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Thread: Problem With Stability Theorem

  1. #1
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    Problem With Stability Theorem

    Okay, I have the following autonomous differential equation:

    $\displaystyle dx/dt = f(x) = x^2-5x+6$

    I found the steady states of the equation of be x*1=2 and x*2=3

    When I sub them into $\displaystyle f'(x^*) = 2x-5$, I get $\displaystyle f'(x^*1) = -1 $ and $\displaystyle f'(x^*2)= 1$.

    Now I was told that the stability theorem states that if $\displaystyle f'(x^*) < 0$, then x* is stable and when $\displaystyle f'(x^*) > 0$, x* is unstable, however, when I create my phase line diagram is shows that x*2 is stable.

    Could someone shed some light on this?

    Thanks,

    JS
    Last edited by jschlarb; Feb 7th 2009 at 11:10 AM.
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  2. #2
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    Quote Originally Posted by jschlarb View Post
    Okay, I have the following autonomous differential equation:

    $\displaystyle dx/dt = f(x) = x^2-5x+6$

    I found the steady states of the equation of be x*1=2 and x*2=3

    When I sub them into $\displaystyle f'(x^*) = 2x-5$, I get $\displaystyle f'(x^*1) = -1 $ and $\displaystyle f'(x^*2)= 1$.

    Now I was told that the stability theorem states that if $\displaystyle f'(x^*) < 0$, then x* is stable and when $\displaystyle f'(x^*) > 0$, x* is unstable, however, when I create my phase line diagram is shows that x*2 is stable.

    Could someone shed some light on this?

    Thanks,

    JS
    Here's the direction field- it shows that $\displaystyle x = 2$ is stable and $\displaystyle x = 3$ unstable.
    Attached Thumbnails Attached Thumbnails Problem With Stability Theorem-dirfield.jpg  
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  3. #3
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    I think I found my problem: When I made a general shape of the graph, I made it so it curved down instead of curve up :S.

    Thanks!
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  4. #4
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    I also have another part of the question I need help with:

    It asks to use separation of variable to determine the equation explicitly, however when I integrate it, I get an equation that doesn't make sense:

    $\displaystyle 1=Ke^t$

    I separated the variables to get an equation with which I needed to use integration by parts. I determined that A=1 and B=-1 to which i got the final equation to be

    $\displaystyle ln|x-2|-ln|x-3|+c = t+c$

    Taking the e function to everything and simplifying got me that. Could there have been a mistake in my integration (I apologize, I'm not too familiar with the math functions on this site so I wasn't able to show you what I did).
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  5. #5
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    Quote Originally Posted by jschlarb View Post
    I also have another part of the question I need help with:

    It asks to use separation of variable to determine the equation explicitly, however when I integrate it, I get an equation that doesn't make sense:

    $\displaystyle 1=Ke^t$

    I separated the variables to get an equation with which I needed to use integration by parts. I determined that A=1 and B=-1 to which i got the final equation to be

    $\displaystyle ln|x-2|-ln|x-3|+c = t+c$

    Taking the e function to everything and simplifying got me that. Could there have been a mistake in my integration (I apologize, I'm not too familiar with the math functions on this site so I wasn't able to show you what I did).
    No - you're good to go. Put the constants together and call them $\displaystyle \ln c$ (it cleans things up a bit.) Then solve for x

    $\displaystyle \ln \left| \frac{x-2}{x-3}\right| = \ln c + t$

    and solving for x gives

    $\displaystyle x = \frac{3 c e^t - 2}{ c e^t - 1}$

    You can see if $\displaystyle c = 0$ you have your first critical point whereas if you let $\displaystyle c = \frac{1}{\bar{c}}$ then
    $\displaystyle x = \frac{3e^t - 2\bar{c}}{ e^t - \bar{c}}$

    and letting $\displaystyle \bar{c} = 0$ gives your second.
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