求解复合超几何微分方程边值问题解的一种新方法

A New Method for Solving the Boundary Value Problem of Compound Hypergeometric Differential Equation

摘要: 针对复合超几何微分方程的边值问题，研究其求解过程并分析解的结构，发现：首先利用左、右区超几何方程的任意一组线性无关的解构造引解函数，然后利用右区引解函数和右边值条件的系数构造右相似核函数，左区引解函数和交界处衔接条件的系数构造左相似核函数，最后将左相似核函数与左边值条件的系数组合得到该边值问题解的相似结构，并总结出求解该类边值问题的相似构造法。

Abstract: In the boundary value problem of the compound hypergeometric differential equation, first of all, the leading function is constructed by any two linearly independent solutions of the definite solution equation in the left and right regions. Then, right similar kernel function is obtained through leading function and coefficients of the boundary condition in right area. Furthermore, with the combination of leading function in left area and coefficients of junction conditions, left similar kernel function can also get. Finally, combining left similar kernel function with coefficients of the boundary condition in left area, it is easy to get a similar structure of the solutions in this boundary value problem. Overall, we conclude that a new method to solve this kind of boundary value problem. It is called similar construction method.

1. 引言

超几何(超比)方程或Gauss方程，它是具有 $0,1,\infty$ 三个正则奇点的Fuchs型原型方程，其解超比函数或Gauss函数，是一类重要的特殊函数 [1]。我们知道：凡是具有三个正则奇点的Fuchs型方程的解都可以用超比函数表达 [1]。且在数学物理中，富克斯方程(Fuchs Equation)有很广的应用背景，所以研究超几何方程问题很是重要。

自2013年起，研究了在 $x=1$ 处的欧拉超几何微分方程的第一、二类边值问题 [2]、在无穷大处的欧拉超几何微分方程的第一、二类边值问题 [3]、在 $x=0$ 处的欧拉超几何微分方程的第三类边值问题 [4] 和超几何方程的一类非齐次边值问题 [5] 等，得到了这些边值问题解的相似结构。

本文研究以下复合超几何微分方程的边值问题：

$\{\begin{array}{l}x\left(1-x\right){y^{″}}_1+\left[c_1-\left(a_1+b_1+1\right)x\right]{y^{\prime }}_1+a_1{b}_1{y}_1=0,\alpha <x<\theta \\ x\left(1-x\right){y^{″}}_2+\left[c_2-\left(a_2+b_2+1\right)x\right]{y^{\prime }}_2+a_2{b}_2{y}_2=0,\theta <x<\beta \\ {\left[e{y}_1+\left(1+ef\right){y^{\prime }}_1\right]|}_{x=\alpha }=Q\\ {y_1|}_{x=\theta }={\lambda y_2|}_{x=\theta }\\ {{y^{\prime }}_1|}_{x=\theta }={\mu {y^{\prime }}_2|}_{x=\theta }\\ {\left[G{y}_2+H{y^{\prime }}_2\right]|}_{x=\beta }=0\end{array}$ (1)

其中 $a_i,b_i,c_i,\left(i=1,2\right),\alpha ,\theta ,\beta ,\lambda ,\mu ,e,f,G,H,Q$ 均为实数， $c_i,a_i-b_i,c_i-a_i-b_i,\left(i=1,2\right)$ 不是正整数， $\left|x\right|<1,Q>0,\beta >\alpha >0,G^2+H^2\ne 0$。

2. 预备知识

超几何方程

$x\left(1-x\right){y^{″}}_i+\left[c_i-\left(a_i+b_i+1\right)x\right]{y^{\prime }}_i+a_i{b}_i{y}_i=0,\left(i=1,2;c_i\in N^{*}\right)$ (2)

在奇点 $x=0$ 处的通解 [1] 为

$y_i=C_{i1}F\left(a_i,b_i,c_i,x\right)+C_{i2}{x}^{1-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right),$ (3)

