Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

硼掺杂对(2,2)单壁碳纳米管电子结构影响研究

周宝艳

周宝艳. 硼掺杂对(2,2)单壁碳纳米管电子结构影响研究[J]. 华南师范大学学报(自然科学版), 2013, 45(1). DOI: 10.6054/j.jscnun.2012.12.010
引用本文: 周宝艳. 硼掺杂对(2,2)单壁碳纳米管电子结构影响研究[J]. 华南师范大学学报(自然科学版), 2013, 45(1). DOI: 10.6054/j.jscnun.2012.12.010
ZHOU Bao-Yan. THE STUDY ON THE EFFECTS OF THE ELECTRONIC STRUCTURE OF BORON-DOPED(2,2) SINGLE-WALLED CARBON NANOTUBES[J]. Journal of South China Normal University (Natural Science Edition), 2013, 45(1). DOI: 10.6054/j.jscnun.2012.12.010
Citation: ZHOU Bao-Yan. THE STUDY ON THE EFFECTS OF THE ELECTRONIC STRUCTURE OF BORON-DOPED(2,2) SINGLE-WALLED CARBON NANOTUBES[J]. Journal of South China Normal University (Natural Science Edition), 2013, 45(1). DOI: 10.6054/j.jscnun.2012.12.010

硼掺杂对(2,2)单壁碳纳米管电子结构影响研究

基金项目: 

福建省自然科学基金

详细信息
    通讯作者:

    周宝艳

  • 中图分类号: 

    O471.5

THE STUDY ON THE EFFECTS OF THE ELECTRONIC STRUCTURE OF BORON-DOPED(2,2) SINGLE-WALLED CARBON NANOTUBES

More Information
    Corresponding author:

    ZHOU Bao-Yan

  • 摘要: 基于密度泛函理论下的第一性原理,计算了不同B掺杂浓度的SWCNT的几何形状结构、杂质的形成能、能带结构和态密度(DFT),研究了B掺杂对(2,2)单壁碳纳米管(SWCNT)电子结构的影响.B杂质的引入使SWCNT的管径增大,杂质的形成能为负值,表明(2,2)SWCNT进行掺B的过程为放热反应,B原子以替位形式掺入碳纳米管中是可行的.掺杂B后,SWCNT的费米能级向价带迁移,使(2,2)SWCNT转变为n型半导体.
    Abstract: We studied boron -doped (2,2) single-walled carbon nanotubes(SWCNT) by using the first-principle method based on density functional theory. In this paper,Geometry structure, formation energy, band structure and density of states(DFT) are calculated. The results show the diameter of SWCNT will be increased when boron is adulterated, the process of boron doping SWCNT will release energy, it is feasible to substitute a carbon atom with a boron atom in SWCNT. It is also found that Fermi energy move to valence band, and B-doped carbon nanotubes is n type semiconductor.
  • 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:

    {uttΔu=|v|p((x,t)Rn×(0,T)),vttΔv=|u|q((x,t)Rn×(0,T)),(u,ut,v,vt)(0,x)=ε(u0,u1,v0,v1)(x)(xRn),

    其中, qp>1,n1,ε>0,Δ是拉普拉斯算子。该二阶半线性波动系统的临界曲线为

    αW:=max

    \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。

    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)

    \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)

    \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)

    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。

    本文的主要结果如下:

    定理1  设p, q>1p, 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} 。

    \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}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\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UV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}

    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)

    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)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)

    \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)

    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}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)

    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)

    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}

    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)} 。

    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}

    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}

    Y_{2}(n, p, q)>0时, t的指数是正的。

    \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, 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)} 。

    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]t>E_{2} \varepsilon^{-1 / Y_{1}(n, p, q)}时, 有t \geqslant R\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]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)}\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。

计量
  • 文章访问数:  1291
  • HTML全文浏览量:  73
  • PDF下载量:  882
  • 被引次数: 0
出版历程
  • 收稿日期:  2012-03-05
  • 修回日期:  2012-04-14
  • 刊出日期:  2013-01-24

目录

/

返回文章
返回