用向量分析推導(dǎo)歐拉恒等式與復(fù)變?nèi)呛瘮?shù)

2021-04-26 00:40 作者:中國大黃鴨鴨 0人讀過 | 我要投稿

第一節(jié)　基礎(chǔ)知識

1?微基分

具體內(nèi)容略，2333~~~

【up主】還有，上面正確的寫法應(yīng)該是“積微成著”的“積”，不是“搞基搞比利”的“基”，2333~

覺得看著枯燥？不如買一包薯?xiàng)l吧

矢函數(shù)？？屎函數(shù)？？

第二節(jié)　定義復(fù)變?nèi)呛瘮?shù)

　　在推導(dǎo)復(fù)變?nèi)呛瘮?shù)之前，我們先討論一下三角函數(shù)到底是什么。我想下面這張圖，應(yīng)該已經(jīng)很好地定義了 $%5Csin%20%CE%B8$ 和 $%5Ccos%20%CE%B8$ 。下圖主要依賴于幾何思維，而非代數(shù)思維，而三角函數(shù)的出處就是幾何。因此它對定義復(fù)變?nèi)呛瘮?shù)，掌握了絕對發(fā)言權(quán)，自然也比解析延拓更有發(fā)言權(quán)。

　　根據(jù)基本的微分幾何常識，可知 $%7C%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%7C%3D1$ 恒成立。積分式中的“ $%C2%B1$ ”，是因?yàn)橛?jì)算向量模長時(shí)，會求出 $2$ 個(gè)根，而此處 $%CE%B8$ 可能等于正根，也可能等于負(fù)根。

接下來的內(nèi)容，就是為了檢驗(yàn)紅字部分是否恒成立。

聽說鴨子的陰莖是螺旋形的，貌似可以用三角函數(shù)描述

第三節(jié)　證明圖中 $%5Cvec%20r%20%5Cleft(%201%20%5Cright)%3D%5Cleft(%20%5Ccos%20%CE%B8%20%2C%20%5Csin%20%CE%B8%20%5Cright)$

　　注：本節(jié)暫時(shí)只考慮 $%CE%B8%E2%88%88%5Cmathbb%20R$ 的情況。復(fù)變?nèi)呛瘮?shù)的推導(dǎo)過程在第四節(jié)。

$%E2%88%B5%5Cfrac%7Bd%20%5Cvec%20r%7D%7Bdt%7D%3D%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r$

$%E2%88%B4%5Cfrac%7Bd%5E2%20%5Cvec%20r%7D%7Bdt%5E2%7D%3D%CE%B8%5E2%20%5Cvec%20k%E2%88%A7%20%5Cleft(%20k%E2%88%A7%5Cvec%20r%20%5Cright)%3D%CE%B8%5E2%20%5Cleft(%20%5Cleft%3C%5Cvec%20r%2C%5Cvec%20k%20%5Cright%3E%20%5Cvec%20k%20-%5Cleft%3C%5Cvec%20k%2C%5Cvec%20k%20%5Cright%3E%20%5Cvec%20r%20%5Cright)%3D-%CE%B8%5E2%20%5Cvec%20r$

　　記 $y%3D%5Cvec%20r_n%2Cn%E2%88%88%5Cmathbb%20Z_%2B$ ，則：

$%5Cfrac%7Bd%5E2%20y%7D%7Bdt%5E2%7D%3D-%CE%B8%5E2y%5C%20%E2%91%B4%5C%20%5C%20$

　　此時(shí)定義 $p%3Dp(t)%3D%5Cfrac%7Bdy%7D%7Bdt%7D%5C%20%E2%91%B5%5C%20%5C%20$ ，則

$%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7Bdp%7D%7Bdx%7D%3D%5Cfrac%7Bdp%7D%7Bdy%7D%5Cfrac%7Bdy%7D%7Bdx%7D%3Dp%5Cfrac%7Bdp%7D%7Bdy%7D%5C%20%E2%91%B6%5C%20%5C%20$

