Math 5120
Homework 6.1 Solutions
1. (a) To find the minimal wave speed for the spread of rabies (which implies that R0 > 1 from class), we
need to classify the fixed points of the system
1
u′ = − R0 uv
c
1
′
ln u − 1 .
v =c v+u−
R0
The steady states are (1, 0) and (u2 , 0), where 0 < u2 < 1. Recall that a traveling wave solution in
the PDE corresponds to a solution in the ODE that starts at one fixed point and ends at the other,
satisfying the boundary conditions. In this case, the solution should start at (1, 0) and end at (u1 , 0),
where u(+∞) = u1 = u2 . The Jacobian for this system is
#
"
1
1
− c R0 v − c R0 u
.
J(u, v) =
c
c 1 − R10
For J(u1 , 0), the trace is c and the determinant is R0 u1 − 1. Since the trace is positive, and we
require the traveling wave solution to approach (u1 , 0), this point has to be a saddle. Therefore,
the
√
c±
c2 −4(R −1)
0
determinant must be negative, and so u1 < R10 . The eigenvalues of J(1, 0) are λ1,2 =
.
2
This is either an unstable spiral or an unstable node. Since v represents a population, it must be the
2
case that
√ v ≥ 0. To ensure this, (1, 0) must be a node, which means that c − 4(R0 −∗ 1) ≥√0. Thus,
c ≥ 2 R0 − 1 must be true, so that the minimal wave speed for the spread of rabies is c = 2 R0 − 1.
(b) To convert the wave speed into dimensional coordinates, rewrite the dimensionless wave variable z in
terms of the dimensional x and t, and construct the dimensional wave variable, say ζ = x + κt, where
κ will be the dimensional wave speed.
z = ξ + cτ
r
p
γ
x + 2γ R0 − 1t
=
D
s
p
2γ D(R0 − 1)
D
⇒
z =x+
t
√
γ
γ
p
= x + 2 Dγ(R0 − 1)t
s
βN
−1
= x + 2 Dγ
γ
p
= x + 2 D(βN − γ)
⇒ ζ = x + κt,
p
where κ = 2 D(βN − γ) is the dimensional wave speed. If γ = 1, β = 0.8, N = 2 and D = 1060,
√ 80
then it would take the epidemics 80
≈ 1.6 years to reach a location 80 km away.
κ =
2
D(βN −γ)
(c) We can prevent the epidemic from happening by making R0 < 1. R0 can be made smaller by decreasing
N = S + I + R. Realistically, this can happen by vaccinating susceptible foxes, which would essentially
limit potential infections. Another way would be to identify infected foxes and quickly remove them
from the population permanently, which would increase γ and therefore decrease R0 . The former
strategy is probably more feasible.
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Homework 6.2 Solutions
1. (a) If the muskrat density at radius R at time t is given by
M
R2
u(R, t) =
,
exp αt −
4πDt
4Dt
then the total number of muskrats that are farther than R2 (t) from the origin at time t is
N (t) =
Z
0
2π
Z
∞
u(R, t)R dR dθ
R2 (t)
2π Z ∞
R2
M
R dR dθ
exp αt −
=
4Dt
0
R2 (t) 4πDt
Z
Z
w M eαt 2π ∞ 1
=
dw dθ
exp −
4πDt 0
4Dt
R22 (t) 2
Z
R22 (t)
M eαt 2π
dθ
exp −
=
2π
4Dt
0
2π
M eαt
R22 (t)
=
θ
exp −
2π
4Dt
0
R22 (t)
= M exp αt −
.
4Dt
Z
(b) We want to find R2 (t) satisfying N (t) ≤ m ≤ M .
R22 (t)
m
R22 (t)
≤ m ⇒ exp αt −
≤
M exp αt −
4Dt
4Dt
M
m
m
R2 (t)
≤ ln
⇒ αt − 2
⇒ 4Dαt2 − R22 (t) ≤ 4D ln
t
4Dt
M
M
m
√
t ≪ αt2 , so R22 (t) ≈ 4Dαt2 and R2 (t) ≈ 4αDt, making the speed of wave
For large t, ln
√ M
propagation 4αD.
(c) If α = 0, there is no traveling wave solution, and so the speed of propagation in either case is zero.
√
2. (a) The wave front speed is 2 αD ≈ 3.3 meters/year.
(b) At the wave front speed in part (a), the total distance covered in 20, 000 years would be about 66 km,
which is significantly less than the actual distance of 1000 km. The oaks spread much faster than the
model predicted.
(c) Other factors that could contribute to the faster invasion by oaks in Britain are weather and other
external influences (e.g. animals) that cause the acorns to be carried farther than the estimated
dispersal distance.
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