In this section, we show the existence and multiplicity of positive solutions for (1.1). Let \(q_{0}=\min_{t\in[1, T]_{\mathbb{Z}}}q(t)\) and define
$$\begin{aligned} \phi(r)=\max \bigl\{ \tilde{f}(t,z): t\in[1,T]_{\mathbb{Z}}, z\in [0,r] \bigr\} , \quad\mbox{for } r>0, \end{aligned}$$
and
$$\begin{aligned} \psi(r)= \min \biggl\{ \tilde{f}(t,z): t\in[1, T]_{\mathbb{Z}}, z\in \biggl[ \biggl(r-\frac{1}{\mu} \biggr)q_{0}, r \biggr] \biggr\} , \quad\mbox{for }r>\frac {1}{\mu}. \end{aligned}$$
Theorem 3.1
Assume that (C1) and (C2) hold. Suppose that there exist
\(r, R>0\)
such that
\(\frac{1}{\mu}< r< R\)
and
- (C3):
-
\(\phi(r)\leq\frac{r}{\max_{t\in[1,T]_{\mathbb {Z}}}\sum_{s=1}^{T}G(t,s)}\)
and
\(\psi(R)\geq \frac{R}{\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\).
Then problem (1.1) has at least one positive solution. In addition, if
- (C4):
-
\(f(t,z)\leq0\)
for all
\(t\in[1,T]_{\mathbb{Z}}\)
and
\(z>0\)
sufficiently large, and
- (C5):
-
\(L<\frac{1}{\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\),
then problem (1.1) has at least two positive solutions.
Proof
Assume (C3) holds. We first prove by Lemma 2.5 that problem (2.3) has at least one positive solution v. In the Banach space \(\mathbb{E}=\{v: [0,T+1]_{\mathbb{Z}}\to\mathbb {R}\}\) endowed with the norm \(\|v\|=\max_{[0,T+1]_{\mathbb{Z}}}|v(t)|\), define
$$\begin{aligned} Fv(t)=\sum_{s=1}^{T} G(t,s) \tilde{f}\bigl(s,v(s)-u_{0}(s)\bigr). \end{aligned}$$
(3.1)
Then by (C1), \(F: \mathbb{E}\to\mathbb{E}\) is completely continuous. Define the cone
$$\begin{aligned} P=\bigl\{ v\in\mathbb{E}: v(t)\geq q(t)\|v\|, t \in[1,T]_{\mathbb{Z}} \bigr\} . \end{aligned}$$
(3.2)
By Lemma 2.2, \(F(P)\subset P\). Thus, a fixed point of F in P is a positive solution of problem (2.3).
Let
$$\begin{aligned} \Omega_{1}=\bigl\{ v\in\mathbb{E}: \|v\|< r\bigr\} ,\qquad \Omega_{2}=\bigl\{ v\in\mathbb{E}: \|v\|< R\bigr\} . \end{aligned}$$
(3.3)
For \(v\in P\cap\partial\Omega_{1}\), we have \(v(s)-u_{0}(s)\geq q(s)\|v\|-u_{0}(s)\geq(\mu r-1)u_{0}(s)> 0\), \(s\in[1,T]_{\mathbb{Z}}\). It follows that \(\tilde{f}(s, v(s)-u_{0}(s))\leq\phi(r)\), \(s\in[1,T]_{\mathbb{Z}}\). Notice that \(G(0, s)=G(1,s)\) and \(G(T,s)=G(T+1,s)\), \(s\in[1,T]_{\mathbb {Z}}\). Then, by (C3), we have
$$\begin{aligned}[b] \|Fv\|&=\max_{t\in[0,T+1]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \tilde{f}\bigl(s,v(s)-u_{0}(s)\bigr) \\ &\leq \max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \phi (r) \\ &\leq r. \end{aligned} $$
That is, \(\|Fv\|\leq\|v\|\) for \(v\in P\cap\partial\Omega_{1}\).
