (0,1), \\ \psi_{2}''(t)+a_{2}(t)=0 & t \in (0,1), \\ \psi_{1}(0)=0=\psi_{2}(0), & \\ \psi_{1}(1)- \int_{0}^{1} g_{1}(s)\psi_{2}(s)\,ds=0, & \\ \psi_{2}(1)- \int_{0}^{1} g_{2}(s)\psi_{1}(s)\,ds=0. & \end{cases}\displaystyle \end{aligned}$$
We recall that \(p_{\lambda, \mu }(t)= (\rho \lambda +\rho_{1}\mu)t\) and \(q_{\lambda, \mu }(t)=(\rho_{2}\lambda +\rho \mu)t\) and denote \(\phi_{1}(t)= 1+(\rho \nu_{1}+ \rho_{1}\nu_{2})t\) and \(\phi_{2}(t)= 1+( \rho \nu_{2}+ \rho_{2}\nu_{1})t\). Then let \(\alpha = \|\psi_{1}\|_{ \infty }+\|\psi_{2}\|_{\infty }+\rho +\rho_{1}+\rho_{2} +\rho \nu_{2} +\rho_{2}\nu_{1}+\rho \nu_{1}+\rho_{1}\nu_{2} +2\). By (H1) there exist \(\tilde{\lambda }\approx 0\), \(\tilde{\mu }\approx 0\), and \(\tilde{\delta }\approx 0\), sufficiently small, such that
$$\begin{aligned}& f_{1}\bigl( (\tilde{\lambda }+\tilde{\mu } +\tilde{ \delta } ) \Vert \psi_{1} \Vert _{\infty }+p_{\tilde{\lambda },\tilde{\mu }}(1)+ \tilde{\delta }\phi _{1}(1), (\tilde{\lambda }+\tilde{\mu } +\tilde{ \delta } ) \Vert \psi_{2} \Vert _{\infty }+q_{\tilde{\lambda },\tilde{\mu }}(1)+ \tilde{\delta } \phi_{2}(1)\bigr) \\& \quad \leq \frac{1}{\alpha } \bigl( (\tilde{\lambda }+\tilde{\mu } + \tilde{ \delta } ) \Vert \psi_{1} \Vert _{\infty }+p_{\tilde{\lambda }, \tilde{\mu }}(1)+ \tilde{\delta }\phi_{1}(1) \\& \qquad {} + (\tilde{\lambda }+\tilde{\mu } +\tilde{\delta } ) \Vert \psi_{2} \Vert _{\infty }+q_{\tilde{\lambda },\tilde{\mu }}(1)+\tilde{\delta }\phi _{2}(1) \bigr). \end{aligned}$$
(12)
Now we define \(Z_{1}(t)=(\tilde{\lambda }+\tilde{\mu }+ \tilde{\delta })\psi_{1}+\tilde{ \delta }\phi_{1}(t)\) and \(Z_{2}(t)=( \tilde{\lambda }+\tilde{\mu }+\tilde{\delta })\psi_{2} + \tilde{\delta }\phi_{2}(t)\). Then, from (H) and (12) we have \(Z_{1}(0)=\tilde{\delta }\),
