数学物理学报, 2023, 43(3): 771-784

曲率控制细胞和组织生长演化模型的Cauchy问题

王增桂,

聊城大学数学科学学院 山东聊城 252059

Cauchy Problem for the Evolution of Cells and Tissue During Curvature-Controlled Growth

Wang Zenggui,

School of Mathematical Sciences, Liaocheng University, Shandong Liaocheng 252059

通讯作者: *王增桂, E-mail: wangzenggui@lcu.edu.cn

收稿日期: 2022-03-22   修回日期: 2023-01-12  

基金资助: 山东省自然科学基金(ZR2021MA084)
聊城大学科研基金(318012025)
聊城大学强特色智能科学与技术学科基金(319462208)

Received: 2022-03-22   Revised: 2023-01-12  

Fund supported: Natural Science Foundation of Shandong Province(ZR2021MA084)
Scientific Research Foundation of Liaocheng University(318012025)
Discipline with Strong Characteristics of Liaocheng University-Intelligent Science and Technology(319462208)

摘要

该文研究了一类由曲率控制细胞和组织生长演化的Cauchy问题, 根据支撑函数的定义, 将拟线性退化的演化方程转化成一类非齐次拟线性双曲方程组. 进一步通过对拟线性双曲方程组的解的先验估计, 证明了该双曲曲率流Cauchy问题经典解的生命跨度.

关键词： 曲率控制下细胞和组织的演化; 非齐次拟线性双曲方程组; 先验估计; 生命跨度

Abstract

In this paper, We consider Cauchy problem for the evolution of cells and tissue during curvature-controlled growth. By the definition of Riemann invariants, the evolution equation can be rewritten as a non-homogeneous quasilinear hyperbolic system. the lifespan of classical solution to the initial value problem is given by a priori estimation of the solution of the quasilinear hyperbolic system.

Keywords： The evolution of cells and tissue during curvature-controlled growth; Non-homogeneous quasilinear hyperbolic system; Priori estimation; Lifespan

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本文引用格式

王增桂. 曲率控制细胞和组织生长演化模型的Cauchy问题[J]. 数学物理学报, 2023, 43(3): 771-784

Wang Zenggui. Cauchy Problem for the Evolution of Cells and Tissue During Curvature-Controlled Growth[J]. Acta Mathematica Scientia, 2023, 43(3): 771-784

1 引言

最近, Alias 与 Buenzli[1-2] 提出了一种基于细胞和组织生长的数学模型, 组织合成细胞表面演化的法向加速度线性地依赖于平均曲率. 假设二维空间中的组织界面$F(t)$ 用显式的前沿跟踪参数$\gamma(s,t)$来描述, 其中$s$是一个任意的一维参数, $t$是时间. 如果参数化$\gamma(s,t)$正交, 即时间线$t\mapsto\gamma(s,t)$始终垂直于界面, 控制组织边界和细胞密度的演化方程为

$\begin{equation}\label{eq:1.1}\gamma_{t}=v{\vec{n}}=k\rho{\vec{n}},\end{equation}$
$\begin{equation}\label{eq:1.2}\rho_{t}=-k \rho^{2}\kappa+D\rho_{\ell\ell}-A\rho,\end{equation}$

其中下标表示偏导数, $v$对应于界面的法向速度, $k$是细胞分泌率(每单位时间每个细胞分泌的新组织的体积), $\rho$是组织合成细胞表面密度, $\ell$为弧长, $d\ell=gds$, $g=|\gamma_{s}|$, ${\vec{t}}=\frac{\gamma_{s}}{|\gamma_{s}|}$ 为 $F(t)$的单位切向量, ${\vec{n}}$ 是垂直于组织基质的单位外法向量, 而$ \kappa={\vec{t}}\cdot{\vec{n}}_{l}$是有符号曲率, 当$\kappa>0$时, 组织基质为凸面; 当$k<0$时, 组织基质为凹面. 方程(1.2)中$D\rho_{\ell\ell}$ 表示平行于具有扩散率D的界面的单元扩散, 而$-A\rho$表示以速率A耗散减少的活性细胞数量. 如果分泌速率$k$为常数, 方程(1.2)等价于

$\begin{equation}\label{eq:1.3}v_{t}=-v^{2}\kappa+Dv_{\ell\ell}-Av,\end{equation}$

其中, $v_{t}$对应于界面的法向加速度$\gamma_{tt}\cdot{\vec{n}}$. 因此, 在右侧的第一项使此流成为双曲曲率流. Alias 与 Buenzli[3]利用水平集的方法研究了曲率控制增长时细胞和组织的演化, 进一步采用三种不同的数值方法进行数值模拟.

双曲曲率流是曲线或曲面运动的热点研究内容之一. 为了描述氦晶体融化时表现出来的固态原子仅在他们融化并进入液态时才运动的现象,Gurtin与 Podio-Guidugli[4] 研究了一类满足双曲型偏微分方程的平面曲线演化问题.Rostein等[5] 进一步分析了在双曲晶体曲率作用下的闭多项式曲线运动,并给出了数值分析.丘[6]提出振动膜(或曲面运动)以平均曲率向量场为加速度场的双曲几何流方程. Lefloch 与Smoczyk[7] 提出了一类法向双曲平均曲率流, 证明了曲线在有限时间内产生奇点;同时还证明了在一维图情形下演化方程弱解的整体存在性. 李与王[8] 利用拟线性双曲方程组的局部解理论研究了文献[7]中的法向双曲平均曲率流Cauchy问题经典解的生命跨度. Lefloch与闫[27] 证明了（奇异）自相似解的存在性及其 基于Nash-Moser迭代的加权的Sobolev空间的稳定性. 何、孔与刘[10] 研究了丘[6]提出的双曲平均曲率流. 孔、刘与王[11]研究了法向加速度为平均曲率的平面曲线演化问题, 首先证明了平面曲线的保凸性, 其次分析了在初始速度非负时, 平面曲线演化的渐近性态. 孔与王[12] 根据双曲方程的特征值理论分析了平面曲线在双曲平均曲率流作用下奇性的形成问题. 何、黄与邢[13]研究了该曲率流的自相似解. 在双曲曲线收缩流的基础上, 王[14⇓-16] 研究了带有外力场的双曲平均曲率流的演化问题及其奇性分析. 王[17]对Minkowski 空间中的双曲平均曲率流作用下类空曲线Cauchy问题经典解的生命跨度进行了研究. 毛[18]研究了强迫双曲平均曲率流, 给出了解的局部存在唯一性, 推广了文献[10,12] 中相应的结果. 毛等[19]研究了双曲逆平均曲率流, 假设初始紧光滑超曲面是平均凸与星形的, 证明了双曲逆平均曲率流的短时间存在性. 毛等[20] 研究了一类双曲曲线流, 曲线法向加速度为平均曲率的幂函数, 证明了解的局部存在唯一性. 王[21]利用拟线性双曲方程组的局部解理论, 研究了一类双曲逆平均曲率流初值问题经典解的生命跨度.

曹与沃[22]研究了一类凸超曲面的保法向双曲曲率流, 进一步在文献[23]中, 沃等人提出并研究了双曲仿射曲线流, 在此基础上, 王[24] 对带有耗散项的双曲仿射曲线流进行了分析和研究. Notz[25]研究了Riemannian流形中闭超曲面的运动几何方程, 在球形曲面情况下该方程可以描述一个移动肥皂泡的非理想化模型. 闫[26] 引进了一个拟线性退化的双曲方程描述在球对称位势中闭曲面的运动, 基于一个新的Nash-Morse迭代方法证明了这种Cauchy问题大时间的存在性和具有小初值的稳定性.

