Convergence of flows of uniformly converging vector fields

Question

My questions is about the flows of certain "well behaved" sequence of vector fields. Let $X^n: U \subset R^n \rightarrow R^n$ be a sequence of smooth vector fields which

i. Are equicontinous and equibounded

ii. $X^n$(x) converges fastly and uniformly (with "speed" $||X^{n+k}(p)-X^k(p)||< r^{k}$ for r<1 in $C^{0}$ manner) to some Holder contiunous vector field X(x)

My question is, does the flows of $X^n$ starting at a point $x \in M$ converge to a flow of X (for bounded time if needed). Now I know that flows of X need not be unique since it is just continuous but it will have flows by Peano curve theorem. And I can only prove the convergence if I have that $X^n$ are equlipschitz which is not the case because the limit is not Lipschitz for certain. What I can do is for bounded time, only find a subsequence among the integral curves which converge to an integral curve of X using Arzela Ascoli theorems (you can also show integral curves are equi-lipschitz for bounded time).

It is hard to give a counter example to this too because the vector fields satisfy many properties.

(p.s: I will give points to the first person to give a counter example to give a necessary condition or to point me in the right direction or otherwise if not I will just pick one best answer).

(p.s2: I know that if the convergence is in $C^1$ topology for instance you can show this through gronwall inequality (for fixed time). But all the references I could find or all the proofs I could give is always requires the derivatives to be uniformly Lipschitz or the limit to be Lipschitz or convergence to be in $C^1$ or higher topology)

Thanks

Answer 1

One-dimensional counterexample. Let $U=(-1,1)$, $X(x)=x^{1/3}$, $$X^n(x)=(x^2+\epsilon_n^2)^{-1/3}(x+(-1)^n \epsilon_n) \tag1$$ where you can choose $\epsilon_n>0$ converging to $0$ as fast as you want. Consider the solution curve of $X^n$ beginning at $0$, denoted $x_n(t)$. If $n$ is even, then $x_n'(0)$ is positive, which makes $x_n$ increasing for all times (solutions of autonomous equations don't turn back). If $n$ is odd, then $x_n'(0)<0$ and $x_n$ is decreasing for all times.

Thus, the only way for the sequence $x_n(1)$ to have a limit is to converge to $0$. But in fact, $x_n(1)>1/3$ for even $n$. Indeed, assuming $n$ is even, the value of $ x_n(1)$ is determined from the equation $$\int_0^{x_n(1)} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx=1 \tag2$$ Since $$\begin{split} &\int_0^{1/3} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx \\ &= \int_0^{\epsilon_n} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx+ \int_{\epsilon_n}^{1/3} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx \\ &\le \int_0^{\epsilon_n} (2 \epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx+ \int_{0}^{1/3} (2x^2 )^{1/3}(x)^{-1} \,dx \\& = (2\epsilon^2)^{1/3}\log 2+3^{1/3}\cdot 2^{-2/3} \end{split}$$ is less than $1$ for small $\epsilon$, it follows that $x_n(1)>1/3$. $\Box$

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Of course, both subsequences $x_{2n}$ and $x_{2n+1}$ converge to (different) solution curves of $X$. But this is something you already know from Arzelà-Ascoli.