APPENDIX D DIFFERENTIAL AND DIFFERENCE EQUATIONS

Transcription

1 APPENDIX D DIFFERENTIAL AND DIFFERENCE EQUATIONS Differential and difference equations playa key role in the solution of most queueing models. In this appendix we review some of the fundamentals concerning these types of equations. 0.1 Ordinary Differential Equations A differential equation is an equation involving a function and its derivatives. An example of such an equation might be d 2 y dy 3 x 3 dx x dx - x Y = 6e, (D.I) where y is a function of x, that is, y = y(x). The problem is to determine the most general y(x) that satisfies (D.I). Prior to discussing methods of solution to such equations, we first discuss the nomenclature involved with categorizing differential equations. Fundamentals of Queueing Theory, Fourth Edition. By D. Gross, J. F. Shortie, J. M. Thompson, and C. M. Harris Copyright 2008 John Wiley & Sons, Inc. 467

2 468 DIFFERENTIAL AND DIFFERENCE EQUATIONS Classification A differential equation is called ordinary if it involves only total (as opposed to partial) derivatives. Differential equations are further categorized by order and degree. Thus a differential equation of the form d"y d"-ly dy ao(x)-d + a1(x)-d an-1(x)-d + an(x)y = f(x) (D.2) ~ ~- x is called a linear ordinary differential equation of order n. The order refers to the highest derivative in the equation, while the degree (linear in this case) refers to the exponent on the dependent variable y and its derivatives. When the coefficients an (x) are independent of x, the equation is said to be constant coefficients. If the right-hand side of (D.2) is zero, then the equation is called homogeneous. Thus the equation d"y d"-ly dy ao-+al an-l- + any = 0 dxn dxn- 1 dx is a linear, homogeneous differential equation of order n with constant coefficients. The descriptor "ordinary" is understood and generally omitted unless one is dealing simultaneously with ordinary and partial differential equations Solutions Discussion in this appendix is restricted to solutions of linear ordinary differential equations. Solution techniques for nonlinear differential equations are extremely complex, and furthermore, the types of differential equations that arise from our interest in queueing analyses are usually linear. Consider the following linear differential equation of second order with constant coefficients, namely, y" + 3y' + 2y = 6e x, (D.3) where the prime notation is now used to denote differentiation. One solution to (D.3) is (D.4) which can be verified by substitution. This is referred to as a particular solution to (D.3). Another solution to (D.4) is (D.S) where C1 is any constant. This solution can also be verified by substitution. It contains the particular solution of (D.4) and is a more general solution. We desire the most general solution to any differential equation, which we refer to simply as the general solution. It turns out that the general solution of (D.3) is given by (D.6)

3 ORDINARY DIFFERENTIAL EQUATIONS 469 Any particular solution can be obtained by specifying the arbitrary constants C1 and C2 For example, the particular solution given by (D.4) results from (D.6) when C1 = C2 = O. The number of arbitrary constants appearing in a general solution of a linear ordinary differential equation can be shown to be equal to the order n. Since (D.3) is of order two, two constants appear in the general solution given by (D.6). Another way oflooking at the solution given by (D.6) is to first consider solutions to a homogeneous equation obtained from (D.3) by setting the right-hand side to zero. The homogeneous equation then becomes linear: y" + 3y' + 2y = O. (D.7) We note that C 1e-x and C2e-2x are both solutions to (D.7). Also, ex is a solution to the original nonhomogeneous equation (D.3), so that the general solution consists of a linear combination of all solutions to the homogeneous equation (the general solution to the homogeneous equation) plus a particular solution to the nonhomogeneous equation. It can be proved that for a linear ordinary differential equation of order n there are n solutions to the homogeneous equation, so that the general solution is comprised of a linear combination of the n solutions (thus yielding n arbitrary constants) plus a particular solution to the nonhomogeneous equation. See, for example, Rainville and Bedient (1969). To determine the constants of a general solution, that is, which particular solution is desired, one must utilize boundary conditions. A boundary condition is a condition on the function y(x) for a specific x, and results from the model which the differential equation represents. For the equation given by (D.3), suppose one knows from the physical situation that generated (D.3) that both the function and its derivative must be zero when x is zero, that is, y(o) = y' (0) = o. Using these conditions in (D.6) yields two equations in two unknowns, namely, 0= C1 + C2 + 1, o = -C1-2C2 + 1, which result in C1 = -3 and C2 = 2, giving the particular solution and the general solution as We see then, for an nth order equation, n boundary conditions are required to obtain a particular solution from the general solution. Thus the fundamental approach presented here in solving differential equations is to first find the general solution and then, using the boundary conditions, find the particular solution desired. Emphasis in this appendix is on finding general solutions.

