#! https://zhuanlan.zhihu.com/p/627068881
积分变换法
笔记源代码(https://github.com/cjyyx/notes/tree/main/%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0/%E6%95%B0%E7%90%86%E6%96%B9%E6%B3%95)
前置知识
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 \(若 $f_{1}(t) \fallingdotseq \bar{f}_{1}(p), f_{2}(t) \fallingdotseq \bar{f}_{2}(p)$ ，则\)
\[f_{1}(t) * f_{2}(t) \fallingdotseq \bar{f}_{1}(p) \bar{f}_{2}(p)\]
误差函数
误差函数
\[\mathrm{erf} (x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} \mathrm{e}^{-z^{2}} \mathrm{~d} z\]余误差函数
\[\mathrm{erfc} (x) = 1 - \mathrm{erf} (x)\]二阶非齐次线性方程的通解
有二阶非齐次线性方程
\[y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{2}(x) y=f(x)\] \[\text{设 } y_{1}(x), y_{2}(x) \text{ 是对应的二阶齐次线性方程}\] \[y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{2}(x) y=0\]的基本解组，则该二阶非齐次线性方程的通解为
\[y(x)=c_{1} y_{1}(x)+c_{2} y_{2}(x)+\int_{x_{0}}^{x} \frac{y_{1}(x)\left(-y_{2}(\xi)\right)+y_{2}(x) y_{1}(\xi)}{y_{1}(\xi) y_{2}^{\prime}(\xi)-y_{2}(\xi) y_{1}^{\prime}(\xi)} f(\xi) d \xi\]傅里叶变换法
对于无界空间的定解问题，傅里叶变换是一种很适用的求解方法。
无限长弦的自由振动
\[\left\{\begin{aligned} & u_{t}-a^{2} u_{x x}=0 \quad (-\infty<x<\infty) \\ & \left.u\right|_{t=0}=\varphi(x),\quad \left.u_{t}\right|_{t=0}=\psi(x) \end{aligned}\right.\]傅里叶变换
\[\left\{\begin{aligned} & U^{\prime \prime}+k^{2} a^{2} U=0 \\ & \left.U\right|_{t=0}=\Phi(k),\left.\quad U^{\prime}\right|_{t=0}=\Psi(k) \end{aligned}\right.\]解为
\[\begin{aligned} U(t, k)= & \frac{1}{2} \Phi(k) \mathrm{e}^{\mathrm{i} k a t}+\frac{1}{2 a} \frac{1}{\mathrm{i} k} \Psi(k) \mathrm{e}^{\mathrm{i} k a t} \\ & +\frac{1}{2} \Phi(k) \mathrm{e}^{-\mathrm{i} k a t}-\frac{1}{2 a} \frac{1}{\mathrm{i} k} \Psi(k) \mathrm{e}^{-\mathrm{i} k a t} \end{aligned}\]逆傅里叶变换，得达朗贝尔公式
\[u(x, t)=\frac{1}{2}[\varphi(x+a t)+\varphi(x-a t)]+\frac{1}{2 a} \int_{x-a t}^{x+a t} \psi(\xi) \mathrm{d} \xi\]无限长细杆的热传导问题
\[\left\{\begin{aligned} & u_{t}-a^{2} u_{x x}=0 \quad(-\infty<x<\infty) \\ & \left.u\right|_{t=0}=\varphi(x) \end{aligned}\right.\]傅里叶变换
\[\left\{\begin{aligned} & U^{\prime}+k^{2} a^{2} U=0 \\ & \left.U\right|_{t=0}=\Phi(k) \end{aligned}\right.\]解为
\[U(t, k)=\Phi(k) \mathrm{e}^{-k^{2} a^{2} t}\]逆傅里叶变换，得
\[u(x, t)=\int_{-\infty}^{\infty} \varphi(\xi)\left[\dfrac{1}{2 a \sqrt{\pi t}} \exp\{-\dfrac{(x-\xi)^{2}}{4 a^{2} t}\}\right] \mathrm{d} \xi\]三维无界空间的自由振动
