【銀蛇出品】數(shù)學(xué)漫談13——簡單三角級(jí)數(shù)的和函數(shù)及其在解PDE中的應(yīng)用

2022-08-24 13:27 作者:山舞_銀蛇 0人讀過 | 我要投稿

前置知識(shí)：冪級(jí)數(shù)、Fourier級(jí)數(shù)、復(fù)數(shù)的基本運(yùn)算

前言：在大多數(shù)時(shí)候，我們更關(guān)注一個(gè)函數(shù)的Fourier級(jí)數(shù)，分別去研究基波和諧波分量，而很少關(guān)注給定的一個(gè)Fourier級(jí)數(shù)（或更一般的三角級(jí)數(shù)）將收斂到哪個(gè)函數(shù)上．這其中的一個(gè)原因是求級(jí)數(shù)本身就缺少一般通法，多數(shù)時(shí)候需要根據(jù)已知級(jí)數(shù)配湊或者用特殊方法求得．對(duì)于一類具有比較簡單形式的三角級(jí)數(shù)來說，可通過復(fù)數(shù)運(yùn)算求得其和函數(shù)，算是比較通用的一類方法．本文除考察這些三角級(jí)數(shù)的和函數(shù)，還將考察其在求解線性PDE中的應(yīng)用．

關(guān)鍵內(nèi)容：三角級(jí)數(shù)、和函數(shù)、二階線性PDE

??? 求冪級(jí)數(shù)和和函數(shù)其實(shí)并沒有一般的通法，相反相當(dāng)多的特殊函數(shù)就是通過冪級(jí)數(shù)嚴(yán)格定義的．例如冪級(jí)數(shù)

$%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bx%5En%7D%7Bn!%7D%3D1%2Bx%2B%5Cfrac%7Bx%5E2%7D%7B2%7D%2B%5Cfrac%7Bx%5E3%7D%7B6%7D%2B%5Ccdots$

是微分方程y'=y的一個(gè)特解，可求得其和函數(shù)為e^x．在此基礎(chǔ)上，結(jié)合復(fù)數(shù)運(yùn)算可以獲得sin x, cos x等初等函數(shù)的嚴(yán)格定義．再如冪級(jí)數(shù)

$%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D(-1)%5Enx%5En%3D1-x%2Bx%5E2-x%5E3%2B%5Ccdots$

利用等比數(shù)列求和公式知其和函數(shù)為1/(1+x)，再通過變量代換、逐項(xiàng)求導(dǎo)、求原函數(shù)可獲得ln(1+x), arctan x等函數(shù)的嚴(yán)格定義．總之，都是通過求簡單級(jí)數(shù)的和函數(shù)，再通過配湊獲得相對(duì)復(fù)雜級(jí)數(shù)的和函數(shù)，所以這類問題具有較強(qiáng)的技巧性．

??? 下面，我們考察一下三角級(jí)數(shù)的和函數(shù)應(yīng)該如何計(jì)算．前邊兩個(gè)例子的處理手法是十分經(jīng)典的，在繼續(xù)閱讀前請(qǐng)保證熟知這些內(nèi)容．

??? 首先給出我們要研究的三角級(jí)數(shù)形式：

定義1. ?稱形如

$%5Cfrac%7Ba_0%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(a_n%5Ccos%20nx%2Bb_n%5Csin%20nx)$

的函數(shù)項(xiàng)級(jí)數(shù)為三角級(jí)數(shù)，記為式(1)．

??? 根據(jù)De Moivre公式e^inx=cos nx+i sin nx，式(1)中的三角函數(shù)可以寫成復(fù)指數(shù)函數(shù)，進(jìn)而寫成以e^ix為變量的冪級(jí)數(shù)，結(jié)合冪級(jí)數(shù)求和公式便可求得相應(yīng)三角級(jí)數(shù)的和函數(shù)．反過來，我們也可以直接將冪級(jí)數(shù)的變量用e^ix代替，由于余弦、正弦分別對(duì)應(yīng)實(shí)部、虛部（考慮De Moivre公式即可），所以直接對(duì)變量代換后的冪級(jí)數(shù)取實(shí)部、虛部，也就得到了三角級(jí)數(shù)的和函數(shù)．

??? 下面考察幾個(gè)簡單的例子．

例1. 求 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Ccos%20nx%2C%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Csin%20nx$ 的和函數(shù)．

解：考察冪級(jí)數(shù)

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Dz%5En%3D%5Cfrac%7Bz%7D%7B1-z%7D%3DS(z)$

令z=e^ix，則

$%5Cbegin%7Balign*%7D%0AS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%26%3D%5Cfrac%7B1%7D%7B1-%5Ccos%20x-%5Cmathrm%7Bi%7D%5Csin%20x%7D-1%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B1-%5Ccos%20x%2B%5Cmathrm%7Bi%7D%5Csin%20x%7D%7B2-2%5Ccos%20x%7D-1%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B-1%2B%5Ccos%20x%2B%5Cmathrm%7Bi%7D%5Csin%20x%7D%7B2(1-%5Ccos%20x)%7D%0A%5Cend%7Balign*%7D$

