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x2 3
R = d1 ln |t| ? ? arctan ? d2 + ln |x3 + C1 | + C2 .
x1 2
A8 = J12 , J23 , J14 , J23 , J24 , J34
Ansatz:
1
? = ? t + f (?), R = g(?), ? = (t ? 2?)2 + 2 x2 + x2 + x2 .
1 2 3
2
Solution (in implicit form):

? = ? 1 t + 1 (t ? 2?)2 + 2(x2 + x2 + x2 ) + C1 ,
1 2 3
2 2
3
R = ? ln (t ? 2?)2 + 2(x2 + x2 + x2 ) + C2 .
1 2 3
4
Some exact solutions of a conformally invariant nonlinear Schr?dinger equation
o 61

A9 = J01 , J02 , J03 , J12 , J13 , J23
Ansatz:
x2 + x2 + x2
1 1
? = ? ? t.
? = f (?) + 1 2 3
, R = g(?),
4t 4t 2
Solution:
x2 ? 4C1 t + 8C1 x2 ? 2(t ? 2C1 )2
2
3
R = ? ln
?= , + C2 .
4t ? 8C1 t ? 2C1
2
A10 = J12 + P0 + d1 N, P3 + d2 N, P4 (d1 ? 0, d2 > 0)
Ansatz:
1 1 x2 x2
? = f (?) ? t ? v arctan , R = g(?) ? d1 arctan + d2 x3 ,
2 x1 x1
2
? = x2 + x2 .
1 2

Solution:
1 1 x2 ?
? = ? t ? v arctan x2 + x2 ? 1 ? arctan x2 + x2 ? 1 + C1 ,
+ 1 2 1 2
2 x1 2
2
x2 1
R = ?d1 arctan + d2 x3 ? ln |x2 + x2 ? 1| ?
1 2
x1 4
? ?d1 arctan x2 + x2 ? 1 + C2 , ? = ±1.
1 2

A11 = G1 + 2T + dN, P2 + N, P3
Ansatz:
t3 x1 t d
R = g(?) + v t + x2 ,
? = f (?) ? ? = t2 ? 2x1 .
+ ,
6 2 2
Solution:
t3
?2 x1 t
(t ? 2x1 )3/2 ? +
?= + C1 ,
6 6 2
1 ?d d
R = ? (t2 ? 2x1 ) ? v (t2 ? 2x1 )1/2 + v t + x2 + C2 , ? = ±1.
2 2 2
A12 = G1 + 2T + dN, J23 , P2 , P3
Ansatz:
t3 x1 t d
R = g(?) + v t,
? = f (?) ? + ? = t2 ? 2x1 .
,
6 2 2
Solution:
t3
?2 x1 t
? = (t ? 2x1 ) ? +
3/2
+ C1 ,
6 6 2
1 ?d d
R = ? (t2 ? 2x1 ) + v (t2 ? 2x1 )1/2 + v t + C2 , ? = ±1.
2 2 2
62 P. Basarab-Horwath, L.L. Barannyk, W.I. Fushchych

A13 = G1 + d1 N, G2 + P2 + d2 N, P3 + d3 N (d1 , d2 , d3 ? 0)
Ansatz:
x2 x2
+ v v2
1
? = f (?) + ,
4t 2 2( 2t + 1)
d1 x1 d2 x2
R = g(?) + v + v + d3 x3 , ? = t.
2t 2t + 1
Solution:
x2 x2
+ v v2
1
?= + C1 ,
4t 2 2( 2t + 1)
1v
d1 x1 d2 x2 1
R= v +v + d3 x3 ? ln |t| ? | 2t + 1| + C2 .
2 2
2t 2t + 1

