Difference Equations to Differential Equations — Section 1.5: Nonlinear Difference Equations
Difference Equations
to
Differential Equations
Section 1.5
Nonlinear Difference Equations
In Section 1.4 we discussed the difference equation
xn+1 = αxn ,
(1.5.1)
n = 0, 1, 2, . . ., as a model for either growth or decay and we saw that its solution is given
by
xn = αn x0 ,
n = 0, 1, 2, . . .. Now
( 0,
lim αn =
n→∞
0 < α < 1,
1,
α = 1,
∞, α > 1,
(1.5.2)
from which it follows that if {xn } is a solution of (1.5.1) with x0 > 0, then
( 0,
lim xn = x0 lim αn =
n→∞
n→∞
0 < α < 1,
x0 , α = 1,
∞, α > 1.
(1.5.3)
These limiting values are consistent with our radioactive decay example since, in that case,
0 < α < 1 and we would expect the amount of a radioactive element to decline toward
0 over time. The case 0 < α < 1 also may make sense for a population model if the
population is declining and heading toward extinction. However, the unbounded growth
indefinitely into the future implied by the case α > 1 is very unlikely for a population
model: eventually ecological or even sociological problems come to the forefront, such as
when the population begins to overreach the resources available to it, and the rate of
growth of the population changes. Even for bacteria growing in a Petri dish, diminishing
food and space eventually cause a change in the rate of growth. Hence the equation
xn+1 = αxn ,
(1.5.4)
for n = 0, 1, 2, . . . and α > 1, called the uninhibited, or natural, growth model, although
often accurate as a model of population growth over short periods of time, is usually too
simplistic for predictions over long time spans.
The inhibited growth model
Suppose we wish to model the growth of a certain population which, without ecological
constraints, would grow at a rate of 100β% per unit of time. That is, if xn represents the
1
Copyright c by Dan Sloughter 2000
2
Nonlinear Difference Equations
Section 1.5
size of the population after n units of time and there are no constraints on the size of the
population, then
xn+1 − xn = βxn
(1.5.5)
for n = 0, 1, 2, . . .. However, suppose that, because of the limitation of resources, the
population will begin to decline if it ever has more than M individuals. We call M the
carrying capacity of the available resources, the maximum population which is sustainable
over time. Then it would be reasonable to modify our model by forcing the amount of
increase over a unit of time to decrease as the size of the population approaches M and
to become negative if the size of the population ever exceeds M . One way to accomplish
this is to multiply the term βxn in (1.5.5) by
M − xn
,
M
a ratio which is close to 1 when xn is small, close to 0 when xn is close to M , and negative
when xn exceeds M . This leads us to the difference equation
M − xn
xn+1 − xn = βxn
,
M
n = 0, 1, 2, . . ., or, equivalently,
xn+1 = xn +
β
xn (M − xn ),
M
(1.5.6)
n = 0, 1, 2, . . ., which we call the inhibited growth model, also known as the discrete logistic
equation. This is an example of a nonlinear difference equation because if we multiply
β 2
out the right-hand side of the equation we have a quadratic term, namely, M
xn . Such
equations are, in general, far more difficult to solve than the linear difference equations
we considered in Section 1.4; in fact, many nonlinear difference equations are not solvable
in terms of the elementary functions of calculus. Hence we will not consider any methods
for solving such equations, relying instead on computing specific solutions by iterating the
equation using a calculator or, preferably, a computer.
Example Suppose a population of owls, currently numbering 100, has a natural growth
rate of 4%, but, because of the limited resources of their natural habitat, can sustain a
population of no more than 500. If we let xn represent the size of the population n years
from now, then, using the inhibited growth model, we should have
xn+1 = xn +
0.04
xn (500 − xn ) = xn + 0.00008xn (500 − xn )
500
for n = 0, 1, 2, . . .. Using this equation we are able to compute, for example, the predicted
size of the population for the next 10 years:
Year
Population
0
1
2
3
4
5
6
7
8
9
10
100.0 103.2 106.5 109.8 113.3 116.8 120.3 124.0 127.8 131.5 135.4
Section 1.5
Nonlinear Difference Equations
3
500
400
300
200
100
20
40
60
80
100
120
140
Figure 1.5.1 Inhibited population growth with β = 0.04
Here, and in subsequent tables, we have rounded our results to the first decimal place. It is
interesting to compare these results to the corresponding results for the uninhibited growth
model. If we let yn be the predicted population n years from now using the uninhibited
growth model, then we would have
yn+1 = yn + 0.04yn = 1.04yn ,
n = 0, 1, 2, . . ., which has the exact solution
yn = 100(1.04)n
for n = 0, 1, 2, . . .. From this model we obtain the following predicted population sizes:
Year
Population
0
1
2
3
4
5
6
7
8
9
10
100.0 104.0 108.2 112.5 117.0 121.7 126.5 131.6 136.9 142.3 148.0
As we would expect, the population is growing more slowly under the inhibited population
growth model than under the uninhibited model. Moreover, this difference will become
more pronounced over time. For example, after 150 years we would have x150 = 495.4 and
y150 = 35, 892, illustrating how the inhibited growth model is constrained by the carrying
capacity of 500 while the uninhibited growth model will have unbounded growth. Figures
1.5.1 and 1.5.2 provide a graphical comparison of the two models for n = 0, 1, 2, . . . , 150.
