脉冲微分方程理论的出现使得对理论物理、化学、生物技术和人口动力学等学科中的某些过程和现象的精确模拟成为可能.近年来, 由于分数阶微分方程在反常扩散、多孔介质力学、非牛顿流体力学、粘弹性力学、软物质力学、生物医学以及控制系统等领域的广泛应用, 其研究越来越受到人们的关注.不论在理论还是应用中, 人们主要研究分数阶微分方程的定性性质, 研究方向众多, 如解的存在性以及带有初值条件或边界条件的分数阶微分方程解的稳定性, 其相关定性理论研究迅速发展, 并得到了一些研究成果.偏微分方程解的振动问题已有一些学者通过研究得到了一些结论^[1]-[3].近年来, 分数阶微分方程的振动性问题受到广泛关注, 一些学者研究带阻尼项的分数阶微分方程的振动性^[4]-[10], 并陆续有很好的研究成果发表^[11]-[17].但关于脉冲分数阶微分方程的振动性研究却很少, 本篇论文借鉴现有的一些研究结果以及方法来进一步讨论脉冲分数阶偏微分方程的性质.

本文将讨论如下方程

$ \begin{equation} \begin{cases} &D_{+, t}^{\alpha}\left(r(t)D_{+, t}^{\alpha}u(x, t)\right)+p(t)D_{+, t}^{\alpha}u(x, t) = a(t)h(u)\Delta u(x, t)\\ & +\sum\limits_{i = 1}^{j}a_{i}(t)h_{i}(u(x, t-\tau_{i}))\Delta u(x, t-\tau_{i})-m(x, t, u(x, t))-q(x, t)F(u(x, t)), \quad t\neq t_{k}, \\ &D_{+, t}^{\alpha}u(x, t_{k}^{+})-D_{+, t}^{\alpha}u(x, t_{k}^{-}) = \alpha_{k}D_{+, t}^{\alpha}u(x, t_{k}), \ k = 1, 2, 3, \cdots, (x, t)\in\Omega\times R_{+}\equiv E, \\ &u(x, t_{k}^{+})-u(x, t_{k}^{-}) = \beta_{k}u(x, t_{k}), \ k = 1, 2, 3, \cdots, (x, t)\in\Omega\times R_{+}\equiv E, \end{cases} \end{equation} $

(1.1)

边界条件

$ \begin{align} \frac{\partial u(x, t)}{\partial N} = 0, \qquad (x, t)\in\partial\Omega\times R_{+}, \qquad t\neq t_{k}, \end{align} $

(1.2)

其中$ \alpha\in(0, 1) $, $ \Delta $为拉普拉斯算子, $ \Omega $是$ R^{n} $内的有界域, $ \partial\Omega $充分光滑, $ \overline{\Omega} = \Omega\cup\partial\Omega $, $ N $为$ \partial\Omega $的单位外法向量, $ 0 < t_{1} < \cdots < t_{i} < \cdots $且$ \lim\limits_{i\to\infty}t_{i} = +\infty $.

如下是本文的基本假设

(H1) $ a(t), p(t), a_{i}(t)\in C(R_{+}; R_{+}) $, $ r(t)\in C^{\alpha}(R_{+}; R_{+}) $且$ \tau_{i}\geq0 $是常数, $ i\in I_{j} = \{1, 2\cdots, j\} $; $ \alpha_{k} > -1, \; \beta_{k} > -1 $.

(H2) $ q(x, t)\in C(\overline{E}; R_{+}) $, $ q(t) = \min\limits_{x\in\overline{\Omega}}q(x, t) $; $ h'(u), h_{i}^{'}(u), F(u)\in C(R; R) $, 对于$ x\neq0 $, 存在一个正常数$ c $, 使得$ \frac{F(x)}{x}\geq c > 0 $; $ u\neq0, \; uh'(u) > 0, \; uh_{i}^{'}(u) > 0 $.

