一类具有非线性项的弱耦合半线性双波动系统全局解的非存在性

欧阳柏平; 侯春娟

doi:10.6054/j.jscnun.2023041

一类具有非线性项的弱耦合半线性双波动系统全局解的非存在性

欧阳柏平^,,
侯春娟

广州华商学院数据科学学院，广州 511300

基金项目:

广东省基础与应用基础研究基金省市联合基金项目 2021A1515111048

广东省普通高校重点项目(自然科学) 2019KZDXM042

广州华商学院校内项目 2020HSDS01

详细信息

通讯作者:
欧阳柏平, Email: oytengfei79@tom.com

中图分类号: O175.4
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出版历程
- 收稿日期: 2022-10-21
- 网络出版日期: 2023-08-25
- 刊出日期: 2023-06-24

Nonexistence of Global Solutions to a Class of Weakly Coupled Semilinear Double-Wave System with Nonlinear Terms

OUYANG Baiping^,,
HOU Chunjuan

College of Data Science, Guangzhou Huashang College, Guangzhou 511300, China

摘要

摘要: 考虑了一类非线性项的弱耦合半线性双波动系统在次临界情况下解的爆破问题：首先，引入若干时变泛函，结合微分不等式方法，得到了该泛函的迭代框架和第一下界；然后，运用迭代技巧和切片方法，证明了该双波动系统柯西问题解的爆破，并推出了其解的生命跨度上界。
- 非线性项 /
- 弱耦合半线性双波动系统 /
- 爆破
Abstract: Blow-up of solutions to a class of weakly coupled semilinear double-wave system with nonlinear terms in the subcritical case is considered. By introducing some time-dependent functional associated with differential inequality methods, an iteration frame and the first lower bound of solutions are obtained. Then, blow-up of solutions to the Cauchy problem is proved via the iteration technique and slicing methods. Meanwhile, the upper bound of the lifespan for solutions is derived.
- nonlinear term /
- weakly coupled semilinear double-wave system /
- blow-up

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近几十年来, 有关非线性项下弱耦合半线性波动方程和波动系统柯西问题解的全局存在性及爆破问题成为了学者们关注的热点。部分学者^[1-7]研究了以下二阶半线性波动系统解的爆破问题:

$\left\{\begin{array}{l}u_{t t}-\Delta u=|v|^{p} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ v_{t t}-\Delta v=|u|^{q} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ \left(u, u_{t}, v, v_{t}\right)(0, x)=\varepsilon\left(u_{0}, u_{1}, v_{0}, v_{1}\right)(x) \quad\left(x \in \mathbb{R}^{n}\right), \end{array}\right.$

其中, $q 、p>1, n \geqslant 1, \varepsilon>0, \Delta$ 是拉普拉斯算子。该二阶半线性波动系统的临界曲线为

$\alpha_{W}:=\max \left\{\frac{4+3 q+p^{-1}}{p q-1}, \frac{4+3 p+q^{-1}}{p q-1}\right\}=\frac{n-1}{2} 。$

当 $\alpha_{W}<(n-1) / 2$ 时, 在初始数据满足一定的约束条件下存在全局解; 当 $\alpha_{W} \geqslant(n-1) / 2$ 时, 其解爆破。

CHEN和REISSIG^[8]考虑了如下具有阻尼项的弱耦合半线性波动系统解的爆破问题:

$\left\{\begin{array}{l}u_{t t}-\Delta u+u_{t}=|v|^{p} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ v_{t t}-\Delta v=|u|^{q} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ \left(u, u_{t}, v, v_{t}\right)(0, x)=\varepsilon\left(u_{0}, u_{1}, v_{0}, v_{1}\right)(x) \quad\left(x \in \mathbb{R}^{n}\right), \end{array}\right.$

其中, $q 、p>1, \varepsilon>0$ 。该文应用迭代技巧推出了初始数据满足一定条件时解的爆破及其生命跨度估计。

高阶半线性波动方程解的爆破已有很多研究成果^[9-16], 如: CHEN和PALMIERI^[9]讨论了以下半线性Moore-Gibson-Thompson (MGT) 方程解的柯西问题:

$\left\{\begin{array}{l} \beta u_{t t t}+u_{t t}-\Delta u-\beta \Delta u_{t}=|u|^{p} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ \left(u, u_{t}, u_{t t}\right)(0, x)=\varepsilon\left(u_{0}, u_{1}, u_{2}\right)(x) \quad\left(x \in \mathbb{R}^{n}\right), \end{array}\right.$

其中, $\varepsilon>0, p>1, \beta 、\varepsilon>0$ 。该文应用迭代方法和切片方法, 分别得到了在次临界、临界情况下解的全局非存在性和生命跨度估计。

双波动模型是基本波动方程的一个推广。特别地, 如果将 $\operatorname{div}$ 算子作用于弹性波方程 $u_{t t}-a^{2} \Delta u+$ $\left(b^{2}-a^{2}\right) \nabla \operatorname{div} u=0$ , 则可得到一类双波动方程。

本文考虑如下具有非线性项的弱耦合半线性双波动系统柯西问题解的爆破现象:

$\left\{\begin{array}{l} \left(\partial_{t}^{2}-\Delta\right)^{2} u=|v|^{p} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ \left(\partial_{t}^{2}-\Delta\right)^{2} v=|u|^{q} \quad\left((x, t) \in \mathbb{R}^{n} \times(0, T)\right), \\ \left(u, u_{t}, u_{t t}, u_{t t t}, v, v_{t}, v_{t t}, v_{t t t}\right)(0, x)= \\ \quad \varepsilon\left(u_{0}, u_{1}, u_{2}, u_{3}, v_{0}, v_{1}, v_{2}, v_{3}\right)(x) \quad\left(x \in \mathbb{R}^{n}\right), \end{array}\right.$

(1)

其中, $q 、p>1, \varepsilon>0, \Delta$ 是拉普拉斯算子, $\left(\partial_{t}^{2}-\Delta\right)^{2} u=$ $u_{t t t t}-2 \Delta u_{t t}+\Delta^{2} u_{\text {。 }}$

