This document summarizes a lesson on higher dimensional difference equations. It discusses: 1) Linear systems described by equations of the form y(k+1) = Ay(k) and their solutions involving eigenvalues and eigenvectors of A. 2) Qualitative analysis of diagonal systems based on the magnitudes of the eigenvalues determining behaviors like attraction, repulsion, or saddle points. 3) The nonlinear case where equilibria y* are found as solutions to g(y*)=y* and stability is determined by eigenvalues of the Jacobian matrix Dg(y*) evaluated at the equilibria.

1. Lesson 31
First Order, Higher Dimensional Difference
Equations
Math 20
April 30, 2007
Announcements
PS 12 due Wednesday, May 2
MT III Friday, May 4 in SC Hall A
Final Exam: Friday, May 25 at 9:15am, Boylston 110 (Fong
Auditorium)

2. Recap
Higher dimensional linear systems
Examples
Markov Chains
Population Dynamics
Solution
Qualitative Analysis
Diagonal systems
Examples
Higher dimensional nonlinear

3. one-dimensional linear difference equations
Fact
The solution to the inhomogeneous difference equation
yk+1 = ayk + b
(with a = 1) has solution
b b
yk = ak y0 − +
1−a 1−a
Please try not to memorize this. When a and b have actual
values, it’s either to follow this process:
1. Start with ak times an undetermined parameter c (this
satisfies the homogenized equation)
2. Find the equilibrium value y∗ .
3. Add the two and pick c to match y0 when k = 0.

4. Nonlinear equations
sl
op
e
=
−
Fact 1
slo
The equilibrium pe
=g y0
point y∗ of the
(y
∗)
nonlinear
y2
difference
equation
yk+1 = g(yk ) is y1
stable if
1
|g (yk )| < 1. =
e
op
sl

5. Recap
Higher dimensional linear systems
Examples
Markov Chains
Population Dynamics
Solution
Qualitative Analysis
Diagonal systems
Examples
Higher dimensional nonlinear

6. Let’s kick it up a notch and look at the multivariable, linear,
homogeneous difference equation
y(k + 1) = Ay(k )
(we move the index into parentheses to allow y(k ) to have
coordinates and to avoid writing yk,i .)

7. Skipping class
Example
This example was a Markov chain with transition matrix
0.7 0.8
A=
0.3 0.2
Then the probability of going or skipping on day k satisfies the
equation
p(k + 1) = Ap(k )

8. Example
Female lobsters have more eggs each season the longer they
live. For this reason, it is illegal to keep a lobster that has laid
eggs.
Let yi be the number of lobsters in a fishery which are i years
alive. Then the difference equation might have the simplified
form  
0 100 400 700
0.1 0 0 0
y(k + 1) =   y(k)
 0 0.3 0 0
0 0 0.9 0

9. Mmmm. . . Lobster

10. Formal solution
y(1) = Ay(0)
y(2) = Ay(1) = A2 y(0)
y(3) = Ay(2) = A3 y(0)
So

11. Formal solution
y(1) = Ay(0)
y(2) = Ay(1) = A2 y(0)
y(3) = Ay(2) = A3 y(0)
So
Fact
The solution to the homogeneous system of linear difference
equations y(k + 1) = Ay(k) is
y(k) = Ak y(0)

12. Flop count
To multiply two n × n matrices takes n3 (n − 1) additions or
multiplications (flop=floating point operation)

13. Flop count
To multiply two n × n matrices takes n3 (n − 1) additions or
multiplications (flop=floating point operation)
So finding Ak takes about n4k flops!

14. Now what?
Suppose v is an eigenvector of A with eigenvalue λ . Then the
solution to the problem
y(k + 1) = Ay(k), y(0) = v
is

15. Now what?
Suppose v is an eigenvector of A with eigenvalue λ . Then the
solution to the problem
y(k + 1) = Ay(k), y(0) = v
is
y(k ) = λ k v

16. Now what?
Suppose v is an eigenvector of A with eigenvalue λ . Then the
solution to the problem
y(k + 1) = Ay(k), y(0) = v
is
y(k ) = λ k v
Suppose
y(0) = c1 v1 + c2 v2 + · · · + cm vm