于是

$%5Cbegin%7Balign*%7D%0A%26%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Ccos%20nx%3D%5Cmathrm%7BRe%7D%5BS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%5D%3D-%5Cfrac%7B1%7D%7B2%7D%2C%20x%5Cne%202n%5Cpi%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5B6pt%5D%0A%26%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Csin%20nx%3D%5Cmathrm%7BIm%7D%5BS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Ccot%5Cfrac%7Bx%7D%7B2%7D%2C%20x%5Cne%202n%5Cpi%2C%20n%5Cin%5Cmathbb%7BN%7D%0A%5Cend%7Balign*%7D$

這其中用到了三角公式 $%5Ccot%20%5Cfrac%7Bx%7D%7B2%7D%3D%5Cfrac%7B%5Csin%20x%7D%7B1-%5Ccos%20x%7D$ ，后邊還會(huì)多次用到類似的三角公式，其證明不難，留給讀者自行證明．

例2. 求 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Ccos%20nx%7D%7Bn%7D%2C%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn%7D$ 的和函數(shù)．

解：考察冪級(jí)數(shù)

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bz%5En%7D%7Bn%7D%3D%5Cint_0%5Ez%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7Dz%5En%5C%2C%5Cmathrm%7Bd%7Dz%20%3D%5Cint_0%5Ez%5Cfrac%7B%5Cmathrm%7Bd%7Dz%7D%7B1-z%7D%3D-%5Cln(1-z)%3DS(z)$

令z=e^ix，則

$%5Cbegin%7Balign*%7D%0AS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%26%3D-%5Cln(1-%5Ccos%20x-%5Cmathrm%7Bi%7D%5Csin%20x)%5C%5C%5B6pt%5D%0A%26%3D-%5Cln%5Csqrt%7B(1-%5Ccos%20x)%5E2%2B%5Csin%5E2%20x%7D%2B%5Cmathrm%7Bi%7D%5Carctan%5Cfrac%7B%5Csin%20x%7D%7B1-%5Ccos%20x%7D%5C%5C%5B6pt%5D%0A%26%3D-%5Cfrac%7B1%7D%7B2%7D%5Cln(2-2%5Ccos%20x)%2B%5Cmathrm%7Bi%7D%5Carctan%5Cleft(%5Ccot%5Cfrac%7Bx%7D%7B2%7D%5Cright)%0A%5Cend%7Balign*%7D$

此處需要考察arctan(cot x/2)：

$%5Cbegin%7Balign*%7D%0A%5Carctan%5Cleft(%5Ccot%5Cfrac%7Bx%7D%7B2%7D%5Cright)%26%3D%5Carctan%5Cleft(1%5CBig%2F%5Ctan%5Cfrac%7Bx%7D%7B2%7D%5Cright)%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Carctan%5Cleft(%5Ctan%5Cfrac%7Bx%7D%7B2%7D%5Cright)%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Cfrac%7Bx%2Bn%5Cpi%7D%7B2%7D%2C%20n%5Cin%5Cmathbb%7BN%7D%0A%5Cend%7Balign*%7D$

對(duì)前邊的和函數(shù)進(jìn)一步化簡得

$%5Cbegin%7Balign*%7D%0AS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%3D-%5Cln%5Cleft%7C2%5Csin%5Cfrac%7Bx%7D%7B2%7D%5Cright%7C%2B%5Cmathrm%7Bi%7D%5Cfrac%7B-x%2Bn%5Cpi%7D%7B2%7D%2Cn%5Cin%5Cmathbb%7BN%7D%0A%5Cend%7Balign*%7D$

于是

$%5Cbegin%7Balign*%7D%0A%26%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Ccos%20nx%7D%7Bn%7D%3D%5Cmathrm%7BRe%7D%5BS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%5D%3D-%5Cln%5Cleft%7C2%5Csin%5Cfrac%7Bx%7D%7B2%7D%5Cright%7C%2C%20x%5Cne%202n%5Cpi%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5B6pt%5D%0A%26%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn%7D%3D%5Cmathrm%7BIm%7D%5BS(%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D)%5D%3D%5Cfrac%7B-x%2Bn%5Cpi%7D%7B2%7D%2C%20x%5Cne%202n%5Cpi%2C%20n%5Cin%5Cmathbb%7BN%7D%0A%5Cend%7Balign*%7D$

可以通過逐項(xiàng)求導(dǎo)得到例1的結(jié)果，證明解的正確性．

??? 作 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn%7D$ 的圖像，如圖1所示

圖1 三角級(jí)數(shù)\sum \sin(nx)/n的和函數(shù)

其曲線形狀是一個(gè)鋸齒波．

??? 類似的，通過上述手法結(jié)合逐項(xiàng)積分，可以求得以下三角級(jí)數(shù)的和函數(shù)，不再一一給出過程．

問題1. 以下三角級(jí)數(shù)會(huì)收斂到其和函數(shù)上，請(qǐng)證明：

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Ccos%20nx%7D%7Bn%5E2%7D%3D%5Cfrac%7B3x%5E2-6%5Cpi%20x%2B2%5Cpi%5E2%7D%7B12%7D%2Cx%5Cin%5B0%2C2%5Cpi%5D$

（提示：對(duì) $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn%7D$ 逐項(xiàng)積分，并利用 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5E2%7D%3D%5Cfrac%7B%5Cpi%5E2%7D%7B6%7D$ ）

