Math 5110/6830
Homework 3.1 Solutions
1. (a)
Mt+1 = Mt − g(Mt )Mt + S = Mt −
Mtp
Mt + S
K p + Mtp
with K = 2 and S = 1 gives
Mt+1 = Mt −
Mtp
Mt + 1.
2p + Mtp
A fixed point for this equation therefore satisfies
M∗ = M∗ −
M ∗p
M ∗ + 1.
2p + M ∗ p
With M ∗ = 2,
?
2 =2−
?
2 =3−
2p
2p
·2+1
+ 2p
1
·2
2
X
2 = 2.
Because 2p2+2p = 21 is true independently of p, we know that M ∗ = 2 is an equilibrium point
for any value of p.
p
(b) Recall that a fixed point M ∗ = f (M ∗ ) is stable when |f ′ (M ∗ )| < 1, where
f (Mt ) = Mt −
Mtp
Mtp+1
M
+
S
=
M
−
+ 1.
t
t
K p + Mtp
2p + Mtp
Differentiating, we obtain
f ′ (Mt ) = 1 −
(p + 1)Mtp (2p + Mtp ) − Mtp+1 (pMtp−1 )
.
(2p + Mtp )2
For M ∗ = 2,
f ′ (2) = 1 −
=1−
=1−
=1−
=1−
=1−
If 2−p
4 < 1, then
(p + 1)2p (2p + 2p ) − 2p+1 (p · 2p−1 )
(2p + 2p )2
p
p+1
(p + 1)2 · 2
− p · 22p
(2p+1 )2
(p + 1)22p+1 − p · 22p
22p+2
2p+1
2
(p + 1 − p · 2−1 )
22p+1 · 2
p + 1 − 2p
2
2−p
1 p+2
4 − (p + 2)
=
=
2
2
4
4
1
⇒
−4 <
2−p
<
4
2−p<
⇒
−6 <
−p <
⇒
6>
−1 <
1
4
2
p > −2.
M∗
Thus when −2 < p < 6, the fixed point
= 2 is stable. It is unstable when p < −2
or p > 6. The solution oscillates toward the equilibrium when two conditions are satisfied:
when −2 < p < 6 and when f ′ (2) < 0, i.e. when p > 2. Thus, stable oscillations occur
when 2 < p < 6.
(c) Choose p = 7 and M0 = 3.
4
Mt+1=f(Mt)
Mt+1=Mt
3.5
cobweb
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
2. (a) Given an+1 = an + k(M − an )(an − m) with M = 5000, m = 100 and k = 0.0001, we have
an+1 = an + 0.0001(5000 − an )(an − 100).
A fixed point a∗ satisfies
a∗ = a∗ + 0.0001(5000 − a∗ )(a∗ − 100)
⇒
0 = 0.0001(5000 − a∗ )(a∗ − 100)
So, a∗ = 5000 and a∗ = 100 are the fixed points of this model.
Stability of a fixed point a∗ requires that |f ′ (a∗ )| < 1, where f (an ) = an + 0.0001(5000 −
an )(an − 100). Differentiation gives
f ′ (an ) = 1 + 0.0001[−1(an − 100) + (5000 − an )]
= 1 + 0.0001(5100 − 2an )
= 1.51 − 0.0002an
When a∗ = 5000, f ′ (5000) = 1.51−1 = 0.51. Since |f ′ (5000)| < 1, this fixed point is stable.
2
When a∗ = 100, f ′ (100) = 1.51 − 0.02 = 1.49. Since |f ′ (100)| > 1, this fixed point is
unstable.
Cobweb and sketch of solution for a0 = 50
250
50
an+1=f(an)
200
an+1=an
cobweb
150
0
100
−50
n
50
a
0
−50
−100
−100
−150
−150
−200
−250
−100
−50
0
50
100
150
−200
200
0
0.5
1
1.5
2
n
2.5
3
3.5
4
For a0 = 50, the population of whales decreases without bound, which is not biologically
possible.
