Stability of a Scalar Difference Equation¶

Proposition¶

Consider the linear difference equation of order $k$ with constant coefficients and constant right-hand side

x_{t+k}+a_{k-1}x_{t+k-1}+\cdots+a_0x_t=b,

with characteristic polynomial $p$ and $p(1)\ne0$ $0$ , so that $x^*=b/p(1)$ is the unique equilibrium. The equilibrium is called (globally) asymptotically stable if every solution converges to $x^*$ , whatever the initial values. This holds if and only if every characteristic root $r$ satisfies $\lvert r\rvert<1$ $1$ .

First order, $x_{t+1}=a x_t+b$ with $a\ne1$ $1$ : asymptotically stable if and only if $\lvert a\rvert<1$ $1$ .
Second order, $x_{t+2}+a_1x_{t+1}+a_0x_t=b$ $b$ : asymptotically stable if and only if $\lvert a_0\rvert<1$ and $\lvert a_1\rvert<1+a_0$ $a_{0}$ .

Interpretation¶

The deviation $x_t-x^*$ solves the homogeneous equation, so it is a combination of terms $t^j r^t$ . Each term tends to $0$ exactly when the modulus $\lvert r\rvert$ is below $1$ ; the polynomial factor $t^j$ from a repeated root cannot prevent decay. One root with $\lvert r\rvert\ge1$ is enough to spoil convergence for some initial values.

$The time path of x_{t+1} = −0.8x_t + 9 from x_0 = 0 alternates 0, 9, 1.8, … around the line x* = 5 inside a dashed envelope that shrinks like 0.8^t.$

The figure shows the first example below, $x_{t+1}=-0.8x_t+9$ with $x_0=0$ $0$ . Because $a=-0.8$ is negative, $x_t$ jumps to alternate sides of $x^*=5$ $5$ ; because $\lvert a\rvert<1$ $1$ , the distance $\lvert x_t-x^*\rvert=5(0.8)^t$ shrinks every period, so the path stays inside the dashed envelope $5\pm5(0.8)^t$ and converges to $5$ .

Move $a$ across $\pm1$ and follow the cobweb: each period multiplies the distance to $x^*$ by $|a|$ , so every starting value $x_0$ converges when $|a|<1$ and moves away when $|a|>1$ $1$ .

The definition quantifies over all initial values. An unstable equation can still have particular solutions that converge, namely those whose coefficients on the large roots happen to be zero.

In Math V, asymptotic stability (all solutions converge) requires all roots strictly inside the unit circle. The equation is called stable when all solutions stay bounded: all roots satisfy $\lvert r\rvert\le1$ and those with $\lvert r\rvert=1$ are simple. Such roots give solutions that neither grow nor die out, so the equation is stable but not asymptotically stable. Compare the continuous-time criterion for the stability of an equilibrium of an ordinary differential equation, where decay requires $f'(x^*)<0$ (negative real parts for linear systems) rather than a modulus below $1$ .

Key Results¶

For the first-order equation, $x_t=x^*+a^t(x_0-x^*)$ $x^{*})$ . For $0<a<1$ the approach is monotone, for $-1<a<0$ it is a damped oscillation, for $\lvert a\rvert>1$ the path diverges unless $x_0=x^*$ $x^{*}$ , and for $a=-1$ it is a two-cycle.
A pair of simple nonreal roots $\rho \mathrm{e}^{\pm\mathrm{i}\theta}$ gives oscillations with amplitude proportional to $\rho^t$ : damped if $\rho<1$ $1$ , constant if $\rho=1$ $1$ , explosive if $\rho>1$ $1$ .
Necessary conditions for asymptotic stability: $\lvert a_0\rvert<1$ (since $\lvert a_0\rvert=\prod_i\lvert r_i\rvert$ $\prod_{i} ∣ r_{i} ∣$ ) and $p(1)>0$ $0$ .
The speed of convergence is set by the largest root modulus $\max_i\lvert r_i\rvert$ .
If $p(1)=0$ $0$ , $r=1$ is a root, there is no unique equilibrium and the equation is not asymptotically stable.

Examples¶

Damped oscillation. $x_{t+1}=-0.8x_t+9$ has $x^*=9/1.8=5$ and $x_t=5+(-0.8)^t(x_0-5)$ $5)$ . From $x_0=0$ $0$ : $x_1=9$ $9$ , $x_2=1.8$ $1.8$ , approaching $5$ from alternating sides. Asymptotically stable, since $\lvert-0.8\rvert<1$ $1$ .

Asymptotically stable with complex roots. $x_{t+2}-x_{t+1}+0.5x_t=0$ $0$ : here $a_0=0.5$ $0.5$ , $a_1=-1$ $- 1$ , and $\lvert a_0\rvert<1$ $1$ , $\lvert a_1\rvert=1<1.5$ $1.5$ . Directly: $r=\tfrac{1\pm\mathrm{i}}{2}$ with $\lvert r\rvert=\tfrac{\sqrt{2}}{2}\approx0.707<1$ $1$ .

Unstable, yet some solutions converge. $x_{t+2}-2.5x_{t+1}+x_t=0$ has roots $2$ and $0.5$ ; the test fails because $\lvert a_0\rvert=1$ $1$ . The general solution $c_1 2^t+c_2(0.5)^t$ diverges unless $c_1=0$ $0$ ; for instance $x_0=2$ $2$ , $x_1=1$ gives $x_t=2(0.5)^t\to0$ $0$ .

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 2

Aliases: stability of a linear recurrence, asymptotic stability of a difference equation Keywords: |a| < 1, all characteristic roots inside the unit circle, |a_0| < 1 and |a_1| < 1 + a_0, global asymptotic stability, damped oscillation Symbols: product, rho, square root, theta

Last updated: 17 October 2026 21:53