电磁场数值分析与计算课程笔记
无PPT，按照板书整理，若有错误敬请指正。
目录
1.电磁场数值分析与计算01-场论
2.电磁场数值分析与计算02-Maxwell方程组
3.电磁场数值分析与计算03-电磁场数值分析的定解问题
4.电磁场数值分析与计算04-边界条件
5.电磁场数值分析与计算05-有限元方法介绍
6.电磁场数值分析与计算06-2D有限元分析
插值函数
插值函数又称形状函数、权重函数(Interpolation/Shape/Weighting Function)，指的是在离散数据的基础上补插连续函数，使得这条连续曲线通过全部给定的离散数据点。
1D空间的插值
一维一阶插值
给定两个点及其对应的物理量\((x_1,\Phi_1)\quad(x_2,\Phi_2)\quad(x_1<x_2)\)，求\(\Phi(x_1<x<x_2)\)构造线性插值函数\(\tilde \Phi(x)=ax+b\)
\(x=x_1 \quad \tilde \Phi(x_1)=\Phi_1 \quad \Rightarrow \quad ax_1+b=\Phi_1\)
\(x=x_2 \quad \tilde \Phi(x_2)=\Phi_2 \quad \Rightarrow \quad ax_2+b=\Phi_2\)
使用克莱姆法则求解方程组：
\(\left[ {\begin{array}{*{20}{c}} x_1&1 \\ x_2&1 \end{array}}\right] \left[ {\begin{array}{*{20}{c}} a \\ b \end{array}}\right] = \left[ {\begin{array}{*{20}{c}} \Phi_1 \\ \Phi_2 \end{array}}\right]\)
\(a=\frac{\left[ {\begin{array}{*{20}{c}} \Phi_1&1 \\ \Phi_2&1 \end{array}}\right]}{\left[ {\begin{array}{*{20}{c}} x_1&1 \\ x_2&1 \end{array}}\right]}=\frac{\Phi_1-\Phi_2}{x_1-x_2} \quad b=\frac{\left[ {\begin{array}{*{20}{c}} x_1&\Phi_1 \\ x_2&\Phi_2 \end{array}}\right]}{\left[ {\begin{array}{*{20}{c}} x_1&1 \\ x_2&1 \end{array}}\right]}=\frac{\Phi_2 x_1 - \Phi_1 x_2}{x_1-x_2}\)
\(\Rightarrow \tilde \Phi(x) = \frac{\Phi_1-\Phi_2}{x_1-x_2}x+\frac{\Phi_2 x_1 - \Phi_1 x_2}{x_1-x_2}=N_1(x)\Phi_1+N_2(x)\Phi_2=\frac{x-x_2}{x_1-x_2}\Phi_1+\frac{x-x_1}{x_2-x_1}\Phi_2\)
其中，线性插值函数\(N_1(x)=\frac{x-x_2}{x_1-x_2} \quad N_2(x)=\frac{x-x_1}{x_2-x_1}\)
\(N_1(x) \left\{ {\begin{array}{*{20}{c}} N_1(x_1)=1 \\ N_1(x_2)=0 \end{array}}\right. \quad N_2(x) \left\{ {\begin{array}{*{20}{c}} N_2(x_1)=0 \\ N_2(x_2)=1 \end{array}}\right.\)
图1
一维一阶线性插值函数
一维二阶插值
给定两个点及其对应的物理量\((x_1,\Phi_1)\quad(x_2,\Phi_2)\quad(x_3,\Phi_3)\)
构造线性插值函数\(\tilde
\Phi(x)=ax^2+bx+c=N_1(x)\Phi_1+N_2(x)\Phi_2+N_3(x)\Phi_3\)
\(x=x_1 \quad \tilde \Phi(x_1)=\Phi_1 \quad
\Rightarrow \quad ax_1^2+bx_1+c=\Phi_1\)
\(x=x_2 \quad \tilde \Phi(x_2)=\Phi_2 \quad
\Rightarrow \quad ax_2^2+bx_2+c=\Phi_1\)
