﻿ m维各向异性热传导方程的奇异内边界问题研究 Study of Singular Internal Boundary Problems for m Dimensional Anisotropic Heat Conduction Equations

Pure Mathematics
Vol.08 No.01(2018), Article ID:23645,24 pages
10.12677/PM.2018.81011

Study of Singular Internal Boundary Problems for m Dimensional Anisotropic Heat Conduction Equations

Xiaoqing Wu

College of Science, Southwest Petroleum University, Chengdu Sichuan

Received: Jan. 3rd, 2018; accepted: Jan. 20th, 2018; published: Jan. 31st, 2018

ABSTRACT

In this paper, first of all, the mathematical model I is established on anisotropic heat conduction equation in m dimension infinite domain. Mathematical model I: seek $\left\{w\left(x,t\right);x\left(t\right)\right\}$ , make it satisfy to

$\left\{\begin{array}{l}\frac{\partial w}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}w}{\partial {x}_{k}\partial {x}_{j}}+\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 (I)

The free term of the equation is $\gamma \left(t\right)\delta \left(x-x\left(t\right)\right)$ , among them, $\delta \left(x-x\left(t\right)\right)$ is m dimensional Diracfunction, $\gamma \left(t\right)$ is strength function of singular source. The exact solution $\left\{w\left(x,t\right);x\left(t\right)\right\}$ of the mathematical model I is obtained by the matrix theory and the generalized eigenfunction method, under the condition that the matrix $A={\left({a}_{kj}\right)}_{m×m}$ is real symmetric positive definite matrix and $A=B{B}^{T}$ . The matrix B is a lower triangular matrix, whose main diagonal element is positive. The singular internal boundary is demonstrated as $x\left(t\right)=tB\upsilon +\chi ,T>t>0$ . We establish the free boundary problem IIa and problem IIb on homogeneous heat conduction equation. The problem IIa is free boundary problem in region ${Ε}_{-}\left(t\right)$ . The problem IIb is free boundary problem in region ${Ε}_{+}\left(t\right)$ . It is also solves these two questions of problem IIa and IIb. These two free boundaries are the same $x\left(t\right)=tB\upsilon +\chi ,T>t>0$ .

Keywords:Heat Conduction Equation, Multidimensional (m Dimensional), Anisotropy, Free Boundary Problem, Real Symmetric Positive Definite Matrix

m维各向异性热传导方程的奇异内边界问题研究

$\left\{\begin{array}{l}\frac{\partial w}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}w}{\partial {x}_{k}\partial {x}_{j}}+\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 (I)

1. 引言

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

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

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

$\left\{\begin{array}{l}\frac{\partial w}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}w}{\partial {x}_{k}\partial {x}_{j}}+\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),\text{\hspace{0.17em}}x\in {R}^{m},\text{\hspace{0.17em}}x\left(t\right)\in {R}^{m},\text{\hspace{0.17em}}0 (I)

2. 符号说明

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

m维列向量的集合 ${R}^{m}\triangleq \left\{x|x={\left({x}_{1},\cdots ,{x}_{m}\right)}^{\mathrm{T}},{x}_{j}\in R\right\}$

$\forall x={\left({x}_{1},\cdots ,{x}_{m}\right)}^{\mathrm{T}}\in {R}^{m},\text{\hspace{0.17em}}y={\left({y}_{1},\cdots ,{y}_{m}\right)}^{\mathrm{T}}\in {R}^{m},\text{\hspace{0.17em}}〈x,y〉\triangleq \underset{k=1}{\overset{m}{\sum }}{x}_{k}{y}_{k},\text{\hspace{0.17em}}{‖x‖}^{2}\triangleq \underset{k=1}{\overset{m}{\sum }}{x}_{k}^{2};$

$x\left(t\right)\triangleq {\left({x}_{1}\left(t\right),\cdots ,{x}_{m}\left(t\right)\right)}^{\mathrm{T}}$

$x={\left({x}_{1},\cdots ,{x}_{m}\right)}^{\mathrm{T}}>0⇔{x}_{l}>0,l\in \left\{1,\cdots ,m\right\}$

$x={\left({x}_{1},\cdots ,{x}_{m}\right)}^{\mathrm{T}}\ge 0⇔{x}_{l}\ge 0,l\in \left\{1,\cdots ,m\right\}$

$-\infty

$x\left(t\right)

