HW1_Solutions
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University of California, Los Angeles *
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Course
30B
Subject
Aerospace Engineering
Date
Apr 3, 2024
Type
Pages
17
Uploaded by JudgeRiver13513
Problem 4.1. Further Exercise 1
Exercise 4.1.FE 1
Is a saddle point an attractor? Justify your answer.
For something to be an attractor, it has to attract other trajectories in space.
However, for a
saddle point, it generally does not.
The trajectories can approach the point from an angle, and
then almost always run away in the long run.
Thus, a saddle point is NOT
an attractor.
Problem 4.1. Further Exercise 2
Exercise 4.1.FE 2
Does a trajectory that approaches a limit cycle attractor ever reach the
attractor? Explain.
So let’s say that the trajectory will actually reach the attractor. Let’s think about this...
1. If the trajectory reaches the attractor, that means that the attractor is actually a part of the
trajectory.
2. However, we know that all trajectories will approach the attractor. So if #1 is true, it means
that the attractor will belong to ALL trajectories.
3. If #2 is true, then the trajectories will cross each other at least at the region of the limit
cycle attractor.
4. However, we know that trajectories can NEVER cross.
As a result, the limit cycle attractor can never belong to any trajectory. In other words,
A trajectory that approaches a limit cycle attractor will NEVER
reach the
attractor.
Problem 4.1. Further Exercise 4
Exercise 4.1.FE 4
Describe jet lag and recovery from it in dynamical terms.
The body controls sleep cycle via the circadian rhythm, which oscillates throughout the day.
After a long flight, the circadian rhythm witnesses a drastic switch, where the levels of biomarkers
do not math with the (expected) normal levels at certain hours. However, the body is not able to
adapt immediately since it takes time to control levels of biomarkers in the entire body.
Rather, the system slowly adjusts and approaches the stable limit cycle where the circadian rhythm
is normally at.
1
Problem 4.1. Further Exercise 6
Exercise 4.1.FE 6
Suppose a 2D system has a stable equilibrium point that is located
somewhere outside a limit cycle.
Can a trajectory starting inside the limit cycle reach this
point? Justify your answer.
(Hint: It may help to draw the situation.)
As we can see, if a trajectory starts inside the limit cycle, it will try to approach the limit cycle but
can never reach it (see
Further Exercise 2
). However, if it can never reach the cycle, it can never
cross it to reach the stable equilibrium point on the other side of the cycle. Thus,
A trajectory starting inside the limit cycle CANNOT
reach this point.
Problem 4.2.3
Exercise 4.2.3
Verify this assertion.
We are looking to verify the assertion:
“This negative feedback model will not oscillate, no matter
how steep the feedback.”
.
To do this, we will need to look at the trajectories and/or time series for the system when it has:
(1) a low value of
n
(not-steep feedback), and (2) a high value of
n
(steep feedback).
There are a few different ways you can accomplish this: animation, interactive, assign-directly, etc.
We will show the codes for the most simple method, which is to assign a definite value to
n
.
Recommended Initial Condition:
H
= 0
.
25 and
G
= 2
2
1
var(
‘H, G’
)
# define H and G as symbolic CoCalc variables
2
n =
5
# assign a value to n
3
time
list = srange(
0
,
100
,
0.1
)
# define the list of time points to obtain solutions
4
5
Hprime(H, G) =
1
/ (
1
+ Gˆn ) -
0.2
* H
# define H’
6
Gprime(H, G) = H -
0.2
* G
# define G’
7
8
sol = desolve
odeint( [ Hprime(H, G), Gprime(H, G)], dvars = [H, G], ics = [
0.25
,
2
],
times = time
list)
# simulate the model
9
10
list
plot( sol )
# plot the trajectory
Small
n
Value (e.g.
n
= 5
)
Large
n
Value (e.g.
n
= 50
)
As observed, regardless of the steepness of the feedback (i.e. how large the value of
n
is), the system
always has one stable equilibrium point, instead of a stable oscillation/limit cycle attractor.
