DOI:10.12677/PM.2018.81011,PDF,HTML,XML,被引量下载: 3,111浏览: 8,896
作者:吴小庆：西南石油大学理学院，四川 成都
关键词:热传导方程；多维(m维)；各向异性；自由边界问题；实对称正定矩阵；Heat Conduction Equation；Multidimensional (m Dimensional)；Anisotropy；Free Boundary Problem；Real Symmetric Positive Definite Matrix

文章引用：吴小庆. m维各向异性热传导方程的奇异内边界问题研究[J]. 理论数学, 2018, 8(1): 75-98. https://doi.org/10.12677/PM.2018.81011

1. 引言

一个偏微分方程的定解问题，若其定解区域的部分边界是待定的，它和定解问题的解彼此相关且必

须同时确定。这类定解问题，人们称之为自由边界问题，其待定边界称为自由边界。所有自由边界问题都是非线性问题。自由边界问题的研究有着广泛的实际背景。在渗流力学、等离子物理、塑性力学、射流等方面都提出了各种不同形式的定常和不定常自由边界问题 [1] - [12] 。为了研究m维各向异性热传导方程的自由边界问题，我们建立起相应的m维各向异性热传导方程的奇异内边界问题。本文假设在区域

$\Omega \triangleq \left\{\left(x,t\right)|x\in R^m,t\in \left(0,T\right)\right\}$

存在无重点的一条奇异内边界 $\Gamma \triangleq \left\{\left(x,t\right)|x=x\left(t\right),t\in \left(0,T\right),\forall t_1\ne t_2,x\left(t_1\right)\ne x\left(t_2\right)\right\}$；向量函数 $x\left(t\right)\in R^m$把 $R^m$分成 ${\overline{{\rm E}}}_{-}\left(t\right)$和 ${\overline{{\rm E}}}_{+}\left(t\right)$两部分， $\forall t\in \left(0,T\right)$，满足条件 ${\overline{{\rm E}}}_{-}\left(t\right)\cup {\overline{{\rm E}}}_{+}\left(t\right)=R^m$， ${\overline{{\rm E}}}_{-}\left(t\right)\cap {\overline{{\rm E}}}_{+}\left(t\right)=x\left(t\right)$； $\Gamma \subset \Omega$把区域 $\Omega$分成两个区域 ${\Omega }_{-}$和 ${\Omega }_{+}$， $\Omega ={\Omega }_{-}\cup \Gamma \cup {\Omega }_{+}$， ${\Omega }_{-}\cap {\Omega }_{+}=\Phi$(空集)；在 ${\Omega }_{-}$和 ${\Omega }_{+}$两个区域都分别满足齐次的m维各向异性热传导方程，在奇异内边界 $\Gamma$上热传导方程具有一定奇性，在整个区域 $\Omega$用具有特殊自由项

$\gamma \left(t\right)\delta (x-x(\; t\; )\; )$

的非齐次m维各向异性热传导方程来描述奇异内边界上的奇性， $\delta \left(x-x\left(t\right)\right)$为m维狄拉克d-函数， $\gamma \left(t\right)$为奇异源强度函数。其中奇异内边界 $\Gamma$是待定的，它和问题的解函数 $w\left(x,t\right)$彼此相关且必须同时确定。

显然寻求 $\left\{w\left(x,t\right),\Gamma \right\}$与寻求 $\left\{w\left(x,t\right),x\left(t\right)\right\}$是等价的。我们同时求 $\left\{w\left(x,t\right),x\left(t\right)\right\}$，使其满足数学模型I。称数学模型I为奇异内边界问题。即有

数学模型I (m维无穷区域各向异性非齐次热传导方程齐次初始条件的奇异内边界问题)：

求 $\left\{w\left(x,t\right),x\left(t\right)\right\}$，使其满足

$\{\begin{array}{l}\frac{\partial w}{\partial t}=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}w}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial w}{\partial x_k}}-rw+\gamma \left(t\right)\delta \left(x-x\left(t\right)\right),x\in R^m,x\left(t\right)\in R^m,0<t<T\\ w\left(x,0\right)=0,x\in R^m\\ w\left(x\left(t\right),t\right)=\underset{x\in R^m}{\max }w\left(x,t\right),0<t<T\\ \underset{x\to -\infty }{\lim }\left|w\right|<\infty ,\underset{x\to \infty }{\lim }\left|w\right|<\infty \end{array}$(I)

应用矩阵理论和广义特征函数法，在条件 $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B\in R^{m\times m}$为正线下三角矩阵)下，获得了数学模型I的充分光滑的精确解 $\left\{w\left(x,t\right);x\left(t\right)\right\}$。同时获得了m维各向异性齐次热传导方程的自由边界问题II_a和问题II_b的充分光滑的精确解。问题II_a和问题II_b具有公共的自由边界 $x\left(t\right)=tB\upsilon +\chi ,T>t>0$。称m维向量函数表达式 $$为最佳热源边界。表达式 $x\left(t\right)=tB\upsilon +\chi ,T>t>0$满足条件 $\frac{dx(t)}{dt}=B\upsilon$；其中 $\upsilon \triangleq {\left({\upsilon }_1,\cdots ,{\upsilon }_m\right)}^{\mathrm{T}}$， ${\upsilon }_p={\displaystyle \underset{k=1}{\overset{p}{\sum }}\left(q_k-r\right)c_{kp}},p=1,\cdots ,m$。公式表明m维向量 $B\upsilon$由m维热传导方程中出现的所有参数 $a_{kj},q_j,r$唯一确定。

2. 符号说明

m为某正整数； $B\in R^{m\times m}$为m阶实矩阵， $B^{\mathrm{T}}$表示 $B$的转置矩阵；

实数集 $R\triangleq \left\{s|s\in \left(-\infty ,\infty \right)\right\}$；

m维列向量的集合 $R^m\triangleq \left\{x|x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}},x_j\in R\right\}$；

$\forall x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}\in R^m,y={\left(y_1,\cdots ,y_m\right)}^{\mathrm{T}}\in R^m,\langle x,y\rangle \triangleq {\displaystyle \underset{k=1}{\overset{m}{\sum }}{x}_k{y}_k},{\Vert x\Vert }^2\triangleq {\displaystyle \underset{k=1}{\overset{m}{\sum }}{x}_k^2};$

向量函数：

$x\left(t\right)\triangleq {\left(x_1\left(t\right),\cdots ,x_m\left(t\right)\right)}^{\mathrm{T}}$；

$x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}>0\iff x_l>0,l\in \left\{1,\cdots ,m\right\}$；

$x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}\ge 0\iff x_l\ge 0,l\in \left\{1,\cdots ,m\right\}$；

$-\infty <x<x\left(t\right)\iff -\infty <x_j<x_j\left(t\right),j=1,\cdots ,m$；

$x\left(t\right)<x<\infty \iff x_j\left(t\right)<x_j<\infty ,j=1,\cdots ,m$；

${\mathrm{E}}_{-}\left(t\right)\triangleq \left\{x|-\infty <x<x\left(t\right)\right\},{\mathrm{E}}_{+}\left(t\right)\triangleq \left\{x|x\left(t\right)<x<\infty \right\}$

${\bar{E}}_{-}\left(t\right)\triangleq \left\{x|-\infty <x\le x\left(t\right)\right\},{\bar{E}}_{+}\left(t\right)\triangleq \left\{x|x\left(t\right)\le x<\infty \right\}$

$\Gamma \triangleq \left\{\left(x,t\right)|x=x\left(t\right),t\in \left(0,T\right)\right\}$；

$\Omega \triangleq \left\{\left(x,t\right)|x\in R^m,t\in \left(0,T\right)\right\}$；

${\Omega }_{-}\triangleq \left\{\left(x,t\right)|x\in {{\rm E}}_{-}\left(t\right),t\in \left(0,T\right)\right\},{\Omega }_{+}\triangleq \left\{\left(x,t\right)|x\in {{\rm E}}_{+}\left(t\right),t\in \left(0,T\right)\right\}$；

m维开区间：

$\left(x,y\right)\triangleq \left\{s|s={\left(s_1,\cdots ,s_m\right)}^{\mathrm{T}},x_j<s_j<y_j,j=1,\cdots ,m\right\}$；

m维闭区间：

$\left[x,y\right]\triangleq \left\{s|s={\left(s_1,\cdots ,s_m\right)}^{\mathrm{T}},x_j\le s_j\le y_j,j=1,\cdots ,m\right\}$；

多元(m元)函数 $\phi \left(x\right)\triangleq \phi \left(x_1,\cdots ,x_m\right)$； $\phi :x\in R^m\to R$的连续映射，是定义域是 $R^m$，值域在 $R$中的函数。函数的支集 $\mathrm{supp}\phi \triangleq \left\{x|\phi \left(x\right)\ne 0\right\}$，支集的闭包 $\overline{\mathrm{supp}\phi }\triangleq \overline{\left\{x|\phi \left(x\right)\ne 0\right\}}$；

记函数集合

${\Lambda }_{\left(x,y\right)}\triangleq \left\{\phi |\phi \in C\left(R^m\right),\forall s\in R^m,\phi =\phi \left(s\right)\ge 0,且\overline{\mathrm{supp}\phi }\subset \left(x,y\right)\subset R^m\right\},x\in R^m,y\in R^m$。

3. 主要结果

3.1. m维无穷区域各向异性非齐次热传导方程非齐次初始条件的初值问题的求解

定解问题I (m维无穷区域各向异性非齐次热传导方程非齐次初始条件的初值问题)：

$\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}u}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial u}{\partial x_k}}-ru+\gamma \left(t\right)\delta \left(x-x\left(t\right)\right),x\in R^m,x\left(t\right)\in R^m,0<t<T(1)\\ u\left(x,0\right)=\phi \left(x\right),x\in R^m(2)\\ \underset{x\to -\infty }{\lim }\left|u\right|<\infty ,\underset{x\to \infty }{\lim }\left|u\right|<\infty (\; 3\; )\end{array}$

