Determinants and Their Applications in Mathematical Physics

(Chris Devlin) #1
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:

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