Problem 4.2.4
Exercise 4.2.4
In the delay differential equation
Y
′
(
t
) = 16
Y
(
t
−
2) + 8
Y
(
t
), what does
Y
(
t
−
2) refer to? What does
Y
(
t
) refer to?
We have:
•
Y
(
t
)
≡
the value of
Y
at time
t
(i.e. right now)
•
Y
(
t
−
2)
≡
the value of
Y
at time
t
−
2 (i.e.
two
time units
prior
)
Problem 4.2.5
3
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Exercise 4.2.5
What aspect of the function does
V
max
control?
Let’s graph
V
max
·
X
n
1+
X
n
for a few different values of
V
max
. We also pick:
n
= 5.
X
Ventilation
Rate
V
max
= 1
V
max
= 2
V
max
= 3
As we observed, the
V
max
controls the maximum value for the Ventilation Rate,
and graphically, it controls the height of the function.
Problem 4.2.6
Exercise 4.2.6
Let
X
′
= 6
−
16
·
X
(
t
−
0
.
2)
5
1 +
X
(
t
−
0
.
5)
5
·
X
Assume that for all
t
≤
0,
X
(
t
) = 0
.
5. Use Euler’s method with a step size of 0.1 to approximate
X
(0
.
3).
A Note on Our Notation:
“
X
(
t
= 3)
” vs “
X
3
”
In this case, the first
notation means the value of
X
at
t
= 3.
Meanwhile, the second
notation means the value of
X
on its third iteration
. This may or may
not mean at
t
= 3. If the step size is ∆
t
= 1, then by the third step, you will reach
t
= 3,
hence
X
(
t
= 3) =
X
3
for ∆
t
= 1. However, let’s say that ∆
t
= 0
.
01, then by the third step,
you only reach
t
= 0
.
03, and clearly
X
3
=
X
(
t
= 0
.
03)
̸
=
X
(
t
= 3) for ∆
t
̸
= 1.
To reach
t
= 0
.
3 in step size of ∆
t
= 0
.
1, we need to take 3 steps
.
We also have the following equations:
X
′
(
t
) = 6
−
16
·
X
(
t
−
0
.
2)
5
1 +
X
(
t
−
0
.
5)
5
·
X
(
t
)
(1)
X
t
new
=
X
t
old
+
X
′
(
t
old
)
·
∆
t
(2)
4
Step #0: Find
X
at
t
= 0
As
t
= 0
≤
0:
X
(0) = 0
.
5
Step #1: Find
X
at
t
= 0
.
1
We have
t
new
= 0
.
1 and
t
old
= 0. It is important to remember that we are evaluating
X
′
at
t
old
(as
seen in Equation (2)).
X
(
t
−
0
.
2)
X
(
t
−
0
.
5)
X
≡
X
old
X
(0
−
0
.
2) =
X
(
−
0
.
2) = 0
.
5
X
(0
−
0
.
5) =
X
(
−
0
.
5) = 0
.
5
X
old
= 0
.
5
Inputting to Equations (1) and (2), we have:
X
(0
.
1) =
X
(0) +
X
′
(0)
·
0
.
1 = 0
.
5 +
"
6
−
16
·
(0
.
5)
5
1 + (0
.
5)
5
·
(0
.
5)
#
·
(0
.
1) = 1
.
07576
5
Step #2: Find
X
at
t
= 0
.
2
We have
t
new
= 0
.
2 and
t
old
= 0
.
1. It is important to remember that we are evaluating
X
′
at
t
old
(as seen in Equation (2)).
X
(
t
−
0
.
2)
X
(
t
−
0
.
5)
X
≡
X
old
X
(0
.
1
−
0
.
2) =
X
(
−
0
.
1) = 0
.
5
X
(0
.
1
−
0
.
5) =
X
(
−
0
.
4) = 0
.
5
X
old
= 1
.
07576
Inputting to Equations (1) and (2), we have:
X
(0
.
2) =
X
(0
.
1) +
X
′
(0
.
1)
·
0
.
1 = 1
.
07576 +
"
6
−
16
·
(0
.
5)
5
1 + (0
.
5)
5
·
(1
.
07576)
#
·
(0
.
1) = 1
.
62360
Step #3: Find
X
at
t
= 0
.
