4.2 正常点与正则奇点附近的级数解(李家春）
我们首先要回顾一下微分方程教程中已经学过的知识, 即求正常点附近的级数解, 从 \(\S 4.1\) 的讨论, 我们已知解的形式为
\[y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\tag{4.2.1}
\]
只须将 (4.2.1.) 代入微分方程, 再确定系数 \(a_{n}\) 即可.
[ 例 4.2.1] ]
求方程
\[y^{\prime}=2 x y
\]
在 \(x=0\) 附近的级数解.
以式 (4.2.1) 代入 (4.2.2), 并让 \(x\) 的同幂系数相等, 得递推方程
\[y''-xy = 0\\ \sum_{n=1}^{\infty} na_{n}\left(x\right)^{n-1} = 2x\cdot \sum_{n=0}^{\infty} a_{n}\left(x\right)^{n} \\ \sum_{n=1}^{\infty} na_{n}\left(x\right)^{n-1} = 2 \cdot \sum_{n=0}^{\infty} a_{n}\left(x\right)^{n+1} \]对应同次幂\(x\)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, 奇数项系数为零
\[a_{2 n-1}=0
\]
偶数项系数可用 \(a_{0}\) 来表示
\[a_{2 m}=\frac{1}{m !} a_{0}
\]
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
\[y''-xy = 0\tag{4.2.7}
\]
的级数解
Airay方程，全空间内正常点
同样地, 以式 (4.2.1) 代入 (4.2.7), 令同次幂系数相等, 得递推 公式
\[a_{2}=0, \quad a_{n}=\frac{a_{n-3}}{n(n-1)} \quad(n \geq 3)
\]
\[\begin{align}
\sum_{n=2}^{\infty} n\cdot(n-1)a_{n}x^{n-2}
&=x\sum_{n=0}^{\infty} a_{n}x^{n}
\\
\sum_{n=2}^{\infty} n\cdot(n-1)a_{n}x^{n-2}
&=\sum_{n=0}^{\infty} a_{n}x^{n+1}
\end{align}
\]
对应\(x\)同次幂有
\[\begin{array}{l}
2a_2\cdot x^0 &
\\
6a_3 \cdot x^1 & a_0\cdot x^1
\\
\vdots&\vdots
\\
n(n-1)a_n\cdots x^{n-2} & a_{n-3}\cdot x^{n-2}
\end{array}
\]
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
\[\begin{aligned}
&a_{3 m}=\frac{{a}_{0}}{3 m(3 m-1) \cdots 6 \cdot 5 \cdot 3 \cdot 2}=\frac{\Gamma\left(\frac{2}{3}\right)}{9^{m} m ! \Gamma\left(m+\frac{2}{3}\right)} a_{0} \\
&a_{3 m+1}=\frac{a_{1}}{(3 m+1) 3 m \cdots 7 \cdot 6 \cdot 4 \cdot 3}=\frac{\Gamma\left(\frac{4}{3}\right)}{9^{m} m ! \Gamma\left(m+\frac{4}{3}\right)} a_{1}
\end{aligned}
\]
\[a_{3 m}=\frac{{a}_{0}}{3 m(3 m-1) \cdots 6 \cdot 5 \cdot 3 \cdot 2}= \frac{\Gamma\left(\frac{2}{3}\right)}{9^{m} m ! \Gamma\left(m+\frac{2}{3}\right)} a_{0}\\ 其中，3\cdot6\cdot9\dots3m=3^m\cdot1\cdot2\cdot3\dots m=3^m\cdot m! \\另外，2\cdot5 \cdot 8 \cdot 11\dots(3m-1)\\ 提出m个3\\3^m\cdot\frac 2 3\frac 5 3\frac 8 3\dots\frac{3m-1} 3\\ 又因为\Gamma( m+\frac 2 3)=(m -\frac 1 3)\dots \frac 8 3\frac 5 3\frac 2 3\Gamma(\frac 2 3)\\ 所以3^m\cdot\frac 2 3\frac 5 3\frac 8 3\dots\frac{3m-1} 3=3^m\frac{\Gamma( m+\frac 2 3)}{\Gamma(\frac 2 3)} \]
Airy 方程的级数解为
\[y(x)=c_{1} \sum_{m=0}^{\infty} \frac{x^{3 m}}{9^{m} m ! \Gamma\left(m+\frac{2}{3}\right)}+c_{2} \sum_{m=0}^{\infty} \frac{x^{3 m+1}}{9^{m} m ! \Gamma\left(m+\frac{4}{3}\right)}
\]
适当选择 \(c_{1}, c_{2}\) 可以得到常用的 Airy 函数: \(c_{1}=3^{-\frac{2}{3}}, c_{2}=-3^{-\frac{4}{3}}\), \(y(x)=A i(x)\), 当 \(c_{1}=3^{-\frac{1}{6}}, c_{2}=3^{-\frac{5}{6}}, y(x)=B i(x)\).