本文主要对$\gamma(s,t)$为严格凸闭曲线的情况进行研究, 特别地, 当$D=A=0$时, 演化方程方程(1.1)、方程(1.3)的Cauchy问题, 即

$\begin{equation}\label{eq:1.4}\left\{\begin{array}{ll} \gamma_{t }=v{\vec{n}},\\ v_{t}=-v^{2}\kappa,\\ \gamma(s,0)=\gamma_{0}(s),\quad v(s,0)=v_{0}(s)>0.\end{array}\right.\end{equation}$

主要结果如下.

${\bf定理1.1}$ 假设$\gamma_{0}$为光滑闭曲线, 曲率$k_{0}>0$, $v_{0}(s)\in C^{1}(S^{1})$, 则初值问题(1.4)局部解的生命跨度的下界$\underline{\delta}$为

$\begin{equation} \label{eq:1.5}\underline{\delta}=\min\left\{ \frac{1}{8\|{\varphi}\|+2}, \frac{A_{0}(\|{\varphi}\|)+A_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\|}{B_{0}(\|{\varphi}\|)+B_{1}(\|{\varphi}\|)\| \dot{{\varphi}}\|+B_{2}(\|{\varphi}\|)\|\dot{{\varphi}}\|^2}, \frac{1}{C_{0}(\|{\varphi}\|)+C_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\|}\right\}.\end{equation} $

其中, ${{\varphi}}=({\varphi}_{1},\cdots,{\varphi}_{5})^{T}=\big(\frac{\tilde{v_{0}}-\tilde{v_{0}}'}{h''+h}, \frac{-\tilde{v_{0}}-\tilde{v_{0}}'}{h''+h},h, \tilde{v_{0}},h'\big)^{T}. $

$\begin{eqnarray*} && A_{0}(\|{\varphi}\|)=8\|{\varphi}\|^2+3\|{\varphi}\|, A_{1}(\|{\varphi}\|)=2\|{\varphi}\|+1, A_{2}(\|{\varphi}\|)=8\|{\varphi}\|^2+2\|{\varphi}\|,\\ &&A_{3}(\|{\varphi}\|)=(64\|{\varphi}\|^2+4)(8\|{\varphi}\|^{2}+6\|{\varphi}\|),B_{0}(\|{\varphi}\|)= A_{1}(\|{\varphi}\|)A_{3}(\|{\varphi}\|)+A_{2}(\|{\varphi}\|), \\ &&B_{1}(\|{\varphi}\|)=2A_{1}(\|{\varphi}\|)^2(64\|{\varphi}\|^2+4)+ 4A_{1}(\|{\varphi}\|)(8\|{\varphi}\|^{2}+6\|{\varphi}\|), B_{2}(\|{\varphi}\|)=8A_{1}(\|{\varphi}\|)^{2}.\\ &&C_{0}(\|{\varphi}\|)=12(1+2\|{\varphi}\|)^{2}(8\|{\varphi}\|+1)^2, C_{1}(\|{\varphi}\|)=12(1+2\|{\varphi}\|)^{2}.\end{eqnarray*}$

论文结构如下. 在第2节, 根据严格凸曲线的支撑函数, 曲线演化方程(1.4)被约化成曲线支撑函数满足的一个双曲型Monge-Ampére方程, 进一步给出其Riemann不变量满足的拟线性双曲方程组, 得到解的局部存在唯一性定理.在第3节, 利用双曲拟线性双曲方程组的局部解理论, 给出定理1.1的证明.

2 解的局部存在唯一性定理

假设严格凸闭曲线$\gamma(s,t)$满足(1.4), $\theta$是曲线$\gamma$的单位外法向角, 则单位外法向量${\vec{n}}$和单位切向量${\vec{t}}$分别为

${\vec{n}}=(\cos \theta,\sin \theta),\quad {\vec{t}}=(-\sin \theta, \cos \theta).$

由Frenet公式可得$\frac{\partial\theta}{\partial \ell}=\kappa$. 令

$\widetilde{\gamma}(\theta,\tau)=\gamma(s(\theta,\tau), t(\theta,\tau))=\gamma(s(\theta,\tau), t(\theta,\tau)),$

其中$t(\theta,\tau)=\tau$.则曲线$\gamma$ 的支撑函数为

$S(\theta,\tau)=(\widetilde{\gamma}(\theta,\tau),{\vec{n}}).$

求$S$关于$\theta$的一阶偏导数

$ S_{\theta}(\theta,\tau)=(\widetilde{\gamma}_{\theta}(\theta,\tau),{\vec{n}})+(\widetilde{\gamma}(\theta,\tau),{\vec{n}}_{\theta}) =(\widetilde{\gamma}(\theta,\tau),{\vec{t}}). $

则曲线$\tilde{\gamma}$可表示为

$\widetilde{\gamma}=S{\vec{n}}+S_{\theta}{\vec{t}}.$

易得曲线$\tilde{\gamma}$的平均曲率为

$k=\frac{1}{S_{\theta\theta}+S}.$

由

$\frac{\partial\widetilde{\gamma}}{\partial \tau}=\frac{\partial \gamma} {\partial u}\frac{\partial u}{\partial\tau}+\frac{\partial \gamma}{\partial t}.$

得支撑函数$S$关于$\tau$的一阶偏导数满足

$ S_{\tau}=\left(\frac{\partial\widetilde{\gamma}}{\partial \tau},-{\vec{n}}\right) =\left(\frac{\partial \gamma}{\partial u}\frac{\partial u}{\partial \tau}+\frac{\partial \gamma}{\partial t},-{\vec{n}}\right)=\left(\frac{\partial\gamma}{\partial t},-{\vec{n}}\right) $

及

$ S_{\tau\tau} =\left (\frac{\partial^{2}\gamma}{\partial u\partial t}\frac{\partial u}{\partial \tau}+\frac{\partial^{2} \gamma}{\partial t^{2}},-{\vec{n}}\right) =\left(\frac{\partial^{2}\gamma}{\partial u\partial t}\frac{\partial u}{\partial \tau},-{\vec{n}}\right)-\left|\frac{\partial \gamma}{\partial {t}}\right|^2k =\frac{\partial u}{\partial \tau}\left(\frac{\partial^{2}\gamma}{\partial u\partial t},-{\vec{n}}\right)-S_{\tau}^2k. $

由$S_{\theta}=(\widetilde{\gamma},{\vec{t}})$ 知

$S_{\tau\theta}=\Big|\frac{\partial \gamma}{\partial u}\Big|\frac{\partial u}{\partial\tau}.$

又由

$S_{\theta\tau}=\left(\frac{\partial^{2}\gamma}{\partial u\partial t} \frac{\partial u}{\partial \theta},-{\vec{n}}\right)+ \left(\frac{\partial \gamma}{\partial t},-\frac{\partial{\vec{n}}}{\partial\theta}\right) =\left(\frac{\partial^{2}\gamma}{\partial u\partial t} \frac{\partial u}{\partial s}\frac{\partial s}{\partial \theta},-{\vec{n}}\right) =\frac{1}{k\Big|\frac{\partial \gamma}{\partial u}\Big|}\left(\frac{\partial^{2}\gamma}{\partial u\partial t},-{\vec{n}}\right) $

