Oscillation of a class of second-order linear impulsive differential equations

Abstract

In this paper, we investigate the oscillation of a class of second-order linear impulsive differential equations of the form

$$\{\begin{array}{c}{(a(t)[x^{\prime }(t)+\lambda x(t)])}^{\prime }+p(t)x(t)=0,t\ge t_0,t\ne t_k,\hfill \\ x(t_k^{+})=b_{k}x(t_k),x^{\prime }(t_k^{+})=c_k{x}^{\prime }(t_k),k=1,2,\dots .\hfill \end{array}$$

By using the equivalence transformation and the associated Riccati techniques, some interesting results are obtained.

MSC:34A37, 34C10.

1 Introduction

Impulsive differential equations are recognized as adequate mathematical models for studying evolution processes that are subject to abrupt changes in their states at certain moments. Many applications in physics, biology, control theory, economics, applied sciences and engineering exhibit impulse effects (see [1–4]). In recent years, the study of the oscillation of all solutions of impulsive differential equations have been the subject of many research works. See, for example, [5–11] and the references cited therein.

In this article, we consider the second-order linear impulsive differential equation of the form

$$\{\begin{array}{c}{(a(t)[x^{\prime }(t)+\lambda x(t)])}^{\prime }+p(t)x(t)=0,t\ge t_0,t\ne t_k,\hfill \\ x(t_k^{+})=b_{k}x(t_k),x^{\prime }(t_k^{+})=c_k{x}^{\prime }(t_k),k=1,2,\dots ,\hfill \end{array}$$

(1.1)

where $0\le t_0<t_1<\cdots <t_k\to \mathrm{\infty }$, ${lim}_{k\to \mathrm{\infty }}{t}_k=+\mathrm{\infty }$, $a(t)\in C([t_0,\mathrm{\infty }),(0,\mathrm{\infty }))$ and $p(t)\in C([t_0,\mathrm{\infty }),\mathbb{R})$, $\{b_k\}$ and $\{c_k\}$ are two known sequences of positive real numbers, λ is a real number, and

$$\begin{array}{c}{x}^{\prime }(t_k)=x^{\prime }\left(t_k^{-}\right)=\underset{h\to 0^{-}}{lim}\frac{x(t_k+h)-x(t_k)}{h},\hfill \\ x^{\prime }\left(t_k^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{x(t_k+h)-x(t_k^{+})}{h}.\hfill \end{array}$$

Let $J\subset \mathbb{R}$ be an interval and $PC(J,\mathbb{R})=\{x:J\to \mathbb{R}:x(t)\text{ be continuous everywhere except}\text{at some }{t}_k\text{ at which }x(t_k^{+})\text{ and }x(t_k^{-})\text{ exist and }x(t_k^{-})=x(t_k)\}$.

A function $x\in PC([t_0,\mathrm{\infty }),\mathbb{R})$ is said to be a solution of Eq. (1.1) if

• (i) $x(t)$ satisfies ${(a(t)[x^{\prime }(t)+\lambda x(t)])}^{\prime }+p(t)x(t)=0$ for $t\in [t_0,\mathrm{\infty })$ and $t\ne t_k$,

• (ii) $x(t_k^{+})=b_{k}x(t_k)$, $x^{\prime }(t_k^{+})=c_k{x}^{\prime }(t_k)$ for each $t_k$, and $x(t)$ and $x^{\prime }(t)$ are left continuous for each $t_k$, $k=1,2,\dots$ .

Definition 1.1 A nontrivial solution of Eq. (1.1) is said to be nonoscillatory if the solution is eventually positive or eventually negative. Otherwise, it is said to be oscillatory. Eq. (1.1) is said to be oscillatory if all solutions are oscillatory.

If $\lambda =0$, then Eq. (1.1) reduces to the second-order linear differential equation with impulses

$$\{\begin{array}{c}{(a(t)x^{\prime }(t))}^{\prime }+p(t)x(t)=0,t\ge t_0,t\ne t_k,\hfill \\ x(t_k^{+})=b_{k}x(t_k),x^{\prime }(t_k^{+})=c_k{x}^{\prime }(t_k),k=1,2,\dots .\hfill \end{array}$$

(1.2)

In [12] Luo et al. and [13] Guo et al. gave some excellent results on the oscillation and nonoscillation of Eq. (1.2) by using associated Riccati techniques and an equivalence transformation. Moreover, Luo et al. showed that the oscillation of Eq. (1.2) can be caused by impulsive perturbations, though the corresponding equation without impulses admits a nonoscillatory solution.