\(x=x_3 \quad \tilde \Phi(x_3)=\Phi_3 \quad
\Rightarrow \quad ax_3^2+bx_3+c=\Phi_1\)
使用克莱姆法则求解方程组：
\(\left[ {\begin{array}{*{20}{c}}
x_1^2&x_1&1 \\ x_2^2&x_2&1 \\ x_3^2&x_3&1
\end{array}}\right] \left[ {\begin{array}{*{20}{c}} a \\ b \\c
\end{array}}\right] = \left[ {\begin{array}{*{20}{c}} \Phi_1 \\ \Phi_2
\\ \Phi_3 \end{array}}\right]\)
（求解过程及其结果略）
线性插值函数：
\(N_1(x_1)=1 \quad N_2(x_1)=0 \quad
N_3(x_1)=0\)
\(N_1(x_2)=0 \quad N_2(x_2)=1 \quad
N_3(x_2)=0\)
\(N_1(x_3)=0 \quad N_2(x_3)=0 \quad
N_3(x_3)=1\)
\(N_1(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}
\quad N_2(x)=\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)} \quad
N_3(x)=\frac{(x-x_1)(x-x_2)}{(x_3-x_2)(x_1-x_2)} \quad\)
一维N阶插值
给定\(n+1\)个点及其对应的物理量\((x_1,\Phi_1)\quad(x_2,\Phi_2)\cdots(x_{n+1},\Phi_{n+1})\)
线性插值函数：
\(N_1(x)=\frac{(x-x_2)(x-x_3)\cdots(x-x_n)(x-x_{n+1})}{(x_1-x_2)(x_1-x_3)\cdots(x_1-x_n)(x_1-x_{n+1})}\)
\(N_i(x)=\frac{(x-x_1)(x-x_2)\cdots(x-x_{i-1})(x-x_{i+1})\cdots(x-x_n)(x-x_{n+1})}{(x_i-x_1)(x_i-x_2)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)(x_i-x_{n+1})}\)
性质：
\(N_k(x_i)=\left\{ {\begin{array}{*{20}{c}}
1&i=k \\ 0&i \ne k \end{array}}\right. \quad \sum\limits_{i =
1}^n N_k(x_i)=1\)
分母是\((x-x_k)\)的连乘（\(k \ne i\)）；分子是\((x_i-x_k)\)的连乘（\(k \ne i\)）
2D空间的插值
二维一阶插值
图2
二维一阶插值
使用三角形进行剖分，每个三角形为一个单元(element)，其中包含三个顶点(node)，分别为\((x_1,y_1,\Phi_1)\quad(x_2,y_2,\Phi_2)\quad(x_3,y_3,\Phi_3)\)
\(ax_1+by_1+c=\Phi_1\)
\(ax_2+by_2+c=\Phi_2\)
\(ax_3+by_3+c=\Phi_3\)
使用克莱姆法则求解方程组，求解过程及其结果略。
其中算式分母项\(D=\left[
{\begin{array}{*{20}{c}} x_1&y_1&c \\ x_2&y_2&c \\
x_3&y_3&c
\end{array}}\right]=2\Delta\)，其大小为所剖分的三角形面积的2倍。
线性插值函数展开为形状函数：\(\tilde
\Phi(x)=\sum\limits_1^3 N_k(x,y)\Phi_k\)
\(N_1(x,y)=\frac{1}{2\Delta}(p_1 x+q_1
y+r_1)\)
\(N_2(x,y)=\frac{1}{2\Delta}(p_2 x+q_2
y+r_2)\)
\(N_3(x,y)=\frac{1}{2\Delta}(p_3 x+q_3
y+r_3)\)
\(\left\{ {\begin{array}{*{20}{c}}
N_1(x_1,y_1)=1 \\ N_1(x_2,y_2)=0 \\ N_1(x_3,y_3)=0 \end{array}}\right.