本文采用迭代思路和切片方法对问题(1) 进行探讨, 避开了由于无界乘子的引人而使得Kato引理难以应用的问题。首先, 构造若干能量泛函, 得到其迭代框架和第一下界; 然后, 运用迭代技巧推导出问题(1) 全局解的非存在性以及生命跨度估计。

1. 预备知识

首先给出方程组(1) 的柯西问题弱解的定义:

定义1 设 $\left(u_{0}, u_{1}, u_{2}, u_{3}, v_{0}, v_{1}, v_{2}, v_{3}\right) \in$ $\left(H^{3}\left(\mathbb{R}^{n}\right) \times H^{2}\left(\mathbb{R}^{n}\right) \times H^{1}\left(\mathbb{R}^{n}\right) \times L^{2}\left(\mathbb{R}^{n}\right)\right) \times\left(H^{3}\left(\mathbb{R}^{n}\right) \times\right.$ $\left.H^{2}\left(\mathbb{R}^{n}\right) \times H^{1}\left(\mathbb{R}^{n}\right) \times L^{2}\left(\mathbb{R}^{n}\right)\right)$ 。称 $(u, v)$ 为问题(1) 在 $[0, T)$ 上能量弱解, 如果

$\begin{aligned} u \in & C\left([0, T), H^{3}\left(\mathbb{R}^{n}\right)\right) \cap C^{1}\left([0, T), H^{2}\left(\mathbb{R}^{n}\right)\right) \cap \\ & C^{2}\left([0, T), H^{1}\left(\mathbb{R}^{n}\right)\right) \cap C^{3}\left([0, T), L^{2}\left(\mathbb{R}^{n}\right)\right) \cap \\ & L_{\mathrm{loc}}^{q}\left([0, T) \times \mathbb{R}^{n}\right), \\ v \in & C\left([0, T), H^{3}\left(\mathbb{R}^{n}\right)\right) \cap C^{1}\left([0, T), H^{2}\left(\mathbb{R}^{n}\right)\right) \cap \\ & C^{2}\left([0, T), H^{1}\left(\mathbb{R}^{n}\right)\right) \cap C^{3}\left([0, T), L^{2}\left(\mathbb{R}^{n}\right) \cap\right. \\ & L_{\mathrm{loc}}^{p}\left([0, T) \times \mathbb{R}^{n}\right) \end{aligned}$

满足

$\begin{array}{*{20}{l}} {\int_{{\mathbb{R}^n}} {{u_{ttt}}} (t, x)\varphi (t, x){\text{d}}x - \int_{{\mathbb{R}^n}} {{u_{ttt}}} (0, x)\varphi (0, x){\text{d}}x - } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} {{u_{ttt}}} } (s, x){\varphi _t}(s, x){\text{d}}x{\text{d}}s + 2\int_0^t {\int_{{\mathbb{R}^n}} {\left( {\nabla {u_{tt}}(s, x) \times } \right.} } } \\ \ \ \ \ {\nabla \varphi (s, x)){\text{d}}x{\text{d}}s + \int_0^t {\int_{{\mathbb{R}^n}} u } (s, x){\Delta ^2}\varphi (s, x){\text{d}}x{\text{d}}s = } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} | } v(s, x){|^p}\varphi (s, x){\text{d}}x{\text{d}}s} \end{array}$

(2)

和

$\begin{array}{*{20}{l}} {\int_{{\mathbb{R}^n}} {{v_{ttt}}} (t, x)\psi (t, x){\text{d}}x - \int_{{\mathbb{R}^n}} {{v_{ttt}}} (0, x)\psi (0, x){\text{d}}x - } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} {{v_{ttt}}} } (s, x){\psi _t}(s, x){\text{d}}x{\text{d}}s + 2\int_0^t {\int_{{\mathbb{R}^n}} {\left( {\nabla {v_{tt}}(s, x) \times } \right.} } } \\ \ \ \ \ {\nabla \psi (s, x)){\text{d}}x{\text{d}}s + \int_0^t {\int_{{\mathbb{R}^n}} v } (s, x){\Delta ^2}\psi (s, x){\text{d}}x{\text{d}}s = } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} | } u(s, x){|^q}\psi (s, x){\text{d}}x{\text{d}}s, } \end{array}$

(3)

其中, $\varphi(t, x), \psi(t, x) \in C_{0}^{\infty}\left([0, T) \times \mathbb{R}^{n}\right), t \in[0, T)$ 。

对式(3) 和式(4) 应用分部积分, 可得

$\begin{array}{*{20}{l}} {\int_{{\mathbb{R}^n}} {{u_{ttt}}} (t, x)\varphi (t, x){\text{d}}x - \int_{{\mathbb{R}^n}} {{u_{tt}}} (t, x){\varphi _t}(t, x){\text{d}}x + } \\ \ \ \ \ {\int_{{\mathbb{R}^n}} {{u_t}} (t, x){\varphi _{tt}}(t, x){\text{d}}x - \int_{{\mathbb{R}^n}} u (t, x){\varphi _{ttt}}(t, x){\text{d}}x + } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} u } (s, x){\varphi _{tttt}}(s, x){\text{d}}x{\text{d}}s - 2\int_{{\mathbb{R}^n}} {{u_t}} (t, x)\Delta \varphi (t, x){\text{d}}x + } \\ \ \ \ \ {2\int_{{\mathbb{R}^n}} u (t, x)\Delta {\varphi _t}(t, x){\text{d}}x - 2\int_0^t {\int_{{\mathbb{R}^n}} u } (s, x)\Delta {\varphi _{tt}}(s, x){\text{d}}x{\text{d}}s - } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} \Delta } \nabla u(s, x) \cdot \nabla \varphi (s, x){\text{d}}x{\text{d}}s = } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} | } v(s, x){|^p}\varphi (s, x){\text{d}}x{\text{d}}s + \varepsilon \int_{{\mathbb{R}^n}} {{u_3}} (x)\varphi (0, x){\text{d}}x - } \\ \ \ \ \ {\varepsilon \int_{{\mathbb{R}^n}} {{u_2}} (x){\varphi _t}(0, x){\text{d}}x + \varepsilon \int_{{\mathbb{R}^n}} {{u_1}} (x){\varphi _{tt}}(0, x){\text{d}}x - } \\ \ \ \ \ {\varepsilon \int_{{\mathbb{R}^n}} {{u_0}} (x){\varphi _{ttt}}(0, x){\text{d}}x - 2\varepsilon \int_{{\mathbb{R}^n}} {{u_1}} (x)\Delta \varphi (0, x){\text{d}}x + } \\ \ \ \ \ {2\varepsilon \int_{{\mathbb{R}^n}} {{u_0}} (x)\Delta {\varphi _t}(0, x){\text{d}}x} \end{array}$

