RiemannのP関数と超幾何関数で遊んでみようなのだ！

\begin{equation} \frac{d^{2}y}{dx^{2}}+[\frac{1-\alpha-\alpha^{'}}{x-a}+\frac{1-\beta-\beta^{'}}{x-b}+\frac{1-\gamma-\gamma^{'}}{x-c}]\frac{dy}{dx}+[\frac{\alpha\alpha^{'}\left(a-b\right)\left(a-c\right)}{x-a}+\frac{\beta\beta^{'}\left(b-c\right)\left(b-a\right)}{x-b}+\frac{\gamma\gamma^{'}\left(c-a\right)\left(c-b\right)}{x-c}]\frac{y}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}=0 \end{equation}
の$x=a$近傍の級数解を$u=\sum_{n=0}^{\infty}a_{n}\left(x-a\right)^{n+s}$とおくのだ。また下記のことが成り立つので、RiemannのP微分方程式に$\left(x-a\right)^{2}$を両辺にかけて$\left(x-a\right)^{s}$の係数を考えるのだ。\begin{eqnarray} \left\{ \begin{array}{l} \frac{du}{dx}=\sum_{n=0}^{\infty}\left(n+s\right)\left(x-a\right)^{n+s}\\ \frac{d^{2}u}{dx^{2}}=\sum_{n=0}^{\infty}\left(n+s\right)\left(n+s-1\right)\left(x-a\right)^{n+s} \end{array} \right. \end{eqnarray}
\begin{equation} s\left(s-1\right)+\left(1-\alpha-\alpha^{'}\right)s+\alpha\alpha^{'}=\left(s-\alpha\right)\left(s-\alpha^{'}\right)\\=0 \end{equation}
このことから、$x=a$での特性指数は$\alpha,\alpha^{'}$であることが分かる。$x=b,x=c$でも同様なのだ。このことからRiemannのP関数は確定特異点と特性指数を書いたものだとわかるのだ。

$x^{'}=\frac{rx+s}{px+q}$とすると次のように書き直せるのだ。
\begin{eqnarray} x^{'}&=&\frac{\frac{r}{p}\left(px+q\right)+s-\frac{qr}{p}}{px+q}\\ &=&\frac{r}{p}+\frac{s-\frac{qr}{p}}{px+q}\\ &=&\frac{r}{p}+\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{x+\frac{q}{p}} \end{eqnarray}
$x^{'}$で行っていることを細かく分けるのだ。

1. $x_{1}\leftarrow x+\frac{q}{p}\quad \left(平行移動\right)$
2. $x_{2}\leftarrow \frac{1}{x_{1}}\quad \left(反転\right)$
3. $x_{3}\leftarrow \frac{1}{p}\left(s-\frac{qr}{p}\right)x_{2}\quad \left(拡大・回転\right)$
4. $x_{4}\leftarrow x_{3}+\frac{r}{p}\quad \left(平行移動\right)$

1. 平行移動：$x_{1}=x+\delta$とすると、$v\left(x_{1}\right)=u\left(x_{1}-\delta\right),\frac{d}{dx}=\frac{d}{dx_{1}},\frac{d^{2}}{dx^{2}}=\frac{d^{2}}{dx_{1}^{2}}$よりRiemannのPBe分方程式は次のように書けるのだ。\begin{equation} \frac{d^{2}v}{dx_{1}^{2}}+[\frac{1-\alpha-\alpha^{'}}{x_{1}-\left(a+\delta\right)}+\frac{1-\beta-\beta^{'}}{x_{1}-\left(b+\delta\right)}+\frac{1-\gamma-\gamma^{'}}{x_{1}-\left(c+\delta\right)}]\frac{dv}{dx_{1}}+[\frac{\alpha\alpha^{'}\left(a-b\right)\left(a-c\right)}{x_{1}-\left(a+\delta\right)}+\frac{\beta\beta^{'}\left(b-c\right)\left(b-a\right)}{x_{1}-\left(b+\delta\right)}+\frac{\gamma\gamma^{'}\left(c-a\right)\left(c-b\right)}{x_{1}-\left(c+\delta\right)}]\frac{v}{\{x_{1}-\left(a+\delta\right)\}\{x_{1}-\left(b+\delta\right)\}\{x_{1}-\left(c+\delta\right)\}}=0 \end{equation}より、平行移動による解は次のように書けるのだ。\begin{equation} P\begin{Bmatrix}a&b&c\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\rightarrow P\begin{Bmatrix}a+\delta&b+\delta&c+\delta\\\alpha&\beta&\gamma&x+\delta\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix} \end{equation}
