本文主要研究如下具非线性记忆项的高阶阻尼双曲系统的爆破行为

$ \begin{eqnarray} \left\{ \begin{array}{ll} u_{tt}+\left(-\Delta\right)^{m}u +u_t=\frac{1}{\Gamma\left(1-\gamma\right)} \int_0^t \left(t-s\right)^{-\gamma}\left|v\left(s\right)\right|^{p}{\rm d}s, &\left(x, t\right)\in{\Bbb R}^N\times{\Bbb R}_+^1, \\ v_{tt}+\left(-\Delta\right)^{m}v +v_t=\frac{1}{\Gamma\left(1-\delta\right)}\int_0^t \left(t-s\right)^{-\delta}\left|u\left(s\right)\right|^{q}{\rm d}s, &\left(x, t\right)\in{\Bbb R}^N\times{\Bbb R}_+^1, \\ u\left(x, 0\right) =u_0\left(x\right), u_t\left(x, 0\right) =u_1\left(x\right), &x\in {\Bbb R}^N, \\ v\left(x, 0\right) =v_0\left(x\right), v_t\left(x, 0\right) =v_1\left(x\right), &x\in {\Bbb R}^N, \end{array}\right. \end{eqnarray} $

(1.1)

其中$m$, $N\in{\Bbb Z}^+$, $\gamma$, $\delta\in \left(0, 1\right)$, $p$, $q\in \left(1, +\infty\right)$, $\Gamma$表示Euler-Gamma函数, $\left(u_i, v_i\right)\in W^{1-i, 2m}\left( {\Bbb R}^N\right)\times W^{1-i, 2m}\left( {\Bbb R}^N\right)$, 且$\mbox{supp}\left(u_i, v_i\right)\subset B_\rho\equiv \left\{x\in {\Bbb R}^N: \left|x\right|<\rho\right\}$, 其中$\rho$为一合适的正常数, $i=0, 1$.

近些年来, 很多作者对具阻尼项的双曲方程整体解的存在性与非存在性进行了研究. Li和Zhou^[1], Todorova和Yordanov^[2], Zhang^[3]研究了如下初值问题解的整体存在性与有限时刻爆破

$ \begin{eqnarray} \left\{ \begin{array}{llll} u_{tt}-\Delta u +u_t=\left|u\right|^{p}, &\left(x, t\right)\in{\Bbb R}^N\times{\Bbb R}_+^1, \\ u\left(x, 0\right) =u_0\left(x\right), u_t\left(x, 0\right) =u_1\left(x\right), &x\in {\Bbb R}^N, \end{array} \right . \end{eqnarray} $

(1.2)

得到了临界指数$p_f=1 + \frac{2}{N}$.

Fino^[4]研究了问题

$ \begin{eqnarray} \left\{ \begin{array}{ll} u_{tt}-\Delta u +u_t= \int_0^t \left(t-s\right)^{-\gamma}\left|u\left(s\right)\right|^{p}{\rm d}s, &\left(x, t\right)\in{\Bbb R}^N\times{\Bbb R}_+^1, \\[2mm] u\left(x, 0\right) =u_0\left(x\right), u_t\left(x, 0\right) =u_1\left(x\right), &x\in {\Bbb R}^N, \end{array}\right. \end{eqnarray} $

(1.3)

其中$\gamma\in\left(0, 1\right)$, $p\in\left(1, +\infty\right)$.当空间维数$N\in\left\{1, 2, 3\right\}$时, Fino讨论了小初值解的整体存在性和渐近行为; 对任意的$N\in {\Bbb Z}^+$, Fino得到了解的爆破结果. D'Abbicco^[5]在条件$N\in\left\{1, 2, 3, 4, 5\right\}$下研究了问题(1.3)整体解的存在性与非存在性.

许勇强^[6]和Berbiche^[7]分别在条件$m=1$下考察了问题(1.1), 给出了解在有限时刻发生爆破的条件; 同时, 文献[6]中, 作者获得了一维情况下小初值解的整体存在性并给出了解的渐近行为, 而文献[7]中, 作者给出了$N\in\left\{1, 2, 3\right\}$时解的整体存在性结果.