4 470 DIFFERENTIAL AND DIFFERENCE EQUATIONS Separation of Variables The easiest type of differential equation to solve is one for which separation of variables is possible. The general solution can then be obtained by integrating both sides. For example, consider the equation We can write dy 2 y- = 3x +2e x. dx ydy = (3x 2 + 2e X )dx. Integrating both sides and combining the arbitrary constants arising from indefinite integration yields y2-2 -x e x +C. If, in general, we have an equation of the form [even for g(y), u(y) nonlinear] we can separate variables to obtain and the general solution is! dy f(x)g(y) dx = h(x)u(y), g(y) dy - h(x) dx u(y) - f(x), g(y)! h(x) u(y) dy = f(x) dx + C. (D.8) Although the examples thus far have been linear differential equations of the first order, it may also be possible to separate variables in higher-order linear equations. For example, the solution for d 2 y dx 2 = f(x) can be obtained by integrating twice to yield since d 2 y d(dy/dx) dx2 dx and integrating the first time gives a solution dx d! Y f(x)dx + C 1

5 ORDINARY DIFFERENTIAL EQUATIONS 471 EXAMPLE D.I Find the general solution of y" = 6x 2. Integrating once gives and integrating a second time yields y' = 2x 3 + C Linear Differential Equations of First Order The linear differential equation of the first order can be written in general terms as dy dx + a(x)y = f(x). (D.9) If we can determine a function g( x) so that when both sides of (D.9) are multiplied by it, the equation can be put in the form d(gy) - f dx -g, (D. 10) then the solution can be determined by separating variables, that is, the solution becomes or gy = / gfdx+c 1/ gfdx+-. C y=- (D.ll) 9 9 Such a function as 9 is referred to as an integrating factor. We can, for linear first-order differential equations, find 9 as follows. Using the product rule of differentiation, and dividing through by g, (D. 10) can be written as dy +.. dg = f. dx gdx (D.12) For (D.12) to be equivalent to (D.9) we must have Integrating both sides yields 1 dg -- = a(x). gdx lng = / a(x)dx + C1

6 472 DIFFERENTIAL AND DIFFERENCE EQUATIONS or g = ec1ea(x), where A(x) = J a(x)dx. Since we are seeking only a particular g that will yield equivalency for (D.9) and (D.12), we are free to set the constant C1 to any value we desire. It is most convenient to set C1 = O. Hence a suitable integrating factor is g = ea(x). Using (D.l2) in (D.ll) yields the final solution for y, namely, y = e-a(x) J ea(x) f(x)dx + Ce-A(x). (D.13) (D.14) Unfortunately, no such similar method is possible for obtaining solutions to higherorder linear differential equations. We will consider, however, some higher-order equations of specific types Linear Differential Equations with Constant Coefficients The simplest linear equation of higher order is one where the coefficients are independent of x, namely, d"y d"-1 y dy ao- + a an-1- + any = f(x). dxn dxn- 1 dx (D.IS) The approach here is to first find the n solutions to the homogeneous equation aod d"y d"-1 y dy +a1-d an- 1- d +any=o, xn xn X (D.16) and then find a particular solution for the nonhomogeneous equation. The form of (D.16) suggests that the homogeneous solutions are of the form e TX, since the nth derivative is a multiple of the function itself, that is, d"e TX n TX -- =r e. dx Now if e TX is a solution to (D.16), then we have (D.17) which implies for a nontrivial solution (y = e TX =I- 0) that (D.18) Equation (D.18) is called the characteristic or operator equation. The characteristic equation can also be obtained directly by looking at the derivative as an operator, say,

