Space derivative of flow of ODE with monotone source

Question

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 $\Phi(t, \cdot)$?

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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 $f$)?

Answer 1

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.