根据 \(\S 4.1\) 的讨论, 我们可把 \(x_{0}\) 为正则奇点的方程改写成下述 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
\[\begin{aligned}
&p(x)=p_{0}+p_{1}\left(x-x_{0}\right)+\cdots \\
&q(x)=q_{0}+q_{1}\left(x-x_{0}\right)+\cdots
\end{aligned}
\]
为 \(x_{0}\) 的邻域内的解析函数, 我们可假定该方程有 Frobenius 型级数 解.
\[y=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}
\]
且不防有 \(a_{0} \neq 0\), 将上式代入方程 (4.2.12), 得
L是线性算子，\(L y=y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y\)
\(L(f+g)=Lf+Lg\)
正则解其中一个是F型级数，另一个可能是F型级数，也可能是F型级数组合。（至少有一个F型级数解）
\[\begin{aligned}
L y=&y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y\\=& a_{0}\left(x-x_{0}\right)^{\alpha-2}\left[\alpha(\alpha-1)+p_{0} \alpha+q_{0}\right] +\sum_{n=1}^{\infty}\left(x-x_{0}\right)^{\alpha-2+n}\left\{p(\alpha+n) a_{n}\right.\\
&\left.+\sum_{k=0}^{n-1}\left[(\alpha+k) p_{n-k}+q_{n-k}\right] a_{k}\right\}=0
\end{aligned}
\]
\[y''=(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ y'=(n+\alpha)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-1}\\ \frac{p(x)}{x-x_0}=\sum_{n=0}^{\infty}p_n(x-x_0)^{n-1}\\ \frac{q(x)}{(x-x_0)^2}=\sum_{n=0}^{\infty}p_n(x-x_0)^{n-2}\\ Ly=(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\ +\sum_{i=0}^{\infty}p_n(x-x_0)^{i-1}\cdot(n+\alpha)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-1}\\+\sum_{i=0}^{\infty}p_n(x-x_0)^{i-2}\cdot\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n} \]利用待定系数法，x的同幂次放到一起。
双重积分性质
\[\sum_{i=0}^{\infty}a_ix^i\cdot \sum_{j=0}^{\infty}b_jx^j=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_jx^{i+j}=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_jx^{n}=\sum_{n=0}^{\infty}(\sum_{i=0}^{n}a_ib_{n-i})x^n=\sum_{n=0}^{\infty}\sum_{i=0}^{n}a_ib_{n-i}x^n
\]
\[\begin{align}
Ly=&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\
&+\sum_{i=0}^{\infty}p_i(x-x_0)^{i-1}\cdot(n+\alpha)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-1}\\&+\sum_{i=0}^{\infty}q_i(x-x_0)^{i-2}\cdot\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\\
=&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\
&+\sum_{i=0}^{\infty}p_i(x-x_0)^{i-1}\cdot(j+\alpha)\sum_{j=0}^{\infty}a_j(x-x_0)^{j+\alpha-1}\\
&+\sum_{i=0}^{\infty}q_i(x-x_0)^{i-2}\cdot\left(x-x_{0}\right)^{\alpha} \sum_{j=0}^{\infty} a_{j}\left(x-x_{0}\right)^{j}\\
=&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\
