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4
where ? is the arbitrary constant spinor.
Substituting the above results into (3.20) we obtain the following solutions of the
non-linear Dirac equation.
If k = 1
2

1
?1/4
exp ? ?2 ?3 tan?1 (x2 /x3 ) ?
?(x) = x2 + x2
2 3
2
(5.3)
1 ?1
(?3 + a?2 ) ln x2 + x2 + 2a tan?1 (x2 /x3 )
? exp ? i??? 1 + a2
? ?.
2 3
2
1
If k = 2

1
?1/4
exp ? ?2 ?3 tan?1 (x2 /x3 ) ?
?(x) = x2 + x2
2 3
2 (5.4)
(2k?1)/4
?1
? exp 2i?k(1 ? 2k) 1/2k
x2 x2
(??)
? + ?3 ?.
2 3

It is important to note that equation (4.3) can be reduced to the two-dimensional
Dirac equation. This fact can be used for obtaining new non-trivial classes of solutions
of (1.1). If we choose in (4.3), ? = ?(?1 , ?2 ) then
[?3 + ?(?0 ? ?2 )]??1 + 2?1 ??2 = i?(??)1/2k ?. (5.5)
?
Having made a change of variables
1
z1 = ? 1 , z2 = ?2
2
and denoting
?1 = ?3 + ?(?0 ? ?2 ), ?2 = ?1
we obtain
?1 ?z1 + ?2 ?z2 = i?(??)1/2k ?, (5.6)
?
where ?a ?b + ?b ?a = 2gab and a, b = 1, 2.
(i) We look for a solution of (5.6) in the form
(5.7)
?(z) = (?a za f (zb zb ) + ig(zb zb ))?,
274 W.I. Fushchych, R.Z. Zhdanov

where ? is a constant spinor and f , g are unknown scalar functions. Substitution of
(5.7) into (5.6) gives the system of ODE

df 1 1/2k
= ?(??)1/2k g 2 ? ?f 2
f +? ? g,
d? 2
dg 1 1/2k
= ?(??)1/2k g 2 ? ?f 2
? f.
d? 2

The partial solution of this system is given by the formulae (k < 0)

k
1/2
k 2 + |k|
? ?(k+1)/2 ,
f = |k|1/2 ?
?(??)1/2k
?
(5.8)
k
1/2
k 2 + |k|
?1 ?1/2
? ?k/2 .
g = ? 1 + |k| ?
?(??)1/2k
?

(ii) We shall look for a solution of (5.6) in the form

?(z) = ?a za (zb zb )?1 ?(?a za /zb zb ), (5.9)
a, b = 1, 2,

where ? = ?(?) is a four-component spinor, ? = (?a za )/(zb zb ) and k = 1 . It follows
2
from (5.6) that ?(?) satisfies the system of ODE of the form

d? ?
(?a ?a ) = i?(??)?,
d?

whose general solution has the form

?1
?(?) = exp ?i?(??) ?1 + ?2
2 2
(5.10)
? (?a ?a )? ?.

Using formulae (3.18), (5.7)–(5.10) we obtain the following solutions of the nonli-
near Dirac equation (1.1).
If k < 0

1
?1 (?0 ? ?2 )(x0 ? x2 ) [?3 + ?(?0 ? ?2 )] ?
?(x) = exp
2
(5.11)
1
? [x3 + ?(x0 ? x2 )] + ?1 2x1 + (x0 ? x2 )2 f (?) + ig(?) ?,
2

where

1 2
? = [x3 + ?(x0 ? x2 )]2 + 2x1 + (x0 ? x2 )2
4

and f (?), g(?) are defined by (5.8).
On some exact solutions of a system of non-linear differential equations 275

1
If k = 2

1
?1 (?0 ? ?2 )(x0 ? x2 ) [?3 + ?(?0 ? ?2 )] ?
?(x) = exp
2
1
? ?1 ?
? [x3 + ?(x0 ? x2 )] + ?1 2x1 + (x0 ? x2 )2
2
(5.12)
1
?1
? exp ?i?(??) ?1 + ?2 ?1 [?3 + ?(?0 ? ?2 )] + ?2 ?1 ?
2 2
?
2
1
? ?1 ?,
? ?1 [x3 + ?(x0 ? x2 )] + ?2 2x1 + (x0 ? x2 )2
2

where

1 2
? = [x3 + ?(x0 ? x2 )]2 + 2x1 + (x0 ? x2 )2 .
4

Let us point out one of the possible ways of obtaining ungenerable families of
solutions. On applying the procedure of generation of solutions by Lorentz rotations
in the plane (x0 , x1 ) to the solution (5.1) one obtains