其他的一些相关工作可参考文献[8-12].受上述工作的启发, 我们讨论问题(1.1)解的爆破行为.本文结构如下:第二节, 给出Riemann-Liouville分数积分和分数阶导数的一些定义和性质, 并给出问题(1.1)弱解的定义; 第三节, 类似于文献[4, 6, 10], 利用试验函数方法, 给出问题(1.1)弱解的一个爆破结果.

本节我们给出Riemann-Liouville分数积分和分数阶导数的一些定义和性质, 相关细节可参考文献[13-14].

定义2.1^[14]  给定$\alpha\in \left(0, 1\right)$, $f\in L^r\left(0, T\right)$, $1\leq r\leq \infty$, 则$f$的Riemann-Liouville分数积分$J^{1-\alpha}_{0|t}f\left(t\right)$定义为

$ J^{1-\alpha}_{0|t}f\left(t\right)=\frac{1}{\Gamma\left(1-\alpha\right)} \int_0^t \left(t-s\right)^{-\alpha}f\left(s\right){\rm d}s. $

定义2.2^[14]  给定$f\in AC\left[0, T\right]$, 其$\alpha\in \left(0, 1\right)$阶Riemann-Liouville型左边导数$D^\alpha_{0|t}f\left(t\right)$和右边导数$D^\alpha_{t|T}f\left(t\right)$分别定义为

$ D^\alpha_{0|t}f\left(t\right)=\partial _tJ^{1-\alpha}_{0|t}f\left(t\right), $

$ D^\alpha_{t|T}f\left(t\right)=-\frac{1}{\Gamma\left(1-\alpha\right)}\partial _t\int_t^T \left(s-t\right)^{-\alpha}f\left(s\right){\rm d}s. $

引理2.1^[14]  给定$f$, $g\in C\left[0, T\right]$, $\alpha\in \left(0, 1\right)$, 对任意的$t\in \left[0, T\right]$, $D^\alpha_{0|t}f$及$D^\alpha_{t|T}g$存在且连续, 则

$ \begin{equation}\label{2f} \int_0^T \left(D^\alpha_{0|t}f\left(t\right)\right)g\left(t\right){\rm d}t= \int_0^Tf\left(t\right)\left( D^\alpha_{t|T}g\left(t\right)\right){\rm d}t. \end{equation} $

(2.1)

引理2.2^[13]   (1) 对任意的$1\leq r\leq \infty$, 有

$ \begin{equation} D^{\alpha}_{0|t}J^{\alpha}_{0|t}=Id_{L^{r}\left(0, T\right)}\end{equation} $

(2.2)

在$\left[0, T\right]$上几乎处处成立;

(2)  定义$AC^{n+1}\left[0, T\right]=\left\{g: \left[0, T\right]\rightarrow {\Bbb R}, \partial_t^n g\in AC\left[0, T\right]\right\}$, 其中$n$为非负整数.任取$g\in AC^{n+1}\left[0, T\right]$, 则

$ \begin{equation} \left(-1\right)^n \partial_t^n D^\alpha_{t|T} g= D^{n+\alpha}_{t|T}g. \end{equation} $

(2.3)

给定

$ \rho\left(t\right)=\left(1-\frac{t}{T}\right)^l_+=\max\left\{0, \left(1-\frac{t}{T}\right)^l\right\}, $

其中$t\geq 0$, $T>0$, $l\gg1$, 则对任意的$\alpha\in \left(0, 1\right)$, 有

$ \begin{equation} D^{\alpha}_{t|T}\rho\left(t\right)=\frac{\left(1+l-\alpha\right)\Gamma\left(1+l\right)}{\Gamma\left(2+l-\alpha\right)}T^{-l}\left(T-t\right)^{l-\alpha}_+, \end{equation} $

(2.4)