4 The extended subalgebras of the extended
Poincar? algebra AP (1, 4)
?
e
If we add the dilatation operator D to the algebra AP (1, 4)?N , we obtain the algebra
?
AP (1, 4) ? N = Pµ , Jµ? , N, D . In this section we give a list of the subalgebras of
?
AP (1, 4) ? N which are not equivalent to subalgebras of AP (1, 4) ? N , as well as the
corresponding ansatzes and solutions of (4), (5).
A14 = J04 + a1 N, D + a2 N, P3 (a1 ? 0, a2 arbitrary)
Ansatz:
x2 3 x1
R = g(?) + a1 ln |t| ? a1 + a2 + ln |x1 |,
? = 1 f (?), ?= .
t 2 x2
Solution:
x2 1
R = a1 ln |t| + a2 ? a1 + ln |x1 | ? 2(a2 + 1) ln |x2 | + C.
? = 1,
4t 2
A15 = J12 + a1 J04 + a2 N, D + a3 N, P3 (a1 > 0)
Ansatz:
x2 + x2 3 a3 x2
ln x2 + x2 ? a2 arctan ,
?= 1 2
f (?), R = g(?) + + 1 2
t 4 2 x1
x2
? = 2 ln |t| ? ln(x2 + x2 ) + 2a1 arctan .
1 2
x1
Solution:
x2 + x2
1 2
?= ,
4t
x2
R = g 2 ln |t| ? ln x2 + x2 + 2a1 arctan ?
1 2
x1
x2 1
? a2 arctan ? ln x2 + x2 ,
1 2
x1 2
where g is an arbitrary function of one variable.
Some exact solutions of a conformally invariant nonlinear Schr?dinger equation
o 63

A16 = J04 + a1 N, J12 + a2 N, D + a3 N (a1 , a2 ? 0; a3 arbitrary)
Ansatz:
x2 + x2
?= 1 2
f (?),
t
x2 2a1 + 2a3 + 3
R = g(?) + a1 ln |t| ? a2 arctan ? ln(x2 + x2 ),
1 2
x1 4
x2 + x2
? = 1 2 2.
x3
Solution:
x2 + x2
?= 1 2
,
4t
x2 a1 + 1 1 + 2a3
R = a1 ln |t| ? a2 arctan ? ln(x2 + x2 ) ? ln |x3 | + C.
1 2
x1 2 2
A17 = J04 + a1 D + a2 N, J12 + a3 D + a4 N, P3 (a2 + a2 = 0).
1 3
Ansatz:
x2 + x2
?= 1 2
f (?),
t
3a1 + 2a2 3a3 + 2a4 x2
ln |t| ? ?
R = g(?) + arctan
2 2 x1
3a1 + 2a2
? ln x2 + x2 ,
1 2
4
x2
? = a1 ln |?| ? a1 ln |t| + 2a3 arctan ? ln x2 + x2 .
1 2
x1
Solution:
x2 + x2
1 2
?= ,
4t
a1 + 2a2 a1 + 2a2 + 2 a3 + 2a4 x2
ln |t| ? ln(x2 + x2 ) ?
R= arctan + C.
1 2
2 4 2 x1
A18 = J04 + D + aN, J23 , P2 , P3
Ansatz:
3
R = g(?) ? a + ln |x1 |,
? = x2 f (?), ? = t.
1
2
Solution:
x2 3
R = (a + 1) ln |4t + C1 | ? a + ln |x1 | + C2 .
1
?= ,
4t + C1 2
A19 = G1 , J04 + a1 D + a2 N, P3 (a1 = 0, a2 arbitrary)
Ansatz:
x2 x2 3a1 + 2a2 (1?a1 )/a1
R = g(?) ?
2
f (?) + 1 ,
?= ln x2 , ? = tx2 .
4t 4t 2a1
Solution:
x2 + x2 a1 + 2a2 a1 + 2a2 + 2
ln |t| ?
1 2
?= , R= ln x2 + C.
4t 2 2
64 P. Basarab-Horwath, L.L. Barannyk, W.I. Fushchych

A20 = J04 ? D + M + aN, J23 , P2 , P3
Ansatz:
x2
1 3
? = f (?) + v ln |t|, R = g(?) + a ? ln |x1 |, 1
?= .
2 t
22
Solution:
v
x2 ? 4 2t ? |x1 |
v
x2
1 ? 1
? = v ln |t| + (|x1 | + ? x2 ? 4 2t) + v ln
1
+ C1 ,
v
1
t
22 22 2 ? 4 2t + |x |
x1 1
v
x2 ? 4 2t ? |x1 |
v
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