In this section, some results on pmoment stability and oscillation in mean of the secondorder linear differential systems (1) with random impulses (2) are presented. Inspired by [19], we obtain the following lemma, which guarantees {lim}_{k\to +\mathrm{\infty}}{\xi}_{k}=+\mathrm{\infty} with probability 1.
Lemma 3.1 Assume that the condition (H) holds, then {lim}_{k\to +\mathrm{\infty}}{\xi}_{k}=+\mathrm{\infty} with probability 1.
Proof For a given nonnegative random variable τ and a constant c>0, we define
{\tau}^{(c)}:=\{\begin{array}{ll}\tau ,& \text{if}\tau \le c,\\ 0,& \text{otherwise}.\end{array}
By Kolmogorov’s three series theorem and Kolmogorov’s zeroorone laws [20], {\sum}_{k=1}^{+\mathrm{\infty}}{\tau}_{k}=+\mathrm{\infty} almost surely if and only if at least one of the series {\sum}_{k=1}^{+\mathrm{\infty}}\mathbb{P}({\tau}_{k}>c) and {\sum}_{k=1}^{+\mathrm{\infty}}\mathbb{E}({\tau}_{k}^{(c)}) diverges for some c>0 or any c>0.
From the CDF of the Erlang distribution, we obtain that \mathbb{P}({\tau}_{k}>c)=1\mathbb{P}({\tau}_{k}\le c)=1F(c;{m}_{k},\lambda ) is increasing in k, together with \mathbb{P}({\tau}_{1}>c)>0, we conclude that {\sum}_{k=1}^{+\mathrm{\infty}}\mathbb{P}({\tau}_{k}>c) diverges. Thus, we know that {\sum}_{k=1}^{+\mathrm{\infty}}{\tau}_{k}=+\mathrm{\infty} almost surely. The proof is complete. □
Lemma 3.2 (see [11])
y(t) is a solution of the system (1) with (2) if and only if
y(t)=\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}[1+{b}_{i}]\cdot {\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t))x(t),
where x(t) is a solution of the system (3) with the same initial conditions of the system (1) with (2), and \mathcal{X} is the index function, i.e.,
{\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t)=\{\begin{array}{ll}1,& \mathit{\text{if}}{\xi}_{k}\le t{\xi}_{k+1},\\ 0,& \mathit{\text{otherwise}}.\end{array}
Here and in the sequel, we assume that a product equals unity if the number of factors is equal to zero.
Theorem 3.1 Let the condition (H) hold. Further assume that there are a finite number of {b}_{k} such that {b}_{k}<1. If there exists a T\in {\mathbb{R}}_{\tau} such that
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})
does not change its sign for all t\ge T, then all solutions of the system (1) with (2) are oscillatory in mean if and only if all solutions of the system (3) are oscillatory. Here and in the following, {z}_{0}\equiv t{t}_{0}.
Proof Let y(t) be any sample path solution of the system (1) with (2), then it follows from Lemma 3.2 that
y(t)=\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot {\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t))x(t).
Since there are a finite number of {b}_{k} such that {b}_{k}<1, there exists a K such that {\prod}_{i=1}^{k}(1+{b}_{i}) are either nonnegative or nonpositive for all k\ge K. Hence, by either monotone convergence or Tonelli’s theorem [20],
\begin{array}{rl}\mathbb{E}y(t)& =\mathbb{E}\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot {\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t))x(t)\\ =\sum _{k=0}^{+\mathrm{\infty}}\mathbb{E}(\prod _{i=1}^{k}(1+{b}_{i})\cdot {\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t))x(t)\\ =\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \mathbb{E}{\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t))x(t).\end{array}
Further,
\begin{array}{rl}\mathbb{E}{\mathcal{X}}_{[{\xi}_{k},{\xi}_{k+1})}(t)& =\mathbb{P}({\tau}_{1}+{\tau}_{2}+\cdots +{\tau}_{k}\le {z}_{0}<{\tau}_{1}+{\tau}_{2}+\cdots +{\tau}_{k+1})\\ =\mathbb{P}({\tau}_{1}+{\tau}_{2}+\cdots +{\tau}_{k}\le {z}_{0})\mathbb{P}({\tau}_{1}+{\tau}_{2}+\cdots +{\tau}_{k+1}\le {z}_{0})\\ =\sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{e}^{\lambda {z}_{0}}{(\lambda {z}_{0})}^{n}}{n!}.\end{array}
So,
\begin{array}{rl}\mathbb{E}y(t)& =\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{e}^{\lambda {z}_{0}}{(\lambda {z}_{0})}^{n}}{n!})x(t)\\ ={e}^{\lambda {z}_{0}}x(t)\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!}).\end{array}
Since
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})
does not change its sign for all t\ge T,
{e}^{\lambda {z}_{0}}x(t)\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})
does not change its sign for all t\ge T, too. Hence, \mathbb{E}y(t) has the same sign as x(t) for all t\ge T. That is, all solutions of the system (1) with (2) are oscillatory in mean if and only if all solutions of the system (3) are oscillatory. The proof is complete. □
Theorem 3.2 Let the condition (H) hold and further assume that there are a finite number of {b}_{k} such that {b}_{k}<1. Then all solutions of the system (1) with (2) are oscillatory in mean if and only if all solutions of the system (3) are oscillatory.