(4)

和

$\begin{array}{*{20}{l}} {\int_{{\mathbb{R}^n}} {{v_{ttt}}} (t, x)\psi (t, x){\text{d}}x - \int_{{\mathbb{R}^n}} {{v_{tt}}} (t, x){\psi _t}(t, x){\text{d}}x + } \\ \ \ \ \ {\int_{{\mathbb{R}^n}} {{v_t}} (t, x){\psi _{tt}}(t, x){\text{d}}x - \int_{{\mathbb{R}^n}} v (t, x){\psi _{ttt}}(t, x){\text{d}}x + } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} v } (s, x){\psi _{tttt}}(s, x){\text{d}}x{\text{d}}s - 2\int_{{\mathbb{R}^n}} {{v_t}} (t, x)\Delta \psi (t, x){\text{d}}x + } \\ \ \ \ \ {2\int_{{\mathbb{R}^n}} v (t, x)\Delta {\psi _t}(t, x){\text{d}}x - 2\int_0^t {\int_{{\mathbb{R}^n}} v } (s, x)\Delta {\psi _{tt}}(s, x){\text{d}}x{\text{d}}s - } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} \Delta } \nabla v(s, x) \cdot \nabla \psi (s, x){\text{d}}x{\text{d}}s = } \\ \ \ \ \ {\int_0^t {\int_{{\mathbb{R}^n}} | } u(s, x){|^q}\psi (s, x){\text{d}}x{\text{d}}s + \varepsilon \int_{{\mathbb{R}^n}} {{v_3}} (x)\psi (0, x){\text{d}}x - } \\ \ \ \ \ {\varepsilon \int_{{\mathbb{R}^n}} {{v_2}} (x){\psi _t}(0, x){\text{d}}x + \varepsilon \int_{{\mathbb{R}^n}} {{v_1}} (x){\psi _{tt}}(0, x){\text{d}}x - } \\ \ \ \ \ {\varepsilon \int_{{\mathbb{R}^n}} {{v_0}} (x){\psi _{ttt}}(0, x){\text{d}}x - 2\varepsilon \int_{{\mathbb{R}^n}} {{v_1}} (x)\Delta \psi (0, x){\text{d}}x + } \\ \ \ \ \ {2\varepsilon \int_{{\mathbb{R}^n}} {{v_0}} (x)\Delta {\psi _t}(0, x){\text{d}}x} \end{array}。$

(5)

令 $t \rightarrow T$ , 则 $(u, v)$ 满足方程组(1) 给出的弱解的定义。

下面给出本文定理证明所需引理。

引理1^[10] 函数 ${\mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x)}$ 定义如下:

$\mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x)=\left\{\begin{array}{l} \mathrm{e}^{x}+\mathrm{e}^{-x} \quad(n=1), \\ \int_{\mathbb{S}^{n-1}} \mathrm{e}^{x \cdot \omega} \mathrm{d} \sigma_{\omega} \quad(n \geqslant 2), \end{array}\right.$

其中, $x \in \mathbb{R}^{n}, \mathbb{S}^{n-1}$ 为n-1维球面。 $\mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x)$ 为正光滑函数, 且 $\Delta \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x)=\mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x), \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \sim|x|^{-(n-1) / 2} \mathrm{e}^{|x|} \quad(|x| \rightarrow \infty)$ 。

引理2^[11] 函数 $\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}$ 的渐近性如下:

$\int_{B_{R+t}} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x)^{p^{\prime}} \mathrm{d} x \leqslant c_{1}(R+t)^{(n-1)\left(1-p^{\prime} / 2\right)},$

(6)

其中, $c_{1}>0, \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}=\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x)=\mathrm{e}^{-t} \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x), \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x)$ 的定义见引理 $1, p^{\prime}=p(p-1), B_{R+t}$ 表示以原点为中心 $R+t$ 为半径的球。

2. 主要结论

本文的主要结果如下:

定理1 设 $p, q>1$ 且 $p, q<(n+4) /(n-4)$ , 如果 $n>4$ , 则

$\max \left\{\frac{4+3 q+p^{-1}}{p q-1}, \frac{4+3 p+q^{-1}}{p q-1}\right\}>\frac{n-1}{2} 。$

设 $\left(u_{0}, u_{1}, u_{2}, u_{3}, v_{0}, v_{1}, v_{2}, v_{3}\right) \in\left(H^{3}\left(\mathbb{R}^{n}\right) \times H^{2}\left(\mathbb{R}^{n}\right) \times\right.$ $\left.H^{1}\left(\mathbb{R}^{n}\right) \times L^{2}\left(\mathbb{R}^{n}\right)\right)^{2}$ , 其中 $u_{i}, v_{i} \quad(i=0, 1, 2, 3)$ 是不恒为0的具有非负紧支集的函数, 包含在半径为 $R$ 的球 $B_{R}$ 中。特别地, 假设 $u_{3}(x)+u_{2}(x)>u_{1}(x)+u_{0}(x)$ , $v_{3}(x)+v_{2}(x)>v_{1}(x)+v_{0}(x)$ , 如果 $(u, v)$ 是方程组(1) 的解, 其生命跨度 $T(\varepsilon)$ 满足

$\operatorname{supp} u(t, \cdot), \operatorname{supp} v(t, \cdot) \subset B_{t+R} \quad(t \in(0, T)),$

