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Consider the system of differential equations $$\dot{x}(t) = f(t,x)$$ where $x \in \mathbb{R}^n$ and $f$ is a $C^{\infty}$ function. Suppose that every trajectory which begins in the closed unit ball tends to zero as $t \rightarrow \infty$. Does it follow that there exists a ball around the origin $\mathcal{B}$ such that every trajectory that begins in the unit ball stays in $\mathcal{B}$?

My natural inclination is to think the answer ought to be yes and one should be able to show this by picking a convergent subsequence somehow. I'm having trouble making this work, however.

Here is where I am stuck. Supposing $x_i(t)$ is a trajectory that begins from $x_i(0)$ at $t=0$ and has distance at least $i$ away from the origin at some later time; and supposing $\lim_{i \rightarrow \infty} x_i(0) = x$; then I can't see how to derive a contradiction between the fact that the trajectory beginning from $x$ approaches zero, and is consequently bounded, and the the fact that $||x_i(t_i)||_2 \geq i$ for some $t_i$. If each $t_i$ was below some $T$ a contradiction is easy to obtain, but how to deal with $t_i$'s which blow up?

Finally, this is not homework.

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There's a distinct difference between the open and closed unit ball here. If the ball is open, ask what could happen to a trajectory starting on the boundary. If the ball is closed (hence compact) ask whether maximum distance from 0 is a nice function of initial position. – Bob Pego Mar 23 '12 at 17:29
Thanks for your comment. I edited the question to say explicitly that the unit ball should be closed. – robinson Mar 23 '12 at 20:17
@BobPego - It sounds like you know the solution, and, if that is indeed the case, would you consider posting an answer with a sketch of the argument? Thank you. – robinson Mar 23 '12 at 20:19
For the closed ball, the argument I was thinking of has sprung a leak. (The idea was, for fixed $x_0$ there is $T_0$ so $x(t,x_0)$ is small for $t>T_0$. Then for $x_1$ near $x_0$, $x(t,x_1)$ is close to $x(t,x_0)$ for $t\in[0,T_0]$. But we know nothing now for $t>T_0$!) Your subsequence idea looks more promising. Choose $(t_n,x_n)$ so $|x(t_n,x_n)|>n$, say and look at $t_*=\liminf t_n$. But I have to run just now.. – Bob Pego Mar 23 '12 at 21:50
Right, but then in the case of $t_{*}=+\infty$ I am stuck :( – robinson Mar 24 '12 at 0:24
up vote 2 down vote accepted

In fact, the forward image $x(t,{\mathcal{B}})$ need not be uniformly bounded in ${\mathbb R}^n$ for $n>1$. We can modify any given stable vector field, say $x'=-x$, by messing with a small slice of phase space for a fixed interval of time, forcing trajectories in this slice to take large excursions. During these excursions, other trajectories can be frozen by using a smooth cutoff in time. And no trajectory ever gets messed with more than once, so ultimately every trajectory approaches the origin.

The details: Fix a smooth bump function $u:{\mathbb R}\to[0,2]$ such that $u(t)>0$ for $0<t<1$ and $u(t)=0$ otherwise, and $\int_0^1 u(t)\,dt = 1$. Let $c(t)=\sum_{k\in\mathbb Z} u(t-2k-1)$. Then $c$ is $2$-periodic and $c(t)=0$ for $t\in[2k,2k+1]$, and every solution of the system $$x'=-c(t)x$$ approaches 0 as $t\to\infty$, with $x(t+2)=e^{-1} x(t)$ for all $t$.

Now consider a modified system of the form $$ x_1' = -c(t)x_1+ \sum_{k=1}^\infty k\, u'(t-2k)v_k(x_2),\quad x_j'= -c(t)x_j \quad (j>1).$$ We choose $v_k$ to be a non-negative cutoff function supported in $[a_k,b_k]\subset (0,1)$, of the form $$ v_k(y)= u\left(\frac{y-a_k}{b_k-a_k}\right). $$ For $t\in[2k,2k+1]$, $x_2(t)$ is constant, and $x_1(t)-x_1(2k)=k\,u(t-2k)v_k(x_2).$ Thus, for $x_2(t)\in (a_k,b_k)$ at time $t=2k$, $x_1(t)$ makes an excursion of order $k$, and returns to its starting value at time $t=2k+1$.

Finally, we choose $a_k, b_k\to0$ as $k\to \infty$ with $b_k<e^{-2}a_k$. Then for any solution of the system, $x_2(2k)\in(a_k,b_k)$ for at most one $k$, hence the solution $x(t)\to0$ as $t\to\infty$. On the other hand, solutions with $x_2(2k)\in(a_k,b_k)$ correspond to initial data with $x_2(0)\in (e^ka_k,e^kb_k)$. Since $e^kb_k\to0$, the forward image $x(t,{\mathcal B})$ is not bounded uniformly in $t$.

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Thanks for what looks like a fantastic answer. Please give me a few days to think about what you wrote before I accept it. – robinson Mar 26 '12 at 0:23

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