许多自然科学和社会科学中的实际问题可以由偏微分方程来描述,例如固体的颤动,人口的增长,疾病的传播,电磁波的放射等.对这些现象的研究,人们总假设状态的变化和系统参数的改变是连续的.但是,我们容易想到某些突然出现的自然现象,比如电路系统中开关的闭合、机械钟摆的运动、电流中的脉冲、种群生态系统的定时捕捞或补给、药剂学中的定时给药的过程、地震、海啸及烈性传染病SARS等灾难的出现使得所述的系统产生了跳跃,也就是出现了脉冲.脉冲现象在现代科技各领域的实际问题中是普遍存在的,其数学模型往往可归结为脉冲微分系统.脉冲微分系统最突出的特点是能够充分考虑到瞬时突变现象对状态的影响,能更深刻、更精确地反映事物的变化规律.近年来,人们发现这类系统在航天技术、信息科学、控制系统、通讯、生命科学、医学,经济领域均有重要应用.大量的学者对脉冲微分方程的定性理论(如脉冲常微分方程或偏微分方程解的存在性、唯一性、稳定性及振动性等方面)产生了极大的兴趣,取得了一系列丰硕的成果^{[1, 2]}.

另一方面,振动理论是微分方程定性理论的一个重要的分支,也是近年来定性理论研究中一个比较活跃的方向.在自然界、工业领域、日常生活及社会生活中,都会碰到各种形式的振动现象,如机械的振动,无线电中的电磁震荡等.广泛的应用背景使得微分方程振动理论的研究在近三十年取得了丰硕的成果^{[3, 4]},但具有时滞和脉冲的偏微分方程研究却进展缓慢,这主要是由于脉冲和时滞同时出现,使得系统呈现出新的特征,给研究带来了困难.因此关于脉冲时滞偏微分系统的定性理论之一的振动性问题有待进一步的探讨,需建立其特有的理论体系.

本文研究如下非线性脉冲时滞双曲型方程组

$\begin{align} & \frac{{{\partial }^{2}}}{\partial {{t}^{2}}}[{{u}_{i}}(x,t)+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t){{u}_{i}}(x,t-{{\mu }_{s}})]+\sum\limits_{r=1}^{n}{{{b}_{ir}}}(x,t){{f}_{ir}}({{u}_{r}}(x,t-\sigma))\\ &={{a}_{i}}(t){{h}_{i}}({{u}_{i}}(x,t))\Delta {{u}_{i}}(x,t)+\sum\limits_{j=1}^{m}{{{a}_{ij}}}(t){{h}_{ij}}({{u}_{i}}(x,t-{{\tau }_{j}}))\Delta {{u}_{i}}(x,t-{{\tau }_{j}}),\ \\ & \ t\ne {{t}_{k}},(x,t)\in \Omega \times {{\mathbb{R}}^{+}}=G,\\ \end{align}$

(1.1)

$\begin{equation}\label{eq:a2} u_i(x,t^+_k)=p_{ki}(x,t_k,u_i(x,t_k)),\ \ k=1,2,\cdots, \end{equation}$

(1.2)

$\begin{equation}\label{eq:a3} u_{it}(x,t^+_k)=q_{ki}(x,t_k,u_{it}(x,t_k)),\ \ k=1,2,\cdots \end{equation}$

(1.3)

在两类边界条件

$\frac{\partial {{u}_{i}}(x,t)}{\partial \nu }+{{\beta }_{i}}(x,t){{u}_{i}}(x,t)=0,\;\;(x,t)\in \partial \Omega \times {{\mathbb{R}}^{+}},$

(1.4)

${{u}_{i}}(x,t)=0,\;\;(x,t)\in \partial \Omega \times {{\mathbb{R}}^{+}},\ i\in {{I}_{n}}=\{1,2,\cdots,n\}$

(1.5)

及初始条件

$\begin{equation}\label{eq:a6} u_i(x,t)=\Phi_i(x,t),\ \frac{\partial u_i(x,t)}{\partial t}=\Psi_i(x,t),(x,t)\in\Omega\times[-\delta,0] \end{equation}$

(1.6)

下的振动性.其中$\ \Omega\subset {\Bbb R}^N$表示具有逐片光滑边界$\partial\Omega$的有界区域,$\nu$表示边界$\partial\Omega$的单位外法向量,$u_{it}(x,t)=\frac{\partial u_i(x,t)}{\partial t},\ {\Bbb R}^+=[0,+\infty),$ $ \beta_i(x,t)\in PC(\partial\Omega\times {\Bbb R}^+,{\Bbb R}^+),PC$表示以$t=t_k$ $(k\in I_\infty)$为第一类间断点且在该处左连续的分段连续函数类.$\delta=\max\{\mu_s,\: \sigma,\ \tau_j\},\ \Phi_i(x,t)\in C^2(\Omega\times[-\delta,0],{\Bbb R}),\ \Psi_i(x,t)\in C^1(\Omega\times[-\delta,0],{\Bbb R}),\ s\in I_l,j\in I_m,i\in I_n$.

