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?(i) ?(i)
? exp[?(S03 + ?S12 ) ln(x0 + x3 )]


Let us substitute ans?tze (4.1) from table 1 info (3.15). As a result we obtain the
a
following reduced ODEs:
??
?2 ? + iR? = 0,
(1)
??
?0 ? + iR? = 0,
(2)
??
?
(?0 + ?3 )? + iR? = 0,
(3)
? ?(i) ??
? ? ? ?
?(?0 + ?3 )S ? + [?(?0 + ?3 ) + (?0 + ?3 )]? + iR? = 0,
(4) 03
? ?(i) ??
?
?(?0 + ?3 )S03 ?+ ?2 ? + iR? = 0,
(5)
1 ? ?(i) ? ?
? ?2 S03 ?1 ? + iR? = 0,
(6)
?
1 ? ?(i) ??
? ?
? ?2 S03 ? + [?(?0 + ?3 )e?/? ? ?2 ]? + iR? = 0,
(7)
?
v
1 ? ?(i)
v ?1 S12 ? + 2 ? ?2 ? + iR? = 0,
?
(8)
?
1 ? ?(i) ??
?0 S12 ? + ?3 ? + iR? = 0,
(9)
?
1 ? ?(i) ??
? ?3 S12 ? + ?0 ? + iR? = 0,
(10)
?
1? ? ?(i) ??
?
? (?0 ? ?3 )S12 ? + (?0 + ?3 )? + iR? = 0,
(11)
2
1 ? ?(i) ?(i) ??
?
? ?1 (S01 + S31 )? + (?0 + ?3 )? + iR? = 0,
(12)
?
1? ? ?(i) ?(i) ??
?
(?2 ? ??1 )(S01 + S31 )? + (?0 + ?3 )? + iR? = 0,
(13)
??
? ?(i) ?(i) ??
?
??2 (S + S )? + (?0 + ?3 )? + iR? = 0,
(14) 01 31

?(i) ?(i) ? ??
?
(S01 + S31 )(?0 + ?3 )? + 2?1 ? + iR? = 0,
(15)
(i) (i) ??
? ? ? ? ?
(S01 + S31 )(?0 + ?3 )? + 2(?2 ? ??1 )? + iR? = 0,
(16)
332 W.I. Fushchych, W.M. Shtelen, S.V. Spichak

v
1 ? ?(i) ?(i)
? v ?1 (S03 + ?S12 )? + 2 ? ?2 ? + iR? = 0,
??
(17)
??
? ?(i) ?(i) ??
? ? ? ?
?(?0 + ?3 )(S03 + ?S12 )? + [?(?0 + ?3 ) + (?0 ? ?3 )]? + iR? = 0,
(18)
1 ? ?(i) ?(i) ? ?(i) ?(i) ??
?
? [?1 (S01 + S31 ) + ?2 (S02 + S32 )]? + (?0 + ?3 )? + iR? = 0,
(19)
?
[?(? + ?) ? ?]?1 {[??2 ? (? + ?)?1 ](S + S ) +
(i) (i)
? ?
(20) 01 31

? ?(i) ?(i) ??
? ?
+ (?1 ? ? ?2 )(S02 + S32 )}? + (?0 + ?3 )? + iR? = 0,
[(??1 + (? + ?)?1 ?2 )? ?1 (S01 + S31 ) ?
?(i) ?(i)
? ?
(21)
? (?2 (? + ?)?1 (S02 + S32 )]? + (?0 + ?3 )? + iR? = 0,
?(i) ?(i) ??
? ?
(4.4)
?
?2 ?(i) ?(i)
1 ? 1 ?(i) ?(i) ??
?
? ? (S01 + S31 ) ? (S + S32 ) ? + (?0 + ?3 )? + iR? = 0,
(22)
? + 1 02
?
? ?(i) ? ?(i) ?(i) ??
? ? ? ?
?[(?0 + ?3 )S03 + ?1 (S31 + S31 )]? + [?(?0 + ?3 ) + ?0 ? ?3 ]? + iR? = 0,
(23)
? ?(i) ? ?(i) ?(i) ??
? ? ?
?[(?0 + ?3 )S03 + ?1 (S31 + S31 )]? + [?2 ? ?(?0 + ?3 )]? + iR? = 0,
(24)
1 ?(i) ?
2S03 (?0 + ?3 )? ?1 + S12 (?0 ? ?3 ) ? + (?0 + ?3 )? + iR? = 0,
?(i) ? ??
? ? ?
(25)
2
? ?(i) ?(i) ?(i) ? ?(i) ?(i) ?
?
?[(?0 + ?3 )S03 + (S01 + S31 )?1 + (S02 + S32 )?2 ]? +
(26)
??
? ? ?
+ [(?0 + ?3 )? + (?0 ? ?3 )]? + iR? = 0.