2. 拡大・回転：$x_{2}=px$とすると$v\left(x_{2}\right)=u\left(\frac{x_{2}}{p}\right),\frac{d}{dx}=p\frac{d}{dx_{2}},\frac{d^{2}}{dx^{2}}=p^{2}\frac{d^{2}}{dx_{2}^{2}}$より、RiemannのP微分方程式は次のように書き直せるのだ。\begin{eqnarray} &&p^{2}\frac{d^{2}v}{dx_{2}^{2}}+p[\frac{1-\alpha-\alpha^{'}}{\frac{x_{2}}{p}-a}+\frac{1-\beta-\beta^{'}}{\frac{x_{2}}{p}-b}+\frac{1-\gamma-\gamma^{'}}{\frac{x_{2}}{p}-c}]\frac{dv}{dx_{2}}+[\frac{\alpha\alpha^{'}\left(a-b\right)\left(a-c\right)}{\frac{x_{2}}{p}-a}+\frac{\beta\beta^{'}\left(b-c\right)\left(b-a\right)}{\frac{x_{2}}{p}-b}+\frac{\gamma\gamma^{'}\left(c-a\right)\left(c-b\right)}{\frac{x_{2}}{p}-x}]\frac{v}{\{\frac{x_{2}}{p}-a\}\{\frac{x_{2}}{p}-b\}\{\frac{x_{2}}{p}-c\}}=0\\ &&\frac{d^{2}v}{dx_{2}^{2}}+[\frac{1-\alpha-\alpha^{'}}{x_{2}-pa}+\frac{1-\beta-\beta^{'}}{x_{2}-pb}+\frac{1-\gamma-\gamma^{'}}{x_{2}-pc}]\frac{dv}{dx_{2}}+[\frac{\alpha\alpha^{'}\left(pa-pb\right)\left(pa-pc\right)}{x_{2}-pa}+\frac{\beta\beta^{'}\left(pb-pc\right)\left(pb-pa\right)}{x_{2}-pb}+\frac{1-\gamma-\gamma^{'}}{x_{2}-pc}]\frac{v}{\left(x_2-pa\right)\left(x_{2}-pb\right)\left(x_{2}-pc\right)}=0 \end{eqnarray}より回転・拡大によるRiemannのP関数は次のように変換されるのだ。\begin{equation} P\begin{Bmatrix}a&b&c\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\rightarrow P\begin{Bmatrix}pa&pb&pc\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix} \end{equation}
3. 反転の場合：$x_{3}=\frac{1}{x}$とすると次のように書けるのだ。$v\left(x_{3}\right)=u\left(\frac{1}{x_{3}}\right),\frac{d}{dx}=-x_{3}^{2}\frac{d}{dx_{3}},\frac{d^{2}}{dx^{2}}=x_{3}^{4}\frac{d^{2}}{dx_{3}^{2}}+2x_{3}^{3}\frac{d}{dx_{3}}$この事を用いるとRiemannのP微分方程式は下記のように変換されることが分かるのだ。\begin{eqnarray} &&x_{3}^{4}\frac{d^{2}v}{dx_{3}^{2}}+2x_{3}^{3}\frac{dv}{dx_{3}}-x_{3}^{2}[\frac{1-\alpha-\alpha^{'}}{\frac{1}{x_{3}}-a}+\frac{1-\beta-\beta^{'}}{\frac{1}{x_{3}}-b}+\frac{1-\gamma-\gamma^{'}}{\frac{1}{x_{3}}-c}]\frac{dv}{dx_{3}}+[\frac{\alpha\alpha^{'}\left(a-b\right)\left(a-c\right)}{\frac{1}{x_{3}}-a}+\frac{\beta\beta^{'}\left(b-c\right)\left(b-a\right)}{\frac{1}{x_{3}}-b}+\frac{\gamma\gamma^{'}\left(c-a\right)\left(c-b\right)}{\frac{1}{x_{3}}-c}]\frac{v}{\left(\frac{1}{x_{3}}-a\right)\left(\frac{1}{x_{3}}-b\right)\left(\frac{1}{x_{3}}-c\right)}=0\\ &&x_{3}\frac{d^{2}v}{dx_{3}^{2}}+[2-\frac{1-\alpha-\alpha^{'}}{1-ax_{3}}-\frac{1-\beta-\beta^{'}}{1-bx_{3}}-\frac{1-\gamma-\gamma^{'}}{1-cx_{3}}]\frac{dv}{dx_{3}}+[\frac{\alpha\alpha^{'}\left(a-b\right)\left(a-c\right)}{1-ax_{3}}+\frac{\beta\beta^{'}\left(b-c\right)\left(b-a\right)}{1-bx_{3}}+\frac{\gamma\gamma^{'}\left(c-a\right)\left(c-b\right)}{1-cx_{3}}]\frac{x_{3}v}{\left(1-ax_{3}\right)\left(1-bx_{3}\right)\left(1-cx_{3}\right)}=0 \end{eqnarray}ここで$A=1-\alpha-\alpha^{'},B=1-\beta-\beta^{'},C=1-\gamma-\gamma^{'},D=\alpha\alpha^{'},E=\beta\beta^{'},F=\gamma\gamma^{'}$とおくと次のように書けるのだ。