其中 $F\left(a_i,b_i,c_i,x\right),x^{1-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right)$ 是方程(2)的两个线性无关的解， $C_{i1},C_{i2}$ 是任意常数。

根据超几何函数的微分性质 [1]

$\frac{\text{d}}{\text{d}x}F\left(a_i,b_i,c_i,x\right)=\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,x\right),$ (4)

可以求出

$\begin{array}{c}{y^{\prime }}_i=C_{i1}\left[\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,x\right)\right]\\ +C_{i2}[\left(1-c_i\right)x^{-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right)\begin{array}{c}\\ \end{array}\\ +x^{1-c_i}\frac{\left(a_i+1-c_i\right)\left(b_i+1-c_i\right)}{2-c_i}F\left(a_i+2-c_i,b_i+2-c_i,3-c_i,x\right)]\end{array}$ (5)

为了求解复合超几何微分方程的边值问题，我们构造引解函数

$\begin{array}{c}{\phi }_{0,0}^i\left(x,\xi \right)\\ =F\left(a_i,b_i,c_i,x\right){\xi }^{1-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,\xi \right)\\ -F\left(a_i,b_i,c_i,\xi \right)x^{1-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right)\end{array}$ (6)

且

$\begin{array}{c}{\phi }_{1,0}^i\left(x,\xi \right)\\ =\frac{\partial {\phi }_{0,0}^i\left(x,\xi \right)}{\partial x}\\ =\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,x\right){\xi }^{1-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,\xi \right)\\ -F\left(a_i,b_i,c_i,\xi \right)\left(1-c_i\right)x^{-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right)\\ -F\left(a_i,b_i,c_i,\xi \right)x^{1-c_i}\frac{\left(a_i+1-c_i\right)\left(b_i+1-c_i\right)}{2-c_i}F\left(a_i+2-c_i,b_i+2-c_i,3-c_i,x\right),\end{array}$ (7)

$\begin{array}{c}{\phi }_{0,1}^i\left(x,\xi \right)\\ =\frac{\partial {\phi }_{0,0}^i\left(x,\xi \right)}{\partial \xi }\\ =F\left(a_i,b_i,c_i,x\right)\left(1-c_i\right){\xi }^{-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,\xi \right)\\ +F\left(a_i,b_i,c_i,x\right){\xi }^{1-c_i}\frac{\left(a_i+1-c_i\right)\left(b_i+1-c_i\right)}{2-c_i}F\left(a_i+2-c_i,b_i+2-c_i,3-c_i,\xi \right)\\ -\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,\xi \right)x^{1-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right),\end{array}$ (8)

$\begin{array}{c}{\phi }_{1,1}^i\left(x,\xi \right)\\ =\frac{{\partial }^2{\phi }_{0,0}^i\left(x,\xi \right)}{\partial x\partial \xi }=\frac{{\partial }^2{\phi }_{0,0}^i\left(x,\xi \right)}{\partial \xi \partial x}\\ =\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,x\right)\left(1-c_i\right){\xi }^{-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,\xi \right)\\ +\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,x\right){\xi }^{1-c_i}\frac{\left(a_i+1-c_i\right)\left(b_i+1-c_i\right)}{2-c_i}F\left(a_i+2-c_i,b_i+2-c_i,3-c_i,\xi \right)\\ -\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,\xi \right)\left(1-c_i\right)x^{-c_i}F\left(a_i+1-c_i,b_i+1-c_i,2-c_i,x\right)\\ -\frac{a_i{b}_i}{c_i}F\left(a_i+1,b_i+1,c_i+1,\xi \right)x^{1-c_i}\frac{\left(a_i+1-c_i\right)\left(b_i+1-c_i\right)}{2-c_i}F\left(a_i+2-c_i,b_i+2-c_i,3-c_i,x\right).\end{array}$ (9)

