1 Introduction
We obtain sufficient conditions for the existence of oscillatory solutions of impulsive linear differential equations of the form
$$ \begin{align}\begin{cases} (a(t)y^{\prime})^{\prime}+b(t)y=0,& t\neq \tau_{k},\\[3pt] \Delta y+a_{k}y=0, & t=\tau_{k},\\[3pt] \Delta a(t)y^{\prime}+b_{k}y+c_{k} y^{\prime}=0, &t=\tau_{k}. \end{cases} \end{align} $$
The functions
$a(t)>0$
and
$b(t)$
are assumed to be piecewise left continuous on
$[t_{0},\infty )$
for some
$t_{0}\geq 0$
;
$\{a_{k}\}$
,
$\{b_{k}\}$
and
$\{c_{k}\}$
are real sequences;
$\tau _{k+1}> \tau _{k}$
for all
$k=1,2,\ldots $
;
$\lim _{k\to \infty }\tau _{k}=\infty $
and
$\Delta \varphi (\tau _{k})=\varphi (\tau _{k}^{+})-\varphi (\tau _{k}^{-})$
with
$ \varphi (\tau _{k}^{\pm })=\lim _{t\to \tau _{k}^{\pm }}\varphi (t) $
. As the impulse effects contain both the solution and its derivative, they are said to be of mixed type. By separated impulse effects, we mean that the impulse effects contain either only the solution or only the derivative of the solution, for example,
$\Delta y+a_{k}y=0$
and
$ \Delta a(t)y^{\prime }+c_{k} y^{\prime }=0$
, where
$t=\tau _{k}$
.
For the sake of brevity, the notation
$\underline {n}(t):=\inf \{k: \tau _{k}\geq t\}$
,
$\overline {n}(t):={\rm sup} \{k: \tau _{k}< t\}$
and
$\displaystyle \omega _{k}:= ({1-c_{k}/a(\tau _{k})})/({1-a_{k}})$
is used. The following hypotheses are assumed throughout the paper:
-
(H1)
$1-c_{k}/a(\tau _{k})>0,\ 1-a_{k}>0$
and
$b_{k}\leq 0$
,
$k=1,2,\ldots $
; -
(H2) there exists a function
$f(t): [t_{0},\infty )\to (0,\infty )$
such that
$f^{\prime }(t)$
exists on
$[t_{0},\infty )$
and where
$$ \begin{align*} g(t):=b(t)+\alpha(t)f^{\prime}(t)+\frac{(\alpha(t)f(t))^{2}}{a(t)}\geq 0,\end{align*} $$
$$ \begin{align*}\alpha(t):=\begin{cases} \displaystyle-\sum\limits_{k={\underline{n}(t_{0})}}^{\overline{n}(t)} \dfrac{b_{k}}{f(\tau_{k})(1-c_{k}/a(\tau_{k}))}\prod\limits_{i=k}^{\overline{n}(t)}{\omega_{i}} &\text{if}\ b_{k}\neq 0,\\[22pt] \displaystyle\prod\limits_{k={\underline{n}(t_{0})}}^{\overline{n}(t)}{\omega_{k}} &\text{if}\ b_{k}= 0. \end{cases}\end{align*} $$
Differential equations containing impulse effects are practical tools to represent many evolutionary processes such as biological models, physical phenomena and engineering problems, and the corresponding theory is quite rich. Their qualitative theory has been investigated deeply by many researchers (see the famous books [Reference Bainov and Simeonov4, Reference Lakshmikantham, Bainov and Simeonov6]). It is well known that impulse effects can cause radical changes in the structure of the solution of a differential equation. For example, a nonoscillatory unforced differential equation may turn out to be oscillatory under impulsive conditions [Reference Akgöl and Zafer2, Reference Luo and Shen7, Reference Sugie9–Reference Wen, Zeng, Peng and Huang11]. Since it is not easy to make a prediction, it is crucial to study the long-time behaviour of impulsive differential equations, in particular, their oscillatory properties. We refer to [Reference Agarwal, Karakoc and Zafer1] for an excellent survey on the oscillation of differential equations under impulse effects and the papers [Reference Bainov, Domshlak and Simeonov3, Reference Özbekler and Zafer8, Reference Sugie9] regarding self-adjoint impulsive differential equations with continuous solutions, namely, equations derived by setting
$a_{k}=0=c_{k}$
in (1.1). There are only a few studies of their counterparts having discontinuous solutions (see [Reference Guo and Xu5, Reference Luo and Shen7], where differential equations with separated impulse effects were considered). To the best of our knowledge, the only paper dealing with oscillation of equations with mixed impulse effects of the form (1.1) is [Reference Akgöl and Zafer2], in which a Leighton-type oscillation theorem is produced.
2 Main results
We start with some auxiliary lemmas.
