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Question:

Let $f: \Omega = \mathbb R \times \mathbb R^n \to \mathbb R^n$ continuous with $f(t,x) = f(x)$, locally Lipschitz such that $|f| \leq M$ in $\Omega$. Show that for every $t \in \mathbb R$ $$\begin{align}\varphi_t : \mathbb R^n &\to \mathbb R^n\\x_0 &\mapsto \varphi(t, x_0)\end{align}$$ is a homeomorphism, where $\varphi (t,x_0)$ is a unique solution defined in $\mathbb R$ for the IPV $$x' = f(x) ,\,\, x(0) = x_0 \tag{*}$$

Attempt:

  • $\varphi_t$ is injective

As $\varphi (t,x_0)$ is unique for each $x_0$, if $x_0 \neq y_0$ then $\varphi_t (x_0) = \varphi (t,x_0) \neq \varphi (t,y_0) = \varphi_t (y_0) $.

  • $\varphi_t$ is surjective

$\varphi_t$ is injective and defined from $\mathbb R^n$ to $\mathbb R^n$

  • $\varphi_t$ is continuous

Suppose it isn't. Then there exists a sequence of points $x_n$ in $\mathbb R^n$ such that $x_n \to x_0$ and $\varphi (t,x_n)\not\to \varphi(t,x_0)$. Consider $\varphi_n(\tau) = \varphi(\tau,x_n)$, where, $\tau \in [0,t]$. Then

$$|\varphi'_n(\tau)| = |\varphi'(\tau, x_n)| = |f(\tau, \varphi(\tau, x_n))| \leq M$$

Then the set $E =\{\varphi_n : [0,1] \to \mathbb R^n\}$ is equicontinous. Plus, by the Mean Value Theorem

$$|\varphi_n(\tau) - \varphi_n (\sigma)| \leq \sup_{\theta \in [0,1]} |\varphi'(\theta)| |\tau - \sigma| \leq M |\tau - \sigma|$$

Thus $(\varphi_n)$ is uniformly bounded. By the Arzelá-Ascoli Theorem there exists a subsequence $(\varphi_{n_k})$ converging to $\hat\varphi_0 \neq \varphi_0$ given by

$$\varphi_{n_k} = x_{n_k} + \int_0^{\tau} f(s, \varphi_{n_k}(s)) ds$$

Letting $k\to \infty$ we have

$$\hat\varphi_{0} = x_{0} + \int_0^{\tau} f(s, \hat\varphi_{0}(s)) ds$$

which is solution to $(*)$ and $\hat\varphi_0 \neq \varphi_0$, yielding a contradiction.

  • $\varphi_t^{-1}$ is continuous

  • $\phi_{t + s} = \phi_t \circ \phi_s$, for any given $t,s \in \mathbb R$.

Note: We still don't have the Continous Dependence on solutions Theorem

I'm having trouble with the last two parts.

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    $\begingroup$ The penultimate point follows from the last. Note that $\varphi_0 = \operatorname{id}_{\mathbb{R}^n}$. $\endgroup$ – Daniel Fischer Sep 16 '15 at 21:30
  • $\begingroup$ So, is it $$\varphi_t^{-1} (\varphi (t, x_0)) = id_{\mathbb R^n} = \varphi_0$$ $\endgroup$ – Aaron Maroja Sep 16 '15 at 21:40
  • $\begingroup$ Using the last bullet point, what is $\varphi_t^{-1}$? $\endgroup$ – Daniel Fischer Sep 16 '15 at 21:41
  • $\begingroup$ Continuous. If what have done on the last bullet is correct. $\endgroup$ – Aaron Maroja Sep 16 '15 at 21:42
  • $\begingroup$ No, I mean, can you give a different expression for $\varphi_t^{-1}$ using the last bullet point and $\varphi_0 = \operatorname{id}$? $\endgroup$ – Daniel Fischer Sep 16 '15 at 21:44
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There are two points you need to spend some work showing,

  1. for all $t\in \mathbb{R}$, the map $\varphi_t$ is continuous, and
  2. for all $s,t\in \mathbb{R}$, we have $\varphi_{t+s} = \varphi_t \circ \varphi_s$.

Together with the obvious (sorry) $\varphi_0 = \operatorname{id}_{\mathbb{R}^n}$, all the other points follow automatically.

The property 2. shows that for all $t\in \mathbb{R}$, we have

$$\varphi_t \circ \varphi_{-t} = \varphi_{t + (-t)} = \varphi_0 = \varphi_{-t} \circ \varphi_t,$$

thus $\varphi_t$ is a bijection whose inverse is $\varphi_{-t}$.

Hence, if property 1. is also shown, it follows that $\varphi_t^{-1}$ is continuous, and hence $\varphi_t$ a homeomorphism, since $\varphi_t^{-1} = \varphi_{-t}$ is also a map whose continuity is asserted in 1.

You have already shown the continuity of $\varphi_t$, but strictly speaking only for $t \geqslant 0$. For the case $t < 0$, the same argument works, but needs some small changes in notation. I would, however, prefer a direct argument rather than your proof by contradiction, but that's a matter of personal taste. The essence of the argument is in any case that the uniform limit of the $\varphi_{n_k}$ is again a solution of the differential equation $x' = f(x)$ - or its equivalent integral form - and so the convergence $\varphi_{n_k}(0) \to x_0$ shows that $\varphi(t,x_0) = \lim \varphi_{n_k}(t)$.

So it remains to tackle property 2. Fix an arbitrary $x_0 \in \mathbb{R}^n$, and $s,t \in \mathbb{R}$. Let $\psi \colon \mathbb{R} \to \mathbb{R}^n$ be the solution to the IVP $(\ast)$. (From the question text, it seems you already know that it has a unique solution that is defined on all of $\mathbb{R}$.)

By definition, we have

  • $\varphi_s(x_0) = \psi(s)$, and
  • $\varphi_{t+s}(x_0) = \psi(t+s)$.

But since the differential equation is autonomous, the function

$$\eta \colon \tau \mapsto \psi(\tau+s)$$

satisfies $\eta'(\tau) = f(\eta(\tau))$, hence is a solution to the IVP $x' = f(x),\, x(0) = \eta(0) = \psi(s) = \varphi_s(x_0)$. Therefore, we have $\varphi_t(\varphi_s(x_0)) = \eta(t) = \psi(t+s)$, i.e.

$$\varphi_t(\varphi_s(x_0)) = \varphi_{t+s}(x_0).$$

Since $x_0$ was arbitrary, it follows that indeed $\varphi_{t+s} = \varphi_t \circ \varphi_s$.

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  • $\begingroup$ Thanks for clarifying. Yes, showing that $(*)$ has an unique solution for all $\mathbb R$ is part $(a)$ of the problem. Also, everything on the book is done for $t \geq 0$. $\endgroup$ – Aaron Maroja Sep 16 '15 at 22:44
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    $\begingroup$ But we need the $t < 0$ case for the inverse. So even if the book only cares about $t \geqslant 0$, for this it is better to also consider negative $t$. $\endgroup$ – Daniel Fischer Sep 16 '15 at 22:53
  • $\begingroup$ I see, you're right. $\endgroup$ – Aaron Maroja Sep 16 '15 at 22:54

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