Homework #4 Solutions
Mathematical Modeling, Spring 2017
Dr. Doreen De Leon
1. For each of the following first-order difference equations:
(a) solve the equation;
(b) find the equilibrium solution(s); and
(c) analyze the stability of all equilibrium solutions.
Do not use Maple for this problem.
(1) yn = 5yn−1 , y0 = 10
2
(2) yn = yn−1 , y0 = 81
3
(3) yn = 0.1yn−1 + 3.2, y0 = 1.3
(4) yn = yn−1 − 100, y0 = 10
(5) yn = −3yn−1 + 4, y0 = 4
Solution:
(1) (a) The general solution is yn = 10(5n ), n ∈ N.
(b) The equilibrium solution is found by solving y = 5y, giving y = 0.
(c) Since lim yn = ∞, y = 0 is unstable.
n→∞
n
2
(2) (a) The general solution is yn = 81
, n ∈ N.
3
2
(b) The equilibrium solution is found by solving y = y, giving y = 0.
3
(c) Since lim yn = 0, y = 0 is stable.
n→∞
(3) (a) The general solution is
n
yn = 1.3(0.1) + 3.2
or
yn = 1.3(0.1)n +
1 − (0.1)n
1 − 0.1
,
32
(1 − (0.1)n ), n ∈ N.
9
(b) The equilibrium solution is found by solving y = 0.1y + 3.2, giving y =
1
32
.
9
(c) Since
32
= y,
9
lim yn =
n→∞
the equilibrium is stable.
(4) (a) The general solution is
yn = 10 − 100n, n ∈ N.
(b) There is no equilibrium solution, since y = y − 100 has no solution.
(c) Not applicable.
(5) (a) The general solution is
n
yn = 4(−3) + 4
1 − (−3)n
4
,
or
yn = 4(−3)n + 1 − (−3)n ,
which simplifies to
yn = 3(−3)n + 1, n ∈ N.
(b) The equilibrium solution is found by solving y = −3y + 4, giving y = 1¿
(c) Since lim yn = −∞, the equilibrium is unstable.
n→∞
2. Problem 5.2.7. Consider the simple interest formula Sn = (1+np)S0 and the compound
interest formula Sn = (1 + p/r)n S0 . There are three options to earn interest. Company
A offers simple interest at a rate of 6%. Company B offers compound interest at a 4%
rate with a conversion period of one month. Company C offers compound interest at
a 4% rate with a conversion period of three months.
(a) Calculate for the three cases the amount on deposit after 5, 10, 15, and 20 years
for any principal S0 .
(b) Which interest offer maximizes the amount on deposit after 5, 10, 15, and 20
years?
Solution:
(a) Note that for Companies B and C, n is the number of conversion periods, which
is kr, where k is the number of years. For Company A, n = k.
k
5
10
15
20
Company A Company B
1.3S0
1.221S0
1.6S0
1.491S0
1.9S0
1.820S0
2.2S0
2.222S0
Company C
1.220S0
1.489S0
1.817S0
2.217S0
Table 1: Amount of money on deposit for Companies A, B, and C after k years.
2
(b) The amount on deposit is maximized after 5, 10, and 15 years by Company A (i.e.,
simple interest). However, after 20 years, the amount on deposit is maximized by
Company B (i.e., interest at 4% with monthly conversion).
3. Cipro is an antibiotic taken to combat many infections, including anthrax. Cipro is
filtered from the blood by the kidneys. Each 24-hour period, the kidneys filter out
about one-third of the Cipro that was in the blood at the beginning of the 24-hour
period.
(a) Write a difference equation to model the situation where a patient is given only
one 500-mg dose of Cipro. If the level of the drug must be maintained at 250 mg,
how long before the drug is no longer effective?
(b) Now, assume that the patient must take an additional 500 mg per day. Use
a difference equation to determine how much blood the patient has in his/her
bloodstream after 5 days.
(c) For the situation in part (b), what is the equilibrium value of the level of Cipro
in the bloodstream? Is this level effective?
Solution:
(a) Let yn be the amount of drug in the patient after n 24-hour periods. Then yn
satisfies
1
yn = yn−1 − yn−1 ,
3
or
2
yn = yn−1 , y0 = 500.
3
For any n, yn is given by
n
2
, n = 1, 2, . . .
yn = 500
3
To determine when the drug stops becoming effective, we must determine n so
that
n
2
250 = 500
,
3
or
n
2
1
= .
3
2
Taking the natural logarithm of both sides and solving gives
n=
ln(1/2)
= 1.7.
ln(2/3)
Since n must be a positive integer, n = 2, so after two days the drug is no longer
effective.
3
(b) In this case, the difference equation is given by
2
yn = 500 + yn−1 , y0 = 500.
3
Then for all n,
n
2
yn = 500
+ 500
3
or
n !
