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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!

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  • $\begingroup$ Hey Wenguang, welcome to MO. (a) Did you mean a family of 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, 2018 at 5:55
  • $\begingroup$ If we let $K_t(x_0)$ stand for the set of the values at $t > 0$ of all possible solutions taking the value $x_0$ at time zero then, under the assumption that $b$ is continuous, the mapping $$x_0 \mapsto K_t(x_0)$$ is, for $t$ fixed, upper hemi-continuous (in the modern terminology). Jean Duchon's example below shows that its need not be continuous in the Hausdorff topology. For more, see G. R. Sell's paper On the fundamental theory of ordinary differential equations, Journal of Differential Equations 1 (1965), 370-392. $\endgroup$
    – user539887
    Mar 29, 2018 at 8:01

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Suppose e.g. $b(x) = \sqrt{x}$ for $0 \le x \le 4$. If $0 < x_0 < 4$, the solution with $x(0) = x_0$ is $$x_t(x_0) = \left(\frac{t+2\sqrt{x_0}}{2}\right)^2$$ which is unique in $-\sqrt{x_0} \le t \le 4 - 2\sqrt{x_0}$. But at a given $t > 0$, this is $t^2/4 + t \sqrt{x_0} + x_0^2$, which is not a Lipschitz function of $x_0 \in (0,4)$.

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Take $b(x)=|x|^{-1/2}x$ (truncated for large $|x|$ for boundedness). Then, for small $t>0$, one has $x_t(x_0)>ct^2$ for $x_0>0$, but $x_t(x_0)<-ct^2$ for $x_0<0$, so that $x_t$ is discontinuous, whatever $x_t(0)$ (which can be anywhere in $[-ct^2,+ct^2]$).

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