140 6. FUNCTIONAL SEPARABILITY
e reader can verify that (6.113) arises on solving
u
xy
D F .u/u
x
u
y
; u
xz
D F .u/u
x
u
z
; u
yz
D F .u/u
y
u
z
; (6.114)
where F D f
00
=f
0
. We wish to make (6.112) and (6.114) compatible. By differentiating
(6.112) and (6.114) with respect to x; y; and z, we calculate all third-order derivatives. Re-
quiring that these be compatible with each other gives rise to
2
F
0
C F
2
u
xx
F u
2
x
F
00
C 2FF
0
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q D 0;
(6.115a)
2
F
0
C F
2
u
yy
F u
2
y
F
00
C 2FF
0
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q D 0;
(6.115b)
2
F
0
C F
2
u
zz
F u
2
z
F
00
C 2FF
0
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q D 0;
(6.115c)
and the first-order condition
3F
00
C 8FF
0
C 2F
3
u
2
x
C u
2
y
C u
2
z
3Q
00
C 3FQ
0
C
7F
0
C 4F
2
Q D 0: (6.116)
We can solve (6.115) for u
xx
; u
yy
; and u
zz
, provided that F
0
C F
2
¤ 0. As F
0
C F
2
D 0 is a
special case, we consider this first. is gives F D u
1
, with the constant of integration omitted.
Equation (6.115) reduce to
u
2
Q
00
uQ
0
C Q D 0: (6.117)
is integrates to
Q D q
1
u C q
2
u ln juj; (6.118)
where q
1
and q
2
are arbitrary constants, with q
2
¤ 0 as we are interested in NLPDEs. However,
as we can scale u in our original PDE (6.112) with this particular Q, we can set q
1
D 0 without
loss of generality. Since F D f
00
=f
0
, then from (6.113) we obtain
u D e
ACBCC
: (6.119)
Substitution of (6.119) into (6.112) with (6.118) gives
A
00
C A
02
C B
00
C B
02
C C
00
C C
02
e
ACBCC
D q
2
e
ACBCC
.
A C B C C
/
; (6.120)
which we see separates into
A
00
C A
02
D q
2
A C a; B
00
C B
02
D q
2
B C b; C
00
C C
02
D q
2
C C c; (6.121)
where a; b and c are constants such that a C b C c D 0. We can further transfer the constants
a; b; and c to the solution (6.119), allowing us to set a D b D c D 0 and the ODEs in (6.121)