我们假设下列条件均成立:

${\rm \widetilde{H}}_1)$ $0<t_1<t_2<\cdots <t_k<\cdots,\lim\limits_{k\rightarrow\infty}t_k=\infty.\ \mu_s,\ \sigma,\tau_j$均为非负常数,$s\in I_l,j\in I_m.$

${\rm \widetilde{H}}_2)$ $a_i(t),a_{ij}(t)\in PC({\Bbb R}^+,{\Bbb R}^+),b_{ir}(x,t)\in PC(\overline{\Omega}\times{\Bbb R}^+,{\Bbb R}),$ $b_{ii}(x,t)>0,$ $ c_i(t)\in C^2({\Bbb R}^+,$ ${\Bbb R}^+),$ $ \sum\limits^l_{s=1}c_s(t)<1,r\in I_n,j\in I_m,i\in I_n.$

${\rm \widetilde{H}}_3)$ $f_{ir}(u_r)\in PC({\Bbb R},{\Bbb R})$,当$u_r\neq0$时,$f_{ir}(u_r)/u_r\geq C_{ir},C_{ir}$为正常数;$f_{ir}(-u_r)=-f_{ir}(u_r),$ $r\in I_n,i\in I_n.$

${\rm \widetilde{H}}_4)$ $h_i(u_i),h_{ij}(u_i)\in C^1({\Bbb R},{\Bbb R})$,当$u\neq 0$时,$ h_i(u_i)> 0,$ $h_{ij}(u_i)> 0;$ $ u_ih'_i(u_i)\geq 0,$ $ u_ih'_{ij}(u_i)\geq 0,$ $ h_i(0)=0,\ h_{ij}(0)=0,$ $ j\in I_m,i\in I_n.$

${\rm \widetilde{H}}_5)$ $p_{ki}(x,t_k,u_i(x,t_k)),q_{ki}(x,t_k,u_{it}(x,t_k))\in PC(\bar{\Omega}\times{\Bbb R}^+\times{\Bbb R},{\Bbb R}),$存在正常数$\overline{a}_{ki},$ $\underline{a}_{ki},$ $ \overline{b}_{ki},$ $ \underline{b}_{ki},$使得$\xi\neq 0,\eta\neq 0,k\in I_{\infty},$

$$ \min\limits_{i\in I_n}\{\underline{a}_{ki}\}=\underline{a}_k\leq\underline{a}_{ki}\leq\frac{p_{ki}(x,t_k,\xi)}{\xi}\leq\overline{a}_{ki}\leq \overline{a}_k=\max\limits_{i\in I_n}\{\overline{a}_{ki}\}, $$ $$ \min\limits_{i\in I_n}\{\underline{b}_{ki} \}=\underline{b}_k\leq\underline{b}_{ki}\leq\frac{q_{ki}(x,t_k,\eta)}{\eta}\leq\overline{b}_{ki}\leq \overline{b}_k=\max\limits_{i\in I_n}\{\overline{b}_{ki}\}, $$

其中$\overline{b}_k\leq\underline{a}_k$.

构造序列$\{\bar{t}_k\}=\{t_k\}\cup\{t_{ks}\}\cup\{t_{k\sigma}\}\cup\{t_{kj}\}$,其中$t_{ks}=t_k+\mu_s,\ t_{k\sigma}=t_k+\sigma,\ t_{kj}=t_k+\tau_j$,且$\bar{t}_k<\bar{t}_{k+1},\ s\in I_l,\ j\in I_m,\ k\in I_\infty.$

引入以下记号

$$ b_{ii}(t)=\min\limits_{x\in\overline\Omega}{b_{ii}(x,t)},\:\:\: \overline{b_{ir}(t)}=\max\limits_{x\in\overline\Omega}{|b_{ir}(x,t)|}, $$ $$ B(t)=\min\limits_{i\in I_n}\bigg\{b_{ii}(t)C_{ii}-\sum\limits_{r\neq i} \overline{b_{ri}(t)}C_{ri}\bigg\}>0,\ r\in I_n,\ i\in I_n. $$ $$ z(t)=U(t)+\sum\limits^l_{s=1}c_s(t)U(t-\mu_s),\ U(t)=\sum\limits_{i=1}^nU_i(t), $$ $$ U_i(t)=\int_\Omega Z_i(x,t){\rm d}x,\: Z_i(x,t)=\gamma_iu_i(x,t),\ \gamma_i={\rm sgn} u_i(x,t). $$

定义1.1 称向量值函数$u(x,t)=\{u_1(x,t),\ u_2(x,t),\ \cdots,\ u_n(x,t)\}$是问题(1.1)-(1.4)或问题(1.1)-(1.3),(1.5)的解,如果以下条件成立:

1) 当$-\delta\leq t\leq0$时,$u_i(x,t)=\Phi_i(x,t),\ \frac{\partial u_i(x,t)}{\partial t}=\Psi_i(x,t)$;

2) 当$0\leq t\leq \bar{t}_1=t_1$时,$u_i(x,t)$与问题(1.1)-(1.4)或问题(1.1)-(1.3),(1.5)的解一致;

3) 当$\bar{t}_k<t\leq\bar{t}_{k+1},\ \bar{t}_k\in\{t_k\}\setminus(\{t_{ks}\}\cup\{t_{k\sigma}\}\cup\{t_{kj}\})$时,$u_i(x,t)$与问题(1.1)-(1.4)或问题(1.1)-(1.3),(1.5)的解一致;

4) 当$\bar{t}_k<t\leq\bar{t}_{k+1},\ \bar{t}_k\in\{t_{ks}\}\cup\{t_{k\sigma}\}\cup\{t_{kj}\}$时,$u_i(x,t)$满足(1.4)或(1.5)并且与下列问题的解一致

