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0593_C14_fm Page 507 Tuesday, May 7, 2002 6:56 AM

Stability 507

B
k
m F(t)
frictionless
FIGURE 14.7.1
Undamped linear mass–spring oscillator. x

these criteria form a stronger stability test than that used in the foregoing sections. Spe-
cifically, if the Routh–Hurwitz criteria are satisfied, the dynamical system will return to
the equilibrium position after a disturbance. In the foregoing sections, however, we simply
required that a small disturbance away from equilibrium remain small or bounded.
To illustrate these concepts consider a system governed by the differential equation:

2
dx dx
a + a + ax = 0 (14.7.11)
0 2 1 2
dt dt
Let the solution x have the form of Eq. (14.6.46). That is,

x = Xe t λ (14.7.12)

Then, we immediately obtain:

a λ + a λ + a = 0 (14.7.13)
2
0 1 2
The Routh–Hurwitz criteria for stability are then:

a > 0, a > 0, and a > 0 (14.7.14)
0 1 2
To interpret this result, consider again the undamped linear mass–spring oscillator
depicted in Figure 14.7.1. Recall that the governing equation for this system is (see Eq.
(13.3.1)):

+
˙˙
mx kx = 0 (14.7.15)
where, as before, m is the mass, x is the displacement, and k is the spring modulus. Then,
by seeking a solution in the form of Eq. (14.7.12), we immediately obtain:

mλ += 0 (14.7.16)
2
k
By comparison with Eq. (14.7.13), see that the coefficients a , a , and a are:
0 1 2
a = m, a = 0, and a = k (14.7.17)
0 1 2

The first and third expressions of Eq. (14.7.17) satisfy Eq. (14.7.14), but the second
expression violates the requirement a > 0. Thus, the equilibrium position x = 0 of the
1
mass–spring oscillator is unstable according to the Routh–Hurwitz criteria. Intuitively,
however, we would expect the equilibrium position to be stable because a small disturbance