则存在一个正常数 $\varepsilon_{0}$ , 使得当 $\varepsilon \in\left(0, \varepsilon_{0}\right]$ 时, $(u, v)$ 在有限时间爆破，其生命跨度的上界估计为

$T(\varepsilon) \leqslant \widetilde{C} \varepsilon^{-\frac{1}{\max \left\{Y_{1}(n, p, q), Y_{2}(n, p, q)\right\}}, }$

其中, $\widetilde{C}$ 独立于 $\varepsilon$ , 且

$\begin{aligned} & Y_{1}(n, p, q)=\frac{4+3 q+p^{-1}}{p q-1}-\frac{n-1}{2}, \\ & Y_{2}(n, p, q)=\frac{4+3 p+q^{-1}}{p q-1}-\frac{n-1}{2} 。 \end{aligned}$

证明定义如下泛函

$U(t)=\int_{\mathbb{R}^{n}} u(t, x) \mathrm{d} x, V(t)=\int_{\mathbb{R}^{n}} v(t, x) \mathrm{d} x 。$

(7)

由波动方程有限传播速度和定理条件, 可知当 $\left(u_{0}\right.$ , $\left.u_{1}, u_{2}, u_{3}, v_{0}, v_{1}, v_{2}, v_{3}\right) \in\left(H^{3}\left(\mathbb{R}^{n}\right) \times H^{2}\left(\mathbb{R}^{n}\right) \times H^{1}\left(\mathbb{R}^{n}\right) \times\right.$ $\left.L^{2}\left(\mathbb{R}^{n}\right)\right)^{2}$ 在 $B_{R}$ 中具有紧支集时, 问题(1) 的局部弱解属于其能量空间并且在 $B_{R+t}$ 中具有紧支集。

在式(4)、(5) 中, 选取 $\varphi \equiv 1$ 和 $\psi \equiv 1, \{(s, x) \in$ $\left.[0, t] \times \mathbb{R}^{n}:|x| \leqslant R+s\right\}$ , 有

$\int_{\mathbb{R}^{n}} u_{t t t}(t, x) \mathrm{d} x-\int_{\mathbb{R}^{n}} u_{t t t}(0, x) \mathrm{d} x=\int_{0}^{t} \int_{\mathbb{R}^{n}}|v(s, x)|^{p} \mathrm{~d} x \mathrm{~d} s,$

(8)

$\int_{\mathbb{R}^{n}} v_{t t t}(t, x) \mathrm{d} x-\int_{\mathbb{R}^{n}} v_{t t t}(0, x) \mathrm{d} x=\int_{0}^{t} \int_{\mathbb{R}^{n}}|u(s, x)|^{q} \mathrm{~d} x \mathrm{~d} s 。$

(9)

结合式(7)~(9), 得到

$U^{\prime \prime \prime}(t)=U^{\prime \prime \prime}(0)+\int_0^t \int_{\mathbb{R}^n}|v(s, x)|^p \mathrm{~d} x \mathrm{~d} s,$

(10)

$V^{\prime \prime \prime}(t)=V^{\prime \prime \prime}(0)+\int_0^t \int_{\mathbb{R}^n}|u(s, x)|^q \mathrm{~d} x \mathrm{~d} s_{\circ}$

(11)

对式(10)关于 $t$ 积分3次, 可得

$\begin{gathered} U(t)=U(0)+U^{\prime}(0) t+\frac{1}{2} U^{\prime \prime}(0) t^2+\frac{1}{6} U^{\prime \prime \prime}(0) t^3+ \\ \ \ \ \ \ \int_0^t \int_0^s \int_0^\tau \int_0^\sigma \int_{\mathbb{R}^n}|v(\eta, x)|^p \mathrm{~d} x \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant 0 。 \end{gathered}$

(12)

由Hölder不等式及 $\operatorname{supp} v(t, \cdot) \subset B_{t+R}(t \in(0, T))$ , 可推出

$\int_{\mathbb{R}^{n}}|v(\eta, x)|^{p} \mathrm{~d} x \geqslant C_{1}(R+\eta)^{-n(p-1)}(V(\eta))^{p},$

(13)

其中 $C_{1}>0$ 。

将式(13)代入式(12), 可得

$\begin{aligned} U(t) & \geqslant \int_0^t \int_0^s \int_0^\tau \int_0^\sigma \int_{\mathbb{R}^n}|v(\eta, x)|^p \mathrm{~d} x \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant \\ & C_1 \int_0^t \int_0^s \int_0^\tau \int_0^\sigma(R+\eta)^{-n(p-1)}(V(\eta))^p \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s 。 \end{aligned}$

(14)

同样, 对式(11) 在 $[0, t]$ 上积分3次, 整理得到

$\begin{aligned} V(t) & \geqslant \int_0^t \int_0^s \int_0^\tau \int_0^\sigma \int_{\mathbb{R}^n}|u(\eta, x)|^q \mathrm{~d} x \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant \\ & C_2 \int_0^t \int_0^s \int_0^\tau \int_0^\sigma(R+\eta)^{-n(q-1)}(U(\eta))^q \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s, \end{aligned}$

(15)

其中 $C_{2}>0$ 。

式(14)、(15) 提供了迭代框架。下面推导 $U$ 和 $V$ 的下界序列及其第一下界。由引理2, 有

$\left(\partial_{t}^{2}-\Delta\right)^{2} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}=0 \text { 。 }$

定义泛函 $U_{0}(t) 、V_{0}(t)$ 为:

$\begin{aligned} & U_{0}(t)=\int_{\mathbb{R}^{n}} u(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x, \\ & V_{0}(t)=\int_{\mathbb{R}^{n}} v(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x 。 \end{aligned}$

(16)