　　將 $%5C%20%E2%91%B6%5C%20%5C%20$ 代入 $%5C%20%20%E2%91%B4%5C%20%5C%20$ ，得

$p%5Cfrac%7Bdp%7D%7Bdy%7D%3D-%CE%B8%5E2y$

　　等式兩邊乘 $dy$ 并求積分，得

$%5Cint%20pdp%3D-%CE%B8%5E2%5Cint%20ydy$

　　化簡，得

$p%5E2%3DC_1%5E2-%CE%B8%5E2y%5E2$

　　上式開平方并求倒數(shù)，得

$%5Cfrac%201p%3D%5Cfrac%201%7B%5Csqrt%7BC_1%5E2-%CE%B8%5E2y%5E2%7D%7D$

　　將 $%5C%20%E2%91%B5%5C%20%5C%20$ 代入，得

$%5Cfrac%20%7Bdt%7D%7Bdy%7D%3D%5Cfrac%201%7B%5Csqrt%7BC_1%5E2-%CE%B8%5E2y%5E2%7D%7D$

　　等式兩邊乘 $dy$ 并求積分，得

$%5Cint%20dt%20%3D%20%5Cint%20%5Cfrac%20%7Bdy%7D%7B%5Csqrt%7BC_1%5E2-%CE%B8%5E2y%5E2%7D%7D%3D%5Cfrac%201%7BC_1%7D%5Cint%20%5Cfrac%20%7Bdy%7D%7B%5Csqrt%7B1-%20%5Cleft(%20%5Cfrac%7B%CE%B8y%7D%7BC_1%7D%20%5Cright)%5E2%20%7D%7D$

　　記 $%5Csin%20%CF%89%3D%5Cfrac%7B%CE%B8y%7D%7BC_1%7D$ ，于是 $%5Ccos%20%CF%89%3D%5Csqrt%7B1-%20%5Cleft(%20%5Cfrac%7B%CE%B8y%7D%7BC_1%7D%20%5Cright)%5E2%20%7D$ 。則：

$C_1t%20%3D%5Cint%20%5Cfrac%20%7Bd%20%5Csin%20%CF%89%7D%7B%5Ccos%20%CF%89%7D%20%3D%20%5Cint%20%5Cfrac%20%7B%5Ccos%20%CF%89%7D%7B%5Ccos%20%CF%89%7Dd%CF%89%3D%CF%89%2BC_2%3D%20%5Csin%5E%7B-1%7D%20%5Cfrac%7B%CE%B8y%7D%7BC_1%7D%20%2BC_2$

$%E2%88%B4y%3D%20%5Cfrac%20%7BC_1%20%5Csin%20%5Cleft(%20C_1t-C_2%20%5Cright)%7D%CE%B8$

小心，別把西瓜汁蹭到衣服上~~

　　此時(shí)可以基本確定：

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Cfrac%20%7BC_1%20%5Csin%20%5Cleft(%20C_1t-C_2%20%5Cright)%7D%CE%B8%2C%5Cfrac%20%7BC_3%20%5Csin%20%5Cleft(%20C_3%20t-C_4%20%5Cright)%7D%CE%B8%20%5Cright%5D%5C%20%E2%91%B7%5C%20%5C%20$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Cfrac%20%7BC_1%5E2%20%5Ccos%20%5Cleft(%20C_1%20t-C_2%20%5Cright)%20%7D%CE%B8%2C%5Cfrac%20%7BC_3%5E2%20%5Ccos%20%5Cleft(%20C_3%20t-C_4%20%5Cright)%20%7D%CE%B8%20%5Cright%5D$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5C%7B%20%5Cfrac%20%7BC_1%5E2%20%5Csin%20%5Cleft%5B%2090%C2%B0-%20%5Cleft(%20C_1t-C_2%20%5Cright)%20%5Cright%5D%7D%CE%B8%2C%5Cfrac%20%7BC_3%5E2%20%5Csin%20%5Cleft%5B%2090%C2%B0-%20%5Cleft(%20C_3t-C_4%20%5Cright)%20%5Cright%5D%20%7D%CE%B8%20%5Cright%5C%7D$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Cfrac%20%7BC_1%5E2%20%5Csin%20%5Cleft(%2090%C2%B0%2BC_2-C_1t%20%20%5Cright)%20%7D%CE%B8%2C%5Cfrac%20%7BC_3%5E2%20%5Csin%20%5Cleft(%2090%C2%B0%2BC_4-C_3t%20%5Cright)%20%7D%CE%B8%20%5Cright%5D%5C%20%E2%91%B8%5C%20%5C%20$

