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. 49
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?2 1
? , where ?2 ? R\{0}. Then the form of ? 2 is given by formula
ii. M =
?2
0
(4.27), and
? ?1 ?1 ? ?1 ?1 ?1
? 1 = ?11 ? (?12 ? C0 ?22 ?2 )?2 ? C0 (?21 ? ?22 ?2 )?2 ?2 +
? ? ?
? ?1 ?1
+ (?21 ? ?22 ?2 )?2 ? + C3 ?21 (?) + C4 ?22 (?) ? Ci ? i (?),
?
where
?1
? j (?) = D2 ? 2?? (?) ? C0 ?2j (?)
2j
if D2 = 0,
1 1
? 1 (?) = 1 ? 2 e ? 2 (?) = 1 ? 3 e
2 C0 ? 2 C0 ? if D2 = 0.
,
2 6

?1 ??2
? , where ?i ? R, ?2 = 0. Then
iii. M =
?2 ?1
? 1 = (?i ?i )?1 (?21 ?1 + ?22 ?2 )? + (?i ?i )?1 (?11 ?1 + ?12 ?2 ) ?
? ? ? ?
? C0 (?i ?i )?2 ?21 (?2 ? ?1 ) ? ?22 2?1 ?2 + Cn ?1n (?),
2 2
? ?
? 2 = (?i ?i )?1 (??21 ?2 + ?22 ?1 )? + (?i ?i )?1 (??11 ?2 + ?12 ?1 ) ?
? ? ? ?
? C0 (?i ?i )?2 ?21 2?1 ?2 + ?22 (?2 ? ?1 ) + Cn ?2n (?),
2 2
? ?
where n = 1, 4,
(C0 ? 4?1 )2 + (4?2 )2 ,
2
?=
1 |?2 |
2(? + C0 ? 4?1 ), 2(? ? C0 + 4?1 ),
1 2 2
?1 = ?2 = 4 ?2
4
?11 (?) = ?22 (?) = exp ( 1 C0 ? ?1 )? cos ?2 ?,
2
?? (?) = ? (?) = exp ( 1 C0 ? ?1 )? sin ?2 ?,
21 12
2
1
13 24
? (?) = ? (?) = exp ( 2 C0 + ?1 )? cos ?2 ?,
?23 (?) = ??14 (?) = exp ( 1 C0 + ?1 )? sin ?2 ?.
2

If ? = 0, the last equation of system (4.12) implies that ? 3 = ?? (translating ?,
the integration constant can be made to vanish). The other equations of system (4.12)
can be written in the form
h = ??1 a2 ?2 (?)d? ? 1 ? 2 ? 2 ,
2
(4.29)
??? ? ???? + µij ? = ?1i + ?2i ?,
i i j


where ?11 = c11 , ?21 = c21 , ?12 = c12 , ?22 = c22 + ?2 a2 ?. The form of the general
solution of system (4.29) depends on the Jordan form of the matrix M = {µij }. Now,
let us transform the dependent variables
?i = eij ? j ,
where the constants eij are determined by means of the system of linear algebraic
equations
eij µjk = µij ejk
? (i, j, k = 1, 2)
?
with the condition det{eij } = 0. Here M = {?ij } is the real Jordan form of the
µ
i
matrix M. The new unknown functions ? have to satisfy the following system
??? ? ???? + µij ? j = ?1i + ?2i ?,
i i
(4.30)
? ? ?
Symmetry reduction and exact solutions of the Navier–Stokes equations 199

?
where ?1i = eij ?1j , ?2i = eij ?2j . Depending on the form of M, we consider the
? ?
following cases:
?
A. det M = 0 (this is equivalent to the condition det M = 0).
0?
? , where ? ? {0; 1}. Then
i. M =
00
2 2 2
1 1 1
e 2 ?? d? ? ? ?1 ?22 ? + ?12 e? 2 ?? d? d?,
? 2 = C1 + C2 e 2 ?? (4.31)
? ?
2 2 2
1 1 1
e 2 ?? d? ? ? ?1 ?21 ? + e? 2 ?? (?11 ? ?? 2 )d? d?.
? 1 = C3 + C4 e 2 ??
? ?