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn%5E2%7D%3D-%5Cint_%7B0%7D%5E%7Bx%7D%5Cln%5Cleft%7C2%5Csin%5Cfrac%7Bz%7D%7B2%7D%5Cright%7C%5C%2C%5Cmathrm%7Bd%7Dz%2Cx%5Cin%5B0%2C2%5Cpi%5D$

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn%5E3%7D%3D%5Cfrac%7Bx%5E3-3%5Cpi%20x%5E2%2B2%5Cpi%5E2%20x%7D%7B12%7D%2Cx%5Cin%5B0%2C2%5Cpi%5D$

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7B%5Ccos%20nx%7D%7Bn%7D%3D-%5Cln%5Cleft(2%5Ccos%5Cfrac%7Bx%7D%7B2%7D%5Cright)%2Cx%5Cin(-%5Cpi%2C%5Cpi)$

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7B%5Csin%20nx%7D%7Bn%7D%3D-%5Cfrac%7Bx%7D%7B2%7D%2Cx%5Cin(-%5Cpi%2C%5Cpi)$

$%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Ccos%20(2n%2B1)x%7D%7B2n%2B1%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cln%5Cleft%7C%5Ctan%5Cfrac%7Bx%7D%7B2%7D%5Cright%7C%2Cx%5Cin(-%5Cpi%2C%5Cpi)$

$%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20(2n%2B1)x%7D%7B2n%2B1%7D%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cpi%2F4%2C%26x%5Cin(0%2C%5Cpi)%5C%5C%0A-%5Cpi%2F4%2C%26x%5Cin(-%5Cpi%2C0)%0A%5Cend%7Barray%7D%5Cright.$

??? 問題1中，級(jí)數(shù) $%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20(2n%2B1)x%7D%7B2n%2B1%7D$ 的圖像顯然具有矩形波的形式．

問題2. 求 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Ccos%20nx%7D%7Bn!%7D%2C%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn!%7D$ 的和函數(shù)．

答案：

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Ccos%20nx%7D%7Bn!%7D%3D%5Cmathrm%7Be%7D%5E%7B%5Ccos%20x%7D%5Ccdot%5Ccos(%5Csin%20x)-1$

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%20nx%7D%7Bn!%7D%3D%5Cmathrm%7Be%7D%5E%7B%5Ccos%20x%7D%5Ccdot%5Csin(%5Csin%20x)$

??? 接下來我們考察一個(gè)形式比較特殊的三角級(jí)數(shù)：

例3. 求 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7B%5Ccos%20nx%7D%7Bn%5E2%2Bu%5E2%7D%2C%20u%5Cin%5Cmathbb%7BR%7D$ 的和函數(shù)．

解：考察冪級(jí)數(shù) $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7Bz%5En%7D%7Bn%5E2%2Bu%5E2%7D$ ，記其和函數(shù)為 $S(z)$ ．

??? 等式兩側(cè)對(duì)z求導(dǎo)，再乘以z，得

$zS'(z)%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7Bnz%5En%7D%7Bn%5E2%2Bu%5E2%7D$

等式兩側(cè)再對(duì)z求導(dǎo)，再乘以z，得

$zS'(z)%2Bz%5E2S''(z)%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7Bn%5E2z%5En%7D%7Bn%5E2%2Bu%5E2%7D$

于是疊加得到微分方程

$z%5E2S''%2BzS'%2Bu%5E2S%3D%5Cfrac%7B1%7D%7B1%2Bz%7D-1$

然而該微分方程的求解是十分困難的．事實(shí)上它并沒有初等形式的解．

??? 于是考慮直接作變換z=e^ix，則根據(jù)鏈?zhǔn)椒▌t有

$%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dz%7D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B%5Cmathrm%7Bd%7Dz%7D%3D-%5Cmathrm%7Bi%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dx%7D$

并將和函數(shù)S(z)寫成 $S(z)%3DU(x)%2B%5Cmathrm%7Bi%7DV(x)$ ．于是上述方程變形為

$%5Cleft(u%5E2-%5Cfrac%7B%5Cmathrm%7Bd%7D%5E2%7D%7B%5Cmathrm%7Bd%7Dx%5E2%7D%5Cright)(U%2B%5Cmathrm%7Bi%7DV)%3D-%5Cfrac%7B1%7D%7B2%7D%5Cleft(1%2B%5Cmathrm%7Bi%7D%5Ctan%5Cfrac%7Bx%7D%7B2%7D%5Cright)$

除了虛數(shù)單位i以外，其余部分均是實(shí)數(shù)，因此它等價(jià)于下述兩個(gè)方程

$%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D%0A%5Cdfrac%7B%5Cmathrm%7Bd%7D%5E2%20U%7D%7B%5Cmathrm%7Bd%7Dx%5E2%7D-u%5E2U%3D%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5B3pt%5D%0A%5Cdfrac%7B%5Cmathrm%7Bd%7D%5E2%20V%7D%7B%5Cmathrm%7Bd%7Dx%5E2%7D-u%5E2V%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctan%5Cdfrac%7Bx%7D%7B2%7D%0A%5Cend%7Barray%7D%5Cright.$