$\begin{align} & \frac{{{\partial }^{2}}}{\partial {{t}^{2}}}[{{u}_{i}}(x,{{t}^{+}})+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t){{u}_{i}}(x,{{(t-{{\mu }_{s}})}^{+}})]+\sum\limits_{r=1}^{n}{{{b}_{ir}}}(x,t){{f}_{ir}}({{u}_{r}}(x,{{(t-\sigma)}^{+}}))\\ &={{a}_{i}}(t){{h}_{i}}({{u}_{i}}(x,{{t}^{+}}))\Delta {{u}_{i}}(x,{{t}^{+}})+\sum\limits_{j=1}^{m}{{{a}_{ij}}}(t){{h}_{ij}}({{u}_{i}}(x,{{(t-{{\tau }_{j}})}^{+}}))\Delta {{u}_{i}}(x,{{(t-{{\tau }_{j}})}^{+}}),\ \\ & \ t\ne {{t}_{k}},(x,t)\in \Omega \times {{\mathbb{R}}^{+}}=G,\\ \end{align}$

$$ u_i(x,\bar{t}^+_k)=u_i(x,\bar{t}_k),\ \ u_{it}(x,\bar{t}^+_k)=u_{it}(x,\bar{t}_k), \:\bar{t}_k\in(\{t_{ks}\}\cup\{t_{k\sigma}\}\cup\{t_{kj}\})\setminus\{t_k\} $$ 或者

$$ u_i(x,\bar{t}^+_k)=p_{k_s}(x,\bar{t}_k,u_i(x,\bar{t}_k)),\ \ u_{it}(x,\bar{t}^+_k)=q_{k_s}(x,\bar{t}_k,u_{it}(x,\bar{t}_k)), $$ $$ \bar{t}_k\in(\{t_{ks}\}\cup\{t_{k\sigma}\}\cup\{t_{kj}\})\cap\{t_k\}, $$ 其中$k_s$由等式$\bar{t}_k=t_{k_s}$确定.

定义1.2 称向量值函数$u(x,t)=\{u_1(x,t),\ u_2(x,t),\ \cdots,\ u_n(x,t)\}$的非平凡分量$u_i(x,t)$ $(i\in I_n)$在区域$G$内非振动,如果它是最终正或最终负;否则称为振动的.

定义1.3 称问题(1.1)-(1.4)或问题(1.1)-(1.3),(1.5)的解向量$u(x,t)=\{u_1(x,t),$ $ u_2(x,t),$ $ \cdots,u_n(x,t)\}$在区域$G$内振动,如果至少有一个非平凡的分量是振动的.

以下将给出一个重要的引理,用来估计一阶脉冲微分不等式解的范围.我们仅给出引理的叙述,有兴趣的读者可以参考文献[1].

引理1.1 假设以下条件成立:

${\rm A}_1)$ 序列$\{t_k\}$满足$0\le {{t}_{0}}<{{t}_{1}}<{{t}_{2}}<\cdots ,\ \underset{k\to \infty }{\mathop{\lim }}\,{{t}_{k}}=\infty $;

${\rm A}_2)$ $m(t)\in PC^1({\Bbb R}^+,{\Bbb R})$在点列$\{t_k\}\:(k=1,2,\cdots)$处左连续;

${\rm A}_3)$ $k=1,2,\cdots $且$t\geq t_0$时,

$$ m'(t)\leq p(t)m(t)+q(t),\ \ t\neq t_k, $$ $$ m(t^+_k)\leq d_km(t_k)+e_k $$ 成立,其中$p(t),q(t)\in C({\Bbb R}^+,{\Bbb R}),\ d_k\geq0,\ e_k$是常数.则

$\begin{array}{l} m(t) \le m({t_0})\prod\limits_{{t_0} < {t_k} < t} {{d_k}} \exp \left( {\int_{{t_0}}^t p (s){\rm{d}}s} \right) + \int_{{t_0}}^t {\prod\limits_{s < {t_k} < t} {{d_k}} } \exp \left( {\int_s^t p (r){\rm{d}}r} \right)q(s){\rm{d}}s\\ \;\;\;\;\;\;\;\;\;\; + \;\sum\limits_{{t_0} < {t_k} < t} {\prod\limits_{{t_k} < {t_j} < t} {{d_j}} } \exp \left( {\int_{{t_k}}^t p (s){\rm{d}}s} \right){e_k}. \end{array}$

引理2.1 假设向量值函数$u(x,t)=\{u_1(x,t),\ u_2(x,t),\ \cdots,\ u_n(x,t)\}$是问题(1.1)-(1.4)在区域G上的一个非振动解,则函数$z(t)$满足以下脉冲微分不等式

$ \begin{equation}\label{eq:b1} z''(t)+\bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)B(t)z(t-\sigma)\leq0,\ \ t\neq t_k, \end{equation} $

(2.1)

$\begin{equation}\label{eq:b2} \underline{a}_k\leq\frac{z(t_k^+)}{z(t_k)}\leq\overline{a}_k,\end{equation}$

(2.2)

$ \begin{equation}\label{eq:b3} \underline{b}_k\leq\frac{z'(t_k^+)}{z'(t_k)}\leq\overline{b}_k,\ \ k\in I_{\infty}.\end{equation} $

(2.3)

证 假设$u(x,t)=\{u_1(x,t),\ u_2(x,t),\ \cdots,\ u_n(x,t)\}$是问题(1.1)-(1.4)在区域$G$上的一个非振动解.不失一般性,假定存在$\overline{t_0}>0,$使得当$(x,t)\in\Omega\times[\overline{t_0},+\infty)$时,${\rm sgn}u_i(x,t)\neq 0,i \in I_n$.

