常微分方程/超几何方程的Frobenius解法

在下面，我们使用以费迪南德·格奥尔格·弗罗贝尼乌斯命名的弗罗贝尼乌斯方法来求解被称为超几何微分方程的二阶微分方程。这是一种使用微分方程的级数解的方法，我们假设解采用级数的形式。对于复杂的常微分方程，这通常是我们使用的方法。

超几何微分方程的解非常重要。例如，可以证明勒让德微分方程是超几何微分方程的特例。因此，通过求解超几何微分方程，人们可以在进行必要的代换后，直接比较它的解以得到勒让德微分方程的解。有关更多详细信息，请查看超几何微分方程

我们将证明这个方程有三个奇点，即在x = 0，x = 1 和无穷远处。然而，由于这些将被证明是正则奇点，我们将能够假设一个级数形式的解。由于这是一个二阶微分方程，我们必须有两个线性无关的解。

然而，问题是我们假设的解可能是或不是独立的，或者更糟的是，甚至可能没有定义（取决于方程参数的值）。这就是为什么我们将研究参数的不同情况并相应地修改我们假设的解。

方程

在所有奇点附近求解超几何方程

$x(1-x)y''+\left\{\gamma -(1+\alpha +\beta )x\right\}y'-\alpha \beta y=0$

x = 0 附近的解

令

${\begin{aligned}P_{0}(x)=-\alpha \beta ,&&P_{1}(x)=\gamma -(1+\alpha +\beta )x,&&P_{2}(x)=x(1-x)\end{aligned}}$

那么

$P_{2}(0)=0,P_{2}(1)=0.\,$

因此，x = 0 和x = 1 是奇点。让我们从x = 0 开始。要查看它是否为正则奇点，我们研究以下极限

${\begin{aligned}\lim _{x\to a}{\frac {(x-a)P_{1}(x)}{P_{2}(x)}}&=\lim _{x\to 0}{\frac {(x-0)(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\lim _{x\to 0}{\frac {x(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\gamma \\\lim _{x\to a}{\frac {(x-a)^{2}P_{0}(x)}{P_{2}(x)}}&=\lim _{x\to 0}{\frac {(x-0)^{2}(-\alpha \beta )}{x(1-x)}}=\lim _{x\to 0}{\frac {x^{2}(-\alpha \beta )}{x(1-x)}}=0\end{aligned}}$

因此，两个极限都存在，并且 *x* = 0 是一个正则奇点。所以，我们假设解的形式为

$y=\sum _{r=0}^{\infty }a_{r}x^{r+c}$

其中 *a*₀ ≠ 0。所以，

${\begin{aligned}y'=\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}&&y''=\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}.\end{aligned}}$

将这些代入超几何方程，得到

${\begin{aligned}&x\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}-x^{2}\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\&\quad -(1+\alpha +\beta )x\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-\alpha \beta \sum _{r=0}^{\infty }a_{r}x^{r+c}=0\end{aligned}}$

也就是，

${\begin{aligned}&\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\&\quad -(1+\alpha +\beta )\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c}-\alpha \beta \sum _{r=0}^{\infty }a_{r}x^{r+c}=0\end{aligned}}$

为了简化这个方程式，我们需要所有指数都相同，都等于 r + c - 1，即最小的指数。因此，我们如下转换指数

${\begin{aligned}&\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=1}^{\infty }a_{r-1}(r+c-1)(r+c-2)x^{r+c-1}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\&\quad -(1+\alpha +\beta )\sum _{r=1}^{\infty }a_{r-1}(r+c-1)x^{r+c-1}-\alpha \beta \sum _{r=1}^{\infty }a_{r-1}x^{r+c-1}=0\end{aligned}}$

因此，分离从 0 开始的和的第一个项，我们得到

${\begin{aligned}&a_{0}(c(c-1)+\gamma c)x^{c-1}+\sum _{r=1}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=1}^{\infty }a_{r-1}(r+c-1)(r+c-2)x^{r+c-1}\\&\quad +\gamma \sum _{r=1}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )\sum _{r=1}^{\infty }a_{r-1}(r+c-1)x^{r+c-1}-\alpha \beta \sum _{r=1}^{\infty }a_{r-1}x^{r+c-1}=0\end{aligned}}$

现在，根据 x 的所有幂的线性独立性，也就是函数 1、x、x² 等的线性独立性，x^k 的系数对于所有的 k 都为零。因此，根据第一项，我们有

$a_{0}(c(c-1)+\gamma c)=0\,$

这是指标方程。因为 a₀ ≠ 0，我们有

$c(c-1+\gamma )=0.\,$

因此，

${\begin{aligned}c_{1}=0&&c_{2}=1-\gamma \end{aligned}}$

此外，从其余项中，我们有

${\begin{aligned}&((r+c)(r+c-1)+\gamma (r+c))a_{r}\\&\quad +(-(r+c-1)(r+c-2)-(1+\alpha +\beta )(r+c-1)-\alpha \beta )a_{r-1}=0\end{aligned}}$

因此，

${\begin{aligned}a_{r}&={\frac {(r+c-1)(r+c-2)+(1+\alpha +\beta )(r+c-1)+\alpha \beta }{(r+c)(r+c-1)+\gamma (r+c)}}a_{r-1}\\&={\frac {(r+c-1)(r+c+\alpha +\beta -1)+\alpha \beta }{(r+c)(r+c+\gamma -1)}}a_{r-1}\end{aligned}}$

