由于分数阶微积分非常适合于刻画具有记忆和遗传性质的材料和过程, 所以许多重要的数学模型均是由含有分数阶导数的微分方程描述的, 这些性质在经典模型中常常是被忽略的.如今, 分数阶微分方程越来越多地被用来描述光学、热学系统、流变学、流体力学系统、信号处理、系统辨识、控制和机器人及其他应用领域中的问题^[1-5].

常微分方程、偏微分方程和脉冲偏微分方程的振动性质在过去已有许多研究成果^[6-12].在分数阶微分方程的研究中, 振动性同样具有十分重要的意义.近年来, 已经有许多学者对分数阶常微分方程和分数阶偏微分方程的振动性质进行了研究^[13-20].然而, 目前为止, 对具有若干时滞的带脉冲的分数阶偏微分方程振动性质的研究依然很少. 2017年Raheem A和Maqbul M利用微分不等式等方法研究了一类带脉冲和强迫项的分数阶偏微分方程在Robin和Dirichlet边界条件下解的振动性^[21]. Qu Zhuo^[22]研究了具有多个时滞的带强迫项和脉冲项的分数阶偏微分方程解的振动性.

本文中, 我们在改进的Riemann-Liouville分数阶定义下, 研究一类分数阶脉冲时滞偏微分方程在两类不同的边界条件下解的强迫振动性质.

$ \begin{equation} \left\{ \begin{aligned} &D_{+, t} ^{2\alpha} u(t, x)+p(t)D_{+, t} ^\alpha u(t, x) = a(t)h(u)\Delta u+\sum\limits_{i = 1}^m a_i (t)h_i (u(t-\tau_i, x))\Delta u(t-\tau_i, x)\\ &\; \; \; \; -\sum\limits_{j = 1}^n q_j(t, x)f_j(u(t-\delta_j, x))-g(t, x), t \neq t_k, (t, x)\in R_+ \times \Omega, \\ &D_{+, t} ^\alpha u(t_k^+, x)-D_{+, t} ^\alpha u(t_k ^-, x) = \sigma(t_k, x)D_{+, t} ^\alpha u(t_k, x), \\ &u(t_k^+, x)-u(t_k ^-, x) = \theta(t_k, x)u(t_k, x), \; \; t = t_k, (t, x)\in R_+ \times \Omega, k = 1, 2\cdots. \end{aligned} \right. \end{equation} $

(1.1)

满足边界条件

$ \begin{equation} \frac{\partial u(t, x)}{\partial n} = 0, \; \; (t, x)\in R_+ \times \partial\Omega, t\neq t_k, \end{equation} $

(1.2)

或

$ \begin{equation} u(t, x) = 0, \; \; (t, x)\in R_+ \times \partial\Omega, t\neq t_k. \end{equation} $

(1.3)

其中$ \alpha \in (0, 1) $是常数, $ D_{+, t}^{\alpha} $表示改进的Riemann-Liouville正半轴分数阶微积分定义下函数$ u(t, x) $关于变量$ t $的$ \alpha $阶微分, $ \Omega $是在$ R^n $上的有界域, 其边界$ \partial \Omega $是光滑的, 且$ \overline\Omega = \Omega \cup \partial \Omega $; $ R_+ = (0, +\infty) $, $ \Delta $为拉普拉斯算子, $ n $为边界$ \partial \Omega $的外法线方向向量, $ \tau_j, \delta_j $为非负常数: $ i = 1, 2, 3, \cdots, m $且$ j = 1, 2, 3, \cdots, n $; $ p(t), a(t), a_i (t)\in PC[R_+, R_+] $, $ \lim\limits_{k\rightarrow +\infty} t_k = +\infty $, $ 0<t_1< \cdots <t_k <\cdots $且强迫项$ g(t, x)\in PC[R_+ \times \overline\Omega, R] $, $ PC $表示以$ t = t_k, k = 1, 2, \cdots $为第一类间断点且在该处左连续的分段连续函数类, 问题(1.1)和(1.2)(或(1.1)和(1.3))的解$ u(t, x) $和它的分数阶导数$ D_{+, t} ^\alpha u(t, x) $是以$ t = t_k $为第一类间断点且在该处左连续的分段连续函数, 即

$ \begin{eqnarray*} u(t_k^-, x) = u(t_k, x);\quad D_{+, t} ^\alpha u(t_k^-, x) = D_{+, t} ^\alpha u(t_k, x). \end{eqnarray*} $

以下为本文的基本假设:

(H1) $ f_j:R \rightarrow R $为连续函数, 且当$ u\neq0 $存在正常数$ k_j $, 有$ f_j(u)/u\geq k_j >0 $;

(H2) $ q_j(t, x)\in PC[R_+ \times \bar{\Omega}, R_+] $, $ q(t) = \min\limits_{1\leq j\leq n}\min\limits_{x\in \bar{\Omega}}q_j(t, x) $;

(H3) $ h(u), h_i(u)\in C(R, R); uh'(u)\geq 0, uh'_i(u)\geq 0; $

(H4) $ \sigma(t_k, x)\leq \alpha_k, \theta(t_k, x)\leq\theta_k\leq0, \alpha_k, \theta_k $是常数;

(H5) $ g(t, x)\in PC[R_+ \times \bar{\Omega}) $.