Lemma 2.1. Let
$$ \begin{align}x(t):=y(t)\exp\bigg\{{-}\int\nolimits_{t_{0}}^{t} \frac{\alpha(s)f(s)}{a(s)}\,{d}s\bigg\},\end{align} $$
where
$y(t)$
is a solution of (1.1). Then,
$x(t)$
is a solution of the differential equation with separated impulse effects:
$$ \begin{align} \begin{cases} (a(t)x^{\prime})^{\prime}+2\alpha(t)f(t)x^{\prime}+g(t)x=0, \quad & t\neq \tau_{k},\\[3pt] \Delta x+a_{k} x=0,\quad & t=\tau_{k},\\[3pt] \Delta(a(t)x^{\prime})+c_{k}x^{\prime}=0,\quad & t=\tau_{k}. \end{cases} \end{align} $$
Proof. Let
$y(t)$
be a solution of (1.1). For
$t\neq \tau _{k}$
,
$$ \begin{align*} a(t)y^{\prime}(t)=[a(t)x^{\prime}(t)+\alpha(t)f(t)x(t)]\exp\bigg\{\int\nolimits_{t_{0}}^{t} \frac{\alpha(s)f(s)}{a(s)}\,{d}s\bigg\} \end{align*} $$
and
$$ \begin{align*}(a(t)y^{\prime}(t))^{\prime}+b(t)y(t)= [(a(t)x^{\prime}(t))^{\prime} + 2\alpha(t)f(t)x^{\prime}(t)+g(t)x(t)]\exp\bigg\{\int\nolimits_{t_{0}}^{t} \frac{\alpha(s)f(s)}{a(s)}\,{d}s\bigg\},\end{align*} $$
which clearly implies that
$$ \begin{align} (a(t)x^{\prime}(t))^{\prime} + 2\alpha(t)f(t)x^{\prime}(t)+g(t)x(t)=0, \quad t\neq \tau_{k}. \end{align} $$
For
$t=\tau _{k}$
,
$k=1,2,\ldots $
, it is easy to see that
$$ \begin{align} \Delta x\vert_{t=\tau_{k}}+a_{k} x(\tau_{k}) = [\Delta y\vert_{t=\tau_{k}}+a_{k} y(\tau_{k}) ]\exp\bigg\{{-}\int\nolimits_{t_{0}}^{\tau_{k}} \frac{\alpha(s)f(s)}{a(s)}\,{d}s\bigg\} =0. \end{align} $$
Noting that
$$ \begin{align*}\alpha(\tau_{k}^{+})=\omega_{k}\alpha(\tau_{k})-\frac{b_{k}}{f(\tau_{k})(1-a_{k})},\end{align*} $$
we see that
$$ \begin{align} & \Delta( a(t)x^{\prime})\vert_{t=\tau_{k}}+c_{k}x^{\prime}(\tau_{k}) \nonumber \\[6pt] & = \bigg(\Delta( a(t)y^{\prime}- \alpha(t) f(t) y) \vert_{t=\tau_{k}} +c_{k} \bigg[ y^{\prime}(\tau_{k})- \frac{\alpha(\tau_{k}) f(\tau_{k}) y(\tau_{k})}{a(\tau_{k})}\bigg]\bigg) \exp\bigg\{{-}\int\nolimits_{t_{0}}^{\tau_{k}} \frac{\alpha(s)f(s)}{a(s)}\,{d}s\bigg\} \nonumber \\[6pt] & = y(\tau_{k}) \bigg(\!-b_{k}-f(\tau_{k})\bigg[(1-a_{k})\alpha(\tau_{k}+)-\alpha(\tau_{k})+\frac{c_{k}\alpha(\tau_{k})}{a(\tau_{k})}\bigg]\bigg) \exp\bigg\{{-}\int\nolimits_{t_{0}}^{\tau_{k}} \frac{\alpha(s)f(s)}{a(s)}\,{d}s\bigg\} \nonumber\\[6pt] & =0. \end{align} $$
Thus, from (2.3)–(2.5), we conclude that
$x(t)$
is a solution of (2.2).
Lemma 2.2. Let
$x(t)$
be a nonoscillatory solution of (2.2). If
$$ \begin{align} \displaystyle \lim_{n\to\infty} \frac{1}{a(\tau_{n})}\sum\limits_{i=0}^{n-1} \prod\limits_{j=0}^{i}\frac{1} {\omega_{j}} \int\nolimits_{\tau_{i}}^{\tau_{i+1}} g(t)\,{d}t=\infty, \end{align} $$
then
$x(t)x^{\prime }(t)$
is ultimately negative.
Proof. Suppose that
$x(t)$
is ultimately positive. First, we will show that
$x^{\prime }(t)$
is nonoscillatory. We assume on the contrary that
$x^{\prime }(t)$
is oscillatory. Then, there is some
$k\in \mathbb {N}$
and
$t_{a}\in (\tau _{k},\tau _{k+1}]$
such that
$x^{\prime }(t_{a})=0$
. Thus, in view of (2.2),
$$ \begin{align} a(t_{a})x^{\prime\prime}(t_{a})=-g(t_{a})x(t_{a})<0, \end{align} $$
which implies that there is some interval
$(t_{a}, t_{a}+\delta )$
,
$\delta>0$
, in which
$x^{\prime }(t)$
is decreasing. Hence,
$$ \begin{align} x^{\prime}(t)<0, \quad t\in (t_{a}, t_{a}+\delta).\end{align} $$
Now, assume that
$t_{a}$
is the first root and
$x^{\prime }$
has another root in the same interval, that is, there is some
$t_{b}\in (t_{a}, \tau _{k+1})$
such that
$x^{\prime }(t_{b})=0$
. From (2.8), this implies that
$x^{\prime \prime }(t_{b})\geq 0$
. However, from (2.7), we see that
$a(t_{b})x^{\prime \prime }(t_{b})<0$
which leads to a contradiction. Hence,
$x^{\prime }(t)$
cannot have a root in
$(t_{a}, \tau _{k+1})$
, that is,
$x^{\prime }(t)<0$
there. This implies that
$$ \begin{align}x^{\prime}(\tau_{k+1}^{+})=(1-{c_{k+1}}/{a(\tau_{k+1})})x^{\prime}(\tau_{k+1})<0.\end{align} $$
If we again suppose that there is some
$t_{c}$
such that
$x^{\prime }(t_{c})=0$
, from (2.7), we obtain
$x^{\prime \prime }(t_{c})<0$
which implies
$x^{\prime }(t)<0$
,
$t\in (t_{c}-\delta , t_{c}+\delta )$
, contradicting
$x^{\prime }(t_{c})=0$
. Hence,
$x^{\prime }(t)<0$
on
$(\tau _{k+1}, \tau _{k+2}]$
. By similar arguments, it can be seen that
$$ \begin{align*}x^{\prime}(t)<0, \quad t\in (\tau_{k+i}, \tau_{k+i+1}], \ i\in\mathbb{N}.\end{align*} $$
Fix some
$T\geq t_{0}$
and let
$\tau _{k}\geq T$
. If
$x^{\prime }(t)<0$
on
$(\tau _{k}, \tau _{k+1}]$
, from (2.9),
$x^{\prime }(\tau _{k+1}^{+})<0$
and, by the above discussion,
$x^{\prime }(t)<0$
on
$(\tau _{k+i}, \tau _{k+i+1}]$
, for all
$i\in \mathbb {N}$
. Thus,
$x^{\prime }(t)<0$
on
$[T,\infty )$
.
