4.3勒让德方程正常点附近邻域上的求解

在 z = 0 为中心的区域上求解勒让德方程

\[(1 - z^2)u''(z) - 2zu'(z) + l(l + 1)u(z) = 0 \]

化为标准形式 \(u''(z) - \frac{2z}{1 - z^2}u'(z) + \frac{l(l+1)}{1 - z^2}u(z) = 0\)

\[p(z) = -\frac{2z}{1 - z^2}, \quad q(z) = \frac{l(l+1)}{1 - z^2} \]

\(z_0 = 0\) ———— 方程的正常点
\(z_1 = 1, \quad z_2 = -1\) ——— 方程的孤立奇点
\(|z| < 1\) ———— 相应的解析圆域

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求解：

用$$u(z) = \sum_{k=0}^{\infty} a_k z^k$$展开

可得到$$a_{k+2} = \frac{k(k+1) - l(l+1)}{(k+1)(k+2)} a_k = \frac{(k-l)(k+l+1)}{(k+1)(k+2)} a_k$$

\[u(z) = a_0 u_0(z) + a_1 u_1(z) \]

具体过程见附录

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附录：

\[u(z) = \sum_{k=0}^{\infty} a_k z^k \rightarrow (1 - z^2)u''(z) - 2zu'(z) + l(l + 1)u(z) = 0 \]

\[\sum_{k=0}^{\infty} k(k - 1)a_k z^{k-2} - \sum_{k=0}^{\infty} k(k - 1)a_k z^k - 2\sum_{k=0}^{\infty} ka_k z^k + l(l + 1) \sum_{k=0}^{\infty} a_k z^k = 0 \]

\[\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} z^k - \sum_{k=0}^{\infty} k(k-1)a_k z^k - 2\sum_{k=0}^{\infty} ka_k z^k + l(l + 1) \sum_{k=0}^{\infty} a_k z^k = 0 \]

由各幂次项系数必为 0，最终就得到

\[a_{k+2} = \frac{k(k+1) - l(l+1)}{(k+1)(k+2)} a_k \]

化简使其更容易递推：$$= \frac{(k-l)(k+l+1)}{(k+1)(k+2)} a_k$$