因为$Z_i(x,t)=\gamma_iu_i(x,t),\gamma_i={\rm sgn}u_i(x,t),$则$ Z_i(x,t)>0.$根据条件${\rm \widetilde{H}}_1)$,知必存在$t_0\geq \overline{t_0}$,使得当$(x,t)\in \Omega\times[t_0,+\infty)$时,$Z_i(x,t-\mu_s)>0,Z_r(x,t-\sigma)>0,Z_i(x,t-\tau_j)>0,\ s\in I_l,r\in I_n,j\in I_m,\ i\in I_n$.

当$t\geq t_0,\ t\neq t_k(k\in I_{\infty})$时,(1.1)式两边同时在区域$\Omega$上关于变量$x$积分,得

$\begin{align} & \frac{{{\text{d}}^{2}}}{\text{d}{{t}^{2}}}[\int_{\Omega }{{{Z}_{i}}}(x,t)\text{d}x+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)\int_{\Omega }{{{Z}_{i}}}(x,t-{{\mu }_{s}})\text{d}x] \\ &+\sum\limits_{r=1}^{n}{\frac{{{\gamma }_{i}}}{{{\gamma }_{r}}}}\int_{\Omega }{{{b}_{ir}}}(x,t){{f}_{ir}}({{Z}_{r}}(x,t-\sigma))\text{d}x \\ &={{a}_{i}}(t)\int_{\Omega }{{{h}_{i}}}({{u}_{i}}(x,t))\Delta {{Z}_{i}}(x,t)\text{d}x \\ &+\sum\limits_{j=1}^{m}{{{a}_{ij}}}(t)\int_{\Omega }{{{h}_{ij}}}({{u}_{i}}(x,t-{{\tau }_{j}}))\Delta {{Z}_{i}}(x,t-{{\tau }_{j}})\text{d}x,\ i\in {{I}_{n}}.\\ \end{align}$

应用Green公式、边界条件(1.4)及条件${\rm \widetilde{H}}_4)$,得

$\begin{align} & \int_{\Omega }{{{h}_{i}}}({{u}_{i}}(x,t))\Delta {{Z}_{i}}(x,t)\text{d}x=\int_{\partial \Omega }{{{h}_{i}}}({{u}_{i}})\frac{\partial {{Z}_{i}}}{\partial \nu }\text{d}S-\int_{\Omega }{\nabla }{{h}_{i}}({{u}_{i}})\cdot \nabla {{Z}_{i}}\text{d}x \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-{{\gamma }_{i}}\int_{\partial \Omega }{{{\beta }_{i}}}(x,t){{u}_{i}}{{h}_{i}}({{u}_{i}})\text{d}S-{{\gamma }_{i}}\int_{\Omega }{{{h}_{{{i}'}}}}({{u}_{i}})|\text{grad}u{{|}^{2}}\text{d}x \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \le 0,\\ \end{align}$

$$ \int_\Omega h_{ij}(u_i(x,t-\tau_{j}))\Delta Z_i(x,t-\tau_{j}){\rm d}x\leq0,\ j\in I_m,i\in I_n, $$ 这里${\rm d}S$为边界$\partial\Omega$的面积元素.由条件${\rm \widetilde{H}}_3)$,可得

$$ \int_{\Omega}b_{ir}(x,t)f_{ir}(Z_r(x,t-\sigma)){\rm d}x\geq C_{ir}\int_{\Omega}b_{ir}(x,t)Z_r(x,t-\sigma){\rm d}x,\ r\in I_n,i\in I_n. $$

因$\:U_i(t)=\int_{\Omega}Z_i(x,t){\rm d}x$,很明显$U_i(t)>0$,且

$\begin{equation}\label{eq:b4} \bigg[U_i(t)+\sum\limits^l_{s=1}c_s(t)U_i(t-\mu_s)\bigg]''+b_{ii}(t)C_{ii}U_i(t-\sigma)-\sum\limits_{r\neq i}\overline{b_{ir}(t)}C_{ir}U_r(t-\sigma)\leq0,\ i\in I_n. \end{equation}$

(2.4)

因$\:U(t)=\sum\limits_{i=1}^n U_i(t),$有$U(t)>0,t\geq t_0.$将(2.4)式中的$n$个不等式垂直相加,可得

$\begin{align} & 0\ge {{[U(t)+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)U(t-{{\mu }_{s}})]}^{\prime \prime }}+\sum\limits_{i=1}^{n}{[}{{b}_{ii}}(t){{C}_{ii}}{{U}_{i}}(t-\sigma)-\sum\limits_{r\ne i}{\overline{{{b}_{ir}}(t)}}{{C}_{ir}}{{U}_{r}}(t-\sigma)] \\ & ={{[U(t)+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)U(t-{{\mu }_{s}})]}^{\prime \prime }}+\sum\limits_{i=1}^{n}{[}{{b}_{ii}}(t){{C}_{ii}}-\sum\limits_{r\ne i}{\overline{{{b}_{ri}}(t)}}{{C}_{ri}}]{{U}_{i}}(t-\sigma)\\ & \ge {{[U(t)+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)U(t-{{\mu }_{s}})]}^{\prime \prime }}+\underset{i\in {{I}_{n}}}{\mathop{\min }}\,[{{b}_{ii}}(t){{C}_{ii}}-\sum\limits_{r\ne i}{\overline{{{b}_{ri}}(t)}}{{C}_{ri}}]\sum\limits_{i=1}^{n}{{{U}_{i}}}(t-\sigma)\\ &={{[U(t)+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)U(t-{{\mu }_{s}})]}^{\prime \prime }}+B(t)U(t-\sigma),\\ \end{align}$