If $a(t)\equiv 1$ and $\lambda \ne 0$, then Eq. (1.1) reduces to the impulsive Langevin equation of the form

$$\{\begin{array}{c}\frac{d}{dt}(\frac{d}{dt}+\lambda )x(t)+p(t)x(t)=0,t\ge t_0,t\ne t_k,\hfill \\ x(t_k^{+})=b_{k}x(t_k),x^{\prime }(t_k^{+})=c_k{x}^{\prime }(t_k),k=1,2,\dots .\hfill \end{array}$$

(1.3)

The Langevin equation (first formulated by Langevin in 1908) is found to be an effective tool to describe the evolution of physical phenomena in fluctuating environments. For more details of the Langevin equation without impulses and its applications, we refer the reader to [14].

If $\lambda =0$ and $b_k=c_k=1$ for all $k=1,2,\dots$ , then Eq. (1.1) reduces to the self-adjoint second-order differential equation

$${(a(t)x(t))}^{\prime }+p(t)x(t)=0,t>t_0.$$

(1.4)

There are many good results on the oscillation and nonoscillation of Eq. (1.4); see, for example, [15–18].

2 Main results

Now we are in the position to establish the main result.

Lemma 2.1 If the second-order differential equation

$$\begin{array}{r}{[\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)a(t)\{y^{\prime }(t)+\lambda (1-\frac{2}{a(t)}\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k))y(t)\}]}^{\prime }\\ +\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)[p(t)+\frac{{\lambda }^2}{a(t)}{(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k))}^2\\ -{\lambda }^2\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k)]y(t)=0,t>T,\end{array}$$

(2.1)

is oscillatory, then Eq. (1.1) is oscillatory, where $d_k=c_k/b_k$, $k=1,2,\dots$ .

Proof For the sake of contradiction, suppose that Eq. (1.1) has an eventually positive solution $x(t)$. Then there exists a constant $T\ge t_0$ such that $x(t)>0$ for $t\ge T$.

We let

$$u(t)=\frac{a(t){(e^{\lambda t}x(t))}^{\prime }}{e^{\lambda t}x(t)}=\frac{a(t)(x^{\prime }(t)+\lambda x(t))}{x(t)},t\ge T.$$

Then

$$\begin{array}{rl}{u}^{\prime }(t)=& \frac{e^{\lambda t}x(t){[a(t){(e^{\lambda t}x(t))}^{\prime }]}^{\prime }-a(t){[{(e^{\lambda t}x(t))}^{\prime }]}^2}{{(e^{\lambda t}x(t))}^2}\\ =& \frac{{[a(t){(e^{\lambda t}x(t))}^{\prime }]}^{\prime }}{e^{\lambda t}x(t)}-\frac{a(t){[{(e^{\lambda t}x(t))}^{\prime }]}^2}{{(e^{\lambda t}x(t))}^2}\\ =& \frac{{(e^{\lambda t}[a(t)(x^{\prime }(t)+\lambda x(t))])}^{\prime }}{e^{\lambda t}x(t)}-\frac{a(t){[{(e^{\lambda t}x(t))}^{\prime }]}^2}{{(e^{\lambda t}x(t))}^2}\\ =& \frac{{(a(t)(x^{\prime }(t)+\lambda x(t)))}^{\prime }}{x(t)}+\lambda \frac{a(t)(x^{\prime }(t)+\lambda x(t))}{x(t)}\\ -\frac{a(t){[{(e^{\lambda t}x(t))}^{\prime }]}^2}{{(e^{\lambda t}x(t))}^2}\\ =& -p(t)+\lambda u(t)-\frac{u^2(t)}{a(t)}\end{array}$$

can be obtained. Therefore,

$$u^{\prime }(t)-\lambda u(t)+\frac{u^2(t)}{a(t)}+p(t)=0,t\ge T,t\ne t_k.$$

(2.2)

