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for any given fixed p 2 , p 1 would move to the f = 0 partial equilibrium curve. This curve would, therefore, give the effects of p 2 on the equilibrium values of p 1 . In other words, for a system controlled by the first equation of [link] if p 2 is given, the f = 0 curve will give the equilibrium value p s u b 1 approaches. .ppIn fact, however, p 2 is not fixed, but must obey the second equation in [link] . If this equation is examined separately, we have a second curve calledthe partial equilibrium curve for p 2 given by

g ( p 1 , p 2 ) = 0

A similar analysis of this equation shows the effects of p 1 on the equilibrium values of p 2 , and can be visualized by the following illustration of a g = 0 curve.

If these two curves are considered simultaneously, then not only are the singular points determined by the intersections, but the stabilityof the points and the nature and direction of the trajectories can be estimited by the signs of p ˙ 1 and p ˙ 2 in the various regions.For these illustrated curves of f = 0 and g = 0 , we have
This determines the singular point, and the directions show that it is stable.Applying this to the Lotka-Volterra competition model of [link] for the partial equilibrium curves gives

x ˙ = a ( x - x y ) = 0

or

y = 1

and

y ˙ = c ( y - x y ) = 0

or

x = 1

In the phase plane, these are

Another tool that is very useful and is related to the preceding discussion is the method of isoclines. [link] [link] Here we find curves in the phase plane where all the trajectories that cross that curve have the same slope.The partial equilibrium curves are two isoclines. The f = 0 curve implies from [link] that the slope of all trajectories along that curve is zero.The slope of all trajectories along the g = 0 curve is infinite. If we find the isocline for a slope of m , this is done from [link] by setting

f g = m

For the competition model with a = c = 1 , we have

x - x y y - x y = m

Solving for x as a function of y gives

x = m y 1 + ( m - 1 ) y

The m = 0 isocline is y = 1 . The m = isocline is x = 1 . The m = 1 isocline is

x = y

and m = - 1 gives

x = - y ( 1 - 2 y ) .

In the phase plane the isocline looks like

Note how the isoclines aid one in sketching or visualizing the phase plane solution trajectories.

This should be enough detail on this approach to allow application to the various two-variable models that can be so interesting.

  • Competition Models

We will now return to the competition model of [link] and examine it in more detail.Consider a situation where the uninhibited growth rate of population p 1 is 10%. This implies a = 0 . 1 in [link] . Assume that the negative effects of p 2 are such that 100 membersof p 2 cancel the positive effects of one member of p 1 .>From the first equation of [link] , we have

p ˙ 1 = a p 1 - b p 1 p 2 = ( a - b p 2 ) p 1

If a = 0 . 1 , then b = 0 . 001 . We also assume that p 2 has the same self-growth rate, and p 1 affects p 2 in the same way that p 2 affects p 1 . This gives c = 0 . 1 and d = 0 . 001 . The model becomes

p ˙ 1 = 0 . 1 p 1 - 0 . 001 p 1 p 2
p ˙ 2 = 0 . 1 p 2 - 0 . 001 p 1 p s b 2 .

Using Euler's method to convert these differential equations to difference equations, we see that

p 1 ( n + 1 ) = p 1 ( n ) + T a p 1 ( n ) - T b p 1 ( n ) p 2 ( n )
p 2 ( n + 1 ) = p 2 ( n ) + T c p 2 ( n ) - T d p 1 ( n ) p 2 ( n )

These were programmed on a Tektronix 31 programmable calculator with an automatic plotter to give the phase plane output shown in Figure G.The trajectories were generated by running the simulation with various initial populations.For example, the lowest trajectory was run with an initial population of p 1 = 25 and p 2 = 50 . The next one used p 1 = 30 and p 2 = 35 .

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Source:  OpenStax, Dynamics of social systems. OpenStax CNX. Aug 07, 2015 Download for free at https://legacy.cnx.org/content/col10587/1.9
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