Mathematics Asked by 0xbadf00d on July 24, 2020

Let $dinmathbb N$, $uin C^{0,:1}(mathbb R^d,mathbb R^d)$ and $c:=|u|_{C^{0,:1}(mathbb R^d,:mathbb R^d)}$ (the semi-norm given by the Lipschitz constant).

I would like to show that there is a $tau>0$ such that $$T_t(x):=x+tu(x);;;text{for }xinmathbb R^d$$

satisfies

- $T_t$ is a bijection for all $tin[0,tau]$;
- $T,T^{-1}in C^{0,:1}(mathbb R^d,C^0([0,tau],mathbb R^d))$.

It’s easy to see that $$left|T(x)-T(y)right|_{C^0([0,:tau],:mathbb R^d)}le(1+tau c)left|x-yright|;;;text{for all }x,yinmathbb R^d.$$ Assuming $taule(2c)^{-1}$ (**I’ve read at other places that we need to assume $taulemin(1,(2c)^{-1})$ but I don’t understand why**), we can show that $T_t$ is bijective for all $tin[0,tau]$:

In fact, let $tin[0,tau]$, $yinmathbb R^d$ and $$f(x):=y-(T_t(x)-x);;;text{for }xinmathbb R^d.$$ Then begin{equation}begin{split}left|f(x_1)-f(x_2)right|&=left|(x_1-x_2)-(T_t(x_1)-T_t(x_2))right|\&le tcleft|x_1-x_2right|end{split}tag2end{equation} for all $x_1,x_2inmathbb R^d.$ If $taule(2c)^{-1}$, then $tcle2^{-1}<1$ and hence $f$ is a strict contraction so that, by the Banach fixed-point theorem, there is a unique $xinmathbb R^d$ with $f(x)=x$, which is equivalent to $y=T_t(x)$.

From the reverse triangle inequality and $(2)$ we see that $$left|x_1-x_2right|le2left|T_t(x_1)-T_t(x_2)right|tag3$$ for all $tin[0,tau]$ and $x_1,x_2inmathbb R^d$ and hence $$left|T_t^{-1}(x)-T_t^{-1}(y)right|le2left|T_t(T_t^{-1}(x))-T_t(T_t^{-1}(y))right|=2left|x-yright|tag4$$ for all $tin[0,tau]$ and $x,yinmathbb R^d$.

How can we show the remaining claims? And, if $uin C^1(mathbb R^d,mathbb R^d)$, under which condition can we show that ${rm D}T_t=operatorname{id}_{mathbb R^d}+t{rm D}u$ has a nonnegative determinant?

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