APM2614 Assignment 3 (Accurate Solutions) 01 August 2025

APM2614

ASSIGNMENT 03

Fixed Closing Date: 01 August 2025

, Assignment 03 Solutions



Question 1
Consider the following plane autonomous system:
(
ẋ = x − y
ẏ = x + y − 2xy

• Find the nature of all the singular points.
• 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
Solving the system:
x−y =0 ⇒ x=y
2
x + y − 2xy = 0 ⇒ x + x − 2x = 0 ⇒ 2x(1 − x) = 0 ⇒ x = 0 or x = 1
Hence, the singular points are:
(0, 0) and (1, 1)

Step 2: Linearization via Jacobian
Define:
f (x, y) = x − y, g(x, y) = x + y − 2xy
Compute the Jacobian matrix:
" #
∂f ∂f  
∂x ∂y 1 −1
J(x, y) = ∂g ∂g =
∂x ∂y
1 − 2y 1 − 2x

At (0, 0):
 
1 −1
J(0, 0) =
1 1
Find eigenvalues from:

1−λ −1
det(J − λI) = = (1 − λ)2 + 1 = 0 ⇒ (1 − λ)2 = −1 ⇒ λ = 1 ± i
1 1−λ

Conclusion: Complex eigenvalues with positive real part Unstable spiral point.




1

, At (1, 1):
 
1 −1
J(1, 1) =
−1 −1


1−λ −1 √
det(J − λI) = = (1 − λ)(−1 − λ) − 1 = −(1 − λ2 ) − 1 = λ2 − 2 ⇒ λ = ± 2
−1 −1 − λ

Conclusion: Real eigenvalues of opposite signs Saddle point.

Step 3: Isoclines
• Vertical isocline: ẋ = 0 ⇒ x = y
• Horizontal isocline: ẏ = 0 ⇒ x + y − 2xy = 0 ⇒ y = x
2x−1


Final Answer (Question 1)
• Singular points: (0, 0) and (1, 1)
• (0, 0): Unstable spiral point

• (1, 1): Saddle point
• Isoclines:
– Vertical: y = x
x
– Horizontal: y = 2x−1




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