(H3) $ m\in C(\overline{E}\times R; R) $,

$ m(x, t, \eta) = \left\{ \begin{array}{rcl} \geq0, & &{\eta\in(0, +\infty)}, \\ \leq0, & &{\eta\in(-\infty, 0)}. \end{array}\right. $

(H4) $ u(x, t) $和它的分数阶导数$ D_{+, t}^{\alpha}u(x, t) $是分段连续函数, $ t = t_{k}, k = 1, 2, 3, \cdots $为第一类间断点, 且在$ t = t_{k} $处左连续, 即$ D_{+, t}^{\alpha}u(x, t_{k}^{-}) = D_{+, t}^{\alpha}u(x, t_{k}), u(x, t_{k}^{-}) = u(x, t_{k}). $

定义1.1 ^[18]    $ f:R_{+}\rightarrow R $, 阶数为$ \alpha > 0 $的Riemann-Liouville左侧分数阶积分定义如下

$ \begin{align} (I_{0+}^{\alpha}f)(t) = \frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-v)^{\alpha-1}f(v)dv. \end{align} $

(1.3)

上式在$ R_{+} $是逐点定义的, $ \Gamma $是gamma函数.

定义1.2 ^[19]   修正后的Riemann-Liouville分数阶导数定义如下

$ \begin{equation} \begin{cases} &D_{t}^{\alpha}f(t) = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{0}^{t}{(t-\xi)^{-\alpha}}(f(\xi)-f(0))d\xi , \;0<\alpha<1, \\ &D_{t}^{\alpha}f(t) = \left(f^{n}(t)\right)^{\alpha-n}, \;1\leq n\leq\alpha<n+1. \end{cases} \end{equation} $

(1.4)

下面给出关于$ \alpha $阶修正后的Riemann-Liouville分数阶导数的一些计算公式

$ \begin{eqnarray*} D_{t}^{\alpha}t^{r}& = &\frac{\Gamma(1+\alpha)}{\Gamma(1+r-\alpha)}t^{r-\alpha}, \\ D_{t}^{\alpha}\left(f(t)h(t)\right)& = &h(t)D_{t}^{\alpha}f(t)+f(t)D_{t}^{\alpha}h(t), \\ D_{t}^{\alpha}f[h(t)]& = &f'_{h}D_{t}^{\alpha}h(t) = D_{h}^{\alpha}f[h(t)](h'(t))^{\alpha}. \end{eqnarray*} $

及一些在本文的证明中要用到的记号

$ U(t) = \int_{\Omega}u(x, t)dx, \quad R(t) = I_{+}^{\alpha}\left(\frac{p(t)}{r(t)}\right), \quad\xi = \frac{t^{\alpha}}{\Gamma(1+\alpha)}, \quad \widetilde{r}(\xi) = r(t), $

$ \quad \widetilde{p}(\xi) = p(t), \quad\widetilde{q}(\xi) = q(t), \quad\widetilde{\psi}(\xi) = \psi(t), \quad\widetilde{R}(\xi) = R(t). $

定理2.1   如果下列脉冲分数阶微分不等式

$ \begin{equation} \begin{cases} &D_{+, t}^{\alpha}\left[r(t)D_{+, t}^{\alpha}U(t)\right]+p(t)D_{+, t}^{\alpha}U(t)\leq-cq(t)U(t)\\ &D_{+, t}^{\alpha}U(t^{+}) = (1+\alpha_{k})D_{+, t}^{\alpha}U(t), \qquad k = 1, 2, 3, \cdots, \\ &U(t^{+}) = (1+\beta_{k})U(t), \qquad k = 1, 2, 3, \cdots \end{cases} \end{equation} $

(2.1)

没有最终正解, 且下列脉冲分数阶微分不等式

$ \begin{equation} \begin{cases} &D_{+, t}^{\alpha}\left[r(t)D_{+, t}^{\alpha}U(t)\right]+p(t)D_{+, t}^{\alpha}U(t)\geq-c q(t)U(t)\\ &D_{+, t}^{\alpha}U(t^{+}) = (1+\alpha_{k})D_{+, t}^{\alpha}U(t), \qquad k = 1, 2, 3, \cdots, \\ &U(t^{+}) = (1+\beta_{k})U(t), \qquad k = 1, 2, 3, \cdots \end{cases} \end{equation} $

(2.2)

没有最终负解, 那么方程(1.1)和(1.2)的每个非平凡解$ u(x, t) $在$ E $内都是振动的.

