In this section we present sufficient conditions on the functions f and g such that problem (S)–(BC) has positive solutions with respect to a cone.
We present the assumptions that we shall use in the sequel.
-
\((H1)\)
:
-
\(a_{i}\ge 0\), \(\xi _{i}\in \mathbb{N}\) for all \(i=\overline{1,p}\), \(1\le \xi _{1}<\cdots <\xi _{p}\le N-1\),
\(b_{i}\ge 0\), \(\eta _{i}\in \mathbb{N}\) for all \(i=\overline{1,q}\), \(1\le \eta _{1}<\cdots <\eta _{q}\le N-1\),
\(c_{i}\ge 0\), \(\zeta _{i}\in \mathbb{N}\) for all \(i=\overline{1,r}\), \(1\le \zeta _{1}<\cdots <\zeta _{r}\le N-1\),
\(d_{i}\ge 0\), \(\rho _{i}\in \mathbb{N}\) for all \(i=\overline{1,l}\), \(1\le \rho _{1}<\cdots <\rho _{l}\le N-1\), and
\(\sum_{i=1}^{p}a_{i}<1\), \(\sum_{i=1}^{q}b_{i}<1\), \(\sum_{i=1}^{r}c_{i}<1\), \(\sum_{i=1}^{l}d_{i}<1\).
-
\((H2)\)
:
-
The functions \(f, g:\{1,\ldots ,N-1\}\times \mathbb{R} _{+}\times \mathbb{R}_{+}\to \mathbb{R}_{+}\) are continuous, (\(\mathbb{R}_{+}=[0,\infty )\)).
-
\((H3)\)
:
-
There exist functions \(a, b\in C(\mathbb{R}_{+}, \mathbb{R}_{+})\) such that
-
(a)
\(a(\cdot )\) is concave and strictly increasing on \(\mathbb{R}_{+}\) with \(a(0)=0\);
-
(b)
$$\textstyle\begin{cases} f_{0}^{i}= \liminf_{v\to 0+} \frac{f(n,u,v)}{a(v)}\in (0,\infty ], \\ \quad \text{uniformly with respect to } (n,u)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, \quad \text{and} \\ g_{0}^{i}= \liminf_{u\to 0+} \frac{g(n,u,v)}{b(u)}\in (0,\infty ], \\ \quad \text{uniformly with respect to } (n,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}; \end{cases} $$
-
(c)
\(\lim_{u\to 0+} \frac{a(Cb(u))}{u}=\infty \) exists for any constant \(C>0\).
-
\((H4)\)
:
-
There exist \(\alpha _{1}, \alpha _{2}>0\) with \(\alpha _{1}\alpha _{2}\le 1\) such that
$$\textstyle\begin{cases} f_{\infty }^{s}= \limsup_{v\to \infty } \frac{f(n,u,v)}{v^{\alpha _{1}}}\in [0,\infty ), \\ \quad \text{uniformly with respect to } (n,u)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, \quad \text{and} \\ g_{\infty }^{s}= \lim_{u\to \infty } \frac{g(n,u,v)}{u^{\alpha _{2}}}=0 \\ \quad \text{exists uniformly with respect to } (n,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}. \end{cases} $$
-
\((H5)\)
:
-
There exist the functions \(c, d\in C(\mathbb{R}_{+}, \mathbb{R}_{+})\) such that
-
(a)
\(c(\cdot )\) is concave and strictly increasing on \(\mathbb{R}_{+}\);
-
(b)
$$\textstyle\begin{cases} f_{\infty }^{i}= \liminf_{v\to \infty } \frac{f(n,u,v)}{c(v)}\in (0,\infty ], \\ \quad \text{uniformly with respect to } (n,u)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, \quad \text{and} \\ g_{\infty }^{i}= \liminf_{u\to \infty } \frac{g(n,u,v)}{d(u)}\in (0,\infty ], \\ \quad \text{uniformly with respect to } (n,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}; \end{cases} $$
-
(c)
\(\lim_{u\to \infty } \frac{c(Cd(u))}{u}=\infty \) exists for any constant \(C>0\).
-
\((H6)\)
:
-
There exist \(\beta _{1}, \beta _{2}>0\) with \(\beta _{1} \beta _{2}\ge 1\) such that
$$\textstyle\begin{cases} f_{0}^{s}= \limsup_{v\to 0+} \frac{f(n,u,v)}{v^{\beta _{1}}}\in [0,\infty ), \\ \quad \text{uniformly with respect to } (n,u)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, \quad \text{and} \\ g_{0}^{s}= \lim_{u\to 0+} \frac{g(n,u,v)}{u^{\beta _{2}}}=0 \\ \quad \text{exists uniformly with respect to } (n,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}. \end{cases} $$
-
\((H7)\)
:
-
The functions \(f(n,u,v)\) and \(g(n,u,v)\) are nondecreasing with respect to u and v, and there exists \(N_{0}>0\) such that
$$f(n,N_{0},N_{0})< \frac{3N_{0}}{(N^{2}-1)\max \{A,B\}} \quad \text{and} \quad g(n,N_{0},N _{0})< \frac{3N_{0}}{(N^{2}-1)\max \{A,B\}} $$
for all \(n\in \{1,\ldots ,N-1\}\).
By using the Green functions \(G_{1}\) and \(G_{2}\) from Sect. 2, our problem (S)–(BC) can be written equivalently as the following system:
$$ \textstyle\begin{cases} u_{n}= \sum_{i=1}^{N-1}G_{1}(n,i)f(i,u_{i},v_{i}), \quad n=\overline{0,N}, \\ v_{n}= \sum_{i=1}^{N-1}G_{2}(n,i)g(i,u_{i},v_{i}), \quad n=\overline{0,N}. \end{cases} $$
(7)
Then \((u,v)=((u_{n})_{n=\overline{0,N}},(v_{n})_{n=\overline{0,N}})\) is a solution of problem (S)–(BC) if and only if \((u,v)\) is a solution of system (7).
