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leads to a simple differential equation for ?(t) and equation (6) determines ?(t) up to
a constant. Hence, the maximal symmetry algebra is H, Pj , Jjk , Gj , Q, D, A , where
?1
A = e2?2 t R?R ? (12)
?S .
?2
6. Finally, if u and v vanish identically, the maximal Lie symmetry algebra is
H, Pj , Jjk , Gj , Q, D , I , the direct sum of the Schr?dinger algebra (though with
o
a different representative D of the generator of dilations) with a one-dimensional
algebra generated by I, where
(13)
D = 2t?t + xk ?xk ,

(14)
I = R?R .

The invariance under I reflects real homogeneity of the equation (1); together with Q
it generates complex rescalings of ?.
2.2. n > 3. In all cases the algebras remain symmetry algebras for arbitrary di-
mension n. We believe that they are still maximal, but we have no proof of maximality
for arbitrary n. The algebras of the cases 1–3 and 6 have been studied in [18], and
the finite transformations they generate are well known. The structure of the algebra
of case 4 was investigated in [12, 19, 20].
As for the generators B and A, they generate the following finite transformations:

? > g B ?, g B ?(x, t) = exp( (1 ? i2?1 t))?(x, t),
?1
? > g A ?, 1?i
g A ?(x, t) = exp e2?2 t ?(x, t),
?2

3. Reduction and exact solutions
Using the operators of symmetry we will construct ans?tze reducing equation (1)
a
to a system of ordinary differential equations (ODEs). The algebras of the cases 1–3
and 6 are subalgebras of the maximal symmetry algebra of the linear Schr?dinger o
equation; their structure was studied in detail and corresponding ans?tze are well
a
known. Thus we concentrate on the cases 4 and 5, and particularly on the reduction
by those subalgebras containing the “new” generators A and B. The solutions obtained
in this way might reflect the nonlinear structure of equation (1) with f (?) := (?1 +
i?2 ) ln ? + i?3 . We consider mainly the case of three spatial variables, n = 3.
3.1. Case 4: f (?) := ?1 ln ? + i?3 ; or u(R2 ) = ?1 ln(R2 ), v(R2 ) = ?3
1. B + G1 , G2 , G3 . The ansatz
m x2
x1
+ g(t) + i ?2?1 x1 + (15)
?(x, t) = exp + h(t)
t 2t
reduces equation (1) to the system
dg 2 ?1 31 1
?
= + 2d 2 + ?3 ,
dt m 2t t
2 ?2
dh 1
=? ? 2?1 g(t).
1
+
2m t2
dt m
Symmetries and reductions of nonlinear Schr?dinger equations
o 397

Having solved this system we find the solution
m x2
1
?(x, t) = t exp ?3 t + (x1 ? 2d) + c1 + i ?2?1 x1 + ? ? 1 ? 3 t2 ?
k
t 2t
?1 1
? 2?1 kt ln t + 2?1 k ? c1 ? t + 4d?1 ln t ? + c2 ,
m 2m t

where k := 2 m 1 ? 3 and c1 , c2 are real constants.
?
2
2. B + ?H, J12 + ?P3 , ? ? R=0 , ? ? R. For ?3 = 0, the ansatz
t ?1
+ g(?1 , ?2 ) + i ? t2 + h(?1 ?2 ) (16)
?(x, t) = exp ,
? ?
1
? ?x3 , reduces equation (1) to the
x2
with ?1 := (x2 + x2 ) 2 and ?2 := arctan
1 2 x1
system
?2 ?2
h1
+ 2g1 h1 + 2g2 h2 1 + 2 ?
h11 + h22 1 + +
2
?1 ?1 ?1
2
?2
2md ? g1
? 2 2 2
g11 + g22 1 + 2 + + 2g1 + 2g1 + 2g2 1 + 2 =
?1 ?1 ?1
2m 1
?3 ?
= ,
?
?2 ?2
g1
+ g1 + g2 1 + 2 ?
2 2
g11 + g22 1 + 2 +
?1 ?1 ?1
2
4m?1 ?
? g ? h2 ? h2 1 + 2 = 0,
1 2
?1
where subscripts denote derivatives, i.e. g1 := ?g/??1 , etc.
3. B + ?H + ?G1 , J23 , ? ? R=0 , ? ? R. The ansatz
?1 2 m? 2 3
t m?
x1 t ? t ? , (17)
?(x, t) = exp + g(?1 , ?2 ) + i t + h(?1 ?2 )
3 ?2
? ? ?
?t2 1
? x1 and ?2 := (x2 + x2 ) 2 reduces equation (1) to the system
with ?1 := 2 3
2?

