Modulus of continuity of flow for non-Lipschitz vector fields satisfies Osgood condition

Question

An Osgood modulus of continuity is an increasing function $\omega:(0,1]\to(0,1]$ such that $\int_0^1\frac{dt}{\omega(t)}=\infty$.

We say a vector field $X$ satisfies Osgood condition with modulus $\omega$ if locally there is a and $|X(x)-X(y)|\le\omega(|x-y|)$ (here I don't want a constant).

When $X$ satisfies Osgood condition, we know the ODE has uniqueness hence the flow $\mathfrak F_X(t,x)$ of $X$, given by $\begin{cases}\frac d{dt}\mathfrak F_X(t;x)=X(\mathfrak F_X(t;x))\\\mathfrak F_X(t;x)=x\end{cases}$ is well-defined locally in a small time interval.

My question is, given a Osgood modulus $\omega$, is there any estimate under modulus of continuity, that for any Osgood vector fields $X$ of modulus $\omega$, what is the modulus of continuity for $\mathfrak F_X$?

In DiPerna-Lions people mostly treat the cases where the vector field has bounded divergence, and the flow is not a regular Lagrange flow. But this is not true for general Osgood vector fields. I found a paper here of Clop the talk about the general Osgood vector fields but the regularity result seems still rough if we focus on some "smoother" modulus.

Consider a log-Lipschitz ODE $x'=x\log|x|$, $\begin{cases}\frac d{dt}\phi(t;s)=\phi\log|\phi|\\\phi(0;s)=s\end{cases}$, then $\phi(t;s)=|s|^{e^t}\operatorname{sgn}s$. This flow is not regular Lagrange.

Here globally $\phi$ is $\bigcup_{\alpha>0}C^\alpha$. And locally near 0 we have for any $\epsilon>0$ there is a neighborhood $0\in U_\epsilon\subset\mathbb R\times\mathbb R$, such that $\phi$ is $C^{1-\epsilon}(U_\epsilon)$.

For general log-Lipschitz vector field, is the flow (global or local) regularity in this case sharp?

I think in Saric's paper Here, for Zygmund vector field in 1-dim, the global result is $\bigcup_{\alpha>0}C^\alpha$ holds (so-called a quasisymmetric flow). What about the higher dimension?

If we consider log-log-Lipschitz case: Consider $x'=x\log|x|\log|\log|x||$, $\begin{cases}\frac d{dt}\phi(t;s)=\phi\log|\phi|\log|\log|\phi||\\\phi(0;s)=s\end{cases}$, then near 0 we have $\phi(t;s)=|s|^{|\log s|^{e^t-1}}\operatorname{sgn}s$. It's not even locally Holder near 0.

It's likely that the modulus of the flow depends on the asymtoptic behavior of $\int_\delta^1\frac{dt}{\omega(t)}$ as $\delta\to0$.

Answer 1

Let $X:\mathbb{R}^n\to\mathbb{R}^n$ a vector field with modulus of continuity $\omega:[0,\infty)\to[0,\infty)$ (we have no reason here to bound the domain of $\omega$ to $(0,1]$) that satisfies the Osgood condition $\int_0^1{ds\over\omega(s)}=+\infty$. Being uniformly continuous on $\mathbb{R}^n$, the vector field $X$ has a linear growth, and its flow $\mathfrak{F}_X(t,x)$ is defined for all $(t,x)\in\mathbb{R}\times\mathbb{R}^n$.

By the Osgood condition, there is a well defined function $\delta:[0,+\infty)\times[0,+\infty)\to[0,+\infty) $ defined by

$$\int_r^{\delta(t,r)}{ds\over\omega(s)}=t$$ for all $r\ge0$; the Osgood condition also implies $\delta(t,r)=o(1)$ as $r\to0$.

Moreover, for any $x$, $y$ in $\mathbb{R}^n$, and for any $t\in\mathbb{R}$ we have

$$\|\mathfrak{F}_X(t,x)-\mathfrak{F}_X(t,y)\|\le \delta(|t|,\|x-y\|).$$
Therefore e.g. the flow map at time $|t|\le T$ are all equicontinuous with m.o.c. $\delta(T,\cdot)$. (If we want a m.o.c. for $\mathfrak{F}_X$ in the pair $(t,x)$, to bound $\|\mathfrak{F}_X(t,x)-\mathfrak{F}_X(u,y)\|$ we just need to add a term bounding $\|\mathfrak{F}_X(t,x)-\mathfrak{F}_X(u,x)\|$, that is the m.o.c. for the single orbit generated from $x$).

The reason is that $\delta$ solves the Cauchy problem $$\cases{\partial_t\delta(t,r)=\omega(\delta(t,r))\\ \delta(0,r)=r}$$ so the above bound follows by the usual ODE comparison arguments with super-solutions, starting from

$$\|\mathfrak{F}_X(t,x)-\mathfrak{F}_X(t,y)\|\le\|x-y\|+\big|\int_0^t\big\|X(\mathfrak{F}_X(s,x))-X(\mathfrak{F}_X(s,y))\big\|ds\big|.$$