$$\begin{aligned}& Z_{1}''(t)+a_{1}(t)f_{1} \bigl(Z_{1}(t)+p_{\tilde{\lambda },\tilde{\mu }}(t), Z_{2}(t)+q_{\tilde{\lambda },\tilde{\mu }}(t) \bigr) \\& \quad = -(\tilde{\lambda }+\tilde{\mu }+\tilde{\delta })a_{1}(t)+a_{1}(t)f _{1}\bigl(Z_{1}(t)+p_{\tilde{\lambda },\tilde{\mu }}(t),Z_{2}(t)+q_{ \tilde{\lambda },\tilde{\mu }}(t) \bigr) \\& \quad \leq a_{1}(t) \bigl\{ f_{1}\bigl( (\tilde{\lambda }+ \tilde{\mu } + \tilde{\delta } ) \Vert \psi_{1} \Vert _{\infty }+p_{\tilde{\lambda }, \tilde{\mu }}(1)+\tilde{\delta }\phi_{1}(1), \\& \qquad {} (\tilde{\lambda }+\tilde{\mu } +\tilde{\delta } ) \Vert \psi_{2} \Vert _{\infty }+q_{\tilde{\lambda },\tilde{\mu }}(1)+\tilde{\delta } \phi_{2}(1)\bigr) -( \tilde{\lambda }+\tilde{\mu }+\tilde{\delta } ) \bigr\} \\& \quad \leq a_{1}(t) \biggl[ \frac{1}{\alpha } \bigl( (\tilde{\lambda }+ \tilde{\mu } +\tilde{\delta } ) \Vert \psi_{1} \Vert _{\infty }+p_{ \tilde{\lambda },\tilde{\mu }}(1)+\tilde{\delta }\phi_{1}(1) \\& \qquad {} + (\tilde{\lambda }+\tilde{\mu } +\tilde{\delta } ) \Vert \psi_{2} \Vert _{\infty }+q_{\tilde{\lambda },\tilde{\mu }}(1)+ \tilde{ \delta }\phi_{2}(1) \bigr) - (\tilde{\lambda }+\tilde{\mu }+ \tilde{ \delta }) \biggr] \\& \quad \leq 0, \end{aligned}$$
and
$$\begin{aligned} Z_{1}(1)-\int_{0}^{1} g_{1}(s)Z_{2}(s)\,ds =\tilde{\delta }\phi_{1}(1) - \tilde{\delta } \int_{0}^{1} g_{1}(s) \phi_{2}(s)\,ds =\tilde{\delta } \geq \delta. \end{aligned}$$
Similarly, we can show that \(Z_{2}\) satisfies the same inequalities. This shows that \((Z_{1},Z_{2})\) is a supersolution of (5) at \((\tilde{\lambda },\tilde{\mu })\). On the other hand, \((0,0)\) is a strict subsolution such that \((0,0)\leq (Z_{1},Z_{2})\) in \([0,1]\). Hence, by Theorem 2.1, (5) has a positive solution at \((\tilde{\lambda },\tilde{\mu })\), and then Lemma 3.2 completes the proof. □
Lemma 3.4
Assume (H2). There exists
\(M>0\)
such that
\(\|(w_{1},w_{2}) \|\leq M\)
for all possible solutions
\((w_{1},w_{2})\)
of (5).
Proof
As \((w_{1},w_{2})\) is a solution of (5), we find
$$\begin{aligned} \Vert w_{1} \Vert _{\infty } \geq& w_{1}(j_{*}) \\ =&\delta \bigl[1+(\rho \nu_{1}+\rho_{1} \nu_{2})j_{*}\bigr] \\ &{} + \int_{0}^{1} \biggl( G(j_{*},s) + \rho_{1} j_{*} \int_{0}^{1} G(\tau, s)g_{2}(\tau)\,d\tau \biggr) a_{1}(s)f_{1}(w_{1}+p_{\lambda, \mu },w _{2}+q_{\lambda, \mu }) \\ &{} + \biggl( \rho j_{*} \int_{0}^{1} G(\tau, s)g_{1}(\tau)\,d\tau \biggr) a _{2}(s)f_{2}(w_{1}+p_{\lambda, \mu },w_{2}+q_{\lambda, \mu }) \,ds \\ \geq& \int_{j_{*}}^{j^{*}} \biggl( \rho j_{*} \int_{j_{*}}^{j^{*}} G( \tau, s)g_{1}(\tau)\,d \tau \biggr) a_{2}(s)f_{2}(w_{1}+p_{\lambda, \mu },w_{2}+q_{\lambda, \mu }) \,ds \\ \geq& \rho j_{*} \int_{j_{*}}^{j^{*}} g_{1}(s) \biggl( \int_{j_{*}}^{s} \tau (1-s) a_{2}( \tau)f_{2}(w_{1}+p_{\lambda, \mu },w_{2}+q_{\lambda, \mu }) \,d\tau \\ &{} + \int_{s}^{j^{*}} s (1-\tau) a_{2}( \tau)f_{2}(w_{1}+p_{ \lambda, \mu },w_{2}+q_{\lambda, \mu } )\,d\tau \biggr)\,ds \\ \geq& \rho j_{*} \int_{j_{*}}^{j^{*}} g_{1}(s)j_{*} \bigl(1-j^{*}\bigr) \biggl( \int_{j_{*}}^{j^{*}} a_{2}(\tau)f_{2}(w_{1}+p_{\lambda, \mu },w_{2}+q _{\lambda, \mu })\,d\tau \biggr)\,ds \\ \geq& \rho j_{*}^{2}\bigl(1-j^{*}\bigr) \biggl( \int_{j_{*}}^{j^{*}} g_{1}(s)\,ds \biggr) \biggl( \int_{j_{*}}^{j^{*}} a_{2}(s)\,ds \biggr) f_{2}\bigl(\gamma \Vert w_{1} \Vert _{\infty },\gamma \Vert w_{2} \Vert _{\infty }\bigr). \end{aligned}$$
Hence we obtain
$$\begin{aligned} \Vert w_{1} \Vert _{\infty }+ \Vert w_{2} \Vert _{\infty } \geq \gamma^{2} \rho j_{*} \biggl( \int_{j_{*}}^{j^{*}} g_{1}(s)\,ds \biggr) \biggl( \int_{j_{*}} ^{j^{*}} a_{2}(s)\,ds \biggr) f_{2}\bigl(\gamma \Vert w_{1} \Vert _{\infty }, \gamma \Vert w_{2} \Vert _{\infty }\bigr), \end{aligned}$$
which implies that
$$ \bigl( \gamma^{3} \rho j_{*} \bigr)^{-1} \biggl( \int_{j_{*}}^{j^{*}} g_{1}(s)\,ds \biggr) ^{-1} \biggl( \int_{j_{*}}^{j^{*}} a_{2}(s)\,ds \biggr) ^{-1} \geq \frac{ f_{2}(\gamma \Vert w_{1} \Vert _{\infty }, \gamma \Vert w_{2} \Vert _{\infty })}{\gamma \Vert w_{1} \Vert _{\infty }+\gamma \Vert w_{2} \Vert _{\infty }}. $$
(13)
If \(\|w_{1}\|_{\infty }\rightarrow \infty \), then (13) contradict to (H2). Similarly, we can show that \(\|w_{2}\|_{\infty }\) is bounded. □
Define \(\mathcal{S}=\{(\lambda, \mu)\in [0,\infty)^{2} \setminus \{(0,0)\} : ( \text{5) has a positive solution at } (\lambda, \mu) \}\). Then \(\mathcal{S}\neq\emptyset \) by Lemma 3.3.
Lemma 3.5
Assume (H) and (H2). Then
\((\mathcal{S}, \leq)\)
is bounded above.