We consider the Banach space \(X=\mathbb{R}^{N+1}=\{u=(u_{n})_{n= \overline{0,N}}, u_{i}\in \mathbb{R}, i=\overline{0,N}\}\) with the maximum norm \(\|\cdot \|\), \(\|u\|=\max_{i=\overline{0,N}}|u_{i}|\), and the Banach space \(Y=X\times X\) with the norm \(\|(u,v)\|_{Y}=\|u\|+\|v \|\). We define the cones
$$ \begin{gathered} P_{1}=\biggl\{ u\in X, u=(u_{n})_{n=\overline{0,N}}, u_{n}\ge \frac{1}{A}k(n) \Vert u \Vert , \forall n=\overline{0,N}\biggr\} \subset X, \\ P_{2}=\biggl\{ v\in X, v=(v_{n})_{n=\overline{0,N}}, v_{n}\ge \frac{1}{B}k(n) \Vert v \Vert , \forall n= \overline{0,N}\biggr\} \subset X, \end{gathered} $$
and \(P=P_{1}\times P_{2}\subset Y\).
We introduce the operators \(Q_{1}, Q_{2}:Y\to X\) and \(Q:Y\to Y\) defined by
$$ \begin{gathered} Q_{1}(u,v)=\bigl(Q_{1n}(u,v) \bigr)_{n=\overline{0,N}}, \qquad Q_{2}(u,v)=\bigl(Q_{2n}(u,v) \bigr)_{n=\overline{0,N}}, \\ Q_{1n}(u,v)= \sum_{i=1}^{N-1}G_{1}(n,i)f(i,u_{i},v_{i}), \quad n=\overline{0,N}, \\ Q_{2n}(u,v)= \sum_{i=1}^{N-1}G_{2}(n,i)g(i,u_{i},v_{i}), \quad n=\overline{0,N}, \\ Q(u,v)=\bigl(Q_{1}(u,v),Q_{2}(u,v)\bigr), \quad (u,v)= \bigl((u_{n})_{n=\overline{0,N}},(v_{n})_{n=\overline{0,N}}\bigr)\in Y. \end{gathered} $$
The pair \((u,v)\) is a solution of problem (S)–(BC) if and only if \((u,v)\) is a fixed point of operator Q in the space Y. So, we will investigate the existence of fixed points of operator Q. Under assumptions \((H1)\) and \((H2)\) and by using Lemma 2.4, we can easily prove that \(Q(P)\subset P\) and the operator \(Q:P\to P\) is completely continuous.
Theorem 3.1
Assume that
\((H1)\), \((H2)\), \((H3)\), and
\((H4)\)
hold. Then problem (S)–(BC) has at least one positive solution.
Proof
By \((H3)\), there exist \(C_{1}>0\), \(C_{2}>0\), and a sufficiently small \(r_{1}>0\) such that
$$ \begin{gathered} f(n,u,v)\ge C_{1} a(v), \quad \forall (n,u)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, v\in [0,r_{1}], \\ g(n,u,v)\ge C_{2} b(u), \quad \forall (n,v)\in \{1,\ldots ,N-1\} \times \mathbb{R}_{+}, u\in [0,r_{1}], \end{gathered} $$
(8)
and
$$ a\bigl(C_{3}b(u)\bigr)\ge \frac{72C_{3}\max \{A,B\}N^{3} u}{C_{1}C_{2}(N^{2}-1)^{2}}, \quad \forall u\in [0,r_{1}], $$
(9)
where \(C_{3}=\max \{\frac{(N-1)C_{2}}{N}h(j), j= \overline{1,N-1} \}\).
We will show that \((Q_{1}(u,v),Q_{2}(u,v))\not \le (u,v)\) for all \((u,v)\in \partial B_{r_{1}}\cap P\). We suppose that there exists \((u,v)\in \partial B_{r_{1}}\cap P\), that is, \(\|(u,v)\|_{Y}=r_{1}\), such that \((Q_{1}(u,v),Q_{2}(u,v))\le (u,v)\). Then \(u\ge Q_{1}(u,v)\) and \(v\ge Q_{2}(u,v)\). By using the monotonicity and concavity of \(a(\cdot )\), the Jensen inequality, Lemma 2.3, relations (8) and (9), we obtain
$$\begin{aligned} u_{n}&\ge Q_{1n}(u,v)= \sum_{i=1}^{N-1}G_{1}(n,i)f(i,u_{i},v_{i}) \ge C_{1} \sum_{i=1}^{N-1}h(i)k(n)a(v_{i}) \\ &\ge C_{1} k(1) \sum_{i=1}^{N-1}h(i)a(v_{i}) = \frac{C_{1}}{N} \sum_{i=1}^{N-1}h(i)a \Biggl( \sum_{j=1}^{N-1}G_{2}(i,j)g(j,u_{j},v_{j}) \Biggr) \\ &\ge \frac{C_{1}}{N} \sum_{i=1}^{N-1}h(i)a \Biggl(C_{2} \sum_{j=1}^{N-1}G_{2}(i,j)b(u_{j}) \Biggr) \ge \frac{C_{1}}{N} \sum_{i=1}^{N-1}h(i)a \Biggl(C_{2} \sum_{j=1}^{N-1}h(j)k(i)b(u_{j}) \Biggr) \\ &\ge \frac{C_{1}}{N} \sum_{i=1}^{N-1}h(i)a \Biggl(C_{2} k(1) \sum_{j=1}^{N-1}h(j)b(u_{j}) \Biggr) = \frac{C_{1}}{N} \Biggl( \sum_{i=1}^{N-1}h(i) \Biggr)a \Biggl( \frac{C_{2}}{N} \sum_{j=1}^{N-1}h(j)b(u_{j}) \Biggr) \\ &\ge \frac{C_{1}(N^{2}-1)}{6N(N-1)} \sum_{j=1}^{N-1}a \biggl( \frac{(N-1)C_{2}}{N}h(j)b(u_{j}) \biggr) \\ &= \frac{C_{1}(N+1)}{6N} \sum_{j=1}^{N-1}a \biggl(\frac{(N-1)C_{2}}{NC_{3}}h(j)\cdot C_{3}b(u _{j}) \biggr) \\ &\ge \frac{C_{1}(N+1)}{6N} \sum_{j=1}^{N-1} \frac{(N-1)C_{2}}{NC_{3}}h(j)a\bigl(C_{3}b(u_{j})\bigr) \\ &\ge \frac{C_{1}C_{2}(N^{2}-1)}{6N^{2}C_{3}} \sum_{j=1}^{N-1}h(j) \frac{72 C_{3} \max \{A,B\}N^{3}}{C_{1}C_{2}(N^{2}-1)^{2}}u_{j} \\ &\ge \frac{12N\max \{A,B\}}{N^{2}-1} \sum_{j=1}^{N-1}h(j) \frac{1}{A}k(j) \Vert u \Vert \\ &\ge \frac{12 N\max \{A,B\}}{N^{2}-1} \sum _{j=1}^{N-1}h(j)\frac{1}{AN} \Vert u \Vert \\ &\ge 2 \Vert u \Vert , \quad \forall n=\overline{1,N-1}. \end{aligned}$$
