我們首先要回顧一下微分方程教程中已經學過的知識, 即求正常點附近的級數解, 從 \(\S 4.1\) 的討論, 我們已知解的形式為
\[y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\tag{4.2.1} \]
只須將 (4.2.1.) 代入微分方程, 再確定系數 \(a_{n}\) 即可.
[ 例 4.2.1] ]
求方程
\[y^{\prime}=2 x y \]
在 \(x=0\) 附近的級數解.
以式 (4.2.1) 代入 (4.2.2), 並讓 \(x\) 的同冪系數相等, 得遞推方程
\[y''-xy = 0\\ \sum_{n=1}^{\infty} na_{n}\left(x\right)^{n-1} = 2x\cdot \sum_{n=0}^{\infty} a_{n}\left(x\right)^{n} \\ \sum_{n=1}^{\infty} na_{n}\left(x\right)^{n-1} = 2 \cdot \sum_{n=0}^{\infty} a_{n}\left(x\right)^{n+1} \]對應同次冪\(x\)有
\[\begin{array}{l} a_1\cdot x^0 & \\ 2a_2\cdot x^1 & 2a_0\cdot x^1 \\ 3a_3\cdot x^2 & 2a_1\cdot x^2 \\ \vdots & \vdots \\ na_{n}\cdot x^{n-1} & 2a_{n-2} \cdot x^{n-1} \\ (n+1)a_{n+1}\cdot x^{n}& 2a_{n-1}\cdot x^{n} \\ (n+2)a_{n+2}\cdot x^{n+1} \qquad& 2a_{n}\cdot x^{n+1} \\ \vdots&\vdots \end{array} \]即
\[a_1 = 2a_1 \qquad na_n = 2a_{n-2} \\\downarrow\\ a_{1}=0,\qquad a_{n}=\frac{2}{n} a_{n-2} (n \geq 2) \]
\[a_{1}=0, \quad a_{n}=\frac{2}{n} a_{n-2} \quad(n \geq 2) \]
所以, 奇數項系數為零
\[a_{2 n-1}=0 \]
偶數項系數可用 \(a_{0}\) 來表示
\[a_{2 m}=\frac{1}{m !} a_{0} \]
因此，有解
\[y=a_{0} \sum_{m=0}^{\infty} \frac{1}{m !} x^{2 m}=a_{0} e^{x^{2}} \]
例4.2.2
求Airay方程
\[y''-xy = 0\tag{4.2.7} \]
的級數解
Airay方程，全空間內正常點
同樣地, 以式 (4.2.1) 代入 (4.2.7), 令同次冪系數相等, 得遞推 公式
\[a_{2}=0, \quad a_{n}=\frac{a_{n-3}}{n(n-1)} \quad(n \geq 3) \]
\[\begin{align} \sum_{n=2}^{\infty} n\cdot(n-1)a_{n}x^{n-2} &=x\sum_{n=0}^{\infty} a_{n}x^{n} \\ \sum_{n=2}^{\infty} n\cdot(n-1)a_{n}x^{n-2} &=\sum_{n=0}^{\infty} a_{n}x^{n+1} \end{align} \]
對應\(x\)同次冪有
\[\begin{array}{l} 2a_2\cdot x^0 & \\ 6a_3 \cdot x^1 & a_0\cdot x^1 \\ \vdots&\vdots \\ n(n-1)a_n\cdots x^{n-2} & a_{n-3}\cdot x^{n-2} \end{array} \]
即
\[a_2 = 0 \qquad a_n = \frac{a_{n-3}}{n(n-1)}\\ \]
\[a_0\rightarrow a_3\rightarrow a_6\rightarrow a_{\dots} \]
\[a_1\rightarrow a_4\rightarrow a_7\rightarrow a_{\dots} \]
\[a_2\rightarrow a_5\rightarrow a_8\rightarrow a_{\dots} \]
所以