即

$\begin{equation}\label{eq:b5} \bigg [U(t)+\sum\limits^l_{s=1}c_s(t)U(t-\mu_s)\bigg]''+B(t)U(t-\sigma)\leq 0,\ t\geq t_0. \end{equation}$

(2.5)

因$z(t)=U(t)+\sum\limits^l_{s=1}c_s(t)U(t-\mu_s),$则

$$ z(t)>0,\quad z(t)\geq U(t). $$

由(2.5)式知$z''(t)\leq 0$且

$\begin{equation}\label{eq:b6} z'(t)>0. \end{equation}$

(2.6)

事实上,如果(2.6)式不成立,则存在$t_1\geq t_0$,满足$z'(t_1)<0.$因为$z'(t)$单调减,则

$$ z(t)-z(t_1)=\int^t_{t_1}z'(s){\rm d}s\leq \int^t_{t_1}z'(t_1){\rm d}s=z'(t_1)(t-t_1), $$

且$\lim\limits_{t\rightarrow+\infty}z(t)=-\infty,$此结论与$z(t)>0$矛盾,所以(2.6)式成立.

由(2.5)式可得

$$ z''(t)+B(t)U(t-\sigma)\leq0,\ t\geq t_0. $$

因为

$\begin{align} & U(t)=z(t)-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)U(t-{{\mu }_{s}})\\ &=z(t)-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)[z(t-{{\mu }_{s}})-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t-{{\mu }_{s}})U(t-2{{\mu }_{s}})] \\ &=z(t)-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)z(t-{{\mu }_{s}})+\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)[\sum\limits_{s=1}^{l}{{{c}_{s}}}(t-{{\mu }_{s}})U(t-2{{\mu }_{s}})] \\ & \ge z(t)-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t)z(t-{{\mu }_{s}})\ge(1-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t))z(t),\\ \end{align}$

所以

$$ U(t-\sigma)\geq \bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)z(t-\sigma). $$

于是,我们得到

$$ z''(t)+\bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)B(t)z(t-\sigma)\leq0,\ t\geq t_0. $$

当$t\geq t_0,t=t_k(k\in I_{\infty})$时,有

$\begin{array}{*{35}{l}} {{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}}}_{k}}\le \frac{{{Z}_{i}}(x,t_{k}^{+})}{{{Z}_{i}}(x,{{t}_{k}})}=\frac{{{\gamma }_{i}}{{u}_{i}}(x,t_{k}^{+})}{{{\gamma }_{i}}{{u}_{i}}(x,{{t}_{k}})}=\frac{{{p}_{ki}}(x,{{t}_{k}},{{u}_{i}}(x,{{t}_{k}}))}{{{u}_{i}}(x,{{t}_{k}})}\le {{{\bar{a}}}_{k}},\\ {{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}}}_{k}}\le \frac{{{Z}_{it}}(x,t_{k}^{+})}{{{Z}_{it}}(x,{{t}_{k}})}=\frac{{{\gamma }_{i}}{{u}_{it}}(x,t_{k}^{+})}{{{\gamma }_{i}}{{u}_{it}}(x,{{t}_{k}})}=\frac{{{q}_{ki}}(x,{{t}_{k}},{{u}_{it}}(x,{{t}_{k}}))}{{{u}_{it}}(x,{{t}_{k}})}\le {{{\bar{b}}}_{k}}.\\ \end{array}$

(2.7)

由(2.7)式可得

$$ \underline{a}_k\leq \frac{U_i(t_k^+)}{U_i(t_k)}\leq\overline{a}_k,\ \ \ \ \underline{a}_k\leq \frac{U(t_k^+)}{U(t_k)}\leq\overline{a}_k, $$ $$ \underline{b}_k\leq \frac{{U_i}'(t_k^+)}{{U_i}'(t_k)}\leq\overline{b}_k,\ \ \ \ \underline{b}_k\leq \frac{U'(t_k^+)}{U'(t_k)}\leq\overline{b}_k, $$

且

$$ \underline{a}_k\leq \frac{z(t_k^+)}{z(t_k)}\leq\overline{a}_k,\ \ \ \ \underline{b}_k\leq \frac{z'(t_k^+)}{z'(t_k)}\leq\overline{b}_k. $$

所以$z(t)$是脉冲微分不等式(1.1)-(1.4)的正解.证毕.

