Page 184 - Numerical methods for chemical engineering

170 4 Initial value problems

We can show that (4.91) is indeed a solution, as
d t 2 t 3 t 4 [0]
˙ x = I + At + AA + AAA + AAAA +· · · x
dt 2! 3! 4!
t t [0]
2 3
= A + tAA + AAA + AAAA +· · · x
2! 3!
t t [0] At [0]
2 3
= A I + At + AA + AAA +· · · x = Ae x = Ax(t) (4.93)
2! 3!

Stability of the steady state of a linear system
Obviously, ˙x = Ax has a steady state at x s = 0 where ˙x = 0, but is it a stable steady state?

Definition A steady state of a dynamic system is one in which the time derivatives of
each state variable are zero. A steady state x s is stable, if following every infinitesimal
perturbation away from x s , the system returns to x s . A steady state x s is unstable,ifany
infinitesimal perturbation causes the system to move away from x s . A steady state for which
a perturbation neither grows nor decays with time is said to be neutrally stable. Stability is
a property of the particular steady state and not of the differential equation.

We now determine the stability of x s = 0 for ˙x = Ax. Let us assume that A is diago-
N
nalizable, so that any v ∈ can be written as the linear combination
v = c 1 w [1] + c 2 w [2] +· · · + c N w [N] c j ∈ C
(4.94)
[ j] [ j]
Aw = λ j w λ j ∈ C
[0]
To determine the stability of x s = 0, we compute the response, starting at x = ε, and
examine whether x(t) returns to x s = 0 or it diverges. That is, if the system is stable, we
must have

At
lim x(t) = lim e ε = 0 (4.95)
t→∞ t→∞
At
We expand e At and ε = c 1 w [1] + c 2 w [2] +···+ c N w [N] to write x(t) = e ε as

t [ j]
2 N
x(t) = I + At + AA + ··· c j w
2!
j=1
N 2
t [ j]
= c j I + At + AA +· · · w
2!
j=1
N 2 N
t 2 [ j] λ j t [ j]
= c j 1 + tλ j + λ +· · · w = c j e w (4.96)
j
2!
j=1 j=1
In general, the eigenvalues of A are complex
a j = Re(λ j ) b j = Im(λ j ) (4.97)
λ j = a j + ib j