1 1
?2 (x) = exp ? ??0 ?1 exp ?1 (?0 ? ?2 )(x0 ? x2 ) ?
2 2
1
? exp ? i?(??)1/2k ?1 2x1 + (x0 ? x2 )2 ?,
?
2
x0 = x0 cosh ? + x1 sinh ?, x1 = x1 cosh ? + x0 sinh ?, x2 = x2 , x3 = x3 .

Let us rewrite this expression in the equivalent form

1 1
?2 (x) = exp ? ??0 ?1 exp ?1 (?0 ? ?2 )(x0 cosh ? + x1 sinh ? ? x2 ) ?
2 2
1 1 1
? exp ??0 ?1 exp ? ??0 ?1 ? i?(??)1/2k ?1 ?
?
2 2 2

? 2x1 cosh ? + 2x0 sinh ? + (x0 cosh ? + x1 sinh ? ? x2 )2 ?

1 1
? exp ??0 ?1 exp ? ??0 ?1 ?.
2 2

On taking into consideration the identities
?
? ?0 cosh ? + ?1 sinh ?, ? = 0,
1 1
exp ? ??0 ?1 ?? exp ?1 cosh ? + ?0 sinh ?, ? = 1,
??0 ?1 =
?
2 2
?? , ? = 2, 3,
276 W.I. Fushchych, R.Z. Zhdanov

we obtain the following expression:

1
(?1 cosh ? + ?0 sinh ?)(?0 sinh ? + ?1 cosh ? ? ?2 ) ?
?2 (x) = exp
2

? (x0 cosh ? + x1 sinh ? ? x2 ) ?

1
? exp ? i?(? ? )1/2k (?1 cosh ? + ?0 sinh ?) ?
?
2

? 2x1 cosh ? + 2x0 sinh ? + (x0 cosh ? + x1 sinh ? ? x2 )2 ?,

where ? = exp ? 2 ??0 ?1 ?.
1

?
Using rest transformations from O(1, 3) ? P(1, 3) in the same way one can find a
family of solutions of equation (1.1) of the form
1 1
(?a)(?b)bx exp ? i?(??)1/2k (?a) 2ax + (bx)2 (5.13)
?(x) = exp ? ?,
2 2
where parameters aµ , bµ satisfy the conditions
aa = ?1, ?a = ?µ aµ , bx = bµ xµ , ab = aµ bµ .
bb = ab = 0,
Applying the formula for generating solutions by scale transformations
?2 (x) = e?k? ?1 (x ), xµ = e? xµ , ? = const
one can obtain
1 1
?(?a)(?b)bx exp ? i?(??)1/2k (?a) 2ax + ?(bx)2 ?. (5.14)
?(x) = exp ?
2 2
At last, generating from (5.14) new solutions by the group of translations, we obtain
an ungenerable family of solutions of the non-linear Dirac equation (1.1).
(i) k ? R1 , k = 0,
1 1
?(?a)(?b)bz exp ? i?(??)1/2k (?a) 2az + ?(bz)2
?(x) = exp ? ?,
2 2
zµ = xµ + ?µ , ?a = ?µ aµ , bz = bµ z µ , az = aµ z µ ,
where ? is an arbitrary constant spinor and ?, ?µ , aµ , bµ are constants satisfying the
following constraints:
aa = ?1, (5.15)
bb = 0, ab = 0.
The same procedure when applied to (5.3), (5.4), (5.11) and (5.12) gives ungene-
rable families of the form
(ii) k ? R1 , k = 0, 1 ,
2

1
?1/4
exp ? (?a)(?b) tan?1 (az/bz) ?
?(x) = (az)2 + (bz)2
2 (5.16)
2 (2k?1)/4k
?1
? exp i2?k(2k ? 1) 1/2k 2
(??)
? (?b) (az) + (bz) ?,
On some exact solutions of a system of non-linear differential equations 277

where aa = ?1, bb = ?1, ab = 0, zµ = xµ + ?µ , ?µ being arbitrary constants, and ?
is an arbitrary constant spinor.
(iii) k = 1 ,
2
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