Throughout the rest of this paper, we always assume that the points of impulse \(t_{k}\) are right-dense for each \(k=1,2,\ldots,m\).
We define
$$\begin{aligned} PC =&\bigl\{ x\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}\rightarrow\mathbb {R}:x_{k}\in C(J_{k},\mathbb{R}), k=0,1,2,\ldots,m, \text{and there exist} \\ &x\bigl(t_{k}^{+}\bigr)\text{ and }x\bigl(t_{k}^{-} \bigr)\text{ with }x\bigl(t_{k}^{-} \bigr)=x(t_{k}), k=1,2,\ldots,m\bigr\} , \end{aligned}$$
where \(x_{k}\) is the restriction of x to \(J_{k}= \left .( t_{k},t_{k+1} ] \right ._{\mathbb{T}}\subset \left .(0,\sigma(T)] \right ._{\mathbb{T}}\), \(k=1,2,\ldots,m\), and \(J_{0}=[0,t_{1}]_{\mathbb{T}}\), \(t_{m+1}=\sigma(T)\).
Let
$$X= \bigl\{ x:x\in PC, x(0)=x\bigl(\sigma(T)\bigr) \bigr\} $$
with the norm \(\Vert x\Vert =\sup_{ t\in[0,\sigma (T)]_{\mathbb{T}}}\vert x(t)\vert \), then X is a Banach space.
Lemma 2.1
[18]
Suppose
\(h:[0,T]_{\mathbb{T}}\rightarrow \mathbb{R}\)
is rd-continuous, then
x
is a solution of
$$x(t)=\int_{0}^{\sigma(T)}G(t,s)h(s)\triangle s+\sum _{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k})\bigr),\quad t\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}, $$
where
$$ G(t,s)= \left\{ \begin{array}{l@{\quad}l} \frac{e_{p}(s,t)e_{p}(\sigma(T),0)}{e_{p}(\sigma(T),0)-1}, & 0\leq s\leq t\leq \sigma(T), \\ \frac{e_{p}(s,t)}{e_{p}(\sigma(T),0)-1}, & 0\leq t< s\leq\sigma(T), \end{array} \right. $$
if and only if
x
is a solution of the boundary value problem
$$\left \{ \begin{array}{l} x^{\bigtriangleup}(t)+p(t)x(\sigma(t))=h(t),\quad t\in J, t\neq t_{k}, \\ x(t_{k}^{+})-x(t_{k}^{-})=I_{k}(x(t_{k}^{-})),\quad k=1,2,\ldots,m, \\ x(0)=x(\sigma(T)). \end{array} \right . $$
Lemma 2.2
Let
\(G(t,s)\)
be defined as in Lemma
2.1, then
$$A=\frac{1}{e_{p}(\sigma(T),0)-1}\leq G(t,s)\leq\frac{e_{p}(\sigma(T),0)}{e_{p}(\sigma(T),0)-1}=B\quad \textit{for all }t,s \in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}. $$
Remark 2.1
Let \(G(t,s)\) be defined as in Lemma 2.1, then \(\int_{0}^{\sigma(T)}G(t,s)p(s)\triangle s=1\).
Let
$$K= \bigl\{ x\in X:x(t)\geq\delta \Vert x\Vert , t\in \bigl[0,\sigma (T) \bigr]_{\mathbb{T}} \bigr\} , $$
where \(\delta=\frac{A}{B}\in(0,1)\). It is not difficult to verify that K is a cone in X.