引理2.2 假设$z(t)$是脉冲微分不等式(2.1)-(2.3)的正(负)解,即假设存在$T\geq t_0$使得当$t\geq T$时,$z(t)>0(z(t)<0)$.如果

$\mathop {\lim }\limits_{t \to + \infty } \int_{{t_0}}^t {\prod\limits_{{t_0} < {t_k} < \eta } {\frac{{{{}_k}}}{{{{\bar b}_k}}}} } F(\eta ){\rm{d}}\eta = + \infty $

(2.8)

成立,则当$t\in[T,t_l]\bigcup\Big(\bigcup\limits^{+\infty}_{k=l}(t_k,t_{k+1}]\Big)$时,$z'(t)\geq0 (z'(t)\leq 0)$,其中$l=\min\{k:t_k\geq T\}$.

证 不妨假设$z(t)$是脉冲微分不等式(2.1)-(2.3)的正解,则当$t\geq T>t_0$时,$z(t)>0$.我们首先证明对任意的$k\geq l$,都有

$\begin{equation}\label{eq:b9} z'(t_k)\geq 0. \end{equation}$

(2.9)

如果(2.9)式不成立,则存在某个$j$,使得$j\geq l$且$z'(t_j)<0.$由(2.3)式,可得

$$ z'(t_j^+)\leq\underline{b}_j z'(t_j)<0. $$

设$z'(t_j^+)=-\beta<0.$由(2.1)式知,$z'(t)$在区间$(t_{j+i-1},t_{j+i}](i=1,2,\cdots)$内单调递减,所以

$$ z'(t_{j+1})\leq z'(t_j^+)=-\beta<0, $$ $$ z'(t_{j+2})\leq z'(t_{j+1}^+)\leq\underline{b}_{j+1}z'(t_{j+1})\leq-\underline{b}_{j+1}\beta<0, $$ $$ z'(t_{j+3})\leq z'(t_{j+2}^+)\leq\underline{b}_{j+2}z'(t_{j+2})\leq-\underline{b}_{j+2}\underline{b}_{j+1}\beta<0. $$

由归纳法可得,对任意的正整数$n\geq 2$,有

$$ z'(t_{j+n})\leq-\prod\limits_{i=1}^{n-1}\underline{b}_{j+i}\beta<0. $$

考虑以下脉冲微分不等式

$$ z''(t)\leq 0,t>t_j,t\neq t_k, $$ $$ z'(t_k^+)\leq \underline{b}_kz'(t_k),k=j+1,j+2,\cdots. $$

令$m(t)=z'(t)$,则

$$ m'(t)\leq 0,t>t_j,t\neq t_k, $$ $$ m(t_k^+)\leq \underline{b}_km(t_k),k=j+1,j+2,\cdots. $$

根据引理1.1,可得

$m(t)\leq m(t_j^+)\prod\limits_{t_j<t_k<t}\underline{b}_k,$

即

$z'(t)\leq z'(t_j^+)\prod\limits_{t_j<t_k<t}\underline{b}_k.$

根据(2.2)式,可得

$$ z(t_k^+)\leq \overline{a}_kz(t_k),k=j+1,j+2,\cdots. $$

再次运用引理1.1,可得

$\begin{align} & z(t)\le z(t_{j}^{+})\prod\limits_{{{t}_{j}}<{{t}_{k}}<t}{{{{\bar{a}}}_{k}}}+\int_{t_{j}^{+}}^{t}{\prod\limits_{s<{{t}_{k}}<t}{{{{\bar{a}}}_{k}}}}({z}'(t_{j}^{+})\prod\limits_{{{t}_{j}}<{{t}_{k}}<s}{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}}}_{k}}})\text{d}s \\ & \ \ \ \ \ \ =\prod\limits_{{{t}_{j}}<{{t}_{k}}<t}{{{{\bar{a}}}_{k}}}[z(t_{j}^{+})-\beta \int_{t_{j}^{+}}^{t}{\prod\limits_{{{t}_{j}}<{{t}_{k}}<s}{\frac{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}}}_{k}}}{{{{\bar{a}}}_{k}}}}}\text{d}s],\\ \end{align}$

(2.10)

由条件(2.8),可得$\lim\limits_{t\rightarrow+\infty}z(t)=-\infty$,此结论与$z(t)>0$矛盾,所以$z'(t_k)\geq 0\:(k\geq l)$.

另外由(2.3)式可推知$z'(t_k^+)\geq \underline{b}_kz'(t_k)\geq 0\:(k\geq l)$.因为$z'(t)$在区间$(t_k,t_{k+1}]\:(k\geq l)$上单调减,显然在此区间上$z'(t)\geq z'(t_{k+1})>0$,当$t\in [T,t_l]$上$z'(t)\geq z'(t_l)>0$.证毕.

定理2.1 如果条件(2.8)与以下条件

$\begin{equation}\label{eq:b11} \lim\limits_{t\rightarrow+\infty}\int^t_{t_0} \prod\limits_{t_0<t_k<\eta}\frac{\underline{a}_k}{\overline{b}_k}F(\eta){\rm d}\eta=+\infty \end{equation}$

(2.11)

均成立,则问题(1.1)-(1.4)的每一个解在区域$G$内振动,其中

$$ F(t)=\bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)B(t). $$

证 用反证法.假设向量值函数$u(x,t)=\{u_1(x,t),u_2(x,t),\cdots,u_n(x,t)\}$是问题(1.1)-(1.4)的一个非振动解.不失一般性,假定存在$\overline{t_0}>0,$使得当$(x,t)\in\Omega\times[\overline{t_0},+\infty),$ $ {\rm sgn}u_i(x,t)\neq0,$ $ i \in I_n$.

