MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – Pietro Majer May 29 '10 at 18:26
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 $\phi$. – user7807 Jul 22 '10 at 21:44
The smooth dependence of a solution of an ODE on initial conditions is a standard topic discussed in introductory textbooks. Voting to close. – Michael Renardy May 6 '12 at 1:59

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.


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.