但是

${\begin{aligned}&(r+c-1)(r+c+\alpha +\beta -1)+\alpha \beta \\&\quad =(r+c-1)(r+c+\alpha -1)+(r+c-1)\beta +\alpha \beta \\&\quad =(r+c-1)(r+c+\alpha -1)+\beta (r+c+\alpha -1)\end{aligned}}$

因此，我们得到递推关系

$a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c+\gamma -1)}}a_{r-1},{\text{ for }}r\geq 1.$

现在让我们通过用a₀而不是a_r − 1表示a_r来简化这个关系。从递推关系（注意：下面，形式为 (u)_r 的表达式指的是Pochhammer符号）。

${\begin{aligned}a_{1}&={\frac {(c+\alpha )(c+\beta )}{(c+1)(c+\gamma )}}a_{0}\\a_{2}&={\frac {(c+\alpha +1)(c+\beta +1)}{(c+2)(c+\gamma +1)}}a_{1}={\frac {(c+\alpha +1)(c+\alpha )(c+\beta +1)}{(c+2)(c+1)(c+\gamma )(c+\gamma +1)}}a_{0}={\frac {(c+\alpha )_{2}(c+\beta )_{2}}{(c+1)_{2}(c+\gamma )_{2}}}a_{0}\\a_{3}&={\frac {(c+\alpha +2)(c+\beta +2)}{(c+3)(c+\gamma +2)}}a_{2}={\frac {(c+\alpha )_{2}(c+\alpha +2)(c+\beta )_{2}(c+\beta +2}{(c+1)_{2}(c+3)(c+\gamma )_{2}(c+\gamma +2)}}a_{0}\\&={\frac {(c+\alpha )_{3}(c+\beta )_{3}}{(c+1)_{3}(c+\gamma )_{3}}}a_{0}\end{aligned}}$

我们可以看到,

$a_{r}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}a_{0},{\text{ for }}r\geq 0$

因此，我们假设的解的形式为

$y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c}.$

现在我们准备研究对应于c₁ − c₂ = γ − 1 不同情况的解（需要注意的是，这简化为研究参数γ的性质：它是否是一个整数）。

对解的分析，以两个根的差 γ − 1 为基础

γ 不是整数

然后 y₁ = y|_{c = 0} 以及 y₂ = y|_{c = 1 − γ}。由于

$y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c},$

我们有

${\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(\gamma )_{r}}}x^{r}\\&=a_{0}\cdot {{}_{2}F_{1}}(\alpha ,\beta ;\gamma ;x)\\y_{2}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1-\gamma +1)_{r}(1-\gamma +\gamma )_{r}}}x^{r+1-\gamma }=a_{0}x^{1-\gamma }\sum _{r=0}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(2-\gamma )_{r}}}x^{r}\\&=a_{0}x^{1-\gamma }{{}_{2}F_{1}}(\alpha -\gamma +1,\beta -\gamma +1;2-\gamma ;x)\end{aligned}}$

因此，$y=A'y_{1}+B'y_{2}.$ 令 *A*′ *a*₀ = *a* 且 *B*′ *a*₀ = *B*。 然后

$y=A{{}_{2}F_{1}}(\alpha ,\beta ;\gamma ;x)+Bx^{1-\gamma }{{}_{2}F_{1}}(\alpha -\gamma +1,\beta -\gamma +1;2-\gamma ;x)\,}$

γ = 1

然后 *y*₁ = *y*|_{*c* = 0}。 由于 γ = 1，我们有

$y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r+c}.$

因此，

${\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r}}}x^{r}=a_{0}{{}_{2}F_{1}}(\alpha ,\beta ;1;x)\\y_{2}&=\left.{\frac {\partial y}{\partial c}}\right|_{c=0}.\end{aligned}}$

为了计算这个导数，设

$M_{r}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}.$

那么

$\ln(M_{r})=\ln \left({\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\right)=\ln(c+\alpha )_{r}+\ln(c+\beta )_{r}-2\ln(c+1)_{r}$

但是

$\ln(c+\alpha )_{r}=\ln {\bigl (}(c+\alpha )(c+\alpha +1)\cdots (c+\alpha +r-1){\bigr )}=\sum _{k=0}^{r-1}\ln(c+\alpha +k).$

因此，

${\begin{aligned}\ln(M_{r})&=\sum _{k=0}^{r-1}\ln(c+\alpha +k)+\sum _{k=0}^{r-1}\ln(c+\beta +k)-2\sum _{k=0}^{r-1}\ln(c+1+k)\\&=\sum _{k=0}^{r-1}{\bigl (}\ln(c+\alpha +k)+\ln(c+\beta +k)-2\ln(c+1+k){\bigr )}\end{aligned}}$

对等式的两边关于c求导，我们得到

${\frac {1}{M_{r}}}{\frac {\partial M_{r}}{\partial c}}=\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right).$

因此，

${\frac {\partial M_{r}}{\partial c}}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right).$

现在，

$y=a_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}=a_{0}x^{c}\sum _{r=0}^{\infty }M_{r}x^{r}.$

因此，

${\begin{aligned}{\frac {\partial y}{\partial c}}&=a_{0}x^{c}\ln(x)\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}\\&\quad +a_{0}x^{c}\sum _{r=0}^{\infty }\left({\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\left\{\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right)\right\}\right)x^{r}\\&=a_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r})^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right)\right)x^{r}.\end{aligned}}$

