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? ?? ? ?3 + (a2 ?2 + ?4 ? a2 )?3 = 0,
13
?? 2
1
?? = ?3 .

??1 ?1 ? ?1 + ?1 ?1 ? ? ?4 ?2 ?2 ? 3? ?1 ?1 + ?2 ? ?2 ?2 + ? ?1 h? = 0,
3. ? ?? ?
?1 2
?? ?? ? ??? + ? ?? ? ?2 ? ? + a1 = 0,
12 2 21
(6.10)
??1 ?3 ? ?3 + a2 ? ?2 ?2 ? ? ?1 ?3 + ?5 = 0,
? ?? ?
1 1
2? + ??? = ?3 .

??1 ?1 ? ?1 + ?1 ?1 ? ? ?4 ?2 ?2 ? 3? ?1 ?1 + ?2 ? ?2 ?2 + ? ?1 h? = 0,
4. ? ?? ?
?1 2
?? ?? ? ??? + ? ?? ? ?2 ? ? + a1 = 0,
12 2 21
(6.11)
??1 ?3 ? ?3 + a2 ? ?2 ?2 ?3 ? ? ?1 ?3 + (?4 ? a2 ? ?2 )?3 = 0,
? ?? ? 2
2?1 + ??1 = ?3 .
?


(?2 ? a?1 )?1 ? (1 + a2 )?1 ? ?1 ?1 ? ?2 ?2 ? ah? ? 2h = 0,
5. ? ??
(? ? a? )?? ? (1 + a )?2 ? 2(a?2 + ?1 ) + h? = 0,
2 1 2 2
?? ? ?
(6.12)
(? ? a? )?? ? (1 + a )??? + 2? ? ? 4?3 + 4a?3 + ?5 = 0,
2 1 3 2 3 13
?
?? ? a?? = 0.
2 1



(?2 ? a?1 )?1 ? (1 + a2 )?1 ? ?1 ?1 ? ?2 ?2 ? ah? ? 2h = 0,
6. ? ??
(? ? a? )?? ? (1 + a )?2 ? 2(a?2 + ?1 ) + h? = 0,
2 1 2 2
?? ? ?
(6.13)
(? ? a? )?? ? (1 + a )??? + a1 ? = 0,
2 1 3 2 3 1

?2 ? a?1 = 0.
? ?


(?2 ? a?1 )?1 ? (1 + a2 )?1 ? ?1 ?1 ? ?2 ?2 ? ah? ? 2h = 0,
7. ? ??
(? ? a? )?? ? (1 + a )?2 ? 2(a?2 + ?1 ) + h? = 0,
2 1 2 2
?? ? ?
(6.14)
(? ? a? )?? ? (1 + a )??? + a1 ? ? ? a2 ?3 + 2aa1 ?3 = 0,
2 1 3 2 3 13
?
1
?2 ? a?1 = 0.
? ?


Numeration of reduced systems (6.8)–(6.14) corresponds to that of the ansatzes
in Table 2. Let us integrate systems (6.8)–(6.14) in such cases when it is possible.
Below, in this section, Ck = const (k = 1, 6).
Symmetry reduction and exact solutions of the Navier–Stokes equations 217

1. We failed to integrate system (6.8) in the general case, but we managed to find
the following particular solutions:
?1 = ?6?(? + C3 , 1 (4 ? 2C1 ), C2 ) ? 2,
a) 3
? = ? = 0, h = 2?1 + C1 ;
2 3

?1 = ?6C1 e2C1 ? ?(eC1 ? + C3 , 0, C2 ) + 3C1 ? 2,
2 2
b)
?2 = 5C1 , ?3 = 0,
h = ?12C1 e2C1 ? ?(eC1 ? + C3 , 0, C2 ) ? 2 ? 13 C1 ? 9 C1 ;
2 2 4
2 4
?1 = C1 , ?2 = C2 , ?3 = 0, h = ? 1 (C1 + C2 ).2 2
c) 2

Here ?(?, ?1 , ?2 ) is the Weierstrass function that satisfies the equation (see [19]):

(?? )2 = 4?3 ? ?1 ? ? ?2 . (6.15)

2. If ?3 = 0, the last equation of (6.9) implies that ?1 = C1 . It follows from the
other equations of (6.9) that
?1
?2 = C3 + C2 eC1 ? ? (a1 C1 ? ?2 )?,
?1 ?1
h = C6 ? ?2 C3 ? ? ?2 C2 C1 eC1 ? + 1 ?2 (a1 C1 ? ?2 )? 2
2

if C1 = 0, and
?2 = C3 + C2 ? + 1 a1 ? 2 ,
2
h = C6 ? ?2 C3 ? ? 1 ?2 C2 ? 2 ? 1 ?2 a1 ? 3
2 6

if C1 = 0. The function ?3 satisfies the equation

?3 ? C1 ?3 + (a2 ? ?4 ? a2 ?2 )?3 = 0. (6.16)
?? ? 2

We solve equation (6.16) for the following cases:
A. C2 = a1 ? ?2 C1 = 0:
? 1C ? 1/2 1/2
? e 2 1 C4 eµ ? + C5 e?µ ? , µ > 0,
?
1
?3 = e 2 C1 ? C4 + C5 ? , µ = 0,
?1
?
e 2 C1 ? C4 cos((?µ)1/2 ?) + C5 sin((?µ)1/2 ?) , µ < 0,

where µ = 1 C1 ? a2 + ?4 + a2 C3 .
2
2
4
B. C1 = a1 = 0, C2 = 0 ([19]):
2
?3 = ? 1/2 Z1/3 1/2 3/2
3 (?a2 C2 ) ? ,
where ? = ? + (C3 a2 ? a2 ? ?4 )/(a2 C2 ). Here Z? (? ) is the general solution of the
2
Bessel equation (4.22).
C. C1 = 0, a1 = 0 ([19]):

