Let
$$\begin{aligned}& \kappa _{3}=\frac{\beta }{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} g(t) ( \log t)^{\beta -1}(1- \log t)\frac{dt}{t},\\& \kappa _{4}=\frac{\beta }{d _{g,h}\varGamma (2\beta +1)} \int _{1}^{e} h(t) (\log t)^{\beta -1}(1- \log t)\frac{dt}{t}, \\& \kappa _{5}=\frac{\beta (\beta -1)}{d_{g,h}\varGamma (2\beta +1)} \int _{1} ^{e} g(t)\frac{dt}{t},\qquad \kappa _{6}=\frac{\beta (\beta -1)}{d_{g,h} \varGamma (2\beta +1)} \int _{1}^{e} h(t)\frac{dt}{t}. \end{aligned}$$
Now we list our assumptions for the nonlinearities \(f_{i}\) (\(i=1,2\)).
-
(H3)
There are \(a_{1i}, b_{1i}\ge 0\) (\(i=1,2\)) and \(l_{1},l_{2}>0\) such that
$$ \begin{aligned}[b] & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}< 1,\qquad b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}< 1, \\ &\det \begin{pmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a _{11}\kappa _{3} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\ge \begin{pmatrix} a_{11}x+b_{11}y-l_{1} \\ a_{12}x+b_{12}y-l_{2} \end{pmatrix}, \quad \forall (t,x,y)\in [1,e]\times \mathbb{R}^{+} \times \mathbb{R}^{+}. \end{aligned} $$
-
(H4)
There are \(a_{2i}, b_{2i}\ge 0\) (\(i=1,2\)) and \(r_{1}>0\) such that
$$ \begin{aligned}[b] & (\kappa _{2}+\kappa _{5}d_{h})a_{21}+\kappa _{6}a_{22}< 1,\qquad (\kappa _{2}+d _{g}\kappa _{6})b_{22}+\kappa _{5}b_{21}< 1, \\ & \det \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d_{g}\kappa _{6})b_{22}- \kappa _{5}b_{21} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\le \begin{pmatrix} a_{21}x+b_{21}y \\ a_{22}x+b_{22}y \end{pmatrix},\quad \forall (t,x,y)\in [1,e]\times [0,r_{1}] \times [0,r_{1}]. \end{aligned} $$
-
(H5)
There are \(a_{3i}, b_{3i}\ge 0\) (\(i=1,2\)) and \(r_{2}>0\) such that
$$ \begin{aligned}[b] & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}< 1,\qquad b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}< 1, \\ & \det \begin{pmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a _{31}\kappa _{3} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\ge \begin{pmatrix} a_{31}x+b_{31}y \\ a_{32}x+b_{32}y \end{pmatrix},\quad \forall (t,x,y)\in [1,e]\times [0,r_{2}] \times [0,r_{2}]. \end{aligned} $$
-
(H6)
There are \(a_{4i}, b_{4i}\ge 0\) (\(i=1,2\)) and \(l_{3},l_{4}>0\) such that
$$ \begin{aligned}[b] & (\kappa _{2}+\kappa _{5}d_{h})a_{41}+\kappa _{6} a_{42}< 1,\qquad (\kappa _{2}+d _{g}\kappa _{6})b_{42}+\kappa _{5}b_{41}< 1, \\ & \det \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d_{g}\kappa _{6})b_{42}- \kappa _{5}b_{41} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\le \begin{pmatrix} a_{41}x+b_{41}y+l_{3} \\ a_{42}x+b_{42}y+l_{4} \end{pmatrix}, \quad \forall (t,x,y)\in [1,e]\times \mathbb{R}^{+}\times \mathbb{R}^{+}. \end{aligned} $$
Let \(B_{\rho }:=\{u\in E:\|u\|<\rho \}\) for \(\rho >0\) in the sequel.
Theorem 3.1
Suppose that (H1)–(H4) hold. Then (1.1) has a positive solution.