3
We have
t
new
= 0
.
3 and
t
old
= 0
.
2. It is important to remember that we are evaluating
X
′
at
t
old
(as seen in Equation (2)).
X
(
t
−
0
.
2)
X
(
t
−
0
.
5)
X
≡
X
old
X
(0
.
2
−
0
.
2) =
X
(0) = 0
.
5
X
(0
.
2
−
0
.
5) =
X
(
−
0
.
3) = 0
.
5
X
old
= 1
.
62360
Inputting to Equations (1) and (2), we have:
X
(0
.
3) =
X
(0
.
2) +
X
′
(0
.
2)
·
0
.
1 = 1
.
62360 +
"
6
−
16
·
(0
.
5)
5
1 + (0
.
5)
5
·
(1
.
62360)
#
·
(0
.
1) =
2
.
14488
In Summary, we have:
Step
Starting
X
n
X
(
t
−
0
.
2)
X
(
t
−
0
.
5)
Compute
X
′
Final
X
n
+1
0
-
-
-
-
X
(0) = 0
.
5
1
X
0
= 0
.
5
0.5
0.5
X
′
= 5
.
75758
X
1
= 1
.
07576
2
X
1
= 1
.
07576
0.5
0.5
X
′
= 5
.
47842
X
2
= 1
.
62360
3
X
2
= 1
.
62360
0.5
0.5
X
′
= 5
.
21280
X
3
= 2
.
14488
Problem 4.2. Further Exercise 1
Exercise 4.2.FE 1
Some people have difficulty maintaining a stable weight. Instead, they
gain a lot of weight, go on a diet, lose the weight, but then gain it back.
This pattern is
sometimes referred to as yo-yo dieting.
6
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a) What kind of feedback loop is involved in this situation?
b) Use your understanding of feedback loops and oscillations to suggest what might help
such a person to stabilize their weight.
Part a:
What kind of feedback loop is involved in this situation?
In this case we have:
Increase in weight
→
decrease/change eating habits (go on a diet)
→
Lost weight
In other words, an increase in the value of a variable ultimately leads to its own decrease. Thus,
this is a
NEGATIVE FEEDBACK LOOP
.
Schematically, we have:
Body Weight
Eating Habits
−
+
Part b:
Use your understanding of feedback loops and oscillations to suggest what
might help such a person to stabilize their weight.
The oscillations are caused by:
1. High sensitivity (steep reaction by the person): too much weight
→
drastic diet
→
too low
weight
→
eat a lot
→
too much weight
→
repeat
2. Time delay by the body because it takes time for body weight to change
It is not as feasible to control the time delay since it is a physiological process. Thus, the person
can only stabilize their weight by not changing their diet too drastically. Just adjust a little. Let
the weight stabilize around a new value. If that is still not the desired weight, then make additional
small change to the diet.
Problem 4.2. Further Exercise 2
Exercise 4.2.FE 2
While traveling, you find yourself in a hotel room in which using the
thermostat leads to large oscillations in the room’s temperature. The thermostat responds to
the room’s air temperature by turning on an air conditioner on the other side of the room if
7
the temperature near the thermostat gets too warm. Similarly, when the temperature near the
thermostat gets cold, the air conditioner switches off. What could the builder of the hotel have
done to prevent the oscillations you are experiencing?
Oscillations are caused by: (1) Steep reaction (high sensitivity), and (2) Time delay
In this case, the reaction is steep (on/off) but it is not controllable (as it is a part of the mechanical
design of the air conditioner to simply be on/off, instead of automatically adjusting how cold it
wants the air to be).
→
Thus, we can only control the time delay.
To reduce drastic oscillations, we need to reduce time delay. In this case, it can be argued that the
time delay is at maximum because the distance between the air conditioner and the thermostat is
the greatest. Thus, the distance the air has to travel is also the greatest.
To reduce time delay, they can put the thermostat closer to the AC (or AC
closer to the thermostat, either way works).
Problem 4.2. Further Exercise 4
Exercise 4.2.FE 4
Meerkats are highly social small carnivores that live in southern Africa.
They rely on each other to raise their young.