式(3) 中令 $\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}=\varphi$ , 易得

$\int_{\mathbb{R}^{n}} u_{t t t}(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x-\int_{0}^{t} \int_{\mathbb{R}^{n}} u_{t t t}(s, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}_{t}(s, x) \mathrm{d} x \mathrm{~d} s+\\ \ \ \ \ \ \ \ \ \ \ \ 2 \int_{0}^{t} \int_{\mathbb{R}^{n}} \nabla u_{t t}(s, x) \cdot \nabla \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(s, x) \mathrm{d} x \mathrm{~d} s+\\ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{t} \int_{\mathbb{R}^{n}} u(s, x) \Delta^{2} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(s, x) \mathrm{d} x \mathrm{~d} s=\\ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{t} \int_{\mathbb{R}^{n}}|v(s, x)|^{p} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(s, x) \mathrm{d} x \mathrm{~d} s+\\ \ \ \ \ \ \ \ \ \ \ \ \int_{\mathbb{R}^{n}} u_{t t t}(0, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(0, x) \mathrm{d} x,$

(17)

其中, $\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \in C_{0}^{\infty}\left([0, T) \times \mathbb{R}^{n}\right), t \in[0, T)$ 。

对于式(17), 进一步由分部积分和 $\mathit{\boldsymbol{ \boldsymbol{\varPsi} }}$ 的性质, 得到

$\int_{\mathbb{R}^{n}} u_{t t t}(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x+\int_{\mathbb{R}^{n}} u_{t t}(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x- \\ \ \ \ \ \ \ \ \ \int_{\mathbb{R}^{n}} u_{t}(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x-\int_{\mathbb{R}^{n}} u(t, x) \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(t, x) \mathrm{d} x= \\ \ \ \ \ \ \ \ \ \ \varepsilon I+\int_{0}^{t} \int_{\mathbb{R}^{n}}|v(s, x)|^{p} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(s, x) \mathrm{d} x \mathrm{~d} s,$

(18)

其中

$\begin{aligned} & I=I\left[u_{0}, u_{1}, u_{2}, u_{3}\right]= \\ & \quad \int_{\mathbb{R}^{n}}\left(u_{3}(x)+u_{2}(x)-u_{1}(x)-u_{0}(x)\right) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x_{\circ} \end{aligned}$

结合式(16) 和式(18), 有

$\begin{aligned} & U_{0}^{\prime \prime \prime}(T)+4 U_{0}^{\prime \prime}(T)+4 U_{0}^{\prime}(T)= \\ & \ \ \ \ \ \quad \varepsilon I+\int_{0}^{t} \int_{\mathbb{R}^{n}}|v(s, x)|^{p} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(s, x) \mathrm{d} x \mathrm{~d} s \geqslant \varepsilon I_{\circ} \end{aligned}$

令 $G(T)=U_{0}^{\prime \prime}(t)+2 U_{0}^{\prime}(t)$ , 有

$G^{\prime}(t)+2 G(t) \geqslant \varepsilon I_{\circ}$

(19)

对式(19) 积分, 有

$G(t) \geqslant\left(G(0)-\frac{1}{2} \varepsilon I\right) \mathrm{e}^{-2 t}+\frac{1}{2} \varepsilon I,$

(20)

从而可得

$U_{0}^{\prime \prime}(t)+2 U_{0}^{\prime}(t) \geqslant\left(G(0)-\frac{1}{2} \varepsilon I\right) \mathrm{e}^{-2 t}+\frac{1}{2} \varepsilon I_{\circ}$

(21)

对式(21) 关于 $t$ 积分2次, 得到

$U_{0}(T) \geqslant \int_{\mathbb{R}^{n}} u_{0}(x) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x+\\ \ \ \ \frac{\varepsilon}{4}\left(1-\mathrm{e}^{-2 t}\right) \int_{\mathbb{R}^{n}}\left(3 u_{1}(x)-2 u_{0}(x)-u_{3}(x)\right) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x+\\ \ \ \ \frac{\varepsilon}{4} t \int_{\mathbb{R}^{n}}\left(u_{3}(x)+u_{2}(x)-u_{1}(x)-u_{0}(x)\right) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x+\\ \ \ \ \frac{\varepsilon}{4} t \mathrm{e}^{-2 t} \int_{\mathbb{R}^{n}}\left(u_{3}(x)+u_{0}(x)-u_{1}(x)-u_{2}(x)\right) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x \geqslant\\ \ \ \ \int_{\mathbb{R}^{n}} u_{0}(x) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x+\frac{\varepsilon \delta}{4} \int_{\mathbb{R}^{n}}\left(2 u_{1}(x)-u_{0}(x)-u_{2}(x)\right) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x+\\ \ \ \ \frac{\varepsilon}{4} t \int_{\mathbb{R}^{n}}\left(u_{3}(x)+u_{2}(x)-u_{1}(x)-u_{0}(x)\right) \mathit{\boldsymbol{ \boldsymbol{\varPhi} }}(x) \mathrm{d} x \geqslant C_{0} \varepsilon,$

其中 $C_{0}>0$ 。

类似地, 可得

$V_{0}(t) \geqslant \widetilde{C}_{0} \varepsilon,$

(23)

其中 $\widetilde{C}_{0}>0$ 。

由Hölder不等式、式(6)和式(23), 有

$\begin{aligned} & \int_{\mathbb{R}^{n}}|v(s, x)|^{p} \mathrm{~d} x \geqslant\\ & \ \ \ \ \ \ \left(\int_{\mathbb{R}^{n}}|v(s, x)| \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}(s, x) \mathrm{d} x\right)^{p}\left(\int_{B_{R+s}} \mathit{\boldsymbol{ \boldsymbol{\varPsi} }}^{\frac{p}{p-1}}(s, x) \mathrm{d} x\right)^{-(p-1)} \geqslant \\ & \ \ \ \ \ \ c_{1}^{1-p} \widetilde{C}_{0}^{p} \varepsilon^{p}(R+s)^{(n-1)-(n-1) p / 2} 。 \end{aligned}$

(24)

联立式(14) 和式(24), 可得

$U(t) \geqslant\\ \ \ \ \ \ \ \ \ c_{1}^{1-p} \widetilde{C}_{0}^{p} \varepsilon^{p} \int_{0}^{t} \int_{0}^{s} \int_{0}^{\tau} \int_{0}^{\sigma}(R+\eta)^{(n-1)-(n-1) p / 2} \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant \\ \ \ \ \ \ \ \frac{c_{1}^{1-p} \widetilde{C}_{0}^{p} \varepsilon^{p}}{n(n+1)(n+2)(n+3)}(R+t)^{-(n-1) p / 2} t^{n+3} 。$

(25)

类似地, 由式(15) 和式(22), 有

$V(t) \geqslant\\ \ \ \ \ \ \ \ \ c_{1}^{1-q} C_{0}^{q} \varepsilon^{q} \int_{0}^{t} \int_{0}^{s} \int_{0}^{\tau} \int_{0}^{\sigma}(R+\eta)^{(n-1)-(n-1) q / 2} \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant\\ \ \ \ \ \ \ \frac{c_{1}^{1-q} C_{0}^{q} \varepsilon^{q}}{n(n+1)(n+2)(n+3)}(R+t)^{-(n-1) q / 2} t^{n+3} \text { 。 }$

(26)

取 $D_{1}=\frac{c_{1}^{1-p} \widetilde{C}_{0}^{p} \varepsilon^{p}}{n(n+1)(n+2)(n+3)}, \alpha_{1}=\frac{(n-1) p}{2}, \beta_{1}=n+3$ , $Q_{1}=\frac{c_{1}^{1-q} C_{0}^{q} \varepsilon^{q}}{n(n+1)(n+2)(n+3)}, a_{1}=\frac{(n-1) q}{2}, b_{1}=n+3$ 。式(25) 和式(26) 可分别简写为

$U(t) \geqslant D_{1}(R+t)^{-\alpha_{1}} t^{\beta_{1}},$

(27)

$V(t) \geqslant Q_{1}(R+t)^{-a_{1}} t^{b_{1}}。$

(28)

下面构造 $U(t)$ 和 $V(t)$ 的迭代序列。具体地说, 设

$U(t) \geqslant D_{j}(R+t)^{-\alpha_{j}} t^{\beta_{j}},$

(29)

$V(t) \geqslant Q_{j}(R+t)^{-a_{j}} t^{b_{j}},$

(30)

其中, $\left\{D_{j}\right\}_{j \in \mathbb{N}} 、\left\{Q_{j}\right\}_{j \in \mathbb{N}} 、\left\{\alpha_{j}\right\}_{j \in \mathbb{N}} 、\left\{a_{j}\right\}_{j \in \mathbb{N}} 、\left\{\beta_{j}\right\}_{j \in \mathbb{N}} 、\left\{b_{j}\right\}_{j \in \mathbb{N}}$ 均为非负实序列。

由式(27) 和式(28) 可知, $j=1$ 时, 式(29) 和式(30) 成立。假设 $j>1$ 时, 式(29)、(30) 成立, 下证式(29)、(30) 对 $j+1$ 也成立。

由式(14)、(15)、(29)、(30), 可得

$U(t) \geqslant C_{1} Q_{j}^{p} \int_{0}^{t} \int_{0}^{s} \int_{0}^{\tau} \int_{0}^{\sigma}(R+\eta)^{-n(p-1)-p a_{j}} \eta^{p b_{j}} \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant \\ \ \ \ \ \ \frac{C_{1} Q_{j}^{p}}{\left(p b_{j}+1\right)\left(p b_{j}+2\right)\left(p b_{j}+3\right)\left(p b_{j}+4\right)}(R+t)^{-n(p-1)-p a_{j}} t^{p b_{j}+4},$

(31)

$V(t) \geqslant C_{2} D_{j}^{q} \int_{0}^{t} \int_{0}^{s} \int_{0}^{\tau} \int_{0}^{\sigma}(R+\eta)^{-n(q-1)-q \alpha_{j}} \eta^{q \beta_{j}} \mathrm{~d} \eta \mathrm{d} \sigma \mathrm{d} \tau \mathrm{d} s \geqslant \\ \frac{C_{2} D_{j}^{q}}{\left(p \beta_{j}+1\right)\left(p \beta_{j}+2\right)\left(p \beta_{j}+3\right)\left(p \beta_{j}+4\right)}(R+t)^{-n(q-1)-q \alpha_{j}} t^{q \beta_{j}+4} \text { 。 }$

(32)

设

$\begin{gathered} D_{j+1}=\frac{C_{1} Q_{j}^{p}}{\left(p b_{j}+1\right)\left(p b_{j}+2\right)\left(p b_{j}+3\right)\left(p b_{j}+4\right)}, \\ \ \ \ \ \ \ \ \alpha_{j+1}=n(p-1)+p a_{j}, \beta_{j+1}=p b_{j}+4, \end{gathered}$

(33)

$Q_{j+1}=\frac{C_{2} D_{j}^{q}}{\left(p \beta_{j}+1\right)\left(p \beta_{j}+2\right)\left(p \beta_{j}+3\right)\left(p \beta_{j}+4\right)}, \\ \ \ \ \ \ \ \ \ \ \ \ a_{j+1}=n(q-1)+q \alpha_{j}, b_{j+1}=q \beta_{j}+4_{\circ}$

(34)

于是, 由式(31)~(34) 可知式(29) 和式(30) 对 $j+1$ 成立。

下面对 $\alpha_{j} 、\beta_{j} 、a_{j} 、b_{j}$ 进行估计。设 $j$ 为奇正整数, 由式(33)、(34) 可得

$\begin{gathered} \alpha_{j}=n(p-1)+p\left(n(q-1)+q \alpha_{j-2}\right)=\cdots= \\ \left(n+\frac{(n-1) p}{2}\right)(p q)^{(j-1) / 2}-n, \\ a_{j}=n(q-1)+q n(p-1)+p q a_{j-2}=\cdots= \\ \left(n+\frac{(n-1) q}{2}\right)(p q)^{(j-1) / 2}-n_{\circ} \end{gathered}$

(35)

同样可得

$\begin{aligned} \beta_{j}=4+4 p+p q \beta_{j-2}=\cdots=\left(\frac{4+4 p}{p q-1}+n+3\right)(p q)^{(j-1) / 2}-\frac{4+4 p}{p q-1} , \\ b_{j}=4+4 q+p q b_{j-2}=\cdots=\left(\frac{4+4 q}{p q-1}+n+3\right)(p q)^{(j-1) / 2}-\frac{4+4 q}{p q-1}。 \end{aligned}$

