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Thread: Why bounded ode solution tends to a constant

  1. #1
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    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 $\displaystyle c$
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Why bounded ode solution tends to a constant

    Quote Originally Posted by xinglongdada View Post
    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

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Why bounded ode solution tends to a constant

    Quote Originally Posted by xinglongdada View Post
    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 $\displaystyle c$
    Chisigma provided you an excellent answer. It would be interesting to transcribe the exact formulation of the problem.
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  4. #4
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    Re: Why bounded ode solution tends to a constant

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