及

$ \begin{equation} D^{\alpha+1}_{t|T}\rho\left(t\right)=\frac{\left(1+l-\alpha\right)\left(l-\alpha\right)\Gamma\left(1+l\right)}{\Gamma\left(2+l-\alpha\right)}T^{-l}\left(T-t\right)^{l-\alpha-1}_+. \end{equation} $

(2.5)

由定义2.1, 我们可将问题(1.1)中的方程重写成如下形式

$ \begin{eqnarray} \left\{ \begin{array}{ll} u_{tt}+\left(-\Delta\right)^{m}u +u_t=J^{1-\gamma}_{0|t}\left(\left|v\right|^{p}\right)\left(t\right), &\left(x, t\right)\in{\Bbb R}^N\times{\Bbb R}_+^1, \\ v_{tt}+\left(-\Delta\right)^{m}v +v_t=J^{1-\delta}_{0|t}\left(\left|u\right|^{q}\right)\left(t\right), &\left(x, t\right)\in{\Bbb R}^N\times{\Bbb R}_+^1. \end{array}\right. \end{eqnarray} $

(2.6)

本节的最后, 我们给出问题(1.1)弱解的定义.

定义2.3  给定$T>0$及初值$( u_i, v_i)\in (L_{\rm loc}^1({\Bbb R}^N))^2$ $(i=0, 1)$.若$( u, v )\in L_{\rm loc}^q ({\Bbb R}^N\times [0, T))\times L_{\rm loc}^p({\Bbb R}^N\times [0, T))$满足

$ \begin{eqnarray}\label{2a} &&\iint_{{\Bbb R}^{N}\times\left(0, T\right)} J^{1-\gamma}_{0|t}\left(\left|v\right|^{p}\right)\varphi {\rm d}x {\rm d}t+\int_{{\Bbb R}^{N}}u_1\left(x\right)\varphi \left(x, 0\right){\rm d}x\\ &&+ \int_{{\Bbb R}^{N}}u_0\left(x\right)\varphi \left(x, 0\right){\rm d}x-\int_{{\Bbb R}^{N}}u_0\left(x\right)\varphi_t \left(x, 0\right){\rm d}x\\ &=&\iint_{{\Bbb R}^{N}\times\left(0, T\right)}u\varphi_{tt}{\rm d}x{\rm d}t +\iint_{{\Bbb R}^{N}\times\left(0, T\right)} u\left(-\Delta\right)^{m}\varphi {\rm d}x{\rm d}t- \iint_{{\Bbb R}^{N}\times\left(0, T\right)}u\varphi_t {\rm d}x {\rm d}t, \end{eqnarray} $

(2.7)

及

$ \begin{eqnarray} &&\iint_{{\Bbb R}^{N}\times\left(0, T\right)} J^{1-\delta}_{0|t}\left(\left|u\right|^{q}\right)\psi {\rm d}x {\rm d}t+\int_{{\Bbb R}^{N}}v_1\left(x\right)\psi \left(x, 0\right){\rm d}x\\ &&+\int_{{\Bbb R}^{N}}v_0\left(x\right)\psi \left(x, 0\right){\rm d}x-\int_{{\Bbb R}^{N}}v_0\left(x\right)\psi_t \left(x, 0\right){\rm d}x\\ &=&\iint_{{\Bbb R}^{N}\times\left(0, T\right)}v\psi_{tt}{\rm d}x{\rm d}t +\iint_{{\Bbb R}^{N}\times\left(0, T\right)} v\left(-\Delta\right)^{m}\psi {\rm d}x{\rm d}t- \iint_{{\Bbb R}^{N}\times\left(0, T\right)}v\psi_t {\rm d}x{\rm d}t, \end{eqnarray} $

(2.8)

则称$\left( u, v \right)$为问题(1.1)的一个弱解, 其中试验函数$\varphi$, $\psi\in C^{2, \infty}_0\left([0, T)\times {\Bbb R}^N\right)$满足