对于 *c* = 0，我们得到

$y_{2}=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)x^{r}.$

因此，y = C′ y₁ + D′ y₂。令 C′ a₀ = C 且 D′ a₀ = D。然后

$y=C{{}_{2}F_{1}}(\alpha ,\beta ;1;x)+D\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)x^{r}$

γ 是一个整数，且 γ ≠ 1

γ ≤ 0

从递推关系

$a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c+\gamma -1)}}a_{r-1},$

我们可以看到，当 c = 0（较小的根）时，a_{1 − γ} → ∞。因此，我们必须进行替换 a₀ = b₀ (c - c_i)，其中 c_i 是使我们的解发散的根。因此，我们取 a₀ = b₀ c，并且我们的假设解采用新形式

$y_{b}=b_{0}x^{c}\sum _{r=0}^{\infty }{\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r}$

然后 y₁ = y_b|_{c = 0}。我们可以看到，在

${\frac {c(c+\alpha )_{1-\gamma }(c+\beta )_{1-\gamma }}{(c+1)_{1-\gamma }(c+\gamma )_{1-\gamma }}}x^{1-\gamma }$

之前的所有的项都消失了，因为分子中有 c。然而，从这一项开始，分子中的 c 就消失了。为了看到这一点，请注意

$(c+\gamma )_{1-\gamma }=(c+\gamma )(c+\gamma +1)\cdots c.$

因此，我们的解决方案形式为

${\begin{aligned}y_{1}&=b_{0}\left({\frac {(\alpha )_{1-\gamma }(\beta )_{1-\gamma }}{(1)_{1-\gamma }(\gamma )_{-\gamma }}}x^{1-\gamma }+{\frac {(\alpha )_{2-\gamma }(\beta )_{2-\gamma }}{(1)_{2-\gamma }(\gamma )_{-\gamma }(1)}}x^{2-\gamma }+{\frac {(\alpha )_{3-\gamma }(\beta )_{3-\gamma }}{(1)_{3-\gamma }(\gamma )_{-\gamma }(1)(2)}}x^{3-\gamma }+\cdots \right)\\&={\frac {b_{0}}{(\gamma )_{-\gamma }}}\sum _{r=1-\gamma }^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r+\gamma -1}}}x^{r}.\end{aligned}}$

现在，

$y_{2}=\left.{\frac {\partial y_{b}}{\partial c}}\right|_{c=1-\gamma }.$

为了计算这个导数，设

$M_{r}={\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}.$

然后，遵循 上一情况 中的方法，我们得到

${\frac {\partial M_{r}}{\partial c}}={\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}\left\{{\frac {1}{c}}+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {1}{c+1+k}}-{\frac {1}{c+\gamma +k}}\right)\right\}.$

现在，

$y_{b}=b_{0}\sum {r=0}^{\infty }{\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c}=b_{0}x^{c}\sum _{r=0}^{\infty }M_{r}x^{r}.$

因此，

${\begin{aligned}{\frac {\partial y}{\partial c}}&=b_{0}x^{c}\ln(x)\sum _{r=0}^{\infty }{\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r}\\&\quad +b_{0}x^{c}\sum _{r=0}^{\infty }{\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}\left\{{\frac {1}{c}}+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {1}{c+1+k}}-{\frac {1}{c+\gamma +k}}\right)\right\}x^{r}\end{aligned}}$

因此，

${\begin{aligned}{\frac {\partial y}{\partial c}}=b_{0}x^{c}\sum _{r=0}^{\infty }&{\frac {c(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}{\Biggl (}\ln x+{\frac {1}{c}}+\\&\quad +\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {1}{c+1+k}}-{\frac {1}{c+\gamma +k}}\right){\Biggr )}x^{r}.\end{aligned}}$

当c = 1- γ 时，我们得到y₂。因此，y = E′ y₁ + F′ y₂。令E′ b₀ = E 且F′ b₀ = F。那么

${\begin{aligned}y&={\frac {E}{(\gamma )_{-\gamma }}}\sum _{r=1-\gamma }^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r+\gamma -1}}}x^{r}\\&\quad {\begin{aligned}{}+Fx^{1-\gamma }\sum _{r=0}^{\infty }&{\frac {(1-\gamma )(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(2-\gamma )_{r}(1)_{r}}}{\Biggl (}\ln x+{\frac {1}{1-\gamma }}+\\&+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k+1-\gamma }}+{\frac {1}{\beta +k+1-\gamma }}-{\frac {1}{2+k-\gamma }}-{\frac {1}{1+k}}\right){\Biggr )}x^{r}.\end{aligned}}\end{aligned}}$

γ > 1

从递推关系

$a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c+\gamma -1)}}a_{r-1},$

我们可以看到当 *c* = 1 - γ（较小的根）时，*a*_{γ − 1} → ∞。因此，我们必须进行替换 *a*₀ = *b*₀(*c* − *c*_*i*)，其中 *c*_*i* 是我们的解为无穷大的根。因此，我们取 *a*₀ = *b*₀(*c* + γ - 1)，我们的假设解将采用新的形式

$y_{b}=b_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r}.$

然后 *y*₁ = *y*_*b*|_{*c* = 1 - γ}。所有之前的项

${\frac {(c+\gamma -1)(c+\alpha )_{\gamma -1}(c+\beta )_{\gamma -1}}{(c+1)_{\gamma -1}(c+\gamma )_{\gamma -1}}}x^{\gamma -1}$

