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On page $23$ Milnor states:

Let $\varphi$ : $\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function which satisfies $$\begin{cases} \varphi(x) > 0, & {\rm for}\,\|x\| < 1 \\ \varphi(x) = 0, &{\rm for}\,\|x\| \geq 1\end{cases}$$ Given any fixed unit vector $c \in \mathbb{S} ^{n-1} $, consider the differential equations $$\frac{dx_i}{dt} = c_{i} \cdot \varphi(x_1,x_2,\dots,x_n); \hspace{10 mm} i = 1, \dots , n.$$ For any $ \hat x \in \mathbb{R}^n$ these equations have a unique solution $x = x(t)$, defined for all real numbers which satisfies the initial condition $$x(0)= \hat x.$$ We will use the notation $x(t) = F_{t}( \hat x)$ for this solution. Then clearly
1) $F_{t}$ is defined for all $t$ and $ \hat x $ and depends smoothly on $t$ and $ \hat x $,
2) $F_{0}( \hat x ) = \hat x $,
3) $F_{s+t}( \hat x ) = F_{s} \circ F_{t}( \hat x )$.

2) is quite clear, however I don't understand why 1) and 3) are valid. Could anybody explain that?

Moreover on page $24$ Milnor states that

clearly, with suitable choice of $c$ and $t$, the diffeomorphism $F_{t}$ will carry the origin to any desired point in the open unit ball.

Is that really 'clear'? Wouldn't it be necessary to show that $x(t) \rightarrow c$ when $t \rightarrow \infty $? If it is, how?

Thanks in advance.

Source: http://www.maths.ed.ac.uk/~aar/papers/milnortop.pdf

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  • $\begingroup$ Because $\phi$ has compact support and is smooth, $\phi'$ is bounded, hence uniformly Lipschitz. This gives existence and uniqueness. The system is time invariant, so time shifted solutions are also solutions. The function $F$ gives the solution at time $t$ starting from $x$. I need to think if there is a quick answer to smoothness with respect to initial condition. $\endgroup$ – copper.hat May 20 '15 at 0:55
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To answer your "Moreover" part: The path $x(t)$ starts at $0$ and travels in a straight line in the direction $c$. The limit of $x(t)$ as $t \to \infty$ must exist, because $x(t)$ is traveling in one direction on the straight line given by the vector $c$. Let's call the limit $x_\infty$. Now, you can imagine intuitively that the path must slow down as $t \to \infty$. By continuity of $\phi$ this should mean that $\phi$ is $0$ at $x_\infty$. This means that $x_\infty$ must be at the boundary of the circle.

More rigorously, Suppose that $|x_\infty| < 1$. There must be a sequence of times $t_n$ ($t_n \to \infty$ as $n \to \infty$) such that $$\lim_{n \to \infty} |x'(t_n)| = 0.$$ (See edit 2 below.) Since (by the differential equation) $|x'(t_n)| = |\phi(x(t_n))||c|$ we have $$0 = \lim_{n \to \infty} |x'(t_n)| = \lim_{n \to \infty}|\phi(x(t_n))||c| = \phi(x_\infty)|c|$$ and therefore $\phi(x_\infty) = 0$, so $x_\infty$ is on the boundary of the circle.

From this picture, you see that $x(t)$ sweeps out the whole line segment $$\{rc : 0 < r < \frac{1}{|c|} \}$$

Edit: let me answer your first part. (1) The existence, uniqueness, and smoothness of the solution is a fact you may learn in an ODE course. There are notes here: math.ucsb.edu/~sideris/Math243-F09/notes-09-10.pdf In particular, you want:

--Corollary 5.1.1 (Smooth dependence on initial conditions)

--Theorem 2.3.1 (Existence and Uniqueness)

It is a bit unfair to say 'clearly'.

(3) The third property essentially follows from uniqueness. In words, the property says that "flowing for time $s+t$ is the same as flowing for time $t$, then flowing for time $s$." To see this let $\hat x$ and $t$ be given. Notice that $$y(s) = F_{s+t}(\hat x)$$ and $$z(s) = F_s \circ F_t(\hat x)$$ both satisfy the same equation and initial conditions: $$\frac{dy(s)}{ds} = \phi(y(s))c \qquad y(0) = F_t(\hat x)$$ $$\frac{dz(s)}{ds} = \phi(z(s))c \qquad z(0) = F_t(\hat x)$$ so by uniqueness they must be the same.

Edit 2: Clarification of the speed going to 0. Suppose there is no such sequence of times. This would mean that $|x'(t)|$ stays large after some point in time. Specifically, for some time $T$ and small $\delta$, for all $t > T$, we have $|x'(t)| > \delta$. Thus the total distance traveled in the path $x(t)$: $$\int_0^\infty |x'(t)|dt$$ is infinite. That contradicts the fact that $x(t)$ starts at the origin, travels in the direction given by $c$, and does not go farther than distance 1 from the origin.

Edit 3: Clarification clarification. Let's assume that $|c| = 1$, otherwise you just have to normalize some things.

Thinking about the path of the origin, we're looking at the ODE $$\frac{dx(t)}{dt} = \varphi(x(t))c, \qquad x(0) = 0.$$ Since $x(t)$ is always a multiple of $c$ this is equivalent to the one dimensional ODE $$\frac{dr(t)}{dt} = \psi(r(t)), \qquad r(0) = 0$$ where $$x(t) = r(t)c, \qquad \psi(r) = \varphi(rc).$$ Since $\psi(r)\geq 0$, the solution $r(t)$ is increasing in time. Furthermore $r(t)$ is bounded by $1$, because $\psi(r) = 0$ for $r\geq 1$. Now, $$r(T) = \int_0^T r'(t) dt$$ Since the left hand side is bounded as $T \to \infty$, and since the integrand is nonnegative, we have $$\int_0^\infty r'(t) dt < \infty$$ which tells us $r'(t)$ must go to 0 as $t \to \infty$. So, $$0 = \lim_{t \to \infty}r'(t) = \lim_{t \to \infty}\psi(r(t)) = \psi(\lim_{t \to \infty}r(t))$$ which tells us $\lim_{t \to \infty}r(t)=1$.

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  • $\begingroup$ Thank you, great explanations. Could you clarify why $\lim_{n \to \infty} |x'(t_n)| = 0$ in the first part? $\endgroup$ – Dave2k13 May 21 '15 at 10:36
  • $\begingroup$ I put an edit. The answer is kind of based on intuition, but you can probably turn it into a calculation involving integrals. The key thing for that would be the fact that $\left< x(t), c \right> =|c|^2 \int_0^t \phi(x(s)) ds $ $\endgroup$ – Tim Carson May 21 '15 at 20:47
  • $\begingroup$ I've thought about it, but I can't come up with anything useful. Could you give it a try to write a complete formal solution for the "moreover" part? $\endgroup$ – Dave2k13 May 26 '15 at 18:33
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    $\begingroup$ I added it again. Hopefully putting it as a 1d ode makes it easier. $\endgroup$ – Tim Carson May 27 '15 at 19:10

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