Second-order difference equations

9.2 Second-order difference equations

A general second-order difference equation takes the form

x_t+2 = f (t, x_t, x_t+1).

As for a first-order equation, a second-order equation has a unique solution: by successive calculation we can see that given x₀ and x₁ there exists a uniquely determined value of x_t for all t ≥ 2. (Note that for a second-order equation we need two starting values, x₀ and x₁, rather than one.)

Second order linear equations

Consider the equation

x_t+2 + a_tx_t+1 + b_tx_t = c_t.

We solve this equation by first considering the associated homogeneous equation

x_t+2 + a_tx_t+1 + b_tx_t = 0.

Suppose that we find two solutions of this equation, u_t and v_t. Then

u_t+2 + a_tu_t+1 + b_tu_t = 0

and

v_t+2 + a_tv_t+1 + b_tv_t = 0.

Hence for any constants A and B we have

(Au_t+2 + Aa_tu_t+1 + Ab_tu_t) + (Bv_t+2 + Ba_tv_t+1 + Bb_tv_t) = 0,

(Au_t+2 + Bv_t+2) + (a_tAu_t+1 + a_tBv_t+1) + (b_tAu_t + b_tBv_t) = 0.

That is, Au_t + Bv_t is also a solution of the equation.

This solution has two arbitrary constants in it, so it seems it might be a general solution. But in order to be a general solution, it must be that the two solutions are really independent; for example, they can't be proportional to each other. The solutions are linearly independent if

u₀ v₀

u₁ v₁

≠ 0.

If this condition is satisfied, then Au_t + Bv_t is a general solution of the homogeneous equation.

Now return to the original equation. Suppose that u*_t is a solution of this equation. Let x_t be an arbitrary solution of the original equation. Then x_t − u*_t is a solution of the homogeneous equation. Thus x_t − u*_t = Au_t + Bv_t for some values of A and B, or

x_t = Au_t + Bv_t + u*_t.

That is, any solution of the original equation is the sum of a some solution of this equation and a solution of the homogeneous equation.

Example
Consider the equation

x_t+2 − 5x_t+1 + 6x_t = 2t − 3.

The associated homogeneous equation is

x_t+2 − 5x_t+1 + 6x_t = 0.

Two solutions of this equation are u_t = 2^t and v_t = 3^t. (Shortly we'll see how to find such solutions.) These are linearly independent:

1 1

2 3

= 1.

To find a solution of the original equation we can guess that it takes the form u*_t = at + b. In order for u*_t to be a solution we need

a(t+2) + b − 5[a(t+1) + b] + 6(at + b) = 2t−3.

Equating coefficients, we have a = 1 and b = 0. Thus u*_t = t is a solution.
We conclude that the general solution of the equation is

x_t = A2^t + B3^t + t.

Second-order linear equations with constant coefficients

A second-order linear equation with constant coefficients takes the form

x_t+2 + ax_t+1 + bx_t = c_t.

The strategy for solving such an equation is very similar to the strategy for solving a second-order linear differential equations with constant coefficients. We first consider the associated homogeneous equation

x_t+2 + ax_t+1 + bx_t = 0.

The homogeneous equation

We need to find two solutions of the homogeneous equation

x_t+2 + ax_t+1 + bx_t = 0.

We can guess that a solution takes the form u_t = m^t. In order for u_t to be a solution, we need

m^t(m² + am + b) = 0

or, if m ≠ 0,

m² + am + b = 0.

This is the characteristic equation of the difference equation. Its solutions are

−(1/2)a ± √((1/4)a² − b).

We can distinguish three cases:

Distinct real roots: If a² > 4b, the characteristic equation has distinct real roots, and the general solution of the homogeneous equation is
Am₁^t + Bm₂^t,
where m₁ and m₂ are the two roots.
Repeated real root: If a² = 4b, then the characteristic equation has a single root, and the general solution of the homogeneous equation is
(A + Bt)m^t,
where m = −(1/2)a is the root.
Complex roots: If a² < 4b, then the characteristic equation has complex roots, and the general solution of the homogeneous equation is
Ar^t cos(θt + ω),
where A and ω are constants, r = √b, and cos θ = −a/(2√b), or, alternatively,
C₁r^t cos(θt) + C₂r^t sin(θt),
where C₁ = A cos ω and C₂ = −A sin ω (using the formula that cos(x+y) = (cos x)(cos y) − (sin x)(sin y).