Use the following assumptions to model the
number of adult meerkats,
M
, in a population. You can invent parameters as necessary.
−
The per capita rate at which meerkats give birth to babies who survive to adulthood
is a steep sigmoid function of the adult population, with higher reproductive success at
higher populations.
−
Meerkats die of natural causes at a constant per capita rate
d
.
−
Meerkats are preyed upon by eagles and jackals. These predators have many other prey,
so their population does not depend on the meerkat population.
−
The rate at which jackals prey on meerkats is a nonsigmoid saturating function of the
meerkat population.
−
The rate at which eagles prey on meerkats is a sigmoid function of the meerkat population.
The sigmoid is not very steep.
Please note that for this problem, we will not add the signs to the contributions. We will add the
signs when evaluating the inflow/outflow at the very end.
Part a:
The per capita rate at which meerkats give birth to babies who survive
to adulthood is a steep sigmoid function of the adult population, with higher
reproductive success at higher populations.
For this one, it is up to interpretation, so this answer is acceptable:
8
M
′
M
=
k
·
M
n
(
a
1
)
n
+
M
n
→
M
′
=
k
·
M
n
(
a
1
)
n
+
M
n
·
M
However, one can argue that the number of adults NOW is dependent on the number of babies
born
τ
years back.
And we know that the number of babies born THEN is dependent on the number of adults AT
THAT TIME (
t
−
τ
), not at this moment. As a result, the contribution from birth to
M
′
is:
M
′
M
(
t
−
τ
)
=
k
·
[
M
(
t
−
τ
)]
n
(
a
1
)
n
+ [
M
(
t
−
τ
)]
n
⇐⇒
M
′
=
k
·
[
M
(
t
−
τ
)]
n
(
a
1
)
n
+ [
M
(
t
−
τ
)]
n
·
M
(
t
−
τ
)
(3)
We will utilize this latter version of Equation (3).
Part b:
Meerkats die of natural causes at a constant per capita rate
d
.
The contribution from death to
M
′
is:
M
′
M
=
d
⇐⇒
M
′
=
d
·
M
(4)
Part c:
Meerkats are preyed upon by eagles and jackals.
These predators have
many other prey, so their population does not depend on the meerkat population.
See Equations from Part d and e
Part d:
The rate at which jackals prey on meerkats is a nonsigmoid saturating
function of the meerkat population.
The contribution from Predation [From Jackals] to
M
′
is:
M
′
=
c
1
·
M
a
2
+
M
·
J
(5)
Part e:
The rate at which eagles prey on meerkats is a sigmoid function of the
meerkat population. The sigmoid is not very steep.
The contribution from Predation [From Eagles] to
M
′
is:
M
′
=
c
2
·
M
s
(
a
3
)
s
+
M
s
·
E
(s is small)
(6)
9
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Putting Things Together:
By combining Equations (3) - (6), we have:
M
′
=
[Sum of Inputs]
−
[Sum of Outputs]
=
[Birth]
−
[Death + Predation from Jackals + Predation from Eagles]
=
[Birth]
−
[Death]
−
[Predation from Jackals]
−
[Predation from Eagles]
M
′
(
t
) =
k
·
[
M
(
t
−
τ
)]
n
(
a
1
)
n
+ [
M
(
t
−
τ
)]
n
·
M
(
t
−
τ
)
−
d
·
M
(
t
)
−
c
1
·
M
(
t
)
a
2
+
M
(
t
)
·
J
(
t
)
−
c
2
·
[
M
(
t
)]
s
(
a
3
)
s
+ [
M
(
t
)]
s
·
E
(
t
)
for certain constants
a
1
,
a
2
,
a
3
,
c
1
,
c
2
,
n
,
s
and
τ
Problem 4.2. Further Exercise 5
Exercise 4.2.FE 5
The garibaldi is a large orange fish that lives off the coast of California
and Baja California. Use the assumptions below to write a differential equation for the size of
an
adult
garibaldi population.
−
The number of adults joining a population is the number of eggs laid times the fraction
that hatch times the fraction that survive to adulthood.
−
Garibaldis lay eggs at a constant per capita rate,
b
.