定理3.1.1 (定解问题I的精确解)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)；

2) $x\left(t\right)={\left(x_1\left(t\right),\cdots ,x_m\left(t\right)\right)}^{\mathrm{T}}\in R^m$， $x_k\left(t\right)$为充分光滑的单调函数， $k=1,\cdots ,m$；

3) $\gamma \left(t\right)\in C\left(\left[0,\infty \right)\right)$；

4) $\phi \left(x\right)\in C\left(R^m\right),0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；

则定解问题I有充分光滑的精确解

$u\left(x,t\right)=V\left(x,t\right)+W\left(x,t\right)$(4)

其中

$V\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}\text{d}\xi }$(5)

$W\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}\text{d}\tau }$(6)

定解问题I的求解过程：

记偏微分算子

$L=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^2}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial }{\partial x_k}-r}$(7)

先考虑m维线性常系数偏微分方程在无界区域 $R^m$的特征值问题I

$LE=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}E}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial E}{\partial x_k}-rE}=-\lambda E$(8)

为求解特征值问题I和定解问题I我们建立了引理3.1.1~引理3.1.6。

引理3.1.1：设 $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵，则存在正线下三角矩阵 $B={\left(b_{kj}\right)}_{m\times m}\in R^{m\times m}$满足 $A=B{B}^{\mathrm{T}}$且分解是唯一的；且有

1) 正线下三角矩阵 $B$的行列式 $\left|B\right|={\displaystyle \underset{j=1}{\overset{m}{\prod }}{b}_{jj}}>0$；

2) 记 ${\left(B^{\mathrm{T}}\right)}^{-1}\triangleq C={\left(c_{kj}\right)}_{m\times m}$，则 $C$为正线上三角矩阵， $\left|C\right|={\displaystyle \underset{j=1}{\overset{m}{\prod }}{c}_{jj}}={\displaystyle \underset{j=1}{\overset{m}{\prod }}{b}_{jj}^{-1}}>0,c_{kj}=0,k>j,j=1,\cdots ,m-1$；

3) 记 $I\left(k;x\right)\triangleq {\displaystyle \underset{n=k}{\overset{m}{\sum }}{c}_{kn}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{c}_{jn}{x}_j}},k=1,\cdots ,m$， $x={\left(x_1,\cdots ,x_m\right)}^{\text{T}}\in R^m$；则当 $x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}>0$，有 $I\left(k;x\right)>0,k=1,\cdots ,m$；当 $x={\left(x_1,\cdots ,x_m\right)}^{\text{T}}<0$，有 $I\left(k;x\right)<0,k=1,\cdots ,m$。

3)' 当 $x={\left[x_1,\cdots ,x_m\right]}^{\mathrm{T}}<0$，有 $I\left(k;x\right)=\left[x_1,\cdots ,x_m\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]<0,k\in \left\{1,\cdots ,m\right\}$；当 $x={\left[x_1,\cdots ,x_m\right]}^{\mathrm{T}}>0$，有 $I\left(k;x\right)=\left[x_1,\cdots ,x_m\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]>0,k\in \left\{1,\cdots ,m\right\}$。

证明：由矩阵理论的结论 [13] 即知存在正线下三角矩阵 $B={\left(b_{kj}\right)}_{m\times m}\in R^{m\times m}$满足 $A=B{B}^{\mathrm{T}}$且分解是唯一的。由 $A=B{B}^{\mathrm{T}}$有 $\left|A\right|={\left|B\right|}^2$。

正线下三角矩阵 $B={\left(b_{kj}\right)}_{m\times m}$，正线下三角矩阵 $B$的行列式 $\left|B\right|={\displaystyle \underset{j=1}{\overset{m}{\prod }}{b}_{jj}}>0$，且 $b_{kj}=0,k<j,j=2,\cdots ,m$； $B$的转置矩阵 $B^{\mathrm{T}}$为正线上三角矩阵。 $B^{\mathrm{T}}$的逆矩阵 $C$为正线上三角矩阵。 $\left|C\right|={\displaystyle \underset{j=1}{\overset{m}{\prod }}{c}_{jj}}={\displaystyle \underset{j=1}{\overset{m}{\prod }}{b}_{jj}^{-1}}>0,c_{kj}=0,k>j,j=1,\cdots ,m-1$。

下证3) 由于 $C$为正线上三角矩阵，即有 ${\displaystyle \underset{n=k}{\overset{m}{\sum }}{c}_{kn}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{c}_{jn}{x}_j}}={\displaystyle \underset{n=1}{\overset{m}{\sum }}{c}_{kn}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{c}_{jn}{x}_j}}$和 ${\left[{\displaystyle \underset{j=1}{\overset{1}{\sum }}{c}_{j1}{x}_j},{\displaystyle \underset{j=1}{\overset{2}{\sum }}{c}_{j2}{x}_j}\cdots ,{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}{x}_j}\right]}^{\mathrm{T}}={\left[{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}{x}_j},{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j2}{x}_j}\cdots ,{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}{x}_j}\right]}^{\mathrm{T}}$；从而有

$\begin{array}{c}I\left(k;x\right)\triangleq {\displaystyle \underset{n=k}{\overset{m}{\sum }}{c}_{kn}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{c}_{jn}{x}_j}}={\displaystyle \underset{n=1}{\overset{m}{\sum }}{c}_{kn}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{c}_{jn}{x}_j}}\\ =\left[c_{k1},\cdots ,c_{kk}\cdots ,c_{km}\right]{\left[{\displaystyle \underset{j=1}{\overset{1}{\sum }}{c}_{j1}{x}_j},{\displaystyle \underset{j=1}{\overset{2}{\sum }}{c}_{j2}{x}_j}\cdots ,{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}{x}_j}\right]}^{\mathrm{T}}\\ =\left[c_{k1},\cdots ,c_{kk}\cdots ,c_{km}\right]{\left[{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}{x}_j},{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j2}{x}_j}\cdots ,{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}{x}_j}\right]}^{\mathrm{T}}\end{array}$

$x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}$为列向量，应用分块矩阵的乘法运算即有

${\left[{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}{x}_j},{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j2}{x}_j}\cdots ,{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}{x}_j}\right]}^{\mathrm{T}}=C^{\mathrm{T}}x$(9)

从而有

$I\left(k;x\right)=\left[c_{k1},c_{k2}\cdots ,c_{km}\right]C^{\mathrm{T}}x,k\in \left\{1,\cdots ,m\right\}$(10)

由于 $C{C}^{\mathrm{T}}=\left({\left(B^{\mathrm{T}}\right)}^{-1}\right){\left({\left(B^{\mathrm{T}}\right)}^{-1}\right)}^{\mathrm{T}}=\left({\left(B^{\mathrm{T}}\right)}^{-1}\right)\left(B^{-1}\right)={\left(B{B}^{\mathrm{T}}\right)}^{-1}=A^{-1}$即有

$\left[\begin{array}{c}I\left(1;x\right)\\ I\left(2;x\right)\\ ⋮\\ ⋮\\ I\left(m;x\right)\end{array}\right]=C{C}^{\mathrm{T}}x=A^{-1}x$(11)

$A={\left(a_{kj}\right)}_{m\times m}$为正定矩阵，则 $A^{-1}$为正定矩阵， $x^{\mathrm{T}}{A}^{-1}x$为正定二次齐式，从而有

$x^{\mathrm{T}}\left[\begin{array}{c}I\left(1;x\right)\\ I\left(2;x\right)\\ ⋮\\ ⋮\\ I\left(m;x\right)\end{array}\right]=x^{\mathrm{T}}{A}^{-1}x>0,x\ne 0$(12)

${\displaystyle \underset{j=1}{\overset{m}{\sum }}{x}_{j}I\left(j;x\right)}>0,x\ne 0$(13)

由 $x$的任意性，分别令 $x=y_p{\epsilon }_p,y_p\ne 0,p=1,\cdots ,m$，

其中 ${\epsilon }_1={\left(1,0,0,\cdots ,0\right)}^{\mathrm{T}},{\epsilon }_2={\left(0,1,0,\cdots ,0\right)}^{\mathrm{T}},\cdots ,{\epsilon }_m={\left(0,0,\cdots ,0,1\right)}^{\mathrm{T}}$。

由(13)式即有

$y_p{\displaystyle \underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{y}_p}>0,p=1,\cdots ,m$

即

${\displaystyle \underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{y}_p^2}>0,y_p\ne 0,p=1,\cdots ,m$(14)

再记 $y_p^2\triangleq x_p,p=1,\cdots ,m$；有 $x_p>0,p=1,\cdots ,m$和

${\displaystyle \underset{p=1}{\overset{m}{\sum }}{\displaystyle \underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{y}_p^2}}={\displaystyle \underset{p=1}{\overset{m}{\sum }}{\displaystyle \underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{x}_p}}=I\left(k;x\right),k=1,\cdots ,m$(15)

由(14)，(15)两式即有：当 $x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}$， $x_j>0,j=1,\cdots ,m$，有 $I\left(k;x\right)>0,k=1,\cdots ,m$；显然也有：

当 $x={\left(x_1,\cdots ,x_m\right)}^{\mathrm{T}}$， $x_j<0,j=1,\cdots ,m$，有 $I\left(k;x\right)<0,k=1,\cdots ,m$。由(10)式：

