First Order Difference Equations 123

Ify₀is given, then successive terms of the solution can be found from Eq. (3). Thus y₁=f(y₀),

and

y2=f(y1)=f[f(y0)].

The quantity f[f(y0)]is called the second iterate of the difference equation and is sometimes denoted byf²(y₀). Similarly, the third iteratey₃is given by

y₃=f(y₂)=f{f[f(y₀)]} =f³(y₀), and so on. In general, thenth iteratey_nis

y_n=f(y_n−1)=fⁿ(y₀).

This procedure is referred to as iterating the difference equation. It is often of pri- mary interest to determine the behavior of y_n as n→ ∞. In particular, does y_n approach a limit, and if so, what is it?

Solutions for whichy_nhas the same value for allnare calledequilibrium solutions.

They are frequently of special importance,just as in the study of differential equations.

If equilibrium solutions exist, you can find them by settingy_n+1equal toy_nin Eq. (3) and solving the resulting equation

y_n=f(y_n) (4)

fory_n.

Linear Equations. Suppose that the population of a certain species in a given region in yearn+1, denoted byy_n+1, is a positive multipleρ_nof the populationy_nin yearn;

that is,

yn+1=ρ_nyn, n=0, 1, 2,. . . . (5) Note that the reproduction rate ρ_n may differ from year to year. The difference equation (5) is linear and can easily be solved by iteration. We obtain

y1=ρ0y0,

y₂=ρ₁y₁=ρ₁ρ₀y₀, and, in general,

y_n=ρ_n−1· · ·ρ₀y₀, n=1, 2,. . . . (6) Thus, if the initial population y₀ is given, then the population of each succeeding generation is determined by Eq. (6). Although for a population problemρ_nis intrin- sically positive, the solution (6) is also valid ifρ_nis negative for some or all values of n. Note, however, that ifρ_nis zero for somen, theny_n+1and all succeeding values of yare zero; in other words, the species has become extinct.

If the reproduction rateρ_n has the same valueρfor eachn, then the difference equation (5) becomes

y_n+1=ρy_n (7)

and its solution is

y_n=ρⁿy₀. (8)

124 Chapter 2. First Order Differential Equations

Equation (7) also has an equilibrium solution, namely,y_n=0 for alln, corresponding to the initial valuey₀=0.The limiting behavior of y_n is easy to determine from Eq. (8). In fact,

n→∞lim y_n=

⎧⎪

⎨

⎪⎩

0, if|ρ|<1;

y0, ifρ=1;

does not exist, otherwise.

(9) In other words, the equilibrium solutiony_n =0 is asymptotically stable for|ρ|<1 and unstable for|ρ|>1.

Now we will modify the population model represented by Eq. (5) to include the effect of immigration or emigration. Ifb_n is the net increase in population in year ndue to immigration, then the population in year n+1 is the sum of the part of the population resulting from natural reproduction and the part due to immigration.

Thus

y_n+1=ρy_n+b_n, n=0, 1, 2,. . ., (10) where we are now assuming that the reproduction rateρis constant. We can solve Eq. (10) by iteration in the same manner as before. We have

y₁=ρy₀+b₀, y₂=ρ(ρy₀+b₀)+b₁=ρ²y₀+ρb₀+b₁, y3=ρ(ρ²y0+ρb0+b1)+b2=ρ³y0+ρ²b0+ρb1+b2, and so forth. In general, we obtain

y_n=ρⁿy₀+ρⁿ⁻¹b₀+ · · · +ρb_n−2+b_n−1=ρⁿy₀+

n−1

4

j=0

ρ^n−1−jb_j. (11) Note that the first term on the right side of Eq. (11) represents the descendants of the original population, while the other terms represent the population in yearn resulting from immigration in all preceding years.

In the special case whereb_n =b̸=0 for alln, the difference equation is

y_n+1=ρy_n+b, (12)

and from Eq. (11) its solution is

y_n =ρⁿy₀+(1+ρ+ρ²+ · · · +ρⁿ⁻¹)b. (13) Ifρ̸=1, we can write this solution in the more compact form

y_n =ρⁿy₀+ 1−ρⁿ

1−ρb, (14)

where again the two terms on the right side are the effects of the original population and of immigration, respectively. Rewriting Eq. (14) as

y_n =ρⁿ '

y₀− b 1−ρ

( + b

1−ρ (15)

makes the long-time behavior of y_n more evident. It follows from Eq. (15) that y_n →b/(1−ρ) if |ρ|<1. If |ρ|>1 or if ρ=−1 then y_n has no limit unless y₀=b/(1−ρ). The quantity b/(1−ρ), for ρ̸=1, is an equilibrium solution of

2.9 First Order Difference Equations 125

Eq. (12), as can readily be seen directly from that equation. Of course, Eq. (14) is not valid forρ=1. To deal with that case, we must return to Eq. (13) and letρ=1 there. It follows that

y_n =y₀+nb, (16)

so in this caseynbecomes unbounded asn→ ∞.

