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1. Introduction

In the present work using ideas and methods of S. Lie (see [12, 2]) we have

constructed large classes of exact solutions of the non-linear Dirac equation

?

?µ pµ + ?(??)1/2k ?(x) = 0, (1.1)

k = 0,

where ?µ are 4 ? 4 Dirac matrices, pµ = igµ? ?/?x? , ? = ? † ?0 , x = (x0 , x1 , x2 , x3 ),

?

? is a four-component spinor and k, ? are parameters, and of the system of eight

non-linear equations,

(?µ pµ + ?1 ?µ Aµ + m1 )?(x) = 0,

(1.2)

?

p? p? Aµ ? pµ p? A? = exp(??µ ?) + Aµ (m2 + ?2 A? A? ),

where Aµ (x) is the vector potential of the electromagnetic field and e, ?1 , ?2 , m1 , m2

are constants. If we choose m2 = ?2 = 0, then system (1.2) coincides with equations

of the classical electrodynamics describing interaction of electromagnetic and spinor

fields.

To construct multiparameter families of exact solutions of (1.1) and (1.2) we

essentially use their symmetry properties and the ansatz

(1.3)

?(x) = A(x)?(?) + B(x)

suggested by Fushchych [3, 4] and effectively realised by Fushchych and Shtelen [6, 7]

and Fushchych and Serov [5] for a number of non-linear wave equations. A(x) is a 4?4

matrix and B(x) is a four-component spinor, algorithms for their construction being

cited below, and ?(?) is the column vector, components of which depend in general

on three invariant variables ? = {?1 , ?2 , ?3 } (for more details see Fushchych [3, 4]).

Later we shall consider the case when B(x) = 0.

On using finite transformations it is established that equation (1.1) is invariant

?

under the extended Poincar? group P(1, 3), i.e. under the Poincar? group P(1, 3)

e e

supplemented by a group of scale transformations.

?

Basis elements of the Lie algebra AP(1, 3) have the form

Pµ = pµ , Jµ? = xµ p? ? x? pµ + Sµ? ,

(1.4)

D = xµ pµ ? ik, Sµ? = (i/4)(?µ ?? ? ?? ?µ ), µ, ? = 0, 3.

J. Phys. A: Math. Gen., 1987, 20, P. 4173–4190.

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

A general scheme for constructing solutions of the system (1.1) (solutions of the

system (1.2) are obtained in an analogous way) is as follows. We look for solutions of

?

equation (1.1) which are invariant under the subgroup of the group P(1, 3) generated

?

by linear combination of all basis elements of AP(1, 3)

Q = C µ? Jµ? + C 00 D + C µ Pµ , (1.5)

where C µ? , C 00 , C µ are constants and µ, ? = 0, 3.

The matrix A(x) is a solution of the following system of partial differential equa-

tions (PDE):

(1.6)

QA(x) = 0.

Invariant variables are the first integrals of the Euler–Lagrange system of ordinary

differential equations (ODE)

dx0 dxa

(1.7)

=a , a = 1, 3,

? 0 (x) ? (x)

where ? µ = C µ? x? + C 00 xµ + C µ .

If one knows an explicit form of the matrix A(x) then after substituting (1.3) into

the corresponding equation we shall obtain an equation for a spinor ?(?) depending

on three invariant variables {?1 , ?2 , ?3 } only. This means that ansatz (1.3) with the

chosen matrix A(x) provides separation of variables in equation (1.1). Solutions of the

corresponding equation for ?(?) being substituted in (1.3) yield the solutions of the

initial equation.

To realise this scheme it is necessary first of all to construct in an explicit form

matrices A(x) satisfying (1.6). So one has to solve the first-order linear system of 16

PDE with variable coefficients. It is rather difficult to solve such a system by standard

methods, which is why we use the following trick. The operator Q is transformed into

another operator

Q = W QW ?1 (1.8)

with the help of the invertible operator

W ?1 (x, p) = exp(???), (1.9)

W (x, p) = exp(??),

where

? = ?µ? Jµ? + ?00 D + ?µ Pµ . (1.10)

Transformation W is so chosen that operator Q is as simple as possible. This

purpose can always be achieved because of the Poincar? invariance of system (1.1).

e

From the physical point of view this means that the non-linear Dirac equation is

solved in the fixed reference system. The construction of the solutions which do not

depend on the reference system (ungenerable solutions) is the next step.