7 ORDINARY DIFFERENTIAL EQUATIONS 473 D, so that Dy D2y = D(Dy) dy dx' d2y dx2' Hence (D.16) can be written as Dny = D(Dn-1y) ~y dxn (aodn + a1dn an-1d + an)y = 0, where the characteristic equation is in terms of D instead of r. Denoting the n roots of the characteristic equation by rl, r2,..., r n, we can write and hence theoretically the roots can be found by factorization. * If the n roots are distinct, we then have n solutions etix(i = 1,2,..., n) of the homogeneous equation (D.16). The most general solution of (D.16) is then If the roots are not all distinct, we have less than n solutions. To find the missing solution we can proceed as follows. Suppose that rl is a double root of the characteristic equation. Then we have Observing that a(r - rl)2 _ 2( _ ) ar - r rl, (D.19) we find that the partial derivative with respect to r evaluated at r = rl also vanishes, so that if e T1X is a solution, so too is ae TX / a rlt=tl = xe T1X To verify that xe T1X is a solution consider solutions of the form xe TX Putting this in for y in (D.16) yields Since we can write * Depending on the characteristic equation that results, factorization could be impossible and numerical methods might be required.

8 474 DIFFERENTIAL AND DIFFERENCE EQUATIONS and changing the order of differentiation gives Hence we can write a ((r etx lax n ) a(aetx lax) aetx ao or an-l or + an 8r = o. or :r [(aorn + alrn an-lr + an) etx ] = O. But we have said that the characteristic equation factors into n - 1 roots as given in (D.19) so that :r {[(r - rl)2(r - r2) (r - rn-d] etx } = O. This equation does hold for r = Tl, since the partial with respect to r vanishes at that point. Thus the two solutions for a double root rl are This can be generalized to roots of multiplicity k; that is, if rl has multiplicity k, the solution associated with rl is C1eT1X + C2xeT1X + C3x2eTIX Ckxk-leTlX. When we have multiple roots, if factorization is not possible and we must resort to numerical methods, we might only be able to find (say) n - k distinct roots to the characteristic equation. To find which root (or roots) have multiplicity we can simply take partial derivatives of the characteristic equation and check for which root (or roots) vanish. The roots for which only the first partial derivative of the characteristic equation vanishes have multiplicity two. If a root causes the first, second,..., kth partial derivatives to vanish, it is of multiplicity k + 1. EXAMPLE D.2 Find the general solution for The characteristic equation is D3-4D = 0, which factors into D(D + 2)(D - 2) = 0; hence the roots are rl = 0, r2 = -2, and r3 = +2. The general solution is y = C1 + C2e-2x + C3e2x.

9 ORDINARY DIFFERENTIAL EQUATIONS 475 EXAMPLE D.3 Find the general solution for The characteristic equation is which factors into D3-4D2 + 5D - 2 = 0, (D - 2)(D - 1)2 = o. Thus the roots are Tl = 2 and T2 = T3 = 1, and we have Had we not been able to factor the characteristic equation but had determined that 2 and 1 were all distinct roots, we know that since the characteristic equation is cubic, one root must be double. To find which root it is, we take the partial derivative of the characteristic equation, which gives 3D 2-8D + 5, and evaluating D = 2 and D = 1 yields and so that root D = 1 is the double root. 3(2)2-8(2) + 5 = 1 3(1)2 + 8(1) + 5 = 0 It remains now to discuss the determination of a particular solution for the nonhomogeneous linear differential equation with constant coefficients. There are four methods for finding a particular solution to the nonhomogeneous equation: (1) undetermined coefficients, (2) variation of parameters, (3) differential operators, and (4) Laplace transforms. We briefly discuss the first and third methods here. Laplace transforms are also presented in Appendix C Undetermined Coefficients If the right-hand side of the differential equation given in (D.15) is of the form xm (m is an integer), sin(bx), cos(bx), e bx, and/or products of two or more such functions, we can employ the method of undetermined coefficients to find a particular solution. We first define a family of a function f (x) and its derivatives. The functions specified above are functions with a finite number of derivatives for which the function and its