可得

$S_{\theta\tau}^2\kappa=\Big|\frac{\partial \gamma}{\partial u}\Big|\frac{\partial u}{\partial\tau}\cdot \frac{1}{\Big|\frac{\partial \gamma}{\partial u}\Big|}\left(\frac{\partial^{2}\gamma}{\partial u\partial t},-{\vec{n}}\right)=\frac{\partial u}{\partial \tau}\left(\frac{\partial^{2}\gamma}{\partial u\partial t},-{\vec{n}}\right). $

从而曲线的支撑函数$S$满足方程

$\begin{equation}\label{eq:2.1} S_{\tau\tau}=\left(\frac{\partial^{2}\gamma}{\partial u\partial t}\frac{\partial u}{\partial \tau},-{\vec{n}}\right)-S_{\tau}^2k=S_{\theta\tau}^{2}k-S_{\tau}^{2}k =\frac{S_{\theta\tau}^{2}-S_{\tau}^{2}}{S_{\theta\theta}+S}, \end{equation}$

即

$\begin{equation}\label{eq:2.2}SS_{\tau\tau}+S_{\tau\tau}S_{\theta\theta}-S_{\theta\tau}^{2}+S_{\tau}^{2}=0.\end{equation}$

因此方程(1.4)可简化为

$\begin{equation}\label{eq:2.3}\left\{\begin{array}{ll} SS_{\tau\tau}+S_{\tau\tau}S_{\theta\theta}-S_{\theta\tau}^{2}+S_{\tau}^{2}=0,\\ S(\theta, 0)=h(\theta),\\ S_{\tau}(\theta,0)=\widetilde{v_{0}}(\theta),\end{array}\right. \end{equation}$

其中$h$是$\gamma_{0}$的支撑函数, 且初始速度$\widetilde{v_{0}}>0 $. 易知$S\equiv h(\theta)$是(2.3)的静态解, 此时, $\widetilde{v_{0}}(\theta)=0$. 本文我们只考虑$S_{\tau}>0$的情况, 该方程为双曲型Monge-Ampére 方程. 事实上, 由

$S_{\tau}^2+S\cdot S_{\tau\tau}+0\cdot S_{\theta\tau}+0\cdot S_{\theta\theta}+1\cdot (S_{\tau\tau}S_{\theta\theta}-S_{\theta\tau}^{2})=0$

可知

$A=S_{\tau}^{2},\ B=S,\ C=0,\ D=0, \ E=1.$

则

$\bigtriangleup^{2}(\tau,\theta,S,S_{\tau},S_{\theta})=C^{2}-4BD+4AE=4S_{\tau}^{2}>0.$

下文中将$\tau$记为$t$, 设 $S(\theta,t)\in C^{3}\tilde{D}$是方程(2.3)的解. 则易知函数

$\begin{equation}\label{eq:2.4}r=\frac{C+\Delta-2S_{\theta t}}{2(S_{\theta\theta}+S)}= \frac{S_{t}-S_{\theta t}}{S_{\theta\theta}+S}, \ \ s=\frac{C-\Delta-2S_{\theta t}}{2(S_{\theta\theta}+S)}=\frac{-S_{t}-S_{\theta t}}{S_{\theta\theta}+S}\end{equation}$

是方程(2.3)的Rimeann不变量. 设 $p=S_{t}$, $q=S_{\theta}$, 则方程(2.3) 转化成Riemann不变量形式下的拟线性双曲方程组的Cauchy问题

$\begin{equation}\label{eq:2.5}\left\{\begin{array}{ll} r_{t}+sr_{\theta}=-r\left(r-s\right),\\ s_{t}+rs_{\theta}=-s(r-s),\\ S_{t}+rS_{\theta}=p+rq, \\ p_{t}+sp_{\theta}=sp,\\ q_{t}+rq_{\theta}=p-rS,\end{array}\right.\end{equation}$

对应初值为

$\begin{equation}\label{eq:2.6}\left\{\begin{array}{ll} r(0,\theta)=\frac{\tilde{v_{0}}(\theta)-\tilde{v_{0}}'(\theta)}{h''(\theta)+h(\theta)},\\[3mm] s(0,\theta)=\frac{-\tilde{v_{0}}(\theta)-\tilde{v_{0}}'(\theta)}{h''(\theta)+h(\theta)},\\ S(0,\theta)=h(\theta),\\ p(0,\theta)=\tilde{v_{0}}(\theta),\\ q(0,\theta)=h'(\theta).\end{array}\right. \end{equation}$

若令${u}=(r,s,S,p,q)^{T}$, 其中$u_{1}=r,u_{2}=s,u_{3}=S,u_{4}=p,u_{5}=q$.

$\Lambda=\left[\begin{array}{ccccc} \lambda_{1}&0&0&0&0\\ 0&~\lambda_{2}~&0&~0~&0\\ 0&0&\lambda_{3}&0&0\\ 0&0& 0&\lambda_{4}&0\\ 0&0&0&0&\lambda_{5}\end{array}\right] =\left[\begin{array}{ccccc} s&0&0&0&0\\ 0&~r~&0&~0~&0\\ 0&0&r&0&0\\ 0&0& 0&s&0\\ 0&0&0&0&r\end{array}\right],$
${\mu}=(\mu_{1},\cdots,\mu_{5})^{T }=\left(-r\left(r-s\right),-s\left(r-s\right),p+rq,s p,p-rS\right)^{T}, $

则方程(2.5)及初值(2.6)可改写为

$\begin{equation}\label{eq:2.7}\left\{\begin{array}{ll} \sum_{j=1}^{5}\delta_{ij}\left(\frac{\partial u_{j}}{\partial t}+\lambda_{i}\frac{\partial u_{j}}{\partial \theta}\right)=\mu_{i},i=1,\cdots,5,\\ u_{i}(0,\theta)=\varphi_{i}(\theta), i=1,\cdots,5, \end{array}\right.\end{equation}$
其中, ${\varphi}=(\varphi_{1},\cdots,\varphi_{5})^{T}=\big(\frac{\tilde{v_{0}}(\theta)-\tilde{v_{0}}'(\theta)}{h''(\theta)+h(\theta)}, \frac{-\tilde{v_{0}}(\theta)-\tilde{v_{0}}'(\theta)}{h''(\theta)+h(\theta)},h(\theta), \tilde{v_{0}}(\theta),h'(\theta)\big)^{T}$.

首先假设对任意向量${u}=(u_{1},\cdots,u_{5})^{T}\in {\Bbb R} ^{5}$, 定义

$|{u}|=\max_{1\leq i\leq 5}|u_{i}|,$

对于向量函数${u}={u}(t,\theta)=(u_{1}(t,\theta),\cdots,u_{5}(t,\theta))^{T}$ 的模$\|{u}\|$定义为

$\|{u}\|=\sup_{(t,\theta)\in D}|{u(t,\theta)}|,$

而如果向量函数${u}(t,\theta)$是连续可微的, 还可定义$\|{u}\|_{1}$为

$\|{u}\|_{1}=\|{u}\|+\left\|\frac{\partial {u}}{\partial t}\right\|+\left\|\frac{\partial {u}}{\partial \theta}\right\|.$

类似地, 可定义任一有限的函数集$G$的模 $\|G\|, \|G\|_{1}$.