Proof According to Theorem 3.1, we only need to prove that there exists a T\in {\mathbb{R}}_{\tau} such that
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})
(4)
does not change its sign for all t\ge T.
In the following, we will discuss the sign of (4) in two cases respectively.
Case I. Assume that there are a finite number of {b}_{k} such that {b}_{k}<1 and no {b}_{i}=1. Then there exists a finite set \stackrel{\u02c6}{N}=\{{k}_{i}:{k}_{i}\in \mathbb{N}\text{satisfying}{k}_{1}{k}_{2}\cdots {k}_{n},\text{where}n\text{is finite}\} such that {b}_{k}>1 for all k\in \mathbb{N}\mathrm{\setminus}\stackrel{\u02c6}{N} and {b}_{k}<1 for all k\in \stackrel{\u02c6}{N}.

(a)
If n is odd,
\prod _{i=1}^{k}(1+{b}_{i})<0\phantom{\rule{1em}{0ex}}\text{for all}k\ge {k}_{n}.
So,
\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+\cdots +{m}_{k}}^{{m}_{1}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!}<0\phantom{\rule{1em}{0ex}}\text{for all}k\ge {k}_{n}\text{and}t{t}_{0}.
(5)
For any fixed k, 0\le k<{k}_{n},
\begin{array}{r}\prod _{i=1}^{{k}_{n}}1+{b}_{i}\frac{{(\lambda {z}_{0})}^{{m}_{1}+\cdots +{m}_{{k}_{n}}+j1}}{({m}_{1}+\cdots +{m}_{{k}_{n}}+j1)!}\\ \phantom{\rule{1em}{0ex}}\ge \prod _{i=k+1}^{{k}_{n}}1+{b}_{i}\frac{{(\lambda {z}_{0})}^{{m}_{k+1}+\cdots +{m}_{{k}_{n}}}}{({m}_{1}+\cdots +{m}_{{k}_{n}}+j1)!}\cdot \prod _{i=1}^{k}1+{b}_{i}\frac{{(\lambda {z}_{0})}^{{m}_{1}+\cdots +{m}_{k}+j1}}{({m}_{1}+\cdots +{m}_{k}+j1)!}\\ \phantom{\rule{1em}{0ex}}\ge \prod _{i=k+1}^{{k}_{n}}1+{b}_{i}\frac{{(\lambda {z}_{0})}^{{m}_{k+1}+\cdots +{m}_{{k}_{n}}}}{({m}_{1}+\cdots +{m}_{{k}_{n}}+{m}_{k+1}1)!}\cdot \prod _{i=1}^{k}1+{b}_{i}\frac{{(\lambda {z}_{0})}^{{m}_{1}+\cdots +{m}_{k}+j1}}{({m}_{1}+\cdots +{m}_{k}+j1)!}\end{array}
holds for j=1,2,\dots ,{m}_{k+1}.
In fact,
\prod _{i=k+1}^{{k}_{n}}1+{b}_{i}\frac{{(\lambda {z}_{0})}^{{m}_{k+1}+\cdots +{m}_{{k}_{n}}}}{({m}_{1}+\cdots +{m}_{{k}_{n}}+{m}_{k+1}1)!}={\beta}_{k}\cdot {z}_{0}^{{m}_{k+1}+\cdots +{m}_{{k}_{n}}}\ge {k}_{n}1
holds for all t\ge {t}_{0}+{T}_{k}, where {T}_{k}=max\{1,\frac{{k}_{n}1}{{\beta}_{k}}\}, and
{\beta}_{k}=\prod _{i=k+1}^{{k}_{n}}1+{b}_{i}\frac{{\lambda}^{{m}_{k+1}+\cdots +{m}_{{k}_{n}}}}{({m}_{1}+\cdots +{m}_{{k}_{n}}+{m}_{k+1}1)!}
is a positive constant.
