3.3. GENERALIZED KDV EQUATION 57
Eliminating F
p
from (3.56g) and (3.56h) gives
B D AF F
2
; (3.57)
and returning to either (3.56g) and (3.56h) gives
.
A 2F
/
.F A/F
p
C F
q
D 0; (3.58)
leading to two cases: (i) A D 2F and (ii) A D
FF
p
C F
q
F
p
, but each case ultimately leads to the
same result and thus, we will only pursue the first case. On setting A D 2F in equations (3.56d),
(3.56e) and (3.56f), we obtain
F
2
F
pp
2FF
pq
C F
qq
C 4FF
2
p
4F
p
F
q
D 0; (3.59a)
F
3
F
pp
2F
2
F
pq
C FF
qq
C 3F
2
F
2
p
2FF
p
F
q
F
2
q
D 0; (3.59b)
F
4
F
pp
2F
3
F
pa
C F
2
F
qq
C 2F
3
F
2
p
2FF
2
q
D 0; (3.59c)
and on the elimination of the second-order derivatives we obtain
F
q
FF
p
D 0; (3.60)
and with this shows that (3.59) is identically satisfied. As F satisfies (3.60), both (3.56b) and
(3.56c) reduce
C
q
F C
p
D 0: (3.61)
With F and C satisfying (3.59) and (3.61), and the last equation in (3.56), (3.56a) becomes
C C
p
pC
v
qF C
v
F
p
C C
2
F
pp
D 0: (3.62)
If we perform the operation @
q
F @
p
on (3.62) and use (3.60) and (3.61), we obtain
2CF
p
C
p
.p C qF /F
p
C
v
C 3C
2
F
pp
D 0: (3.63)
From (3.62) and (3.63), (upon subtraction), we have
C
F
p
C
p
C 2CF
pp
D 0: (3.64)
If C D 0 then we have
u D F .p; q/; v
xx
C 2F v
xy
C F
2
v
yy
D 0; F
q
FF
p
D 0: (3.65)
If C ¤ 0 then the compatibility of F
p
C
p
C 2CF
pp
D 0 with (3.60) and (3.61) leads to
F
pp
D 0; (3.66)