please have a look at the attached picture ...and help me understand what to do next...

thank you for whatever help anyone can provide ...

plz note , i know how to linearize at the equillibrium point which is ( x= m/n , y =a/b )

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- Jan 10th 2009, 12:58 AMcrbnumerical analysis of competing species??
please have a look at the attached picture ...and help me understand what to do next...

thank you for whatever help anyone can provide ...

plz note , i know how to linearize at the equillibrium point which is ( x= m/n , y =a/b ) - Jan 10th 2009, 05:34 AMHallsofIvy
you can change to the "scaled variables" Y= y/(a/b) and X= x/(m/n) by solving those for x and y: x= (m/n)X and y= (a/b)Y. Then the differential equations become

$\displaystyle \frac{dx}{dt}= \frac{m}{n}\frac{dX}{dt}= (a- b(\frac{a}{b})Y)\frac{m}{n}X$

or

$\displaystyle \frac{dX}{dt}= a(1- Y)X$

and

$\displaystyle \frac{dy}{dt}= \frac{a}{b}\frac{dY}{dt}= (m- n(\frac{m}{n}X)\frac{a}{b}Y$

or

$\displaystyle \frac{dY}{dt}= m(1- X)Y$

which have equilibrium points at (0, 0) and (1, 1).

You then get $\displaystyle \frac{dY}{dX}= \frac{m(1- X)Y}{a(1-Y)X}$

which can be separated as

$\displaystyle a\frac{Y}{1-Y}dY= m\frac{X}{1-X}dX$

Integrate that to get g(Y)= f(X).