Conversely, if
$x^{\prime }(t)>0$
on
$(\tau _{k}, \tau _{k+1}]$
, then
$x^{\prime }(\tau _{k+1}^{+})(1-{c_{k+1}}/{a(\tau _{k+1})})x^{\prime }(\tau _{k+1})>0$
. However, we know that
$x^{\prime }(t)$
has no root in
$(\tau _{k+i}, \tau _{k+i+1}]$
, for all
$n\in \mathbb {N}$
. Thus,
$x^{\prime }(t)>0$
on
$[T,\infty )$
. Hence,
$x^{\prime }(t)$
is nonoscillatory.
Our next aim is to show that
$x^{\prime }(t)$
is ultimately negative, that is, there is some
$T_{*}\geq t_{0}$
such that
$x^{\prime }(t)<0$
for
$t\geq T_{*}$
. Suppose on the contrary, there exists
$k\in \mathbb {N}$
such that
$x^{\prime }(\tau _{k})>0$
for
$\tau _{k}\geq T_{*}$
. Then,
$$ \begin{align*}x^{\prime}(\tau_{k}^{+})=(1-c_{k}/a(\tau_{k}))x^{\prime}(\tau_{k})>0,\end{align*} $$
and so
$x^{\prime }(t)>0$
for
$t\geq \tau _{k}$
. From (H1),
$\alpha (t)> 0$
on
$[t_{0},\infty )$
. Thus, we can write
$$ \begin{align} (a(t)x^{\prime}(t))^{\prime}=-2\alpha(t)f(t)x^{\prime}(t)-g(t)x(t)< -g(t)x(t)\leq 0,\end{align} $$
which shows that
$a(t)x^{\prime }(t)$
is decreasing on each interval
$[\tau _{k+i-1},\tau _{k+i}), \ i\in \mathbb {N}$
. Now, we need to prove that, for
$n\geq 1$
,
$$ \begin{align} x^{\prime}(\tau_{k+n})\leq &\frac{1}{ a(\tau_{k+n})}\prod\limits_{j=0}^{n-1} \bigg(1-\frac{c_{k+j}}{a(\tau_{k+j})}\bigg) \bigg\{ a(\tau_{k})x^{\prime}(\tau_{k}) -x(\tau_{k})\sum\limits_{i=0}^{n-1}\prod\limits_{j=0}^{i} \omega_{k+j} \int\nolimits_{\tau_{k+i}}^{\tau_{k+i+1}} g(t)\, {d}t \bigg\}. \end{align} $$
Integrating (2.10) on
$(\tau _{i}, \tau _{i+1}]$
,
$$ \begin{align*} a(\tau_{k+1})x^{\prime}(\tau_{k+1})\leq a(\tau_{k}^{+})x^{\prime}(\tau_{k}^{+})-\int\nolimits_{\tau_{k}}^{\tau_{k+1}} g(t)x(t)\, {d}t. \end{align*} $$
Since
$x^{\prime }(t)>0$
for
$t\geq \tau _{k}$
, it follows that x is increasing on
$(\tau _{k}, \tau _{k+1}]$
. Hence,
$$ \begin{align} a(\tau_{k+1})x^{\prime}(\tau_{k+1}) & \leq (a(\tau_{k})-c_{k})x^{\prime}(\tau_{k})-x(\tau_{k}^{+})\int\nolimits_{\tau_{k}}^{\tau_{k+1}} g(t)\, {d}t \nonumber \\[6pt] & = \bigg(1-\frac{c_{k}}{a(\tau_{k})}\bigg)\bigg\{a(\tau_{k})x^{\prime}(\tau_{k})-\frac{1}{\omega_{k}} x(\tau_{k})\int\nolimits_{\tau_{k}}^{\tau_{k+1}} g(t)\, {d}t\bigg\}. \end{align} $$
Integrating (2.10) on
$(\tau _{k+1}, \tau _{k+2}]$
and using (2.12),
$$ \begin{align*} & a(\tau_{k+2})x^{\prime}(\tau_{k+2}) \nonumber \\[3pt] &\quad \leq (a(\tau_{k+1})-c_{k+1})x^{\prime}(\tau_{k+1})-x(\tau_{k+1}^{+})\int\nolimits_{\tau_{k+1}}^{\tau_{k+2}} g(t)\, {d}t \\[6pt] &\quad \leq \bigg(1-\frac{c_{k+1}}{a(\tau_{k+1})}\bigg)\bigg\{\bigg(1-\frac{c_{k}}{a(\tau_{k})}\bigg)\bigg[a(\tau_{k})x^{\prime}(\tau_{k}) -\frac{1}{\omega_{k}} x(\tau_{k}) \int\nolimits_{\tau_{k}}^{\tau_{k+1}} g(t)\, {d}t\bigg]\\[6pt] &\qquad- \frac{1}{\omega_{k+1}} x(\tau_{k}^{+}) \int\nolimits_{\tau_{k+1}}^{\tau_{k+2}} g(t)\, {d}t\bigg\}\\[6pt] &\quad = \bigg(1-\frac{c_{k+1}}{a(\tau_{k+1})}\bigg)\bigg(1-\frac{c_{k}}{a(\tau_{k})}\bigg)\bigg\{a(\tau_{k})x^{\prime}(\tau_{k}) - x(\tau_{k})\bigg[ \frac{1}{\omega_{k}}\int\nolimits_{\tau_{k}}^{\tau_{k+1}} g(t)\, {d}t \\[6pt] &\qquad+\frac{1}{\omega_{k}\omega_{k+1}} \int\nolimits_{\tau_{k+1}}^{\tau_{k+2}} g(t)\, {d}t \bigg]\bigg\}. \end{align*} $$
Now, suppose that (2.11) holds for
$n=N$
. Then, for
$t\in (\tau _{k+N}, \tau _{k+N+1}]$
,
$$ \begin{align*} & a(\tau_{k+N+1})x^{\prime}(\tau_{k+N+1}) \nonumber \\[6pt] & \quad \leq (a(\tau_{k+N})-c_{k+N})x^{\prime}(\tau_{k+N})-x(\tau_{k+N}^{+})\int\nolimits_{\tau_{k+N}}^{\tau_{k+N+1}} g(t)\, {d}t \nonumber \\[6pt] & \quad \leq \bigg(1-\frac{c_{k+N}}{a(\tau_{k+N})}\bigg)\bigg\{ \prod\limits_{j=0}^{N-1} \bigg(1-\frac{c_{k+j}}{a(\tau_{k+j})}\bigg) \bigg[ a(\tau_{k})x^{\prime}(\tau_{k}) \nonumber \\[6pt] & \qquad -x(\tau_{k})\sum\limits_{i=0}^{N-1}\prod\limits_{j=0}^{i} \frac{1}{\omega_{k+j}} \int\nolimits_{\tau_{k+i}}^{\tau_{k+i+1}} g(t)\, {d}t \bigg] -\frac{1}{\omega_{k+N}} x(\tau_{k+N}) \int\nolimits_{\tau_{k+N}}^{\tau_{k+N+1}} g(t)\, {d}t\bigg\} \\[6pt] & \quad \leq \prod\limits_{j=0}^{N} \bigg(1-\frac{c_{k+j}}{a(\tau_{k+j})}\bigg) \bigg\{a(\tau_{k})x^{\prime}(\tau_{k}) -x(\tau_{k})\sum\limits_{i=0}^{N}\prod\limits_{j=0}^{i} \frac{1}{\omega_{k+j} } \int\nolimits_{\tau_{k+i}}^{\tau_{k+i+1}} g(t)\, {d}t \bigg\}, \nonumber \end{align*} $$
where, in the last line, the estimate
$x(\tau _{k+N})\geq x(\tau _{k+N-1}^{+})=(1-a_{k+N-1})x(\tau _{k+N-1})$
is used N times. Thus, by induction on n, we see that (2.11) holds for any
$n\geq 1$
.
If we take the limit of both sides of (2.11) as
$n\to \infty $
, from (2.6), we see that
$x^{\prime }(\tau _{k+n})<0$
for sufficiently large values of n. However, this contradicts the assumption that
$x^{\prime }(t)>0$
for
$t\geq \tau _{k}$
. Hence,
$x^{\prime }(\tau _{k})<0$
for
$\tau _{k}\geq T_{*}$
. Since
$x^{\prime }(t)$
has a constant sign for
$t\geq \tau _{k}$
, it follows that
$x^{\prime }(t)<0$
for all
$t\neq \tau _{k+n}$
,
$t\geq T_{*}$
.
If
$x(t)$
is ultimately negative, by repeating all the steps of the proof, it can be shown that there is some
$T^{*}\geq t_{0}$
such that
$x^{\prime }(t)>0$
for
$t\geq T^{*}$
. Thus, the proof is complete.
Theorem 2.3. Suppose that (2.6) holds, and
$$ \begin{align} \displaystyle \limsup_{n\to\infty} \sum\limits_{i=0}^{n-1} \prod\limits_{j=0}^{i} \omega_{j}\int\nolimits_{\tau_{i}}^{\tau_{i+1}} \mu(s,t_{0})\,{d}s=\infty, \end{align} $$
where
$$ \begin{align*}\mu(t,s):=\exp\bigg\{{-}2\int\nolimits_{s}^{t} \frac{f(r)\alpha(r)}{a(r)}\,{d}r\bigg\}.\end{align*} $$
Then, (2.2) is oscillatory.