&+\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_jp_i\cdot(j+\alpha)(x-x_0)^{i+j+\alpha-2}\\
&+\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_jq_i\cdot(x-x_0)^{i+j+\alpha-2}\\
=&(n+\alpha)(n+\alpha-1)\sum_{n=0}^{\infty}a_n(x-x_0)^{n+\alpha-2}\\
&+\sum_{n=0}^{\infty}\left(\sum_{i=0}^{\infty}a_{n-i}p_i\cdot(n-i+\alpha)\right)(x-x_0)^{n+\alpha-2}\\
&+\sum_{n=0}^{\infty}\left(\sum_{i=0}^{\infty}a_{n-i}q_i\right)(x-x_0)^{n+\alpha-2}
\end{align}
\]
n=0时，
\[\begin{align}
&\alpha(\alpha-1)a_0(x-x_0)^{\alpha-2}+p_0a_0\alpha(x-x_0)^{\alpha-2}+q_0a_0(x-x_0)^{\alpha-2}\\
=&a_0(x-x_0)^{\alpha-2}[\alpha(\alpha-1)+p_0\alpha+q_0]\\
=&a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)}
\end{align}
\]
以后，n从1到无穷
\[\sum_{n=1}^{\infty}\left[(n+\alpha)(n+\alpha-1)a_n+\sum_{i=0}^{n}p_ia_{n-i}(n-i+\alpha)+\sum_{i=0}^{n}q_ia_{n-i}\right](x-x_0)^{n+\alpha-2}\\
\]
令n-i=k，
\[\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[(n+\alpha)(n+\alpha-1)a_n+\sum_{k=0}^{n}p_{n-k}a_{k}(k+\alpha)+\sum_{k=0}^{n}q_{n-k}a_{k}\right]
\]
中括号里，
- an的系数
\[(n+\alpha)(n+\alpha-1)+p_0(n+\alpha)+q_0=\mathscr{P}(\mathcal{n+\alpha})
\]
- ak的系数
\[\sum_{k=0}^{n-1}p_{n-k}(k+\alpha)+\sum_{k=0}^{n-1}q_{n-k}=\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})
\]
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\((x-x_0)^{\dots}\)的系数为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不能为0，所以\(\mathscr{P}(\alpha)=0\)
为0没有意义。如果为零，提出一个(x-x_0)，a_1还是充当a_0的位置。
\[\begin{cases}
\mathscr{P(\alpha)}=0&(4.2.17)\\
a_n(\mathscr{P}(\mathcal{n+\alpha}))=-a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})&(4.2.18)
\end{cases}
\]
第二式是递推式,\(知道a_0,推出a_1.知道a_0,a_1,推出a_3,\dots\)
因为\(\mathscr{P(\alpha)}=0\),即\(\alpha(\alpha-1)+p_0\alpha+q_0=0\)（特征方程）,解出\(\alpha\)
这样就可以得到F型级数解\(y=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\)。
- 例：n=1时,k从0求和到0，即只有k=0项
\[a_1\mathscr{P}(\mathcal{1+\alpha})=-a_0(p_{1}\alpha+q_{1})\\
a_1=-\frac{a_0(p_{1}\alpha+q_{1})}{\mathscr{P}(\mathcal{1+\alpha})}
\]
a_n的系数\(\mathscr{P}(\mathcal{n+\alpha})\)如果是0，递推就中断了。
特征方程的根不等, 且 \(\alpha_{1}-\alpha_{2}\) 不等于整数. 这时, 递推公 式 (4.2.18) 中
\[p\left(\alpha_{i}+n\right) \neq 0 \quad(i=1,2 ; n=1,2, \cdots)
\]
这样\(\mathscr{P}(\mathcal{n+\alpha})\)不是是0，递推不会中断。
所以, 对 \(\alpha=\alpha_{1}, \alpha_{2}\), 我们可完全确定 Frobenius 型级数的系数, 方 程 (4.2.12) 有两个 Frobenius 型级数解.
