Math 5120
Homework 5 Solutions
1. (a) For the system
v
cvN
cvN0
dG
=− G−
+
dz
D
D
D
dN
kN G
=−
,
dz
v
the steady states of the form (G∗ , N ∗ ) are P1 = (cN0 , 0) and P2 = (0, N0 ). The Jacobian
v
0
and λ2 = − ckN
evaluated at P1 has eigenvalues λ1 = − D
v . Assuming v > 0, P1 is a
stable node. The Jacobian evaluated at P2 has a √
negative trace and a negative determinant,
−v+
v2 +4DckN
0
making this point a saddle. Here, λ1,2 =
. The G-nullcline is given by
2D
1
N = − c G + N0 , and the N -nullclines are G = 0 and N = 0.
Nullcines for G (green) and N (black); steady states are red
N
N0
0
0
G
cN0
(b) The solution of the ODE system that corresponds to the traveling wave in the PDE is that
which connects the two steady states, leaving from (0, N0 ) and approaching (cN0 , 0). This
is the only solution that will satisfy the boundary conditions. As mentioned in part (a), v
must be positive in order for such a solution to exist.
2. (a) A predator-prey system in which both species move randomly in a 1d setting may be described
by
∂2u
∂u
= Du 2 ± αu − γuv
∂t
∂x
∂2v
∂v
= Dv 2 ± βu + δuv,
∂t
∂x
where u represents the prey, v the predator, and all parameters are positive.
(b) A predator-prey system in which the predator (v) moves toward higher prey (u) densities and
the prey moves toward lower predator densities may be described by
∂2u
∂
∂v
∂u
= Du 2 ± αu − γuv +
ku u
∂t
∂x
∂x
∂x
2
∂ v
∂
∂u
∂v
= Dv 2 ± βu + δuv −
kv v
.
∂t
∂x
∂x
∂x
1
(c) A pair of reacting and diffusing chemicals such that species 1 activates the formation of both
substances and species 2 inhibits the formation of both substances may be described by
∂ 2 s1
∂s1
= D1 2 + α1 s1 − β1 s2
∂t
∂x
∂ 2 s2
∂s2
= D2 2 + α2 s1 − β2 s2 ,
∂t
∂x
where si represents the amount of species i for i = 1, 2. All parameters are assumed to be
positive.
3. The original equation for bacterial chemotaxis is
∂B
∂
∂c
∂
∂B
=−
χB
+
µ
.
∂t
∂x
∂x
∂x
∂x
(a) If the chemotactic sensitivity increases linearly with the chemical concentration, then the
sensitivity term χ will be proportional to c. A modified equation may be
∂
∂c
∂
∂B
∂B
=−
ξcB
+
µ
.
∂t
∂x
∂x
∂x
∂x
(b) If random motion decreases as the cell density increases, this can be represented by the
inclusion of a decreasing function of B in the diffusion term. An example may be the
following:
µ ∂B
∂
∂c
∂
∂B
=−
χB
+
.
∂t
∂x
∂x
∂x B ∂x
(c) If the cell population increases logistically with the carrying capacity proportional to the
concentration of the chemical, then we can add a logistic term to account for this behavior.
A modified equation can be
∂B
∂
∂c
∂
∂B
B
=−
χB
+
µ
+ rB 1 −
.
∂t
∂x
∂x
∂x
∂x
kc
4. Given the model
u
,
ut = Duxx + ku 1 −
K
consider a traveling wave solution such that u(x, t) = ϕ(x − ct). Then with k = 1 and K = 1,
the above PDE becomes the following ODE:
−cϕ′ = Dϕ′′ + ϕ(1 − ϕ),
where ϕ′ denotes the derivative with respect to wave variable z = x − ct. To determine the wave
speed, rewrite the second-order ODE as a system two first-order ODEs.
ϕ′ = ψ
ψ′ = −
c
1
ψ − ϕ(1 − ϕ)
D
D
The steady states (ϕ∗ , ψ ∗ ) of this system are P1 = (0,
√0) and P2 = (1, 0). The eigenvalues
1
of the Jacobian evaluated at P1 are λ1,2 = 2D (−c ± c2 − 4D). At P2 , the eigenvalues are
√
1
(−c ± c2 + 4D). Therefore, P2 is a saddle and P1 is either a stable node or a stable
λ1,2 = 2D
2
spiral for c > 0. Note that if P1 is a stable spiral, then ϕ will oscillate between negative and
positive values near the fixed point. Therefore, we require that P1 be a stable node in order for a
biologically relevant traveling wave solution to exist in the full system.
Subject to this condition,
√
it must be the case that c2 − 4D ≥ 0, which implies that c ≥√2 D. Thus, the minimum wave
speed that will support traveling waves in this model is c∗ = 2 D.
(a) For D = 100, the minimum wave speed is c∗ = 20, and so it will take at least 5 months for
the dingo population to reach the farm 100 miles away.
√
(b) If, instead, D = 50, the minimum wave speed becomes c∗ = 10 2, and it will take at least
7 months for the dingo population to reach the farm 100 miles away.
3