Let $f:[t_0,t_1] \times \mathbb{R} \longrightarrow{\mathbb{R}}$ a continuous function. Suppose that $f$ is a decreasing function on $x$, how can I prove that for every $x_0\in \mathbb{R}$ the problem \begin{cases} x'=f(t,x) \\ x(t_0)=x_0 \end{cases} has an unique solution?
I suppose that I should use Picard's theorem, but I dont know how to take advantage of $f$ being a decreasing function.
On
We have to prove two things, the existence and uniqueness of a solution.
Before starting to prove anything we have to make a really important observation. Since $f(t,x)$ is decreasing on x $$f(t,x_1)\geq f(t,x_2)\hspace{2mm}\forall t\in [t_0,t_1], \forall x_1\leq x_2$$ we get that $$(x_2-x_1)(f(t,x_2)-f(t,x_1))\leq0$$
Existence: Since $f(t,x)$ is a continous function (by hypothesis), we can use Peano's theorem which garantees the existence of a solution. Let's call this solution $\varphi(t)$.
Uniqueness: We first suppose that there existed another solution $\psi(t)$ for the same initial value problem. Let's define another function $$g(t)=(\varphi(t)-\psi(t))^2\geq0, \hspace{2mm}\forall t\in [t_0,t_1]$$ Now $$g'(t)=2(\varphi(t)-\psi(t))(\varphi'(t)-\psi'(t))=2(\varphi(t)-\psi(t))(f(t,\varphi(t))-f(t,\psi(t)))\leq0$$ for all $t\in [t_0,t_1]$.
So $g(t)$ is a decreasing function on $[t_0,t_1]$, which means that $$g(t)\leq g(t_0)=0$$ since we where assuming that $\psi(t)$ was also another solution to the IVP. At the end, we get $$g(t)\leq0 \text{ and }g(t)\geq0\hspace{2mm}\forall t\in[t_0,t_1]$$ Which means $g(t)=0$ for all $t$ in $[t_0,t_1]$, i.e. $\varphi(t)=\psi(t)$ for all $t$ in $[t_0,t_1]$.
Suppose $x_1$ and $x_2$ are two solutions on $[t_0,t_1]$. Let $h(t)=(x_1(t)-x_2(t))^2$ and prove that $h$ is decreasing (that is, $h'\le0$.)