10 476 DIFFERENTIAL AND DIFFERENCE EQUATIONS Table D.I Functions and Their Families Function xm sin bx cos bx e bx Family xm,xm-1,xm-2,...,x2,x,1 sin bx, cos bx cos bx, sin bx e bx derivatives are linearly independent. Table D.llists the families of the aforementioned functions. The family of a function consisting of a product of n terms of this type consists of all possible products of the family members of each of the n terms. For example, the family of x2 cos x is x 2 cos x, X cos x, cos x, x2 sin x, x sin x, and sin x. The method works as follows in three steps: 1. Assuming f (x) is a linear combination of functions or products of functions given in Table D.l, construct the family for each, eliminating families that are included in other families. 2. If any family has a member that is also a solution to the homogeneous equation, replace that family by a new one, obtained by multiplying the original family by x (or the lowest power of x necessary) so that the new family has no members that are also solutions to the homogeneous equation. 3. The particular solution is assumed to be a linear combination of all members of the constructed families. The constants of the linear combination are then found by substituting this particular solution into the differential equation. EXAMPLE D.4 Find the general solution for ylll - y' = 2x + 1-4cosx + 2e x. The general homogeneous solution can be found from previous methods to be y = C1 + C2e x + C3e- x. The families for the right-hand side function are, respectively, {x, I}, {I }, {cos x, sinx}, {ex}. Since {I} is included in {x, I}, we omitthis. Furthermore, since 1 and ex are in the homogeneous solution, their families are replaced by { x2, x} and { xe X }, respectively. Then the resulting terms to be used are { 2. X} X,x,Cosx,Slnx,xe

11 ORDINARY DIFFERENTIAL EQUATIONS 477 and the particular solution is of the form YP = AX2 + Bx + Ccosx + Dsinx + Exe x. Substituting Yp into the differential equation yields C sinx - D cos x + E(xeX + 3eX) - [2Ax + B - Csinx + Dcosx + E(xeX + ex)] = 2x cos x + 2e x, or simplifying we get -2Ax - B + 2Csinx - 2Dcosx + 2Ee x = 2x + 1-4cosx + 2e x. Matching coefficients of like terms yields Hence the particular solution is and the general solution becomes A = -1,B = -1,C = O,D = 2,E = 1. YP = _x2 - X + 2 sin x + xe x Differential Operators We illustrate the use of differential operators on the same equations used in the previous example. The equation can be written in operator notation as y"' - Y' = 2x cos x + 2e x where g, as before, is the right-hand side. This can be factored as D(D + 1)(D - l)y = g. We let hence we have YI = (D + 1) (D - 1) Yi (D.20)

12 478 DIFFERENTIAL AND DIFFERENCE EQUATIONS or Solving directly by integration gives Yl dyl -=g. dx J gdx + C1 J (2x + 1-4cosx + 2e X )dx + C1 x 2 + X - 4sinx + 2e x + C1. Substituting Yl into (D.20) yields the differential equation We next let and get or (D + 1) (D - 1) Y = x 2 + X - 4 sin x + 2e x + C1. Y2 = (D -1)y (D+l)Y2 =x2+x-4sinx++2e x +C1 dy2 2. x C dx + Y2 = X + X - 4 sm x + 2e + 1 (D.21) (D.22) Equation (D.22) is now a first-order equation that can be solved by the solution previously derived in Section D.1.4 and given by (D.14), which yields Y2 = e- x J ex (x2 + x - 4sinx + 2e x + C 1 ) dx + C2e- x = x 2 - X + 1-2sinx + 2cosx + ex + C1 + C2e- x. Now substituting Y2 into (D.21) yields another first-order equation (D -1) Y = x 2 - X + 1-2sinx + 2cosx + ex + C1 + C2e- x. Again using the solution for first-order equations we get Y = ex J e- x (x2 - x + 1-2sinx + 2cosx + ex + C1 + C2e-X) dx + C3e x X C C2 -x C X = -x - x - + smx + xe Te + 3e, which agrees with our previous solution upon redefining the arbitrary constants. This method of using operators applies only for equations with constant coefficients. Essentially, any equation with constant coefficients can be written in operator notation as f(d)y = g(x). If a function f- 1 (D) can be found where

