6.8 The Kadomtsev–Petviashvili Equation 279new formulae for the derivatives of logA:
v=Dx(logA)=∑
r,sA
rs
−∑
rλr, (6.8.15)Dy(logA)=−∑
r,s(br−cs)Ars
+∑
rλrμr, (6.8.16)Dt(logA)=− 4∑
r,s(b2
r−brcs+c2
s)Ars+4
∑
rλr(b2
r
−brcr+c2
r). (6.8.17)
Equations (6.8.16) and (6.8.17) are not applied below but have been
included for their interest.
Eliminating the sum common to (6.8.6) and (6.8.12), the sum commonto (6.8.7) and (6.8.13), and the sum common to (6.8.8) and (6.8.14), we
find new formulas for the derivatives ofA
ij
:Dx(Aij
)=(bi+cj)Aij
−∑
r,sA
is
Arj
,Dy(Aij
)=−(b2
i
−c2
j)A
ij
+∑
r,s(br−cs)Ais
Arj
,Dt(Aij
)=−4(b3
i
+c3
j)A
ij
+4∑
r,s(b2
r
−brcs+c2
s)A
is
Arj
.(6.8.18)Define functionshij,Hij, andHijas follows:
hij=n
∑r=1n
∑s=1bi
r
cj
sA
rs
,Hij=hij+hji=Hji,Hij=hij−hji=−Hji. (6.8.19)The derivatives of these functions are found by applying (6.8.18):
Dx(hij)=∑
r,sbi
rcj
s[
(br+cs)Ars
−∑
p,qA
rq
Aps]
=
∑
r,sbi
rcj
s(br+cs)Ars
−∑
r,qbi
rArq∑
p,scj
sAps=hi+1,j+hi,j+1−hi 0 h 0 j,which is a nonlinear differential recurrence relation. Similarly,
Dy(hij)=hi 0 h 1 j−hi 1 h 0 j−hi+2,j+hi,j+2,Dt(hij)=4(hi 0 h 2 j−hi 1 h 1 j+hi 2 h 0 j−hi+3,j−hi,j+3),Dx(Hij)=Hi+1,j+Hi,j+1−hi 0 h 0 j−h 0 ihj 0 ,Dy(Hij)=(hi 0 h 1 j+h 0 ihj 1 )−(hi 1 h 0 j+h 1 ihj 0 )