Unicity of solution of ODE when the function is continuos and non-decreasing on $x$

Question

I'm studying Ordinary Differential Equations using the book of Viana and Spinar, and I came a cross with a question that I can't formalize (Ex 2.12):

Let $F: U \subset \mathbb{R}^2 \rightarrow \mathbb{R} $ be a continuous function in the open set $U$ and suppose that the function is non-decreasing in $x$, that is, $F(t,x_1) \geq F(t,x_2)$ for every $t$ and $x_1 \leq x_2$. Given $(t_0,x_0) \in U$ and $\gamma_1$, $\gamma_2$ two solutions of $x' = F(t,x)$ with initial condition $\gamma_1(t_0) = x_0 = \gamma_2(t_0)$, then $\gamma_1(t) = \gamma_2(t)$ for every $t \geq t_0$ .

I have an intuition regarding this problem, but I'm having some problems to write the solution properly. My attempt was this:

By the sake of contradiction, suppose that we can find a $s_0 \geq t_0$ such that $\gamma_1(s) > \gamma_2(s)$ for every $s \in (t_0,s_0]$. Then, the supposition on the function tell us that $\gamma'_1(s) \leq \gamma'_2(s)$ for all $s \in (t_0,s_0]$ which is already a contradiction, because the functions are equal on $t_0$.

This argument has a weak point, which is we may not be able to define such a interval $(t_0,s_0]$ such that (letting $g(t) = \gamma_1(t) - \gamma_2(t)$) we have $g(0) = 0$ and $g(t) \geq 0$ for all $t \in (t_0,s_0]$ because the zeros can accumulate.

Can anyone give me a hint?

Answer 1

Let $x \mapsto F(t,x)$ be non-increasing (see the comment of Gonçalo ). For $t \in [t_0,\omega)$ (as long as both solutions exist) you then have $$ [(\gamma_1-\gamma_2)^2]'(t)=2(\gamma_1(t)-\gamma_2(t))(\gamma_1'(t)-\gamma_2'(t)) $$ $$ =2(\gamma_1(t)-\gamma_2(t))(F(t,\gamma_1(t))-F(t,\gamma_2(t))) \le 0, $$ hence $(\gamma_1-\gamma_2)^2$ is non-increasing on $[t_0,\omega)$, and with $(\gamma_1(t_0)-\gamma_2(t_0))^2=0$ this yields $(\gamma_1-\gamma_2)^2=0$ on $[t_0,\omega)$.