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

${\stackrel{¯}{E}}_{-}\left(t\right)\triangleq \left\{x|-\infty

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

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

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

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：

$\left[x,y\right]\triangleq \left\{s|s={\left({s}_{1},\cdots ,{s}_{m}\right)}^{\mathrm{T}},{x}_{j}\le {s}_{j}\le {y}_{j},j=1,\cdots ,m\right\}$

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

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

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}u}{\partial {x}_{k}\partial {x}_{j}}+\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

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \left(x\right)\in C\left({R}^{m}\right),\text{\hspace{0.17em}}0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {‖{B}^{-1}x‖}^{2}}{2T}},\text{\hspace{0.17em}}0\le \theta <1$

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

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

$W\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\text{d}\tau$ (6)

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

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

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

2) 记 ${\left({B}^{\mathrm{T}}\right)}^{-1}\triangleq C={\left({c}_{kj}\right)}_{m×m}$ ，则 $C$ 为正线上三角矩阵， $|C|=\underset{j=1}{\overset{m}{\prod }}{c}_{jj}=\underset{j=1}{\overset{m}{\prod }}{b}_{jj}^{-1}>0,\text{\hspace{0.17em}}{c}_{kj}=0,\text{\hspace{0.17em}}k>j,\text{\hspace{0.17em}}j=1,\cdots ,m-1$

3) 记 $I\left(k;x\right)\triangleq \underset{n=k}{\overset{m}{\sum }}{c}_{kn}\underset{j=1}{\overset{n}{\sum }}{c}_{jn}{x}_{j},\text{\hspace{0.17em}}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 $I\left(k;x\right)=\left[{x}_{1},\cdots ,{x}_{m}\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ {c}_{km}\end{array}\right]<0,\text{\hspace{0.17em}}k\in \left\{1,\cdots ,m\right\}$ ；当 $x={\left[{x}_{1},\cdots ,{x}_{m}\right]}^{\mathrm{T}}>0$ ，有 $I\left(k;x\right)=\left[{x}_{1},\cdots ,{x}_{m}\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ {c}_{km}\end{array}\right]>0,\text{\hspace{0.17em}}k\in \left\{1,\cdots ,m\right\}$

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

$x={\left({x}_{1},\cdots ,{x}_{m}\right)}^{\mathrm{T}}$ 为列向量，应用分块矩阵的乘法运算即有

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

$I\left(k;x\right)=\left[{c}_{k1},{c}_{k2}\cdots ,{c}_{km}\right]{C}^{\mathrm{T}}x,\text{\hspace{0.17em}}k\in \left\{1,\cdots ,m\right\}$ (10)

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

$A={\left({a}_{kj}\right)}_{m×m}$ 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

${x}^{\mathrm{T}}\left[\begin{array}{c}I\left(1;x\right)\\ I\left(2;x\right)\\ ⋮\\ ⋮\\ I\left(m;x\right)\end{array}\right]={x}^{\mathrm{T}}{A}^{-1}x>0,\text{\hspace{0.17em}}x\ne 0$ (12)

$\underset{j=1}{\overset{m}{\sum }}{x}_{j}I\left(j;x\right)>0,\text{\hspace{0.17em}}x\ne 0$ (13)

$x$ 的任意性，分别令 $x={y}_{p}{\epsilon }_{p},{y}_{p}\ne 0,p=1,\cdots ,m$

${y}_{p}\underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{y}_{p}>0,\text{\hspace{0.17em}}p=1,\cdots ,m$

$\underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{y}_{p}^{2}>0,\text{\hspace{0.17em}}{y}_{p}\ne 0,\text{\hspace{0.17em}}p=1,\cdots ,m$ (14)

$\underset{p=1}{\overset{m}{\sum }}\underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{y}_{p}^{2}=\underset{p=1}{\overset{m}{\sum }}\underset{n=1}{\overset{m}{\sum }}{c}_{kn}{c}_{pn}{x}_{p}=I\left(k;x\right),\text{\hspace{0.17em}}k=1,\cdots ,m$ (15)

$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：

$\begin{array}{l}I\left(k;x\right)=\left[{c}_{k1},\cdots ,{c}_{km}\right]{C}^{\mathrm{T}}x\in {R}^{1×1},\text{\hspace{0.17em}}k=1,\cdots ,m;\\ I\left(k;x\right)={\left[I\left(k;x\right)\right]}^{\mathrm{T}}={x}^{\mathrm{T}}C{\left[{c}_{k1},\cdots ,{c}_{km}\right]}^{\mathrm{T}}=\left[{x}_{1},\cdots ,{x}_{m}\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ {c}_{km}\end{array}\right]\end{array}$ (16)

