<< Ïðåäûäóùàÿ ñòð. 51(èç 122 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>
(l · l)?t
1
(5.12)
t
t

The vector-functions m1 and l are chosen in such a way that one can find a fundamen-
tal set of solutions for equation (5.12). For example, let m ? mt = 0 ? t ? (t0 , t1 ). Let
us introduce the notation m := m1 and put l = ?(t)m? mt , where ? ? C ? ((t0 , t1 ), R),
?(t) = 0 ?t ? (t0 , t1 ). Then

|m|?2 dt m + ? m ? mt ,
m · l = 0, mt · l ? m · lt = 0, m2 = ?
k = ? m ? (m ? mt ), ? = (?)2 |m|2 |m ? mt |?2 ,
|m|?2 dt n2 + (?)2 |m ? mt |?2 m,
n2 = ?|m|2 m ? mt , n1 =
?11 (t) = (?)?2 |m ? mt |?2 dt, ?21 (t) = 1 ? ?11 |m|?2 dt,
?12 (t) = 1, ?22 (t) = ? |m|?2 dt,
?10 (t) = 2 ((m ? mt ) · mtt )|m ? mt |?2 + ? ?1 |m|?4 dt ? ?2 |m ? mt |?2 dt,
?20 (t) = ??10 (t) |m|?2 dt + 2 ? ?1 |m|?4 dt.

Consider the following cases: m ? mt ? 0, i.e., m = ?(t)a, where ?(t) ?
?
C ((t0 , t1 ), R), ?(t) = 0 ?t ? (t0 , t1 ), a = const, and |a| = 1. Let us put

l(t) = ? 1 (t)b + ? 2 (t)c,

where ? 1 , ? 2 ? C ? ((t0 , t1 ), R), (? 1 (t), ? 2 (t)) = (0, 0) ? t ? (t0 , t1 ), b = const, |b| = 1,
a · b = 0, and c = a ? b. Then
??2 dt a + ? 1 b + ? 2 c,
m2 = ? ? k = ?? 1 c ? ?? 2 b,
??2 dt n2 + ?? i ? i a,
? = (?)2 ? i ? i , n2 = (?)2 (? 1 b + ? 2 c), n1 =
?11 = (? i ? i )?1 dt, ?21 = 1 ? ?11 ??2 dt, ?12 = 1, ?22 = ? ??2 dt,
?10 = 2 (?t ? 1 ? ? 2 ?t )??1 (? i ? i )?1 dt, ?20 = ??10 ??2 dt.
2 1
Symmetry reduction and exact solutions of the Navier–Stokes equations 205

Note 5.4 In formulas (5.1) and (5.5) solutions of the NSEs (1.1) are expressed in
terms of solutions of the decomposed system of two linear one-dimensional heat
equations (LOHEs) that have the form:
i i
(5.13)
g? = g?? .

The Lie symmetry of the LOHE are known. Large sets of its exact solutions were
constructed [27, 3]. The Q-conditional symmetries of LOHE were investigated in [14].
Moreover, being decomposed system (5.13) admits transformations of the form

g 1 (? , ? ) = F 1 (?, ?, g 1 (?, ?)), ? = G1 (?, ?), ? = H 1 (?, ?),
?
g 2 (? , ? ) = F 2 (?, ?, g 2 (?, ?)), ? = G2 (?, ?), ? = H 2 (?, ?),
?

where (G1 , H 1 ) = (G2 , H 2 ), i.e. the independent variables can be transformed in
the functions g 1 and g 2 in different ways. A similar statement is true for system
(5.19)–(5.20) (see below) if ? = 0.

Note 5.5 It can be proved that an arbitrary Navier–Stokes field (u, p), where

u = w(t, ?) + (k i (t) · x)li (t)

with k i , li ? C ? ((t0 , t1 ), R3 ), k 1 ? k 2 = 0, and ? = (k 1 ? k 2 ) · x, is equivalent to
either a solution from family (5.1) or a solution from family (5.5). The equivalence
transformation is generated by R(m) and Z(?).