本文中, 为方便起见, 引入以下符号:

$ \begin{eqnarray*} &&U(t) = \int_\Omega u(t, x)dx, \; \; G(t) = \int_\Omega g(t, x)dx, \\ &&\tilde{U}(\xi) = U(t), \; \; \tilde{G}(\xi) = G(t), \; \; \tilde{V}(\xi) = V(t), \\ &&\xi = \frac{t^\alpha}{\Gamma(1+\alpha)}, \; \; \xi_l = \frac{t_l^\alpha}{\Gamma(1+\alpha)}, \; \; \tilde{p(\xi)} = p(t), \; \; \tilde{q(\xi)} = q(t). \end{eqnarray*} $

定义1.1^[9]  问题(1.1)与(1.2)或((1.1)与(1.3))的一个非零解$ u(t, x) $在区域$ G $内是非振动的, 如果存在一个常数$ \tau\geq 0 $, 使得在$ (t, x)\in [\tau, +\infty)\times \Omega $, 始终有$ u(t, x)<0 $或者$ u(t, x)>0 $, 否则称振动的.

定义1.2^[1]  在Riemann-Liouville正半轴上的分数阶积分定义下, 关于函数$ x(t):R^+ \rightarrow R $在正半轴上逐点定义的$ \alpha $阶积分表示如下

$ \begin{eqnarray*} (I_+ ^\alpha x)(t): = \frac{1}{\Gamma(\alpha)}\int_0 ^t (t-v)^{\alpha-1}x(v)dv, \; \; t>0. \end{eqnarray*} $

其中$ \alpha\in(0, 1) $; $ \Gamma $表示Gamma函数, 并定义$ \Gamma(\alpha) = \int_0 ^{+\infty} s^{\alpha-1} e^{-s}ds $.

定义1.3^[16]  改进的Riemann-Liouville分数阶$ \alpha $阶导数定义

$ \begin{equation*} D_{+, t} ^\alpha u(t, x) = \left\{ \begin{aligned} &\frac{1}{\Gamma(1-\alpha)}\frac{\partial}{\partial t}\int_0 ^t (t-v)^{-\alpha}u(v, x)dv, \; \; 0<\alpha<1, \\ &(f^{(n)}(t))^{(\alpha-n)}, n \leq \alpha \leq n+1, n \geq 1, \end{aligned} \right. \end{equation*} $

其中$ \alpha\in(0, 1) $; $ \Gamma $表示Gamma函数.

下面给出改进的Riemann-Liouville分数阶定义下一些重要的相关性质^[16]

$ \begin{equation} D_{t} ^\alpha t^r = \frac{\Gamma(1+r)}{\Gamma(1+r-\alpha)}t^{r-\alpha} \end{equation} $

(1.4)

$ \begin{equation} D_{t} ^\alpha (m(t)n(t)) = n(t)D_{t} ^\alpha m(t)+m(t)D_{t} ^\alpha n(t) \end{equation} $

(1.5)

$ \begin{equation} D_{t} ^\alpha m[n(t)] = m_n'[n(t)]D_{t} ^\alpha n(t) = D_{n} ^\alpha m[n(t)](n'(t))^\alpha \end{equation} $

(1.6)

引理1.4^[12]  设$ \lambda $是一常数, 则对如下问题

$ \begin{equation*} \left\{ \begin{split} &\Delta w(x)+\lambda w(x) = 0, \; \; x\in \Omega, \\ &w(x) = 0, \; \; x\in \partial \Omega. \end{split} \right. \end{equation*} $

其最小特征值$ \lambda_0 $为正, 且相应的特征函数$ \phi_0(x) $在$ x\in \Omega $也为正.

