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Thread: Phase Line & Limits

  1. #1
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    Phase Line & Limits

    I understand how to create a horizontal phase line of a function...
    $\displaystyle \frac{dx}{dt}=5x^2(x+5)(x-3)^3$

    --->-----(-5)-----<------(0)------<------(3)----->----


    ...but I don't really understand what it means. What does the horizontal line represent and what do the arrows represent?

    And what procedure would I use for answering the question:

    The solution of the above equation with $\displaystyle x(0)=-4$ is defined on the interval $\displaystyle a<t<b $, find
    $\displaystyle lim_{t\to a+} x(t)= $
    $\displaystyle lim_{t\to b-} x(t)= $

    Is there a way to find the limit analytically? I'm guessing not since we're using the phase line to do it.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The function $\displaystyle v:\mathbb{R}\to \mathbb{R},\;v(x)=5x^2(x+5)(x-3)^3$ represents a velocity field. That is, the equation $\displaystyle x'(t)=v(x)$ is equivalent to find the movement $\displaystyle x=x(t)$ of a particle in the real line such that its velocity in any instant of time $\displaystyle t$ is exactly $\displaystyle v(x)$ . The arrows represent the movement sense. Hope this helps.
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