# Proving a differential equation has a unique solution, where $f(t,x)$ is decreasing.

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.

• Use that every contraction mapping is uniformly continuous and apply Picard's theorem Oct 24 '18 at 12:59

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, $$h'\le0$$.)
• And since you assumed that $x_1$ and $x_2$ are two solutions, that should get you to a contradiction? I don't understand the reasoning in this answer Oct 24 '18 at 19:13
• $h\ge0$, $h(0)=0$, $h$ decreasing. There are not many possibilities for $h$. Oct 24 '18 at 19:50
• Differentiate $h$, substitute $x’_i$ by $f(t,x_i)$ and use that $f(t,x)$ is decreasing in $x$ to deduce that $h’\le0$. Oct 25 '18 at 13:27