For \(v\in P\cap\partial\Omega_{2}\), we have by Lemma 2.3, for \(s\in[1, T]\),
$$\begin{aligned} R\geq v(s)-u_{0}(s)\geq q(s)\|v\|-\frac{q(s)}{\mu} \geq \biggl(R-\frac {1}{\mu} \biggr)q_{0}. \end{aligned}$$
(3.4)
This implies \(\tilde{f}(s, v(s)-u_{0}(s))\geq\psi(R)\) for \(s\in[1, T]\), \(v\in P\cap\partial\Omega_{2}\). Then, by (C3), we have, for \(v\in P\cap \partial\Omega_{2}\),
$$\begin{aligned} \|Fv\| =&\max_{t\in[0,T+1]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \tilde{f}\bigl(s,v(s)-u_{0}(s)\bigr) \\ \geq& \max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \psi (R) \\ \geq& R. \end{aligned}$$
That is, \(\|Fv\|\geq\|v\|\) for \(v\in P\cap\partial\Omega_{2}\).
Therefore, by Lemma 2.5, F has a fixed point \(v_{1}\in P\) satisfying \(r\leq\|v_{1}\|\leq R\), which is a positive solution of problem (2.3). By Lemma 2.3, \(u_{1}(t)=v_{1}(t)-u_{0}(t)\geq q(t)\|v_{1}\|-u_{0}(t)\geq(\mu r-1)u_{0}(t)>0\), \(t\in[1,T]_{\mathbb{Z}}\). Therefore, by Lemma 2.4, \(u_{1}\) is a positive solution of problem (1.1).
Now, let (C4) and (C5) also hold. We prove that problem (1.1) has a distinct second positive solution \(u_{2}(t)\). By (C4), there exists \(D>0\) such that, for \(z>D\),
$$\tilde{f}(t,z)=f(t,z)+Lz+h(t)\leq Lz+h(t),\quad t\in[1,T]_{\mathbb{Z}}. $$
By (C5), we can choose \(R_{\infty}>\max \{R, \frac{D}{q_{0}}+\frac{1}{\mu} \}\) such that
$$\begin{aligned} L+\frac{\bar{h}}{R_{\infty}}\leq\frac{1}{\max_{t\in [1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}, \end{aligned}$$
(3.5)
where \(\bar{h}=\max_{t\in[1,T]_{\mathbb{Z}}}h(t)\). Let \(\Omega_{3}=\{ v\in\mathbb{E}: \|v\|< R_{\infty}\}\). For \(v\in P\cap\partial\Omega_{3}\), similar to (3.4), we have
$$\begin{aligned} v(t)-u_{0}(t)\geq \biggl(R_{\infty}- \frac{1}{\mu} \biggr)q_{0}>D,\quad t\in [1,T]_{\mathbb{Z}}, \end{aligned}$$
(3.6)
which implies that
$$\tilde{f}\bigl(t, v(t)-u_{0}(t)\bigr)\leq L\bigl(v(t)-u_{0}(t) \bigr)+h(t)\leq LR_{\infty}+\bar {h},\quad t\in[1,T]_{\mathbb{Z}}. $$
Thus, by (3.5), we have, for \(v\in P\cap\partial\Omega_{3}\),
$$\begin{aligned}[b] \|Fv\|&= \max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \tilde {f}\bigl(s,v(s)-u_{0}(s)\bigr) \\ &\leq (LR_{\infty}+\bar{h})\max_{t\in[1,T]_{\mathbb{Z}}}\sum _{s=1}^{T}G(t,s) \\ &\leq R_{\infty}. \end{aligned} $$
That is, \(\|Fv\|\leq\|v\|\) for \(v\in P\cap\partial\Omega_{3}\).