\[\left\{\begin{aligned} & u_{t}-a^{2} \Delta_{3} u=0 \\ & \left.u\right|_{t=0}=\varphi(\boldsymbol{r}),\quad \left.u_{t}\right|_{t=0}=\psi(\boldsymbol{r}) \end{aligned}\right.\]其解为泊松公式
\[u(\boldsymbol{r}, t)=\frac{1}{4 \pi a} \frac{\partial}{\partial t} \iint_{S_{a t}^{r}} \frac{\varphi\left(\boldsymbol{r}^{\prime}\right)}{a t} \mathrm{~d} S^{\prime}+\frac{1}{4 \pi a} \iint_{S_{a t}^{r}} \frac{\psi\left(\boldsymbol{r}^{\prime}\right)}{a t} \mathrm{~d} S^{\prime}\] \[\text{其中 } S_{a t}^{r} \text{ 是球心为 } \boldsymbol{r} \text{，半径为 } a t \text{ 的球面；} \mathrm{d} S^{\prime} \text{ 是球面 } S_{a t}^{r} \text{ 的面积元。}\]三维无界空间的受迫振动
\[\left\{\begin{aligned} & u_{t t}-a^{2} \Delta_{3} u=f(\boldsymbol{r}, t) \\ & \left.u\right|_{t=0}=0,\quad \left.u_{t}\right|_{t=0}=0 \end{aligned}\right.\]其解为推迟势
\[u(\boldsymbol{r}, t)=\frac{1}{4 \pi a^{2}} \iiint_{T_{a t}^{r}} \frac{f\left(\boldsymbol{r}^{\prime}, t-\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right| / a\right)}{\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right|} \mathrm{d} V^{\prime}\] \[\text{其中 } T_{a t}^{r} \text{ 是球心为 } \boldsymbol{r} \text{，半径为 } a t \text{ 的球体；} \mathrm{d} V^{\prime} \text{ 是球体 } T_{a t}^{r} \text{ 的体积元。}\]二维无界空间的自由振动
\[\left\{\begin{aligned} & u_{t}-a^{2} \Delta_{2} u=0 \\ & \left.u\right|_{t=0}=\varphi(x, y),\quad \left.u_{t}\right|_{t=0}=\psi(x, y) \end{aligned}\right.\] \[\text{二维空间中的波动其实还是在三维空间中传播的波动，只是该波动跟坐标 } z \text{ 无关而已。} \\ \text{因此三维波动的泊松公式，消除了坐标 } z \text{，就成为二维波动的公式，这称为降维法。}\] \[\text{对于二维问题，球面 } S_{a t}^{r} \text{ 上的积分应代之以 } xy \text{ 平面的圆 } \Sigma_{a t}^{x, y} \text{ 上的积分。}\]
泊松公式在二维问题中成为
\[\begin{aligned} u(x, y, t)= & \frac{1}{2 \pi a} \frac{\partial}{\partial t} \iint_{\Sigma_{a t}^{x, y}} \frac{\varphi\left(x^{\prime}, y^{\prime}\right)}{\sqrt{a^{2} t^{2}-\left(x^{\prime}-x\right)^{2}-\left(y^{\prime}-y\right)^{2}}} \mathrm{~d} x^{\prime} \mathrm{d} y^{\prime} \\ & +\frac{1}{2 \pi a} \iint_{\Sigma_{a t}^{x, y}} \frac{\psi\left(x^{\prime}, y^{\prime}\right)}{\sqrt{a^{2} t^{2}-\left(x^{\prime}-x\right)^{2}-\left(y^{\prime}-y\right)^{2}}} \mathrm{~d} x^{\prime} \mathrm{d} y^{\prime} \end{aligned}\]拉普拉斯变换法
拉普拉斯变换法方法适于求解初值问题，不管方程及边界条件是否为齐次的。