Solutions to Attendance Quiz for Lecture 17 of Dr. Z.’s Dynamical Models in Biology class

Name: Dr. Z.

1. Find all the stable equilibrium points of the 2D continuous dynamical system given by the differential equations

\frac{dx}{dt} = 1 − \frac{2x}{1 + y} ,

\frac{dy}{dt} = 1 − \frac{2y}{1 + x} .

Sol. of 1: To find the equilibrium points we have to solve

1 − \frac{2x}{1 + y} = 0 ,

1 − \frac{2y}{1 + x} = 0 .

Simplifying (you do it!) you get a system of two equations with unknowns x and y, whose solution is

(x, y) = (1, 1) .

So the set of equilibrium points is {(1, 1)} . Next we take the Jacobian of the transformation from R^{2} to R^{2}

(x, y) → (1 − \frac{2x}{1 + y} , 1 − \frac{2y}{1 + x}) .

J(x, y) = [ \begin{matrix}−frac21 + y & \frac{2x}{(1+y)^{2}} \\ \frac{2y}{(1+x)^{2}} & −frac21 + x \end{matrix} ]

Plugging-in x = 1, y = 1, we get

J(1, 1) = [ \begin{matrix}−frac21 + 1 & \frac{2·1}{(1+1)^{2}} \\ \frac{2·1}{(1+1)^{2}} & −frac21 + 1 \end{matrix} ]

[ \begin{matrix}−1 & \frac{1}{2} \\ \frac{1}{2} & −1 \end{matrix} ] .

The characteristic equation is:

(1 + λ)^{2} − \frac{1}{4} = 0

(1 + λ)^{2} − (\frac{1}{2})^{2} = 0

(\frac{1}{2} + λ)(\frac{3}{2} + λ) = 0

The eigenvalues are −\frac{1}{2} and −\frac{3}{2} since they are both negative (in particular, they have a negative real parts) this is stable.

Ans.: So the set of stable equilibrium points is {(1, 1)} .