数学物理学报, 2023, 43(1): 169-180

非线性记忆项的Euler-Poisson-Darboux-Tricomi方程解的爆破

欧阳柏平,

广州华商学院 应用数学系 广州 511300

Blow-up of Solutions to the Euler-Poisson-Darboux-Tricomi Equation with a Nonlinear Memory Term

Ouyang Baiping,

Department of Apllied Mathematics, Guangzhou Huashang College, Guangzhou 511300

收稿日期: 2021-10-26   修回日期: 2022-08-25  

基金资助: 广东省普通高校自然科学重点项目(2019KZDXM042)
广东省普通高校创新团队项目(2020WCXTD008)
广州华商学院校内项目(2020HSDS01)
广州华商学院校内项目(2021HSKT01)

Received: 2021-10-26   Revised: 2022-08-25  

Fund supported: Key Projects of Universities in Guangdong Province (NATURAL SCIENCE)(2019KZDXM042)
Innovation Team Project in Colleges and Universities of Guangdong Province(2020WCXTD008)
Science Foundation of Guangzhou Huashang College(2020HSDS01)
Science Foundation of Guangzhou Huashang College(2021HSKT01)

作者简介 About authors

欧阳柏平,E-mail:oytengfei79@tom.com

摘要

研究了具有非线性记忆项的Euler-Poisson-Darboux-Tricomi方程在次临界情况下解的爆破现象. 利用泛函分析方法结合修正的Bessel方程推出了其柯西问题解的迭代框架和第一下界, 然后通过迭代技巧, 获得了其解的全局非存在性以及解的生命跨度上界估计.

关键词： 非线性记忆项; Euler-Poisson-Darboux-Tricomi方程; 爆破

Abstract

Blow-up phenomenon of solutions to the Euler-Poisson-Darboux-Tricomi equation with a nonlinear memory term in the subcritical case is studied. By using functional methods associated with a modified Bessel equation, an iteration frame and the first lower bound are derived. Then, nonexistence of global solutions to the Cauchy problem for the Euler-Poisson-Darboux-Tricomi equation and an upper bound estimate of solutions for the lifespan are obtained via the iteration technique.

Keywords： Nonlinear memory term; Euler-Poisson-Darboux-Tricomi equation; Blow-up

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本文引用格式

欧阳柏平. 非线性记忆项的Euler-Poisson-Darboux-Tricomi方程解的爆破[J]. 数学物理学报, 2023, 43(1): 169-180

Ouyang Baiping. Blow-up of Solutions to the Euler-Poisson-Darboux-Tricomi Equation with a Nonlinear Memory Term[J]. Acta Mathematica Scientia, 2023, 43(1): 169-180

1 引言

本文讨论如下具有非线性记忆项的Euler-Poisson-Darboux-Tricomi方程柯西问题解的爆破

$\begin{matrix}\label{1} \left\{\begin{array}{ll} u_{tt}-t^{2l}\Delta u+\mu t^{-1}u_t+\nu^2t^{-2}u=C_{\gamma}\int_1^t (t-s)^{-\gamma} |u(s,x)|^p {\rm d}s,& (x,t)\in \Bbb R^n \times (1,T), \\ (u,u_t)(1,x)= \varepsilon (u_0,u_1)(x),& x\in \Bbb R^n, \end{array}\right. \end{matrix}$

其中 $ C_{\gamma}=\frac{1}{\Gamma(1-\gamma)},\ \gamma\in (0,1),\ p>1,\ \varepsilon>0,\ l>-1,$$ \Delta $为拉普拉斯算子, $ \mu, \nu^2$为非负实数,$\Gamma $为欧拉积分.

(1.1)式中如果$\mu=\nu=0$, 文献[1⇓⇓⇓⇓⇓⇓⇓-9]研究了下面具有非线性项的半线性广义Tricomi方程

$ \left\{\begin{array}{ll} u_{tt}-t^{2l}\Delta u=|u|^p,& (x,t)\in \Bbb R^n \times (0,T), \\ (u,u_t)(0,x)= \varepsilon (u_0,u_1)(x),& x\in \Bbb R^n, \end{array}\right. $

其中 $l>0, p>1$.

当$n=1 $时, 其临界指数$p_{crit}(1,l)=1+\frac{2}{l}$. 当$n\geq 2$时, 其临界指数

$p_{crit}(n,l)=\frac{n+1+(n-2)l+\sqrt{((n+1)(l+1)-3l)^2+8(l+1)(nl+n-1)}}{2(nl+n-1)}. $

当$1<p<p_{crit}(n,l)$ 时, 在初始值的合适假设下, 其柯西问题的解在有限时间内爆破, $p>p_{crit}(n,l)$时, 对于小初值情况存在全局解.

文献[10]研究了下面混合非线性项的半线性广义Tricomi方程解的爆破问题

$ \left\{\begin{array}{ll} u_{tt}-t^{2l}\Delta u=|u_t|^p+|u|^q,& (x,t)\in \Bbb R^n \times (0,T), \\ (u,u_t)(0,x)= \varepsilon (u_0,u_1)(x),& x\in \Bbb R^n, \end{array}\right. $

其中 $l>0, p,q>1$.

作者基于迭代的方法在初始数据满足一定的条件下, 推出了其柯西问题能量解在有限时间内爆破, 同时得到了其能量解的生命跨度上界估计

$T(\varepsilon)\leq C\varepsilon^{-\frac{p(q-1)}{\Theta(n,l,p,q)}}.$

与已有的结果相比, 作者扩大了其爆破的区域.

(1.1)式中如果$l=-k,\mu \geq 0, \nu=0$, 则为具有非线性记忆项的Einstein-de Sitter时空上半线性波动方程. 文献[11]对如下非线性项的Einstein-de Sitter时空上半线性波动方程解的爆破进行了研究

$ \left\{\begin{array}{ll} u_{tt}-t^{-2k}\Delta u+\mu t^{-1}u_t=|u|^p,& (x,t)\in \Bbb R^n \times (1,T), \\ (u,u_t)(1,x)= \varepsilon (u_0,u_1)(x),& x\in \Bbb R^n, \end{array}\right. $

其中 $ p>1, k\in [0,1), \varepsilon>0,$$ \Delta$为拉普拉斯算子,$ \mu$为非负实数.

