ñòð. 51 |

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:

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