13 ORDINARY DIFFERENTIAL EQUATIONS 479 the solution to the equation is The material referenced above deals with determining such inverse differential operations Reduction of Order Leaving the topic of particular solutions and returning now to the topic of solutions in general, if one homogeneous solution of a linear differential equation of order n is known, the remainder of the solution can be determined by solving a new linear differential equation of order n -1 in much the same way one can reduce the degree of an algebraic equation when one root is known. Consider the following second-order equation: (D.23) where Y and 9 are functions of x and the coefficients al and a2 may be also. Suppose one solution to the homogeneous equation can be found from inspection and we denote it by Yl (x), that is, (D.24) Then if we let then Y is a solution to (D.23) if Y = YlV, or Simplifying we obtain YlV" + 2y~v' + y~v + al (YlV' + vyd + a2ylv = g. (D.25) But since Yl is a homogeneous solution, using (D.24) and (D.25) gives " (2 I ) I YlV + Yl+alYl v =g. Letting u = v' we get the following first-order equation involving u, namely, (D.26) which can be solved by (D.14) of Section D.1.4. Finally, we can get v from v' = u by integration and the general solution Y = Yl V results.

14 480 DIFFERENTIAL AND DIFFERENCE EQUATIONS EXAMPLE D.S Consider the equation y" - y = x. From inspection we can see that one solution to the homogeneous equation is Thus using this in (D.26) we have the first-order equation or which by (D.l4) yields u' + 2u = xe- x, Integrating u we obtain v as V = -xe and finally (redefining the constant C1 ) -x C1-2x - -e + C2 2 ' Systems of Linear Differential Equations This section considers systems of simultaneous linear differential equations with constant coefficients. To begin, consider the following system of two equations in two unknowns: These can be rewritten using operator notation as (D2-1) Y1-2Y2 = g1, -3Y1 + (D2-2) Y2 = g2. (D.27) Using Cramer's rule the solution yields the following two differential equations of a single variable: (D2-1) \ -3-2 \ \91 (D2-2) Y1 = g2-2 \ (D2-2),

15 or rewriting (D2-1) I -3 ORDINARY DIFFERENTIAL EQUATIONS I _1(D 2-1) gil (D2-2) Y g2 ' (D4-3D2-4)YI = (D2-2)gl + 2g2, (D4-3D2-4)Y2 = (D2 - l)g2 + 3gl. Since gl and g2 are known functions of x, the differentiation implied by the operator can be performed and the right-hand side represented by two known function, hl(x) and h2(x), yielding (D4-3D2-4)YI = hl(x), (D4-3D2-4)Y2 = h2(x). Both equations have identical characteristic equations (which can always be obtained from the determinant of the left-hand side of the system of equations) so the general solution to the homogeneous equations are of the same form. Denoting the four roots to the characteristic equation by rl, r2, r3, and r4, we have the homogeneous solutions YI = CIe T1X + C2e T2X + C3e T3X + C4 e T4X Y2 = C5e T1X + C6e T2X + C7e T3X + C8e T4X (D.28) While there are eight constants, they are not all independent and their relationships can be obtained by substitution of (D.28) in either equation of (D.27) with the right-hand side equal to zero, yielding c - r~ -IC 6-2 2, C - r~ 8-- -IC 2-4 Equations (D.28) are the homogeneous solutions to (D.27). To obtain particular solutions, one can use the method of undetermined coefficients. EXAMPLE D.6 Solve the following for YI and Y2: Y~ - 2YI + 2y~ = 2-4e 2x, 2y~ - 3YI + 3y~ - Y2 = o. (D.29) Considering first the homogeneous solutions, we rewrite the equations in operator notation as (D - 2)YI + 2DY2 = 0, (2D - 3)YI + (3D - I)Y2 = 0, (D.30)