On the other hand, we have

$$u\left(t_k^{+}\right)=\frac{a(t_k^{+})(x^{\prime }(t_k^{+})+\lambda x(t_k^{+}))}{x(t_k^{+})}=\frac{a(t_k)(c_k{x}^{\prime }(t_k)+\lambda b_{k}x(t_k))}{b_{k}x(t_k)}.$$

Let $e_k=\lambda (1-d_k)a(t_k)$, then we get

$$u\left(t_k^{+}\right)=d_{k}u(t_k)+e_k,t_k\ge T,k=1,2,\dots .$$

(2.3)

Now, we define

$$v(t)=\prod _{T\le t_k<t}{d}_k^{-1}[u(t)-\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k],t>T.$$

(2.4)

Then, for $t_n>T$, we get that

$$\begin{array}{rl}v\left(t_n^{+}\right)& =\prod _{T\le t_k\le t_n}{d}_k^{-1}[u\left(t_n^{+}\right)-\sum _{T\le t_k\le t_n}\prod _{t_k<t_j\le t_n}{d}_j{e}_k]\\ =\prod _{T\le t_k\le t_n}{d}_k^{-1}[d_{n}u(t_n)+e_n-\sum _{T\le t_k<t_n}\prod _{t_k<t_j<t_n}{d}_j{d}_n{e}_k-e_n]\\ =\prod _{T\le t_k<t_n}{d}_k^{-1}[u(t_n)-\sum _{T\le t_k<t_n}\prod _{t_k<t_j<t_n}{d}_j{e}_k]\\ =v(t_n).\end{array}$$

Therefore, $v(t)$ is continuous on $(T,\mathrm{\infty })$.

We have

$$\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)\times \left(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k\right)=\sum _{T\le t_k<t}\prod _{T\le t_j\le t_k}{d}_j^{-1}{e}_k.$$

Then, for $t>T$, $t\ne t_n$ and from (2.2), we get that

$$\begin{array}{rl}{v}^{\prime }(t)=& \underset{h\to 0}{lim}(\prod _{T\le t_k<t+h}{d}_k^{-1}[u(t+h)-\sum _{T\le t_k<t+h}\prod _{t_k<t_j<t+h}{d}_j{e}_k]\\ -\prod _{T\le t_k<t}{d}_k^{-1}[u(t)-\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k])/h\\ =& \underset{h\to 0}{lim}\frac{{\prod }_{T\le t_k<t+h}{d}_k^{-1}u(t+h)-{\prod }_{T\le t_k<t}{d}_k^{-1}u(t)}{h}\\ =& \left(\prod _{T\le t_k<t}{d}_k^{-1}\right)\underset{h\to 0}{lim}\frac{u(t+h)-u(t)}{h}\\ =& \left(\prod _{T\le t_k<t}{d}_k^{-1}\right)u^{\prime }(t)=\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)[\lambda u(t)-\frac{u^2(t)}{a(t)}-p(t)]\\ =& \left(\prod _{T\le t_k<t}{d}_k^{-1}\right)\{\lambda [v(t)\left(\prod _{T\le t_k<t}{d}_k\right)+\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k]\\ -\frac{1}{a(t)}{[v(t)\left(\prod _{T\le t_k<t}{d}_k\right)+\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k]}^2-p(t)\}\\ =& \lambda v(t)+\lambda \sum _{T\le t_k<t}\prod _{T\le t_j\le t_k}{d}_j^{-1}{e}_k-\left(\prod _{T\le t_k<t}{d}_k\right)\frac{v^2(t)}{a(t)}\\ -2\frac{v(t)}{a(t)}\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k-\frac{1}{a(t)}\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)\\ \times {\left(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k\right)}^2-\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)p(t).\end{array}$$

The left-hand and the right-hand derivatives of $v(t)$ at $t=t_n$ are given by

$$\begin{array}{c}{v}^{\prime }\left(t_n^{-}\right)=\left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)u^{\prime }\left(t_n^{-}\right)=\left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)u^{\prime }(t_n),\hfill \\ v^{\prime }\left(t_n^{+}\right)=\left(\prod _{T\le t_k\le t_n}{d}_k^{-1}\right)u^{\prime }\left(t_n^{+}\right)=\left(\prod _{T\le t_k\le t_n}{d}_k^{-1}\right)d_n{u}^{\prime }(t_n)=\left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)u^{\prime }(t_n).\hfill \end{array}$$