- 特征方程有等根, \(\alpha_{1}=\alpha_{2}=\alpha\) 这时, 由于
\[p(\alpha+n) \neq 0
\]
我们可以得到, 但仅可得到一个 Frobenius 型级数解. 现在要设法 找第二个线性独立解, 我们已知
\[y_{1}=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}
\]
暂且取 \(\beta\) 略不同于 \(\alpha\) 以及
\(\beta\)是\(\alpha\)的的变分，微小的偏移
\(\widetilde{a_{n}}\)是\(\beta\)通过递推式得到
\[\tilde{y}_{1}=\left(x-x_{0}\right)^{\beta} \sum_{n=0}^{\infty} \widetilde{a_{n}}\left(x-x_{0}\right)^{n}
\]
我们使系数 \(a_{n}(\beta)\) 满足递推关系 (4.2.18), 那么
\(a_{n}(\beta)=\widetilde{a_{n}}\)
\(\widetilde{a_{n}}\)带入（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\(\beta\)通过递推式得到，必然会使中括号里为0，Ly第二项为零。
只剩第一项
\[L \tilde{y}_{1}=a_{0}\left(x-x_{0}\right)^{\beta-2} \mathscr{P(\beta)}
\]
两边对 \(\beta\) 微商, 再令 \(\beta=\alpha\)
\[\begin{array}{l} \left(\frac{\partial }{\partial \beta}L\tilde y_{1}\right)\bigg|_{\beta=\alpha}\\=\left[a_{0}(\beta-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\beta)+a_0\left(x-x_{0}\right)^{\beta-2} \mathscr P'(\beta)\right]\bigg|_{\beta=\alpha}\\ =a_{0}\left[(\alpha-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\alpha)+\left(x-x_{0}\right)^{\alpha-2} \mathscr P'(\alpha)\right]=0\\\because L是线性的\\ \therefore(\frac{\partial L\tilde y_{1}}{\partial \beta}\bigg|_{\beta=\alpha})= L(\frac{\partial \tilde y_{1}}{\partial \beta}\bigg|_{\beta=\alpha})=0，所以\frac{\partial \tilde y_{1}}{\partial \beta}\bigg|_{\beta=\alpha}是解 \end{array} \]
\[L\left(\frac{\partial y_{1}}{\partial \alpha}\right)=a_{0}\left[(\alpha-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\alpha)+\left(x-x_{0}\right)^{\alpha-2} \mathscr P'(\alpha)\right]=0
\]
这是因为特征方程有重根 \(\alpha\) 之故. 这说明方程 (4.2.12) 的另一个与 \(y_{1}\) 线性独立的解为
\[y_{2}=\frac{\partial y_{1}}{\partial \alpha}=y_{1} \ln \left(x-x_{0}\right)+\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty}\left(\frac{\partial a_{n}}{\partial \alpha}\right)\left(x-x_{0}\right)^{n}
\]
\[\begin{align} y_{1}&=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\\ y_2&=\frac{\partial y_1}{\partial\alpha}\\&=\frac{\partial(x-x_0)^{\alpha}}{\partial\alpha}\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} \frac{\partial a_{n}}{\partial \alpha}\left(x-x_{0}\right)^{n}\\ &=\frac{\partial}{\partial \alpha}[e^{\alpha\ln{(x-x_0)}}]+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n}\\ &=\ln{(x-x_0)}e^{\alpha\ln{(x-x_0)}}+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n}\\ &=\ln{(x-x_0)}(x-x_0)^{\alpha}+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n}\\ &=\ln{(x-x_0)}y_1+\left(x-x_{0}\right)^{\alpha}\sum_{n=0}^{\infty} c_n\left(x-x_{0}\right)^{n} \end {align} \]
- 两根之差为整数, 若 \(\alpha_{1}=\alpha_{2}+N, N>0\). 这时, 我们可以 由
\[\mathscr P\left(\alpha_{1}+n\right) \neq 0
\]
得到一个 Frobenius 型级数解.
\[y_{1}=\left(x-x_{0}\right)^{\alpha_{1}} \sum_{n=0}^{\infty} a_{1 n}\left(x-x_{0}\right)^{n}
\]
\(\alpha_1\)有对应解，\(\alpha_2\)会使递推式中断，没有解。
但由于
\[\mathscr P\left(\alpha_{2}+N\right)=0
\]
除非 (4.2.16) 中第二项
\[\sum_{k=0}^{N-1}\left[\left(\alpha_{2}+k\right) p_{N-k}+q_{N-k}\right] a_{k}=0
\]
对特征指数 \(\alpha_{2}\), 一般不能得到 Frobenius 型级数解. 仿照有等根的 情况, 还是取 \(\alpha\) 略不同于 \(\alpha_{1}\), 系数 \(a_{n}(\alpha)\) 满足递推关系的形式解.