$\begin{array}{l}I\left(k;x\right)=\left[c_{k1},\cdots ,c_{km}\right]C^{\mathrm{T}}x\in R^{1\times 1},k=1,\cdots ,m;\\ I\left(k;x\right)={\left[I\left(k;x\right)\right]}^{\mathrm{T}}=x^{\mathrm{T}}C{\left[c_{k1},\cdots ,c_{km}\right]}^{\mathrm{T}}=\left[x_1,\cdots ,x_m\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]\end{array}$(16)

结论也可表述为：3)' 当 $x={\left[x_1,\cdots ,x_m\right]}^{\mathrm{T}}<0$，有 $I\left(k;x\right)=\left[x_1,\cdots ,x_m\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]<0,k\in \left\{1,\cdots ,m\right\}$；当

$x={\left[x_1,\cdots ,x_m\right]}^{\mathrm{T}}>0$，有 $I\left(k;x\right)=\left[x_1,\cdots ,x_m\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]>0,k\in \left\{1,\cdots ,m\right\}$。引理证毕。

记 $x\in R^m,y\in R^m$，作 $R^m$到 $R^m$的线性变换

$x=By$(17)

记

$E\left(x\right)=E\left(By\right)\triangleq Y\left(y\right)$(18)

引理3.1.2：若 $x=By$，则有

${\nabla }_y=B^{\mathrm{T}}{\nabla }_x$(19)

其中

${\nabla }_y\triangleq \left[\begin{array}{c}\frac{\partial }{\partial y_1}\\ ⋮\\ \frac{\partial }{\partial y_m}\end{array}\right]$(20)

为向量偏微分算子。

证明：由(17)式即有

$\left[\begin{array}{c}{x}_1\\ ⋮\\ x_m\end{array}\right]=\left[\begin{array}{c}{\displaystyle \underset{p=1}{\overset{m}{\sum }}{b}_{1p}}{y}_p\\ ⋮\\ {\displaystyle \underset{p=1}{\overset{m}{\sum }}{b}_{mp}}{y}_p\end{array}\right]$(21)

由(18)式即有

$E\left(x\right)=E\left(x_1,\cdots ,x_m\right)=E\left({\displaystyle \underset{p=1}{\overset{m}{\sum }}{b}_{1p}}{y}_p,\cdots ,{\displaystyle \underset{p=1}{\overset{m}{\sum }}{b}_{mp}}{y}_p\right)=Y\left(y\right)$(22)

由复合函数的求导法则，即有

$\left[\begin{array}{c}\frac{\partial }{\partial y_1}\\ ⋮\\ \frac{\partial }{\partial y_m}\end{array}\right]Y\left(y\right)=\left[\begin{array}{c}{b}_{11},\cdots ,b_{m1}\\ ⋮\\ b_{1m},\cdots ,b_{mm}\end{array}\right]\left[\begin{array}{c}\frac{\partial }{\partial x_1}\\ ⋮\\ \frac{\partial }{\partial x_m}\end{array}\right]E\left(x\right)=B^{\mathrm{T}}\left[\begin{array}{c}\frac{\partial }{\partial x_1}\\ ⋮\\ \frac{\partial }{\partial x_m}\end{array}\right]E\left(x\right)$(23)

即(19)式成立。引理证毕。

m维二阶线性偏微方程在无界区域 $R^m$的特征值问题II

${\displaystyle \underset{k=1}{\overset{m}{\sum }}\left[\frac{1}{2}\frac{{\partial }^{2}Y\left(y\right)}{\partial y_k{}^2}-{\upsilon }_k\frac{\partial Y\left(y\right)}{\partial y_k}+\left({\lambda }_k-\frac{r}{m}\right)Y\left(y\right)\right]}=0,y={\left(y_1,\cdots ,y_m\right)}^{\mathrm{T}}\in R^m$(24)

其中

$\lambda ={\displaystyle \underset{k=1}{\overset{m}{\sum }}{\lambda }_k}$(25)

${\upsilon }_k\triangleq {\displaystyle \underset{j=1}{\overset{k}{\sum }}\left(q_j-r\right)c_{jk}},k=1,\cdots ,m$(26)

引理3.1.3：若 $x=By$，则特征值问题I中方程(8)与特征值问题II中方程(24)等价。

证明：由 $A=B{B}^{\mathrm{T}}$，有

${\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\left(\frac{\partial }{\partial x_k}\right)\left(\frac{\partial }{\partial x_j}\right)E\left(x\right)}={\left[\begin{array}{c}\frac{\partial }{\partial x_1}\\ ⋮\\ \frac{\partial }{\partial x_m}\end{array}\right]}^{\mathrm{T}}A\left[\begin{array}{c}\frac{\partial }{\partial x_1}\\ ⋮\\ \frac{\partial }{\partial x_m}\end{array}\right]E\left(x\right)={\left(B^{\mathrm{T}}{\nabla }_x\right)}^{\mathrm{T}}\left(B^{\mathrm{T}}{\nabla }_x\right)E\left(x\right)={\left({\nabla }_y\right)}^{\mathrm{T}}\left({\nabla }_y\right)Y\left(y\right)$(27)

记 $C\triangleq {\left(B^{\mathrm{T}}\right)}^{-1}$，由引理3.1.1矩阵 $C$为正线上三角矩阵。

由(18)式有

$C\left[\begin{array}{c}\frac{\partial }{\partial y_1}\\ ⋮\\ \frac{\partial }{\partial y_m}\end{array}\right]=\left[\begin{array}{c}\frac{\partial }{\partial x_1}\\ ⋮\\ \frac{\partial }{\partial x_m}\end{array}\right]$(28)

由矩阵乘法

$\left[\begin{array}{c}{\displaystyle \underset{p=1}{\overset{m}{\sum }}{c}_{1p}}\frac{\partial }{\partial y_p}\\ ⋮\\ {\displaystyle \underset{p=1}{\overset{m}{\sum }}{c}_{mp}}\frac{\partial }{\partial y_p}\end{array}\right]=\left[\begin{array}{c}\frac{\partial }{\partial x_1}\\ ⋮\\ \frac{\partial }{\partial x_m}\end{array}\right]$(29)

从而

$\frac{\partial }{\partial x_k}={\displaystyle \underset{p=1}{\overset{m}{\sum }}{c}_{kp}}\frac{\partial }{\partial y_p}$(30)

${\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial E\left(x\right)}{\partial x_k}}={\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right){\displaystyle \underset{p=1}{\overset{m}{\sum }}{c}_{kp}}\frac{\partial Y\left(y\right)}{\partial y_p}}={\displaystyle \underset{p=1}{\overset{m}{\sum }}\left[{\displaystyle \underset{k=0}{\overset{m}{\sum }}{c}_{kp}\left(r-q_k\right)}\right]\frac{\partial Y\left(y\right)}{\partial y_p}}$(31)

由于 $C$为正线上三角矩阵，有

${\displaystyle \underset{k=0}{\overset{m}{\sum }}{c}_{kp}\left(r-q_k\right)}={\displaystyle \underset{k=0}{\overset{p}{\sum }}{c}_{kp}\left(r-q_k\right)}$(32)

记

${\upsilon }_p\triangleq {\displaystyle \underset{k=1}{\overset{p}{\sum }}\left(q_k-r\right)c_{kp}},p=1,\cdots ,m$(33)

即有

${\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial E\left(s\right)}{\partial x_k}}=-{\displaystyle \underset{p=1}{\overset{m}{\sum }}{\upsilon }_p}\frac{\partial Y\left(y\right)}{\partial y_p}$(34)

由(27)，(34)两式即知方程(8)与方程(24)等价。引理证毕。

引理3.1.4：特征值问题II的特征值

(35)

所对应的特征函数为

$Y\left(y\right)=Y_{\beta }\left(y\right)={\text{e}}^{{\displaystyle \underset{k=1}{\overset{m}{\sum }}{\alpha }_k{y}_k}}{\text{e}}^{i{\displaystyle \underset{k=1}{\overset{m}{\sum }}{\beta }_k{y}_k}}$(36)

其中 $\upsilon \triangleq {\left({\upsilon }_1,\cdots ,{\upsilon }_m\right)}^{\mathrm{T}}$， ${\upsilon }_p={\displaystyle \underset{k=1}{\overset{p}{\sum }}\left(q_k-r\right)c_{kp}},p=1,\cdots ,m$。

证明：由分离变量法容易求解特征值问题II：

令

$Y\left(y\right)={\displaystyle \underset{k=1}{\overset{m}{\prod }}{Y}_k\left(y_k\right)}$(37)

则

$\frac{1}{2}\frac{{\text{d}}^2{Y}_k\left(y_k\right)}{\text{d}{y}_k^2}-{\upsilon }_k\frac{\text{d}{Y}_k\left(y_k\right)}{\text{d}{y}_k}+\left({\lambda }_k-\frac{r}{m}\right)Y_k\left(y_k\right)=0$(38)

$Y_k\left(y_k\right)={\text{e}}^{{\alpha }_k{y}_k}$(39)

$\frac{1}{2}{\alpha }_k^2-{\upsilon }_k{\alpha }_k+\left({\lambda }_k-\frac{r}{m}\right)=0$(40)

${\alpha }_k={\upsilon }_k\pm \sqrt{{\upsilon }_k^2-2\left({\lambda }_k-\frac{r}{m}\right)}$(41)

${\alpha }_k={\upsilon }_k\pm i\left|{\beta }_k\right|$(42)

$Y_k={\text{e}}^{\left[{\upsilon }_k\pm \sqrt{{\upsilon }_k^2-2\left({\lambda }_k-\frac{r}{m}\right)}\right]y_k}={\text{e}}^{{\upsilon }_k{y}_k}{\text{e}}^{\pm i{y}_k\sqrt{2\left({\lambda }_k-\frac{r}{m}\right)-{\upsilon }_k^2}}={\text{e}}^{{\upsilon }_k{y}_k}{\text{e}}^{i{y}_k{\beta }_k},{\beta }_k\in R$(43)