Proof
Let \(S_{1}=\{(u,v)\in P\times P: (u,v)=A(u,v)+\lambda (\varphi _{1},\varphi _{1}), \forall \lambda \ge 0\}\), where \(\varphi _{1}\) is a fixed element in \(P_{0}\). We claim that \(S_{1}\) is a bounded set in \(P\times P\). Note if there exists \((u,v)\in S_{1}\) such that
$$ u(t)=A_{1}(u,v) (t)+\lambda \varphi _{1}(t), \qquad v(t)=A_{2}(u,v) (t)+ \lambda \varphi _{1}(t)\quad \text{for }t\in [1,e], $$
(3.1)
then this, together with Lemma 2.6, implies that
$$ u,v\in P_{0}. $$
(3.2)
From (3.1) we have
$$ u(t)\ge A_{1}(u,v) (t), \qquad v(t)\ge A_{2}(u,v) (t)\quad \text{for }t\in [1,e]. $$
(3.3)
From the definitions of \(A_{i}\) (\(i=1,2\)), multiplying by \(\mu (t)\) and integrating from 1 to e, Lemmas 2.4 and 2.5 enable us to obtain
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \left ( \textstyle\begin{array}{l} \int _{1}^{e} \mu (t) (\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s)) \frac{ds}{s}\\ \quad {}+\frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s)) \frac{ds}{s} \\ \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s} )\frac{dt}{t} \\ \int _{1}^{e} \mu (t) ( \int _{1}^{e} G_{1}(t,s)f _{2}(s,u(s),v(s))\frac{ds}{s}\\ \quad {}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f _{2}(s,u(s),v(s))\frac{ds}{s} \\ \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s} ) \frac{dt}{t} \end{array}\displaystyle \right ) \\ &\quad \ge \begin{pmatrix} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t}+\kappa _{4} \int _{1}^{e} \mu (t)f_{2}(t,u(t),v(t)) \frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)f _{2}(t,u(t),v(t))\frac{dt}{t}+ \kappa _{3}\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t} \end{pmatrix}. \end{aligned} $$
(3.4)
Combining this with (H3), we have
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \left ( \textstyle\begin{array}{l} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)(a_{11}u(t)+b_{11}v(t)-l _{1})\frac{dt}{t}\\ \quad {}+\kappa _{4} \int _{1}^{e} \mu (t)(a_{12}u(t)+b_{12}v(t)-l _{2})\frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)(a_{12}u(t)+b_{12}v(t)-l_{2})\frac{dt}{t}\\ \quad {}+ \kappa _{3}\int _{1} ^{e} \mu (t)(a_{11}u(t)+b_{11}v(t)-l_{1})\frac{dt}{t} \end{array}\displaystyle \right ), \end{aligned} $$
(3.5)
and
$$ \begin{aligned}[b] & \begin{pmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a _{11}\kappa _{3} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix} \\ &\quad \le \begin{pmatrix} ((\kappa _{1}+\kappa _{3}d_{h})l_{1}+\kappa _{4}l_{2})\int _{1}^{e} \mu (t)\frac{dt}{t} \\ ((\kappa _{1}+d_{g}\kappa _{4})l_{2}+\kappa _{3}l_{1})\int _{1}^{e} \mu (t)\frac{dt}{t} \end{pmatrix}= \begin{pmatrix} \frac{(\kappa _{1}+\kappa _{3}d_{h})l_{1}+\kappa _{4}l_{2}}{\varGamma ( \beta +2)} \\ \frac{(\kappa _{1}+d_{g}\kappa _{4})l_{2}+\kappa _{3}l _{1}}{\varGamma (\beta +2)} \end{pmatrix}. \end{aligned} $$
Solving this matrix inequality, we have
Hence, there exist \(M_{1}>0\), \(M_{2}>0\) such that
$$ \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix} \le \begin{pmatrix} M_{1} \\ M_{2} \end{pmatrix}. $$
Note (3.2), and we find
$$ \begin{pmatrix} \Vert v \Vert \\ \Vert u \Vert \end{pmatrix}\le \begin{pmatrix} \frac{M_{1}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \\ \frac{M_{2}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \end{pmatrix}. $$
This proves that \(S_{1}\) is bounded in \(P\times P\). As a result, if we choose \(R_{1}> \{r_{1},\frac{M_{1}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )} , \frac{M_{2}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )} \}\) (\(r_{1}\) is defined by (H4)), then we have