\[\begin{aligned} &a_{3 m}=\frac{{a}_{0}}{3 m(3 m-1) \cdots 6 \cdot 5 \cdot 3 \cdot 2}=\frac{\Gamma\left(\frac{2}{3}\right)}{9^{m} m ! \Gamma\left(m+\frac{2}{3}\right)} a_{0} \\ &a_{3 m+1}=\frac{a_{1}}{(3 m+1) 3 m \cdots 7 \cdot 6 \cdot 4 \cdot 3}=\frac{\Gamma\left(\frac{4}{3}\right)}{9^{m} m ! \Gamma\left(m+\frac{4}{3}\right)} a_{1} \end{aligned} \]
\[a_{3 m}=\frac{{a}_{0}}{3 m(3 m-1) \cdots 6 \cdot 5 \cdot 3 \cdot 2}= \frac{\Gamma\left(\frac{2}{3}\right)}{9^{m} m ! \Gamma\left(m+\frac{2}{3}\right)} a_{0}\\ 其中，3\cdot6\cdot9\dots3m=3^m\cdot1\cdot2\cdot3\dots m=3^m\cdot m! \\另外，2\cdot5 \cdot 8 \cdot 11\dots(3m-1)\\ 提出m個3\\3^m\cdot\frac 2 3\frac 5 3\frac 8 3\dots\frac{3m-1} 3\\ 又因為\Gamma( m+\frac 2 3)=(m -\frac 1 3)\dots \frac 8 3\frac 5 3\frac 2 3\Gamma(\frac 2 3)\\ 所以3^m\cdot\frac 2 3\frac 5 3\frac 8 3\dots\frac{3m-1} 3=3^m\frac{\Gamma( m+\frac 2 3)}{\Gamma(\frac 2 3)} \]
Airy 方程的級數解為
\[y(x)=c_{1} \sum_{m=0}^{\infty} \frac{x^{3 m}}{9^{m} m ! \Gamma\left(m+\frac{2}{3}\right)}+c_{2} \sum_{m=0}^{\infty} \frac{x^{3 m+1}}{9^{m} m ! \Gamma\left(m+\frac{4}{3}\right)} \]
適當選擇 \(c_{1}, c_{2}\) 可以得到常用的 Airy 函數: \(c_{1}=3^{-\frac{2}{3}}, c_{2}=-3^{-\frac{4}{3}}\), \(y(x)=A i(x)\), 當 \(c_{1}=3^{-\frac{1}{6}}, c_{2}=3^{-\frac{5}{6}}, y(x)=B i(x)\).
根據 \(\S 4.1\) 的討論, 我們可把 \(x_{0}\) 為正則奇點的方程改寫成下述 形式
\[L y=y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y=0\tag{4.1.12} \]
\[y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0\\ y^{\prime \prime}+\frac{p(x)(x-x_{0})}{x-x_{0}} y^{\prime}+\frac{q(x)(x-x_{0})^2}{\left(x-x_{0}\right)^{2}} y=0\\ y^{\prime \prime}+\frac{p(x)^*}{x-x_{0}} y^{\prime}+\frac{q(x)^*}{\left(x-x_{0}\right)^{2}} y=0\\ y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y=0\\ p^*,q^*是解析的 \]
式中
\[\begin{aligned} &p(x)=p_{0}+p_{1}\left(x-x_{0}\right)+\cdots \\ &q(x)=q_{0}+q_{1}\left(x-x_{0}\right)+\cdots \end{aligned} \]
為 \(x_{0}\) 的鄰域內的解析函數, 我們可假定該方程有 Frobenius 型級數 解.
