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(4)
?aµ = ?aµ (x), ?aµ = ?
c=1

The ansatz for Aµ can be searched for in the form
3
Aa (x) Qab (x)B? (?(x)),
b
(5)
=
µ µ?
c=1

where B? (?) are arbitrary smooth and the functions ?(x), Qab (x) satisfy the system
b
µ?
of PDE
?aµ (x)?xµ = 0,
3
(6)
(?aµ ? bc ?µ ? Raµ? )Qcd = 0.
bc
??
c=1

Here, ? bc is the Kronecker symbol, a, b, d = 1, 3, ? = 0, 3.
On the set of solutions of equations (1), the following representation of the Poincar?
e
algebra is realized:
3
? ?
Jµ? = xµ P? ? x? Pµ + ? Aa
µ
Aa µ, ? = 0, 3. (7)
Pµ = ? , ,
µ ?
a? ?Aaµ
?A
a=1

b
Consequently, relations (4) hold true. Moreover, expression for ?aµ has the form

?aµ = Raµ? (x)Ab ,
b
(8)
a, b = 1, 3, µ = 0, 3.
?

That is why formulae (5), (6) can be rewritten in a simpler way. Namely, an ansatz
for the vector-potential of the Yang–Mills field A(x) invariant under a subalgebra of
the algebra AP (1, 3) with basis operators (7) should be searched for in the form

Aa (x) = Qµ? (x)B? (?(x)),
a
(9)
µ
Reduction of the self-dual Yang–Mills equations 475

a
where B? (?) are arbitrary smooth functions and functions ?(x), Qµ? (x) satisfy the
system of PDE
?aµ (x)?xµ = 0,
(10)
?a? (x)?? Qµ? ? Raµ? (x)Q?? = 0,
where a = 1, 3, µ, ? = 0, 3.
Thus, to get a complete description of P (1, 3)-inequivalent ansatzes invariant under
three-dimensional subalgebras of the Poincar? algebra, one has to integrate over-
e
determined system of PDE (10) for each subalgebra. Let us note that compatibility of
equations (10) is guaranteed by the fact that the operators X1 , X2 , X3 form a Lie
algebra.
Bellow, we adduce a complete list of C(1, 3)-inequivalent three-dimensional subal-
gebras of the Poincar? algebra AP (1, 3) following [4]:
e
L1 = P0 , P1 , P2 , L2 = P1 , P2 , P3 ,
L3 = P0 + P3 , P1 , P2 , L4 = J03 + ?J12 , P1 , P2 ,
L5 = J03 , P0 + P3 , P1 , L6 = J03 + P1 , P0 , P3 ,
L7 = J03 + P1 , P0 + P3 , P1 , L8 = J12 + ?J03 , P0 , P3 ,
L9 = J12 + P0 , P1 , P2 , L10 = J12 + P3 , P1 , P2 ,
L11 = J12 + P0 ? P3 , P1 , P2 , L12 = G1 , P0 + P3 , P2 + ?P1 ,
L14 = G1 + P0 ? P3 , P0 + P3 , P2 ,
L13 = G1 + P2 , P0 + P3 , P1 ,
L15 = G1 + P0 ? P3 , P1 + ?P2 , P0 + P3 , L16 = J12 , J03 , P0 + P3 ,
L17 = G1 + P2 , G2 ? P1 + ?P2 , P0 + P3 , L18 = G1 , J03 , P2 ,
L19 = J03 , G1 , P0 + P3 , L20 = J03 + P2 , G1 , P0 + P3 ,
L21 = G1 , J03 + P1 + ?P2 , P0 + P3 , L22 = G1 , G2 , J03 + ?J12 ,
L23 = G1 , P0 + P3 , P1 , L24 = J12 , P1 , P2 ,
L25 = J03 , P0 , P3 , L26 = J01 , J02 , J12 ,
L27 = J12 , J23 , J13 ,
Here, Gi = J0i ? Ji3 (i = 1, 2), ? ? R.
Let us consider, as an example, the procedure of construction of ansatz (9) invari-
ant under subalgebra L4 (? = 0). In this case, system (10) reads
(11a)
?x1 = ?x2 = 0, x0 ?x3 + x3 ?x0 = 0,
x0 Qx3 + x3 Qx0 ? SQ = 0, (11b)
Qx1 = Qx2 = 0,
3
where Q = Qµ? (x) = 0,
µ,?

