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

164 CHAPTER 8. BOUNDARY VALUE PROBLEMS
where A(t) = f y (t, y k (t)) and
For the boundary condition we apply the chain rule
where
If we define
then
Writing y = y k+1 gives
q(t) = f(t, y k (t)) − A(t)y k (t) (8.84)
0 = g + g u (y k (a), y k (b))∆y(a) + g v (y k (a), y k (b))∆y(b) (8.85)
= g + B k a(y k+1 (a) − y k (a)) + B k b (yk+1 (b) − y k (b)) (8.86)
B k a = ∂g
∂u (yk (a), y k (b)); B k b = ∂g
∂v (yk (a), y k (b)) (8.87)
d = −g(y k (a), y k (b)) + B 0 y k (a) + B b y k (b) (8.88)
B a y k+1 (a) + B b y k+1 (b) = d (8.89)
y ′ = A(t)y + q(t) (8.90)
B a y(a) + B b y(b) = d (8.91)
Example 8.4. Calculate A(t) and q(t) for the linear example 8.1, where
y ′ = d ( ) (
)
u v
=
dt v u/t 2 − v/t
Solution. Since
B a =
( 1 0
0 0
(
f(t, u, v) =
) ( 0 0
; B b =
1 0
v
u/t 2 − v/t
the Jacobian is given by
A(t) = df
dy = ∂(f ( )
1, f 2 )
∂(u, v) = ∂f1 /∂u ∂f 1 /∂v
=
∂f 2 /∂u ∂f 2 /∂v
)
)
( 0 1
1/t 2 −1/t
)
(8.92)
(8.93)
(8.94)
(8.95)
where the f 1 and f 2 denote the first and second vector components of f. Similarly,
as expected.
q(t) = f k − Ay k (8.96)
(
v
=
k ) ( ) ( )
0 1 u
k
u k /t 2 − v k −
/t 1/t 2 −1/t v k (8.97)
( ) 0
=
(8.98)
0
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007