$x={\left[{x}_{1},\cdots ,{x}_{m}\right]}^{\mathrm{T}}>0$ ，有 $I\left(k;x\right)=\left[{x}_{1},\cdots ,{x}_{m}\right]C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ {c}_{km}\end{array}\right]>0,\text{\hspace{0.17em}}k\in \left\{1,\cdots ,m\right\}$ 。引理证毕。

$x\in {R}^{m},y\in {R}^{m}$ ，作 ${R}^{m}$${R}^{m}$ 的线性变换

$x=By$ (17)

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

${\nabla }_{y}={B}^{\mathrm{T}}{\nabla }_{x}$ (19)

${\nabla }_{y}\triangleq \left[\begin{array}{c}\frac{\partial }{\partial {y}_{1}}\\ ⋮\\ \frac{\partial }{\partial {y}_{m}}\end{array}\right]$ (20)

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

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

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

m维二阶线性偏微方程在无界区域 ${R}^{m}$ 的特征值问题II

$\underset{k=1}{\overset{m}{\sum }}\left[\frac{1}{2}\frac{{\partial }^{2}Y\left(y\right)}{\partial {y}_{k}{}^{2}}-{\upsilon }_{k}\frac{\partial Y\left(y\right)}{\partial {y}_{k}}+\left({\lambda }_{k}-\frac{r}{m}\right)Y\left(y\right)\right]=0,\text{\hspace{0.17em}}y={\left({y}_{1},\cdots ,{y}_{m}\right)}^{\mathrm{T}}\in {R}^{m}$ (24)

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

${\upsilon }_{k}\triangleq \underset{j=1}{\overset{k}{\sum }}\left({q}_{j}-r\right){c}_{jk},\text{\hspace{0.17em}}k=1,\cdots ,m$ (26)

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

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

$C\left[\begin{array}{c}\frac{\partial }{\partial {y}_{1}}\\ ⋮\\ \frac{\partial }{\partial {y}_{m}}\end{array}\right]=\left[\begin{array}{c}\frac{\partial }{\partial {x}_{1}}\\ ⋮\\ \frac{\partial }{\partial {x}_{m}}\end{array}\right]$ (28)

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

$\frac{\partial }{\partial {x}_{k}}=\underset{p=1}{\overset{m}{\sum }}{c}_{kp}\frac{\partial }{\partial {y}_{p}}$ (30)

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

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

${\upsilon }_{p}\triangleq \underset{k=1}{\overset{p}{\sum }}\left({q}_{k}-r\right){c}_{kp},\text{\hspace{0.17em}}p=1,\cdots ,m$ (33)

$\underset{k=1}{\overset{m}{\sum }}\left(r-{q}_{k}\right)\frac{\partial E\left(s\right)}{\partial {x}_{k}}=-\underset{p=1}{\overset{m}{\sum }}{\upsilon }_{p}\frac{\partial Y\left(y\right)}{\partial {y}_{p}}$ (34)

(35)

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

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

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

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

${\alpha }_{k}={\upsilon }_{k}±i|{\beta }_{k}|$ (42)

${Y}_{k}={\text{e}}^{\left[{\upsilon }_{k}±\sqrt{{\upsilon }_{k}^{2}-2\left({\lambda }_{k}-\frac{r}{m}\right)}\right]{y}_{k}}={\text{e}}^{{\upsilon }_{k}{y}_{k}}{\text{e}}^{±i{y}_{k}\sqrt{2\left({\lambda }_{k}-\frac{r}{m}\right)-{\upsilon }_{k}^{2}}}={\text{e}}^{{\upsilon }_{k}{y}_{k}}{\text{e}}^{i{y}_{k}{\beta }_{k}},\text{\hspace{0.17em}}{\beta }_{k}\in R$ (43)

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

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

$\lambda ={\lambda }_{\beta }=\underset{k=1}{\overset{m}{\sum }}{\lambda }_{{\beta }_{k}}=\underset{k=1}{\overset{m}{\sum }}\frac{{\beta }_{k}^{2}+{\upsilon }_{k}^{2}+\frac{2r}{m}}{2},\text{\hspace{0.17em}}\beta ={\left({\beta }_{1},\cdots ,{\beta }_{m}\right)}^{\mathrm{T}}\in {R}^{m}$ (46)