引理1.5^[21]  假设以下不等式成立

$ \begin{equation*} \begin{split} &\omega'(t)\leq g_1(t)\omega(t)+g_2(t), \; \; t\neq t_k, t\geq \mu, \\ &\omega(t_k^+)\leq(1+a_k)\omega(t_k), \; \; k = 1, 2, \cdots, \end{split} \end{equation*} $

其中$ 0<t_1<t_2<\cdots<t_k<\cdots $, 且$ {\lim\limits_{k\rightarrow\infty}}t_k = +\infty.\; \; \omega\in PC^1 [R_+, R], \; \; g_1, g_2\in [R_+, R] $, 并且$ a_k $是常数.那么,

$ \begin{equation*} \begin{split} &\omega(t)\leq \omega(t_0)\prod\limits_{t_0<t_l<t} (1+\alpha_k)\text{exp}(\int_{t_0}^t g_1(s)ds)+\int_{t_0} ^t \prod\limits_{s<t_l<t} (1+\alpha_k)\text{exp}(\int_s ^t g_1(\sigma)d\sigma)g_2(s)ds, t\geq \mu. \end{split} \end{equation*} $

定理2.1   假设条件(H1)-(H5)成立, 如果分数阶脉冲微分不等式

$ \begin{equation} \left \{ \begin{aligned} &D_{+, t} ^{2\alpha} U(t)+p(t)D_{+, t} ^{\alpha} U(t)\leq -G(t), \\ &D_{+, t} ^\alpha U(t_k^+)\leq (1+\alpha_k)D_{+, t} ^\alpha U(t_k), \\ &U(t_k^+)\leq (1+\theta_k)U(t_k)\; \; , k = 1, 2, \cdots\\ \end{aligned} \right. \end{equation} $

(2.1)

没有最终正解, 并且分数阶脉冲微分不等式

$ \begin{equation} \left \{ \begin{aligned} &D_{+, t} ^{2\alpha} U(t)+p(t)D_{+, t} ^{\alpha} U(t)\geq -G(t), \\ &D_{+, t} ^\alpha U(t_k^+)\geq (1+\alpha_k)D_{+, t} ^\alpha U(t_k), \\ &\; \; U(t_k^+)\geq (1+\theta_k)U(t_k), k = 1, 2, \cdots\\ \end{aligned} \right. \end{equation} $

(2.2)

没有最终负解, 那么问题(1.1)与(1.2)的每一个非零解$ u(t, x) $在G内都是振动的.

证  (用反证法)假设$ u(t, x) $是问题(1.1)与(1.2)的一个非振动解.不失一般性, 我们假设$ u(t, x) $是问题(1.1)与(1.2)的最终正解, 即存在$ \mu>0 $, 当$ (t, x)\in [\mu, +\infty) \times \Omega $时, 使得$ u(t, x)>0, u(t-\tau_i, x)>0 $.

(Ⅰ)当$ t\neq t_k $时, 对(1.1)的第一个方程, 两边同时对$ x $在有界域$ \Omega $上积分得到:

$ \begin{equation} \begin{split} &D_{+, t} ^{2\alpha}\int_\Omega u(t, x)dx+p(t)D_{+, t} ^{\alpha}\int_\Omega u(t, x)dx\\ = &a(t)\int_\Omega h(u(t, x))\Delta udx+\int_\Omega \sum\limits_{i = 1}^m a_i (t)h_i (u(t-\tau_i, x))\Delta u(t-\tau_i, x)dx-\\ &\sum\limits_{j = 1}^n \int_\Omega q_j (t, x) f_j(u(t-\delta_j, x))dx-\int_\Omega g(t, x)dx, t \neq t_k, (t, x)\in R_+ \times \Omega.\\ \end{split} \end{equation} $

(2.3)

根据Green公式, 结合边界条件(1.2)和假设(H3)得到：

$ \begin{equation} \begin{split} &\; \; \; \; \int_\Omega h(u)\Delta u(t, x)dx = \int_{\partial \Omega}h(u)\frac{\partial u(t, x)}{\partial n}dx-\int_\Omega h'(u)|\text{grad} u|^2 dx \\ & = -\int_\Omega h'(u)|\text{grad} u|^2 dx\leq 0, t\geq t_0. \end{split} \end{equation} $

(2.4)

同理可得

$ \begin{equation} \begin{split} \int_\Omega h_i(u(t-\tau_i), x))\Delta u(t-\tau_i, x)dx\leq 0. \end{split} \end{equation} $

(2.5)

根据条件(H1)和(H2)得到

$ \begin{equation} \begin{split} \sum\limits_{j = 1}^n \int_\Omega q_j (t, x)f_j(u(t-\delta_j, x))dx \geq \sum\limits_{j = 1}^n k_j q (t)\int_\Omega u(t-\delta_j, x)dx\geq0, \; \; t\geq t_0.\\ \end{split} \end{equation} $

(2.6)

结合(2.3)–(2.6)可得

$ \begin{equation} D_{+, t} ^{2\alpha} U(t)+p(t)D_{+, t} ^{\alpha} U(t)\leq -G(t), \; \; t\geq t_0.\\ \end{equation} $