$$ (u,v)\neq A(u,v)+\lambda (\varphi _{1},\varphi _{1}), \quad \text{for } (u,v) \in \partial B_{R_{1}}\cap (P\times P), \forall \lambda \ge 0. $$
From Lemma 2.7 we have
$$ i\bigl(A, B_{R_{1}}\cap (P \times P),P \times P\bigr) = 0. $$
(3.6)
Next we claim that
$$ (u,v)\neq \lambda A(u,v),\quad \text{for }(u,v)\in \partial B_{r_{1}} \cap (P\times P), \forall \lambda \in [0,1], $$
(3.7)
where \(r_{1}\) is defined by (H4). Suppose (3.7) is not true. Then there exist \((u,v)\in \partial B_{r_{1}}\cap (P\times P)\) and \(\lambda \in [0,1]\) such that \((u,v) = \lambda A(u,v)\), which implies that
$$ u(t)\le A_{1}(u,v) (t),\qquad v(t)\le A_{2}(u,v) (t)\quad \text{for } t\in [1,e]. $$
(3.8)
Multiplying by \(\mu (t)\) and integrating from 1 to e, Lemmas 2.4 and 2.5 enable us to obtain
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix} \\ &\quad \le \left ( \textstyle\begin{array}{l} \int _{1}^{e} \mu (t) (\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s)) \frac{ds}{s}\\ \quad {}+\frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s)) \frac{ds}{s} \\ \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s} )\frac{dt}{t} \\ \int _{1}^{e} \mu (t) ( \int _{1}^{e} G_{1}(t,s)f _{2}(s,u(s),v(s))\frac{ds}{s}\\ \quad {}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f _{2}(s,u(s),v(s))\frac{ds}{s} \\ \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s} ) \frac{dt}{t} \end{array}\displaystyle \right ) \\ &\quad \le \begin{pmatrix} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t}+\kappa _{6} \int _{1}^{e} \mu (t)f_{2}(t,u(t),v(t)) \frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)f _{2}(t,u(t),v(t))\frac{dt}{t}+ \kappa _{5}\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t} \end{pmatrix}. \end{aligned} $$
(3.9)
Substituting (H4) into this matrix inequality, we obtain
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \le \begin{pmatrix} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)(a_{21}u(t)+b_{21}v(t)) \frac{dt}{t}+\kappa _{6} \int _{1}^{e} \mu (t)(a_{22}u(t)+b_{22}v(t)) \frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)(a _{22}u(t)+b_{22}v(t))\frac{dt}{t}+ \kappa _{5}\int _{1}^{e} \mu (t)(a _{21}u(t)+b_{21}v(t))\frac{dt}{t} \end{pmatrix}. \end{aligned} $$
(3.10)
Consequently, we get
$$ \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{22}-\kappa _{5}b_{21} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$
Therefore, (H4) implies that
Hence,
$$ \int _{1}^{e} u(t)\mu (t)\frac{dt}{t}=0, \qquad \int _{1}^{e} v(t)\mu (t) \frac{dt}{t}=0. $$
Note that \(\mu (t)\not \equiv 0\) for \(t\in [1,e]\), so \(u(t)=v(t) \equiv 0\), \(t\in [1,e]\), which implies that \(\|u\|=\|v\|=0\), contradicting \((u,v)\in \partial B_{r_{1}}\cap (P\times P)\). As a result, (3.7) holds. From Lemma 2.8 we have
$$ i\bigl(A, B_{r_{1}}\cap (P \times P),P \times P\bigr) = 1. $$
(3.11)
From (3.6) and (3.11) we have
$$ \begin{aligned} &i\bigl(A,(B_{R_{1}}\backslash \overline{B}_{r_{1}})\cap (P \times P), P \times P\bigr)\\ &\quad =i\bigl(A, B_{R_{1}}\cap (P \times P),P \times P\bigr)-i\bigl(A, B_{r_{1}} \cap (P \times P),P \times P\bigr)=0-1=-1. \end{aligned} $$
Therefore the operator A has at least one fixed point on \((B_{R_{1}} \backslash \overline{B}_{r_{1}})\cap (P\times P)\). Equivalently, (1.1) has at least one positive solution. This completes the proof. □
Theorem 3.2
Suppose that (H1)–(H2), (H5)–(H6) hold. Then (1.1) has a positive solution.
