Homework 5.1 Solutions
Math 5110/6830
1. (a) For
dx
dt
= x − x3
Phase-line diagram:
Fixed points are x∗ = 0 and x∗ = ±1. Stability analytically (you can get this from your phase-line
diagram too). Let
f (x) = x − x3
f 0 (x) = 1 − 3x2
f 0 (0) = 1
f 0 (1) = −2
f 0 (−1) = −2
Then, x∗ = 0 is unstable and both x = 1 and x = −1 are stable.
Sketch of x(t):
3
2
x(t)
1
0
−1
−2
−3
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
To solve ẋ = x − x3 , we can use separation of variables.
Z
x(t)
x0
Z
x(t)
x0
1
dx = dt
x − x3
Z t
1
dx̄ =
dt̄
x̄ − x̄3
0
1
dx̄ = t
x̄(1 − x̄)(1 + x̄)
Using partial fractions:
1
1
1
2
2
+
−
dx̄
x̄ 1 − x̄ 1 + x̄
x0
x(t)
1
1
ln(x̄) − ln(1 − x̄) − ln(1 + x̄)
2
2
x0
x(t)
x̄
ln √
1 − x̄2 x0
x(t)
x̄
ln √
1 − x̄2 x0
!
x
x0
ln √
− ln p
1 − x2
1 − x20
1/2
x2
ln
1 − x2
x2
1
ln
2
1 − x2
x2
ln
1 − x2
x2
1 − x2
Z
x(t)
2
x
x2
1+
x20 2t
e
1 − x20
= t
= t
= t
= t
= t
=
=
=
=
=
2
1/2
x20
1 − x20
x20
1
t + ln
2
1 − x20
x20
2t + ln
1 − x20
x20 2t
e
1 − x20
x20 2t
e (1 − x2 )
1 − x20
x20 2t
e
1 − x20
= t + ln
x20
e2t
1−x20
x
=
x(t)
=
x0 et
p
(1 − x20 + x20 e2t )
x20
2t
1 + 1−x
2e
0
v
u
2
x0
u
e2t
1−x20
u
x(t) = t x20
2t
1 + 1−x
2e
0
s
x20 e2t
x(t) =
(1 − x20 + x20 e2t )
(b) For
dx
dt
= e−x sin(x)
Phase-line diagram:
Fixed points are x∗ = nπ. Stability analytically (you can get this from your phase-line diagram too).
Let
f (x)
f 0 (x)
= e−x sin(x)
= −e−x sin(x) + e−x cos(x)
Then, x∗ = nπ is unstable for n even and stable for n odd.
Sketch of x(t):
This cannot be solve analytically.
2. To find an equation for this phase-line diagram, we can notice that we will have roots x = −1, x = 0 and
x = 2. So, the equation will take the form
dx
dt
= x(x + 1)(x − 2)
But, to get the correct stability, we can check concavity. We need the function to be concave up at x = −1,
concave down at x = 0 and concave up at x = 2:
f (x) = x(x + 1)(x − 2)
f 0 (x) = 3x2 − 2x − 2
f 00 (x) = 6x − 2
f 00 (−1) = −8
So, this fails. Lets think about it though. If we want this equation to be concave up at x = −1 then the
2
term (x + 1) should be squared. Let’s try dx
dt = x(x + 1) (x − 2):
f 00 (x) = 12x2 − 6
f 00 (−1) = 6
f 00 (0) = −6
f 00 (2) = 42
Then, this works! So, an equation to match the phase-line diagram is:
dx
dt
= x(x + 1)2 (x − 2)
3. We can use separation of variables for this equation:
dN
dt
Z
N (t)
N0
1
dN
N
1
dN̄
N̄
N (t)
ln(N̄ )N0
= K(t)N
= K(t)dt
Z t
=
K(t̄)dt̄
0
t
Z
=
K(t̄)dt̄
0
Z t
ln(N (t)) − ln(N0 ) =
K(t̄)dt̄
0
Z t
N (t)
=
ln
K(t̄)dt̄
N0
0
Rt
N (t)
= e( 0 K(t̄)dt̄)
N0
Rt
N (t) = N e( 0 K(t̄)dt̄)
0
4. Solution using separation of variables:
1
N 1−
Z
N (t)
N̄ 1 −
= rN
dN
= rdt
dN̄
=
Z
1
N0
N
K
dN
dt
N̄
K
1−
N
K
t
rdt̄
0
Using partial fractions:
Z
N (t)
N0
"
1
1
+ K N̄
N̄
1− K
#
Z
dN̄
t
rdt̄
=
0
Now integrate:
ln
N (t)
N̄
ln N̄ − ln 1 −
K N0
!N (t)
N̄
ln
N̄
1− K
N0
!
!