1 − 32
,
1 − 23
n
2
yn = −1000
+ 1500, n = 1, 2, . . . .
3
After 5 days, the patient will have
5
2
y5 = −1000
+ 1500 = 1368,
3
or 1,368 mg of drug in his/her bloodstream.
(c) The equilibrium level is determined by solving
2
y = 500 + y,
3
giving
y = 1500.
The equilibrium level of 1,500 mg is above the minimum 250 mg, and so is definitely effective. The only question that remains is if that level of drug is safe.
4. For each of the following difference equations:
(a) solve the equation; and
(b) determine the long-term behavior of the solution (i.e., the behavior of the solution
as n → ∞).
Do not use Maple for this problem.
(1) yn = 2yn−2 − 1, y0 = 1, y1 = −1
1
1
(2) yn = yn−1 − yn−2 + , y0 = 1, y1 = −1
4
4
(3) yn = 2yn−1 − 2yn−2 − 1, y0 = 1, y1 = −1
Solution:
(1) First, solve the homogeneous equation yn = 2yn−2 . The characteristic equation is
r2 − 2 = 0,
√
giving r = ± 2. Therefore, the homogeneous solution is
√
√
sn = c1 (− 2)n + c2 ( 2)n .
4
Since 1 − A − B = 1 − 2 = −1 6= 0, the particular solution is
zn =
−1
= 1.
−1
The general solution is given by
√
√
yn = sn + zn = c1 (− 2)n + c2 ( 2)n + 1.
We use the initial conditions to find c1 and c2 .
y0 = c1 + c2 + 1 =⇒ 1 = c1 + c2 + 1, or c1 + c2 = 0.
And,
√
√
√
√
√
√
y1 = c1 (− 2)+c2 ( 2)+1 =⇒ −1 = c1 (− 2)+c2 2+1, or c1 (− 2)+c2 ( 2) = 2.
The first equation tells us that c1 = −c2 , so from the second equation we obtain
1
1
c1 = √ and c2 = − √ , and so yn is given by
2
2
√
√
yn = (− 2)n−1 − ( 2)n−1 + 1, n = 1, 2, . . . .
We see that limn→∞ yn does not exist.
(2) First, solve the homogeneous equation. The characteristic equation is
r2 − r −
1
= 0,
4
giving
2
1
= 0.
r−
2
So, r = 1/2, which is a repeated real root. Therefore, the homogeneous solution is
n
n
1
1
+ c2 n
.
s n = c1
2
2
Since 1 − A − B =
1
4
6= 0, the particular solution is given by
zn =
1
4
1
4
= 1.
The general solution is given by
n
n
1
1
yn = sn + zn = c1
+ c2 n
+ 1.
2
2
We use the initial conditions to find c1 and c2 . We have
y0 = c1 + 1 =⇒ 1 = c1 + 1, or c1 = 0.
5
And,
1
1
y 1 = c1
+ c2
+ 1,
2
2
giving
1
−1 = 0 + c2
+ 1.
2
Solving for c2 gives
c2 = −4.
Therefore,
n
1
+ 1, n = 1, 2, . . . .
yn = −4n
2
As n → ∞, yn → ∞.
(3) First, solve the homogeneous equation. The characteristic equation is
r2 − 2r + 2 = 0.
The roots are
r = 1 ± i.
Therefore, the solution takes the form
sn = Rn (c1 cos(nθ) + c2 sin(nθ)),
where
R=
p
√
12 + (−1)2 = 2
and
tan(θ) = 1 =⇒ θ =
π
.
4
So,
nπ nπ √
+ c2 sin
.
sn = ( 2)n c1 cos
4
4
To find zn , since 1 − A − B = 1 − 2 − (−2) 6= 0,
zn =
−1
= −1.
1−2+2
The general solution is then
nπ nπ √
yn = sn + zn = ( 2)n c1 cos
+ c2 sin
− 1.
4
4
To find c1 and c2 , we use the initial conditions.
1 = y0 = c1 − 1 =⇒ c1 = 2,
and
√
1
1
−1 = y1 = 2 c1 √ + c2 √
= 2 + c2 =⇒ c2 = −3.
2
2
Therefore,
nπ nπ √ n
yn = ( 2) 2 cos
− 3 sin
− 1.
4
4
Then limn→∞ yn does not exist.
6
5. Problem 5.3.4. (See textbook for problem statement)
Solution:
(a) Given y0 = 1 and after 1 month that pair of rabbits will have one pair of rabbits,
so y1 = 2 (the original pair plus the one just born). Then
y3 = 2 + 1 = 3,
since the first pair of rabbits will have a pair of rabbits (but the second pair
won’t). Then
y4 = 3 + 2 = 5
y5 = 5 + 3 = 8.