\begin{eqnarray} &&A+B+C=2\\ &&x_{3}\frac{d^{2}v}{dx_{3}^{2}}+[A+B+C-\frac{A}{1-ax_{3}}-\frac{B}{1-bx_{3}}-\frac{C}{1-cx_{3}}]\frac{dv}{dx_{3}}+[\frac{D\left(a-b\right)\left(a-c\right)}{1-ax_{3}}+\frac{E\left(b-c\right)\left(b-a\right)}{1-bx_{3}}+\frac{F\left(c-1\right)\left(c-b\right)}{1-cx_{3}}]\frac{x_{3}v}{\left(1-ax_{3}\right)\left(1-bx_{3}\right)\left(1-cx_{3}\right)}\\ &&=x_{3}\frac{d^{2}v}{dx_{3}^{2}}-x_{3}[\frac{aA}{1-ax_{3}}+\frac{bB}{1-bx_{3}}+\frac{cC}{1-cx_{3}}]\frac{dv}{dx_{3}}+[\frac{D\left(a-b\right)\left(a-c\right)}{1-ax_{3}}+\frac{E\left(b-c\right)\left(b-a\right)}{1-bx_{3}}+\frac{F\left(c-1\right)\left(c-b\right)}{1-cx_{3}}]\frac{x_{3}v}{\left(1-ax_{3}\right)\left(1-bx_{3}\right)\left(1-cx_{3}\right)}\\ &&=0\\ &&\frac{d^{2}v}{dx_{3}^{2}}+[\frac{A}{x_{3}-\frac{1}{a}}+\frac{B}{x_{3}-\frac{1}{b}}+\frac{C}{x_{3}-\frac{1}{c}}]\frac{dv}{dx_{3}}+[\frac{D\left(\frac{1}{a}-\frac{1}{b}\right)\left(\frac{1}{a}-\frac{1}{c}\right)}{x_{3}-\frac{1}{a}}+\frac{E\left(\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}-\frac{1}{a}\right)}{x_{3}-\frac{1}{b}}+\frac{F\left(\frac{1}{c}-\frac{1}{a}\right)\left(\frac{1}{c}-\frac{1}{b}\right)}{x_{3}-\frac{1}{c}}]\frac{v}{\left(x_{3}-\frac{1}{a}\right)\left(x_{3}-\frac{1}{b}\right)\left(x_{3}-\frac{1}{c}\right)}\\ &&=0 \end{eqnarray}
   この計算より、反転によってRiemannのP関数は下記のように変形されることが示されたのだ。\begin{equation} P\begin{Bmatrix}a&b&c\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\rightarrow P\begin{Bmatrix}\frac{1}{a}&\frac{1}{b}&\frac{1}{c}\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix} \end{equation}
4. まとめると、$x^{'}=\frac{rx+s}{px+q}$なる変換についてRiemannのP関数は下記のようになるのだ。\begin{eqnarray} P\begin{Bmatrix}a&b&c\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}&\rightarrow& P\begin{Bmatrix}a+\frac{q}{p}&b+\frac{q}{p}&c+\frac{q}{p}\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\\ &\rightarrow&P\begin{Bmatrix}\frac{1}{a+\frac{q}{p}}&\frac{1}{b+\frac{q}{p}}&\frac{1}{c+\frac{q}{p}}\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\\ &\rightarrow&P\begin{Bmatrix}\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{a+\frac{q}{p}}&\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{b+\frac{q}{p}}&\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{c+\frac{q}{p}}\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\\ &\rightarrow&P\begin{Bmatrix}\frac{r}{p}+\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{a+\frac{q}{p}}&\frac{r}{p}+\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{b+\frac{q}{p}}&\frac{r}{p}+\frac{\frac{1}{p}\left(s-\frac{qr}{p}\right)}{c+\frac{q}{p}}\\\alpha&\beta&\gamma&\frac{rx+s}{px+q}\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix} \end{eqnarray}よって、$x^{'}=T\left(x\right)=\frac{rx+s}{px+q}$とすれば下記のように変換されたことが分かったのだ。\begin{equation} P\begin{Bmatrix}a&b&c\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix}\rightarrow P\begin{Bmatrix}T\left(a\right)&T\left(b\right)&T\left(c\right)\\\alpha&\beta&\gamma&T\left(x\right)\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix} \end{equation}