0 0 0 1
0 0 0 0
S= .
0 0 0 0
1 0 0 0
The first integral of system (11a) has the form ? = x2 ? x2 . Next, from first two
0 3
equations of system (11b), it follows that Q = Q(x0 , x3 ). Since S is a constant matrix,
solutions of the third equation from (l1b) can be looked for in the form (see, e.g. [6])
Q = exp{f (x0 , x3 )S}.
476 R.Z. Zhdanov, V.I. Lahno, W.I. Fushchych

By substituting this expression into (11b), we get
(x0 fx3 , x3 fx0 ? 1) exp{f S} = 0,
where f = ln(x0 + x3 ).
Consequently, a particular solution of equations (lib) can be chosen in the following
way:
Q = exp{ln(x0 + x3 )S}.
By using evident identity S = S 3 , we obtain the equality
Q = I + S sh(ln(x0 + x3 )) + S 2 (ch(ln(x0 + x3 )) ? 1), (12)
where I is a unit (4 ? 4)-matrix.
By substituting the obtained expressions into formula (9), we get an ansatz for
Aµ (x) which is invariant under the algebra L4
Aa = B0 (x2 ? x2 ) ch(ln(x0 + x3 )) + B3 (x2 ? x2 ) sh(ln(x0 + x3 )),
a a
0 0 3 0 3
Aa = B1 (x2 ? x2 ), Aa = B2 (x2 ? x2 ),
a a
(13)
1 0 3 2 0 3
A3 = B3 (x0 ? x3 ) ch(ln(x0 ? x3 )) + B0 (x2 ? x2 ) sh(ln(x2 ? x2 )),
a a2 2 2 2 a
a = 1, 3.
0 3 0 3

The above ansatz has such an unpleasant feature as an asymmetric dependence
on independent variables xµ . To remove this asymmetry, one has to use a solution
generation procedure [7]. As a result, we arrive at the following representation of the
Poincar? invariant ansatz for the vector-potential of the Yang–Mills field:
e
Aµ (x) = Qµ? (x)B ? (?) = {(aµ a? ? dµ d? ) ch ?0 + (dµ a? ? d? aµ ) sh ?0 +
+ 2kµ [(?1 cos ?3 + ?2 sin ?3 )b? + (?2 cos ?3 ? ?1 sin ?3 )c? +
+ (?1 + ?2 )k? exp(??0 )] + (bµ c? ? b? cµ ) sin ?3 ? (cµ c? ? bµ b? ) cos ?3 ?
2 2

? 2(?1 bµ + ?2 cµ )k? exp(??0 )]}B ? (?).
Here, aµ , bµ , cµ , dµ are arbitrary constants satisfying the following equalities:
aµ aµ = ?bµ bµ = ?cµ cµ = ?dµ dµ = 1,
aµ bµ = aµ cµ = aµ dµ = bµ cµ = bµ dµ = cµ dµ = 0,
kµ = aµ + dµ , Qµ , ? are some functionals of x whose explicit form depends on the
choice of the algebra AP (1, 3), µ = 0, 3. Below, we adduce a complete list of functions
Qµ , µ = 0, 3, ? co corresponding to three-dimensional subalgebras of the Poincar? e
algebra (7).
L1 :
?µ = 0, ? = dx;
L2 :
?µ = 0, ? = ax;
L3 :
?µ = 0, ? = ax + dx;
?0 = ? ln |ax + dx|, ?1 = ?2 ?3 = a ln |ax + dx|,
L4 : = 0,
? = (ax)2 ? (dx)2 ;
L5 : ?0 = ? ln |ax + dx|, ?1 = ?2 = ?3 = 0, ? = cx;
L6 : ?0 = bx, ?1 = ?2 = ?3 = 0, ? = cx;
? = ?bx + ln |ax + dx|;
L7 : ?0 = bx, ?1 = ?2 = ?3 = 0,
Reduction of the self-dual Yang–Mills equations 477