证   设$ u(x, t) $是方程(1.1)和(1.2)的一个非振动解.不妨设存在$ t_{0}\geq0 $, 使得$ u(x, t) > 0 $, $ u(x, t-\tau_{i}) > 0 $, $ (x, t)\in\Omega\times[t_{0}, +\infty) $.

当$ t\neq t_{k} $.将方程(1.1)的第一个式子关于$ x $在$ \Omega $内积分, 得到

$ \begin{align} &D_{+, t}^{\alpha}\int_{\Omega}\left(r(t)D_{+, t}^{\alpha}u(x, t)\right)dx+p(t)\int_{\Omega}D_{+, t}^{\alpha}u(x, t)dx\\ = &a(t)\int_{\Omega}h(u)\Delta u(x, t)dx+\sum\limits_{i = 1}^{j}a_{i}(t)\int_{\Omega}h_{i}(u(x, t-\tau_{i}))\Delta u(x, t-\tau_{i})dx\\ &-\int_{\Omega}m(x, t, u(x, t))dx-\int_{\Omega}q(x, t)F(u(x, t))dx. \end{align} $

(2.3)

由Green公式、边值条件(1.2)及条件(H2), 容易得到

$ \begin{eqnarray} &&\int_{\Omega}h(u)\Delta u(x, t)dx\ = \int_{\partial\Omega}h(u)\frac{\partial u(x, t)}{\partial N}dS-\int_{\Omega}h'(u)\mid \text {grad} \;u\mid^{2}dx\\ & = &-\int_{\Omega}h'(u)\mid \text {grad}\; u\mid^{2}dx \leq0, \quad\qquad\qquad\qquad\qquad\qquad\qquad t\geq t_{0}, \end{eqnarray} $

(2.4)

$ \begin{eqnarray} &&\int_{\Omega}h_{i}(u(x, t-\tau_{i}))\Delta u(x, t-\tau_{i})dx \\ & = &-\int_{\Omega}h_{i}'(u(x, t-\tau_{i}))\mid \text {grad}\;u(x, t-\tau_{i})\mid^{2}dx\leq0, \qquad\qquad \;t\geq t_{0}, \end{eqnarray} $

(2.5)

及

$ \begin{eqnarray} &&\int_{\Omega}m(x, t, u(x, t))dx\geq0, \qquad \qquad \qquad \qquad \qquad \qquad \qquad \, \, \, t\geq t_{0}, \end{eqnarray} $

(2.6)

$ \begin{eqnarray} &&\int_{\Omega}q(x, t)F(u(x, t))dx\geq\int_{\Omega}cq(t)u(x, t)dx\geq cq(t)U(t), \quad\, \, \, \, t\geq t_{0}. \end{eqnarray} $

(2.7)

由(2.3)–(2.7)式, 得到

$ \begin{align} D_{+, t}^{\alpha}\left[r(t)D_{+, t}^{\alpha}U(t)\right]+p(t)D_{+, t}^{\alpha}U(t)\leq-cq(t)U(t), \qquad\qquad\quad t\geq t_{0}. \end{align} $

(2.8)

当$ t = t_{k} $.分别对方程(1.1)的第二个式子和第三个式子关于$ x $在$ \Omega $内积分, 得到

$ \begin{align} D_{+, t}^{\alpha}U(t_{k}^{+}) = D_{+, t}^{\alpha}\int_{\Omega}u(x, t_{k}^{+})dx = (1+\alpha_{k})D_{+, t}^{\alpha}\int_{\Omega}u(x, t_{k})dx = (1+\alpha_{k})D_{+, t}^{\alpha}U(t_{k}), \\ \;k = 1, 2, 3, \cdots. \end{align} $

(2.9)

$ \begin{align} U(t_{k}^{+}) = \int_{\Omega}u(x, t_{k}^{+})dx = (1+\beta_{k})\int_{\Omega}u(x, t_{k})dx = (1+\beta_{k})U(t_{k}), \;k = 1, 2, 3, \cdots. \end{align} $

(2.10)

由(2.8)–(2.10)式知, $ U(t) = \int_{\Omega}u(x, t)dx $是脉冲分数阶微分不等式(2.1)的最终正解, 这与假设矛盾.

