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

CHAPTER 9. DIFFERENTIAL ALGEBRAIC EQUATIONS 179
2. We could solve 9.157 for E and eliminate E completely from the system, giving
a system of three explicit ODE’s:
A ′ = −αA(1 − X) + βX (9.139)
B ′ = γX (9.140)
X ′ = αA(1 − X) − (γ + β)X (9.141)
Then the value of E can be computed from the constraint E = 1 − X after
X is known. This is the method that is generally used to solve a semi-explicit
system where we are able to solve for the constraints and reduce the system
to an ODE.
3. We can treat the system as a full implicit DAE as in 9.143 and use Newton’s
method to extract y n at each step. This is the the only procedure that can be
used for a fully implicit system, that is, a DAE where it is not clear how to
separate the constraints from the ODEs, and so we will illustrate the procedure
in detail.
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Algorithm 9.1. Fully Implicit DAE with Backward Euler
To solve the fully implicit differential algebraic equation
At each time step t n , solve
F
F(t, y, y ′ ) (9.142)
(
t n , y n , 1 )
h (y n − y n−1 ) = 0 (9.143)
for y n using Newton’s method
y k+1
n
where the Jacobian J is
= y k n − J −1 F
(
t n , y n , 1 )
h (y n − y n−1 )
J = 1 ∂F
h ∂y ′ + ∂F
∂y
In general it is more efficient to calculate updates by solving
(9.144)
(9.145)
J∆y = F (9.146)
for ∆y, e.g., by Gaussian elimination on a sparse system, and updating
y k+1
n = y k n + ∆y (9.147)
instead of inverting J.
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c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge