数理方法考前必背

特殊函数

勒让德多项式

前几个勒让德多项式

\[\begin{aligned} P_{0}(x) & = 1 \\ P_{1}(x) & = x=\cos \theta \\ P_{2}(x) & = \dfrac{1}{2}\left(3 x^{2}-1\right)=\dfrac{1}{4}(3 \cos 2 \theta+1) \\ P_{3}(x) & = \dfrac{1}{2}\left(5 x^{3}-3 x\right)=\dfrac{1}{8}(5 \cos 3 \theta+3 \cos \theta) \end{aligned} \]

勒让德多项式的正交关系

\[\int_{-1}^{+1} \mathrm{P}_{k}(x) \mathrm{P}_{l}(x) \mathrm{d} x=0 \quad(k \neq l) \]

\[\int_{0}^{\pi} \mathrm{P}_{k}(\cos \theta) \mathrm{P}_{l}(\cos \theta) \sin \theta \mathrm{d} \theta=0 \quad(k \neq l) \]

勒让德多项式的模

\[N_{l}^{2}=\int_{-1}^{+1}\left[\mathrm{P}_{l}(x)\right]^{2} \mathrm{~d} x =\dfrac{2}{2 l+1} \]

\[N_{l}=\sqrt{\dfrac{2}{2 l+1}} \quad(l=0,1,2, \cdots) \]

连带勒让德函数

\[\mathrm{P}_{l}^{m}(x)=\left(1-x^{2}\right)^{\frac{m}{2}} \mathrm{P}_{l}^{[m]}(x) \quad (m=0,1,2, \cdots,l) \]

函数表

\[\begin{aligned} P_{1}^{1}(x) & =\left(1-x^{2}\right)^{\frac{1}{2}}=\sin \theta \\ P_{2}^{1}(x) & =3\left(1-x^{2}\right)^{\frac{1}{2}} x=\frac{3}{2} \sin 2 \theta=3 \sin \theta \cos \theta \\ P_{2}^{2}(x) & =3\left(1-x^{2}\right)=\frac{3}{2}(1-\cos 2 \theta)=3 \sin ^{2} \theta \\ P_{3}^{1}(x) & =\frac{3}{2}\left(1-x^{2}\right)^{\frac{1}{2}}\left(5 x^{2}-1\right)=\frac{3}{8}(\sin \theta+5 \sin 3 \theta)=6 \sin \theta-\frac{15}{2} \sin ^{3} \theta \\ P_{3}^{2}(x) & =15\left(1-x^{2}\right) x=\frac{15}{4}(\cos \theta-\cos 3 \theta)=15 \sin ^{2} \theta \cos \theta \\ P_{3}^{3}(x) & =15\left(1-x^{2}\right)^{\frac{3}{2}}=\frac{15}{4}(3 \sin \theta-\sin 3 \theta)=15 \sin ^{3} \theta \end{aligned} \]

连带勒让德函数的正交关系

\[\int_{-1}^{+1} \mathrm{P}_{k}^{m}(x) \mathrm{P}_{l}^{m}(x) \mathrm{d} x=0 \quad(k \neq l) \]

\[\int_{0}^{\pi} \mathrm{P}_{k}^{m}(\cos \theta) \mathrm{P}_{l}^{m}(\cos \theta) \sin \theta \mathrm{d} \theta=0 \quad(k \neq l) \]

连带勒让德函数的模

\[\left(N_{l}^{m}\right)^{2}=\int_{-1}^{+1}\left[\mathrm{P}_{l}^{m}(x)\right]^{2} \mathrm{~d} x= \dfrac{(l+m) ! 2}{(l-m) !(2 l+1)} \]

\[N_{l}^{m}=\sqrt{\dfrac{(l+m) ! 2}{(l-m) !(2 l+1)}} \]

球坐标系和柱坐标系下的分离变数法

拉普拉斯方程

\[\Delta u=0 \]

球坐标

\[\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial u}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial u}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2} u}{\partial \varphi^{2}}=0 \]

分离变数

\[u(r, \theta, \varphi)=R(r) \mathrm{Y}(\theta, \varphi) \]

分解为

\[\frac{\mathrm{d}}{\mathrm{d} r}\left(r^{2} \frac{\mathrm{d} R}{\mathrm{~d} r}\right)-l(l+1) R=0 \]

\[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial \mathrm{Y}}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2} \mathrm{Y}}{\partial \varphi^{2}}+l(l+1) \mathrm{Y}=0 \]

柱坐标

\[\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial u}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} u}{\partial \varphi^{2}}+\frac{\partial^{2} u}{\partial z^{2}}=0 \]

分离变数

\[u(\rho, \varphi, z)=R(\rho) \Phi(\varphi) Z(z) \]