由(2.6)式可知, ${\varphi}=(\varphi_{1},\cdots,\varphi_{5})^{T}$ 是$D$ 上有界的,且${\varphi}\in C^{1}(D)$, 则存在正常数$M$, 使得$\|{\varphi}\|<M$. 记$R(\delta_{0})=\{(t,\theta)|(t,\theta)\in [\delta_{0}]\times [2\pi]\}$, 并将其扩展区域定义为$E(\delta_{0})=\{(t,\theta,{u})\}|(t,\theta)\in R(\delta_{0}),|{u}|\leq M\}$. 进一步假设$\lambda_{i},\frac{\partial\lambda_{i}}{\partial \theta},\frac{\partial\lambda_{i}}{\partial u_{j}},\mu_{i}, \frac{\partial\mu_{i}}{\partial \theta},\frac{\partial\mu_{i}}{\partial u_{j}}$是$E(\delta_{0})$上的连续函数, 那么由标准的双曲型偏微分方程理论, 存在常数$0<\delta_{*}\leq\delta_{0}$, 使得方程组(2.7)的Cauchy问题在$R(\delta_{*})$上存在唯一的连续可微解$u(t,\theta)$, 即定理2.1.

${\bf定理2.1}$ (局部存在唯一性) 假设$\gamma_{0}$是严格凸闭曲线, $v_{0}(s)\in C^{1}(S^{1})$, 则存在$T>0$ 与一簇严格闭凸曲线$\gamma(\cdot,t)$, 使得当$t\in [0,T)$时, $\gamma(\cdot,t)$满足(1.4)式.

3 定理1.1的证明

在本节中, 根据拟线性双曲方程组的Cauchy问题的局部解理论研究局部解的生命跨度.

第一步, 线性化拟线性双曲方程组(2.7). 为了进行解的第二先验估计, 选取适当的正常数$M_{1}$, 使得

$\|{\varphi}\|+(1+M)\|\dot{{\varphi}}\|+2M^{2}+M<M_{1}.$

对于任意${\nu}\in\sum(\delta|M,M_{1})$, 引入下列函数集合

$\sum(\delta|M,M_{1})=\{{\nu}(t,\theta)|{\nu}=(\nu_{1},\cdots,\nu_{5})^{T}\in C^{1}[R(\delta)],\|{\nu}\|\leq M, \|{{\nu}}\|_{1}\leq M_{1}\}.$

根据一阶线性偏微分方程组

$\begin{equation}\label{eq:3.1}\left\{\begin{array}{ll} \sum_{j=1}^{5}\delta_{lj}\left(\frac{\partial u_{j}}{\partial t}+ \tilde{\lambda}_{l}(t,\theta)\frac{\partial u_{j}}{\partial \theta}\right)=\tilde{\mu}_{l}(t,\theta),l=1,\cdots,5,\\ u_{l}(0,\theta)=\varphi_{l}(\theta), \theta\in [2\pi], l=1,\cdots,5.\end{array}\right. \end{equation}$
其中 $\tilde{\lambda}_{l}(t,\theta)=\lambda_{l}(t,\theta,{\nu}(t,\theta))$, $\tilde{\mu}_{l}(t,\theta)=\mu_{l}(t,\theta,{\nu})$, 即 ${\tilde{\lambda}}=(\nu_{2},\nu_{1},\nu_{1},\nu_{2},\nu_{1})^{T}$;
$ {\tilde{\mu}}={\mu}(t,\theta,{\nu})=(-\nu_{1}(\nu_{1}-\nu_{2}), -\nu_{2}(\nu_{1}-\nu_{2}), \nu_{4}+\nu_{1}\nu_{5}, \nu_{2}\nu_{4}, \nu_{4}-\nu_{1}\nu_{3})^{T}. $

定义函数集$G^{*}=\left\{\lambda_{l},\mu_{l}, \frac{\partial\lambda_{l}}{\partial \theta},\frac{\partial\lambda_{l}} {\partial u_{j}},\frac{\partial\mu_{l}}{\partial \theta},\frac{\partial\mu_{l}}{\partial u_{j}},l,j=1,\cdots,5\right\}$, 其中$\frac{\partial\lambda_{l}}{\partial \theta}=\frac{\partial\mu_{l}}{\partial \theta}=0$$ (l=1,\cdots,5,)$,有

$\left(\frac{\partial \lambda_{l}}{\partial u_{j}}\right)= \left[\begin{array}{ccccc} 0& ~1~&1&~0~&1\\ 1& 0&0&1&0\\ 0&0&0&0&0\\ 0& 0& 0&0&0\\ 0&0&0&0&0\end{array}\right],$
$\left(\frac{\partial \mu_{l}}{\partial u_{j}}\right)= \left[\begin{array}{ccccc} \nu_{2}-2\nu_{1}& -\nu_{2}&\nu_{5}&0&-\nu_{3}\\ \nu_{1}&~ 2\nu_{2}-\nu_{1}~&0&~\nu_{4}~&0\\ 0&0&0&0&-\nu_{1}\\ 0& 0& 1&\nu_{2}&1\\ 0&0&\nu_{1}&0&0\end{array}\right].$

则由

$\max_{l=1,\cdots,5}|\lambda_{l}|=\max\{|\nu_{1}|,|\nu_{2}|\}\leq M,$
$\max_{l=1,\cdots,5}|\mu_{l}|=\max\left\{2M^{2}, M+M^{2}\right\}\leq 2M^{2}+M, $
$\max_{l=1,\cdots,5}\left|\frac{\partial\lambda_{l}}{\partial \theta}\right|=0,\quad \max_{l,j=1,\cdots,5}\left|\frac{\partial\lambda_{l}}{\partial u_{j}}\right|=1, \max_{l=1,\cdots,5}\left|\frac{\partial\mu_{l}}{\partial \theta}\right|=0,$
$ \max_{l,j=1,\cdots,5}\left|\frac{\partial\mu_{l}}{\partial u_{j}}(t,\theta,\nu)\right|\leq\max\left\{3M,1\right\}\leq3M+1,$

可知, $\|G^*\|$为

$\|G^*\|= \max\left\{2M^2+M, 3M+1\right\}.$

令函数集合$G_{2}[{\nu}]=\{\tilde{\lambda}_{l}(t,\theta),\frac{\partial \tilde{\lambda}_{l}}{\partial \theta}(t,\theta,),\tilde{\mu}_{l}(t,\theta),\frac{\partial \tilde{\mu}_{l}}{\partial \theta}(t,\theta)\}$, 则由

$\frac{\partial\tilde{ \lambda}_{l}}{\partial \theta}(t,\theta)=\frac{\partial \lambda_{l}}{\partial \theta}(t,\theta,{\nu})+\sum_{j=1}^{5} \frac{\partial \lambda_{l}}{\partial u_{j}}(t,\theta,{\nu})\frac{\partial \nu_{j}}{\partial \theta},\quad l=1,\cdots,5,$
$\frac{\partial\tilde{ \mu}_{l}}{\partial \theta}(t,\theta)=\frac{\partial \mu_{l}}{\partial \theta}(t,\theta,{\nu})+\sum_{j=1}^{5} \frac{\partial \mu_{l}}{\partial u_{j}}(t,\theta,{\nu})\frac{\partial \nu_{j}}{\partial \theta},\quad l=1,\cdots,5, $

可知

$\frac{\partial\tilde{\lambda}_{l}}{\partial \theta}\leq M_{1}, \frac{\partial\tilde{\mu}_{l}}{\partial \theta}\leq\max\{4M,2M+1\}M_{1}\leq(4M+1)M_{1}.$