其中后一個(gè)方程是沒有簡單初等解的，而我們要求的是前一個(gè)方程．

??? 考慮函數(shù)的奇偶性，前一個(gè)方程的通解為

$U(x)%3DC%5Ccosh(ux)-%5Cfrac%7B1%7D%7B2u%5E2%7D$

然而求出待定系數(shù)C是極困難的，這需要我們知道級(jí)數(shù) $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7Bn%5E2%2Bu%5E2%7D$ 的值，而這個(gè)級(jí)數(shù)并不好求．但我們觀察到解的形式與e^ux有關(guān)，所以我們暫時(shí)擱置例3，不妨考察一下e^ux的Fourier級(jí)數(shù)的形式．

例4. 求 $%5Cmathrm%7Be%7D%5E%7Bux%7D$ 的Fourier級(jí)數(shù)．

解：記

$%5Cmathrm%7Be%7D%5E%7Bux%7D%3D%5Cfrac%7Ba_0%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(a_n%5Ccos%20nx%2Bb_n%5Csin%20nx)$

則其系數(shù)滿足

$a_n%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%7B%5Cpi%7D%5Cmathrm%7Be%7D%5E%7Bux%7D%5Ccos(nx)%5C%2C%5Cmathrm%7Bd%7Dx$

$b_n%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%7B%5Cpi%7D%5Cmathrm%7Be%7D%5E%7Bux%7D%5Csin(nx)%5C%2C%5Cmathrm%7Bd%7Dx$

于是

$%5Cbegin%7Balign*%7D%0Aa_n%2B%5Cmathrm%7Bi%7Db_n%26%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%7B%5Cpi%7D%5Cmathrm%7Be%7D%5E%7Bux%7D(%5Ccos(nx)%2B%5Cmathrm%7Bi%7D%5Csin(nx))%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%7B%5Cpi%7D%5Cmathrm%7Be%7D%5E%7B(u%2B%5Cmathrm%7Bi%7Dn)x%7D%5C%2C%5Cmathrm%7Bd%7Dx%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B1%7D%7B%5Cpi(u%2B%5Cmathrm%7Bi%7Dn)%7D%5Cleft(%5Cmathrm%7Be%7D%5E%7Bu%5Cpi%7D(-1)%5En-%5Cmathrm%7Be%7D%5E%7B-u%5Cpi%7D(-1)%5En%5Cright)%5C%5C%5B6pt%5D%0A%26%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Cfrac%7B(-1)%5En(a-%5Cmathrm%7Bi%7Dn)%7D%7Bn%5E2%2Ba%5E2%7D%5Csinh(u%5Cpi)%0A%5Cend%7Balign*%7D$

分別取實(shí)部、虛部即得系數(shù)．于是

$%5Cmathrm%7Be%7D%5E%7Bux%7D%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Csinh(u%5Cpi)%5Cleft(%5Cfrac%7B1%7D%7B2u%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7Bn%5E2%2Bu%5E2%7D(u%5Ccos%20nx-n%5Csin%20nx)%5Cright)$

??? 實(shí)際上e^ux的Fourier級(jí)數(shù)展開式中已經(jīng)包含了例3中問題的形式，考慮到

$%5Cmathrm%7Be%7D%5E%7B-ux%7D%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Csinh(u%5Cpi)%5Cleft(%5Cfrac%7B1%7D%7B2u%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7Bn%5E2%2Bu%5E2%7D(u%5Ccos%20nx%2Bn%5Csin%20nx)%5Cright)$

二者相加除以2即得

$%5Ccosh(ux)%3D%5Cfrac%7B2%7D%7B%5Cpi%7D%5Csinh(u%5Cpi)%5Cleft(%5Cfrac%7B1%7D%7B2u%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7Bn%5E2%2Bu%5E2%7D(u%5Ccos%20nx)%5Cright)$

這就得到了例3的解，即

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7B%5Ccos%20nx%7D%7Bn%5E2%2Bu%5E2%7D%3D%5Cfrac%7B%5Cpi%5Ccosh(ux)%7D%7B2u%5Csinh(u%5Cpi)%7D-%5Cfrac%7B1%7D%7B2u%5E2%7D$

同時(shí)我們也知道了

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7Bn%5E2%2Bu%5E2%7D%3D%5Cfrac%7B%5Cpi%7D%7B2u%5Csinh(u%5Cpi)%7D-%5Cfrac%7B1%7D%7B2u%5E2%7D$

以及

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5E2%2Bu%5E2%7D%3D%5Cfrac%7B%5Cpi%7D%7B2u%5Ctanh(u%5Cpi)%7D-%5Cfrac%7B1%7D%7B2u%5E2%7D$

??? 同理我們也能求出

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7Bn%5Csin%20nx%7D%7Bn%5E2%2Bu%5E2%7D%3D-%5Cfrac%7B%5Cpi%5Csinh(ux)%7D%7B2%5Csinh(u%5Cpi)%7D$

??? 下面我們考慮在求解線性PDE時(shí)，利用分離變量法求解獲得的級(jí)數(shù)解將收斂到何種和函數(shù)上．

例5. 考慮均質(zhì)直桿的縱向振動(dòng)．設(shè)有一長為l，楊氏模量為E的各向同性的均質(zhì)輕直桿（輕的含義是不計(jì)桿本身的重力），其截面尺寸為A相對(duì)于長度足夠小，使其一端固定，另一端自由．今在自由端施加一沿干軸向的拉力P，使其平衡，隨后撤去拉力P，則桿會(huì)開始作縱向振動(dòng)．根據(jù)振動(dòng)力學(xué)中的力學(xué)原理，我們可以將該問題歸納為求解下述定解問題：