$ \varphi\left(x, T\right)=\psi\left(x, T\right)=0, \varphi_t\left(x, T\right)=\psi_t\left(x, T\right)=0. $

定理3.1  假设

$ \int_{{\Bbb R}^N} u_i\left(x\right){\rm d}x>0, \int_{{\Bbb R}^N} v_i\left(x\right){\rm d}x>0, i=0, 1. $

则当

$ \frac{N\left(pq-1\right)}{2m}\leq \max\left\{pq\left(1-\gamma\right)+p\left(2-\delta\right)+1, pq\left(1-\delta\right)+q\left(2-\gamma\right)+1 \right\} $

或

$ p\leq\frac{1}{\delta} \mbox{ 且 } q\leq\frac{1}{\gamma} $

时, 问题(1.1)的弱解会在有限时刻爆破.

证  反证.假设$\left(u\left(x, t\right), v\left(x, t\right)\right)$为问题(1.1)的一非平凡整体弱解.令$\Phi\left(s\right)\in C_0^{\infty}\left({\Bbb R}^1_+\right)$满足

$ \Phi\left(s\right) =\left\{\begin{array}{ll} 1, & 0\leq s\leq 1, \\ \searrow, & 1\leq s\leq 2, \\ 0, & s\geq 2 . \end{array}\right. $

令

$ \phi_1\left(x\right)=\Phi^ {l_1}\left(\frac{\left|x\right|}{\left(B^{-1}T\right)^{\frac{1}{2m}}}\right), \phi_2\left(t\right)=\left(1-\frac{t}{T}\right)^{l_2}_+, $

其中$B>0$为待定常数, $T$, $l_1$, $l_2\gg1$.

再令$\sigma\left(x, t\right)=\phi_1\left(x\right)\phi_2\left(t\right)$, 在定义2.3中取试验函数

$ \varphi\left(x, t\right)=D^{1-\gamma}_{t|T}\sigma\left(x, t\right), \psi\left(x, t\right)=D^{1-\delta}_{t|T}\sigma\left(x, t\right), $

同时利用(2.1), (2.3), (2.4)和(2.5)式, 得

$ \begin{eqnarray} &&\iint_{{\Bbb R}^{N}\times\left(0, T\right)} D^{1-\gamma}_{0|t}J^{1-\gamma}_{0|t}\left(\left|v\right|^{p}\right)\sigma {\rm d}x{\rm d}t+CT^{\gamma-1}\int_{{\Bbb R}^{N}}u_1\left(x\right)\phi_1\left(x\right){\rm d}x \\ &&+ C\left(T^{\gamma-1}+T^{\gamma-2}\right)\int_{{\Bbb R}^{N}}u_0\left(x\right)\phi_1\left(x\right){\rm d}x \\ &=&C\iint_{{\Bbb R}^{N}\times\left(0, T\right)}u\left(D^{3-\gamma}_{t|T}\sigma+D^{2-\gamma}_{t|T}\sigma\right) {\rm d}x{\rm d}t +\iint_{{\Bbb R}^{N}\times\left(0, T\right)} u\left(-\Delta\right)^{m}\phi_1D^{1-\gamma}_{t|T}\phi_2 {\rm d}x{\rm d}t, \end{eqnarray} $

(3.1)

及

$ \begin{eqnarray} &&\iint_{{\Bbb R}^{N}\times\left(0, T\right)} D^{1-\delta}_{0|t}J^{1-\delta}_{0|t}\left(\left|u\right|^{q}\right)\sigma {\rm d}x{\rm d}t+CT^{\delta-1}\int_{{\Bbb R}^{N}}v_1\left(x\right)\phi_1\left(x\right){\rm d}x\\ &&+ C\left(T^{\delta-1}+T^{\delta-2}\right)\int_{{\Bbb R}^{N}}v_0\left(x\right)\phi_1\left(x\right){\rm d}x \\ &=&C\iint_{{\Bbb R}^{N}\times\left(0, T\right)}v\left(D^{3-\delta}_{t|T}\sigma+D^{2-\delta}_{t|T}\sigma\right) {\rm d}x{\rm d}t +\iint_{{\Bbb R}^{N}\times\left(0, T\right)} v\left(-\Delta\right)^{m}\phi_2D^{1-\delta}_{t|T}\phi_1 {\rm d}x{\rm d}t, \end{eqnarray} $