\begin{equation} \left(\frac{x-a}{x-b}\right)^{\mu}\left(\frac{x-c}{x-b}\right)^{\nu}P\begin{Bmatrix}a&b&c\\\alpha&\beta&\gamma&x\\\alpha^{'}&\beta^{'}&\gamma^{'}\end{Bmatrix} =P\begin{Bmatrix}a&b&c\\\alpha+\mu&\beta-\mu-\nu&\gamma+\nu&x\\\alpha^{'}+\mu&\beta^{'}-\mu-\nu&\gamma^{'}+\nu\end{Bmatrix}\end{equation}
が成り立つ。
実際、$x=a$での近傍の解で特性指数$\alpha$の解を$u=\sum_{n=0}^{\infty}a_{n}\left(x-a\right)^{n+\alpha}$とおくのだ。また$v=\left(\frac{x-a}{x-b}\right)^{\mu}\left(\frac{x-c}{x-b}\right)^{\nu}u$と書くのだ。
\begin{eqnarray} v&=&\left(\frac{x-a}{x-b}\right)^{\mu}\left(\frac{x-c}{x-b}\right)^{\nu}\sum_{n=0}^{\infty}a_{n}\left(x-a\right)^{n+\alpha}\\ &=&\frac{\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\sum_{n=0}^{\infty}a_{n}\left(x-a\right)^{n+\alpha+\mu} \end{eqnarray}
また次の様に書けるのだ。
\begin{eqnarray} \frac{\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}&=&\frac{\left(a-c\right)^{\nu}}{\left(a-b\right)^{\mu+\nu}}\sum_{m=0}^{\infty}\begin{pmatrix}\nu\\m\end{pmatrix}\left(x-a\right)^{m}\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(x-a\right)^{n}\\ &=&\frac{\left(a-c\right)^{\nu}}{\left(a-b\right)^{\mu+\nu}}\sum_{m=0}^{\infty}\left(x-a\right)^{m}\sum_{n=0}^{m}\left(-1\right)^{n}\begin{pmatrix}\nu\\m-n\end{pmatrix} \end{eqnarray}このことから、$v$がRiemannのP微分方程式とすれば、$x=a$の近傍で、特性指数$\alpha+\mu$の解であることが分かるのだ。
同様にして、$v$がRiemannのP微分方程式であるとすれば
\begin{equation} v=P\begin{Bmatrix}a&b&c\\\alpha+\mu&\beta-\mu-\nu&\gamma+\nu&x\\\alpha^{'}+\mu&\beta^{'}-\mu-\nu&\gamma^{'}+\nu\end{Bmatrix} \end{equation}
また$v$は次のRiemannのP微分方程式を満たすことを示すのだ。
\begin{equation} \frac{d^{2}v}{dx^{2}}+[\frac{1-\alpha-\alpha^{'}-2\mu}{x-a}+\frac{1-\beta-\beta^{'}+2\left(\mu+\nu\right)}{x-b}+\frac{1-\gamma-\gamma^{'}-2\nu}{x-c}]\frac{dv}{dx}+[\frac{\left(\alpha+\mu\right)\left(\alpha^{'}+\mu\right)\left(a-b\right)\left(a-c\right)}{x-a}+\frac{\left(\beta-\mu-\nu\right)\left(\beta^{'}-\mu-\nu\right)\left(b-c\right)\left(b-1\right)}{x-b}+\frac{\left(\gamma+\nu\right)\left(\gamma^{'}+\nu\right)\left(c-a\right)\left(c-b\right)}{x-c}]\frac{v}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}=0 \end{equation}
これを満たすことを確認するために下記の事実を用いるのだ。
\begin{eqnarray} \frac{dv}{dx}&=&\frac{d}{dx}\left(\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}u\right)\\ &=&\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{du}{dx}+\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)u\right) \end{eqnarray}
