Page 186 - Numerical methods for chemical engineering
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172 4 Initial value problems
Stability of a steady state of a nonlinear system
We now generalize the stability results to nonlinear systems, ˙x = f (x; Θ), with a steady
state x s , where f (x s ; Θ) = 0 . x s is stable if lim t→∞ x(t) − x s = 0 following any
infinitesimal perturbation ε away from x s . If we only perturb the system slightly, we can
represent each function near x s as
N N
∂ f j ∂ f j
f j (x s + ε) ≈ f j (x s ) + ε k = ε k (4.102)
∂x k ∂x k
k=1 k=1
x s x s
Defining the Jacobian matrix, whose elements are functions of x and Θ,
(∂ f 1 /∂x 1 ) (∂ f 1 /∂x 2 ) ... (∂ f 1 /∂x N )
(∂ f 2 /∂x 1 ) (∂ f 2 /∂x 2 ) ... (∂ f 2 /∂x N )
∂ f m
. . .
J mn (x; Θ) =
. . .
J(x; Θ) =
. . . ∂x n (x;Θ)
(∂ f N /∂x 1 )(∂ f N /∂x 2 ) ... (∂ f N /∂x N )
(4.103)
the function vector in the vicinity of the steady state is approximately
N
f j (x s + ε) ≈ J jk (x s ; Θ)ε k ⇒ f (x s + ε; Θ) ≈ J(x s ; Θ)ε (4.104)
k=1
As x s is fixed, d(x s + ε)/dt = ˙ε, and the dynamics near x s are described by
˙ ε = J(x s ; Θ)ε (4.105)
Thus, we can apply the stability analysis presented above, using
[k] [k]
J(x s ; Θ)w = λ k w (4.106)
The steady state x s of ˙x = f (x; Θ)is stable with respect to infinitesimal perturbations if
every eigenvalue of the Jacobian matrix J(x s ;Θ) has a real part less than zero; Re(λ k ) < 0,
for all k.
If any eigenvalue of J(x s ;Θ) has a positive real part, x s is unstable with respect to
infinitesimal perturbations; Re(λ k ) > 0 for any k.
If all eigenvalues of J(x s ; Θ) have nonnegative real parts, but there is at least one
with a zero real part, x s is neutrally stable with respect to infinitesimal perturbations;
Re(λ k ) ≤ 0 for all k, Re(λ m ) = 0 for some m.
Note that in each of the statements above, we have included the restriction “with respect
to infinitesimal perturbations.” The system may very well respond unstably to large, finite
perturbations even if all eigenvalues of the Jacobian have negative real parts.
Example. Stability of steady states for nonlinear ODE systems
Consider the system of two nonlinear ODEs
2 2
˙ x 1 =−2(x 1 − 1) − 2(x 1 − 1) + (x 2 − 1) =−2x + 2x 1 + x 2 − 1 = f 1
1
2 2 (4.107)
˙ x 2 =−(x 1 − 1) − 3(x 2 − 1) − 4(x 2 − 1) =−x 1 − 3x + 2x 2 + 2 = f 2
2