In each case the solutions are linearly independent:

In the first case the solutions u_t = m₁^t and v_t = m₂^t are linearly independent:

u₀ v₀

u₁ v₁

=

1 1

m₁ m₂

= m₂ − m₁ ≠ 0
In the second case the solutions u_t = m^t and v_t = tm^t are linearly independent:

1 0

m m

= m ≠ 0
In the third case the solutions u_t = r^t cos(θt) and v_t = r^t sin(θt) are linearly independent:

1 0

r cos θ r sin θ

= r sin θ = √b√(1 − cos²θ) = √b√(1 − a²/4b) = √(b − (1/4)a²) > 0.

In the third case, when the characteristic equation has complex root, the solution oscillates. Ar^t is the amplitude (which depends on the initial conditions) at time t, and 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

x_t+2 + x_t+1 − 2x_t = 0.

The roots of the characteristic equation are 1 and −2 (real and distinct). Thus the solution is

x_t = A + B(−2)^t.

Example
Consider the equation

x_t+2 + 6x_t+1 + 9x_t = 0.

The roots of the characteristic equation are −3 (real and repeated). Thus the solution is

x_t = (A + Bt)(−3)^t.

Example
Consider the equation

x_t+2 − x_t+1 + x_t = 0.

The roots of the characteristic equation are complex. We have r = 1 and cos θ = 1/2, so θ = (1/3)π. So the general solution is

x_t = Acos((1/3)πt + ω).

The frequency is (π/3)/2π = 1/6 and the growth factor is 1, so the oscillations are undamped.

The original (nonhomogeneous) equation

To find the general solution of the original equation

x_t+2 + ax_t+1 + bx_t = c_t

we need to find one of its solutions. Suppose that b ≠ 0.

The form of a solution depends on c_t.

Suppose that c_t = c for all t. Then x_t = C is a solution if C = c/(1 + a + b) and if 1 + a + b ≠ 0. (If 1 + a + b = 0 then try x_t = Ct; if that doesn't work try x_t = Ct².)

More generally, if c_t is a linear combination of terms of the form q^t, t^m, cos(pt), and sin(pt) (for some constants q, p, and m), and products of such terms, then the method of undetermined coefficients can be used, with a trial solution of the form suggested by c_t, as illustrated in the following examples. If c_t happens to satisfy the homogeneous equation---and hence is part of the general solution already---then a different approach must be taken, which I do not discuss.

Example
Consider the equation

x_t+2 − 5x_t+1 + 6x_t = 4^t + t² + 3.

The general solution of the associated homogeneous equation is A2^t + B3^t (found by examining the characteristic equation, as before).
To find a particular solution, try

x_t = C4^t + Dt² + Et + F .

Substituting this potential solution into the equation and equating coefficients we find that

C = 1/2, D = 1/2, E = 3/2, and F = 4

gives a solution.
Thus the general solution of the equation is

x_t = A2^t + B3^t + (1/2)4^t + (1/2)t² + (3/2)t + 4.

Stability

As for differential equations, we say that a system is stable if its long-run behavior is not sensitive to the initial conditions.

Consider the second-order equation

x_t+2 + ax_t+1 + bx_t = c_t.

Write the general solution as

x_t = Au_t + Bv_t + u_t*,

where A and B are determined by the initial conditions. This solution is globally asymptotically stable (or simply stable) if the first two terms approach 0 as t → ∞, for all values of A and B. In this case, for any initial conditions, the solution of the equation approaches the particular solution u_t*.

If the first two terms approach zero for all A and B, then u_t and v_t must approach zero. (To see that u_t must approach zero, take A = 1 and B = 0; to see that v_t must approach 0, take A = 0 and B = 1.) A necessary and sufficient condition for this to be so is that the moduli of the roots of the characteristic equation be both less than 1. (The modulus of a complex number α + βi is +√(α² + β²), which is the absolute value of number if the number is real.)

There are two cases:

If the characteristic equation has complex roots then the modulus of each root is √b (the roots are α ± βi, where α = −a/2 and β = √(b − (1/4)a²)). So for stability need b < 1.
If the characteristic equation has real roots then the modulus of each root is its absolute value. So for stability we need the absolute values of each root to be less than 1, or |−a/2 + √(a²/4 − b)| < 1 and |−a/2 − √(a²/4 − b)| < 1.

We can show (see the argument in Sysdaeter and Hammond (1995) on page 756, and problem 10 on page 758) that these conditions are equivalent to the conditions |a| < 1 + b and b < 1. Thus in summary we have:

The solution is stable if and only if modulus of each root of characteristic equation is less than 1. In terms of the coefficients of the terms in the equation, the solution is stable if and only if |a| < 1 + b and b < 1.

Exercises