APPM 2360 HW 11 Solutions
Fall 2014
6.4.2: Consider
#
x0 =
0 1
−1 0
#
x
Computing eigenvalues,
(−λ)(−λ) − (−1) = λ2 + 1 ≡ 0
and so λ1,2 = ±i, and accordingly α = 0. Therefore, the equilibrium point at the origin is a center
point and is neutrally stable.
6.4.5: Consider
#
x0 =
2 1
3 4
#
x
Computing eigenvalues,
(2 − λ)(4 − λ) − 3 = λ2 − 6λ + 5
= (λ − 5)(λ − 1) ≡ 0
and so λ1,2 = 5, 1. Since λ1 , λ2 > 0, then the equilibrium point at the origin is a repelling node/node
source, which is an unstable fixed point.
6.4.8: Consider
mẍ + bẋ + kx = 0,
m, b, k > 0
b
k
k
b
Then let y = ẋ, so y 0 = ẍ = − ẋ − x = − x − y. Solving for nullclines, we derive that (0, 0) is an
m
m
m
m
equilibrium point. Therefore,
# " 0
1
#
ẋ
x
ẋ =
=
k
b
ẏ
y
−
−
m
m
Computing the eigenvalues of the above matrix,
−λ
1
b
k
b
k
= −λ − − λ +
−
m
m
− −λ m
m
Simplifying, we obtain,
λ=
−b/m ±
p
(b/m)2 − 4k/m
2
Considering cases, if
√
ˆ 0 < b < 4mk, then (0, 0) is underdamped.
√
ˆ b = 4mk, then (0, 0) is critically damped.
√
ˆ b > 4mk, then (0, 0) is overdamped.
1
6.4.10: Consider the system,
#
x0 =
0 0
1 1
#
x
(a): Computing the eigenvalues,
(1 − λ)(−λ) = λ(1 − λ) ≡ 0
and so λ1,2 = 0, 1.
x02
(b): For vertical nullclines, x01 = 0 = 0, so there is no nullcline. For horizontal nullclines,
= −x1 +x2 = 0 and therefore nullclines at x1 = x2 , or all points on the line x = y are equilibrium points.
(c): Consider
(A − 0I) =
0 0 0
−1 1 0
which becomes
−x1 + x2 = 0
=⇒ x1 = x2 = s
and therefore
v# = s
1
1
1
Similarly,
−1 0 0
−1 0 0
(A − I) =
which becomes
−x1 = 0
=⇒ x2 = s
and therefore
v# = s
1
0
1
So the general solution is
#
x =c
1
1
1
t
+ c2 e
(d): Direction never changes in the x direction,
2
0
1
6.4.17: Consider the system
#
x0 =
k 0
0 −1
#
x
Solving for eigenvalues,
k−λ
0
0
−1 − λ
= (k − λ)(−1 − λ) ≡ 0
Then the eigenvalues are λ1 = −1 and λ2 = k.
(a): If k < −1, then λ2 < λ1 < 0, so our equilibrium point is an asymptotically stable attracting node.
(b): If k = −1, then λ2 = λ1 < 0, and therefore our equilibrium point is an attracting star
node, stable.
(c): If −1 < k < 0, then we have λ1 < λ2 < 0, and our equilibrium point is an asymptotically stable attracting node.
(d): If k = 0, then λ2 = 0 and there is a row of nonisolated fixed points.
(e): If k > 0, then λ2 > 0 and so λ1 < 0 < λ2 . Therefore, our equilibrium point is an unstable saddle point.
6.4.22: Consider T r(A) 6= 0, so depending on sign, either attracting or repelling. Then
(T r(A))2 − 4|A| < 0
which implies complex eigenvalues, indicating a spiral.
6.4.27: Consider T r(A) < 0, and |A| > 0, then our solution is asymptotically stable, since
eigenvalues are both real and negative.
7.1.18: Consider the following system,
x0 = y − x2 + 1
y 0 = y + x2 − 1
Then solving for the equilibrium point, x0 = 0 = y − x2 + 1, and y 0 = 0 = y + x2 − 1, indicating vertical
nullclines at y = x2 −1 and horizontal nullclines at y = 1−x2 . Therefore, equilibrium points exist at (±1, 0).
