Solutions to Homework Set 6
1) Flexible rod again
: This is a somewhat tedious, but ultimately rewarding exercise.
a) Look back at your solution for homework set 4, where you showed that
Z
1
0
u
*
v
(4)
dx
= [
u
*
v
(3)

(
u
0
)
*
v
00
+ (
u
00
)
*
v
0

(
u
(3)
)
*
v
]
1
0
+
Z
1
0
(
u
(4)
)
*
v dx.
From this we see that taking
v
=
v
00
= 0 at the endpoints as the boundary conditions
for
L
requires
u
=
u
00
= 0 at the endpoints for the as the boundary conditions for
L
†
.
The operator is therefore selfadjoint.
b) There are no zero modes for
L
. The solution is therefore unique, and
F
has no condi
tions to satisfy
c) From
∂
4
x
G
(
x, y
) =
δ
(
x

y
) we see that we need
G
,
G
0
and
G
00
to be continuous at
x
=
y
, and
G
000
must obey the jump condition
G
000
(
y
+
, y
)

G
000
(
y

, y
) = 1
.
The Green function, considered as a function of
x
, must also satisfy the boundary
conditions at
x
= 0
,
1.
d) We satisfy the boundary conditions at the ends by writing
G
(
x, y
) =
(
ax
+
bx
2
,
0
< x < y <
1,
c
(
x

1) +
d
(
x

1)
3
,
0
< y < x <
1
.
The jump condition, the continuity at
x
=
y
of
G
00
,
G
0
, and
G
(in that order) then
read:
6
d

6
b
=
1
,
6
d
(
y

1)
=
6
by,
c
+ 3
d
(
y

1)
2
=
a
+ 3
by
2
,
c
(
y

1) +
d
(
y

1)
3
=
ay
+
by
3
.
The first two equations quickly give
d
=
y/
6 and
b
= (
y

1)
/
6. Rather more effort
produces
c
=
y
(
y

1)(
y
+ 1)
/
6 and
a
=
y
(
y

1)(
y

2)
/
6. Thus
G
(
x, y
) =
(
1
6
[
y
(
y

1)(
y

2)
x
+ (
y

1)
x
3
]
,
0
< x < y <
1,
1
6
[
y
(
y

1)(
y
+ 1)(
x

1) +
y
(
x

1)
3
]
,
0
< y < x <
1
.
e) To see that that the Green function does indeed obey
G
(
x, y
) =
G
(
y, x
) it helps to
rearrange the second alternative so as to expose the powers of
y
:
G
(
x, y
) =
(
1
6
[
y
(
y

1)(
y

2)
x
+ (
y

1)
x
3
]
,
0
< x < y <
1,
1
6
[
x
(
x

1)(
x

2)
y
+ (
x

1)
y
3
]
,
0
< y < x <
1
.
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 Fall '07
 Stone
 Work, UK, Boundary value problem, Boundary conditions, Dirichlet boundary condition, Green's function

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