根据以上引解函数的定义，有

${\phi }_{0,0}^i\left(x,\xi \right)=-{\phi }_{0,0}^i\left(\xi ,x\right),{\phi }_{1,0}^i\left(x,\xi \right)=-{\phi }_{0,1}^i\left(\xi ,x\right),{\phi }_{1,1}^i\left(x,\xi \right)={\phi }_{1,1}^i\left(\xi ,x\right).$ (10)

3. 主要定理

定理若复合超几何微分方程的边值问题(1)在奇点 $x=0$ 处有唯一解，那么其左区( $\alpha <x<\theta$ )解有如下相似结构：

$y_1=Q\frac{1}{e+\frac{1}{f+\Phi \left(\alpha \right)}}\frac{1}{f+\Phi \left(\alpha \right)}\Phi \left(x\right),$ (11)

右区( $\theta <x<\beta$ )解为：

$y_2=Q\frac{1}{e+\frac{1}{f+\Phi \left(\alpha \right)}}\frac{1}{f+\Phi \left(\alpha \right)}\frac{{\phi }_{0,1}^1\left(\theta ,\theta \right)}{\lambda {\Phi }^{*}\left(\theta \right){\phi }_{1,1}^1\left(\alpha ,\theta \right)-\mu {\phi }_{1,0}^1\left(\alpha ,\theta \right)}{\Phi }^{*}\left(x\right),$ (12)

其中 ${\Phi }^{\text{*}}\left(x\right)$ 是右相似核函数，且

${\Phi }^{\text{*}}\left(x\right)=\frac{G{\phi }_{0,0}^2\left(x,\beta \right)+H{\phi }_{0,1}^2\left(x,\beta \right)}{G{\phi }_{1,0}^2\left(\theta ,\beta \right)+H{\phi }_{1,1}^2\left(\theta ,\beta \right)},\theta <x<\beta ;$ (13)

$\Phi \left(x\right)$ 是左相似核函数，且

$\Phi \left(x\right)=\frac{\lambda {\Phi }^{\text{*}}\left(\theta \right){\phi }_{0,1}^1\left(x,\theta \right)-\mu {\phi }_{0,0}^1\left(x,\theta \right)}{\lambda {\Phi }^{\text{*}}\left(\theta \right){\phi }_{1,1}^1\left(\alpha ,\theta \right)-\mu {\phi }_{1,0}^1\left(\alpha ,\theta \right)},\alpha <x<\theta ;$ (14)

${\phi }_{m,n}^i\left(x,\xi \right),\left(i=1,2;m,n=0,1\right)$ 是引解函数。

证明：由预备知识可知超几何方程的通解为(3)式，且其导数为(5)式。则根据左边值条件 ${\left[e{y}_1+\left(1+ef\right){y^{\prime }}_1\right]|}_{x=\alpha }=Q$ 有

$\begin{array}{l}{C}_{11}\left[eF\left(a_1,b_1,c_1,\alpha \right)+\left(1+ef\right)\frac{a_1{b}_1}{c_1}F\left(a_1+1,b_1+1,c_1+1,\alpha \right)\right]\\ +C_{12}\{e{\alpha }^{1-c_1}F\left(a_1+1-c_1,b_1+1-c_1,2-c_1,\alpha \right)\begin{array}{c}\\ \end{array}\\ +\left(1+ef\right)[\left(1-c_1\right){\alpha }^{-c_1}F\left(a_1+1-c_1,b_1+1-c_1,2-c_1,\alpha \right)\\ +{\alpha }^{1-c_1}\frac{\left(a_1+1-c_1\right)\left(b_1+1-c_1\right)}{2-c_1}F\left(a_1+2-c_1,b_1+2-c_1,3-c_1,\alpha \right)]\}=Q\end{array}$ (15)

由衔接条件 ${y_1|}_{x=\theta }={\lambda y_2|}_{x=\theta },{{y^{\prime }}_1|}_{x=\theta }={\mu {y^{\prime }}_2|}_{x=\theta }$ 有