\quad \left\{ {\begin{array}{*{20}{c}} N_2(x_1,y_1)=0 \\ N_2(x_2,y_2)=1
\\ N_2(x_3,y_3)=0 \end{array}}\right. \quad \left\{
{\begin{array}{*{20}{c}} N_3(x_1,y_1)=0 \\ N_3(x_2,y_2)=0 \\
N_3(x_3,y_3)=1 \end{array}}\right.\)
\(p_1=y_2-y_3 \quad p_2=y_3-y_1 \quad
p_3=y_1-y_2\)
\(q_1=x_3-x_2 \quad q_2=x_1-x_3 \quad
q_3=x_2-x_1\)
\(r_1=x_2 y_3-x_3 y_2 \quad r_2=x_3 y_1-x_1
y_3 \quad r_3=x_1 y_2-x_2 y_1\)
三角形的三点按照逆时针标记序号
p的下标为逆时针；q的下标为顺时针；r的下标为逆时针-顺时针
二维二阶插值
图3
二维二阶插值
在二维一阶插值的基础上，再取三个边的中点\((x_4,y_4,\Phi_4)\quad(x_5,y_5,\Phi_5)\quad(x_6,y_6,\Phi_6)\)
同样列出方程组并使用克莱姆法则求解之，求解过程及其结果略
形状函数：\(\tilde \Phi(x)=a_1+a_2 x+a_3 y+a_4
x^2+a_5 xy+a_6 y^2=\sum\limits_{j=1}^6 N_j(x,y)\Phi_j\)
性质：\(N_j(x_k,y_k)=\left\{
{\begin{array}{*{20}{c}} 1&j=k \\ 0&j \ne k \end{array}}\right.
\quad \sum\limits_{j = 1}^n N_j(x_k,y_k)=1\)
\(L_1(x,y)=\frac{1}{2\Delta}(p_1 x+q_1 y+r_1 )
\quad p_1=y_2-y_3 \quad q_1=x_3-x_2 \quad r_1=x_2 y_3-x_3
y_2\)
\(L_2(x,y)=\frac{1}{2\Delta}(p_2 x+q_2 y+r_2 )
\quad p_2=y_3-y_1 \quad q_2=x_1-x_3 \quad r_2=x_3 y_1-x_1
y_3\)
\(L_3(x,y)=\frac{1}{2\Delta}(p_3 x+q_3 y+r_3 )
\quad p_3=y_1-y_2 \quad q_3=x_2-x_1 \quad r_3=x_1 y_2-x_2
y_1\)
\(N_1(x,y)=(2L_1-1)L_1 \quad N_4(x,y)=4L_1
L_2\)
\(N_2(x,y)=(2L_2-1)L_2 \quad N_5(x,y)=4L_2
L_3\)
\(N_3(x,y)=(2L_3-1)L_3 \quad N_6(x,y)=4L_3
L_1\)
\(L(x,y)\)与二维一阶插值的\(N(x,y)\)一致
二维N阶插值
图4
帕斯卡三角形
二维N阶插值的形状函数展开式可以使用帕斯卡三角形（杨辉三角）推算出。
例如二维三阶插值的形状函数为：
\(\tilde \Phi(x)=a_1+a_2 x+a_3 y+a_4 x^2+a_5
xy+a_6 y^2+a_7 x^3+a_8 x^2 y+a_9 xy^2+a_{10} y^3=\sum\limits_{j=1}^{10}
N_j(x,y)\Phi_j\)
二维边值问题
由二阶偏微分方程定义的边值问题
\(-\frac{\partial}{\partial x}(\alpha_x
\frac{\partial \Phi}{\partial x})-\frac{\partial}{\partial y}(\alpha_y
\frac{\partial \Phi}{\partial y})+\beta\Phi=f \quad (x,y \in
\Omega)\)
\(\Phi\)为未知函数，\(\alpha_x,\alpha_y,\beta\)为与求解域中材料特性有关的系数，\(f\)为激励（已知）。
\({\bf{\nabla}}^2 \Phi=0\)，即\(\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}=0 \quad (\alpha_x=1,\alpha_y=1,\beta=0,f=0)\)
边界条件：
