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1
? 2 C1 (C5 ? ? + C6 ? ?? ),
3
(6.23)
?= µ = 0, ? = 0,
?1
?
? 2 C1 (C5 + C6 ln ?), µ = 0, ? = 0,

where µ = ??4 , ? = 1 (C1 ? 4a2 + 4a2 C2 )1/2 . Here and below Z? (? ) is the general
2
2
2
solution of the Bessel equation (4.22).
C. C3 = 0, C1 = 0: ?3 is determined by (6.23), where
?1
µ = 1 a2 C1 (a1 ? ?2 C1 ) ? ?4 , ? = 1 (C1 ? 4a2 + 4a2 C2 )1/2 .
2
2
2 2

D. C1 = a1 = 0: ?3 is determined by (6.23), where
µ = ? 1 a2 C3 ? ?4 , ? = (?a2 + a2 C2 )1/2 .
2
2

E. C3 = 0, C1 ? {0; ?2}, a2 (a1 ? ?2 C1 ) ? 2?4 C1 = 0:
1 1
?3 = ? 2 C1 Z? (µ? 1+ 2 C1 ),

where µ = 2C3 (C1 + 2)?3/2 , ? = (C1 + 2)?1 (C1 ? 4a2 + 4a2 C2 )1/2 .
1/2 2
2

F. C1 = ?2, C3 = 0, a2 (a1 + 2?2 ) + 4?4 = 0 ([19]):

?3 = ? ?1 ? 1/2 Z1/3 ( 2 C3 ? 3/2 ),
1/2
3
?1
where ? = ln ? + C3 (a2 ? a2 C2 ? 1).
2

G. C1 = 2, C3 < 0, 1 ? a2 + a2 C2 ? 0:
2

?3 = W (?, µ, 1 (?C3 )1/2 ? 2 ),
2

where ? = 1 (?C3 )?1/2 (?4?4 +a2 ?2?2 a2 ), µ = 1 (1?a2 +a2 C2 )1/2 . Here W (?, µ, ? )
2 2
8 2
is the general solution of the Whittaker equation (4.21).
5–7. Identical corollaries of system (6.12), (6.13), and (6.14) are the equations
?2 = a?1 + C1 , (6.24)

h = a(1 + a2 )?1 + (2 + 2a2 ? aC1 )?1 + C2 , (6.25)
?
220 W.I. Fushchych, R.O. Popovych

(1 + a2 )?1 + (4a ? C1 )?1 + ?1 ?1 + 4?1 + (1 + a2 )?1 (C1 + 2C2 ) = 0. (6.26)
2
?? ?


We found the following solutions of (6.26):
A. If (1 + a2 )?1 (C1 + 2C2 ) < 4:
2


1/2
?1 = 4 ? (1 + a2 )?1 (C1 + 2C2 ) ? 2.
2
(6.27)

B. If C1 = 4a:

4 (C 2 + 2C2 )
?
?1 = ?6? + C4 , ? 1 ? 2. (6.28)
, C3
3(1 + a2 )
(1 + a2 )1/2 3

Here and below ?(?, ?1 , ?2 ) is the Weierstrass function satisfying equation (6.15). If
C2 = 2 ? 6a2 and C3 = 0, a particular case of (6.28) is the function

?1 = ?6(1 + a2 )? 2 ? 2 (6.29)

(the constant C4 is considered to vanish).
= 4a, (1 + a2 )?1 (C1 + 2C2 ) ? 4 = ?9µ4 :
2
C. If 1

?1 = ?6µ2 e?2? ?(e?? + C4 , 0, C3 ) + 3µ2 ? 2, (6.30)

where ? = (1 + a2 )?1/2 µ?, µ = 1 (4a ? C1 )(1 + a2 )?1/2 . If C3 = 0, a paticular case of
5
(6.30) is the function

?1 = ?6µ2 e?2? (e?? + C4 )?2 + 3µ2 ? 2, (6.31)

where the constant C4 is considered not to vanish.
The function ?3 has to be found for systems (6.12), (6.13), and (6.14) individually.
5. The function ?3 satisfy the equation

(1 + a2 )?3 ? (C1 + 4a)?3 ? (2?1 ? 4)?3 ? ?5 = 0.
?? ?

If ?1 is determined by (6.27), we obtain

?3 = exp 1 (1 + a2 )?1 (C1 + 4a)? ?
?2 ?
? C5 exp(? 1/2 ?) + C6 exp(?? 1/2 ?), ?>0 ?
? ?
? 1/2 1/2 +
C cos((??) ?) + C6 sin((??) ?), ? < 0
?5 ?
? ?
C + C6 ?, ?=0
?5 ?
? ??5 (2?1 ? 4)?1 , 2?1 ? 4 = 0 ?
? ?
?1
??5 (4a + C1 ) ?, 2? ? 4 = 0, C1 + 4a = 0
1
+ ,
?1 ?
? ?
2 ?1 2
? , 2?1 ? 4 = 0, C1 + 4a = 0
2 ?5 (1 + a )

where ? = 1 (1 + a2 )?2 (C1 + 4a)2 ? (1 + a2 )?1 (4 ? 2?1 ).
4
6. In this case ?3 satisfy the equation

(1 + a2 )?3 ? C1 ?3 = a1 ?1 .
?? ?
Symmetry reduction and exact solutions of the Navier–Stokes equations 221

