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Ad(?R(m))Jab = Jab ? ?R(m) + ?2 Z(ma mb ? ma mb ),
? tt tt
Ad(?R(m))R(n) = R(n) + ?Z(mtt · n ? m · ntt ), Ad(?Jab )R(m) = R(m), ?
Ad(?Jab )Jcd = Jcd cos ? + [Jab , Jcd ] sin ? (a, b) = (c, d) = (b, a) ,
where
ma = mb , mb = ?ma , mc = 0, a = b = c = a,
? ? ?
d ? {a; b}.
md = md cos ? + md sin ?, mc = mc , a = b = c = a,
? ? ?
Four adjoint actions are better found by means of integrating a system of form (A.2).
As a result we obtain that
(A.5)
Cases where adjoint actions coincide with the identical mapping are omitted.

Note A.2 If Z(?(t)) ? A(N S)[C ? ((t0 , t1 ), R)] with ?? < t0 or t1 < +?, the
operators Z(?(t)) and Z(?(t + ?)) (Z(?(t)) and Z(e2? ?(te2? ))) that belong to the
different algebras
A(N S)[C ? ((t0 , t1 ), R)] and A(N S)[C ? ((t0 ? ?, t1 ? ?), R)]
(A(N S)[C ? ((t0 , t1 ), R)] and A(N S)[C ? ((t0 e?2? , t1 e?2? ), R)])
respectively. An analogous statement is true for the operator R(m). Equivalence of
subalgebras in Theorems A.1 and A.2 is also meant in this sense.
Symmetry reduction and exact solutions of the Navier–Stokes equations 231

Note A.3 Besides the adjoint representations of operators (1.2) we make use of di-
screte transformation (1.6) for classifying the subalgebras of A(N S),

To prove the theorem of this section, the following obvious lemma is used.

Lemma A.1 Let N ? N.
A. If ? ? C N ((t0 , t1 ), R), then ? ? ? C N ((t0 , t1 ), R) : 2t?t + 2? = ?.
B. If ? ? C N ((t0 , t1 ), R), then ? ? ? C N ((t0 , t1 ), R) : 2t?t ? ? = ?.
C. If mi ? C N ((t0 , t1 ), R) and a ? R, then ? li ? C N ((t0 , t1 ), R) :
2tlt ? l1 + al2 = m1 , 2tlt ? l2 ? al1 = m2 .
1 2

A.2 One-dimensional subalgebras
Theorem A.1 A complete set of A(N S)-inequivalent one-dimensional subalgebras
of A(N S) is exhausted by the following algebras:
1. A1 (?) = D + 2?J12 , where ? ? 0.
1

2. A1 (?) = ?t + ?J12 , where ? ? {0; 1}.
2

3. A1 (?, ?) = J12 + R(0, 0, ?(t)) + Z(?(t)) with smooth functions ? and ?.
3
Algebras A1 (?, ?) and A1 (?, ?) are equivalent if ? ?, ? ? R, ? ? ? C ? ((t0 , t1 ), R):
3? ?
3

? (t) = e?? ?(t), ?(t) = e2? (?(t) + ?tt (t)?(t) ? ?(t)?tt (t)),
?? ?? (A.6)
where t = te?2? + ?.
?
4. A1 (m, ?) = R(m(t)) + Z(?(t)) with smooth functions m and ?: (m, ?) ?
4
(0, 0). Algebras A1 (m, ?) and A1 (m, ?) are equivalent if ? ?, ? ? R, ? C = 0, ? B ?
4??
4
O(3), ? l ? C ? ((t0 , t1 ), R3 ):

m(t) = Ce?? B m(t), ?(t) = Ce2? ?(t) + ltt (t) · m(t) ? mtt (t) · l(t) ,
?? ?? (A.7)

where t = te?2? + ?.
?

Proof. Consider an arbitrary one-dimensional subalgebra generated by
V = a1 D + a2 ?t + a3 J12 + a4 J23 + a5 J31 + R(m) + Z(?).
The coefficients a4 and a5 are omitted below since they always can be made to vanish
If a1 = 0 we get a1 = 1 by means of a change of basis. Next, step-by-step we
?
make a2 , m, and ? vanish by means of the adjoint representations Ad(? 1 a2 a?1 ?t ),
1
2

l ? C ? ((t0 + 1 a2 a?1 , t1 + 1 a2 a?1 ), R3 ),
1 1
2 2
? ? C ? ((t0 + 1 a2 a?1 , t1 + 1 a2 a?1 ), R),
1 1
2 2

and l, ? are solutions of the equations
1
2tlt ? l + a3 a?1 (l2 , ?l1 , 0)T = m, 2t?t + 2? = ? + (ltt · m ? l · mtt )
? ? ? ?
1
2
232 W.I. Fushchych, R.O. Popovych

