MATH 23 HOMEWORK #2 PART B SOLUTIONS
Problem 2.5.7 (Semistable Equilibrium Solutions). Sometimes a constant equilibrium
solution has the property that solutions lying on one side of the equilibrium solution
tend to approach it, whereas solutions lying on the other side depart from it. In this case
the equilibrium solution is said to be semistable.
(a) Consider the equation
dy/dt = k(1 − y)2 ,
(1)
where k is a positive constant. Show that y = 1 is the only critical point, with the
corresponding equilibrium solution y(t) = 1.
(b) Sketch f ( y) versus y. Show that y is increasing as a function of t for y < 1 and
also for y > 1. The phase line has upward-pointing arrows both below and above
y = 1. Thus solutions below the equilibrium solution approach it, and those above
it grow farther away. Therefore, y(t) = 1 is semistable.
(c) Solve Eq. (1) subject to the initial condition y(0) = y0 and confirm the conclusions
reached in part (b).
Solution.
(a) If y is an equilibrium solution, then dy/dt = 0 identically, so
0=
dy
= k (1 − y)2
dt
hence y(t) = 1 identically.
(b)
f ( y) =
dy
dt
k
y
1
2
Since dy/dt = k(1 − y)2 > 0 for y 6= 1, then y is increasing for y < 1 and y > 1.
(c) Separating the variables, if y 6= 1 we have
kt + C =
Z
k dt =
Z
dy
=−
(1 − y)2
1
Z
du
1
1
= =
2
u
1−y
u
where we have made the substitution u = 1 − y, du = −dy. Then
1
1
=⇒ y = 1 −
kt + C
kt + C
and the initial condition gives
1−y =
y0 = y(0) = 1 −
1
1
1
=⇒
= 1 − y0 =⇒ C =
.
C
C
1 − y0
Thus
y = 1−
1
1 − y0
= 1−
1
k (1 − y0 )t + 1
kt + 1− y0
y=1
or
since we previously excluded this case.
Since
1
k (1 − y0 )
1
then y(t) has a vertical asymptote at t = −
and a horizontal asymptote
k (1 − y0 )
at y = 1. If y > 1, then
0 = k(1 − y0 )t + 1 =⇒ t = −
1 < y = 1−
1
1
1
=⇒ 0 < −
=⇒
< 0.
1
1
kt + 1− y0
kt + 1− y0
kt + 1−1y0
Then the numerator and denominator have different signs, so the denominator is
negative:
kt +
1
1
1
< 0 =⇒ kt < −
=⇒ t < −
1 − y0
1 − y0
k (1 − y0 )
since k > 0. Then we are to the left of the vertical asymptote, so y(t) diverges to
1
±∞ as t → −
. On the other hand, if y > 1, then we are to the right of the
k (1 − y0 )
vertical asymptote and y(t) → 1 as t → ∞ since y(t) has a horizontal asymptote
at y = 1. (See below for a plot of y(t) for the values k = 1, y0 = 2.)
4
y
2
−3 −2 −1
1
−2
−4
Problem 2.5.17.
2
2
3
4
t
(a) Solve the Gompertz equation
dy/dt = ry ln(K / y)
subject to the initial condition y(0) = y0 . Hint: You may wish to let u = ln( y/K )l
(b) For the data given in Example 1 in the text (r = 0.71 per year, K = 80.5 × 106 kg,
y0 /K = 0.25), use the Gompertz model to find the predicted value of y(2).
(c) For the same data as in part (b), use the Gompertz model to find the time τ at
which y(τ ) = 0.75K.
Solution.
(a) If y 6= 0, we can make the suggested substitution u = ln( y/K ). Then
y = Keu , so
du
1 dy
K dy
K dy
dy
dy
du
=
=
= u
= e−u
=⇒
= eu .
dt
y/K dt
y dt
Ke dt
dt
dt
dt
Then the Gompertz equation becomes
dy
du
du
=
= ry ln(K / y) = rKeu (−u) =⇒
= −rKu .
dt
dt
dt
Note that u = ln( y/K ) = 0 ⇐⇒ y = K. Then for y 6= K we can separate
variables:
Z
Z
du
ln |u| =
= − rK dt = −rKt + C0
u
and exponentiating yields
eu
ln( y/K ) = u = Ce−rKt =⇒ y/K = eCe
−rKt
=⇒ y = KeCe
−rKt
.