$\begin{array}{l}{C}_{11}F\left(a_1,b_1,c_1,\theta \right)+C_{12}{\theta }^{1-c_1}F\left(a_1+1-c_1,b_1+1-c_1,2-c_1,\theta \right)\\ -C_{21}\lambda F\left(a_2,b_2,c_2,\theta \right)-C_{22}\lambda {\theta }^{1-c_2}F\left(a_2+1-c_2,b_2+1-c_2,2-c_2,\theta \right)=0\end{array}$ (16)

$\begin{array}{l}{C}_{11}\left[\frac{a_1{b}_1}{c_1}F\left(a_1+1,b_1+1,c_1+1,\theta \right)\right]-C_{21}\mu \left[\frac{a_2{b}_2}{c_2}F\left(a_2+1,b_2+1,c_2+1,\theta \right)\right]\\ +C_{12}[\left(1-c_1\right){\theta }^{-c_1}F\left(a_1+1-c_1,b_1+1-c_1,2-c_1,\theta \right)\begin{array}{c}\\ \end{array}\\ +{\theta }^{1-c_1}\frac{\left(a_1+1-c_1\right)\left(b_1+1-c_1\right)}{2-c_1}F\left(a_1+2-c_1,b_1+2-c_1,3-c_1,\theta \right)]\\ -C_{22}\mu [\left(1-c_2\right){\theta }^{-c_2}F\left(a_2+1-c_2,b_2+1-c_2,2-c_2,\theta \right)\begin{array}{c}\\ \end{array}\\ +{\theta }^{1-c_2}\frac{\left(a_2+1-c_2\right)\left(b_2+1-c_2\right)}{2-c_2}F\left(a_2+2-c_2,b_2+2-c_2,3-c_2,\theta \right)]=0\end{array}$ (17)

由右边值条件 ${\left[G{y}_2+H{y^{\prime }}_2\right]|}_{x=\beta }=0$ 有

$\begin{array}{l}{C}_{21}\left\{GF\left(a_2,b_2,c_2,\beta \right)+H\left[\frac{a_2{b}_2}{c_2}F\left(a_2+1,b_2+1,c_2+1,\beta \right)\right]\right\}\\ +C_{22}\{G{\beta }^{1-c_2}F\left(a_2+1-c_2,b_2+1-c_2,2-c_2,\beta \right)\begin{array}{c}\\ \end{array}\\ +H[\left(1-c_2\right){\beta }^{-c_2}F\left(a_2+1-c_2,b_2+1-c_2,2-c_2,\beta \right)\begin{array}{c}\\ \end{array}\\ +{\beta }^{1-c_2}\frac{\left(a_2+1-c_2\right)\left(b_2+1-c_2\right)}{2-c_2}F\left(a_2+2-c_2,b_2+2-c_2,3-c_2,\beta \right)]\}=0\end{array}$ (18)

联立方程(15)、(16)、(17)、(18)，利用(6)~(10)式化简可得

$\begin{array}{l}D=\mu \left[e{\phi }_{0,0}^1\left(\alpha ,\theta \right)+\left(1+ef\right){\phi }_{1,0}^1\left(\alpha ,\theta \right)\right]\left[G{\phi }_{0,1}^2\left(\beta ,\theta \right)+H{\phi }_{1,1}^2\left(\beta ,\theta \right)\right]\\ -\lambda \left[e{\phi }_{0,1}^1\left(\alpha ,\theta \right)+\left(1+ef\right){\phi }_{1,1}^1\left(\alpha ,\theta \right)\right]\left[G{\phi }_{0,0}^2\left(\beta ,\theta \right)+H{\phi }_{1,0}^2\left(\beta ,\theta \right)\right]\end{array}$ (19)