\begin{eqnarray} \frac{d^{2}v}{dx^{2}}&=&\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{d^{2}u}{dx^{2}}+\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)\frac{du}{dx}\right)\\ &+&\mu\frac{\left(x-a\right)^{\mu-1}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{du}{dx}+\left(\frac{\mu-1}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)u\right)\\ &-&\left(\mu+\nu\right)\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu+1}}\left(\frac{du}{dx}+\left(\frac{\mu}{x-a}-\frac{\mu+\nu+1}{x-b}+\frac{\nu}{x-c}\right)u\right)\\ &+&\nu\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu-1}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{du}{dx}+\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu-1}{x-c}\right)u\right)\\ &=&\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\frac{d^{2}u}{dx^{2}}\\ &+&2\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)\frac{du}{dx}\\ &+&\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{\mu\left(\mu-1\right)}{\left(x-a\right)^{2}}-\frac{\mu\left(\mu+\nu\right)}{\left(x-a\right)\left(x-b\right)}+\frac{\mu\nu}{\left(x-a\right)\left(x-c\right)}-\frac{\mu\left(\mu+\nu\right)}{\left(x-a\right)\left(x-b\right)}+\frac{\left(\mu+\nu\right)\left(\mu+\nu+1\right)}{\left(x-b\right)^{2}}-\frac{\nu\left(\mu+\nu\right)}{\left(x-b\right)\left(x-c\right)}+\frac{\mu\nu}{\left(x-a\right)\left(x-c\right)}-\frac{\nu\left(\mu+\nu\right)}{\left(x-b\right)\left(x-c\right)}+\frac{\nu\left(\nu-1\right)}{\left(x-c\right)^{2}}\right)u\\ &=&\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\frac{d^{2}u}{dx^{2}}\\ &+&2\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)\frac{du}{dx}\\ &+&\frac{\left(x-a\right)^{\mu}\left(x-c\right)^{\nu}}{\left(x-b\right)^{\mu+\nu}}\left(\frac{\mu\left(\mu-1\right)}{\left(x-a\right)^{2}}+\frac{\left(\mu+\nu\right)\left(\mu+\nu+1\right)}{\left(x-b\right)^{2}}+\frac{\nu\left(\nu-1\right)}{\left(x-c\right)^{2}}-2\left(\frac{\mu\left(\mu+\nu\right)}{\left(x-a\right)\left(x-b\right)}-\frac{\mu\nu}{\left(x-a\right)\left(x-c\right)}+\frac{\nu\left(\mu+\nu\right)}{\left(x-b\right)\left(x-c\right)}\right)\right)u \end{eqnarray}
これらの事実を用いてRiemannのP微分方程式に代入すると下記の様になるのだ。
\begin{eqnarray} &&\frac{d^{2}u}{dx^{2}}\\ &&+2\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)\frac{du}{dx}\\ &&+\left(\frac{\mu\left(\mu-1\right)}{\left(x-a\right)^{2}}+\frac{\left(\mu+\nu\right)\left(\mu+\nu+1\right)}{\left(x-b\right)^{2}}+\frac{\nu\left(\nu-1\right)}{\left(x-c\right)^{2}}-2\left(\frac{\mu\left(\mu+\nu\right)}{\left(x-a\right)\left(x-b\right)}-\frac{\mu\nu}{\left(x-a\right)\left(x-c\right)}+\frac{\nu\left(\mu+\nu\right)}{\left(x-b\right)\left(x-c\right)}\right)\right)u\\ &&+[\frac{1-\alpha-\alpha^{'}-2\mu}{x-a}+\frac{1-\beta-\beta^{'}+2\left(\mu+\nu\right)}{x-b}+\frac{1-\gamma-\gamma^{'}-2\nu}{x-c}]\\ &&\times \left(\frac{du}{dx}+\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)u\right)\\ &&+[\frac{\left(\alpha+\mu\right)\left(\alpha^{'}+\mu\right)\left(a-b\right)\left(a-c\right)}{x-a}+\frac{\left(\beta-\mu-\nu\right)\left(\beta^{'}-\mu-\nu\right)\left(b-c\right)\left(b-1\right)}{x-b}+\frac{\left(\gamma+\nu\right)\left(\gamma^{'}+\nu\right)\left(c-a\right)\left(c-b\right)}{x-c}]\frac{u}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}\\ &&=\frac{d^{2}u}{dx^{2}}\\ &&+[\frac{1-\alpha-\alpha^{'}}{x-a}+\frac{1-\beta-\beta^{'}}{x-b}+\frac{1-\gamma-\gamma^{'}}{x-c}]\frac{du}{dx}\\ &&+\left(\frac{\mu\left(\mu-1\right)}{\left(x-a\right)^{2}}+\frac{\left(\mu+\nu\right)\left(\mu+\nu+1\right)}{\left(x-b\right)^{2}}+\frac{\nu\left(\nu-1\right)}{\left(x-c\right)^{2}}-2\left(\frac{\mu\left(\mu+\nu\right)}{\left(x-a\right)\left(x-b\right)}-\frac{\mu\nu}{\left(x-a\right)\left(x-c\right)}+\frac{\nu\left(\mu+\nu\right)}{\left(x-b\right)\left(x-c\right)}\right)\right)u\\ &&+[\frac{1-\alpha-\alpha^{'}-2\mu}{x-a}+\frac{1-\beta-\beta^{'}+2\left(\mu+\nu\right)}{x-b}+\frac{1-\gamma-\gamma^{'}-2\nu}{x-c}]\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)u\\ &&+[\frac{\left(\alpha+\mu\right)\left(\alpha^{'}+\mu\right)\left(a-b\right)\left(a-c\right)}{x-a}+\frac{\left(\beta-\mu-\nu\right)\left(\beta^{'}-\mu-\nu\right)\left(b-c\right)\left(b-1\right)}{x-b}+\frac{\left(\gamma+\nu\right)\left(\gamma^{'}+\nu\right)\left(c-a\right)\left(c-b\right)}{x-c}]\frac{u}{\left(x-a\right)\left(x-b\right)\left(x-c\right)} \end{eqnarray}
よって次の事を示せばよいのだ。
\begin{eqnarray} &&\left(\frac{\mu\left(\mu-1\right)}{\left(x-a\right)^{2}}+\frac{\left(\mu+\nu\right)\left(\mu+\nu+1\right)}{\left(x-b\right)^{2}}+\frac{\nu\left(\nu-1\right)}{\left(x-c\right)^{2}}-2\left(\frac{\mu\left(\mu+\nu\right)}{\left(x-a\right)\left(x-b\right)}-\frac{\mu\nu}{\left(x-a\right)\left(x-c\right)}+\frac{\nu\left(\mu+\nu\right)}{\left(x-b\right)\left(x-c\right)}\right)\right)u\\ &&+[\frac{1-\alpha-\alpha^{'}-2\mu}{x-a}+\frac{1-\beta-\beta^{'}+2\left(\mu+\nu\right)}{x-b}+\frac{1-\gamma-\gamma^{'}-2\nu}{x-c}]\left(\frac{\mu}{x-a}-\frac{\mu+\nu}{x-b}+\frac{\nu}{x-c}\right)u\\ &&+[\frac{\mu\left(\alpha+\alpha^{'}+\mu\right)\left(a-b\right)\left(a-c\right)}{x-a}+\frac{-\left(\mu+\nu\right)\left(\beta+\beta^{'}-\mu-\nu\right)\left(b-c\right)\left(b-a\right)}{x-b}+\frac{\nu\left(\gamma+\gamma^{'}+\nu\right)\left(c-a\right)\left(c-b\right)}{x-c}]\frac{u}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}\\ &&=0 \end{eqnarray}
\begin{eqnarray} &&\mu\left(\mu-1\right)\left(x-b\right)^{2}\left(x-c\right)^{2}\\ &&+\left(\mu+\nu\right)\left(\mu+\nu+1\right)\left(x-a\right)^{2}\left(x-c\right)^{2}\\ &&+\nu\left(\nu-1\right)\left(x-a\right)^{2}\left(x-b\right)^{2}\\ &&-2\left(\mu\left(\mu+\nu\right)\left(x-c\right)-\mu\nu\left(x-b\right)+\nu\left(\mu+\nu\right)\left(x-a\right)\right)\left(x-a\right)\left(x-b\right)\left(x-c\right)\\ &&+\left(\left(1-\alpha-\alpha^{'}-2\mu\right)\left(x-b\right)\left(x-c\right)+\left(1-\beta-\beta^{'}+2\mu+2\nu\right)\left(x-a\right)\left(x-c\right)+\left(1-\gamma-\gamma^{'}-2\nu\right)\left(x-a\right)\left(x-b\right)\right)\\ &&\times \left(\left(\mu a-\left(\mu+\nu\right)b+\nu c\right)x+\mu bc -\left(\mu+\nu\right)ca+\nu ab\right)\\ &&+\mu\left(\alpha+\alpha^{'}+\mu\right)\left(a-b\right)\left(a-c\right)\left(x-b\right)\left(x-c\right)\\ &&-\left(\mu+\nu\right)\left(\beta+\beta^{'}-\mu-\nu\right)\left(b-c\right)\left(b-a\right)\left(x-a\right)\left(x-c\right)\\ &&+\nu\left(\gamma+\gamma^{'}+\nu\right)\left(c-a\right)\left(c-b\right)\left(x-a\right)\left(x-b\right)\\ &&=0 \end{eqnarray}