利用迭代技巧和相关泛函分析方法, 作者分别得到了在临界和次临界情况下解的爆破情况, 同时还进一步推出了其生命跨度的上界估计.

文献[12]研究了非线性项的Euler-Poisson-Darboux-Tricomi方程柯西问题解的爆破

$ \left\{\begin{array}{ll} u_{tt}-t^{2l}\Delta u+\mu t^{-1}u_t+\nu^2t^{-2}u=|u|^p,& (x,t)\in \Bbb R^n \times (1,T), \\ (u,u_t)(1,x)= \varepsilon (u_0,u_1)(x),& x\in \Bbb R^n, \end{array}\right. $

其中$ p>1,\varepsilon>0,l>-1, \Delta$为拉普拉斯算子,$ \mu, \nu^2$为非负实数.

在对初始数据一定的约束下, 作者利用迭代方法得到了上式方程其柯西问题解的爆破结果.

本文在文献[12]的基础上研究具有非线性记忆项的Euler-Poisson-Darboux-Tricomi方程解的爆破现象, 主要是探讨非线性记忆项对方程解的非局部影响. 由问题(1.1)等式右边的积分项可知, 其表示系统记忆的信息包含过去的历史信息, 而且$(t-s)^{-\gamma}$ 是衰减的, 这表明越近的信息对系统影响更大.

目前应用泛函分析和迭代技巧处理一类高阶波动方程解的爆破问题取得一些成果[13⇓⇓⇓⇓⇓⇓⇓⇓-22], 本文拟采用其中的办法, 通过引入若干个泛函来构造迭代框架以及第一下界进行研究. 然而与文献[13⇓⇓⇓⇓⇓⇓⇓⇓-22] 不同的是, 问题(1.1)会出现关于时间和空间的积分, 此时需要借助修正的Bessel方程来进行处理.

文中记$w_l(t)=\frac{t^{l+1}}{l+1}$为$t^l$的原函数. 另外, 取$A_l(t)=w_l(t)-w_l(1)$.

2 主要结果

先定义问题(1.1)的能量解.

定义 2.1 设 $(u_0,u_1)\in L_{\rm loc}^1(\Bbb R^n)\times L_{\rm loc}^1(\Bbb R^n)$, 使得 supp $u_0,$ supp $u_1 \subset B_R, R>0$. $u$ 是问题(1.1)在 $[1,T)$ 上的能量解, 如果

$ u\in {\cal C}([1,T),W_{\rm loc}^{1,1}(\Bbb R^n))\cap {\cal C}^1([1,T),L_{\rm loc}^1(\Bbb R^n))\cap L_{\rm loc}^p((1,T)\times \Bbb R^n), \mbox{supp}\ u\subset B_{R+A_l(t)}, t\in (1,T), $

同时下面的积分关系成立

$\begin{matrix}\label{2} &&\int_{\Bbb R^n} u_{t}(t,x)\varphi(t,x) {\rm d}x-\int^t_1\int_{\Bbb R^n}u_{t}(s,x)\varphi_s(s,x){\rm d}x{\rm d}s +\int^t_1\int_{\Bbb R^n}s^{2l}\nabla u(s,x)\nabla\varphi(s,x) {\rm d}x{\rm d}s \\ &&+\mu\int^t_1\int_{\Bbb R^n}s^{-1}u_t(s,x)\varphi(s,x){\rm d}x{\rm d}s+\int^t_1\int_{\Bbb R^n}\nu^{2}s^{-2}u(s,x)\varphi(s,x){\rm d}x{\rm d}s \\ &= &C_{\gamma}\int_1^t \int_{\Bbb R^n}\varphi(s,x)\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s+\varepsilon \int_{\Bbb R^n}u_1(x)\varphi(1,x){\rm d}x, \end{matrix}$

其中 $\varphi(t,x)\in {\cal C}^\infty_0([1,T)\times \Bbb R^n), t\in (1,T), u(1,\cdot)=\varepsilon u_0 \in L_{\rm loc}^1(\Bbb R^n).$

由式(2.1), 应用分部积分, 整理得

$\begin{matrix}\label{3} &&\int_{\Bbb R^n} u_{t}(t,x)\varphi(t,x) {\rm d}x-\int_{\Bbb R^n} u(t,x)\varphi_t(t,x) {\rm d}x+\int^t_1\int_{\Bbb R^n}u(s,x)\varphi_{ss}(s,x){\rm d}x{\rm d}s \\ &&-\int^t_1\int_{\Bbb R^n}s^{2l} u(s,x)\Delta\varphi(s,x) {\rm d}x{\rm d}s+\mu\int_{\Bbb R^n}t^{-1}u(t,x)\varphi(t,x){\rm d}x \\ &&-\mu\int^t_1\int_{\Bbb R^n}s^{-1}u(s,x)\varphi_s(s,x){\rm d}x{\rm d}s+\int^t_1\int_{\Bbb R^n}(\mu+\nu^{2})s^{-2}u(s,x)\varphi(s,x){\rm d}x{\rm d}s \\ &=& C_{\gamma}\int_1^t \int_{\Bbb R^n}\varphi(s,x)\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s+\varepsilon \int_{\Bbb R^n}u_1(x)\varphi(1,x){\rm d}x \\ &&-\varepsilon \int_{\Bbb R^n}u_0(x)\varphi_t(1,x){\rm d}x+\varepsilon \mu\int_{\Bbb R^n}u_0(x)\varphi(1,x){\rm d}x. \end{matrix}$

令 $t\rightarrow T$, 有 $u$ 满足问题(1.1)弱解的定义.