16 482 DIFFERENTIAL AND DIFFERENCE EQUATIONS and the characteristic equation is then which upon expanding yields (D - 2) I (2D - 3) I 2D (3D-I) =0, _D2_D+2=0. Factoring gives the two roots as 1 and - 2 so that the homogeneous solutions are Yl = Clex + C2e-2x, Y2 = C3ex + C4e-2x. To determine the relationship among the constants, we substitute the above in either equation of (D.30) to get and thus To obtain the particular solution, we use the method of undetermined coefficients. The family to be considered is {I, e2x } and we proceed as follows: Substituting into (D.29) gives Yl,p = A + Be2x, Y2,p = C + De2x. 2Be2x - 2A - 2Be2x + 4De2x = 2-4e 2x, 4Be2x - 3A - 3Be2x + 6De2x - C - De2x = 0, or upon simplifying we have - 2A + 4De2x = 2-4e2x, - 3A - C + (B + 5D)e2X = 0. Now equating coefficients of like terms yields -2A = 2, 4D = -4, -3A - C = 0, B + 5D = 0, which finally gives A=-I, B=5, C=3, D=-I,

17 DIFFERENCE EQUATIONS 483 and the general solutions are Yl = Clex + C2e-2x e2x, Y2 = C l ex - C2e-2x + 3 _ e2x 2. The procedure, of course, generalizes to systems of size greater than two. If we have n simultaneous equations, the characteristic equation is obtained from evaluating an n x n determinant Summary In solving ordinary linear differential equations, the first approach should be to determine whether the variables are separable. If they are, the general solution can be obtained directly by integration as discussed in Section D.l.3. If separation of variables is not possible, but the equation is first order, the solution can be obtained from (D. 14) as derived in Section D.l.4. For higher-order equations with constant coefficients, the general solution to the homogeneous equation can be obtained by finding the roots of the characteristic equation (Section D.l.5) and then finding the particular solution via undetermined coefficients (Section D.l.6). Use of operators (Section D.l.7) can also be employed to determine general solutions for nonhomogeneous linear equations with constant coefficients. If one (or more) solutions to the homogeneous equation are known, the order of the equation can be reduced (Section D.l.8), thereby yielding equations of lower order, which may be solved more readily. Finally, in Section D.1.9 solutions of systems of simultaneous linear differential equations with constant coefficients are discussed. 0.2 Difference Equations Consider a function of an independent variable x, where x is now a discrete variable; that is, it can take only integer values. Then the function exists only at discrete points (integer values of x) and we denote this type of function by Yx instead of y(x). The first finite difference of Yx is given as the second finite difference as and the nth finite difference as tly == Yx+l - Yx, tl 2 y = tl(tly) = (Yx+2 - Yx+l) - (YX+l - Yx) = Yx+2-2Yx+l + Yx,

18 484 DIFFERENTIAL AND DIFFERENCE EQUATIONS We define an operator D to be Dyx = Yx+l, D2yx = D(Dyx) = Yx+2, One can easily see the relationship between ~ and D as Linear Difference Equations with Constant Coefficients An equation involving Yx of the type (D.31) is called a linear difference equation of order n with constant coefficients. We shall not treat here the case where the coefficients are also dependent on x. One can see many similarities between difference equations and differential equations and, indeed, the solution techniques are often quite similar. The technique for solving (D.31) is very much like that used for linear differential equations with constant coefficients. In fact, it can be shown that a general solution of (D.31) consists of a linear combination of all solutions to the homogeneous equation (gx replaced by zero) plus a particular solution to (D.31). Also, for nth degree equation, there are n arbitrary constants associated with the homogeneous solution, which in any particular case can be found from n boundary conditions. To find the solution to the homogeneous equation we proceed in a manner similar to Section D.1.S. We first rewrite (D.3l) using operator notation to get The homogeneous solutions are of the form rx (as opposed to etx for differential equations), where r is a root to the characteristic equation To see this, we let Yx = rx in (D.31) and get whereupon factoring out rx we have But since r is a root to the characteristic equation, the left-hand equals zero.