图5
二维边值问题
在\(\Gamma_1\)上，\(\Phi=P\)（常量）
在\(\Gamma_2\)上，\((\alpha_x \frac{\partial \Phi}{\partial x}
{\bf{i}}+\alpha_y \frac{\partial \Phi}{\partial y} {\bf{j}})\cdot
{\bf{n}}+\gamma\Phi=Q\)
在\(\Gamma_d\)上，\(\left\{ {\begin{array}{*{20}{c}} \Phi^+=\Phi^- \\
(\alpha_x^+ \frac{\partial \Phi^+}{\partial x} {\bf{i}}+\alpha_y^+
\frac{\partial \Phi^+}{\partial y} {\bf{j}})\cdot {\bf{n}}=(\alpha_x^-
\frac{\partial \Phi^-}{\partial x} {\bf{i}}+\alpha_y^- \frac{\partial
\Phi^-}{\partial y} {\bf{j}})\cdot {\bf{n}}
\end{array}}\right.\)
（场矢量的法向连续）
区域离散
剖分原则：
①单元之间不能有重叠
②单元之间通过定点连接
③避免出现狭长单元
④单元的数量对存储计算及存储精度均有影响
图6
区域离散与编号
连接矩阵：
| e | n(1,e) | n(2,e) | n(3,e) |
|---|---|---|---|
| (1) | 2 | 4 | 1 |
| (2) | 5 | 4 | 2 |
| (3) | 3 | 5 | 2 |
| (4) | 5 | 6 | 4 |
其中\(e\)表示单元，\(n(i,e)\)中的\(i\)表示节点，单元内的节点按照逆时针排序，排序不唯一，如节点\((1)\)可排序为241、124、412。
部分提供的数据：
(1)各节点坐标：\((x_1,y_1),\cdots,(x_i,y_i)\)
(2)材料属性：\({\begin{array}{*{20}{c}} e_1
\Rightarrow \alpha_{x_1} \quad \alpha_{y_1} \quad \beta_1 \quad f_1 \\
e_2 \Rightarrow \alpha_{x_2} \quad \alpha_{y_2} \quad \beta_2 \quad f_2
\\ e_3 \Rightarrow \alpha_{x_3} \quad \alpha_{y_3} \quad \beta_3 \quad
f_3 \\ e_4 \Rightarrow \alpha_{x_4} \quad \alpha_{y_4} \quad \beta_4
\quad f_4 \end{array}}\)
(3)\(\Gamma_1\)上各节点的\(\Phi\)值
(4)\(\Gamma_2\)上每一段上的\(\gamma\)和\(q\)
单元的插值
图7 单元的插值
\(N_k^e(x,y)=\frac{1}{2\Delta^e}(p_k^e+q_k^e+r_k^e) \quad k=1,2,3\)
构建方程系统（基于伽辽金法）
（边值问题、求解区域边界条件、区域的离散及连接矩阵见上文）
余量\(r=-\frac{\partial}{\partial x}(\alpha_x
\frac{\partial \Phi}{\partial x})-\frac{\partial}{\partial y}(\alpha_y
\frac{\partial \Phi}{\partial y})+\beta\Phi-f\)
对于第\(e\)个单元的加权余量
\(R_i^e=\iint\limits_{\Omega ^e} N_i^e rdxdy
\quad i=1,2,3\cdots\)
\(R_i^e=\iint\limits_{\Omega ^e} N_i^e
[-\frac{\partial}{\partial x}(\alpha_x \frac{\partial \Phi^e}{\partial
x})-\frac{\partial}{\partial y}(\alpha_y \frac{\partial \Phi^e}{\partial
y})+\beta\Phi^e-f] dxdy\)
\(=\iint\limits_{\Omega