Thus,
holds for j=1,2,\dots ,{m}_{k+1}. From (6), we obtain that
\prod _{i=1}^{{k}_{n}}(1+{b}_{i})\cdot \sum _{n={m}_{1}+\cdots +{m}_{{k}_{n}}}^{{m}_{1}+\cdots +{m}_{{k}_{n}+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!}\ge ({k}_{n}1)\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+\cdots +{m}_{k}}^{{m}_{1}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!}.
(7)
From (5) and (7), it follows that
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})<0
for t\ge {t}_{0}+T, where T={max}_{0\le k<{k}_{n}}{T}_{k}.

(b)
If n is even, similar to the procedure of (a) in Case I, we can prove that
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})>0
holds for all t\ge {t}_{0}+T.
From (a), (b), we know that
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})
does not change its sign for all t\ge {t}_{0}+T.
Case II. Assume there are a finite number of {b}_{k} such that {b}_{k}<1 and at least a {b}_{i}=1. Then let m=min\{i\in \mathbb{N}:{b}_{i}=1\}, and let {b}_{k}>1 for all k\in \{1,2,\dots ,m1\}\mathrm{\setminus}\stackrel{\u02c6}{N} and {b}_{i}<1 for all i\in \stackrel{\u02c6}{N}, where \stackrel{\u02c6}{N}=\{{k}_{1},{k}_{2},\dots ,{k}_{n}\} satisfying {k}_{1}<{k}_{2}<\cdots <{k}_{n}. Without loss of generality, we assume {k}_{n}<m. Then
\prod _{i=1}^{k}(1+{b}_{i})=0\phantom{\rule{1em}{0ex}}\text{for all}k\ge m.
Thus,
\begin{array}{r}\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})\\ \phantom{\rule{1em}{0ex}}=\sum _{k=0}^{m1}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!}).\end{array}
Similar to the proof of Case I, we can prove that
\sum _{k=0}^{m1}(\prod _{i=1}^{k}(1+{b}_{i})\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})
does not change its sign for all t\ge {t}_{0}+T.
In summary, by Case I, Case II and Theorem 3.1, all solutions of the system (1) with (2) are oscillatory in mean if and only if all solutions of the system (3) are oscillatory. The proof is complete. □
Remark 3.1 Theorem 3.2 is a generalization of Theorem 2 in [11] since the condition (H) can degenerate to the condition (C) in [11].
Theorem 3.3 Let the condition (H) hold. If there exists a constant \alpha >0 such that
\sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}{1+{b}_{i}}^{p}\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})\le \alpha {e}^{\lambda {z}_{0}}
holds for all t\ge {t}_{0}, then the system (1) with (2) is (uniformly, asymptotically, uniformly asymptotically, etc.) pmoment stable if and only if the system (3) is stable correspondingly.
Proof Let y(t) be any solution of the system (1) with (2). Similar to the proof of Theorem 3.1, we obtain that
\mathbb{E}{y(t)}^{p}={e}^{\lambda {z}_{0}}\cdot \sum _{k=0}^{+\mathrm{\infty}}(\prod _{i=1}^{k}{1+{b}_{i}}^{p}\cdot \sum _{n={m}_{1}+{m}_{2}+\cdots +{m}_{k}}^{{m}_{1}+{m}_{2}+\cdots +{m}_{k+1}1}\frac{{(\lambda {z}_{0})}^{n}}{n!})\cdot {x(t)}^{p}.
By assumption, we obtain that
\mathbb{E}{y(t)}^{p}\le \alpha {x(t)}^{p}.
(8)
So, if the trivial solution of the system (3) is stable, then for any \u03f5>0, there exists a \delta >0 such that
{x}_{0}<\sqrt[p]{\delta}\phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}x(t)<\sqrt[p]{\frac{\u03f5}{\alpha}}\phantom{\rule{1em}{0ex}}\text{for all}t\ge {t}_{0}.
From {x}_{0}={y}_{0} and (8), we obtain that
{{y}_{0}}^{p}<\delta \phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}\mathbb{E}{y(t)}^{p}<\u03f5\phantom{\rule{1em}{0ex}}\text{for all}t\ge {t}_{0},
which means that the trivial solution of the system (1) with (2) is pmoment stable.
The remaining proof is similar to the proof above, so we omit it. The proof is complete. □
Remark 3.2 If {b}_{k}({\tau}_{k})\equiv {b}_{k} is finite for all k=1,2,\dots , when the condition (H) degenerates to the condition (C) in [11], Theorem 3.3 degenerates to Theorem 3 in [11].