若$ u(x, t) $是脉冲分数阶微分方程(1.1)和(1.2)的最终负解.用类似方法, 容易得到$ U(t) = \int_{\Omega}u(x, t)dx $是脉冲分数阶微分不等式(2.2)的最终负解, 这与假设矛盾.定理得证.

引理2.2 ^[18]  若$ G(t) = \int_{0}^{t}(t-v)^{-\alpha}U(v)dv, \; \; \alpha\in(0, 1), \; \; t > 0, $则$ G'(t) = \Gamma(1-\alpha)(D_{+}^{\alpha}U)(t), \; \; \alpha\in(0, 1), \; \; t > 0. $

引理2.3   若$ 0 < \alpha < 1 $, 则$ (D_{+}^{\alpha}I_{0+}^{\alpha}f)(x) = f(x). $

证   由定义1.1和1.2可得

$ \begin{array}{l} (D_{+}^{\alpha}I_{0+}^{\alpha}f)(x) & = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_{0}^{x}\frac{dt}{(x-t)^{\alpha}}(\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{f(s)ds}{(t-s)^{1-\alpha}}-I_{0+}^{\alpha}f(0))\nonumber\\ & = \frac{1}{\Gamma(\alpha)\Gamma(1-\alpha)}\frac{d}{dx}\int_{0}^{x}\frac{dt}{(x-t)^{\alpha}}\int_{0}^{t}\frac{f(s)}{(t-s)^{1-\alpha}}ds\nonumber\\ & = \frac{1}{\Gamma(\alpha)\Gamma(1-\alpha)}\frac{d}{dx}\int_{0}^{x}f(s)ds\int_{s}^{x}(x-t)^{-\alpha}(t-s)^{\alpha-1}dt. \end{array} $

令$ t = s+\mu(x-s) $, 利用Beta函数的定义得

$ \begin{array}{l} \int_{s}^{x}(x-t)^{-\alpha}(t-s)^{\alpha-1}dt = \int_{0}^{1}\mu^{\alpha-1}(1-\mu)^{-\alpha}d\mu = B(\alpha, 1-\alpha), \\ (D_{+}^{\alpha}I_{0+}^{\alpha}f)(x) = \frac{B(\alpha, 1-\alpha)}{\Gamma(\alpha)\Gamma(1-\alpha)}\frac{d}{dx}\int_{0}^{x}f(s)ds = f(x). \end{array} $

引理2.4 ^[20]   假设$ w\in PC^{1}[R_{+}, R], $

$ w'(t)\leq g_{1}(t)w(t)+g_{2}(t), \qquad t\neq t_{k}, \quad t\geq t_{0}, $

$ w(t_{k}^{+})\leq (1+\delta_{k})w(t_{k}), \qquad k = 1, 2, 3, \cdots, $

其中$ g_{1}, g_{2}\in C[R_{+}, R] $, $ \delta_{k} $是常数, $ PC^{1}[R_{+}, R] = \{x(t):R_{+}\to R, \; x(t)\mbox{在除}\; t = t_{k}, k = 1, 2, \cdots\mbox{以外的点连续可微}, x(t_{k}^{+}), x(t_{k}^{-}), x'(t_{k}^{+}), x'(t_{k}^{-}) \mbox{存在, 且}\; x(t_{k}) = x(t_{k}^{-}), x'(t_{k}) = x'(t_{k}^{-})\} $.则

$ \begin{array}{l} w(t)\leq& w(t_{0})\prod\limits_{t_{0}<t_{k}<t}(1+\delta_{k})\exp(\int_{t_{0}}^{t}g_{1}(s)ds)\\ &+\int_{t_{0}}^{t}\prod\limits_{s<t_{k}<t}(1+\delta_{k})\exp(\int_{s}^{t}g_{1}(\sigma)d\sigma)g_{2}(s)ds, \quad\; \; t\geq t_{0}. \end{array} $