$%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20t%5E2%7D%3Da%5E2%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20x%5E2%7D%2C%260%3Cx%3Cl%2Ct%3E0%5C%5C%5B3pt%5D%0Au(x%2C0)%3D%5Cdfrac%7BPx%7D%7BEA%7D%2C%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20t%7D(x%2C0)%3D0%2C%260%5Cleqslant%20x%5Cleqslant%20l%5C%5C%5B3pt%5D%0Au(0%2Ct)%3D0%2C%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D(l%2Ct)%3D0%2C%26t%5Cgeqslant%200%0A%5Cend%7Barray%7D%5Cright.$

這是一個(gè)典型的雙曲型方程．我們關(guān)心級(jí)數(shù)收斂到何種和函數(shù)上，所以略去具體的求解過程，直接給出其Fourier級(jí)數(shù)解為

$u(x%2Ct)%3D%5Cfrac%7B8Pl%7D%7B%5Cpi%5E2%20EA%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7B(2n%2B1)%5E2%7D%5Csin%5Cleft(%5Cfrac%7B(2n%2B1)%5Cpi%7D%7B2l%7D%20x%5Cright)%5Ccos%5Cleft(%5Cfrac%7B(2n%2B1)%5Cpi%7D%7B2l%7D%20at%5Cright)$

利用積化和差公式，可將求和部分寫成兩個(gè)三角級(jí)數(shù)加和的形式，即

$u(x%2Ct)%3D%5Cfrac%7B4Pl%7D%7B%5Cpi%5E2%20EA%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7B(2n%2B1)%5E2%7D%5Cleft(%5Csin%5Cleft(%5Cfrac%7B(2n%2B1)%5Cpi%7D%7B2l%7D%20(x%2Bat)%5Cright)%2B%5Csin%5Cleft(%5Cfrac%7B(2n%2B1)%5Cpi%7D%7B2l%7D%20(x-at)%5Cright)%5Cright)$

于是求該級(jí)數(shù)的和函數(shù)問題被歸結(jié)為求解下述級(jí)數(shù)的和函數(shù)

$%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7B%5Csin%20(2n%2B1)x%7D%7B(2n%2B1)%5E2%7D$

直接給出其結(jié)果為

$%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cfrac%7B%5Csin%20(2n%2B1)x%7D%7B(2n%2B1)%5E2%7D%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cdfrac%7B%5Cpi%7D%7B4%7D(x-2n%5Cpi)%2C%26%5Cleft(-%5Cdfrac%7B%5Cpi%7D%7B2%7D%2B2n%5Cpi%2C%5Cdfrac%7B%5Cpi%7D%7B2%7D%2B2n%5Cpi%5Cright)%5C%5C%5B3pt%5D%0A%5Cdfrac%7B%5Cpi%7D%7B4%7D(%5Cpi-x-2n%5Cpi)%2C%26%5Cleft(%5Cdfrac%7B%5Cpi%7D%7B2%7D%2B2n%5Cpi%2C%20%5Cdfrac%7B3%5Cpi%7D%7B2%7D%2B2n%5Cpi%5Cright)%0A%5Cend%7Barray%7D%5Cright.$

根據(jù)其區(qū)間，可以將直桿在一個(gè)振動(dòng)周期內(nèi)的級(jí)數(shù)解化簡為

$u(x%2Ct)%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cvarepsilon%20x%2C%260%3Ct%3C%5Cdfrac%7Bl-x%7D%7Ba%7D%5C%5C%5B3pt%5D%0A%5Cvarepsilon%20(l-at)%2C%26t%3E%5Cdfrac%7Bl-x%7D%7Ba%7D%2Ct%3E%5Cdfrac%7Bl%2Bx%7D%7Ba%7D%5C%5C%5B3pt%5D%0A-%5Cvarepsilon%20x%2C%26t%3C%5Cdfrac%7Bl%2Bx%7D%7Ba%7D%2Ct%3C%5Cdfrac%7B3l-x%7D%7Ba%7D%5C%5C%5B3pt%5D%0A-%5Cvarepsilon%20(3l-at)%2C%26t%3E%5Cdfrac%7B3l-x%7D%7Ba%7D%2Ct%3E%5Cdfrac%7B3l%2Bx%7D%7Ba%7D%5C%5C%5B3pt%5D%0A%5Cvarepsilon%20x%2C%26t%3C%5Cdfrac%7B3l%2Bx%7D%7Ba%7D%2Ct%3C%5Cdfrac%7B4l%7D%7Ba%7D%5C%5C%5B3pt%5D%0A%5Cend%7Barray%7D%5Cright.%20%2Cx%5Cin%5B0%2Cl%5D$

其中 $%5Cvarepsilon%3D%5Cfrac%7BP%7D%7BEA%7D$ ．特別的，我們可以考察自由端的位移

$u(x%2Ct)%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cvarepsilon%20(l-at)%2C%260%3Ct%3C%5Cdfrac%7B2l%7D%7Ba%7D%5C%5C%5B3pt%5D%0A-%5Cvarepsilon%20(3l-at)%2C%26%5Cdfrac%7B2l%7D%7Ba%7D%3Ct%3C%5Cdfrac%7B4l%7D%7Ba%7D%5C%5C%5B3pt%5D%0A%5Cend%7Barray%7D%5Cright.$

