142 6. FUNCTIONAL SEPARABILITY
Eliminating Q
000
between (6.128) and (6.129) gives two factors: one is F
0
C F
2
, which we as-
sume is not zero; the second is
3F
0
C F
2
Q
00
3F
00
C 11FF
0
C 3F
3
Q
0
C
5FF
00
7F
02
C 7F
2
F
0
C 2F
4
Q D 0:
(6.130)
It is left as an exercise to the reader to show that if (6.130) is satisfied, then (6.128) and (6.129)
are automatically satisfied. To solve (6.130) for a given Q would seem like a daunting task—
solving a highly nonlinear second order ODE for F . However, the ODE is linear for Q and it
is this road we will take. We will first introduce the substitution F D 3S
0
=S. e motivation for
this is in the first term 3F
0
C F
2
. is gives
S
2
S
00
Q
00
S
2
S
000
C 8SS
0
S
00
Q
0
C
5SS
0
S
000
7SS
002
C 20S
02
S
00
Q D 0: (6.131)
One exact solution of (6.131) would then allow us to reduce the order of the ODE. Seeking a
solution of the form Q D Q.S; S
0
/, we are led to the solution Q D S
5
S
0
; letting Q D S
5
S
0
P
where P D P .u/ gives
SS
0
S
00
P
00
SS
0
S
000
2SS
002
2S
02
S
00
P
0
D 0; (6.132)
from which we integrate, giving
P D p
1
C p
2
Z
S
00
S
2
S
002
du; (6.133)
where p
1
and p
2
are constant. Now we will relate this to f . Since F D
3S
0
S
D
f
00
f
0
, then
S
3
D
1
f
0
, which allows (6.133) to be integrated, giving
P D p
1
C p
2
3
f
02
f
00
2f
: (6.134)
We note that f
00
¤ 0 as f
00
D 0 gives Q D 0. is, in turn, gives
Q D
p
1
3
f
00
f
03
p
2
3
f
00
f
03
3
f
0
2
f
00
2f
; (6.135)
or
Q D q
1
f
00
f
03
C q
2
2ff
00
f
03
3
f
0
; (6.136)
where q
1
and q
2
are constant. Now that we know the form of Q, we go after the forms of A; B;
and C . Recall that f .u/ D A.x/ C B.y/ C C.z/, so
u
x
D
A
0
f
0
; u
y
D
B
0
f
0
; u
z
D
C
0
f
0
; (6.137)
6. FUNCTIONAL SEPARABILITY 143
and
u
xx
D
A
00
f
0
f
00
A
02
f
03
; u
yy
D
B
00
f
0
f
00
B
02
f
03
; u
zz
D
C
00
f
0
f
00
C
02
f
03
: (6.138)
Substitution of these into (6.112) with Q given in (6.136) gives
A
00
C B
00
C C
00
f
0
A
02
C B
02
C C
02
f
00
f
03
D q
1
f
00
f
03
C q
2
2ff
00
f
03
3
f
0
; (6.139)
or
A
00
C B
00
C C
00
C 3q
2
D
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g; (6.140)
where g D f
00
=f
02
. If we differentiate (6.140) with respect to x, y; and z, we obtain
A
000
D
2A
0
A
00
C 2q
2
A
0
g C
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g
0
A
0
f
0
; (6.141a)
B
000
D
2B
0
B
00
C 2q
2
B
0
g C
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g
0
B
0
f
0
; (6.141b)
C
000
D
2C
0
C
00
C 2q
2
C
0
g C
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g
0
C
0
f
0
: (6.141c)
Further cross differentiation of (6.141) and canceling the common term A
0
B
0
=f
0
gives
2
A
00
C B
00
C 2q
2
g
0
C
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g
0
f
0
0
D 0; (6.142a)
2
A
00
C C
00
C 2q
2
g
0
C
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g
0
f
0
0
D 0; (6.142b)
2
B
00
C C
00
C 2q
2
g
0
C
A
02
C B
02
C C
02
C q
1
C 2q
2
.A C B C C /
g
0
f
0
0
D 0: (6.142c)
Upon subtraction, (6.142) gives rise to
A
00
B
00
g
0
D 0;
A
00
C
00
g
0
D 0;
B
00
C
00
g
0
D 0: (6.143)
is leads to
A
00
D B
00
D C
00
D 2 (constant); (6.144)
as g
0
D 0 gives that F
0
C F
2
D 0 which we assume to be nonzero. If D 0 then A, B; and C
are linear, so from (6.113) solutions are of the form
f .u/ D ax Cby Ccz Cd; (6.145)
where a d are constant and it’s relatively straightforward to deduce that (6.112) will admit
solutions of this type, provided that Q is of the form
Q D
a
2
C b
2
C c
2
f
00
f
03
: (6.146)
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