(3.2)

其中$C$为不依赖于$T$的常数.注意到

$ \int_{{\Bbb R}^N} u_i\left(x\right){\rm d}x>0, \quad \int_{{\Bbb R}^N} v_i\left(x\right){\rm d}x>0, i=0, 1, $

则由式(2.2), 可得估计

$ \begin{eqnarray} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t &\leq& C\underbrace{\iint_{{\Bbb R}^{N}\times\left(0, T\right)}u\phi_1\left(D^{3-\gamma}_{t|T}\phi_2+D^{2-\gamma}_{t|T}\phi_2\right) {\rm d}x{\rm d}t}_{I_1 }\\ && +\underbrace{\iint_{{\Bbb R}^{N}\times\left(0, T\right)} u\left(-\Delta\right)^{m}\phi_1D^{1-\gamma}_{t|T}\phi_2{\rm d}x{\rm d}t}_{I_2}, \end{eqnarray} $

(3.3)

及

$ \begin{eqnarray} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t &\leq& C\underbrace{\iint_{{\Bbb R}^{N}\times\left(0, T\right)}v\phi_1\left(D^{3-\delta}_{t|T}\phi_2+D^{2-\delta}_{t|T}\phi_2\right) {\rm d}x{\rm d}t}_{I_3 }\\ && +\underbrace{\iint_{{\Bbb R}^{N}\times\left(0, T\right)} v\left(-\Delta\right)^{m}\phi_1D^{1-\delta}_{t|T}\phi_2 {\rm d}x{\rm d}t}_{I_4}. \end{eqnarray} $

(3.4)

由Hölder不等式, 有

$ \begin{eqnarray} I_{1}&\leq& \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t\right)^{\frac{1}{q}}\\ &&\times \underbrace{\left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)}\phi_1\phi_2^{-\frac{1}{q-1}}\left(\left|D_{t|T}^{3-\gamma}\phi_2\right| +\left|D_{t|T}^{2-\gamma}\phi_2\right|\right)^{\frac{q}{q-1}}{\rm d}x{\rm d}t\right)^{\frac{q-1}{q}}}_{K_1}, \end{eqnarray} $

(3.5)

及

$ \begin{equation} I_{2}\leq \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t\right)^{\frac{1}{q}}\underbrace{\left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)}\sigma^{-\frac{1}{q-1}}\left|\Delta^{m}\phi_1D^{1-\gamma}_{t|T}\phi_2 \right|^{\frac{q}{q-1}}{\rm d}x{\rm d}t\right)^{\frac{q-1}{q}}}_{K_2}. \end{equation} $

(3.6)

联立(3.3), (3.5)和(3.6)式, 得

$ \begin{equation} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t\leq\left(K_1+K_2\right) \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t\right)^{\frac{1}{q}}. \end{equation} $

(3.7)

类似地，我们有

$ \begin{equation} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t \leq\left(K_3+K_4\right) \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t\right)^{\frac{1}{p}}, \end{equation} $

(3.8)

这里

$ K_3=\left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)}\phi_1\phi_2^{-\frac{1}{p-1}}\left(\left|D_{t|T}^{3-\delta}\phi_2\right| +\left|D_{t|T}^{2-\delta}\phi_2\right|\right)^{\frac{p}{p-1}}{\rm d}x{\rm d}t\right)^{\frac{p-1}{p}}, $

$ K_4=\left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)}\sigma^{-\frac{1}{p-1}}\left|\Delta^{m}\phi_1D^{1-\delta}_{t|T}\phi_2 \right|^{\frac{p}{p-1}}{\rm d}x{\rm d}t\right)^{\frac{p-1}{p}}. $

