868
Appendix: All Solutions to the Problems
(2) T = −
d
2
dx 2
p(x)
d
2
dx 2
, D T : {u(x)|u(x) ∈ C
4
[0, l]}, with p(x) ∈ C
2
[0, l], and
greater than zero.
Solution
(1) Let v(x)|v(x) ∈ C
2
[0, l], make the inner product
(T u, v) = −
l
0
(x
2 u
)
vdx = − x
2 u
v
l
0
+
l
0
x
2 u
v
dx
= − x
2 u
v
l
0
+
l
0
x
2 v
du = − x
2 u
v
l
0
+ x
2 v
u
l
0
−
l
0
(x
2 v
)
udx
= x
2
(v
u − u
v)
l
0
+ (T v, u)
It can be seen that in order to make T become the symmetric operator, there
should be
x
2
(v
u − u
v)
l
0
= 0
(2) Let v(x)|v(x) ∈ C
4
[0, l], make the inner product
(T u, v) = −
l
0
( pu
)
vdx = −( pu
)
v
l
0
+
l
0
( pu
)
v
dx
= −( pu
)
v
l
0
+ ( pu
)v
l
0
−
l
0
( pu
)v
dx
= −( pu
)
v
l
0
+ ( pu
)v
l
0
− ( pv
)u
l
0
+
l
0
( pv
)
u
dx
= −( pu
)
v
l
0
+ p(u
v
− v
u
)
l
0
+ ( pv
)
u
l
0
−
l
0
( pv
)
udx
= −( pu
)
v
l
0
+ p(u
v
− v
u
)
l
0
+ ( pv
)
u
l
0
+ (T v, u)
It can be seen that in order to make T become the symmetric operator, there
should be
[u( pv
)
− v( pu
)
+ p(u
v
− v
u
)]
l
0
= 0
7.4 Let the domain of definition of the operator T =
d
dx
be D T : {u(x)|u(x) ∈
C
1
[x 0 , x 1 ], y(x 0 ) = y(x 1 ) = 0}, find the conjugate operator of T .
Solution Let v(x)|v(x) ∈ C
1
[x 0 , x 1 ], make the inner product