${E}_{\beta }\left(x\right)={\text{e}}^{〈\upsilon ,{B}^{-1}x〉}{\text{e}}^{i〈\beta ,{B}^{-1}x〉}$ (47)

${Y}_{\beta }\left(y\right)={\text{e}}^{\underset{k=1}{\overset{m}{\sum }}{\upsilon }_{k}{y}_{k}}{\text{e}}^{i\underset{k=1}{\overset{m}{\sum }}{\beta }_{k}{y}_{k}},\text{\hspace{0.17em}}\upsilon \in {R}^{m},\beta \in {R}^{m}$ (48)

$y={\left({y}_{1},\cdots ,{y}_{m}\right)}^{\mathrm{T}}$

$y={B}^{-1}x$

${Y}_{\beta }\left(y\right)={\text{e}}^{〈\upsilon ,y〉}{\text{e}}^{i〈\beta ,y〉}={\text{e}}^{〈\upsilon ,{B}^{-1}x〉}{\text{e}}^{i〈\beta ,{B}^{-1}x〉}={E}_{\beta }\left(x\right)$ (49)

${\int }_{{R}^{m}}{E}_{\beta }\left(x\right){\stackrel{¯}{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)\text{d}x={\left(2\text{π}\right)}^{m}|B|\delta \left({\beta }^{\prime }-\beta \right),\text{\hspace{0.17em}}{\beta }^{\prime }\in {R}^{m},\text{\hspace{0.17em}}\beta \in {R}^{m}$ (50)

${\int }_{{R}^{m}}{E}_{\beta }\left(x\right){\stackrel{¯}{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)\text{d}x={\int }_{{R}^{m}}{\text{e}}^{i〈\beta ,{B}^{-1}x〉}{\text{e}}^{-i〈{\beta }^{\prime },{B}^{-1}x〉}\text{d}x={\int }_{{R}^{m}}{\text{e}}^{i〈\left(\beta -{\beta }^{\prime }\right),{B}^{-1}x〉}\text{d}x$ (51)

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

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

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

$|\frac{\partial \left({x}_{1},\cdots ,{x}_{m}\right)}{\partial \left({y}_{1},\cdots ,{y}_{m}\right)}|=|B|\ne 0,\text{\hspace{0.17em}}x\in {R}^{m}$ (55)

$\begin{array}{l}{\int }_{{R}^{m}}{E}_{\beta }\left(x\right){\stackrel{¯}{E}}_{{\beta }^{\prime }}\left(x\right)\rho \left(x\right)\text{d}x={\int }_{{R}^{m}}{\text{e}}^{i〈\beta ,{B}^{-1}x〉}{\text{e}}^{-i〈{\beta }^{\prime },{B}^{-1}x〉}\text{d}x\\ ={\int }_{{R}^{m}}{\text{e}}^{i〈\left(\beta -{\beta }^{\prime }\right),{B}^{-1}x〉}\text{d}x={\int }_{{R}^{m}}{\text{e}}^{i〈\left(\beta -{\beta }^{\prime }\right),y〉}|\frac{\partial \left({x}_{1},\cdots ,{x}_{m}\right)}{\partial \left({y}_{1},\cdots ,{y}_{m}\right)}|\text{d}y\\ ={\int }_{{R}^{m}}{\text{e}}^{i〈\left(\beta -{\beta }^{\prime }\right),y〉}|B|\text{d}y=|B|{\left(2\text{π}\right)}^{m}\delta \left({\beta }^{\prime }-\beta \right),\text{\hspace{0.17em}}{\beta }^{\prime }\in {R}^{m},\text{\hspace{0.17em}}\beta \in {R}^{m}\end{array}$ (56)

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

(58)

${U}_{\beta }\left(t\right)=\frac{1}{{\left(2\text{π}\right)}^{m}|B|}{\int }_{{R}^{m}}u\left(x,t\right){\stackrel{¯}{E}}_{\beta }\left(x\right)\rho \left(x\right)\text{d}x,\text{\hspace{0.17em}}\beta \in {R}^{m}$ (59)

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

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

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

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

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

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

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

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

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

$\left\{\begin{array}{l}{{U}^{\prime }}_{\beta }\left(t\right)+{\lambda }_{\beta }{U}_{\beta }\left(t\right)-{f}_{\beta }\left(t\right)=0,\text{\hspace{0.17em}}t>0\\ {U}_{\beta }\left(0\right)={\phi }_{\beta }\end{array}$ (69)

