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How to solve the following ODE?

$$\dot{x} = \frac{f(x)}{\|f(x)\|},$$

where $x : \mathbb{R} \to \mathbb{R}^n$, i.e., $x(t)$ is the trajectory. The right-hand side $f : \mathbb{R}^n \to \mathbb{R}^n$ is a continuously differentiable function with respect to $x$. $\|\cdot\|$ is any vector norm.

I think the right-hand side of the ODE is not Lipschitz continuous. For example, let's take $n=1$ and $f(x)=-x$, then the right-hand side $-x/|x|$ is not even continuous. In this case, the theorem of existence and uniqueness of a solution cannot be applied. Then how to analyze the existence of the solution? Is there finite escape time?

Rodrigo de Azevedo
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winston
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    You could separate into $\frac{dx}{ds}=f(x(s))$, $\frac{dt}{ds}=|f(x(s))|$. Then in the example $f(x)=x$ you get $x(s)=x_0e^s$ and $t(s)=|x_0|(e^s-1)$ and eliminating $e^s$ gives $x=x_0+\frac{x_0}{|x_0|}t$ for $t>-|x_0|$. – Lutz Lehmann Dec 11 '18 at 11:20
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    The example $\dot x=x/|x|$ is not interesting as the solutions run away from the problematic points. It is more interesting to consider $\dot x=-x/|x|$ with the sliding mode at $x=0$. One way to approach existence is to consider differential inclusions, Filippov solutions etc – A.Γ. Dec 11 '18 at 11:30
  • @A.Γ. Thank you for your suggestions. I have edited my post. Do you know any paper that analyzes this kind of problem? – winston Dec 11 '18 at 12:04
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    Here is a tutorial on the topic I just come across as one of the links when goggle "filippov solution". – A.Γ. Dec 11 '18 at 13:36
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    Finite escape time seems not possible as the growth rate is bounded by $|dx/dt|\leq 1$. For $n=1$, generalized (fillipov) solutions are needed if the surface $f(x)=0$ 'attracts' the solution (such as $\dot{x}=-sign(x)$). – RTJ Dec 14 '18 at 10:36
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    "Is there finite escape time?" Not in the sense that there would exist some $t_$ such that $x(t)$ is defined for every $t$ in $[0,t_)$ and $x(t)\to\infty$ when $t\to t_$. But it may happen that $x(t)$ is well defined for $t$ in $[0,t_)$ and that $x(t)\to x_$ when $t\to t_$, for some $x_$ such that $f(x_)=0$. Then the solution $x(t)$ cannot be defined on any wider interval $[0,t^)$ with $t^>t_*$. – Did Dec 14 '18 at 12:04
  • The right-hand side is not only discontinuous, but possibly ill-defined on an arbitrarily large set. What do you mean by a solution? – Michał Miśkiewicz Mar 12 '21 at 09:17

1 Answers1

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A qualitative approach (liberally assuming $f$ does not vanish when necessary) is to solve the equation up to a time change and then reparameterize (which, after writing what follows, I realize is exactly what Lutz Lehmann suggests doing in a comment). To be more specific, consider instead

$$\dfrac{dy}{dt}(t)=f(y(t)).$$

$f$ is $C^1$ (in particular locally Lipschitz), so the existence and uniqueness theorem guarantees that for any $y_0\in \mathbb{R}^n$, there is a unique $y\in C^1(\mathbb{R},\mathbb{R}^n)$ satisfying

$$\dfrac{dy}{dt}(t)=f(y(t))\text{ and } y(0)=y_0.$$

Let $x_0\in \mathbb{R}^n$ and let $y$ be the unique solution as above with $y(0)=x_0$. The arclength of $y$ is

$$s(t)=\int_0^t \left\vert\dfrac{dy}{dt}(\tau)\right\vert\, d\tau = \int_0^t \left\vert f(y(\tau))\right\vert\, d\tau.$$

If $f(x_0)\neq0$, since $f$ is continuous, for some time $T\in\mathbb{R}_{>0}$ we have that $s:([0,T],0)\to(\mathbb{R},0)$ is strictly increasing (and is $C^2$), so that its inverse $t:([0,S],0)\to ([0,T],0)$ is well defined (and is $C^2$) (here $S=s(T)$). Take $[x:s\mapsto y\circ t(s)] \in C^1([0,S],\mathbb{R}^n)$. Then

$$\dfrac{dx}{ds}(s)=\dfrac{dy}{dt}(t(s))\,\dfrac{dt}{ds}(s)=\dfrac{f(y(t(s)))}{\dfrac{ds}{dt}(t(s))}=\dfrac{f(y(t(s)))}{\vert f(y(t(s)))\vert}=\dfrac{f(x(s))}{\vert f(x(s))\vert}$$

Rem.1: One can generalize this to arbitrary initial times with a little more care.

Rem.2: $f$ not vanishing at an initial position $x_0$ means that it does not vanish in a neighborhood of $x_0$, which means that one can similarly extend the solution in some finite negative time.

Rem.3: Note that the trajectories of the ODE with $y$ as the unknown partition $\mathbb{R}^n$ (this is not quite a foliation since we don't know if $f$ has zeros). Since the trajectories of both ODE's are the same up to a reparameterization, we have the phase portrait of the ODE with $x$ as the unknown.

Alp Uzman
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