So, \(\|u\|\ge \max_{n=\overline{1,N-1}}u_{n}\ge 2\|u\|\), and then
$$ \Vert u \Vert =0. $$
(10)
In a similar manner, we deduce
$$\begin{aligned} a(v_{i})&\ge a\bigl(Q_{2i}(u,v)\bigr)=a \Biggl( \sum _{j=1}^{N-1}G_{2}(i,j)g(j,u_{j},v_{j}) \Biggr) \\ &\ge \frac{1}{N-1} \sum_{j=1}^{N-1}a \bigl((N-1)G_{2}(i,j)g(j,u_{j},v_{j}) \bigr) \\ &\ge \frac{1}{N-1} \sum_{j=1}^{N-1}a \bigl((N-1)h(j)k(i)C_{2} b(u_{j}) \bigr) \\ &\ge \frac{1}{N-1} \sum_{j=1}^{N-1}a \biggl( \frac{C_{2}(N-1)}{N}h(j)b(u_{j}) \biggr) \\ &= \frac{1}{N-1} \sum_{j=1}^{N-1}a \biggl(\frac{C_{2}(N-1)}{NC_{3}}h(j)\cdot C_{3}b(u _{j}) \biggr) \\ &\ge \frac{1}{N-1} \sum_{j=1}^{N-1} \frac{C_{2}(N-1)}{NC_{3}}h(j)a\bigl(C_{3} b(u_{j})\bigr) \\ &\ge \frac{C_{2}}{NC_{3}} \sum_{j=1}^{N-1}h(j) \frac{72 C_{3} \max \{A,B\}N^{3}}{C_{1}C_{2}(N^{2}-1)^{2}}u_{j} \\ &= \frac{72 N^{2}\max \{A,B\}}{(N^{2}-1)^{2}C_{1}} \sum_{j=1}^{N-1}h(j) \Biggl( \sum_{\theta =1}^{N-1}G_{1}(j, \theta )f(\theta ,u_{\theta },v_{\theta }) \Biggr) \\ &\ge \frac{72 N^{2}\max \{A,B\}}{(N^{2}-1)^{2}C_{1}} \sum_{j=1}^{N-1}h(j) \Biggl( \sum_{\theta =1}^{N-1}h(\theta )k(j)C_{1}a(v_{\theta }) \Biggr) \\ &\ge \frac{72 N\max \{A,B\}}{(N^{2}-1)^{2}} \Biggl( \sum_{j=1}^{N-1}h(j) \Biggr) \Biggl( \sum_{\theta =1}^{N-1}h(\theta )a(v_{\theta }) \Biggr) \\ &= \frac{12 N\max \{A,B\}}{N^{2}-1} \sum_{\theta =1}^{N-1}h( \theta )a(v_{\theta }) \\ &\ge \frac{12 N\max \{A,B\}}{N^{2}-1} \sum_{\theta =1}^{N-1}h( \theta )a \biggl(\frac{1}{B}k(\theta ) \Vert v \Vert \biggr) \\ &\ge \frac{12 N\max \{A,B\}}{N^{2}-1} \Biggl( \sum_{\theta =1}^{N-1}h( \theta ) \Biggr)a \biggl(\frac{1}{BN} \Vert v \Vert \biggr) \\ &\ge 2 N\max \{A,B\} \frac{1}{BN}a\bigl( \Vert v \Vert \bigr)\\ &\ge 2 a \bigl( \Vert v \Vert \bigr), \quad \forall i=\overline{1,N-1}. \end{aligned}$$
Then we conclude that \(a(\|v\|)=a(\sup_{i=\overline{0,N}}v_{i})\ge a(v _{1})\ge 2 a(\|v\|)\), and hence \(a(\|v\|)=0\). By \((H3)\)(a), we obtain
$$ \Vert v \Vert =0. $$
(11)
Therefore, by (10) and (11), we deduce that \(\|(u,v)\|_{Y}=0\), which is a contradiction. Hence \((Q_{1}(u,v),Q_{2}(u,v)) \not \le (u,v)\) for all \((u,v)\in \partial B_{r_{1}}\cap P\). By Theorem 2.3(b), we conclude that the fixed point index
$$ i(Q,B_{r_{1}}\cap P,P)=0. $$
(12)
On the other hand, by \((H4)\) we deduce that there exist \(C_{4}>0\), \(C_{5}>0\), and \(C_{6}>0\) such that
$$ \begin{gathered} f(n,u,v)\le C_{4}v^{\alpha _{1}}+C_{5}, \quad \forall (n,u,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+} \times \mathbb{R}_{+}, \\ g(n,u,v)\le \varepsilon _{1} u^{\alpha _{2}}+C_{6}, \quad \forall (n,u,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}\times \mathbb{R}_{+}, \end{gathered} $$
(13)
with
$$\varepsilon _{1}=\min \biggl\{ \frac{6}{B(N^{2}-1)} \biggl(\frac{3}{4AC _{4}(N^{2}-1)} \biggr)^{\alpha _{2}},\frac{6^{\alpha _{2}+1}}{8B(AC_{4})^{ \alpha _{2}}(N^{2}-1)^{\alpha _{2}+1}} \biggr\} . $$
Then, by (13), we have
$$ \begin{gathered} Q_{1n}(u,v)= \sum _{i=1}^{N-1}G_{1}(n,i)f(i,u_{i},v_{i}) \le \sum_{i=1}^{N-1}A h(i) \bigl(C_{4} v_{i}^{\alpha _{1}}+C_{5}\bigr) \\ \hphantom{Q_{1n}(u,v)}=AC_{4} \sum_{i=1}^{N-1}h(i)v_{i}^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6}, \quad \forall n=\overline{0,N}, \\ Q_{2n}(u,v)= \sum_{i=1}^{N-1}G_{2}(n,i)g(i,u_{i},v_{i}) \le \sum_{i=1}^{N-1}B h(i) \bigl(\varepsilon _{1} u_{i}^{\alpha _{2}}+C_{6}\bigr) \\ \hphantom{Q_{2n}(u,v)}=B\varepsilon _{1} \sum_{i=1}^{N-1}h(i)u_{i}^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6}, \quad \forall n=\overline{0,N}. \end{gathered} $$