${\beta }_k=\pm \sqrt{2\left({\lambda }_k-\frac{r}{m}\right)-{\upsilon }_k^2},{\beta }_k\in R$(44)

${\lambda }_k=\frac{{\beta }_k^2+{\upsilon }_k^2}{2}+\frac{r}{m}$(45)

由(45)式，(25)式即有(35)式成立。由(37)式，(43)式即有(36)式成立。引理证毕。

由(17)和(18)式换回原变量即得特征值问题I的特征函数。于是得到

引理3.1.5：特征值问题I的特征值

$\lambda ={\lambda }_{\beta }={\displaystyle \underset{k=1}{\overset{m}{\sum }}{\lambda }_{{\beta }_k}}={\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\beta }_k^2+{\upsilon }_k^2+\frac{2r}{m}}{2}},\beta ={\left({\beta }_1,\cdots ,{\beta }_m\right)}^{\mathrm{T}}\in R^m$(46)

所对应的特征函数为

$E_{\beta }\left(x\right)={\text{e}}^{\langle \upsilon ,B^{-1}x\rangle }{\text{e}}^{i\langle \beta ,B^{-1}x\rangle }$(47)

其中 $\upsilon \triangleq {\left({\upsilon }_1,\cdots ,{\upsilon }_m\right)}^{\mathrm{T}}$， ${\upsilon }_p={\displaystyle \underset{k=1}{\overset{p}{\sum }}\left(q_k-r\right)c_{kp}},p=1,\cdots ,m$。

证明：

$Y_{\beta }\left(y\right)={\text{e}}^{{\displaystyle \underset{k=1}{\overset{m}{\sum }}{\upsilon }_k{y}_k}}{\text{e}}^{i{\displaystyle \underset{k=1}{\overset{m}{\sum }}{\beta }_k{y}_k}},\upsilon \in R^m,\beta \in R^m$(48)

$y={\left(y_1,\cdots ,y_m\right)}^{\mathrm{T}}$

$y=B^{-1}x$

$Y_{\beta }\left(y\right)={\text{e}}^{\langle \upsilon ,y\rangle }{\text{e}}^{i\langle \beta ,y\rangle }={\text{e}}^{\langle \upsilon ,B^{-1}x\rangle }{\text{e}}^{i\langle \beta ,B^{-1}x\rangle }=E_{\beta }\left(x\right)$(49)

引理证毕。

引理3.1.6：特征值问题I的特征函数系 $E_{\beta }\left(x\right)={\text{e}}^{\langle \upsilon ,B^{-1}x\rangle }{\text{e}}^{i\langle \beta ,B^{-1}x\rangle }$是无界区域 $R^m$带权函数 $\rho \left(x\right)={\text{e}}^{-2\langle \upsilon ,B^{-1}x\rangle }$的完备正交系；正交关系即

${\displaystyle {\int }_{R^m}{E}_{\beta }\left(x\right){\bar{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)\text{d}x}={\left(2\text{π}\right)}^m\left|B\right|\delta \left({\beta }^{\prime }-\beta \right),{\beta }^{\prime }\in R^m,\beta \in R^m$(50)

证明：由于

${\displaystyle {\int }_{R^m}{E}_{\beta }\left(x\right){\bar{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)\text{d}x}={\displaystyle {\int }_{R^m}{\text{e}}^{i\langle \beta ,B^{-1}x\rangle }{\text{e}}^{-i\langle {\beta }^{\prime },B^{-1}x\rangle }\text{d}x}={\displaystyle {\int }_{R^m}{\text{e}}^{i\langle \left(\beta -{\beta }^{\prime }\right),B^{-1}x\rangle }\text{d}x}$(51)

变量替换

$B^{-1}x=y$(52)

则

$x=By=\left[\begin{array}{ccc}{b}_{11}& \cdots & b_{1m}\\ ⋮& \ddots & ⋮\\ b_{m1}& \cdots & b_{mm}\end{array}\right]\left[\begin{array}{c}{y}_1\\ ⋮\\ y_m\end{array}\right]=\left[\begin{array}{c}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{b}_{1j}}{y}_j\\ ⋮\\ {\displaystyle \underset{j=1}{\overset{m}{\sum }}{b}_{mj}}{y}_j\end{array}\right]$(53)

$\left|\frac{\partial \left(x_1,\cdots ,x_m\right)}{\partial \left(y_1,\cdots ,y_m\right)}\right|=\left|\begin{array}{ccc}\frac{\partial x_1}{\partial y_1}& \cdots & \frac{\partial x_1}{\partial y_m}\\ ⋮& \ddots & ⋮\\ \frac{\partial x_m}{\partial y_1}& \cdots & \frac{\partial x_m}{\partial y_m}\end{array}\right|=\left|\begin{array}{ccc}{b}_{11}& \cdots & b_{1m}\\ ⋮& \ddots & ⋮\\ b_{m1}& \cdots & b_{mm}\end{array}\right|=\left|B\right|$(54)

即有变量替换的雅可比行列式

$\left|\frac{\partial \left(x_1,\cdots ,x_m\right)}{\partial \left(y_1,\cdots ,y_m\right)}\right|=\left|B\right|\ne 0,x\in R^m$(55)

由多重积分变量替换公式，即有

$\begin{array}{l}{\displaystyle {\int }_{R^m}{E}_{\beta }\left(x\right){\bar{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)\text{d}x}={\displaystyle {\int }_{R^m}{\text{e}}^{i\langle \beta ,B^{-1}x\rangle }{\text{e}}^{-i\langle {\beta }^{\prime },B^{-1}x\rangle }\text{d}x}\\ ={\displaystyle {\int }_{R^m}{\text{e}}^{i\langle \left(\beta -{\beta }^{\prime }\right),B^{-1}x\rangle }\text{d}x}={\displaystyle {\int }_{R^m}{\text{e}}^{i\langle \left(\beta -{\beta }^{\prime }\right),y\rangle }}\left|\frac{\partial \left(x_1,\cdots ,x_m\right)}{\partial \left(y_1,\cdots ,y_m\right)}\right|\text{d}y\\ ={\displaystyle {\int }_{R^m}{\text{e}}^{i\langle \left(\beta -{\beta }^{\prime }\right),y\rangle }}\left|B\right|\text{d}y=\left|B\right|{\left(2\text{π}\right)}^m\delta \left({\beta }^{\prime }-\beta \right),{\beta }^{\prime }\in R^m,\beta \in R^m\end{array}$(56)

引理证毕。

由引理3.1.5与引理3.1.6的结论可以引入广义特征函数法 [11] [12] 求解定解问题I。

不妨设解 $u\in C\left(R^m\times \left[0,T\right)\right)$，将其表为特征函数的积分形式

$u\left(x,t\right)={\displaystyle {\int }_{R^m}{U}_{\beta }\left(t\right)E_{\beta }\left(x\right)\text{d}\beta }$(57)

将上式两边乘以 ${\bar{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)$再关于变量 $x$在 $R^m$积分，利用正交关系(50)则有

(58)

得到

$U_{\beta }\left(t\right)=\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_{R^m}u\left(x,t\right){\bar{E}}_{\beta }\left(x\right)\rho \left(x\right)\text{d}x},\beta \in R^m$(59)

将方程中的自由项 $f\left(x,t\right)$也表为特征函数的积分形式

$f\left(x,t\right)=\gamma \left(t\right)\delta \left(x-x\left(t\right)\right)={\displaystyle {\int }_{R^m}{f}_{\beta }}\left(t\right)E_{\beta }\left(x\right)\text{d}\beta$(60)

由(59)即有

$f_{\beta }\left(t\right)=\frac{1}{{\left(\text{2π}\right)}^m\left|B\right|}{\displaystyle {\int }_{R^m}\gamma \left(t\right)\delta \left(x-x\left(t\right)\right){\bar{E}}_{\beta }\left(x\right)\rho \left(x\right)\text{d}x}$(61)

应用d-函数的积分性质即

$\begin{array}{c}{f}_{\beta }\left(t\right)=\frac{\gamma \left(t\right)}{{\left(2\text{π}\right)}^m\left|B\right|}{\bar{E}}_{\beta }\left(x\left(t\right)\right)\rho \left(x\left(t\right)\right)=\frac{\gamma \left(t\right)}{{\left(2\text{π}\right)}^m\left|B\right|}{\text{e}}^{\langle \upsilon ,B^{-1}x\left(t\right)\rangle }{\text{e}}^{-i\langle \beta ,B^{-1}x\left(t\right)\rangle }{\text{e}}^{-2\langle \upsilon ,B^{-1}x\left(t\right)\rangle }\\ =\frac{\gamma \left(t\right)}{{\left(2\text{π}\right)}^m\left|B\right|}{\text{e}}^{-i\langle \beta ,B^{-1}x\left(t\right)\rangle }{\text{e}}^{-\langle \upsilon ,B^{-1}x\left(t\right)\rangle }\end{array}$(62)

含参变量积分与算子 $L$的运算交换次序即有

$Lu\left(x,t\right)={\displaystyle {\int }_{R^m}{U}_{\beta }\left(t\right)L{E}_{\beta }\left(x\right)\text{d}\beta }=-{\displaystyle {\int }_{R^m}{U}_{\beta }\left(t\right){\lambda }_{\beta }{E}_{\beta }\left(x\right)\text{d}\beta }$(63)

且有

$\frac{\partial u}{\partial t}\left(x,t\right)={\displaystyle {\int }_{R^m}{U^{\prime }}_{\beta }}\left(t\right)E_{\beta }\left(x\right)\text{d}\beta$(64)