由于分子中的 *c* + γ - 1 而消失。然而，从这一项开始，分子中的 *c* + γ - 1 消失。要看到这一点，请注意

$(c+1)_{\gamma -1}=(c+1)(c+2)\cdots (c+\gamma -1).$

因此，我们的解决方案形式为

${\begin{aligned}y_{1}&=b_{0}x^{1-\gamma }\left({\frac {(\alpha +1-\gamma )_{\gamma -1}(\beta +1-\gamma )_{\gamma -1}}{(2-\gamma )_{\gamma -2}(1)_{\gamma -1}}}x^{\gamma -1}+{\frac {(\alpha +1-\gamma )_{\gamma }(\beta +1-\gamma )_{\gamma }}{(2-\gamma )_{\gamma -2}(1)(1)_{\gamma }}}x^{\gamma }+\cdots \right)\\&={\frac {b_{0}}{(2-\gamma )_{\gamma -2}}}x^{1-\gamma }\sum _{r=\gamma -1}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+1-\gamma }}}x^{r}.\end{aligned}}$

现在，

$y_{2}=\left.{\frac {\partial y_{b}}{\partial c}}\right|_{c=0}.$

为了计算这个导数，设

$M_{r}={\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}.$

然后，根据 上面第二种情况 中的方法，

${\begin{aligned}{\frac {\partial M_{r}}{\partial c}}&={\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}{\Biggl (}{\frac {1}{c+\gamma -1}}+\\&\qquad +\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {1}{c+1+k}}-{\frac {1}{c+\gamma +k}}\right){\Biggr )}\end{aligned}}$

现在，

$y_{b}=b_{0}\sum _{r=0}^{\infty }{\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c}=b_{0}x^{c}\sum _{r=0}^{\infty }M_{r}x^{r}.$

因此，

${\begin{aligned}{\frac {\partial y}{\partial c}}&=b_{0}x^{c}\ln(x)\sum _{r=0}^{\infty }{\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r}+\\&\qquad {\begin{aligned}{}+b_{0}x^{c}\sum _{r=0}^{\infty }&{\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}{\Biggl (}{\frac {1}{c+\gamma -1}}+\\&+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {1}{c+1+k}}-{\frac {1}{c+\gamma +k}}\right){\Biggr )}x^{r}\end{aligned}}\\&{\begin{aligned}{}=b_{0}x^{c}\sum _{r=0}^{\infty }&{\frac {(c+\gamma -1)(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}{\Biggl (}\ln x+{\frac {1}{c+\gamma -1}}+\\&+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {1}{c+1+k}}-{\frac {1}{c+\gamma +k}}\right){\Biggr )}x^{r}.\end{aligned}}\end{aligned}}$

当c = 0时，我们得到y₂。因此，y = G′y₁ + H′y₂。令G′b₀ = E，H′b₀ = F，则

${\begin{aligned}y={\frac {G}{(2-\gamma )_{\gamma -2}}}&x^{1-\gamma }\sum _{r=\gamma -1}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+1-\gamma }}}x^{r}\\&{\begin{aligned}{}+H\sum _{r=0}^{\infty }&{\frac {(1-\gamma )(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(2-\gamma )_{r}(1)_{r}}}{\Biggl (}\ln x+{\frac {1}{\gamma -1}}+\\&+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {1}{1+k}}-{\frac {1}{\gamma +k}}\right){\Biggr )}x^{r}.\end{aligned}}\end{aligned}}$

关于x = 1 的解

现在我们研究奇点x = 1。为了查看它是否是正则奇点，

${\begin{aligned}&\lim _{x\to a}{\frac {(x-a)P_{1}(x)}{P_{2}(x)}}=\lim _{x\to 1}{\frac {(x-1)(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}\\&\quad =\lim _{x\to 1}{\frac {-(\gamma -(1+\alpha +\beta )x)}{x}}=1+\alpha +\beta -\gamma \\&\lim _{x\to a}{\frac {(x-a)^{2}P_{0}(x)}{P_{2}(x)}}=\lim _{x\to 1}{\frac {(x-1)^{2}(-\alpha \beta )}{x(1-x)}}=\lim _{x\to 1}{\frac {(x-1)\alpha \beta }{x}}=0\end{aligned}}$

因此，两个极限都存在，x = 1 是一个正则奇点。现在，我们不假设解的形式为

$y=\sum _{r=0}^{\infty }a_{r}(x-1)^{r+c},$

我们将尝试用点 *x* = 0 的解来表达这种情况的解。我们按如下步骤进行：我们有超几何方程

$x(1-x)y''+(\gamma -(1+\alpha +\beta )x)y'-\alpha \beta y=0.\,$

令 *z* = 1 - *x*。那么

${\begin{aligned}&{\frac {dy}{dx}}={\frac {dy}{dz}}\times {\frac {dz}{dx}}=-{\frac {dy}{dz}}=-y'\\&{\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)={\frac {d}{dx}}\left(-{\frac {dy}{dz}}\right)={\frac {d}{dz}}\left(-{\frac {dy}{dz}}\right)\times {\frac {dz}{dx}}={\frac {d^{2}y}{dz^{2}}}=y''\end{aligned}}$

因此，方程的形式变为

$z(1-z)y''+(\alpha +\beta -\gamma +1-(1+\alpha +\beta )z)y'-\alpha \beta y=0.\,$

由于 *z* = 1 - *x*，超几何方程在 *x* = 1 处的解与该方程在 *z* = 0 处的解相同。但是，*z* = 0 处的解与我们为点 *x* = 0 获得的解相同，如果我们将每个 γ 替换为 α + β - γ + 1。因此，为了获得解，我们只需在先前的结果中进行此替换。还要注意，对于 *x* = 0，*c*₁ = 0 且 *c*₂ = 1 - γ。因此，在我们的例子中，*c*₁ = 0，而 *c*₂ = γ - α - β。现在让我们写出解。需要注意的是，在以下内容中，我们将每个 *z* 替换为 1 - *x*。