Proof
We claim that there exist \(\bar{ \lambda },\bar{\mu }>0\) such that (5) has no solution for \(\lambda > \bar{\lambda }\) or \(\mu > \bar{ \mu }\). Suppose on the contrary that there exists a sequence \((\lambda_{n},\mu_{n})\) such that either \(\lambda_{n} \rightarrow \infty \) or \(\mu_{n} \rightarrow \infty \) and (5) has a positive solution \((w_{1n},w_{2n})\) at \(\lambda =\lambda_{n}\) and \(\mu =\mu_{n}\). Without loss of generality, we assume that \(\lambda _{n} \rightarrow \infty \). First, we observe from (H2) that there exists \(R_{f_{2}} \) sufficiently large such that
$$ f_{2}(u,v) \geq u+v \quad \text{for all } u+v \geq R_{f_{2}}. $$
(14)
Now we choose \(\lambda_{n}\) sufficiently large such that
$$\begin{aligned} \gamma \bigl( \Vert w_{1n} \Vert _{\infty }+ \Vert w_{2n} \Vert _{\infty }\bigr) +j_{*}(\rho +\rho _{2}) \lambda_{n} + j_{*}(\rho_{1}+ \rho) \mu_{n} \geq R_{f_{2}}. \end{aligned}$$
Then, for such \(\lambda_{n}\), from (14) it follows that
$$\begin{aligned}& f_{2} \bigl(\gamma \Vert w_{1n} \Vert _{\infty }+ (\rho \lambda_{n} +\rho_{1}\mu _{n})j_{*}, \gamma \Vert w_{2n} \Vert _{\infty }+ (\rho_{2}\lambda_{n} +\rho \mu_{n})j_{*}\bigr) \\& \quad \geq \gamma \bigl( \Vert w_{1n} \Vert _{\infty }+ \Vert w_{2n} \Vert _{\infty }\bigr) +j_{*}(\rho + \rho_{2}) \lambda_{n} + j_{*}(\rho_{1}+ \rho) \mu_{n}. \end{aligned}$$
(15)
Since \(w_{1n}\) is a solution of (5), using (3), (11), and (15), we have
$$\begin{aligned} \Vert w_{1n} \Vert _{\infty } \geq& w_{1n} (j_{*}) \\ =&\delta \bigl[1+(\rho \nu_{1}+\rho_{1} \nu_{2}) j_{*}\bigr] \\ &{} + \int_{0}^{1} H_{2} ( j_{*}, s) a_{1}(s)f_{1}(w_{1n}+p_{\lambda_{n}, \mu_{n}},w_{2n}+q_{\lambda_{n}, \mu_{n}}) \\ &{} + K_{1} ( j_{*}, s) a_{2}(s)f_{2}(w_{1n}+p_{\lambda_{n}, \mu_{n}},w _{2n}+q_{\lambda_{n}, \mu_{n}})\,ds \\ \geq& \int_{j_{*}}^{j^{*}} K_{1} ( j_{*}, s) a_{2}(s)f_{2}\bigl(\gamma \Vert w _{1n} \Vert _{\infty }+p_{\lambda_{n}, \mu_{n}}(j_{*}), \gamma \Vert w_{2n} \Vert _{\infty }+q_{\lambda_{n}, \mu_{n}}(j_{*})\bigr) \,ds \\ \geq& \bigl( \gamma \bigl( \Vert w_{1n} \Vert _{\infty }+ \Vert w_{2n} \Vert _{\infty }\bigr) +j_{*}( \rho + \rho_{2}) \lambda_{n} \\ &{} + j_{*}(\rho_{1}+\rho) \mu_{n} \bigr) \int_{j_{*}}^{j^{*}} \biggl( \rho j_{*} \int_{0}^{1} G(\tau, s)g_{1}(\tau)\,d\tau \biggr) a _{2}(s)\,ds, \end{aligned}$$
which implies that \(\|w_{1n}\|_{\infty }\rightarrow \infty \), a contradiction to Lemma 3.4. Hence, we conclude that there exist \(\bar{\lambda }, \bar{\mu }>0\) such that (5) has no solution for \(\lambda > \bar{\lambda }\) or \(\mu > \bar{\mu }\). □
The following Lemmas 3.6, 3.7, and 3.8 can be proved by the same ideas as in Lemma 3.5, Lemma 3.6, and Theorem 3.1 in [9], respectively.
Lemma 3.6
Assume (H) and (H2). Then every chain in
\(\mathcal{S}\)
has a unique supremum in
\(\mathcal{S}\).
