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x1 ? ? ln(x0 + x3 ) 1
(?0 + ?3 )?1 exp ?0 ?3 ln(x0 + x3 ) ?
?5 (x) = exp
2(x0 + x3 ) 2
? exp [?2 (?0 + ?3 ) + i?(??)1/2k (?2 ? ?(?0 + ?3 ))][x2 ? ? ln(x0 + x3 )] ?;
?
x2 + 2??x1 1
??0 ? ?1 ?4 ? i?? ?,
?6 (x) = exp ?0 ?3
2? 2?
where ? ? R1 ; ? and ? being determined by formulae (3.40) and (3.41);
2??x3 ? x0 1
??0 ? ?3 ?4 ? i?? ?,
?7 (x) = exp ?1 ?2
2? 2?
where ? ? R1 ; ? and ? being determined by formulae (3.44);
x3 + 2??x0 1
?0 ?4 ? i?? ?,
?8 (x) = exp ?1 ?2 ??3 +
2? 2?
where ? ? R1 ; ? and ? being determined by formulae (3.46);
1
[x3 ? x0 + 4?(x0 + x3 )]?1 ?2 ?
?9 (x) = exp
4
? [4?(?0 + ?3 )?4 + (?0 ? ?3 )?4 ? 4i?? ]?,
? and ? being determined by formulae (3.48);
1? ?
? (?1 ?1 + ?2 ?2 ) + ?3 ?4 (?0 + ?3 ) ?
?10 (x) = exp
2
? exp[i?(??)1/2k ?1 (x1 + ?1 )]?,
?
where ?1 , ?2 , ?3 are arbitrary smooth functions of x0 + x3 .
Symmetry and exact solutions of nonlinear spinor equations 603

k ? R1 , k = 1/2:
1
?0 ?3 ln(x0 + x3 ) ?(x2 ? x2 ),
?11 (x) = exp 0 3
2
?(?) being determined by formulae (3.52)–(3.54) with m = 1;
?12 (x) = [(x1 + ?1 )2 + (x2 + ?2 )2 ]?1/4 ?
1? ?
? exp ? (?1 ?1 + ?2 ?2 ) + ?3 ?4 (?0 + ?3 ) ?
2
1 x1 + ?1
? exp ? ?1 ?2 arctg ?
2 x2 + ?2
2i?k
? exp (??)1/2k ?2 [(x1 + ?1 )2 + (x2 + ?2 )2 ](2k?1)/4k ?,
?
1 ? 2k
where ?1 , ?2 , ?3 are arbitrary smooth functions of x0 + x3 ;
1 1 x2
?13 (x) = (x2 + x2 )?1/4 exp ? ?0 ?1 ln(x0 ? x1 ) ? ?2 ?3 arctg ?
2 3
4 2 x3
? [f1 + ?3 f2 + (?0 + ?1 )f3 + ?3 (?0 + ?1 )f4 ]?,
where fi = fi (x2 + x2 ) are determined by formulae (3.60);
2 3
k ? R1 , k = 1:
x1 1
(?0 + ?3 )?1 exp ?0 ?3 ln(x0 + x3 ) ?(x2 ? x2 ? x2 ),
?14 (x) = exp 0 1 3
2(x0 + x3 ) 2
?(?) being given by formulae (3.52)–(3.54) with m = 2;
k ? R1 , k = 3/2:
?0 + ?3 1
(?1 x1 + ?2 x2 ) exp ?0 ?3 ln(x0 + x3 ) ?(x · x),
?15 (x) = exp
2(x0 + x3 ) 2
?(?) being given by formulae (3.52)–(3.54) with m = 3;
k = 1/2:
1
?0 ?3 ln(x0 + x3 ) ?(x2 ? x2 ),
?16 (x) = exp 0 3
2
?(?) being given by formulae (3.56);
1 x2
?17 (x) = (x2 + x2 )?1/4 exp ? ?2 ?3 arctg ?
2 3
2 x3
i? x2
? exp (??)(?3 + a?2 ) ln(x2 + x2 ) + 2a arctg
? ?;
2 3
2(1 + a2 ) x3
k = 1:
1?
?18 (x) = ??1 exp ?
? (?1 ?1 + ?2 ?2 ) + ?3 ?4 ?
0
2
1?
? ?0 ??1 (?1 (x1 + ?1 ) + ?2 (x2 + ?2 ) (?0 + ?3 ) ?
0
2
i?
? exp (??)1/2 ?1 (x1 + ?1 ) ?;
?
?0
604 W.I. Fushchych, R.Z. Zhdanov