本文有下面的结论.

定理 2.1 设 $l>-1, \mu, \nu^2\geq 0$ 且 $1<p< p_0(n,p,l,\mu,\gamma)$, 其中 $p_0(n,p,l,\mu,\gamma)=\frac{5-2\gamma+\mu+(n-2)(l+1)+\sqrt{(5-2\gamma+\mu+(n-2)(l+1))^2+8(l+1)(\mu+l+(n-1)(l+1))} } {2\mu+2l+2(n-1)(l+1)}. $

设$(u_0,u_1)\in H^1(\Bbb R^n)\times L^2(\Bbb R^n)$ 是非负的紧支集函数, 其支集包含于$B_R(R>0)$. 若$u\in {\cal C}([1,T),W_{\rm loc}^{1,1}(\Bbb R^n))\cap {\cal C}^1([1,T),L_{\rm loc}^1(\Bbb R^n))\cap L_{\rm loc}^p((1,T)\times \Bbb R^n) $ 为问题(1.1)的能量解,满足$u_1(x)+(\frac{\mu-1}{2}-\delta)u_0(x)\geq 0,\delta^2=\frac{(\mu-1)^2-4\nu^2}{4}, \delta\geq 0.$ 其生命跨度记作$T(\varepsilon)$, 则存在一个正常数$\varepsilon_0$, 使得当$\varepsilon\in (0,\varepsilon_0]$ 时 $u$ 在有限时间内爆破, 其生命跨度上界估计为

$ T(\varepsilon)\leq \check{C}\varepsilon^{-\frac{2p(p-1)}{\Theta(p,n,l,\gamma,\mu)}}, $

其中 $\breve{C}$ 为不依赖于 $\varepsilon$ 的正常数, 且

$\Theta(p,n,l,\gamma,\mu)=2(l+1)+(5-2\gamma+\mu+(n-2)(l+1))p-p^2(\mu+l+(n-1)(l+1)). $

3 定理的证明

研究如下方程的正解

$\begin{matrix}\label{4} \Phi_{ss}+(\mu+\nu^2)s^{-2}\Phi=s^{2l}\Delta \Phi+\mu s^{-1}\Phi_s. \end{matrix}$

设函数$\phi(x)$ (参见文献[21])定义如下

$ \phi(x) \doteq \left\{\begin{array}{ll} {\rm e}^{x}+{\rm e}^{-x},&\ \, n=1,\\ \int_{{\Bbb S}^{n-1}} {\rm e}^{x\cdot \omega}{\rm d} \sigma_\omega,& \ \, n\geq 2, \end{array}\right. $

其中 ${\Bbb S}^{n-1}$表示$n-1$维球面, $\phi(x)$ 是正光滑函数, 有如下性质

$ \Delta \phi(x) =\phi(x), $
$\phi (x) \sim |x|^{-\frac{n-1}{2}} {\rm e}^{|x|}, \qquad \mbox{当} \,\, |x|\to \infty. $

设$\Phi=\Phi(s,x)=\lambda(s)\phi(x)$, 结合(3.1)式, 可推得

$\begin{matrix}\label{5} \lambda''+(\mu+\nu^2)s^{-2}\lambda=s^{2l}\lambda+\mu s^{-1}\lambda'. \end{matrix}$

作变量代换, 设$\eta=w_l(s)$, 有

$\begin{matrix}\label{6} \frac{{\rm d}\lambda}{{\rm d}s}s^{-l}=\frac{{\rm d}\lambda}{{\rm d}\eta}, \frac{{\rm d}^2\lambda}{{\rm d}s^2}=\frac{{\rm d}^2\lambda}{{\rm d}\eta^2}s^{2l}+\frac{{\rm d}\lambda}{{\rm d}\eta}ls^{l-1}. \end{matrix}$

联立(3.2)-(3.3)式, 整理可得

$\begin{matrix}\label{7} \eta^2\frac{{\rm d}^2\lambda}{{\rm d}\eta^2}-\frac{(l-\mu)\eta {\rm d}\lambda}{(l+1){\rm d}\eta}+(\frac{\mu+\nu^2}{(l+1)^2}-\eta^2)\lambda=0. \end{matrix}$

设$\lambda(\eta)=\eta^m \xi(\eta)$, 结合(3.4)式, 得到

$\begin{matrix}\label{8} \eta^l\frac{{\rm d}^2\xi}{{\rm d}\eta^2}+(2m+\frac{l-\mu}{l+1})\eta\frac{{\rm d}\xi}{{\rm d}\eta}+(m(m-1)+\frac{(l-\mu)m}{l+1}+(\frac{\mu+\nu^2}{(l+1)^2}-\eta^2))\xi=0. \end{matrix}$

(3.5)式中, 取 $m=\frac{1+\mu}{2(l+1)}$, 得

$\begin{matrix}\label{9} \eta^2\frac{{\rm d}^2\xi}{{\rm d}\eta^2}+\eta\frac{{\rm d}\xi}{{\rm d}\eta}-(\frac{(\mu-1)^2-4\nu^2}{4(l+1)^2}+\eta^2)\xi=0. \end{matrix}$

(3.2)-(3.6)式表明,(3.2)式的解 $\lambda$亦为(3.6)式的解.

