98
V. Hussin et al.
where a 0 , d 0 and d 2 remain arbitrary real parameters. Clearly the solution (24) is
obtained when d 0 = 0.
Now, consider a 0 = 0. We then get different subcases ( = ±1)
– d 2 = 0, d 0 = 0
β 11 (x) = a 2 x 2 + 2 x ,
β 22 (x) = d 2 x 2 − 2 x ,
– d 0 = 0, d 2 = 0
β 11 (x) = a 2 x 2 + 0 x ,
β 22 (x) = 2 x + d 0 .
3.3 The Case of Z 3
In this case we have det Z
†
3 Z 3 =
1 + |x| 2 3 , i.e. r = 3 and κ =
2
3 . With the
solution W 3 as in (14), the condition (9) becomes a third degree polynomial in x − .
Equating the coefficients of different powers of x − to zero we obtain the following
equations:
2x 3
+
√
2β
12 + 5β
22
− x 2
+
3β
11 + 8
√
2β
12 + 6
√
2β
21 + 40β
22
+6x +
3β
11 + 2
√
2β 12 + 6
√
2β
21 + 10β 22
− 36β 11 − 72
√
2β 21 = 0 , (26)
x 2
+
−β
11 + 8
√
2β
12 − 4
√
2β
21 + 6x + β
22 + 4β
22
−x +
4β
11 + 24
√
2β 12 − 8
√
2β
21 + 48β 22
+ 28β 11 + 16
√
2β 21 = 0 , (27)
x +
β
11 − 2
√
2x + β
12 + 8
√
2β
12 − 2
√
2β
21 + 2x + β
22 + 16β
22
−10β
11 + 12
√
2β 12 − 4
√
2β
21 + 12β 22 = 0 ,
(28)
−3β
11 + 4
√
2x + β
12 + 8
√
2β
12 + 2x + β
22 + 4β
22 = 0 ,
(29)
whose solution gives the final form of β 11 and β 22 as
β 11 (x + ) =
3x +
√
2
β 12 (x + ) +
1
√
2
β 21 (x + ),
(30)
β 22 (x + ) =
√
2
4
β 12 (x + ) +
3
√
2
4x +
β 21 (x + ).
(31)
Introducing (30) and (31) into the condition (10) we obtain
1 + 3|x| 2
β 21 − x +
1 + |x| 2
∂ + β 21
2
|x| 4
1 + |x| 2
2
= f (x + ) + g(x − ) ,
(32)