1?
?1/2
[(x1 + ?1 )2 + (x2 + ?2 )2 ]?1/4 exp ?
? (?1 ?1 + ?2 ?2 ) + ?3 ?4 ?
?19 (x) = ?0
2
1?
? ?0 ??1 (?1 (x1 + ?1 ) + ?2 (x2 + ?2 ) (?0 + ?3 ) ?
0
2
1 x1 + ?1
? exp ? ?1 ?2 arctg ?
2 x2 + ?2
?1/2
? exp ?2i?(??)1/2 ?2 [(x1 + ?1 )2 + (x2 + ?2 )2 ]1/4 ?0
? ?;

where ?0 , ?2 , ?3 are arbitrary smooth functions x0 + x3 ;
?0 x0 ? ?1 x1 ? ?2 x2
exp[i?(??)1/2 ?0 x0 (x2 ? x2 ? x2 )?1 ]?;
?20 (x) = ?
(x2 ? x2 ? x2 )3/2 0 1 2
0 1 2
?0 x0 ? ?1 x1 ? ?2 x2
exp[?i?(??)1/2 ?1 x1 (x2 ? x2 ? x2 )?1 ]?;
?21 (x) = ?
2 ? x2 ? x2 )3/2 0 1 2
(x0 1 2
?0 x0 ? ?1 x1 ? ?2 x2 2 1 x1
(x1 + x2 )?1/4 exp ? ?1 ?2 arctg ?
?22 (x) = 2 ? x2 ? x2 2
x0 2 x2
1 2
? exp[2i?(??)1/2 ?2 (x2 + x2 )1/4 (x2 ? x2 ? x2 )?1/2 ]?;
? 1 2 0 1 2
?·x
?23 (x) = 2 3/2 exp[?i?(??)1/2 ?1 x1 (x2 )?1 ]?;
?
(x )
?·x 2 1 x1
(x1 + x2 )?1/4 exp ? ?1 ?2 arctg ?
?24 (x) = 2
x2 2 x2
? exp[2i?(??)1/2 ?2 (x2 + x2 )1/4 (x2 )?1/2 ]?;
? 1 2
x1
(?0 + ?3 )?1 ?
?25 (x) = exp
2(x0 + x3 )
1
? exp ?0 ?3 ln(x0 + x3 ) ?(x2 ? x2 ? x2 ),
0 1 3
2
?(?) being determined by formulae (3.55) or (3.57) with m = 2;
1 1 x2
?0 ?1 ln(x0 + x1 ) ? ?2 ?3 arctg ?
?26 (x) = exp
2 2 x3
? (x2 + x2 )?1/4 [f1 + ?3 f2 + (?0 + ?1 )f3 + ?3 (?0 + ?1 )f4 ]?,
2 3

fi = fi (x2 + x2 ) being given by formulae (3.61);
2 3
k = 3/2:
1
(?0 + ?3 )(?1 x1 + ?2 x2 ) ?
?27 (x) = exp
2(x0 + x3 )
1
? exp ?0 ?3 ln(x0 + x3 ) ?(x · x),
2
where ?(?) is determined by formulae (3.55) or (3.57) with m = 3.
Besides, in ref. [24] two other classes of exact solutions were obtained, essentially
using ansatz (3.23) and the Heisenberg ansatz [14],
k < 0:
1
?28 (x) = exp ?1 (?0 ? ?2 )(x0 ? x2 ) (?3 + ?(?0 ? ?2 ))(x3 + ?(x0 ? x2 )) +
2
1
+ ?1 (2x1 + (x0 ? x2 )2 ) f (?) + ig(?) ?;
2
Symmetry and exact solutions of nonlinear spinor equations 605

