Let $\Phi$ be the flow (defined as in page 6 of [this paper](http://www.numdam.org/article/JEDP_2004____A1_0.pdf)) of the ODE
$$\begin{cases}
\frac{d}{dt}\Phi(x,t) = f(\Phi(x,t),t) \quad t >0 \\
\Phi(x,0) = x \quad x \in \mathbb{R}.
\end{cases}$$



> 
> Is it true that if $f$ is monotone in the first variable then $\Phi$ is Lipschitz? 
> 
> 
> 



Flow of ODE with monotone source
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
\dot{x}-\dot{y}= f(x,t)-f(y,t).
$$


Multiplying both sides by $x-y$ we deduce


$$
(\dot{x}-\dot{y})(x-y) =\big(f(x,t)-f(y,t)\big)(x-y)\leq 0,
$$ 
where the last equality holds because $f$ is decreasing.


Hence
$$
\frac{1}{2}\frac{d}{dt}\big(x-y)^2\leq 0.
$$
Thus the function $t\mapsto \big( x(t)-y(t)\big)^2 $ is decreasing so
$$
\big(x(t)-y(t)\big)^2\leq \big( x(0)-y(0)\big)^2,\;\;\forall t\geq 0,
$$
i.e.,
$$
\Big(\Phi(x\_0,t)-\Phi(y\_0,t)\Big)^2\leq \Big(x\_0-y\_0\Big)^2,\;\;\forall t\geq 0.
$$
In other words, for $t\geq 0$, $\Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.