The same model also provides a framework for solving many problems of a finan- cial character. For such problems,y_n is the account balance in thenth time period, ρ_n=1+r_n, where r_n is the interest rate for that period, and b_n is the amount deposited or withdrawn. The following example is typical.

E X A M P L E

1

A recent college graduate takes out a $10,000 loan to purchase a car. If the interest rate is 12%, what monthly payment is required to pay off the loan in 4 years?

The relevant difference equation is Eq. (12), wherey_nis the loan balance outstanding in the nth month,ρ=1+r, whereris the interest rate per month andbis the effect of the monthly payment. Note thatρ=1.01, corresponding to a monthly interest rate of 1%. Since payments reduce the loan balance,bmust be negative; the actual payment is|b|.

The solution of the difference equation (12) with this value forρand the initial condition y0=10,000 is given by Eq. (15); that is,

yn=(1.01)ⁿ(10,000+100b)−100b. (17)

The value ofbneeded to pay off the loan in 4 years is found by settingy₄₈=0 and solving forb. This gives

b=−100 (1.01)⁴⁸

(1.01)⁴⁸−1 =−263.34. (18)

The total amount paid on the loan is 48 times|b|, or $12,640.32. Of this amount, $10,000 is repayment of the principal and the remaining $2640.32 is interest.

Nonlinear Equations. Nonlinear difference equations are much more complicated and have much more varied solutions than linear equations. We will restrict our attention to a single equation, the logistic difference equation

y_n+1=ρy_n) 1−y_n

k

*, (19)

which is analogous to the logistic differential equation dy

dt =ry) 1− y

K

* (20)

that was discussed in Section 2.5. Note that if the derivative dy/dt in Eq. (20) is replaced by the difference (y_n+1−y_n)/h, then Eq. (20) reduces to Eq. (19) with ρ=1+hrandk=(1+hr)K/hr. To simplify Eq. (19) a little more, we can scale the variabley_nby introducing the new variableu_n =y_n/k. Then Eq. (19) becomes

un+1=ρun(1−un), (21) whereρis a positive parameter.

126 Chapter 2. First Order Differential Equations

We begin our investigation of Eq. (21) by seeking the equilibrium,or constant,solu- tions. These can be found by settingu_n+1equal tou_nin Eq. (21), which corresponds to settingdy/dtequal to zero in Eq. (20). The resulting equation is

un =ρun−ρu²_n, (22)

so it follows that the equilibrium solutions of Eq. (21) are u_n =0, u_n= ρ−1

ρ . (23)

The next question is whether the equilibrium solutions are asymptotically stable or unstable. That is, for an initial condition near one of the equilibrium solutions, does the resulting solution sequence approach or depart from the equilibrium solution?

One way to examine this question is by approximating Eq. (21) by a linear equation in the neighborhood of an equilibrium solution. For example, near the equilibrium solutionu_n=0, the quantityu²_nis small compared tou_nitself, so we assume that we can neglect the quadratic term in Eq. (21) in comparison with the linear terms. This leaves us with the linear difference equation

u_n+1=ρu_n, (24)

which is presumably a good approximation to Eq. (21) foru_n sufficiently near zero.

However, Eq. (24) is the same as Eq. (7), and we have already concluded, in Eq. (9), that u_n→0 as n→ ∞ if and only if |ρ|<1, or (since ρ must be positive) for 0<ρ<1. Thus the equilibrium solutionu_n=0 is asymptotically stable for the linear approximation (24) for this set ofρvalues, so we conclude that it is also asymptoti- cally stable for the full nonlinear equation (21). This conclusion is correct, although our argument is not complete. What is lacking is a theorem stating that the solutions of the nonlinear equation (21) resemble those of the linear equation (24) near the equilibrium solutionu_n=0. We will not take time to discuss this issue here; the same question is treated for differential equations in Section 9.3.

