# 问题及解答

## [Def] 施瓦茨导数(Schwarzian derivative)

Posted by haifeng on 2024-04-21 10:05:40 last update 2024-05-06 14:27:57 | Edit | Answers (2)

$\{f,z\}:=\frac{f'''(z)}{f'(z)}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2$

$\biggl(\frac{f''(z)}{f'(z)}\biggr)'=\frac{f'''(z)f'(z)-f''(z)f''(z)}{(f'(z))^2}=\frac{f'''(z)}{f'(z)}-\biggl(\frac{f''(z)}{f'(z)}\biggr)^2,$

$\{f,z\}=\biggl(\frac{f''(z)}{f'(z)}\biggr)'-\frac{1}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2.$

$g(z)=\frac{af(z)+b}{cf(z)+d},$

$\frac{g'''(z)}{g'(z)}-\frac{3}{2}\biggl(\frac{g''(z)}{g'(z)}\biggr)^2=\frac{f'''(z)}{f'(z)}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2.$

$\{f,t\}=\varphi^{-2}\cdot\Bigl(\{f,z\}+2A^2-2A'\Bigr)$

References:

1

Posted by haifeng on 2024-04-21 10:54:30

$g(z)=\frac{af(z)+b}{cf(z)+d},\quad ad-bc\neq 0.$

$g'(z)=\frac{af'(z)(cf(z)+d)-(af(z)+b)cf'(z)}{(cf(z)+d)^2}=\frac{(ad-bc)f'(z)}{(cf(z)+d)^2}.$

$\begin{split} g''&=\biggl[\frac{(ad-bc)f'}{(cf+d)^2}\biggr]'=(ad-bc)\cdot\frac{f''\cdot(cf+d)^2-f'\cdot 2(cf+d)cf'}{(cf+d)^4}\\ &=(ad-bc)\cdot\frac{(cf+d)f''-2c(f')^2}{(cf+d)^3}, \end{split}$

$\begin{split} g'''&=(ad-bc)\cdot\biggl[\frac{(cf+d)f''-2c(f')^2}{(cf+d)^3}\biggr]'\\ &=(ad-bc)\cdot\frac{\bigl[(cf+d)f''-2c(f')^2\bigr]'\cdot(cf+d)^3-\bigl[(cf+d)f''-2c(f')^2\bigr]\cdot 3(cf+d)^2\cdot cf'}{(cf+d)^6}\\ &=\frac{ad-bc}{(cf+d)^4}\biggl[\Bigl(cf'\cdot f''+(cf+d)f'''-4cf' f''\Bigr)(cf+d)-\Bigl((cf+d)f''-2c(f')^2\Bigr)\cdot 3cf'\biggr]\\ &=\frac{ad-bc}{(cf+d)^4}\biggl[\Bigl((cf+d)f'''-3cf' f''\Bigr)(cf+d)-3c(cf+d)f' f''+6c^2(f')^3\biggr]\\ &=\frac{ad-bc}{(cf+d)^4}\biggl[(cf+d)^2 f'''-6cf' f''\cdot(cf+d)+6c^2(f')^3\biggr] \end{split}$

$\begin{split} \frac{g'''}{g'}&=\frac{ad-bc}{(cf+d)^4}\biggl[(cf+d)^2 f'''-6cf' f''\cdot(cf+d)+6c^2(f')^3\biggr]\cdot\frac{(cf+d)^2}{(ad-bc)f'}\\ &=\frac{1}{(cf+d)^2}\biggl[(cf+d)^2\frac{f'''}{f'}-6c(cf+d)f''+6c^2(f')^2\biggr]\\ &=\frac{f'''}{f'}-\frac{6cf''}{cf+d}+\frac{6c^2(f')^2}{(cf+d)^2} \end{split}$

$\begin{split} \frac{3}{2}\biggl(\frac{g''}{g'}\biggr)^2&=\frac{3}{2}\biggl[(ad-bc)\cdot\frac{(cf+d)f''-2c(f')^2}{(cf+d)^3}\cdot\frac{(cf+d)^2}{(ad-bc)f'}\biggr]^2\\ &=\frac{3}{2}\biggl[\frac{(cf+d)f''-2c(f')^2}{(cf+d)f'}\biggr]^2\\ &=\frac{3}{2}\biggl[\frac{f''}{f'}-\frac{2cf'}{cf+d}\biggr]^2\\ &=\frac{3}{2}\biggl[\biggl(\frac{f''}{f'}\biggr)^2-\frac{4cf''}{cf+d}+\frac{4c(f')^2}{(cf+d)^2}\biggr]\\ &=\frac{3}{2}\biggl(\frac{f''}{f'}\biggr)^2-\frac{6cf''}{cf+d}+\frac{6c(f')^2}{(cf+d)^2}, \end{split}$

$\frac{g'''}{g'}-\frac{3}{2}\biggl(\frac{g''}{g'}\biggr)^2=\frac{f'''}{f'}-\frac{3}{2}\biggl(\frac{f''}{f'}\biggr)^2.$

