Notes on Solving Difference Equations

Notes on Solving Difference Equations Yulei Luo SEF of HKU September 13, 2012 Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 1 / 36

Discrete-time, Differences, and Difference Equations The note is largely based on Fundamental Methods of Mathematical Economics (by Alpha C. Chiang and Kevin Wainwright, 4th edition, 2005). When time is taken to be a discrete variable, so that the variable t is allowed to take integer values only, the concept of the derivative will no longer appropriate (it involves infinitesimal changes, dt) and the change in variables must be described by so called differences ( t). Accordingly, the techniques of difference equations need to be developed. We may describe the pattern of change of y by the following difference equations: where y t+1 = y t+1 y t. y t+1 = 2 (1) or y t+1 = 0.1y t (2) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 2 / 36

Solving a FO Difference Equation Iterative Method. For the FO case, the difference equation describes the pattern of y between two consecutive periods only. Given an initial value y 0, a time path can be obtained by iteration. Consider y t+1 y t = 2 with y 0 = 15, y 1 = y 0 + 2 y 2 = y 1 + 2 = y 0 + 2 (2) and in general, for any period t, y t = y 0 + t (2) = 15 + 2t. (3) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 3 / 36

. Consider y t+1 = 0.9y t with y 0. By iteration, y 1 = 0.9y 0, y 2 = (0.9) 2 y 0 y t = (0.9) t y 0 (4) Consider the following homogeneous difference equation ( n ) t my t+1 ny t = 0 = y t+1 = y0, (5) m which can be written as a more general form y t = Ab t. (6) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 4 / 36

General Method to Solve FO Difference Equation Suppose that we are solving the FO DE: y t+1 + ay t = c (7) The general solution is the sum of the two components: a particular solution y p (which is any solution of the above DE) and a complementary function (CF) y c. Let s first consider the CF. We try a solution of the form y t = Ab t, Ab t+1 + aab t = 0 = b + a = 0 or b = a, which means that the CF should be y c = A ( a) t (8) Consider now the particular solution. We try the simplest form y t = k, k + ak = c = k = c 1 + a = the PI is y p = c 1 + a (a = 1). (9) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 5 / 36

(conti.) If it happens that a = 1, try another solution form y t = kt, k (t + 1) + akt = c c = k = t + 1 + at = c. y p = ct (a = 1). (10) The general solution is then y t = A ( a) t + c 1 + a or y t = A ( a) t + ct (a = 1) (11) Using the initial condition y t = y 0 when t = 0, we can easily determine the definite solution: y 0 = A + c 1 + a = A = y 0 c (a = 1) or 1 + a y 0 = A + c 0 = A = y 0 (a = 1) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 6 / 36

The Dynamic Stability of Equilibrium In the discrete-time case, the dynamic stability depends on the Ab t term. The dynamic stability of equilibrium depends on whether or not the CF (Ab t ) will tend to zero as t. We can divide the range of b into seven distinct regions: see Figure 17.1. { Nonoscillatory if b > 0 Oscillatory if b < 0 ; { Divergent if b > 1 Convergent if b < 1 The role of A. First, it can produce a scale effect without changing the basic configuration of the time path. Second, the sign of A can affect the shape of the path: a negative A can produce a mirror effect as well as a scale effect. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 7 / 36

The Cobweb Model A variant of the market model: it treats Q s as a function not of the current price but of the price of the preceding time period, that is, the supply function is lagged or delayed. Q s,t = S (P t 1 ) (12) When this function interacts with a demand function of the form Q d,t = D (P t ),interesting price dynamics will appear. Assuming linear supply and demand functions, and the market equilibrium implies Q s,t = Q d,t (13) Q d,t = α βp t (α, β > 0) (14) Q s,t = γ + δp t 1 (γ, δ > 0). (15) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 8 / 36

