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

72 CHAPTER 4. IMPROVING ON EULER’S METHOD
The error after n steps is then
e n (t) = |y(t) − ŷ(t)| = δe λ(t−t 0)
1 − e −nλh
∣ e λh − 1 ∣ (4.25)
For |nλh| t 0 + nh, the e λ(t−t 0) factor still
dominates the fraction, and hence the error still tends towards zero.
The analogous vector test equation is written as
y ′ = Ay, y(0) = y 0 (4.27)
where A is a square matrix of constants. The solution is given by the matrix
exponential
y = e At y 0 (4.28)
where the matrix exponential is defined by its Taylor Series,
∞∑
e At t k A k
=
k!
k=0
(4.29)
If the matrix A is diagonalizable, or equivalently, has n linearly independent eigenvectors
v 1 , . . . , v n , with corresponding eigenvalues λ 1 , . . . , λ n , then
where
and
e At = UEU −1 (4.30)
E = diag(e λ 1t , . . . , e λnt ) (4.31)
U = (v 1 , . . . , v n ) (4.32)
Example 4.1. Solve the initial value problem
( ) (
y ′ 0 1
1
= Ay, A = , y(0) =
−2 0
0)
(4.33)
Solution. A is diagonalizable with eigenvalues ±i √ 2 and corresponding eigenvectors
( ) ( )
√ i
2
− √i
and 2
(4.34)
1 1
We have then
y = e At y 0 =
=
=
(
− √i
i
2
1 1
√
2
(
e −i √ )
2t +e i√ 2t
2
i ei√ 2t −e −i√ 2t
√
2
( cos
(√
2t
)
− √ 2 sin (√ 2t ) )
) (
) (
e i√ 2t i
0
0 e −i√ 2t
√ 1
2 2
− √ i 1
2 2
)
(1
0
)
(4.35)
(4.36)
(4.37)
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007