ñòð. 135 |

the same principal part.

The problem of classification of all inequivalent subalgebras of the Poincar? algebra

e

P (, 3) was solved in refs. [43–45]. Integration of the PDE (3.8), (3.9) is carried out

by standard methods but the calculations are rather cumbersome. We give here the

final result in table 1.

As an example we consider the case Q1 = J03 , Q2 = P1 , Q3 = P2 , i.e.,

(x0 p3 ? x3 p0 )? = 0, (3.12)

p1 ? = p2 ? = 0;

1

x0 p3 ? x3 p0 + i?0 ?3 A(x) = 0, (3.13)

p1 A(x) = p2 A(x) = 0.

2

From the last two equations of the system (3.12) it follows that ? = ?(x0 , x3 ).

Substituting this result into the first equation one obtains that ?(x0 , x3 ) is a first

integral of the Euler–Lagrange system

dx0 dx3

= ,

x3 x0

which can be chosen in the form ? = x2 ? x2 .

0 3

A solution of the system (3.13) is looked for in the form

A(x) = exp[?0 ?3 f (x)]

whence it follows that the scalar function f (x) satisfies the following equation:

1

x0 fx3 + x3 fx0 = ,

2

whose particular solution has the form

1

f (x) = ln(x0 + x3 ).

2

Symmetry and exact solutions of nonlinear spinor equations 585

Table 1. P (1, 3)-invariant anz?tze for the spinor field

a

No. Algebra A(x) ?(x)

1 P0 , P1 , P2 I x3

2 P1 , P2 , P3 I x0

P0 + P3 , P1 , P2 x0 + x3

3 I

x2 ? x2

1

exp ? ? ln(x0 + x3 )

4 J03 , P1 , P2 203 0 3

1

J03 , P0 + P3 , P1 exp ? ? ln(x0 + x3 )

5 x2

203

J03 + ?P2 , P0 , P3 exp x2

6 ?? x1

2? 0 3

J03 + ?P2 , P0 + P3 , P1 exp ? ln(x0 + x3 ) ? x2

x2

7 ??

2? 0 3

x2 + x2

exp ? 1 ?1 ?2 arctg x1

8 J12 , P0 , P3 1 2

2 x2

J12 + ?P0 , P1 , P2 exp ? 2? ?1 ?2

x0

9 x3

10 J12 + ?P3 , P1 , P2 exp x3

?? x0

2? 1 2

1

J12 + P0 + P3 , P1 , P2 exp (x3 ? x0 )?1 ?2 x0 + x3

11 4

12 G1 , P0 + P3 , P2 exp (?0 + ?3 )?1 x0 + x3

x1

2(x0 +x3 )

?x1 ?x2

13 G1 , P0 + P3 , P1 + ?P2 exp (?0 + ?3 )?1 x0 + x3

2?(x0 +x3 )

14 G1 + P2 , P0 + P3 , P1 exp (?0 + ?3 )?1 x0 + x3

x2

2

2x1 + (x0 + x3 )2

exp ? 1 (x0 + x3 )(?0 + ?3 )?1

15 G1 + P0 , P0 + P3 , P2 2

exp ? 1 (x0 + x3 )(?0 + ?3 )?1

16 G1 + P0 , P1 + ?P2 , 2(x2 ? ?x1 ) ?

2

? ?(x0 + x3 )2

P0 + P3

x2 + x2

1

17 J03 + ?J12 , P0 , P3 exp ? 2? (?0 ?3 + ??1 ?2 ) arctg x1

1 2

x2

x2 ? x2

1

18 J03 + ?J12 , P1 , P2 exp (?0 ?3 + ??1 ?2 ) ln(x0 + x3 ) 0 3

2

?0 +?3

19 G1 , G2 , P0 + P3 exp (?1 x1 + ?2 x2 ) x0 + x3

2(x0 +x3 )

?0 +?3

20 G1 + P2 , exp ? x0 + x3

2[(x0 +x3 )(x0 +x3 +?)??]

G2 + ?P1 + ?P2 , ? {?1 [(x0 + x3 + ?)x1 ? ?x2 ] +

P0 + P3 + ?2 [(x0 + x3 )x2 ? x1 ]}

21 G1 , G2 + P1 + ?P2 , exp (?0 + ?3 )?1 + x0 + x3

x1

2(x0 +x3 )

P0 + P3 + ?

x2

2(x0 +x3 )(x0 +x3 +?)

? (?0 + ?3 )[?2 (x0 + x3 ) ? ?1 ]

22 G1 , G2 + P2 , exp (?0 + ?3 ) + x0 + x3

x1

?

2(x0 +x3 ) 1

P0 + P3 + x2

?

2(x0 +x3 +1) 2

x2 ? x2 ? x2

exp (?0 + ?3 )?1 ?

x1

23 G1 , J03 , P2 0 1 3

2(x0 +x3 )

1

? exp 2 ?0 ?3 ln(x0 + x3 )

x1 ?? ln(x0 +x3 )

24 J03 + ?P1 + ?P2 , exp (?0 + ?3 )?1 ? x2 ? ? ln(x0 + x3 )

2(x0 +x3 )

exp 1 ?0 ?3 ln(x0 + x3 )

G1 , P 0 + P 3 ? 2

?0 +?3

25 J12 + P0 + P3 , G1 , G2 exp (?1 x1 + ?2 x2 ) ? x0 + x3

2(x0 +x3 )

? exp ? 4(xx·x 3 ) ?1 ?2

0 +x

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

Table 1 (Continued)

No. Algebra A(x) ?(x)

?0 +?3

26 J03 + ?J12 , G1 , G2 exp (?1 x1 + ?2 x2 ) ? x·x

2(x0 +x3 )

exp 1 (?0 ?3 + ??1 ?2 ) ln(x0 +

? x3 )

2

Notation:

?