(2.7)

(Ⅱ)当$ t = t_k $时, 对(1.1)的第二、三个方程, 两边同时对$ x $在有界域$ \Omega $上积分，并结合条件(H4)得到

$ \begin{equation} \begin{split} D_{+, t} ^\alpha \int_\Omega u(t_k^+, x)dx = &D_{+, t} ^\alpha U(t_k^+, x) \leq(1+\alpha_k)D_{+, t} ^\alpha\int_\Omega u(t_k, x)dx = (1+\alpha_k)D_{+, t}^\alpha U(t_k), \\ U(t_k^+, x) = &\int_\Omega u(t_k^+, x)dx\leq(1+\theta_k)\int_\Omega u(t_k, x)dx\\ = &(1+\theta_k)U(t_k, x)\; \qquad\qquad\qquad\qquad\qquad\qquad\; k = 1, 2, 3, \cdots\\ \end{split} \end{equation} $

(2.8)

因此, 由脉冲微分不等式(2.7)–(2.8)可知函数$ U(t) = \int_\Omega u(t, x)dx $是分数阶脉冲微分不等式(2.1)的最终正解.与假设条件相矛盾.

另一方面, 如果$ u(t, x) $是问题(1.1)与(1.2)在$ G $内的最终负解, 即存在$ \mu>0 $, 当$ (t, x)\in [\mu, +\infty) \times \Omega $时, 使得$ u(t, x)<0, u(t-\tau_i, x)<0 $.

(Ⅲ)当$ t\neq t_k $时, 由(2.3)式根据Green公式, 结合边界条件(1.2)和假设(H3)得：

$ \begin{equation} \begin{split} \int_\Omega h(u)\Delta u(t, x)dx = &\int_{\partial \Omega}h(u)\frac{\partial u(t, x)}{\partial n}dx-\int_\Omega h'(u)|\text{grad} u|^2 dx \\ = &-\int_\Omega h'(u)|\text{grad} u|^2 dx \geq 0, t \geq t_0.\\ \end{split} \end{equation} $

(2.9)

同理可得

$ \begin{equation} \begin{split} \int_\Omega h_i(u(t-\tau_i, x))\Delta u(t-\tau_i, x)dx\geq 0, t \geq t_0.\\ \end{split} \end{equation} $

(2.10)

根据条件(H1)和(H2)得到

$ \begin{equation} \begin{split} \sum\limits_{j = 1}^n \int_\Omega q_j (t, x)f_j(u(t-\delta_j, x))dx \leq \sum\limits_{j = 1}^n k_j q (t)\int_\Omega u(t-\delta_j, x)dx\leq0, \; \; t\geq t_0.\\ \end{split} \end{equation} $

(2.11)

结合(2.3), (2.9)–(2.11)可得

$ \begin{equation} D_{+, t} ^{2\alpha} U(t)+p(t)D_{+, t} ^{\alpha} U(t)\geq -G(t), \; \; t\geq t_0.\\ \end{equation} $

(2.12)

(Ⅳ)当$ t = t_k $时, 对(1.1)的第二、三个方程, 两边同时对$ x $在有界域$ \Omega $上积分，并结合条件(H4)得到：

$ \begin{equation} \begin{split} D_{+, t} ^\alpha \int_\Omega u(t_k^+, x)dx& = D_{+, t} ^\alpha U(t_k^+) \geq(1+\alpha_k)D_{+, t} ^\alpha\int_\Omega u(t_k, x)dx = (1+\alpha_k)D_{+, t}^\alpha U(t_k), \\ U(t_k^+)& = \int_\Omega u(t_k^+, x)dx\geq(1+\theta_k)\int_\Omega u(t_k, x)dx\\ & = (1+\theta_k)U(t_k, )\; \qquad\qquad\qquad\qquad\qquad\qquad\; k = 1, 2, 3, \cdots\\ \end{split} \end{equation} $

(2.13)

因此, 由脉冲微分不等式(2.12)–(2.13)可知函数$ U(t) = \int_\Omega u(t, x)dx $是分数阶脉冲微分不等式(2.2)的最终负解, 与假设条件相矛盾.证毕.

定理2.2   假设$ h(u) = h_i(u) = 1 $, 且条件(H1)–(H2)、(H4)–(H5)成立, 如果分数阶脉冲微分不等式(2.1)没有最终正解, 并且(2.2)没有最终负解, 那么问题(1.1)与(1.3)的每个非零解$ u(t, x) $在$ G $内是振动的.

