A modification of the predatorprey equations is given by x

A modification of the predator-prey equations is given by x\' = x(1 - x) - axy/x + 1 y\' = y (1 - y), where a > 0 is a parameter. (a) Find all equilibrium points and classify them. (b) Sketch the nullclines and the phase portraits for different values of a. (c) Describe any bifurcations that occur as a varies.

Solution

dx/dt=x(1-x)-axy/x+1 dy/dt=y(1-y)

dx/dt=x-x²-axy/x+1 for a=1

dx/dt=x-x²-xy/x+1

To find the equillibrium points we solve the system of equations

x-x²-xy/x+1=0 y(1-y)=0

x(x+1)-x²(x+1)-xy=0 y-y²=0

x²+x-x³-x²-xy=0 y=y² implies y=0 or y=1

x-x³-xy=0

putting y=0 we get x=x³ putting y=1 we get x³=0

Hence the equillibrium points are (0,0),(1,0),(0,1)

so we have three equilibrium points, all of which are biologically meaningful. The equilibrium point (0, 0) corresponds to the constant solution x(t)=0 y(t)=0 and the meaning of this equilibrium point is that if we start with no prey and no predators, there will never be any prey or predators (no spontaneous generation) since our model does not take immigration into account. The equilibrium point (1, 0) corresponds to the constant solution x(t)=1 y(t)=0 and the meaning of this equilibrium point is that if we start with 1 prey and no predators, there will always be 1 prey and no predators. This makes sense because 1 is the carrying capacity of the prey population in the logistic differential equation

for a=2

we have dx/dt=x-x²-2xy/x+1

solving the system of linear equations

x(x+1)-x²(x+1)-2xy=0

x²+x-x³-x²-2xy=0

-x³+x-2xy=0

x(-x²+1-2y)=0

x=0 or2y=1-X²

the equillibrium points is(0,0) ,(-1,1),

(-1,1) is not possible dolution

so the eqiyillibrium point is (0,0).