THE STUDY ON THE EFFECTS OF THE ELECTRONIC STRUCTURE OF BORON-DOPED(2,2) SINGLE-WALLED CARBON NANOTUBES
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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.
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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:
{utt−Δu=|v|p((x,t)∈Rn×(0,T)),vtt−Δv=|u|q((x,t)∈Rn×(0,T)),(u,ut,v,vt)(0,x)=ε(u0,u1,v0,v1)(x)(x∈Rn), 其中, q、p>1,n⩾1,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 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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