\[y=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n} \]
且不防有 \(a_{0} \neq 0\), 將上式代入方程 (4.2.12), 得
L是線性算子，\(L y=y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y\)
\(L(f+g)=Lf+Lg\)
正則解其中一個是F型級數，另一個可能是F型級數，也可能是F型級數組合。（至少有一個F型級數解）
\[\begin{aligned} L y=&y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y\\=& a_{0}\left(x-x_{0}\right)^{\alpha-2}\left[\alpha(\alpha-1)+p_{0} \alpha+q_{0}\right] +\sum_{n=1}^{\infty}\left(x-x_{0}\right)^{\alpha-2+n}\left\{p(\alpha+n) a_{n}\right.\\ &\left.+\sum_{k=0}^{n-1}\left[(\alpha+k) p_{n-k}+q_{n-k}\right] a_{k}\right\}=0 \end{aligned} \]
\[y''=(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ y'=(n+\alpha)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-1}\\ \frac{p(x)}{x-x_0}=\sum_{n=0}^{\infty}p_n(x-x_0)^{n-1}\\ \frac{q(x)}{(x-x_0)^2}=\sum_{n=0}^{\infty}p_n(x-x_0)^{n-2}\\ Ly=(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ +\sum_{i=0}^{\infty}p_n(x-x_0)^{i-1}\cdot(n+\alpha)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-1}\\+\sum_{i=0}^{\infty}p_n(x-x_0)^{i-2}\cdot\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n} \]利用待定系數法，x的同冪次放到一起。
雙重積分性質
\[\sum_{i=0}^{\infty}a_ix^i\cdot \sum_{j=0}^{\infty}b_jx^j=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_jx^{i+j}=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_jx^{n}=\sum_{n=0}^{\infty}(\sum_{i=0}^{n}a_ib_{n-i})x^n=\sum_{n=0}^{\infty}\sum_{i=0}^{n}a_ib_{n-i}x^n \]
\[\begin{align} Ly=&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ &+\sum_{i=0}^{\infty}p_i(x-x_0)^{i-1}\cdot(n+\alpha)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-1}\\&+\sum_{i=0}^{\infty}q_i(x-x_0)^{i-2}\cdot\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\\ =&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ &+\sum_{i=0}^{\infty}p_i(x-x_0)^{i-1}\cdot(j+\alpha)\sum_{j=0}^{\infty}a_j(x-x_0)^{j+\alpha-1}\\ &+\sum_{i=0}^{\infty}q_i(x-x_0)^{i-2}\cdot\left(x-x_{0}\right)^{\alpha} \sum_{j=0}^{\infty} a_{j}\left(x-x_{0}\right)^{j}\\ =&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ &+\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_jp_i\cdot(j+\alpha)(x-x_0)^{i+j+\alpha-2}\\ &+\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_jq_i\cdot(x-x_0)^{i+j+\alpha-2}\\ =&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ &+\sum_{n=0}^{\infty}\left(\sum_{i=0}^{\infty}a_{n-i}p_i\cdot(n-i+\alpha)\right)(x-x_0)^{n+\alpha-2}\\ &+\sum_{n=0}^{\infty}\left(\sum_{i=0}^{\infty}a_{n-i}q_i\right)(x-x_0)^{n+\alpha-2} \end{align} \]
n=0時，
\[\begin{align} &\alpha(\alpha-1)a_0(x-x_0)^{\alpha-2}+p_0a_0\alpha(x-x_0)^{\alpha-2}+q_0a_0(x-x_0)^{\alpha-2}\\ =&a_0(x-x_0)^{\alpha-2}[\alpha(\alpha-1)+p_0\alpha+q_0]\\ =&a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)} \end{align} \]