N (t)
N0
− ln
1 − NK0
1 − NK(t)


1 − NK0 N (t)

ln  1 − NK(t) N0
= rt
= rt
= rt
= rt
K − N0
N (t) = N0 ert
K − N (t)
(K − N0 )N (t) = N0 ert (K − N (t))
(K − N0 + N0 ert )N (t) = N0 Kert
N0 Kert
N (t) =
K − N0 + N0 ert
N0 ert
N (t) =
N0 rt
1 + K (e − 1)
Solution using the substitution x = N1 :
With x = N1 , or equivalently N = x1 , we have:
dN
dt
= −
1 dx
x2 dt
Substituting this into the logistic eqn:
−
1 dx
x2 dt
dx
dt
1
1 1
1−
x
Kx
1
= r
−x
K
= r
Now separate variables and integrate:
Z t
1
dx̄
=
rdt̄
1
x0
0
K − x̄
x(t)
1
− ln
− x̄
= rt
K
x0
1
1
− ln
− x(t) + ln
− x0
= rt
K
K
1
− x0
ln 1K
= rt
K − x(t)
1 − Kx0
= ert
1 − Kx(t)
1 − Kx(t) = (1 − Kx0 )e−rt
1
x(t) =
[1 − (1 − Kx0 )e−rt ]
K
Z
x(t)
Returning to the original variables:
1
N (t)
=
N (t)
=
=
=
=
5. Phase-line diagram:
Solution:
1
K
1− 1−
e−rt
K
N0
K
1 − 1 − NK0 e−rt
Kert
ert − 1 +
K
N0
rt
N0 e
− NK0 + 1
N0 rt
K e
N0 ert
1 + NK0 (ert − 1)
Homework 5.2 Solutions
Math 5110/6830
1. (a) Fixed points satisfy:
dx
=0 =
dt
2
1 − e−x
The only fixed point is x∗ = 0. To determine its stability:
f (x)
=
2
1 − e−x
2
f 0 (x) = 2xe−x
f 0 (0) = 0
From this, we cannot determine the stability of x∗ = 0. Therefore, we will need to do this graphically:
As you can see, this point is half-stable.
(b) Fixed points satisfy:
dx
=0 =
dt
ln(x)
The only fixed point is x∗ = 1. To determine its stability:
f (x)
=
ln(x)
1
f 0 (x) =
x
f 0 (1) = 1
And, x∗ = 1 is unstable.
(c) Fixed points satisfy:
dx
=0
dt
= x(1 − x)(2 − x)
The fixed points are x∗ = 0, x∗ = 1, and x∗ = 2. To determine their stability:
f (x)
f 0 (x)
f 0 (0)
f 0 (1)
f 0 (2)
=
=
=
=
=
x(1 − x)(2 − x)
(1 − x)(2 − x) − x(2 − x) − x(1 − x)
2
−1
2
And, x∗ = 0 is unstable, x∗ = 1 is stable and x∗ = 2 is unstable.
(d) Fixed points satisfy:
dx
= 0 = ax − x3 = x(a − x2 )
dt
√
√
The fixed points are x∗ = 0, x∗ = − a, and x∗ = a. To determine their stability, let
f (x)
= ax − x3
Then, for a > 0
f 0 (x) = a − 3x2
f 0 (0) = a
√
f 0 (− a) = −2a
√
f 0 ( a) = −2a
√
√
For a > 0, x∗ =
0 is unstable, x∗ = − √
a is stable and x∗ = a is stable. But for a < 0, x∗ = 0 is
√
stable, x∗ = − a is unstable and x∗ = a is unstable. And, for a = 0, there is only one fixed point
x∗ = 0 which we need to do a graphical stability analysis for:
Looks stable to me.
2. (a) To determine stability, let
= −xc
and
f 0 (x) = −cxc−1
f (x)
Then, for c = 1 this is stable. However, we need to do this graphically for other values of c: ????
(b) To find an equation for time here, we can separate variables and solve this DE:
dx
dt
= −xc
Z x(t)
Z t
1
dx̄ =
−dt̄
x̄c
x0
0
Z x(t)
Z t
x̄−c dx̄ =
−dt̄
x0
0
−c+1
x̄
−c + 1
x(t)
= −t
x0
x−c+1
0
x(t)−c+1
−
−c + 1
−c + 1
x(t)−c+1
= −t
= x−c+1
− t(−c + 1)
0
−c+1
1
x(t) = x0
− t(−c + 1) −c+1
Letting x0 = 1:
x(t)
=
1
[1 − t(−c + 1)] −c+1
Now, to find the time it takes til x(t) = 0:
1
x(t) = 0 = [1 − t(−c + 1)] −c+1
0 = 1 − t(−c + 1)
1
t =
1−c
3. To show that there are infinite solutions to ẋ = x1/3 , we can find at least 2 solutions that pass through the
same point. Notice that x(t) = 0 is one solution that passes through the point (0, 0). The other solution,
found by separation of variables, is x(t) = 23 x3/2 . This solution also passes through the point (0, 0).