由(2)式即有

$\phi \left(x\right)=u\left(x,0\right)={\displaystyle {\int }_{R^m}{U}_{\beta }}\left(0\right)E_{\beta }\left(x\right)\text{d}\beta$(65)

${\phi }_{\beta }=U_{\beta }\left(0\right)=\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(x\right){\bar{E}}_{\beta }\left(x\right)\rho \left(x\right)\text{d}x}$(66)

将(60) (63) (64)式代入方程(1)即有

${\displaystyle {\int }_{R^m}\left[{U^{\prime }}_{\beta }\left(t\right)+{\lambda }_{\beta }{U}_{\beta }\left(t\right)-f_{\beta }\left(t\right)\right]E_{\beta }\left(x\right)\text{d}\beta }=0$(67)

由特征函数系的完备正交性即有

${U^{\prime }}_{\beta }\left(t\right)+{\lambda }_{\beta }{U}_{\beta }\left(t\right)-f_{\beta }\left(t\right)=0$(68)

从而得到非齐次常微分方程的初值问题

$\{\begin{array}{l}{U^{\prime }}_{\beta }\left(t\right)+{\lambda }_{\beta }{U}_{\beta }\left(t\right)-f_{\beta }\left(t\right)=0,t>0\\ U_{\beta }\left(0\right)={\phi }_{\beta }\end{array}$(69)

用常数变易法得到非齐次常微分方程的初值问题的解为

$U_{\beta }\left(t\right)={\phi }_{\beta }{\text{e}}^{-{\lambda }_{\beta }t}+{\displaystyle {\int }_0^t{f}_{\beta }\left(\tau \right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau }$(70)

$u\left(x,t\right)={\displaystyle {\int }_{R^m}{U}_{\beta }}\left(t\right)E_{\beta }\left(x\right)\text{d}\beta ={\displaystyle {\int }_{R^m}{\phi }_{\beta }{\text{e}}^{-{\lambda }_{\beta }t}}{E}_{\beta }\left(x\right)\text{d}\beta +{\displaystyle {\int }_{R^m}\left[{\displaystyle {\int }_0^t{f}_{\beta }\left(\tau \right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau }\right]E_{\beta }\left(x\right)\text{d}\beta }$(71)

$u\left(x,t\right)=V\left(x,t\right)+w\left(x,t\right)$(72)

$V\left(x,t\right)={\displaystyle {\int }_{R^m}{\phi }_{\beta }{\text{e}}^{-{\lambda }_{\beta }t}}{E}_{\beta }\left(x\right)\text{d}\beta$(73)

$W\left(x,t\right)={\displaystyle {\int }_{R^m}\left[{\displaystyle {\int }_0^t{f}_{\beta }\left(\tau \right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau }\right]E_{\beta }\left(x\right)\text{d}\beta }$(74)

将(62)代入方程(74)即有

$\begin{array}{l}W\left(x,t\right)={\displaystyle {\int }_{R^m}\left[{\displaystyle {\int }_0^t\frac{\gamma \left(\tau \right)}{{\left(2\text{π}\right)}^m\left|B\right|}{\bar{E}}_{\beta }\left(x\left(\tau \right)\right)\rho \left(x\left(\tau \right)\right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau }\right]E_{\beta }\left(x\right)\text{d}\beta }\\ ={\displaystyle {\int }_0^t\frac{\gamma \left(\tau \right)}{{\left(2\text{π}\right)}^m\left|B\right|}{\text{e}}^{\langle \upsilon ,B^{-1}\left(x-x\left(\tau \right)\right)\rangle }{\text{e}}^{-{\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_k^2+\frac{2r}{m}}{2}}\left(t-\tau \right)}}\text{d}\tau {\displaystyle {\int }_{R^m}{\text{e}}^{-{\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\beta }_k^2}{2}}\left(t-\tau \right)}}{\text{e}}^{i\langle \beta ,B^{-1}\left(x-x\left(\tau \right)\right)\rangle }\text{d}\beta \\ =\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_0^t{\text{e}}^{\langle \upsilon ,B^{-1}\left(x-x\left(\tau \right)\right)\rangle }\gamma \left(\tau \right){\text{e}}^{-{\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_k^2+\frac{2r}{m}}{2}}\left(t-\tau \right)}}\text{d}\tau {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\displaystyle {\int }_{-\infty }^{\infty }{\text{e}}^{-\frac{{\beta }_k^2}{2}\left(t-\tau \right)}}}{\text{e}}^{i{\beta }_k{\left[B^{-1}\left(x-x\left(\tau \right)\right)\right]}_k}\text{d}{\beta }_k\end{array}$

$\begin{array}{l}=\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_0^t[\gamma \left(\tau \right){\text{e}}^{-{\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_k^2+\frac{2r}{m}}{2}}\left(t-\tau \right)}}\text{d}\tau {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\left(\frac{2\text{π}}{t-\tau }\right)}^{\frac{1}{2}}}{\text{e}}^{-\frac{{\left\{{\left[B^{-1}\left(x-x\left(\tau \right)\right)\right]}_k\right\}}^2}{2\left(t-\tau \right)}}{\text{e}}^{{\upsilon }_k{\left[B^{-1}\left(x-x\left(\tau \right)\right)\right]}_k}\\ =\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_0^t\gamma \left(\tau \right){\text{e}}^{-{\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_k^2+\frac{2r}{m}}{2}}\left(t-\tau \right)}}\text{d}\tau {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\left(\frac{2\text{π}}{t-\tau }\right)}^{\frac{1}{2}}}{\text{e}}^{-\frac{{\left\{{\left[B^{-1}\left(x-x\left(\tau \right)\right)\right]}_k\right\}}^2}{2\left(t-\tau \right)}}{\text{e}}^{{\upsilon }_k{\left[B^{-1}\left(x-x\left(\tau \right)\right)\right]}_k}\\ =\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_0^t\gamma \left(\tau \right){\text{e}}^{-{\displaystyle \underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_k^2+\frac{2r}{m}}{2}}\left(t-\tau \right)}}\text{d}\tau {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\left(\frac{2\text{π}}{t-\tau }\right)}^{\frac{1}{2}}}{\text{e}}^{-\frac{{\left\{{\left[B^{-1}\left(x-x\left(\tau \right)\right)\right]}_k-\left(t-\tau \right){\upsilon }_k\right\}}^2}{2\left(t-\tau \right)}}{\text{e}}^{\frac{\left(t-\tau \right){\upsilon }_k^2}{2}}\end{array}$

由于 ${\displaystyle \underset{k=1}{\overset{m}{\sum }}{\left[{\left(B^{-1}x\right)}_k-{\left(B^{-1}x\left(\tau \right)\right)}_k-\left(t-\tau \right){\upsilon }_k\right]}^2}={\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2$

即知

$W\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}}\text{d}\tau$(75)

将(66)式代入(73)式即有

$\begin{array}{c}V\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right){\text{e}}^{-2\langle \upsilon ,B^{-1}\xi \rangle }\text{d}\xi {\displaystyle \underset{R^m}{\int }{\text{e}}^{\langle \upsilon ,B^{-1}\xi \rangle }{\text{e}}^{-i\langle \beta ,B^{-1}\xi -x\rangle }}{\text{e}}^{-{\lambda }_{\beta }t}}{\text{e}}^{\langle \upsilon ,B^{-1}\xi \rangle }\text{d}\beta \\ =\frac{1}{{\left(2\text{π}\right)}^m\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right)\text{d}\xi {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\text{e}}^{-\left[\frac{{\upsilon }_k^2}{2}+\frac{r}{m}\right]t}}{\displaystyle {\int }_{-\infty }^{\infty }{\text{e}}^{-\frac{{\beta }_k^2}{2}t}}}{\text{e}}^{i{\beta }_k{\left[B^{-1}\left(x-\xi \right)\right]}_k}\text{d}{\beta }_k\end{array}$(76)

$\begin{array}{c}V\left(x,t\right)=\frac{1}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right)\text{d}\xi {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\text{e}}^{-\left[\frac{{\upsilon }_k^2}{2}+\frac{r}{m}\right]t}{\text{e}}^{-\frac{{\left\{{\left[B^{-1}\left(x-\xi \right)\right]}_k\right\}}^2}{2t}}}}\\ =\frac{1}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right)\text{d}\xi {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\text{e}}^{-\left[\frac{{\upsilon }_k^2}{2}+\frac{r}{m}\right]t}{\text{e}}^{-\frac{{\left\{{\left[B^{-1}\left(x-\xi \right)\right]}_k\right\}}^2-2{\left[B^{-1}\left(x-\xi \right)\right]}_{k}t{\upsilon }_k+{\left[t{\upsilon }_k\right]}^2-{\left[t{\upsilon }_k\right]}^2}{2t}}}}\\ =\frac{1}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right)\text{d}\xi {\displaystyle \underset{k=1}{\overset{m}{\prod }}{\text{e}}^{-\left[\frac{{\upsilon }_k^2}{2}+\frac{r}{m}\right]t}{\text{e}}^{-\frac{{\left\{{\left[B^{-1}\left(x-\xi \right)\right]}_k-t{\upsilon }_k\right\}}^2-{\left[t{\upsilon }_k\right]}^2}{2t}}}}\\ =\frac{e^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right){\text{e}}^{-\frac{{\displaystyle \underset{k=1}{\overset{m}{\sum }}{\left[{\left(B^{-1}x\right)}_k-{\left(B^{-1}\xi \right)}_k-t{\upsilon }_k\right]}^2}}{2t}}\text{d}\xi }\end{array}$