可見，桿端位移的形式符合三角波的形式．

例6. 考慮如下熱傳導(dǎo)的定解問題：

$%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20t%7D%3Dk%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20x%5E2%7D%2C%260%3Cx%3Cl%2Ct%3E0%5C%5C%5B3pt%5D%0Au(x%2C0)%3D%5Cdfrac%7Bx(l-x)%7D%7Bl%5E2%7D%2C%260%5Cleqslant%20x%5Cleqslant%20l%5C%5C%5B3pt%5D%0Au(0%2Ct)%3D%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D(l%2Ct)%3D0%2C%26t%5Cgeqslant%200%0A%5Cend%7Barray%7D%5Cright.$

這是一個(gè)拋物型方程，直接給出該問題的Fourier級(jí)數(shù)解為

$u(x%2Ct)%3D%5Cfrac%7B2%7D%7B%5Cpi%5E3%7D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2-%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5Cpi(-1)%5En%7D%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E3%7D%5Cexp%5Cleft(-%5Cfrac%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E2%5Cpi%5E2%7D%7Bl%5E2%7Dkt%5Cright)%5Csin%5Cfrac%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5Cpi%7D%7Bl%7Dx$

然而該問題沒辦法直接用前邊的方法求出解析表達(dá)式，這是因?yàn)閮H用普通的復(fù)數(shù)的冪級(jí)數(shù)無法表示含有 $%5Cmathrm%7Be%7D%5E%7B-n%5E2%7D$ 項(xiàng)的級(jí)數(shù)．為此，需要引入Jacobi $%5Cvartheta$ 函數(shù)：

定義2. 用級(jí)數(shù)定義以下四個(gè)特殊函數(shù)，稱之為Jacobi $%5Ccolor%7Bred%7D%7B%5Cvartheta%7D$ 函數(shù)：

$%5Cvartheta_1(v%3B%5Ctau)%3D2%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7D%5Ctau%5Cpi%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E2%7D%5Csin(2n%2B1)%5Cpi%20v$

$%5Cvartheta_2(v%3B%5Ctau)%3D2%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7D%5Ctau%5Cpi%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E2%7D%5Ccos(2n%2B1)%5Cpi%20v$

$%5Cvartheta_3(v%3B%5Ctau)%3D1%2B2%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7D%5Cpi%5Ctau%20n%5E2%7D%5Ccos(2n%5Cpi%20v)$

$%5Cvartheta_4(v%3B%5Ctau)%3D1%2B2%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7D%5Cpi%5Ctau%20n%5E2%7D%5Ccos(2n%5Cpi%20v)$

我們將v視為變量，將τ僅視為取固定值的參數(shù)．這幾個(gè)符號(hào)是Jacobi引入的，關(guān)于這些函數(shù)的更多性質(zhì)我們不做探討，可參考王竹溪、郭敦仁所著《特殊函數(shù)概論》．

??? 為計(jì)算方便，可以適當(dāng)?shù)厝》e分下限，使結(jié)果具有比較簡單的形式：

$%5Cbegin%7Balign*%7D%0A%26%5Cint_%7B0%7D%5E%7B%5Cfrac%7Bx%7D%7B2l%7D%7D%5Cint_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7Bw%7D%5Cvartheta_1%5Cleft(v%3B%5Cfrac%7B%5Cmathrm%7Bi%7D%5Cpi%20kt%7D%7Bl%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dv%5Cmathrm%7Bd%7Dw%5C%5C%0A%3D%26-%5Cfrac%7B2%7D%7B%5Cpi%5E2%7D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B(-1)%5En%7D%7B%5Cleft(2n%2B1%5Cright)%5E2%7D%0A%5Cexp%5Cleft(-%5Cfrac%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E2%5Cpi%5E2%7D%7Bl%5E2%7Dkt%5Cright)%5Csin%5Cfrac%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5Cpi%7D%7Bl%7Dx%0A%5Cend%7Balign*%7D$

$%5Cbegin%7Balign*%7D%0A%26%5Cint_%7B0%7D%5E%7B%5Cfrac%7Bx%7D%7B2l%7D%7D%5Cint_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7Bu%7D%5Cint_%7B0%7D%5E%7Bw%7D%5Cvartheta_2%5Cleft(v%3B%5Cfrac%7B%5Cmathrm%7Bi%7D%5Cpi%20kt%7D%7Bl%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dv%5Cmathrm%7Bd%7Dw%5Cmathrm%7Bd%7Du%5C%5C%0A%3D%26-%5Cfrac%7B2%7D%7B%5Cpi%5E3%7D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7B%5Cleft(2n%2B1%5Cright)%5E3%7D%0A%5Cexp%5Cleft(-%5Cfrac%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E2%5Cpi%5E2%7D%7Bl%5E2%7Dkt%5Cright)%5Csin%5Cfrac%7B%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5Cpi%7D%7Bl%7Dx%0A%5Cend%7Balign*%7D$