On some exact solutions of a system of non-linear differential equations 263

2. Construction of the matrix A(x)

Before proceeding with a direct solution of the system (1.6) let us simplify it using

the method described in the introduction. To do this we need the Campbell–Hausdorff

formula

?

?k

{Q1 , Q2 }k ,

exp(?Q1 )Q2 exp(??Q1 ) =

k! (2.1)

k=0

{Q1 , Q2 }0 = Q2 , {Q1 , Q2 }n = [Q1 , {Q1 , Q2 }n?1 ],

where Q1 , Q2 are operators and [A, B] = AB ? BA.

A fundamental role is played by the following lemma.

Lemma. The operator Q = C µ? Jµ? = Ak Mk + Bl Nl , where Mk = ? 1 ?klm Jlm ,

2

Nk = J0k , by a transformation Q > Q = V QV ?1 , where V = exp(?µ? Jµ? , can be

reduced to one of the following forms:

2

(A · B)2 + A2 ? B 2

(i) Q = ?J01 + ?J23 , = 0,

A · B = A2 ? B 2 = 0.

(ii) Q = ?(J01 + J12 ),

Proof. Let us introduce new operators

Ka = (i/2)(Ma ? iNa ),

Ja = (i/2)(Ma + iNa ), a = 1, 3.

One can easily check that the following commutational relations hold:

(2.2)

[Ja , Jb ] = i?abc Jc , [Ka , Kb ] = i?abc Kc , [Ja , Kb ] = 0

so Q = ak Jk + bl Kl , where ak = ?Bk ? iAk and bl = Bl ? iAl .

Using (2.1) and (2.2) one obtains

?

1/2 2 1/2

V1 QV1?1 = a2 a2 a2 a2 a2

Q= + + J1 + + + a3 K1 = ?J01 + ?J23 ,

1 2 3 1 2

where

?1/2

V1 = exp ?i tan?1 (a2 /a3 )J1 exp i tan?1 a1 a2 + a2 + ?/2 J2 ?

2 3

(2.3)

?1/2

?1 ?1

? exp ?i tan b1 b 2 b2

(b2 /b3 )K1 exp i tan + + ?/2 K2 .

2 3

It is evident that these formulae lose their validity in the case

a2 + a2 + a2 = 0 ? A2 = B 2 , A · B = 0.

1 2 3

Therefore one can use this approach only in case (i). Let us now consider case

(ii). It follows from (2.1) that

exp(?Ma )Ak Mk exp(??Ma ) =

(2.4)

= Ak Mk cos ? + Aa Ma (1 ? cos ?) + ?akl Ak Ml sin ?

(no summation is performed over a),

exp(?Ma )Bl Nl exp(??Ma ) =

(2.5)

= Bl Nl cos ? + Ba Na (1 ? cos ?) + ?akl Bk Nl sin ?

(no summation is performed over a).

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

Using identities (2.4) and (2.5), one can be convinced that the following equality

holds:

Q = V2 QV2?1 = V2 (Ak Mk + Bl Nl )V2?1 = ?|A| sgnA3 (J01 + J12 ),

where

1/2

V2 = exp tan?1 (A1 /A2 )M3 exp tan?1 /A3 M1 ?

A2 + A2

1 2

tan?1 [B3 |A|/(B2 A1 ? B1 A2 )] + ??(B1 A2 ? B2 A1 ) M3

? exp ,

1, x ? 0, 1, x > 0,

sgn x = ?(x) =

?1, x < 0, 0, x ? 0.

This completes the proof. Let us prove the main statement.

