The rectification theorem says that near a regular point every vector field $v$ is standard, that is, there exists a local coordinate system such that $v=\frac{\partial }{\partial x\_1}$.


In the textbooks (e.g. of Arnold and Hartmann) the vector field is assumed to be $C^r$ with $r\ge 1$ and the local coordinate system such that $v=\frac{\partial }{\partial x\_1}$ is also of smoothness $C^r$.


Why the smoothness of the coordinate system is not $C^{r+1}$? What (counter)example shows this?



Smoothness of coordinates in the rectification theorem for ODE
In dimension $1$, it's true that a flowboxing change of coordinates for a $C^r$ vector field is $C^{r+1}$, but this is no longer true in dimensions greater than $1$.


Basically, the reason is this: If $V$ is a $C^r$ vector field on $\mathbb{R}^2$ and $V(0,0)\not=0$, then there exist local $C^r$ coordinates $(x^1,x^2)$ centered on $(0,0)$ such that $V = \partial/\partial x^1$. Any other set $(y^1,y^2)$ of such $C^r$ 'flowbox' coordinates is locally of the form
$$
(y^1,y^2) = \bigl(x^1 + f(x^2), \ g(x^2)\bigr)
$$
for some $C^r$ functions $f$ and $g$ *of one variable*, that satisfy $f(0)=g(0)=0$. Generally, if $(x^1, x^2)$ is only $C^r$, there will not exist such functions $f$ and $g$ that will make $y^1$ and $y^2$ be $C^{r+1}$.


For a specific example, let $h:\mathbb{R}\to\mathbb{R}$ be a $C^r$ function that is not $C^{r+1}$ and satisfies $h(0)\not=0$ and consider the vector field $V$ defined on a neighborhood of $(0,0)$ in the $uv$-plane by
$$
V := \frac{1}{h(v)}\,\frac{\partial}{\partial u} + 0\,\frac{\partial}{\partial v}.
$$  

The $C^r$ local flowbox coordinates for $V$ on a neighborhood of $(0,0)$ are of the form
$$
(x^1,x^2) = \bigl(\ u\,h(v)+f(v),\ g(v)\ \bigr)
$$
where $f$ and $g$ are $C^r$ functions. Clearly, it is not possible to choose $f$ and $g$ so that $x^1$ will be of class $C^{r+1}$.


