Math 5120
Homework 4.1 Solutions
1. (a) The level curves of the Hamiltonian for the spruce budworm model show that (0, 0) is a
center.
Hamiltonian and level curves
1.5
4
1
0.5
2
0
v
H(u,v)
3
1
−0.5
0
−1
2
1
−1.5
2
1
0
0
−1
−2
−2
−1
v
−2
−2
−1.5
−1
−0.5
u
0
u
0.5
1
1.5
(b) The approximate steady state solution of the full PDE model for D = 0.25, µ = 1 and l = 2
is satisfied by the conditions u(0) = 0 and v(0) = 0.7499 ≈ 0.75. This yields u(2) = 0, as
desired.
0.45
0.4
0.35
0.3
u(x)
0.25
0.2
0.15
0.1
0.05
0
0
0.2
0.4
0.6
0.8
1
x
1.2
1.4
1.6
1.8
2
(c) For larger µ covering the same distance (l = 2), v(0), and consequently the maximum value
of u(x), must also increase. If the population is changing more quickly, then u(x) will grow
faster, reach a maximum value, and then decrease as x approaches l. Since u(x) is increasing
faster for larger µ, this in turn requires that v(x) be higher initially in order to compensate,
particularly if a density of u(x) = 0 is to be attained within the same amount of space, that
is at l = 2. For example, if µ = 1.25, then v(0) must be 1.026–up from 0.75–in order for
u(2) = 0 to be satisfied.
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Homework 4.2 Solutions
1. (a) The system of ODEs that describes the steady-state solution of ut = uxx + u(1− u)(u− 1/2)
is
u′ = v
v ′ = −u(1 − u)(u − 1/2),
where u′ and v ′ are derivatives with respect to x. The corresponding boundary conditions
are v(0) = v(l) = 0.
(b) The Jacobian of the linearized system is
0
J(u, v) =
3u2 − 3u +
1
2
1
0
The equilibria of this system and corresponding stability are as follows:
q
• P1 = (0, 0): The eigenvalues of J(0, 0) are λ1,2 = ± 12 , and so this point is a saddle.
• P2 = ( 12 , 0): The eigenvalues of J( 12 , 0) are λ1,2 = ± 21 i, and so this point is a possible
center. (This can be verified by looking at the level curves of the Hamiltonian found in
part (c)).
q
• P3 = (1, 0): The eigenvalues of J(1, 0) are λ1,2 = ± 12 , and so this point is also a
saddle.
∂H
′ ′
(c) Recall that a Hamiltonian is some function H(u, v) satisfying ∇H = ∂H
,
∂u ∂v = (−v , u ).
To determine H(u, v), we integrate.
3
1
∂H
= −v ′ = −u3 + u2 − u
∂u
2
2
1
1
1
⇒ H(u, v) = − u4 + u3 − u2 + c1 (v)
4
2
4
1
∂H
′
= u = v ⇒ H(u, v) = v 2 + c2 (u)
∂v
2
1 2 1 4 1 3 1 2
⇒ H(u, v) = v − u + u − u
2
4
2
4
0.5
0.4
0.3
0.2
v
0.1
0
−0.1
−0.2
−0.3
−0.4
−0.5
−0.5
0
0.5
u
2
1
1.5
(d) The (u, v) = (u, ux ) plane has the phase portrait shown below.
0.8
0.6
0.4
v
0.2
0
−0.2
−0.4
−0.6
−0.8
−0.5
0
0.5
u
1
1.5
(u0, v0) = (0.45, 0)
(u0, v0) = (0.55, 0)
0≤ x≤ 2π
0≤ x≤ 2π
0.54
0.54
0.52
0.52
u(x)
u(x)
(e) The possible steady-state solutions satisfying the homogeneous Neumann boundary conditions for 0 ≤ x ≤ l = 2π are shown as functions of x below.
0.5
0.5
0.48
0.48
0.46
0.46
0.02
−0.005
0.015
−0.01
v(x)
v(x)
0
0.01
0.005
0
−0.015
−0.02
0
2
4
6
x
0
2
4
6
x
(f) These steady-state solutions reflect the membrane potential along the length of the axon.
3