One of the most common applications
of non-linear systems of ordinary differential equations
are the so-called predator-prey or a Lotka-Volterra systems.
These systems arise when two species interact,
one as the prey and one as the predator.
These equations also see applications
in economics and in chemical reactions.
In biology, this system of equations
explains the natural periodic variations of populations
of different species in nature.
Before the application of differential equations
these periodic variations in the population,
baffled biologists.
We will keep with the classic example
of hares and foxes in a forest,
it is the easiest to understand.
x equals a number of hares, the prey
and y equals the number of foxes, the predator.
When there are a lot of hares,
there is plenty of food for the foxes,
so the fox population grows.
However, when the fox population grows,
the foxes eat more hares, so when there are lots of foxes
the hare population should go down and vice versa.
The Lotka-Volterra model proposes
that this behavior is described
by the following system of differential equations.
x prime equals the quantity a minus by times x
and y prime equals the quantity cx minus d times y,
where a, b, c and d are some parameters
that describe the interaction of the foxes and hares.
In this model these are all positive numbers.
Now it's analyze the idea behind this model.
The model is a slightly more complicated idea
based on the exponential population model.
So given the two differential equations, if we expand,
we have x prime equals ax minus byx
and y prime equals cxy minus dy.
The hares are expected to simply grow exponentially
in the absence of foxes.
That is where the ax terms comes in.
The growth in population is proportional to
the population itself.
So looking at our equation for x prime, if y is 0.
Again, notice how we have just x prime equals ax.
We are assuming the hares always find enough food
and have enough space to reproduce.
However, there is another component.
The -byx,
that is the population also is decreasing proportional to
the number of foxes.
Together we write the equation
as x prime equals the quantity a minus by times x.
So it is like exponential growth or decay,
but the constant depends on the number of foxes.
The foxes need food, the hares to reproduce.
The more food, the bigger the rate of growth,
hence the cyx term in the equation for y prime.
On the other hand, there are natural deaths
in the fox population and hence the term -dy
and now also get a specific example.
Let's consider the system x prime equals
the quantity 0.4 minus 0.01y times x
and y prime equals the quantity 0.003x
minus 0.3 times y.
Below we have a graph of the phase portrait for the system
as well as a graph of x and y.
Let's look at this on the next slide.
On the left again
we have the phase portrait for the predator-prey model.
In this example, it makes sense to also plot x and y
as graphs with respect to time, which we see on the right.
Therefore, the second graph,
again, the graph on the right
is the graph of x and y on the vertical axis
against time on the horizontal axis.
The particular solution graphed
was with initial conditions of 20 foxes and 50 hares.
Looking at the vertical axis, we can now tell
that the red graph is the graph for the foxes
and the blue graph is the graph for the hares.
And now going back to our specific system
let's determine the critical points, classify them
and also determine the equation of the trajectories.
Recall, find the critical points
need to determine the order repairs
were both x prime and y prime equal to 0,
which results in the system shown here.
Notice both systems are satisfied
with an x and y are both 0.
The system is also satisfied when x is 100 and y is 40.
Giving us a second critical point of 100,40.
Notice that 0.003x minus 0.3 is 0,
when x is 100
and 0.4 minus 0.01y is 0,
when Y is 40.
The critical 0,0
is when there are zero hares and zero foxes.
The critical point 100,40
is when there are 100 hares and 40 foxes.
For the next step, we determine the Jacobian matrix
where f of x,y is equal to x prime
and g of x,y is equal to y prime.
For the entries in the Jacobian matrix
the partial of f with respect to x is 0.4 minus 0.01y.
The partial of f with respect to y is -0.01x.
In the second row, the partial of g with respect to x
is 0.003y
and the partial g with respect to y
is 0.003x minus 0.3.
And now we determine the Jacobian matrix
at the critical points
and then determine the corresponding eigenvalues.
Recall the types of eigenvalues in most cases
help us classify the critical point.
At the critical 0,0 the Jacobian has entries
0.4, 0, 0, -0.3.
Recall to find the corresponding eigenvalues
we set up and solve the equation.
The determinant of the difference
of J and lambda I equals 0
which in this case results in lambda.
The eigenvalues equal -0.3 and 0.4.
Recall, because we have two real eigenvalues
with opposite sign,
the critical point 0,0 is a saddle.
This should make sense.
If you start with some foxes with no hares
then the foxes would grow extinct,
that is you would approach the origin.
If you start with no foxes in a few hares
then the hares would just keep multiplying without check
and so, you would go away from the origin
which is verified analyzing the vector field.
And now we can see are the critical point 100,40
where the Jacobian matrix has entries,
0, -1, 0.12, 0.
In this case, the eigenvalues are lambda equals
plus or minus i times the square root of 0.12.
Notice the eigenvalues are purely imaginary
which is the case where we cannot quite classify
the critical point using only linearization.
We could have a stable center, a spiral sink
or a spiral source.
That is the equilibrium could be asymptotically stable,
a stable or unstable.
Of course, the previous phase portrait
which I've shown below
seems to imply we have a stable center,
but we can't just trust the picture.
Perhaps the oscillations are getting larger and larger
but only very slowly.
Of course, this would be a bad result
as it would imply something will go wrong
with our population sooner or later.
We also only graphed a specific example
with specific trajectories,
so how can we be sure that we are in a stable situation
at the critical point 100,40?
As we said before
in the case of purely imaginary eigenvalues
we have to do a bit more work.
Let's see if we can find the equations of the trajectories.
To do this, we'll consider dy/dx.
Recall dy/dx is equal to dy/dt divided by dx/dt
which we have resulting in the following quotient.
And we can solve this differential equation
using separation of variables.
If we separate the variables,
the result is the quantity 0.4 minus 0.01y
divided by ydy
equals the quantity 0.003x minus 0.3
divided by xdx.
Next, we integrate both sides of the equation
and the result is 0.04 natural log absolute value of y
minus 0.01y equals 0.003x
minus 0.3 natural log absolute value of x plus C.
Let's go ahead and solve this for C.
And now we have the equation for the trajectories,
given a particular value of C, we can graph a trajectory.
So now let's graph some of the trajectories.
I've used Desmos for the graph on the right.
Notice on the left I used C equals
1.8, 1.9, 2, 2.1, and 2.15.
Notice the trajectories are closed loops
around the critical point 100,40.
We now know the critical point 100,40 is a center
and is a stable equilibrium point.
We do not have to worry about the foxes or hares
going extinct or their populations exploding.
I hope you found this helpful.