Proof. Suppose on the contrary that
$x(t)$
is a nonoscillatory solution of (2.2). If we assume
$x(t)$
is ultimately positive, namely, there exists some
$T\geq t_{0}$
such that
$x(t)>0$
for
$t\geq T$
, in view of Lemma 2.2,
$x^{\prime }(t)<0$
for
$t\geq T$
and
$t\neq \tau _{k}$
. Since
$g(t)>0$
, from (2.2),
$$ \begin{align*}(a(t)x^{\prime}(t))^{\prime}+2\alpha(t)f(t)x^{\prime}(t)<0, \quad t\geq T, \ t\neq \tau_{k},\end{align*} $$
that is,
$$ \begin{align} \frac{(a(t)x^{\prime}(t))^{\prime}}{a(t)x^{\prime}(t)}+\frac{2\alpha(t)f(t)}{a(t)}>0. \end{align} $$
Define
${\tau _{k}}:=\min \{\tau _{j}: \tau _{j}\geq T\}$
. For
$t\in (\tau _{k}, \tau _{k+1}]$
, integration of (2.14) yields
$$ \begin{align*} \ln \bigg(\frac{a(t)x^{\prime}(t)}{a(\tau_{k}^{+})x^{\prime}(\tau_{k}^{+})}\bigg)+2 \int\nolimits_{\tau_{k}}^{t} \frac{\alpha(r)f(r)}{a(r)}\, {d}r>0.\end{align*} $$
Since
$x^{\prime }(t)<0$
for
$t\geq T$
, this implies that
$$ \begin{align} x^{\prime}(t)< x^{\prime}(\tau_{k}^{+})\mu(\tau_{k}, t)=\bigg(1-\frac{c_{k}}{a(\tau_{k})}\bigg)x^{\prime}(\tau_{k})\mu(\tau_{k},t), \quad t\in (\tau_{k}, \tau_{k+1}].\end{align} $$
Setting
$t=\tau _{k+1}$
,
$$ \begin{align}x^{\prime}(\tau_{k+1})< \bigg(1-\frac{c_{k}}{a(\tau_{k})}\bigg)x^{\prime}(\tau_{k})\mu(\tau_{k},\tau_{k+1}). \end{align} $$
Integrating (2.15) on
$(\tau _{k}, \tau _{k+1}]$
,
$$ \begin{align}x(\tau_{k+1})< (1-a_{k})x(\tau_{k})+ \bigg(1-\frac{c_{k}}{a(\tau_{k})}\bigg)x^{\prime}(\tau_{k})\int\nolimits_{\tau_{k}}^{\tau_{k+1}} \mu(\tau_{k},s)\, {d}s.\end{align} $$
Now, we apply the same procedure on
$(\tau _{k+1}, \tau _{k+2}]$
and similarly obtain
$$ \begin{align}x(\tau_{k+2})< (1-a_{k+1})x(\tau_{k+1})+ \bigg(1-\frac{c_{k+1}}{a(\tau_{k+1})}\bigg)x^{\prime}(\tau_{k+1})\int\nolimits_{\tau_{k+1}}^{\tau_{k+2}} \mu(\tau_{k+1},s)\, {d}s.\end{align} $$
Observe that
$\mu (\tau _{k},\tau _{k+1})\mu (\tau _{k+1},s)=\mu (\tau _ k,s) $
. Thus, using (2.16) and (2.17) in (2.18),
$$ \begin{align*} & x(\tau_{k+2}) \nonumber \\[3pt] & \ \ \ < (1-a_{k})(1-a_{k+1})\bigg\{ x(\tau_{k})+ x^{\prime}(\tau_{k})\bigg[\omega_{k} \int\nolimits_{\tau_{k}}^{\tau_{k+1}} \mu(\tau_{k},s)\, {d}s+\omega_{k} \omega_{k+1}\int\nolimits_{\tau_{k+1}}^{\tau_{k+2}} \mu(\tau_{k},s)\, {d}s\bigg]\bigg\}. \end{align*} $$
Suppose that
$$ \begin{align} x(\tau_{k+n})< \prod\limits_{j=0}^{n-1}(1-a_{k+j})\bigg\{ x(\tau_{k})+ x^{\prime}(\tau_{k})\sum\limits_{i=0}^{n-1}\prod\limits_{j=0}^{i} \omega_{k+j}\int\nolimits_{\tau_{k+i}}^{\tau_{k+i+1}} \mu(\tau_{k},s)\, {d}s\bigg\} \end{align} $$
for
$n=N$
. Then, in a similar way to the proof of (2.18), it can be shown that (2.19) holds for
$n=N+1$
. Thus, by induction, the inequality (2.19) is true for any
$n\geq 1$
.
Applying (2.13) in (2.19) leads to the contradiction that
$x(\tau _{n})<0$
for sufficiently large values of n. Hence,
$x(t)$
is oscillatory.
Theorem 2.4. Suppose that (2.6) holds and that there exists a continuous function
$h(t): [t_{0},\infty )\to (0,\infty )$
such that
$h^{\prime }(t)$
exists on
$[t_{0},\infty )$
and
$a(t)h^{\prime }(t)\geq 2\alpha (t)f(t)h(t)$
. If
$$ \begin{align} \displaystyle \limsup_{n\to\infty} \sum\limits_{i=1}^{n} \prod\limits_{j=1}^{i-1} \omega_{j}\int\nolimits_{\tau_{i-1}}^{\tau_{i}} h(s)g(s)\,{d}s=\infty \end{align} $$
and
$$ \begin{align} \limsup_{n\to\infty} \int\nolimits_{\tau_{i}}^{\tau_{i+1}} \frac{{d}t}{a(t)h(t)}\geq 1, \end{align} $$
then (2.2) is oscillatory.