以后，n從1到無窮
\[\sum_{n=1}^{\infty}\left[(n+\alpha)(n+\alpha-1)a_n+\sum_{i=0}^{n}p_ia_{n-i}(n-i+\alpha)+\sum_{i=0}^{n}q_ia_{n-i}\right](x-x_0)^{n+\alpha-2}\\ \]
令n-i=k，
\[\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[(n+\alpha)(n+\alpha-1)a_n+\sum_{k=0}^{n}p_{n-k}a_{k}(k+\alpha)+\sum_{k=0}^{n}q_{n-k}a_{k}\right] \]
中括號里，
- an的系數
\[(n+\alpha)(n+\alpha-1)+p_0(n+\alpha)+q_0=\mathscr{P}(\mathcal{n+\alpha}) \]
- ak的系數
\[\sum_{k=0}^{n-1}p_{n-k}(k+\alpha)+\sum_{k=0}^{n-1}q_{n-k}=\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k}) \]
所以
\[Ly=a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)}+\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[a_n(\mathscr{P}(\mathcal{n+\alpha}))+a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\right]=0\quad(4.2.16) \]
也就是要求
\[\begin{cases} a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)}=0\\ \sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[a_n(\mathscr{P}(\mathcal{n+\alpha}))+a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\right]=0 \end{cases} \]
也就是要求\((x-x_0)^{\dots}\)的系數為0
\[\begin{cases} a_0\mathscr{P(\alpha)}=0\\ a_n(\mathscr{P}(\mathcal{n+\alpha}))+a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})=0 \end{cases} \]
a_0不能為0，所以\(\mathscr{P}(\alpha)=0\)
為0沒有意義。如果為零，提出一個(x-x_0)，a_1還是充當a_0的位置。
\[\begin{cases} \mathscr{P(\alpha)}=0&(4.2.17)\\ a_n(\mathscr{P}(\mathcal{n+\alpha}))=-a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})&(4.2.18) \end{cases} \]
第二式是遞推式,\(知道a_0,推出a_1.知道a_0,a_1,推出a_3,\dots\)
因為\(\mathscr{P(\alpha)}=0\),即\(\alpha(\alpha-1)+p_0\alpha+q_0=0\)（特征方程）,解出\(\alpha\)
這樣就可以得到F型級數解\(y=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\)。
- 例：n=1時,k從0求和到0，即只有k=0項
\[a_1\mathscr{P}(\mathcal{1+\alpha})=-a_0(p_{1}\alpha+q_{1})\\ a_1=-\frac{a_0(p_{1}\alpha+q_{1})}{\mathscr{P}(\mathcal{1+\alpha})} \]
a_n的系數\(\mathscr{P}(\mathcal{n+\alpha})\)如果是0，遞推就中斷了。
特征方程的根不等, 且 \(\alpha_{1}-\alpha_{2}\) 不等於整數. 這時, 遞推公 式 (4.2.18) 中
\[p\left(\alpha_{i}+n\right) \neq 0 \quad(i=1,2 ; n=1,2, \cdots) \]
這樣\(\mathscr{P}(\mathcal{n+\alpha})\)不是是0，遞推不會中斷。
所以, 對 \(\alpha=\alpha_{1}, \alpha_{2}\), 我們可完全確定 Frobenius 型級數的系數, 方 程 (4.2.12) 有兩個 Frobenius 型級數解.
- 特征方程有等根, \(\alpha_{1}=\alpha_{2}=\alpha\) 這時, 由於
\[p(\alpha+n) \neq 0 \]
我們可以得到, 但僅可得到一個 Frobenius 型級數解. 現在要設法 找第二個線性獨立解, 我們已知
\[y_{1}=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n} \]
暫且取 \(\beta\) 略不同於 \(\alpha\) 以及
\(\beta\)是\(\alpha\)的的變分，微小的偏移
\(\widetilde{a_{n}}\)是\(\beta\)通過遞推式得到
\[\tilde{y}_{1}=\left(x-x_{0}\right)^{\beta} \sum_{n=0}^{\infty} \widetilde{a_{n}}\left(x-x_{0}\right)^{n} \]
我們使系數 \(a_{n}(\beta)\) 滿足遞推關系 (4.2.18), 那么
\(a_{n}(\beta)=\widetilde{a_{n}}\)
\(\widetilde{a_{n}}\)帶入（4.2.16）