Proof
We use similar methods as in Theorem 3.1 to prove this theorem. We first claim that
$$ (u,v)\neq A(u,v)+\lambda (\varphi _{2},\varphi _{2}),\quad \text{for } (u,v) \in \partial B_{r_{2}}\cap (P \times P), \forall \lambda \ge 0, $$
(3.12)
where \(\varphi _{2}\in P\) is a given element. Suppose the claim is not true. Then there exist \((u,v)\in \partial B_{r_{2}}\cap (P\times P)\) and \(\lambda \ge 0\) such that \((u,v) =A(u,v)+\lambda (\varphi _{2},\varphi _{2})\), which implies that
$$ u(t)\ge A_{1}(u,v) (t),\qquad v(t)\ge A_{2}(u,v) (t) \quad \text{for } t\in [1,e]. $$
Similar to (3.4), (3.5), from (H5) we obtain
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \begin{pmatrix} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)(a_{31}u(t)+b_{31}v(t)) \frac{dt}{t}+\kappa _{4} \int _{1}^{e} \mu (t)(a_{32}u(t)+b_{32}v(t)) \frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)(a _{32}u(t)+b_{32}v(t))\frac{dt}{t}+ \kappa _{3}\int _{1}^{e} \mu (t)(a _{31}u(t)+b_{31}v(t))\frac{dt}{t} \end{pmatrix}, \end{aligned} $$
and
$$ \begin{aligned}[b] & \begin{pmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a _{31}\kappa _{3} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \end{aligned} $$
Thus \(u(t)=v(t)\equiv 0\) for \(t\in [1,e]\), and \(\|u\|=\|v\|=0\), which contradicts \((u,v)\in \partial B_{r_{2}}\cap {(P\times P)}\). Consequently, (3.12) holds, and from Lemma 2.7 we have
$$ i\bigl(A, B_{r_{2}}\cap (P \times P),P \times P\bigr) = 0. $$
(3.13)
Let \(S_{2}=\{(u,v)\in P\times P: (u,v)=\lambda A(u,v), \forall \lambda \in [0,1] \}\). Now we prove that \(S_{2}\) is bounded in \(P\times P\). Note if there exists \((u,v)\in S_{2}\), then
$$ u(t)\le A_{1}(u,v) (t),\qquad v(t)\le A_{2}(u,v) (t) \quad \text{for } t\in [1,e], $$
and similar to (3.9), (3.10), and by (H6) we have
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix} \\ &\quad \le \left ( \textstyle\begin{array}{l} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)(a_{41}u(t)+b_{41}v(t)+l _{3})\frac{dt}{t}\\ \quad {}+\kappa _{6} \int _{1}^{e} \mu (t)(a_{42}u(t)+b_{42}v(t)+l _{4})\frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)(a_{42}u(t)+b_{42}v(t)+l_{4})\frac{dt}{t}\\ \quad {}+ \kappa _{5}\int _{1} ^{e} \mu (t)(a_{41}u(t)+b_{41}v(t)+l_{3}))\frac{dt}{t} \end{array}\displaystyle \right ). \end{aligned} $$
Thus
$$ \begin{aligned} & \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{42}-\kappa _{5}b_{41} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \le \begin{pmatrix} \frac{(\kappa _{2}+\kappa _{5}d_{h})l_{3}+\kappa _{6}l_{4}}{\varGamma ( \beta +2)} \\ \frac{(\kappa _{2}+d_{g}\kappa _{6})l_{4}+\kappa _{5}l _{3}}{\varGamma (\beta +2)} \end{pmatrix}. \end{aligned} $$
Solving this matrix inequality, we have
Hence, there exist \(M_{3}>0\), \(M_{4}>0\) such that
$$ \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t) \mu (t)\frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} M_{3} \\ M_{4} \end{pmatrix}. $$
Note that \((u,v)\in S_{2}\), and from Lemma 2.6, we find \(u,v\in P_{0}\). Thus, we obtain
$$ \begin{pmatrix} \Vert u \Vert \\ \Vert v \Vert \end{pmatrix}\le \begin{pmatrix} \frac{M_{3}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \\ \frac{M_{4}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \end{pmatrix}. $$
This proves that \(S_{2}\) is bounded in \(P\times P\). As a result, if we take \(R_{2}> \{r_{2},\frac{M_{3}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )}, \frac{M_{4}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )} \}\) (\(r_{2}\) is defined by (H5)), we conclude that
$$ (u,v)\neq \lambda A(u,v),\quad \text{for }(u,v)\in \partial B_{R_{2}} \cap (P\times P), \forall \lambda \in [0,1]. $$
(3.14)
From Lemma 2.8 we have