Plotting the solution,
3
Hence, (−1, 0) is an unstable spiral, and (1, 0) is an unstable saddle.
7.1.25: Consider the equation
ẍ + (ẋ2 − 1)ẋ + x = 0
(a): Let ẋ = y so that
ẏ = ẍ = (1 − y 2 )y − x
Accordingly,
ẋ = 0 = y
ẏ = 0 = (1 − y 2 )y − x
implying that x = y − y 3 . Therefore, equilibrium at (0, 0). Plotting the solution,
(c): (0, 0) is an unstable repelling spiral, and x(t) = 0 is an unstable solution.
(d): Trajectories around the origin are periodic.
4
7.1.31: Consider the system
ẋ = 2xy
ẏ = y 2 − x2 − 1
Then the vertical nullclines correspond to 2xy = 0, and the horizontal nullclines correspond to y 2 = x2 + 1.
Therefore, the equilibrium points are located at (0, 1) unstable, (0, −1) stable.
Plotting the solution,
7.2.4: Consider the following system,
x0 = y
y 0 = − sin(x) − y
Solving for nullclines, then we observe that y = 0 and y = − sin(x), and accordingly we have equilibrium
points at (0, 0) and (±nπ, 0).
Computing the Jacobian, then
J(x, y) =
0
1
− cos(x) −1
The Jacobian at J(0, 0) is
J(0, 0) =
0
1
−1 −1
Solving for eigenvalues of the above matrix, then we obtain
√
3
1
i
λ1,2 = − ±
2
2
and hence (0, 0) is an attracting spiral.
The Jacobian at (±nπ, 0) is
J(±nπ, 0) =
where the eigenvalues are
0 1
1 −1
√
1
5
λ1,2 = − ±
2
2
and hence (nπ, 0) is an unstable saddle for n ∈ 2N + 1, and is an attracting spiral for n = 2N.
Plotting the solution,
5
7.2.8: Consider the following system,
x0 = x − 3y + 2xy
y 0 = 4x − 6y − xy
Solving for nullclines, we set
x − 3y + 2xy = 0
4x − 6y − xy = 0
2 2
Then we find equilibrium points at (0, 0) and
,
by solving the above.
3 5
Computing the Jacobian,
J(x, y) =
1 + 2y −3 + 2x
4 − y −6 − x
Evaluating at (0, 0), then
J(0, 0) =
1 −3
4 −6
with corresponding eigenvalues λ1,2 = −2, −3. Hence, (0, 0) is an attracting node.
2 2
Similarly, evaluating the Jacobian at
,
provides
3 5


9
5
2 2
 5 −3 
,
=  18
J
20 
3 5
−
5
3
√
1
5
2 2
with corresponding eigenvalues λ1,2 = − ±
, and hence
,
is an unstable saddle. Plotting
2
2
3 5
the solution,
6
7.2.11: Consider the following equation,
ẍ + ẋ + x + x3 = 0
Let y = x0 , and then y 0 = −y − x − x3 . Solving for nullclines yields our equilibrium point at (0, 0).
Computing the Jacobian,
J(x, y) =
0
1
−1 − 3x2 −1
Evaluating at (0, 0), then we arrive at
J(0, 0) =
with corresponding eigenvalues λ1,2
the solution,
0
1
−1 −1
√
1
3
= − ±i
, and hence (0, 0) is an attracting spiral. Plotting
2
2
7.2.19: Consider the following equation,
ẍ + ẋ + ẋ3 + x = 0
Let y = ẋ, then y 0 = ẍ = −y − y 3 − x. SOlving for nullclines, we derive that (0, 0) is an equilibrium point.
Computing the Jacobian,
J(x, y) =
0
1
−1 −1 − 3y 2
7
Evaluating the Jacobian at (0, 0), then
J(0, 0) =
0
1
−1 −1
√
1
with corresponding eigenvalues λ1,2 = − ± i 3, and hence (0, 0) is an attracting spiral. Plotting
2
the solution,
8