Math 5110/6830
Homework 6.1 Solutions
1. (a) The fixed points are x = −1, x = 0 and x = 1. From the graph, we see that x = −1 and
x = 1 are both stable, while x = 0 is unstable.
Vector field (left) and x(t) for different initial conditions (right)
1.5
2
1.5
1
1
0.5
x(t)
ẋ
0.5
0
0
−0.5
−0.5
−1
−1
−1.5
−2
−1.5
−1
−0.5
0
0.5
x
1
−1.5
1.5
0
0.5
1
1.5
2
t
2.5
3
3.5
4
(b) The fixed points are at kπ for k ∈ Z. For k even or 0, the fixed point is unstable. For k
odd, the fixed point is stable.
ẋ from x = −4π to x = π in its entirety
4
x 10
9
8
7
6
ẋ
5
4
3
2
1
0
−12
−10
−8
−6
x
−4
−2
0
2
4
4
3
3
2
2
1
1
x(t)
ẋ
Vector field (close-up; left) and x(t) for different initial conditions (right)
0
−3π
−2π
0
−π
0
−1
−1
−2
−2
−3
−3
−4
−4
−12
−10
−8
−6
x
−4
−2
0
2
0
2
4
6
8
10
t
12
14
16
2. An equation that is consistent with the given phase portrait is ẋ = x(x − 2)(x + 1)2 .
1
18
20
4
3
2
ẋ
1
0
−1
−2
−3
−4
−1.5
−1
−0.5
0
0.5
1
x
1.5
2
2.5
3. Separating variables gives
dN
N 1−
N
K
= rdt
⇒
K
dN = rdt.
N (K − N )
K
Using partial fraction decomposition, we find that N (K−N
) =
1
1
+
dN = rdt
N
K −N
Z 1
1
+
dN = rt + c1
N
K −N
ln N − ln |K − N | = rt + c1
N = rt + c1
ln K −N
1
N
+
1
K−N ,
which gives
N0
N
= cert ⇒ c =
K −N
K − N0
N
N0
=
ert .
K −N
K − N0
Solving this for N yields the solution N (t) =
KN0 ert
.
K + N0 (ert − 1)
4. The fixed point of this equation is N = 1b , and it is stable. Note that Ṅ is undefined for N ≤ 0,
1
so N = 0 is not a fixed point. In the following pictures, a = 1 and b = 20
.
Vector field (left) and N (t) for different initial conditions (right)
8
25
6
20
4
15
N(t)
dN/dt
2
0
10
1
b
−2
5
−4
−6
0
5
10
15
20
0
25
N
2
0
1
2
3
4
5
t
6
7
8
9
10
Homework 6.2 Solutions
2
1. (a) The fixed points satisfy 1 − e−x = 0, to which x = 0 is the solution. Let f (x) = ẋ. Then
f 0 (0) = 0, which gives us no stability information. So, we need to analyze this graphically.
Based on the graph below, this fixed point is semistable (unstable).
1
0.9
0.8
0.7
ẋ
0.6
0.5
0.4
0.3
0.2
0.1
0
−2
−1.5
−1
−0.5
0
x
0.5
1
1.5
2
(b) The fixed points satisfy ln x = 0, to which x = 1 is the solution. Let f (x) = ẋ. Then
f 0 (1) = 1 > 0, and therefore the fixed point is unstable.
(c) The fixed points satisfy x(1 − x)(2 − x) = 0, to which x = 0, x = 1 and x = 2 are solutions.
Let f (x) = ẋ.
For x = 0: f 0 (0) = 2 > 0, and so x = 0 is an unstable fixed point.
For x = 1: f 0 (1) = −1 < 0, and so x = 1 is a stable fixed point.
For x = 2: f 0 (2) = 2 > 0, and so x = 2 is an unstable fixed point.
√
√
(d) The fixed points satisfy ax − x3 = 0, to which x = 0, x = a and x = − a are solutions.
Note that there is only one fixed point for a ≤ 0. In general, f 0 (x) = a − 3x2 , where
f (x) = ẋ. The following table outlines the stability of the fixed points for various a values.
f 0 (x∗ )
−2a
a
−2a
a<0
only 1 FP
stable
only 1 FP
a=0
only 1 FP
stable (see graph)
only 1 FP
Case where a = 0
1
0.8
0.6
0.4
0.2
ẋ
x∗
√
− a
0
√
a
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
3
0
x
0.5
1
a>0
stable
unstable
stable
Bifurcation Diagram for a
4
3
2
x*
1
0
−1
−2
−3
−4
−10
−8
−6
−4
−2
4
0
a
2
4
6
8
10