$$ i\bigl(A, B_{R_{2}}\cap (P \times P),P \times P\bigr) = 1. $$
(3.15)
From (3.13) and (3.15) we have
$$ \begin{aligned} &i\bigl(A,(B_{R_{2}}\backslash \overline{B}_{r_{2}})\cap (P \times P), P \times P\bigr)\\ &\quad =i\bigl(A, B_{R_{2}}\cap (P \times P),P \times P\bigr)-i\bigl(A, B_{r_{2}} \cap (P \times P),P \times P\bigr)=1-0=1. \end{aligned} $$
Therefore the operator A has at least one fixed point on \((B_{R_{2}} \backslash \overline{B}_{r_{2}})\cap (P\times P)\). Equivalently, (1.1) has at least one positive solution. This completes the proof. □
Example 3.3
Let \(\beta =2.5\), \(h(t)=g(t)=\log t\) for \(t\in [1,e]\). Then \(d_{h}=d_{g}=\int _{1}^{e} (\log t)^{\beta } \frac{dt}{t}=\frac{2}{7}\), \(d_{g,h}=1-\int _{1}^{e} h(t) (\log t)^{ \beta -1} \frac{dt}{t}\cdot \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}=1-\frac{4}{49}=\frac{45}{49}\). This implies that (H2) holds. Next, we calculate \(\kappa _{i}\) (\(i=1,2,3,4,5,6\)) as follows:
$$\begin{aligned}& \kappa _{1}=\frac{\beta ^{2}\varGamma (\beta )}{\varGamma (2\beta +2)}=\frac{2.5^{2} \varGamma (2.5)}{\varGamma (7)}\approx 0.01154,\\& \kappa _{2}=\frac{\beta -1}{ \varGamma (\beta +2)}=\frac{1.5}{\varGamma (4.5)} \approx 0.129, \\& \begin{aligned} \kappa _{3}&=\kappa _{4}=\frac{\beta }{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} (\log t) (\log t)^{\beta -1}(1-\log t)\frac{dt}{t}\\ &=\frac{2.5}{ \frac{45}{49}\varGamma (6)} \int _{1}^{e} (\log t)^{2.5}(1-\log t) \frac{dt}{t}\approx 0.00144, \end{aligned} \\& \kappa _{5}=\kappa _{6}= \frac{\beta (\beta -1)}{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} ( \log t)\frac{dt}{t}= \frac{2.5\times 1.5}{\frac{45}{49}\varGamma (6)} \int _{1}^{e} (\log t)\frac{dt}{t} \approx 0.017. \end{aligned}$$
Case 1. Let \(a_{11}=10\), \(a_{12}=600\), \(b_{11}=630\), \(b_{12}=7\), \(a_{21}=3\), \(a_{22}=4\), \(b_{21}=3\), \(b_{22}=2\). Then we have
$$\begin{aligned}& a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12} \kappa _{4}=10\times 0.012+600 \times 0.00144< 1,\\& b_{12}(\kappa _{1}+d_{g}\kappa _{4})+b_{11}\kappa _{3}= 7\times 0.012+63 0 \times 0.00144< 1, \\& (\kappa _{2}+\kappa _{5}d_{h})a_{21}+ \kappa _{6}a_{22}=0.134\times 3+0.017 \times 4 < 1,\\& ( \kappa _{2}+d_{g}\kappa _{6})b_{22}+ \kappa _{5}b_{21}=0.134 \times 2+0.017\times 3< 1, \\& \begin{vmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g}\kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a_{11}\kappa _{3} \end{vmatrix}= \begin{vmatrix} 7.57 & -0.016 \\ -0.009 & 7.21 \end{vmatrix}>0, \end{aligned}$$
and
$$ \begin{vmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{22}-\kappa _{5}b_{21} \end{vmatrix}= \begin{vmatrix} 0.53 & -0.436 \\ -0.587 & 0.681 \end{vmatrix}>0. $$
Let \(f_{1}(t,x,y)= (10x+630 y)^{\gamma _{1}}\), \(f_{2}(t,x,y)=(600x+7 y)^{ \gamma _{2}} \) for \(t\in [1,e]\), \(x,y\in \mathbb{R}^{+}\), \(\gamma _{1}, \gamma _{2}>1\). Then we have
$$\begin{aligned}& \liminf_{a_{11}x+b_{11}y\to +\infty }\frac{f_{1}(t,x,y)}{a_{11}x+b _{11}y}=\liminf _{10x+630y\to +\infty } \frac{(10x+630 y)^{\gamma _{1}}}{10x+630y}=+\infty ,\\& \quad \text{uniformly on }t\in [1,e], \\& \liminf_{a_{12}x+b_{12}y\to +\infty }\frac{f_{2}(t,x,y)}{a_{12}x+b _{12}y}=\liminf _{600x+7 y\to +\infty }\frac{(600x+7 y)^{\gamma _{2}}}{600x+7 y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \limsup_{a_{21}x+b_{21}y\to 0^{+}} \frac{f_{1}(t,x,y)}{a_{21}x+b_{21}y}=\limsup _{3x+3y\to 0^{+}} \frac{(10x+630 y)^{\gamma _{1}}}{3x+3y}=0,\quad \text{uniformly on }t \in [1,e], \end{aligned}$$
and
$$ \limsup_{a_{22}x+b_{22}y\to 0^{+}} \frac{f_{2}(t,x,y)}{a_{22}x+b_{22}y}=\limsup _{4x+2y\to 0^{+}} \frac{(600x+7 y)^{\gamma _{2}}}{4x+2y}=0,\quad \text{uniformly on }t\in [1,e]. $$
As a result, (H3)–(H4) hold.