19 DIFFERENCE EQUATIONS 485 Since the characteristic equation has n roots, the general solution to the homogeneous equation is Yx = C1rf + C2r~ Cnr~. Multiple roots can be handled in a manner analogous to differential equations in that for a root of multiplicity k, the first k -1 derivatives of the characteristic equation with respect to D must vanish and the k solutions are of the form rx, xrx, x(x -l)rx,..., x (x - 1)... (x - k + 1) rx, since in taking the i th derivative or rx one obtains x(x - 1) (x - i + l)rxr-i and the r-i can be absorbed in the arbitrary constant. To find a particular solution to (D.3I), the method of undetermined coefficients can be employed. We illustrate the procedures on the following example. EXAMPLE D.7 Consider the difference equation Yx+2 + 6Yx+1 + 9yx = 16x2. The homogeneous equation in operator notation is (D2 + 6D + 9) Yx = 0, and the solution to the characteristic equation has two roots at -3. Hence the solution is Yx = C 1 (_3)X + C2x( _3)x. To find the particular solution, the family of x2 gives terms {x2, x, I}. Therefore Yx,P = Ax2 + Bx + C, and substituting this into the original equation gives or A(x + 2)2 + B(x + 2) + C + 6[A(x + 1)2 + B(x + 1) + CJ + 9[Ax2 + Bx + CJ = 16x2, 16Ax2 + (16A + 16B)x + loa + 8B + 16C = 16x2. Equating like coefficients yields the conditions or finally 16A = 16, 16A + 16B = 0, loa + 8B + 16C = 0, Thus the particular solution is and the general solution becomes A=l, B=-l, C=-~. 2 1 Yx,P = X - X - 8' Yx = C1( _3)X + C2x( _3)X + x2 - X - ~.

20 486 DIFFERENTIAL AND DIFFERENCE EQUATIONS Systems of Linear Difference Equations The solution to systems of difference equations is analogous to the procedure used in Section D.1.9 for differential equations. One first writes the equation in operator notation, finds the characteristic equation using the determinant ofthe left-hand side "coefficients," solves for the roots, and obtains the homogeneous solution as linear combinations of ri (instead of eti,x as for differential equations). The number of constants are then reduced as before by substituting the homogeneous solutions into the homogeneous equations. Then a particular solution can be found (if the equations are nonhomogeneous) by the method of undetermined coefficients. EXAMPLE D.S Consider the following system of difference equations to be solved for Y and z. Yx+1-3yx + Zx+1-3zx = 2, 2Yx+l - 5yx + 3zx+l - 3zx = 6(4)x. (D.32) We first obtain the homogeneous solutions by solving the characteristic equation obtained after writing in operator notation. The characteristic equation is (D - 3) I (2D - 5) (D - 3) I (3D - 3) = 0, which upon calculating the determinant yields D2 - D - 6 = o. The roots can be found by factoring to be 3 and - 2. Thus the homogeneous solutions are Yx = C1(3)X + C2 ( _2)X, Zx = C 3 (3)X + C 4 ( _2)x. To reduce the number of arbitrary constants we substitute the above in the original equations of (D.32) with the right-hand side set to zero. This yields the relations and hence the homogeneous solutions become Yx = C1 (3)X + C2 ( _2)X, Zx = - ~1 (3y - C 2 ( _2)x. To obtain the particular solution, we employ undetermined coefficients. The family of the first right-hand side of (D.32) is {1} and the family of the second

21 DIFFERENCE EQUATIONS 487 is {4X}. Thus we have Substituting into (D.32) we get Yx,P = A + B(4)X, zx,p = C + D(4)x. A + 4B(4)X - 3A - 3B(4)'E + C + 4D(4)X - 3C - 3D(4)X = 2, 2A + SB(4)X - 5A - 5B(4)X + 3C + 12D(4)X - 3C - 3D(4)X = 6(4)x. Upon simplification we obtain Equating like coefficients yields -2A - 2C + (B + D)(4)X = 2, -3A + (3B + 9D)(4)X = 6(4)x. -2A - 2C = 2, B + D = 0, -3A = 0, 3B + 9D = 6, which gives A=O, B=-l, C=-l, D=l. The general solution is then Yx = C1(3)X + C 2 ( _2)X - (4)X, Zx = - C 1 (3)"' - C 2 ( _2)X (4)x. 6 This method of finding particular solutions for systems of equations through the use of undetermined coefficients does not always work. For example, one can verify that undetermined coefficients do not yield a particular solution of the following set of equations: Yx+1-2yx + 2zx = 2, 2Yx+1-3yx + 3Zx+1 - Zx = 6( 4)x. For such cases, other methods are necessary: However, further detailed treatment of finding particular solutions is not necessary since differential and difference equations encountered in queueing theory are, for the most part, homogeneous.