Therefore, by Lemma 2.5, F has a fixed point \(v_{2}\in P\) such that \(R\leq\|v_{2}\|\leq R_{\infty}\). By Lemma 2.4, \(u_{2}(t)=v_{2}(t)-u_{0}(t)\) is a second positive solution of problem (1.1). The proof is complete. □
Corollary 3.2
Let (C1) and (C2) hold. Assume that there exist
\(\frac{1}{\mu}< r< R\)
such that (C3) hold. Then, if
\(\lim_{z\to\infty}\frac {f(t,z)}{z}=0\)
uniformly for
\(t\in[1,T]_{\mathbb{Z}}\)
and
\(L<\frac{1}{\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\), problem (1.1) has at least two positive solutions.
Proof
Let F, P, and \(\Omega_{1}\), \(\Omega_{2}\) be defined as (3.1), (3.2), and (3.3), respectively. From the proof of Theorem 3.1, we know by (C3) that F has a fixed point \(v_{1}\) such that \(r\leq\|v_{1}\|\leq R\) and \(u_{1}(t)=v_{1}(t)-u_{0}(t)\) is a positive solution of (1.1). Now, we prove that F has a second fixed point \(v_{2}\in P\).
Take \(\epsilon>0\) sufficiently small such that \(L+\epsilon\leq\frac {1}{\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\). Note that \(\lim_{z\to\infty}\frac{f(t,z)}{z}=0\) implies
$$\lim_{z\to\infty}\frac{\tilde{f}(t,z)}{z}=L $$
uniformly for \(t\in[1,T]_{\mathbb{Z}}\). Then there exists \(D>0\) such that, for \(z>D\), \(\tilde{f}(t,z)\leq(L+\epsilon)z\) holds for all \(t\in [1,T]_{\mathbb{Z}}\). Choose \(R_{\infty}=\max \{R, \frac{D}{q_{0}}+\frac{1}{\mu} \}+1\) and \(\Omega_{3}=\{v\in\mathbb{E}: \|v\|< R_{\infty}\}\). For \(v\in P\cap\partial\Omega_{3}\), we see that (3.6) holds and hence \(\tilde{f}(t, v(t)-u_{0}(t))\leq(L+\epsilon)(v(t)-u_{0}(t))\leq (L+\epsilon)R_{\infty}\) for \(t\in[1,T]_{\mathbb{Z}}\). Thus, for \(v\in P\cap\partial\Omega_{3}\),
$$\begin{aligned} \|Fv\| =& \max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \tilde {f}\bigl(s,v(s)-u_{0}(s)\bigr) \\ \leq& (L+\epsilon)R_{\infty}\max_{t\in[1,T]_{\mathbb{Z}}}\sum _{s=1}^{T}G(t,s) \\ \leq& R_{\infty}. \end{aligned}$$
That is, \(\|Fv\|\leq\|v\|\) for \(v\in P\cap\partial\Omega_{3}\). Therefore, by Lemma 2.5, F has a fixed point \(v_{2}\in P\) such that \(R\leq\|v_{2}\|\leq R_{\infty}\), and hence \(u_{2}(t)=v_{2}(t)-u_{0}(t)\) is a second positive solution of problem (1.1). The proof is complete. □
Theorem 3.3
Assume that (C1) and (C2) hold. Suppose that there exist
\(r, R>0\)
such that
\(\frac{1}{\mu}< r< R\)
and
- (C3)∗
:
-
\(\phi(R)\leq\frac{R}{\max_{t\in[1,T]_{\mathbb {Z}}}\sum_{s=1}^{T}G(t,s)}\)
and
\(\psi(r)\geq \frac{r}{\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\).
Then problem (1.1) has at least one positive solution. In addition, if
- (C4)∗
:
-
\(f(t,z)\geq0\)
for all
\(t\in[1,T]_{\mathbb{Z}}\)
and
\(z>0\)
sufficiently large, and
- (C5)∗
:
-
\(L> \frac{2}{q_{0}\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\),
then problem (1.1) has at least two positive solutions.