(conti.) In equilibrium, the model can be reduced to the following FO DE βp t + δp t 1 = α + δ = P t+1 + δ β P t = α + δ β (16) Consequently, we have P t = ( P 0 α + γ ) ( δ ) t + α + γ β + δ β β + δ. (17) The particular integral P = α+γ β+δ is the intertemporal equilibrium price of the model. We can rewrite the price dynamics as follows P t = ( P 0 P ) ( δ β ) t + P. (18) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 9 / 36

(conti.) P 0 P can have both the scale effect and the mirror effect on the price dynamics. Given our model specification (δ, β > 0), we can deduce an oscillatory time path because δ β < 0. That s why we call the model the Cobweb model. The model has three possibilities of oscillation patterns: Explosive if δ > β Uniform if δ = β Damped if δ < β See Figure 17.2 in CW. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 10 / 36

Nonlinear Difference Equations-The Qualitative-Graphic Approach When nonlinearity occurs in the case of FO DE models, we can use the graphic approach (Phase diagram) to analyze the properties of the DE. Consider the following nonlinear DEs y t+1 + y 3 t = 5 or y t+1 + sin y t ln y t = 3 = y t+1 = f (y t ) when y t+1 and y t are plotted against each other, the resulting diagram is a phase diagram and the curve corresponding to f is a phase line. See Figure 17.4. The first two phase lines, f 1 and f 2, are characterized by positive slopes f 1 (0, 1) and f 2 > 1 and the remaining two, f 3 and f 4, are negatively sloped f 3 ( 1, 0) and f 4 < 1 Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 11 / 36

(conti.) For the phase line f 1, the iterative process leads from y 0 to y in a steady path, without oscillation. For the phase line f 2 (whose slope is greater than 1), a divergent path appears. For phase lines, f 3 and f 4, the slopes are negative. The oscillatory time paths appear. Summary: The algebraic sign of the slope of the phase line determines whether there will be oscillation, and the absolution value of its slope governs the question of convergence. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 12 / 36

Second Order Difference Equation A second-order difference equation involves the second difference of y : 2 y t+2 = ( y t+2 ) = (y t+2 y t+1 ) where is the first difference. = (y t+2 y t+1 ) (y t+1 y t ) = y t+2 2y t+1 + y t, (19) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 13 / 36

SO Linear DEs with Constant Coeffi cients and Constant Term A simple variety of SO equation takes the form y t+2 + a 1 y t+1 + a 2 y t = c (20) We first discuss particular solution. As usual, try the simplest solution form y t = k, which means that y p = k = c 1 + a 1 + a 2 (1 + a 1 + a 2 = 0) (21) In case a 1 + a 2 = 1, try another solution form y t = kt, which means that y p = kt = c a 1 + 2 t (22) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 14 / 36

(conti.) We next discuss the complementary function which is the solution of the reduced homogenous equation (c = 0). As in the FO DE case, try the following solution form y t = Ab t = (23) Ab t+2 + a 1 Ab t+1 + a 2 Ab t = 0 = (24) b 2 + a 1 b + a 2 = 0 (25) This quadratic characteristic equation have two roots: b 1, b 2 = a 1 ± a1 2 4a 2 2 and both should appear in the general solution of the reduced DE. There are three possibilities. (26) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 15 / 36

Case 1 (distinct real roots) When a 2 1 4a 2 > 0, the CF can be written as y c = A 1 b t 1 + A 2 b t 2. (27) Example: Consider which means that b 1 = 1, b 2 = 2, y t+2 + y t+1 2y t = 12, (28) y t = A 1 + A 2 ( 2) t + 4t (29) where A 1 and A 2 can be determined by two initial conditions y 0 = 4 and y 1 = 5 : 4 = A 1 + A 2 and 5 = A 1 + A 2 ( 2) + 4 = A 1 = 3 and A 2 = 1. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 16 / 36

