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Consider the ODE $$ \begin{cases} \partial_t\Phi(t,x) = f(t,\Phi(t,x)), &\ t>0, \ x \in \mathbb R \\ \Phi(0,x) = x, & x \in \mathbb R \end{cases} $$ where $f$ is function which is a non-increasing in the second variable (without other assumptions on regularity).

Then $\Phi$ exists and is Lipschitz with respect to space (Flow of ODE with monotone source).

How can one compute the a.e. space derivative of this Lipschitz flow $\Phi(t, \cdot)$?


Remark. Note that, if $f$ was Lipschitz, we would get that the space derivative of the flow $\partial_x \Phi$ satisfies

$$\partial_t \partial_x \Phi = \partial_x f(t,\Phi(t,x))\partial_x \Phi.$$

To reiterate, the question of this post is the following:

In general, how can we compute $\partial_x \Phi(t,\cdot)$ if we only assume that $f$ is function which is a non-increasing in the second variable (without other assumptions on regularity of $f$)?

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  • $\begingroup$ To quote Help Center: "You should only ask practical, answerable questions based on actual problems that you face. Chatty, open-ended questions diminish the usefulness of our site and push other questions off the front page." And apparently that was the reason why your question was downvoted (not by me). $\endgroup$ – user539887 Jun 11 at 7:27
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    $\begingroup$ @user539887 I'm asking how to compute the derivative of the flow of one ODE under more general assumptions than those that are on standard books. What is not practical or not answerable about it? $\endgroup$ – Jay Jun 11 at 8:44
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    $\begingroup$ <a href="mathoverflow.net/questions/328182/… of ODE with monotone source</a> contains the answer. $\endgroup$ – Bazin Jun 11 at 12:31
  • $\begingroup$ @Bazin There uniqueness is proved. There is no computation of the space derivative. $\endgroup$ – Jay Jun 11 at 12:32
  • $\begingroup$ After your edit of 7 minutes ago the question becomes much better: to me, it seems legitimate now. $\endgroup$ – user539887 Jun 11 at 12:40
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Let me deal with the autonomous case for simplicity : $\dot x= f(x)$. Then following the reference mentioned in my comment, we have for two solutions $x_1,x_2$ with respective initial data $y_1, y_2$, $$ \dot x_2-\dot x_1= f(x_2)-f(x_1)\Longrightarrow \frac{d}{dt}\Vert x_2-x_1\Vert^2=2(f(x_2)-f(x_1))(x_2-x_1)\le 0, $$ since $f$ is non-increasing, so that for $t\ge 0$, $$ \Vert x_2(t,y_2)-x_1(t,y_1)\Vert^2\le \Vert x_2(0,y_2)-x_1(0,y_1)\Vert^2=\Vert y_2-y_1\Vert^2, $$ proving that the flow is $1-$Lipschitz continuous. Let $\phi$ be the flow defined by $$ \partial_t\phi(t,y)= f(\phi(t,y)), \quad \phi(0,y)=y. \tag{1}$$ The $1D$ autonomous situation can be exploited: separating the variables we get $$ \frac{dx}{f(x)}=dt $$ and if $f$ is say continuous and does not vanish at $y$, we may define a $C^1$ function $F$ such that $F'=1/f$ and get that $F$ is invertible locally $$ F(\phi(t,y))-F(y)=t,\quad \phi(t,y)=F^{-1}(t+F(y)), $$ and with the last formula you get an explicit expression for the derivative of $\phi$ wrt $y$. Note that going back to (1), denoting $\psi(t,y)=\partial_y\phi$, you get that $\psi$ should satisfy formally a linear ODE, but the chain rule may fail since a priori $f'$ is a (negative) measure but $\phi'$ is only bounded measurable.

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  • $\begingroup$ Yes, I understand that the flow is Lipschitz, hence differentiable a.e. The question of this post is: How does one compute the space derivative? $\endgroup$ – Jay Jun 13 at 17:34
  • $\begingroup$ I have added a little bit more. $\endgroup$ – Bazin Jun 13 at 20:23

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