从而, $\|G_{2}[{\nu}]\|$为

$ \|G_{2}[{\nu}]\|=\max\left\{2M^2+M, (4M+1)M_{1}\right\}.$

首先, 证明Cauchy问题(2.7)解的第一限先验估计式. 定义

$\frac{d}{d_{l}\varsigma}=\frac{\partial}{\partial\varsigma}+\lambda_{l}\frac{\partial}{\partial\xi}.$

为沿第$l$族特征线的微商$\frac{d}{d_{l}\varsigma}$. 则由第一相关积分,

$\begin{equation}\label{eq:3.2}u_{k}(t,\theta)=u_{k}^{0}(t,\theta)+\int_{0}^{t}\mu_{k}(\tau,x_{k}(\varsigma;t,\theta)){\rm d}\varsigma, k=1,\cdots,5,\end{equation} $

其中$u_{k}^{0}(t,\theta)=\varphi_{k}(x_{k}(0;t,\theta)),k=1,\cdots,5.$ 当$0\leq t\leq\delta$时

$\|{u}\|\leq\|{\varphi}\|+(2M^{2}+M)\delta.$

下证Cauchy问题(2.7)解的第二先验估计式, 即解的一阶导数的上界. 记

$\begin{equation}\label{eq:3.3}\left\{\begin{array}{ll} {v}=(v_{1}(t,\theta),\cdots,v_{5}(t,\theta),)^{T}=\left(\frac{\partial u_{1}(t,\theta)}{\partial t},\cdots,\frac{\partial u_{5}(t,\theta)}{\partial t}\right)^{T},\\[3mm] {w}=(w_{1}(t,\theta),\cdots,w_{5}(t,\theta),)^{T}=\left(\frac{\partial u_{1}(t,\theta)}{\partial \theta},\cdots,\frac{\partial u_{5}(t,\theta)}{\partial \theta}\right)^{T}.\end{array}\right.\end{equation}$

第二相关积分为

$\begin{equation}\label{eq:3.4}\left\{\begin{array}{ll} v_{i}(t,\theta)=v_{i}^{0}(t,\theta)-\lambda_{i}\int_{0}^{t}\bar{\mu}_{i}(\tau,x_{i}(\tau;t,\theta)){\rm d}\tau, \quad&(i=1,\cdots,5),\\ v_{i}^{0}(t,\theta)=\mu_{i}(t,\theta)-\lambda_{l}(t,\theta)\dot{\varphi}_{i}(x_{i}(0;t,\theta)),\quad&(i=1,\cdots,5).\end{array}\right.\end{equation}$
$\begin{equation}\label{eq:3.5} \left\{\begin{array}{ll} w_{i}(t,\theta)=w_{i}^{0}(t,\theta)+\int_{0}^{t}\bar{\mu}_{i}(\tau,x_{i}(\tau;t,\theta)){\rm d}\tau, \quad&(i=1,\cdots,5),\\[2mm] w_{i}^{0}(t,\theta)=\dot{\varphi}_{i}(x_{i}(0;t,\theta)),\quad&(i=1,\cdots,5).\end{array}\right.\end{equation}$

其中

$\bar{\mu}_{i}(t,\theta)=\frac{\partial \mu_{i}(t,\theta)}{\partial \theta}-\frac{\partial \lambda_{i}}{\partial \theta}w_{i},\quad (i=1,\cdots,5).$

则在$R(\delta)$上, 可得

$\max_{i=1,\cdots,5}|w_{i}^{0}(t,\theta)|\leq\|\dot{{\varphi}\|}, \max_{i=1,\cdots,5}|v_{i}^{0}(t,\theta)|\leq M\|\dot{{\varphi}\|}+2M^{2}+M.$

令

$v(\tau)=\sup_{i=1,\cdots,5}\{|v_{i}(t,\theta)|,(t,\theta)\in R(\tau)\},\quad w(\tau)=\sup_{i=1,\cdots,5}\{|w_{i}(t,\theta)|,(t,\theta)\in R(\tau)\},$

则成立

$\|v\|=\sup_{0\leq \tau\leq\delta}v(\tau)=v(\delta),\quad\|w\|=\sup_{0\leq\tau\leq\delta}w(\tau)=w(\delta).$

由(3.5)式, 当$0\leq t\leq\delta$时

$w(t)\leq\|\dot{{\varphi}}\|+(4M+1)M_{1}\delta+M_{1}\int_{0}^{t}w(\tau){\rm d}\tau, $

由Gronwall不等式

$w(t)\leq\left[\|\dot{{\varphi}}\|+\left(4M+1\right)M_{1}\delta\right]e^{M_{1}t},$

从而有

$\begin{equation}\label{eq:3.6}\|{w}\|\leq\left[\|\dot{{\varphi}}\|+\left(4M+1\right)M_{1}\delta\right]e^{M_{1}\delta}.\end{equation}$

由

$ v(t)\leq M\|\dot{{\varphi}}\|+2M^{2}+M +\left(4M+1\right)MM_{1}\delta+MM_{1} \int_{0}^{t}w(\tau){\rm d}\tau,$

将$w(t)\leq\left[\|\dot{{\varphi}}\|+\left(4M+1\right)M_{1}\delta\right]e^{M_{1}t}$ 代入上式,得

$v(t)\leq M\dot{\|{\varphi}\|}+2M^{2}+M +M\left[\|\dot{{\varphi}}\|+\left(4M+1\right)M_{1}\delta\right]e^{M_{1}t},$

从而有

$\begin{equation}\label{eq:3.7}\|{v}\|\leq2M^{2}+M+ M\left[\|\dot{{\varphi}}\|+\left(4M+1\right)M_{1}\delta\right]e^{M_{1}\delta},\end{equation} $

于是由$\|{u}\|_{1}$定义可知

$\begin{matrix}\label{eq:3.8} \|{u}\|_{1}&=&\|{u}\|+\|{v}\|+\|{w}\|\\ &\leq&\|{\varphi}\|+(2M^{2}+M)(1+\delta) +(M+1)\left[\|\dot{{\varphi}}\|+\left(4M+1\right)M_{1}\delta\right]e^{M_{1}\delta},\end{matrix}$

假设$\delta\leq\frac{1}{M_{1}}$, 由Taylor展开式

$e^{M_{1}\delta}<1+M_{1}\delta e^{M_{1}\delta}<1+3M_{1}\delta,$

代入(3.8)式, 并简化$\delta^{2}$项,

$\begin{equation}\label{eq:3.9}\|{u}\|_{1}\leq\|{\varphi}\|+\left(2M^{2}+M\right)(1+\delta)+(M+1)\|\dot{{\varphi}}\| +(M+1)(16M^2+4\|\dot{{\varphi}}\|+4)M_{1}\delta.\end{equation}$

下面将给出Cauchy问题(2.7)的第三估计式, 即估计解的一阶偏导数的连续性模. 设$\psi(t,\theta)$是定义在区域$R$上的函数, $\psi(t,\theta)$在$R$上的连续性模定义为非负值函数

$\rho(\eta)=\rho(\eta|\psi)=\sup\limits_{|t'-t''|\leq\eta<\infty\atop \|\theta'-\theta''|\leq\eta<\infty} \{|\psi(t',\theta')-\psi(t'',\theta'')|,\quad(t',\theta'),(t'',\theta'')\in D\}.$

类似地, 可分别定义向量函数${\psi}=(\psi_{1}(t,\theta),\cdots,\psi_{n}(t,\theta))^{T}$ 或一般的有限个函数的集合$\Psi=\{\psi_{i}\}$ 的连续性模为