Case 2 (repeated real roots) When a 2 1 4a 2 = 0, the CF can be written as y c = A 3 b t + A 4 tb t. (30) Example: Consider which means that b 1 = b 2 = 3, y t+2 + 6y t+1 + 9y t = 4, (31) y t = A 3 ( 3) t + A 4 t ( 3) t + 1 4 (32) where A 1 and A 2 can be determined by two initial conditions y 0 and y 1. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 17 / 36

Case 3 (complex roots) When a 2 1 4a 2 < 0, b 1, b 2 = h ± vi where h = a 1 a 2 2 and v = 1 +4a 2 2. The CF is y c = A 1 b t 1 + A 2 b t 2 = A 1 (h + vi) t + A 2 (h vi) t. (33) De Moivre theorem implies that (h ± vi) t = R t (cos θt ± i sin θt) where R = h 2 + v 2 = a 2, cos θ = h R, sin θ = v R (34) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 18 / 36

(conti.) The CF can be rewritten as y c = A 1 R t (cos θt + i sin θt) + A 2 R t (cos θt i sin θt) (35) = R t [(A 1 + A 2 ) cos θt + (A 1 A 2 ) i sin θt] = R t (A 5 cos θt + A 6 i sin θt) (36) where R and θ can be determined once h and v become known. Example: Consider y t+2 + 1 4 y t = 5,which means that h = 0, v = 1 2, R = y c = ( ) 1 2 0 + = 1 2 2, cos θ = 0, sin θ = 1, θ = π = (37) 2 ( ) 1 t ( A 5 cos π 2 2 t + A 6i sin π ) 2 t. (38) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 19 / 36

The Convergence of Time Path The convergence of time path y is determined by the two characteristic roots of the SO DE. In Case 1 if b 1 > 1 and b 2 > 1, then both components in the CF will be explosive and y c must be divergent. if b 1 < 1 and b 2 < 1, then both components in the CF will converge to 0 as t goes to infinity, as will y c also. if b 1 > 1 and b 2 < 1, then A 2 b t 2 tend to converge to 0, while A 1b t 1 tends to deviate further from 0 and will eventually render the path divergent. Call the root with higher absolute value the dominant root since this root sets the tone of the time path. A time path will converge iff the dominant root is less than 1 in absolute value. The non-dominant root also affects the time path, at least in the beginning periods. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 20 / 36

In Case 2 (repeated roots), for the term A 4 tb t, if b > 1, the b t term will be explosive. and the multiplicative t term also serves to intensify the explosiveness as t increases. if b < 1, the b t term will be converge. and the multiplicative t will offset the convergence as t increases. It turns out the damping force b t of will eventually dominant the exploding force t. Hence, the basic requirement for convergence is still that the root be less than 1 in absolution value. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 21 / 36

In Case 3 (complex roots), The term A 5 cos θt + A 6 i sin θt produces a fluctuation pattern of a periodic nature. Since time is discrete, the resulting path displays a sort of stepped fluctuation. The term R t determines the convergence of y : determines whether the stepped fluctuation is to be intensified or mitigated as t increases. Hence, the basic requirement for convergence is still that the root be less than 1 in absolution value. The fluctuation can be gradually narrowed down iff R < 1 (Note that R is just the absolute value of the complex roots h ± vi). Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 22 / 36

Difference Equations System So far our dynamic analysis has focused on a single difference equation. However, some economic models may include a system of simultaneous dynamic equations in which several variables need to be handled. Hence, the solution method to solve such dynamic system need to be introduced. The dynamic system with several dynamic equations and several variables can be equivalent with a single higher order equation with a single variable. Hence, the solution of a dynamic system would still include a set of PI and CF, and the dynamic stability of the system would still depend on the absolution values (for difference equation system) of the characteristic roots in the CF. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 23 / 36

The Transformation of a Higher-order Dynamic Equation In particular, a SO difference equation can be rewritten as two simultaneous FOC equations in two variables. Consider the following example: y t+2 + a 1 y t+1 + a 2 y t = c (39) If we introduce an artificial new variable x, defined as x t = y t+1, we can then express the original SO equation by the following two FO DE x t+1 + a 1 x t + a 2 y t = c (40) y t+1 x t = 0 (41) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 24 / 36

