6.9 The Benjamin–Ono Equation 285To prove (6.9.17), perform the row operations (6.9.24) on Pn+1,n+2and apply (6.9.7). To prove (6.9.18), perform the same row operations on
P
∗
n+1,n+2
, apply the third equation in (6.9.8), and transpose the result.To prove (6.9.19), note thatQ+Qn+2,n+2=∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
c 1 1c 2 1··· ···[bij]n ··· ······ ···cn 1−c 1 −c 2 ··· −cn 00− 1 − 1 ··· − 101∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
n+2. (6.9.25)
The row operations
R
′
i
=Ri−Rn+2, 1 ≤i≤n,leave a single nonzero element in the last column. The result appears after
applying the second equation in (6.9.7).
To prove (6.9.20), note thatQ−Qn+2,n+2can be expressed as a deter-minant similar to (6.9.25) but with the element 1 in the bottom right-hand
corner replaced by−1. The row operations
R
′
i=Ri+Rn+2,^1 ≤i≤n,leave a single nonzero element in the last column. The result appears after
applying the second equation of (6.9.8) and transposing the result.
6.9.3 Proof of the Main Theorem
Denote the left-hand side of (6.9.1) byF. Then, it is required to prove that
F= 0. Applying (6.9.3), (6.9.5), (6.9.11), and (6.9.17),
Ax=∑
r∂A
∂θr∂θr∂x=ω∑
rcrArr (6.9.26)=ω∑
r∑
scrArs=ωPn+1,n+2=ωQn+1,n+2. (6.9.27)Taking the complex conjugate of (6.9.27) and referring to (6.9.18),
A
∗
x
=−ωP∗
n+1,n+2=(−1)n
ωQn+2,n+1.