于是該定解問題的解可以寫成

$%5Cbegin%7Balign*%7D%0Au(x%2Ct)%3D%264%5Cint_%7B0%7D%5E%7B%5Cfrac%7Bx%7D%7B2l%7D%7D%5Cint_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7Bw%7D%5Cvartheta_1%5Cleft(v%3B%5Cfrac%7B%5Cmathrm%7Bi%7D%5Cpi%20kt%7D%7Bl%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dv%5Cmathrm%7Bd%7Dw%5C%5C%26-16%5Cint_%7B0%7D%5E%7B%5Cfrac%7Bx%7D%7B2l%7D%7D%5Cint_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7Bu%7D%5Cint_%7B0%7D%5E%7Bw%7D%5Cvartheta_2%5Cleft(v%3B%5Cfrac%7B%5Cmathrm%7Bi%7D%5Cpi%20kt%7D%7Bl%5E2%7D%5Cright)%5Cmathrm%7Bd%7Dv%5Cmathrm%7Bd%7Dw%5Cmathrm%7Bd%7Du%0A%5Cend%7Balign*%7D$

至于Jacobi $%5Cvartheta$ 函數(shù)的近似取值，可通過查表或數(shù)值計(jì)算的手段獲得．

問題3. 考慮均質(zhì)直桿在簡支邊界條件下的橫向振動(dòng)問題，該問題可抽象為如下四階方程的定解問題：

$%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20t%5E2%7D%2Ba%5E2%5Cdfrac%7B%5Cpartial%5E4%20u%7D%7B%5Cpartial%20x%5E4%7D%3D0%2C%260%3Cx%3Cl%2Ct%3E0%5C%5C%5B3pt%5D%0Au(x%2C0)%3Du_0%5Cdfrac%7Bx(3l%5E2-4x%5E2)%7D%7Bl%5E3%7D%2C%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20t%7D(x%2C0)%3D0%2C%260%5Cleqslant%20x%5Cleqslant%20l%5C%5C%5B3pt%5D%0Au(0%2Ct)%3Du(l%2Ct)%3D%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20x%5E2%7D(0%2Ct)%3D%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20x%5E2%7D(l%2Ct)%3D0%2C%26t%5Cgeqslant%200%0A%5Cend%7Barray%7D%5Cright.$

直接給出該問題的Fourier級(jí)數(shù)解為

$u(x%2Ct)%3D2u_0%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D(-1)%5En%5Cleft(%5Cfrac%7B1%7D%7Bn%5Cpi%7D-%5Cfrac%7B24%7D%7B(n%5Cpi)%5E3%7D%5Cright)%5Ccos%20%5Cleft(%5Cfrac%7Bn%5Cpi%7D%7Bl%7D%5Cright)%5E2%20at%5Csin%5Cfrac%7Bn%5Cpi%7D%7Bl%7Dx$

試用Jacobi $%5Cvartheta$ 函數(shù)表出該解．

答案：此處我們只使用積分號(hào) $%5Cint$ 表示求一次原函數(shù)，并指定求原函數(shù)時(shí)附加的積分常數(shù)為0

$u(x%2Ct)%3Du_0%5Cleft(%5Cint-48%5Ciiint%5Cright)%5Cleft(%5Cvartheta_4%5Cleft(%5Cfrac%7Bx%7D%7B2l%7D%3B%5Cfrac%7Ba%5Cpi%20t%7D%7Bl%5E2%7D%5Cright)%2B%5Cvartheta_4%5Cleft(%5Cfrac%7Bx%7D%7B2l%7D%3B-%5Cfrac%7Ba%5Cpi%20t%7D%7Bl%5E2%7D%5Cright)-2%5Cright)$

??? 最后給出一類筆者未能解決的問題，即二維調(diào)和方程或重調(diào)和方程的三角級(jí)數(shù)解將收斂到何種特殊的函數(shù)上．下面簡要介紹問題求解過程，供讀者進(jìn)行思考和求解．

問題5. 考慮如下定解問題

$%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%0A%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20x%5E2%7D%2B%5Cdfrac%7B%5Cpartial%5E2%20u%7D%7B%5Cpartial%20y%5E2%7D%3D0%2C%260%3Cx%3Ca%2C0%3Cy%3Cb%5C%5C%5B3pt%5D%0Au(x%2C0)%3D%5Cdfrac%7Bx%7D%7Ba%7D%2Cu(x%2Cb)%3D0%2C%260%5Cleqslant%20x%5Cleqslant%20a%5C%5C%5B3pt%5D%0A%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D(0%2Cy)%3D%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D(a%2Cy)%3D0%2C%260%5Cleqslant%20y%5Cleqslant%20b%0A%5Cend%7Barray%7D%5Cright.$

這是一個(gè)橢圓型方程，求解該問題可以設(shè)解的形式為二重三角級(jí)數(shù)，即

$u_%7Bmn%7D(x%2Cy)%3Da_%7Bmn%7D%5Csin%5Cleft(%5Cfrac%7Bm%5Cpi%20x%7D%7Ba%7D%2B%5Cvarphi_m%5Cright)%5Csin%5Cleft(%5Cfrac%7Bn%5Cpi%20y%7D%7Bb%7D%2B%5Cvarphi_n%5Cright)$

再對(duì)指標(biāo) $m$ 或 $n$ 求和，將另一個(gè)指標(biāo)暫時(shí)視為常量，以獲得一重的三角級(jí)數(shù)．也可以直接利用分離變量法求解獲得以下結(jié)果

