# Why bounded ode solution tends to a constant

• Aug 6th 2011, 08:07 PM
Why bounded ode solution tends to a constant
Dear all, consider now a solution
$\displaystyle x(t):\ [0,\infty)\to R$
to an ode
$\displaystyle x'=f(x)$
If $\displaystyle x(t)$ is bounded. Show $\displaystyle \lim_{t\to\infty}x(t)=c$
for some constant $\displaystyle c$
• Aug 7th 2011, 03:07 AM
chisigma
Re: Why bounded ode solution tends to a constant
Quote:

Dear all, consider now a solution
$\displaystyle x(t):\ [0,\infty)\to R$
to an ode
$\displaystyle x'=f(x)$
If $\displaystyle x(t)$ is bounded. Show $\displaystyle \lim_{t\to\infty}x(t)=c$
for some constant $\displaystyle c$

A counterxample: $\displaystyle x(t)= \sin t$ is bounded and obeys to ODE $\displaystyle x^{'}= \sqrt{1-x^{2}}$ , but the $\displaystyle \lim_{t \rightarrow \infty} x(t)$ doesn't exsist...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Aug 7th 2011, 11:04 AM
FernandoRevilla
Re: Why bounded ode solution tends to a constant
Quote:

Dear all, consider now a solution $\displaystyle x(t):\ [0,\infty)\to R$ to an ode $\displaystyle x'=f(x)$ . If $\displaystyle x(t)$ is bounded. Show $\displaystyle \lim_{t\to\infty}x(t)=c$ for some constant $\displaystyle c$