(b) The difference equation modeling the above is
yn = yn−1 + yn−2 .
(c) The characteristic equation for the difference equation is
r2 − r − 1 = 0.
The roots of this equation are
r1,2
√
1± 5
=
,
2
and so the general solution is given by
√ !n
1+ 5
y n = c1
+ c2
2
√ !n
1− 5
.
2
Using the initial conditions, we may solve for c1 and c2 . We have
1 = y0 = c1 + c2 ,
and
1 = y1 = c1
√ !
1+ 5
+ c2
2
√ !
1− 5
.
2
Solving gives c1 = 1 and c2 = −1. So, the solution is
√ !n
√ !n
1+ 5
1− 5
yn =
−
.
2
2
(d) After one year, there are
y12 = 321.99,
or 322 rabbits.
7
6. Problem 5.4.1. Consider the following nonlinear first-order difference equations, where
n = 1, 2, . . . . Find the equilibrium points and determine whether the equilibrium states
are stable or not.
1
1 + yn−1
yn−1
(b) yn =
1 + yn−1
(a) yn =
2
)
(c) yn = yn−1 (1 − yn−1
Solution:
(a) The equilibrium solution is obtained by solving
y=
1
,
1+y
or
y 2 + y − 1 = 0.
The solution is given by
√
−1 ± 5
.
y=
2
To determine the stability of these equilibria, we need to find f 0 (y).
f 0 (y) = −
Then
0
f
−1 +
so y =
2
√
5
√ !
5 = 1
1+
√ 2
−1+ 5
2
=
1
√ 2
1+ 5
2
< 1,
is stable. And
0
f
√
−1 +
2
1
.
(1 + y)2
−1 −
2
√ !
5 = 1
√
1+
−1− 5
2
2 = −1 +
since (1 − 5) ≈ 1.52 < 4. Therefore, y =
2
(b) The equilibrium solution is obtained by solving
2
y=
y
,
1+y
or
y(1 + y) = y,
giving
y 2 = 0.
8
√
5
1
√ 2
1− 5
2
> 1,
is not stable.
Therefore, there is only one equilibrium solution, y = 0. To determine its stability,
we need to find f 0 (y).
1
.
f 0 (y) =
(1 + y)2
Then,
f 0 (0) = 1.
In this case, the stability of y = 0 cannot be determined using this approach.
Instead, we will look at some solutions using an arbitrary initial value y0 . The
first few solutions are
y0
y1 =
1 + y0
y1
y2 =
1 + y1
=
1
y0
1+y0
y0
+ 1+y
0
y0
1 + 2y0
y2
y3 =
1 + y2
=
=
=
1
y0
1+2y0
y0
+ 1+2y
0
y0
.
1 + 3y0
It appears that the general solution is given by
y0
yn =
.
1 + ny0
As n → ∞, yn → 0, so y = 0 is stable.
(c) The equilibrium solution is obtained by solving
y = y(1 − y 2 ),
giving
y = y − y 3 =⇒ y 3 = 0.
Therefore, the only equilibrium solution is y = 0. To determine its stability, we
will find f 0 (y).
f 0 (y) = 1 − 3y 2 .
Again, we have the problem that f 0 (0) = 1 and the linearization approach does
not give us an answer. Instead, we will again look at some solutions using an
arbitrary initial value y0 . The first few solutions are
y1 = y0 − y03
y2 = y1 − y13
= y0 − y03 − (y0 − y0 )3
= y09 − 3y07 + 3y05 − 2y03 + y0 .
It seems that if y0 > 1, yn → ∞ as n → ∞. Therefore, y = 0 is not stable.
9
7. Problem 5.4.2. The Ricker equation yn = αyn−1 exp(−βyn−1 ) is often used for the
modeling of fish populations. In this equation, α is the maximal growth rate of the
organism, and β represents the inhibition of growth due to overpopulation (EdelsteinKeshet 2005).
(a) Calculate the nonzero equilibrium solution to this equation.
(b) Show under which conditions the equilibrium is stable.
Solution:
(a) The equilibrium solutions are determined by solving
y = αye−βy .
Then
y(1 − αe−βy ) = 0,
and assuming y 6= 0,
e−βy =
1
,
α
giving
y=
ln(α)
.
β
This is only defined if α > 0.
(b) To determine the stability of this equilibrium solution, we need to determine f 0 (y).
f 0 (y) =
d
(αye−βy ) = αe−βy − αβye−βy .
dy
Then
|f 0 (y)| = αe−βy |1 − βy|,
and
0 ln(α) ln(α)
−
ln
α
f
= αe
1 − β
= |1 − ln(α)|.
β
β
So,
0 ln(α) f
= |1 − ln(α)| < 1
β
only if
0 < ln(α) < 2,
or if
1 < α < e2 .
10