For convenience, we denote
$$\begin{aligned}& f^{0}=\lim_{ x\rightarrow0^{+}}\sup\max_{t\in [ 0,T ] _{\mathbb{T}}} \frac{f(t,x)}{p(t)x},\qquad I^{0}(k)=\lim_{ x\rightarrow 0^{+}}\sup \frac{I_{k}(x)}{x}; \\& f_{0}=\lim_{ x\rightarrow0^{+}}\inf\min_{t\in [ 0,T ] _{\mathbb{T}}} \frac{f(t,x)}{p(t)x},\qquad I_{0}(k)=\lim_{ x\rightarrow 0^{+}}\inf \frac{I_{k}(x)}{x}; \\& f^{\infty}=\lim_{ x\rightarrow\infty}\sup\max_{t\in [ 0,T ] _{\mathbb{T}}} \frac{f(t,x)}{p(t)x},\qquad I^{\infty}(k)=\lim_{ x\rightarrow\infty}\sup \frac{I_{k}(x)}{x}; \\& f_{\infty}=\lim_{ x\rightarrow\infty}\inf\min_{t\in [ 0,T ] _{\mathbb{T}}} \frac{f(t,x)}{p(t)x},\qquad I_{\infty}(k)=\lim_{ x\rightarrow\infty}\inf \frac{I_{k}(x)}{x}; \\& f^{u}=\sup_{x\in[\delta u,u]}\max_{t\in [ 0,T ] _{\mathbb {T}}} \frac{f(t,x)}{p(t)x},\qquad I^{u}(k)=\sup_{x\in[\delta u,u]} \frac{I_{k}(x)}{x}; \\& f_{u}=\inf_{x\in[\delta u,u]}\min_{t\in [ 0,T ] _{\mathbb {T}}} \frac{f(t,x)}{p(t)x},\qquad I_{u}(k)=\inf_{x\in[\delta u,u]} \frac{I_{k}(x)}{x}. \end{aligned}$$
Now we state our main results.
Theorem 2.1
Suppose there exist
\(0<\alpha<\beta\)
such that
$$f(t,x)\geq0\quad \textit{and}\quad I_{k}(x)\geq0,\quad t\in [ 0,T ] _{\mathbb{T}}, x\in[\delta\alpha,\beta]. $$
Then the problem (1.1) has at least one positive solution if one of the following two conditions holds:
-
(i)
\(f_{\alpha}+A\sum_{k=1}^{m}I_{\alpha}(k)\geq1\), \(f^{\beta}+B\sum_{k=1}^{m}I^{\beta}(k)\leq1\);
-
(ii)
\(f^{\alpha}+B\sum_{k=1}^{m}I^{\alpha}(k)\leq1\), \(f_{\beta}+A\sum_{k=1}^{m}I_{\beta}(k)\geq1\).
Proof
We define an operator \(\Phi:K\rightarrow X\) by
$$\Phi(x) (t)=\int_{0}^{\sigma(T)}G(t,s)f\bigl(s,x\bigl( \sigma(s)\bigr)\bigr)\triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(u(t_{k})\bigr),\quad t\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}. $$
It is obvious that fixed points of Φ are solutions of the problem (1.1). Similar to [18], \(\Phi:K\rightarrow X\) is completely continuous.
Define the open sets
$$\begin{aligned}& \Omega_{1} = \bigl\{ x\in X:\Vert x\Vert < \alpha\bigr\} , \\& \Omega_{2} = \bigl\{ x\in X:\Vert x\Vert <\beta\bigr\} . \end{aligned}$$
Firstly, we claim that \(\Phi:K\cap(\overline{\Omega}_{2}\setminus \Omega _{1})\rightarrow K\).
In fact, for any \(x\in K\cap(\overline{\Omega}_{2}\setminus\Omega _{1})\), we have \(\delta\alpha\leq x\leq\beta\), by Lemma 2.2
$$\Vert \Phi x\Vert \leq B \Biggl[ \int_{0}^{\sigma(T)}f \bigl(s,x\bigl(\sigma (s)\bigr)\bigr)\triangle s+\sum _{k=1}^{m}I_{k}\bigl(x(t_{k}) \bigr) \Biggr] $$
and
$$\begin{aligned} ( \Phi x ) (t) =&\int_{0}^{\sigma(T)}G(t,s)h_{x}(s) \triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k})\bigr) \\ \geq&A \Biggl[ \int_{0}^{\sigma(T)}f\bigl(s,x\bigl( \sigma(s)\bigr)\bigr)\triangle s+\sum_{k=1}^{m}I_{k} \bigl(x(t_{k})\bigr) \Biggr] . \end{aligned}$$
So
$$( \Phi x ) (t)\geq\frac{A}{B}\Vert \Phi x\Vert =\delta \Vert \Phi x \Vert ,\quad \textit{i.e.},\Phi x\in K. $$
Therefore, \(\Phi:K\cap(\overline{\Omega}_{2}\setminus\Omega _{1})\rightarrow K\).
