Competition system

**alexmahone** · December 25th 2011, 07:19 AM

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

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

$a_1, a_2, b_1, b_2, c_1, c_2$ 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 $c_1c_2<b_1b_2$ 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?

**CaptainBlack** · December 25th 2011, 09:56 PM

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 $c_1c_2<b_1b_2$ 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

**alexmahone** · December 25th 2011, 10:10 PM

Originally Posted by CaptainBlack

Consider a series expansion about the critical point.

CB

But there is no series in my question!

**CaptainBlack** · December 25th 2011, 11:37 PM

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

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