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

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 $$\delta\to0$$.

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

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

• Just curious, is there any reference about this argument. It's a great proof! – yaoliding Apr 14 '19 at 17:00