Secondly, we prove the result provided condition (i) holds.
Since
$$f(t, x)\geq f_{\alpha}p(t)x,\qquad I_{k}(x)\geq I_{\alpha}(k)x,\quad k=1,2,\ldots,m, x\in[\delta\alpha,\alpha]. $$
Let \(e\equiv1\), then \(e\in K\). We assert that
$$ x\neq\Phi x+\lambda e\quad \text{for }x\in K\cap\partial\Omega_{1} \text{ and }\lambda>0. $$
(2.1)
If not, there would exist \(x_{0}\in K\cap\partial\Omega_{1}\) and \(\lambda _{0}>0 \) such that \(x_{0}=\Phi x_{0}+\lambda_{0}e\).
Since \(x_{0}\in K\cap\partial\Omega_{1}\), then \(\delta\alpha=\delta \Vert x_{0}\Vert \leq x_{0}(t)\leq\alpha\). Let \(\mu=\min_{t\in[0,\sigma (T)]_{\mathbb{T}}}x_{0}(t)\), then for any \(t\in[0,\sigma(T)]_{\mathbb{T}}\), by the first inequality of (i) we have
$$\begin{aligned} x_{0}(t) =& ( \Phi x_{0} ) (t)+\lambda_{0} \\ =&\int_{0}^{\sigma(T)}G(t,s)f\bigl(s, x_{0} \bigl(\sigma(s)\bigr)\bigr)\triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x_{0}(t_{k})\bigr)+\lambda_{0} \\ \geq&\int_{0}^{\sigma(T)}G(t,s)f_{\alpha}p(s)x_{0}\bigl(\sigma(s)\bigr)\triangle s+A\sum _{k=1}^{m}I_{\alpha}(k)x(t_{k})+ \lambda_{0} \\ \geq&\mu \Biggl[ f_{\alpha}+A\sum_{k=1}^{m}I_{\alpha}(k) \Biggr] +\lambda _{0} \\ \geq&\mu+\lambda_{0}. \end{aligned}$$
This implies that \(\mu\geq\mu+\lambda_{0}\), and this is a contradiction. Therefore (2.1) holds.
On the other hand, by using the second inequality of (i), we have
$$f(t, x)\leq \Biggl[ 1-B\sum_{k=1}^{m}I^{\beta}(k) \Biggr] p(t)x,\quad t\in[0,T]_{\mathbb{T}}, x\in [ \delta\beta,\beta ] . $$
We assert that
$$ \Vert \Phi x\Vert \leq \Vert x\Vert \quad \text{for }x\in K\cap \partial \Omega_{2}. $$
(2.2)
In fact, for any \(x\in K\cap\partial\Omega_{2}\), then \(\delta\beta =\delta \Vert x\Vert \leq x(t)\leq\beta\), we have
$$\begin{aligned} ( \Phi x ) (t) =&\int_{0}^{\sigma(T)}G(t,s)f\bigl(s, x \bigl(\sigma (s)\bigr)\bigr)\triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k})\bigr) \\ \leq& \Biggl( 1-B\sum_{k=1}^{m}I^{\beta}(k) \Biggr) \int_{0}^{\sigma (T)}G(t,s)p(s)x\bigl(\sigma(s) \bigr)\triangle s+B\sum_{k=1}^{m}I^{\beta}(k)x(t_{k}) \\ \leq& \Biggl[ \Biggl( 1-B\sum_{k=1}^{m}I^{\beta}(k) \Biggr) \int_{0}^{\sigma (T)}G(t,s)p(s)\triangle s+B\sum _{k=1}^{m}I^{\beta}(k) \Biggr] \Vert x\Vert \\ =&\Vert x\Vert . \end{aligned}$$
Therefore, \(\Vert \Phi x\Vert \leq \Vert x\Vert \).
It follows from Remark 1.1, (2.1), and (2.2) that Φ has a fixed point \(x\in K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\).