令$Z_i(x,t)=\gamma_iu_i(x,t),\gamma_i={\rm sgn}u_i(x,t),$则$Z_i(x,t)>0.$由条件${\rm \widetilde{H}}_1)$知,存在$t_0\geq \overline{t_0}$,使得当$(x,t)\in \Omega\times[t_0,+\infty),$ $ Z_i(x,t-\mu_s)>0,$ $ Z_r(x,t-\sigma)>0,$ $Z_i(x,t-\tau_{j})>0,$ $ s\in I_l,r\in I_n,$ $ j\in I_m,i\in I_n$.由引理$2.1$可知,$z(t)$是脉冲微分不等式(2.1)-(2.3)的正解.

当$t\geq t_0,\ t\neq t_k(k\in I_{\infty})$时,定义

$\begin{equation}\label{eq:b12} w(t)=\frac{z'(t)}{z(t)},\ \ t\geq t_0. \end{equation}$

(2.12)

由引理$2.2$知,$w(t)\geq0,\ t\geq t_0$.不妨假设$z(t_0)=1$,由(2.12)式得

$\begin{equation}\label{eq:b13} z(t)=\exp\left(\int^t_{t_0}w(s){\rm d}s\right), \end{equation}$

(2.13)

$\begin{equation}\label{eq:b14} z'(t)=w(t)\exp\left(\int^t_{t_0}w(s){\rm d}s\right), \end{equation}$

(2.14)

$\begin{equation}\label{eq:b15} z''(t)=w^2(t)\exp\left(\int^t_{t_0}w(s){\rm d}s\right)+w'(t)\exp\left(\int^t_{t_0}w(s){\rm d}s\right). \end{equation}$

(2.15)

将式(2.13)-(2.15)代入(2.1)式,得

$\begin{align} & {{w}^{2}}(t)\exp \left(\int_{{{t}_{0}}}^{t}{w}(s)\text{d}s \right)+{w}'(t)\exp \left(\int_{{{t}_{0}}}^{t}{w}(s)\text{d}s \right)\\ &+(1-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t-\sigma))B(t)\exp \left(\int_{{{t}_{0}}}^{t-\sigma }{w}(s)\text{d}s \right)\le 0,\\ \end{align}$

化简得

$\begin{equation}\label{eq:b16} w^2(t)+w'(t)+\bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)B(t)\exp \bigg(-\int^t_{t-\sigma}w(s){\rm d}s\bigg)\leq 0. \end{equation}$

(2.16)

根据(2.16)式及条件$\overline{b}_k\leq\underline{a}_k$,易知$w(t)$是单调减函数,则当$t\geq t_0$时,$w(t)\leq w(t_0)$且

$\begin{equation}\label{eq:b17} w^2(t)+w'(t)+\exp(-\sigma w(t_0))\bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)B(t)\leq0. \end{equation}$

(2.17)

由(2.17)式可得到

$$ w'(t)+\exp(-\sigma w(t_0))\bigg(1-\sum\limits^l_{s=1}c_s(t-\sigma)\bigg)B(t)\leq0. $$

由式(2.2),(2.3)和(2.12)式,可得

$$ w(t^+_k)=\frac{z'(t^+_k)}{z(t^+_k)}\leq\frac{\overline{b}_kz'(t_k)}{\underline{a}_kz(t_k)}=\frac{\overline{b}_k}{\underline{a}_k}w(t_k),\ k\in I_{\infty}. $$

于是,我们得到了如下脉冲微分不等式

$\begin{align} & {w}'(t)\le -\exp(-\sigma w({{t}_{0}}))(1-\sum\limits_{s=1}^{l}{{{c}_{s}}}(t-\sigma))B(t)\\ &=-\exp(-\sigma w({{t}_{0}}))F(t),\ \ t\ne {{t}_{k}},\\ & w(t_{k}^{+})\le \frac{{{{\bar{b}}}_{k}}}{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}}}_{k}}}w({{t}_{k}}).\\ \end{align}$

应用引理1.1,结合条件(2.11),可得

$\begin{align} & w(t)\le w({{t}_{0}})\prod\limits_{{{t}_{0}}<{{t}_{k}}<t}{\frac{{{{\bar{b}}}_{k}}}{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}}}_{k}}}}-\exp(-\sigma w({{t}_{0}}))\int_{{{t}_{0}}}^{t}{\prod\limits_{\eta <{{t}_{k}}<t}{\frac{{{{\bar{b}}}_{k}}}{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}}}_{k}}}}}F(\eta)\text{d}\eta \\ & \ \ \ \ \ \ \ =\prod\limits_{{{t}_{0}}<{{t}_{k}}<t}{\frac{{{{\bar{b}}}_{k}}}{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}}}_{k}}}}[w({{t}_{0}})-\exp(-\sigma w({{t}_{0}}))\int_{{{t}_{0}}}^{t}{\prod\limits_{{{t}_{0}}<{{t}_{k}}<\eta }{\frac{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}}}_{k}}}{{{{\bar{b}}}_{k}}}}}F(\eta)\text{d}\eta] \\ & \ \ \ \ \ <0,\\ \end{align}$

与$w(t)\geq 0$矛盾.证毕.

推论2.1 当边界条件换为(1.5)式时,引理2.1,引理2.2及定理2.1中的结论仍成立,故问题(1.1)-(1.3),(1.5)的每一个解都是振动的.

