查看“数学物理方法”的源代码

* [https://www.bilibili.com/video/BV1Fe4y1u7Lx/ B站网课视频]

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=== 复变函数 ===
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====复变函数可导的充要条件====
<math>f(z)=u(x,y)+iv(x,y)</math>可导的充要条件为：偏导<math>\frac{\partial u}{\partial x}</math>、<math>\frac{\partial u}{\partial y}</math>、<math>\frac{\partial v}{\partial x}</math>、<math>\frac{\partial v}{\partial y}</math>存在、连续，且满足柯西-黎曼条件。
====柯西-黎曼条件====
#笛卡尔系 ：<math>\begin{align}\left\{\begin{array}{l}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\\
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
\end{array}\right.\end{align}</math>
#极坐标系：<math>\begin{align}\left\{\begin{array}{l}
\frac{\partial u}{\partial \rho}=\frac{1}{\rho}\frac{\partial v}{\partial \varphi},\\
\frac{1}{\rho}\frac{\partial u}{\partial \varphi}=-\frac{\partial v}{\partial \rho}.
\end{array}\right.\end{align}</math>
====已知实部（虚部）求虚部（实部）====
一般解法：曲线积分法。

=== 复变函数积分 ===
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=== 幂级数展开 ===
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=== 留数定理 ===
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=== 傅里叶变换 ===
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==== 傅里叶展开小结 ====
<math>\begin{align}f(x)=\left\{\begin{array}{ll}
\sum_{n=1}^\infty B_k\sin\frac{n\pi x}{l},&~~~(f(0)=f(l)=0)\\
\sum_{n=0}^\infty B_k\sin\frac{(n+1/2)\pi x}{l},&~~~(f(0)=f'(l)=0)\\
\sum_{n=0}^\infty A_k\cos\frac{(n+1/2)n\pi x}{l},&~~~(f'(0)=f(l)=0)\\
\sum_{n=0}^\infty A_k\cos\frac{n\pi x}{l}.&~~~(f'(0)=f'(l)=0)
\end{array}\right.\end{align}</math>

==== 常用公式 ====
傅里叶展开的时候，经常涉及到形如<math>\int_0^1\xi^n\sin\lambda \xi{\rm d}\xi</math>和<math>\int_0^1\xi^n\cos\lambda\xi{\rm d}\xi</math>的积分。可以利用<math>\xi^ne^{i\lambda\xi}=\frac{1}{i^n}\frac{{\rm d}^n}{{\rm d}\lambda^n}e^{i\lambda\xi}</math>，将其化简。最终结果为，

<math>\begin{align}
\int_0^1\xi^n\sin\lambda\xi{\rm d}\xi=\left\{\begin{array}{ll}
(-1)^{k+1}\frac{{\rm d}^{2k+1}}{{\rm d}\lambda^{2k+1}}\frac{\sin\lambda}{\lambda},&(n=2k+1)\\
(-1)^{k+1}\frac{{\rm d}^{2k}}{{\rm d}\lambda^{2k}}\frac{\cos\lambda-1}{\lambda},&(n=2k)
\end{array}\right.~~~(k=0,1,2,\ldots)\end{align}</math>
<math>\begin{align}
\int_0^1\xi^n\cos\lambda\xi{\rm d}\xi=\left\{\begin{array}{ll}
(-1)^{k+1}\frac{{\rm d}^{2k+1}}{{\rm d}\lambda^{2k+1}}\frac{\cos\lambda-1}{\lambda},&(n=2k+1)\\
(-1)^{k}\frac{{\rm d}^{2k}}{{\rm d}\lambda^{2k}}\frac{\sin\lambda}{\lambda},&(n=2k)
\end{array}\right.~~~(k=0,1,2,\ldots)
\end{align}</math>

=== 拉普拉斯变换 ===
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<math>\bar f(p)=\int_0^\infty f(t)e^{-pt}{\rm d}t</math>， <math>~~~~~~({\rm Re}p>0</math>；<math>{\rm Re}p</math>足够大，使得变换收敛<math>)</math>

<math>{\cal L}[t^ne^{st}]=\frac{n!}{(p-s)^{n+1}},</math>

<math>{\cal L}[t^nf(t)]=(-1)^n\frac{{\rm d}^n}{{\rm d}p^n}\bar f(p).</math>

=== 数学物理定解问题 ===
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=== 分离变数法 ===
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=== 级数解法、本征值问题 ===
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=== 球函数 ===
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==== 轴对称情形 ====
<math>u(r,\theta)=\sum_{l=0}^\infty\left(A_l r^l+\frac{B_l}{r^{l+1}}\right) {\rm P}_l(\cos\theta) </math>
==== 一般情形 ====
<math>\begin{align}
u(r,\theta,\varphi)=&\sum_{m=0}^\infty \sum_{l=m}^\infty\left(A_l^m r^l+\frac{B_l^m}{r^{l+1}}\right) {\rm P}_l^m(\cos\theta)\cos m\varphi\\
&+\sum_{m=1}^\infty \sum_{l=m}^\infty\left(C_l^m r^l+\frac{D_l^m}{r^{l+1}}\right) {\rm P}_l^m(\cos\theta)\sin m\varphi
 \end{align}</math>

=== 柱函数 ===
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[[文件:教学-数理-柱函数-1.png|800px]]
[[文件:教学-数理-柱函数-2.png|800px]]

=== 格林函数法 ===
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==== 主要思想 ====
<math>  \begin{align}
    \left\{
    \begin{array}{l}
      \Delta u=f({\bf r}),\\
      \left.\left(\alpha \frac{\partial u}{\partial n}+\beta u\right)\right|_\Sigma=\varphi(M),
    \end{array}
\right.
  \end{align}
</math>

转化为求解格林函数边值问题

第一类边界条件：<math> \begin{align}&\left\{\begin{array}{l}
      \Delta G=\delta({\bf r}-{\bf r}_0),\\
      G|_\Sigma=0,
    \end{array}
\right.&&u({\bf r})=\iiint_TG({\bf r},{\bf r}_0)f({\bf r}_0){\rm d} V_0+\iint_{\Sigma}\varphi({\bf r_0})\frac{\partial G({\bf r},{\bf r}_0)}{\partial n_0}{\rm d} S_0,
\end{align}</math>

第二类边界条件：<math> \begin{align}
\left\{
      \begin{array}{l}
      \Delta G=\delta({\bf r}-{\bf r}_0)-\frac{1}{V_T},\\
      \frac{\partial G}{\partial n}|_\Sigma=0,
      \end{array}
\right.&&u({\bf r})=\iiint_TG({\bf r},{\bf r}_0)f({\bf r}_0){\rm d} V_0-\iint_{\Sigma}\varphi({\bf r_0})G({\bf r},{\bf r}_0){\rm d} S_0,\end{align}</math>

第三类边界条件：
<math> \begin{align}
\left\{
      \begin{array}{l}
      \Delta G=\delta({\bf r}-{\bf r}_0),\\
      \left(\alpha \frac{\partial G}{\partial n}+\beta G\right)|_\Sigma=0,
      \end{array}
\right.&&u({\bf r})=\iiint_TG({\bf r},{\bf r}_0)f({\bf r}_0){\rm d} V_0-\frac{1}{\alpha}\iint_{\Sigma}\varphi({\bf r_0})G({\bf r},{\bf r}_0){\rm d} S_0,
\end{align}</math>

=== 积分变换法 ===
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=== 保角变换法 ===
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=== 非线性数学物理问题简介 ===
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