Hence, for $t=t_n$, we have

$$\begin{array}{rcl}{v}^{\prime }(t_n)& =& \left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)u^{\prime }(t_n)=\left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)\underset{t\to t_n^{-}}{lim}{u}^{\prime }(t)\\ =& \left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)\underset{t\to t_n^{-}}{lim}[\lambda u(t)-\frac{u^2(t)}{a(t)}-p(t)]\\ =& \left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)[\lambda u(t_n)-\frac{u^2(t_n)}{a(t_n)}-p(t_n)]\\ =& \left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)\{\lambda [v(t_n)\left(\prod _{T\le t_k<t_n}{d}_k\right)+\sum _{T\le t_k<t_n}\prod _{t_k<t_j<t_n}{d}_j{e}_k]\\ -\frac{1}{a(t_n)}{[v(t_n)\left(\prod _{T\le t_k<t_n}{d}_k\right)+\sum _{T\le t_k<t_n}\prod _{t_k<t_j<t_n}{d}_j{e}_k]}^2-p(t_n)\}\\ =& \lambda v(t_n)+\lambda \sum _{T\le t_k<t_n}\prod _{T\le t_j\le t_k}{d}_j^{-1}{e}_k-\left(\prod _{T\le t_k<t_n}{d}_k\right)\frac{v^2(t_n)}{a(t_n)}\\ -2\frac{v(t_n)}{a(t_n)}\sum _{T\le t_k<t_n}\prod _{t_k<t_j<t_n}{d}_j{e}_k-\frac{1}{a(t_n)}\left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)\\ \times {\left(\sum _{T\le t_k<t_n}\prod _{t_k<t_j<t_n}{d}_j{e}_k\right)}^2-\left(\prod _{T\le t_k<t_n}{d}_k^{-1}\right)p(t_n).\end{array}$$

Thus,

$$\begin{array}{r}{v}^{\prime }(t)+\left(\prod _{T\le t_k<t}{d}_k\right)\frac{v^2(t)}{a(t)}-\lambda v(t)+2\frac{v(t)}{a(t)}\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k\\ -\lambda \sum _{T\le t_k<t}\prod _{T\le t_j\le t_k}{d}_j^{-1}{e}_k+\frac{1}{a(t)}\left(\prod _{T\le t_k<t}{d}_k^{-1}\right){\left(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k\right)}^2\\ +\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)p(t)=0,t>T.\end{array}$$

(2.5)

We set

$$w(t)=exp\left\{{\int }_T^t[\left(\prod _{T\le t_k<s}{d}_k\right)\frac{v(s)}{a(s)}-\lambda +\frac{2}{a(s)}\sum _{T\le t_k<s}\prod _{t_k<t_j<s}{d}_j{e}_k]ds\right\},t>T.$$

Then, $w(t)>0$ for $t>T$ and

$$w^{\prime }(t)=w(t)[\left(\prod _{T\le t_k<t}{d}_k\right)\frac{v(t)}{a(t)}-\lambda +\frac{2}{a(t)}\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k].$$

From (2.5), we obtain

$$\begin{array}{r}{[\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)a(t)w^{\prime }(t)+\lambda \left(\prod _{T\le t_k<t}{d}_k^{-1}\right)a(t)w(t)-2\sum _{T\le t_k<t}\prod _{T\le t_j\le t_k}{d}_j^{-1}{e}_{k}w(t)]}^{\prime }\\ =w(t)v^{\prime }(t)+v(t)w^{\prime }(t)\\ =w(t)[v^{\prime }(t)+\left(\prod _{T\le t_k<t}{d}_k\right)\frac{v^2(t)}{a(t)}-\lambda v(t)+2\frac{v(t)}{a(t)}\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k]\\ =w(t)[\lambda \sum _{T\le t_k<t}\prod _{T\le t_j\le t_k}{d}_j^{-1}{e}_k-\frac{1}{a(t)}\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)\\ \times {\left(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j{e}_k\right)}^2-\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)p(t)].\end{array}$$