Therefore,
?1
?3 = C5 + C6 exp (1 + a2 )?1 C1 ? + a1 C1 ?1 (?)d? +

+ exp (1 + a2 )?1 C1 ? exp ?(1 + a2 )?1 C1 ? ?1 (?)d?

for C1 = 0, and
?3 = C5 + C6 ? + a1 (1 + a2 )?1 ? ?1 (?)d? ? ??1 (?)d?
for C1 = 0.
7. The function ?3 satisfy the equation
(1 + a2 )?3 ? (C1 + 2a1 a)?3 + (a2 ? a1 ?1 )?3 = 0. (6.32)
?? ? 1

A. If ?1 is determined by (6.27), it follows that
?3 = exp 1 (1 + a2 )?1 (C1 + 2a1 a)? ?
?2 ?
? C5 exp(? 1/2 ?) + C6 exp(?? 1/2 ?), ?>0 ?
? C cos((??)1/2 ?) + C6 sin((??)1/2 ?), ? < 0 ,
?5 ?
C5 + C6 ?, ?=0

where ? = 1 (1 + a2 )?2 (C1 + 2a1 a)2 ? (1 + a2 )?1 (a2 ? a1 ?1 ).
1
4
B. If C1 = 4a, that is, ?1 is determined by (6.27), we obtain
?3 = exp a(a1 + 2)(1 + a2 )?1 ? ?(? ),

where ? = (1 + a2 )?1/2 ? + C4 . Here the function ? = ?(? ) is the general solution of
of the following Lame equation ([19]):
?? ? + 6a1 ?(? ) + a2 + 2a1 ? a2 (2 + a1 )2 (1 + a2 )?1 ? = 0
1

with the Weierstrass function
?(? ) = ? ?, 1 4 ? (1 + a2 )?1 (C1 + 2C2 ) , C3 .
2
3

Consider the particular case when C2 = 2 ? 6a2 and C3 = 0 additionally, i.e., ?1
can be given in form (6.29). Depending on the values of a and a1 , we obtain the
following expression for ?3 :
Case 1. a1 = ?2, a1 = 2a2 :
1/2
(2 + a1 )(a1 ? 2a2 )
a(2 + a1 )
?3 = |?|1/2 exp ? Z? ?,
1 + a2 1 + a2

where ? = ( 1 ? 6a1 )1/2 .
4
Case 2. a1 = ?2: ?3 = C5 ? 4 + C6 ? ?3 .
Case 3. a1 = 2a2 :
v
Case 3.1. 48a2 < 1: ?3 = |?|1/2 e2a? C5 ? ? + C6 ? ?? , where ? = 1 1 ? 48a2 .
2
v
Case 3.2. 48a2 = 1, that is, a = ± 12 3: ?3 = |?|1/2 (C5 + C6 ln ?).
1
222 W.I. Fushchych, R.O. Popovych

Case 3.3. 48a2 > 1: ?3 = |?|1/2 e2a? C5 cos(? ln ?) + C6 sin(? ln ?) , where
v
? = 1 48a2 ? 1.
2

C. Let the conditions

(1 + a2 )?1 (C1 + 2C2 ) ? 4 = ?9µ4
2
C1 = 4a,

be satisfied, that is, let ?1 be determined by (6.30). Transforming the variables in
equation (6.32) by the formulas:

?3 = ? ?1/2 exp + 2aa1 )(1 + a2 )?1 ? ?(? ),
1
2 (C1

? = exp ?µ(1 + a2 )?1/2 ? ,

we obtain the following equation in the function ? = ?(? ):

? 2 ?? ? + 6a1 ? 2 ?(? + C4 , 0, C3 ) + ? ? = 0, (6.33)

where ? = µ?2 a2 + 2a1 ? 1 (1 + a2 )?1 (C1 + 2aa1 )2 ? 3a1 + 1 . If ? = 0, equation
2
1 4 4
(6.33) is a Lame equation.
In the particular case when ?1 is determined by (6.31), equation (6.33) has the
form:

? 2 (? + C4 )2 ?? ? + 6a1 ? 2 + ?(? + C4 )2 ? = 0. (6.34)

By means of the following transformation of variables:
?1
? = |?|?1 |? ? 1|?2 ?(?), ? = ?C4 ?,

where ?1 (?1 ? 1) + ? = 0 and ?2 (?2 ? 1) + 6a1 = 0, equation (6.34) is reduced to a
hypergeometric equation of the form (see [19]):

?(? ? 1)??? + 2(?1 + ?2 )? ? 2?1 )?? + 2?1 ?2 ? = 0.

If ? = 0, equation (6.34) implies that

(? + C4 )2 ?? ? + 6a1 ? = 0.

Therefore,

? = C5 |? + C4 |1/2?? + C6 |? + C4 |1/2+?

where ? = ( 1 ? 6a1 )1/2 ,
1
if a1 < 24 , 4

? = |? + C4 |1/2 C5 + C6 ln |? + C4 |
1
if a1 = and
24 ,

? = |? + C4 |1/2 C5 cos(? ln |? + C4 |) + C6 sin(? ln |? + C4 |)

where ? = (6a1 ? 1 )1/2 .
1
if a1 > 24 , 4
Symmetry reduction and exact solutions of the Navier–Stokes equations 223

7 Exact solutions of system (2.9)
Among the reduced systems from Section 2, only particular cases of system (2.9) have
Lie symmetry operators that are not induced by elements from A(N S). Therefore,
Lie reductions of the other systems from Section 2 give only solutions that can be

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