(36)

当 $j$ , $j-1$ 为偶正整数时，为正奇数, 故有

$\begin{aligned} & \beta_{j}=p b_{j-1}+4=q^{-1}\left(\frac{4+4 q}{p q-1}+n+3\right)(p q)^{j / 2}-\frac{4+4 p}{p q-1}, \\ & b_{j}=q \beta_{j-1}+4=p^{-1}\left(\frac{4+4 p}{p q-1}+n+3\right)(p q)^{j / 2}-\frac{4+4 q}{p q-1} 。 \end{aligned}$

(37)

由式(36)、(37), 可知

(1) 当 $j$ 为奇正整数时,

$\beta_{j} \leqslant B_{0}(p q)^{(j-1) / 2}, b_{j} \leqslant \widetilde{B}_{0}(p q)^{(j-1) / 2} ;$

(2) 当 $j$ 为偶正整数时,

$\beta_{j} \leqslant B_{0}(p q)^{j / 2}, b_{j} \leqslant \widetilde{B}_{0}(p q)^{j / 2},$

其中, $B_{0}=B_{0}(n, p, q), \widetilde{B}_{0}=\widetilde{B}_{0}(n, p, q)$ , 均为与 $j$ 无关的正数。

下面估计 $D_{j}$ 和 $Q_{j}$ 。结合式(33) (37), 有

$D_{j}=\frac{C_{1} Q_{j-1}^{p}}{\left(p b_{j-1}+1\right)\left(p b_{j-1}+2\right)\left(p b_{j-1}+3\right)\left(p b_{j-1}+4\right)} \geqslant \\ \ \ \ \ \ \frac{C_{1} Q_{j-1}^{p}}{\beta_{j}^{4}} \geqslant C_{1} B_{0}^{-4} Q_{j-1}^{p}(p q)^{-2 j}, \\ Q_{j}=\frac{C_{2} D_{j-1}^{q}}{\left(p \beta_{j-1}+1\right)\left(p \beta_{j-1}+2\right)\left(p \beta_{j-1}+3\right)\left(p \beta_{j-1}+4\right)} \geqslant \\ \ \ \ \ \ \frac{C_{2} D_{j-1}^{q}}{b_{j}^{4}} \geqslant C_{2} \widetilde{B}_{0}^{-4} D_{j-1}^{q}(p q)^{-2 j} 。$

由此可得

$\begin{aligned} D_{j} \geqslant & C_{1} B_{0}^{-4} C_{2}^{p} \widetilde{B}_{0}^{-4 p} D_{j-2}^{p q}(p q)^{-2 p(j-1)-2 j}= \\ & E_{0} D_{j-2}^{p q}(p q)^{-2 p(j-1)-2 j}, \end{aligned}$

(38)

$\begin{aligned} Q_{j} \geqslant & C_{2} \widetilde{B}_{0}^{-4} C_{1}^{q} B_{0}^{-4 q} Q_{j-2}^{p q}(p q)^{-2 q(j-1)-2 j}= \\ & \widetilde{E}_{0} Q_{j-2}^{p q}(p q)^{-2 q(j-1)-2 j} \text { 。 } \end{aligned}$

(39)

当 $j$ 为奇正整数时, 对式(38) 两边取对数, 由递推关系, 得到

$\log D_{j} \geqslant \log E_{0}+p q \log D_{j-2}-(2 p(j-1)+2 j) \log (p q) \geqslant \\ \ \ \ \ \ \ \ \ \ \ \cdots \geqslant(p q)^{\frac{j-1}{2}}\left[\frac{\log E_{0}}{p q-1}+\log D_{1}+\right. \\ \ \ \ \ \ \ \ \ \ \ \left.\frac{2 p(p q-1)-(2 p+2)(3 p q-1)}{(p q-1)^{2}} \log (p q)\right]-\\ \ \ \ \ \ \ \ \ \ \ \frac{\log E_{0}+2 p \log (p q)}{p q-1}+\frac{(2 p+2) \log (p q)}{(p q-1)^{2}} \times\\ \ \ \ \ \ \ \ \ \ \ [2 p q+(p q-1) j]。$

(40)

设 $j_{0}$ 为满足下式的最小正整数:

$j_{0} \geqslant \frac{\log E_{0}+2 p \log (p q)}{(2 p+2) \log (p q)}-\frac{2 p q}{p q-1},$

则式(40) 可化为

$\log D_{j} \geqslant\\ \ \ (p q)^{(j-1) / 2} \log \left(E_{0}^{1 /(p q-1)} D_{1}(p q)^{(2 p(p q-1)-(2 p+2)(3 p q-1)) /(p q-1)^{2}}\right)= \\ \ \ (p q)^{(j-1) / 2} \log \left(E_{1} \varepsilon^{p}\right),$

(41)

其中, $E_{1}=E_{1}(n, p, q), j \geqslant j_{0}$ 。

由式(39), 类似可得

$\log Q_{j} \geqslant\\ \ \ (p q)^{(j-1) / 2} \log \left(\widetilde{E}_{0}^{1 /(p q-1)} Q_{1}(p q)^{(2 q(p q-1)-(2 q+2)(3 p q-1)) /(p q-1)^{2}}\right)= \\ \ \ (p q)^{(j-1) / 2} \log \left(\widetilde{E}_{1} \varepsilon^{q}\right),$

(42)

其中, $\widetilde{E}_{1}=\widetilde{E}_{1}(n, p, q) ; j \geqslant j_{1}, j_{1}$ 为正整数, 且 $j_{1} \geqslant$ $\frac{\log \widetilde{E}_{0}+2 q \log (p q)}{(2 q+2) \log (p q)}-\frac{2 p q}{p q-1}$ 。

当 $j$ 为奇正整数且 $j \geqslant \max \left\{j_{0}, j_{1}\right\}$ 时, 由式(29)、(35)、(36)、(41), 有