Therefore,

$$\begin{array}{r}{[\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)a(t)\{w^{\prime }(t)+\lambda (1-\frac{2}{a(t)}\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k))w(t)\}]}^{\prime }\\ +\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)[p(t)+\frac{{\lambda }^2}{a(t)}{(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k))}^2\\ -{\lambda }^2\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k)]w(t)=0,t>T.\end{array}$$

This implies that $w(t)$ is an eventually positive solution of Eq. (2.1) which is a contradiction. A similar argument can be used to prove that Eq. (2.1) cannot have an eventually negative solution. Therefore, from Definition 1.1, the solution of Eq. (2.1) is oscillatory. The proof is complete. □

Lemma 2.2 (Leighton type oscillation criteria)

Assume that the functions $g(t),q(t)\in PC([t_0,\mathrm{\infty }),\mathbb{R})$ and $h(t)\in PC([t_0,\mathrm{\infty }),(0,\mathrm{\infty }))$.

If

$${\int }_T^{\mathrm{\infty }}g(s)e^{-{\int }_T^{s}q(\sigma )d\sigma }ds=\mathrm{\infty }\mathit{\text{and}}{\int }_T^{\mathrm{\infty }}\frac{1}{h(s)}{e}^{{\int }_T^{s}q(\sigma )d\sigma }ds=\mathrm{\infty },$$

then

$${(h(t)[y^{\prime }(t)+q(t)y(t)])}^{\prime }+g(t)y(t)=0$$

(2.6)

is oscillatory.

Proof Let $y(t)$ be a nonoscillatory solution of the Eq. (2.6). Without loss of generality, we assume that there exists a $T\ge t_0$ such that $y(t)>0$ for $t\ge T$. We define

$$u(t)=\frac{h(t){(e^{{\int }_T^{t}q(\sigma )d\sigma }y(t))}^{\prime }}{e^{{\int }_T^{t}q(\sigma )d\sigma }y(t)},t\ge T.$$

Then the equation

$$u^{\prime }(t)-q(t)u(t)+\frac{u^2(t)}{h(t)}+g(t)=0$$

(2.7)

has a solution $u(t)$ on $[T,\mathrm{\infty })$. It is easy to see that the solution of Eq. (2.7) satisfies the following equation:

$$\begin{array}{rl}u(t)=& e^{{\int }_T^{t}q(\sigma )d\sigma }u(T)-e^{{\int }_T^{t}q(\sigma )d\sigma }{\int }_T^t\frac{u^2(s)}{h(s)}{e}^{-{\int }_T^{s}q(\sigma )d\sigma }ds\\ -e^{{\int }_T^{t}q(\sigma )d\sigma }{\int }_T^{t}g(s)e^{-{\int }_T^{s}q(\sigma )d\sigma }ds.\end{array}$$

(2.8)

Since ${\int }_T^{\mathrm{\infty }}g(s)e^{-{\int }_T^{s}q(\sigma )d\sigma }ds=\mathrm{\infty }$, then there exists $\tau >T$ such that

$$u(T)-{\int }_T^{t}g(s)e^{-{\int }_T^{s}q(\sigma )d\sigma }ds<0$$

for all t in $[\tau ,\mathrm{\infty })$. Hence, from (2.8), it follows that

$$u(t)<-e^{{\int }_T^{t}q(s)ds}{\int }_T^t\frac{u^2(s)}{h(s)}{e}^{-{\int }_T^{s}q(\sigma )d\sigma }ds,t\in [\tau ,\mathrm{\infty }).$$

Let

$$r(t)={\int }_T^t\frac{u^2(s)}{h(s)}{e}^{-{\int }_T^{s}q(\sigma )d\sigma }ds,t\in [\tau ,\mathrm{\infty }).$$

Then $u(t)<-r(t)e^{{\int }_T^{t}q(\sigma )d\sigma }$ and

$$r^{\prime }(t)=\frac{u^2(t)}{h(t)}{e}^{-{\int }_T^{t}q(\sigma )d\sigma }>\frac{r^2(t)}{h(t)}{e}^{{\int }_T^{t}q(\sigma )d\sigma },t\in [\tau ,\mathrm{\infty }).$$

(2.9)