with m(t) = a?1 m(t ? 1 a2 a?1 ) and ?(t) = a?1 ?(t ? 1 a2 a?1 ). Such l and ? exist in
? ?
1 1 1 1
2 2
virtue of Lemma A.1. As a result we obtain the algebra A1 (?), where 2? = a3 a?1 .
1 1
In case ? < 0 additionally one has to apply transformation (1.6) with b = 1.
If a1 = 0 and a2 = 0, we make a2 = 1 by means of a change of basis. Next, step-
?
by-step we make m and ? vanish by means of the adjoint representations Ad(R(l))
and Ad(Z(?)), where l ? C ? ((t0 , t1 ), R3 ), ? ? C ? ((t0 , t1 ), R), and
1
a2 lt + a3 (l2 , ?l1 , 0)T = m, a2 ?t = ? + (ltt · m ? l · mtt ).
2
1
If a3 = 0 we obtain the algebra A2 (0) at once. If a3 = 0, using the adjoint repre-
sentation Ad(?D) and transformation (1.6) (in case of need), we obtain the algebra
A1 (1).
2
If a1 = a2 = 0 and a3 = 0, after a change of basis and applying the adjoint
representation Ad(R(?a?1 m2 , a?1 m1 , 0)) we get the algebra A1 (?, ?), where ? =
?
3
3 3
?1 3 ?1 ?2
a3 m and ? = a3 ? + a3 (mtt m ? m mtt ). Equivalence relation (A.6) is generated
1 2 12
?
If a1 = a2 = a3 = 0, at once we get the algebra A1 (m, ?). Equivalence relation
4

A.3 Two-dimensional subalgebras
Theorem A.2 A complete set of A(N S)-inequivalent two-dimensional subalgebras
of A(N S) is exhausted by the following algebras:
1. A2 (?) = ?t , D + ?J12 , where ? ? 0.
1

2. A2 (?, ?) = D, J12 + R(0, 0, ?|t|1/2 ) + Z(?t?1 ) , where ? ? 0, ? ? 0.
2

3. A2 (?, ?) = ?t , J12 + R(0, 0, ?) + Z(?) , where ? ? {0; 1}, ? ? 0 if ? = 1 and
3
? ? {0; 1} if ? = 0.
4. A2 (?, ?, µ, ?, ?) = D + 2?J12 , R |t|?+1/2 (? cos ?, ? sin ?, µ) + Z(?|t|??1 ) ,
4
where ? = ? ln |t|, ? > 0, µ ? 0, ? ? 0, µ2 + ? 2 = 1, ?? = 0, and ? ? 0.
5. A2 (?, ?) = D, R(0, 0, |t|?+1/2 ) + Z(?|t|??1 ) , where ?? = 0 and ? ? 0.
5

6. A2 (?, µ, ?, ?) = ?t + J12 , R(?e?t cos t, ?e?t sin t, µe?t ) + Z(?e?t ) , where µ ? 0,
6
? ? 0, µ2 + ? 2 = 1, ?? = 0, and ? ? 0.
7. A2 (?, ?) = ?t , R(0, 0, e?t ) + Z(?e?t ) , where ? ? {?1; 0; 1}, ?? = 0, and ? ? 0.
7

8. A2 (?, ? 1 , ?, ? 2 ) = J12 + R(0, 0, ?) + Z(? 1 ), R(0, 0, ?) + Z(? 2 ) with smooth
8
functions (of t) ?, ?, and ? i : (?, ? 2 ) ? (0, 0) and ?tt ? ? ??tt ? 0. Algebras
?? ??
A2 (?, ? 1 , ?, ? 2 ) and A2 (?, ? 1 , ?, ? 2 ) are equivalent if ? C1 = 0, ? ?, ?, C2 ? R,
8 8
? ? ? C ? ((t0 , t1 ), R):
?(t) = e? (?(t) + C2 ?(t)), ?(t) = C1 e?? ?(t),
?? ??
??
? 1 (t) = e2? ? 1 (t) + ?tt (t)?(t) ? ?(t)?tt (t) +
(A.8)
+ C2 (? 2 (t) + ?tt (t)?(t) ? ?(t)?tt (t)) ,
??
? 2 (t) = C1 e2? (? 2 (t) + ?tt (t)?(t) ? ?(t)?tt (t)),
where t = te?2? + ?.
?
Symmetry reduction and exact solutions of the Navier–Stokes equations 233

9. A2 (m1 , ?1 , m2 , ?2 ) = R(m1 (t)) + Z(?1 (t)), R(m2 (t)) + Z(?2 (t)) with smooth
9
functions mi and ?i :
m1 · m2 ? m1 · m2 = 0, rank (m1 , ?1 ), (m2 , ?2 ) = 2.
tt tt