定理2.5   假设存在$ t^{*}\geq0 $,

$ \begin{align} \int_{t^{*}}^{\infty}\frac{1}{r(s)e^{R(s)}}ds = +\infty, \end{align} $

(2.11)

且

$ \begin{align} \lim\limits_{\xi\to\infty}\int_{t^{*}}^{\xi}\prod\limits_{t^{*}<\xi_{k}<s}\frac{1+\beta_{k}}{1+\alpha_{k}}\widetilde{\psi}(s)ds = +\infty, \end{align} $

(2.12)

其中$ \widetilde{\psi}(s) = ce^{\widetilde{R}(s)}\widetilde{q}(s) $.那么方程(1.1)–(1.2)每个非平凡解$ u(x, t) $在$ E $内振动.

证   用反证法.假设$ U(t) $是脉冲分数阶微分不等式(2.1)的非振动解.不失一般性, 假设$ U(t) $是脉冲分数阶微分不等式(2.1)的最终正解, 存在$ t^{*}\geq0 $, 使得$ U(t) > 0, U(t-\tau_{i}) > 0, G(t) > 0 $, $ t\geq t^{*} $.由(2.1)式及引理2.3, 有

$ \begin{align} D_{+}^{\alpha}\left[e^{R(t)}r(t)D_{+}^{\alpha}U(t)\right] & = e^{R(t)}D_{+}^{\alpha}\left[r(t)D_{+}^{\alpha}U(t)\right]+r(t)D_{+}^{\alpha}U(t)D_{+}^{\alpha}e^{R(t)}\\ & = e^{R(t)}D_{+}^{\alpha}\left[r(t)D_{+}^{\alpha}U(t)\right]+r(t)D_{+}^{\alpha}U(t)e^{R(t)}D_{+}^{\alpha}I_{0+}^{\alpha}\left(\frac{p(t)}{r(t)}\right)\\ & = e^{R(t)}D_{+}^{\alpha}\left[r(t)D_{+}^{\alpha}U(t)\right]+e^{R(t)}p(t)D_{+}^{\alpha}U(t)\\ & = e^{R(t)}\left[D_{+}^{\alpha}[r(t)D_{+}^{\alpha}U(t)]+p(t)D_{+}^{\alpha}U(t)\right]\\ &<-ce^{R(t)}q(t)U(t) <0. \end{align} $

(2.13)

由上式知$ e^{R(t)}r(t)D_{+}^{\alpha}U(t) $在$ t\geq t^{*} $上是减函数.不妨设$ D_{+}^{\alpha}U(t) > 0 $, $ t\in[t^{*}, \infty) $.否则, 存在$ T\in[t^{*}, \infty) $, 使得$ D_{+}^{\alpha}U(T) < 0 $, $ e^{R(t)}r(t)D_{+}^{\alpha}U(t)\leq e^{R(T)}r(T)D_{+}^{\alpha}U(T) = c_{1} < 0, $其中$ c_{1} $在$ t\in[T, \infty) $是一个常数.根据引理2.2, 有

$ \begin{align} \frac{G'(t)}{\Gamma(1-\alpha)} = D_{+}^{\alpha}U(t)\leq\frac{c_{1}}{e^{R(t)}r(t)}. \end{align} $

(2.14)

对上述不等式从$ T $到$ t $积分, 得

$ \begin{eqnarray} &&\int_{T}^{t}\frac{G'(s)}{\Gamma(1-\alpha)}ds\leq\int_{T}^{t}\frac{c_{1}}{e^{R(s)}r(s)}ds, \end{eqnarray} $

(2.15)

$ \begin{eqnarray} &&G(t)\leq G(T)+\Gamma(1-\alpha)c_{1}\int_{T}^{t}\frac{1}{e^{R(s)}r(s)}ds. \end{eqnarray} $

(2.16)

当$ t\to+\infty $, $ \lim\limits_{t\to\infty}G(t)\leq-\infty $, 这与$ G(t) > 0 $矛盾.因此$ D_{+}^{\alpha}U(t) > 0 $, $ t > T $.

