Mathematical methods for economic theory

Martin J. Osborne

9.2 Second-order difference equations

Definition
A second-order difference equation is an equation
xt+2 = f(txtxt+1),
where f is a function of three variables. A solution of the second-order difference equation xt+2 = f(txtxt+1) is a function x of a single variable whose domain is the set of integers such that xt+2 = f(txtxt+1) for every integer t, where xt denotes the value of x at t.
As for a first-order difference equation, we can find a solution of a second-order difference equation by successive calculation. The only difference is that for a second-order equation we need the values of x for two values of t, rather than one, to get the process started. Given xt and xt+1 for some value of t, we use the equation to find xt+2, and then use the equation again for xt+1 and xt+2 to find xt+3, and so forth. At each step, the new value of x is uniquely determined, so we have the following result.
Proposition
For every pair of numbers x0 and x1, every second-order difference equation xt+2 = f(txtxt+1) has a unique solution in which the value of x is x0 at 0 and x1 at 1.

Second-order linear difference equations with constant coefficients

Definition
A linear second-order difference equation with constant coefficients is a second-order difference equation that may be written in the form
xt+2 + axt+1 + bxt = ct,
where a, b, and ct for each value of t, are numbers. The equation is homogeneous if ct = 0 for all t.
The method for finding a solution of a linear second-order difference equation follows the lines of the method for finding a solution of a linear second-order differential equation. Suppose that x and y are solutions of the equation
xt+2 + axt+1 + bxt = ct,
which we refer to subsequently as the “original” equation. Define zt = yt − xt. Then zt+2 + azt+1 + bzt = [yt+2 + ayt+1 + byt] − [xt+2 + axt+1 + bxt] = ct − ct = 0. That is, z is a solution of the homogeneous equation
xt+2 + axt+1 + bxt = 0.
Conversely, let x be a solution of the original equation, let z be a solution of the homogeneous equation, and define yt = xt + zt for all t. Then y is a solution of the original equation.

Thus we have the following result.

Proposition
Let x be a solution of the linear second-order ordinary difference equation with constant coefficients
xt+2 + axt+1 + bxt = ct.
Then y is a solution of this equation if and only if yt = xt + zt for all t, where z is a solution of the (homogeneous) equation
xt+2 + axt+1 + bxt = 0.
An implication of this result is that we can find the set of all solutions of the original equation by finding the set of all solutions of the homogeneous equation and a single solution of the original equation, as described in the following procedure.
Procedure for finding general solution of linear second-order difference equation with constant coefficients
The solutions of the difference equation
xt+2 + axt+1 + bxt = ct
may be found as follows.
  1. Find the solutions of the associated homogeneous equation xt+2 + axt+1 + bxt = 0.
  2. Find a single solution of the original equation xt+2 + axt+1 + bxt = ct.
  3. Add together the solutions found in steps 1 and 2.

1. Find the solutions of a homogeneous equation

You might guess that the homogeneous equation
xt+2 + axt+1 + bxt = 0
has a solution of the form xt = mt. For this function to be a solution, we need
mt+2 + amt+1 + bmt = 0
for all t, or
mt(m2 + am + b) = 0
or, if m ≠ 0,
m2 + am + b = 0.
This equation is the characteristic equation of the difference equation. Its solutions are
−(1/2)a ± √((1/4)a2 − b).
We have the following result, analogous to a result for homogeneous second-order differential equations.
Proposition  
Consider the homogeneous linear second-order difference equation with constant coefficients
xt+2 + axt+1 + bxt = 0.
The set of solutions of this equation depends on the character of the roots of the characteristic equation m2 + am + b = 0 as follows.
Distinct real roots
If a2 > 4b, in which case the characteristic equation has distinct real roots, say m1 and m2, every solution of the equation has the form
Amt
1
 + Bmt
2
for numbers A and B.
Single real root
If a2 = 4b, in which case the characteristic equation has a single root, every solution of the equation has the form
(A + Bt)mt
for numbers A and B, where m = −(1/2)a is the root.
Complex roots
If a2 < 4b, in which case the characteristic equation has complex roots, every solution of the equation has the form
Art cos(θt + ω)
for numbers A and ω, where r = √b and cos θ = −a/(2√b). This solution may alternatively be expressed as
C1rt cos(θt) + C2rt sin(θt),
where C1 = A cos ω and C2 = −A sin ω (using the formula cos(x+y) = (cos x)(cos y) − (sin x)(sin y).
Source  
For a proof, see Goldberg (1958), pp. 134–142.
In the third case, in which the characteristic equation has complex roots, the solutions oscillate. We use the following terminology: Art is the amplitude of the oscillations at time t, r is growth factor, θ/2π is the frequency of the oscillations, and ω is the phase (which depends on the initial conditions). If |r| < 1 then the oscillations are damped; if |r| > 1 then they are explosive.
Example
Consider the equation
xt+2 + xt+1 − 2xt = 0.
The roots of the characteristic equation are 1 and −2 (real and distinct). Thus every solution has the form
xt = A + B(−2)t.
Example
Consider the equation
xt+2 + 6xt+1 + 9xt = 0.
The single root of the characteristic equation is −3. Thus every solution has the form
xt = (A + Bt)(−3)t.
Example
Consider the equation
xt+2 − xt+1 + xt = 0.
The roots of the characteristic equation are complex. We have r = 1 and cos θ = 1/2, so θ = (1/3)π. So every solution has the form
xt = Acos((1/3)πt + ω).
The frequency of the oscillations is (π/3)/2π = 1/6 and the growth factor is 1, so the oscillations are neither damped nor explosive.

