Consider the following ODE: $$ x′(t)=b(x(t)),\quad x(0)=x_0. $$ If $b$ is bounded and Holder continuous, then the Cauchy-Peano theorem ensures that there exists a solution to the above equation (but in general not unique). The question is:

is it possible that there always exists a solution $x_t(x_0)$ which depends Lipschitz continuous in $x_0$? Or $x_0\to x_t(x_0)$ is one-to-one?

Many thanks for the answers!

familyof solutions $x_t (x_0)$ such that they are Lipschitzin $x_0$ for all $t\geq 0$? (b) How can $x_0 \to x_t (x_0)$ be one-to-one if there's no uniqueness? $\endgroup$ – Amir Sagiv Mar 28 '18 at 5:55On the fundamental theory of ordinary differential equations, Journal of Differential Equations1(1965), 370-392. $\endgroup$ – user539887 Mar 29 '18 at 8:01