46206 Operational Calculus的教學與學習分享
Operational Calculus的教學與學習分享

### 1. 傅立葉轉換基本性質探討

1.1. 暫態 (Transient state) 的 $\cos(t)$ 與 $\sin(t)$ 函數在 $t\in(0,\pi)$ 才有值的微分轉換關係是否成立？

\begin{align} f(t)&=\left\{\begin{array}{ccl} \cos(t),&&0\le t\le \pi\\[5pt] 0,&&\hbox{otherwise,} \end{array}\right.\label{1} \end{align} \begin{align} g(t)&=\left\{\begin{array}{ccl} \sin(t),&&0\le t\le \pi\\[5pt] 0,&&\hbox{otherwise,} \end{array}\right.\label{2} \end{align}

$f(t)$ 和 $g(t)$ 圖形分別如圖一與圖二所示。

\begin{align} F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t} dt=\frac{i\omega(e^{-i\omega \pi}+1)}{(1-\omega^2)}.\label{3} \end{align}

\begin{align} G(\omega)=\int_{-\infty}^{\infty}g(t)e^{-i\omega t} dt=\frac{(e^{-i\omega \pi}+1)}{(1-\omega^2)}.\label{4} \end{align}

\begin{align} g'(t)&=f(t),\label{5}\\ f'(t)&=-g(t).\label{6} \end{align}

\begin{align} {\cal F}\{g'(t)\}&={\cal F}\{f(t)\},\label{7}\\ {\cal F}\{f'(t)\}&={\cal F}\{-g(t)\}.\label{8} \end{align}

\begin{align} i\omega G(\omega)&=F(\omega),\label{9}\\ \hbox{與 } i\omega F(\omega)&=-G(\omega).\label{10} \end{align}

\begin{align} i\omega G(\omega)&=i\omega\frac{(e^{-i\omega \pi}+1)}{(1-\omega^2)}=\frac{i\omega(e^{-i\omega \pi}+1)}{(1-\omega^2)}=F(\omega),\label{11}\\ i\omega F(\omega)&=-\omega^2\frac{(e^{-i\omega \pi}+1)}{(1-\omega^2)}\not=\frac{(e^{-i\omega \pi}+1)}{(1-\omega^2)}=-G(\omega).\label{12} \end{align}

$f(t)$ 經由微分後可得

\begin{align} f'(t)=(\delta(t)-\delta(t-\pi))\cos(t)-g(t),\label{13} \end{align}

$g(t)$ 經由微分後可得

\begin{align} g'(t)=(\delta(t)-\delta(t-\pi))\sin(t)+f(t),\label{14} \end{align}

\begin{align} i\omega F(\omega)&=\int_{-\infty}^\infty (\delta(t)-\delta(t-\pi))\cos(t)e^{-i\omega t} dt-G(\omega),\label{15}\\ \hbox{與 } i\omega G(\omega)&=\int_{-\infty}^\infty (\delta(t)-\delta(t-\pi))\sin(t)e^{-i\omega t} dt+F(\omega).\label{16} \end{align}

\begin{align} \int_{-\infty}^\infty \delta(t-a)k(t)dt=k(a),\label{17} \end{align}

\begin{align} i\omega G(\omega)=\sin(0)e^{-i\omega 0}-\sin(\pi)e^{-i\omega \pi}+F(\omega),\label{18} \end{align}

\begin{align} i\omega G(\omega)=F(\omega),\label{19} \end{align}

\begin{align} i\omega F(\omega)=\cos(0)e^{-i\omega 0}-\cos(\pi)e^{-i\omega \pi}-G(\omega).\label{20} \end{align}

\begin{align} i\omega F(\omega)=1-(-1)e^{-i\omega \pi}-G(\omega).\label{21} \end{align}

\begin{align} g'(t)=f(t),\quad -\infty \lt t \lt \infty\label{22} \end{align}

\begin{align} f'(t)\not=-g(t),\quad -\infty \lt t \lt \infty.\label{23} \end{align}

1.2. 簡諧函數 $\cos(t)$ 可以看成是 $\sin(t)$ 的微分, 也可以看成 $-\sin(t-\frac\pi 2)$, 其所對應的傅立葉轉換是否相同, 並做探討。

\begin{align} f_c(t)=&\,\cos(t),\quad -\infty \lt t \lt \infty,\label{24}\\ g_s(t)=&\,\sin(t),\quad -\infty \lt t \lt \infty.\label{25} \end{align}