Lemma 3.7
Assume (H) and (H2). Then there exists
\(\mu^{*} \in [\tilde{\mu }, \bar{\mu }]\)
such that (5) has a positive solution at
\((0,\mu)\)
for all
\(0< \mu \leq \mu^{*}\)
and no solution at
\((0,\mu)\)
for all
\(\mu > \mu^{*}\). Similarly, there exists
\(\lambda^{*} \in [\tilde{\lambda }, \bar{\lambda }]\)
such that (5) has a positive solution at
\((\lambda,0)\)
for all
\(0< \lambda \leq \lambda^{*}\)
and no solution at
\((\lambda,0)\)
for all
\(\lambda > \lambda^{*}\).
Lemma 3.8
Assume (H1) and (H2). Then there exists a continuous curve
\(\Gamma_{\delta }\)
splitting
\([0,\infty)^{2} \setminus \{ (0,0) \}\)
into two disjoint subsets
\(\Theta_{\delta, 1}\)
and
\(\Theta_{\delta, 2}\)
such that (5) has at least one positive solution if
\((\lambda, \mu) \in \Theta_{1} \cup \Gamma_{\delta }\)
and no positive solution if
\((\lambda, \mu) \in \Theta_{\delta, 2}\).
Notice that by the sub- and supersolution theorem we can show that if \(0< \bar{\delta }< \delta\), then \(\Theta_{\delta, 1} \cup \Gamma_{\delta }\subset \Theta_{\bar{\delta }, 1} \cup \Gamma_{\bar{\delta }}\). Let \(\Theta = \bigcup_{\delta >0} ( \Theta_{\delta,1} \cup \Gamma_{\delta })\). Then \(\Theta \subset \Theta_{0, 1}\). Now we prove our main theorem, Theorem 1.1, by taking \(\Theta_{1}= \Theta_{0,1}\), \(\Theta_{2}=\Theta_{0,2}\), and \(\Gamma = \Gamma_{0}\) in Theorem 3.1.
Theorem 3.1
Assume (H), (H1), and (H2). Then there exists a continuous curve
\(\Gamma_{0}\)
splitting
\([0,\infty)^{2} \setminus \{ (0,0) \}\)
into two disjoint subsets
\(\Theta_{0, 1}\), \(\Theta_{0, 2}\)
and a subset
\(\Theta \subset \Theta_{0,1}\)
such that (4) has at least two positive solutions for
\((\lambda, \mu) \in \Theta\), at least one positive solution for
\((\lambda, \mu) \in (\Theta_{0,1} \setminus \Theta)\cup \Gamma_{0} \), and no positive solution for
\((\lambda, \mu) \in \Theta_{0, 2}\).
Proof
We prove that (4) has a positive second solution for \((\lambda, \mu) \in \Theta\). If \((\lambda, \mu) \in \Theta\), then there exists \(\delta >0 \) such that \((\lambda, \mu) \in ( \Theta_{\delta,1} \cup \Gamma_{\delta })\). Now we let \((w_{\delta,1}, w_{\delta,2})\) be a positive solution of (5) at \((\lambda, \mu)\) and define \(\Omega = \{(w_{1},w_{2}) \in X: -\epsilon < w_{1}(t) < w_{\delta, 1}(t), -\epsilon < w_{2}(t)< w_{\delta, 2}(t), t \in [0,1] \}\). Then Ω is a bounded open set in X such that \(0 \in \Omega\). Here we denote
$$ \bigl(A_{\lambda, \mu }(w_{1},w_{2}) (t), B_{\lambda, \mu } (w_{1},w_{2}) (t)\bigr):= \bigl(A_{\lambda, \mu,0}(w_{1},w_{2}) (t), B_{\lambda, \mu,0} (w _{1},w_{2}) (t)\bigr) $$