分析解关于两个根之差 γ − α − β 的表达式

γ − α − β 不是整数

${\begin{aligned}y&=A\cdot {{}_{2}F_{1}}(\alpha ,\beta ;\alpha +\beta -\gamma +1;1-x)\\&\quad +B(1-x)^{\gamma -\alpha -\beta }{{}_{2}F_{1}}(\gamma -\alpha ,\gamma -\beta ;\gamma -\alpha -\beta +1;1-x)\end{aligned}}$

γ − α − β = 0

${\begin{aligned}y&=C\cdot {{}_{2}F_{1}}(\alpha ,\beta ;1;1-x)\\&\quad +D\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln(1-x)+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)(1-x)^{r}\end{aligned}}$

γ − α − β 是整数，且 γ − α − β ≠ 0

γ − α − β > 0

${\begin{aligned}y&={\frac {E}{(\alpha +\beta -\gamma +1)_{\gamma -\alpha -\beta -1}}}\sum _{r=1-\gamma }^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r+\alpha +\beta -\gamma }}}(1-x)^{r}+{}\\&\quad {\begin{aligned}{}+F(1-x)^{\gamma -\alpha -\beta }\sum _{r=0}^{\infty }&{\frac {(\gamma -\alpha -\beta )(\gamma -\beta )_{r}(\gamma -\alpha )_{r}}{(1+\gamma -\alpha -\beta )_{r}(1)_{r}}}{\Biggl (}\ln(1-x)+{\frac {1}{\gamma -\alpha -\beta }}+{}\\&+\sum _{k=0}^{r-1}\left({\frac {1}{k+\gamma -\beta }}+{\frac {1}{k+\gamma -\alpha }}-{\frac {1}{1+k+\gamma -\alpha -\beta }}-{\frac {1}{1+k}}\right){\Biggr )}(1-x)^{r}\end{aligned}}\end{aligned}}$

γ − α − β < 0

${\begin{aligned}y&={\frac {G}{(1+\gamma -\alpha -\beta )_{\alpha +\beta -\gamma -1}}}(1-x)^{\gamma -\alpha -\beta }\sum _{r=\alpha +\beta -\gamma }^{\infty }{\frac {(\gamma -\beta )_{r}(\gamma -\alpha )_{r}}{(1)_{r}(1)_{r+\gamma -\alpha -\beta }}}(1-x)^{r}+{}\\&\quad {\begin{aligned}{}+H\sum _{r=0}^{\infty }&{\frac {(\gamma -\alpha -\beta )(\gamma -\beta )_{r}(\gamma -\alpha )_{r}}{(1+\gamma -\alpha -\beta )_{r}(1)_{r}}}{\Biggl (}\ln(1-x)+{\frac {1}{\alpha +\beta -\gamma }}+{}\\&+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {1}{1+k}}-{\frac {1}{\alpha +\beta -\gamma +1+k}}\right){\Biggr )}(1-x)^{r}\end{aligned}}\end{aligned}}$

无穷远处解

最后，我们研究当 *x* → ∞ 时的奇点。由于我们无法直接研究，我们令 *x* = *s*⁻¹。然后当 *x* → ∞ 时，方程的解与当 *s* = 0 时，修改后的方程的解相同。我们有

${\begin{aligned}&x(1-x)y''+\left\{\gamma -(1+\alpha +\beta )x\right\}y'-\alpha \beta y=0\\&{\frac {dy}{dx}}={\frac {dy}{ds}}\times {\frac {ds}{dx}}=-s^{^{2}}\times {\frac {dy}{ds}}=-s^{^{2}}y'\\&{\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)={\frac {d}{dx}}\left(-s^{^{2}}\times {\frac {dy}{ds}}\right)={\frac {d}{ds}}\left(-s^{^{2}}\times {\frac {dy}{ds}}\right)\times {\frac {ds}{dx}}\\&\left((-2s)\times {\frac {dy}{ds}}+(-s^{2}){\frac {d^{2}y}{ds^{2}}}\right)\times (-s^{2})=2s^{3}y'+s^{4}y''\end{aligned}}$

因此，该方程将采用新的形式

${\frac {1}{s}}\left(1-{\frac {1}{s}}\right)\left(2s^{3}y'+s^{4}y''\right)+\left(\gamma -(1+\alpha +\beta ){\frac {1}{s}}\right)(-s^{2}y')-\alpha \beta y=0$

可简化为

$(s^{3}-s^{2})y''+{\bigl (}(2-\gamma )s^{2}+(\alpha +\beta -1)s{\bigr )}y'-\alpha \beta y=0.$

令

$P_{0}(s)=-\alpha \beta ,\qquad P_{1}(s)=((2-\gamma )s^{2}+(\alpha +\beta -1)s),\qquad P_{2}(s)=(s^{3}-s^{2}).$