^e}[\alpha_x\frac{\partial N_i^e}{\partial x}\frac{\partial
\Phi^e}{\partial x}+\alpha_y\frac{\partial N_i^e}{\partial
y}\frac{\partial \Phi^e}{\partial y}+\beta
N_i^e\Phi^e]dxdy-\iint\limits_{\Omega ^e}N_i^e
fdxdy-\oint\limits_{\Gamma^e}N_i^e{\bf{D}}\cdot{\bf{n^e}}d\Gamma\)
（推导过程略，因为不考）
\(\Gamma^e\)—\(\Omega^e\)的边界，\(\bf{n^e}\)—\(\Gamma^e\)的单位法向量，\({\bf{D}}=(\alpha_x\frac{\partial\Phi}{\partial
x}{\bf{i}}+\alpha_y\frac{\partial\Phi}{\partial
y}{\bf{i}})\)
单元中的\(\Phi^e\)用插值函数表示\(\Phi^e=N_1^e \Phi_1^e+N_2^e \Phi_2^e+N_3^e
\Phi_3^e=\sum\limits_{j=1}^{3}N_j^e \Phi_j^e\)
又由\(\frac{\partial \Phi^e}{\partial
x}=\sum\limits_{j=1}^{3} \Phi_j^e\frac{\partial N_j^e}{\partial x} \quad
\frac{\partial \Phi^e}{\partial y}=\sum\limits_{j=1}^{3}
\Phi_j^e\frac{\partial N_j^e}{\partial y} \quad \frac{\partial
N_j^e}{\partial x}=\frac{1}{2\Delta^e}p_j^e \quad \frac{\partial
N_j^e}{\partial y}=\frac{1}{2\Delta^e}q_j^e\)
\(\Rightarrow R_i^e=\iint\limits_{\Omega
^e}\sum\limits_{j=1}^{3}\Phi_j^e(\alpha_x\frac{\partial N_i^e}{\partial
x}\frac{\partial N_j^e}{\partial x}+\alpha_y\frac{\partial
N_i^e}{\partial y}\frac{\partial N_j^e}{\partial y}+\beta N_i^e
N_j^e)dxdy-\iint\limits_{\Omega ^e}N_i^e
fdxdy-\oint\limits_{\Gamma^e}N_i^e{\bf{D}}\cdot{\bf{n^e}}d\Gamma\)
\(i=1\)时有\(R_1^e=\iint\limits_{\Omega
^e}\Phi_1^e(\alpha_x\frac{\partial N_1^e}{\partial x}\frac{\partial
N_1^e}{\partial x}+\alpha_y\frac{\partial N_1^e}{\partial
y}\frac{\partial N_1^e}{\partial y}+\beta N_1^e
N_1^e)dxdy\)
\(+\iint\limits_{\Omega
^e}\Phi_2^e(\alpha_x\frac{\partial N_1^e}{\partial x}\frac{\partial
N_2^e}{\partial x}+\alpha_y\frac{\partial N_1^e}{\partial
y}\frac{\partial N_2^e}{\partial y}+\beta N_1^e
N_2^e)dxdy\)
\(+\iint\limits_{\Omega
^e}\Phi_3^e(\alpha_x\frac{\partial N_1^e}{\partial x}\frac{\partial
N_3^e}{\partial x}+\alpha_y\frac{\partial N_1^e}{\partial
y}\frac{\partial N_3^e}{\partial y}+\beta N_1^e
N_3^e)dxdy-\iint\limits_{\Omega ^e}N_1^e