Integrating (2.9) from $\tau >T$ to ∞, we obtain

$$-\frac{1}{r(\mathrm{\infty })}+\frac{1}{r(\tau )}>{\int }_{\tau }^{\mathrm{\infty }}\frac{1}{h(s)}{e}^{{\int }_T^{s}q(\sigma )d\sigma }ds.$$

Hence,

$${\int }_{\tau }^{\mathrm{\infty }}\frac{1}{h(s)}{e}^{{\int }_T^{s}q(\sigma )d\sigma }ds<\frac{1}{r(\tau )}<\mathrm{\infty },$$

which is a contradiction. Thus, the solution $y(t)$ is oscillatory. The proof is complete. □

Theorem 2.3 Assume that

$$\begin{array}{r}{\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)[p(t)+\frac{{\lambda }^2}{a(t)}{(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k))}^2\\ -{\lambda }^2\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k)]exp\{-\lambda {\int }_T^t(1-\frac{2}{a(s)}\\ \times \sum _{T\le t_k<s}\prod _{t_k<t_j<s}{d}_j(1-d_k)a(t_k)\left)ds\right\}dt=\mathrm{\infty }\end{array}$$

(2.10)

and

$$\begin{array}{r}{\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<t}{d}_k\right)\frac{1}{a(t)}\\ \times exp\left\{\lambda {\int }_T^t(1-\frac{2}{a(s)}\times \sum _{T\le t_k<s}\prod _{t_k<t_j<s}{d}_j(1-d_k)a(t_k))ds\right\}dt=\mathrm{\infty },\end{array}$$

(2.11)

where $d_k=c_k/b_k$, $k=1,2,\dots$ . Then Eq. (1.1) is oscillatory.

If $b_k=c_k$, $k=1,2,\dots$ , then $d_k=1$, $k=1,2,\dots$ and (1.1) becomes

$$\{\begin{array}{c}{(a(t)[x^{\prime }(t)+\lambda x(t)])}^{\prime }+p(t)x(t)=0,t\ge t_0,t\ne t_k,\hfill \\ x(t_k^{+})=b_{k}x(t_k),x^{\prime }(t_k^{+})=b_k{x}^{\prime }(t_k),k=1,2,\dots .\hfill \end{array}$$

(2.12)

Theorem 2.4 Eq. (2.12) is oscillatory if and only if

$${(a(t)[y^{\prime }(t)+\lambda y(t)])}^{\prime }+p(t)y(t)=0,t\ge t_0,$$

(2.13)

is oscillatory.

Proof From Lemma 2.1, we only need to prove that if Eq. (2.12) is oscillatory, then Eq. (2.13) is oscillatory.

Without loss of generality, we suppose that $y(t)$ is an eventually positive solution of (2.13) such that $y(t)>0$ for $t\ge T\ge t_0$. Set

$$x(t)=\left(\prod _{T\le t_k<t}{b}_k\right)y(t),t>T.$$

Then, for $t>T$, we have $x(t)>0$, and for $t_n>T$,

$$x\left(t_n^{+}\right)=\left(\prod _{T\le t_k\le t_n}{b}_k\right)y\left(t_n^{+}\right)=b_n\left(\prod _{T\le t_k<t_n}{b}_k\right)y(t_n)=b_{n}x(t_n).$$

Moreover, for $t\ne t_n>T$, we have

$$x^{\prime }(t)=\left(\prod _{T\le t_k<t}{b}_k\right)y^{\prime }(t),$$

and

$$x^{\prime }\left(t_n^{+}\right)=\left(\prod _{T\le t_k\le t_n}{b}_k\right)y^{\prime }\left(t_n^{+}\right)=b_n\left(\prod _{T\le t_k<t_n}{b}_k\right)y^{\prime }(t_n)=b_n{x}^{\prime }(t_n)=c_n{x}^{\prime }(t_n).$$

Now we have for $t\ne t_n$

$$\begin{array}{rl}{(a(t)[x^{\prime }(t)+\lambda x(t)])}^{\prime }& ={(a(t)[\left(\prod _{T\le t_k<t}{b}_k\right)y^{\prime }(t)+\lambda \left(\prod _{T\le t_k<t}{b}_k\right)y(t)])}^{\prime }\\ =\left(\prod _{T\le t_k<t}{b}_k\right){(a(t)[y^{\prime }(t)+\lambda y(t)])}^{\prime }\\ =-\left(\prod _{T\le t_k<t}{b}_k\right)p(t)y(t)=-p(t)x(t).\end{array}$$

