1.3 Solutions to Dynamical Systems
Dynamical systems are typically described implicitly by providing a rule for the change
of the system from time t to time 𝑡 + 1. Given the evolution rule, dynamical systems are
straight-forward to evolve but it can be difficult to summarily describe their general
behavior.
Understanding the ‘general behavior’ of a dynamical system is a vague phrase, but might
include any of the following:



Writing down an explicit description (a formula for the nth term of the sequence
that only requires n as an input)
An understanding of the long-term behavior of the dynamical system for an
initial condition a0 (can always be achieved by evolving the system numerically
or considering the explicit description in the limit as 𝑛 → ∞)
Identifying the set of fixed points (equilibrium values) and classifying their
stability
Analytic Methods for Investigating the Behavior of a Dynamical System
First, we’ll look at analytic methods for understanding the behavior of dynamical systems
and then in the next class project we’ll combine these with numerical methods.
Analytic methods include finding the explicit description of a dynamical system, which
we can do for all linear dynamical systems, or finding the fixed points, which can be
accomplished more generally.
Analytic Methods for Finding the Explicit Description of Linear Dynamical Systems
I. Of a Geometric Sequence
II. Of an Arithmetic Sequence
III. Of a Mixed Sequence
HW1.3 (1bef)
Analytic Methods for Finding the Fixed Points of Linear Dynamical Systems
I. Solving for fixed points by solving ∆𝑎𝑛 = 0
II. Discussing stability
HW1.3 (2bcegh)
I. Of a Geometric Sequence
an1  ra n , a0  a0
General Solution:
Example:
a0  a0
a1  ra0
an1  3an , a0  1
a2

an
 ra1  r ra0   r 2 a0
The solution (explicit description) is
a n  3 n a0  3 n .
 r n a0
II. Of an Arithmetic Sequence
an1  an  b, a0  a0
General Solution:
a0  a0
a1  a0  b
a2  a1  b 

an
 a0  nb
Example:
an1  an  1.1, a0  2
a0  b   b  a0  2b
The solution (explicit description) is
an  2  1.1n .
III. Of a Mixed Sequence
an1  ran  b, a0  a0
General Solution:
a0  a0
a1  ra0  b
a2  ra1  b  r ra0  b   b  r 2 a0  rb  2b
a3


 ra 2  b  r r 2 a0  rb  2b  b


 r 3 a0  r 2b  rb  b

an1  r n1a0  r n  r n1    r  1 b

Sn
To find an explicit description for Sn:
S n  r n  r n 1    1
rS n  r r n  r n 1    1  r n 1  r n    r
rS n  r n 1  r n    r   r n 1  S n  1
rS n  S n  r n 1  1
S n (r  1)  r n 1  1
r n 1  1
S 
r 1
n
So we have:
an 1  r
 r n 1  1 
br n 1
b
b 
b

n 1
b  r a0 
a0  

 r n 1  a0 

r 1 r 1
r 1 r 1

 r 1 
n 1
or more simply:
a n  k1r n  k 2 where k1, k2 are constants
And we can solve a system of equations to find these constants.
Example:
an1  2an  1, a0  3
(the sequence has terms 3, 5, 9, 17, …)
a n  k1 2  k 2 and we can solve for the constants using:
n
a0  k1 2 0  k 2  k1  k 2  3 

  k1  2, k 2  1
1
a1  k1 2  k 2  2k1  k 2  5

n
so a n  2  2  1 .
Compare the common method of shifting a sequence in order to find the
explicit description:
To find an explicit description for Sn:
S n  r n  r n 1    1
rS n  r r n  r n 1    1
rS n  r n 1  r n    r
rS n  r n 1  r n    r 
rS n  r n 1  S n  1
rS n  S n  r n 1  1
S n (r  1)  r n 1  1
r n 1  1
S 
r 1
n
To express 0.22.. as a fraction:
0. 2  x
10  0. 2  10 x
2.2  10 x
2  0. 2  10 x
2  x  10x
2  9x
2
x
9