(14)
We consider now the functions \(\widetilde{p}, \widetilde{q}: \mathbb{R}_{+}\to \mathbb{R}_{+}\) defined by
$$ \begin{gathered} \widetilde{p}(w)= \frac{AC_{4}(N^{2}-1)}{6} \biggl[ \biggl( \frac{3w}{4AC_{4} (N^{2}-1)} \biggr)^{\alpha _{2}}+ \frac{BC_{6}(N^{2}-1)}{6} \biggr]^{\alpha _{1}}+ \frac{AC_{5}(N^{2}-1)}{6}, \\ \widetilde{q}(w)= \frac{6^{\alpha _{2}}}{8(AC_{4}(N^{2}-1))^{\alpha _{2}}} \biggl( \frac{AC_{4}(N^{2}-1)}{6}w^{\alpha _{1}}+ \frac{AC_{5}(N^{2}-1)}{6} \biggr)^{\alpha _{2}}+ \frac{BC_{6}(N^{2}-1)}{6}. \end{gathered} $$
Because
$$\lim_{w\to \infty } \frac{\widetilde{p}(w)}{w}= \lim_{w\to \infty } \frac{\widetilde{q}(w)}{w}= \textstyle\begin{cases} 0, & \text{if } \alpha _{1}\alpha _{2}< 1, \\ 1/8, & \text{if } \alpha _{1}\alpha _{2}=1, \end{cases} $$
we conclude that there exists \(R_{1}>r_{1}\) such that
$$ \widetilde{p}(w)\le \frac{1}{4}w, \qquad \widetilde{q}(w) \le \frac{1}{4}w, \quad \forall w\ge R_{1}. $$
(15)
We will show that \((u,v)\not \le (Q_{1}(u,v),Q_{2}(u,v))\) for all \((u,v)\in \partial B_{R_{1}}\cap P\). We suppose that there exists \((u,v)\in \partial B_{R_{1}}\cap P\), that is, \(\|(u,v)\|_{Y}=R_{1}\), such that \((u,v)\le (Q_{1}(u,v),Q_{2}(u,v))\). So, by (14), we obtain
$$ \begin{gathered} u_{n}\le Q_{1n}(u,v)\le AC_{4} \sum_{i=1}^{N-1}h(i)v_{i}^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6}, \quad \forall n=\overline{0,N}, \\ v_{n}\le Q_{2n}(u,v)\le B\varepsilon _{1} \sum _{i=1}^{N-1}h(i)u_{i}^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6}, \quad \forall n=\overline{0,N}. \end{gathered} $$
Then, for all \(n=\overline{0,N}\), we deduce
$$\begin{aligned} u_{n}&\le AC_{4} \sum_{i=1}^{N-1}h(i) \Biggl(B\varepsilon _{1} \sum_{j=1}^{N-1}h(j)u_{j}^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \Biggr) ^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \\ &=AC_{4} \frac{N^{2}-1}{6} \Biggl(B\varepsilon _{1} \sum _{j=1}^{N-1}h(j)u_{j}^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \Biggr)^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \\ &\le AC_{4} \frac{N^{2}-1}{6} \Biggl(B\varepsilon _{1} \sum_{j=1}^{N-1}h(j) \Vert u \Vert ^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \Biggr)^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \\ &=AC_{4} \frac{N^{2}-1}{6} \biggl(B\varepsilon _{1} \frac{N^{2}-1}{6} \Vert u \Vert ^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \biggr)^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \\ &\le AC_{4} \frac{N^{2}-1}{6} \biggl[ \biggl( \frac{3 \Vert u \Vert }{4AC_{4}(N^{2}-1)} \biggr)^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \biggr]^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \\ &\le AC_{4} \frac{N^{2}-1}{6} \biggl[ \biggl( \frac{3 \Vert (u,v) \Vert _{Y}}{4AC_{4}(N^{2}-1)} \biggr)^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \biggr]^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6}, \end{aligned}$$
(16)
and
$$ \begin{aligned}[b] v_{n}&\le B\varepsilon _{1} \sum_{i=1}^{N-1}h(i)u_{i}^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \\ &\le B\varepsilon _{1} \sum_{i=1}^{N-1}h(i) \Biggl(AC_{4} \sum_{j=1}^{N-1}h(j)v_{j}^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \Biggr)^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \\ &\le B\varepsilon _{1} \frac{N^{2}-1}{6} \biggl(AC_{4} \frac{N^{2}-1}{6} \Vert v \Vert ^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \biggr)^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \\ &\le \frac{6^{\alpha _{2}}}{8(AC_{4}(N^{2}-1))^{\alpha _{2}}} \biggl(AC_{4} \frac{N^{2}-1}{6} \Vert v \Vert ^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \biggr)^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6} \\ &\le \frac{6^{\alpha _{2}}}{8(AC_{4}(N^{2}-1))^{\alpha _{2}}} \biggl(AC_{4} \frac{N^{2}-1}{6} \bigl\Vert (u,v) \bigr\Vert _{Y}^{\alpha _{1}}+AC_{5} \frac{N^{2}-1}{6} \biggr)^{\alpha _{2}}+BC_{6} \frac{N^{2}-1}{6}. \end{aligned} $$