\[\tilde{y}_{1}=\left(x-x_{0}\right)^{\alpha} \sum_{n=1}^{\infty} a_{n}\left(x-x_{0}\right)^{n}
\]
\[构造\\ \tilde y_1=(x-x_0)^{\alpha}\sum_{n=0}^{\infty}\tilde {a_n}(x-x_0)^n \]
代入方程 (4.2.12), 两边微商, 再令 \(\alpha=\alpha_{1}\), 得
\[L\left[\left.\frac{\partial \widetilde{y}_{1}}{\partial \alpha}\right|_{\alpha_{1}}\right]=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right) \neq 0
\]
\[\begin{array}{l} L\left(\frac{\partial y_{1}}{\partial \alpha}\right)=a_{0}\left[(\alpha-2)\left(x-x_{0}\right)^{\alpha-3} \mathscr P(\alpha)+\left(x-x_{0}\right)^{\alpha-2} \mathscr P'(\alpha)\right]=0\\ \because\mathscr P(\alpha_1)=0\\ \therefore L\left(\frac{\partial y_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_1}=a_0(x-x_{0})^{\alpha_1-2} \mathscr P'(\alpha_1)\\=a_0(x-x_{0})^{\alpha_2+N-2} \mathscr P'(\alpha_2+N)\ne0①\\ \therefore\left(\frac{\partial y_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_1}不是方程的解\\ \alpha_1=\alpha_2+N\\ ①是非齐次方程，非齐次方程的通解等于它对应齐次方程通解加上非齐次的特解。 \end{array} \]
这是因为 \(\alpha_{2}\) 并不是特征方程的重根的缘故, 我们再求非齐次方程
\[\begin{aligned}
L(y) &=y^{\prime \prime}+\frac{p(x)}{x-x_{0}} y^{\prime}+\frac{q(x)}{\left(x-x_{0}\right)^{2}} y \\
&=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} p^{\prime}\left(\alpha_{2}+N\right)
\end{aligned}
\]
的 Frobenius 型级数解.
假设它的一个特解如下
\[\bar{y}=\left(x-x_{0}\right)^{\alpha_{2}} \sum_{n=0}^{\infty} c_{n}\left(x-x_{0}\right)^{n}
\]
把\(\bar y\)代入Ly
仿照
\[Ly=a_0(x-x_0)^{\alpha-2}\mathscr{P(\alpha)}\\+\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha-2}\left[a_n(\mathscr{P}(\mathcal{n+\alpha}))+a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\right]\quad(4.2.16)\\
\]
写出
\[\begin{array}{l}
L\bar y=c_0(x-x_0)^{\alpha_2-2}\mathscr P(\alpha_2)\\+\sum_{n=1}^{\infty}(x-x_0)^{n+\alpha_2-2}\left[c_n(\mathscr{P}(\mathcal{n+\alpha_2}))+c_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha_2)+q_{n-k})\right]\\
=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right)\\
\because \mathscr P(\alpha_2)=0\\
\therefore \sum_{n=1}^{\infty}(x-x_0)^{n+\alpha_2-2}\left[c_n(\mathscr{P}(\mathcal{n+\alpha_2}))+c_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha_2)+q_{n-k})\right]\\
=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right)
\\需要幂次相同的系数相同。n=N时，
\\(x-x_0)^{N+\alpha_2-2}\left[c_N\mathscr{P}(N+\alpha_2)+c_k\sum_{k=0}^{N-1}(p_{N-k}(k+\alpha_2)+q_{N-k})\right]\\=a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} \mathscr P^{\prime}\left(\alpha_{2}+N\right)\\
\left[c_N\mathscr{P}(N+\alpha_2)+c_k\sum_{k=0}^{N-1}(p_{N-k}(k+\alpha_2)+q_{N-k})\right]=a_0\mathscr P^{\prime}\left(\alpha_{2}+N\right)\\
\end{array}
\]
这样一来, 确定 \(c_{N}\) 的递推关系为
\[\begin{aligned}
\mathscr P\left(\alpha_{2}+N\right) c_{N}=&-\sum_{k=0}^{N-1}\left[(\alpha+k) p_{N-k}+q_{N-k}\right) c_{k} \\
&+a_{0} \mathscr P^{\prime}\left(\alpha_{n}+N\right)\quad(4.2.32)
\end{aligned}
