APM2614
ASSIGNMENT 03
Fixed Closing Date: 01 August 2025
,Solution to Question 1
Problem Statement
Consider the following plane autonomous system:
ẋ = x − y
ẏ = x + y − 2xy.
Find the nature of all the singular points and sketch the phase plane diagram with the trajectories and all
the isoclines.
Step 1: Find the Singular Points
Singular points occur where ẋ = 0 and ẏ = 0. Set the equations to zero:
x − y = 0 =⇒ y = x, (1)
x + y − 2xy = 0. (2)
Substitute y = x from equation (1) into equation (2):
x + x − 2x · x = 2x − 2x2 = 2x(1 − x) = 0.
Solve for x:
x=0 or x = 1.
Using y = x:
• If x = 0, then y = 0, giving the singular point (0, 0).
• If x = 1, then y = 1, giving the singular point (1, 1).
Thus, the singular points are (0, 0) and (1, 1).
Step 2: Linearize the System
To determine the nature of the singular points, compute the Jacobian matrix of the system:
f (x, y) = x − y, g(x, y) = x + y − 2xy.
The Jacobian matrix is: " # " #
∂f ∂f
∂x ∂y 1 −1
J= ∂g ∂g
= .
∂x ∂y 1 − 2y 1 − 2x
Evaluate J at each singular point.
1
, Step 3: Analyze the Singular Point at (0, 0)
At (0, 0): " # " #
1 −1 1 −1
J(0, 0) = = .
1−2·0 1−2·0 1 1
Find the eigenvalues by solving det(J − λI) = 0:
" #
1−λ −1
J − λI = .
1 1−λ
det(J − λI) = (1 − λ)(1 − λ) − (−1)(1) = (1 − λ)2 + 1 = λ2 − 2λ + 2.
Solve the quadratic equation: √ √
2± 4−8 2 ± −4
λ= = = 1 ± i.
2 2
The eigenvalues are λ = 1 ± i, which are complex with positive real part (Re(λ) = 1 > 0). Thus, (0, 0) is an
unstable spiral point, as trajectories spiral outward.
Step 4: Analyze the Singular Point at (1, 1)
At (1, 1): " # " #
1 −1 1 −1
J(1, 1) = = .
1−2·1 1−2·1 −1 −1
Find the eigenvalues: " #
1−λ −1
J − λI = .
−1 −1 − λ
det(J − λI) = (1 − λ)(−1 − λ) − (−1)(−1) = (1 − λ)(−1 − λ) − 1.
= λ2 − (1 − 1)λ − 1 − 1 = λ2 + 2.
√
λ2 = −2 =⇒ λ = ± 2i.
√
The eigenvalues are λ = ± 2i, which are purely imaginary. This suggests a center at (1, 1), where
trajectories form closed orbits.
Step 5: Compute the Isoclines
dy ẏ
Isoclines are curves where the slope dx = ẋ is constant:
dy x + y − 2xy
= .
dx x−y
• Horizontal isocline (ẋ = 0):
x − y = 0 =⇒ y = x.
• Vertical isocline (ẏ = 0):
x
x + y − 2xy = 0 =⇒ x(1 − 2y) + y = 0 =⇒ y(1 − 2x) = −x =⇒ y = .
2x − 1
2
ASSIGNMENT 03
Fixed Closing Date: 01 August 2025
,Solution to Question 1
Problem Statement
Consider the following plane autonomous system:
ẋ = x − y
ẏ = x + y − 2xy.
Find the nature of all the singular points and sketch the phase plane diagram with the trajectories and all
the isoclines.
Step 1: Find the Singular Points
Singular points occur where ẋ = 0 and ẏ = 0. Set the equations to zero:
x − y = 0 =⇒ y = x, (1)
x + y − 2xy = 0. (2)
Substitute y = x from equation (1) into equation (2):
x + x − 2x · x = 2x − 2x2 = 2x(1 − x) = 0.
Solve for x:
x=0 or x = 1.
Using y = x:
• If x = 0, then y = 0, giving the singular point (0, 0).
• If x = 1, then y = 1, giving the singular point (1, 1).
Thus, the singular points are (0, 0) and (1, 1).
Step 2: Linearize the System
To determine the nature of the singular points, compute the Jacobian matrix of the system:
f (x, y) = x − y, g(x, y) = x + y − 2xy.
The Jacobian matrix is: " # " #
∂f ∂f
∂x ∂y 1 −1
J= ∂g ∂g
= .
∂x ∂y 1 − 2y 1 − 2x
Evaluate J at each singular point.
1
, Step 3: Analyze the Singular Point at (0, 0)
At (0, 0): " # " #
1 −1 1 −1
J(0, 0) = = .
1−2·0 1−2·0 1 1
Find the eigenvalues by solving det(J − λI) = 0:
" #
1−λ −1
J − λI = .
1 1−λ
det(J − λI) = (1 − λ)(1 − λ) − (−1)(1) = (1 − λ)2 + 1 = λ2 − 2λ + 2.
Solve the quadratic equation: √ √
2± 4−8 2 ± −4
λ= = = 1 ± i.
2 2
The eigenvalues are λ = 1 ± i, which are complex with positive real part (Re(λ) = 1 > 0). Thus, (0, 0) is an
unstable spiral point, as trajectories spiral outward.
Step 4: Analyze the Singular Point at (1, 1)
At (1, 1): " # " #
1 −1 1 −1
J(1, 1) = = .
1−2·1 1−2·1 −1 −1
Find the eigenvalues: " #
1−λ −1
J − λI = .
−1 −1 − λ
det(J − λI) = (1 − λ)(−1 − λ) − (−1)(−1) = (1 − λ)(−1 − λ) − 1.
= λ2 − (1 − 1)λ − 1 − 1 = λ2 + 2.
√
λ2 = −2 =⇒ λ = ± 2i.
√
The eigenvalues are λ = ± 2i, which are purely imaginary. This suggests a center at (1, 1), where
trajectories form closed orbits.
Step 5: Compute the Isoclines
dy ẏ
Isoclines are curves where the slope dx = ẋ is constant:
dy x + y − 2xy
= .
dx x−y
• Horizontal isocline (ẋ = 0):
x − y = 0 =⇒ y = x.
• Vertical isocline (ẏ = 0):
x
x + y − 2xy = 0 =⇒ x(1 − 2y) + y = 0 =⇒ y(1 − 2x) = −x =⇒ y = .
2x − 1
2