−
Because garibaldis sometimes eat their own eggs, the fraction of eggs that hatch is a
declining sigmoid function of the adult population.
−
Larval garibaldis float as plankton before becoming adults and joining a population.
Thus, the number of individuals joining a population is proportional to the number that
hatched six years earlier, with proportionality constant
r
.
−
Adult garibaldis die at a constant per capita rate
d
.
Please note that for this problem, we will not add the signs to the contributions. We will add the
signs when evaluating the inflow/outflow at the very end.
Part 1:
Contribution from Birth
•
The number of adults joining a population is the number of eggs laid times
the fraction that hatch times the fraction that survive to adulthood.
•
Garibaldis lay eggs at a constant per capita rate,
b
.
10
•
Because garibaldis sometimes eat their own eggs, the fraction of eggs that
hatch is a declining sigmoid function of the adult population.
•
Larval garibaldis float as plankton before becoming adults and joining a
population. Thus, the number of individuals joining a population is
proportional to the number that hatched six years earlier, with proportionality
constant
r
.
We have the following diagram:
Number of
Eggs (
G
1
)
Hatched
Eggs (
G
2
)
Survived to
Adulthood (
G
3
)
Time:
t
−
6
Time:
t
−
6
Time:
t
We know that the number of “new adults joining a population” at time
t
is dependent on the
number of eggs hatched six years earlier (as seen in Bullet #4), which in turn is dependent on the
number of eggs produced AT THAT TIME.
11
Let’s start with the following statement:
Bullet #1:
The number of adults joining a population is the number of eggs laid times
the fraction that hatch times the fraction that survive to adulthood.
We have our first equation:
G
′
=
G
3
= (# of eggs)
·
(Fraction Hatching)
·
(Fraction Survived)
(7)
Bullet #2:
Garibaldis lay eggs at a constant per capita rate,
b
We saw from the diagram that the number of eggs is from the time
t
−
6.
In turn, the number of eggs laid is dependent on the population of Garibaldis at time
t
−
6 (i.e.
per-capita of the population at
t
−
6). Thus,
1
G
(
t
−
6)
·
G
1
=
b
⇐⇒
G
1
= (# of Eggs) =
b
·
G
(
t
−
6)
(8)
Bullet #3:
Because garibaldis sometimes eat their own eggs, the fraction of eggs that
hatch is a declining sigmoid function of the adult population.
We saw from the diagram that the hatching occurs at time
t
−
6. Thus,
(Fraction Hatching) =
k
a
n
+ [
G
(
t
−
6)]
n
(9)
Note that this is NOT the same as
G
2
because
G
2
=
(Fraction Hatching)
·
G
1
Note that the numerator does NOT have to be 1 because if
G
(
t
−
6) = 0
, it is only required that:
k
a
n
= 1
for some constants
a
,
n
, and
k
Bullet #4:
Larval garibaldis float as plankton before becoming adults and joining a
population.
Thus, the number of individuals joining a population is proportional to
the number that hatched six years earlier, with proportionality constant
r
.
We have:
(Fraction Survived) =
r
(10)
Putting Things Together:
Let’s plug in Equations (8) - (10) into Equation (7), we have the contribution from birth to
G
′
:
G
′
=
(
b
·
G
(
t
−
6)
)
·
(
k
a
n
+
G
(
t
−
6)
n
)
·
(
r
)
(11)
12
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Part 2:
Contribution from Death
We know that:
Bullet #5:
Adult garibaldis die at a constant per capita rate
d
.
The contribution of death is in real time (i.e. at time
t
), thus
1
G
(
t
)
·
G
′
=
d
⇐⇒
G
′
=
d
·
G
(
t
)
(12)
Part 3:
Putting Things Together
By combining Equations (11) and (12), we have:
G
′
=
[Sum of Inputs]
−
[Sum of Outputs]
=
[Birth]
−
[Death]
G
′
(
t
) =
(
b
·
G
(
t
−
6)
)
·
(
k
a
n
+
G
(
t
−
6)
n
)
·
(
r
)
−
d
·
G
(
t
)
for some constants
b
,
a
,
k
,
n
,
r
, and
d
Problem 4.3.1
Exercise 4.3.1
Explain what each term in this model means and why it has the algebraic
form (for example,
SP
2
) that it does.
