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Thread: Competition system

  1. #1
    MHF Contributor alexmahone's Avatar
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    Competition system

    $\displaystyle \frac{dx}{dt}=a_1x-b_1x^2-c_1xy=x(a_1-b_1x-c_1y)$

    $\displaystyle \frac{dy}{dt}=a_2y-b_2y^2-c_2xy=y(a_2-b_2y-c_2x)$

    $\displaystyle a_1, a_2, b_1, b_2, c_1, c_2$ are positive constants.

    Consider the nonzero critical point $\displaystyle (x_E, y_E)$, which is the solution of the system $\displaystyle a_1-b_1x-c_1y=0$ and $\displaystyle a_2-b_2x-c_2y=0$.

    The stability of $\displaystyle (x_E, y_E)$ depends on whether $\displaystyle c_1c_2<b_1b_2$ or $\displaystyle c_1c_2>b_1b_2$. Could someone please direct me to a proof of this or guide me through it if it is not too computationally tedious?
    Last edited by alexmahone; Dec 25th 2011 at 07:30 AM.
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  2. #2
    Grand Panjandrum
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    Re: Competition system

    Quote Originally Posted by alexmahone View Post
    $\displaystyle \frac{dx}{dt}=a_1x-b_1x^2-c_1xy=x(a_1-b_1x-c_1y)$

    $\displaystyle \frac{dy}{dt}=a_2y-b_2y^2-c_2xy=y(a_2-b_2y-c_2x)$

    $\displaystyle a_1, a_2, b_1, b_2, c_1, c_2$ are positive constants.

    Consider the nonzero critical point $\displaystyle (x_E, y_E)$, which is the solution of the system $\displaystyle a_1-b_1x-c_1y=0$ and $\displaystyle a_2-b_2x-c_2y=0$.

    The stability of $\displaystyle (x_E, y_E)$ depends on whether $\displaystyle c_1c_2<b_1b_2$ or $\displaystyle c_1c_2>b_1b_2$. Could someone please direct me to a proof of this or guide me through it if it is not too computationally tedious?
    Consider a series expansion about the critical point.

    CB
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Competition system

    Quote Originally Posted by CaptainBlack View Post
    Consider a series expansion about the critical point.

    CB
    But there is no series in my question!
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  4. #4
    Grand Panjandrum
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    Re: Competition system

    Quote Originally Posted by alexmahone View Post
    But there is no series in my question!
    Let $\displaystyle (x_0,y_0)$ be the critical point. Put $\displaystyle (x,y)=(x_0+\varepsilon, y_0+\delta)$. Now consider the solution to the system as a power series in $\displaystyle \varepsilon$ and $\displaystyle \delta$ (and truncate after the first non-zero (lowest order) non-constant terms)

    Then if: $\displaystyle \partial_t\varepsilon \ne 0$ and $\displaystyle \partial_t\delta \ne 0$ you have a stable critical point iff:

    $\displaystyle \partial_t\varepsilon < 0$

    and

    $\displaystyle \partial_t\delta < 0$

    CB
    Last edited by CaptainBlack; Dec 25th 2011 at 11:51 PM.
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