Math 5110/6830
Homework 1 Solutions
1. (a) If In+1 = (1 − p 1 )In and I0 is the initial number of infected individuals, then given I0 ,
2
we can solve the equation recursively for In . Since I1 = (1 − p 1 )I0 , it follows that
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I2 = (1 − p 1 )I1 = (1 − p 1 )2 I0 and I3 = (1 − p 1 )3 I0 . Continuing in this fashion, we
2
2
2
obtain In = (1 − p 1 )n I0 = (1 − α∆t)n I0 as the solution to the discrete model.
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Assume for now that α > 0 and that 0 < 1 − α∆t < 1. Then (1 − α∆t)n < 1, which implies
that as time goes on, i.e. as n gets bigger, the number of infected people is going to decay.
(b) To solve the continuous model I 0 (t) = −αI(t), we first let I(0) = I0 . If you don’t immediately see the solution to this equation, rewrite the model as dI
dt = −αI and solve by
separating the variables. When we do this, we get the following:
dI
= −αdt
I
Z
Z
1
dI = −α dt
I
ln I = −αt + c1
eln I = e−αt+c1
I(t) = ce−αt ,
where c = ec1 . Now, to solve for c, we use the fact that I(0) = I0 : I(0) = ce−α·0 = c = I0 ,
and therefore I(t) = I0 e−αt .
(c) For α = 0.3 and ∆t = 0.5, the discrete solution undershoots the continuous solution.
∆ t = 0.5
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Number of infected individuals
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Time, days
Varying ∆t gives a closer approximation of the discrete solution to the continuous solution for
smaller values of ∆t, while the discrete solutions plotted with larger values of ∆t are farther
away from the continuous solution. When α = 0.3, ∆t = 18 gives the best agreement while
∆t = 10 gives the worst agreement. This is because the discrete model breaks down when
the probability of recovery, p∆t , exceeds 1, which occurs when ∆t > α1 . This is unrealistic,
as we obtain negative numbers of infected individuals with the discrete model under this
condition.
1
∆ t = 0.25
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Number of infected individuals
Number of infected individuals
∆ t = 0.125
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Time, days
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Time, days
∆t=1
∆ t = 10
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Number of infected individuals
Number of infected individuals
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0
−50
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−150
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Time, days
Time, days
(d) With ∆t = 0.5, as α decreases, the number of infected individuals decreases more slowly,
whereas when α increases, In and I(t) decay more quickly. The solutions decay more slowly
for smaller values of α because 1−p∆t increases as α decreases, which means that (1−p∆t )n
is going to be greater for smaller α, thus increasing In . Similarly, −αt is less negative for
smaller α, thus making e−αt greater.
In blue: α = .3; In red: α = 0.2
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Number of infected individuals
Number of infected individuals
In blue: α = .3; In red: α = 0.1
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Time, days
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Time, days
2
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In blue: α = .3; In red: α = 0.5
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Number of infected individuals
Number of infected individuals
In blue: α = .3; In red: α = 0.4
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Time, days
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Time, days
2. (a) Choosing a time unit of ∆t = 1 minute, the corresponding probability of cell division after
time ∆t is p∆t = g · ∆t, where g is the rate at which a cell divides. So the probability of a
cell dividing after just 1 minute is p1 = g · 1 = g.
(b) At time t + ∆t, the amount of cells in the dish will be the number at time t plus the number
that divided after ∆t minutes, which is given by the probability of division times the amount
at time t. Let B(t) be the number of cells in the petri dish at time t. Then the discrete-time
model relating the amount of cells time t and t + ∆t is B(t + ∆t) = B(t) + p∆t B(t) .
(c) Define the growth rate to be g, as above. Then since p∆t = g · ∆t, we have B(t + ∆t) =
B(t) + p∆t B(t) = B(t) + g∆tB(t). To derive the continuous-time model, subtract B(t)
from both sides, divide by ∆t, and take the limit as ∆t goes to 0:
B(t + ∆t) − B(t) = g∆tB(t)
B(t + ∆t) − B(t)
g∆tB(t)
⇒
=
= gB(t)
∆t
∆t
B(t + ∆t) − B(t)
⇒
lim
= lim gB(t) = gB(t)
∆t→0
∆t→0
∆t
⇒
B 0 (t) = gB(t)
(d) To solve both models, first let B0 be the initial number of cells in the petri dish.
Discrete-time model: Defining Bn = B(n · ∆t), we can write the discrete model as
Bn+1 = (1 + p∆t )Bn . Then B1 = (1 + p1 )B0 , B2 = (1 + p1 )2 B0 , B3 = (1 + p1 )3 B0 , and so
Bn = (1 + p1 )n B0 = (1 + g)n B0 . Since each cell divides into two identical copies of itself
1
min−1 , and the solution to the discrete
every 10 minutes, the growth
rate g is therefore 10
n
1
model is Bn = 1 + 10
B0 .
Continuous-time model: The model B 0 (t) = gB(t) can be solved using separation of
1
variables (see 1(b)). The continuous solution is B(t) = B0 egt = B0 e 10 t .
Comparison: There is close agreement between the discrete and continuous solutions for
smaller values of n and t. The solutions begin to diverge for larger values, with the solution
to the discrete model increasing more slowly than that of the continuous model.
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Number of cells in dish
200
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Time, minutes
20
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