1
$\begingroup$

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

$\endgroup$
2
+50
$\begingroup$

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 $x_n(1)>1/3$. $\Box$


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.

$\endgroup$
  • $\begingroup$ Thank you for the example, I will try to generalize it into 2 dimensions and simulate the flows numerically to construct a nice instructive counter examples. I have some follow up questions if you dont mind. 1- You said we know sequences converge by Arzela-Ascoli. But what we actually directly know is that they have convergent subsequences right? For instance might it be even that the original sequence has a subsequence whose flow does not converge? I think so. $\endgroup$ – Sina May 22 '13 at 13:31
  • $\begingroup$ 2- If we know that a certain subsequence (indexed by I) of flows starting at $x_0$ converge, then considering the flows of same subsequence vectorfields $X^n$ with $n \in I$ starting from a nearby point y, is it possible to say their flows also converge? I tried to use smoothness of solutions with respect to initial conditions but was not able to get it. $\endgroup$ – Sina May 22 '13 at 13:35
  • $\begingroup$ Finally a remark if we knew that the final vector field had unique integral curves (albeit being continuous) we know that every subsequence has a further subsequence which converges. And we can show they must converge to the integral curve of the limit vector field, therefore all the sequence must converge. Right? $\endgroup$ – Sina May 22 '13 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.