(17)
By using (16), (17), and (15), we conclude that \(u_{n}\le \frac{1}{4}\|(u,v)\|_{Y}\) and \(v_{n}\le \frac{1}{4}\|(u,v) \|_{Y}\) for all \(n=\overline{0,N}\). Therefore we obtain that \(\|(u,v)\|_{Y}\le \frac{1}{2}\|(u,v)\|_{Y}\), and so \(\|(u,v)\|_{Y}=0\), which is a contradiction because \(\|(u,v)\|_{Y}=R_{1}>0\). So, \((u,v)\not \le (Q_{1}(u,v),Q_{2}(u,v))\) for all \((u,v)\in \partial B _{R_{1}}\cap P\). By Theorem 2.3(a), we deduce that the fixed point index
$$ i(Q,B_{R_{1}}\cap P,P)=1. $$
(18)
Because Q has no fixed points on \(\partial B_{r_{1}}\cup \partial B _{R_{1}}\), by (12) and (18), we conclude that
$$ i\bigl(Q,(B_{R_{1}}\setminus \overline{B}_{r_{1}})\cap P,P \bigr)=i(Q,B_{R_{1}} \cap P,P)-i(Q,B_{r_{1}}\cap P,P)=1. $$
So the operator Q has at least one fixed point \((u^{1},v^{1})\in (B _{R_{1}}\setminus \overline{B}_{r_{1}})\cap P\), with \(r_{1}<\|(u^{1},v ^{1})\|_{Y}<R_{1}\), that is, \(\|u^{1}\|>0\) or \(\|v^{1}\|>0\). Because \(u^{1}\in P_{1}\) and \(v^{1}\in P_{2}\), we obtain \(u^{1}_{n}>0\) for all \(n=\overline{1,N}\) or \(v_{n}^{1}>0\) for all \(n=\overline{1,N}\). □
Theorem 3.2
Assume that
\((H1)\), \((H2)\), \((H5)\), and
\((H6)\)
hold. Then problem (S)–(BC) has at least one positive solution.
Proof
By \((H5)\) there exist \(C_{i}>0\), \(i=7,\ldots ,11\), such that
$$ \begin{gathered}[b] f(n,u,v)\ge C_{7} c(v)-C_{8}, \qquad g(n,u,v)\ge C_{9} d(u)-C_{10}, \\ \quad \forall (n,u,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+} \times \mathbb{R}_{+}, \end{gathered} $$
(19)
and
$$ c\bigl(C_{12}d(u)\bigr)\ge \frac{72 C_{12}N^{3}\max \{A,B\}u}{C_{7}C_{9}(N^{2}-1)^{2}}-C_{11}, \quad \forall u\in \mathbb{R}_{+}, $$
(20)
where \(C_{12}=\max \{\frac{C_{9}(N-1)}{N}h(i), i=\overline{1,N-1} \}>0\). Then we obtain
$$ \begin{gathered} Q_{1n}(u,v)= \sum _{i=1}^{N-1}G_{1}(n,i)f(i,u_{i},v_{i}) \ge \sum_{i=1}^{N-1}G_{1}(n,i) \bigl(C_{7} c(v_{i})-C_{8}\bigr) \\ \hphantom{Q_{1n}(u,v)}\ge \sum_{i=1}^{N-1}h(i)k(n) \bigl(C_{7}c(v_{i})-C_{8}\bigr) \ge \sum _{i=1}^{N-1}h(i)k(1) \bigl(C_{7} c(v_{i})-C_{8}\bigr) \\ \hphantom{Q_{1n}(u,v)}= \frac{1}{N} \sum_{i=1}^{N-1}h(i) \bigl(C_{7}c(v_{i})-C_{8}\bigr), \quad \forall n=\overline{1,N-1}, \\ Q_{2n}(u,v)= \sum_{i=1}^{N-1}G_{2}(n,i)g(i,u_{i},v_{i}) \ge \sum_{i=1}^{N-1}G_{2}(n,i) \bigl(C_{9} d(u_{i})-C_{10}\bigr) \\ \hphantom{Q_{2n}(u,v)}\ge \sum_{i=1}^{N-1}h(i)k(n) \bigl(C_{9}d(u_{i})-C_{10}\bigr) \ge \sum _{i=1}^{N-1}h(i)k(1) \bigl(C_{9} d(u_{i})-C_{10}\bigr) \\ \hphantom{Q_{2n}(u,v)}= \frac{1}{N} \sum_{i=1}^{N-1}h(i) \bigl(C_{9}d(u_{i})-C_{10}\bigr), \quad \forall n=\overline{1,N-1}. \end{gathered} $$
(21)
We will prove that the set \(U=\{(u,v)\in P, (u,v)=Q(u,v)+\lambda (\varphi ^{1},\varphi ^{2}), \lambda \ge 0\}\) is bounded, where \((\varphi ^{1},\varphi ^{2})\in P\setminus \{(0,0)\}\). Indeed, \((u,v)\in U\) implies that \(u\ge Q_{1}(u,v)\), \(v\ge Q_{2}(u,v)\) for some \(\varphi ^{1}, \varphi ^{2}\ge 0\). By (21), we obtain
$$\begin{aligned}& u_{n}\ge Q_{1n}(u,v)\ge \frac{C_{7}}{N} \sum_{i=1}^{N-1}h(i)c(v_{i})-C_{13}, \quad \forall n=\overline{1,N-1}, \end{aligned}$$
(22)
$$\begin{aligned}& v_{n}\ge Q_{2n}(u,v)\ge \frac{C_{9}}{N} \sum_{i=1}^{N-1}h(i)d(u_{i})-C_{14}, \quad \forall n=\overline{1,N-1}, \end{aligned}$$
(23)
where \(C_{13}=C_{8}(N^{2}-1)/(6N)\), \(C_{14}=C_{10}(N^{2}-1)/(6N)\).
