Let $\;x'(t)=F(t,x(t))$ be an ODE such that $F(t,x)$ is monotonically decreasing in respect to the variable $x$, for any constant $t$. Let $x_1, x_2$ be two solutions and define: $\; x_s(t):=x_1(t)-x_2(t)$. Prove that for every $a<b$ the following inequality holds: $|x_s(a)| \geq |x_s(b)|$
My attempt: let $t \in (a,b)$. We have: $$x'_s(t)=x'_1(t)-x'_2(t)=F(t,x_1(t)-F(t,x_2(t))$$
If $x_1(t) \geq x_2(t)$ then since $F$ is monotonic $x'_s(t) \leq 0$ and thus $x_s(t)=|x_s(t)|$ is not increasing.
If $x_1(t) < x_2(t)$ then since $F$ is monotonic $x'_s(t) \geq 0$ and thus $- x_s(t)=|x_s(t)|$ is not increasing.
And in conclusion $|x_s(t)|$ is not increasing in the interval $(a,b)$ and thus $|x_s(a)| \geq |x_s(b)|$.
This is the general idea, but my problem is that "not increasing" is not a pointwise property, that is: if I want to use the derivative in order to show that the function is not increasing in an interval, the derivative should be non positive in every point of that interval. But at some points, $|x_s(t)|$ is not differentiable! those are the points where $x_s(t)=0$. How can I formally explain that in my proof?
Consider $(x_s(t))^2$ instead of $|x_s(t)|$. Then $$ \frac{d}{dx}\,(x_s(t))^2=2(x_1-x_2)(f(t,x_1)-f(t,x_2)\le0. $$