Mathsdifferential Equations.pdf

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Section 1.4
33
Difference Equations
At this point almost all of our sequences have had explicit formulas for their terms. That is, we have looked
mainly at sequences for which we could write the
nth
term as
a
n
=
f
(
n
) for some known function
f.
For example, if
n
+1
-
a
n
= --------------
n
2
+ 3
11
101
-
-
then it is an easy matter to compute explicitly, say,
a
10
= -------- or
a
100
= -------------- . In such cases we are able to com-
103
10003
pute any given term in the sequence without reference to any other terms in the sequence. However, it is often the case
in applications that we do not begin with an explicit formula for the terms of a sequence; rather, we may know only
some relationship between the various terms. For example, we may know an equation which expresses the value of
a
n
as a function of the terms
a
1
,
a
2
, …,
a
n
– 1
. Such an equation is called a
difference equation.
If we can find a function
f
such that
a
n
=
f
(
n
) , then we will have
solved
the difference equation. In this section we will consider a class of dif-
ference equations that are solvable in this sense; in the next section we will discuss an example where an explicit solu-
tion is not possible.
Example
Suppose a certain population of owls is growing at the rate of 2% per year. If we let
x
0
represent the size of
the initial population of owls and
x
n
the number of owls
n
years later,
n
= 1
,
2
,
3
, …
, then
x
n
+ 1
=
x
n
+ 0.02x
n
= 1.02x
n
(1.4.1)
for
n
= 0
,
1
,
2
, …
. That is, the number of owls in any given year is the number of owls in the previous year plus 2%
of that number. Equation
(1.4.1)
is an example of a difference equation; it says that the number of owls in a given year
is determined by the number of owls in the previous year. Hence we know the value of a specific
x
n
only if we know
the value of
x
n
– 1
. In particular, to get the sequence started we would have to know the value of
x
0
. For example, if
initially we have a population of
x
0
= 100 owls and we want to know what the population will be after 4 years, we
may compute
x
1
= 1.02x
0
=
(
1.02
) (
100
)
= 102
x
2
= 1.02x
1
=
(
1.02
) (
102
)
= 104.04
x
3
= 1.02x
2
=
(
1.02
) (
104.04
)
= 106.1208
x
4
= 1.02x
3
=
(
1.02
) (
106.1208
)
= 108.243216.
Thus we would expect about 108 owls in the population after 4 years.
Note that we may work backwards to find
x
4
explicitly in terms of
x
0
:
34
Section 1.4
x
4
= 1.02x
3
=
(
1.02
)
1.02x
2
=
(
1.02
) (
1.02
)
1.02x
1
=
(
1.02
) (
1.02
) (
1.02
)
1.02x
0
= 1.02
4
x
0
.
This is interesting because it indicates that we can compute
x
4
without reference to the values of
x
1
,
x
2
, and
x
3
, pro-
vided, of course, that we know the value of
x
0
. If we do this in general, then we have solved the difference equation
x
n
+ 1
= 1.02x
n
. Namely, we have
x
n
= 1.02x
n
– 1
= 1.02
2
x
n
– 2
= 1.02
3
x
n
– 3
:
= 1.02
n
x
0
.
For example, if
x
0
= 100 as above, then we can compute
x
20
= 1.02
20
(
100
) ≅
149
owls, or even
x
100
= 1.02
100
(
100
) ≅
724
owls, without having to compute any intermediate values..
(1.4.2)
700
600
500
400
300
200
100
0
20
40
60
80
100
Figure 1.4.1 Plot of
(
n
,
x
n
)
where
x
n
= 1.02
n
(
100
)
Section 1.4
35
For a geometric feeling of how the population is changing with time, Figure 1.4.1 shows a plot of the points
(
n
,
x
n
)
for
n
= 0
,
1
,
2
, …,
100 . Of course, whether or not our model will provide an accurate prediction of the owl
population 100 or 200 years into the future is an entirely different question. Frequently, a simple population model
like this will be valid only for a short span of time during which the rate of growth of population remains stable.
By replacing 1.02 with an arbitrary constant
α
in
(1.4.2),
we arrive at the general result that the solution of
the difference equation
x
n
+ 1
=
αx
n
,
n
= 0
,
1
,
2
, …
is given by
x
n
=
α
n
x
0
,
n
= 0
,
1
,
2
, …
.
(1.4.4)
(1.4.3)
Note that this difference equation, and its solution, are useful whenever we are interested in a sequence of numbers
where the
(
n
+ 1
)
st term is a constant proportion of the
nth
term. Our first example, where a population was
assumed to grow at a constant rate, is a common example of this type of behavior. Another common example is when
a quantity decreases at a constant rate over time. This behavior is discussed in the next example in the context of
radioactive decay.
Example
Radium is a radioactive element which decays at a rate of 1% every 25 years. This means that the amount
left at the beginning of any given year is equal to the amount at the beginning of the previous year minus 1% of that
amount. That is, if
x
0
is the initial amount of radium and
x
n
is the amount of radium still remaining after 25n years,
then
x
n
+ 1
=
x
n
– 0.01x
n
= 0.99x
n
(1.4.5)
for
n
= 0
,
1
,
2
, …
. Since this is a difference equation of the form of
(1.4.3)
with
α
= 0.99 , we know that the solu-
tion is of the form
(1.4.4).
Namely,
x
n
= 0.99
n
x
0
for
n
= 0
,
1
,
2
, …
. For example, the amount left after 100 years is given by
x
4
= 0.99
4
x
0
= 0.9606x
0
,
where we have rounded the answer to four decimal places. That is, approximately 96% of the initial amount of
radium will be left after 100 years. A plot of the amount of radium left versus number of years, assuming an initial
amount of 500 grams, is given in Figure 1.4.2.
The half-life of a radioactive element is the number of years required for one-half of an initial amount to
decay. In this example, if 25N is the half-life of radium and
x
0
is the initial amount, then we would have
1
--
x
0
= 0.99
N
x
0
,
-
2
(1.4.6)
36
Section 1.4
600
500
400
300
200
100
0
2000
4000
6000
8000
10000
Figure 1.4.2 Amount of radium versus number of years
which implies
1
1
1
-
-
-- = 0.99
N
⇒
log
--
= log
10
(
0.99
N
) ⇒
log
--
=
N
log
10
(
0.99
)
.
-
10
2
10
2
2
Solving for
N
here would give us
1
-
log
--
10
2
----------------------------- = 68.98 ,
-
log
10
(
0.99
)
rounding to two decimal places. Since we are working with sequences,
N
must be an integer. Hence we will round this
result to the nearest integer and use
N
= 69 . Recalling that we are working with 25 year units of time, this shows
that the half-life of radium is approximately
(
25
) (
69
)
= 1725 years. For example, this means that if we started
with an initial amount of 100 grams of radium, after 1725 years we would still have 50 grams left. It would then take
an additional 1725 years until the remaining amount would be reduced to 25 grams.
Although we have stated the results of the preceding example in discrete time units, namely, units of 25
years each, later we will see that the results hold for continuous time as well. In other words, although the difference
equation
(1.4.5)
has been set up for nonnegative integer values of
n,
the solution
(1.4.6)
is valid for arbitrary nonneg-
ative values of
n.
But we will hold off discussion of these ideas until we discuss differential equations in Chapter 6.
It is interesting to compare the plots in Figures 1.4.1 and 1.4.2. The first is an example of
exponential
growth,
whereas the second is an example of
exponential decay.
In the first, the steepness of the graph increases with
time; in the second, the graph flattens out over time, showing that the rate of decay is decreasing. The difference
equation
(1.4.3)
will always lead to the first behavior when
α >
1 and to the second when
α <
1 .
Section 1.4
37
First-order linear difference equations
A difference equation of the form
x
n
+ 1
=
αx
n
+
β,
n
= 0
,
1
,
2
, …
,
(1.4.7)
where
α
and
β
are constants, is called a
first-order linear difference equation.
Note that the difference equation
(1.4.3)
is of this form with
β
= 0 . A procedure analogous to the method we used to solve
(1.4.3)
will enable us to
solve this equation as well. Namely,
x
1
=
αx
0
+
β
implies that
x
2
=
αx
1
+
β
=
α ( αx
0
+
β )
+
β
=
α
2
x
0
+
αβ
+
β
=
α
2
x
0
+
β ( α
+ 1
)
.
Similarly,
x
3
=
αx
2
+
β
=
α  α
2
x
0
+
β ( α
+ 1
)
.
+
β
=
α
3
x
0
+
α
2
β
+
αβ
+
β
=
α
3
x
0
+
β ( α
2
+
α
+ 1
)
and
x
4
=
αx
3
+
β
=
α  α
3
x
0
+
β ( α
2
+
α
+ 1
) 
+
β
=
α
4
x
0
+
α
3
β
+
α
2
β
+
αβ
+
β
=
α
4
x
0
+
β ( α
3
+
α
2
+
α
+ 1
)
.
Continuing in this fashion for
n
steps would give us
x
n
=
α
n
x
0
+
β ( α
n
– 1
+
α
n
– 2
+
…
+
α
2
+
α
+ 1
)
for
n
= 0
,
1
,
2
, …
. Now in our discussion of geometric series in Section 1.3 we saw that
1+
α
+
α
2
+
…
+
α
n
– 2
α
n
– 1
1 –
α
n
= --------------
-
1–
α
(1.4.8)
+

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