By the monotonicity and concavity of \(c(\cdot )\) and the Jensen inequality, inequality (23) implies that
$$\begin{aligned} c(v_{n}+C_{14})& \ge c \Biggl( \frac{C_{9}}{N} \sum_{i=1}^{N-1}h(i)d(u_{i}) \Biggr) \\ &\ge \frac{1}{N-1} \sum_{i=1}^{N-1}c \biggl( \frac{C_{9}(N-1)}{N}h(i)d(u_{i}) \biggr) \\ &= \frac{1}{N-1} \sum_{i=1}^{N-1}c \biggl( \frac{C_{9}(N-1)}{NC_{12}}h(i)\cdot C_{12}d(u_{i}) \biggr) \\ &\ge \frac{1}{N-1} \sum_{i=1}^{N-1} \frac{C_{9}(N-1)}{NC_{12}}h(i)c \bigl(C_{12}d(u_{i}) \bigr) \\ &= \frac{C_{9}}{NC_{12}} \sum_{i=1}^{N-1}h(i)c \bigl(C_{12}d(u_{i})\bigr), \quad \forall n= \overline{1,N-1}. \end{aligned}$$
(24)
Since \(c(v_{n})\ge c(v_{n}+C_{14})-c(C_{14})\), by relations (22), (23), and (24), we deduce
$$ \begin{aligned} u_{n}&\ge \frac{C_{7}}{N} \sum _{i=1}^{N-1}h(i)c(v_{i})-C_{13} \\ &\ge \frac{C_{7}}{N} \sum_{i=1}^{N-1}h(i) \bigl[c(v_{i}+C_{14})-c(C_{14})\bigr]-C_{13} \\ &= \frac{C_{7}}{N} \sum_{i=1}^{N-1}h(i)c(v_{i}+C_{14})-C_{15} \\ &\ge \frac{C_{7}}{N} \sum_{i=1}^{N-1}h(i) \Biggl[ \frac{C_{9}}{NC_{12}} \sum_{j=1}^{N-1}h(j)c \bigl(C_{12}d(u_{j})\bigr) \Biggr]-C_{15} \\ &= \frac{C_{7}C_{9}(N^{2}-1)}{6N^{2}C_{12}} \sum_{j=1}^{N-1}h(j)c \bigl(C_{12}d(u_{j})\bigr)-C_{15} \\ &\ge \frac{C_{7}C_{9}(N^{2}-1)}{6N^{2}C_{12}} \sum_{j=1}^{N-1}h(j) \biggl( \frac{72 C_{12}N^{3}\max \{A,B\}}{C_{7}C_{9} (N^{2}-1)^{2}}u_{j}-C _{11} \biggr)-C_{15} \\ &= \frac{12N\max \{A,B\}}{N^{2}-1} \sum_{j=1}^{N-1}h(j)u_{j}-C_{16} \ge 2 \Vert u \Vert -C_{16}, \quad \forall n=\overline{1,N-1}, \end{aligned} $$
where \(C_{15}=\frac{C_{7}c(C_{14})(N^{2}-1)}{6N}+C_{13}\), \(C_{16}=\frac{C _{7}C_{9}C_{11}(N^{2}-1)^{2}}{36N^{2}C_{12}}+C_{15}\).
Therefore \(\|u\|\ge u_{1}\ge 2\|u\|-C_{16}\), and then
$$ \Vert u \Vert \le C_{16}. $$
(25)
Since \(c(v_{n})\ge c (\frac{1}{B}k(n)\|v\| )\ge c (\frac{1}{BN} \|v\| )\ge \frac{1}{BN}c(\|v\|)\) for all \(n=\overline{1,N-1}\), then by relations (19), (22), (23), and (24), we obtain
$$\begin{aligned} c(v_{n})&\ge c(v_{n}+C_{14})-c(C_{14}) \\ &\ge \frac{C_{9}}{NC_{12}} \sum_{i=1}^{N-1}h(i)c \bigl(C_{12}d(u_{i})\bigr)-c(C_{14}) \\ &\ge \frac{C_{9}}{NC_{12}} \sum_{i=1}^{N-1}h(i) \biggl( \frac{72C_{12}N^{3}\max \{A,B\}}{C_{7}C_{9}(N^{2}-1)^{2}}u_{i}-C_{11} \biggr)-c(C _{14}) \\ &= \frac{72 N^{2}\max \{A,B\}}{C_{7}(N^{2}-1)^{2}} \sum_{i=1}^{N-1}h(i)u_{i}-C_{17} \\ &\ge \frac{72 N^{2}\max \{A,B\}}{C_{7}(N^{2}-1)^{2}} \sum_{i=1}^{N-1}h(i) \Biggl( \frac{C_{7}}{N} \sum_{j=1}^{N-1}h(j)c(v_{j})-C_{13} \Biggr)-C_{17} \\ &= \frac{12N\max \{A,B\}}{N^{2}-1} \sum_{j=1}^{N-1}h(j)c(v_{j})-C_{18} \\ &\ge \frac{12 N\max \{A,B\}}{N^{2}-1} \sum_{j=1}^{N-1}h(j) \frac{1}{BN}c\bigl( \Vert v \Vert \bigr)-C_{18} \\ &\ge 2c\bigl( \Vert v \Vert \bigr)-C_{18}, \quad \forall n= \overline{1,N-1}, \end{aligned}$$
where \(C_{17}=\frac{C_{9}C_{11}(N^{2}-1)}{6NC_{12}}+c(C_{14})\), \(C_{18}=\frac{12C_{13}N^{2}\max \{A,B\}}{C_{7}(N^{2}-1)}+C_{17}\).
