Is it possible to define the flow of a continous function? Let $f$ $\in$ $\mathcal{C}^{1}(\mathbb{R}^n,\mathbb{R}^n)$  and consider the ODE:
\begin{align*}
y'&=f(y)\\
y(0)&=y_0
\end{align*}
Once $f$ is a $\mathcal{C}^1$ function we can define the flow $\varphi$ of $f$. The function $\varphi$ satisfies the following conditions:


*

*$\varphi:D\rightarrow \mathbb{R}^n $

*$\varphi(p,0)=p \hspace{0.1cm}$; $\forall$ $p$ $\in$ $\mathbb{R}^n$

*$ \frac{d}{dt}\varphi(p,t) = f(\varphi(p,t)) $, $\forall$ $(p,t)$ $\in$ $D$
where $D$ the "definition domain of $f$", i.e, $D =\{(x,s)\in \mathbb{R}^{n+1};$ $s$ $\in$ $(\omega_x^-,\omega_x^+)$}, being that $(\omega_x^-,\omega_x^+)$ the maximal domain of solution of the ODE
\begin{align*}
y'&=f(y)\\
y(0)&=x
\end{align*}
It's possible to prove that $D$ is an open set and $\varphi$ is a $\mathcal{C}^1$ function.
Now, I would like to know if it is possible to define the flow of $f$ supposing $f$ just a continuous function, I know that maybe this is impossible because in this case, we do not have a solution uniqueness of the differential equation.

My Question: If $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is a continous function, is possible to define a continous function $\varphi:D\rightarrow \mathbb{R}^n$ (where $D\supset\mathbb{R}^n \times \{0\}$ is an open subset of $\mathbb{R}^{n+1}$) such that, $\varphi(p,0) = p$ and $\frac{d}{dt}\varphi(p,s) = f(\varphi(p,s))?$

 A: It is possible to define a multifunction (that is, a function from some subset of $\mathbb{R} \times \mathbb{R}^n$ into the set of compact connected subsets of $\mathbb{R}^n$).  
But perhaps you are asking whether one can choose in a continuous way one solution for each initial value.  This is impossible, as the following example shows.
Let 
$$
f(y) = \begin{cases}
2\sqrt{y} & \text{ for } y \ge 0,
\\
0 & \text{ for } y < 0.
\end{cases}
$$
Since for each $p > 0$ we have uniqueness of solutions, your function must have the property
$$
\varphi(p,s) = (\sqrt{p} + s)^2, \quad p > 0, \ s \in \mathbb{R}.
$$
Similarly,
$$
\varphi(p,s) = p, \quad p < 0, \ s \in \mathbb{R}.
$$
We thus have
$$
\lim\limits_{p\to 0^-} \varphi(p, 1) = 0 \ne 1 = \lim\limits_{p\to 0^+} \varphi(p, 1).
$$
Incidentally, notice that $\lim\limits_{p\to 0^-} \varphi(p, \cdot)$ is the minimal solution of the IVP
$$
\begin{cases}
y' = f(y) \\
y(0) = 0,
\end{cases}
$$
and $\lim\limits_{p\to 0^+} \varphi(p, \cdot)$ is the maximal solution of the above IVP.