由 $\delta^2=\frac{(\mu-1)^2-4\nu^2}{4}, \delta\geq 0$, 则(3.6)式化为

$\begin{matrix}\label{10} \eta^2\frac{{\rm d}^2\xi}{{\rm d}\eta^2}+\eta\frac{{\rm d}\xi}{{\rm d}\eta}-(\frac{\delta^2}{(l+1)^2}+\eta^2)\xi=0. \end{matrix}$

选取第二类Bessel函数$K_{\frac{\delta}{l+1}}(\eta)$作为(3.7)式的解. 于是, 如果忽略一个正常数因子, 有

$\begin{matrix}\label{11} \lambda(s)=s^{\frac{\mu+1}{2}}K_{\frac{\delta}{l+1}}(w_l(s)). \end{matrix}$

由此, 可得

$\begin{matrix}\label{12} \Phi(s,x)=\lambda(s)\phi(x)=s^{\frac{\mu+1}{2}}K_{\frac{\delta}{l+1}}(w_l(s))\phi(x). \end{matrix}$

接下来, 将通过引入辅助泛函得到需要的迭代框架. 为此, 设

$\begin{matrix}\label{13} F(t)=\int_{\Bbb R^n} u(t,x){\rm d}x. \end{matrix}$

(2.2)式中, 取 $\varphi=\varphi(s,x)\equiv 1$,其中 $(s,x)\in [t]\times \Bbb R^n$ 且$ |x|\leq R+A_l(s)$, 得到

$\begin{matrix}\label{14} &&\int_{\Bbb R^n} u_{t}(t,x){\rm d}x+\mu\int^t_1\int_{\Bbb R^n}s^{-1}u_t(s,x){\rm d}x{\rm d}s+\int^t_1\int_{\Bbb R^n}\nu^{2}s^{-2}u(s,x){\rm d}x{\rm d}s \\ &= &C_{\gamma}\int_1^t \int_{\Bbb R^n}\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s+\varepsilon \int_{\Bbb R^n}u_1(x){\rm d}x. \end{matrix}$

结合(3.10)和(3.11)式, 有

$\begin{matrix}\label{15} &&F'(t)+\mu\int^t_1\int_{\Bbb R^n}s^{-1}u_t(s,x){\rm d}x{\rm d}s+\int^t_1\int_{\Bbb R^n}\nu^{2}s^{-2}u(s,x){\rm d}x{\rm d}s \\ &=& C_{\gamma}\int_1^t \int_{\Bbb R^n}\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s+F'(1). \end{matrix}$

(3.12)式对$t$求导, 可推出

$\begin{matrix}\label{16} F''(t)+\mu t^{-1}F'(t)+\nu^{2}t^{-2}F(t)= C_{\gamma} \int_{\Bbb R^n}\int_1^t(t-s)^{-\gamma} |u(s,x)|^p {\rm d}s{\rm d}x. \end{matrix}$

对(3.13)式进一步化简, 整理可得

$\begin{matrix}\label{17} t^{-(q_2+1)}\frac{{\rm d}}{{\rm d}t}(t^{q_2+1-q_1}\frac{{\rm d}(t^{q_1}F(t))}{{\rm d}t})= C_{\gamma} \int_{\Bbb R^n}\int_1^t(t-s)^{-\gamma} |u(s,x)|^p {\rm d}s{\rm d}x, \end{matrix}$

其中$q_1, q_2$ 满足$q_1 q_2=\nu^2, q_1+ q_2+1=\mu$, 特别地, 取$q_2=\frac{\mu-1}{2}+\delta, q_1=\frac{\mu-1}{2}-\delta $.

对(3.14)式在 $[t]$上积分,有

$\begin{matrix}\label{18} F(t)&=&t^{-q_1}F(1)+t^{-q_1}\int_1^t F'(1)s^{-(q_2+1-q_1)}{\rm d}s \\ &&+ C_{\gamma}t^{-q_1} \int_1^t s^{-(q_2+1-q_1)}\int_1^s \tau^{q_2+1}\int_{\Bbb R^n}\int_1^\tau(\tau-\sigma)^{-\gamma} |u(\sigma,x)|^p {\rm d}\sigma {\rm d}x {\rm d}\tau {\rm d}s \\ &\geq & C_{\gamma}t^{-q_1} \int_1^t s^{-(q_2+1-q_1)}\int_1^s \tau^{q_2+1}\int_{\Bbb R^n}\int_1^\tau(\tau-\sigma)^{-\gamma} |u(\sigma,x)|^p {\rm d}\sigma {\rm d}x {\rm d}\tau {\rm d}s. \end{matrix}$

结合定理条件以及Hölder不等式, 可得

$\begin{matrix}\label{19} \int_{\Bbb R^n} |u(\sigma,x)|^p {\rm d}x\geq c_0(R+A_l(\sigma))^{-n(p-1)}(F(\sigma))^p, \end{matrix}$

其中 $c_0=c_0(n,p,l)>0. $

由(3.15)和(3.16)式, 可推出

$\begin{matrix}\label{20} F(t)&\geq & C_{\gamma}c_0t^{-q_1} \int_1^t s^{-(q_2+1-q_1)}\int_1^s \tau^{q_2+1}\int_1^\tau(\tau-\sigma)^{-\gamma} (R+A_l(\sigma))^{-n(p-1)}(F(\sigma))^p {\rm d}\sigma {\rm d}\tau {\rm d}s \\ &\geq & C_{\gamma}\widetilde{c}_0t^{-q_1} \int_1^t s^{-(q_2+1-q_1)}\int_1^s \tau^{q_2+1}\int_1^\tau(\tau-\sigma)^{-\gamma} (\sigma+1)^{-n(l+1)(p-1)}(F(\sigma))^p {\rm d}\sigma {\rm d}\tau {\rm d}s \\ &\geq & C_{\gamma}\widetilde{c}t^{-(q_2+1+\gamma)-n(l+1)(p-1)} \int_1^t \int_1^s \tau^{q_2+1}\int_1^\tau(F(\sigma))^p {\rm d}\sigma {\rm d}\tau {\rm d}s, \end{matrix}$

其中$\widetilde{c}_0,\widetilde{c}$为正常数, $t\geq 1$.