Proof. Suppose on the contrary that
$x(t)$
is a nonoscillatory solution of (2.2). We may assume
$x(t)>0$
for
$t\geq T$
for some
$T\geq t_{0}$
. Then, by Lemma 2.2,
$x^{\prime }(t)<0$
for
$t\geq T$
. Define
${\tau _{k}}:=\min \{\tau _{j}: \tau _{j}\geq T\}$
. Multiplying the first line of (2.2) by
$h(t)/x(t)$
, and then integrating it on
$(\tau _{k},t]$
, for
$t\in (\tau _{k}, \tau _{k+1}]$
,
$$ \begin{align} \nonumber \frac{h(t)a(t)x^{\prime}(t)}{x(t)}&- \frac{h(\tau_{k})a(\tau_{k})x^{\prime}(\tau_{k}^{+})}{x(\tau_{k}^{+})}+\int\nolimits_{\tau_{k}}^{t} a(s)h(s)\bigg(\frac{x^{\prime}(s)}{x(s)}\bigg)^{2}\,{d}s \\[6pt] & -\int\nolimits_{\tau_{k}}^{t} [a(s)h^{\prime}(s)-2\alpha(s)f(s)h(s)]\frac{x^{\prime}(s)}{x(s)}\,{d}s +\int\nolimits_{\tau_{k}}^{t} h(s)g(s)\,{d}s=0, \end{align} $$
which implies that
$$ \begin{align*} \frac{h(t)a(t)x^{\prime}(t)}{x(t)}<\frac{h(\tau_{k})a(\tau_{k})x^{\prime}(\tau_{k}^{+})}{x(\tau_{k}^{+})}-\int\nolimits_{\tau_{k}}^{t} h(s)g(s)\,{d}s,\end{align*} $$
and so
$$ \begin{align*} \frac{h(\tau_{k+1})a(\tau_{k+1})x^{\prime}(\tau_{k+1})}{x(\tau_{k+1})}<\omega_{k}\frac{h(\tau_{k})a(\tau_{k})x^{\prime}(\tau_{k})}{x(\tau_{k})}-\int\nolimits_{\tau_{k}}^{\tau_{k+1}} h(s)g(s)\,{d}s.\end{align*} $$
For
$t\in (\tau _{k+1}, \tau _{k+2}]$
, similarly, we can show
$$ \begin{align*} \frac{h(t)a(t)x^{\prime}(t)}{x(t)} & < \frac{h(\tau_{k+1})a(\tau_{k+1})x^{\prime}(\tau_{k+1}^{+})}{x(\tau_{k+1}^{+})}-\int\nolimits_{\tau_{k+1}}^{t} h(s)g(s)\,{d}s\\[6pt] & < \omega_{k+1}\bigg\{\omega_{k}\frac{h(\tau_{k})a(\tau_{k})x^{\prime}(\tau_{k})}{x(\tau_{k})}-\int\nolimits_{\tau_{k}}^{\tau_{k+1}} h(s)g(s)\,{d}s\bigg\}-\int\nolimits_{\tau_{k+1}}^{t} h(s)g(s)\,{d}s.\end{align*} $$
The last inequality holds for
$t=\tau _{k+2}$
. Using similar arguments and applying induction on n, it is not hard to prove that for
$n\geq 1$
,
$$ \begin{align*} \frac{h(\tau_{k+n})a(\tau_{k+n})x^{\prime}(\tau_{k+n})}{x(\tau_{k+n})}< \frac{h(\tau_{k})a(\tau_{k})x^{\prime}(\tau_{k})}{x(\tau_{k})}\prod\limits_{j=0}^{n-1}\omega_{k+j} -\sum_{i=1}^{n-1} \prod\limits_{j=i}^{n-1}\omega_{k+j} \int\nolimits_{\tau_{k+i-1}}^{\tau_{k+i}} h(s)g(s)\, {d}s. \end{align*} $$
However, from Lemma 2.2,
$x(t)x^{\prime }(t)<0$
. So, using (2.20),
$$ \begin{align} \frac{h(\tau_{n})a(\tau_{n})x^{\prime}(\tau_{n})}{x(\tau_{n})}< -\sum_{i=1}^{n-1} \prod\limits_{j=i}^{n-1}\omega_{j} \int\nolimits_{\tau_{i-1}}^{\tau_{i}} h(s)g(s)\, {d}s\to-\infty, \quad\mbox{as } n\to\infty. \end{align} $$
Now, from (2.22),
$$ \begin{align*} \frac{h(t)a(t)x^{\prime}(t)}{x(t)}+\int\nolimits_{\tau_{k}}^{t} a(s)h(s)\bigg(\frac{x^{\prime}(s)}{x(s)}\bigg)^{2}\,{d}s <\frac{h(\tau_{k})a(\tau_{k})x^{\prime}(\tau_{k}^{+})}{x(\tau_{k}^{+})}-\int\nolimits_{\tau_{k}}^{t} h(s)g(s)\,{d}s \end{align*} $$
and from (2.23), this implies that there is a sufficiently large
$\tau _{\ell }$
so that
$$ \begin{align*} \frac{h(t)a(t)x^{\prime}(t)}{x(t)}+\int\nolimits_{\tau_{\ell}}^{t} a(s)h(s)\bigg(\frac{x^{\prime}(s)}{x(s)}\bigg)^{2}\, {d}s\leq -1, \end{align*} $$
for
$t\geq \tau _{\ell }$
. Hence,
$$ \begin{align} \frac{h(t)a(t)x^{\prime}(t)}{x(t)}\leq -1-\int\nolimits_{\tau_{\ell}}^{t} a(s)h(s)\bigg(\frac{x^{\prime}(s)}{x(s)}\bigg)^{2}\, {d}s. \end{align} $$
Since
$x^{\prime }(t)<0$
,
$$ \begin{align*} \frac{x^{\prime}(t)}{x(t)}\leq a(t)h(t)\bigg(\frac{x^{\prime}(t)}{x(t)}\bigg)^{2} \bigg(1+\int\nolimits_{\tau_{\ell}}^{t} \bigg(\frac{x^{\prime}(s)}{x(s)}\bigg)^{2}\, {d}s\bigg)^{-1}. \end{align*} $$
Integrating the last inequality on
$(\tau _{\ell },t]$
yields
$$ \begin{align*} \ln\bigg( \frac{x^{\prime}(\tau_{\ell}^{+})}{x(t)}\bigg)\leq \ln \bigg(1+\int\nolimits_{\tau_{\ell}}^{t} \bigg(\frac{x^{\prime}(s)}{x(s)}\bigg)^{2}\, {d}s\bigg). \end{align*} $$
From (2.24), it follows that
$$ \begin{align*} \frac{x^{\prime}(\tau_{\ell}^{+})}{x(t)}\leq - a(t)h(t)\frac{x^{\prime}(t)}{x(t)}, \quad\mbox{that is,}\quad x^{\prime}(t)\leq - \frac{x^{\prime}(\tau_{\ell}^{+})}{a(t)h(t)}.\end{align*} $$
Now, we integrate the last expression on
$(\tau _{\ell },\tau _{\ell +1}]$
to obtain
$$ \begin{align*} x(\tau_{\ell +1})\leq x(\tau_{\ell}^{+})\bigg(1-\int\nolimits_{\tau_{\ell }}^{\tau_{\ell +1}}\frac{{d}t}{a(t)h(t)}\bigg).\end{align*} $$
Thus, using the hypothesis (2.21), it is easy to see
$$ \begin{align*}\limsup_{\ell\to\infty} x(\tau_{\ell +1})\leq 0,\end{align*} $$
which leads to a contradiction because of the assumption that
$x(t)$
is ultimately positive. Hence,
$x(t)$
is oscillatory.
