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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.

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  • $\begingroup$ Use that every contraction mapping is uniformly continuous and apply Picard's theorem $\endgroup$
    – Vasya
    Oct 24 '18 at 12:59
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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, $h'\le0$.)

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  • $\begingroup$ 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 $\endgroup$ Oct 24 '18 at 19:13
  • $\begingroup$ $h\ge0$, $h(0)=0$, $h$ decreasing. There are not many possibilities for $h$. $\endgroup$ Oct 24 '18 at 19:50
  • $\begingroup$ I'm sorry but I don't follow that you are trying to say, could you elaborate? $\endgroup$ Oct 24 '18 at 23:29
  • $\begingroup$ Differentiate $h$, substitute $x’_i$ by $f(t,x_i)$ and use that $f(t,x)$ is decreasing in $x$ to deduce that $h’\le0$. $\endgroup$ Oct 25 '18 at 13:27

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