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Suppose we have a $C^k$ vector field $v$ and let $\phi_t$ be the corresponding flow. I have estimates on $v$ and its derivatives: $|v| < C_0$, $|Dv| < C_1$, $|D^2v| < C_2$, ... $|D^kv| < C_k$. My question is: which estimates can be derived for the flow $\phi_t(x)$ as a function of $x$ and its derivatives wrt $x$?

I was thinking about this: suppose $t$ is small then we have $$ \phi_t(x)=x+tv(x)+\frac 1 2 t^2 Dv(x)v(x)+... $$ so apparently I should have nontrivial estimates in x which involve all the derivatives of $v$... is this reasonable?

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    $\begingroup$ Hi! Have you tried the Gronwall lemma? Also, the derivative wrt x satisfy their usual linear ODE, so you can use the G. lemma also there. $\endgroup$
    – Pietro Majer
    Commented May 29, 2010 at 18:26
  • $\begingroup$ Do you want to obtain estimates of the form $| \phi_t(x) | < M$, where $M$ does not depend on $t$? If so, then no: let $v(x) = 1$ on the real line is a counterexample (in this case $\phi_t(x) = x + t$). The above comment by Pietro allows us to obtain estimates of the form $| \phi_t(x) | <Ce^ {M t}$, where $M$ and $C$ do not depend on $x$, and similarly for (spatial) derivatives of $\phi$. $\endgroup$
    – user7807
    Commented Jul 22, 2010 at 21:44
  • $\begingroup$ The smooth dependence of a solution of an ODE on initial conditions is a standard topic discussed in introductory textbooks. Voting to close. $\endgroup$
    – Michael Renardy
    Commented May 6, 2012 at 1:59

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Let $X=\sum_{1\le j\le n}a_j(x)\partial_{x_j}$ be a Lipschitz-continuous vector field on some open subset of $\mathbb R^n$. The flow is then Lipschitz-continuous: it is a consequence of Gronwall's inequality. In fact, with $$ \dot \Phi(t,y)=X(\Phi(t,y)),\quad \Phi(0,y)=y, $$ we have $ \Phi(t,y_1)-\Phi(t,y_2)=y_1-y_2+\int_0^t\Bigl(X(\Phi(s,y_1))-X(\Phi(s,y_2))\Bigr) ds $ and consequently for $t\ge 0$ $$ \vert \Phi(t,y_1)-\Phi(t,y_2)\vert\le \vert y_1-y_2 \vert+ \int_0^t L\vert \Phi(s,y_1))-\Phi(s,y_2)\vert ds=R(t). $$ As a result, we get $ \dot R(t)\le L R(t),\quad R(0)=\vert y_1-y_2 \vert $ so that $$\vert \Phi(t,y_1)-\Phi(t,y_2)\vert\le R(t)\le \vert y_1-y_2 \vert e^{tL}. $$ When the vector field is $C^1$, the flow is also $C^1$ with respect to $x$, but the proof is not so simple (the previous argument is somehow a first step). The Birkhoff-Rota book on ODE provides a nice proof.

Bazin.

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