Algebras A2 (m1 , ?1 , m2 , ?2 ) and A2 (m1 , ?1 , m2 , ?2 ) are equivalent if ? ?, ? ? R,
9? ???
9
? {aij }i,j=1,2 : det{aij } = 0, ? B ? O(3), ? l ? C ? ((t0 , t1 ), R3 ):

mi (t) = e?? aij B mj (t),
??
(A.9)
?i (t) = e2? aij ?j (t) + ltt (t) · mj (t) ? l(t) · mj (t) ,
?? tt

where t = te?2? + ?.
?
10. A2 (?, ?) = D + ?J12 , Z(|t|? ) , where ? ? 0, ? ? R.
10

11. A2 (?) = ?t + J12 , Z(e?t ) , where ? ? R.
11

12. A2 (?) = ?t , Z(e?t ) , where ? ? {?1; 0; 1}.
12

The proof of Theorem A.2 is analogous to that of Theorem A.1. Let us take an
arbitrary two-dimensional subalgebra generated by two linearly independent operators
of the form
V i = ai D + ai ?t + ai J12 + ai J23 + ai J31 + R(mi ) + Z(?i ),
1 2 3 4 5

where ai = const (n = 1, 5) and [V 1 , V 2 ] ? V 1 , V 2 . Considering the different
n
possible cases we try to simplify V i by means of adjoint representation as much as
possible. Here we do not present the proof of Theorem A.2 as it is too cumbersome.

A.4 Three-dimensional subalgebras
We also constructed a complete set of A(N S)-inequivalent three-dimensional subal-
gebras. It contains 52 classes of algebras. By means of 22 classes from this set one
can obtain ansatzes of codimension three for the Navier–Stokes field. Here we only
give 8 superclasses that arise from unification of some of these classes:
1. A3 = D, ?t , J12 .
1

2. A3 = D + ?J12 , ?t , R(0, 0, 1) , where ? ? 0. Here and below ?, ?, ?1 , ?2 , µ,
2
?, and aij are real constants.
3. A3 (?, ?, ?1 , ?2 ) = D, J12 + ? R(0, 0, |t|1/2 ln |t|) + Z(?2 |t|?1 ln |t|) + Z(?1 |t|?1 ),
3
R(0, 0, |t|?+1/2 ) + Z(?2 |t|??1 ) , where ?? = 0, ?1 ? 0, ? ? 0, and ??2 = 0.
4. A3 (?, ?, ?1 , ?2 ) = ?t , J12 +Z(?1 )+? R(0, 0, t)+Z(?2 t) , R(0, 0, e?t )+Z(?2 e?t ) ,
4
where ?? = 0, ??2 = 0, and, if ? = 0, the constants ?, ?1 , and ?2 satisfy one of the
following conditions:
? = 1, ?1 ? 0; ? = 0, ?1 = 1, ?2 ? 0; ? = ?1 = 0, ?2 ? {0; 1}.

5. A3 (?, m1 , m2 , ?1 , ?2 ) = D + 2?J12 , R(m1 ) + Z(?1 ), R(m2 ) + Z(?2 ) , where
5
? ? 0, rank(m1 , m2 ) = 2,
tmi ? 1 mi + ?(mi2 , ?mi1 , 0)T = aij mj ,
t 2
t?t + ?i = aij ?j , aij = const,
i
234 W.I. Fushchych, R.O. Popovych

(a11 + a22 ) a21 m1 · m1 + (a22 ? a11 )m1 · m2 ? a12 m2 · m2 +
(A.10)
+ 2?(m12 m21 ? m11 m22 ) = 0.
This superclass contains eight inequivalent classes of subalgebras that can be obtained
from it by means of a change of basis and the adjoint actions
if ? > 0 (? = 0) respectively. Here the functions n and ? satisfy the following
equations:
tnt ? 1 n + ?(n2 , ?n1 , 0)T = bi mi ,
2
t?t + ? = bi ?i + 1 t(nttt · n ? ntt · nt ) + ntt · n + ?(n1 n2 ? n1 n2 ).
tt tt
2

6. A3 (?, m1 , m2 , ?1 , ?2 ) = ?t + ?J12 , R(m1 ) + Z(?1 ), R(m2 ) + Z(?2 ) , where
6
? ? {0; 1}, rank(m1 , m2 ) = 2,
mi ? ?(mi2 , ?mi1 , 0)T = aij mj , t?i = aij ?j ,
t t

and aij are constants satisfying (A.10). This superclass contains eight inequivalent
classes of subalgebras that can be obtained from it by means of a change of basis and