Now, we can easily establish the following oscillation criteria for (1.1).
In view of (2.1) and Lemma 2.1, it can be seen that Theorems 2.5 and 2.6 follow directly from Theorems 2.3 and 2.4, respectively.
3 Examples
In this section, we describe some examples to illustrate our results. For each example, the graphs show the discontinuities in a small interval and the oscillating behaviour on a larger interval.
Example 3.1. Consider the impulsive differential equation
$$ \begin{align} \begin{cases}\bigg(\dfrac{1}{t^{2}} y^{\prime}\bigg)^{\prime}+\dfrac{t+2}{(t+1)^{2}}y=0, & t\neq k, t\geq 2, \\[14pt] \Delta y+\displaystyle\frac{k+3}{2(k+2)}y=0,\ \ \Delta \bigg(\dfrac{1}{t^{2}} y^{\prime}\bigg)-\frac{3}{4k^{3}(k+1)(k+2)}y+\frac{1}{2k^{2}}y^{\prime}=0, & t=k, \ k>2.\end{cases} \end{align} $$
Let
$t_{0}=2$
. Clearly,
$$ \begin{align*}\tau_{k}=k, \ a(t)=\frac{1}{t^{2}}, \ b(t)=\frac{t+2}{(t+1)^{2}}, \ a_{k}=\frac{k+3}{2(k+2)}, \ b_{k}=-\frac{3}{4k^{3}(k+1)(k+2)},\ c_{k}=\frac{1}{2k^{2}},\end{align*} $$
and so,
$1-a_{k}=(k+1)/(2(k+2))>0$
and
$1-c_{k}/a(\tau _{k})= 1/2$
. Thus, (H1) holds. If we choose
$f(t)=3/(2t^{3}(t+1))$
, from
$\omega _{k}=(k+2)/(k+1)$
,
$$ \begin{align*}\alpha(t)=\sum\limits_{k=2}^{\overline{n}(t)} \frac{1}{k+2}\prod\limits_{i=k}^{\overline{n}(t)}\frac{i+2}{i+1}= ({\overline{n}(t)+2})\sum\limits_{k=2}^{\overline{n}(t)}\frac{1}{(k+2)(k+1)}=\frac{\overline{n}(t)-1}{3} .\end{align*} $$
For
$t\in (i, i+1]$
,
$t\geq 2$
, we have
$\overline {n}(t)=i+1$
and
$\alpha (t)=i/3$
, which implies that
$$ \begin{align*} g(t)&=b(t)+\alpha(t)f^{\prime}(t)+\frac{(\alpha(t)f(t))^{2}}{a(t)}\\[3pt] &> \frac{t+2}{(t+1)^{2}}-\frac{4t+3}{2t^{3}(t+1)^{2}}+\frac{(t-1)^{2}}{4t^{4}(t+1)^{2}}>\frac{1}{t+1}>0. \end{align*} $$
Thus, (H2) also holds and
$$ \begin{align*} \frac{1}{a(\tau_{n})}\sum\limits_{i=0}^{n-1} \prod\limits_{j=0}^{i}\frac{1} {\omega_{j}} \int\nolimits_{\tau_{i}}^{\tau_{i+1}} g(t)\,{d}t> n^{2}\sum\limits_{i=0}^{n-1} \frac{1}{i+2}\ln \bigg(\frac{i+2}{i+1}\bigg). \end{align*} $$
If we take the limit of both sides, we see that (2.6) is satisfied. However, since
$\alpha (t)<{t}/{3}$
for
$t\geq 2$
, we have the estimate
$$ \begin{align*}\mu(s,t_{0})=\exp\bigg\{{-}2\int\nolimits_{2}^{s} \frac{f(r)\alpha(r)}{a(r)}\,{d}r\bigg\}>\exp\bigg\{{-}\int\nolimits_{2}^{s} \frac{1}{r+1}\,{d}r\bigg\}=\frac{3}{s+1}.\end{align*} $$
Hence, from
$$ \begin{align*} \sum\limits_{i=0}^{n-1} \prod\limits_{j=0}^{i} \omega_{j}\int\nolimits_{\tau_{i}}^{\tau_{i+1}} \mu(s,t_{0})\,{d}s>3\sum\limits_{i=0}^{n-1}(i+2)\int\nolimits_{i}^{i+1}\frac{1}{s+1}\, {d}s= 3\sum\limits_{i=0}^{n-1}(i+2) \ln\bigg(\frac{i+2}{i+1}\bigg),\end{align*} $$
we see that (2.13) also holds. Thus, by Theorem 2.5, (3.1) is oscillatory. This conclusion is illustrated in Figure 1.