正如我们所说，我们只研究当 s = 0 时解。我们可以看到，这是一个奇点，因为 P₂(0) = 0。为了判断它是否是正则奇点，

${\begin{aligned}&{\underset {s\to a}{\mathop {\lim } }}\,{\frac {\left(s-a\right)P_{1}(s)}{P_{2}(s)}}={\underset {s\to 0}{\mathop {\lim } }}\,{\frac {\left(s-0\right)((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{(s^{3}-s^{2})}}={\underset {s\to 0}{\mathop {\lim } }}\,{\frac {((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{s^{2}-s}}\\&={\underset {s\to 0}{\mathop {\lim } }}\,{\frac {((2-\gamma )s+(\alpha +\beta -1))}{s-1}}=1-\alpha -\beta {\text{ }}{\text{. }}\\&{\underset {s\to a}{\mathop {\lim } }}\,{\frac {\left(s-a\right)^{2}P_{0}(s)}{P_{2}(s)}}={\underset {s\to 0}{\mathop {\lim } }}\,{\frac {\left(s-0\right)^{2}\left(-\alpha \beta \right)}{(s^{3}-s^{2})}}={\underset {x\to 0}{\mathop {\lim } }}\,{\frac {\left(-\alpha \beta \right)}{s-1}}=\alpha \beta {\text{ }}{\text{.}}\end{aligned}}$

因此，这两个极限都存在，并且 s = 0 是一个正则奇点。因此，我们假设解的形式为

$y=\sum \limits _{r=0}^{\infty }{a_{r}s^{r+c}}$

其中 a₀ ≠ 0。

因此，

$y'=\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}$ 和 $y''=\sum \limits _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}}.$

代入修正后的超几何方程，我们得到

${\begin{aligned}&(s^{3}-s^{2})y''+((2-\gamma )s^{2}+(\alpha +\beta -1)s)y'-\alpha \beta y=0\\&s^{3}\sum \limits _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}}-s^{2}\sum \limits _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c-2}}\\&+(2-\gamma )s^{2}\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}+(\alpha +\beta -1)s\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}-\alpha \beta \sum \limits _{r=0}^{\infty }{a_{r}s^{r+c}}=0\end{aligned}}$

即

${\begin{aligned}\sum \limits _{r=0}^{\infty }&{a_{r}(r+c)(r+c-1)s^{r+c+1}}-\sum \limits _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}\\&+(2-\gamma )\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c+1}}+(\alpha +\beta -1)\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum \limits _{r=0}^{\infty }{a_{r}s^{r+c}}=0\end{aligned}}$

为了简化这个方程，我们需要所有幂都相同，等于r + c，即最小幂。 因此，我们如下更改索引

${\begin{aligned}&\sum \limits _{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}}-\sum \limits _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}\\&+(2-\gamma )\sum \limits _{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}}+(\alpha +\beta -1)\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum \limits _{r=0}^{\infty }{a_{r}s^{r+c}}=0\end{aligned}}$

因此，分离从 0 开始的和的第一个项，我们得到

${\begin{aligned}&a_{0}\left\{-(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right\}s^{c}+\sum \limits _{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}}\\&-\sum \limits _{r=1}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}+(2-\gamma )\sum \limits _{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}}\\&+(\alpha +\beta -1)\sum \limits _{r=1}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum \limits _{r=1}^{\infty }{a_{r}s^{r+c}}=0\\\end{aligned}}$

现在，从所有 s 的幂的线性无关性（即函数 1，s，s²，...，s^k 的系数对于所有 k 都消失）。因此，从第一项我们有

$a_{0}\left\{-(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right\}=0$

这是指标方程。因为 a₀ ≠ 0，我们有

$(c)(-c+1+\alpha +\beta -1)-\alpha \beta )=0.\,$

因此，c₁ = α 且 c₂ = β。

此外，从其余的项我们有

${\begin{aligned}&\left\{(r+c-1)(r+c-2)+(2-\gamma )(r+c-1)\right\}a_{r-1}\\&+\left\{-(r+c)(r+c-1)+(\alpha +\beta -1)(r+c)-\alpha \beta \right\}a_{r}=0\end{aligned}}$

因此，

${\begin{aligned}&a_{r}=-{\frac {\left\{(r+c-1)(r+c-2)+(2-\gamma )(r+c-1)\right\}}{\left\{-(r+c)(r+c-1)+(\alpha +\beta -1)(r+c)-\alpha \beta \right\}}}a_{r-1}\\&{\text{ }}={\frac {\left\{(r+c-1)(r+c-\gamma )\right\}}{\left\{(r+c)(r+c-\alpha -\beta )+\alpha \beta \right\}}}a_{r-1}\end{aligned}}$

但是

${\begin{aligned}(r+c)(r+c-\alpha -\beta )+\alpha \beta &=(r+c-\alpha )(r+c)-\beta (r+c)+\alpha \beta \\&=(r+c-\alpha )(r+c)-\beta (r+c-\alpha ).\end{aligned}}$

因此，我们得到递推关系

$a_{r}={\frac {(r+c-1)(r+c-\gamma )}{(r+c-\alpha )(r+c-\beta )}}a_{r-1},\,\forall r\geq 1$

现在，让我们通过用 *a*₀ 而不是 *a*_{*r* − 1} 来表示 *a*_r 来简化这个关系。从递推关系，

${\begin{aligned}&a_{1}={\frac {(c)(c+1-\gamma )}{(c+1-\alpha )(c+1-\beta )}}a_{0}\\&a_{2}={\frac {(c+1)(c+2-\gamma )}{(c+2-\alpha )(c+2-\beta )}}a_{1}={\frac {(c+1)(c)(c+2-\gamma )(c+1-\gamma )}{(c+2-\alpha )(c+1-\alpha )(c+2-\beta )(c+1-\beta )}}a_{0}\\&={\text{ }}{\frac {(c)_{2}(c+1-\gamma )_{2}}{(c+1-\alpha )_{2}(c+1-\beta )_{2}}}a_{0}\end{aligned}}$