We are looking at this model:
S
′
=
V
0
−
cSP
2
P
′
=
cSP
2
−
kP
We have:
•
V
0
: the production rate of fructose-6-phosphate (from the two-step conversion of glucose not
featured in reaction scheme)
•
cSP
2
: the rate at which one molecule of fructose-6-phosphate is lost (and conversely, one
molecule of ADP is formed)
◦
SP
2
: the probability that ONE molecule of fructose-6-phosphate and TWO molecules of
ADP will meet each other AT THE SAME TIME (
P
2
because there are two molecules
of ADP)
◦
c
:
the probability constant that tells us how likely one successful encounter between
three molecules (1 fructose-6-phosphate + 2 ADPs) will result in a successful forward
reaction
13
•
kP
: the rate of degradation of ADP
◦
k
: the per-capita rate of degradation of ADP
Problem 4.3.2
Exercise 4.3.2
Why can
h
act as a half-saturation density?
In other words, what is the
consumption rate when
N
=
h
, and what does this mean biologically?
At
N
=
h
,
f
(
N
=
h
) =
C
max
·
h
h
+
h
=
C
max
·
h
2
·
h
=
C
max
2
Thus, mathematically, we see that at
N
=
h
, the rate of consumption of the predator is at half of
its maximum rate.
Biologically, it can act as a switch between low and high stable values:
•
Above h, an increase in N will has less effect in the rate of consumption. Thus, if N increases
at a really fast rate (i.e.
big ∆
N
), the predators’ consumption rate cannot keep up (i.e.
small
∆
f
), thus the number of preys increases to a higher value, as shown in
blue
below.
•
Below h, an increase in N will cause a big increase in the rate of consumption.
Thus if N
increases even slightly (i.e.
small ∆
N
), the predators can eat a lot more (high increase in
consumption rate, i.e.
big ∆
f
), thus the number of preys decreases to a lower value, as shown
in
red
below.
N
f
h
∆
N
∆
f
∆
N
∆
f
0
14
Problem 4.3.3
Exercise 4.3.3
Find the equilibria for this model using the parameter values in Figure
4.34
.
(Hint: Work with the second equation first.)
We have the following equations:
N
′
=
r
1
·
N
·
1
−
N
k
−
w
·
N
d
+
N
·
P
(13)
P
′
=
r
2
·
P
·
1
−
j
·
P
N
(14)
Let’s simplify Equation (13) a little more, by factoring out the
N
. Equation (13) becomes:
N
′
=
r
1
·
N
·
1
−
N
k
−
w
·
N
d
+
N
·
P
=
r
1
·
N
·
1
−
N
k
−
w
d
+
N
·
N
·
P
⇐⇒
N
′
=
N
·
"
r
1
·
1
−
N
k
−
w
d
+
N
·
P
#
(15)
We know that the equilibria of the system occur when
N
′
= 0 AND
P
′
= 0.
Thus, we need to
set BOTH
Equations (14) and (15) to be equal to 0 (recall that Equation (15) is a variation of
Equation (13)). We can follow the hint by solving for
P
using
P
′
= 0 first, and then plug the
P
value into the equation for
N
′
= 0 to find the value for
N
.
Using Equation (14), we have:
P
′
= 0
⇐⇒
r
2
·
P
·
1
−
j
·
P
N
= 0
⇐⇒
P
= 0
OR
1
−
j
·
P
N
= 0
⇐⇒
P
= 0
OR
j
·
P
N
= 1
⇐⇒
P
= 0
OR
j
·
P
=
N
⇐⇒
P
= 0
OR
P
=
1
j
·
N
As we can see, there are two values of
P
that will make
P
′
= 0. Now, let’s use
each
of them, and
find the corresponding
N
value in the solution using Equation (15).
It is also important to note that for
P
′
= 0,
P
′
must
be defined. In other words,
N
̸
= 0.