$f_c(t)$ 和 $g_s(t)$ 函數圖分別如圖三與圖四所示。

\begin{align} \int_{-\infty}^\infty \delta(t)e^{-i\omega t} dt=1,\label{26} \end{align}

1 的傅立葉反轉換等於 $\delta(t)$, 即

\begin{align} \frac 1{2\pi}\int_{-\infty}^\infty 1e^{i\omega t}d\omega=\delta(t).\label{27} \end{align}

\begin{align} \int_{-\infty}^\infty 1e^{-i\omega \tau} d\omega=2\pi\delta(-\tau)=2\pi \delta(\tau).\label{28} \end{align}

\eqref{28} 式可改寫成

\begin{align} \int_{-\infty}^\infty 1e^{-i\omega t} dt=2\pi\delta(\omega),\label{29} \end{align}

\begin{align} \int_{-\infty}^\infty e^{it}1e^{-i\omega t} dt=2\pi\delta(\omega-1),\label{30} \end{align}

\begin{align} \int_{-\infty}^\infty e^{-it}1e^{-i\omega t} dt=2\pi\delta(\omega+1).\label{31} \end{align}

\begin{align} F_c(\omega)=&\,{\cal F}\{f_c(t)\}=\int_{-\infty}^\infty \dfrac{e^{it}+e^{-it}}{2}e^{-i\omega t} dt=\pi\big(\delta(\omega-1)+\delta(\omega+1)\big),\label{32}\\ \hbox{與 } G_s(\omega)=&\,{\cal F}\{g_s(t)\}=\int_{-\infty}^\infty \dfrac{e^{it}-e^{-it}}{2i}e^{-i\omega t} dt=\frac \pi i\big(\delta(\omega-1)-\delta(\omega+1)\big).\label{33} \end{align}

\begin{align} F_c(\omega)=&\,\pi\big(\delta(\omega-1)+\delta(\omega+1)\big),\label{34}\\ \hbox{與 } G_s(\omega)=&\,\frac \pi i\big(\delta(\omega-1)-\delta(\omega+1)\big).\label{35} \end{align}

\begin{align} {\cal F}\Big\{\frac{d\sin(t)}{dt}\Big\}=&\,\omega\pi\big(\delta(\omega-1)-\delta(\omega+1)\big),\label{36}\\ \hbox{與 } {\cal F}\Big\{-\sin\Big(t-\frac \pi 2\Big)\Big\}=&\,-e^{i\omega \frac \pi 2}\frac \pi i\big(\delta(\omega-1)-\delta(\omega+1)\big).\label{37} \end{align}

\begin{align} {\cal F}\Big\{-\sin\Big(t-\frac \pi 2\Big)\Big\}=\Big(i\cos(\omega\frac \pi 2)+\sin(\omega \frac\pi 2)\Big)\pi\big(\delta(\omega-1)- \delta(\omega+1)\big).\label{38} \end{align}

1.3. 高斯函數 $e^{-t^2}$ 的傅立葉轉換與機率密度分布函數探討

\begin{align} p(t)=&\,te^{-t^2},\label{39}\\ q(t)=&\,e^{-t^2},\label{40} \end{align}

$p(t)$ 與 $q(t)$ 兩函數在時間域存在如下關係：

\begin{align} tq(t)=&\,p(t),\label{41}\\ q'(t)=&\,-2p(t).\label{42} \end{align}

\begin{align} iQ'(\omega)=&\,P(\omega),\label{43}\\ i\omega Q(\omega)=&\,-2P(\omega),\label{44} \end{align}