${U}_{\beta }\left(t\right)={\phi }_{\beta }{\text{e}}^{-{\lambda }_{\beta }t}+{\int }_{0}^{t}{f}_{\beta }\left(\tau \right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau$ (70)

$u\left(x,t\right)={\int }_{{R}^{m}}{U}_{\beta }\left(t\right){E}_{\beta }\left(x\right)\text{d}\beta ={\int }_{{R}^{m}}{\phi }_{\beta }{\text{e}}^{-{\lambda }_{\beta }t}{E}_{\beta }\left(x\right)\text{d}\beta +{\int }_{{R}^{m}}\left[{\int }_{0}^{t}{f}_{\beta }\left(\tau \right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau \right]{E}_{\beta }\left(x\right)\text{d}\beta$ (71)

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

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

$W\left(x,t\right)={\int }_{{R}^{m}}\left[{\int }_{0}^{t}{f}_{\beta }\left(\tau \right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau \right]{E}_{\beta }\left(x\right)\text{d}\beta$ (74)

$\begin{array}{l}W\left(x,t\right)={\int }_{{R}^{m}}\left[{\int }_{0}^{t}\frac{\gamma \left(\tau \right)}{{\left(2\text{π}\right)}^{m}|B|}{\stackrel{¯}{E}}_{\beta }\left(x\left(\tau \right)\right)\rho \left(x\left(\tau \right)\right){\text{e}}^{-{\lambda }_{\beta }\left(t-\tau \right)}\text{d}\tau \right]{E}_{\beta }\left(x\right)\text{d}\beta \\ ={\int }_{0}^{t}\frac{\gamma \left(\tau \right)}{{\left(2\text{π}\right)}^{m}|B|}{\text{e}}^{〈\upsilon ,{B}^{-1}\left(x-x\left(\tau \right)\right)〉}{\text{e}}^{-\underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_{k}^{2}+\frac{2r}{m}}{2}\left(t-\tau \right)}\text{d}\tau {\int }_{{R}^{m}}{\text{e}}^{-\underset{k=1}{\overset{m}{\sum }}\frac{{\beta }_{k}^{2}}{2}\left(t-\tau \right)}{\text{e}}^{i〈\beta ,{B}^{-1}\left(x-x\left(\tau \right)\right)〉}\text{d}\beta \\ =\frac{1}{{\left(2\text{π}\right)}^{m}|B|}{\int }_{0}^{t}{\text{e}}^{〈\upsilon ,{B}^{-1}\left(x-x\left(\tau \right)\right)〉}\gamma \left(\tau \right){\text{e}}^{-\underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_{k}^{2}+\frac{2r}{m}}{2}\left(t-\tau \right)}\text{d}\tau \underset{k=1}{\overset{m}{\prod }}{\int }_{-\infty }^{\infty }{\text{e}}^{-\frac{{\beta }_{k}^{2}}{2}\left(t-\tau \right)}{\text{e}}^{i{\beta }_{k}{\left[{B}^{-1}\left(x-x\left(\tau \right)\right)\right]}_{k}}\text{d}{\beta }_{k}\end{array}$

$\begin{array}{l}=\frac{1}{{\left(2\text{π}\right)}^{m}|B|}{\int }_{0}^{t}\left[\gamma \left(\tau \right){\text{e}}^{-\underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_{k}^{2}+\frac{2r}{m}}{2}\left(t-\tau \right)}\text{d}\tau \underset{k=1}{\overset{m}{\prod }}{\left(\frac{2\text{π}}{t-\tau }\right)}^{\frac{1}{2}}{\text{e}}^{-\frac{{\left\{{\left[{B}^{-1}\left(x-x\left(\tau \right)\right)\right]}_{k}\right\}}^{2}}{2\left(t-\tau \right)}}{\text{e}}^{{\upsilon }_{k}{\left[{B}^{-1}\left(x-x\left(\tau \right)\right)\right]}_{k}}\\ =\frac{1}{{\left(2\text{π}\right)}^{m}|B|}{\int }_{0}^{t}\gamma \left(\tau \right){\text{e}}^{-\underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_{k}^{2}+\frac{2r}{m}}{2}\left(t-\tau \right)}\text{d}\tau \underset{k=1}{\overset{m}{\prod }}{\left(\frac{2\text{π}}{t-\tau }\right)}^{\frac{1}{2}}{\text{e}}^{-\frac{{\left\{{\left[{B}^{-1}\left(x-x\left(\tau \right)\right)\right]}_{k}\right\}}^{2}}{2\left(t-\tau \right)}}{\text{e}}^{{\upsilon }_{k}{\left[{B}^{-1}\left(x-x\left(\tau \right)\right)\right]}_{k}}\\ =\frac{1}{{\left(2\text{π}\right)}^{m}|B|}{\int }_{0}^{t}\gamma \left(\tau \right){\text{e}}^{-\underset{k=1}{\overset{m}{\sum }}\frac{{\upsilon }_{k}^{2}+\frac{2r}{m}}{2}\left(t-\tau \right)}\text{d}\tau \underset{k=1}{\overset{m}{\prod }}{\left(\frac{2\text{π}}{t-\tau }\right)}^{\frac{1}{2}}{\text{e}}^{-\frac{{\left\{{\left[{B}^{-1}\left(x-x\left(\tau \right)\right)\right]}_{k}-\left(t-\tau \right){\upsilon }_{k}\right\}}^{2}}{2\left(t-\tau \right)}}{\text{e}}^{\frac{\left(t-\tau \right){\upsilon }_{k}^{2}}{2}}\end{array}$

