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4 2

1
(3.24)
(vii) ?(x) = exp ?0 ?1 ln(x0 + x1 ) ?(?),
2

1
(viii) ?(x) = exp ? ?2 ?3 tan?1 (x2 /x3 ) ?(?), (3.25)
2

(ix) ?(x) = x?k ?(?), (3.26)
0

(3.27)
(x) ?(x) = ?(?),

(3.28)
(xi) ?(x) = ?(?),

(3.29)
(xii) ?(x) = ?(?).
?
The problem of finding all the ans?tze for P(1, 3)-invariant solutions is therefore
a
completely solved. The second step of the algorithm — the reduction of the Dirac
equation — will be realised in the next section.
270 W.I. Fushchych, R.Z. Zhdanov

4. Reduction of the non-linear Dirac equation (1.1)
It was pointed out above that substitution of ansatz (1.3) into (1.1) results in a
reduction by one of a number of independent variables. This means that the equation
obtained will depend on the three independent variables ?1 , ?2 , ?3 . Omitting cum-
bersome calculations we write down resulting systems of PDE:
(i) k(?2 ? ?0 )? + (?0 ? ?2 ) ?1 + a?2 ?2 ?3 +
22

+ (?0 + ?2 )?2 ? 2a?1 ?1 ?3 ?2 ? 2?3 ?1 ?2 ??1 +
2 2
(4.1)
+ (?0 ? ?2 )?2 ? ?3 ?2 ??2 +
2

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

1 ?1
(?0 ? ?2 )?1 ? + (?0 ? ?2 )??1 + ?3 ??2 +
(4.2)
2
?1
+ (?0 + ?2 )?1 + (?0 ? ?2 )?3 ?1 ??3 = i?(??)1/2k ?,
?

(ii) [?3 + ?(?0 ? ?2 )]??1 + 2?1 ??2 +
(4.3)
3
+ (2?2 ) + (?0 ? ?2 )?2 )??3 = i?(??)1/2k ?,
?
2
1 ?1
? ?4 (?0 ? ?2 )? + (?0 ? ?2 )??1 + (?0 + ?2 )?1 ? 2? ?1 ?1 ?3 +
(iii)
(4.4)
2
?1
?2 2
+ (?0 ? ?2 ) ? ?3 + ?2 ?1 ??2 + (??1 ? ?3 ?1 )??3 = i?(??) 1/2k
? ?,

1
(1 ? 2k)?3 ? + (?1 ?3 )1/2 (?0 ? ?1 ?1 )??1 + 2(?3 + a?2 )??2 +
(iv)
2 (4.5)
1/2 ?1/2
+ 2?3 ? (?0 + ?1 ?1 )?3 ?1 ?3 ??3 = i?(??)1/2k ?,
?

?1/2(a+1) 1/2(a+1)
(v) ?k ?0 cosh ln ?1 ? ?1 sinh ln ?1 +
1 1
?1/2(a+1)
+ (a + 1)?1 (?0 + ?1 )?1
1/2
??
+ ?3 ?2
2 2
(4.6)
1/2(a+1) 1/2(a+1)
? 2(a + 1)?1 ?0 cosh ln ?1 ? ?1 sinh ln ?1 ?? 1 +
1/2(a+1) 1/2(a+1) 1/2
? ?1 sinh ln ?1 ? ?3 ?2
+ 2 ?0 cosh ln ?1 ? 2 ?? 2 +
1/2
+ 2(a?2 + b?3 )?2 ??3 = i?(??)1/2k ?,
?

1/2 1/2
?k ?0 cosh ln ?1 ? ?1 sinh ln ?1 +
1 1
1/2 1/2 1/2
+ (?0 ? ?1 )?1 + ?3 ?2 ? + (?0 + ?1 )?1 ??1 +
4 2 (4.7)
1/2 1/2 1/2
? ?1 sinh ln ?1 ? 2?3 ?2
+ 2?2 ?0 cosh ln ?1 ?? 2 +
1/2
+ 2(b?3 ? ?2 )?2 ??3 = i?(??)1/2k ?,
?

1
[(1 ? 2k)(?0 + ?1 ) + ?3 ?2 ]? + 2[(? ? 1)?0 + (? + 1)?1 ]?1 ??1 +
(vi)
2 (4.8)
1/2 1/2
+ 2?2 ?0 + ?1 ? ?2 ?3 ??2 + 2(?2 + b?3 )?2 ??3 = i?(??)1/2k ?, ?
On some exact solutions of a system of non-linear differential equations 271

1
(?0 + ?1 )? + [?0 (?1 + 1) + ?1 (?1 ? 1)]??1 +
(vii)
2 (4.9)
+ (?0 + ?1 ? ?2 )??2 + ?3 ??3 = i?(??)1/2k ?,
?

1 ?1/2 ?1/2
1/2
(viii) ?1 ? + 2?1 ?3 ??1 + ?1 ?2 + ?1 ?0 + ?2 ?1 ??2 +
(4.10)
2
+ (?2 ?0 + ?1 ?1 )??3 = i?(??)1/2k ?,
?

(ix) ?k?0 ? + (?a ? ?a ?0 )??a = i?(??)1/2k ?, (4.11)
?