分解为

\[\Phi^{\prime \prime}+\lambda \Phi=0 \]

\[Z^{\prime \prime}-\mu Z=0 \]

\[\frac{\mathrm{d}^{2} R}{\mathrm{~d} \rho^{2}}+\frac{1}{\rho} \frac{\mathrm{d} R}{\mathrm{~d} \rho}+\left(\mu-\frac{m^{2}}{\rho^{2}}\right) R=0 \]

亥姆霍兹方程

\[\Delta v+k^{2} v=0 \quad ( k \neq 0 ) \]

球坐标

分离变数

\[v(r, \theta, \varphi)=R(r) Y(\theta, \varphi) \]

得球函数方程

\[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial \mathrm{Y}}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2} \mathrm{Y}}{\partial \varphi^{2}}+l(l+1) \mathrm{Y}=0 \]

和 \(l\) 阶球贝塞尔方程

\[\frac{\mathrm{d}}{\mathrm{d} r}\left(r^{2} \frac{\mathrm{d} R}{\mathrm{~d} r}\right)+\left[k^{2} r^{2}-l(l+1)\right] R=0 \]

柱坐标

分离变数

\[v(\rho, \varphi, z)=R(\rho) \Phi(\varphi) Z(z) \]

分解为

\[\Phi^{\prime \prime}+\lambda \Phi=0 \]

\[Z^{\prime \prime}+\nu^{2} Z=0 \]

\[\frac{\mathrm{d}^{2} R}{\mathrm{~d} \rho^{2}}+\frac{1}{\rho} \frac{\mathrm{d} R}{\mathrm{~d} \rho}+\left(k^{2}-\nu^{2}-\frac{\lambda}{\rho^{2}}\right) R=0 \]

格林函数法

第一边值问题的泊松方程的格林函数法

\[\Delta u=f(\boldsymbol{r}) \quad(\boldsymbol{r} \in T) \]

\[u(\boldsymbol{r})=\varphi(\boldsymbol{r}) \quad(\boldsymbol{r} \in \Sigma) \]

则

\[u\left(\boldsymbol{r}_{0}\right)=\iiint_{T} \mathrm{G}\left(\boldsymbol{r}, \boldsymbol{r}_{0}\right) f(\boldsymbol{r}) \mathrm{d} V+\iint_{\Sigma} \varphi(\boldsymbol{r}) \frac{\partial \mathrm{G}\left(\boldsymbol{r}, \boldsymbol{r}_{0}\right)}{\partial n} \mathrm{~d} S \]

其中，\(\mathrm{G}\) 称为泊松方程第一边值问题的格林函数

\[\Delta \mathrm{G}\left(r, r_{0}\right)=\delta\left(r-r_{0}\right), \quad \mathrm{G}(\boldsymbol{r})=0 \quad(\boldsymbol{r} \in \Sigma) \]

用电像法求格林函数

球域以内的第一边值问题

\[\left\{\begin{aligned} & \nabla^{2} \mathrm{G}=\delta\left(r-r_{0}\right) \\ & \left.\mathrm{G}\right|_{r=R}=0 \end{aligned}\right. \]

其解为

\[\begin{aligned} \mathrm{G}\left(\boldsymbol{r}, \boldsymbol{r}_{0}\right) & =-\frac{1}{4 \pi} \frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{0}\right|}+\frac{R}{r_{0}} \frac{1}{4 \pi} \frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{1}\right|} \\ & =-\frac{1}{4 \pi} \frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{0}\right|}+\frac{R}{r_{0}} \frac{1}{4 \pi} \frac{1}{\left|\boldsymbol{r}-\dfrac{R^{2}}{r_{0}^{2}} \boldsymbol{r}_{0}\right|} \end{aligned} \]

无穷大平面上凸起的半球的第一边值问题

像电荷已在图中标出，其解为

\[G\left(\boldsymbol{r}, \boldsymbol{r}_{0}\right)=-\frac{1}{4 \pi}\frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{0}\right|}+\frac{1}{4 \pi}\frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{(1)}^{*}\right|}+\frac{1}{4 \pi}\frac{R}{r_{0}}\frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{(2)}^{*}\right|}-\frac{1}{4 \pi}\frac{R}{r_{0}}\frac{1}{\left|\boldsymbol{r}-\boldsymbol{r}_{(3)}^{*}\right|} \]

其中

\[\left\{\begin{aligned} \boldsymbol{r}_{0} & =\left(x_{0}, y_{0}, z_{0}\right) \\ \boldsymbol{r}_{(1)}^{*} & =\left(x_{0}, y_{0},-z_{0}\right) \\ \boldsymbol{r}_{(2)}^{*} & =\frac{R^{2}}{r_{0}}\left(x_{0}, y_{0}, z_{0}\right) \\ \boldsymbol{r}_{(3)}^{*} & =\frac{R^{2}}{r_{0}}\left(x_{0}, y_{0},-z_{0}\right) \end{aligned}\right. \]

