21.04.2015 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

CHAPTER 9. DIFFERENTIAL ALGEBRAIC EQUATIONS 171<br />

<strong>The</strong>orem 9.1. Let Ay ′ + By = f be a linear constant coefficient DAE with regular<br />

pencil λA + B. <strong>The</strong>n there exists nonsingular matrices P, Q, such that<br />

where u = Q −1 y,<br />

P AQ =<br />

P AQu ′ + P BQu = P f (9.50)<br />

( ) I 0<br />

; P BQ =<br />

0 N<br />

( ) C 0<br />

0 I<br />

(9.51)<br />

and N is a matrix of nilpotency k, for some integer k ≥ 0. If N = 0 define k = 1; if<br />

A is nonsignular, P AQ = I and k = 0. If det(λA + B) is constant then P AQ = n,<br />

P BQ = I.<br />

Definition 9.6. <strong>The</strong> index of the pencil λA + B is the degree of nilpotency in<br />

equation 9.51.<br />

<strong>The</strong>orem 9.2. Let Ay ′ + By = f be a linear constant coefficient DAE with regular<br />

pencil λA + B. <strong>The</strong>n the index of the pencil equals the index of the DAE.<br />

This theorem says that there exist nonsingular matrices P , Q, such that we can<br />

write<br />

( ) ( ) ( ( ) ( I 0 u<br />

′ C 0 u g<br />

0 N v ′ +<br />

=<br />

(9.52)<br />

0 I)<br />

v h)<br />

where<br />

Qy =<br />

( ( u g<br />

, P f =<br />

v)<br />

h)<br />

(9.53)<br />

<strong>The</strong>n<br />

u ′ + Cu = g (9.54)<br />

Nv ′ + v = h (9.55)<br />

and N is k-nilpotent, where k is the index of the DAE. <strong>Equation</strong> 9.54 has a unique<br />

solution that is fully determined by its initial conditions (this is the existence theorem<br />

for ordinary differential equations). <strong>Equation</strong> 9.55 also has a unique solution,<br />

but does not require initial conditions, because the matrix N is nilpotent (N k =0).<br />

To see this let D be the derivative operator; then 9.55 becomes<br />

But since<br />

NDv + v = h (9.56)<br />

(ND + I)v = h (9.57)<br />

v = (ND + I) −1 h (9.58)<br />

(I + ND) −1 = I − ND + N 2 D 2 − N 3 D 3 + N 4 D 4 + · · · + N k−1 D k−1 (9.59)<br />

c○2007, B.E.<strong>Shapiro</strong><br />

Last revised: May 23, 2007<br />

Math 582B, Spring 2007<br />

California State University Northridge

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!