5.2 Investigation of system (3.13)–(3.16)
Consider system 8 from Subsection 3.2, i.e., equations (3.13)–(3.16). Equation (3.16)
immediately gives
?2
w1 = ? 1 ?t ??1 + (? ? 1)z2 , (5.14)
2

where ? = ?(t) is an arbitrary smooth function of z1 = t. Substituting (5.14) into
remaining equations (5.13)–(5.15), we get
?1 ?3 ?3
(?t ??1 )t ? 1 (?t ??1 )2 z2 ? ?t z2 ? (? ? 1)2 z2 + (w2 ? ?)2 z2 , (5.15)
1
q2 = 2 2

?1
w1 ? w22 + ?z2 ? 1 ?t ??1 z2 w2 = 0,
2 2 2
(5.16)
2

?1 ?2
w1 ? w22 + ?z2 ? 1 ?t ??1 z2 w2 + ?(w2 ? ?)z2 = 0.
3 3 3
(5.17)
2

Recall that ? = ?(t) and ? = ?(t) are arbitrary smooth functions of t; ? ? {0; 1}.
After the change of the independent variables

|?(t)|dt, z = |?(t)|1/2 z2 (5.18)
?=

in equations (5.16) and (5.17), we obtain a linear system of a simpler form:

w? ? wzz + ? (? )z ?1 wz = 0,
2 2 2
(5.19)
?

w? ? wzz + (?(? ) ? 2)z ?1 wz + ?(w2 ? ?(? ))z ?2 = 0,
3 3 3
(5.20)
? ?
206 W.I. Fushchych, R.O. Popovych

where ? (? ) = ?(t) and ?(? ) = ?(t). Equation (5.15) implies
? ?
(?t ??1 )t ? 1 (?t ??1 )2 z2 ? ?t ln |z2 | ?
1 2
q= 4 2
(5.21)
?2 ?3
? 1 (? ? 1)2 z2 + (w2 (?, z) ? ?(? ))2 z2 dz2 .
?
2

Formulas (5.14), (5.18)–(5.21), and ansatz (3.8) determine a solution of the NSEs
(1.1).
If ? = 0 system (5.19)–(5.20) is decomposed and consists of two translational
linear equations of the general form
f? + ? (? )z ?1 fz ? fzz = 0, (5.22)
?
where ? = ? (? = ? ? 2) for equation (5.19) ((5.20)). Tilde over ? is omitted below.
? ?? ?
Let us investigate symmetry properties of equation (5.22) and construct some of its
exact solutions.

Theorem 5.1 The MIA of (5.22) is given by the following algebras
if ?(? ) = const;
a) L1 = f ?f , g(?, z)?f
? if ?(? ) = const, ? ? {0; ?2};
b) L2 = ?? , D, ?, f ?f , g(?, z)?f
L3 = ?? , D, ?, ?z + 1 ?z ?1 f ?f , G = 2? ?? ? (z ? ?z ?1 ? )f ?f , f ?f ,
?
c) 2
? ? {0; ?2}.
if
g(?, z)?f
?
Here D = 2? ?? + z?z , ? = 4? 2 ?? + 4? z?z ? (z 2 + 2(1 ? ?)? )f ?f ; g = g(?, z) is an
arbitrary solution of (5.22).

When ? = 0, equation (5.22) is the heat equation, and, when ? = ?2, it is reduced
?
to the heat equation by means of the change f = zf .
For the case ? = const equation (5.22) can be reduced by inequivalent one-
dimensional subalgebras of L2 . We construct the following solutions:
For the subalgebra ?? + af ?f , where a ? {?1; 0; 1}, it follows that
= e?? z ? (C1 J? (z) + C2 Y? (z)) if a = ?1,
f
= e? z ? (C1 I? (z) + C2 K? (z)) if a = 1,
f
= C1 z ?+1 + C2 if a = 0 and ? = ?1,
f
= C1 ln z + C2 if a = 0 and ? = ?1.
f
Here J? and Y? are the Bessel functions of a real variable, whereas I? and K? are
the Bessel functions of an imaginary variable, and ? = 1 (? + 1).
2
?
For the subalgebra D + 2af ?f , where a ? R, it follows that
1 1
f = |? |a e? 2 ? |?| 2 (??1) W ? 1) ? a, 1 (? + 1), ?
1
4 (? 4

with ? = 1 z 2 ? ?1 . Here W (?, µ, ?) is the general solution of the Whittaker equation
4

4? 2 W?? = (? 2 ? 4?? + 4µ2 ? 1)W.