? 0
? . Then the form of ? 2 is given by formula (4.31), and
ii. M =
0 0
2 2
1 1
e 2 ?? d? ? ? ?1 e 2 ?? + ? ?1 ?11 +
? 1 = C3 ? + C4 ? ?
2 2
1 1
+ ? ?1 ?21 ?? e 2 ?? ?1 (?)d? ? e 2 ?? ?1 (?) ,
?
2
1
e? 2 ?? d?.
where ?1 (?) =
?1 0
? , where ?1 ? R\{0; ?}. Then ? 2 is determited by (4.31), and
iii. M =
00
1
the form of ? is given by (4.33) (see below).
?
B. det M = 0, det{?ij ? ??ij } = 0 (this is equivalent to the conditions det M = 0,
µ
det{µij ? ??ij } = 0; here ?ij is the Kronecker symbol).
? ?
? , where ? ? {0; 1}. Then
i. M =
0 ?
2 2
1 1
? 2 = C1 ? + C2 ? e 2 ?? d? ? ? ?1 e 2 ?? + ? ?1 ?12 + vspace1mm
?
(4.32)
2 2
1 1
+ ? ?1 ?22 ?? e 2 ?? ?1 (?)d? ? e 2 ?? ?1 (?) ,
?
2 2
1 1
e 2 ?? d? ? ? ?1 e 2 ?? + ? ?1 ?11 +
? 1 = C3 ? + C4 ? ?
2 2
1 1
e 2 ?? ?2 (?)d? ? e 2 ?? ?2 (?) + ? ?1 (?21 ? ? ?? 2 ),
+ ?? ?
2 2
1 1
e? 2 ?? d?, ?2 (?) = ? ?1 e? 2 ?? (?21 ? ??? )d?.
where ?1 (?) = 2
?
?1 0
? , where ?1 ? R\{0; ?}. In this case ? 2 is determined by
ii. M =
0?
(4.32), and the form of ? 1 is given by (4.33) (see below).
?
C. det M = 0, det{?ij ? ??ij } = 0 (this is equivalent to the condition det M = 0,
µ
det{µij ? ??ij } = 0: here ?ij is the Kronecker symbol).
?1 0
? , where ?i ? R\{0; ?}. Then
i. M =
?2
0
2
1
?1
? 1 = ?1 ?11 + (?1 ? ?)?1 ?21 ? + |?|?1/2 e 4 ?? ?
? ?
(4.33)
?1 ?1
? C3 M 2 ?1 ? 2 ?1 ? + 1 , ? 1 , 1 ?? 2 ,
1
+ 1 , 1 , 1 ?? 2 + C4 M 1
442 4 42
200 W.I. Fushchych, R.O. Popovych
2
1
?1
? 2 = ?2 ?12 + (?2 ? ?)?1 ?22 ? + |?|?1/2 e 4 ?? ?
? ?
(4.34)
?1 ?1
? C1 M 2 ?2 ? 2 ?2 ? + 1 , ? 1 , 1 ?? 2 ,
1
+ 1 , 1 , 1 ?? 2 + C2 M 1
442 4 42

where M (?, µ, ? ) is the Whittaker function:
1 1
M (?, µ, ? ) = ? 2 +µ e? 2 ? 1 F1 + µ ? ?, 2µ + 1, ? ,
1 (4.35)
2

and 1 F1 (a, b, ? ) is the degenerate hypergeometric function defined by means of the
series:
?
a(a + 1) . . . (a + n ? 1) ? n
1 F1 (a, b, ? ) = 1 + ,
b(b + 1) . . . (b + n ? 1) n!
n=1

b = 0, ?1, ?2, . . ..
?1 ??2
? , where ?i ? R, ?2 = 0. Then
ii. M =
?2 ?1

? 1 = (?j ?j )?1 (?1 ?11 + ?2 ?12 ) + ((?1 ? ?)2 + ?2 )?1 ((?1 ? ?)?21 + ?2 ?22 )? +
2
? ? ? ?
+ C1 Re ? 1 (?) ? C2 Im ? 1 (?) + C3 Re ? 2 (?) ? C4 Im ? 2 (?),
? 2 = (?j ?j )?1 (??2 ?11 + ?1 ?12 ) +
? ?
+ ((?1 ? ?)2 + ?2 )?1 (??2 ?21 + (?1 ? ?)?22 )? +
2
? ?
+ C1 Im ? 1 (?) + C2 Re ? 1 (?) + C3 Im ? 2 (?) + C4 Re ? 2 (?),
where
+ ?2 i)? ?1 + 1 , 1 , 1 ?? 2 ,
1
? 1 (?) = M 2 (?1 442
?1
+ ?2 i)? + 1 , ? 1 , 1 ?? 2 , i2 = ?1.
1
? 2 (?) = M 2 (?1 4 42


?2 1
? , where ?2 ? R\{0; ?}. Here the form of ? 2 is given by
iii. M =
?2
0
(4.34), and
? ?1 ?1
? 1 = (?11 ? ?12 ?2 )?2 + ?21 ? ?22 (?2 ? ?)?1 (?2 ? ?)?1 ? +
? ? ?
2
1
+ |?|?1/2 e 4 ?? C3 ?1 (? ) + C4 ?2 (? ) ? ? ?1 ?1 (? ) ? ?1 ?2 (? )Ci ?i (? )d? +

+ ? ?1 ?2 (? ) ? ?1 ?1 (? )Ci ?i (? )d? ,

where ? = 1 ?? 2 ,
2
?1 ?1
2 ?2 ? 2 ?2 ? + 1, ?1, ? .
1
+ 1, 1, ? , 1
?1 (? ) = M ?2 (? ) = M
44 4 4

Note 4.3 The general solution of the equation
??? ? ???? ? (n + 1)?? = 0,
where n is an integer and n ? 0, is determined by the formula
?2
dn 1 ??2 dn 1 ??2
2
1
2 ??
?= e2 C1 + C2 e e2 d? .

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. 49
( 122 .)



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