为了得到 $F(t)$的第一下界, 引入下面泛函

$\begin{matrix}\label{21} U(t)=\int_{\Bbb R^n}u(t,x)\Phi(t,x){\rm d}x. \end{matrix}$

(2.2)式中, 取$\Phi=\varphi $, $\Phi$的定义见(3.9)式, 由波方程的有限传播速度可知, $u$ 具有紧支集, 因此 $\Phi$ 的支集条件可以去掉, 有

$\begin{matrix}\label{22} &&\int_{\Bbb R^n} u_{t}(t,x)\Phi(t,x) {\rm d}x-\int_{\Bbb R^n} u(t,x)\Phi_t(t,x) {\rm d}x+\int^t_1\int_{\Bbb R^n}u(s,x)\Phi_{ss}(s,x){\rm d}x{\rm d}s \\ &&-\int^t_1\int_{\Bbb R^n}s^{2l} u(s,x)\Delta\Phi(s,x) {\rm d}x{\rm d}s+\mu\int_{\Bbb R^n}t^{-1}u(t,x)\Phi(t,x){\rm d}x \\ &&-\mu\int^t_1\int_{\Bbb R^n}s^{-1}u(s,x)\Phi_s(s,x){\rm d}x{\rm d}s+\int^t_1\int_{\Bbb R^n}(\mu+\nu^{2})s^{-2}u(s,x)\Phi(s,x){\rm d}x{\rm d}s \\ &=& C_{\gamma}\int_1^t \int_{\Bbb R^n}\Phi(s,x)\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s+\varepsilon \int_{\Bbb R^n}u_1(x)\Phi(1,x){\rm d}x \\ &&-\varepsilon \int_{\Bbb R^n}u_0(x)\Phi_t(1,x){\rm d}x+\varepsilon \mu\int_{\Bbb R^n}u_0(x)\Phi(1,x){\rm d}x. \end{matrix}$

联立(3.1)和(3.19)式, 得到

$\begin{matrix}\label{23} &&\int_{\Bbb R^n} u_{t}(t,x)\Phi(t,x) {\rm d}x-\int_{\Bbb R^n} u(t,x)\Phi_t(t,x) {\rm d}x+\mu\int_{\Bbb R^n}t^{-1}u(t,x)\Phi(t,x){\rm d}x \\ &=& C_{\gamma}\int_1^t \int_{\Bbb R^n}\Phi(s,x)\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s+\varepsilon \int_{\Bbb R^n}u_1(x)\Phi(1,x){\rm d}x \\ &&-\varepsilon \int_{\Bbb R^n}u_0(x)\Phi_t(1,x){\rm d}x+\varepsilon \mu\int_{\Bbb R^n}u_0(x)\Phi(1,x){\rm d}x. \end{matrix}$

由(3.18)式, 易得

$\begin{matrix}\label{24} U'(t)=\int_{\Bbb R^n}u_t(t,x)\Phi(t,x){\rm d}x+\frac{\lambda'(t)}{\lambda(t)}U(t). \end{matrix}$

结合(3.18)-(3.21)式, 可推出

$\begin{matrix}\label{25} &&U'(t)-\frac{2\lambda'(t)}{\lambda(t)}U(t)+\mu t^{-1}U(t) \\ &=& C_{\gamma}\int_1^t \int_{\Bbb R^n}\Phi(s,x)\int_1^s(s-\tau)^{-\gamma} |u(\tau,x)|^p {\rm d}\tau {\rm d}x{\rm d}s +\varepsilon \int_{\Bbb R^n}u_1(x)\lambda(1)\phi(x){\rm d}x \\ &&+\varepsilon \int_{\Bbb R^n}(\mu \lambda(1)-\lambda'(1))u_0(x)\phi(x){\rm d}x. \end{matrix}$

由第二类修正Bessel函数有关递推关系式

$K'_{\chi}(z)=-K_{\chi+1}(z)+\frac{\chi}{z}K_{\chi}(z), $

结合(3.8)式, 得到

$\begin{matrix}\label{26} \lambda'(s)&=&\frac{\mu+1}{2}s^{\frac{\mu-1}{2}}K_{\frac{\delta}{l+1}}(w_l(s)) +s^{\frac{\mu-1}{2}}K'_{\frac{\delta}{l+1}}(w_l(s))w'_l(s) \\ &=&(\frac{\mu+1}{2}+\delta)s^{\frac{\mu-1}{2}}K_{\frac{\delta}{l+1}}(w_l(s)) -s^{\frac{\mu+1}{2}+l}K_{\frac{\delta}{l+1}+1}(w_l(s)). \end{matrix}$

由(3.8)和(3.22)-(3.23)式以及 $K_{\chi}(z) (\chi\in R, z>0)$的非负性, 有

$\begin{matrix}\label{27} && \int_{\Bbb R^n}u_1(x)\lambda(1)\phi(x){\rm d}x+ \int_{\Bbb R^n}(\mu \lambda(1)-\lambda'(1))u_0(x)\phi(x){\rm d}x \\ &=&\int_{\Bbb R^n}K_{\frac{\delta}{l+1}+1}(1)u_0(x)\phi(x){\rm d}x+ \int_{\Bbb R^n}K_{\frac{\delta}{l+1}}(1)(u_1(x) +(\frac{\mu-1}{2}-\delta)u_0(x))\phi(x){\rm d}x \\ &>&0. \end{matrix}$

设

$I=I[u_0,u_1]= \int_{\Bbb R^n}u_1(x)\lambda(1)\phi(x){\rm d}x+ \int_{\Bbb R^n}(\mu \lambda(1)-\lambda'(1))u_0(x)\phi(x){\rm d}x, $

由(3.22)式, 有

$\begin{matrix}\label{28} U'(t)-\frac{2\lambda'(t)}{\lambda(t)}U(t)+\mu t^{-1}U(t)\geq \varepsilon I. \end{matrix}$

对(3.25)式进一步整理, 得到

$\begin{matrix}\label{29} \frac{\lambda^2(t)}{t^\mu}\frac{{\rm d}}{{\rm d}t}(\frac{t^\mu}{\lambda^2(t)}U(t)) \geq \varepsilon I. \end{matrix}$