Case 2. Let \(a_{31}=8\), \(a_{32}=620\), \(b_{31}=630\), \(b_{32}=7\), \(a_{41}=3\), \(a_{42}=4\), \(b_{41}=3\), \(b_{42}=2\). Then we have
$$\begin{aligned}& a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32} \kappa _{4}=8\times 0.012+620 \times 0.00144< 1,\\& b_{32}(\kappa _{1}+d_{g}\kappa _{4})+b_{31}\kappa _{3}= 7\times 0.012+63 0 \times 0.00144< 1, \\& (\kappa _{2}+\kappa _{5}d_{h})a_{41}+ \kappa _{6}a_{42}=0.134\times 3+0.017 \times 4 < 1,\\& ( \kappa _{2}+d_{g}\kappa _{6})b_{42}+ \kappa _{5}b_{41}=0.134 \times 2+0.017\times 3< 1, \\& \begin{vmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g}\kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a_{31}\kappa _{3} \end{vmatrix}= \begin{vmatrix} 7.57 & -0.0112 \\ -0.009 & 7.45 \end{vmatrix}>0, \end{aligned}$$
and
$$ \begin{vmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{42}-\kappa _{5}b_{41} \end{vmatrix}= \begin{vmatrix} 0.53 & -0.436 \\ -0.587 & 0.681 \end{vmatrix}>0. $$
Let \(f_{1}(t,x,y)= (8x+630 y)^{\gamma _{3}}\), \(f_{2}(t,x,y)=(620x+7 y)^{ \gamma _{4}} \) for \(t\in [1,e]\), \(x,y\in \mathbb{R}^{+}\), \(\gamma _{3}, \gamma _{4}\in (0,1)\). Then we have
$$\begin{aligned}& \liminf_{a_{31}x+b_{31}y\to 0^{+}} \frac{f_{1}(t,x,y)}{a_{31}x+b_{31}y}=\liminf _{8x+630y\to 0^{+}}\frac{(8x+630 y)^{\gamma _{3}}}{8x+630y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \liminf_{a_{32}x+b_{32}y\to 0^{+}} \frac{f_{2}(t,x,y)}{a_{32}x+b_{32}y}=\liminf _{620x+7 y\to 0^{+}}\frac{(620x+7 y)^{\gamma _{4}}}{620x+7 y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \limsup_{a_{41}x+b_{41}y\to +\infty } \frac{f_{1}(t,x,y)}{a_{41}x+b _{41}y}=\limsup _{3x+3y\to +\infty } \frac{(8x+630 y)^{\gamma _{3}}}{3x+3y}=0,\quad \text{uniformly on }t\in [1,e], \end{aligned}$$
and
$$ \limsup_{a_{42}x+b_{42}y\to +\infty } \frac{f_{2}(t,x,y)}{a_{42}x+b _{42}y}=\limsup _{4x+2y\to +\infty } \frac{(620x+7 y)^{\gamma _{4}}}{4x+2y}=0,\quad \text{uniformly on }t\in [1,e]. $$
As a result, (H5)–(H6) hold.