274 6. Applications of Determinants in Mathematical Physics
into the KdV equation yields
ut+6uux+uxxx=2Dx(
F
w
2)
, (6.7.56)
where
F=wwxt−wxwt+3w2
xx−^4 wxwxxx+wwxxxx.Hence, the KdV equation will be satisfied if
F=0. (6.7.57)
Theorem 6.15. The KdV equation in the form (6.7.56) and (6.7.57) is
satisfied by the Wronskianwdefined as follows:
w=∣
∣Dj−^1
x (ψi)∣
∣
n,
where
ψi= exp(
1
4
b2
iz)
φi,φi=pie1 / 2
i +qie− 1 / 2
i ,ei= exp(−bix+b3
it).zis independent ofxandtbut is otherwise arbitrary.bi,pi, andqiare
constants.
Whenz=0,pi=λi, andqi=μi, thenψi=φiandw=Wso that thistheorem differs little from Theorem 6.14 but the proof of Theorem 6.15
which follows is direct and independent of the proofs of Theorems 6.13 and
6.14. It uses the column vector notation and applies the Jacobi identity.
Proof. Since
(Dt+4D3
x)φi=0,it follows that
(Dt+4D3
x)ψi=0. (6.7.58)Also
(Dz−D2
x
)ψi=0. (6.7.59)Since each row ofwcontains the factor exp
(
1
4b
2
i
z)
,
w=eBz
W,where
W=
∣
∣Dj−^1
x
(φi)∣
∣
nand is independent ofzand
B=
1
4∑
ib2
i