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本帖最后由 hbghlyj 于 2022-11-1 22:15 编辑 对于三阶可微的函数$f$,定义$S(f)=\left({\frac {f''(x)}{f'(x)}}\right)'-{\frac {1}{2}}\left({f''(x) \over f'(x)}\right)^{2}$称为$f$的Schwarzian derivative.
⑴若$S(f)=0$,则$f(x)=\frac {ax+b}{cx+d},ad-bc\ne0$.
证明:设$g(x)=\frac {f''(x)}{f'(x)}=(\ln f'(x))'$,题中方程即$g'=\frac12g^2$.
分离变量,积分得$\int2g^{-2}~\mathrm dg=\int\mathrm dx$,故$-2g^{-1}=x+C_1$,$g=-\frac2{x+C_1}$.
所以$\ln f'(x)=\int g(x)~\mathrm dx=-2\ln(x+C_1)+\ln C_2⇒f'(x)=C_2(x+C_1)^{-2}$,于是$f(x)={\frac {ax+b}{cx+d}}$.
⑵$S(f)=S(g)$,如果$g(x)=\frac{af(x)+b}{cf(x)+d},ad-bc\ne0$.Apostol, Mathematical Analysis, Ex 5.7 e
证明:设$h(x)=\frac {ax+b}{cx+d}$,则$g=h○f$,$g'(x)=h'(f(x))·f'(x)$,$g''(x)=h''(f(x))·(f'(x))^2+h'(f(x))·f''(x)$.
$\left(g''(x) \over g'(x)\right)'=\left({\frac {h''(f(x))·f'(x)^2+h'(f(x))·f''(x)}{h'(f(x))·f'(x)}}\right)'$$=\frac{-f''(x)^2 h'(f(x))^2+f'(x)^4 \left(-h''(f(x))^2\right)+h^{(3)}(f(x)) f'(x)^4 h'(f(x))+f^{(3)}(x) f'(x) h'(f(x))^2+f'(x)^2 f''(x) h'(f(x)) h''(f(x))}{f'(x)^2 h'(f(x))^2}$
使用⑴,我们有$h^{(3)}(x)=\frac32·{h''(x)^2\over h'(x)}$.把上式中的$h^{(3)}$换掉,变为$\left(g''(x) \over g'(x)\right)'=\frac{-2 f''(x)^2 h'(f(x))^2+f'(x)^4 h''(f(x))^2+2 f^{(3)}(x) f'(x) h'(f(x))^2+2 f'(x)^2 f''(x) h'(f(x)) h''(f(x))}{2 f'(x)^2 h'(f(x))^2}$
$S(g)=\left({\frac {g''(x)}{g'(x)}}\right)'-{\frac {1}{2}}\left({g''(x) \over g'(x)}\right)^{2}=\frac{2 f^{(3)}(x) f'(x)-3 f''(x)^2}{2 f'(x)^2}=S(f)$.证毕. |
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