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are constructed.

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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