 轉換前 ($t$ 域) 轉換後 ($\omega$ 域) $tq(t)=p(t)$ $iQ'(\omega)=P(\omega)$ $q'(t)=-2p(t)$ $i\omega Q(\omega)=-2P(\omega)$

\begin{align} Q'(\omega)=&\,\frac {-1}2\omega Q(\omega),\label{45}\\ \hbox{解得 $Q(\omega)$} Q(\omega)=&\,ce^{\frac{-1}{4}\omega^2}\qquad \hbox{($c$ 為常數)}.\label{46} \end{align}

\begin{align} Q(\omega)=\int_{-\infty}^\infty e^{-t^2}e^{-i\omega t} dt.\label{47} \end{align}

\begin{align} c=\sqrt{\pi},\label{48} \end{align}

\begin{align} Q(\omega)=\sqrt{\pi}e^{-\frac 14\omega^2}.\label{49} \end{align}

\begin{align} {\cal F}\bigg\{\frac 1{\sqrt{2\pi}\sigma}e^{-\frac{(t-\mu)^2}{2\sigma^2}}\bigg\}=e^{-\mu i\omega}e^{\frac{-\omega^2\sigma^2}{2}},\label{50} \end{align}

\begin{align} f(t)=\frac 1{\sqrt{2\pi}\sigma}e^{-\frac{(t-\mu)^2}{2\sigma^2}}.\label{51} \end{align}

\begin{align} F(\omega)=\int_{-\infty}^\infty \frac 1{\sqrt{2\pi}\sigma}e^{-\frac{(t-\mu)^2}{2\sigma^2}}e^{-i\omega t} dt= \frac 1{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{-\big(\frac{(t-\mu)^2}{2\sigma^2}+i\omega t\big)}dt.\label{52} \end{align}

\begin{align} u=\frac{t-\mu}{\sqrt 2\sigma},\qquad du=\frac 1{\sqrt 2\sigma}dt,\label{53} \end{align}

\begin{align} F(\omega)=\frac 1{\sqrt{\pi}}e^{-\mu i\omega}\int_{-\infty}^\infty e^{-(u^2+i\omega \sqrt{2}\sigma u)} du,\label{54} \end{align}

\begin{align} F(\omega)=\frac 1{\sqrt{\pi}}e^{-\mu i\omega}e^{\frac{-\omega^2\sigma^2}2} \int_{-\infty}^\infty e^{-\big(u+\frac{i\omega}{\sqrt 2}\sigma\big)^2}du.\label{55} \end{align}

\begin{align} \int_{-\infty}^\infty e^{-\big(u+\frac{i\omega}{\sqrt 2}\sigma\big)^2}du=\int_{-\infty}^\infty e^{-v^2}dv=\sqrt{\pi}\label{56} \end{align}

\begin{align} \int_{L_1}f(z)dz=\int_{-R_2}^{R_1} e^{-u^2}du.\label{57} \end{align}

\begin{align} \int_{L_3}f(z)dz=\int_{R_1}^{-R_2} e^{-\big(u+\frac{i\omega}{\sqrt 2}\sigma\big)^2}du.\label{58} \end{align}

\begin{align} \int_{L_2}f(z)dz=\int_0^1 e^{-\big(R_1+\frac{i\omega t}{\sqrt 2}\sigma\big)^2}dt=\int_0^1 e^{-R_1^2}e^{\frac{\omega^2 t^2\sigma^2}2} e^{-i\sqrt 2 R_1\omega t}dt,\label{59} \end{align}

\begin{align} |f(z)|=e^{-R_1^2}e^{\frac{\omega^2 t^2\sigma^2}2}\le e^{-R_1^2}e^{\frac{\omega^2 \sigma^2}2}=M_R. \label{60} \end{align}

\begin{align} \left|\int_c f(z)dz\right|\le ML,\label{61} \end{align}

\begin{align} \left|\int_{L_2} f(z)dz\right|\le \frac{i\omega}{\sqrt 2}\sigma\cdot e^{-R_1^2}e^{\frac{\omega^2 \sigma^2}2}.\label{62} \end{align}

\begin{align} \lim_{R_1\to\infty} \frac{i\omega}{\sqrt 2}\sigma\cdot e^{-R_1^2}e^{\frac{\omega^2 \sigma^2}2}=0,\label{63} \end{align}

\begin{align} \lim_{R_1\to\infty} \int_{L_2}f(z)dz=0.\label{64} \end{align}

\begin{align} \lim_{R_2\to\infty} \int_{L_4}f(z)dz=0.\label{65} \end{align}

\begin{align} \int_{-\infty}^\infty e^{-u^2}du=\int_{-\infty}^\infty e^{-\big(u+\frac{i\omega}{\sqrt 2}\sigma\big)^2}du.\label{66} \end{align}

\begin{align} F(\omega)=\frac{1}{\sqrt\pi}e^{-\mu i\omega}e^{\frac{-\omega^2\sigma^2}{2}}\int_{-\infty}^\infty e^{-u^2}du.\label{67} \end{align}