$u(x%2Cy)%3D%5Cfrac%7Bb-y%7D%7B2b%7D%2B%5Cfrac%7B2%7D%7B%5Cpi%5E2%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1-(-1)%5En%7D%7Bn%5E2%7D%5Cfrac%7B%5Csinh%5Cfrac%7Bn%5Cpi%7D%7Ba%7D(y-b)%7D%7B%5Csinh%5Cfrac%7Bn%5Cpi%20b%7D%7Ba%7D%7D%5Ccos%5Cfrac%7Bn%5Cpi%7D%7Ba%7Dx$

可以證明，二重三角級(jí)數(shù)解與分離變量求得的級(jí)數(shù)解是一致的．然而若要做進(jìn)一步化簡，勢(shì)必要考慮形如 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csinh%20n(x-a)%7D%7B%5Csinh%20na%7D$ 的級(jí)數(shù)如何求和，這是十分復(fù)雜的，筆者也沒有查到相關(guān)的特殊函數(shù)．

問題6. 考慮均質(zhì)薄板的靜力平衡問題．彈性力學(xué)中的平面薄板的撓曲問題滿足方程

$D%5CDelta%5E2w%3Dq$

其中D是給定常數(shù)，q=q(x,y)是給定函數(shù)，Δ是Laplace算子．這是一個(gè)四階常系數(shù)線性偏微分方程，若假設(shè)所考慮的 $q%3Dq_0$ 為常數(shù)，薄板是矩形的，且四邊簡支（位移w=0，力矩 $%5Cfrac%7B%5Cpartial%5E2%20w%7D%7B%5Cpartial%20x%5E2%7D$ 或 $%5Cfrac%7B%5Cpartial%5E2%20w%7D%7B%5Cpartial%20y%5E2%7D%3D0$ ），則可假設(shè)位移具有二重三角級(jí)數(shù)的形式．經(jīng)計(jì)算可得二重三角級(jí)數(shù)解的形式為：

$w(x%2Cy)%3D%5Cfrac%7B16q_0%7D%7B%5Cpi%5E6%20D%7D%5Csum_%7Bm%3D1%2C3%2C5%2C%5Ccdots%7D%5E%7B%5Cinfty%7D%5Csum_%7Bn%3D1%2C3%2C5%2C%5Ccdots%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Csin%5Cfrac%7Bm%5Cpi%20x%7D%7Ba%7D%5Csin%5Cfrac%7Bn%5Cpi%20y%7D%7Bb%7D%7D%7Bmn%5Cleft(%5Cfrac%7Bm%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7Bn%5E2%7D%7Bb%5E2%7D%5Cright)%5E2%7D$

??? 此外，或者通過對(duì)一個(gè)指標(biāo)求和，或者通過假設(shè)解具有

$w(x%2Cy)%3D%5Csum_%7Bm%3D1%2C3%2C5%2C%5Ccdots%7D%5E%7B%5Cinfty%7DY(y)%5Csin%5Cfrac%7Bm%5Cpi%20x%7D%7Ba%7D$

的形式，可求得一重三角級(jí)數(shù)解具有以下形式

$w(x%2Cy)%3D%5Cfrac%7B4q_0a%5E4%7D%7B%5Cpi%5E5%20D%7D%5Csum_%7Bm%3D1%2C3%2C5%2C%5Ccdots%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bm%5E5%7D%5Cleft(1-%5Cfrac%7B2%2B%5Cfrac%7Bm%5Cpi%20b%7D%7B2a%7D%5Ctanh%5Cfrac%7Bm%5Cpi%20b%7D%7B2a%7D%7D%7B2%5Ccosh%5Cfrac%7Bm%5Cpi%20b%7D%7B2a%7D%7D%5Ccosh%5Cfrac%7Bm%5Cpi%7D%7Ba%7Dy%2B%5Cfrac%7B%5Cfrac%7Bm%5Cpi%20b%7D%7B2a%7D%7D%7B2%5Csinh%5Cfrac%7Bm%5Cpi%20b%7D%7B2a%7D%7D%5Cfrac%7B2y%7D%7Bb%7D%5Csinh%5Cfrac%7Bm%5Cpi%7D%7B2a%7Dy%5Cright)%5Csin%5Cfrac%7Bm%5Cpi%7D%7Ba%7Dx$

至于進(jìn)一步能否將該級(jí)數(shù)化簡成某種特殊函數(shù)，歡迎大家做進(jìn)一步討論．

參考資料：

[1] (德)E. Zeidler等著．?dāng)?shù)學(xué)指南——實(shí)用數(shù)學(xué)手冊(cè)[M]．李文林等譯．北京：科學(xué)出版社，2012．

[2] 李莉，王峰．?dāng)?shù)學(xué)物理方程(第2版) [M]．哈爾濱：哈爾濱工業(yè)大學(xué)出版社，2016．

[3] 王竹溪，郭敦仁．特殊函數(shù)概論[M]．北京：北京大學(xué)出版社，2000．

[4]?于開平，鄒經(jīng)湘．結(jié)構(gòu)動(dòng)力學(xué)(第3版) [M]．哈爾濱：哈爾濱工業(yè)大學(xué)出版社，2015．

[5] 徐芝綸．彈性力學(xué)(第5版，下冊(cè)) [M]．北京：高等教育出版社，2016．

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