由于

${\displaystyle \underset{k=1}{\overset{m}{\sum }}{\left[{\left(B^{-1}x\right)}_k-{\left(B^{-1}\xi \right)}_k-t{\upsilon }_k\right]}^2}={\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2$

即有

$V\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}\text{d}\xi }$(78)

即定解问题I的解中 $$的表达式(78)式即知(5)式成立，解中 $W\left(x,t\right)$的表达式(75)式即(6)式成立。定理3.1.1证毕。

附注3.1.1：在定理3.1.1的条件4)下，易证(5)式右端积分在 $\bar{\Omega }$绝对一致收敛，则所表示的 $V$在 $\bar{\Omega }$连

续，即 $V\in C(\bar{\Omega })$；且易证 $\frac{\partial V}{\partial x}\in C(\Omega ),\frac{\partial V}{\partial t}\in C(\Omega )$。

3.2. m维无穷区域各向异性热传导方程的奇异内边界问题

数学模型I (m维无穷区域各向异性非齐次热传导方程齐次初始条件的奇异内边界问题)：

求 $\left\{w\left(x,t\right);x\left(t\right)\right\}$，使其满足

$\{\begin{array}{l}\frac{\partial w}{\partial t}=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}w}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial w}{\partial x_k}}-rw+\gamma \left(t\right)\delta \left(x-x\left(t\right)\right),x\in R^m,x\left(t\right)\in R^m,0<t<T(79)\\ w\left(x,0\right)=0,x\in R^m(80)\\ w\left(x\left(t\right),t\right)=\underset{x\in R^m}{\max }w\left(x,t\right),0<t<T(81)\\ \underset{x\to -\infty }{\lim }\left|w\right|<\infty ,\underset{x\to \infty }{\lim }\left|w\right|<\infty (\; 82\; )\end{array}$

定理3.2.1 (数学模型I的精确解)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)；

2) $x\left(t\right)={\left(x_1\left(t\right),\cdots ,x_m\left(t\right)\right)}^{\mathrm{T}}\in R^m$， $x_k\left(t\right)$为充分光滑的单调函数， $k=1,\cdots ,m$；

3) $\gamma \left(t\right)\in C\left(\left[0,T\right)\right),\gamma \left(t\right)\ge 0$；

则数学模型I的充分光滑的精确解为 $\left\{w\left(x,t\right);x\left(t\right)\right\}$：

$\{\begin{array}{l}w\left(x,t\right)=\frac{1}{{\left(\text{2π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}}\text{d}\tau (83)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 84\; )\end{array}$

证明：定解问题I的精确解 $W\left(x,t\right)$表达式(6)即有(83)式满足(79)式，(80)式，(82)式三式。再令

$\begin{array}{l}{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2\\ =\langle B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ,B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \rangle \\ \triangleq J\left(x,t;x\left(\tau \right),\tau \right)\triangleq J\end{array}$(85)

下面由引理3.2.1和引理3.2.2完成定理3.2.1的证明；也可以由引理3.2.3和引理3.2.4完成定理3.2.1的证明。记

$B^{-1}x\left(t\right)-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \equiv 0,\forall t\ge \tau \ge 0$(86)

引理3.2.1：(86)式等价于(84)式。

证明：由(86)即有恒等式

$B^{-1}x\left(t\right)-t\upsilon \equiv B^{-1}x\left(\tau \right)-\tau \upsilon ,\forall t\ge \tau \ge 0$(87)

令 $\tau =0$即得

$B^{-1}x\left(t\right)-t\upsilon \equiv B^{-1}x(\; 0\; )$

在上式两端左乘 $B$矩阵即有

$x\left(t\right)-tB\upsilon \equiv x\left(0\right)\triangleq \chi$(88)

即得(84)式。反之(84)式成立， $\forall t\ge \tau \ge 0$，有 $x\left(t\right)-tB\upsilon \equiv \chi$和 $x\left(\tau \right)-\tau B\upsilon \equiv \chi$两式，两式相减则有恒等式 $x\left(t\right)-x\left(\tau \right)-\left(t-\tau \right)B\upsilon \equiv 0,\forall t\ge \tau \ge 0$；在上式两端左乘 $B^{-1}$矩阵即有(86)式成立。

引理3.2.2：若(83)式，(84)式成立；则有(81)式成立。

证明：若(84)式成立，则有(86)式成立，即有

$B^{-1}x\left(t\right)-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \equiv 0,\forall t\ge \tau \ge 0$

${\text{e}}^{-\frac{{\Vert B^{-1}x\left(t\right)-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}\equiv 1\ge {\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}\ge 0,\forall x\in R^m,\forall t\ge \tau \ge 0$(89)

由(83)式，故有(81)式成立。证毕。

由引理3.2.1和引理3.2.2即知(83) (84)式满足(81)式，从而满足数学模型I。证毕。

引理3.2.3：当(83)和(84)式成立时，则有

(a) $\frac{\partial w}{\partial x_k}>0,\left(x,t\right)\in {\Omega }_{-},k=1,\cdots ,m$；(b) $\frac{\partial w}{\partial x_k}<0,\left(x,t\right)\in {\Omega }_{+},k=1,\cdots ,m$。

证明：由于

$J={\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2={\Vert C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2$

$\begin{array}{c}\frac{\partial J}{\partial x_l}=\frac{\partial \langle C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon ,C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon \rangle }{\partial x_l}\\ =2\langle \frac{\partial C^{\mathrm{T}}x}{\partial x_l},C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon \rangle \end{array}$(90)

$C^{\mathrm{T}}x=\left[\begin{array}{ccc}{c}_{11}& \cdots & c_{m1}\\ ⋮& \ddots & ⋮\\ c_{1m}& \cdots & c_{mm}\end{array}\right]\left[\begin{array}{c}{x}_1\\ ⋮\\ x_m\end{array}\right]=\left[\begin{array}{c}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}}{x}_j\\ ⋮\\ {\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}}{x}_j\end{array}\right]$(91)

$\frac{\partial C^{\mathrm{T}}x}{\partial x_l}=\frac{\partial }{\partial x_l}\left[\begin{array}{c}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}}{x}_j\\ ⋮\\ {\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}}{x}_j\end{array}\right]=\left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right]$(92)

$\begin{array}{c}\frac{\partial J}{\partial x_l}=2\langle \left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right],C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon \rangle \\ =2{\left[C^{\mathrm{T}}\left(x-x\left(\tau \right)-\left(t-\tau \right)B\upsilon \right)\right]}^{\mathrm{T}}\left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right]\\ =2{\left[x-x\left(\tau \right)-\left(t-\tau \right)B\upsilon \right]}^{\mathrm{T}}C\left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right]\end{array}$(93)

由(93)即知，(84)式成立的充要条件是

${\frac{\partial J}{\partial x_l}|}_{x=x\left(t\right)}\equiv 0,l=1,2,\cdots ,m$(94)

又由

$\frac{\partial w}{\partial x_l}\left(x,t\right)=\frac{-1}{2{\left(\text{2π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}-1}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}}\frac{\partial J}{\partial x_l}\text{d}\tau$(95)

$\frac{\partial J}{\partial x_l}=2\langle \left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right],C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon \rangle$(96)

$\frac{\partial J}{\partial x_l}=2{\left(C^{\mathrm{T}}x-C^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon \right)}^{\mathrm{T}}\left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right]=2{\left(x-x\left(\tau \right)-\left(t-\tau \right)B\upsilon \right)}^{\mathrm{T}}C\left[\begin{array}{c}{c}_{l1}\\ ⋮\\ c_{lm}\end{array}\right]$(97)

(a) $\left(x,t\right)\in {\Omega }_{-},-\infty <x<x\left(t\right),\forall t>0$

$x-x\left(\tau \right)-\left(t-\tau \right)B\upsilon <x\left(t\right)-x\left(\tau \right)-\left(t-\tau \right)B\upsilon \equiv 0$

即有

$x-x\left(\tau \right)-\left(t-\tau \right)B\upsilon <0$(98)

由引理3.1.1中3)'的结论即得 $\frac{\partial J}{\partial x_l}<0,l=1,2,\cdots ,m$；再由(95)式即有

$\forall \left(x,t\right)\in {\Omega }_{-},\frac{\partial w}{\partial x_l}\left(x\left(t\right),t\right)>0,l=1,2,\cdots ,m$(99)

从而有

$w\left(x\left(t\right),t\right)=\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }w\left(x,t\right),0<t<T$(100)

同理

(b) $\left(x,t\right)\in {\Omega }_{+},\frac{\partial w}{\partial x_l}\left(x\left(t\right),t\right)<0,l=1,2,\cdots ,m$(101)

从而有

$w\left(x\left(t\right),t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\max }w\left(x,t\right),0<t<T$(102)

引理3.2.3证毕。

记

$\frac{\partial w}{\partial x_l}\left(x\left(t\right),t\right)=0,l=1,2,\cdots ,m,\forall t\in \left(0,T\right)$(103)

引理3.2.4：由(83)式确定的 $w\left(x,t\right)$，使得(84)式与(103)式等价。

证明：由(95)式即知，当且仅当(94)式成立有(103)式成立。从而当且仅当(84)式成立时有(103)式成立。

引理3.2.4证毕。

当(83)和(84)式成立时，有引理3.2.3，引理3.2.4成立，从而有(103)式，(100)式，(102)式成立，推出(81)式成立。由此也得到定理3.2.1的证明。证毕。

附注3.2.1：定理3.2.1中解 $w\left(x,t\right)$由(83)式给出，它依赖于奇异源强度函数 $\gamma \left(t\right)>0$的确定。由(83)式，(86)式即知

$w\left(x\left(t\right),t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}}\text{d}\tau \triangleq \mu \left(t\right)$(104)