我们可以看到,

$a_{r}={\frac {(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}a_{0}{\text{ }}\forall {\text{r}}\geq {\text{0}}$

因此，我们假设的解的形式为

$y=a_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}\right)}$

现在我们可以研究对应于 c₁ − c₂ = α − β 不同情况下的解。

根据两个根之差 α - β 分析解

α − β 不是整数

那么 y₁ = y|_{c = α} 和 y₂ = y|_{c = β}。 由于

$y=a_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}\right)}$,

我们有

${\begin{aligned}y_{1}&=a_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(\alpha +1-\beta )_{r}}}s^{r+\alpha }\right)}=a_{0}s_{2}^{\alpha }F_{1}(\alpha ,{\text{ }}\alpha +1-\gamma ;{\text{ }}\alpha +1-\beta ;{\text{ s)}}\\y_{2}&=a_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(\beta )_{r}(\beta +1-\gamma )_{r}}{(\beta +1-\alpha )_{r}(1)_{r}}}s^{r+\beta }\right)=a_{0}s_{2}^{\beta }F_{1}(\beta ,{\text{ }}\beta +1-\gamma ;{\text{ }}\beta +1-\alpha ;{\text{ s)}}}\end{aligned}}$

因此，y = A′y₁ + B′y₂。令 A′a₀ = A 和 B′a₀ = B。然后，注意到 s = x^-1，

${\begin{aligned}&y=Ax_{2}^{-\alpha }F_{1}(\alpha ,{\text{ }}\alpha +1-\gamma ;{\text{ }}\alpha +1-\beta ;{\text{ x}}^{-1})+Bx_{2}^{-\beta }F_{1}(\beta ,{\text{ }}\beta +1-\gamma ;{\text{ }}\beta +1-\alpha ;{\text{ x}}^{-1})\\\end{aligned}}$

α − β = 0

然后 y₁ = y|_{c = α}。由于 α = β，我们有

$y=a_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}s^{r+c}\right)}$

因此，

${\begin{aligned}&y_{1}=a_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r}}}s^{r+\alpha }\right)}=a_{0}s_{2}^{\alpha }F_{1}(\alpha ,{\text{ }}\alpha +1-\gamma ;{\text{ 1; s)}}\\&y_{2}={\frac {\partial y}{\partial c}}{\text{ at }}c=\alpha \end{aligned}}$

为了计算这个导数，设

${\begin{aligned}&M_{r}={\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\end{aligned}}$

然后使用上述γ = 1情况下的方法，我们可以得到

${\begin{aligned}&{\frac {\partial M_{r}}{\partial c}}={\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\left\{\sum \limits _{k=0}^{r-1}{\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {2}{c+1-\alpha +k}}\right)}\right\}\\\end{aligned}}$

现在，

${\begin{aligned}y=a_{0}s^{c}\sum \limits _{r=0}^{\infty }{\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}s^{r}\right)}=a_{0}s^{c}\sum \limits _{r=0}^{\infty }{M_{r}s^{r}}\end{aligned}}$

因此

${\begin{aligned}&=a_{0}s^{c}{\text{ ln(s)}}\sum \limits _{r=0}^{\infty }{\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}s^{r}\right)}\\&{\text{ }}+a_{0}s^{c}\sum \limits _{r=0}^{\infty }{\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\left\{\sum \limits _{k=0}^{r-1}{\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {2}{c+1-\alpha +k}}\right)}\right\}s^{r}\right)}\\\end{aligned}}$

因此，

${\begin{aligned}&{\frac {\partial y}{\partial c}}=a_{0}s^{c}\sum \limits _{r=0}^{\infty }{\left(\left({\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\right)\left(\ln s+\sum \limits _{k=0}^{r-1}{\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {2}{c+1-\alpha +k}}\right)}\right)s^{r}\right)}\\\end{aligned}}$

当 *c* = α 时，我们得到

${\begin{aligned}&y_{2}=a_{0}s^{\alpha }\sum \limits _{r=0}^{\infty }{\left(\left({\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{\left((1)_{r}\right)^{2}}}\right)\left(\ln s+\sum \limits _{k=0}^{r-1}{\left({\frac {1}{\alpha +k}}+{\frac {1}{\alpha +1-\gamma +k}}-{\frac {2}{1+k}}\right)}\right)s^{r}\right)}\\&\end{aligned}}$

因此，*y* = *C*′*y*₁ + *D*′*y*₂。令 *C*′*a*₀ = *C* 且 *D*′*a*₀ = *D*。注意到 *s* = *x*^-1，

$y=Cx_{2}^{-\alpha }F_{1}(\alpha ,\alpha +1-\gamma ;1;x^{-1})$

$+Dx^{-\alpha }\sum \limits _{r=0}^{\infty }{\left(\left({\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{\left((1)_{r}\right)^{2}}}\right)\left(\ln x^{-1}+\sum \limits _{k=0}^{r-1}{\left({\frac {1}{\alpha +k}}+{\frac {1}{\alpha +1-\gamma +k}}-{\frac {2}{1+k}}\right)}\right)x^{-r}\right)}$

α − β 是整数，且 α − β ≠ 0

α − β > 0

从递推关系

$a_{r}={\frac {(r+c-1)(r+c-\gamma )}{(r+c-\alpha )(r+c-\beta )}}a_{r-1}$

我们可以看到，当c = β（较小的根）时，a_{α - β} → ∞。因此，我们必须进行替换a₀ = b₀(c − c_i)，其中c_i 是使我们的解无限的根。因此，我们取a₀ = b₀(c − β)，我们假设的解将采用新的形式

