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. 14
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?2
This shows that the equation (2) is invariant with respect to the operator C.
Note 1. Solutions of the equation (1) can be generated by means of transformations [1]:
x0 > x0 = x0 , xa > xa = xa , u > u = exp{?(1 ? 2ibx0 )}u
which are generated by the operator B.
From the commutation relations for the operator C
[C, P0 ] = d1 C, [C, Pa ] = [C, Jab ] = [C, Q] = [C, Ga ] = 0
and for the operator B
d 1 , d2 ? R
[B, P0 ] = d2 Q, [B, Pa ] = [B, Jab ] = [B, Q] = [B, Ga ] = 0,
it follows that the algebras AG3 (1, n) and AG4 (1, n) differ.
58 W.I. Fushchych, V.I. Chopyk

3. Lie reduction by number of independent variables. In this paper we sys-
tematically use symmetry properties of equations (1) and (2) to find their exact
solutions. The method of finding exact solutions of differential equations is based on
Lie’s ideas of invariant solutions and it is described in full detail in [4, 5].
In this section we describe the some ansatzes of codimension 1 and 2

u = f (x0 , x)?(?1 , ?2 ) exp{g(x0 , x) + ?(?1 , ?2 )},

where the functions f , g and new variables ?i = ?i (x0 , x) are determined by means
of operators of subalgebras of AG3 (1, n) and AG4 (1, n).
Let us consider some subalgebras of AG3 (1, n), which reduce the equation (1) to
system of differential equations with one and two independent variables.
1) B + ?P0 , Jab , ? = 0. The ansatz and corresponding systems of reduced equati-
ons has the form:
x0 b
?(?) exp i ? x2 + ?(?) ? ? R, (8)
u = exp , ? = 0,
?0
?

where ? = (x2 )1/2 , x2 = x2 + · · · + x2 and
n
1

n?1
1
? + 2??? + ??? + ??
?? ? ? = 0,
?
? ?
?? + ?(n ? 1)? ?1 ? ? ???2 = 2b? ln ?,
? ?
2 2
where ? = ?? , ? = ?? , ? = ??? , ? = ??? .
?? ??
? ?2 ? ?2
? ?
2) B + ?P0 , J12 + ?P3 , ?, ? = 0, ?, ? ? R

x0 b
?(?1 , ?2 ) exp i ? x2 + ?(?1 , ?2 ) ? ? R, (9)
u = exp , ? = 0,
?0
?

where
x2 x3
?.
?1 = (x2 + x2 )1/2 , ?2 = arctg
1 2
x1 ?
The system of reduced equations has the form (for the case n = 3)
?2
??1 ? + 2??1 ?1 + 2??2 ?2 (?1 + ? ?2 ) + ???11 +
?2 ?1
+ ???22 (?1 + ? ?2 ) + ???1 ?1 = 0,
?2 ?1 ?2
??11 + ??22 (?1 + ? ?2 ) + ???1 ? ???2 + ??2 (?1 + ? ?2 ) = 2b? ln ?.
1 2

3) The ansatz

x3
x0 x0 x1 b
? x2 ? 0 + ?(?1 , ?2 ) (10)
u = exp ?(?1 , ?2 ) exp i
? 0 6??
? ?

when n = 3 reduces equation (1) to the system:
?1
2??1 ?1 + 2??2 ?2 + ??1 ? + ??(?2 ?2 + ?11 + ?22 ) = 0,
(11)
?1
??11 + ??22 + ??2 ?2 ? ??(?2 + ?2 ) = 2b? ln ? ? (2??)?1 ??1 ,
1 2
Symmetry analysis and ansatzes for the Schr?dinger equations
o 59

where
?x2
? x1 , ?2 = (x2 + x2 )1/2 ,
0
?1 = ? = 0,
2 3
?
?? ??
?i = , ?i = , i = 1, 2.
??i ??i
Solving the system of reduced equations (11) one can following partial solution of the
equation (1)

x2 x0
? da xa + c1 +
0
u = exp + 1
8??b ?
(12)
x0 2b x0
+i ? ? x2 + da xa + da xa + c2 ,
0
?2 3
6?? ?

where da , ci , ? ? R, k = 1, 2, 3, a = 1, n and da satisfy the following conditions:
k k

1 1 1
, da d a = 2 , da d a = ?
da d a = ,
11 12 13
8b2 ?2 ? 8? b 2??
? b 1
d a d a = 2 , da d a = ? , da d a = ? 2bc1 .
22 23 33
16??b2
4? ?
It is easy to see that the exact solution (12) of the nonlinear equation (1) is
non-analytical by b.
Note 2. The ansatzes (7)–(9) follows from the fact that the equation (1) is invariant
to the operator B.
Let us adduce some examples of reduction of equation (2).
Example 1. C + ?P3 , J12 . The ansatz

1
exp(2?2 x0 )x3 ?(?1 , ?2 ) ?
u = exp
?
(13)
?1
? exp i ? exp(2?2 x0 )x3 + ?(?1 , ?2 ) , ? = 0,
??2

where ?1 = x0 , ?2 = (x2 + x2 )1/2 reduces equation (2) (when n = 3) to the system:
1 2