$%E2%88%B5%5Cvec%20r%E2%80%99%20%5Cleft(%20t%20%5Cright)%20%3D%20%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r%20%5Cleft(%20t%20%5Cright)$

$%E2%88%B4%5Cvec%20r%E2%80%99%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20-C_3%20%5Csin%20%5Cleft(%20C_3t-C_4%20%5Cright)%20%EF%BC%8CC_1%20%5Csin%20%5Cleft(%20C_1t-C_2%20%5Cright)%20%5Cright%5D%5C%20%E2%91%B9%5C%20%5C%20$

　　由 $%5C%20%E2%91%B8%5C%20%5C%20$ 與 $%5C%20%E2%91%B9%5C%20%5C%20$ ，得

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_1%3D-C_3%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%26%20%E2%91%BA%5C%5C%0AC_1%3D%5C%20%5C%20%5C%20%5Cfrac%7BC_3%5E2%7D%CE%B8%20%26%20%E2%91%BB%5C%5C%0AC_3%3D-%5Cfrac%7BC_1%5E2%7D%CE%B8%20%20%20%20%20%20%26%20%E2%91%BC%5C%5C%0AC_4%3D-C_2-90%C2%B0%20%20%20%20%20%20%20%20%20%20%20%20%20%20%26%20%E2%91%BD%5C%20%5C%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　將 $%5C%20%E2%91%BA%5C%20%5C%20$ 代入 $%5C%20%E2%91%BB%5C%20%5C%20$ 與 $%5C%20%E2%91%BC%5C%20%5C%20$ ，得

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_3%3D-%5Cfrac%7BC_3%5E2%7D%CE%B8%20%5C%5C%0AC_1%3D%5C%20%5C%20%5C%20%5Cfrac%7BC_1%5E2%7D%CE%B8%20%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　約去公因式，得

$%5C%20%E2%91%BE%5C%20%5C%20$ $%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_1%3D%5C%20%5C%20%5C%20%CE%B8%20%5C%5C%0AC_3%3D-%CE%B8%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　將 $%5C%20%E2%91%BD%5C%20%5C%20$ 和 $%5C%20%E2%91%BE%5C%20%5C%20$ 代入 $%5C%20%E2%91%B7%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Csin%20%5Cleft(%20%CE%B8t-C_2%20%5Cright)%2C%5Csin%20%5Cleft(%20%CE%B8t%2BC_2%2B90%C2%B0%20%5Cright)%20%5Cright%5D%5C%20%E2%91%BF%5C%20%5C%20$

$%E2%88%B5%5Cvec%20r%20%5Cleft(%200%20%5Cright)%20%3D%20%5Cvec%20i$

　　記 $%E2%88%80k%2C%E2%88%80l%20%E2%88%88%20%5Cmathbb%20Z$ ，則

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0A0%CE%B8-C_2%20%20%20%20%20%20%20%20%26%20%3D90%C2%B0(4k%2B1)%20%20%5C%5C%0A0%CE%B8%2BC_2%2B90%C2%B0%20%26%20%3D%2090%C2%B0(4l%2B0)%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　化簡，得

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_2%20%3D%2090%C2%B0%20%5Cleft(%20-4k-1%20%5Cright)%20%5C%5C%0AC_2%20%3D%2090%C2%B0%20%5Cleft(%20%5C%20%5C%20%5C%204l%5C%20-1%20%5Cright)%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cright.$

　　代入 $%5C%20%E2%91%BF%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft%5B%20%5Csin%20%5Cleft(%20%CE%B8t%2B90%C2%B0%20%5Cright)%2C-%5Csin%20%5Cleft(%20-%CE%B8t%20%5Cright)%20%5Cright%5D$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cleft(%20%5Ccos%20%CE%B8t%2C%5Csin%20%CE%B8t%20%5Cright)%5C%20%E2%92%80%5C%20%5C%20$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%201%20%5Cright)%20%3D%20%5Cleft(%20%5Ccos%20%CE%B8%2C%5Csin%20%CE%B8%20%5Cright)$

　　另外，這個(gè)結(jié)論也可以驗(yàn)證——將 $%5C%20%E2%92%80%5C%20%5C%20$ 代入圖中的向量微分方程即可。

第四節(jié) 復(fù)變?nèi)呛瘮?shù)如何計(jì)算？

　　將圖中的 $%CE%B8$ 換成 $i%CE%B8%E2%88%88%5Cmathbb%20I$ ，可得

$%5Cfrac%7Bd%20%5Cvec%20r%7D%7Bdt%7D%3Di%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r$