积分变换法

傅里叶变换

\[F(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(x) \mathrm{e}^{-\mathrm{i} \omega x} \mathrm{~d} x \]

\[f(x)=\int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{\mathrm{i} \omega x} \mathrm{~d} \omega \]

导数定理

\[\mathscr{F}\left[f^{\prime}(x)\right]=\mathrm{i} \omega F(\omega) \]

\[\mathscr{F}\left[f^{(n)}(x)\right]=(\mathrm{i} \omega)^{n} F(\omega) \]

积分定理

\[\mathscr{F}\left[\int^{(x)} f(\xi) \mathrm{d} \xi\right]=\frac{1}{\mathrm{i} \omega} F(\omega) \]

相似性定理

\[\mathscr{ F }[f(a x)]=\frac{1}{a} F\left(\frac{\omega}{a}\right) \]

延迟定理

\[\mathscr{F}\left[f\left(x-x_{0}\right)\right]=\mathrm{e}^{-\mathrm{i} \omega x_{0}} F(\omega) \]

位移定理

\[\mathscr{F}\left[\mathrm{e}^{\mathrm{i} \omega_{0} x} f(x)\right]=F\left(\omega-\omega_{0}\right) \]

卷积定理

\[\mathscr{F}\left[f_{1}(x) * f_{2}(x)\right]=2 \pi F_{1}(\omega) F_{2}(\omega) \]

三重傅里叶变换

\[F(\boldsymbol{k})=\frac{1}{(2 \pi)^{3}} \iiint_{-\infty}^{\infty} f(\boldsymbol{r})\left[\mathrm{e}^{\mathrm{i} k \cdot r}\right]^{*} \mathrm{~d} \boldsymbol{r} \]

\[f(\boldsymbol{r})=\iiint_{-\infty}^{\infty} F(\boldsymbol{k}) \mathrm{e}^{\mathrm{i} k \cdot r} \mathrm{~d} \boldsymbol{k} \]

拉普拉斯变换

\[\bar{f}(p)=\int_{0}^{\infty} f(t) \mathrm{e}^{-p t} \mathrm{~d} t \]

\[f(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \bar{f}(\sigma+\mathrm{i} \omega) \mathrm{e}^{(\sigma+i \omega)} \mathrm{d} \omega = \frac{1}{2 \pi \mathrm{i}} \int_{\sigma-\mathrm{i} \infty}^{\sigma+\mathrm{i} \infty} \bar{f}(p) \mathrm{e}^{\mathrm{i} p} \mathrm{d} p \]

线性定理

若 \(f_{1}(t) \fallingdotseq \bar{f}_{1}(p), f_{2}(t) \fallingdotseq \bar{f}_{2}(p)\) , 则

\[c_{1} f_{1}(t)+c_{2} f_{2}(t) \fallingdotseq c_{1} \bar{f}_{1}(p)+c_{2} \bar{f}_{2}(p) \]

导数定理

\[f^{\prime}(t) \fallingdotseq p \bar{f}(p)-f(0) \]

\[f^{(n)}(t) \fallingdotseq p^{n} \bar{f}(p)-p^{n-1} f(0)-p^{n-2} f^{\prime}(0)-\cdots-p f^{(n-2)}(0)-f^{(n-1)}(0) \]

积分定理

\[\int_{0}^{t} \psi(\tau) \mathrm{d} \tau \fallingdotseq \frac{1}{p} \mathscr{L}[\psi(t)] \]

相似性定理

\[f(a t) \fallingdotseq \frac{1}{a} \bar{f}\left(\frac{p}{a}\right) \]

位移定理

\[\mathrm{e}^{-\lambda t} f(t) \fallingdotseq \bar{f}(p+\lambda) \]

延迟定理

\[f\left(t-t_{0}\right) \fallingdotseq \mathrm{e}^{-p t_{0}} \bar{f} (p) \]

卷积定理

若 \(f_{1}(t) \fallingdotseq \bar{f}_{1}(p), f_{2}(t) \fallingdotseq \bar{f}_{2}(p)\) ，则

\[f_{1}(t) * f_{2}(t) \fallingdotseq \bar{f}_{1}(p) \bar{f}_{2}(p) \]

二阶非齐次线性方程的通解

有二阶非齐次线性方程

\[y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{2}(x) y=f(x) \]

设 \(y_{1}(x), y_{2}(x)\) 是对应的二阶齐次线性方程

\[y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{2}(x) y=0 \]