令

$ w(t) = e^{R(t)}\frac{r(t)D_{+}^{\alpha}U(t)}{U(t)}, $

则$ w(t) > 0 $, 由(2.1)式及引理2.3, 容易得到

$ \begin{align} D_{+}^{\alpha}w(t) & = e^{R(t)}D_{+}^{\alpha}[\frac{r(t)D_{+}^{\alpha}U(t)}{U(t)}]+\frac{r(t)D_{+}^{\alpha}U(t)}{U(t)}D_{+}^{\alpha}e^{R(t)}\\ & = e^{R(t)}\frac{D_{+}^{\alpha}[r(t)D_{+}^{\alpha}U(t)]}{U(t)}-\frac{e^{R(t)}r(t)D_{+}^{\alpha}U(t)D_{+}^{\alpha}U(t)}{U^{2}(t)}+\frac{p(t)}{r(t)}w(t)\\ &\leq e^{R(t)}\frac{-p(t)D_{+, t}^{\alpha}U(t)-c q(t)U(t)}{U(t)}-\frac{w^{2}(t)}{e^{R(t)}r(t)}+\frac{p(t)}{r(t)}w(t)\\ &\leq-ce^{R(t)}q(t)-\frac{p(t)}{r(t)}w(t)-\frac{w^{2}(t)}{e^{R(t)}r(t)}+\frac{p(t)}{r(t)}w(t)\\ &\leq-ce^{R(t)}q(t). \end{align} $

(2.17)

即

$ \begin{align} D_{+}^{\alpha}w(t)\leq-\psi(t), \end{align} $

(2.18)

其中$ \psi(t) = ce^{R(t)}q(t) $.利用(H1)和$ w(t) $的定义式, 不等式(2.1)的第二个式子和第三个式子变为

$ \begin{align} w(t_{k}^{+}) = \frac{1+\alpha_{k}}{1+\beta_{k}}w(t_{k}), \qquad k = 1, 2, 3, \cdots. \end{align} $

(2.19)

令$ \xi = \frac{t^{\alpha}}{\Gamma(1+\alpha)} $, $ \widetilde{w}(\xi) = w(t) $, $ \widetilde{\psi}(\xi) = \psi(t) $, 则$ \widetilde{w}(\xi) > 0 $, 可以得到

$ \begin{align} D_{+}^{\alpha}w(t) = D_{+}^{\alpha}\widetilde{w}(\xi) = \widetilde{w'}(\xi)D_{+}^{\alpha}(\xi) = \widetilde{w'}(\xi), \end{align} $

(2.20)

则(2.18)式变为

$ \begin{align} \widetilde{w'}(\xi)\leq-\widetilde{\psi}(\xi). \end{align} $

(2.21)

由(2.19)–(2.21)式, 容易得到

$ \begin{equation} \begin{cases} \widetilde{w'}(\xi)\leq-\widetilde{\psi}(\xi), \qquad \xi\neq \xi_{k}, \quad \xi\geq t^{*}, \\ \widetilde{w}(\xi_{k}^{+}) = \frac{1+\alpha_{k}}{1+\beta_{k}}\widetilde{w}(\xi_{k}), \qquad k = 1, 2, 3, \cdots.\\ \end{cases} \end{equation} $

(2.22)

根据引理2.4, 得

$ \begin{align} \widetilde{w}(\xi) &\leq\widetilde{w}(t^{*})\prod\limits_{t^{*}<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}-\int_{t^{*}}^{\xi}\prod\limits_{s<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}\widetilde{\psi}(s)ds\\ & = \prod\limits_{t^{*}<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}\left[\widetilde{w}(t^{*})-\int_{t^{*}}^{\xi}\prod\limits_{t^{*}<\xi_{k}<s}\frac{1+\beta_{k}}{1+\alpha_{k}}\widetilde{\psi}(s)ds\right] <0, \end{align} $

(2.23)

这与$ \widetilde{w}(\xi) > 0 $矛盾, 定理2.5的第一部分证毕.

