6.8 The Kadomtsev–Petviashvili Equation 277
−
∑
r
αrA
(n)
r,n− 1
∑
s
βsA
(n)
sn
=
∑
r
∑
s
αrβs
[
AnA
(n)
rs;n− 1 ,n−
∣
∣
∣
∣
∣
A
(n)
r,n− 1 A
(n)
rn
A
(n)
s,n− 1
A
(n)
sn
∣
∣
∣
∣
∣
]
=0,
which completes the proof of the theorem.
Exercise.Prove that
logw=k+ logW,
wherekis independent ofxand, hence, thatwandW yield the same
solution of the KdV equation.
6.8 The Kadomtsev–Petviashvili Equation
6.8.1 The Non-Wronskian Solution
The KP equation is
(ut+6uux+uxxx)x+3uyy=0. (6.8.1)
The substitutionu=2vxtransforms it into
(vt+6v
2
x+vxxx)x+3vyy=0. (6.8.2)
Theorem 6.16. The KP equation in the form (6.8.2) is satisfied by the
function
v=Dx(logA),
where
A=|ars|n,
ars=δrser+
1
br+cs
,
er= exp
[
−(br+cr)x+(b
2
r−c
2
r)y+4(b
3
r+c
3
r)t+εr
]
= exp
[
−λrx+λrμry+4λr(b
2
r−brcr+c
2
r)t+εr
]
,
λr=br+cr,
μr=br−cr.
Theεrare arbitrary constants and thebrandcsare constants such that
br+cs=0, 1 ≤r, s≤n, but are otherwise arbitrary.
Proof. The proof consists of a sequence of relations similar to those
which appear in Section 6.7 on the KdV equation. Those identities which
arise from the double-sum relations (A)–(D) in Section 3.4 are as follows: