6.4 Difference Equations
In a difference equation the unknowns are one or more sequences of numbers, e.g.
x0, x1, x2, x3, …, xn,
y0, y1, y2, y3, …, yn,
In applications we are often observing some system that is changing with time and x and
y represent physical quantities that change with time. Instead of observing x and y at all
times we only observe them at equally spaced discrete times and n represents the nth time
we observe x and y. For example, we might observe the rabbit and fox population in a
region once a year on July 1 and xn and yn are the rabbit and fox populations on July 1 of
year n where n counts years from some starting year.
The difference equations express xn+1 and yn+1 in terms of x0, x1, x2, x3, …, xn and y0, y1, y2,
y3, …, yn. In applications the formulas for xn+1 and yn+1 are usually based on physical
considerations. In this section we shall restrict our attention to the situation where xn+1
and yn+1 are given by linear formulas in terms of xn and yn. In addition to the difference
equations we are often given the values of x0 and y0. These starting values are called
initial conditions.
Example 1.
Consider the difference equations
(1)
xn+1 = xn + 3yn
(2)
yn+1 = 4xn + 2yn
along with the initial conditions x0 = 2 and y0 = 1.
It is always possible to use the difference equations along with the initial conditions to
calculater the values of xn and yn for any particular value of n.
Example 1 (continued). Find x2 and y2 in Example 1.
Substituting n = 1 and x0 = 2 and y0 = 1 into (1) and (2) we get
x1 = x0 + 3y0 = 2 + (3)(1) = 5
y1 = 4x0 + 2y0 = (4)(2) + (2)(1) = 10
x2 = x1 + 3y1 = 5 + (3)(10) = 35
y2 = 4x1 + 2y1 = (4)(5) + (2)(10) = 40
If possible we would like to "solve" the difference equations to get formulas for xn and yn.
In the case where the difference equations are linear we can do this using the powers of
the coefficient matrix. We illustrate this technique with the equations in Example 1.
We begin by writing (1) and (2) in vector form.
6.4 - 1
 xn+1  =  xn + 3yn  =  1 3   xn 
 yn+1 
 4xn + 2yn 
 4 2   yn 
or
(3)
un+1 = Aun
where
x
un =  ynn 
1 3
A =  4 2 
x0
2
The initial conditions can be expressed as  y  =  1  or u0
0
2
=  1 . The equation (3) for n = 0
says
u1 = Au0
For n = 1 the equation (3) says u2 = Au1. Putting in u1 = Au0 for u1 on the right we get
u2 = A(Au0) = A2u0
For n = 2 the equation (3) says u3 = Au2. Putting in u2 = A2u0 for u2 on the right we get
u3 = A(A2u0) = A3u0
Continuing in this fashion we see that
un = Anu0
So
n
n
 xn  =  1 3   x0  =  1 3   2 
 yn 
 4 2   y0 
4 2 1
n
1 3
1 3  5 + 4  (- 2)
In the previous section we saw that  4 2  = 7  4  5n - 4  (- 2)n
n
n
n
n
n
n
 xn  = 1  3  5n + 4  (- 2)n 3  5n - 3  (- 2) n  2 
 yn 
7  4  5 - 4  (- 2)
4  5 + 3  (- 2)  1 
n
n
1
9

5
+
5

(2)
= 7  12  5n - 5  (- 2)n 
So
xn =
9  5n + 5  (- 2)n
7
7
yn =
12  5n - 5  (- 2)n
7
7
6.4 - 2
3  5n - 3  (- 2)n 
.
4  5n + 3  (- 2)n 
So
In general the solutions of homogeneous, constant coefficient linear difference equations
have the form c1(1)n + c2(2)n +  + cm(m)n where 1, 2, …, m, are the eigenvalues of
the coefficient matrix.
Here is a physical situation that gives rise to difference equations.
Example 2. ZipCo has done a study of employee absenteeism. They found that from
one day to the next
20% of those absent one day return to work the next
10% of those at work one day are absent the next
Suppose today 100 employees are absent and 800 employees are at work. Find a formula
for the number of employees absent and at work n days from now.
Let
sn = the number of employees absent on day n
wn = the number of employees at work on day n
The given information translates into the following difference equations.
sn+1 = 0.8sn + 0.1wn
wn+1 = 0.2sn + 0.9wn
along with the initial conditions s0 = 100 and w0 = 800. In vector form one has
 sn+1  =  0.8 0.1   sn 
 wn+1 
 0.2 0.9   wn 
 s0  =  100 
 w0   800 
So
n
 sn  =  0.8 0.1   s0 
 wn 
 0.2 0.9   w0 
0.8 0.1
We need to find the eigenvalues and eigenvectors of A =  0.2 0.9 .
0.8 - 
0.1 
0 = det( A - I ) = 
= (0.8 - )(0.9 - ) – (0.1)(0.2)
 0.2
0.9 -  
= 2 - 1.7 + 0.7 = ( - 1)( - 0.7)
So the eigenvalues are
1 = 1
and
2 = 0.7
x
An eigenvector v =  y  for 1 = 1 satisfies
6.4 - 3
 0  = (A - I)v =  - 0.2 0.1   x 
0
 0.2 - 0.1   y 
So
0 = - 0.2x + 0.1y
0 =
0.2x - 0.1y
x
1
So an eigenvector for 1 = 1 is any vector v =  y  with y = 2x. So any multiple of the vector  2  is an
x
eigenvector for 1 = 1. An eigenvector v =  y  for 2 = 0.7 satisfies
 0  = (A - I)v =  0.1
0
 0.2
0.1   x 
0.2   y 
0 = 0.1x + 0.1y
0 = 0.2x + 0.2y
1
These equations are equivalent to x = - x. So any multiple of the vector  - 1  is an eigenvector for
2 = 0.7. So
-1
n
n
0 1 1
 0.8 0.1  = TDnT -1 =  1 1   1
 0.2 0.9 
 2 - 1   0 (0.7)n   2 - 1 
1
(0.7)n
-1 -1 -1
1 1
(0.7)n
1 1
=  2 - (0.7)n   3   - 2 1  = 3  2 - (0.7)n   2 - 1 
1 1 + 2  (0.7)
= 3  2 - 2  (0.7)n
n
So
 sn  = 1  1 + 2  (0.7)n
 wn 
3  2 - 2  (0.7)
n
1 - (0.7)n 
2 + (0.7)n 
1 - (0.7)n  s0 
2 + (0.7)n  w0 
1 (s + w ) + (2s - w )  (0.7)
= 3  2(s0 + w0 ) - (2s 0- w 0)  (0.7)n 
0
0
0
0
So
1
2s - w
sn = 3 (s0 + w0) +  0 3 0  (0.7)n
2
2s - w
wn = 3 (s0 + w0) -  0 3 0  (0.7)n
In the case that s0 = 100 and w0 = 800 one has
sn = 300 - 200  (0.7)n
wn = 600 + 200  (0.7)n
Note that (0.7)n  0 as n  . So as n   one has
6.4 - 4
n
sn  300
wn  600
in the case where the intial conditions are s0 = 100 and w0 = 800 and
1
sn  3 (s0 + w0)
2
wn  3 (s0 + w0)
in the case of general initial conditions. So in the long run, about 1/3 of the total number
of employees are absent on any particular day and 2/3 are at work.
6.4 - 5