对(3.26)式在 $[t]$上积分, 整理有

$\begin{matrix}\label{30} U(t)&\geq&\frac{U(1)}{\lambda^2(1)}t^{-\mu}\lambda^2(t)+ \varepsilon I t^{-\mu}\lambda^2(t)\int_1^t \frac{s^\mu}{\lambda^2(s)}{\rm d}s \geq \varepsilon I t^{-\mu}\lambda^2(t)\int_1^t \frac{s^\mu}{\lambda^2(s)}{\rm d}s. \end{matrix}$

利用 $K_{\chi}(z) $的渐近性[23], 当$z>0, z\rightarrow\infty$时, 有$K_{\chi}(z)=\sqrt{\frac{\pi}{2z}}{\rm e}^{-z}(1+O(z^{-1})). $ 从而存在 $t_0>1,C_1,C_2>0, s\geq t_0$, 使得

$\begin{matrix}\label{31} C_1{\rm e}^{-2w_l(s)}s^{\mu-l}\leq\lambda^2(s)\leq C_2{\rm e}^{-2w_l(s)}s^{\mu-l}. \end{matrix}$

联立(3.27)-(3.28)式, 可得

$\begin{matrix}\label{32} U(t)&\geq &\frac{\varepsilon IC_1}{C_2}{\rm e}^{-2w_l(t)} t^{-l}\int_{t_0}^t {\rm e}^{2w_l(s)}s^l{\rm d}s \geq \frac{\varepsilon IC_1}{C_2}{\rm e}^{-2w_l(t)} t^{-l}\int_{\frac{t}{2}}^t {\rm e}^{2w_l(s)}s^l{\rm d}s \\ &\geq&\frac{\varepsilon IC_1}{2C_2} t^{-l}(1- {\rm e}^{-\frac{2}{l+1}(1-2^{\frac{1}{l+1}})(2t_0)^{l+1}})=\widetilde{C}\varepsilon t^{-l}, \end{matrix}$

其中 $t\geq 2t_0=t_1$.

运用 Hölder不等式和定理条件, 得到

$\begin{matrix}\label{33} U(t)\leq \left(\int_{\Bbb R^n} |u(t,x)|^p {\rm d}x\right)^{\frac{1}{p}} \bigg(\int_{B_{R+A_l(t)}} \Phi(t,x)^{p'} {\rm d}x\bigg)^{\frac{1}{p'}}, \end{matrix}$

其中 $p'$ 为$p$ 的共轭指数.

从而有

$\begin{matrix}\label{34} \int_{\Bbb R^n} |u(t,x)|^p {\rm d}x\geq U^p(t) \bigg(\int_{B_{R+A_l(t)}} \Phi(t,x)^{p'} {\rm d}x\bigg)^{-(p-1)}. \end{matrix}$

利用文献[24], 可推得

$\begin{matrix}\label{35} \int_{B_{R+A_l(t)}} \Phi(t,x)^{p'} {\rm d}x&=&\lambda^{p'}(t)\int_{B_{R+A_l(t)}} \phi(x)^{p'} {\rm d}x \\ &\leq &C_3\lambda^{p'}(t){\rm e}^{(R+A_l(t))p'}(R+A_l(t))^{n-1-\frac{(n-1)p'}{2}}, \end{matrix}$

其中$C_3>0.$

由(3.32)式, 进一步得到

$\begin{matrix}\label{36} \bigg(\int_{B_{R+A_l(t)}} \Phi(t,x)^{p'} {\rm d}x\bigg)^{p-1}\leq \widetilde{C}_3t^{\frac{(\mu-l)p}{2}}{\rm e}^{(R-w_l(1))p}(R+A_l(t))^{(n-1)(p-1)-\frac{(n-1)p}{2}}, \end{matrix}$

其中$\widetilde{C}_3>0.$

联立(3.29)、(3.31)和(3.33)式, 有

$\begin{matrix}\label{37} \int_{\Bbb R^n} |u(t,x)|^p {\rm d}x\geq \widetilde{\widetilde{C}}_3 \varepsilon^pt^{-\frac{(\mu+l+(n-1)(l+1))p}{2}+(n-1)(l+1)}, \end{matrix}$

其中$\widetilde{\widetilde{C}}_3 >0.$

由(3.15)和(3.34)式, 可得

$\begin{matrix}\label{38} F(t)&\geq & C_{\gamma}\widetilde{\widetilde{C}}_3\varepsilon^pt^{-q_1} \int_{t_1}^t s^{-(q_2+1-q_1)}\int_{t_1}^s \tau^{q_2+1}\int_{t_1}^\tau(\tau-\sigma)^{-\gamma} \sigma^{-\frac{(\mu+l+(n-1)(l+1))p}{2}+(n-1)(l+1)} {\rm d}\sigma {\rm d}\tau {\rm d}s \\ &\geq & C_{\gamma}\widetilde{\widetilde{C}}_3\varepsilon^pt^{-(q_2+1+\gamma+\frac{(\mu+l+(n-1)(l+1))p}{2})} \int_{t_1}^t \int_{t_1}^s \tau^{q_2+1}\int_{t_1}^\tau\sigma^{(n-1)(l+1)} {\rm d}\sigma {\rm d}\tau {\rm d}s \\ &\geq & \frac{C_{\gamma}\widetilde{\widetilde{C}}_3\varepsilon^p t^{-(q_2+1+\gamma+\frac{(\mu+l+(n-1)(l+1))p}{2})}(t-t_1)^{(n-1)(l+1)+q_2+4}}{((n-1)(l+1)+1)((n-1)(l+1)+q_2+3)((n-1)(l+1)+q_2+4)}, \end{matrix}$

其中$t\geq t_1.$

记

$K_0=\frac{C_{\gamma}\widetilde{\widetilde{C}}_3\varepsilon^p} {((n-1)(l+1)+1)((n-1)(l+1)+q_2+3)((n-1)(l+1)+q_2+4)}, $$ \alpha_0=q_2+1+\gamma+\frac{(\mu+l+(n-1)(l+1))p}{2}, \beta_0=(n-1)(l+1)+q_2+4, $