的基本解组，则该二阶非齐次线性方程的通解为

\[y(x)=c_{1} y_{1}(x)+c_{2} y_{2}(x)+\int_{x_{0}}^{x} \frac{y_{1}(x)\left(-y_{2}(\xi)\right)+y_{2}(x) y_{1}(\xi)}{y_{1}(\xi) y_{2}^{\prime}(\xi)-y_{2}(\xi) y_{1}^{\prime}(\xi)} f(\xi) d \xi \]

无限长弦的自由振动

\[u_{t}-a^{2} u_{x x}=0 \quad (-\infty<x<\infty) \]

\[\left.u\right|_{t=0}=\varphi(x),\quad \left.u_{t}\right|_{t=0}=\psi(x) \]

其解为达朗贝尔公式

\[u(x, t)=\frac{1}{2}[\varphi(x+a t)+\varphi(x-a t)]+\frac{1}{2 a} \int_{x-a t}^{x+a t} \psi(\xi) \mathrm{d} \xi \]

三维无界空间的自由振动

\[u_{t}-a^{2} \Delta_{3} u=0 \]

\[\left.u\right|_{t=0}=\varphi(\boldsymbol{r}),\quad \left.u_{t}\right|_{t=0}=\psi(\boldsymbol{r}) \]

其解为泊松公式

\[u(\boldsymbol{r}, t)=\frac{1}{4 \pi a} \frac{\partial}{\partial t} \iint_{S_{a t}^{r}} \frac{\varphi\left(\boldsymbol{r}^{\prime}\right)}{a t} \mathrm{~d} S^{\prime}+\frac{1}{4 \pi a} \iint_{S_{a t}^{r}} \frac{\psi\left(\boldsymbol{r}^{\prime}\right)}{a t} \mathrm{~d} S^{\prime} \]

其中 \(S_{a t}^{r}\) 是球心为 \(\boldsymbol{r}\)，半径为 \(a t\) 的球面；\(\mathrm{d} S^{\prime}\) 是球面 \(S_{a t}^{r}\) 的面积元。

二维无界空间的自由振动

\[u_{t}-a^{2} \Delta_{2} u=0 \]

\[\left.u\right|_{t=0}=\varphi(x, y),\quad \left.u_{t}\right|_{t=0}=\psi(x, y) \]

使用降维法

\[\begin{aligned} \mathrm{d} \sigma^{\prime} & =\mathrm{d} S^{\prime} \cos \theta=\mathrm{d} S^{\prime} \frac{\sqrt{a^{2} t^{2}-\rho^{2}}}{a t} \\ & =\mathrm{d} S^{\prime} \frac{\sqrt{a^{2} t^{2}-\left(x^{\prime}-x\right)^{2}-\left(y^{\prime}-y\right)^{2}}}{a t} \end{aligned} \]

泊松公式在二维问题中成为

\[\begin{aligned} u(x, y, t)= & \frac{1}{2 \pi a} \frac{\partial}{\partial t} \iint_{\Sigma_{a t}^{x, y}} \frac{\varphi\left(x^{\prime}, y^{\prime}\right)}{\sqrt{a^{2} t^{2}-\left(x^{\prime}-x\right)^{2}-\left(y^{\prime}-y\right)^{2}}} \mathrm{~d} x^{\prime} \mathrm{d} y^{\prime} \\ & +\frac{1}{2 \pi a} \iint_{\Sigma_{a t}^{x, y}} \frac{\psi\left(x^{\prime}, y^{\prime}\right)}{\sqrt{a^{2} t^{2}-\left(x^{\prime}-x\right)^{2}-\left(y^{\prime}-y\right)^{2}}} \mathrm{~d} x^{\prime} \mathrm{d} y^{\prime} \end{aligned} \]

无限长细杆的热传导问题

\[u_{t}-a^{2} u_{x x}=0 \quad(-\infty<x<\infty) \]

\[\left.u\right|_{t=0}=\varphi(x) \]

其解为

\[u(x, t)=\int_{-\infty}^{\infty} \varphi(\xi)\left[\dfrac{1}{2 a \sqrt{\pi t}} \exp\{-\dfrac{(x-\xi)^{2}}{4 a^{2} t}\}\right] \mathrm{d} \xi \]

三维无界空间的受迫振动

\[u_{t t}-a^{2} \Delta_{3} u=f(\boldsymbol{r}, t) \]

\[\left.u\right|_{t=0}=0,\quad \left.u_{t}\right|_{t=0}=0 \]

其解为推迟势

\[u(\boldsymbol{r}, t)=\frac{1}{4 \pi a^{2}} \iiint_{T_{a t}^{r}} \frac{f\left(\boldsymbol{r}^{\prime}, t-\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right| / a\right)}{\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right|} \mathrm{d} V^{\prime} \]

其中 \(T_{a t}^{r}\) 是球心为 \(\boldsymbol{r}\)，半径为 \(a t\) 的球体；\(\mathrm{d} V^{\prime}\) 是球体 \(T_{a t}^{r}\) 的体积元。