APM2614 Assignment 3 2026 Applied Dynamical Systems Due 31 July 2026

UNIVERSITY OF SOUTH AFRICA (UNISA)
Department of Mathematical Sciences



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APPLIED DYNAMICAL SYSTEMS
Assignment 03 — Year Module 2026


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Module Code: APM2614

Module Name: Applied Dynamical Systems

Assignment No.: 03

Due Date: 31 July 2026

Semester: Year Module 2026

Unique Number: [Available on myUnisa]




Submitted in partial fulfilment of the requirements for Applied Dynamical Systems
at the University of South Africa.

,UNISA | APM2614 Applied Dynamical Systems — Assignment 03



Question 1

Consider the plane autonomous system


ẋ = y, ẏ = x − x3 .


Find the nature of all the singular points and sketch the phase plane diagram with the trajec-
tories and all the isoclines.


1.1 Singular Points


A singular point satisfies ẋ = 0 and ẏ = 0.

Since ẋ = y, this implies y = 0.

Since ẏ = x − x3 , setting this to zero gives


x − x3 = 0 =⇒ x(1 − x2 ) = 0.


Therefore x = 0 or x = ±1.

The singular points are
(−1, 0), (0, 0), (1, 0).


1.2 Jacobian Matrix


Let f (x, y) = y and g(x, y) = x − x3 . Then

∂f ∂f ∂g ∂g
= 0, = 1, = 1 − 3x2 , = 0.
∂x ∂y ∂x ∂y

Hence  
0 1
J(x, y) =  .
1 − 3x2 0




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,UNISA | APM2614 Applied Dynamical Systems — Assignment 03



1.3 Classification of (0, 0)


At x = 0,  
0 1
J = .
1 0

The characteristic equation is


−λ 1
=0 =⇒ λ2 − 1 = 0 =⇒ λ = ±1.
1 −λ

The eigenvalues have opposite signs, implying that (0, 0) is a saddle point, which is unstable.


1.4 Classification of (1, 0)


At x = 1, 1 − 3(1)2 = −2, so  
0 1
J = .
−2 0

The characteristic equation is

√
λ2 + 2 = 0 =⇒ λ = ±i 2.


The eigenvalues are purely imaginary, implying that (1, 0) is a centre, stable but not asymp-
totically stable.


1.5 Classification of (−1, 0)


At x = −1, 1 − 3(−1)2 = −2, giving the same Jacobian as above:
 
0 1 √
J = , λ = ±i 2.
−2 0

Therefore (−1, 0) is also a centre, stable but not asymptotically stable.




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,UNISA | APM2614 Applied Dynamical Systems — Assignment 03



1.6 Isoclines


Horizontal isoclines occur where ẏ = 0, that is


x − x3 = 0 =⇒ x = −1, x = 0, x = 1.


These are three vertical lines in the phase plane.

Vertical isoclines occur where ẋ = 0, that is y = 0, the x-axis.


1.7 Direction of Motion


Since ẋ = y: if y > 0 then ẋ > 0, so trajectories move right; if y < 0 then ẋ < 0, so trajectories
move left.

Since ẏ = x − x3 , testing each region:


• For x < −1 (e.g. x = −2): ẏ = −2 − (−8) = 6 > 0, trajectories move upward.
• For −1 < x < 0 (e.g. x = −0.5): ẏ = −0.5 − (−0.125) = −0.375 < 0, trajectories move
downward.
• For 0 < x < 1 (e.g. x = 0.5): ẏ = 0.5 − 0.125 = 0.375 > 0, trajectories move upward.
• For x > 1 (e.g. x = 2): ẏ = 2 − 8 = −6 < 0, trajectories move downward.


1.8 Phase Plane Diagram

y




(−1, 0) (0, 0) (1, 0)
x
Centre Saddle Centre




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, UNISA | APM2614 Applied Dynamical Systems — Assignment 03


Key Distinction
The phase portrait shows a saddle at the origin with separatrices along the eigenvector
directions, surrounded by two families of closed orbits (centres) at (−1, 0) and (1, 0).
The dashed vertical lines x = −1, 0, 1 are the horizontal isoclines, and the x-axis is the
vertical isocline.



1.9 Summary Table


Table 1: Classification of singular points
Singular Point Eigenvalues Nature
(0, 0) 1, −1
√ Saddle point (unstable)
(1, 0) ±i√2 Centre (stable)
(−1, 0) ±i 2 Centre (stable)




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