白状するとこれを手計算で行うのはあきらめたのだ(´;ω;｀)。
ただ、どうやってこの式が成り立つかを確認したかは伝えておくのだ。それは下記の様なのだ。

1. pythonによるコーディング：from sympy import symbols, expand, collect

# 定数と変数の定義
x, a, b, c, mu, nu = symbols('x a b c mu nu')
alpha, alpha, beta, beta, gamma, gamma_ = symbols('alpha alpha_ beta beta_ gamma gamma')

# 与えられた式
expr = (mu(mu-1)(x-b)2*(x-c)2
        + (mu+nu)(mu+nu+1)(x-a)2*(x-c)2
        + nu(nu-1)(x-a)2*(x-b)2
        - 2(mu(mu+nu)(x-c) - munu(x-b) + nu(mu+nu)(x-a))(x-a)(x-b)(x-c)
        + ((1-alpha-alpha - 2mu)(x-b)(x-c)
           + (1-beta-beta_ + 2mu + 2nu)(x-a)(x-c)
           + (1-gamma-gamma_ - 2nu)(x-a)(x-b))
        * ((mua - (mu+nu)b + nuc)x + mubc - (mu+nu)ca + nuab)
        + mu(alpha + alpha_ + mu)(a-b)(a-c)(x-b)(x-c)
        - (mu+nu)(beta + beta_ - mu - nu)(b-c)(b-a)(x-a)(x-c)
        + nu(gamma + gamma_ + nu)(c-a)(c-b)(x-a)*(x-b))

# 展開と整理
expanded_expr = expand(expr)

# xの多項式として整理
polynomial = collect(expanded_expr, x)

print(polynomial)
2. 出力された結果をchatgptに入力して整形
3. $\alpha+\alpha^{'}+\beta+\beta^{'}+\gamma+\gamma^{'}=1$であることを用いてなんやかんやするのだ！

1. $x=a$の場合：\begin{eqnarray} &&\mu\left(\mu-1\right)\left(a-b\right)^{2}\left(a-c\right)^{2}\\ &&+\left(1-\alpha-\alpha^{'}-2\mu\right)\mu\left(a-b\right)^{2}\left(a-c\right)^{2}\\ &&+\mu\left(\alpha+\alpha^{'}+\mu\right)\left(a-b\right)^{2}\left(a-c\right)^{2}\\ &&=0 \end{eqnarray}
2. $x=c$の場合:$a,cの対称性より0$
3. $x=b$の場合：\begin{eqnarray} &&\left(\mu+\nu\right)\left(\mu+\nu+1\right)\left(b-a\right)^{2}\left(b-c\right)^{2}\\ &&-\left(\mu+\nu\right)\left(1-\beta-\beta^{'}+2\mu+2\nu\right)\left(b-a\right)^{2}\left(b-c\right)^{2}\\ &&-\left(\mu+\nu\right)\left(\beta+\beta^{'}-\mu-\nu\right)\left(b-a\right)^{2}\left(b-c\right)^{2}\\ &&=0 \end{eqnarray}
4. x=0の場合：\begin{eqnarray} &&\mu\left(\mu-1\right)b^{2}c^{2}\\ &&+\left(\mu+\nu\right)\left(\mu+\nu+1\right)a^{2}c^{2}\\ &&+\nu\left(\nu-1\right)a^{2}b^{2}\\ &&-2\mu\left(\mu+\nu\right)abc^{2}+2\mu\nu ab^{2}c-2\nu\left(\mu+\nu\right)a^{2}bc\\ &&+\left(\left(1-\alpha-\alpha^{'}-2\mu\right)bc+\left(1-\beta-\beta^{'}+2\mu+2\nu\right)ac+\left(1-\gamma-\gamma^{'}-2\nu\right)ab\right)\left(\mu bc-\left(\mu+\nu\right)ca+\nu ab\right)\\ &&+\mu bc\left(\alpha+\alpha^{'}+\mu\right)\left(a-b\right)\left(a-c\right)\\ &&-\left(\mu+\nu\right)\left(\beta+\beta^{'}-\mu-\nu\right)ac\left(b-a\right)\left(b-c\right)\\ &&+\nu ab\left(\gamma+\gamma^{'}+\nu\right)\left(c-a\right)\left(c-b\right)\\ &&=0 \end{eqnarray}