证  (用反证法)假设$ u(t, x) $是问题(1.1)与(1.3)的一个非振动解(其中$ h(u) = h_i(u) = 1 $).不失一般性, 我们假设$ u(t, x) $是问题(1.1)与(1.3)的最终正解, 即存在$ \mu>0 $, 当$ (t, x)\in [\mu, +\infty)\times \Omega $时, 使得$ u(t, x)>0, u(t-\tau_i, x)>0 $.

(Ⅰ)当$ t\neq t_k $时, 对(1.1)的第一个方程, 两边同时乘以$ \phi_0(x) $并对$ x $在有界域$ \Omega $上积分得到：

$ \begin{equation} \begin{split} &D_{+, t} ^{2\alpha}\int_\Omega u(t, x)\phi_0(x)dx+p(t)D_{+, t} ^{\alpha}\int_\Omega u(t, x)\phi_0(x)dx\\ = \; &a(t)\int_\Omega \phi_0(x)\Delta udx +\int_\Omega \sum\limits_{i = 1}^m a_i (t)\phi_0(x)\Delta u(t-\tau_i, x)dx\\ &-\sum\limits_{j = 1}^n \int_\Omega q_j (t, x)\phi_0(x) f_j(u(t-\delta_j, x))dx -\int_\Omega \phi_0(x)g(t, x)dx, \quad t \geq\mu. \end{split} \end{equation} $

(2.14)

根据Green公式、边界条件(1.3)和引理1.4, 得到：

$ \begin{equation} \begin{split} \int_\Omega \phi_0(x)\Delta u(t, x)dx = \int_ \Omega \Delta\phi_0(x) u(t, x)dx = -\lambda_0\int_\Omega u(t, x)\phi_0(x) dx<0, t\geq\mu. \end{split} \end{equation} $

(2.15)

$ \begin{equation} \begin{split} \int_\Omega \phi_0(x)\Delta u(t-\tau_i, x)dx = -\lambda_0\int_\Omega u(t-\tau_i, x)\phi_0(x) dx<0, \qquad \qquad \qquad t\geq\mu. \end{split} \end{equation} $

(2.16)

并且根据条件(H1)–(H2), 得到

$ \begin{equation} \begin{split} \sum\limits_{j = 1}^n \int_\Omega q_j (t, x)f_j(u(t-\delta_j, x))\phi_0(x)dx \geq \sum\limits_{j = 1}^n k_j q (t)\int_\Omega u(t-\delta_j, x)\phi_0(x)dx\geq0, \; \; t\geq \mu. \end{split} \end{equation} $

(2.17)

进一步有

$ \begin{equation} \begin{split} &-\int_\Omega g(t, x)\phi_0(x)dx = -Q(t). \end{split} \end{equation} $

(2.18)

结合(2.14)–(2.18)有

$ \begin{equation} D_{+, t} ^{2\alpha} Y(t)+p(t)D_{+, t} ^{\alpha} Y(t)\leq -Q(t), \; \; t\geq\mu.\\ \end{equation} $

(2.19)

其中$ Y(t) = \int_\Omega u(t, x)\phi_0(x)dx, Q(t) = \int_\Omega g(t, x)\phi_0(x)dx $.

(Ⅱ)当$ t = t_k $时, 对(1.1)的第二、三个方程, 两边同时乘以$ \phi_0(x) $, 并对$ x $在有界域$ \Omega $上积分, 根据条件(H4), 由类似的方法可得

$ \begin{eqnarray} && D_{+, t} ^\alpha Y(t_k^+)\leq(1+\alpha_k)D_{+, t} ^\alpha Y(t_k), \\ && Y(t_k^+) = \int_\Omega u(t_k^+, x)\phi_0(x)dx\leq(1+\theta_k)\int_\Omega u(t_k, x)\phi_0(x)dx = (1+\theta_k)Y(t_k)\; \; k = 1, 2, 3, \cdots \end{eqnarray} $

(2.20)

因此, 由脉冲微分不等式(2.19)–(2.20)可知, 函数$ Y(t) = \int_\Omega u(t, x)\phi_0(x)dx $是分数阶脉冲微分不等式(2.1)的最终正解.与假设条件相矛盾.

另一方面, 如果$ u(t, x) $是问题(1.1)与(1.3)在$ G $内的最终负解, 由相似的方法, 可知函数$ Y(t) = \int_\Omega u(t, x)\phi_0(x)dx $是分数阶脉冲微分不等式(2.2)的最终负解.与假设条件相矛盾.证毕.

