ñòð. 138 |

Symmetry and exact solutions of nonlinear spinor equations 595

1 ?1/2 1/2

(9) iz9 ?2 ?9 + 2iz9 ?2 ?9 = ?2 (?9 , ?9 ),

? ?

2

where ?l = ?l (zl ), zl being determined by formulae (3.18)–(3.21).

Following ref. [24] we obtain via direct reduction of (3.16) to ODE equations of the

form (the number of PDE from which the ODE is obtained is given in parentheses)

1

(1 ? 2k)?3 ? + 2(?3 + a?2 )??2 = ?i?2 (?, ?)

(1) ? (5);

2

1 ?1/2 1/2

)? + 2?3 ?2 ??2 = ?i?2 (?, ?)

(2) (?0 + ?1 + ?3 ?2 ? (8);

2

(3.33)

1

[(1 ? 2k)(?0 + ?1 ) + ?3 ?2 ]? +

(3)

2

1/2 1/2

+ 2?2 (?0 + ?1 ? ?3 ?2 )??2 = ?i?2 (?, ?)

? (9);

?k?0 ? + (?1 ? ?1 ?0 )??1 = ?i?2 (?, ?)

(4) ? (12).

So we have constructed systems of ODE whose solutions, when substituted into

corresponding ansatze, give rise to exact solutions of the initial nonlinear Dirac equati-

on.

3.3. Exact solutions of the nonlinear Dirac–Heisenberg equation. To integrate

eqs. (3.24), (3.32) and (3.33) one can apply various methods. We restrict ourselves to

?

those ODE which can be integrated in quadrature. Let us put ?1 ? ?2 ? ?(??)1/2k ?,

? and k const. Then the PDE (3.10) and (3.14) take the form

?

[?µ pµ ? ?(??)1/2k ]?(x) = 0. (3.34)

The PDE (3.34) was suggested by W. Heisenberg [4, 36] as a possible basic equation

for the unified field theory. According to theorems 2 and 3 it is invariant under the

extended Poincar? group P (1, 3). In the case k = 3/2, eq. (3.34) admits the conformal

e

group C(1, 3). Therefore, to reduce the PDE (3.34) one can apply both the ans?tze of

a

table 1 and of table 2. As a result we obtain eqs. (3.24), (3.22) and (3.33), where

?1 (?, ?) ? ?2 (?, ?) ? ?(?, ?)1/2k ?.

? ? ?

If one multiplies ODE (3) of (3.24) by ?0 + ?3 and uses the identity (?0 + ?3 )2 = 0,

then the following compatibility condition of eq. (3) of (3.24) appears:

(?0 + ?3 )? = 0,

? ?

whence it easily follows that ?? = 0. So ?? = ?? = 0, i.e., the factor (??)1/2k

? ?

determining the nonlinear character of the PDE (3.34) vanishes. Analogous results

hold for eqs. (12)–(14) and (19)–(22) of (3.24). Such solutions are not considered.

ODE (1), (2), (15), (16) and (24) of (3.24) are trivially integrated if one notes

that the condition ?? = const holds. Let us consider, for example, eq. (1). After

?

multiplying it by i?2 one obtains

? = i?(??)1/2k ?2 ?. (3.35)

? ?

The conjugate spinor satisfies the following equation:

? = ?i?(??)1/2k ??2 .

? (3.36)

? ? ?

596 W.I. Fushchych, R.Z. Zhdanov

Multiplying (3.35) by ? and (3.36) by ? we come to the equality

?

(d/d?)(??) ? ?? + ?? = 0,

?

? ? ??

whence it follows that ?? = const. Consequently, the ODE (3.35) is equivalent to a

?

linear one,

? = i?C 1/2k ?2 ?,

? ?? = C,

?

whose general solution has the form ?(?) = exp(i?C 1/2k ?2 ?)?. Since ?(?) =

?

1/2k

?2 ?), then ?? = ?? or ?? = C. Finally, the general solution of

? exp(?i?C

? ? ? ?

(3.35) takes the form

?(?) = exp[i?(??)1/2k ?2 ?]?. (3.37)

?

Hereafter ? is an arbitrary constant spinor. Let us note that, taking into account the

identity (i?2 )2 = 1, expression (3.37) can be rewritten in the following way:

?(?) = {cosh[?(??)1/2k ?] + i?2 sinh[?(??)1/2k ?]}?.

? ?

The general solutions of eqs. (2), (15), (16) and (24) of (3.24) are constructed in

the same way. Omitting intermediate calculations we write down the final result,

?(?) = exp[?i?(??)1/2k ?0 ?]?,

?

1

?(?) = exp i?(??)1/2k ?1 ? ?,

?

2

(3.38)

i?

(??)1/2k (?2 ? ??1 )? ?,

?(?) = exp ?

2)

2(1 + ?

?(?) = exp{[?2 (?0 + ?3 ) + i?(??)1/2k (?2 ? ?(?0 + ?3 ))]?}?.

?

To construct the solution of ODE (6) of (3.24) we use its symmetry properties. Above

it was established that this equation is invariant under the Lie algebra ?? , ?0 ?3 . We

look for the solution which is invariant under the group generated by the operator

Q = ?? ? ??0 ?3 , ? = const, i.e., ?(?) has to satisfy the additional constraint

Q? ? (?? ? ??0 ?3 )? = 0.