Situation #1:
P
= 0
for
P
′
= 0
By plugging
P
= 0 into Equation (15), Equation (15) becomes:
15
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N
′
=
N
·
"
r
1
·
1
−
N
k
−
w
d
+
N
·
0
|
{z
}
0
#
=
N
·
r
1
·
1
−
N
k
(16)
To find
N
′
= 0, we then set Equation (16) to be equal to 0. Thus, we have:
N
′
= 0
⇐⇒
N
·
r
1
·
1
−
N
k
= 0
⇐⇒
N
= 0
OR
1
−
N
k
= 0
⇐⇒
N
= 0
OR
N
k
= 1
⇐⇒
N
= 0
OR
N
=
k
Since
N
̸
= 0, we have one equilibrium point:
N
=
k
and
P
= 0
Situation #2:
P
=
1
j
·
N
for
P
′
= 0
By plugging
P
=
1
j
·
N
into Equation (15), Equation (15) becomes:
N
′
=
N
·
"
r
1
·
1
−
N
k
−
w
d
+
N
·
1
j
·
N
#
(17)
To find
N
′
= 0, we then set Equation (17) to be equal to 0. Thus, we have:
N
′
= 0
⇐⇒
N
·
"
r
1
·
1
−
N
k
−
w
d
+
N
·
1
j
·
N
#
= 0
⇐⇒
N
= 0
OR
r
1
·
1
−
N
k
−
w
d
+
N
·
1
j
·
N
= 0
Since we mentioned that
N
̸
= 0 for
P
′
to be defined, we only will choose the latter case.
Let’s solve this equation for
N
first:
r
1
·
1
−
N
k
−
w
d
+
N
·
1
j
·
N
= 0
⇐⇒
r
1
·
k
−
N
k
−
w
·
N
j
·
(
d
+
N
)
= 0
⇐⇒
r
1
·
(
k
−
N
)
·
j
·
(
d
+
N
)
−
w
·
k
·
N
k
·
j
·
(
d
+
N
)
= 0
⇐⇒
r
1
·
(
k
−
N
)
·
j
·
(
d
+
N
)
−
w
·
k
·
N
= 0
16
⇐⇒
r
1
·
j
·
k
·
d
+
k
·
N
−
d
·
N
−
N
2
−
w
·
k
·
N
= 0
⇐⇒
(
−
r
1
·
j
)
·
N
2
+ (
r
1
·
j
·
k
−
r
1
·
j
·
d
−
w
·
k
)
·
N
+
r
1
·
j
·
k
·
d
= 0
(18)
Using real values of
r
1
= 1,
r
2
= 0
.
1 =
1
10
,
k
= 7,
d
= 1,
j
= 1, and
w
= 0
.
3 =
3
10
, our Equation
(18) becomes:
(
−
1
·
1)
·
N
2
+
1
·
1
·
7
−
1
·
1
·
1
−
3
10
·
7
·
N
+ 1
·
1
·
7
·
1 = 0
⇐⇒ −
N
2
+
39
10
·
N
+ 7 = 0
⇐⇒
10
·
N
2
−
39
·
N
−
70 = 0
This equation has two solutions:
N
=
−
(
−
39)
±
p
(
−
39)
2
−
4
·
(10)
·
(
−
70)
2
·
(10)
=
39
±
√
4321
20
⇐⇒
N
≈
5
.
2367 OR
N
≈ −
1
.
3367
Since
−
1
.
3367
<
0,
N
=
−
1
.
3367 is not within our state space. As a result,
N
≈
5
.
2367.
Substituting
N
≈
5
.
2367 back to equation
P
=
1
j
·
N
, we have:
P
=
1
j
·
N
=
1
1
·
N
=
N
≈
5
.
2367
Therefore, our equilibrium point will be:
N
≈
5
.
2367 and
P
≈
5
.
2367 (you can round down to
whole values if you want)
In summary, our equilibria are:
•
N
=
k
= 7
and
P
= 0
•
N
≈
5
.
2367
and
P
≈
5
.
2367
Note: we don’t need to worry about the fraction
w
·
N
d
+
N
because since
d >
0
and
N >
0
,
d
+
N >
0
,
i.e. no chance of being equal to 0
17