定理2.3   假设条件(H1)–(H5)成立, 如果存在$ \mu_2\geq0 $, 满足

$ \begin{equation} \begin{split} \int_{\mu_2}^\infty \text{exp}(-\int_{t_0}^t p(\sigma)d\sigma)ds = \infty, \end{split} \end{equation} $

(2.21)

存在$ \mu_1\geq0 $, 满足

$ \begin{equation} \begin{split} \lim\limits_{\xi\rightarrow\infty}sup\frac{\int_{\mu_1}^\xi\prod\limits_{s<\xi_l<\xi} (1+\alpha_k) \text{exp}(-\int_{s}^\xi\tilde p(\sigma)d\sigma)\tilde G(s)ds}{\prod\limits_{\mu_1<\xi_l<\xi} (1+\alpha_k)\text{exp}(-\int_{\mu_1}^\xi \tilde p(s)ds) } = \infty, \xi_l = \xi(t_l), k = 1, 2, 3, \cdots\\ \end{split} \end{equation} $

(2.22)

并且

$ \begin{equation} \begin{split} \lim\limits_{\xi\rightarrow\infty}inf\frac{\int_{\mu_1}^\xi \prod\limits_{s<\xi_l<\xi} (1+\alpha_k) \text{exp}(-\int_{\mu_1}^\xi \tilde p(\sigma)d\sigma)\tilde G(s)ds}{\prod\limits_{\mu_1<\xi_l<\xi} (1+\alpha_k)\text{exp}(-\int_{\mu_1}^\xi\tilde p(s)ds) } = -\infty, \xi_l = \xi(t_l), k = 1, 2, 3, \cdots. \end{split} \end{equation} $

(2.23)

那么, 问题(1.1)、(1.2)的每个解在$ G $内是振动的.

证  (反证法)证明该定理, 先证明分数阶脉冲微分不等式(2.1)没有最终正解且分数阶脉冲微分不等式(2.2)没有最终负解.不失一般性, 假设$ U(t) $是分数阶脉冲微分不等式(2.1)的最终正解, 那么存在$ \mu_1\geq0 $使得$ U(t)\geq0 $, $ U(t-\tau)\geq0, t\geq\mu_1 $.令$ V(t) = \text{exp}(\int_{t_0}^t p(s)ds) $, 由性质(1.4)得, $ D_{+, t} ^\alpha \xi(t) = D_{+, t} ^\alpha [\frac{t^\alpha}{\Gamma(1+\alpha)}] = 1 $.再由性质(1.6)得：

$ \begin{equation} \begin{split} &D_{t} ^\alpha U(t) = D_{t} ^\alpha \tilde{U}(\xi) = \tilde{U}'(\xi)D_{t} ^\alpha \xi(t) = \tilde{U}'(\xi), \\ &D_{t} ^{2\alpha} U(t) = D_{t} ^\alpha \tilde{U}'(\xi) = \tilde{U}''D_{t} ^\alpha \xi(t) = \tilde{U}''(\xi). \end{split} \end{equation} $

(2.24)

由式(2.24), 则式(2.19)可以变为

$ \begin{equation} \begin{split} \tilde{U}''(\xi)+\tilde{p}(\xi)\tilde{U}'(\xi)\leq-\tilde{G}(\xi).\\ \end{split} \end{equation} $

(2.25)

故有

$ \begin{equation} \begin{split} [\tilde{U}'(\xi)V(\xi)]' = \tilde{U}''(\xi)V(\xi)+\tilde{U}'(\xi)V(\xi)\tilde{p}(\xi)\leq-\tilde{G}(\xi)V(\xi)<0.\\ \end{split} \end{equation} $

(2.26)

又根据(1.1)的第三个式子, 当$ \xi\in[\xi_1, \infty) $时, $ \tilde{U}'(\xi)V(\xi) $严格单调递减, 并且不变号.由于当$ \xi\in[\xi_1, \infty) $, $ V(\xi)>0 $, 可知$ \tilde{U}'(\xi) $最终不变号.当$ \xi\in[\xi_1, \infty) $时, $ \tilde{U}'(\xi)>0 $, 否则, $ \tilde{U}'(\xi)<0 $, 即存在$ \xi_2\in[\xi_1, \infty) $, 使得$ \tilde{U}'(\xi)<0 $.由于$ \tilde{U}'(\xi)V(\xi) $在$ \xi\in[\xi_1, \infty) $上严格单调递减, 所以当$ \xi\in[\xi_2, \infty) $时, $ \tilde{U}'(\xi)V(\xi)\leq\tilde{U}'(\xi_2)V(\xi_2) = c_1<0 $.故以下不等式成立

$ \begin{equation} \begin{split} \tilde{U}'(\xi)\leq\frac{c_1}{V(\xi)} = c_1 \text{exp} (-\int_{t_0}^\xi p(v)dv)<0, \xi\in[\xi_2, \infty).\\ \end{split} \end{equation} $

(2.27)

上式两边同时除以$ c_1 $并积分得

$ \begin{equation} \begin{split} \int_{\xi_2}^\xi \text{exp} (-\int_{t_0}^s p(v)dv)ds\leq\frac{U(\xi)-\tilde{U}(\xi_2)}{c_1}<\frac{-\tilde{U}(\xi_2)}{c_1}, \xi\in[\xi_2, \infty).\\ \end{split} \end{equation} $

(2.28)

让$ \xi\rightarrow\infty $可得$ \int_{\xi_2}^\xi \text{exp} (-\int_{t_0}^s p(v)dv)ds<\frac{-\tilde{U}(\xi_2)}{c_1}<\infty $, 与式(2.21)矛盾.

