Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 8965

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

$
0
0

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?


Viewing all articles
Browse latest Browse all 8965

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>