$W\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\text{d}\tau$ (75)

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

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

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

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

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

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

$\left\{\begin{array}{l}\frac{\partial w}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}w}{\partial {x}_{k}\partial {x}_{j}}+\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

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\gamma \left(t\right)\in C\left(\left[0,T\right)\right),\gamma \left(t\right)\ge 0$

$\left\{\begin{array}{l}w\left(x,t\right)=\frac{1}{{\left(\text{2π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\text{d}\tau \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(83\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left( 84 \right)\end{array}$

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

${B}^{-1}x\left(t\right)-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon \equiv 0,\text{\hspace{0.17em}}\forall t\ge \tau \ge 0$ (86)

${B}^{-1}x\left(t\right)-t\upsilon \equiv {B}^{-1}x\left(\tau \right)-\tau \upsilon ,\text{\hspace{0.17em}}\forall t\ge \tau \ge 0$ (87)

$\tau =0$ 即得

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

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

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

${\text{e}}^{-\frac{{‖{B}^{-1}x\left(t\right)-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\equiv 1\ge {\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\ge 0,\text{\hspace{0.17em}}\forall x\in {R}^{m},\text{\hspace{0.17em}}\forall t\ge \tau \ge 0$ (89)

(a) $\frac{\partial w}{\partial {x}_{k}}>0,\text{\hspace{0.17em}}\left(x,t\right)\in {\Omega }_{-},\text{\hspace{0.17em}}k=1,\cdots ,m$ ；(b) $\frac{\partial w}{\partial {x}_{k}}<0,\text{\hspace{0.17em}}\left(x,t\right)\in {\Omega }_{+},\text{\hspace{0.17em}}k=1,\cdots ,m$

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

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

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

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

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

${\frac{\partial J}{\partial {x}_{l}}|}_{x=x\left(t\right)}\equiv 0,\text{\hspace{0.17em}}l=1,2,\cdots ,m$ (94)

$\frac{\partial w}{\partial {x}_{l}}\left(x,t\right)=\frac{-1}{2{\left(\text{2π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}-1}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\frac{\partial J}{\partial {x}_{l}}\text{d}\tau$ (95)

$\frac{\partial J}{\partial {x}_{l}}=2〈\left[\begin{array}{c}{c}_{l1}\\ ⋮\\ {c}_{lm}\end{array}\right],{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}x\left(\tau \right)-\left(t-\tau \right)\upsilon 〉$ (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) $\left(x,t\right)\in {\Omega }_{-},\text{\hspace{0.17em}}-\infty 0$

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

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

$\forall \left(x,t\right)\in {\Omega }_{-},\text{\hspace{0.17em}}\frac{\partial w}{\partial {x}_{l}}\left(x\left(t\right),t\right)>0,\text{\hspace{0.17em}}l=1,2,\cdots ,m$ (99)

$w\left(x\left(t\right),t\right)=\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\mathrm{max}}w\left(x,t\right),\text{\hspace{0.17em}}0 (100)

(b) $\left(x,t\right)\in {\Omega }_{+},\text{\hspace{0.17em}}\frac{\partial w}{\partial {x}_{l}}\left(x\left(t\right),t\right)<0,\text{\hspace{0.17em}}l=1,2,\cdots ,m$ (101)

$w\left(x\left(t\right),t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{max}}w\left(x,t\right),\text{\hspace{0.17em}}0 (102)