fdxdy-\oint\limits_{\Gamma^e}N_1^e{\bf{D}}\cdot{\bf{n^e}}d\Gamma\)
（\(R_2^e\)、\(R_3^e\)同理）
写成矩阵形式：
\(\left\{ {\begin{array}{*{20}{c}} R_1^e \\
R_2^e \\ R_3^e \end{array}}\right\}=\left[ {\begin{array}{*{20}{c}}
k_{11}^e&k_{12}^e&k_{13}^e \\ k_{21}^e&k_{22}^e&k_{23}^e
\\ k_{31}^e&k_{32}^e&k_{33}^e \end{array}}\right]\left\{
{\begin{array}{*{20}{c}} \Phi_1^e \\ \Phi_2^e \\ \Phi_3^e
\end{array}}\right\}-\left\{ {\begin{array}{*{20}{c}} b_1^e \\ b_2^e \\
b_3^e \end{array}}\right\}-\left\{ {\begin{array}{*{20}{c}} g_1^e \\
g_2^e \\ g_3^e \end{array}}\right\}\)
矩阵可简化记为：\(\{R^e\}=[k^e]\{\Phi^e\}-\{b^e\}\{g^e\}\)
其中\(k_{ij}^e=\iint\limits_{\Omega
^e}\sum\limits_{j=1}^{3}\Phi_j^e(\alpha_x\frac{\partial N_i^e}{\partial
x}\frac{\partial N_j^e}{\partial x}+\alpha_y\frac{\partial
N_i^e}{\partial y}\frac{\partial N_j^e}{\partial y}+\beta N_i^e
N_j^e)dxdy \quad i=1,2,3 \quad j=1,2,3\)
\(b_i^e=\iint\limits_{\Omega ^e}N_i^e fdxdy
\quad g_i^e=\oint\limits_{\Gamma^e}N_i^e{\bf{D}}\cdot{\bf{n^e}}d\Gamma
\quad i=1,2,3\)
根据公式\(\iint\limits_{\Omega
^e}(N_1^e)^l(N_2^e)^m(N_3^e)^n dxdy=\frac{l!m!n!}{(l+m+n+2)!} \cdot
2\Delta\)
可得\(\iint\limits_{\Omega ^e}N_i
dxdy=\frac{1}{3}\Delta \quad \iint\limits_{\Omega ^e}(N_i)^2
dxdy=\frac{1}{6}\Delta\)
\(\iint\limits_{\Omega ^e}N_i N_j
dxdy=\frac{1}{12}\Delta\)
进而得到\(k_{ij}^e=\)\(\frac{1}{4\Delta^e}(\alpha_x^e p_i^e
p_j^e+\alpha_y^e q_i^e q_j^e)\)\(+\frac{\Delta^e}{12}\beta^e(1+\delta_{ij})\quad
\delta_{ij}=\left\{ {\begin{array}{*{20}{c}} 1 \quad i=j \\ 0 \quad i
\ne j \end{array}}\right. \quad b_i^e=\frac{\Delta f}{3}\)
求图6的\([k]\)矩阵：
对单元1的节点进行标号，按照表1的连接矩阵标号，以逆时针顺序，将节点2、4、1依次标为①、②、③
得到矩阵\([k^{(1)}]=\left[
{\begin{array}{*{20}{c}} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}
\\ k_{21}^{(1)}&k_{22}^{(1)}&k_{23}^{(1)} \\
k_{31}^{(1)}&k_{32}^{(1)}&k_{33}^{(1)}
\end{array}}\right]=\left[ {\begin{array}{*{20}{c}}
k_{22}&k_{24}&k_{21} \\ k_{42}&k_{44}&k_{41} \\
k_{12}&k_{14}&k_{11} \end{array}}\right]\)