例3.1 考虑以下脉冲时滞中立型双曲方程组

$\begin{align} & \frac{{{\partial }^{2}}}{\partial {{t}^{2}}}[{{u}_{2}}(x,t)+\frac{1}{2}{{u}_{2}}(x,t-2\pi)]-{{\text{e}}^{t}}{{u}_{1}}(x,t-\frac{\pi }{2})+2({{t}^{2}}+1){{u}_{2}}(x,t-\frac{\pi }{2})\\ &=2u_{2}^{2}(x,t)\Delta {{u}_{2}}(x,t)+tu_{2}^{2}(x,t-\frac{3\pi }{2})\Delta {{u}_{2}}(x,t-\frac{3\pi }{2}),\ t>1,\ t\ne {{8}^{k}},\\ & {{u}_{2}}(x,{{({{8}^{k}})}^{+}})=7{{u}_{2}}(x,{{8}^{k}}),\\ & {{u}_{2t}}(x,{{({{8}^{k}})}^{+}})={{u}_{2t}}(x,{{8}^{k}}),\ \ k\in {{I}_{\infty }},(x,t)\in(0,\pi)\times {{\mathbb{R}}^{+}},\\ \end{align}$

(3.1)

边界条件为

$\begin{equation}\label{eq:c2} \frac{\partial}{\partial x}u_i(0,t)=\frac{\partial}{\partial x} u_i(\pi,t)=0,\ \ t>1,\ t\neq 8^k,i=1,2, \end{equation}$

(3.2)

其中

$$ n=2,\ l=1,\ m=1,\ N=1,\mu_1=2\pi,\sigma=\frac{\pi}{2}, \tau_1=\frac{3\pi}{2}, $$ $$ c_1(t)=\frac{1}{2},\ a_1(t)=t^2,\:a_2(t)=2,\ h_1(u)=h_2(u)=u^2,\ a_{11}(t)=a_{21}(t)=t, $$ $$ h_{11}(u)=h_{21}(u)=u^2,f_{11}(u)=f_{12}(u)=f_{21}(u)=f_{22}(u)=u, $$ $$ b_{11}(x,t)=3(x^2+1)e^t, b_{12}(x,t)=-2t^2,\ b_{21}(x,t)=-{\rm e}^t,\ b_{22}(x,t)=2(t^2+1). $$

很容易验证条件${\rm \widetilde{H}_1)}-{\rm \widetilde{H}_5)}$及以下条件均成立

$$ b_{11}(t)=3{\rm e}^t,\ b_{22}(t)=2(t^2+1),\ \overline{b_{12}(t)}=2t^2,\ \overline{b_{21}(t)}={\rm e}^t,\ B(t)=2. $$

并且

$\begin{align} & \underset{t\to+\infty }{\mathop{\lim }}\,\int_{{{t}_{0}}}^{t}{\prod\limits_{{{t}_{0}}<{{t}_{k}}<s}{\frac{{{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}}}_{k}}}{{{{\bar{a}}}_{k}}}}}\text{d}s=\int_{1}^{+\infty }{(}\prod\limits_{1<{{t}_{k}}<s}{\frac{1}{7}})\text{d}s \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_{1}^{{{t}_{1}}}{(}\prod\limits_{1<{{t}_{k}}<s}{\frac{1}{7}})\text{d}s+\int_{t_{1}^{+}}^{{{t}_{2}}}{(}\prod\limits_{1<{{t}_{k}}<s}{\frac{1}{7}})\text{d}s \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\int_{t_{2}^{+}}^{{{t}_{3}}}{(}\prod\limits_{1<{{t}_{k}}<s}{\frac{1}{7}})\text{d}s+\cdots \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge \int_{t_{1}^{+}}^{{{t}_{2}}}{\prod\limits_{1<{{t}_{k}}<s}{\frac{1}{7}}}\text{d}s+\int_{t_{2}^{+}}^{{{t}_{3}}}{\prod\limits_{1<{{t}_{k}}<s}{\frac{1}{7}}}\text{d}s+\cdots \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{7}\times({{8}^{2}}-8)+{{(\frac{1}{7})}^{2}}\times({{8}^{3}}-{{8}^{2}})+\cdots \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =7\sum\limits_{n=1}^{+\infty }{{{(\frac{8}{7})}^{n}}}=\infty,\\ \end{align}$

所以(2.8)式成立.另外

$$ \lim\limits_{t\rightarrow +\infty}\int_{t_0}^t\prod\limits_{t_0<t_k<s}\frac{\underline{a}_k}{\overline{b}_k} F(\eta){\rm d}\eta =\int_{1}^{+\infty}\bigg(\prod\limits_{1<t_k<s}\frac{4}{3}\bigg){\rm d}\eta =\infty, $$

于是(2.11)式也成立.根据定理2.1知,问题(3.1),(3.2)的每一个解在区域$(0,\pi)\times{\Bbb R}^+$上是振动的.

注3.1 如果边界条件(3.2)换成如下Dirichlet边界条件

$\begin{equation}\label{eq:c3} u_i(0,t)=u_i(\pi,t)=0,\ \ t>1,\ t\neq 8^k,i=1,2, \end{equation}$

(3.3)

根据推论2.1,问题(3.1),(3.3)的每一个解在区域$(0,\pi)\times{\Bbb R}^+$内是振动的.