Uniqueness of ODE as a consequence of the qualitative behaviour I am interested in uniqueness results for ordinary differential equations. 
Let’s start with a basic example. Consider the initial value problem $\dot{u} = 2\sqrt{|u|}$ with $u(0) =0$. It is well known that this problem has infinitely many solutions. However if we make further assumptions about the behaviour of the solution we can get a unique solution to the right. Assuming for example that $u(t)>0$ for all $t >0$ we deduce that the unique solution is given by $u(t) = t^2$. We see, in some situations further knowledge of the behaviour of the solution is helpful to deduce uniqueness. The one dimensional case is for example discussed in this article.
The special situation i am interested in is an ODE the form $\dot{u} = f(u)$ for a continuous function $f \in C(\mathbb{R}^n;\mathbb{R}^n)$. In this situation one can deduce the existence of solutions for a given initial value $u_0$ but  a priori there is no statement about uniqueness of this solution. In spirit of the introductory example it seems natural to ask if one gets uniqueness under additional assumptions.
Let’s assume that there is compact subset $ K\subset \mathbb{R}^n$ such that the following assumptions hold. 


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*$f$ is locally lipschitz continuous in the interior of $K$.

*Every solution that starts in the interior of $K$ stays there for all time. Especially there is a unique global solution for such initial values.

*Given $u_0 \in K$ there is at least one short time solution of the initial value problem. Every of these solution fulfills $u(t) \in K$ for all $t \in (0,t_+)$. Therefore also for initial values on the boundary of $K$ once again one has a global solution.



Can one deduce that given an initial value on the boundary the solution under above conditions is once again unique. Even tough we do only assume f to be continuous in a neighborhood of this initial value?

Looking at some numerical calculations and the corresponding graphs i think this might be true. 
Some further questions:


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*Do you know about a two dimensional system where one does not have uniqueness of solutions? And where one can rescue uniqueness with similiar assumptions? 

*I do not see how to generalize the results given 1 to the multi dimensional case. Do you have an idea? 

*Does the knowledge of a lyapunov function help?


Some findings:
Here one can find a theorem which is somehow what i was thinking of. So however one needs to verify a not so nice estimate on $f$ and its derivatives. 
 A: Let me give an example to start with.
I was asked (in the exercise) to give a function $f \in C(\mathbb{R})$ such that $x f(x) > 0$ for $x \neq 0$. Such that the problem $\begin{cases}x' + f(x) = 0 \\ x(0) = x'(0) = 0\end{cases}$ doesn't have local uniqueness and I chose $f(x) = sign(x) \sqrt{|x|}$. There is a theorem stablishing that since $(2 \sqrt{x})' = \frac 1 {sign(x) \sqrt{|x|}}$ in a neighbourhood to the right of $0$ then there is no local uniqueness.
But let's go to the example. The system $\begin{cases}x''+f(x) = 0 \\ x(0) = x'(0) = 0\end{cases}$ has a unique global solution. Namely the equilibrium $0$. For that you convert into a system $\begin{cases}x_1' = x_2 \\ x_2' = - sign(x) \sqrt{|x|}\end{cases}$. Then you use as a guiding (Lyapunov) function $V(x_1,x_2) = \frac {x_2^2} 2 + \int_0^x -sign(x) \sqrt{|x|} =\frac {x_2^2} 2 + \frac 2 3 |x|^{2/3}$ which is the canonical choice here called the energy function. Then you use the fact that this is a first integral meaning that $\dot V(x_1,x_2) = \langle \nabla V(x_1,x_2) , f(x_1,x_2) \rangle = 0$ and that the level sets of $V$ are invariant under first integrals. Since $V$ can only be zero when $(x_1,x_2) = (0,0)$ which gives you that you have a real equilibrium solution even if the original field was not differentiable. 