2

Posted by haifeng on 2024-05-06 15:06:53

$A(z)=\frac{1}{2}\frac{\varphi'(z)}{\varphi(z)},$

$A'(z)=\frac{1}{2}\frac{\varphi''(z)\varphi(z)-(\varphi'(z))^2}{\varphi^2(z)}.$

$\begin{split} 2(A^2-A')&=2\biggl[\frac{1}{4}\frac{\varphi'^2}{\varphi^2}-\frac{1}{2}\frac{\varphi''\varphi-\varphi'^2}{\varphi^2}\biggr]\\ &=\frac{1}{2}\frac{\varphi'^2}{\varphi^2}-\frac{\varphi''\varphi-\varphi'^2}{\varphi^2}\\ &=\frac{3\varphi'^2-2\varphi''\varphi}{2\varphi^2}\\ &=\frac{3}{2}\biggl(\frac{\varphi'}{\varphi}\biggr)^2-\frac{\varphi''}{\varphi}\\ &=\frac{3}{2}\biggl(\frac{t''(z)}{t'(z)}\biggr)^2-\frac{t'''(z)}{t'(z)}. \end{split}$

$\begin{split} &\varphi^{-2}\Bigl[\{f,z\}+2A^2-2A'\Bigr]\\ =&\frac{1}{(t'(z))^2}\biggl[\frac{f'''(z)}{f'(z)}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2+\frac{3}{2}\biggl(\frac{t''(z)}{t'(z)}\biggr)^2-\frac{t'''(z)}{t'(z)}\biggr]\\ =&\frac{f'''(z)}{f'(z)}\frac{1}{(t'(z))^2}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2\frac{1}{(t'(z))^2}+\frac{3}{2}\biggl(\frac{t''(z)}{t'(z)}\biggr)^2\frac{1}{(t'(z))^2}-\frac{t'''(z)}{t'(z)}\frac{1}{(t'(z))^2} \end{split}\tag{*}$

$f'(z)=f'(t)\cdot t'(z)$

$f''(z)=(f'(t)\cdot t'(z))'_z=f''(t)\cdot (t'(z))^2+f'(t)\cdot t''(z).$

$\begin{split} f'''(z)&=\bigl[f''(t)\cdot (t'(z))^2+f'(t)\cdot t''(z)\bigr]'_z\\ &=f'''(t)\cdot(t'(z))^3+f''(t)\cdot 2t'(z)t''(z)+f''(t)\cdot t'(z)t''(z)+f'(t)\cdot t'''(z). \end{split}$

$\begin{split} \frac{f'''(z)}{f'(z)}\frac{1}{(t'(z))^2}&=\frac{f'''(t)\cdot(t'(z))^3+f''(t)\cdot 2t'(z)t''(z)+f''(t)\cdot t'(z)t''(z)+f'(t)\cdot t'''(z)}{f'(t)\cdot (t'(z))^3}\\ &=\frac{f'''(t)}{f(t)}+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{t'''(z)}{(t'(z))^3}, \end{split}$

$\begin{split} \frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2\frac{1}{(t'(z))^2}&=\frac{3}{2}\biggl(\frac{f''(t)\cdot (t'(z))^2+f'(t)\cdot t''(z)}{f'(t)\cdot t'(z)}\biggr)^2\cdot\frac{1}{(t'(z))^2}\\ &=\frac{3}{2}\frac{1}{(t'(z))^2}\biggl(\frac{f''(t)}{f'(t)}t'(z)+\frac{t''(z)}{t'(z)}\biggr)^2\\ &=\frac{3}{2}\frac{1}{(t'(z))^2}\biggl[\biggl(\frac{f''(t)}{f'(t)}\biggr)^2(t'(z))^2+2\frac{f''(t)}{f'(t)}t''(z)+\biggl(\frac{t''(z)}{t'(z)}\biggr)^2\biggr]\\ &=\frac{3}{2}\biggl(\frac{f''(t)}{f'(t)}\biggr)^2+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{3}{2}\frac{(t''(z))^2}{(t'(z))^4} \end{split}$

$\begin{split} &\varphi^{-2}\Bigl[\{f,z\}+2A^2-2A'\Bigr]\\ =&\biggl(\frac{f'''(t)}{f(t)}+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{t'''(z)}{(t'(z))^3}\biggr)-\biggl[\frac{3}{2}\biggl(\frac{f''(t)}{f'(t)}\biggr)^2+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{3}{2}\frac{(t''(z))^2}{(t'(z))^4}\biggr]+\frac{3}{2}\frac{(t''(z))^2}{(t'(z))^4}-\frac{t'''(z)}{(t'(z))^3}\\ &=\frac{f'''(t)}{f'(t)}-\frac{3}{2}\biggl(\frac{f''(t)}{f'(t)}\biggr)^2 \end{split}$