\]
适当调整系数 \(a_{0}\), 使它满足使方程 (4.2.32) 右端为零的条件, 即
\(\mathscr P\left(\alpha_{2}+N\right)\)=0
\[a_{0}=\frac{\sum_{k=0}^{N-1}\left[(\alpha+k) p_{N-k}+q_{N-k}\right] c_{k}}{p^{\prime}\left(\alpha_{n}+N\right)}
\]
\[\begin{array}{l} (4.4.32)左边=0，a_0使右边为零，那么c_N可以为任意值。\\ 只不过a_0不是任意的了。通常F型级数a_0是任意的，a_1,a_2,a_3\dots都与a_0有关。\\特解中a_0不是任意的，c_1,c_2,c_3\dots都不是任意的，c_N是任意的。c_{N+1}又不任意了(利用齐次的递推式)。 \end{array} \]
非齐次方程 \((4.70)\) 有 Frobenius 型级数解 \((4.2 .31)\), 那么
\[y_{2}=\left(\frac{\partial \tilde{y}_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_{1}}-\bar{y}
\]
是减不是加的原因
\[\begin{array}{l} L(y_2)=L\left(\frac{\partial \tilde{y}_{1}}{\partial \alpha}\right)\bigg|_{\alpha=\alpha_{1}}-L\bar{y}\\ =a_0(x-x_{0})^{\alpha_1-2} \mathscr P'(\alpha_1)-a_{0}\left(x-x_{0}\right)^{\alpha_{2}+N-2} p^{\prime}\left(\alpha_{2}+N\right)=0\\ 即L(y_2)=0 \end{array} \]
- \(\frac{\partial \tilde{y}_{1}}{\partial \alpha}\bigg|_{\alpha=\alpha_{1}}\)是带\(\ln\)的特解，\(\bar y\)是F型级数的特解，两个解是线性独立，相减是通解。（不理解）🤔
就是原方程的另一个线性独立解, 其一般形式为
\[y_{2}=y_{1}(x) \ln \left(x-x_{0}\right)+\left(x-x_{0}\right)^{\alpha_{2}} \sum_{n=0}^{\infty} d_{n}\left(x-x_{0}\right)^{n}
\]
这就是 Fuchs 所预言的形式.
对于高阶方程, 情况更要复杂一些, 应该如何来进行求解请读 者思考. 对于 \(x_{0}\) 为正则奇点的高阶方程, 其一般形式为
高阶不要求
\[\begin{gathered}
y^{(n)}+\frac{q_{n-1}(x)}{x-x_{0}} y^{(n-1)}+\frac{q_{n-2}(x)}{\left(x-x_{0}\right)^{2}} y^{(n-2)}+\cdots \\
+\frac{q_{0}(x)}{\left(x-x_{0}\right)^{n}} y=0
\end{gathered}
\]
式中 \(q_{0}, q_{1}, \cdots, q_{n-1}\) 在 \(x_{0}\) 邻域内解析, 它的特征指数方程应为
\[\begin{array}{r}
\alpha(\alpha-1) \cdots(\alpha-n+1)+q_{n-1}\left(x_{0}\right) \alpha(\alpha-1) \cdots(\alpha-n+2) \\
+q_{n-2}\left(x_{0}\right) \alpha(\alpha-1) \cdots(\alpha-n+3)+\cdots+q_{0}\left(x_{0}\right)=0
\end{array}
\]
[ 例 4.2.3] 求 Euler 方程的解
\[x^{2} y^{\prime \prime}+p_{0} x y^{\prime}+q_{0} y=0
\]
它可以化为
\[y^{\prime \prime}+\frac{p_{0}}{x} y^{\prime}+\frac{q_{0}}{x^{2}} y=0
\]
x=0是正则奇点
它的特征方程为
\[\alpha(\alpha-1)+p_{0} \alpha+q_{0}=0
\]
特征方程\(\mathscr P(\alpha)= [\alpha(\alpha-1)+p_0\alpha+q_0]\)
当 \(\alpha_{1} \neq \alpha_{2}\) 时, 有解
\[\begin{aligned}
&y_{1}=c_{1} x^{\alpha_{1}} \\
&y_{2}=c_{2} x^{\alpha_{2}}
\end{aligned}
\]
F型级数解\(y=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\)
递推式\(a_n(\mathscr{P}(\mathcal{n+\alpha}))=-a_k\sum_{k=0}^{n-1}(p_{n-k}(k+\alpha)+q_{n-k})\)
n=1时,\(\because p_1,q_1=0,a_1(\mathscr{P}(\mathcal{1+\alpha}))=-a_0(p_{1}\alpha+q_{1})=0,a_1=0\)
同理，\(a_2,a_3,a_{\dots}\)=0.只有a_0不是零。
当 \(\alpha_{1}=\alpha_{2}=\alpha\) 时, 有解
\[\begin{aligned}
&y_{1}=c_{1} x^{\alpha} \\
&y_{2}=c_{2} x^{\alpha} \ln x
\end{aligned}
\]
\(y_{1}=\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\)
\(y_{2}=\frac{\partial y_{1}}{\partial \alpha}=y_{1} \ln \left(x-x_{0}\right)+\left(x-x_{0}\right)^{\alpha} \sum_{n=0}^{\infty}\left(\frac{\partial a_{n}}{\partial \alpha}\right)\left(x-x_{0}\right)^{n}\)
显然, 这是上面所讨论结果的特例.