2. Find a solution of a nonhomogeneous equation

Consider the nonhomogeneous equation
xt+2 + axt+1 + bxt = ct.
If ct = c for all t, then xt = c/(1 + a + b) is a solution if 1 + a + b ≠ 0, xt = ct/(a + 2) is a solution if 1 + a + b = 0 and a ≠ −2, and xt = ct2/2 if 1 + a + b = 0 and a = −2 (in which case b = 1).

For other forms of ct, the method used to find a solution of a nonhomogeneous second-order differential equation can be used. For example, if ct is a linear combination of terms of the form qt, tm, cos(pt), and sin(pt), for constants q, p, and m, and products of such terms, then guess that the equation has a solution that is a linear combination of such terms; substitute such a function into the equation and see whether there are coefficients that generate a solution. If the ct you find happens to satisfy the homogeneous equation, then a different approach must be taken, which I do not discuss.

Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 2t − 3.
To find a solution, guess that there is one of the form at + b. For this function to be a solution, we need
a(t+2) + b − 5[a(t+1) + b] + 6(at + b) = 2t − 3.
Equating the coefficients of t and the constant on each side, we have a = 1 and b = 0. Thus xt = t for all t is a solution.
Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 4t + t2 + 3.
To find a solution, try
xt = C4t + Dt2 + Et + F.
Substituting this potential solution into the equation and equating the coefficients of 4t, t2, and the constant on each side we find that
C = 1/2, D = 1/2, E = 3/2, and F = 4.

Thus a solution of the equation is

xt = (1/2)4t + (1/2)t2 + (3/2)t + 4.

3. Add together the solutions found in steps 1 and 2

Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 2t − 3.
The associated homogeneous equation is
xt+2 − 5xt+1 + 6xt = 0.
The characteristic equation has two real roots, 2 and 3, so by a previous result, every solution of the equation has the form A2t + B3t.

By a previous example, a solution of the original equation is xt = t for all t.

We conclude that every solution of the equation has the form

xt = A2t + B3t + t.
Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 4t + t2 + 3.
As in the previous example, every solution of the associated homogeneous equation has the form A2t + B3t.

By a previous example, a solution of the original equation is

xt = (1/2)4t + (1/2)t2 + (3/2)t + 4.

Thus every solution of the original equation has the form

xt = A2t + B3t + (1/2)4t + (1/2)t2 + (3/2)t + 4.

Equilibrium and stability

The notions of equilibrium and stability for a second-order difference equation are analogous to the ones for a differential equation.
Definition
An equilibrium of the linear second-order difference equation with constant coefficients
xt+2 + axt+1 + bxt = c
is a number x* such that xt = x* for all t is a solution of the equation. An equilibrium x* is stable if every solution of the equation converges to x*.
Now, by an earlier analysis, if 1 + a + b ≠ 0 then a solution of
xt+2 + axt+1 + bxt = c
is xt = c/(1 + a + b) for all t. Thus if 1 + a + b ≠ 0 then c/(1 + a + b) is an equilibrium of the equation. If 1 + a + b = 0 then the equation has no constant solution, and hence no equilibrium.

Also, from a previous result we know that every solution of

xt+2 + axt+1 + bxt = c
is the sum of any given solution of this equation and a solution of the homogeneous equation
xt+2 + axt+1 + bxt = 0.
Thus if 1 + a + b ≠ 0 then every solution of
xt+2 + axt+1 + bxt = c
is the sum of c/(1 + a + b) and a solution of the homogeneous equation
xt+2 + axt+1 + bxt = 0.
So the equilibrium c/(1 + a + b) is stable if and only if every solution of the homogeneous equation converges to zero. We can use the earlier result characterizing the solutions of the homogeneous equation to find conditions under which these solutions converge to zero, as follows. Recall that the characteristic equation of the difference equation is m2 + am + b = 0.
Characteristic equation has distinct real roots
Every solution of the difference equation has the form
Amt
1
 + Bmt
2
for numbers A and B. These solutions converge to zero if and only if |m1| < 1 and |m2| < 1.
Characteristic equation has single real root
Every solution of the difference equation has the form
(A + Bt)mt
for numbers A and B, where m = −(1/2)a is the root. These solutions converge to zero if and only if |m| < 1. (The function tmt converges to zero if and only if |m| < 1.)
Characteristic equation has complex roots
Every solution of the equation has the form
Art cos(θt + ω)
for numbers A and ω, where r = √b and cos θ = −a/(2√b). These solutions converge to zero if and only if r < 1.
We can unify these conditions by defining the modulus of a complex number α + βi to be √(α2 + β2).