联立(3.7)和(3.8)式, 得

$ \begin{equation} \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t\right)^{\frac{pq-1}{pq}}\leq \left(K_1+K_2\right)\left(K_3+K_4\right)^{\frac{1}{q}}, \end{equation} $

(3.9)

$ \begin{equation} \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t\right)^{\frac{pq-1}{pq}}\leq \left(K_1+K_2\right)^{\frac{1}{p}}\left(K_3+K_4\right). \end{equation} $

(3.10)

接下来, 我们对$K_1$, $K_2$, $K_3$, $K_4$进行估计.为此, 作尺度变换

$ \tau=\frac{t}{T}, \xi=\left(\frac{T}{B}\right)^{-\frac{1}{2m}}x, $

则

$ \begin{eqnarray*} K_{1}^{\frac{q}{q-1}}&=&\iint_{{\Bbb R}^{N}\times\left(0, T\right)}\phi_1\left(x\right)\phi_2^{-\frac{1}{q-1}}\left(t\right)\left(\left|D_{t|T}^{3-\gamma}\phi_2\left(t\right)\right| +\left|D_{t|T}^{2-\gamma}\phi_2\left(t\right)\right|\right)^{\frac{q}{q-1}}{\rm d}x{\rm d}t\\ &\leq &C\int_0^T\phi_2^{-\frac{1}{q-1}}\left(t\right)\left(\left|D_{t|T}^{3-\gamma}\phi_2\left(t\right)\right|^{\frac{q}{q-1}} +\left|D_{t|T}^{2-\gamma}\phi_2\left(t\right)\right|^{\frac{q}{q-1}}\right){\rm d}t \int_{\left|x\right|\leq 2\left(\frac{T}{B}\right)^{\frac{1}{2m}}}\phi_1\left(x\right){\rm d}x\\ &=&CB^{-\frac{N}{2m}}T^{\frac{N}{2m}-l_2}\int_0^T \left(T-t\right)^{\frac{q\left(l_2+\gamma-3\right)-l_2}{q-1}}\left[1+\left(T-t\right)^{\frac{q}{q-1}}\right]{\rm d}t\int_{\left|\xi\right|\leq 2}\phi_1\left(\xi\right){\rm d}\xi\\ &=&CB^{-\frac{N}{2m}}T^{1+\frac{N}{2m}-\frac{q\left(2-\gamma\right)}{q-1}}+CB^{-\frac{N}{2m}}T^{1+\frac{N}{2m}-\frac{q\left(3-\gamma\right)}{q-1}}\\ &\leq &CB^{-\frac{N}{2m}}T^{1+\frac{N}{2m}-\frac{q\left(2-\gamma\right)}{q-1}}\left(1+T^{-\frac{q}{q-1}}\right) . \end{eqnarray*} $

注意到$T\gg1$, 则进一步有

$ \begin{equation} K_{1}\leq CB^{-\frac{N\left(q-1\right)}{2mq}}T^{\frac{q-1}{q}\left(1+\frac{N}{2m}\right)-\left(2-\gamma\right)}. \end{equation} $

(3.11)

类似地, 有

$ \begin{equation} K_{3}\leq CB^{-\frac{N\left(p-1\right)}{2mp}}T^{\frac{p-1}{p}\left(1+\frac{N}{2m}\right)-\left(2-\delta\right)}. \end{equation} $

(3.12)