h2 2md g2
? ? 2g1 ? 2g2
2 2
2g1 h1 + 2g2 h2 + h11 + h22 + g11 + g22 + =
?2 ?2
2m 1
?3 ?
= ,
?
2?m2 ?1 g2 4?1 m
? ? g11 ? g22 ? ? g1 ? g2 +
h2 h2 2 2
+ g = 0.
1 2 2? ?2
For ? = 1/?3 , ?3 = 0 and d = /2m we have found the following partial solution of
this system:
? 4m2 d2 )m? 2
2 2
m? m ?1 (
?2 +
g(?1 , ?2 ) = ?1 + 2 + ,
? 4m ? 4m2 d2
2 d2 2 16 3 ?2 ?3
2
2 ??1 1
2md
h(?1 , ?2 ) = f (?1 , ?2 ).
398 W.I. Fushchych, V. Chopyk, P. Nattermann, W. Scherer

The corresponding solution of equation (1) has then the form
m? 2 2
t m? m ?1
t? (x2 + x2 ) +
?(x, t) = exp + x1 + 2
? 4m 1 2
2? 2 d2
? 4? 1 2 ??1
( 2 ? 4m2 d2 )m? 2
2
+2 + +
? 4m2 d2 16 3 ?2 ?3
1
m2 ? 2 ?1 2 m? 2 3 m2 d?
m?
? t? t? 2
+i tx1 + x1 +
?? ? 1
2 2 ? 2 ?1 3 ?2
? ?
2m2 d?1
(x2 + x2 ) + c .
+2
? 4m 2 3
2 d2

4. B + ?H, Jjk , ? ? R=0 . The ansatz
t ?1
+ g(?) + i ? t2 + h(?) (18)
?(x, t) = exp ,
? ?
x2 + x2 + x2 , reduces equation (1) to the system
where ? := 1 2 3
2
d2 h d2 g
2 dh dg dh 2md 2 dg dg 2m 1
? ?3 ?
+ +2 + +2 = ,
d? 2 d? 2
? d? d? d? ? d? d? ?
2 2
d2 g 2 dg dg dh 4m?1
? ?
+ + g = 0.
d? 2 ? d? d? d?
Its partial solution for the case ? = 1/?3 and d = /2m is
3
m?1 x2 +
g(?) = ,
? 4m2 d2
2 2
2m2 d?1
x2 + c,
h(?) = 2
? 4m2 d2
where c is an arbitrary real constant. The corresponding solution of equation (1) has
then the form
t 3
m?1 x2 +
?(x, t) = exp +2 +
? 4m2 d2
? 2
2m2 d?1 ?1
x2 ? t2 + c .
+i 2 ? 4m2 d2 ?
3.2. Case 5: f (?) := (?1 + i?2 ) ln ?; or u(R2 ) = ?1 ln(R2 ), v(R2 ) = ?2 ln(R2 );
?2 = 0.
1. A + ?P1 , G2 , G3 , ? ? R=0 . The ansatz
m x2 + x2
1 2?2 t ?1 2?2 t
x1 + g(t) + i ? 2 3
?(x, t) = exp e e x1 + + h(t)
? ??2 2 t
reduces equation (1) to the system of ODEs
dg 1 1 ?1
? 2?2 g = ? + 2 + 2d e4?2 t ,
dt t ? m?2
?2
dh
= ?2?1 g + 1 ? 1 e4?2 t .
?2
2m?2
dt 2
Symmetries and reductions of nonlinear Schr?dinger equations
o 399

Having solved this system we obtain the following exact solution of equation (1):
1 2?2 t 1 ?1
+ 2d e4?2 t ?
x1 + ce2?2 t +
?(x, t) = exp e 2
? 2?2 ? m?2
m x2 + x2
?1 2?2 t ?1 c 2?2 t ?1
?Ei(?2?2 t)e2?2 t + i ? ? ?
2 3
e x1 + e ln(2?2 t) +
??2 2 t ?2 ?2
?2 4md 4?2 t ?1
1?3 1 ? + Ei(?2?2 t)e2?2 t ,
+ e
2
2?
8m? 2 ?2 ?2
ak xk
exp(ax)
where c is a real constant and Ei(ax) = k!k . This solution
dx = ln x +
x
k=1
is non-analytical in ?2 , and for n = 1 can be written in explicit form.
2. A + ?J12 , G3 , ? ? R=0 . The ansatz

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