How do I prove that any second-order equation with three regular singular points can be transformed into a...












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Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?










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    Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?










    share|cite|improve this question









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      $begingroup$


      Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?










      share|cite|improve this question









      $endgroup$




      Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?







      ordinary-differential-equations hypergeometric-function






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      asked Jan 27 at 2:51









      user638526user638526

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