In this book we will have uses for finite difference equations only in Chaps.8and 9. For that reason we will give only a brief introduction to solution techniques for them. Readers who wish more details can consult one of several texts on difference equations; see, e.g., Goldberg (1986) or Spiegel (1971).
Iff(k) is a real function of time, then thedifference operatorapplied tof is defined as
Δf(k) =f(k+ 1)−f(k). (A.20) The factorial power ofk is defined as
k(n)=k(k−1)(k−2). . .(k−(n−1)). (A.21) It is easy to show that
Δk(n)=nk(n−1). (A.22) Because this formula is similar to the corresponding formula for the derivative d(kn)/dk, the factorial powers of k play an analogous role for finite differences that the ordinary powers of k play for differential calculus.
If f(k) is a real function of time, then the anti-difference operator Δ−1 applied to f is defined as another function g = Δ−1f(k) with the property that
Δg=f(k). (A.23)
One can easily show that
Δ−1k(n)= (1/(n+ 1))k(n+1)+c, (A.24) where c is an arbitrary constant. Equation (A.24) corresponds to the integration formula for powers ofk in calculus.
Note that formulas (A.22) and (A.24) are similar to, respectively, differentiation and integration of the power function kn in calculus. By
A.5. Solutions of Finite Difference Equations 415 analogy with calculus, therefore, we can solve difference equations in- volving polynomials in ordinary powers of k by first rewriting them as polynomials involving factorial powers of k so that (A.22) and (A.24) can be used. We show next how to do this.
A.5.1 Changing Polynomials in Powers of k into Factorial Powers of k
We first give an abbreviated list of formulas that show how to change powers of k into factorial powers ofk:
k0 = k(0) = 1 (by definition), k1 = k(1),
k2 = k(1)+k(2),
k3 = k(1)+ 3k(2)+k(3),
k4 = k(1)+ 7k(2)+ 6k(3)+k(4),
k5 = k(1)+ 15k(2)+ 25k(3)+ 10k(4)+k(5).
The coefficients of the factorial powers on the right-hand sides of these equations are calledStirling numbers of the second kind, after the person who first derived them. This list can be extended by using a more com- plete table of these numbers, which can be found in books on difference equations cited earlier.
Example A.1 Express k4−3k+ 4 in terms of factorial powers.
Solution Using the equations above we have
k4 =k(1)+ 7k(2)+ 6k(3)+k(4), −3k=−3k(1), 4 = 4, so that
k4−3k+ 4 =k(4)+ 6k(3)+ 7k(2)−2k(1)+ 4.
Example A.2 Solve the following difference equation in Example8.7:
Δλk =−k+ 5, λ6= 0.
Solution We first change the right-hand side into factorial powers so that it becomes
Δλk=−k(1)+ 5.
416 A. Solutions of Linear Differential Equations Applying (A.24), we obtain
λk=−(1/2)k(2)+ 5k(1)+c,
where c is a constant. Applying the condition λ6 = 0, we find that c=−15,so that the solution is
λk=−(1/2)k(2)+ 5k(1)−15. (A.25) However, we would like the answer to be in ordinary powers of k.
The way to do that is discussed in the next section.
A.5.2 Changing Factorial Powers of k into Ordinary Powers of k
In order to change factorial powers of k into ordinary powers of k, we make use of the following formulas:
k(1) = k, k(2) = −k+k2, k(3) = 2k−3k2+k3,
k(4) = −6k+ 11k2−6k3+k4,
k(5) = 24k−50k2+ 35k3−10k4+k5.
The coefficients of the factorial powers on the right-hand sides of these equations are called Stirling numbers of the first kind. This list can also be extended by using a more complete table of these numbers, which can be found in books on difference equations.
Solution of Example A.2 Continued By substituting the first two of the above formulas into (A.25), we see that the desired answer is
λk=−(1/2)k2+ (11/2)k−15, (A.26) which is the solution needed for Example 8.7.
Exercises for Appendix A 417 Exercises for Appendix A
E A.1 Show that the solution of Eq. (A.3) is given by (A.4).
E A.2 If A =
⎡
⎢⎣ 3 2 2 3
⎤
⎥⎦, show that Λ =
⎡
⎢⎣ 5 0 0 2
⎤
⎥⎦ and P =
⎡
⎢⎣ 1 1 1 −1
⎤
⎥⎦.
Use (A.15) to solve (A.7) for this data, given thatz(0) =
⎡
⎢⎣ 1 2
⎤
⎥⎦.
E A.3 If A =
⎡
⎢⎣ 3 3 2 4
⎤
⎥⎦, show that Λ =
⎡
⎢⎣ 6 0 0 1
⎤
⎥⎦ and P =
⎡
⎢⎣ 1 3 1 −2
⎤
⎥⎦.
Use (A.15) to solve (A.7) for this data, given thatz(0) =
⎡
⎢⎣ 0 5
⎤
⎥⎦.
E A.4 After you have read Sect.6.1, re-solve the production-inventory example stated in Eqs. (6.1) and (6.2), (ignoring the control constraint (P ≥ 0) by the method of Sect.A.4. The linear two-point boundary value problem is stated in Eqs. (6.6) and (6.7).
Appendix B
Calculus of Variations and Optimal Control Theory
Here we introduce the subject of the calculus of variations by analogy with the classical topic of maximization and minimization in calculus;
see Gelfand and Fomin (1963), Young (1969), and Leitmann (1981) for rigorous treatments of the subject. The problem of the calculus of varia- tions is that of determining a function that maximizes a given functional, the objective function. An analogous problem in calculus is that of de- termining a point at which a specific function, the objective function, is maximum. This, of course, is done by taking the first derivative of the function and equating it to zero. This is what is called the first-order condition for a maximum. A similar procedure will be employed to de- rive the first-order condition for the variational problem. The analogy with classical optimization extends also to the second-order maximiza- tion condition of calculus. Finally, we will show the relationship between the maximum principle of optimal control theory and the necessary con- ditions of the calculus of variations. It is noted that this relationship is similar to the one between the Kuhn-Tucker conditions in mathematical programming and the first-order conditions in classical optimization.
We start with the “simplest” variational problem in the next section.
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420 B. Calculus of Variations and Optimal Control Theory