同理，根据表1得到单元2、单元3、单元4的矩阵为
\([k^{(2)}]=\left[ {\begin{array}{*{20}{c}}
k_{55}&k_{54}&k_{52} \\ k_{45}&k_{44}&k_{42} \\
k_{25}&k_{24}&k_{22} \end{array}}\right] \quad [k^{(3)}]=\left[
{\begin{array}{*{20}{c}} k_{33}&k_{35}&k_{32} \\
k_{53}&k_{55}&k_{52} \\ k_{23}&k_{25}&k_{22}
\end{array}}\right] \quad [k^{(4)}]=\left[ {\begin{array}{*{20}{c}}
k_{55}&k_{56}&k_{54} \\ k_{65}&k_{66}&k_{64} \\
k_{45}&k_{46}&k_{44} \end{array}}\right]\)
矩阵中\(k_{ij}^{(e)}\)表示第\(e\)个单元中的第\(j\)个节点对第\(i\)个节点的影响；
\(k_{ij}\)表示全局系统中的第\(j\)个节点对第\(i\)个节点的影响。
根据上述关于各单元的矩阵得到整个系统的\([k]\)矩阵，计算方法为\([k]=\sum\limits_{e=1}^M [k^{(e)}]\)。
如\([k^{(2)}]\)与\([k^{(4)}]\)上有元素\(k_{54}\)，则对应矩阵\([k]\)的第5行第4列填\(k{54}^{(2)+(4)}\)。
若所有单元对应的矩阵均无\(k_{ij}\)，则\([k]\)对应位置填\(0\)。
矩阵有对称性，即\(k_{ij}=k_{ji}\)，因此由矩阵\([k]\)的第5行第4列填\(k_{54}^{(2)+(4)}\)可知第4行第5列填\(k_{45}^{(2)+(4)}\)。
\([k]=\left[ {\begin{array}{*{20}{c}}
k_{11}^{(1)}&k_{12}^{(1)}&0&k_{14}^{(1)}&0&0 \\
k_{21}^{(1)}&k_{22}^{(1)+(2)+(3)}&k_{23}^{(3)}&k_{24}^{(1)+(2)}&k_{25}^{(2)+(3)}&0
\\ 0&k_{32}^{(3)}&k_{33}^{(3)}&0&k_{35}^{(3)}&0 \\
k_{41}^{(1)}&k_{42}^{(1)+(2)}&0&k_{44}^{(1)+(2)+(4)}&k_{45}^{(2)+(4)}&k_{46}^{(4)}
\\
0&k_{52}^{(2)+(3)}&k_{53}^{(3)}&k_{54}^{(2)+(4)}&k_{55}^{(2)+(3)+(4)}&k_{56}^{(4)}
\\ 0&0&0&k_{64}^{(4)}&k_{65}^{(4)}&k_{66}^{(4)}
\end{array}}\right]\)
\([b]\)与\([g]\)的矩阵中，其元素上标与\([k]\)矩阵对角线上元素商标相同，即
\(\{b\}=\left\{ {\begin{array}{*{20}{c}}
b_{1}^{(1)} \\ b_{2}^{(1)+(2)+(3)} \\ b_{3}^{(3)} \\ b_{4}^{(1)+(2)+(4)}
\\ b_{5}^{(2)+(3)+(4)} \\ b_{6}^{(4)} \end{array}} \right\} \quad
\{g\}=\left\{ {\begin{array}{*{20}{c}} g_{1}^{(1)} \\
g_{2}^{(1)+(2)+(3)} \\ g_{3}^{(3)} \\ g_{4}^{(1)+(2)+(4)} \\
g_{5}^{(2)+(3)+(4)} \\ g_{6}^{(4)} \end{array}} \right\}\)
狄利赫莱边界条件的处理
注：这里没听懂。
假设上文矩阵中节点3、5、6的电磁位\(\Phi\)已知，即\(\Phi_3=p_3,\Phi_5=p_5,\Phi_6=p_6\)，有两种处理方法。
(1)划0置1法
将已知\(\Phi\)的节点对应\([k]\)矩阵中的对角线元素置1，行和列上其他元素置0，得到矩阵
\([k]=\left[ {\begin{array}{*{20}{c}}
k_{11}^{(1)}&k_{12}^{(1)}&0&k_{14}^{(1)}&0&0 \\