In a similar way, we can prove the result by Theorem 1.1 if condition (ii) holds. □
Theorem 2.2
Suppose there exist
\(0<\alpha<\rho<\beta\)
such that
$$f(t,x)\geq0\quad \textit{and}\quad I_{k}(x)\geq0, \quad t\in [ 0,T ] _{\mathbb{T}}, x\in[\delta\alpha,\beta]. $$
Then the problem (1.1) has at least two positive solution if one of the following two conditions holds:
-
(i)
\(f_{\alpha}+A\sum_{k=1}^{m}I_{\alpha}(k)\geq1\), \(f^{\rho}+B\sum_{k=1}^{m}I^{\rho}(k)<1\), \(f_{\beta}+A\sum_{k=1}^{m}I_{\beta}(k)\geq1\);
-
(ii)
\(f^{\alpha}+B\sum_{k=1}^{m}I^{\alpha}(k)\leq1\), \(f_{\rho}+A\sum_{k=1}^{m}I_{\rho}(k)>1\), \(f^{\beta}+B\sum_{k=1}^{m}I^{\beta}(k)\leq1\).
Proof
We only prove the result when condition (i) holds. In a similar way we can obtain the result if condition (ii) holds.
Define \(\Omega_{1}\), \(\Omega_{2}\) as in Theorem 2.1 and define
$$\Omega_{3}=\bigl\{ x\in X:\Vert x\Vert < \rho\bigr\} . $$
Similar to the proof of Theorem 2.1, we can prove that
$$\begin{aligned}& x\neq\Phi x+\lambda e\quad \text{for }x\in K\cap\partial\Omega_{1} \text{ and } \lambda>0, \end{aligned}$$
(2.3)
$$\begin{aligned}& x\neq\Phi x+\lambda e\quad \text{for }x\in K\cap\partial\Omega_{2} \text{ and } \lambda>0, \end{aligned}$$
(2.4)
where \(e\equiv1\in K\), and
$$ \Vert \Phi x\Vert < \Vert x\Vert \quad \text{for }x\in K\cap \partial \Omega_{3}. $$
(2.5)
Thus we can obtain the existence of two positive solutions \(x_{1}\) and \(x_{2}\) by using Theorem 1.1 and Remark 1.1, respectively. It is easy to see that \(\alpha\leq \Vert x_{1}\Vert <\rho<\Vert x_{2}\Vert \leq \beta\). □
Theorem 2.3
Suppose that there exist positive
\(0<\alpha _{1}<\beta _{1}<\alpha_{2}<\beta_{2}<\cdots<\alpha_{n}<\beta_{n}\)
such that
$$f(t,x)\geq0\quad \textit{and} \quad I_{k}(x)\geq0,\quad t\in [ 0,T ] _{\mathbb{T}}, x\in[\delta\alpha_{1},\beta_{n}]. $$
Then the problem (1.1) has at least
n
positive solutions
\(x_{i}\) (\(1\leq i\leq n\)) satisfying
\(\alpha_{i}\leq \Vert x_{i}\Vert \leq \beta_{i}\), \(1\leq i\leq n\), if one of the following two conditions holds:
-
(i)
\(f_{\alpha_{i}}+A\sum_{k=1}^{m}I_{\alpha_{i}}(k)\geq1\), \(f^{\beta _{i}}+B\sum_{k=1}^{m}I^{\beta_{i}}(k)\leq1\);
-
(ii)
\(f^{\alpha_{i}}+B\sum_{k=1}^{m}I^{\alpha_{i}}(k)\leq1\), \(f_{\beta _{i}}+A\sum_{k=1}^{m}I_{\beta_{i}}(k)\geq1\).
Remark 2.2
In Theorem 2.3, assume (i) and (ii) are replaced by
-
(i)
\(f_{\alpha_{i}}+A\sum_{k=1}^{m}I_{\alpha_{i}}(k)>1\), \(f^{\beta _{i}}+B\sum_{k=1}^{m}I^{\beta_{i}}(k)<1\);
-
(ii)
\(f^{\alpha_{i}}+B\sum_{k=1}^{m}I^{\alpha_{i}}(k)<1\), \(f_{\beta _{i}}+A\sum_{k=1}^{m}I_{\beta_{i}}(k)>1\).
Then the problem (1.1) has at least \(2n-1\) positive solutions.