\[Ly=a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)}\\+\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[a_n(\mathscr{P}(\mathcal{n+\alpha}))+a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\right]=0\quad(4.2.16) \]
\(\widetilde{a_{n}}\)是\(\beta\)通過遞推式得到，必然會使中括號里為0，Ly第二項為零。
只剩第一項
\[L \tilde{y}_{1}=a_{0}\left(x-x_{0}\right)^{\beta-2} \mathscr{P(\beta)} \]
兩邊對 \(\beta\) 微商, 再令 \(\beta=\alpha\)
\[\begin{array}{l} \left(\frac{\partial }{\partial \beta}L\tilde y_{1}\right)\bigg|_{\beta=\alpha}\\=\left[a_{0}(\beta-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\beta)+a_0\left(x-x_{0}\right)^{\beta-2} \mathscr P'(\beta)\right]\bigg|_{\beta=\alpha}\\ =a_{0}\left[(\alpha-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\alpha)+\left(x-x_{0}\right)^{\alpha-2} \mathscr P'(\alpha)\right]=0\\\because L是線性的\\ \therefore(\frac{\partial L\tilde y_{1}}{\partial \beta}\bigg|_{\beta=\alpha})= L(\frac{\partial \tilde y_{1}}{\partial \beta}\bigg|_{\beta=\alpha})=0，所以\frac{\partial \tilde y_{1}}{\partial \beta}\bigg|_{\beta=\alpha}是解 \end{array} \]
\[L\left(\frac{\partial y_{1}}{\partial \alpha}\right)=a_{0}\left[(\alpha-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\alpha)+\left(x-x_{0}\right)^{\alpha-2} \mathscr P'(\alpha)\right]=0 \]
這是因為特征方程有重根 \(\alpha\) 之故. 這說明方程 (4.2.12) 的另一個與 \(y_{1}\) 線性獨立的解為
\[y_{2}=\frac{\partial y_{1}}{\partial \alpha}=y_{1} \ln \left(x-x_{0}\right)+\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty}\left(\frac{\partial a_{n}}{\partial \alpha}\right)\left(x-x_{0}\right)^{n} \]
\[\begin{align} y_{1}&=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\\ y_2&=\frac{\partial y_1}{\partial\alpha}\\&=\frac{\partial(x-x_0)^{\alpha}}{\partial\alpha}\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} \frac{\partial a_{n}}{\partial \alpha}\left(x-x_{0}\right)^{n}\\ &=\frac{\partial}{\partial \alpha}[e^{\alpha\ln{(x-x_0)}}]+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n}\\ &=\ln{(x-x_0)}e^{\alpha\ln{(x-x_0)}}+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n}\\ &=\ln{(x-x_0)}(x-x_0)^{\alpha}+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n}\\ &=\ln{(x-x_0)}y_1+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n} \end {align} \]
- 兩根之差為整數, 若 \(\alpha_{1}=\alpha_{2}+N, N>0\). 這時, 我們可以 由
\[\mathscr P\left(\alpha_{1}+n\right) \neq 0 \]
得到一個 Frobenius 型級數解.
\[y_{1}=\left(x-x_{0}\right)^{\alpha_{1}} \sum_{n=0}^{\infty} a_{1 n}\left(x-x_{0}\right)^{n} \]
\(\alpha_1\)有對應解，\(\alpha_2\)會使遞推式中斷，沒有解。
但由於
\[\mathscr P\left(\alpha_{2}+N\right)=0 \]
除非 (4.2.16) 中第二項
\[\sum_{k=0}^{N-1}\left[\left(\alpha_{2}+k\right) p_{N-k}+q_{N-k}\right] a_{k}=0 \]
對特征指數 \(\alpha_{2}\), 一般不能得到 Frobenius 型級數解. 仿照有等根的 情況, 還是取 \(\alpha\) 略不同於 \(\alpha_{1}\), 系數 \(a_{n}(\alpha)\) 滿足遞推關系的形式解.
