Why bounded ode solution tends to a constant

**xinglongdada** · August 6th 2011, 08:07 PM

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

**chisigma** · August 7th 2011, 03:07 AM

Originally Posted by xinglongdada

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

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

Kind regards

$\chi$ $\sigma$

**FernandoRevilla** · August 7th 2011, 11:04 AM

Originally Posted by xinglongdada

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

Chisigma provided you an excellent answer. It would be interesting to transcribe the exact formulation of the problem.

**xinglongdada** · October 16th 2011, 06:32 PM

Indeed, this is a very nice counterexample...Good luck

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