[ 例 4.2.4] 用 Frobenius 方法, 求修正的 Bessel 方程之解.
\[y^{\prime \prime}+\frac{1}{x} y^{\prime}-\left(1+\frac{\nu^{2}}{x^{2}}\right) y=0
\]
它的特征方程为
\[\alpha^{2}-\nu^{2}=0
\]
\(p(x)=1,q(x)=-(x^2+\nu^2)\)
特征方程\(\mathscr P(\alpha)= [\alpha(\alpha-1)+p_0\alpha+q_0]=\alpha^2-\nu^2=0\)
\(\alpha=\pm\nu\)
- 当 \(\nu\) 不为零或整数时
两个根只差为\(2\nu\),有两个F型级数解
\[I_{\pm \nu}(x)=\sum_{n=0}^{\infty} \frac{\left(\frac{x}{2}\right)^{2 n \pm \nu}}{\Gamma(\pm \nu+n+1) n !}
\]
这是两个线性独立解.
- 当 \(\nu\) 为半整数时,
差就是整数(并且是奇数）。
刚好 (4.2.18) 右端为零, 上式仍是方程的解.
右端为零的原因：
\[\begin{cases} p_0=1,&&&p_1,p_2,\dots=0\\ q_0=-\nu^2,&q_1=0,&q_2=-1,&q_3,\dots=0 \end{cases} \]\[\begin{array}{l} a_n(\mathscr{P}(\mathcal{n+\alpha}))=-a_{k} \sum_{k=0}^{n-1}\left(p_{n-k}(k+\alpha)+q_{n-k}\right) \\ =-\left[\left[-\nu \cdot p_{n}+q_{n}\right] a_{0}\right. \\ +\left[(-\nu+1) p_{n-1}+q_{n-1}\right] a_{1} \\ +\left[(-\nu+2) p_{n-2}+q_{n-2}\right] a_{2} \\ +\cdots \\ +\left[(-\nu+n-2) p_{2}+q_{2}\right] a_{n-2} \\ \left.\left.+\left[(-\nu+n-1) p_{1}+q_{1}\right] a_{n-1}\right]\right]① \\ =a_{n-2}\\ \therefore a_n(\mathscr{P}(\mathcal{n+\alpha}))=a_{n-2}② \end{array} \]上式中的p,q系数只有 非零，为-1。所以①式倒数第二项留下来了。
\[\begin{array}{l} \therefore n\ne\nu-(-\nu)时，递推式左边不为零，由a_0推出a_2,a_4,\dots(偶数项存在)；\\ a_1由a_0推出，a_1\mathscr P(1+\alpha)=-a_0(p_1+q_1)=0,a_1=0\\ \therefore 由②式，a_3,a_5,\dots=0(奇数项不存在) \end{array} \]\(设\nu-(-\nu)=m，则m必为奇数\)
\(p(\alpha+m) a_{m}=-\sum_{k=0}^{m-1}\left[(\alpha+k) p_{m-k}+q_{m-k}\right] a_{k}=a_{m-2}\)，
左边是零但是右边也为零,使递推式满足。
- 当 \(\nu=0\) 时, 有重根, 第一个解为
\[I_{0}(x)=\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2} x\right)^{2 n}}{(n !)^{2}}
\]
为求出第二个解, 先令 \(\alpha \neq 0\), 将 Frobenius 型级数代入方程, 得到 递推关系, \(a_{0} \neq 0\).
\[\begin{aligned}
&a_{2 n-1}(\alpha)=0 \\
&a_{2 n}(\alpha)=\frac{a_{0}}{(\alpha+2 n)^{2}(\alpha+2 n-2)^{2} \cdots(\alpha+2)^{2}}
\end{aligned}
\]
由此导出 \(b_{0} \neq 0\).
\[\begin{aligned}
&b_{2 n-1}=0 \\
&b_{2 n}=\left.\frac{\partial a_{2 n}}{\partial \alpha}\right|_{\alpha=0}=-\frac{a_{0}}{2^{2 n}(n !)^{2}}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)
\end{aligned}
\]