(x) (?0 + ?1 )??1 + ?2 ??2 + ?3 ??3 = i?(??)1/2k ?, (4.12)
?

(xi) ?a ??a = i?(??)1/2k ?, (4.13)
?

(xii) ?0 ??1 + ?2 ??2 + ?3 ??3 = i?(??)1/2k ?, (4.14)
?

wher ??a = ??/??a and a = 1, 3.
A partial solution of one of the equations (4.1)–(4.14) through formulae (3.16)–
(3.29) gives a partial solution of the non-linear Dirac equation. To obtain a partial
solution of the reduced equation one can again apply the reduction procedure. But
it demands a knowledge of the symmetry of equations (4.1)–(4.14). Investigation of
symmetrical properties of equations in question is a very interesting problem (for
example, equation (4.12) possesses an infinite-parameter symmetry group) and it will
be considered in a future paper. We shall perform the direct reduction (if it is possible)
of systems (4.1)–(4.14) to systems of ODE.
Let us suppose that in (4.1) ? = ?(?2 ). It follows that

k(?2 ? ?0 )? + ?2 (?0 ? ?2 ? ?2 ?3 )??2 = i?(??)1/2k ?. (4.15)
?

Similarly, if one chooses ? = ?(?3 ) then

k(?2 ? ?0 )? + [(?2 ? ?0 )(1 + ?3 ) + a?1 )??3 = i?(??)1/2k ?. (4.16)
?

(4.15) and (4.16) are non-linear systems of ODE.
Equation (4.2) gives the following system of ODE:
1 ?1
(?0 ? ?2 )??1 + ?1 (?0 ? ?2 )? = i?(??)1/2k ?. (4.17)
?
2
From (4.3) it follows that

[?3 + ?(?0 ? ?2 )]??1 = i?(??)1/2k ?, (4.18)
?

2?1 ??2 = i?(??)1/2k ?. (4.19)
?

Systems (4.4) and (4.5) can be reduced to the systems of ODE of the form

2?(?0 ? ?2 )??1 + (?0 ? ?2 )?4 ? = 2i?(??)1/2k ?, (4.20)
?

1
(1 ? 2k)?3 ? + 2(?3 + a?2 )??2 = i?(??)1/2k ?. (4.21)
?
2
272 W.I. Fushchych, R.Z. Zhdanov

We did not succeed in reducing systems (4.6)–(4.8) to ODE. From (4.9) one can
obtain three systems of ODE:
1
(?0 + ?1 )? + [(?0 + ?1 )?1 + ?0 ? ?1 ]??1 = i?(??)1/2k ?, (4.22)
?
2
1
(?0 + ?1 )? + (?0 + ?1 ? ?2 )]??2 = i?(??)1/2k ?, (4.23)
?
2
1
(?0 + ?1 )? + ?3 ??3 = i?(??)1/2k ?. (4.24)
?
2
Equation (4.10) gives the system
1 ?1/2 1/2
+ 2?3 ?1 ??1 = i?(??)1/2k ?. (4.25)
?3 ?1 ?
2
Equations (4.11)–(4.14) are reduced to the following systems of ODE:
?k?0 ? + (?a ? ?a ?0 )??a = i?(??)1/2k ?, (4.26)
?

(?0 + ?1 )??1 = i?(??)1/2k ?, (4.27)
?

?a ??a = i?(??)1/2k ?, (4.28)
?

?0 ??1 = i?(??)1/2k ? (4.29)
?

(no summation is carried over a).
Symmetry properties of the non-linear Dirac equation therefore enable us to reduce
the problem of finding its partial solution to an essentially simpler one of integration
of systems of ode (4.15)–(4.29). To solve these systems one can apply various methods
including numerical ones.

5. Construction of exact solutions of the non-linear Dirac equation (1.1)
We shall consider only systems of ODE solvable in quadratures, but we shall not
consider cases which give already known solutions. The general solution of (4.19) has
the form
?(?2 ) = exp[?(i?/2)(??)1/2k ?1 ?2 ]?,
?
where ? is an arbitrary constant spinor.
Substituting the above result into (3.18), we obtain a solution of the initial equation
(1.1):
1
(?0 ? ?2 )(x0 ? x2 ) ?
?(x) = exp
2 (5.1)
? exp ?(i?/2)(??) ?1 2x1 + (x0 ? x2 )
1/2k 2
? ?.
1
Let us next consider equation (4.21). Under k = its general solution has the
2
form
1 ?1
?(?2 ) = exp ? i??? 1 + a2 (5.2)
? (?3 + a?2 )?2 ?,
2
where ? is a constant spinor.
On some exact solutions of a system of non-linear differential equations 273

Under k = 1 , a = 0, we did not succeed in integrating the corresponding equation.
2
If a = 0, then making a change of variables we obtain
1
(2k ? 1)?2 ?(?2 ),
?(?2 ) = exp
4
1
(1 ? 2k)k ?1 ?2 ?3 ??2 = i?(??)1/2k ?.
?
2 exp
4
The general solution of the last equation is given by the formula
1
? = exp (2i?k)(1 ? 2k)?1 (??)1/2k exp (2k ? 1)k ?1 ?2 ?3 ?,
?
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