1. $abc^{2}$の係数について：\begin{equation} \left(abc^{2}の係数\right)=\nu\left(\alpha+\alpha^{'}+\beta+\beta^{'}+\gamma+\gamma^{'}-1\right)=0 \end{equation}
2. $a,c$の対称性より$\left(a^{2}bcの係数\right)=0$
3. $ab^{2}c$の係数について：\begin{equation} \left(ab^{2}cの係数\right)=-\left(\mu+\nu\right)\left(\alpha+\alpha^{'}+\beta+\beta^{'}+\gamma+\gamma^{'}-1\right)=0 \end{equation}
4. 以上より、証明が完了したのだ！

$T\left(z\right)=\frac{rz+s}{pz+q}$とすると
\begin{equation} v\left(z\right)=u\left(T\left(z\right)\right)=P\begin{Bmatrix}\frac{s}{q}&\frac{r+s}{p+q}&\frac{r}{p}\\0&0&a&T\left(z\right)\\1-c&c-a-b&b\end{Bmatrix} \end{equation}
\begin{equation} w\left(z\right)=\frac{\left(z-\frac{s}{q}\right)^{\mu}\left(z-\frac{r}{p}\right)^{\nu}}{\left(z-\frac{r+s}{p+q}\right)^{\mu+\nu}}v\left(z\right)=P\begin{Bmatrix}\frac{s}{q}&\frac{r+s}{p+q}&\frac{r}{p}\\\mu&-\mu-\nu&a+\nu&T\left(z\right)\\1-c+\mu&c-a-b-\mu-nu&b+\nu\end{Bmatrix} \end{equation}
\begin{equation} \frac{\left(z-\frac{s}{q}\right)^{\mu}\left(z-\frac{r}{p}\right)^{\nu}}{\left(z-\frac{r+s}{p+q}\right)^{\mu+\nu}}F\left(a,b;c;\frac{rz+s}{pz+q}\right)=P\begin{Bmatrix}\frac{s}{q}&\frac{r+s}{p+q}&\frac{r}{p}\\\mu&-\mu-\nu&a+\nu&T\left(z\right)\\1-c+\mu&c-a-b-\mu-\nu&b+\nu\end{Bmatrix} \end{equation}
ここで次の様に定めると
\begin{eqnarray} \left\{ \begin{array}{l} \mu=c-1\\ \nu=-b \end{array} \right. \end{eqnarray}
次のような結果が得られるのだ。
\begin{eqnarray} \frac{\left(z-\frac{s}{q}\right)^{c-1}\left(z-\frac{r}{p}\right)^{-b}}{\left(z-\frac{r+s}{p+q}\right)^{c-b-1}}F\left(a,b;c;\frac{rz+s}{pz+q}\right)=P\begin{Bmatrix}\frac{s}{q}&\frac{r+s}{p+q}&\frac{r}{p}\\c-1&b-c+1&a-b&\frac{rz+s}{pz+q}\\0&1-a&0\end{Bmatrix} \end{eqnarray}
ここで次の事実を用いるのだ。
$T\left(z\right)=\frac{rz+s}{pz+q}$とし、$T\left(a\right)=0,T\left(b\right)=1,T\left(c\right)=\infty$となるようになる$p,q,r,s$は次の様に書けるのだ。
\begin{eqnarray} \left\{ \begin{array}{l} p=\frac{b-a}{b-c}\\ q=-\frac{b-a}{b-c}c\\ r=1\\ s=-a\\ \end{array} \right. \end{eqnarray}
そこで、$a=\frac{s}{q},b=\frac{r}{p},c=\frac{r+s}{p+q}$とおき、次のような変換を考えるのだ。
\begin{eqnarray} U\left(z\right)=\frac{\left(b-c\right)\left(z-a\right)}{\left(b-a\right)\left(z+c\right)}&=&\frac{\left(\frac{r}{p}-\frac{r+s}{p+q}\right)\left(z-\frac{s}{q}\right)}{\left(\frac{r}{p}-\frac{s}{q}\right)\left(z-\frac{r+s}{p+q}\right)}\\ &=&\frac{p\left(qr-ps\right)\left(qz-s\right)}{\left(qr-ps\right)\left(\left(p+q\right)z-\left(r+s\right)\right)}\\ &=&\frac{p\left(qz-s\right)}{\left(p+q\right)z-\left(r+s\right)} \end{eqnarray}
この$U$を用いると最終的に次のような結論を得るのだ。
\begin{eqnarray} \frac{\left(U\left(z\right)-\frac{s}{q}\right)^{c-1}\left(U\left(z\right)-\frac{r}{p}\right)^{-b}}{\left(U\left(z\right)-\frac{r+s}{p+q}\right)^{c-b-1}}F\left(a,b;,c;U\left(\frac{rz+s}{pz+q}\right)\right)&=&P\begin{Bmatrix}0&1&\infty\\0&0&1-a&U\left(\frac{rz+s}{pz+q}\right)\\c-1&a-b&b-c+1\end{Bmatrix}\\ &=&F(1-a,b-c+1;2-c;U\left(\frac{rz+s}{pz+q}\right)) \end{eqnarray}
長くなったのでまとめるのだ！