\[\tilde{y}_{1}=\left(x-x_{0}\right)^{\alpha} \sum_{n=1}^{\infty} a_{n}\left(x-x_{0}\right)^{n} \]
\[構造\\ \tilde y_1=(x-x_0)^{\alpha}\sum_{n=0}^{\infty}\tilde {a_n}(x-x_0)^n \]
代入方程 (4.2.12), 兩邊微商, 再令 \(\alpha=\alpha_{1}\), 得
\[L\left[\left.\frac{\partial \widetilde{y}_{1}}{\partial \alpha}\right|_{\alpha_{1}}\right]=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right) \neq 0 \]
\[\begin{array}{l} L\left(\frac{\partial y_{1}}{\partial \alpha}\right)=a_{0}\left[(\alpha-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\alpha)+\left(x-x_{0}\right)^{\alpha-2} \mathscr P'(\alpha)\right]=0\\ \because\mathscr P(\alpha_1)=0\\ \therefore L\left(\frac{\partial y_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_1}=a_0(x-x_{0})^{\alpha_1-2} \mathscr P'(\alpha_1)\\=a_0(x-x_{0})^{\alpha_2+N-2} \mathscr P'(\alpha_2+N)\ne0①\\ \therefore\left(\frac{\partial y_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_1}不是方程的解\\ \alpha_1=\alpha_2+N\\ ①是非齊次方程，非齊次方程的通解等於它對應齊次方程通解加上非齊次的特解。 \end{array} \]
這是因為 \(\alpha_{2}\) 並不是特征方程的重根的緣故, 我們再求非齊次方程
\[\begin{aligned} L(y) &=y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y \\ &=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} p^{\prime}\left(\alpha_{2}+N\right) \end{aligned} \]
的 Frobenius 型級數解.
假設它的一個特解如下
\[\bar{y}=\left(x-x_{0}\right)^{\alpha_{2}} \sum_{n=0}^{\infty} c_{n}\left(x-x_{0}\right)^{n} \]
把\(\bar y\)代入Ly
仿照
\[Ly=a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)}\\+\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[a_n(\mathscr{P}(\mathcal{n+\alpha}))+a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\right]\quad(4.2.16)\\ \]
寫出
\[\begin{array}{l} L\bar y=c_0(x-x_0)^{\alpha_2-2}\mathscr P(\alpha_2)\\+\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha_2-2}\left[c_n(\mathscr{P}(\mathcal{n+\alpha_2}))+c_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha_2)+q_{n-k})\right]\\ =a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right)\\ \because \mathscr P(\alpha_2)=0\\ \therefore \sum_{n=1}^{\infty}(x-x_0)^{n+\alpha_2-2}\left[c_n(\mathscr{P}(\mathcal{n+\alpha_2}))+c_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha_2)+q_{n-k})\right]\\ =a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right) \\需要冪次相同的系數相同。n=N時， \\(x-x_0)^{N+\alpha_2-2}\left[c_N\mathscr{P}(N+\alpha_2)+c_k\sum_{k=0}^{N-1}(p_{N-k}(k+\alpha_2)+q_{N-k})\right]\\=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right)\\ \left[c_N\mathscr{P}(N+\alpha_2)+c_k\sum_{k=0}^{N-1}(p_{N-k}(k+\alpha_2)+q_{N-k})\right]=a_0\mathscr P^{\prime}\left(\alpha_{2}+N\right)\\ \end{array} \]
這樣一來, 確定 \(c_{N}\) 的遞推關系為
\[\begin{aligned} \mathscr P\left(\alpha_{2}+N\right) c_{N}=&-\sum_{k=0}^{N-1}\left[(\alpha+k) p_{N-k}+q_{N-k}\right) c_{k} \\ &+a_{0} \mathscr P^{\prime}\left(\alpha_{n}+N\right)\quad(4.2.32) \end{aligned} \]
適當調整系數 \(a_{0}\), 使它滿足使方程 (4.2.32) 右端為零的條件, 即
\(\mathscr P\left(\alpha_{2}+N\right)\)=0
\[a_{0}=\frac{\sum_{k=0}^{N-1}\left[(\alpha+k) p_{N-k}+q_{N-k}\right] c_{k}}{p^{\prime}\left(\alpha_{n}+N\right)} \]
\[\begin{array}{l} (4.4.32)左邊=0，a_0使右邊為零，那么c_N可以為任意值。\\ 只不過a_0不是任意的了。通常F型級數a_0是任意的，a_1,a_2,a_3\dots都與a_0有關。\\特解中a_0不是任意的，c_1,c_2,c_3\dots都不是任意的，c_N是任意的。c_{N+1}又不任意了(利用齊次的遞推式)。 \end{array} \]
非齊次方程 \((4.70)\) 有 Frobenius 型級數解 \((4.2 .31)\), 那么
\[y_{2}=\left(\frac{\partial \tilde{y}_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_{1}}-\bar{y} \]
是減不是加的原因
\[\begin{array}{l} L(y_2)=L\left(\frac{\partial \tilde{y}_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_{1}}-L\bar{y}\\ =a_0(x-x_{0})^{\alpha_1-2} \mathscr P'(\alpha_1)-a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} p^{\prime}\left(\alpha_{2}+N\right)=0\\ 即L(y_2)=0 \end{array} \]
- \(\frac{\partial \tilde{y}_{1}}{\partial \alpha}\bigg|_{\alpha=\alpha_{1}}\)是帶\(\ln\)的特解，\(\bar y\)是F型級數的特解，兩個解是線性獨立，相減是通解。（不理解）🤔
就是原方程的另一個線性獨立解, 其一般形式為
\[y_{2}=y_{1}(x) \ln \left(x-x_{0}\right)+\left(x-x_{0}\right)^{\alpha_{2}} \sum_{n=0}^{\infty} d_{n}\left(x-x_{0}\right)^{n} \]
這就是 Fuchs 所預言的形式.
