If $x'(t) \leq f(x(t))$, then $x(t)\leq y(t)$ for which $y'(t)=f(y(t))$?

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be some Lipschitz function, and assume $x(t)$ satisfies $$x'(t) \leq f(x(t))\ {\rm with }\ x(0) = a$$ Let $y(t)$ be the unique solution to $y'(t) = f(y(t))$ with $y(0)=a$, does it then hold that $$x(t) \leq y(t)$$ for all $t\geq 0$?

Answer 1

Since $x'(0)\leq f(a)=y'(0)$, then $x(t)\leq y(t)$ for small $t>0$.

If $x(t_0)=y(t_0)$ for some $t_0>0$, then $x'(t_0)\leq f(x(t_0)) = y'(t_0)$ so that $x(t_0+t)\leq y(t_0+t)$ for small $t>0$.

[add] If $0<x'(0)=y'(0)$ and $x(t) > y(t)$ for all $ 0< t\leq \varepsilon$, then there is $t_1<t_2$ s.t. $x(t_1)=y(t_2)$.

Hence $$ x'(t_1) < f(x(t_1))=y'(t_2) $$ which is a contradiction.

Answer 2

If you consider $h(t):=x(t)-y(t)$ you have that

$|h’(t)|=|x’(t)-y’(t)|=|x’(t)-f(y(t))|$

$\leq |f(x(t))-f(y(t))|\leq |x(t)-y(t)|=|h(t)|$

So

$\frac{|h’(t)|}{|h(t)|}\leq 1$

And for each $\epsilon>0$

$log(|h(t)|)-log(|h(\epsilon)|)\leq t$

so

$|h(t)|\leq |h(\epsilon)|e^t$ and

$|x(t)|\leq |y(t)| + |h(\epsilon)|e^t$

So for $\epsilon\to 0$, by continuity of $h$, you have that

$ |x(t)|\leq |y(t)|+ $|h(0)|e^t$ =|y(t)|$

So

$|x(t)|<|y(t)|$