Enumerations (1)–(26) in (4.4) correspond to those of the ans?tze in table 1; the
a
??
dot denotes differentiation with respect to the corresponding ? and R = F (??, ?M ?).
Below we obtain some solutions of reduces ODEs (4.4) in the case of a non-
?(2) ?(1) ?
standard representation of AP (1, 3) realized by matrices Sµ? = Sµ? + Qµ? (see (3.4),
?(1)
(3.10) and (3.11)). The cases with Sµ? are analogous to those considered in [3, 4, 5,
6].
First of all we note that the condition of compatability for equations (3), (12)–(14),
??
(19) and (22) in (4.4) results in R ? F (??, ?M ?) = 0 and therefore such cases are
rather trivial.
Consider equation (5) in (4.4), choosing
?
? ? ??,
R = ??1/2k , (4.5)

where ?, k = 0 are arbitrary real constants. From equation (5) we find as a corollary
(or condition of compatibility)

d2 ? 2?k 1+1/2k 1
k=? ,
= 4??1/2k (4.6)
? + c0 ,
2
d? 1 + 2k 2

c0 is an arbitrary real constant. A particular solution of (4.6) is
?2k
2?
c? (4.7)
?(?) = ? ,
1 + 2k
On the connection between solutions of Dirac and Maxwell equations 333

c is an arbitrary real constant. Let us go back to equation (5) in (4.4). Using (4.7)
we obtain a linear ODE and its general solution has the form

i(1 ? 2k) ? 2 2?? 1 + 2k ? 0 ? 3 ?
? = exp ? ? ln c ? exp ? (? + ? )S03 ?
2 1 + 2k 4?
[c ? 2??/(1 + 2k)]?2k
[c ? 2??/(1 + 2k)]2k+2
?
? ?2 ? ? (4.8)
2k + 2 2k
[c ? 2??/(1 + 2k)]2k+2 1
+i + ?,
2k[c ? 2??/(1 + 2k)]2k
2k + 2

where ? is an arbitrary 16-component constant column satisfying the conditions

2?k
?? ??? ?? ? ??
?(?0 + ?3 )Q03 ?2 ? = ?i?(?0 + ?3 ), (4.9)
?? = 0,
? ?Q03 ? = .
1 + 2k
Let us write down the general solution of equation (5) in (4.4) in the case of the
spinor representation (i = 1). It can be found without difficulty and has the form

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

where ? is an arbitrary 16-component column.
Consider equation (15) in (4.4). In this case, by analogy with (5) in (4.4) consi-
dered above, we find

d2 ? 2?k 1+1/2k 1
k=?
= ??1/2k ? + c0 ,
2
d? 1 + 2k 2

and then

1? ?
? ?
?(?) = exp[i?1 ?(?)] exp ?1 (Q01 + Q31 )(?0 + ?3 ) ?
2
(4.11)
? ?
?
? cosh ?(y)dy + i?1 sinh ?(y)dy ?,

where

?? = 0,
?
2?k
?? ? ? ? ? ?? ? ? ?
??1 (Q01 + Q31 )(?0 + ?3 )? = i?(Q01 + Q31 )(?0 + ?3 )? = ,
1 + 2k
??
?(?) = ?(1 + 2k) ln c ? .
1 + 2k

Analogously, in the case of equation (16) in (4.4) we have

d2 ? ? 2?k 1+1/2k
?1/2k
= ? +c
2 2
d? 1+? 1 + 2k
334 W.I. Fushchych, W.M. Shtelen, S.V. Spichak

and
i?(?) ? 2 ?
(? ? ??1 ) ?
?(?) = exp 2)
2(1 + ?
1 ? ? ? ?
? exp (Q01 + Q31 )(?0 + ?3 ) ? (4.12)
2)
2(1 + ?
? ?
?(y) ?(y)
cosh v sinh v
? ?
(?1 ? ??2 ) ?2
dy + i 1+ dy ?,
1 + ?2 1 + ?2
where
??
v

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