$\frac{\partial w}{\partial {x}_{l}}\left(x\left(t\right),t\right)=0,\text{\hspace{0.17em}}l=1,2,\cdots ,m,\text{\hspace{0.17em}}\forall t\in \left(0,T\right)$ (103)

$w\left(x\left(t\right),t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}\text{d}\tau \triangleq \mu \left(t\right)$ (104)

${\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}\text{d}\tau ={\left(\text{2π}\right)}^{\frac{m}{2}}|B|\mu \left(t\right)$ (105)

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

$\left\{\begin{array}{l}\frac{\partial V}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}V}{\partial {x}_{k}\partial {x}_{j}}+\underset{k=1}{\overset{m}{\sum }}\left(r-{q}_{k}\right)\frac{\partial V}{\partial {x}_{k}}-rV,\text{\hspace{0.17em}}x\in {R}^{m},\text{\hspace{0.17em}}0

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

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

$\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}}{\mathrm{max}}V\left(x,t\right)=V\left(x\left(t\right),t\right),0

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

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

$\frac{\partial {V}_{a}}{\partial {x}_{k}}\left(x,t\right)>0,\left(x,t\right)\in {\Omega }_{-},k\in \left\{1,2,\cdots ,m\right\}$ ，满足 $\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\mathrm{max}}{V}_{a}\left(x,t\right)={V}_{a}\left(x\left(t\right),t\right),0

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

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

$\frac{\partial {V}_{b}}{\partial {x}_{k}}\left(x,t\right)<0,\left(x,t\right)\in {\Omega }_{+},k\in \left\{1,2,\cdots ,m\right\}$ ，满足 $\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{max}}{V}_{b}\left(x,t\right)={V}_{b}\left(x\left(t\right),t\right),0

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

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

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

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

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

$I\triangleq {‖{B}^{-1}x-{B}^{-1}\xi -t\upsilon ‖}^{2}={‖{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}\xi -t\upsilon ‖}^{2}=〈{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}\xi -t\upsilon ,{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}\xi -t\upsilon 〉$

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

$\frac{\partial I}{\partial {x}_{k}}=\frac{\partial 〈{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}\xi -t\upsilon ,{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}\xi -t\upsilon 〉}{\partial {x}_{k}}=2〈\frac{\partial {C}^{\mathrm{T}}x}{\partial {x}_{k}},{C}^{\mathrm{T}}x-{C}^{\mathrm{T}}\xi -t\upsilon 〉$ (117)

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

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

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

$\left(x,t\right)\in {\Omega }_{-},-\infty

$x-\xi -tB\upsilon

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

$\frac{\partial I}{\partial {x}_{k}}=2{\left(x-\xi -tB\upsilon \right)}^{\mathrm{T}}C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ {c}_{km}\end{array}\right]<0,\text{\hspace{0.17em}}k\in \left\{1,2,\cdots ,m\right\}$ (122)

$\frac{\partial {V}_{a}}{\partial {x}_{k}}=-\frac{1}{2t}\frac{{\text{e}}^{-rt}}{{\left(2\text{π}t\right)}^{\frac{m}{2}}|B|}{\int }_{{}_{\left(\chi ,\infty \right)}}\phi \left(\xi \right){\text{e}}^{-\frac{1}{2t}{‖{B}^{-1}x-{B}^{-1}\xi -\upsilon t‖}^{2}}\frac{\partial I}{\partial {x}_{k}}\text{d}\xi >0,\text{\hspace{0.17em}}k\in \left\{1,2,\cdots ,m\right\}$ (123)

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

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

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

$\frac{\partial I}{\partial {x}_{k}}=2{\left(x-\xi -tB\upsilon \right)}^{\mathrm{T}}C\left[\begin{array}{c}{c}_{k1}\\ ⋮\\ {c}_{km}\end{array}\right]>0,\text{\hspace{0.17em}}k\in \left\{1,2,\cdots ,m\right\}$ (126)

$\frac{\partial {V}_{b}}{\partial {x}_{k}}\left(x,t\right)<0,\text{\hspace{0.17em}}\left(x,t\right)\in {\Omega }_{+},\text{\hspace{0.17em}}k\in \left\{1,2,\cdots ,m\right\}$ (127)

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

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}u}{\partial {x}_{k}\partial {x}_{j}}+\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),\text{\hspace{0.17em}}x\in {R}^{m},\text{\hspace{0.17em}}x\left(t\right)\in {R}^{m},0