$\rho(\eta|{\psi})=\max_{i=1,\cdots,n}\rho(\eta|\psi_{i}), \rho(\eta|\Psi)=\max_{\psi_{i}\in \Psi}\rho(\eta|\psi_{i}).$

第$i$条特征曲线$x_i=x_{i}(\varsigma;t,\theta)$满足

$\left\{\begin{array}{ll} \frac{dx_{i}}{{\rm d}\varsigma}=\lambda_{i}(\varsigma,\xi), \\ x_{i}(t;t,\theta)=\theta.\end{array}\right.$

有

$\begin{equation}\label{eq:3.10}\left\{\begin{array}{ll} \frac{\partial x_{i}}{\partial \theta}(\varsigma;t,\theta)=e^{\int_{t}^{\varsigma}\frac{\partial\lambda_{i}}{\partial\xi}(\varsigma_{1},x_{l}(\varsigma_{1};t,\theta)){\rm d}\varsigma}_{1},\\[3mm] \frac{\partial x_{i}}{\partial t}(\varsigma;t,\theta)=-\lambda_{l}e^{\int_{t}^{\varsigma}\frac{\partial\lambda_{i}}{\partial\xi}(\varsigma_{1},x_{l}(\varsigma_{1};t,\theta)){\rm d}\varsigma_{1}},\end{array}\right. \end{equation} $

从而, 对于$i=1,\cdots,5$, 我们有

$\begin{equation}\label{eq:3.11}\left|\frac{\partial x_{i}}{\partial \theta}\right|\leq \exp\left(\left\| \frac{\partial\lambda_{i}}{\partial \theta}\right\|t\right)\leq 1 +\left\|\frac{\partial\lambda_{i}}{\partial \theta}\right\|\delta \exp\left(\left\| \frac{\partial\lambda_{i}}{\partial \theta}\right\|\right)\delta)\leq1+3M_{1}\delta,\end{equation} $
$\begin{equation} \label{eq:3.12}\left|\frac{\partial x_{i}}{\partial t}\right|\leq \|\lambda_{i}\|\exp \left(\left\| \frac{\partial\lambda_{i}}{\partial \theta}\right\|t\right)\leq(1+3M_{1}\delta)M.\end{equation}$

因此

$\rho(\eta|w_{i}^{0}(t,\theta))=\rho(\eta|\dot{\varphi}(x_{i}(0;t,\theta)))\leq (1+M)(1+3M_{1}\delta)\rho(\eta|\dot{\varphi}_{i}). $

由

$\begin{array}{ll}\rho(\eta|v_{i}^{0}(t,\theta))&=\rho(\eta|\mu_{i}-\lambda_{i}\dot{\varphi}(x_{i}(0;t,\theta)))\\ &\leq\rho(\eta|\mu_{i})+\|\lambda_{i}\|\rho(\eta|\dot{\varphi}((x_{i}(0;t,\theta))))+\|\dot{\varphi}_{i}\|\rho(\eta|\lambda_{i}),\end{array}$

记$\chi(\eta)=\max\{\rho(\eta|\mu_{i}),\rho(\eta|\lambda_{i})\}$, 则

$\begin{equation}\label{eq:3.13}\rho(\eta|v_{i}^{0}(t,\theta))\leq M(1+M)(1+3M_{1}\delta)\rho(\eta|\dot{\varphi}_{i})+(1+\|\dot{\varphi}\|)\chi(\eta).\end{equation}$

综上可得

$\rho(\eta|{w}^{0})\leq (1+M)(1+3M_{1}\delta)\rho(\eta|\dot{{\varphi}}), $
$\rho(\eta|{v}^{0})\leq M(1+M)(1+3M_{1}\delta)\rho(\eta|\dot{{\varphi}})+(1+\|\dot{{\varphi}}\|)\chi(\eta). $

上两式相加

$\begin{equation}\label{eq:3.14}\rho(\eta|{w}^{0})+\rho(\eta|{v}^{0})\leq(1+M)^{2}(1+3M_{1}\delta)\rho(\eta|\dot{{\varphi}})+(1+\|\dot{{\varphi}}\|)\chi(\eta).\end{equation}$

令

$\rho(\tau,\eta|{w})=\sup\limits_{ \stackrel{i=1,\cdots,5} {\stackrel{(t',\theta'),(t'',\theta'')\in R(M)} {\scriptscriptstyle |t'-t''|\leq\eta,|\theta'-\theta''|\leq\eta} }} |w_{i}(t',\theta')-w_{i}(t'',\theta'')|, $
$\rho(\tau,\eta|{v})=\sup\limits_{ \stackrel{i=1,\cdots,5} {\stackrel{(t',\theta'),(t'',\theta'')\in R(M)} {\scriptscriptstyle |t'-t''|\leq\eta,|\theta'-\theta''|\leq\eta}}} |v_{i}(t',\theta')-v_{i}(t'',\theta'')|, $

由第二相关积分可以估计$\rho(\tau,\eta|{w}), \rho(\tau,\eta|{v})$. 令

$\chi_{1}(\eta)=\max_{i=1,\cdots,5}\rho\left(\tau,\eta|\frac{\partial\mu_{i}}{\partial \theta}\right), \rho\left(\tau,\eta|\frac{\partial\lambda_{i}}{\partial \theta}\right).$

当$\bar{t}\leq\delta$时

$ \rho(\bar{t},\eta|\bar{\mu}_{i}\left(\tau,x_{i}(\tau;t,\theta)\right))\leq 4(1+M)\rho(\tau,\eta|\bar{\mu}_{i}(\tau,\xi)), $
$\begin{eqnarray*} \rho(\tau,\eta|\bar{\mu}_{i}(\tau,\xi))&=&\rho\left(\tau,\eta\Big|\frac{\partial\mu_{i}(\tau,\xi)}{\partial \theta}- \frac{\partial\lambda_{i}(\tau,\xi)}{\partial \theta}w_{i}(\tau,\xi)\right)\\ &\leq&\rho\left(\tau,\eta\Big|\frac{\partial\mu_{i}}{\partial \theta}\right)+\left\|\frac{\partial\lambda_{i}}{\partial \theta}\right\|\rho\left(\tau,\eta|w_{i}(\tau,\xi)\right) +\|w_{i}\|\rho\left(\tau,\eta\Big|\frac{\partial\lambda_{i}}{\partial \theta}(\tau,\xi)\right)\\ &\leq&M_{1}\rho(\tau,\eta|{w})+(1+\|{w}\|)\chi_{1}(\eta)\\ &\leq&M_{1}\rho(\tau,\eta|{w})+\left(3\|\dot{{\varphi}}\|+12M+3\right)\chi_{1}(\eta), \end{eqnarray*} $
$\rho(\bar{t},\eta|\bar{\mu}_{i}(\tau,x_{i}(\tau;t,\theta)))\leq 4(1+M)\left[M_{1}\rho(\tau,\eta|{w}) +\left(3\|\dot{{\varphi}}\|+12M+3\right)\chi_{1}(\eta)\right].$

由$\bar{\mu}_{i}=\frac{\partial\tilde{\mu}_{i}}{\partial \theta}-\frac{\partial\tilde{\lambda}_{i}}{\partial \theta}w_{i}, i=1,\cdots,5.$ 得

$ \|\bar{\mu}_{i}\|\leq\left\|\frac{\partial\tilde{\mu}_{i}}{\partial \theta}\right\| +\left\|\frac{\partial\tilde{\lambda}_{i}}{\partial \theta}\right\|\cdot\|w_{i}\| \leq\left(3\|\dot{{\varphi}}\|+16M+4\right)M_{1}, $