另一方面, 可对$K_2$进行如下估计

$ \begin{eqnarray} K_{2}^{\frac{q}{q-1}}&=&\iint_{{\Bbb R}^{N}\times\left(0, T\right)}\phi_1^{-\frac{1}{q-1}}\left(x\right)\phi_2^{-\frac{1}{q-1}}\left(t\right)\left|\Delta ^m \phi _1\left(x\right)D_{t|T}^{1-\gamma}\phi_2\left(t\right)\right|^{\frac{q}{q-1}}{\rm d}x{\rm d}t\\ &=&\int_0^T\phi_2^{-\frac{1}{q-1}}\left(t\right)\left|D_{t|T}^{1-\gamma}\phi_2\left(t\right)\right|^{\frac{q}{q-1}}{\rm d}t \int_{\left|x\right|\leq 2\left(\frac{T}{B}\right)^{\frac{1}{2m}}}\phi_1^{-\frac{1}{q-1}}\left|\Delta ^m \phi _1\left(x\right)\right|{\rm d}x\\ &=&B^{\frac{q}{q-1}-\frac{N}{2m}}T^{\frac{N}{2m}-\frac{q}{q-1}-l_2}\int_0^T \left(T-t\right)^{\frac{q\left(l_2+\gamma-1\right)-l_2}{q-1}}{\rm d}t \int_{\left|\xi\right|\leq 2}\phi_1^{-\frac{1}{q-1}}\left|\Delta_{\xi} ^m \phi _1\left(\xi\right)\right|{\rm d}\xi\\ &\leq& CB^{\frac{q}{q-1}-\frac{N}{2m}}T^{1+\frac{N}{2m}-\frac{q\left(2-\gamma\right)}{q-1}}. \end{eqnarray} $

(3.13)

类似地, 对$K_4$, 有

$ \begin{equation} K_{4}^{\frac{p}{p-1}}\leq CB^{\frac{p}{p-1}-\frac{N}{2m}}T^{1+\frac{N}{2m}-\frac{p\left(2-\delta\right)}{p-1}}. \end{equation} $

(3.14)

联立(3.11)和(3.13)式, 得

$ \begin{equation} K_{1}+K_{2}=C\left(1+B\right)B^{-\frac{N\left(q-1\right)}{2mq}}T^{\frac{q-1}{q}\left(1+\frac{N}{2m}\right)+\gamma-2}. \end{equation} $

(3.15)

联立(3.12)和(3.14)式, 得

$ \begin{equation} K_{3}+K_{4}=C\left(1+B\right)B^{-\frac{N\left(p-1\right)}{2mp}}T^{\left(1+\frac{N}{2m}\right)\frac{p-1}{p}+\delta-2}. \end{equation} $

(3.16)

至此, (3.9), (3.10), (3.15)和(3.16)式蕴含

$ \begin{eqnarray} \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t\right)^{1-\frac{1}{pq}}&\leq& \left(K_1+K_2\right)\left(K_3+K_4\right)^{\frac{1}{q}}\\ &=&C\left(1+B\right)^{1+\frac{1}{q}}B^{-\frac{N}{2q}\left(\frac{q-1}{m}+\frac{p-1}{mp}\right)} T^{\theta_1}, \end{eqnarray} $

(3.17)

及

$ \begin{eqnarray} \left(\iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t\right)^{1-\frac{1}{pq}}&\leq& \left(K_1+K_2\right)^{\frac{1}{p}}\left(K_3+K_4\right)\\ &=&C\left(1+B\right)^{1+\frac{1}{p}}B^{-\frac{N}{2p}\left(\frac{p-1}{m}+\frac{q-1}{mq}\right)} T^{\theta_2}, \end{eqnarray} $

(3.18)

其中

$ \theta_1=\frac{q-1}{q}\left(1+\frac{N}{2m}\right)+\frac{1}{q}\left[\frac{p-1}{p}\left(1+\frac{N}{2m}\right)+\delta-2\right]+\gamma-2, $

$ \theta_2=\frac{p-1}{p}\left(1+\frac{N}{2m}\right)+\frac{1}{p}\left[\frac{q-1}{q}\left(1+\frac{N}{2m}\right)+\gamma-2\right]+\delta-2. $

接下来, 我们将分3种情形来完成定理的证明.在情形1和情形2中, 不妨假定

$ \max\left\{pq\left(1-\gamma\right)+p\left(2-\delta\right)+1, pq\left(1-\delta\right)+q\left(2-\gamma\right)+1 \right\}=pq\left(1-\gamma\right)+p\left(2-\delta\right)+1. $