Therefore,

$${(a(t)[x^{\prime }(t)+\lambda x(t)])}^{\prime }+p(t)x(t)=0,t\ne t_n,t>T.$$

We get that $x(t)$ is an eventually positive solution of (2.12), a contradiction, and so the proof is complete. □

Corollary 2.5 Assume that

$${\int }_T^{\mathrm{\infty }}p(t)e^{-\lambda t}dt=\mathrm{\infty }$$

(2.14)

and

$${\int }_T^{\mathrm{\infty }}\frac{e^{\lambda t}}{a(t)}dt=\mathrm{\infty }.$$

(2.15)

Then Eq. (2.12) is oscillatory.

3 Some examples

In this section, we illustrate our results with two examples.

Example 3.1 Consider the following impulsive Langevin equation:

$$\{\begin{array}{c}\frac{d}{dt}(\frac{d}{dt}+\frac{2}{3})x(t)+5^{t^3}x(t)=0,t>0,t\ne k,\hfill \\ x(k^{+})=\frac{k}{k+1}x(k),x^{\prime }(k^{+})=x^{\prime }(k),k=1,2,\dots .\hfill \end{array}$$

(3.1)

Set $d_k=\frac{k+1}{k}$, $\lambda =\frac{2}{3}$, $a(t)=a(t_k)\equiv 1$ and $p(t)=5^{t^3}$. If $T\in (m,m+1]$ for some integer $m\ge 0$, then we get

$$\left(\prod _{T\le t_k<[t]+1}\frac{k}{k+1}\right)=\frac{m+1}{m+2}\cdot \frac{m+2}{m+3}\cdots \frac{[t]}{[t]+1}=\frac{m+1}{[t]+1},$$

and

$$\begin{array}{r}\sum _{T\le t_k<[t]}\prod _{t_k<t_j<[t]}\left(\frac{j+1}{j}\right)\left(\frac{1}{k}\right)\\ =\frac{1}{m+1}\cdot \frac{m+3}{m+2}\cdot \frac{m+4}{m+3}\cdots \frac{[t]}{[t]-1}\\ +\frac{1}{m+2}\cdot \frac{m+4}{m+3}\cdot \frac{m+5}{m+4}\cdots \frac{[t]}{[t]-1}+\cdots +\frac{1}{[t]-2}\cdot \frac{[t]}{[t]-1}+\frac{1}{[t]-1}\\ =[t](\frac{1}{m+1}\cdot \frac{1}{m+2}+\frac{1}{m+2}\cdot \frac{1}{m+3}+\cdots +\frac{1}{[t]-2}\cdot \frac{1}{[t]-1})+\frac{1}{[t]-1}\\ =[t](\frac{1}{m+1}-\frac{1}{[t]-1})+\frac{1}{[t]-1}\\ =\frac{[t]}{m+1}-1,\end{array}$$

where $[\cdot ]$ denotes the greatest integer function.