若边值函数 $\mu \left(t\right)$为已知函数，则

${\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}}\text{d}\tau ={\left(\text{2π}\right)}^{\frac{m}{2}}\left|B\right|\mu \left(t\right)$(105)

是关于奇异源强度函数 $\gamma \left(t\right)$作为未知函数的第一类Volterra积分方程。

附注3.2.2：奇异源强度函数 $\gamma \left(t\right)$也可以是广义函数，此时积分方程为

${\displaystyle {\int }_{0^{-}}^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}}\text{d}\tau ={\left(\text{2π}\right)}^{\frac{m}{2}}\left|B\right|\mu \left(t\right)$(105)'

例如，若边值函数为 $\mu \left(t\right)={\left(\text{2π}\right)}^{-\frac{m}{2}}{\left|B\right|}^{-1}{t}^{-\frac{m}{2}}{\text{e}}^{-rt}$，容易验证 $\gamma \left(t\right)=\delta \left(t\right),t\ge 0$是满足积分方程(105)'的解。

定解问题II (m维无穷区域各向异性齐次热传导方程非齐次初始条件的初值问题)：

$\{\begin{array}{l}\frac{\partial V}{\partial t}=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}V}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial V}{\partial x_k}}-rV,x\in R^m,0<t<T(106)\\ V\left(x,0\right)=\phi \left(x\right),x\in R^m(107)\\ \underset{x\to -\infty }{\lim }\left|V\right|<\infty ,\underset{x\to \infty }{\lim }\left|V\right|<\infty (\; 108\; )\end{array}$

由定理3.1.1即知定解问题II的解由(5)式给出。

定理3.2.2 (定解问题II的解的性质定理)：若 $x\left(t\right)$由(84)式表出；则

2) 当 $\phi \left(x\right)=\delta \left(\chi -x\right),\chi \in R^m$；定解问题II的充分光滑的精确解可表为

$V\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{1}{2t}{\Vert B^{-1}x-B^{-1}\chi -t\upsilon \Vert }^2}$(109)

且 $\frac{\partial V}{\partial x_k}>0,\left(x,t\right)\in {\Omega }_{-}$； $\frac{\partial V}{\partial x_k}<0,\left(x,t\right)\in {\Omega }_{+},k=1,\cdots ,m$，满足 $\underset{x\in R^m}{\max }V\left(x,t\right)=V\left(x\left(t\right),t\right),0<t<T$。

2) 当 $\phi \in {\Lambda }_{\left(\chi ,\infty \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；定解问题II的充分光滑的精确解可表为

$V_a\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(\chi ,\infty \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}}\text{d}\xi$(110)

且 $\frac{\partial V_a}{\partial x_k}\left(x,t\right)>0,\left(x,t\right)\in {\Omega }_{-},k\in \left\{1,2,\cdots ,m\right\}$，满足 $\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }{V}_a\left(x,t\right)=V_a\left(x\left(t\right),t\right),0<t<T$；

3) 当 $\phi \in {\Lambda }_{\left(-\infty ,\chi \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；定解问题II的充分光滑的精确解可表为

$V_b\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(-\infty ,\chi \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}}\text{d}\xi$(111)

且 $\frac{\partial V_b}{\partial x_k}\left(x,t\right)<0,\left(x,t\right)\in {\Omega }_{+},k\in \left\{1,2,\cdots ,m\right\}$，满足 $\underset{x\in \left[x\left(t\right),\infty \right)}{\max }{V}_b\left(x,t\right)=V_b\left(x\left(t\right),t\right),0<t<T$。

证明：1) 当 $\phi \left(x\right)=\delta \left(\chi -x\right),\chi \in R^m$；且 $x\left(t\right)=tB\upsilon +\chi ,t>0$则由(5)式，定解问题II的解可表为

$\begin{array}{c}V\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{R^m}\delta \left(\chi -\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}\text{d}\xi }\\ =\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{1}{2t}{\Vert B^{-1}x-B^{-1}\chi -t\upsilon \Vert }^2}\end{array}$(112)

$0\equiv {\Vert B^{-1}x\left(t\right)-B^{-1}\chi -t\upsilon \Vert }^2\le {\Vert B^{-1}x-B^{-1}\chi -t\upsilon \Vert }^2$(113)

${\text{e}}^{-\frac{1}{2t}{\Vert B^{-1}x-B^{-1}\chi -t\upsilon \Vert }^2}\le {\text{e}}^{-\frac{1}{2t}{\Vert B^{-1}x\left(t\right)-B^{-1}\chi -t\upsilon \Vert }^2}$(114)

$V\left(x,t\right)\le V\left(x\left(t\right),t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}$(115)

满足 $\underset{x\in R^m}{\max }V\left(x,t\right)=V\left(x\left(t\right),t\right),0<t<T$。

与下面2)中的证明方法类似可证： $\frac{\partial V}{\partial x_k}>0,\left(x,t\right)\in {\Omega }_{-}$； $\frac{\partial V}{\partial x_k}<0,\left(x,t\right)\in {\Omega }_{+},k=1,\cdots ,m$。

2) 当 $\phi \in {\Lambda }_{\left(\chi ,\infty \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；定解问题II的充分光滑的精确解 $V_a\left(x,t\right)$可由(110)表出，记

$I\triangleq {\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2={\Vert C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \Vert }^2=\langle C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon ,C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \rangle$

由(110)有

$\frac{\partial V_a}{\partial x_k}=-\frac{1}{2t}\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(\chi ,\infty \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{1}{2t}{\Vert C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \Vert }^2}\frac{\partial I}{\partial x_k}\text{d}\xi }$(116)

$\frac{\partial I}{\partial x_k}=\frac{\partial \langle C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon ,C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \rangle }{\partial x_k}=2\langle \frac{\partial C^{\mathrm{T}}x}{\partial x_k},C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \rangle$(117)

$C^{\mathrm{T}}x=\left[\begin{array}{ccc}{c}_{11}& \cdots & c_{m1}\\ ⋮& \ddots & ⋮\\ c_{1m}& \cdots & c_{mm}\end{array}\right]\left[\begin{array}{c}{x}_1\\ ⋮\\ x_m\end{array}\right]=\left[\begin{array}{c}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}}{x}_j\\ ⋮\\ {\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}}{x}_j\end{array}\right]$(118)

$\frac{\partial C^{\mathrm{T}}x}{\partial x_k}=\frac{\partial }{\partial x_k}\left[\begin{array}{c}{\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{j1}}{x}_j\\ ⋮\\ {\displaystyle \underset{j=1}{\overset{m}{\sum }}{c}_{jm}}{x}_j\end{array}\right]=\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]$(119)

$\begin{array}{c}\frac{\partial I}{\partial x_k}=2\langle \frac{\partial C^{\mathrm{T}}x}{\partial x_k},C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \rangle =2\langle \left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right],C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \rangle \\ =2{\left(C^{\mathrm{T}}x-C^{\mathrm{T}}\xi -t\upsilon \right)}^{\mathrm{T}}\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]\end{array}$(120)

由 $\left(x,t\right)\in {\Omega }_{-},-\infty <x<x\left(t\right);\xi \in \left(\chi ,\infty \right),\chi <\xi <\infty$

有 $x-\xi -tB\upsilon <x\left(t\right)-\chi -tB\upsilon \equiv 0$

即

$x-\xi -B\upsilon t<0$(121)

再由引理3.1.1中3)'的结论即得

$\frac{\partial I}{\partial x_k}=2{\left(x-\xi -tB\upsilon \right)}^{\mathrm{T}}C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]<0,k\in \left\{1,2,\cdots ,m\right\}$(122)

从而

$\frac{\partial V_a}{\partial x_k}=-\frac{1}{2t}\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(\chi ,\infty \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{1}{2t}{\Vert B^{-1}x-B^{-1}\xi -\upsilon t\Vert }^2}\frac{\partial I}{\partial x_k}\text{d}\xi }>0,k\in \left\{1,2,\cdots ,m\right\}$(123)

即有

$\frac{\partial V_a}{\partial x_k}\left(x,t\right)>0,\left(x,t\right)\in {\Omega }_{-},k\in \left\{1,2,\cdots ,m\right\}$(124)

故满足 $\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }{V}_a\left(x,t\right)=V_a\left(x\left(t\right),t\right),0<t<T$；

3)当 $\phi \in {\Lambda }_{\left(-\infty ,\chi \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；定解问题II的充分光滑的精确解 $V_b\left(x,t\right)$可由(111)式表出。当

$\left(x,t\right)\in {\Omega }_{+},x\left(t\right)<x<\infty ;\xi \in \left(-\infty ,\chi \right),-\infty <\xi <\chi$;

$x-\xi -tB\upsilon >x\left(t\right)-\chi -tB\upsilon \equiv 0$(125)

与2)中证明类似，由引理3.1.1中3)'的结论即得

$\frac{\partial I}{\partial x_k}=2{\left(x-\xi -tB\upsilon \right)}^{\mathrm{T}}C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ c_{km}\end{array}\right]>0,k\in \left\{1,2,\cdots ,m\right\}$(126)

从而

$\frac{\partial V_b}{\partial x_k}\left(x,t\right)<0,\left(x,t\right)\in {\Omega }_{+},k\in \left\{1,2,\cdots ,m\right\}$(127)

故满足 $\underset{x\in \left[x\left(t\right),\infty \right)}{\max }{V}_b\left(x,t\right)=V_b\left(x\left(t\right),t\right),0<t<T$。证毕。

数学模型II (m维无穷区域各向异性非齐次热传导方程非齐次初始条件的奇异内边界问题)：

求 $\left\{u\left(x,t\right);x\left(t\right)\right\}$使其满足

定理3.2.3 (数学模型II奇异内边界问题解的存在定理)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)，