若$ U(t) $是不等式(2.2)的一个非振动解.不妨设$ U(t) $是脉冲分数阶微分不等式(2.2)的最终负解, 则$ G(t) < 0, t\in [t^{*}, \infty) $.用类似方法, 容易得到$ D_{+}^{\alpha}U(t) < 0 $, $ t\geq t^{*} $.令$ w(t) = -e^{R(t)}\frac{r(t)D_{+}^{\alpha}U(t)}{U(t)} $, 则$ w < 0 $.根据(2.2)式, 可以得到 $ D_{+}^{\alpha}w(t)\geq c e^{R(t)}q(t) = \psi(t) $和$ w(t_{k}^{+}) = \frac{1+\alpha_{k}}{1+\beta_{k}}w(t_{k}), k = 1, 2, 3, \cdots. $即

$ \begin{equation} \begin{cases} \widetilde{w'}(\xi)\geq\widetilde{\psi}(\xi), \qquad \xi\neq \xi_{k}, \quad \xi\geq t^{*}, \\ \widetilde{w}(\xi_{k}^{+}) = \frac{1+\alpha_{k}}{1+\beta_{k}}\widetilde{w}(\xi_{k}), \qquad k = 1, 2, 3, \cdots.\\ \end{cases} \end{equation} $

(2.24)

若$ \widetilde{w}(\xi) = -\widetilde{v}(\xi) $

$ \begin{equation} \begin{cases} \widetilde{v'}(\xi)\leq-\widetilde{\psi}(\xi), \qquad \xi\neq \xi_{k}, \quad \xi\geq t^{*}, \\ \widetilde{v}(\xi_{k}^{+}) = \frac{1+\alpha_{k}}{1+\beta_{k}}\widetilde{v}(\xi_{k}), \qquad k = 1, 2, 3, \cdots.\\ \end{cases} \end{equation} $

(2.25)

由引理2.4, 得

$ \begin{align} \widetilde{v}(\xi)\leq \widetilde{v}(t^{*})\prod\limits_{t^{*}<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}-\int_{t^{*}}^{\xi}\prod\limits_{s<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}\widetilde{\psi}(s)ds, \end{align} $

(2.26)

则

$ \begin{align} \widetilde{w}(\xi) &\geq\widetilde{w}(t^{*})\prod\limits_{t^{*}<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}+\int_{t^{*}}^{\xi}\prod\limits_{s<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}\widetilde{\psi}(s)ds\\ & = \prod\limits_{t^{*}<\xi_{k}<\xi}\frac{1+\alpha_{k}}{1+\beta_{k}}[\widetilde{w}(t^{*})+\int_{t^{*}}^{\xi}\prod\limits_{t^{*}<\xi_{k}<s}\frac{1+\beta_{k}}{1+\alpha_{k}}\widetilde{\psi}(s)ds] >0, \end{align} $

(2.27)

这与结论$ \widetilde{w}(\xi) < 0 $矛盾.定理2.5证毕.

例1   考虑如下问题

$ \begin{equation} \begin{cases} &D_{+, t}^{\frac{1}{2}}[e^{-t-\frac{2}{3\sqrt{\pi}}t^{\frac{3}{2}}}D_{+, t}^{\frac{1}{2}}u(x, t)]+\frac{1}{2}te^{-t-\frac{2}{3\sqrt{\pi}}t^{\frac{3}{2}}}D_{+, t}^{\frac{1}{2}}u(x, t)\\ & = e^{t}u^{2}\Delta u(x, t)+t^{\frac{2}{3}}\Delta u(x, t-\frac{\pi}{3})-\frac{u^{3}(x, t)}{1+t^{2}+x^{2}}-(x^{2}+t^{2})u(x, t), \quad t\neq t_{k}, \\ &D_{+, t}^{\frac{1}{2}}u(x, t_{k}^{+})-D_{+, t}^{\frac{1}{2}}u(x, t_{k}^{-}) = D_{+, t}^{\frac{1}{2}}u(x, t_{k}), k = 1, 2, 3, \cdots, \, (x, t)\in \Omega\times R_{+}\equiv E, \\ &u(x, t_{k}^{+})-u(x, t_{k}^{-}) = u(x, t_{k}), \ k = 1, 2, 3, \cdots, \, (x, t)\in \Omega\times R_{+}\equiv E, \end{cases} \end{equation} $

(3.1)