Then \(c(\|v\|)\ge c(v_{1})\ge 2c(\|v\|)-C_{18}\), and so \(c(\|v\|) \le C_{18}\). By \((H5)\)(a) and (c), we deduce that \(\lim_{v\to \infty }c(v)=\infty \). Thus there exists \(C_{19}>0\) such that
$$ \Vert v \Vert \le C_{19}. $$
(26)
By (25) and (26), we conclude that \(\|(u,v)\|_{Y} \le C_{16}+C_{19}\) for all \((u,v)\in U\). That is the set U is bounded. Then there exists a sufficiently large \(R_{2}>0\) such that \((u,v) \neq Q(u,v)+\lambda (\varphi ^{1},\varphi ^{2})\) for all \((u,v)\in \partial B_{R_{2}}\cap P\) and \(\lambda \ge 0\). By Theorem 2.2 we deduce that
$$ i(Q,B_{R_{2}}\cap P,P)=0. $$
(27)
On the other hand, by \((H6)\) there exist \(C_{20}>0\) and a sufficiently small \(r_{2}>0\), (\(r_{2}< R_{2}\), \(r_{2}\le 1\)) such that
$$ \begin{gathered} f(n,u,v)\le C_{20}v^{\beta _{1}}, \quad \forall (n,u)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, v\in [0,r_{2}], \\ g(n,u,v)\le \varepsilon _{2}u^{\beta _{2}}, \quad \forall (n,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}, u\in [0,r_{2}], \end{gathered} $$
(28)
where \(\varepsilon _{2}= (2AB^{\beta _{1}}C_{20} (\frac{N^{2}-1}{6} ) ^{\beta _{1}+1} )^{-1/\beta _{1}}>0\).
We will show that \((u,v)\not \le Q(u,v)\) for all \((u,v)\in \partial B _{r_{2}}\cap P\). We suppose that there exists \((u,v)\in \partial B _{r_{2}}\cap P\), that is, \(\|(u,v)\|_{Y}=r_{2}\le 1\), such that \((u,v)\le (Q_{1}(u,v),Q_{2}(u,v))\), or \(u\le Q_{1}(u,v)\) and \(v\le Q_{2}(u,v)\). Then by (28) we obtain
$$ \begin{aligned} u_{n}&\le Q_{1n}(u,v)= \sum _{i=1}^{N-1}G_{1}(n,i)f(i,u_{i},v_{i}) \le AC_{20} \sum_{i=1}^{N-1}h(i)v_{i}^{\beta _{1}} \\ &\le AC_{20} \sum_{i=1}^{N-1}h(i) \Biggl( \sum_{j=1}^{N-1}G_{2}(i,j)g(j,u_{j},v_{j}) \Biggr)^{\beta _{1}} \\ &\le AC_{20} \sum_{i=1}^{N-1}h(i) \Biggl(B \sum_{j=1}^{N-1}h(j)\varepsilon _{2} u_{j}^{\beta _{2}} \Biggr)^{\beta _{1}} \\ &\le \frac{AB^{\beta _{1}}C_{20}(N^{2}-1)\varepsilon _{2}^{\beta _{1}}}{6} \Biggl( \sum_{j=1}^{N-1}h(j) \Biggr)^{\beta _{1}} \Vert u \Vert ^{\beta _{1}\beta _{2}} \\ &=AB^{\beta _{1}}C_{20} \biggl( \frac{N^{2}-1}{6} \biggr)^{\beta _{1}+1}\varepsilon _{2}^{\beta _{1}} \Vert u \Vert ^{\beta _{1}\beta _{2}} \\ &\le AB^{\beta _{1}}C_{20}\varepsilon _{2}^{\beta _{1}} \biggl( \frac{N^{2}-1}{6} \biggr)^{\beta _{1}+1} \Vert u \Vert = \frac{1}{2} \Vert u \Vert , \quad \forall n=\overline{0,N}. \end{aligned} $$
Therefore \(\|u\|\le \frac{1}{2}\|u\|\), so
$$ \Vert u \Vert =0. $$
(29)
In addition
$$ \begin{aligned}[b] v_{n}&\le Q_{2n}(u,v)= \sum_{i=1}^{N-1}G_{2}(n,i)g(i,u_{i},v_{i}) \\ &\le B \sum_{i=1}^{N-1}h(i)\varepsilon _{2} u_{i}^{\beta _{2}}\le \frac{B\varepsilon _{2} (N^{2}-1)}{6} \Vert u \Vert ^{\beta _{2}}, \quad \forall n=\overline{0,N}. \end{aligned} $$
(30)
By (29) and (30) we deduce that \(\|v\|=0\), and then \(\|(u,v)\|_{Y}=0\), which is a contradiction because \(\|(u,v)\|_{Y}=r _{2}>0\). Then \((u,v)\not \le Q(u,v)\) for all \((u,v)\in \partial B_{r _{2}}\cap P\). By Theorem 2.3(a), we conclude that
$$ i(Q, B_{r_{2}}\cap P,P)=1. $$
(31)
Because Q has no fixed points on \(\partial B_{r_{2}}\cup \partial B _{R_{2}}\), by (27) and (31), we deduce that
$$ i\bigl(Q,(B_{R_{2}}\setminus \overline{B}_{r_{2}})\cap P,P \bigr)=i(Q,B_{R_{2}} \cap P,P)-i(Q,B_{r_{2}}\cap P,P)=-1. $$
So the operator Q has at least one fixed point \((u^{2},v^{2})\in (B _{R_{2}}\setminus \overline{B}_{r_{2}})\cap P\), with \(r_{2}<\|(u^{2},v ^{2})\|_{Y}<R_{2}\), which is a positive solution for our problem (S)–(BC). □
Theorem 3.3
Assume that assumptions
\((H1)\), \((H2)\), \((H3)\), \((H5)\), and
\((H7)\)
hold. Then problem (S)–(BC) has at least two positive solutions.
