MATH 2250—Final Exam Take-Home Portion
SOLUTIONS
Due Wed., Dec. 17, 2008, 10:30AM (at the exam)
• Do all questions on your own — do not get help from others and do
not work together on these questions.
• You may use the book, your notes, a calculator, and any other materials
you wish, and you may take as much time as you need (up to the due
date).
• If you get stuck, refer to similar examples and questions/answers in the
text book. You may seek help or work together on textbook questions
(but then work out the exam questions on your own!!).
• Show all your work and try not to skip steps.
• This is due at the exam and will not be accepted after. Start early and
plan ahead!
• This will count as 20% of your final exam grade (two out of ten questions total).
1. Find all the critical points for each of the following systems of autonomous differential equations. Then, for each critical point, determine its classification (node, saddle, spiral, or center) and its stability
(linearized, if necessary).
(a)
dR
=J −R
dt
dJ
= −R
dt
Note, if you attended Tuesday’s lecture, this is “romantic but
easily bored Romeo” matched with “tragic Juliet” (you don’t have
to know what I’m talking about to do the problem!).
SOLUTION:
A critical point is a point in the (R, J) plane where both derivatives are equal to zero. Here we have one critical point at (0, 0).
This is a linear system, and in matrix–vector form we have
"
x0 =
−1 1
−1 0
#
x
where I have defined x as the vector (R, J). We can classify the
critical point by finding the eigenvalues of the matrix:
−1 − λ 1
−1
−λ
=0
→
(−1−λ)(−λ)+1 = 0
→
λ2 +λ+1 = 0
√
The roots are λ = (−1 ± 3)/2, which are complex roots with
negative real part. This means we can classify the critical point
as a stable spiral
(b)
dx
= x − xy 3
dt
dy
=x+y+1
dt
SOLUTION:
Any critical points (x∗ , y ∗ ) will satisfy both equations x − xy 3 = 0
and x+y+1 = 0. The first equation can be rewritten x(1−y 3 ) = 0,
which is satisfied if x = 0 or if y = 1. We plug each of these values,
in turn, into the second equation to arrive at two critical points:
(0, −1) and (−2, 1).
We use the Jacobian matrix to analyze the local, linearized stability of each critical point. Taking the appropriate partial derivatives we get
"
J=
1 − y 3 −3xy 2
1
1
#
evaluated at (x, y) = (x∗ , y ∗ )
At the critical point (0, −1) we have
"
J=
2 0
1 1
#
This matrix has two positive eigenvalues, λ = 2, 1, which means
that the point (0, −1) is an unstable node. At the critical point
(−2, 1) we have
"
#
0 6
J=
1 1
Here the eigenvalue equation is λ2 − λ − 6 = 0, which has roots
3 and −2. One eigenvalue is positive and one is negative, which
means the point (−2, 1) is a saddle point, which means it is unstable.
2. Consider a population N of prey and P of predators, governed by the
following differential equations:
N
dN
= rN 1 −
dt
K
− bN P
dP
= bN P − dP
dt
The parameters r, K, b, and d are constants. The biological meaning
of each constant is as follows: r is the intrinsic rate of growth of the
prey population; K is the carrying capacity of the prey in the absence
of predators; b is a rate at which predators consume prey and convert
that food into new predators; d is a death rate for the predators. Use
the following values to answer the questions below: r = 0.1, K = 400,
b = 0.01, and d = 2.
Find all equilibria (critical points) of this system, and explain what
each one represents biologically (for example, extinction of one or both
species, or coexistence). Then perform a linearization at each critical
point and determine its classification and stability. Finally, write a
summary of what the model predicts will happen to the two populations. Include in your summary what would happen to each population
in the absence of the other, as well as what you expect to happen when
they interact at positive initial conditions.
SOLUTION:
A critical point (N ∗ , P ∗ ) would solve both of the following equations
(which I have factored from the above):
N
N r 1−
K
− bP = 0
P (bN − d) = 0
The first equation is solved if
N
r
1−
(i) N = 0 or (ii) P =
b
K
The second equation is solved if
(iii) P = 0 or (iv) N =
d
b
Combining (i) and (iii) leads to the critical point (N ∗ , P ∗ ) = (0, 0)
(extinction of both populations).
Combining (i) and (iv) leads to a contradiction (both cannot be true).
Combining (ii) and (iii) leads to the critical point (N ∗ , P ∗ ) = (K, 0) =
(400, 0) (extinction of the predator population only).
Combining (ii) and (iv) leads to the critical point (N ∗ , P ∗ ) = (d/b, r(1−
d/(bk))/b) = (200, 5) (coexistence of predator and prey).
We use the Jacobian matrix to analyze the local, linearized stability of
each critical point. Taking the appropriate partial derivatives we get
"
J=
.1 − .0005N − .01P −.01N
.01P
.01N − 2
#
evaluated at (N ∗ , P ∗ )
At (0, 0) we have
"
J=
.1 0
0 −2
#
This has eigenvalues .1 and −2. One positive and one negative eigenvalue means (0, 0) is a saddle point and unstable.
At (400, 0) we have
"
J=
−.1 −4
0
2
#
This has eigenvalues −.1 and 2, so (400, 0) is also a saddle point and
unstable.
At (200, 5) we have
"
J=
−.05 −2
.05
0
#
Solving (−.05 − λ)(−λ) + .1 = λ2 + .05λ + .1 = 0 gives us complex
eigenvalues with negative real part, so (200, 5) is a stable spiral.
In the absence of the prey species, the predator equation reduces to
dP/dt = −2P , which means the predator species dies out. In the absence of the predator species, the prey equation reduces to dN/dt =
.1N (1 − N/400), for which N = 400 is a stable equilibrium, so the prey
species would settle at a population of 400. If both species are present,
the evidence above says that we do not expect either species to go extinct, because the critical points (0, 0) and (400, 0) are locally unstable.
The coexistence equilibrium (200, 5) is locally stable, however, so we’d
expect the populations to move with damped oscillations toward those
numbers from positive initial conditions (although it would take further
analysis to prove this is the case for general initial conditions).