and
$$ T_{\lambda, \mu } (w_{1},w_{2}) (t):=T_{\lambda, \mu,0} (w_{1},w _{2}) (t). $$
Then \(T_{\lambda, \mu }: \mathcal{K} \cap \bar{\Omega }\rightarrow \mathcal{K}\) is condensing. Let \((w_{1},w_{2}) \in \mathcal{K} \cap \partial \Omega\). Then there exists \(t_{0} \in [0,1]\) such that either \(w_{1}(t_{0}) =w_{\delta, 1} (t_{0})\) or \(w_{2}(t_{0}) = w_{\delta, 2}(t_{0})\). Suppose that \(w_{1}(t_{0}) =w_{\delta, 1} (t_{0})\). Then by (H)
$$\begin{aligned} A_{\lambda, \mu }(w_{1},w_{2}) (t_{0}) =& \int_{0}^{1} H_{2}(t_{0},s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2}+q_{\lambda,\mu }) \\ &{} +K_{1}(t_{0},s)a_{2}(s)f_{2}(w_{1}+p_{\lambda,\mu },w_{2}+q_{\lambda,\mu }) \,ds \\ < & \delta \bigl[1+ (\rho \nu_{1} +\rho_{1} \nu_{2})t_{0}\bigr] \\ &{} + \int_{0}^{1} H_{2}(t_{0},s) a_{1}(s)f_{1}(w_{\delta,1}+p_{\lambda, \mu }, w_{\delta,2}+q_{\lambda,\mu }) \\ &{} +K_{1}(t_{0},s)a_{2}(s)f_{2}(w_{\delta,1}+p_{\lambda,\mu },w_{ \delta,2}+q_{\lambda,\mu }) \,ds \\ =& w_{\delta,1}(t_{0}) \\ =& w_{1}(t_{0})\leq \nu w_{1}(t_{0}) \end{aligned}$$
for all \(\nu \geq 1\). Thus \(T_{\lambda, \mu } (w_{1},w_{2}) \neq \nu (w_{1},w_{2})\) for all \((w_{1},w_{2}) \in \mathcal{K} \cap \partial \Omega \) and \(\nu \geq 1\). By Lemma 2.1 we conclude
$$ i(T_{\lambda, \mu }, \mathcal{K} \cap \Omega, \mathcal{K})=1. $$
Next, we denote \(\kappa = \min_{J \times J} K_{1}(t,s)\). By (H2) there exists \(R_{f_{2}}>0\) sufficiently large such that
$$\begin{aligned} \textstyle\begin{array}{ll} f_{2}(u,v) \geq \eta (u+v) \quad \text{for } u+v \geq R_{f_{2}}, \end{array}\displaystyle \end{aligned}$$
where \(\eta >0\) can be chosen such that \(\eta \kappa \gamma \int_{j _{*}}^{j^{*}} a_{2}(s)\,ds >1\). Let \(R=\max \{M+1, \frac{1}{\gamma }R _{f_{2}} \}\), where M is defined in Lemma 3.4. Let \(\mathcal{K}_{R} = \{(w_{1},w_{2}) \in \mathcal{K}: \|(w_{1},w_{2})\|< R \}\). Then, by Lemma 3.4, \((w_{1},w_{2}) \neq T_{\lambda, \mu }(w_{1},w_{2})\) for \((w_{1},w_{2}) \in \partial \mathcal{K}_{R}\). Moreover, if \((w_{1},w _{2}) \in \partial \mathcal{K}_{R}\), then
$$ \min_{J} \bigl(w_{1}(t)+w_{2}(t)\bigr) \geq \gamma \bigl( \Vert w_{1} \Vert _{\infty }+ \Vert w _{2} \Vert _{\infty }\bigr) \geq R_{f_{2}} , $$
which implies that \(f_{2}(w_{1}(t),w_{2}(t)) \geq \eta (w_{1}(t)+w _{2}(t))\) for all \(t \in J\). Finally, we find that, for \((w_{1},w_{2}) \in \partial \mathcal{K}_{R}\),
