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

158 CHAPTER 8. BOUNDARY VALUE PROBLEMS
and hence, since q = 0,
⎛
⎜
y(t) = c 1 ⎝
where the parameter vector c = ( c 1
equation 8.16.
Substituting 8.16 into 8.28,
If we define the matrix
then
t 2 + 1
2t
t 2 − 1
2t 2
⎞ ⎛
⎟ ⎜
⎠ + c 2 ⎝
t 2 − 1
2t
t 2 + 1
2t 2
⎞
⎟
⎠ (8.30)
c 2
) T is determined by the boundary conditions,
y(a) = Y (a)c (8.31)
y(b) = Y (b)c − Y (b)
∫ b
a
Y −1 (s)q(s)ds (8.32)
d = B a y(a) + B b y(b) (8.33)
= (B a Y (a) + B b Y (b))c − B b Y (b)
∫ b
a
Y −1 (s)q(s)ds (8.34)
Q = B a Y (a) + B b Y (b) (8.35)
c = Q −1 [d − B b Y (b)
∫ b
a
]
Y −1 (s)q(s)ds
(8.36)
Example 8.3. Solve the boundary value problem considered in section 8.1 using
this formalism.
Solution. We have a = 1 and b = 2, so
( ) 1 0
Y (a) = ; Y (b) =
0 1
( 5/4
) 3/4
3/8 5/8
(8.37)
and hence (see example 8.1),
( ) ( )
1 0 1 0
Q =
+
0 0 0 1
Hence
Q −1 =
( 0 0
1 0
) ( ) 5/4 3/4
=
3/8 5/8
( 1 0
)
−5/3 4/3
and therefore (since q = 0 for this problem)
( )
c = Q −1 1 0
d =
=
−5/3 4/3
( ) 1
−1/3
( ) 1 0
5/4 3/4
(8.38)
(8.39)
(8.40)
The solution of the differential equation is then (we only state the first component
since that is all we are really interested in, as it originated in a 2nd order equation):
u(t) = t2 + 1
2t
− 1 t 2 − 1
= t2 + 2
3 2t 3t
(8.41)
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007