2) $\gamma \left(t\right)\in C\left(\left[0,\infty \right)\right),\gamma \left(t\right)\ge 0$，

3) $\phi \left(x\right)=\delta \left(x-\chi \right)$；

则数学模型II存在充分光滑的精确解 $\left\{u\left(x,t\right),x\left(t\right)\right\}$：

$\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V\left(x,t\right)(132)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 133\; )\end{array}$

其中

$V\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{1}{2t}{\Vert B^{-1}x-B^{-1}\chi -t\upsilon \Vert }^2}$(134)

$w\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}\text{d}\tau }$(135)

证明：由定理3.2.1，定理3.2.2即得。证毕。

推论3.2.1 (数学模型II奇异内边界问题解的存在定理)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)；

2) $\gamma \left(t\right)=\delta \left(t\right),t\ge 0$；

3) $\phi \left(x\right)=\delta \left(x-\chi \right)$；

则数学模型II存在充分光滑的精确解 $\left\{u\left(x,t\right),x\left(t\right)\right\}$：

$\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V\left(x,t\right)=\frac{2t^{-\frac{m}{2}}{\text{e}}^{-rt}}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(t\right)\Vert }^2}{2t}}(136)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 137\; )\end{array}$

其中

$V\left(x,t\right)=\frac{t^{-\frac{m}{2}}{\text{e}}^{-rt}}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(t\right)\Vert }^2}{2t}}$(138)

$w\left(x,t\right)=\frac{t^{-\frac{m}{2}}{\text{e}}^{-rt}}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(t\right)\Vert }^2}{2t}}$(139)

3.3. m维各向异性齐次热传导方程非齐次初始条件的自由边界问题II_a和II_b

问题II_a(m维各向异性齐次热传导方程非齐次初始条件在区域 ${\Omega }_{-}$上的自由边界问题)：

求 $\left\{u\left(x,t\right),x\left(t\right)\right\}$，使其满足

$\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}u}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial u}{\partial x_k}}-ru,x\in \left(-\infty ,x\left(t\right)\right),0<t<T(140)\\ u\left(x,0\right)=\phi \left(x\right),x\in R^m(141)\\ u\left(x\left(t\right),t\right)=\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }u\left(x,t\right),0<t<T(142)\\ \underset{x\to -\infty }{\lim }\left|u\right|<\infty (\; 143\; )\end{array}$

定理3.3.1 (自由边界问题II_a解的存在定理)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)，

2) $\gamma \left(t\right)\in C\left(\left[0,\infty \right)\right),\gamma \left(t\right)\ge 0$，

3) $\phi \in {\Lambda }_{\left(\chi ,\infty \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；

则自由边界问题II_a存在充分光滑的精确解 $\left\{u\left(x,t\right),x\left(t\right)\right\}$：

$\{\begin{array}{l}u\left(x,t\right)=w(x,t)+V_a\left(x,t\right)(144)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 145\; )\end{array}$

其中

$V_a\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(\chi ,\infty \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}}\text{d}\xi$(146)

$w\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}}\text{d}\tau$(147)

证明：由于当 $x\ne x\left(t\right)$时， $\delta \left(x-x\left(t\right)\right)\equiv 0$，故当 $x\in \left(-\infty ,x\left(t\right)\right)$时， $w\left(x,t\right)$满足齐次方程(140)，从而(144)式满足(140) (141) (143)三式；由引理3.2.3证明中(100)式有 $w\left(x\left(t\right),t\right)=\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }w\left(x,t\right),0<t<T$，又定理3.2.2证明了 $V_a\left(x,t\right)$满足 $\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }{V}_a\left(x,t\right)=V_a\left(x\left(t\right),t\right),0<t<T$；即有

$\begin{array}{c}\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }u\left(x,t\right)=\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }w\left(x,t\right)+\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\max }{V}_a\left(x,t\right)\\ =w\left(x\left(t\right),t\right)+V_a\left(x\left(t\right),t\right)=u\left(x\left(t\right),t\right),0<t<T\end{array}$

即得(142)式成立。证毕。

推论3.3.1 (自由边界问题II_a解的存在定理)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)，

2) $\gamma \left(t\right)=\delta \left(t\right),t\ge 0$

3) $\phi \in {\Lambda }_{\left(\chi ,\infty \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；

则自由边界问题II_a存在充分光滑的精确解：

$\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V_a\left(x,t\right)(148)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 149\; )\end{array}$

其中

$V_a\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(\chi ,\infty \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}}\text{d}\xi$(150)

$w\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(t\right)\Vert }^2}{2\left(t-\vartheta \right)}}$(151)

问题II_b(m维各向异性齐次热传导方程非齐次初始条件在区域 ${\Omega }_{+}$上的自由边界问题)：

求 $\left\{u\left(x,t\right),x\left(t\right)\right\}$，使其满足

$\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}{\displaystyle \underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}}\frac{{\partial }^{2}u}{\partial x_k\partial x_j}+{\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(r-q_k\right)\frac{\partial u}{\partial x_k}}-ru,x\in \left(x\left(t\right),\infty \right),0<t<T(152)\\ u\left(x,0\right)=\phi \left(x\right),x\in R^m(153)\\ u\left(x\left(t\right),t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\max }u\left(x,t\right),0<t<T(154)\\ \underset{x\to \infty }{\lim }\left|u\right|<\infty (\; 155\; )\end{array}$

定理3.3.2 (自由边界问题II_b解的存在定理)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)；

2) $\gamma \left(t\right)\in C\left(\left[0,\infty \right)\right),\gamma \left(t\right)\ge 0$，

3) $\phi \in {\Lambda }_{\left(-\infty ,\chi \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；

则自由边界问题II_b存在充分光滑的精确解 $\left\{u\left(x,t\right),x\left(t\right)\right\}$：

$\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V_b\left(x,t\right)(156)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 157\; )\end{array}$

其中

$V_b\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(-\infty ,\chi \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}}\text{d}\xi$(158)

$w\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \Vert }^2}{2\left(t-\tau \right)}}}\text{d}\tau$(159)

证明：由于当时， $\delta \left(x-x\left(t\right)\right)\equiv 0$，故当 $x\in \left(x\left(t\right),\infty \right)$时， $w\left(x,t\right)$满足齐次方程(152)，从而(156)式

满足(152) (153) (155)三式；由引理3.2.3证明中(102)式

$w\left(x\left(t\right),t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\max }w\left(x,t\right),0<t<T$，

定理3.2.2证明了 $V_b\left(x,t\right)$满足

$\underset{x\in \left[x\left(t\right),\infty \right)}{\max }{V}_b\left(x,t\right)=V_b\left(x\left(t\right),t\right),0<t<T$即得

$\begin{array}{c}\underset{x\in \left[x\left(t\right),\infty \right)}{\max }u\left(x,t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\max }w\left(x,t\right)+\underset{x\in \left[x\left(t\right),\infty \right)}{\max }{V}_b\left(x,t\right)\\ =w\left(x\left(t\right),t\right)+V_b\left(x\left(t\right),t\right)=u\left(x\left(t\right),t\right),0<t<T\end{array}$

即得(154)式成立。证毕。

推论3.3.2：(自由边界问题II_b解的存在定理)：若

1) $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)；

2) $\gamma \left(t\right)=\delta \left(t\right),t\ge 0$；

3) $\phi \in {\Lambda }_{\left(-\infty ,\chi \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {\Vert B^{-1}x\Vert }^2}{2T}},0\le \theta <1$；

则自由边界问题II_a存在充分光滑的精确解 $\left\{u\left(x,t\right),x\left(t\right)\right\}$：

$\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V_b\left(x,t\right)(160)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0(\; 161\; )\end{array}$

其中

$V_b\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\displaystyle {\int }_{{}_{\left(-\infty ,\chi \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}\xi -t\upsilon \Vert }^2}{2t}}}\text{d}\xi$(162)

$w\left(x,t\right)=\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}\left|B\right|}{\text{e}}^{-\frac{{\Vert B^{-1}x-B^{-1}x\left(t\right)\Vert }^2}{2t}}$(163)

4. 结论

I) 对于在m维(m为某正整数)无穷区域 $R^m$的抛物型方程的不定常自由边界问题，可以转化为相应的奇异内边界问题来研究。假设在区域 $\Omega \triangleq \left\{\left(x,t\right)|x\in R^m,t\in \left(0,T\right)\right\}$存在一条无重点的奇异内边界

$\Gamma \triangleq \left\{\left(x,t\right)|x=x\left(t\right),t\in \left(0,T\right),\forall t_1\ne t_2,x\left(t_1\right)\ne x\left(t_2\right)\right\}$

建立奇异内边界问题数学模型I。所求得的向量函数 $x\left(t\right)=tB\upsilon +\chi ,T>t>0$就是问题II_a和问题II_b的公共自由边界。

II) 在条件 $A={\left(a_{kj}\right)}_{m\times m}$为m阶实对称正定矩阵， $A=B{B}^{\mathrm{T}}$( $B$为正线下三角矩阵)下；获得了无穷区域 $R^m$的m维各向异性热传导方程的自由边界问题II_a和问题II_b存在充分光滑的精确解。问题II_a和问题II_b具有公共的自由边界： $x\left(t\right)=tB\upsilon +\chi ,T>t>0$，其中 $\upsilon \triangleq {\left({\upsilon }_1,\cdots ,{\upsilon }_m\right)}^{\mathrm{T}}$， ${\upsilon }_p={\displaystyle \underset{k=1}{\overset{p}{\sum }}\left(q_k-r\right)c_{kp}},p=1,\cdots ,m$。公式表明m维向量 $B\upsilon$由m维各向异性的热传导方程中出现的所有参数 $a_{kj},q_j,r$唯一确定。