17. Now what?
Suppose v is an eigenvector of A with eigenvalue λ . Then the
solution to the problem
y(k + 1) = Ay(k), y(0) = v
is
y(k ) = λ k v
Suppose
y(0) = c1 v1 + c2 v2 + · · · + cm vm
Then
Ay(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm
A2 y(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm
2 2 2

18. Now what?
Suppose v is an eigenvector of A with eigenvalue λ . Then the
solution to the problem
y(k + 1) = Ay(k), y(0) = v
is
y(k ) = λ k v
Suppose
y(0) = c1 v1 + c2 v2 + · · · + cm vm
Then
Ay(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm
A2 y(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cm λm vm
2 2 2
If A is diagonalizable, we can take m = n and write any initial
vector as a linear combination of eigenvalues.

19. The big picture
Fact
Let A have a complete system of eigenvalues and eigenvectors
λ1 , λ2 , . . . , λn and v1 , v2 , . . . , vn . Then the solution to the
difference equation y(k + 1) = Ay(k) is
y(k ) = Ak y(0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cn λn vn
k k k
where c1 , c2 , . . . , cn are chosen to make
y(0) = c1 v1 + c2 v2 + · · · + cn vn

20. Recap
Higher dimensional linear systems
Examples
Markov Chains
Population Dynamics
Solution
Qualitative Analysis
Diagonal systems
Examples
Higher dimensional nonlinear

21. Iterating diagonal systems
Consider a 2 × 2 matrix of the form
λ1 0
D=
0 λ2
Then the λ ’s tell the behavior of the system.

22. Picture in terms of eigenvalues
λ1 > λ2 > 1: repulsion away from the origin

23. Picture in terms of eigenvalues
λ1 > λ2 > 1: repulsion away from the origin
1 > λ1 > λ2 > 0: attraction to the origin

24. Picture in terms of eigenvalues
λ1 > λ2 > 1: repulsion away from the origin
1 > λ1 > λ2 > 0: attraction to the origin
λ1 > 1 > λ2 : saddle point

25. Picture in terms of eigenvalues
λ1 > λ2 > 1: repulsion away from the origin
1 > λ1 > λ2 > 0: attraction to the origin
λ1 > 1 > λ2 : saddle point

26. Picture in terms of eigenvalues
λ1 > λ2 > 1: repulsion away from the origin
1 > λ1 > λ2 > 0: attraction to the origin
λ1 > 1 > λ2 : saddle point
For negative eigenvalues just square them and use the above
results.

27. Back to skipping class
Example
If
0.7 0.8
A=
0.3 0.2

28. Back to skipping class
Example
If
0.7 0.8
A=
0.3 0.2
The eigenvectors (in decreasing order of absolute value) are
−1
8/11
1
with eigenvalue 1 and 12 with eigenvalue − 10 .
3/11
2

29. Back to skipping class
Example
If
0.7 0.8
A=
0.3 0.2
The eigenvectors (in decreasing order of absolute value) are
−1
8/11
1
with eigenvalue 1 and 12 with eigenvalue − 10 . So the
3/11
2
8/11
system converges to a multiple of 3 .
/11

30. Back to the lobsters
We had  
0 100 400 700
0.1 0 0 0
A= 
 0 0.3 0 0
0 0 0.9 0
The eigenvalues are 3.80293, −2.84895, −0.476993 +
1.23164i, −0.476993 − 1.23164i and the first eigenvector is
T
0.999716 0.0233099 0.00489153

31. Back to the lobsters
We had  
0 100 400 700
0.1 0 0 0
A= 
 0 0.3 0 0
0 0 0.9 0
The eigenvalues are 3.80293, −2.84895, −0.476993 +
1.23164i, −0.476993 − 1.23164i and the first eigenvector is
T
0.999716 0.0233099 0.00489153
The population will grow despite the increased harvesting!

32. Recap
Higher dimensional linear systems
Examples
Markov Chains
Population Dynamics
Solution
Qualitative Analysis
Diagonal systems
Examples
Higher dimensional nonlinear

33. The nonlinear case
Consider now the nonlinear system
y(k + 1) = g(y(k)).
The process is as it was with the one-dimensional nonlinear:
1. Look for equilibria y∗ with g(y∗ ) = y∗
2. Linearize about the equilibrium using the matrix
∂ gi
A = Dg(y∗ ) =
∂ yj
3. The eigenvalues of A determine the stability of y∗ .