即

$\begin{equation}\label{eq:3.15}\|\bar{{\mu}}\|\leq\left(3\|\dot{{\varphi}}\|+16M+4\right)M_{1}. \end{equation} $

由连续模的性质

$\rho(\eta|\int_{a(t)}^{b(t)}{\psi}(\tau,t){\rm d}\tau)\leq\|{\psi}\|[\rho(\eta|a)+\rho(\eta|b)] +\sup_{t}\int_{a(t)}^{b(t)}\rho(\eta|{\psi}(\tau,t))\tau, $

可得

$\begin{eqnarray*} \rho(\bar{t},\eta|\int_{0}^{t}\bar{\mu}_{i}(\tau,M_{i}(\tau;t,\theta)){\rm d}\tau) &\leq&\|\bar{\mu}_{i}\|\eta +\int_{0}^{\bar{t}}\rho(\tau_{1},\eta|\bar{\mu}_{i}(\tau,M_{i}(\tau;t,\theta))){\rm d}\tau_{1}\\ &\leq&\left(3\|\dot{{\varphi}}\|+16M+4\right)M_{1}\eta+4(1+M)M_{1}\int_{0}^{\bar{t}}\rho(\tau,\eta|w){\rm d}\tau\\ &&+4(1+M)\left(3\|\dot{{\varphi}}\|+12M+3\right)M_{1}\delta \chi_{1}(\eta).\end{eqnarray*}$

因此

$\begin{eqnarray*} \rho(t,\eta|v_{i})+\rho(t,\eta|w_{i})&\leq&\rho(t,\eta|v_{i}^{0})+\rho(t,\eta|w_{i}^{0})+(1+\|\lambda\|) \rho(t,\eta|\int_{0}^{t}\bar{\mu}_{i}{\rm d}\tau)+\|\int_{0}^{t}\bar{\mu}_{i}{\rm d}\tau\|\chi(\eta) \\ &\leq&\left[1+\|\dot{{\varphi}}\|+(1+M_{1}+MM_{1}) \left(3\|\dot{{\varphi}}\|+16M+4\right)M\right]\chi(\eta)\\ &&+4(1+M)^{2}\left[\left(3\|\dot{{\varphi}}\|+12M+3\right)\chi_{1}(\eta)\delta+\rho(\eta|\dot{{\varphi}})\right]\\ &&+4 M_{1}(1+M)^{2}\left[\int_{0}^{t}\rho(\tau,\eta|{w}){\rm d}\tau +\int_{0}^{t}\rho(\tau,\eta|{v}){\rm d}\tau\right],\end{eqnarray*} $

由Gronwall不等式知

$\begin{matrix} \label{eq:3.16} \rho(t,\eta|v)+\rho(t,\eta|w)&\leq&e^{4(1+M)^2}\Big\{\left[1+\|\dot{{\varphi}}\|+(1+M_{1} +MM_{1}) \left(3\|\dot{{\varphi}}\|+16M+4\right)M\right]\chi(\eta)\\ &&+4(1+M)^{2}\left[\rho(\eta|\dot{{\varphi}})+\left(3\|\dot{{\varphi}}\|+12M+3\right)\delta\chi_{1}(\eta)\right]\Big\}.\end{matrix} $

下一步构造非线性算子的一致有界性. 由拟线性双曲方程组的局部解理论, 在区域$R(\delta)$上存在唯一的连续可微解${u}=(u_{1}(t,\theta),\cdots, u_{5}(t,\theta))^{T}$. 记

${u}=T{v}.$

则算子$T$的不动点就是Cauchy问题(2.7)的解, 只须证明$\delta$足够小时,算子$T$有唯一的不动点即可.

首先证明存在$\delta_{1}=\delta_{1}(\|{\varphi}\|,\|{\varphi}\|_{1})>0$ 足够小, 算子$T:\sum(\delta|M,M_{1})\to \sum(\delta|M,M_{1})$, 其中$\delta\leq\delta_{1}$. 在$R(\delta)$上, 由${u}=T{v}$可知

$\label{eq:3.17}\|{u}\|\leq\|{\varphi}\|+\left(2M^{2}+M\right)\delta, $

取$M=2\|{\varphi}\|$,当

$\delta=\frac{1}{8\|{\varphi}\|+2},$

成立

$\|{u}\|\leq M.$

由(3.12)式得

$\begin{eqnarray*}\|{u}\|_{1}&\leq&\|{\varphi}\|+\left(8\|{\varphi}\|^{2}+2\| {\varphi}\|\right)(1+\delta)+(2\|{\varphi}\|+1)\|\dot{{\varphi}}\| \\ && +(2\|{\varphi}\|+1)(64\|{\varphi}\|^2+4\|\dot{{\varphi}}\|+4)M_{1}\delta. \end{eqnarray*}$

取

$ M_{1}=2\|{\varphi}\|+2(M+1)\|\dot{{\varphi}}\|+4M^{2}+2M =8\|{\varphi}\|^{2}+6\|{\varphi}\|+2(2\|{\varphi}\|+1)\|\dot{{\varphi}}\|. $

记

$\begin{eqnarray*} &&A_{0}(\|{\varphi}\|)=8\|{\varphi}\|^2+3\|{\varphi}\|, A_{1}(\|{\varphi}\|)=2\|{\varphi}\|+1, \\ &&A_{2}(\|{\varphi}\|)=8\|{\varphi}\|^2+2\|{\varphi}\|, A_{3}(\|{\varphi}\|)=(64\|\varphi\|^2+4)(8\|{\varphi}\|^{2}+6\|{\varphi}\|), \\ &&B_{0}(\|{\varphi}\|)=A_{1}(\|{\varphi}\|)A_{3}(\|{\varphi}\|)+A_{2}(\|{\varphi}\|),\\ &&B_{1}(\|{\varphi}\|)=2A_{1}(\|{\varphi}\|)^2(64\|{\varphi}\|^2+4)+4A_{1}(\|{\varphi}\|)(8\|{\varphi}\|^{2}+6\|{\varphi}\|), \end{eqnarray*}$

则可取

$\begin{equation}\label{eq:3.18}\delta=\frac{ A_{0}(\|{\varphi}\|)+A_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\|}{B_{0}(\|{\varphi}\|)+B_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\|+8A_{1}(\|{\varphi}\|)^{2}\| \dot{{\varphi}}\|^2}. \end{equation}$

于是, 取

$\begin{equation}\label{eq:3.19}\delta_{1}=\min\left\{\frac{1}{8\|{\varphi}\|+2}, \frac{A_{0}(\|{\varphi}\|)+A_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\|}{B_{0}(\|{\varphi}\|)+B_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\| +8A_{1}(\|{\varphi}\|)^{2}\|\dot{{\varphi}}\|^2}\right\}, \end{equation}$

则算子$T$是一个从函数集合$\sum(\delta|M,M_{1})$ 到自身的变换, 其中$\delta\leq\delta_{1}$.