Note that it appears that
lim xn = 500,
n→∞
while
lim yn = ∞.
n→∞
With the inhibited growth model, if 0 < β < 1 and xn < M , then
β
xn
< 1,
M
4
Nonlinear Difference Equations
Section 1.5
35000
30000
25000
20000
15000
10000
5000
20
40
60
80
100
120
140
Figure 1.5.2 Uninhibited population growth with β = 0.04
so
xn+1 = xn + β
xn
(M − xn ) < xn + (M − xn ) = M
M
(1.5.7)
for n = 0, 1, 2, . . .. Thus if 0 < β < 1, and we start with x0 < M , then xn < M for all n.
Moreover, since
xn
β
> 0,
M
we have
xn+1 = xn + β
xn
(M − xn ) > xn
M
for all n. Hence the sequence {xn } is monotone increasing and bounded, and so must have
a limit. In Problem 8 you will be asked to verify that this limit is in fact M , as appeared
to be the case in the previous example.
If β > 1, it may be the case that there are values of n for which xn > M , in which case
β
xn
(M − xn ) < 0
M
and, as a consequence, xn+1 < xn .
Example Suppose x0 = 100 and M = 500 as in the previous example, but now let
β = 1.5. That is,
xn+1 = xn +
1.5
xn (M − xn ) = xn + 0.003xn (500 − xn )
500
for n = 0, 1, 2, . . .. This equation generates the following values:
Year
Population
0
1
2
3
4
5
6
7
8
9
10
100.0 220.0 404.8 520.4 448.5 505.3 497.2 501.4 499.3 500.3 499.8
Section 1.5
Nonlinear Difference Equations
5
500
400
300
200
100
10
20
30
40
50
Figure 1.5.3 Inhibited population growth with β = 1.5
Note how the values increase rapidly (as we should expect with such a large value for β)
to above the carrying capacity of 500, but then oscillate about 500, with the oscillations
diminishing in size. In fact it may be shown that it is also true in this case that
lim xn = 500.
n→∞
See Figure 1.5.3.
It is possible to show that, for the inhibited growth model of (1.5.6),
lim xn = M
n→∞
whenever 0 < β ≤ 2. However, there are other possible behaviors when β > 2.
Example Suppose x0 = 100 and M = 500, as in the previous examples, but now let
β = 2.3. That is,
xn+1 = xn +
2.3
xn (M − xn ) = xn + 0.0046xn (500 − xn )
500
for n = 0, 1, 2, . . .. This equation generates the following values:
Year
Population
0
1
2
3
4
5
6
7
8
9
10
100.0 284.0 566.2 393.8 586.2 353.8 591.7 342.0 590.6 344.5 590.9
Year
Population
11
12
13
14
15
16
17
18
19
20
343.8 590.8 344.0 590.9 343.9 590.8 343.9 590.8 343.9 590.8
Notice that instead of approaching a single limiting value, the population is settling down
to an oscillation between 344 and 591. We say that the sequence {xn } is approaching a
limiting cycle of period 2, as shown in Figure 1.5.4.
6
Nonlinear Difference Equations
Section 1.5
600
500
400
300
200
100
10
20
30
40
50
Figure 1.5.4 Inhibited population growth with β = 2.3
600
500
400
300
200
100
10
20
30
40
50
Figure 1.5.5 Inhibited population growth with β = 2.48
It is possible to obtain limiting cycles of longer periods by increasing β. For example,
Figure 1.5.5 shows the effect of letting β = 2.48. Note that {xn } appears to be approaching
a limiting cycle of period 4.
With appropriate choices for β and x0 , it is in fact possible for the inhibited growth
model to exhibit limiting cycles of any given period. This is related to the fact that it is
possible for this model to behave chaotically. Intuitively, a sequence is chaotic if it displays
erratic behavior which, although in theory completely determined by a difference equation
such as (1.5.6), is in practice unpredictable because small changes in the initial value
x0 yield strikingly different sequences. For example, Figures 1.5.6 and 1.5.7 illustrate
the differing behavior of the inhibited growth model with β = 2.95, first for an initial
population of 100 and then for an initial population of 101.