所以$ \tilde{U}'(\xi)>0 $.令$ \omega(\xi) = \tilde{U}'(\xi)>0 $, 根据分数阶脉冲微分不等式(2.1)的前两个式子, 可得

$ \begin{equation*} \begin{split} \omega'(\xi)\leq-\tilde{p}(\xi)\omega(\xi)-\tilde{G}(\xi), \quad \xi\geq\xi_0, \xi\neq\xi_k, \quad \omega(t_{k}^{+})\leq(1+\alpha_k)\omega(t_k), \quad k = 1, 2, 3, \cdots\\ \end{split} \end{equation*} $

由引理1.5可得

$ \begin{equation*} \begin{split} \omega(\xi)\leq\omega(\xi_0)\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds) -\int_{\xi_0}^\xi \prod\limits_{s<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{s}^\xi \tilde{p}(\sigma)d\sigma)\tilde{G}(s)ds, \\ \end{split} \end{equation*} $

因此,

$ \begin{equation*} \begin{split} \frac{\omega(\xi)}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds)}\leq \omega(\xi_0) -\frac{\int_{\xi_0}^\xi \prod\limits_{s<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{s}^\xi \tilde{p}(\sigma)d\sigma)\tilde{G}(s)ds}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds)}.\\ \end{split} \end{equation*} $

当$ \xi\rightarrow\infty $, 并由(2.22)式可得

$ \begin{equation*} \begin{split} \lim\limits_{\xi\rightarrow\infty}inf\frac{\omega(\xi)}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi\tilde {p}(s)ds)} = -\infty, \\ \end{split} \end{equation*} $

可知, 与$ \omega(\xi)>0 $矛盾.证毕.

另一方面, 假设分数阶脉冲微分不等式(2.2)有最终负解$ U(t) $, 那么存在$ \tau_1\geq0 $, 使得$ U(t)<0 $, $ U(t-\delta_i)<0 $, $ G(t)<0, t\geq \tau_1 $, $ \tilde{\omega}(\xi) = \tilde{U}'(\xi) $由以上类似的方法, 可得到$ \tilde{\omega}(\xi) = \tilde{U}'(\xi)<0 $, 根据不等式(2.2), 可得

$ \begin{equation*} \begin{split} &\tilde{\omega'}(\xi)\geq-\tilde{p}(\xi)\tilde{\omega}(\xi)-\tilde{G}(\xi), \xi\geq\xi_0, \xi\neq\xi_k, \\ &\tilde{\omega}(t_{k}^{+})\geq(1+\alpha_k)\omega(t_k), k = 1, 2, 3, \ldots.\\ \end{split} \end{equation*} $

令$ \tilde{\omega}(\xi) = -m(\xi) $, 可得

$ \begin{equation*} \begin{split} &m'(\xi)\leq-\tilde{p}(\xi)m(\xi)+\tilde{G}(\xi), \xi\geq\xi_0, \xi\neq\xi_k, \\ &m(t_{k}^{+})\leq(1+\alpha_k)m(t_k), k = 1, 2, 3, \ldots.\\ \end{split} \end{equation*} $

根据引理1.5, 有

$ \begin{equation*} \begin{split} m(\xi)\leq m(\xi_0)\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds) +\int_{\xi_0}^\xi \prod\limits_{s<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{s}^\xi \tilde{p}(\sigma)d\sigma)\tilde{G}(s)ds, \\ \end{split} \end{equation*} $

所以, 有

$ \begin{equation*} \begin{split} \frac{m(\xi)}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds)}\leq m(\xi_0) +\frac{\int_{\xi_0}^\xi \prod\limits_{s<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{s}^\xi \tilde{p}(\sigma)d\sigma)\tilde{G}(s)ds}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds)}, \end{split} \end{equation*} $

则

$ \begin{equation*} \begin{split} \frac{\tilde\omega(\xi)}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds)}\geq \tilde\omega(\xi_0) -\frac{\int_{\xi_0}^\xi \prod\limits_{s<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{s}^\xi \tilde{p}(\sigma)d\sigma)\tilde{G}(s)ds}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi \tilde{p}(s)ds)}. \end{split} \end{equation*} $

当$ \xi\rightarrow\infty $, 可得

$ \begin{equation*} \begin{split} \lim\limits_{\xi\rightarrow\infty}sup\frac{\tilde\omega(\xi)}{\prod\limits_{\xi_0<\xi_1<\xi}(1+\alpha_k)\text{exp} (-\int_{\xi_0}^\xi\tilde {p}(s)ds)} = \infty, \\ \end{split} \end{equation*} $

可知, 与$ \tilde\omega(\xi)<0 $矛盾, 证毕.定理得证.

