EFTA01156014.pdf

Coming together Martin considers two sorts of coming together models. I will start by analyzing Model 2 which is defined by the equations • x,' = au; - (PE ixo+Yx1+$) + • x.' =13x, xn., - (fix, + x„. • = -40y- • = by +I„anx„. (2.1) The proposition that follows summarizes what can be said at this point. By way of notation, the proposition uses x, to denote the unique, positive solution to the cubic equation x(x+1)2 = f3/b. Proposition 1: Suppose that a2 > 2b and a3 - a2 > b. Then there is at most one equilibrium solution to (2.1) with both y > 0 and xi > 0. If, in addition, either a,ab or (x.+I)2 <El oc , 0 )" , then there are no y > 0 and x1 > 0 equilibrium solutions. On the other hand, if the three conditions a2 > 2b and a3 - a2 > b and a1 <b hold, then there exists one and only one equilibrium solution to (2.1) with both y > 0 and xi > 0 solution. As noted by Martin, this definition of $ has y + Eiwnx.= 1 at an equlibrium point. It is also the case that a solution with x1 > 0 has all x„,, positive also. Proof ofProposition 1: An equilibrium point with both y # 0 and xi x 0 is characterized by the conditions: 4) • xi = (1)4(13xi + .2 • x.= a x„., = am' xl where a = ftx,l((3xl+4)). • 4: =b. (2.2) Here, S is shorthand for L., ads,. Note in this regard that the condition on x1 can be derived from the top bullet in (2.1) by noting that En,;_ x, . 4,403x,+40x, (2.3) EFTA01156014  To see what to make of the equilibria, use the third bullet in (2.2) to set fit = b in the second bullet and so write a = ftx,/((tx, +b) and thus xn as a function of xi. This done, then the top bullet in (2.2) defines x, and y as solutions to the equations • Yr = 1 - ari( )n.1 • y = 1 - 112 )(if; +1.02 • (2.4) where ri = MI The equations in (2.4) imply that an equilibrium point obeys (xi + 0 2 = I nko anti ( (x11:1•11) (2.5) By renaming x1 = rix and an = ban, the equation in (2.5) can be written as 0=-(x+1)2+~„m . (2.6) with the constraint that x (x + 1)2 s Tr' to have y *0. The expression in (2.6) can be written as 0 =1, - 1+ (12-2)x + (13-12- 1)x2 + h(x) (2.7) where h(x) > 0 and h'(x) > 0 for x a 0. Moreover, h(x) x3 for x near 0. Note for reference momentarily that the derivative of the right hand side is 12 - 2 + 2(a3 - A2 - 1)x + h'(x) (2.8) This last expression is positive if both a2 - 2 and a3 - a2 - 1 are positive. It follows as a consequence that if this condition holds, then there is at most 1 solution to (2.7). Meanwhile, the right hand side of (2.7) at x = 0 is al - 1. These last two observations lead to the following: • There are no x > 0 solutions to (2.6) if a, a 1 and a2 s 2 and fi3- A2- I a 0. • There exists a unique x > 0 solution to (2.6) it al <1 and a2 z 2 and a3- a2- 1 > 0. (2.9) It remains to be seen when the solution to (2.6) is such that x(x +1)2 < rf l. To this end, let x, denote the one real, positive solution to the cubic equation x(x+ 1)2 -111 = 0. As noted above, if both a2 a 2 and a 3 - a2 - 1 a 0, then the right hand side of (2.10) is an increasing function of x. This being the case, there is exists a y > 0 and x, > 0 equilibrium point if both the second bullet in (2.9) holds and if EFTA01156015  (x.+ 1 ) 2 < Ink° LI((x. .0 )n • (2.10) This solution is unique. On the other hand, if a, a 2 and a3 - a2 - I a 0 and if the inequality in (2.10) is reversed, then there are no y > 0 and xi > 0 solutions to (2.10). EFTA01156016