$\begin{array}{l}{D}_1=Q\{{\theta }^{1-c_1}F\left(a_1+1-c_1,b_1+1-c_1,2-c_1,\theta \right)\times \mu \left[G{\phi }_{0,1}^2\left(\beta ,\theta \right)+H{\phi }_{1,1}^2\left(\beta ,\theta \right)\right]\begin{array}{c}\\ \end{array}\\ -[\left(1-c_1\right){\theta }^{-c_1}F\left(a_1+1-c_1,b_1+1-c_1,2-c_1,\theta \right)+{\theta }^{1-c_1}\frac{\left(a_1+1-c_1\right)\left(b_1+1-c_1\right)}{2-c_1}\\ \begin{array}{c}\\ \end{array}\times F\left(a_1+2-c_1,b_1+2-c_1,3-c_1,\theta \right)]\lambda \left[G{\phi }_{0,0}^2\left(\beta ,\theta \right)+H{\phi }_{1,0}^2\left(\beta ,\theta \right)\right]\}\end{array}$ (20)

$\begin{array}{l}{D}_2=-Q\{F\left(a_1,b_1,c_1,\theta \right)\mu \left[G{\phi }_{0,1}^2\left(\beta ,\theta \right)+H{\phi }_{1,1}^2\left(\beta ,\theta \right)\right]\begin{array}{c}\\ \end{array}\\ -\frac{a_1{b}_1}{c_1}F\left(a_1+1,b_1+1,c_1+1,\theta \right)\lambda \left[G{\phi }_{0,0}^2\left(\beta ,\theta \right)+H{\phi }_{1,0}^2\left(\beta ,\theta \right)\right]\}\end{array}$ (21)

$\begin{array}{c}{D}_3=Q\{G{\beta }^{1-c_2}F\left(a_2+1-c_2,b_2+1-c_2,2-c_2,\beta \right)\begin{array}{c}\\ \end{array}\\ +H[\left(1-c_2\right){\beta }^{-c_2}F\left(a_2+1-c_2,b_2+1-c_2,2-c_2,\beta \right)\begin{array}{c}\\ \end{array}\\ +{\beta }^{1-c_2}\frac{\left(a_2+1-c_2\right)\left(b_2+1-c_2\right)}{2-c_2}F\left(a_2+2-c_2,b_2+2-c_2,3-c_2,\beta \right)]\}{\phi }_{0,1}^1\left(\theta ,\theta \right)\end{array}$ (22)

$D_4=-Q\left\{GF\left(a_2,b_2,c_2,\beta \right)+H\left[\frac{a_2{b}_2}{c_2}F\left(a_2+1,b_2+1,c_2+1,\beta \right)\right]\right\}{\phi }_{0,1}^1\left(\theta ,\theta \right)$ (23)

再根据Cramer法则： $C_{11}=\frac{D_1}{D},C_{12}=\frac{D_2}{D},C_{21}=\frac{D_3}{D},C_{22}=\frac{D_4}{D}$，即可证明复合超几何微分方程的边值

问题(1)的左和右区解分别为(11)和(12)式。

4. 相似构造法

根据以上求解复合超几何微分方程边值问题的过程，归纳出求解边值问题(1)的相似构造法的步骤 [6] 如下：

第一步由超几何方程的任意一组线性无关的解，构造出引解函数 ${\phi }_{m,n}^i\left(x,\xi \right),\left(i=1,2;m,n=0,1\right)$，如(6)~(9)式；

第二步将 ${\phi }_{i,j}^2\left(x,\xi \right),\left(i,j=0,1\right)$ 及右边值条件 ${\left[G{y}_2+H{y^{\prime }}_2\right]|}_{x=\beta }=0$ 的系数组合，生成右相似核函数，即(13)式；

第三步将 ${\phi }_{i.j}^1\left(x,\xi \right),\left(i,j=0,1\right),{\Phi }^{\text{*}}\left(\theta \right)$ 及 $x=\theta$ 处的衔接条件 ${y_1|}_{x=\theta }={\lambda y_2|}_{x=\theta },{{y^{\prime }}_1|}_{x=\theta }={\mu {y^{\prime }}_2|}_{x=\theta }$ 的系数组合，生成左相似核函数，即(14)式；