?3 = (? + C2 a?1 )?1/2 W ?, 1 , ( 1 a1 a2 )?1/2 (? + C2 a?1 )2 ,
1 1
4 2

where ? = 1 ( 1 a1 a2 )?1/2 a2 ? ?4 ? a2 C3 + 1 a2 C3 a?1 . Here W (?, µ, ? ) is the general
2
2 1
42 2
solution of the Whittaker equation (4.21).
218 W.I. Fushchych, R.O. Popovych

D. C1 = 0, C2 = 0, a1 ? ?2 C1 = 0 ([19]):
1 1
?1
?3 = e 2 C1 ? Z? 2C1 (?a2 C2 )1/2 e 2 C1 ? ,
1/2
?1
where ? = C1 C1 + 4(?4 + a2 C3 ? a2 )
2
. Here Z? (? ) is the general solution of the
2
Bessel equation (4.22).
E. C1 = 0, a1 ? ?2 C1 = 0, C2 = 0 ([19]):
1 1/2 3/2
?1
a2 (a1 C1 ? ?2 )
2
?3 = e 2 C1 ? ? 1/2 Z1/3 ? ,
3

?1
where ? = ? + a2 ? 1 C1 ? C3 a2 ? ?4 / a2 (a1 C1 ? ?2 ) . Here Z? (? ) is the general
2
2 4
solution of the Bessel equation (4.22).
If ?3 = 0, then ?1 = ?3 ? (translating ?, the integration constant can be made to
vanish),
2 2 2
1 1 1
e? 2 ?3 ? d? d? + ?2 ?,
?2 = C1 + C2 e 2 ?3 ? d? + a1 e 2 ?3 ?
2 2
1 1
?1
h = C3 ? 1 (?2 + ?3 )? 2 ? ?2 C1 ? ? ?2 C2 ? e 2 ?3 ? d? ? ?3 e 2 ?3 ? ?
2 2
2
2 2 2 2
1 1 1 1
?1 ?1
e? 2 ?3 ? d? d? ? ?3 e 2 ?3 ? e? 2 ?3 ? d? + ?3 ? ,
? ?2 a1 ? e 2 ?3 ?

and the function ?3 satisfies the equation
?3 ? ?3 ??3 + (a2 ? ?4 ? a2 ?2 )?3 = 0. (6.17)
?? ? 2

We managed to find a solution of (6.17) only for the case a1 = C2 = 0, i.e.,
2
1
?2
1/2
?3 = e 4 ?3 ? V ?3 (? + 2a2 ?2 ?3 ), ? ,
?1 2 ?2
where ? = 4?3 ?4 + a2 C1 ? a2 (?2 ?3 + 1) . Here V (?, ?) is the general solution of
2
the Weber equation
4V? ? = (? 2 + ?)V. (6.18)
3. The general solution of system (6.10) has the form:
?1 = C1 ? ?2 + 1 ?3 , (6.19)
2
2
1
? C1 +1 e 4 ?3 ? d? ? 1 ?2 ? 2 +
?2 = C2 + C3 2
(6.20)
2 2
1 1
? ?C1 ?1 e? 4 ?3 ? d? d?,
? C1 +1 e 4 ?3 ?
+ a1

2
1
? C1 ?1 e 4 ?3 ? d? +
?3 = C4 + C5
2 2
1 1
? C1 ?1 e 4 ?3 ? ? 1?C1 e? 4 ?3 ? (?5 + a2 ? ?2 ?2 )d? d?,
+

h = C6 ? 1 ?3 ? 2 ? 1 C1 ? ?2 + (?2 (?))2 ? ?3 d? ? ?2 ? ?1 ?2 (?)d?.
2 2
(6.21)
8 2

4. System (6.11) implies that the functions ?i and h are determined by (6.19)–
(6.21), and the function ?3 satisfies the equation
?3 ? (C1 ?1)? ?1 + 1 ?3 ? ?3 + a2 ? ?2 (a2 ??2 ) ? ?4 ?3 = 0. (6.22)
?? ?
2
Symmetry reduction and exact solutions of the Navier–Stokes equations 219

We managed to solve equation (6.22) in following cases:
A. C3 = a1 = 0, ?3 = 0:
2
1 1
?3 = ? 2 C1 ?1 e 8 ?3 ? W (?, µ, 1 ?3 ? 2 ),
4
?1
where ? = 1 2 ? C1 ? (4?4 + 2?2 a2 )?3 , µ = 1 (C1 ? 4a2 + 4a2 C2 )1/2 . Here
2
2
4 4
W (?, µ, ? ) is the general solution of the Whittaker equation (4.21).
Let ?3 = 0, then
?
C1 = ?2,
? C2 + C3 ln ? + 1 (a1 + 2?2 )? 2 ,
? 4
C2 + 1 C3 ? 2 + 1 a1 ? 2 (ln ? ? 1 ),
?2 = C1 = 0,
? 2 2 2
? ?1
C2 + C3 (C1 + 2)?1 ? C1 +2 ? 1 C1 (a1 ? ?2 C1 )? 2 , C1 = 0, ?2.
2

B. C3 = a1 ? ?2 C1 = 0:
? 1C
? ? 2 1 Z? (µ1/2 ?), µ = 0,
?

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