For the subalgebra ?? + ? + af ?f , where a ? R, it follows that
1
f = (4? 2 + 1) 4 (??1) exp(?? ? + 1 a arctan 2? )?(?)
2
Symmetry reduction and exact solutions of the Navier–Stokes equations 207

with ? = z 2 (4? 2 + 1)?1 . The function ? is a solution of the equation

4???? + 2(1 ? ?)?? + (? ? a)? = 0.

For example if a = 0, then ?(?) = ? µ C1 Jµ ( 1 ?) + C2 Yµ ( 1 ?) , where µ = 1 (? + 1).
2 2 4
Consider equation (5.22), where ? is an arbitrary smooth function of ? .

Theorem 5.2 Equation (5.22) is Q-conditional invariant under the operators

Q1 = ?? + g 1 (?, z)?z + g 2 (?, z)f + g 3 (?, z) ?f (5.23)

if and only if
g? ? ?z ?1 gz + ?z ?2 g 1 ? gzz + 2gz g 1 ? ?? z ?1 + 2gz = 0,
1 1 1 1 2
(5.24)
g? + ?z ?1 gz ? gzz + 2gz g k = 0, k = 2, 3,
k k k 1

and

Q2 = ?z + B(?, z, f )?f (5.25)

if and only if

B? ? ?z ?2 B + ?z ?1 Bz ? Bzz ? 2BBzf ? B 2 Bf f = 0. (5.26)

An arbitrary operator of Q-conditional symmetry of equation (5.22) is equivalent to
either an operator of form (5.23) or an operator of form (5.25).

Theorem 5.2 is proved by means of the method described in [13].

Note 5.6 It can be shown (in a way analogous to one in [13]) that system (5.24) is
reduced to the decomposed linear system

f? + ?z ?1 fz ? fzz = 0
a a a
(5.27)

by means of the following non-local transformation
fzz f 2 ? f 1 fzz
1 2
+ ?z ?1 ,
g =? 1 2
1
fz f ? f 1 fz 2

f 1 f 2 ? fz fzz
12
(5.28)
g 2 = ? zz z ,
fz f 2 ? f 1 fz
1 2

g 3 = fzz ? ?z ?1 fz + g 1 fz ? g 2 f 3 .
3 3 3

Equation (5.26) is reduced, by means of the change

B = ??? /?f , ? = ?(?, z, f )

and the hodograph transformation

y0 = ?, y1 = z, y2 = ?, ? = f,

to the following equation in the function ? = ?(y0 , y1 , y2 ):
?1
?y0 + ?(y0 )y1 ?y1 ? ?y1 y1 = 0.
208 W.I. Fushchych, R.O. Popovych

Therefore, unlike Lie symmetries Q-conditional symmetries of (5.22) are more
extended for an arbitrary smooth function ? = ?(? ). Thus, Theorem 5.2 implies that
equation (5.22) is Q-conditional invariant under the operators
X = ?? + (? ? 1)z ?1 ?z , G = (2? + C)?z ? zf ?f
?z ,
with C = const. Reducing equation (5.22) by means of the operator G, we obtain the
following solution:
f = C2 z 2 ? 2 (?(? ) ? 1)d? + C1 . (5.29)
In generalizing this we can construct solutions of the form
N
T k (? )z 2k , (5.30)
f=
k=0

where the coefficients T k = T k (? ) (k = 0, N ) satisfy the system of ODEs:
 << Ïðåäûäóùàÿ ñòð. 51(èç 122 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>