Cobweb and sketch of solution for a0 = 200
5000
6000
an+1=f(an)
4500
an+1=an
5000
cobweb
4000
3500
4000
n
3000
a
3000
2500
2000
2000
1500
1000
1000
500
0
0
1000
2000
3000
4000
5000
6000
0
0
5
10
15
20
25
n
For a0 = 200, the population of whales will increase to the stable equilibrium at a∗ = 5000.
(b) Observing the dynamics of the whale population over 20 years for varying a0 , notice that the
model breaks down for a0 < 100, which gives rise to negative populations. With a0 > 100,
however, the model stabilizes. As a0 increases, the time it takes to reach the equilibrium
value of a∗ = 5000 gets shorter and shorter.
3
5000
4000
an
3000
2000
1000
0
−1000
0
2
4
6
8
10
n (years)
12
14
16
18
20
Matlab Code
%
k
M
m
Define parameters in the model.
= .0001;
= 5000;
= 100;
N = 20; % total number of years
% Define the given equation as a function with x as the independent variable.
f = @(x) x + k.*(M-x).*(x-m);
year = 0:N;
% This for-loop will set up the initial value of a.
for j=20:40:1060
a(1) = j;
% This for-loop generates the solution for the given a0.
for n=1:N
% Plug the previous value of a_n into the function to get the next
% value.
a(n+1) = f(a(n));
end
% Open figure 1
figure(1)
% Plot the solution and don’t erase it from the figure. "year" will be
% the x-values, "a" will be the y-values.
plot(year,a)
hold all
end
4
% Specify the ranges for x- and y-values in the form [xmin xmax ymin ymax].
axis([0 N -1000 M])
%Label axes
xlabel(’n (years)’)
ylabel(’a_n’)
Note: Any single quotations in the code will not copy correctly into Matlab (a LA TE X issue), so
you will need to retype them after pasting the code into the Editor. Also, in order to erase the figure
after you run this code, but before running another code that requires plotting, type ‘clf’ (without
quotations) in the command window.
5
Homework 3.2 Solutions
1. (a) As can be seen in the pictures below, the form for g(xn ) used in the Beverton-Holt model is
quite different from that used in the logistic model. For most values of r, g(xn ) > 0, and it
decreases nonlinearly to zero for increasing xn . When r < 1, however, three things happen:
K
K
; (2) g is undefined for xn = 1−r
; and (3) g is
(1) g is positive and increasing for xn > 1−r
K
negative and increasing for xn < 1−r . K determines how quickly g decreases (for r > 1) or
increases (for r < 1); smaller values of K produce faster changes in g.
r = 0.5
4
K = 1000
3
2
g(xn)
1
0
−1
−2
−3
−4
0
500
1000
1500
2000
2500
xn
3000
3500
4000
4500
5000
r = 10
r = 100
4
4
K=1
K = 10
K = 1000
K = 5000
3
3
2.5
2.5
2
1.5
2
1.5
1
1
0.5
0.5
0
K=1
K = 10
K = 1000
K = 5000
3.5
g(xn)
n
g(x )
3.5
0
500
1000
1500
2000
2500
x
3000
3500
4000
4500
0
5000
0
n
500
1000
1500
2000
2500
x
3000
3500
4000
4500
5000
n
(b) Fixed points satisfy the equation
x∗ =
x∗ r
,
∗
1 + r−1
K x
to which x∗ = 0 and x∗ = K are solutions.
Stability: Let f (xn ) = g(xn )xn . Stability of x∗ is guaranteed when |f ′ (x∗ )| < 1. Differentiating f (xn ) gives
r−1
r 1 + r−1
r
′
K xn − rxn K
=
f (xn ) =
2 .
2
r−1
1 + r−1
x
1
+
x
n
n
K
K
For x˜ = 0: f ′ (0) = r, and so this fixed point is stable for |r| < 1. Since r must be
positive, it follows that x∗ = 0 is stable for 0 < r < 1 and unstable for r > 1.
6
For x˜ = K:
f ′ (K) =
r
1+
2
r−1
K K
r
(1 + r − 1)2
r
= 2
r
1
= .
r
=
This fixed point is stable for 1r < 1, and since r > 0, this means that x∗ = K is stable for
r > 1 and is unstable for 0 < r < 1.