k = 1/2:
1
?1 (?0 ? ?2 )(x0 ? x2 ) ?
?29 (x) = exp
2
1
? (?3 + ?(?0 ? ?2 ))(x3 + ?(x0 ? x2 )) + ?1 (2x1 + (x0 ? x2 )2 ) ? ?1 ?
2
i?(??)
?
? exp 2 + ? 2 {?1 [?3 + ?(?0 ? ?2 )] + ?2 ?1 } ?
?1 2
1
? ?1 [x3 + ?(x0 ? x2 )] + ?2 [2x1 + (x0 ? x2 )2 ] ? ?1 ?,
2
where
1
? = [x3 + ?(x0 ? x2 )]2 + [2x1 + (x0 ? x2 )2 ]2 ,
4
1/2 k
(1 ? k)
f (?) = |k|?1/2 ? ? ?(k+1)/2 ,
?(??)1/2k
?
k
(1 ? k)1/2
? ?k/2 ,
g(?) = ±(1 ? k) ?
1/2
?(??)1/2k
?
?, ?1 and ?2 are arbitrary constants.
The existence of exact solutions depending on arbitrary functions is connected
with the fact that the additional constraint
(p0 + p3 )?(x) = 0
selects the subset of solutions of the Dirac–Heisenberg equation admitting the infinite-
dimensional algebra (3.17). As established in ref. [29], the large class of Poincar?- e
invariant equations (Bhabha-type equations)
[?µ pµ + m]?(x) = 0, (3.62)
m = const,
possess such a property. In (3.62) ? = {?1 , . . . , ?n }, x = (x0 , x1 , x2 , . . . , xl }, l ? 2,
?µ are n ? n matrices satisfying the conditions
[?? , Sµ? ] = i(gµ? ?? ? g?? ?µ ), Sµ? = i(?µ ?? ? ?? ?µ ),
(3.63)
gµ? = diag (1, ?1, . . . , ?1, ?1), ?, µ, ? = 0, . . . , l.
It is well known that eq. (3.62) is invariant under the Poincar? algebra P (1, l) having
e
basis operators of the form [46]
Jµ? = xµ P? ? x? Pµ + Sµ? .
Pµ = ig µ? ?/?x? ,
We impose on ?(x) the additional constraint
(P0 + Pl )?(x) = 0,
from which an equation for ?(?) = ?(x0 + xl , x1 , . . . , xl?1 ) follows,
? ?
l?1
?i(?0 + ?l )??0 + ?j ??j + m? ?(?) = 0. (3.64)
j=1
606 W.I. Fushchych, R.Z. Zhdanov

Proposition 3. Equation (3.64) is invariant under the infinite-dimensional Lie al-
gebra with the following basis operators:
l?1
1?
?k (?0 )??k + i?k (?0 )(Skl ? S0k ) ,
Q1 = ??0 , Q2 =
2 (3.65)
k=1
Qab = ?a ??b ? ?b ??a + iSab , a, b = 1, . . . , l ? 1,
?
where ??µ = ?/??µ , µ = 0, . . . , l ? 1, ?k = d?k /d?0 , ?k (?) are arbitrary functions.
Proof. For linear equations the following statement holds [31]: An operator Q is the
symmetry operator of the linear equation
L(x)? = 0
iff there exists a matrix R(x) such that
[Q, L] = R(x)L.
We shall prove that
(3.66)
[Q, L] = 0.
If Q ? Q1 , Qab , then the statement is quite evident. Let us consider the case
Q = Q2 . If we shall show that (?0 + ?l )(S0k ? Skl ) = 0, then proposition 3 will be
proved. Choosing k = 1 one has
(?0 + ?l )(S01 ? S1l ) = i(?0 + ?l )(?0 ?1 ? ?1 ?0 ? ?1 ?l + ?l ?1 ) =
= i(?0 ?0 ?1 ? ?0 ?1 ?0 ) + i(?l ?l ?1 ? ?l ?1 ?l ) +
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