Theorem. The operator Q = Ak Mk + Bl Nl + C 00 D + C µ Pµ with the help of trans-

formation (1.8) can be reduced to one of the following forms:

A · B = 0, A2 = B 2 ,

(A)

(2.6)

(i) Q = J01 + J12 + aD,

(ii) Q = J01 + J12 + ?P3 ? P0 , (2.7)

(2.8)

(iii) Q = J01 + J12 + ?P3 ,

2

(A · B)2 + A2 ? B 2

(B) = 0,

(2.9)

(iv) Q = J23 + aD,

(2.10)

(v) Q = J01 + bJ23 + aD,

(2.11)

(vi) Q = J01 + bJ23 + D + ?P0 ,

(2.12)

(vii) Q = J01 + P2 ,

(2.13)

(viii) Q = J23 + ?1 P0 + ?2 P1 ,

(C) A = B = 0,

(2.14)

(ix) Q = D,

(2.15)

(x) Q = P0 + P1 ,

(2.16)

(xi) Q = P0 ,

(2.17)

(xii) Q = P1 .

Proof. If A = 0, B = 0 then it follows from the lemma that there exists an operator

V1 (V2 ) of the form (1.9) such that

under A · B = A2 ? B 2 = 0,

(a)

V1 QV1?1 = ?(J01 + J12 ) + ?D + ?µ Pµ ,

2

under (A · B)2 + A2 ? B 2

(b) = 0,

V2 QV2?1 = ?J01 + ?J23 + ?D + ?µ Pµ .

On some exact solutions of a system of non-linear differential equations 265

It is clear from (1.6) and (1.7) that operators Q and ?Q, ? = 0, generate the same

invariant solutions. One may suppose that ? = 1.

We need the following formulae which are consequences of the Campbell–Haus-

dorff formula:

exp(i?µ Pµ )J?? exp(?i?µ Pµ ) = J?? + (?? P? ? ?? P? ), (2.18)

exp(i?µ Pµ )D exp(?i?µ Pµ ) = D ? ?µ Pµ , (2.19)

exp(i?µ Pµ )P? exp(?i?µ Pµ ) = P? . (2.20)

Let us consider the case (a):

Q > Q = exp(i?µ Pµ )(J01 + J12 + ?D + ?? P? )(exp(?i?µ Pµ ) =

= J01 + J12 + ?D + ?µ Pµ + ?1 P0 ? ?2 P1 ? ?1 P2 ? ??? P? .

Under ? = 0 one can always choose ?? that

Q = J01 + J12 + ?D

and under ? = 0 so that

? ? 0.

Q = J01 + J12 + ?P0 + ?P3 ,

If in the last operator ? = 0, then

= exp(?i ln |?|D)(J01 + J12 + ?P0 + ?P3 ) exp(i ln |?|D) =

Q

= J01 + J12 ? P0 + ?P3 .

If ? = 0 then

Q = J01 + J12 + ?P3 .

Let us now consider case (b). If ? = 0 then on dividing into ? and on transforming

the operator Q according to (2.18)–(2.20) we obtain

Q = exp(i?µ Pµ )(J01 + bJ23 + ?D + ?µ Pµ ) exp(?i?µ Pµ ) =

= J01 + (?1 P0 ? ?0 P1 ) + bJ23 + b(?3 P2 ? ?2 P3 ) + ?D ? ??µ Pµ + ?µ Pµ .

Under ? = ±1, ?2 + b2 = 0 it is always possible to choose ?µ so that

Q = J01 + bJ23 + ?D.

Under ? = ±1 it is possible to choose ?µ so that

Q = J01 + bJ23 + ?D + ?P0 .

Under ? = b = 0 there exist such ?µ that

Q = J01 + P2 .

Under ? = 0 using formulae (2.18)–(2.20) one can check that the operator Q can

be reduced to one of the following forms:

Q = J23 + aD, ? = 0,

Q = J23 + ?1 P0 + ?2 P1 , ? = 0.

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