$U(t) \geqslant \exp \left((p q)^{(j-1) / 2} \log \left(E_{1} \varepsilon^{p}\right)\right) \times\\ (R+t)^{n-(n+(n-1) p / 2)(p q)^{(j-1) / 2}} t^{((4+4 p) /(p q-1)+n+3)(p q)^{(j-1) / 2-(4+4 p) /(p q-1)}=}\\ \exp \left((p q)^{(j-1) / 2}\left(\log \left(E_{1} \varepsilon^{p}\right)-\left(n+\frac{(n-1) p}{2}\right) \log (R+t)+\right.\right.\\ \left.\left.\left(\frac{4+4 p}{p q-1}+n+3\right) \log t\right)\right)(R+t)^{n} t^{-(4+4 p) /(p q-1)} 。$

当 $t \geqslant R$ 时, 有

$U(T) \geqslant \exp \left((p q)^{(j-1) / 2}\left(\log \left(E_{1} \varepsilon^{p} 2^{-(n+(n-1) p / 2)} \times\right.\right.\right.\\ \ \ \ \ \ \ \ \left.\left.\left.t^{(4+4 p) /(p q-1)+n+3-(n+(n-1) p / 2)}\right)\right)\right)(R+t)^{n} t^{-(4+4 p) /(p q-1)} 。$

(43)

式(43) 右边项指数函数中 $t$ 的指数为

$\begin{aligned} & \frac{4+4 p}{p q-1}+n+3-\left(n+\frac{(n-1) p}{2}\right)=p\left(\frac{4+3 q+p^{-1}}{p q-1}-\frac{n-1}{2}\right)= \\ & \ \ \ \ \ \ \ \ \ \ p Y_{1}(n, p, q) \text { 。 } \end{aligned}$

当 $Y_{1}(n, p, q)>0$ 时, $t$ 的指数是正的。

类似的推导, 联立式(30)、(35)、(36)、(42), 得到

$V(t) \geqslant \exp \left((p q)^{(j-1) / 2}\left(\log \left(\widetilde{E}_{1} \varepsilon^{q}\right)-\right.\right.\\ \left.\left.\left(n+\frac{(n-1) q}{2}\right) \log (R+t)+\left(\frac{4+4 q}{p q-1}+n+3\right) \log t\right)\right) \times\\ (R+t)^{n} t^{-(4+4 q) /(p q-1)} \geqslant \exp \left((p q)^{(j-1) / 2}\left(\log \left(\widetilde{E}_{1} \varepsilon^{q} 2^{-(n+(n-1) q / 2)} \times\right.\right.\right.\\ \left.\left.\left.t^{(4+4 q) /(p q-1)+n+3-(n+(n-1) q / 2)}\right)\right)\right)(R+t)^{n} t^{-(4+4 q) /(p q-1)},$

(44)

其中 $t \geqslant R$ 。

此时, 式(44) 右边项指数函数中 $t$ 的指数为

$\begin{aligned} & \frac{4+4 q}{p q-1}+n+3-\left(n+\frac{(n-1) q}{2}\right)=q\left(\frac{4+3 p+q^{-1}}{p q-1}-\frac{n-1}{2}\right)= \\ & \ \ \ \ \ \ \ \ \ q \Upsilon_{2}(n, p, q) \text {。} \end{aligned}$

当 $Y_{2}(n, p, q)>0$ 时, $t$ 的指数是正的。

设 $\varepsilon_{0}=\varepsilon_{0}\left(u_{0}, u_{1}, u_{2}, u_{3}, v_{0}, v_{1}, v_{2}, v_{3}, n, p, q, R\right)>$ 0, 使得

$\varepsilon_{0}^{-1 / Y_{1}(n, p, q)} \geqslant\left(E_{1} 2^{-(n+(n-1) p / 2)}\right)^{1 /\left(p Y_{1}(n, p, q)\right)} 。$

记 $E_2=\left(E_1 2^{-(n+(n-1) p / 2)}\right)^{-1 /\left(p Y_1(n, p, q)\right)}$ 。当 $\varepsilon \in(0, \left.\varepsilon_0\right]$ 和 $t>E_{2} \varepsilon^{-1 / Y_{1}(n, p, q)}$ 时, 有 $t \geqslant R$ 和 $\log \left(E_{1} \varepsilon^{p} \times\right.$ $\left.2^{-(n+(n-1) p / 2)} t^{p Y_{1}(n, p, q)}\right)>0$ 。

式(43) 中, 对 $j \rightarrow \infty$ , 当 $\varepsilon \in\left(0, \varepsilon_{0}\right]$ 和 $t>$ $E_{2} \varepsilon^{-1 / Y_{1}}(n, p, q)$ 时, 可得 $U(t)$ 的下界爆破。

同理, 当 $Y_{2}(n, p, q)>0$ 时, 对于合适的 $\varepsilon_{0}=$ $\varepsilon_{0}\left(u_{0}, u_{1}, u_{2}, u_{3}, v_{0}, v_{1}, v_{2}, v_{3}, n, p, q, R\right)>0$ , 有

$\varepsilon_{0}^{-1 / Y_{2}(n, p, q)} \geqslant\left(\widetilde{E}_{1} 2^{-(n+(n-1) q / 2)}\right)^{1 /\left(q Y_{2}(n, p, q)\right)} 。$

因而, 当 $\varepsilon \in\left(0, \varepsilon_{0}\right] 、t>\widetilde{E}_{2} \varepsilon^{-1 / Y_{2}(n, p, q)}$ 且 $\widetilde{E}_{2}=$ $\left(\widetilde{E}_{1} 2^{-(n+(n-1) q / 2)}\right)^{-1 /\left(q Y_{2}(n, p, q)\right)}$ 时, 令 $j \rightarrow \infty$ , 可推出 $V(t)$ 的下界爆破。

由以上讨论可知, 问题(1) 的全局解不存在, 进一步可得局部解 $(u, v)$ 的生命跨度估计为

$T(\varepsilon) \leqslant \widetilde{C} \varepsilon^{-\frac{1}{\max \left\{Y_{1}(n, p, q), Y_{2}(n, p, q)\right\}}, }$

其中 $\widetilde{C}$ 为正常数。证毕。

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