Suppose $v$ is a (possibly time-dependent) vector field on a compact manifold $M$.

Its flow is a mapping $g: M \times \mathbb{R} \rightarrow M$, where $g$ satisfies the following conditions (written in the local chart):

$g(x, 0) = x$

$\frac{dg}{dt} (x, t) = v(g(x, t), t)$

I would like to find for a fixed $\epsilon$ such $\delta$ that, for any $\delta$-variation of the vector field, the differential flow for time $1$ would not perturb more than on $\epsilon$.

The norm on vector fields is $\sup_{x\in M}(|v(x)|)$ w.r.t. to some arbitrary riemannian metric on $M$, and the distance between mappings $f$, $g$ is defined as $\sup_{x \in M}(dist(f(x), g(x)))$.

While this fact is almost surely standard, I can not figure out if it follows from the continuous dependence of the solution of ODE on parameters. Does anyone know a good reference?

  • 2
    I think that it is a question suitable for MSE rather than for MO. I will give you a hint, however: write the flow as a solution of an integral equation, and apply Grönwall's inequality. But the above holds if $v$ is locally Lipschitz, for example. If $v$ is continuous only, we do not need to have uniqueness. – user539887 May 15 at 11:40
  • I think the statement should hold for any measurable $v$ which is in $\epsilon$-ball in $l^\infty$, provided solution exists. – Asya Rorschach May 15 at 12:27
  • @AsyaRorschach Without uniqueness of solutions the question does not make sense. Even continuity is too weak assumption to guarantee uniqueness, so measurability is way too weak. – Oleg Eroshkin May 15 at 14:45
  • @AsyaRorschach For an example that even under the assumption of continuity one need not select a continuous solution, see Jean Duchon's answer to ODE with Holder drift - Cauchy-Peano theorem. – user539887 May 15 at 15:27
  • @user539887 will estimate still be in $\sup |v|$, or Lipschitz constant also? – Asya Rorschach May 15 at 17:25
up vote 0 down vote accepted

Let me consider two autonomous vector fields $X_1, X_2$ on a compact smooth manifold $\mathcal M$ and assume that the Lipschitz condition is true for both of them. The flow $\psi_j$ of $X_j$ is defined on $\mathbb R\times \mathcal M$ by $$ \dot \psi_j(t,m)=X_j(\psi_j(t,m)), \quad \psi_j(0,m)=m. $$ Now we consider for $m$ in a coordinate chart and $t\ge 0$ small enough $$ \psi_2(t,m)-\psi_1(t,m)=\int_0^t\bigl( X_2(\psi_2(s,m))-X_1(\psi_1(s,m))\bigr) ds, $$ so that \begin{multline} \vert\psi_2(t,m)-\psi_1(t,m)\vert\le \int_0^t\bigl\vert X_2(\psi_2(s,m))-X_1(\psi_1(s,m))\bigr\vert ds \\ \le \int_0^t\bigl\vert X_2(\psi_2(s,m))-X_2(\psi_1(s,m))\bigr\vert ds +\int_0^t\bigl\vert X_2(\psi_1(s,m))-X_1(\psi_1(s,m))\bigr\vert ds \\ \le \int_0^t L_2\vert\psi_2(s,m)-\psi_1(s,m)\vert ds +t\Vert X_2-X_1\Vert=R(t). \end{multline} We have thus with $\rho(t)=\vert\psi_2(t,m)-\psi_1(t,m)\vert$, $$ \dot R(t)=L_2\rho(t)+\Vert X_2-X_1\Vert\le L_2 R(t)+\Vert X_2-X_1\Vert, $$ so that $ \frac{d}{dt}\log\bigl(L_2 R(t)+\Vert X_2-X_1\Vert\bigr)\le L_2 $ and $$ \log\bigl(L_2 R(t)+\Vert X_2-X_1\Vert\bigr)\le \log\bigl(\Vert X_2-X_1\Vert\bigr)+ L_2 t, $$ entailing $L_2 \rho(t)+\Vert X_2-X_1\Vert\le L_2 R(t)+\Vert X_2-X_1\Vert\le e^{L_2 t}\Vert X_2-X_1\Vert $ and $$ \rho(t)\le L_2^{-1}(e^{L_2 t}-1)\Vert X_2-X_1\Vert. $$

  • Thanks for the proof! Could you please explain how do you use that $t$ is arbitrary small? Also, am I right that $X_1$ Lipschitz constant is not used in the estimate? – Asya Rorschach May 16 at 9:16
  • @AsyaRorschach I choose $t$ small enough to stay in the coordinate chart. On the other hand, although I did not use a Lipschitz constant for $X_1$, I was glad to have the Lipschitz assumption on $X_1$ to ensure the existence of a flow. You can make various variations on the above inequalities and control $\psi_2(t,m_2)-\psi_1(t,m_1)$, which would make the role of $X_1, X_2$ more symmetrical. Also, you can get flows with an Osgood-type hypothesis "logarithmically close" to Lipschitz, e.g. for log-Lipschitz vector fields and I believe that the reasoning should be similar to the above arguments. – Bazin May 16 at 19:37

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