Determinants and Their Applications in Mathematical Physics

6.7 The Korteweg–de Vries Equation 267

=0+

1

2

∑

p,r

(b

i+2

p

b

j

r

−b

j+2

p

b

i

r

)A

pr

=

1

2

(φi+2,j−φi,j+2),

which is identical with the right side of (a). This completes the proof of

(a).

Referring to (6.7.8) withr, s→p, qandi, j→r, s,

Dx(φij)=

∑

r

∑

s

b

i

rb

j

sDx(A

rs

)

=

∑

r

∑

s

b

i

r

b

j

s

[

1

2

(br+bs)A

rs

−

∑

p

∑

q

A

rq

A

ps

]

=

1

2

∑

r

∑

s

b

i

r

b

j

s

(br+bs)A

rs

−

∑

q,r

b

i

r

A

rq

∑

p,s

b

j

s

A

ps

=

1

2

(φi+1,j+φi,j+1)−φi 0 φj 0 ,

which proves (b). Part (c) is proved in a similar manner.

Particular cases of (a)–(c) are

φ 00 φ 11 −φ

2

10

=

1

2

(φ 21 −φ 03 ), (6.7.18)

Dx(φ 00 )=φ 10 −φ

2

00

,

Dt(φ 00 )=2φ 00 φ 20 −φ

2

10

−φ 30. (6.7.19)

The preparations for finding the derivatives ofvare now complete. The

formula forvgiven by (6.7.7) can be written

v=φ 00 −constant.

Differentiating with the aid of parts (b) and (c) of the lemma,

vx=φ 10 −φ

2

00 ,

vxx=

1

2

(φ 20 +φ 11 − 6 φ 00 φ 10 +4φ

3

00

),

vxxx=

1

4

(φ 30 +3φ 21 − 8 φ 00 φ 20 − 14 φ

2

10

,

+48φ

2

00 φ^10 −^6 φ^00 φ^11 −^24 φ

4

00 )

vt=2φ 00 φ 20 −φ

2

10 −φ^30. (6.7.20)

Hence, referring to (6.7.18),

4(vt+6v

2

x+vxxx)=3

[

(φ 21 −φ 30 )−2(φ 00 φ 11 −φ

2

10 )

]

=0,

which completes the verification of the first form of solution of the KdV

equation by means of recurrence relations.