$$\begin{aligned} A_{\lambda, \mu }(w_{1},w_{2}) (j_{*}) =& \int_{0}^{1} H_{2}(j_{*},s) a_{1}(s)f_{1}(w_{1}+p_{\lambda,\mu }, w_{2},q_{\lambda,\mu }) \\ &{} +K_{1}(j_{*},s)a_{2}(s)f_{2}(w_{1}+p_{\lambda,\mu },w_{2}+q_{\lambda,\mu }) \,ds \\ \geq& \kappa \int_{j_{*}}^{j^{*}} a_{2}(s)f_{2}(w_{1}+p_{\lambda, \mu },w_{2}+q_{\lambda,\mu }) \,ds \\ \geq& \kappa \eta \int_{j_{*}}^{j^{*}} a_{2}(s) \bigl(w_{1}(s)+w_{2}(s)\bigr)\,ds \\ \geq& \kappa \eta \gamma \bigl\Vert (w_{1},w_{2}) \bigr\Vert \int_{j_{*}}^{j^{*}} a _{2}(s)\,ds \\ >& \bigl\Vert (w_{1},w_{2}) \bigr\Vert . \end{aligned}$$
Therefore \(\|T_{\lambda, \mu }(w_{1},w_{2})\| \geq \|A_{\lambda, \mu }(w_{1},w_{2})\|_{\infty }\geq A_{\lambda, \mu }(w_{1},w_{2})(j _{*}) > \|(w_{1},w_{2})\|\), and by Lemma 2.2 we find
$$ i(T_{\lambda, \mu }, \mathcal{K}_{R}, \mathcal{K})=0. $$
By the additivity of the fixed point index we obtain
$$ 0= i(T_{\lambda, \mu }, \mathcal{K}_{R}, \mathcal{K}) = i(T_{\lambda, \mu }, \mathcal{K}\cap \Omega, \mathcal{K}) + i(T_{\lambda, \mu }, \mathcal{K}_{R}\setminus \overline{\mathcal{K}\cap \Omega }, \mathcal{K}). $$
Since \(i(T_{\lambda, \mu }, \mathcal{K}\cap \Omega, \mathcal{K}) =1\), we find \(i(T_{\lambda, \mu }, \mathcal{K}_{R}\setminus \overline{ \mathcal{K}\cap \Omega }, \mathcal{K})=-1\), which implies that \(T_{\lambda, \mu }\) has a fixed point on \(\mathcal{K}\cap \Omega \) and on \(\mathcal{K}_{R}\setminus \overline{\mathcal{K}\cap \Omega }\). Hence the proof is completed. □
Now, we give a simple example for the main results.
Example 3.2
Consider \(N=3\) and \(r_{0} =1\) for system (1). Then \(\Omega_{e}=\{ x\in (-\infty, \infty)^{3} : |x| \geq 1 \}\). Let \(K_{i} (r)= r^{-\alpha_{i}}\) for \(r\in (1,\infty)\), where \(\alpha _{i} >2\) for \(i =1,2\). Then there exists ν such that \(1<\nu < \min \{\alpha_{1} , \alpha_{2} \} -1\), and hence we get that \(\nu - \alpha_{i} <-1\) and \(\int_{1}^{\infty } r^{\nu } K_{i} (r)\,dr = \int_{1}^{\infty } r^{\nu - \alpha_{i}}\,dr <\infty \) for \(i=1,2\). Let \(l_{i} (r) = r^{-\beta_{i}}\) for \(r\in (1,\infty)\), where \(\beta_{i} >4 \pi +2 \) (\(i=1,2\)). It is easy to see that \(w_{3} \int_{1}^{\infty } r l_{i} (r)\,dr =4\pi \int_{1}^{\infty } r^{1-\beta_{i}}\,dr = \frac{4 \pi }{\beta_{i} -2} < 1\) as \(w_{3} =4\pi\). Now, if \(f_{1} (u,v) = u ^{2} +v^{3}\) and \(f_{2} (u,v)=e^{u+v} -(u+v)\), then (H), (H1), and (H2) hold for \(f_{1}\) and \(f_{2}\). Thus the conclusion of Theorem 1.1 is valid. We note that, for the corresponding radial transformed problem (2) of (1), the conclusion of Theorem 3.1 is also valid.