?0 = ? arctg[bx(cx)?1 ], ?1 = ?2 = 0, ?3 = ? arctg[bx(cx)?1 ],
L8 :
? = (bx)2 + (cx)2 ;
?0 = ?1 = ?2 = 0, ?3 = ?ax, ? = dx;
L9 :
L10 : ?0 = ?1 = ?2 = 0, ?3 = dx, ? = ax;
1
?0 = ?1 = ?2 = 0, ?3 = ? (dx + ax), ? = ax + dx;
L11 :
2
1
?0 = ?2 = ?3 = 0, ?1 = (bx ? ?cx)(ax + dx)?1 , ? = ax + dx;
L12 :
2
1
L13 : ?0 = ?2 = ?3 = 0, ?1 = cx, ? = ax + dx;
2
1
?0 = ?2 = ?3 = 0, ?1 = ? (ax + dx), ? = 4bx ? (ax + dx)2 ;
L14 :
4
1
?0 = ?2 = ?3 = 0, ?1 = ? (ax + dx),
L15 :
4
? = 4(?bx ? cx) ? ?(ax + dx)2 ;
?0 = ? ln |ax + dx|, ?1 = ?2 = 0, ?3 = ? arctg[bx(cx)?1 ],
L16 :
? = (bx)2 + (cx)2 ;
1 cx + (? + ax + dx)bx
L17 : ?0 = 0, ?1 = ,
2 1 + (ax + dx)(? + ax + dx)
bx ? cx(ax + dx)
1
?2 = ? , ?3 = 0, ? = ax + dx;
2 1 + (ax + dx)(? + ax + dx)
1 bx
?0 = ? ln |ax + dx|, ?1 =
L18 : , ?2 = ?3 = 0,
2 ax + dx
? = (ax)2 ? (bx)2 ? (dx)2 ;
1 bx
?0 = ? ln |ax + dx|, ?1 =
L19 : , ?2 = ?3 = 0, ? = cx;
2 ax + dx
1 bx
?0 = ? ln |ax + dx|, ?1 =
L20 : , ?2 = ?3 = 0,
2 ax + dx
? = cx + ln |ax + dx|;
1 ?bx + ln |ax ? dx|
?0 = ? ln |ax + dx|, ?1 = ?
L21 : , ?2 = ?3 = 0,
2 ax + dx
? = cx + ? ln |ax + dx|;
1 bx 1 cx
?0 = ? ln |ax + dx|, ?1 =
L22 : , ?2 = ,
2 ax ? dx 2 ax ? dx
?3 = ? ln |ax + dx|, ? = (ax)2 ? (bx)2 ? (cx)2 ? (dx)2 .

Here, ax = aµ xµ , bx = bµ xµ , cx = cµ xµ , dx = dµ xµ , µ = 0, 3.
Note. Ansatzes invariant under subalgebras L23 , L24 , L25 , L26 , L27 yield so-called
partially-invariant solutions (the term was introduced by L.V. Ovsyannikov [8]) which
cannot be represented in the form (13) and are not considered here.
Substitution of ansatzes (13), (14) into system of PDE (1) demands very cumberso-
me computations. This is why we omit these and adduce only the final result-system
of ordinary differential equations for Bµ (?).
478 R.Z. Zhdanov, V.I. Lahno, W.I. Fushchych

General form of the reduced system is the following:
1
?µ?? T ?? , (14)
Tµ? = µ, ? = 0, 3,
2
where
Tµ? = Gµ (?)Bµ ? G? Bµ ? Hµ?? (?)B ? + eBµ ? B?
and functions Gµ (?), Hµ? (?) are calculated according to the following formulae:
Gµ (?) = Qµ? ?x? ,
Hµ?? (?) = Q? Q??x? Q?? ? Q? ? Q??x? Q?µ .
µ ?

In the above formulae, overdot means differentiation with respect to ?.
Thus, the form of the reduced equations for functions Bµ (?) depends on the
explicit forms of functions Gµ (?), Hµ?? (?). Below, we adduce a list of these functions
corresponding to ansatzes (13), (14).
Gµ = ?dµ , Hµ?? = 0;
L1 :
L2 : Gµ = aµ , Hµ?? = 0;
L3 : Gµ = kµ , Hµ?? = 0;
Gµ = ?[aµ ? dµ + kµ ?],
L4 :
Hµ?? = ??[(aµ d? ? dµ a? )k? + ?(k? (b? cµ ? c? bµ ) ? kµ (b? cµ ? c? b? ))];
Gµ = cµ , Hµ?? = ??(aµ d? ? dµ a? )k? ;
L5 :
Gµ = cµ , Hµ?? = (aµ d? ? a? dµ )b? + (a? d? ? a? d? )bµ ;
L6 :
Gµ = ?bµ + ?kµ , Hµ?? = ?(aµ d? ? a? dµ )b? + (a? d? ? a? d? )bµ ;
L7 :
v
L8 : Gµ = 2cµ ?,
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