The initial condition implies
y0 = y(0) = KeC =⇒ y0 /K = eC =⇒ C = ln( y0 /K )
so
y = K exp (ln( y0 /K ) exp(−rKt)) = K ( y0 /K )e
−rt
.
or the constant solution (which we previously excluded) y = K which clearly satisfies the Gompertz equation.
(b) Using the provided data, we have
y(2) = K exp (ln( y0 /K ) exp(−2rK )) ≈ 5.75797 · 107 kg .
(c) We use Mathematica to find a numerical solution to the equation
0.75K = y(τ ) = K exp (ln( y0 /K ) exp(−rKτ ))
which produces τ ≈ 2.21484 yrs.
Problem 2.5.24. Consider the cohort of individuals born in a given year (t = 0), and let
n(t) be the number of these individuals surviving t years later. Let x(t) be the number of
members of this cohort who have not had smallpox by year t and who are therefore still
susceptible. Let β be the rate at which susceptibles contract smallpox, and let ν be the
rate at which people who contract smallpox die from the disease. Finally, let µ (t) be the
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death rate from all causes other than smallpox. Then dx/dt, the rate at which the number
of susceptibles declines, is given by
dx/dt = −[β + µ (t)] x .
Also
dn/dt = −νβx − µ (t)n .
(a) Let z = x/n, and show that z satisfies the initial value problem
dz/dt = −βz(1 − νz),
z(0) = 1 .
(2)
Observe that the initial value problem (2) does not depend on µ (t).
(b) Find z(t) by solving (2).
(c) Bernoulli estimated that ν = β = 1/8. Using these values, determine the proportion of 20-year-olds who have not had smallpox.
(a) Since z = x/n, we use the quotient rule to find dz/dt:
Solution.
nx0 − xn0
xn0
x
x0
−(β + µ ) x
dz
=
−
− 2 (−νβx − µn)
=
=
2
2
dt
n
n
n
n
n
2
x
x
= −(β + µ ) z + βν 2 + µ = −βz − µz + βνz2 + µz = −βz + βνz2
n
n
= βz(−1 + νz) = −βz(1 − νz) .
Alternatively, we can use the multivariable chain rule: since z = x/n, then
x
d
∂z dx ∂z dn
1
z=
+
= − (β + µ (t)) x + − 2 (−νβx − µ (t)n)
dt
∂x dt
∂n dt
n
n
2
x
x
x
= − (β + µ (t)) + νβ 2 + µ (t) = − z(β + µ ) + νβz2 + µz = − zβ + νβz2
n
n
n
= βz(−1 + νz) = −βz(1 − νz) .
Since x(0) = n(0) (no one is infected at time 0), then z(0) = 1.
(b) If z 6= 0 and z 6= 1/ν, then we can separate variables as follows:
Z
dz
=−
z(1 − νz)
Z
β dt = −βt + C0 .
To integrate the lefthand side, we use partial fraction decomposition:
1
A
B
= +
=⇒ 1 = A(1 − νz) + Bz = ( B − Aν ) z + A
z(1 − νz)
z
1 − νz
=⇒ A = 1 and B − Aν = 0 =⇒ A = 1 and B = ν .
So
Z
dz
=
z(1 − νz)
Z
dz
+
z
Z
z ν
.
dz = ln | z| − ln |1 − νz| = ln 1 − νz
1 − νz 4
Exponentiating, then
z
= Ce−βt =⇒ z = (1 − νz)Ce−βt = Ce−βt − νzCe−βt
1 − νz
=⇒ z(1 + Cνe−βt ) = z + νCe−βt z = Ce−βt
=⇒ z =
Ce−βt
=
1 + Cνe−βt
1
1 βt
Ce
+ν
.
Using the initial condition z(0) = 1, then
1
1
1
1
1 = z(0) = 1
=⇒
+ ν = 1 =⇒
= 1 − ν =⇒ C =
C
C
1−ν
C +ν
so we have
z=
1
1 βt
Ce
=
1
ν + (1 − ν )eβt
+ν
or z = 0 or z = 1/ν, since we excluded these cases and they clearly satisfy (2).
(c) For β = ν = 1/8, we have
1
z=
(1 / 8 ) + (7 / 8 )et/8
so
1
z(20) =
≈ 0.092724104 .
(1/8) + (7/8)e20/8
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