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \left(x\right)=\delta \left(x-\chi \right)$

$\left\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V\left(x,t\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(132\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left( 133 \right)\end{array}$

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

$w\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\text{d}\tau$ (135)

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \left(x\right)=\delta \left(x-\chi \right)$

$\left\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+V\left(x,t\right)=\frac{2{t}^{-\frac{m}{2}}{\text{e}}^{-rt}}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(t\right)‖}^{2}}{2t}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\left(136\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\left( 137 \right)\end{array}$

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

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

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

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}u}{\partial {x}_{k}\partial {x}_{j}}+\underset{k=1}{\overset{m}{\sum }}\left(r-{q}_{k}\right)\frac{\partial u}{\partial {x}_{k}}-ru,\text{\hspace{0.17em}}x\in \left(-\infty ,x\left(t\right)\right),\text{\hspace{0.17em}}0

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \in {\Lambda }_{\left(\chi ,\infty \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {‖{B}^{-1}x‖}^{2}}{2T}},0\le \theta <1$

$\left\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+{V}_{a}\left(x,t\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\left(144\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left( 145 \right)\end{array}$

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

$w\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\text{d}\tau$ (147)

$\begin{array}{c}\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\mathrm{max}}u\left(x,t\right)=\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\mathrm{max}}w\left(x,t\right)+\underset{x\in \left(-\infty ,x\left(t\right)\right]}{\mathrm{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),\text{\hspace{0.17em}}0

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \in {\Lambda }_{\left(\chi ,\infty \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {‖{B}^{-1}x‖}^{2}}{2T}},0\le \theta <1$

$\left\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+{V}_{a}\left(x,t\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(148\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\left( 149 \right)\end{array}$

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

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

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

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}=\frac{1}{2}\underset{k,j=1}{\overset{m}{\sum }}{a}_{kj}\frac{{\partial }^{2}u}{\partial {x}_{k}\partial {x}_{j}}+\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

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \in {\Lambda }_{\left(-\infty ,\chi \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {‖{B}^{-1}x‖}^{2}}{2T}},0\le \theta <1$

$\left\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+{V}_{b}\left(x,t\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(156\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left( 157 \right)\end{array}$

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

$w\left(x,t\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{m}{2}}|B|}{\int }_{0}^{t}{\left(t-\tau \right)}^{-\frac{m}{2}}\gamma \left(\tau \right){\text{e}}^{-r\left(t-\tau \right)}{\text{e}}^{-\frac{{‖{B}^{-1}x-{B}^{-1}x\left(\tau \right)-\left(t-\tau \right)\upsilon ‖}^{2}}{2\left(t-\tau \right)}}\text{d}\tau$ (159)

$w\left(x\left(t\right),t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{max}}w\left(x,t\right),\text{\hspace{0.17em}}0

$\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{max}}{V}_{b}\left(x,t\right)={V}_{b}\left(x\left(t\right),t\right),\text{\hspace{0.17em}}0 即得

$\begin{array}{c}\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{max}}u\left(x,t\right)=\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{max}}w\left(x,t\right)+\underset{x\in \left[x\left(t\right),\infty \right)}{\mathrm{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),\text{\hspace{0.17em}}0

1) $A={\left({a}_{kj}\right)}_{m×m}$ 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) $\phi \in {\Lambda }_{\left(-\infty ,\chi \right)},0\le \phi \left(x\right)\le M{\text{e}}^{\frac{\theta {‖{B}^{-1}x‖}^{2}}{2T}},0\le \theta <1$

$\left\{\begin{array}{l}u\left(x,t\right)=w\left(x,t\right)+{V}_{b}\left(x,t\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(160\right)\\ x\left(t\right)=tB\upsilon +\chi ,T>t>0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left( 161 \right)\end{array}$

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

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

$\Gamma \triangleq \left\{\left(x,t\right)|x=x\left(t\right),t\in \left(0,T\right),\forall {t}_{1}\ne {t}_{2},x\left({t}_{1}\right)\ne x\left({t}_{2}\right)\right\}$

II) 在条件 $A={\left({a}_{kj}\right)}_{m×m}$ 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IIa和问题IIb存在充分光滑的精确解。问题IIa和问题IIb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}=\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。

Study of Singular Internal Boundary Problems for m Dimensional Anisotropic Heat Conduction Equations[J]. 理论数学, 2018, 08(01): 75-98. http://dx.doi.org/10.12677/PM.2018.81011

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