(c)
Cobwebbing for r > 1 (left) and 0 < r < 1 (right)
r = 2.5
r = 0.5
xn+1=f(xn)
5000
xn+1=xn
4500
xn+1=f(xn)
5000
xn+1=xn
4500
cobweb
cobweb
4000
4000
3500
3500
3000
3000
2500
2500
2000
2000
1500
1500
1000
1000
500
0
500
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0
5000
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
(d)
Solutions for r > 1 (left) and 0 < r < 1 (right)
2500
4500
4000
2000
3500
n
1500
2500
x
x
n
3000
2000
1000
1500
1000
500
500
0
0
2
4
6
8
n
10
12
14
16
0
0
2
4
6
8
n
10
12
14
16
As predicted in part (b), r > 1 leads to stability of x∗ = K and instability of x∗ = 0, while
0 < r < 1 leads to the opposite result.
2. (a) Comparison to the logistic model: The Ricker model doesn’t become negative for any value
of xn and decreases nonlinearly.
Comparison to the Beverton-Holt model: As in the BH model, the expression for g(xn ) in
the Ricker model is nonlinear. Also, for r > 1, the model decays nonlinearly to zero and is
7
always positive. Larger values of K cause g to decrease more slowly, as in the BH model.
In contrast to the BH model, the Ricker model decreases to zero for any value of r, not just
those greater than 1. Another difference is that g is extremely sensitive to changes in r, as
smaller increases in r lead to significantly greater changes in g.
r = 0.5
1.8
K = 10
K = 100
K = 1000
K = 5000
1.6
1.4
g(xn)
1.2
1
0.8
0.6
0.4
0.2
0
0
500
1000
1500
2000
2500
xn
3000
3500
4000
4500
5000
4
r=5
150
2.5
K = 10
K = 100
K = 1000
K = 5000
r = 10
x 10
K = 10
K = 100
K = 1000
K = 5000
2
100
g(xn)
g(xn)
1.5
1
50
0.5
0
0
500
1000
1500
2000
2500
x
3000
3500
4000
4500
5000
0
0
500
1000
1500
n
2000
2500
x
3000
3500
4000
4500
5000
n
(b) Fixed points satisfy the equation
x∗
x = x exp r 1 −
.
K
∗
∗
This leads to the following:
x∗
0 = x∗ 1 − exp r 1 −
K
x∗ = 0
x∗
and 1 = exp r 1 −
K
⇒
⇒
⇒
x∗
0=r 1−
K
x∗ = K .
(c) Let f (xn ) = g(xn )xn . Stability of x∗ is guaranteed when |f ′ (x∗ )| < 1. Differentiating f (xn )
gives
h
h
xn i
xn i
r
f ′ (xn ) = exp r(1 −
) + xn exp r(1 −
) ·−
K
K
K
h
r xn i ) · 1 − xn
= exp r(1 −
K
K
˜
′
r
r
For x = 0: f (0) = e , which is stable when |e | < 1. Since er > 0 for all r, |er | = er ,
and so x∗ = 0 is stable when r < 0. Since we’re given that r > 0, this fixed point is always
unstable.
8
For x˜ = K: f ′ (K) = 1 − r, so we require |1 − r| < 1 for stability. This means
⇒
−1 < 1 − r <
1
−2 <
0
⇒
−r <
2>
r > 0.
Therefore, x∗ = K is stable provided 0 < r < 2 and is unstable whenever r > 2.
(d)
Cobweb and sketch of solution for r = 1.8
6000
xn+1=f(xn)
6000
xn+1=xn
cobweb
5000
5000
4000
x
n
4000
3000
3000
2000
2000
1000
1000
0
0
1000
2000
3000
4000
5000
0
6000
0
2
4
6
8
10
12
14
16
18
20
12
14
16
18
20
n
Cobweb and sketch of solution for r = 0.4
7000
xn+1=f(xn)
4500
xn+1=xn
6000
cobweb
4000
5000
3500
3000
x
n
4000
3000
2500
2000
1500
2000
1000
1000
500
0
0
1000
2000
3000
4000
5000
6000
7000
0
0
2
4
6
8
10
n
9