则(3.35)式化为

$\begin{matrix}\label{39} F(t)\geq K_0t^{-\alpha_0}(t-t_1)^{\beta_0},\quad t\geq t_1. \end{matrix}$

接下来, 将通过迭代方法来证明本文的结论. 为此, 设

$\begin{matrix}\label{40} F(t)\geq K_jt^{-\alpha_j}(t-L_jt_1)^{\beta_j}, \end{matrix}$

其中 $t\geq L_jt_1,\{K_j\}_{j\in {\Bbb N}}$,$\{\alpha_j\}_{j\in {\Bbb N}}$,$\{\beta_j\}_{j\in {\Bbb N}}$ 均为非负实数列, $\{L_j\}_{j\in {\Bbb N}}$ 为无限积收敛的部分积序列, 其定义为

$ L_j=\prod\limits_{k=0}^j l_k, l_k=1+p^{-k}, k,j\in {\Bbb N}. $

$\{L_j\}_{j\in {\Bbb N}}$ 的无限积定义为$ \prod\limits_{k=0}^\infty l_k, l_k=1+p^{-k}, k\in {\Bbb N}$.

由(3.36)式可知,(3.37)式对于$j=0$成立. 假如(3.37)式对任意的$j\geq 0$成立, 以下证明对$j+1$也成立.

联立(3.17)和(3.37)式,有

$\begin{matrix}\label{41} F(t)&\geq& C_{\gamma}\widetilde{c}K_j^pt^{-(q_2+1+\gamma)-n(l+1)(p-1)} \int_{L_jt_1}^t \int_{L_jt_1}^s \tau^{q_2+1}\int_{L_jt_1}^\tau \sigma^{-p\alpha_j}(\sigma-L_jt_1)^{p\beta_j} {\rm d}\sigma {\rm d}\tau {\rm d}s \\ &\geq & C_{\gamma}\widetilde{c}K_j^pt^{-(q_2+1+\gamma)-n(l+1)(p-1)-p\alpha_j} \int_{L_jt_1}^t \int_{L_jt_1}^s \tau^{q_2+1}\int_{L_jt_1}^\tau (\sigma-L_jt_1)^{p\beta_j} {\rm d}\sigma {\rm d}\tau {\rm d}s \\ &\geq &\frac{C_{\gamma}\widetilde{c}K_j^pt^{-(q_2+1+\gamma)-n(l+1)(p-1)-p\alpha_j}} {(p\beta_j+1)(p\beta_j+q_2+3)} \int_{\frac{t}{l_{j+1}}}^t(s-L_jt_1)^{p\beta_j+q_2+3} {\rm d}s \\ &\geq & \frac{C_{\gamma}\widetilde{c}K_j^p(l_{j+1}-1)t^{-(q_2+1+\gamma)-n(l+1)(p-1)-p\alpha_j} (t-L_{j+1}t_1)^{p\beta_j+q_2+4}}{(p\beta_j+1)(p\beta_j+q_2+3)l_{j+1}^{p\beta_j+q_2+4}}, \end{matrix}$

其中 $t\geq L_{j+1}t_1.$

令

$\begin{equation}\label{42} K_{j+1}=\frac{C_{\gamma}\widetilde{c}K_j^p(l_{j+1}-1)} {(p\beta_j+1)(p\beta_j+q_2+3)l_{j+1}^{p\beta_j+q_2+4}}, \end{equation}$
$\begin{equation}\label{43} \alpha_{j+1}=(q_2+1+\gamma)+n(l+1)(p-1)+p\alpha_j=p\alpha_j+r, \end{equation}$
$\begin{equation}\label{44} \beta_{j+1}=p\beta_j+q_2+4. \end{equation}$

于是,(3.38)式化为

$\begin{matrix}\label{45} F(t)\geq K_{j+1}t^{-\alpha_{j+1}}(t-L_{j+1}t_1)^{\beta_{j+1}}. \end{matrix}$

(3.42)式说明(3.37)式对于 $j+1$成立. 以下将对$\alpha_j,\beta_j,K_j$ 进行估计.

由(3.40)和(3.41)式以及递推关系, 可推出

$\begin{matrix}\label{46} \alpha_j&=&p\alpha_{j-1}+r=\cdots=r(1+p+p^2+\cdots+p^{j-1})+\alpha_0p^j \\ &=&(\frac{r}{p-1}+\alpha_0)p^j-\frac{r}{p-1}, \end{matrix}$
$\begin{matrix} \label{47} \beta_{j}&=&p\beta_{j-1}+q_2+4=\cdots=(q_2+4)(1+p+p^2+\cdots+p^{j-1})+\beta_0p^j \\ &=&(\frac{q_2+4}{p-1}+\beta_0)p^j-\frac{q_2+4}{p-1}. \end{matrix}$

另外,

$\begin{equation}\label{48} \lim\limits_{j\rightarrow\infty}l_{j}^{\beta_j}=\lim\limits_{j\rightarrow\infty}(1+p^{-j})^{(\frac{q_2+4}{p-1}+\beta_0)p^j-\frac{q_2+4}{p-1}}={\rm e}^{\frac{q_2+4}{p-1}+\beta_0}. \end{equation}$

结合(3.39)和(3.45)式, 可得

$\begin{matrix}\label{49} K_{j}&=&\frac{C_{\gamma}\widetilde{c}K_{j-1}^p(l_{j}-1)} {(p\beta_{j-1}+1)(p\beta_{j-1}+q_2+3)l_{j}^{p\beta_{j-1}+q_2+4}} \\ &\geq &C_{\gamma}\widetilde{c}{\rm e}^{-(\frac{q_2+4}{p-1}+\beta_0)}(\frac{q_2+4}{p-1}+\beta_0)^{-2}p^{-3j}K_{j-1}^p=Dp^{-3j}K_{j-1}^p. \end{matrix}$

