Biology Reference
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average length of the infection is 1
b
. Thus, the average number of
secondary infections produced by the single infective among the S(0)
susceptibles is at most a
b
. The number a
b
S
ð
0
Þ
S
ð
0
Þ
estimates the basic
reproduction number of the infection defined as the average number of
secondary infections that a single infective can produce in a fully
susceptible population.
We next determine the long-term behavior of S(t),R(t), and I(t).
3. The Long-Term Evolution of the Disease for the SIR Model
We first show that according to the SIR model not everyone will catch
the disease.
E XERCISE 2-6
Use dS
dS
=
dt
dt ¼ a
e b
R
ð
t
Þ
dR ¼
S to show that if R(0)
¼
0, then S
ð
t
Þ¼
S
ð
0
Þ
:
dR
=
b
Let lim t !1 R
ð
t
Þ
be denoted by R(
1
). Because R(t) is non-decreasing as a
function of t, we obtain:
a
b
Þ a
b
R
ð
t
N
:
Combined with the result from Exercise 2-6, this yields:
e b
e b
R
ð
t
Þ
N
S
ð
t
Þ¼
S
ð
0
Þ
S
ð
0
Þ
>
0
:
Notice that because S(t) is decreasing as time increases, and R(t)is
increasing, we know that S
ð1Þ ¼
lim t !1
S
ð
t
Þ
and R
ð1Þ ¼
lim t !1
R
ð
t
Þ
exist. Passing to a limit for t , when t
!1
, we obtain
e b
N
S
0, showing that in the long run, a
fraction of the population will never get infected.
ð1Þ ¼
lim t !1 S
ð
t
Þ
S
ð
0
Þ
>
E XERCISE 2-7
We have argued that not everyone will catch the disease. Is this
intuitively plausible? If so, explain why. If not, explain some
discrepancies between the model and reality.
Next, we show that eventually the number of infectives goes to 0. Notice
that because S(
1
) and R(
1
) exist and S(t)
þ
I(t)
þ
R(t)
¼
N for all t,
we obtain that:
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