ñòð. 63 |

The only thing left is to consider the case A = B = 0, i.e. Q = ?D + ?µ Pµ . Using

formulae (2.18)–(2.20) it is easy to be convinced that under ? = 0

exp[(i/?)?µ Pµ ](?D + ?µ Pµ ) exp[?(i/?)?µ Pµ ] = ?D.

If ? = 0 then analysing three possibilities ?µ ?µ = 0, ?µ ?µ > 0, ?µ ?µ < 0 we obtain

operators (2.15)–(2.17). The theorem is proved.

Note 1. When proving the theorem we used only commutational relations of an

?

algebra AP(1, 3) and we did not use its concrete representation.

?

Note 2. It is seen from the proof that P(1, 3)-invariant solutions are exhausted by

solutions generated from ones invariant under operators (2.6)–(2.17) with the help of

?

transformations from P(1, 3).

This theorem essentially simplifies the problem of finding ans?tze because instead

a

of integrating the system (1.6) where Q is an operator of the general form (1.5), it is

enough to find a partial solution of this system with Q having the form (2.6)–(2.17).

For example, let us consider case (2.9). The matrix A(x) is a solution of the

following matrix system of PDE

1

x2 Ax3 ? x3 Ax2 + ?2 ?3 A + axµ Axµ ? akA = 0, (2.21)

2

where Axa = ?A/?xa , a = 0, 3.

We look for a partial solution of (2.21) of the form

(2.22)

A(x) = f (x) exp(g(x)?2 ?3 ).

Substituting (2.22) into (2.21) we obtain

1

x2 fx3 ? x3 fx2 + axµ fxµ ? akf + f x2 gx3 ? x3 gx2 + axµ gxµ + ?2 ?3 ?

2

? exp(g(x)?2 ?3 ) = 0.

A partial solution of the last system is given by formulae

1

?k/2

g(x) = ? tan?1 (x2 /x3 ).

f (x) = x2 + x2 ,

2 3

2

Finally

1 ?k/2

A(x) = exp ? ?2 ?3 tan?1 (x2 /x3 ) x2 + x2 .

2 3

2

In the same way we have obtained matrices A(x) which correspond to operators

(2.6)–(2.17)

1 ?1

(i) a = 0, A(x) = (x0 ? x2 )?k exp a ?1 (?0 ? ?2 ) ln(x0 ? x2 ) , (2.23)

2

1

x1 (x0 ? x2 )?1 ?1 (?0 ? ?2 ) , (2.24)

a = 0, A(x) = exp

2

1

?1 (?2 ? ?0 )(x2 ? x0 ) , (2.25)

(ii) A(x) = exp

2

On some exact solutions of a system of non-linear differential equations 267

1 ?1

? ?1 (?2 ? ?0 )x3 , (2.26)

(iii) A(x) = exp

2

1

?k/2

exp ? ?2 ?3 tan?1 (x2 /x3 ) ,

(iv) A(x) = x2 + x2 (2.27)

2 3

2

?k/2

(v) a = ?1, A(x) = x2 ? x2 ?

0 1

(2.28)

1 1

? exp (a + 1)?1 ?0 ?1 ln(x0 + x1 ) ? ?2 ?3 tan?1 (x2 /x3 ) ,

2 2

?k/2

a = ?1, A(x) = x2 ? x2 ?

0 1

(2.29)

1 1

? exp ? ?0 ?1 ln(x0 ? x1 ) ? ?2 ?3 tan?1 (x2 /x3 ) ,

4 2

(vi) A(x) = (2x0 + 2x1 + ?)?k/2 ?

(2.30)

1 1

? exp ?0 ?1 ln(2x0 + 2x1 + ?) ? tan?1 (x2 /x3 )?2 ?3 ,

4 2

1

(2.31)

(vii) A(x) = exp ?0 ?1 ln(x0 + x1 ) ,

2

1

(viii) A(x) = exp ? ?2 ?3 tan?1 (x2 /x3 ) , (2.32)

2

(ix) A(x) = x?k I, (2.33)

0

(2.34)

(x) A(x) = I,

(2.35)

(xi) A(x) = I,

(2.36)

(xii) A(x) = I,

where I is a unit 4 ? 4 matrix.

3. Ans?tze for the non-linear Dirac equation (1.1)

a

As pointed out in the introduction, to find invariant variables ?1 (x), ?2 (x), ?3 (x)

it is necessary to find all the first integrals of the Euler–Lagrange system of ODE

dxµ

= Cµ? x? + C00 xµ + Cµ . (3.1)

d?

Because of the lemma proved above, one can restrict oneself to the following cases

of the system (3.1):

C01 = ?C12 = 1, C00 = a, rest coefficients are equal to 0,

(i)

C01 = ?C12 = 1, C0 = ?1,

(ii)

C3 = ??, rest coefficients are equal to 0,

C01 = ?C12 = 1, C3 = ??, rest coefficients are equal to 0,

(iii)

C23 = ?1, C00 = a, rest coefficients are equal to 0,

(iv)

C01 = 1, C23 = ?b, C00 = a, rest coefficients are equal to 0,

(v)

C01 = 1, C23 = ?b, C00 = 1,

(vi)

C0 = ?, rest coefficients are equal to 0,

C01 = 1, C2 = ?1, rest coefficients are equal to 0,

(vii)

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

C23 = ?1, C0 = ?1 , C1 = ??2 , rest coefficients are equal to 0,

(viii)

(ix) Cµ? = 0, C00 = 1, Cµ = 0,

= 0, C0 = ?C1 = 1, C2 = C3 = 0,

(x) Cµ? = C00

(xi) Cµ? = C00 = 0, C1 = C2 = C3 = 0, C0 = 1,

= 0, C0 = C2 = C3 = 1, C1 = ?1.