第四步由左边值条件 ${\left[e{y}_1+\left(1+ef\right){y^{\prime }}_1\right]|}_{x=\alpha }=Q$ 的系数和 $\Phi \left(x\right),\Phi \left(\alpha \right)$ 组装可得左区( $\alpha <x<\theta$ )和右区( $\theta <x<\beta$ )解，即(11)和(12)式。

5. 举例

根据相似构造法的步骤，求解下列边值问题( $a_1=\frac{1}{3},b_1=\frac{1}{4},c_1=\frac{1}{2};a_2=\frac{1}{4},b_2=\frac{1}{5},c_2=\frac{1}{2};\alpha =\frac{1}{5},\theta =\frac{1}{3}$， $\beta =\frac{1}{2},e=2,f=3,Q=2,\lambda =\mu =1,G=2,H=3$ )：

$\{\begin{array}{l}x\left(1-x\right){y^{″}}_1+\left(\frac{1}{2}-\frac{19}{12}x\right){y^{\prime }}_1+\frac{1}{12}{y}_1=0,\frac{1}{5}<x<\frac{1}{3}\\ x\left(1-x\right){y^{″}}_2+\left(\frac{1}{2}-\frac{29}{20}x\right){y^{\prime }}_2+\frac{1}{20}{y}_2=0,\frac{1}{3}<x<\frac{1}{2}\\ {\left[2y_1+7{y^{\prime }}_1\right]|}_{x=\frac{1}{5}}=2\\ {y_1|}_{x=\frac{1}{3}}={y_2|}_{x=\frac{1}{3}}\\ {{y^{\prime }}_1|}_{x=\frac{1}{3}}={{y^{\prime }}_2|}_{x=\frac{1}{3}}\\ {\left[2y_2+3{y^{\prime }}_2\right]|}_{x=\frac{1}{2}}=0\end{array}$ (24)

第一步由方程 $x\left(1-x\right){y^{″}}_1+\left(\frac{1}{2}-\frac{19}{12}x\right){y^{\prime }}_1+\frac{1}{12}{y}_1=0,\left(\frac{1}{5}<x<\frac{1}{3}\right)$ 的两个线性无关的解，作引解函数

${\phi }_{0,0}^1\left(x,\xi \right)={\xi }^{\frac{1}{2}}F\left(\frac{1}{3},\frac{1}{4},\frac{1}{2},x\right)F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},\xi \right)-x^{\frac{1}{2}}F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},x\right)F\left(\frac{1}{3},\frac{1}{4},\frac{1}{2},\xi \right)$

$\begin{array}{c}{\phi }_{1,0}^1\left(x,\xi \right)=\frac{1}{6}{\xi }^{\frac{1}{2}}F\left(\frac{4}{3},\frac{5}{4},\frac{3}{2},x\right)F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},\xi \right)-\frac{1}{2}{x}^{-\frac{1}{2}}F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},x\right)F\left(\frac{1}{3},\frac{1}{4},\frac{1}{2},\xi \right)\\ -\frac{5}{12}{x}^{\frac{1}{2}}F\left(\frac{11}{6},\frac{7}{4},\frac{5}{2},x\right)F\left(\frac{1}{3},\frac{1}{4},\frac{1}{2},\xi \right)\end{array}$

$\begin{array}{c}{\phi }_{0,1}^1\left(x,\xi \right)=\frac{1}{2}{\xi }^{-\frac{1}{2}}F\left(\frac{1}{3},\frac{1}{4},\frac{1}{2},x\right)F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},\xi \right)\text{+}\frac{5}{12}{\xi }^{\frac{1}{2}}F\left(\frac{1}{3},\frac{1}{4},\frac{1}{2},x\right)F\left(\frac{11}{6},\frac{7}{4},\frac{5}{2},\xi \right)\\ -\frac{1}{6}{x}^{\frac{1}{2}}F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},x\right)F\left(\frac{4}{3},\frac{5}{4},\frac{3}{2},\xi \right)\end{array}$