$y_{b}=b_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}\right)}$

然后y₁ = y_b|_{c = β}。我们可以看到，在

${\frac {(c-\beta )(c)_{\alpha -\beta }(c+1-\gamma )_{\alpha -\beta }}{(c+1-\alpha )_{\alpha -\beta }(c+1-\beta )_{\alpha -\beta }}}s^{\alpha -\beta }$

之前的所有项都消失了，因为分子中存在c − β。

但是从这一项开始，分子中的c − β 就消失了。为了看到这一点，请注意

$(c+1-\alpha )_{\alpha -\beta }=(c+1-\alpha )(c+2-\alpha )\cdots (c-\beta ).$

因此，我们的解决方案形式为

${\begin{aligned}&y_{1}=b_{0}\left({\frac {(\beta )_{\alpha -\beta }(\beta +1-\gamma )_{\alpha -\beta }}{(\beta +1-\alpha )_{\alpha -\beta -1}(1)_{\alpha -\beta }}}s^{\alpha -\beta }+{\frac {(\beta )_{\alpha -\beta +1}(\beta +1-\gamma )_{\alpha -\beta +1}}{(\beta +1-\alpha )_{\alpha -\beta -1}(1)(1)_{\alpha -\beta +1}}}s^{\alpha -\beta +1}+...\right)\\&={\frac {b_{0}}{(\beta +1-\alpha )_{\alpha -\beta -1}}}\sum \limits _{r=\alpha -\beta }^{\infty }{\left({\frac {(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+\beta -\alpha }}}s^{r}\right)}\\&\\\end{aligned}}$

现在，

$y_{2}=\left.{\frac {\partial y_{b}}{\partial c}}\right|_{c=\alpha }.$

为了计算这个导数，设

$M_{r}={\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}.$

然后使用γ = 1上述情况中的方法，我们得到

${\begin{aligned}{\frac {\partial M_{r}}{\partial c}}&={\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}{\Biggl (}{\frac {1}{c-\beta }}+\\&\qquad +\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {1}{c+1-\alpha +k}}-{\frac {1}{c+1-\beta +k}}\right){\Biggr )}\end{aligned}}$

现在，

$y_{b}=b_{0}\sum \limits _{r=0}^{\infty }{\left({\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}\right)}=b_{0}x^{c}\sum \limits _{r=0}^{\infty }{M_{r}s^{r}}$

因此，

${\begin{aligned}{\frac {\partial y}{\partial c}}&=b_{0}s^{c}\ln(s)\sum _{r=0}^{\infty }{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r}\\&\quad {\begin{aligned}{}+b_{0}s^{c}\sum _{r=0}^{\infty }&{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}{\Biggl (}{\frac {1}{c-\beta }}+\\&+\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {1}{c+1-\alpha +k}}-{\frac {1}{c+1-\beta +k}}\right){\Biggr )}s^{r}\end{aligned}}\end{aligned}}$

因此，

${\begin{aligned}{\frac {\partial y}{\partial c}}=b_{0}s^{c}\sum _{r=0}^{\infty }&{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}{\Biggl (}\ln s+{\frac {1}{c-\beta }}+\\&+\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {1}{c+1-\alpha +k}}-{\frac {1}{c+1-\beta +k}}\right){\Biggr )}s^{r}\end{aligned}}$

当 c = α 时，我们得到 y₂。因此， y = E′y₁ + F′y₂。令 E′b₀ = E 和 F′b₀ = F。注意 s = x^-1，我们得到

${\begin{aligned}y&={\frac {E}{(\beta +1-\alpha )_{\alpha -\beta -1}}}\sum _{r=\alpha -\beta }^{\infty }{\frac {(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+\beta -\alpha }}}x^{-r}\\&\quad {\begin{aligned}{}+Fx^{-\alpha }\sum _{r=0}^{\infty }&{\frac {(\alpha -\beta )(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(\alpha +1-\beta )_{r}}}{\Biggl (}\ln x^{-1}+{\frac {1}{\alpha -\beta }}\\&+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\alpha +1+k-\gamma }}-{\frac {1}{1+k}}-{\frac {1}{\alpha +1+k-\beta }}\right){\Biggr )}x^{-r}\end{aligned}}\end{aligned}}$

α − β < 0

从这里情况的对称性来看，我们发现

${\begin{aligned}y&={\frac {G}{(\alpha +1-\beta )_{\beta -\alpha -1}}}\sum _{r=\beta -\alpha }^{\infty }{\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r+\alpha -\beta }}}x^{-r}\\&\quad {\begin{aligned}{}+Hx^{-\beta }\sum _{r=0}^{\infty }&{\frac {(\beta -\alpha )(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(\beta +1-\alpha )_{r}}}{\Biggl (}\ln x^{-1}+{\frac {1}{\beta -\alpha }}\\&+\sum _{k=0}^{r-1}\left({\frac {1}{\beta +k}}+{\frac {1}{\beta +1+k-\gamma }}-{\frac {1}{1+k}}-{\frac {1}{\beta +1+k-\alpha }}\right){\Biggr )}x^{-r}\end{aligned}}\end{aligned}}$

参考文献

• Ian Sneddon (1966). Special functions of mathematical physics and chemistry. OLIVER B. ISBN 978-0050013342.