Hence,

$$\begin{array}{r}{\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<t}{d}_k^{-1}\right)[p(t)+\frac{{\lambda }^2}{a(t)}{(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k))}^2\\ -{\lambda }^2\sum _{T\le t_k<t}\prod _{t_k<t_j<t}{d}_j(1-d_k)a(t_k)]exp\{-\lambda {\int }_T^t(1-\frac{2}{a(s)}\\ \times \sum _{T\le t_k<s}\prod _{t_k<t_j<s}{d}_j(1-d_k)a(t_k)\left)ds\right\}dt\\ ={\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<t}\frac{k}{k+1}\right)[5^{t^3}+{\left(\frac{2}{3}\right)}^2{\left(\sum _{T\le t_k<t}\prod _{t_k<t_j<t}\left(\frac{j+1}{j}\right)(-\frac{1}{k})\right)}^2\\ -{\left(\frac{2}{3}\right)}^2\sum _{T\le t_k<t}\prod _{t_k<t_j<t}\left(\frac{j+1}{j}\right)(-\frac{1}{k})]exp\{-\frac{2}{3}{\int }_T^t(1-2\\ \times \sum _{T\le t_k<s}\prod _{t_k<t_j<s}\left(\frac{j+1}{j}\right)(-\frac{1}{k})\left)ds\right\}dt\\ \ge {\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<[t]+1}\frac{k}{k+1}\right)[5^{t^3}+{\left(\frac{2}{3}\right)}^2\left(\sum _{T\le t_k<[t]}\prod _{t_k<t_j<[t]}\left(\frac{j+1}{j}\right)\left(\frac{1}{k}\right)\right)\\ \times (\sum _{T\le t_k<[t]}\prod _{t_k<t_j<[t]}\left(\frac{j+1}{j}\right)\left(\frac{1}{k}\right)+1)]exp\{-\frac{2}{3}{\int }_T^t(1+2\\ \times \sum _{T\le t_k<[s]+1}\prod _{t_k<t_j<[s]+1}\left(\frac{j+1}{j}\right)\left(\frac{1}{k}\right)\left)ds\right\}dt\\ \ge {\int }_T^{\mathrm{\infty }}\left(\frac{m+1}{[t]+1}\right)[5^{t^3}+{\left(\frac{2}{3}\right)}^2(\frac{[t]}{m+1}-1)\\ \times \left(\frac{[t]}{m+1}\right)]exp\{-\frac{2}{3}{\int }_T^t(1+2(\frac{[s]+1}{m+1}-1))ds\}dt\\ \ge {\int }_T^{\mathrm{\infty }}\left(\frac{m+1}{t+1}\right)\left(5^{t^3}\right)exp\{-\frac{2}{3}{\int }_T^t(1+2\left(\frac{s+1}{T}\right))ds\}dt\\ =\mathrm{\infty },\end{array}$$

and

$$\begin{array}{r}{\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<t}{d}_k\right)\frac{1}{a(t)}exp\left\{\lambda {\int }_T^t(1-\frac{2}{a(s)}\sum _{T\le t_k<s}\prod _{t_k<t_j<s}{d}_j(1-d_k)a(t_k))ds\right\}dt\\ ={\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<t}\frac{k+1}{k}\right)\\ \times exp\left\{\frac{2}{3}{\int }_T^t(1+2\sum _{T\le t_k<s}\prod _{t_k<t_j<s}\frac{j+1}{j}\cdot \frac{1}{k})ds\right\}dt\\ \ge {\int }_T^{\mathrm{\infty }}\left(\prod _{T\le t_k<[t]}\frac{k+1}{k}\right)exp\{\frac{2}{3}{\int }_T^t(1)ds\}dt\\ ={\int }_T^{\mathrm{\infty }}(\frac{m+2}{m+1}\cdot \frac{m+3}{m+2}\cdots \frac{[t]}{[t]-1})exp\{\frac{2}{3}{\int }_T^t(1)ds\}dt\\ ={\int }_T^{\mathrm{\infty }}\left(\frac{[t]}{m+1}\right)exp\{\frac{2}{3}(t-T)\}dt\\ \ge {\int }_T^{\mathrm{\infty }}\left(\frac{t-1}{m+1}\right)exp\{\frac{2}{3}(t-T)\}dt\\ =\mathrm{\infty }.\end{array}$$

By Theorem 2.3, Eq. (3.1) is oscillatory.

Example 3.2 Consider the equation

$$\{\begin{array}{c}{(\frac{1}{{(t^2+1)}^{\frac{1}{2}}}[x^{\prime }(t)+\frac{3}{4}x(t)])}^{\prime }+{\pi }^{t}x(t)=0,t\ge 0,t\ne t_k,\hfill \\ x(t_k^{+})=b_{k}x(t_k),x^{\prime }(t_k^{+})=b_k{x}^{\prime }(t_k),k=1,2,\dots ,\hfill \end{array}$$

(3.2)

where $\{b_k\}$ is a known sequence of positive real numbers. It is easy to see that

$${\int }_0^{\mathrm{\infty }}{\pi }^t{e}^{-\frac{3}{4}t}dt=\mathrm{\infty }$$

and

$${\int }_0^{\mathrm{\infty }}{(t^2+1)}^{\frac{1}{2}}{e}^{\frac{3}{4}t}dt=\mathrm{\infty }.$$

By Corollary 2.5, Eq. (3.2) is oscillatory.