If the characteristic equation has complex roots, then, using the formula for the roots of a quadratic equation, these roots are −a/2 ± i√(b − a2/4). Thus the modulus of each root is √(a2/4 + b − a2/4) = √b. The modulus of a real number is simply its absolute value (the complex number α + βi is real if and only if β = 0), so the conditions we found for every solution of the homogeneous equation to converge to zero are, in every case, simply that the modulus of each root of the characteristic equation is less than 1.

We have the following result.

Proposition  
For any value of c, an equilibrium of the linear second-order difference equation with constant coefficients xt+2 + axt+1 + bxt = c is stable if and only if the modulus of each root of the characteristic equation m2 + am + b = 0 is less than 1, or, equivalently, if and only if |a| < 1 + b and b < 1.
Proof  
The arguments preceding the statement establish that a necessary and sufficient for the equilibrium to be stable is that the modulus of each root is less than 1.

I now show that this condition is equivalent to the conditions |a| < 1 + b and b < 1.

Note that the second condition is equivalent to the two conditions a < 1 + b, or b > a − 1, and −a < 1 + b, or b > −a − 1. In the following figure, b > a − 1 if b is above the blue line and b > −a − 1 if b is above the red line. Thus the set of values of (ab) satisfying the conditions |a| < 1 + b and b < 1 is the shaded triangle. The characteristic equation has real roots if b lies below the black curve b = a2/4 and complex roots if b lies above this curve. The figure is useful in following the logic of the subsequent slightly intricate algebra.

real complex −2 2 1 −1 a b b = a2/4 b = a − 1 b = −a − 1

First consider the case in which the roots of the characteristic equation are real. Then the roots are −a/2 ± √(a2 − 4b)/2, so the modulus of each root is less than 1 if and only if

−1  <  a/2 + √(a2 − 4b)/2 < 1
−1  <  a/2 − √(a2 − 4b)/2 < 1
or
−2 + a   <  √(a2 − 4b) < 2 + a
−2 + a   <  −√(a2 − 4b) < 2 + a.
Now, the second inequality on the left implies the first inequality on the left, and the first inequality on the right implies the second inequality on the right, so that the modulus of each roots is less than 1 if and only if
√(a2 − 4b)  <  2 + a
−2 + a  <  −√(a2 − 4b).
For the first inequality to hold, we need a > −2 (because the left-hand side is positive) and, squaring both sides,
a2 − 4b < (2 + a)2 = 4 + 4a + a2
or
1 + a + b > 0.
For the second inequality to hold, we need a < 2 (because the right-hand side is negative) and, squaring both sides,
(−2 + a)2 > a2 − 4b
(both sides are negative, so the inequality changes), or
4 − 4a + a2 > a2 − 4b
or
1 − a + b > 0.
In summary, if the roots of the characteristic equation are real, which is true if and only if a2 > 4b, then the modulus of each root is less than 1 if and only if
−2 < a < 2, 1 + a + b > 0, and 1 − a + b > 0.
Now, if b < 1 then the last two inequalities imply −2 < a < 2. Further, −2 < a < 2 and a2 > 4b imply b < 1. Thus if the roots of the characteristic equation are real, then the modulus of each root is less than 1 if and only if
b < 1, 1 + a + b > 0, and 1 − a + b > 0.
Now consider the case in which the roots of the characteristic equation are complex. Then a2 < 4b, so that b > 0, and, as noted previously, the modulus of each root is √b. Thus the modulus of each root is less than 1 if and only if b < 1. Now, a2 < 4b and b < 1 imply that |a| < 2, so that (a − 2)2 > 0. But (a − 2)2 = a2 − 4a + 4, so a2 > 4a − 4, or a2/4 > a − 1. Given a2 < 4b, we have b > a − 1, or 1 − a + b > 0. The inequality |a| < 2 implies also that (a + 2)2 > 0, so that a2 + 4a + 4 > 0, or a2 > −4a − 4, or a2/4 > −a − 1. Thus given a2 < 4b, we have b > −a − 1, or 1 + a + b > 0. Thus if the modulus of each root of the characteristic equation is less than 1, we have
b < 1, 1 + a + b > 0, and 1 − a + b > 0.

In conclusion, whether the roots of the characteristic equation are real or complex, the modulus of each root is less than 1 if and only if

b < 1, 1 + a + b > 0, and 1 − a + b > 0.