利用(3.46)式, 对其两边取对数, 可推出

$\begin{matrix}\label{50} \log K_j&\geq& \log D-3j\log p+p\log K_{j-1}\geq\cdots \\ &\geq&(1+p+p^2+\cdots+p^{j-1})\log D+p^j\log K_0-3(p^{j-1}+2p^{j-2}+\cdots+j)\log p \\ &=&p^j(\frac{\log D}{p-1}+\log K_0-\frac{3p\log p}{(p-1)^2})+\frac{3(p+(p-1)j)\log p}{(p-1)^2}-\frac{\log D}{p-1}, \ \ \forall j\in {\Bbb N}. \end{matrix}$

令 $j_0=j_0(n,p,l,\mu,\gamma)\in {\Bbb N}$, 满足

$ j_0\geq \frac{\log D}{3\log p}-\frac{p}{p-1}. $

于是, 由(3.47)式可得

$\begin{matrix}\label{51} \log K_j\geq p^j(\frac{\log D}{p-1}+\log K_0-\frac{3p\log p}{(p-1)^2})= p^j\log(E_0\varepsilon^p), \end{matrix}$

其中 $E_0=E_0(n,p,l,\mu,\gamma)>0,j \geq j_0.$

由$L_j$定义以及数学分析相关理论, 设$L=\lim\limits_{j\rightarrow\infty}L_j=\prod\limits_{j=0}^\infty l_j\in \Bbb R.$ 又$l_j>1$, 故$j\rightarrow\infty$时, 得到$L_j\rightarrow L$. 另外, 当$j\in \Bbb R$ 和 $t\geq Lt_1$ 时,(3.37)式成立. 也就是,

$\begin{matrix}\label{52} F(t)\geq K_jt^{-\alpha_j}(t-Lt_1)^{\beta_j}. \end{matrix}$

由(3.43)-(3.44)以及(3.48)-(3.49)式, 对于$j\geq j_0, t\geq Lt_1$, 有

$\begin{matrix}\label{53} F(t)&\geq& \exp(p^j\log(E_0\varepsilon^p))t^{-((\frac{r}{p-1}+\alpha_0)p^j-\frac{r}{p-1})} (t-Lt_1)^{(\frac{q_2+4}{p-1}+\beta_0)p^j-\frac{q_2+4}{p-1}} \\ &=&\exp(p^j(\log(E_0\varepsilon^p)-(\frac{r}{p-1}+\alpha_0)\log t+(\frac{q_2+4}{p-1}+\beta_0)\log (t-Lt_1))) \\ &&\times t^{\frac{r}{p-1}}(t-Lt_1)^{-\frac{q_2+4}{p-1}}. \end{matrix}$

当$t\geq 2Lt_1$时, 可得

$\begin{matrix}\label{54} F(t)&\geq& \exp p^j(\log(E_02^{-(\frac{q_2+4}{p-1}+\beta_0)}\varepsilon^pt^{\frac{q_2+4}{p-1}+\beta_0-(\frac{r}{p-1}+\alpha_0)})) t^{\frac{r}{p-1}}(t-Lt_1)^{-\frac{q_2+4}{p-1}} \\ & =& \exp p^j(\log(E_1\varepsilon^pt^{\frac{q_2+4}{p-1}+\beta_0-(\frac{r}{p-1}+\alpha_0)})) t^{\frac{r}{p-1}}(t-Lt_1)^{-\frac{q_2+4}{p-1}}, \end{matrix}$

其中$E_1=E_02^{-(\frac{q_2+4}{p-1}+\beta_0)}.$

(3.51)式指数函数$t$的指数如下

$\begin{matrix}\label{55} & &\frac{q_2+4}{p-1}+\beta_0-(\frac{r}{p-1}+\alpha_0)=\frac{q_2+4-r}{p-1}+\beta_0-\alpha_0 \\ &=&\frac{2(l+1)+(5-2\gamma+\mu+(n-2)(l+1))p-p^2(\mu+l+(n-1)(l+1))}{2(p-1)} \\ &=&\frac{\Theta(n,p,l,\gamma,\mu)}{2(p-1)}. \end{matrix}$

对于$1<p<p_0(n,p,l,\mu,\gamma)$, 有$\Theta(n,p,l,\gamma,\mu)>0.$

固定$\varepsilon_0=\varepsilon_0(u_0,u_1,n,p,l,\gamma,\mu,\nu^2)>0$, 有下式成立

$\varepsilon_0^{-\frac{2p(p-1)}{\Theta(n,p,l,\gamma,\mu)}}\geq 2Lt_1E_1^{\frac{2(p-1)}{\Theta(n,p,l,\gamma,\mu)}}. $

于是, 对于 $\varepsilon\in(0,\varepsilon_0]$ 和$t>E_1^{-\frac{2(p-1)}{\Theta(n,p,l,\gamma,\mu)}} \varepsilon^{-\frac{2p(p-1)}{\Theta(n,p,l,\gamma,\mu)}}$, 可得

$\log(E_1\varepsilon^pt^{\frac{\Theta(n,p,l,\gamma,\mu)}{2(p-1)}})>0. $

因此, $\varepsilon \in (0,\varepsilon_0]$ 和 $t>E_1^{-\frac{2(p-1)}{\Theta(n,p,l,\gamma,\mu)}} \varepsilon^{-\frac{2p(p-1)}{\Theta(n,p,l,\gamma,\mu)}}$ 时, 取$j\rightarrow\infty,$ 可推得(3.51)式中 $F(t)$的下界爆破. 由此得到问题(1.1)不存在全局解. 同时, 进一步可推得$u$的生命跨度

$T(\varepsilon)\leq \check{C}\varepsilon^{-\frac{2p(p-1)}{\Theta(p,n,l,\gamma,\mu)}}, $

其中 $\breve{C}$ 为正常数. 定理2.1得证.

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