# Competition system

• Dec 25th 2011, 08:19 AM
alexmahone
Competition system
$\frac{dx}{dt}=a_1x-b_1x^2-c_1xy=x(a_1-b_1x-c_1y)$

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

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

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

The stability of $(x_E, y_E)$ depends on whether $c_1c_2 or $c_1c_2>b_1b_2$. Could someone please direct me to a proof of this or guide me through it if it is not too computationally tedious?
• Dec 25th 2011, 10:56 PM
CaptainBlack
Re: Competition system
Quote:

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

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

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

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

The stability of $(x_E, y_E)$ depends on whether $c_1c_2 or $c_1c_2>b_1b_2$. 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
• Dec 25th 2011, 11:10 PM
alexmahone
Re: Competition system
Quote:

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

CB

But there is no series in my question!
• Dec 26th 2011, 12:37 AM
CaptainBlack
Re: Competition system
Quote:

Originally Posted by alexmahone
But there is no series in my question!

Let $(x_0,y_0)$ be the critical point. Put $(x,y)=(x_0+\varepsilon, y_0+\delta)$. Now consider the solution to the system as a power series in $\varepsilon$ and $\delta$ (and truncate after the first non-zero (lowest order) non-constant terms)

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

$\partial_t\varepsilon < 0$

and

$\partial_t\delta < 0$

CB