>> Let's talk about
discrete dynamical systems.
So we'll go through an example
of the predator-prey model.
So this is based on exercise
5.6.5 of the textbook.
You can read the
description there.
The spotted owls are basically
going to eat flying squirrels.
So let O sub k be the number
of spotted owls at time k,
where k is measured in months.
And S sub k is going
to be the number
of flying squirrels measured in
thousands at time k. And suppose
that you know that
the number of owls
at time k plus 1 is 0.4
times the number of owls
at time k plus 0.3 times the
number of squirrels in thousands
at time k. And similarly,
the number of squirrels
at time k plus 1 is
negative .5 times the number
of owls plus 1.2 times
the number of squirrels.
All right.
So let's first give
some interpretations
of what these linear
relationships are saying.
The 0.4 times Ok in
the first equation --
what that's saying is that
if there were no squirrels,
so if Sk was 0 then only 40%
of the owls would
survive to the next month.
All right.
So 0.4 times Ok.
The 1.2 times Sk, in this
second equation here,
says that if there
were no owls --
so if Ok was 0, if there
were no owls as predators,
then the squirrel population
would just increase by a factor
of 1.2 so that's a 20%
increase every month.
The 0.3 times Sk in the first
equation says that every 10 --
well, 10,000 squirrels can
support an additional owl.
There'll be an additional
owl that can survive.
And the negative .5 Ok
in the second equation --
this is saying that
every two owls, right,
would be eating 1000
squirrels each month.
So negative .5 times the
number of owls is contributing
to the squirrel population.
So let's set up our
discrete dynamical system.
Let X sub k be this
vector Ok Sk.
Then X sub k plus 1
is A times X sub k,
where this matrix A has
.4 in the top left entry,
.3 in the top right entry,
negative .5 in the bottom left,
and 1.2 in the bottom right.
And so now if we're going to
study this dynamical system,
we need to study this
matrix A. And so we turn
to eigenvalues and eigenvectors.
And so we want to first find
the eigenvalues of this matrix.
And so we've got to compute
its characteristic polynomial.
So it's the determinant
of A minus lambda I.
I compute the determinant
and multiply everything out.
I get this quadratic
polynomial in lambda.
And I'm interested in finding
the roots of that polynomial.
So I can factor it
either by inspection
or by using the quadratic
formula.
And I see it factors as lambda
minus .7 times lambda minus .9
which tells me the
eigenvalues of A are .7 and .9.
So I have these eigenvalues
of .7 and .9.
What does this tell me
about the dynamical system?
Well, one of the things
I notice is that both
of the eigenvalues are
less than 1 in magnitude,
less than 1 in absolute value.
So that tells me the
origin is an attractor.
All right.
So things are attracted
towards the origin.
If I would've had one
eigenvalues bigger than 1
in magnitude and the
other one smaller than 1,
I would have a saddle point.
I have both eigenvalues
bigger than 1 in magnitude,
I would have a repeller.
And so let's next
compute the eigenvectors.
So for the eigenvalue 0.7,
I've got to solve A minus .7
times the identity times X
equals the 0 vector.
I set up my augmented
matrix and row reduce.
I see my pivot there which tells
me X2 is a non-pivot column,
corresponding to a
non-pivot column so it's free.
And X1 is expressible in
terms of the free parameter t
and so I see that in parametric
vector form X is t times 1, 1.
So I have this eigenvector 1, 1.
For the eigenvalue .9,
I'm solving A minus
.9 times the identity,
all times X equals the 0 vector.
Again, I set up the augmented
matrix and row reduce to solve.
I find my pivot here
which tells me X2 is free
so I'll call it t,
some free parameter.
And X1 is 3/5t.
So in parametric vector
form, X is t times 3/5, 1.
And then I have my eigenvector.
So here I'm allowed
to choose any vector,
any non-zero vector
in this space.
I'm going to prefer to deal
with, say, an integer vector,
a vector with integer entries.
And so I'm just going to
choose t to be 5 and I compute
and then I have this
vector 3, 5.
And so to summarize,
I have eigenvalue .7
with corresponding
eigenvector 1, 1.
I have eigenvalue .9 with
corresponding eigenvector 3, 5.
So the direction of
greatest attraction --
this is going to be the line
through the origin
in the vector 1, 1.
And I can see that because
0.7 is the smaller eigenvalue
in magnitude in absolute
value and it's attraction
because that value
is smaller than 1.
There's no direction of
greatest repulsion because both
of the eigenvalues are
less than 1 in magnitude.
So the origin is an attractor
and so there's no direction
of greatest repulsion.
Like so, here is
my eigenvectors.
Eigenvectors, I've drawn
here in blue and red.
And the eigenvalues being less
than 1 means things
flow towards the origin.
So if I had some initial
condition, X naught,
I can express it as a linear
combination of V1 and V2
because V1 and V2, the
eigenvectors, form a basis
for R2 which means any
initial condition, X naught,
can be written as C1
times V1 plus C2 times V2.
But then this tells me
that X sub k is C1 times .7
to the k times V1 plus
C2 times .9 to the k V2.
And so let me just draw in
the conditions on C1 and C2
that correspond to
these regions.
So I have the eigenvectors drawn
in here again in blue and red.
So this top region, top wedge
has C1 less than or equal to 0.
And the bottom one have C1
bigger than or equal to 0.
The bottom region has C2
less than or equal to 0.
And the top two regions have
C2 bigger than or equal to 0.
And so when I'm looking at how
this thing flows in this region,
everything is getting
pulled towards the origin.
As k gets large, it's going
to look more and more like V2
because .9 to the k is going
to be bigger than .7 to the k.
And so these paths are sort
of curving slightly to be more
like V2 as they approach
the origin.
In this region where C1
is less than or equal to 0
and C2 is greater than or
equal to 0, when the value
of C2 is really, really large
and the value of C1 is really,
really small in magnitude,
right,
there might be a little portion
where the path actually goes --
increases in terms
of owls and squirrels
for a little while
before it gets pulled
down because these
exponentials have base --
.7 and .9 are smaller than 1.
All right.
And so similar argument
tells you that in this region
where C1 is bigger than or
equal to 0 and C2 is bigger than
or equal to 0, we also get
pulled towards the origin.
There might be a slight increase
in squirrels and owls based
on these coefficients but
eventually .7 to the k and .9
to be k are going to
shrink fast enough
that they get pulled
to the origin.
And so this is bad news
for our owls and squirrels.
It tells us no matter what the
initial condition is, right,
the initial condition is going
to correspond to some point
in this first quadrant.
No matter what the
initial population is,
eventually the squirrels and
the owls are going to die out.