边界条件为

$ \begin{align} \frac{\partial u(x, t)}{\partial N} = 0, \qquad (x, t)\in\partial\Omega\times R_{+}, \qquad t\neq t_{k}, \end{align} $

(3.2)

证  在(3.1)式中, $ r(t) = e^{-t-\frac{2}{3\sqrt{\pi}}t^{\frac{3}{2}}}, \; p(t) = \frac{1}{2}te^{-t-\frac{2}{3\sqrt{\pi}}t^{\frac{3}{2}}}, \; a(t) = e^{t}, a_{1}(t) = t^{\frac{2}{3}}, \, m(x, t, u) = \frac{u^{3}(x, t)}{1+t^{2}+x^{2}} $, $ h(u) = u^{2} $, $ h_{1}(u) = 1 $, $ q(x, t) = x^{2}+t^{2} $, $ q(t) = \min(x^{2}+t^{2}) = t^{2}, F(u) = u, \frac{F(u)}{u}\geq c = 1 $, $ j = 1 $, $ \alpha_{k} = \beta_{k} = 1 $.那么

$ \begin{array}{l} &R(t) = I_{0+}^{\frac{1}{2}}(\frac{p(t)}{r(t)})\ = \frac{1}{\Gamma(\frac{1}{2})}\int_{0}^{t}(t-v)^{\frac{1}{2}-1}(\frac{v}{2})dv = \frac{1}{\Gamma(\frac{1}{2})}\int_{0}^{t}(-v)d(t-v)^{\frac{1}{2}} = \frac{2}{3\sqrt{\pi}}t^{\frac{3}{2}}, \\ &\int_{t^{*}}^{\infty}\frac{1}{r(s)e^{R(s)}}ds = \int_{t^{*}}^{\infty}\frac{1}{e^{-s-\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}}ds = \int_{t^{*}}^{\infty}e^{s}ds = +\infty, \end{array} $

则

$ \begin{array}{l} \lim\limits_{\xi\to\infty}\int_{t^{*}}^{\xi}e^{\widetilde{R(s)}}\widetilde{q(s)}ds\ & = \lim\limits_{\xi\to\infty}\int_{t^{*}}^{\xi}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}s^{2}ds = \lim\limits_{\xi\to\infty}[\sqrt{\pi}s^{\frac{3}{2}}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}\mid_{t^{*}}^{\xi}-\int_{t^{*}}^{\xi}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}\sqrt{\pi}ds^{\frac{3}{2}}]\nonumber\\ & = \lim\limits_{\xi\to\infty}[\sqrt{\pi}s^{\frac{3}{2}}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}\mid_{t^{*}}^{\xi}-\int_{t^{*}}^{\xi}\frac{3\sqrt{\pi}}{2}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}s^{\frac{1}{2}}ds]\nonumber\\ & = \lim\limits_{\xi\to\infty}[\sqrt{\pi}s^{\frac{3}{2}}e^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}\mid_{t^{*}}^{\xi}-\frac{3\pi}{2}\int_{t^{*}}^{\xi}de^{\frac{2}{3\sqrt{\pi}}s^{\frac{3}{2}}}]\nonumber\\ & = \lim\limits_{\xi\to\infty}[\sqrt{\pi}\xi^{\frac{3}{2}}e^{\frac{2}{3\sqrt{\pi}}\xi^{\frac{3}{2}}}-\frac{3\pi}{2}e^{\frac{2}{3\sqrt{\pi}}\xi^{\frac{3}{2}}}-\sqrt{\pi}(t^{*})^{\frac{3}{2}}e^{\frac{2}{3\sqrt{\pi}}(t^{*})^{\frac{3}{2}}}-\frac{3\pi}{2}e^{\frac{2}{3\sqrt{\pi}}(t^{*})^{\frac{3}{2}}}]\nonumber\\ & = +\infty, \end{array} $

$ \begin{array}{l} \lim\limits_{\xi\to\infty}\int_{t^{*}}^{\xi}\prod\limits_{t^{*}<\xi_{k}<s}\frac{1+\beta_{k}}{1+\alpha_{k}}\widetilde{\psi(s)}ds = +\infty. \end{array} $

则由定理2.5可知, 问题(3.1)和(3.2)的解都是振动的.