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I know how to solve nullclines, the following link is very helpful

http://mcb.berkeley.edu/courses/mcb137/exercises/Nullclines.pdf

However I don't understand how to solve equations that have only variables assuming from the Lotka–Volterra equations.

Take as example the following equations

dx/dt=ax+axy

dy/dt=ax^2+ay

how would I find the nullclines, and the fixed points?

user3424493
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1 Answers1

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You can still proceed in the usual way, but your results will contain unspecified constants $ \ a \ , \ b \ , \ c \ , \ e \ , \ \text{and} \ f \ $ (only $ \ x \ $ and $ \ y \ $ are the variables):

$$ \frac{dx}{dt} \ = \ 0 \ = \ x \ (a - by - ex) \ \ , \ \ \frac{dy}{dt} \ = \ 0 \ = \ y \ (cx - f) \ . $$

So the nullclines are the $ \ x-$ and $ \ y-$ axes (the lines $ \ y = 0 \ $ and $ \ x = 0 \ $ ), the vertical line $ \ x = \frac{f}{c} \ , $ and the line $ \ ex + by = a \ . $ The fixed or equilibrium points are the intersection points of these various lines, where both derivatives are zero. EDIT: There are three points for which that is the case (correcting an oversight I made -- thank you, Amzoti, for alerting me to this gaffe). So some of the coordinates of the points are going to be expressions involving those unspecified constants, rather than particular numbers.

EDIT (3/27) -- Here are a couple of possible arrangements of the nullclines and fixed points (it is assumed here that the quantities represented require that $ \ x \ge 0 \ $ and $ \ y \ge 0 \ $ .

For $ \ a, b, c, e, f \ > \ 0 \ , $ with $ \ \frac{a}{e} > \frac{f}{c} > 1 \ , $ we have (You are not yet asked to work out the "flow" in each region, but that will come up soon...)

enter image description here

By contrast, another possibility is to have $ \ a, c, e, f \ > \ 0 \ \ \text{and} \ \ b < 0 \ $ with $ \ \frac{f}{c} > \frac{a}{e} > 1 \ ; $ here, we have four fixed points, since the sloping nullcline does not intersect $ \ x = 0 \ $ for $ \ y \ge 0 \ . $

enter image description here

colormegone
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  • That's the tricky part: since there are no values assigned to the constants, there are various possible "flow fields" for this system, depending on the relative values of the constants. I will add a couple example graphs in my post shortly to illustrate this. – colormegone Mar 27 '14 at 04:07
  • If it's not met (as in the second graph), there will be a different configuration. I've shown one each with the inequality between $ \ \frac{a}{e} \ $ and $ \ \frac{f}{c} \ $ running either way. Other situations can be produced by changing the sign of various of the constants. – colormegone Mar 28 '14 at 00:06
  • Definitely my mistake: I neglected to check for simultaneous solution! I will edit my post and diagrams accordingly. (I even said both derivatives had to be zero, then didn't act on that...) – colormegone Mar 28 '14 at 00:53
  • I should stop trying to write detailed responses in the "wee hours". I got distracted by the regions of different flow directions and forgot to check my solutions for the pair of equations. – colormegone Mar 28 '14 at 01:06
  • I think I've got this sorted out now. Thank you for spotting my oversights. – colormegone Mar 28 '14 at 01:16