?1
?1 + 2??2 ?2 + ??(?2 ?2 + ?22 ) = 2?2 ? ln ?,
??2 ? exp(4?2 ?1 )(1 ? ?2 ??2 )? + ??22 + ??2 ?2 ? ??1 ? ???2 = 2?1 ? ln ?.
?1
12 2

Example 2. The ansatz

1 ?1
exp(2?2 x0 ) ?(?) exp i ? (14)
u = exp exp(2?2 x0 ) + ?(?) ,
2??2
2?2 ? 2

where ? = (x2 )1/2 reduces (2) when ?2 = 0 to the system of ODE:

2??? + ??? + ??? ?1 (n ? 1)? = 2?2 ? ln ?,
?? ? ?
?? + ?(n ? 1)? ?1 ? + ???2 = 2?1 ? ln ?.
? ?
60 W.I. Fushchych, V.I. Chopyk

Example 3. The ansatz
x2
exp(2?2 x0 ) ?(?1 , ?2 ) ?
u = exp arctg
x1
(15)
x2 + · · · + x2
?1 x2
? exp i n
+3
exp(2?2 x0 ) arctg + ?(?1 , ?2 ) ,
?2 x1 4?x0
where ?1 = x0 , ?2 = (x2 + x2 )1/2 reduces the equation (2) (when n ? 2) to the
1 2
system:
n?2
?1 ?2 ?1
2
?1 + 2? exp(4?2 ?1 ) ??2 + ?2 ?2 ?2 + 2???2 ?2 + ???22 + ?=
?2 2?1
= 2?2 ? ln ?,
? exp(4?2 ?1 )?2 ?(1 ? ?2 ??2 ) + 2??2 (1 + ?2 ) ? ??1 ? ???2 = 2?1 ? ln ?.
?2 ?1
12 2

Example 4. The ansatz
u = exp(exp(2?2 x0 )x1 ?(x0 ) ?
x2 + · · · + x2 (16)
?1
? exp i n
x1 exp(2?2 x0 ) + 2 + ?(x0 ) ,
?2 4?x0
reduces the equation (2) when ?2 = 0 to the system:
n?1
? ? 2??1 ??1 ? exp(4?2 x0 ) +
? = 2?2 ? ln ?,
2
2x0 (17)
? = ? exp(4?2 x0 ) + ??1 ??1 exp(2?2 x0 ) ? 2?1 ln ?.
? 2

The system of equations (17) by means of the change of variables ? = exp ? is
reduced to a linear system of ODE which has the general solution of the form
n?1
??1
2 exp(4?2 x0 ) ? exp(2?2 x0 ) d1 + F (2?2 ) , d1 ? R,
?=
?2 2
2?2
? ?1
exp(4?2 x0 ) 1 ? 21 + (18)
?= (? + 2d1 ) exp(2?2 x0 ) +
4?2 ?2 2?2
+ ?1 (n ? 1) F (2?2 ) exp(2?2 x0 )dx,

where
dx0
F (?) = exp(??x0 ) .
x0
The substitution of (18) into the ansatz (16) gives the following solution of the equa-
tion (2) when ?2 = 0 for n = 1
??1
u = exp (x1 ? d1 ) exp(2?2 x0 ) + exp(4?2 x0 ) +
?2
2
??2 ? 2??2
??1 + 2?1 ?2 d1 ?1
? x1 exp(2?2 x0 ) + 2 1
+i exp(2?2 x0 ) .
2 3
2?2 ?2 4?2
Note 3. The ansatzes (13)–(16) are obtained from the fact that the equation (2) is
invariant with respect to the algebra AG4 (1, n) (as distinct from the equation (1)).
Symmetry analysis and ansatzes for the Schr?dinger equations
o 61

4. Component-wise reduction. The reduction by number of dependent variables
of the equations (1), (2) is possible because of invariance these equations respectively
to the operators B and C.
1) For reduction of the equation (1) by operator B it is necessary to change or
variables:
W = F (x0 , x) ? i(4bx0 )?1 ln(u/u? ), V = ln |u| ? i(4bx0 )?1 ln(u/u? ), (19)
where F is a some real function.
Then the change of variables (19) is constructed, the equation (1) has the form:
F0 ? W0 + V0 + 4?bx0 (Fa ? Wa + Va )(Wa + Fa ) + 2?bx0 (?W ? ?F ) = 0,
?(Fa ? Wa + Va )(Fa ? Wa + Va ) + ?(?F ? ?W + ?V ) ?
? 2bx0 (W0 ? F0 ) ? 4?b2 x2 (Wa ? Fa )(Wa ? Fa ) = 2bV,
0

?2
?F ?W ?V
where Fµ = ?xµ , ?xµ , ?xµ , and the operator B has the form
Wµ = Vµ = ?= ?xa ?xa
?
B = ?W .
The reduction of the equation (1) by operator B is equivalent to the condition
W = 0.
Thus, we can find the solutions of the equation (1) in the form:
u = exp{V (x0 , x) + (1 ? 2ibx0 )F (x0 , x)}, (20)

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