$%E2%88%B4%5Cfrac%7Bd%5E2%20%5Cvec%20r%7D%7Bdt%5E2%7D%3D%CE%B8%5E2%20%5Cvec%20r$

$%E2%88%B4%5Cfrac%7Bd%5Cvec%20r%7D%7Bdt%7D%3D%C2%B1%CE%B8%20%5Cvec%20r$

　　記 $y%3D%5Cvec%20r_n%2Cn%E2%88%88%5Cmathbb%20Z_%2B$ ，則：

$%5Cfrac%7Bdy%7D%7Bdt%7D%3D%C2%B1%CE%B8y$

$%E2%88%B4%5Cfrac%7Bdt%7Dy%5Cfrac%7Bdy%7D%7Bdt%7D%3D%C2%B1%5Cfrac%7Bdt%7Dy%CE%B8y$

$%E2%88%B4%5Cfrac%7Bdy%7Dy%3D%C2%B1%CE%B8dt$

$%E2%88%B4%5Cint%20%5Cfrac%7Bdy%7Dy%3D%C2%B1%CE%B8%5Cint%20dt$

$%E2%88%B4%5Cln%20y%3D%C2%B1%CE%B8t%2BC_0$

$%E2%88%B4y_1%3DC_1e%5E%7B%CE%B8t%7D%2Cy_2%3DC_2e%5E%7B-%CE%B8t%7D$

　　根據(jù)二階常系數(shù)線性齊次微分方程的通解規(guī)則，由于 $y_1%2Cy_2$ 線性無關(guān)，因此 $y%3DAy_1%2BBy_2$ ，即：

$y%3DC_1e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3D%5Cleft(%20C_1e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D%2C%20C_3e%5E%7B%CE%B8t%7D%2BC_4e%5E%7B-%CE%B8t%7D%20%5Cright)$

$%E2%88%B4%5Cvec%20r'%20%5Cleft(%20t%20%5Cright)%3D%CE%B8%5Cleft(%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%2C%20C_3e%5E%7B%CE%B8t%7D-C_4e%5E%7B-%CE%B8t%7D%20%5Cright)%5C%20%E2%92%81%5C%20%5C%20$

$%E2%88%B5%5Cvec%20r%E2%80%99%20%5Cleft(%20t%20%5Cright)%20%3D%20i%CE%B8%5Cvec%20k%E2%88%A7%5Cvec%20r%20%5Cleft(%20t%20%5Cright)$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%20%3D%20%5Cfrac%20i%CE%B8%20%5Cvec%20k%E2%88%A7%5Cvec%20r%E2%80%99$

　　由 $%5C%20%E2%92%81%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3Di%5Cleft(%20-C_3e%5E%7B%CE%B8t%7D%2BC_4e%5E%7B-%CE%B8t%7D%20%2C%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%2C%20%5Cright)$

$%E2%88%B4%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3D%5Cleft%5B%20C_1e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D%2C%20i%20%5Cleft(%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%20%5Cright)%20%5Cright%5D%5C%20%E2%92%82%5C%20%5C%20$

你正在吃的是奧利奧還是丹麥曲奇？

　　再根據(jù) $%5C%20%E2%92%82%5C%20%5C%20$ 可得：

$%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7B1%7D%0AC_1%2BC_2%3D1%5C%5C%0Ai%5Cleft(C_1-C_2%20%5Cright)%3D0%0A%5Cend%7Barray%7D%0A%5Cright.$

　　解得：

$C_1%3DC_2%3D%5Cfrac%2012$

　　代入 $%5C%20%E2%92%82%5C%20%5C%20$ ，得

$%5Cvec%20r%20%5Cleft(%20t%20%5Cright)%3D%5Cleft%5B%20%5Cfrac%2012%20%5Cleft(%20e%5E%7B%CE%B8t%7D%2BC_2e%5E%7B-%CE%B8t%7D%20%5Cright)%2C%20%5Cfrac%20i2%20%5Cleft(%20C_1e%5E%7B%CE%B8t%7D-C_2e%5E%7B-%CE%B8t%7D%20%5Cright)%20%5Cright%5D$