Proof
By using \((H7)\), for any \((u,v)\in \partial B_{N_{0}} \cap P\), we obtain
$$ \begin{gathered} Q_{1n}(u,v)\le A \sum _{i=1}^{N-1}h(i)f(i,N_{0},N_{0})< \frac{3AN_{0}}{(N^{2}-1)\max \{A,B\}} \sum_{i=1}^{N-1}h(i)\le \frac{N_{0}}{2}, \quad \forall n=\overline{0,N}, \\ Q_{2n}(u,v)\le B \sum_{i=1}^{N-1}h(i)g(i,N_{0},N_{0})< \frac{3BN_{0}}{(N^{2}-1)\max \{A,B\}} \sum_{i=1}^{N-1}h(i)\le \frac{N_{0}}{2}, \quad \forall n=\overline{0,N}. \end{gathered} $$
Then we deduce
$$ \bigl\Vert Q(u,v) \bigr\Vert _{Y}= \bigl\Vert Q_{1}(u,v) \bigr\Vert + \bigl\Vert Q_{2}(u,v) \bigr\Vert < N_{0}= \bigl\Vert (u,v) \bigr\Vert _{Y}, \quad \forall (u,v)\in \partial B_{N_{0}}\cap P. $$
Because Q has no fixed points on \(\partial B_{N_{0}}\), by Theorem 2.1 we conclude that
$$ i(Q,B_{N_{0}}\cap P,P)=1. $$
(32)
On the other hand, from \((H3)\) and \((H5)\), and the proofs of Theorems 3.1 and 3.2, we know that there exist a sufficiently \(r_{1}>0\) (\(r_{1}< N_{0}\)) and a sufficiently large \(R_{2}>N_{0}\) such that
$$ i(Q,B_{r_{1}}\cap P,P)=0, \qquad i(Q,B_{R_{2}}\cap P,P)=0. $$
(33)
Because Q has no fixed points on \(\partial B_{r_{1}}\cup \partial B _{R_{2}}\cup \partial B_{N_{0}}\), by relations (32) and (33), we obtain
$$ \begin{gathered} i\bigl(\mathcal{Q},(B_{R_{2}}\setminus \bar{B}_{N_{0}})\cap P,P\bigr)=i( \mathcal{Q},B_{R_{2}}\cap P,P)-i( \mathcal{Q},B_{N_{0}}\cap P,P)=-1, \\ i\bigl(\mathcal{Q}, (B_{N_{0}}\setminus \bar{B}_{r_{1}})\cap P,P \bigr)=i( \mathcal{Q},B_{N_{0}}\cap P,P)-i(\mathcal{Q},B_{r_{1}}\cap P,P)=1. \end{gathered} $$
Then \(\mathcal{Q}\) has at least one fixed point \((u^{1},v^{1})\in (B _{R_{2}}\setminus \bar{B}_{N_{0}})\cap P\) and has at least one fixed point \((u^{2},v^{2})\in (B_{N_{0}}\setminus \bar{B}_{r_{1}})\cap P\). Therefore, problem (S)–(BC) has two distinct positive solutions \((u^{1},v^{1})\), \((u^{2},v^{2})\). □
Remark 3.1
In \((H3)\), if \(a(v)=v^{p}\) with \(p\le 1\) and \(b(u)=u^{q}\) with \(q>0\), the condition from \((H3)\)(c) is satisfied if \(pq<1\). In \((H5)\), if \(c(v)=v^{p}\) with \(p\le 1\), and \(d(u)=u^{q}\) with \(q>0\), the condition from \((H5)\)(c) is satisfied if \(pq>1\).
Examples
-
(1)
We consider \(f(n,u,v)=\frac{n}{n+1}(1+e^{-(u+v)})\) and \(g(n,u,v)=(1+e ^{-n})u^{\theta }\) for \((n,u,v)\in \{1,\ldots ,N-1\}\times \mathbb{R} _{+}\times \mathbb{R}_{+}\). For \(a(v)=v^{p}\) with \(p\le 1\), and \(b(u)=u^{q}\) for \(q>0\) and \(pq<1\), then assumptions \((H3)\) and \((H4)\) are satisfied if \(q>\theta \) and \(\alpha _{2}>\theta \). For example, if \(\theta =\frac{5}{4}\), \(p=\frac{1}{3}\), \(q=\frac{4}{3}\), \(\alpha _{1}=\frac{1}{3}\), and \(\alpha _{2}=3\), we can apply Theorem 3.1, and we deduce that problem (S)–(BC) has at least one positive solution.
-
(2)
We consider \(f(n,u,v)=(1+e^{-u})v^{\theta _{1}}\) and \(g(n,u,v)=(1+e ^{-v})u^{\theta _{2}}\) for \((n,u,v)\in \{1,\ldots ,N-1\}\times \mathbb{R}_{+}\times \mathbb{R}_{+}\). For \(c(v)=v^{p}\) with \(p\le 1\), and \(d(u)=u^{q}\) for \(q>0\) and \(pq>1\), then assumptions \((H5)\) and \((H6)\) are satisfied if \(p<\theta _{1}\), \(q<\theta _{2}\), \(\beta _{1}<\theta _{1}\), and \(\beta _{2}<\theta _{2}\). For example, if \(\theta _{1}=4\), \(\theta _{2}=2\), \(p=\frac{3}{5}\), \(q=\frac{9}{5}\), \(\beta _{1}=3\), and \(\beta _{2}=\frac{1}{3}\), we can apply Theorem 3.2, and we conclude that problem (S)–(BC) has at least one positive solution.