为了算子$T$的等度连续性, 我们考虑$\delta_{2}$的选取. 令

$\tilde{\chi}(\eta)=\max_{l=1,\cdots,5}\{\rho(\eta|\tilde{\lambda}_{l}),\rho(\eta|\tilde{\mu}_{l}),\eta\},$

由连续模的定义与$M_{1}$的构造可知

$\tilde{\chi}(\eta)\leq\max\left\{M_{1}\eta,(3M+1)M_{1}\eta,\eta\right\} =\left(3M+1\right)M_{1}\eta. $

由(3.16)式及

$\rho\left(\tau,\eta|\frac{\partial\tilde{\lambda}_{i}}{\partial \theta}\right)\leq \max\left\{\rho\left(\tau,\eta|\frac{\partial\nu_{1}} {\partial \theta}\right),\rho\left(\tau,\eta|\frac{\partial\nu_{2}} {\partial \theta}\right)\right\}\leq \rho\left(\tau,\eta|\frac{\partial{\nu}} {\partial \theta}\right),$ $\rho\left(\tau,\eta|\frac{\partial\tilde{\mu}_{i}}{\partial \theta}\right)\leq \max\left\{4M\rho\left(\tau,\eta|\frac{\partial{\nu}} {\partial \theta}\right),(2M+1)\rho\left(\tau,\eta|\frac{\partial{\nu}} {\partial \theta}\right) \right\} \leq\left(4M+1\right)\rho\left(\eta|\frac{\partial{\nu}} {\partial \theta}\right),$

可得

$\chi_{1}(\eta)\leq\left(4M+1\right)\rho\left(\eta|\frac{\partial{\nu}} {\partial \theta}\right).$

令

$\begin{eqnarray*} M_{2}(\eta)&=&e^{4(1+M)^2}\Big\{4(1+M)^{2}\rho(\eta|\dot{{\varphi}})\left(4M+1\right) \big[1+\|\dot{{\varphi}}\|\\ &&+(1+M_{1}+MM_{1}) \left(3\|\dot{{\varphi}}\|+16M+4\right)\big]M_{1}\eta \Big\}, \end{eqnarray*}$

则可取

$\delta_{2}=\frac{1}{4(1+M)^{2}\left(3\|\dot{{\varphi}}\|+12M+3\right)\left(4M+1\right)}, $

令 $C_{0}(\|{\varphi}\|)=12(1+2\|{\varphi}\|)^{2}(8\|{\varphi}\|+1)^2$, $C_{1}(\|{\varphi}\|)=12(1+2\|{\varphi}\|)^{2}$, 取

$\begin{equation}\label{eq:3.20}\delta_{2}=\frac{1}{C_{0}(\|{\varphi}\|)+C_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\|}.\end{equation}$

那么当$0\leq\delta\leq\delta_{2}$ 时, 可得

$\lim_{\eta\to 0}M_{2}(\eta)=0$

与

$\rho(\eta|{v})+\rho(\eta|{w})\leq M_{2}(\eta).$

从而取 $\delta_{3}=\min\{\delta_{1},\delta_{2}\}$, 定义

$\sum(\delta_{3})=\sum(\delta_{3}|M,M_{1},M_{2}(\eta))= \Big\{{\nu}(t,\theta)|{\nu}\in\sum(\delta_{3}|M,M_{1}), \rho(\eta\Big|\frac{\partial{\nu} }{\partial \theta})\leq M_{2}(\eta)\Big\}. $

则算子$T$是一个将函数集合$\sum(\delta_{3})$ 映到自身的算子. 由于在$C^{1}$ 模下, $\sum(\delta_{3})$集合一致有界且等度连续, 所以$\sum(\delta_{3})$是准紧的. 而

$\lim_{n\to\infty}{\nu}_{n}={\nu},$

则

$\rho(\eta|{\nu})=\lim_{n\to\infty}\rho(\eta|{\nu}_{n})\le qM_{2}(\eta),\|{\nu}\|\leq M,\|{\nu}\|_{1}\leq M_{1},$

从而$\sum(\delta_{3})$在$C^{1}[R(\delta)]$中闭, 即为紧的, 因此$\sum(\delta_{3})$是完备的. 由$\|{\nu}\|\leq\|{\nu}\|_{1}$可知$C^{1}$模下的收敛性可以推出$C^{0}$ 模下的收敛, 从而$\sum(\delta_{3})$在$C^{0}$ 模下也是完备的. 因此, 只需证明存在$\delta_{*}\leq\delta_{3}$, 使得$T$为$R(\delta_{*})$上的压缩映射, 则不动点问题有解.

令 ${\nu}^{1},{\nu}^{2}\in\sum(\delta_{3})$, 则${u}^{1}=T{\nu}^{1},{u}^{2}=T{\nu}^{2}\in\sum(\delta_{3})$. 设${\nu}^{*}={\nu}^{1}-{\nu}^{2}$, ${u}^{*}={u}^{1}-{u}^{2}$, 且

$\left\{\begin{array}{ll} \tilde{\lambda}_{l}^{\alpha}(t,\theta)=\lambda_{l}^{(\alpha)}(t,\theta,\nu^{\alpha}(t,\theta)),\\ \tilde{\lambda}_{l}^{\alpha}(t,\theta)=\lambda_{l}^{\alpha}(t,\theta,\nu^{\alpha}(t,\theta)), (\alpha=1,2). \end{array}\right. $

则有

$\left\{\begin{array}{ll} \frac{\partial u^{*}_{i}}{\partial t}+\tilde{\lambda}_{i}^{1}(t,\theta)\frac{\partial u^{*}_{i}}{\partial t} =\tilde{h}_{i}(t,\theta),\\ t=0: u^{*}_{i}=0,\quad(i=1,\cdots,5). \end{array}\right. $

其中$\tilde{h}_{i}=-(\tilde{\lambda}_{i}^{1}-\tilde{\lambda}_{i}^{2})\frac{\partial u^{2}_{i}}{\partial \theta} +(\tilde{\mu}^{1}_{i}-\tilde{\mu}^{2}_{i})$, $i=1,\cdots,5$. 在$R(\delta)$上, 计算可得

$\|\tilde{{h}}\|\leq\left(M_{1}+3M+1\right)\|{\nu}^{*}\|. $

由第一先验估计, 可得

$\|{u}^{*}\|\leq\left(M_{1}+3M+1\right)\delta\|{\nu}^{*}\|,$

于是取

$\delta_{4}=\frac{1}{3\left(M_{1}+3M+1\right)},$

则有

$\begin{equation}\label{eq:3.21}\delta_{4}=\frac{1}{3(8\|{\varphi}\|^{2}+12\|{\varphi}\|+(4\|{\varphi}\|+2)\|\dot{{\varphi}}\|+1)},\end{equation}$

那么当选取$\delta_{*}=\min\{\delta_{3},\delta_{4}\}$ 时, 算子$T$ 是$R(\delta_{*})$上的压缩映射.

综上所述,可知 $\delta_{*}=\min\{\delta_{3},\delta_{4}\}$, 又因为$\delta_{3}<\delta_{4}$. 因此,Cauchy问题的局部解的存在区间的下界$\delta_{*}$为

$\begin{equation}\label{eq:3.22}\delta_{*}=\min\left\{ \frac{1}{8\|{\varphi}\|+2}, \frac{A_{0}(\|{\varphi}\|)+A_{1}(\|{\varphi}\|)\|\dot{{\varphi}}}{B_{0}(\|{\varphi}\|)+B_{1}(\|{\varphi}\|)\|\dot{{\varphi}}\| +B_{2}(\|{\varphi}\|)\|\dot{{\varphi}}\|^2}, \frac{1}{C_{0}(\|{\varphi}\|)+C_{1}(\|{\varphi}\|) \|\dot{{\varphi}}\|}\right\}.\end{equation}$

定理1.1得证.

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