The general solution of the above equation is given by the formula

?(?) = exp(??0 ?3 ?)?1 ,

where ?1 is an arbitrary constant spinor. Substituting this expression into the initial

ODE one has

1

??1 ?0 ?3 ? ?1 ?4 exp(??0 ?3 ?)?1 = ?i?? exp(??0 ?3 ?)?1 ,

2?

where ? = (?1 ?1 )1/2k . Multiplying this equality by exp(???0 ?3 ) we come to the

?

system of linear algebraic equations for ?1 ,

1

??2 ? ?1 ?4 ?1 = ?i?? ?1 . (3.39)

2?

Symmetry and exact solutions of nonlinear spinor equations 597

The system (3.39) is diagonalized by the following substitution:

1

??2 ? ?1 ?4 ? i?? ?,

?1 =

2?

whence it follows that

1

??2 ? + ?2 ? 2 ? = 0.

2

4?

Consequently

? = ?(4?2 ? 2 ?2 ? 1)1/2 /2?, ? = ±1. (3.40)

Imposing on ? the condition ? = (?1 ?1 )1/2k one obtains a nonlinear algebraic equation

?

for ? ,

? 2k = 2?2 ? 2 (??) + 2i?? ?(??2 ?4 ?) ? i?? ??1 (??1 ?4 ?). (3.41)

? ? ?

Finally, the particular solution of ODE (6) of (3.34) takes the form

1

??2 ? ?1 ?4 ? i?? ?, (3.42)

?(?) = exp(??0 ?3 ?)

2?

? and ? being determined by formulae (3.40) and (3.41).

An analogous method can be applied to construct solutions of equations (9–11),

the result being

1

??0 ? ?3 ?4 ? i?? ?, (3.43)

?(?) = exp(??1 ?2 ?)

2?

? and ? being determined by the formulae

? = ?(1 ? 4?2 ?2 ? 2 )1/2 /2?,

(3.44)

? 2k = 2?2 ? 2 (??) ? i?? ??1 (??3 ?4 ?) + 2i?? ?(??0 ?4 ?);

? ? ?

1

?0 ?4 ? i?? ?, (3.45)

?(?) = exp(??1 ?2 ?) ??3 +

2?

? and ? being determined by the formulae

? = ?(4?2 ?2 ? 2 + 1)1/2 /2?,

(3.46)

? 2k = 2?2 ? 2 (??) + 2i?? ?(??3 ?4 ?) + i?? ??1 (??0 ?4 ?);

? ? ?

?(?) = exp(??1 ?2 ?)[4?(?0 + ?3 )?4 + (?0 ? ?3 )?4 ? 4i?? ]?, (3.47)

? and ? being determined by the formulae

? = ??2 ? 2 ,

(3.48)

? 2k = 32?2 ? 2 (??) + 8i?? [?(?0 ? ?3 )?4 ?] ? 32i?3 ? 3 [?(?0 + ?3 )?4 ?].

? ? ?

Eq. (8) of (3.24) is, via the change of variables

?(?) = ? ?1/4 ?(?),

598 W.I. Fushchych, R.Z. Zhdanov

reduced to the following ODE:

2i? 1/2 ?2 ? = ?? ?1/4k (??)1/2k ?.

? ?

Multiplying it by 1 i?2 ? ?1/2 we come to an equation of the form

2

1

? = i?? ?(1+2k)/4k (??)1/2k ?,

? ?

2

whose general solution is given by the formulae

2i?k

(??)1/2k ?2 ? (2k?1)/4k ?,

k = 1/2 : ?(?) = exp ?

1 ? 2k

1

k = 1/2 : ?(?) = exp i?(??)?2 ln ? ?.

?

2

So the general solution of ODE (8) has the form

2i?k

?(?) = ? ?1/4 exp (??)1/2k ?2 ? (2k?1)/4k ?,

k = 1/2 : ?

1 ? 2k (3.49)

?(?) = ? ?1/4 exp 1

k = 1/2 : 2 i?(??)?2

? ln ? ?.

Besides we have succeeded in integrating eqs. (4), (23) and (26) of (3.24) (for

? = 0). These ODE can be written in the following way:

1

m(?0 + ?3 )? + [?(?0 + ?3 ) + ?0 ? ?3 ]? = ?i?(??)1/2k ?,

? ?

2

where for m = 1, 2, 3 eqs. (4), (23), (26) of (3.24) are obtained. Multiplying both

parts of the equality by ?(?0 + ?3 ) + ?0 ? ?3 , comes to the ODE

4? ? = ?{m(1 + ?0 ?3 ) + i?(??)1/2k [?(?0 + ?3 ) + ?0 ? ?3 ]}?, (3.50)

? ?

and the equation for the conjugate spinor has the form

4? ? = ??{m(1 ? ?0 ?3 ) ? i?(??)1/2k [?(?0 + ?3 ) + ?0 ? ?3 ]}.

?

? ? ?

Multiplying the first equation by ? and the second by ? one obtains the following

?

relation:

4(?? + ??) = ?2m??,

?

? ?? ?

whence it follows that ?? = c? ?m/2 , C = const. Substitution of the above result into

?

(3.50) gives rise to a linear equation for ?(?),

ñòð. 138 |