Let $\Phi$ be the flow (defined as in page 6 of this paper) 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?


1 Answer 1


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.

  • $\begingroup$ Thank you. How can the argument be made rigorous even when $f$ is not smooth and $\Phi$ is not a classical solution but a regular Lagrangian flow? $\endgroup$
    – user124345
    Apr 16, 2019 at 11:48
  • $\begingroup$ The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces $\endgroup$ Apr 16, 2019 at 12:05
  • $\begingroup$ The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7. $\endgroup$ Apr 16, 2019 at 12:08
  • $\begingroup$ In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $\mathbb{R}\ni x\mapsto f(x)-x\in\mathbb{R}$ is onto. $\endgroup$ Apr 16, 2019 at 14:30
  • $\begingroup$ Thank you. What if $f$ is increasing? $\endgroup$
    – user124345
    Apr 16, 2019 at 23:48

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