情形1    $\frac{N\left(pq-1\right)}{2m}<pq\left(1-\gamma\right)+p\left(2-\delta\right)+1$.经计算, 此时有$\theta_1<0$.取$B=1$, 在式(3.17)两边关于$T\rightarrow \infty$取极限, 得

$ \lim\limits_\limits{T\rightarrow \infty}\int_0^T\int_{\left|x\right|\leq 2T^{\frac{1}{2m}}}v^p\phi {\rm d}x{\rm d}t=0. $

由Lebesgue控制收敛定理, 有

$ \int_0^{\infty}\int_{{\Bbb R}^N}v^p\left(x, t\right){\rm d}x{\rm d}t=0, $

从而

$ \begin{equation} v\equiv 0, \end{equation} $

(3.19)

结合(3.8)和(3.19)式, 可推导出$u\equiv v\equiv 0$, 即$\left(u\left(x, t\right), v\left(x, t\right)\right)$为问题(1.1)的平凡解.矛盾.

情形2  $\frac{N\left(pq-1\right)}{2m}=pq\left(1-\gamma\right)+p\left(2-\delta\right)+1$.此时有$\theta_1=0$.取$B$足够大, 且限定$1 <<B<T$, 使得$T\rightarrow \infty$时$B\nrightarrow \infty$.接下来, 在$\Omega\times \left[0, T\right]$上重新估计$K_2$和$K_4$, 其中

$ \begin{eqnarray*} \Omega=\left\{x\in {\Bbb R}^N : B^{-\frac{1}{2m+1}}T^{\frac{1}{2m+1}}\leq\left|x\right|\leq 2B^{-\frac{1}{2m+1}}T^{\frac{1}{2m+1}}\right\}. \end{eqnarray*} $

类似之前的步骤(亦可见文献[4]), 有

$ \begin{equation} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t\leq CT^{\frac{pq\theta_1}{pq-1}}B^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}}+CL_1^\frac{1}{pq}T^{\theta_1}B^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}} \left(B+B^\frac{1}{q}+B^{\frac{q+1}{q}}\right), \end{equation} $

(3.20)

及

$ \begin{equation} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|u\right|^{q}\sigma {\rm d}x{\rm d}t\leq CT^{\frac{pq\theta_2}{pq-1}}B^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}}+CL_2^\frac{1}{pq}T^{\theta_2}B^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}} \left(B+B^\frac{1}{p}+B^{\frac{p+1}{p}}\right), \end{equation} $

(3.21)

其中

$ L_1=\int_0^T\int_{\Omega} |v^p| \sigma {\rm d}x{\rm d}t, L_2=\int_0^T\int_{\Omega }|u^q| \sigma {\rm d}x{\rm d}t. $

进而, $\theta_1=0$蕴含

$ \begin{equation} \iint_{{\Bbb R}^{N}\times\left(0, T\right)} \left|v\right|^{p}\sigma {\rm d}x{\rm d}t\leq CB^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}} +CL_1^\frac{1}{pq}B^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}} \left(B+B^\frac{1}{q}+B^{\frac{q+1}{q}}\right). \end{equation} $

(3.22)

令$T\rightarrow \infty$, 则由Lebesgue控制收敛定理, 有

$ \int_0^{\infty}\int_{{\Bbb R}^{N}} \left|v\right|^{p}\sigma {\rm d}x {\rm d}t\leq CB^{-\frac{N\left(pq-1\right)}{2m\left(2m+1\right)pq}}. $

再令$B\rightarrow \infty$, 得$v\equiv 0$.进而可推导出$u\equiv v\equiv 0$, 即$\left(u\left(x, t\right), v\left(x, t\right)\right)$为问题(1.1)的平凡解.矛盾.

情形3  $p\leq \frac{1}{\delta}$且$q\leq \frac{1}{\gamma}$.类似于情形1和情形2的讨论, 可以证明问题(1.1)的弱解会在有限时刻爆破.在此, 不再赘述.