Solving Simultaneous Dynamic Equations Suppose that we are given x t+1 + 6x t + 9y t = 4 (42) y t+1 x t = 0 (43) To solve this two-de system, we still need to seek the PI and the CF, and sum them to obtain the desired time paths of the two variables x and y. We first solve for the PI. As usual, try the constant solution: y t+1 = y t = y and x t+1 = x t = x = x = y = 1 4. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 25 / 36

For the CF, try the following function forms x t = mb t and y t = nb t where m and n are arbitrary constants and b represents the characteristic root. Next, we need to find the values of m, n, and b that satisfy the reduced version. Substituting these guessed solutions into the above dynamic system and cancelling out the common term b t gives (b + 6) m + 9n = 0 (44) m + bn = 0 (45) which is a linear homogeneous-equation system in m and n. We can rule out the uninteresting trivial solution (m = n = 0) by requiring that b + 6 9 1 b = b2 + 6b + 9 = 0 (46) This characteristic equation have two roots b (= b 1 = b 2 ) = 3. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 26 / 36

(conti.) Given each b i (i = 1, 2), the above homogeneous equation implies that there will have an infinite number of solutions for (m, n) For this repeated-root case, we have m i = k i n i (47) x t = m 1 ( 3) t + m 2 t ( 3) t, y t = n 1 ( 3) t + n 2 t ( 3) t which must satisfy y t+1 = x t : n 1 ( 3) t+1 + n 2 (t + 1) ( 3) t+1 = m 1 ( 3) t + m 2 t ( 3) t = Setting n 1 = A 3 and n 2 = A 4 gives m 1 = 3 (n 1 + n 2 ), m 2 = 3n 2 x c = 3A 3 ( 3) t 3A 4 (t + 1) ( 3) t (48) y c = A 3 ( 3) t + A 4 t ( 3) t (49) Note that both time paths have the same ( 3) t term, so they both explosive oscillation. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 27 / 36

Matrix Notation We can analyze the above dynamic system by using matrix. The above two-equation system can be written as [ ][ ] [ ][ ] [ ] 1 0 xt+1 6 9 xt 4 + = 0 1 y t+1 1 0 y t 0 }{{}}{{}}{{}}{{}}{{} I u K v d (50) Try a constant PI first, [ ] x (I + K ) = d = y [ x y ] = (I + K ) 1 d = [ 1/4 1/4 ]. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 28 / 36

(conti.) Next, try the CF [ ] [ ] [ mb t+1 m u = nb t+1 = b t+1 m and v = n n [ ] m (bi + K ) = 0 n To avoid the trivial solution, we must have where A i are arbitrary constants. bi + K = 0 = b = 3 = m i = k i n i, where n i = A i, m i = k i A i ] b t = Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 29 / 36

(conti.) With distinct real roots, [ ] [ xc m1 b = 1 t + m 2b t ] [ 2 k1 A y c n 1 b1 t + n 2b2 t = 1 b1 t + k 2A 2 b2 t A 1 b1 t + A 2b2 t ]. (51) With repeated roots, [ xc y c ] [ m1 b = 1 t + m 2tb2 t n 1 b1 t + n 2tb2 t ] (52) The general solution can be written as [ ] [ xt xc = y t y c ] + [ x y ]. (53) Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 30 / 36

Two-Variable Phase Diagram: Discrete-time Case Now we shall discuss the qualitative-graphic (phase-diagram) analysis of a nonlinear difference equation system. Specifically, we focus on the following two-equation system x t+1 x t = f (x t, y t ) (54) y t+1 y t = g (x t, y t ) (55) which is called autonomous system (t is not an explicit argument in f and g). The two-variable phase diagram (PD) can answer the qualitative questions: the location and the dynamic stability of the intertemporal equilibrium. The most crucial task of the PD is to determine the direction of movement of the two variables over time. In the two-variable case, we can also draw the PD in the space of (x, y). Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 31 / 36