對於高階方程, 情況更要復雜一些, 應該如何來進行求解請讀 者思考. 對於 \(x_{0}\) 為正則奇點的高階方程, 其一般形式為
高階不要求
\[\begin{gathered} y^{(n)}+\frac{q_{n-1}(x)}{x-x_{0}} y^{(n-1)}+\frac{q_{n-2}(x)}{\left(x-x_{0}\right)^{2}} y^{(n-2)}+\cdots \\ +\frac{q_{0}(x)}{\left(x-x_{0}\right)^{n}} y=0 \end{gathered} \]
式中 \(q_{0}, q_{1}, \cdots, q_{n-1}\) 在 \(x_{0}\) 鄰域內解析, 它的特征指數方程應為
\[\begin{array}{r} \alpha(\alpha-1) \cdots(\alpha-n+1)+q_{n-1}\left(x_{0}\right) \alpha(\alpha-1) \cdots(\alpha-n+2) \\ +q_{n-2}\left(x_{0}\right) \alpha(\alpha-1) \cdots(\alpha-n+3)+\cdots+q_{0}\left(x_{0}\right)=0 \end{array} \]
[ 例 4.2.3] 求 Euler 方程的解
\[x^{2} y^{\prime \prime}+p_{0} x y^{\prime}+q_{0} y=0 \]
它可以化為
\[y^{\prime \prime}+\frac{p_{0}}{x} y^{\prime}+\frac{q_{0}}{x^{2}} y=0 \]
x=0是正則奇點
它的特征方程為
\[\alpha(\alpha-1)+p_{0} \alpha+q_{0}=0 \]
特征方程\(\mathscr P(\alpha)= [\alpha(\alpha-1)+p_0\alpha+q_0]\)
當 \(\alpha_{1} \neq \alpha_{2}\) 時, 有解
\[\begin{aligned} &y_{1}=c_{1} x^{\alpha_{1}} \\ &y_{2}=c_{2} x^{\alpha_{2}} \end{aligned} \]
F型級數解\(y=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\)
遞推式\(a_n(\mathscr{P}(\mathcal{n+\alpha}))=-a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\)
n=1時,\(\because p_1,q_1=0,a_1(\mathscr{P}(\mathcal{1+\alpha}))=-a_0(p_{1}\alpha+q_{1})=0,a_1=0\)
同理，\(a_2,a_3,a_{\dots}\)=0.只有a_0不是零。
當 \(\alpha_{1}=\alpha_{2}=\alpha\) 時, 有解
\[\begin{aligned} &y_{1}=c_{1} x^{\alpha} \\ &y_{2}=c_{2} x^{\alpha} \ln x \end{aligned} \]
\(y_{1}=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\)
\(y_{2}=\frac{\partial y_{1}}{\partial \alpha}=y_{1} \ln \left(x-x_{0}\right)+\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty}\left(\frac{\partial a_{n}}{\partial \alpha}\right)\left(x-x_{0}\right)^{n}\)
顯然, 這是上面所討論結果的特例.
