The Computable Differential Equation Lecture ... - Bruce E. Shapiro

172 CHAPTER 9. DIFFERENTIAL ALGEBRAIC EQUATIONS
we conclude that
v = (I − ND + · · · + N k−1 D k−1 )h (9.60)
= h − Nh ′ + N 2 h ′′ − · · · + N k−1 h ( k − 1) (9.61)
In other words, equation 9.55 is not a differential equation at all, despite the presence
of the derivative, but is an algebraic equation for v. This is sometimes referred
to as a hidden constraint of the DAE.
Two key points emerge from the preceding discussion. First, we see that DAE’s
may have hidden constraints, namely, part of the solution may be fully determined
independently of any initial conditions. Second, two many initial conditions can
over-determine the system. If too many initial conditions are provided, that is,
more than are necessary to determine u, then there may not be any solution at all.
Theorem 9.3. Let Ay ′ + By = f be a solvable linear constant coefficient DAE with
index k ≥ 1. Then (A + λB) −1 has a pole of order k at λ = 0, and (A + λB) −1 A
has pole of order k − 1 at λ = 0.
Proof. We will give the proof for the first statement. The second is left as an exercise.
Hence
I = Q −1 Q (9.62)
= Q −1 (A + λB) −1 (A + λB)Q (9.63)
= Q −1 (A + λB) −1 P −1 P (A + λB)Q (9.64)
= Q −1 (A + λB) −1 P −1 (P AQ + λP BQ) (9.65)
Q −1 (A + λB) −1 P −1 = (P AQ + λP BQ) −1 (9.66)
(A + λB) −1 = Q(P AQ + λP BQ) −1 P (9.67)
( ) −1 I + λC 0
= Q
P (9.68)
0 N + λI
( )
(I + λC)
−1
0
= Q
0 (N + λI) −1 P (9.69)
Since (this can be verified by multiplying out the product, using the fact that N k =
0)
I = 1 [I − 1 λ λ N + 1 ]
λ 2 N 2 + · · · + (−1)k−1
λ k−1 N k−1 [N + λI] (9.70)
we know that
(N + λI) −1 = 1 λ
∑k−1
i=0
−1 i
λ i N i (9.71)
Hence A + λB has elements that are proportional to 1/λ k . This gives a pole of oder
k when λ → 0.
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007