$\begin{array}{c}{\phi }_{1,1}^1\left(x,\xi \right)=\frac{1}{12}{\xi }^{-\frac{1}{2}}F\left(\frac{4}{3},\frac{5}{4},\frac{3}{2},x\right)F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},\xi \right)+\frac{5}{72}{\xi }^{\frac{1}{2}}F\left(\frac{4}{3},\frac{5}{4},\frac{3}{2},x\right)F\left(\frac{11}{6},\frac{7}{4},\frac{5}{2},\xi \right)\\ -\frac{1}{12}{x}^{-\frac{1}{2}}F\left(\frac{5}{6},\frac{3}{4},\frac{3}{2},x\right)F\left(\frac{4}{3},\frac{5}{4},\frac{3}{2},\xi \right)-\frac{5}{72}{x}^{\frac{1}{2}}F\left(\frac{11}{6},\frac{7}{4},\frac{5}{2},x\right)F\left(\frac{4}{3},\frac{5}{4},\frac{3}{2},\xi \right)\end{array}$

由方程 $x\left(1-x\right){y^{″}}_2+\left(\frac{1}{2}-\frac{29}{20}x\right){y^{\prime }}_2+\frac{1}{20}{y}_2=0,\left(\frac{1}{3}<x<\frac{1}{2}\right)$ 的两个线性无关的解，作引解函数

${\phi }_{0,0}^2\left(x,\xi \right)={\xi }^{\frac{1}{2}}F\left(\frac{1}{4},\frac{1}{5},\frac{1}{2},x\right)F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},\xi \right)-x^{\frac{1}{2}}F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},x\right)F\left(\frac{1}{4},\frac{1}{5},\frac{1}{2},\xi \right)$

$\begin{array}{c}{\phi }_{1,0}^2\left(x,\xi \right)=\frac{1}{10}{\xi }^{\frac{1}{2}}F\left(\frac{5}{4},\frac{6}{5},\frac{3}{2},x\right)F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},\xi \right)-\frac{1}{2}{x}^{-\frac{1}{2}}F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},x\right)F\left(\frac{1}{4},\frac{1}{5},\frac{1}{2},\xi \right)\\ -\frac{7}{20}{x}^{\frac{1}{2}}F\left(\frac{7}{4},\frac{17}{10},\frac{5}{2},x\right)F\left(\frac{1}{4},\frac{1}{5},\frac{1}{2},\xi \right)\end{array}$

$\begin{array}{c}{\phi }_{0,1}^2\left(x,\xi \right)=\frac{1}{2}{\xi }^{-\frac{1}{2}}F\left(\frac{1}{4},\frac{1}{5},\frac{1}{2},x\right)F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},\xi \right)\text{+}\frac{7}{20}{\xi }^{\frac{1}{2}}F\left(\frac{1}{4},\frac{1}{5},\frac{1}{2},x\right)F\left(\frac{7}{4},\frac{17}{10},\frac{5}{2},\xi \right)\\ -\frac{1}{10}{x}^{\frac{1}{2}}F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},x\right)F\left(\frac{5}{4},\frac{6}{5},\frac{3}{2},\xi \right)\end{array}$