Proof
Let F, P, and \(\Omega_{1}\), \(\Omega_{2}\) be defined as (3.1), (3.2), and (3.3), respectively. Consider the fixed point of operator F in the cone P. Similar to the arguments in the proof of Theorem 3.1, if (C3)∗ holds, then we have \(\|Fv\|\geq\|v\|\) for \(v\in P\cap\partial\Omega_{1}\) and \(\|Fv\|\leq \|v\|\) for \(v\in P\cap\partial\Omega_{2}\). Thus, by Lemma 2.5, F has a fixed point \(v_{1}\in P\) satisfying \(r\leq\|v_{1}\|\leq R\), which is a positive solution of problem (2.3) and satisfies \(v_{1}(t)-u_{0}(t)>0\), \(t\in[1,T]_{\mathbb{Z}}\). Therefore, by Lemma 2.4, \(u_{1}=v_{1}(t)-u_{0}(t)\) is a positive solution of problem (1.1).
Now, let (C4)∗ and (C5)∗ hold. By (C4)∗, there exists \(D>0\) such that, for \(z>D\),
$$\tilde{f}(t,z)=f(t,z)+Lz+h(t)\geq Lz+h(t),\quad t\in[1,T]_{\mathbb{Z}}. $$
By (C5)∗, one can choose \(R_{\infty}>R+\max \{\frac{2D}{q_{0}}, \frac{2}{\mu} \}\) such that
$$\begin{aligned} \frac{1}{2}q_{0}L+\frac{h_{0}}{R_{\infty}}\geq \frac{1}{\max_{t\in [1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}, \end{aligned}$$
(3.7)
where \(h_{0}=\min_{t\in[1,T]_{\mathbb{Z}}}h(t)\). Let \(\Omega_{3}=\{v\in\mathbb{E}: \|v\|< R_{\infty}\}\). Then, for \(v\in P\cap\partial\Omega_{3}\), we have
$$\begin{aligned} v(t)-u_{0}(t)\geq \biggl(R_{\infty}-\frac{1}{\mu} \biggr)q_{0}\geq\frac{1}{2} q_{0}R_{\infty}>D,\quad t \in[1,T]_{\mathbb{Z}}, \end{aligned}$$
which implies that
$$\begin{aligned} \tilde{f}\bigl(t, v(t)-u_{0}(t)\bigr)\geq L\bigl(v(t)-u_{0}(t) \bigr)+h(t)\geq\frac {1}{2}Lq_{0}R_{\infty}+h_{0},\quad t\in[1,T]_{\mathbb{Z}}. \end{aligned}$$
Thus, by (3.7), we have, for \(v\in P\cap\partial\Omega_{3}\),
$$\begin{aligned} \|Fv\| =& \max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s) \tilde {f}\bigl(s,v(s)-u_{0}(s)\bigr) \\ \geq& \biggl(\frac{1}{2}Lq_{0}R_{\infty}+h_{0} \biggr)\max_{t\in[1,T]_{\mathbb {Z}}}\sum_{s=1}^{T}G(t,s) \\ \geq& R_{\infty}. \end{aligned}$$
That is, \(\|Fv\|\geq R_{\infty}=\|v\|\) for \(v\in P\cap\partial\Omega_{3}\). Therefore, F has another fixed point \(v_{2}\in P\) such that \(r\leq \|v_{1}\|\leq R\leq\|v_{2}\|\leq R_{\infty}\). By Lemma 2.4, problem (1.1) has two positive solutions \(u_{1}(t)=v_{1}(t)-u_{0}(t)\) and \(u_{2}(t)=v_{2}(t)-u_{0}(t)\). The proof is complete. □
The following result can be obtained directly from Theorem 3.3.
Corollary 3.4
Let (C1) and (C2) hold. Assume that there exist
\(\frac{1}{\mu}< r< R\)
such that (C3)∗
hold. Then, if
\(\lim_{z\to\infty}\frac {f(t,z)}{z}=\infty\)
uniformly for
\(t\in[1,T]_{\mathbb{Z}}\)
and
\(L> \frac{2}{q_{0}\max_{t\in[1,T]_{\mathbb{Z}}}\sum_{s=1}^{T}G(t,s)}\), problem (1.1) has at least two positive solutions.