(xii) Cµ? = C00

Solution of the system (3.1) in cases (i)–(xii) above is carried out in the usual

way, so we write down its first integrals omitting intermediate calculations.

(i) a = 0, ?1 = x2 ? x2 ? x2 x?2 , ?2 = (x0 ? x2 )x?1 ,

0 1 2 3 3

(3.2)

?1

?3 = ax1 (x0 ? x2 ) ? ln(x0 ? x2 ),

a = 0, ?1 = x0 ? x2 , ?2 = x3 , ?3 = x2 ? x2 ? x2 , (3.3)

0 1 2

(ii) ?1 = x3 + ?(x0 ? x2 ), ?2 = 2x1 + (x0 ? x2 )2 ,

(3.4)

?3 = 3x3 + 3x1 (x0 ? x2 ) + (x0 ? x2 )3 ,

(iii) ?1 = x0 ? x2 , ?2 = x2 ? x2 ? x2 , ?3 = ?x1 ? (x0 ? x2 )x3 , (3.5)

0 1 2

(iv) ?1 = x0 x?1 , ?2 = ln x2 + x2 + 2a tan?1 (x2 /x3 ),

2 3

1

(3.6)

?1

2 2

?3 = x2 + x3 (x0 x1 ) ,

(v) a = ?1, ?3 = b ln x2 + x2 + 2a tan?1 (x2 /x3 ),

2 3

(3.7)

?(a+1) ?1

?1 = (x0 + x1 )2a x2 ? x2 , ?2 = x2 ? x2 x2 + x2 ,

0 1 0 1 2 3

?1

a = ?1, ?1 = x0 + x1 , ?2 = x2 ? x2 x2 + x2 ,

0 1 2 3

(3.8)

?1

?3 = b ln x2 + x3 ? 2 tan (x2 /x3 ),

2 2

(vi) ?1 = (2x0 + 2x1 + ?) exp[2? ?1 (x1 ? x0 )],

?1

?2 = (2x0 + 2x1 + ?) x2 + x2 (3.9)

,

2 3

?1

2 2

?3 = b ln x2 + x3 + 2 tan (x2 /x3 ),

(vii) ?1 = x2 ? x2 , ?2 = ln(x0 + x1 ) ? x2 , ?3 = x3 , (3.10)

0 1

(viii) ?1 = x2 + x2 , ?2 = tan?1 (x2 /x3 ) + ?1 x0 + ?2 x1 ,

2 3

(3.11)

?3 = ?2 x0 + ?1 x1 , ??1 ?1 + ?2 ?2 = 1,

(ix) ?a = xa x?1 , a = 1, 3, (3.12)

0

(3.13)

(x) ?1 = x0 + x1 , ?2 = x2 , ?3 = x3 ,

(3.14)

(xi) ?a = xa , a = 1, 3,

(3.15)

(xii) ?1 = x0 , ?2 = x2 , ?3 = x3 .

On some exact solutions of a system of non-linear differential equations 269

Now substituting (2.23 )–(2.36) and (3.2)–(3.15) into (1.3) under B(x) = 0 we

obtain the following set of ans?tze for the non-linear Dirac equation (1.1):

a

1 ?1

(i) ?(x) = (x0 ? x2 )?k exp a ?1 (?2 ? ?0 ) ln(x0 ? x2 ) ?(?), (3.16)

2

1

x1 (x0 ? x2 )?1 ?1 (?2 ? ?0 ) ?(?), (3.17)

?(x) = exp

2

1

?1 (?2 ? ?0 )(x0 ? x2 ) ?(?), (3.18)

(ii) ?(x) = exp

2

1 ?1

? ?1 (?2 ? ?0 )x3 ?(?), (3.19)

(iii) ?(x) = exp

2

1

?k/2

exp ? ?2 ?3 tan?1 (x2 /x3 ) ?(?),

(iv) ?(x) = x2 + x2 (3.20)

2 3

2

?k/2

(v) ?(x) = x2 ? x2 ?

0 1

(3.21)

1 1

? exp (a + 1)?1 ?0 ?1 ln(x0 + x1 ) ? ?2 ?3 tan?1 (x2 /x3 ) ?(?),

2 2

?k/2

?(x) = x2 ? x2 ?

0 1

(3.22)

1 1

? exp ? ?0 ?1 ln(x0 ? x1 ) ? ?2 ?3 tan?1 (x2 /x3 ) ?(?),

4 2

(vi) ?(x) = (2x0 + 2x1 + ?)?k/2 ?

(3.23)

1 1

? exp ?0 ?1 ln(2x0 + 2x1 + ?) ? ?2 ?3 tan?1 (x2 /x3 ) ?(?),

ñòð. 63 |