$\begin{array}{c}{\phi }_{1,1}^2\left(x,\xi \right)=\frac{1}{20}{\xi }^{-\frac{1}{2}}F\left(\frac{5}{4},\frac{6}{5},\frac{3}{2},x\right)F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},\xi \right)\text{+}\frac{7}{200}{\xi }^{\frac{1}{2}}F\left(\frac{5}{4},\frac{6}{5},\frac{3}{2},x\right)F\left(\frac{7}{4},\frac{17}{10},\frac{5}{2},\xi \right)\\ -\frac{1}{20}{x}^{-\frac{1}{2}}F\left(\frac{5}{4},\frac{6}{5},\frac{3}{2},\xi \right)F\left(\frac{3}{4},\frac{7}{10},\frac{3}{2},x\right)-\frac{7}{200}{x}^{\frac{1}{2}}F\left(\frac{7}{4},\frac{17}{10},\frac{5}{2},x\right)F\left(\frac{5}{4},\frac{6}{5},\frac{3}{2},\xi \right)\end{array}$

第二步由 ${\phi }_{i,j}^2\left(x,\xi \right),\left(i,j=0,1\right)$ 及右边值条件 ${\left[2y_2+3{y^{\prime }}_2\right]|}_{x=\frac{1}{2}}=0$ 的系数，生成右相似核函数：

${\Phi }^{\text{*}}\left(x\right)=\frac{2{\phi }_{0,0}^2\left(x,\frac{1}{2}\right)+3{\phi }_{0,1}^2\left(x,\frac{1}{2}\right)}{2{\phi }_{1,0}^2\left(\frac{1}{3},\frac{1}{2}\right)+3{\phi }_{1,1}^2\left(\frac{1}{3},\frac{1}{2}\right)},\left(\frac{1}{3}<x<\frac{1}{2}\right)$ (25)

第三步由 ${\phi }_{i.j}^1\left(x,\xi \right),\left(i,j=0,1\right),{\Phi }^{\text{*}}\left(\theta \right)$ 及交界点的衔接条件 ${y_1|}_{x=\frac{1}{3}}={y_2|}_{x=\frac{1}{3}},{{y^{\prime }}_1|}_{x=\frac{1}{3}}={{y^{\prime }}_2|}_{x=\frac{1}{3}}$ 的系数，生成

左相似核函数：

$\Phi \left(x\right)=\frac{{\Phi }^{\text{*}}\left(\frac{1}{3}\right){\phi }_{0,1}^1\left(x,\frac{1}{3}\right)-{\phi }_{0,0}^1\left(x,\frac{1}{3}\right)}{{\Phi }^{\text{*}}\left(\frac{1}{3}\right){\phi }_{1,1}^1\left(\frac{1}{5},\frac{1}{3}\right)-{\phi }_{1,0}^1\left(\frac{1}{5},\frac{1}{3}\right)},\left(\frac{1}{5}<x<\frac{1}{3}\right)$ (26)

第四步由左边值条件 ${\left[2y_1+7{y^{\prime }}_1\right]|}_{x=\frac{1}{5}}=2$ 的系数和 $\Phi \left(x\right),\Phi \left(\alpha \right)$ 组装可得左区( $\frac{1}{5}<x<\frac{1}{3}$ )解：

$y_1=2\frac{1}{2+\frac{1}{3+\Phi \left(\frac{1}{5}\right)}}\frac{1}{3+\Phi \left(\frac{1}{5}\right)}\Phi \left(x\right)$ (27)

和右区( $\frac{1}{3}<x<\frac{1}{2}$ )解：

$y_2=Q\frac{1}{2+\frac{1}{3+\Phi \left(\frac{1}{5}\right)}}\frac{1}{3+\Phi \left(\frac{1}{5}\right)}\frac{{\phi }_{0,1}^1\left(\frac{1}{3},\frac{1}{3}\right)}{{\Phi }^{\text{*}}\left(\frac{1}{3}\right){\phi }_{1,1}^1\left(\frac{1}{5},\frac{1}{3}\right)-{\phi }_{1,0}^1\left(\frac{1}{5},\frac{1}{3}\right)}{\Phi }^{\text{*}}\left(x\right)$ (28)

致谢

感谢审稿人的审阅及对文章的意见和建议。

基金项目

本项研究由西华大学人才引进项目(编号：Z201076)资助。