定理2.4   在定理2.2的条件下，若存在$ \mu_2\geq0, \mu_1\geq0 $使得(2.21)–(2.23)成立，那么问题(1.1)、(1.3)的每个解在$ G $内是振动的.

例子3.1   考虑如下形式的具有多时滞的分数阶脉冲偏微分方程

$ \begin{equation} \left\{ \begin{split} &D_{+, t} ^{\frac{1}{2}} u(t, x)+\frac{1}{t}D_{+, t} ^{\frac{1}{4}}u(t, x) = e^{-t}u^2 (t, x)\Delta u(t, x) + \frac{t^2}{2} u^2(t-\frac{1}{2}, x)\Delta u(t-\frac{1}{2}, x)\\ &\; \; \; \; -(1+t^2+x^2)u(t-\frac{2\pi}{3}, x)e^{[u(t-\frac{2\pi}{3}, x)]^2} - tcosx, \; \; t\neq t_k , (t, x)\in R_+ \times \Omega = G, \\ &D_{+, t} ^{\frac{1}{4}} u(t_k ^+, x)-D_{+, t} ^{\frac{1}{4}} u(t_k ^-, x) = t_k ^{-3}cosx\cdot D_{+, t} ^{\frac{1}{4}} u(t_k, x), \\ &u(t_k ^+, x)-u(t_k ^-, x) = -t^2\sin x\cdot u(t_k, x)\; \; t = t_k, k = 1, 2, \cdots, \end{split} \right. \end{equation} $

(2.29)

边界条件满足

$ \begin{equation} \begin{split} \frac{\partial u(t, x)}{\partial n} = 0, \; \; (t, x)\in R_+ \times \partial\Omega, t\neq t_k. \end{split} \end{equation} $

(2.30)

其中$ \alpha = \frac{1}{4}, \Omega = (0, \frac{\pi}{2}), m = n = 1, p(t) = \frac{1}{t}, r(t) = \frac{1}{t}. a(t) = e^{-t}, a_1(t) = \frac{t^2}{2}, h(u(t, x)) = u^2(t, x), \tau_1 = \frac{1}{2}, \delta_1 = \frac{2\pi}{3}, $ $ {q_1}(t,x) = {x^2} + {t^2} + 1,{f_1}(u(t,x)) = u(t,x){e^{{u^2}}},g(t,x) = tcosx, $$\sigma ({t_k},x) = t_k^{ - 3}cosx,\theta ({t_k},x) = - {t^2}\sin x,{\alpha _k} = t_k^{ - 3}, $$ {\theta _k} = - {t^2},(t,x) \in {R^ + } \times (0,\frac{\pi }{2}).$计算可得

$ \begin{align*} \int_{\tau_2} ^\infty \text{exp}(-\int_{t_0} ^t p(\sigma) d\sigma)ds = \int_{\tau_2} ^\infty \text{exp}(-\int_{t_0} ^t \frac{1}{\sigma}p(\sigma)d\sigma)ds = \int_{\tau_2} ^\infty \frac{t_0}{t}dt = \infty, \end{align*} $

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow \infty}\frac{\int_{\tau_1}^ t\prod\limits_{s<t_l<t}(1+\alpha_k)\text{exp}(-\int_s^ t p(\sigma) d\sigma)G(s)ds}{\prod\limits_{\tau_1<t_l<t}(1+\alpha_k)\text{exp}(-\int_{\tau_1}^t p(s))ds} = \infty, \end{eqnarray*} $

且有

$ \begin{eqnarray*} \liminf\limits_{t\rightarrow \infty}\frac{\int_{\tau_1}^ t\prod\limits_{s<t_l<t}(1+\alpha_k)\text{exp}(-\int_s^ t p(\sigma) d\sigma)G(s)ds}{\prod\limits_{\tau_1<t_l<t}(1+\alpha_k)\text{exp}(-\int_{\tau_1}^t p(s))ds} = -\infty. \end{eqnarray*} $

因此, 满足定理2.3的所有条件.故问题(3.1)–(3.2)的所有非零解为振动的.