Now consider the other equilibrium solutionu_n =(ρ−1)/ρ. To study solutions in the neighborhood of this point, we write

u_n= ρ−1

ρ +v_n, (25)

where we assume that v_n is small. By substituting from Eq. (25) in Eq. (21) and simplifying the resulting equation, we eventually obtain

v_n+1=(2−ρ)v_n−ρv²_n. (26)

Sincev_nis small, we again neglect the quadratic term in comparison with the linear terms and thereby obtain the linear equation

v_n+1=(2−ρ)v_n. (27)

Referring to Eq. (9) once more, we find thatv_n→0 asn→ ∞for|2−ρ|<1, or in other words for 1<ρ<3. Therefore, we conclude that for this range of values ofρ, the equilibrium solutionu_n =(ρ−1)/ρis asymptotically stable.

Figure 2.9.1 contains the graphs of solutions of Eq. (21) for ρ=0.8,ρ=1.5, andρ=2.8, respectively. Observe that the solution converges to zero for ρ=0.8

2.9 First Order Difference Equations 127

and to the nonzero equilibrium solution forρ=1.5 andρ=2.8. The convergence is monotone forρ=0.8 andρ=1.5 and is oscillatory forρ=2.8. The graphs shown are for particular initial conditions, but the graphs for other initial conditions are similar.

2 4 6 8

(a) 0.8

0.6 0.4 0.2

n u_n

2 4 6 8

(b) 0.8

0.6 0.4 0.2

n u_n

u_n = ¹₃

2 4 6 8

(c) 0.8

0.6 0.4 0.2

u_n = = 0.6429^1.8~

2.8

n u_n

FIGURE 2.9.1 Solutions ofun+1=ρun(1−un): (a)ρ=0.8; (b)ρ=1.5; (c)ρ=2.8.

Another way of displaying the solution of a difference equation is shown in Figure 2.9.2. In each part of this figure, the graphs of the parabola y=ρx(1−x) and of the straight line y=xare shown. The equilibrium solutions correspond to the points of intersection of these two curves. The piecewise linear graph consist- ing of successive vertical and horizontal line segments, sometimes called a stairstep diagram, represents the solution sequence. The sequence starts at the point u₀ on thex-axis. The vertical line segment drawn upward to the parabola atu₀corresponds to the calculation ofρu₀(1−u₀)=u₁.This value is then transferred from they-axis to thex-axis; this step is represented by the horizontal line segment from the parabola to the line y=x. Then the process is repeated over and over again. Clearly, the sequence converges to the origin in Figure 2.9.2aand to the nonzero equilibrium solution in the other two cases.

To summarize our results so far: the difference equation (21) has two equilib- rium solutions, u_n=0 and u_n=(ρ−1)/ρ; the former is asymptotically stable for 0≤ρ<1, and the latter is asymptotically stable for 1<ρ<3. Whenρ=1, the two equilibrium solutions coincide atu=0; this solution can be shown to be asymptoti- cally stable. In Figure 2.9.3 the parameterρis plotted on the horizontal axis anduon the vertical axis. The equilibrium solutionsu=0 andu=(ρ−1)/ρare shown. The intervals in which each one is asymptotically stable are indicated by the solid portions of the curves. There is anexchange of stabilityfrom one equilibrium solution to the other atρ=1.

For ρ>3, neither of the equilibrium solutions is stable, and the solutions of Eq. (21) exhibit increasing complexity asρincreases. Forρsomewhat greater than 3, the sequenceu_nrapidly approaches a steady oscillation of period 2; that is,u_noscil- lates back and forth between two distinct values. Forρ=3.2, a solution is shown in Figure 2.9.4. Forngreater than about 20, the solution alternates between the values 0.5130 and 0.7995. The graph is drawn for the particular initial conditionu₀=0.3, but it is similar for all other initial values between 0 and 1. Figure 2.9.4balso shows the same steady oscillation as a rectangular path that is traversed repeatedly in the clockwise direction.

128 Chapter 2. First Order Differential Equations

(a) 0.8

0.6

0.4

0.2 y

0.2 0.4 0.6 0.8 1

y = x

u₀ = 0.3

x y = x (1 – x)ρ

(b)

0.2 0.4 0.6 0.8 1 x

y = x

u₀ = 0.85

1 3

1

,3

y = x (1 – x)ρ 0.8

0.6

0.4

0.2 y

0.2 0.4 0.6 0.8

(c) 1

y = x

(0.6429..., 0.6429...) y

1 x

u₀ = 0.3 y = x (1 – x)ρ 0.8

0.6

0.4

0.2

FIGURE 2.9.2 Iterates ofun+1=ρun(1−un): (a)ρ=0.8; (b)ρ=1.5; (c)ρ=2.8.

1

0.5

–0.5

1 2 3 ρ

u

u = 0

Asymptotically stable

Unstable u = ( – 1)/ρ ρ

FIGURE 2.9.3 Exchange of stability forun+1=ρun(1−un).

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