\begin{align} F(\omega)=e^{-\mu i\omega}e^{\frac{-\omega^2\sigma^2}{2}}.\label{68} \end{align}

\begin{align} F(\omega)=\,&e^{\frac{\mu^2}{2\sigma^2}}e^{-\big(\frac{\omega\sigma}{\sqrt 2}+\frac{\mu i}{\sqrt 2\sigma}\big)^2}.\label{69}\\ \hbox{再做整理, 可得 } F(\omega)=\,&e^{\frac{\mu^2}{2\sigma^2}}e^{-\frac{(\omega-(-\mu i/\sigma^2))^2}{2(1/\sigma)^2}}.\label{70} \end{align}

\begin{align} F(\omega)=\frac{\sqrt{2\pi}}{\sigma} e^{\frac{\mu^2}{2\sigma^2}}\cdot\frac{1}{\sqrt{2\pi}(1/\sigma)} e^{-\frac{(\omega-(-\mu i/\sigma^2))^2}{2(1/\sigma)^2}}=\frac{\sqrt{2\pi}}{\sigma} e^{\frac{\mu^2}{2\sigma^2}}N\Big(-\frac{\mu i}{\sigma^2},\frac 1{\sigma}\Big); \label{71} \end{align}

\begin{align} F(\omega)=\frac{\sqrt{2\pi}}{\sigma} e^{\frac{\mu^2}{2\sigma^2}}i\cdot\frac{1}{\sqrt{2\pi}(i/\sigma)} e^{\frac{(\omega-(-\mu i/\sigma^2))^2}{2(i/\sigma)^2}} =\frac{\sqrt{2\pi}}{\sigma} e^{\frac{\mu^2}{2\sigma^2}}i\cdot \bar N\Big(\frac{-\mu}{\sigma^2}i,\frac 1{\sigma}i\Big), \label{72} \end{align}

\begin{align} \bar N(\bar \mu,\bar\sigma)=\frac{1}{\sqrt{2\pi}\bar\sigma} e^{\frac{(\omega-\bar\mu)^2}{2\bar\sigma^2}};\label{73} \end{align}

 零次矩 一次矩 二次矩 原始函數 $f_0(t)$ $\int_{-\infty}^{\infty}f_0(t)dt=1$ $\int_{-\infty}^{\infty}tf_0(t)dt=0$ $\int_{-\infty}^{\infty}t^2f_0(t)dt=\sigma^2$ 一次轉換 $f_1(t)$ $\int_{-\infty}^{\infty}f_1(t)dt=\frac{\sqrt{2\pi}}{\sigma}$ $\int_{-\infty}^{\infty}tf_1(t)dt=0$ $\int_{-\infty}^{\infty}t^2f_1(t)dt=\frac{\sqrt{2\pi}}{\sigma^3}$ 二次轉換 $f_2(t)$ $\int_{-\infty}^{\infty}f_2(t)dt=2\pi$ $\int_{-\infty}^{\infty}tf_2(t)dt=0$ $\int_{-\infty}^{\infty}t^2f_2(t)dt=2\pi\sigma^2$

### 2. 拉氏轉換基本性質探討

#### 2.1. $t^n$ 的拉氏轉換

\begin{align} p(t)=\,&t^{n-1},\label{74}\\ q(t)=\,&t^{n}.\label{75} \end{align}

\begin{align} {\cal L}\{t^n\}=\,&Q_n (s)=\frac{-1}n\cdot \big(Q_n (s)+s\cdot Q_n'(s)\big).\label{76} \end{align}

\begin{align} Q_n(s)=\,&k_ns^{-(n+1)}\quad\hbox{($k_n$ 為常數).}\label{77} \end{align}

\begin{align} {\cal L}\{t^{1/2}\}=Q_{1/2} (s)=\int_0^\infty \sqrt{t}\cdot e^{-st} dt,\label{78} \end{align}

\begin{align} Q_{1/2} (1)=k_{1/2}.\label{79} \end{align}

\begin{align} k_{1/2}=\int_0^\infty \sqrt{t}\cdot e^{-t} dt.\label{80} \end{align}

