6.7 The Korteweg–de Vries Equation 265Eliminating the sum common to (6.7.3) and (6.7.5) and the sum common
to (6.7.4) and (6.7.6),
v=Dx(logA)=∑
r∑
sA
rs
−∑
rbr, (6.7.7)Dx(Aij
)=1
2(bi+bj)Aij
−∑
r∑
sA
is
Arj. (6.7.8)
Returning to (A) and (B),
Dt(logA)=∑
rb3
rerArr
, (6.7.9)Dt(Aij
)=−∑
rb3
r
erAir
Arj. (6.7.10)
Now return to (C) and (D) withfr=gr=b
3
r
.
∑
rb3
r
erArr
+∑
r∑
s(b2
r
−brbs+b2
s)A
rs
=∑
rb3
r, (6.7.11)
∑
rb3
r
erAir
Arj
+∑
r∑
s(b2
r
−brbs+b2
s)A
is
Arj=
1
2
(b3
i+b3
j)Aij. (6.7.12)
Eliminating the sum common to (6.7.9) and (6.7.11) and the sum common
to (6.7.10) and (6.7.12),
Dt(logA)=∑
rb3
r−∑
r∑
s(b2
r−brbs+b2
s)Ars
, (6.7.13)Dt(Aij
)=∑
r∑
s(b2
r−brbs+b2
s)Ais
Arj
−1
2
(b3
i+b3
j)Aij
.(6.7.14)The derivativesvxandvtcan be evaluated in a convenient form with the
aid of two functionsψisandφijwhich are defined as follows:
ψis=∑
rbi
rArs
, (6.7.15)φij=∑
sbj
sψis,=
∑
r∑
sbi
rbj
sArs=φji. (6.7.16)They are definitions ofψisandφij.
Lemma.The function φij satisfies the three nonlinear recurrence rela-
tions:
a.φi 0 φj 1 −φj 0 φi 1 =1
2(φi+2,j−φi,j+2),b.Dx(φij)=
1
2(φi+1,j+φi,j+1)−φi 0 φj 0 ,