Qualitative Analysis:
Example 1: Consider the following predator-prey system
The solution curves in the phase plane is as follows:
1
Sketch the R(t)- and F(t)-graphs for the solutions with initial points A, B, and C.
2
Find the longterm behavior of the populations over time.
Answer: First, notice that the point A is an equilibrium point with non-zero coordinates. If we set
we get the following three points: (0,0), (2.5,0), and (1/0.8,1/1.5). Clearly, we have
1
The sketches of the R(t)- and F(t)- graphs for the solutions with initial points A, B, and C respectively,
are given below
2
It is clear that all the solutions on the phase-plane tend to the equilibrium point A when t goes to infinity.
Therefore, in all the cases, we have
,
Example 2: The following depicts the phase portrait of the solution to an autonomous system of differential
equations with initial conditions x(0)=2 and y(0)=1. Sketch the time series for x(t) and y(t).
Answer:
Example 3: Consider the first order system
1.
Graph the nullclines for this system.
2.
Insert direction field "arrows" at the nullclines
3.
Determine the general direction of the direction field in each of the regions cut out by the nullclines.
4.
Sketch possible phase portraits for the two initial conditions indicated by "dots" for as long as the phase
portraits stay within the graph's frame.
5.
Find all the equilibrium points of the system.
Answer:
1.
The x-nullclines can be obtained by setting x'=0, i.e.
The y-nullclines can be obtained by setting y'=0, i.e.
For 2, 3, 4, use the following graph:
5.
The equilibrium points of the system are the intersection points of the x-nullclines and y-nullclines.
Therefore, we get
,
which yields x=2 and
. Hence, we have two equilibrium points
.
Example 4: Consider the system
.
Draw the nullclines and find all the equilibrium points. Use this information to determine the fate of the
solutions corresponding to the following initial conditions
.
Answer: Note that this system is non linear. The x-nullclines are given by
We recognize the two lines x=0 (y-axis) and the line -x-y+1=0.
The y-nullclines are given by
We recognize the line y=0 (x-axis) and the circle
graph below we give the nullclines:
(centered at (0,0) with radius 2). In the
The equilibrium points may be obtained as the intersection between the x-nullclines and y-nullclines. There are
6 points (see the above graph). The solutions corresponding to the given points are drawn in white in the graph
below:
Example 5: Consider the system
Find the equilibrium points and the nullclines. Draw the vector field. Sketch some solutions and specially the
solutions around the equilibrium points.
Answer. The equilibrium points are given by the algebraic system
It is easy to see that we must have x=0 (from the second equation) and therefore y=0. Hence the system has one
equilibrium point (0,0). The x-nullcline are given by the equation
which is the graph of the function
. The y-nullcline are given by the equation
which reduces to the line x=0 (the y-axis). In the picture below, we draw the vector field as well as the
nullclines.
Clearly, the solutions spiral around the equilibrium point (see the picture below)
Notice that the solutions spiral and get closer to a cycle. We can see this better by looking at the graphs of x
versus t as well as the graphs of y versus t.
and
Example 6: Consider the system
Find the equilibrium points and the nullclines. Draw the vector field. Sketch some solutions and specially the
solutions around the equilibrium points.
Answer. The equilibrium points are given by the algebraic system
From the first equation, we get y=0. The second equation gives x=0 or x=1. Hence the equilibrium points are
(0,0) and (1,0). The x-nullcline are given by the equation y = 0 which is the x-axis and the y-nullcline are given
by the equation
, which reduces to the two vertical lines x=0 (the y-axis) and x=1. In the picture
below, we draw the vector field as well as the nullclines.
Clearly, the solutions spiral around the equilibrium point (1,0) and get away from the other equilibrium point
(0,0)(see the picture below)
A closer look at the solutions around the equilibrium point (1,0) gives
Clearly the solutions are close to cycles. The graphs of x versus t as well as the graphs of y versus t illustrate
better this remark:
and