k_{21}^{(1)}&k_{22}^{(1)+(2)+(3)}&0&k_{24}^{(1)+(2)}&0&0
\\ 0&0&1&0&0&0 \\
k_{41}^{(1)}&k_{42}^{(1)+(2)}&0&0&0&0 \\
0&0&0&0&1&0 \\ 0&0&0&0&0&1
\end{array}}\right]\)
\(\left[ {\begin{array}{*{20}{c}}
k_{11}&k_{12}&k_{14} \\ k_{21}&k_{22}&k_{24} \\
k_{41}&k_{42}&k_{44} \end{array}}\right] \left[
{\begin{array}{*{20}{c}} \Phi_1 \\ \Phi_2 \\ \Phi_3
\end{array}}\right]=\left\{ {\begin{array}{*{20}{c}}
b_1-k_{13}p_3-k_{15}p_5-k_{16}p_6 \\ b_2-k_{23}p_3-k_{25}p_5-k_{26}p_6
\\ b_4-k_{43}p_3-k_{45}p_5-k_{46}p_6 \end{array}}\right\}\)
这样就将原来的\(6 \times
6\)矩阵转化为\(3 \times
3\)矩阵，方便了运算。
(2)置大数法
将已知\(\Phi\)的节点对应\([k]\)矩阵中的对角线元素置一个很大的数后带入求解
如取\(k_{33}=10^{70},b_3=p_3 \times
10^{70}\)
区域单元的内边界对\(g\)无贡献，只有边界\(\Gamma\)才对\(g\)有贡献
若节点\(i\)在\(\Gamma\)内部，则对应\(g_i=0\)；
若节点\(i\)在\(\Gamma\)上，则对应\(g_i \ne 0\)。
2D有限元分析的应用
静电场问题
二维：\(-\varepsilon_r
{\bf{\nabla}}\Phi=\frac{\rho}{\varepsilon_0}\)
\(-\frac{\partial}{\partial
x}(\varepsilon_r\frac{\partial \Phi}{\partial
x})-\frac{\partial}{\partial y}(\varepsilon_r\frac{\partial
\Phi}{\partial y})=\frac{\rho}{\varepsilon_0}\)
边界：\(\left\{ {\begin{array}{*{20}{c}}
\Phi=p \\ \frac{\partial \Phi}{\partial n}=0
\end{array}}\right.\)
媒质交界面：\(\left\{ {\begin{array}{*{20}{c}}
\Phi^+=\Phi^- \quad \varepsilon_r^+\frac{\partial \Phi^+}{\partial
n}=\varepsilon_r^-\frac{\partial \Phi^-}{\partial n} \\ E_t^+=E_t^-
\quad D_n^+=D_n^- \end{array}}\right.\)
2D磁场
\({\bf{J}}=J_z{\bf{k}} \quad
{\bf{\nabla}}\times{\bf{H}}={\bf{J}} \Rightarrow
-\frac{1}{\mu_r}{\bf{\nabla}}^2 A_z=\mu_0 J_z{\bf{k}}\)
\({\bf{\nabla}}\times\frac{1}{\mu_r}{\bf{\nabla}}\times
A_z{\bf{k}}=\mu_0 J_z{\bf{k}}\)
\(-\frac{\partial}{\partial
x}(\frac{1}{\mu}\frac{\partial A_z}{\partial
x})-\frac{\partial}{\partial y}(\frac{1}{\mu}\frac{\partial
A_z}{\partial y})=\mu_0 J_z\)
交界面：\(\left\{ {\begin{array}{*{20}{c}}
A_z^+=A_z^- \\ \frac{1}{\mu_r^+}\frac{\partial A_z^+}{\partial
n}=\frac{1}{\mu_r^-}\frac{\partial A_z^-}{\partial n}
\end{array}}\right.\)