Figure 1 Illustrations for Example 3.1.
Example 3.2. Consider the impulsive differential equation
$$ \begin{align} \begin{cases}({e^{-t}}y^{\prime})^{\prime}+(1+e^{-t})y=0, & t\neq k, t\geq 2, \\[3pt] \Delta y-e^{k}y=0,\ \ \Delta ({e^{-t}}y^{\prime})-\dfrac{1}{4(k^{2}-k)}y-y^{\prime}=0, & t=k, k>2.\end{cases} \end{align} $$
Let
$t_{0}=2$
. Clearly,
$$ \begin{align*}\tau_{k}=k,\ a(t)=e^{-t}, \ b(t)=1+e^{-t}, \ a_{k}=-e^{k}, \ b_{k}=-\frac{1}{4(k^{2}-k)}, \ c_{k}=-1.\end{align*} $$
Thus, (H1) holds. Since
$\omega _{k}=1$
, by choosing
$f(t)=e^{-t}$
, we get
$$ \begin{align}\alpha(t)=\frac{1}{4}\sum\limits_{k=2}^{\overline{n}(t)} \frac{e^{k}}{(1+e^{k})(k^{2}-k)}.\end{align} $$
From
$2/5 < e^{k}/(1+e^{k})< 1$
, we may write
${1}/{10t}<\alpha (t)<{1}/{2}$
. Then
$$ \begin{align*}g(t)>1+e^{-t}-\frac{1}{2e^{t}}+\frac{1}{100t^{2}e^{t}}>1,\end{align*} $$
so that
$$ \begin{align*} \frac{1}{a(\tau_{n})}\sum\limits_{i=0}^{n-1} \prod\limits_{j=0}^{i}\frac{1} {\omega_{j}} \int\nolimits_{\tau_{i}}^{\tau_{i+1}} g(t)\,{d}t> e^{n} \frac{(n-1)n}{2}. \end{align*} $$
Taking the limit of both sides, it is easily seen that (2.6) holds.
Now, if we take
$h(t)=e^{t}$
, we can write
$2\alpha (t)f(t)h(t)<1=a(t)h^{\prime }(t)$
, and
$$ \begin{align*}\int\nolimits_{i}^{i+1} \frac{{d}t}{a(t)h(t)}=1,\end{align*} $$
which shows that (2.21) holds. Finally, to check the hypothesis (2.20), we write
$$ \begin{align*} \sum\limits_{i=1}^{n} \prod\limits_{j=1}^{i-1} \omega_{j}\int\nolimits_{\tau_{i-1}}^{\tau_{i}} h(s)g(s)\,{d}s>\sum\limits_{i=1}^{n}\int\nolimits_{i-1}^{i}e^{s}\, {d}s=e^{n}-1.\end{align*} $$
Clearly, the last expression tends to infinity as
$n\to \infty $
. Hence, by means of Theorem 2.6, (3.2) is oscillatory. The oscillation behaviour can also be seen in Figure 2.
Figure 2 Illustrations for Example 3.2.
4 Concluding remarks
The following remark demonstrates the novelty of Theorem 2.5 and hence also Theorem 2.3.
Remark 4.1. As far as we know, the only paper that deals with the oscillation of (1.1) is [Reference Akgöl and Zafer2]. If we attempt to apply the Leighton-type theorem [Reference Akgöl and Zafer2, Theorem 2.1] to (3.1), we compute
$$ \begin{align*} &\int\nolimits_{2}^{t} b(s)\prod\limits_{k=2}^{\overline{n}(s)} \frac{1}{\omega_{k}} \,{d}s +\sum\limits_{k=2}^{\overline{n}(t)} \frac{b_{k}}{1-c_{k}/a(\tau_{k})}\prod\limits_{j=2}^{k} \frac{1}{\omega_{j}}\nonumber \\[6pt] &\quad=3\int\nolimits_{2}^{t} \frac{(s+2)}{(s+1)^{2}(\overline{n}(s)+2)}-\frac{9}{2}\sum\limits_{k=2}^{\overline{n}(t)}\frac{1}{k^{3}(k+1)(k+2)^{2}}, \end{align*} $$
which is finite. Hence, the hypothesis of Theorem 2.1 in [Reference Akgöl and Zafer2] is not satisfied. As we have shown in Example 3.1, our result shows that this system oscillates.
Finally, the next remark shows the usefulness of Theorem 2.6 as an alternative to Theorem 2.5.
Remark 4.2. In Example 3.2, from (3.3), we have
$\alpha (t)=c$
, where
$1/5<c<1/2$
. Thus,
$$ \begin{align*}\mu(s,t_{0})=\exp\bigg\{{-}c\int\nolimits_{2}^{s}\bigg(1-\frac{1}{s}\bigg)\,{d}s\bigg\}=s^{c} e^{-cs},\end{align*} $$
where c is a suitable positive constant. This implies that
$$ \begin{align*} \sum\limits_{i=0}^{n-1} \prod\limits_{j=0}^{i} \omega_{j}\int\nolimits_{\tau_{i}}^{\tau_{i+1}} \mu(s,t_{0})\,{d}s=\sum\limits_{i=0}^{n-1} s^{c} e^{-cs} <\infty. \end{align*} $$
Thus, the hypothesis (2.13) does not hold, and so Theorem 2.5 cannot be applied to the system (3.2).
