There is a function $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ given. This function is non-increasing, so that we have:
$$\forall\,{x_1, x_2 \in \mathbb{R}},\; x_1< x_2 \implies f(t, x_1) - f(t, x_2) \ge 0.$$
There is also a differential equation given with the initial condition:
$$\begin{cases} x'=f(t,x)\\x(t_0) = x_0.\end{cases}$$
Let's consider two solutions of the equation above: $\phi_1, \phi_2$.
To show uniqueness, it should be proved that $\phi_1 \equiv\phi_2$.
This is my attempt:
assume $\phi_1 \not\equiv \phi_2$ and $\phi_1 < \phi_2$.
Because both $\phi_1$ and $\phi_2$ are solutions I can write:
$$\begin{cases} \phi_1'=f(t,\phi_1)\\\phi(t_0) = x_0\end{cases}$$
and also
$$\begin{cases} \phi_2'=f(t,\phi_2)\\\phi_2(t_0) = x_0.\end{cases}$$
Now I can consider this
$$\phi_1' - \phi_2' = f(t,\phi_1) - f(t,\phi_2) \ge 0,$$
thus
$$\phi_1' \ge \phi_2'.$$
By integrating both sides of the equation above I do get
$$\phi_1 - \phi_2 \ge C.$$
However that doesn't lead me to anything useful.
I was trying to show that $\phi_1$ and $\phi_2$ differ by at most a constant and then use Picard's theorem. Unfortunately my attempt failed. I would appreciate any hints or tips.
You get $$ \frac{d}{dt}[ϕ_1(t)-ϕ_2(t)]^2=2[ϕ_1(t)-ϕ_2(t)][f(t,ϕ_1(t))-f(t,ϕ_2(t))]\le 0 $$ by the non-increasing assumption. Thus $$ [ϕ_1(t)-ϕ_2(t)]^2\le[ϕ_1(t_0)-ϕ_2(t_0)]^2~\text{ for }~ t>t_0 $$ with the obvious consequence.