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Math Help - Dynamical Systems

  1. #1
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    Dynamical Systems

    Hi, I am studying dynam systems and have missed some classes and am not sure if i'm on the right track or not,

    Dynamical system;

    dx/dt = rx - ( x/(1 + (x^2)))

    when asked to calc the fixed points for this system i figured to set the RHS to equal 0 and solve for x which gave me;

    x = (+/-) sqrt((1/r) - 1 )

    Have i done this correctly?

    It then asks me to determine the stability of the fixed points and i'm unsure how to proceed.

    Any help will be greatly appreciated am really struggling with this subject.

    Many thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by monster View Post
    Hi, I am studying dynam systems and have missed some classes and am not sure if i'm on the right track or not,

    Dynamical system;

    dx/dt = rx - ( x/(1 + (x^2)))

    when asked to calc the fixed points for this system i figured to set the RHS to equal 0 and solve for x which gave me;

    x = (+/-) sqrt((1/r) - 1 )

    Have i done this correctly?
    yes

    It then asks me to determine the stability of the fixed points and i'm unsure how to proceed.

    Any help will be greatly appreciated am really struggling with this subject.

    Many thanks.
    Let x_0 denote a fixed point, and put x=x_0+\varepsilon, now use \varepsilon as the new variable, and if when the DE is linearised:

    \frac{d\varepsilon}{dt}<0 when \varepsilon>0 and \frac{d\varepsilon}{dt}>0 when \varepsilon<0 the fixed point is stable.

    That is a positive departure from the fixed point gives a negative rate of change of departure, and a negative departude gives a positive rate of change of departure.

    and so on ..

    CB
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  3. #3
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    dont forget the trivial x=0 fixed point
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  4. #4
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    Dynamical systems cont.

    Many thanks for help above!

    Have come across another problem of similar nature where i am again to calculate the fixed points and thier stability but the equation has me stumped.

    dx/dt = (x^3) - ((r+1)x^2 ) + 2rx - r^2

    obviously a cubic but am unsure how to calculate the fixed points of this system.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by monster View Post
    Many thanks for help above!

    Have come across another problem of similar nature where i am again to calculate the fixed points and thier stability but the equation has me stumped.

    dx/dt = (x^3) - ((r+1)x^2 ) + 2rx - r^2

    obviously a cubic but am unsure how to calculate the fixed points of this system.
    By inspection x=r is a root of the cubic:

    x^3-(r+1)x^2+2rx-r^2=0

    and as:

    x^3-(r+1)x^2+2rx-r^2=(x-r)(x^2-x+r)

    this is the only real root and so the only real fixed point.

    CB
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  6. #6
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    Cheers! helps heaps,

    Have gone ahead and completed my original questing and calculated the stability of the fixed points,

    I said that;
    <br />
f(x) = rx - \frac{x}{1+x^2}<br />
    and then that the derivative;
    <br />
f ' (x) = r - \frac{1-x^2}{(1+x^2)^2}<br />
    and i then substituted equation of fixed points in for x,
    which gave a messy function of r which i simplified down to;
    <br />
f ' (x_0) = 2r - 2r^2<br />
    Where x_0denotes fixed points.

    And considering that the question stated r > 0
    i figure that;

    fixed points where 0 < r < 1 are unstable
    fixed points where r > 1 are stable
    and when r=1 fixed points inconclusive?

    I hope i did all the algebra in between steps correctly, but have i gone about this the right way am trying to follow text book but a little hard to understand without examples.

    Many thanks for help, is VERY helpful.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by monster View Post
    Cheers! helps heaps,

    Have gone ahead and completed my original questing and calculated the stability of the fixed points,

    I said that;
    <br />
f(x) = rx - \frac{x}{1+x^2}<br />
    and then that the derivative;
    <br />
f ' (x) = r - \frac{1-x^2}{(1+x^2)^2}<br />
    and i then substituted equation of fixed points in for x,
    which gave a messy function of r which i simplified down to;
    <br />
f ' (x_0) = 2r - 2r^2<br />
    Where x_0denotes fixed points.

    And considering that the question stated r > 0
    i figure that;

    fixed points where 0 < r < 1 are unstable
    fixed points where r > 1 are stable
    and when r=1 fixed points inconclusive?

    I hope i did all the algebra in between steps correctly, but have i gone about this the right way am trying to follow text book but a little hard to understand without examples.

    Many thanks for help, is VERY helpful.
    Putting x=r+\varepsilon we get:

     <br />
\frac{d \varepsilon}{dt}=r^2 \varepsilon + O(\varepsilon^2) <br />

    So if r \ne 0 a small enough positive deviation from the fixed point gives a positive error rate and a small enough negative error gives a negative error rate. Thus the fixed point is unstable for non-zero r, (when r=0 we would have to look at higher order terms in \varepsilon to investigate stability)

    Alternativly solve the linearised equation for \varepsilon :

     <br />
\frac{d \varepsilon}{dt}=r^2 \varepsilon <br />

    to get:

    \varepsilon = \varepsilon_0 e^{r^2 t}

    which for r \ne 0 shows the error grows.

    CB
    Last edited by CaptainBlack; September 16th 2009 at 11:44 PM.
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