(Conti.) In this case, we have two demarcation lines: x t+1 = x t+1 x t = f (x t, y t ) = 0 (56) y t+1 = y t+1 y t = g (x t, y t ) = 0 (57) which interact at point E representing the intertemporal equilibrium ( x t+1 = 0 and y t+1 = 0) and divide the space into 4 regions. (will be specified later.) If the demarcation line can be solved for y in terms of x, we can plot the line in the (x, y) space. Otherwise, we can use the implicit-function theorem to derive: slope of x t+1 = dy dx f / x x t+1 =0 = f / y = f x ; (58) f y slope of y t+1 = dy dx y t+1 =0 = g/ x g/ y = g x g y. (59) Specifically, we assume that f x < 0, f y > 0, g x > 0, g y < 0,which means that both slopes are positive. Further assume that f x fy > g x g y. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 32 / 36

(conti.) The two curves, at any other point, either x or y changes over time according to the signs of x t+1 and y t+1 at that point: d ( x t+1 ) dx = f x < 0, (60) which means that as we move from west to east in the space (as x increases), x t+1 decrease so that the sign of x t+1 must pass through three stages, in the order: +, 0,. Similarly, d ( y t+1 ) dy = g y < 0, (61) which means that as we move from south to north in the space (as y increases), y t+1 decreases so that the sign of y t+1 must pass through three stages, in the order: +, 0,. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 33 / 36

Linearization of a Nonlinearization Difference-Equation System Another qualitative technique of analyzing a nonlinear difference equation system is to examine its linear approximation which is derived by using the Taylor expansion of the system around its intertemporal equilibrium. At the point of expansion (i.e., the IE), the linear approximation has the same equilibrium as the original nonlinear system. In a suffi ciently small neighborhood of E, the linear approximation should have the same general streamline configuration as the original system. As long as we confine our stability analysis to the immediate neighborhood of the IE, the approximated system can include enough information from the original nonlinear system. This analysis is called local stability analysis. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 34 / 36

(Conti.) For the two difference equation system, we have x t+1 = f (x 0, y 0 ) + f x (x 0, y 0 ) (x x 0 ) + f y (x 0, y 0 ) (y y 0 ) y t+1 = g (x 0, y 0 ) + g x (x 0, y 0 ) (x x 0 ) + g y (x 0, y 0 ) (y y 0 ) For purpose of local stability analysis, the above linearization can be put a simpler form. First, the expansion point is the IE, (x, y) and f (x, y) = g (x, y) = 0. We then have another form of linearization x t+1 (1 + f x (x, y)) x f y (x, y) y = f x (x, y) x f y (x, y) y y t+1 g x (x, y) x (1 + g y (x, y)) y = g x (x, y) x g y (x, y) y which means that [ xt+1 x y t+1 y ] [ ] 1 + fx f y g x 1 + g y (x,y ) }{{} J E [ xt x y t y ] = [ 0 0 ]. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 35 / 36

(Conti.) The Jacobian matrix JE in the above reduced system can determine the local stability of the equilibrium. Denote [ ] [ ] 1 + fx f J E = y a b = (62) g x 1 + g y c d (x,y ) The characteristic roots of the reduced linearization is r a b c r d = r 2 (a + d) r + (ad bc) = 0 = trace (J E ) = r 1 + r 2 = a + d = 2 + f x + g y (63) det (J E ) = r 1 r 2 = ad bc = (1 + f x ) (1 + g y ) f y g x = (64) r 1, r 2 = trace (J E ) ± (trace (J E )) 2 4 det (J E ) 2 There are also four cases for the local stability of the above system, but here we only focus on the most popular economic case: The saddle-point case in which r 1 > 1 and r 2 < 1. Luo, Y. (SEF of HKU) Difference Equations September 13, 2012 36 / 36