[ 例 4.2.4] 用 Frobenius 方法, 求修正的 Bessel 方程之解.
\[y^{\prime \prime}+\frac{1}{x} y^{\prime}-\left(1+\frac{\nu^{2}}{x^{2}}\right) y=0 \]
它的特征方程為
\[\alpha^{2}-\nu^{2}=0 \]
\(p(x)=1,q(x)=-(x^2+\nu^2)\)
特征方程\(\mathscr P(\alpha)= [\alpha(\alpha-1)+p_0\alpha+q_0]=\alpha^2-\nu^2=0\)
\(\alpha=\pm\nu\)
- 當 \(\nu\) 不為零或整數時
兩個根只差為\(2\nu\),有兩個F型級數解
\[I_{\pm \nu}(x)=\sum_{n=0}^{\infty} \frac{\left(\frac{x}{2}\right)^{2 n \pm \nu}}{\Gamma(\pm \nu+n+1) n !} \]
這是兩個線性獨立解.
- 當 \(\nu\) 為半整數時,
差就是整數(並且是奇數）。
剛好 (4.2.18) 右端為零, 上式仍是方程的解.
右端為零的原因：
\[\begin{cases} p_0=1,&&&p_1,p_2,\dots=0\\ q_0=-\nu^2,&q_1=0,&q_2=-1,&q_3,\dots=0 \end{cases} \]\[\begin{array}{l} a_n(\mathscr{P}(\mathcal{n+\alpha}))=-a_{k} \sum_{k=0}^{n-1}\left(p_{n-k}(k+\alpha)+q_{n-k}\right) \\ =-\left[\left[-\nu \cdot p_{n}+q_{n}\right] a_{0}\right. \\ +\left[(-\nu+1) p_{n-1}+q_{n-1}\right] a_{1} \\ +\left[(-\nu+2) p_{n-2}+q_{n-2}\right] a_{2} \\ +\cdots \\ +\left[(-\nu+n-2) p_{2}+q_{2}\right] a_{n-2} \\ \left.\left.+\left[(-\nu+n-1) p_{1}+q_{1}\right] a_{n-1}\right]\right]① \\ =a_{n-2}\\ \therefore a_n(\mathscr{P}(\mathcal{n+\alpha}))=a_{n-2}② \end{array} \]上式中的p,q系數只有 非零，為-1。所以①式倒數第二項留下來了。
\[\begin{array}{l} \therefore n\ne\nu-(-\nu)時，遞推式左邊不為零，由a_0推出a_2,a_4,\dots(偶數項存在)；\\ a_1由a_0推出，a_1\mathscr P(1+\alpha)=-a_0(p_1+q_1)=0,a_1=0\\ \therefore 由②式，a_3,a_5,\dots=0(奇數項不存在) \end{array} \]\(設\nu-(-\nu)=m，則m必為奇數\)
\(p(\alpha+m) a_{m}=-\sum_{k=0}^{m-1}\left[(\alpha+k) p_{m-k}+q_{m-k}\right] a_{k}=a_{m-2}\)，
左邊是零但是右邊也為零,使遞推式滿足。
- 當 \(\nu=0\) 時, 有重根, 第一個解為
\[I_{0}(x)=\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2} x\right)^{2 n}}{(n !)^{2}} \]
為求出第二個解, 先令 \(\alpha \neq 0\), 將 Frobenius 型級數代入方程, 得到 遞推關系, \(a_{0} \neq 0\).
\[\begin{aligned} &a_{2 n-1}(\alpha)=0 \\ &a_{2 n}(\alpha)=\frac{a_{0}}{(\alpha+2 n)^{2}(\alpha+2 n-2)^{2} \cdots(\alpha+2)^{2}} \end{aligned} \]
由此導出 \(b_{0} \neq 0\).
\[\begin{aligned} &b_{2 n-1}=0 \\ &b_{2 n}=\left.\frac{\partial a_{2 n}}{\partial \alpha}\right|_{\alpha=0}=-\frac{a_{0}}{2^{2 n}(n !)^{2}}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right) \end{aligned} \]