\begin{align} {\cal L}\{t^{-1/2}\}=Q_{(-1/2)}(s)=\int_0^\infty \frac 1{\sqrt{t}}\cdot e^{-st} dt,\label{81} \end{align}

\begin{align} Q_{(-1/2)}(s)=\int_0^\infty \frac 1v\cdot e^{-sv^2}2vdv=2\int_0^\infty e^{-sv^2}dv.\label{82} \end{align}

\begin{align} Q_{(-1/2)}(1)=k_{(-1/2)}.\label{83} \end{align}

\begin{align} k_{(-1/2)}=2\int_0^\infty e^{-v^2} dv=\sqrt{\pi}.\label{84} \end{align}

\begin{align} Q_{(-1/2)} (s)=k_{(-1/2)} s^{-(-1/2+1)}={\sqrt{\frac\pi s}}.\label{85} \end{align}

\begin{align} Q_{1/2} (s)=-\frac d{ds}\Big(\sqrt{\frac{\pi}s}\Big)=\frac{\sqrt{\pi}}{2s^{3/2}}.\label{86} \end{align}

#### 2.2. 尤拉-柯西(Euler-Cauchy)微分方程的兩次拉氏轉換

\begin{align} at^2 y'' (t)+bty' (t)+cy(t)=0\quad\hbox{(其中 $a,b,c$ 為常數)},\label{87} \end{align}

\begin{align} L\{cy(t)\}=cY(s),\label{88} \end{align}

$by' (t)$ 的轉換依基本轉換性質可得

\begin{align} L\{by'(t)\}=bsY(s)-by(0),\label{89} \end{align}

$ay'' (t)$ 轉換依基本轉換性質可得

\begin{align} L\{ay''(t)\}=as^2Y(s)-asy(0)-ay'(0).\label{90} \end{align}

\begin{align} L\{ty\}=-Y'(s)\label{91} \end{align}

\begin{align} L\{at^2y''(t)\}=as^2Y''(s)+4asY'(s)+2aY(s).\label{92} \end{align}

\begin{align} L\{bty'(t)\}=-bsY'(s)+bY(s).\label{93} \end{align}

\begin{align} L\{at^2y''(t)+bty'(t)+cy(t)\}=as^2Y''(s)+(4a-b)sY'(s)+(2a-b+c)Y(s).\label{94} \end{align}

\begin{align} L\{as^2Y''(s)+(4a-b)sY'(s)+(2a-b+c)Y(s)\}=at^2y''(t)+bty'(t)+cy(t).\label{95} \end{align}

### 參考文獻

Doetsch, G., Introduction to the Theory and Application of Laplace Transform, Springer-verlag New York Heidberg Berlin, 1970. Mikusinski, J., Operational Calculus, Pergmon Press, 1958. Yosida, K., Operational Calculus: A Theory of Hyperfunctions, Applied Mathematics Science Vol. 55, Springer, 1984. 林琦焜。 傅立葉分析與應用。 滄海書局, 2010。 陳正宗。 工程數學講義。 海洋大學, 基隆, 2021。 陳正宗。 工程數學教學經驗談。 工程力學與數學創意教學研討會, 台北, 2004。 陳正宗。 工程數學教學拾趣。 數學傳播季刊, 31(4), 18-37, 2007。 Churchill, R. V., Operational Mathematics, New York: McGraw-Hill Book Co., 1958. Lighthill, M.J., An Introduction to Fourier Analysis and Generalised Functions, Cambridge University Press, 1958. Carslaw, H. S. and Jaeger J. C., Operational Methods in Applied Mathematics, Dover Publications, 1963. Kreyszig, E., Advanced Engineering Mathematics, New York: Wiley, 1988. O'Neil, P.V., Advanced Engineering Mathematics, Boston: Thomson, 1995. Riley, K. F., Hobson M.P., and Bence S.J., Mathematical Methods for Physics and Engineering, Cambridge University Press, 2002. https://www.youtube.com/watch?v=8yp640IzOyM (rigorous) https://www.youtube.com/watch?v=_wWk84m4_x0 (not rigorous) https://www.youtube.com/watch?v=iLQ-E0FA85Q&t=1068s (not rigorous) https://www.youtube.com/watch?v=KxiAoDLRyFc (not rigorous) https://www.youtube.com/watch?v=QmjVdjW23VU (NTOU/MSV new approach). 林琦焜。 private communication, 2021。