"Reparametrizing" a differential system of the first order (Vinograd theorem?)

Question

Consider a continuous function $f:\Omega\subset\mathbb R^n\longrightarrow \mathbb R^n$ such that for every $x\in\Omega$ the Cauchy problem: $$(\ast)\left\{\begin{array} {ll} y'=f(y)\\ y(0)=x \end{array}\right. $$ has a unique maximal solution in some interval $]a_x,b_x[$.

In classroom my professor stated the following result quoting it as the Vinograd theorem:

Suppose that a function $f$ is given as above, and define $g:\mathbb R^n\longrightarrow\mathbb R^n$ such that $$ g(x):=\left\{\begin{array} {ll} \frac{f(x)}{1+||f(x)||}\cdot \frac{\textrm{dist}(x,\partial\Omega)}{1+\textrm{dist}(x,\partial\Omega)}& \textrm{if}\; x\in\Omega\\ 0& \textrm{if}\; x\notin\Omega \end{array}\right. $$ Then, for every $x\in\mathbb R^n$, the Cauchy problem $$\left\{\begin{array} {ll} y'=g(y)\\ y(0)=x \end{array}\right. $$ has a unique global solution, namely defined on the whole $\mathbb R$, such that its support is the same as the support of the solution of the problem $(\ast)$.

My question(s):

1. In literature I can't find any "Vinograd theorem" similar to the above result. Do you know any decent reference?
2. In alternative to the point $1.$ I'd like a (very quick) sketch of the proof. I particular I don't understand why such a modification of the function $f$ fixes the supports of the solutions.

Answer 1

The idea is that the modified dynamics slows down the solutions of the first dynamics when such a solution approaches $\partial\Omega$ and/or accelerates too much, so that the maximal interval of definition $]a_x,b_x[$ is sent to $\mathbb R$ and the paths themselves stay the same.

In other words, the solutions $z_x$ of the second system are related to the solutions $y_x$ of the first system by $$ z_x(t)=y_x(\vartheta_x(t)), $$ for some increasing homeomorphism $\vartheta_x:\mathbb R\to]a_x,b_x[$ such that $\vartheta_x(0)=x$. This is because $g(x)=\lambda(x)f(x)$ for every $x$, for some scalar positive function $\lambda$.

Typo: Replace $\displaystyle\frac{f(x)}{||1+f(x)||}$ by $\displaystyle\frac{f(x)}{1+\|f(x)\|}$.