Gk = J0k + J3k ? (x0 + x3 )pk ? xk (p0 + p3 ) + 1 i(?0 + ?3 )?k , exp{R} ? I + 1

Rn ,

2 n!

n=1

I is the 4 ? 4 matrix, ?, ? ? R1 .

Finally one has

1

?(x) = x0 ? x2 , A(x) = exp ?0 ?3 ln(x0 + x3 ) .

3

2

Other triplets Q1 , Q2 , Q3 from table 1 are treated in an analogous way (we show

in table 1 only algebras Q1 , Q2 , Q3 giving nontrivial ans?tze (3.7)).

a

It is important to note that the ans?tze listed in the table 1 do not exhaust all

a

possible substitutions of the form (3.7) reducing the PDE (3.10) to ODE. Principally

different ans?tze are obtained when the conditions of theorem 4 are valid and some

a

operators Qa are not symmetry operators of eq. (3.10) (conditional invariance).

To investigate the conditional invariance of a differential equation one can also

apply the infinitesimal Lie algorithm [32–34]. However, the determining equations

to be solved are nonlinear (see refs. [39, 40]). To avoid this difficulty the following

method was suggested [24, 30]: firstly, the dimension of the PDE is decreased by

one using its group-theoretical properties and then the maximal symmetry of the

reduced equations is investigated. Under certain circumstances this procedure yields

such operators Qa that the initial equation is conditionally invariant under Qa .

?

We realize the above scheme for the PDE (1.5) invariant under the group P (1, 3),

i.e. for equations of the form

? ?

? ?

?µ pµ ? = ?2 (?, ?) ? [(F1 + F2 ?4 )(??)1/2k ]?. (3.14)

?

P (1, 3)-invariant ans?tze for the spinor field reducing (3.14) to three-dimensional

a

PDE were constructed in refs. [24,29]. The general form is

(3.15)

?(x) = A(x)?(?1 , ?2 , ?3 ),

where ? is a new unknown spinor; the 4 ? 4 matrix A(x) and the scalar functi-

ons ?i (x) are determined from table 2 (each ansatz in table 2 corresponds to some

one-dimensional subalgebra of the algebra AP (1, 3); for more detail see ref. [24]).

Substitution of the ans?tze (3.15), with A(x) and ?(x) as listed in the table 2, into

a

the PDE (3.14) results in a reduction by one of the number of independent variables,

i.e., the equations obtained depend on the three independent variables ?1 , ?2 , ?3

only. Omitting intermediate calculations we write down the reduced equations for

?(?1 , ?2 , ?3 ).

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

22 2 2

(1)

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

2

+ [a?1 + (?2 ? ?0 )(?3 + 1)]??3 = ?i?2 (?, ?);

?

?

Table 2. P (1, 3)-invariant anz?tze for the spinor field

a

No. A(x) ?1 (x) ?2 (x) ?3 (x)

1

1 0 2 0 1 2

(x2 ? x2 )?k exp ? (? ? ?0 ) ln(x0 ? x2 ) (x2 ? x2 ? x2 )x?2 (x0 ? x2 )x?1 ax1 (x0 ? x2 )?1 ? ln(x0 ? x2 )

3 3

2a 1 2

2 x3 0 1 2

exp ? 1 x1 (x0 ? x2 )?1 ?1 (?0 ? ?2 ) x0 ? x2 x2 ? x2 ? x2

2

1

exp

3 (x2 ? x0 )?1 (?2 ? ?0 ) x1 + ?(x0 ? x2 ) 2x1 + (x0 ? x2 )2 3x3 + 3x1 (x0 ? x2 ) + (x0 ? x2 )3

2

1

exp x ? (?

4 1 1 2

? ?0 ) x0 ? x2 x2 ? x2 ? x2 ?x1 ? (x0 ? x2 )x3

2? 3 1 2

x2 x2

ln(x2 + x2 ) + 2 arctg

5 x0 x?1

2 3 2 3 2 3 0 1

(x2 + x2 )?k/2 exp ? 1 ?2 ?3 arctg (x2 + x2 )(x2 ? x2 )?1

1

2 x3 x3

1 x2

?? b ln(x2 + x2 ) + 2a arctg

6 0 1 0 1 0 1 2 3 2 3

(x2 ? x2 )?k/2 exp ln(x0 + x1 ) ? (x2 ? x2 )?a?1 (x0 + x1 )2a (x2 ? x2 )(x2 + x2 )?1

2(a+1) 0 1 x3

x2

? 1 ?2 ?3 arctg

2 x3

x2

7 x0 + x1

0 1 0 1 2 3 2 3

(x2 ? x2 )?k/2 exp ? 1 ?0 ?1 ln(x0 ? x1 ) ? (x2 ? x2 )(x2 + x2 )?1 b ln(x2 + x2 ) ? 2 arctg

4 x3

x2

? 1 ?2 ?3 arctg

2 x3

1 x2 x2

??

exp x2 + x2

8 b ln(x0 + x1 ) + arctg

0 1 2 3

ln(x0 + x1 ) ? 1 ?2 ?3 arctg x2 ? x2

201 2 x3 x3

1

?? b ln(x2 + x2 ) + 2 arctg x2

9 (2x0 + 2x1 + ?)?k/2 exp ln(2x0 + 2 3

(2x0 + 2x1 + ?) ? (2x0 + 2x1 + ?) ?

401 x3

2

x2

(x1 2 3

ñòð. 135 |