Page 282 - Determinants and Their Applications in Mathematical Physics
P. 282
6.7 The Korteweg–de Vries Equation 267
1
b )A
=0 + (b i+2 j j+2 i pr
b − b
2 p r p r
p,r
1
= (φ i+2,j − φ i,j+2 ),
2
which is identical with the right side of (a). This completes the proof of
(a).
Referring to (6.7.8) with r, s → p, q and i, j → r, s,
D x (φ ij )= b b D x (A )
rs
i j
r s
r s
= b b 1 (b r + b s )A rs − A A ps
rq
i j
2
r s
r s p q
1
= b b (b r + b s )A rs − b A rq b A ps
j
i
i j
2 r s r s
r s q,r p,s
1
= (φ i+1,j + φ i,j+1 ) − φ i0 φ j0 ,
2
which proves (b). Part (c) is proved in a similar manner.
Particular cases of (a)–(c) are
1
φ 00 φ 11 − φ 2 = (φ 21 − φ 03 ), (6.7.18)
10 2
2
D x (φ 00 )= φ 10 − φ ,
00
D t (φ 00 )=2φ 00 φ 20 − φ 2 10 − φ 30 . (6.7.19)
The preparations for finding the derivatives of v are now complete. The
formula for v given by (6.7.7) can be written
v = φ 00 − constant.
Differentiating with the aid of parts (b) and (c) of the lemma,
2
v x = φ 10 − φ ,
00
3
1
v xx = (φ 20 + φ 11 − 6φ 00 φ 10 +4φ ),
2 00
2
1
v xxx = (φ 30 +3φ 21 − 8φ 00 φ 20 − 14φ ,
4 10
2
4
+48φ φ 10 − 6φ 00 φ 11 − 24φ )
00
00
v t =2φ 00 φ 20 − φ 2 10 − φ 30 . (6.7.20)
Hence, referring to (6.7.18),
2
2
4(v t +6v + v xxx )=3 (φ 21 − φ 30 ) − 2(φ 00 φ 11 − φ )
10
x
=0,
which completes the verification of the first form of solution of the KdV
equation by means of recurrence relations.