Section 1.5
Nonlinear Difference Equations
7
700
600
500
400
300
200
100
10
20
30
40
50
Figure 1.5.6 Inhibited population growth with β = 2.95 and x0 = 100
700
600
500
400
300
200
100
10
20
30
40
50
Figure 1.5.7 Inhibited population growth with β = 2.95 and x0 = 101
Problems
1. A population of weasels has a natural growth rate of 3% per year. Let wn be the
number of weasels n years from now and suppose there are currently 300 weasels.
(a) Suppose the carrying capacity of the weasel’s habitat is 1000. Using an inhibited
growth model, write a difference equation which describes how the population
changes from year to year.
(b) Using the difference equation from part (a), compute wn for n = 1, 2, . . . , 150.
(c) How many years will it take for the population to double? To triple?
(d) Plot wn versus n for n = 0, 1, 2, . . . , 150. From the plot, guess lim wn .
n→∞
(e) Compare your answers with those to Problem 3 in Section 1.4.
2. Suppose a population of northern pike in a lake in Montana has a natural growth rate
of 4.5% per year, but the lake can support no more than 10,000 pike. Let pn be the
8
Nonlinear Difference Equations
Section 1.5
number of pike n years from now and suppose p0 = 1000.
(a) Use the inhibited growth model to write a difference equation which describes how
the population changes from year to year.
(b) Using the difference equation from part (a), compute pn for n = 1, 2, 3, . . . 50.
(c) How many years will it take for the population to double? To triple?
(d) Plot pn versus n for n = 0, 1, 2, . . . 200. From the plot, guess lim pn .
n→∞
(e) How many years will it take for the population to reach 9500?
3. Do Problem 2 assuming an uninhibited growth model and no restrictions on the number
of pike that the lake can support.
4. Suppose rn represents the number of snowshoe rabbits in a certain National Forest in
Alaska after n years with an initial value of r0 = 5000. Moreover, suppose the forest
can support no more than 10,000 rabbits and {rn }satisfies the inhibited growth model
rn+1 = rn +
β
rn (10, 000 − rn )
10, 000
for n = 0, 1, 2, . . .. For each of the following values for β, plot rn versus n for n =
0, 1, 2, . . . , 100 and comment on the behavior of the sequence, in particular noting any
limiting values or limiting cycles
(a) β = 0.5
(b) β = 1.5
(c) β = 2.4
(d) β = 2.5
(e) β = 2.56
(f) β = 2.9
5. Using an initial value of x0 = 0.5, let {xn } be the sequence which satisfies the difference
equation
xn+1 = µxn (1 − xn ),
n = 0, 1, 2, . . .. Plot xn versus n for the following values of µ and comment on the
behavior of the sequence, in particular noting any limiting values or limiting cycles.
(a) µ = 0.9
(b) µ = 1.0
(c) µ = 1.5
(d) µ = 2.0
(e) µ = 2.5
(f) µ = 3.0
(g) µ = 3.1
(h) µ = 3.5
(i) µ = 3.57
(j) µ = 1 +
(k) µ = 3.99
(l) µ = 4.0
√
8
6. Repeat Problem 5 starting with an initial value of x0 = 0.6.
7. If f is any function defined for real numbers, then the difference equation
xn+1 = f (xn ),
Section 1.5
Nonlinear Difference Equations
9
n = 0, 1, 2, . . . , is called a discrete dynamical system. For any given initial condition
x0 , the sequence {xn } which satisfies this equation is called an orbit of f . Note that
an orbit of f is simply the sequence of points
x0 , f (x0 ), f (f (x0 )), f (f (f (x0 ))), . . . .
For example, the difference equation in Problem 5 is an example of a discrete dynamical
system with f (x) = µx(1 − x). For each of the following, compute 50 terms of the
given orbit and discuss its behavior.
(a) x0 = 10, f (x) = 2x
(c) x0 = 2 f (x) = cos(x)
1
2
(e) x0 = 5, f (x) =
x+
2
x
(g) x0 = 0, f (x) = x2 + 1.0
(i) x0 = 0, f (x) = x2 − 0.8
(k) x0 = 0, f (x) = x2 − 1.9
(b) x0 = 100, f (x) = 0.8x
(d) x0 = 2, f (x) = sin(x)
2x2
(f) x0 = 1, f (x) = 2
3x − 5
(h) x0 = 0, f (x) = x2 − 0.5
(j) x0 = 0, f (x) = x2 − 1.0
(l) x0 = 0, f (x) = x2 − 2.0
8. Assuming that the sequence {xn } satisfying the inhibited growth model equation
xn+1 = xn +
has a limit, show that lim xn = M .
n→∞
β
xn (M − xn )
M