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d? n d? n
Symmetry reduction and exact solutions of the Navier–Stokes equations 201

Note 4.4 If function ? satisfies the equation

??? ? ???? + ?? = 0 (? = ??),

?(?)d? = (? + ?)?1 (??? ? ?? ) + C1 .
then

7. The last equation of system (4.13) is the compatibility condition of the NSEs
(1.1) and ansatz (4.7). Integrating this equation, we obtain that

? 3 = C0 (? i ? i )?1 , C0 = 0.

As ?3 = ??? (? 3 )?1 ?3 = 2?1 ?3 , ?3 = C3 ? i ? i . Then system (4.13) is reduced to the
3
?
equations

?1 = ?1 (?)?1 ? ?2 (?)?2 ,
?
(4.36)
?2 = ?2 (?)?1 + ?1 (?)?2 ,
?

?2 ?1
where ?1 = ?C0 (? i ? i )2 ? ?1 and ?2 = ?2 ? C3 C0 (? i ? i )2 . System (4.36) implies
that

?2 (?)d? ? C2 sin
?1 = exp ?1 (?)d? ?2 (?)d?
C1 cos ,

?2 = exp ?1 (?)d? ?2 (?)d? + C2 cos ?2 (?)d?
C1 sin .

8. Let us apply the trasformation generated by the operator R(k(t)), where

kt = ??1 (nb · k)mb ? ?,
t

to ansatz (4.8). As a result we obtain an ansatz of the same form, where the functi-
?
ons ? and h are replaced by the new functions ? and h:
?

? = ? ? ??1 (na · k)ma + kt = 0,
? t
h = h ? ? (mtt · k)(na · k) + 1 ??2 (mb · ma )(na · k)(nb · k).
?1
? a
tt
2

?
Let us make h vanish by means of the transformation generated by the operator
?
Z(?h(t)). Therefore, the functions ?a and h can be considered to vanish. The equation
(na · ma ) = 0 is the compatibility condition of ansatz (4.8) and the NSEs (1.1).
t

Note 4.5 The solutions of the NSEs obtained by means of ansatzes 5–8 are equivalent
to either solutions (5.1) or solutions (5.5).

5 Reduction of the Navier–Stokes equations
to linear systems of PDEs
Let us show that non-linear systems 8 and 9, from Subsection 3.2, are reduced to
linear systems of PDEs.
202 W.I. Fushchych, R.O. Popovych

5.1 Investigation of system (3.17)–(3.18)
Consider system 9 from Subsection 3.2, i.e., equations (3.17) and (3.18). Equation
(3.18) integrates with respect to z2 to the following expression:

k · w = ?(t).

Here ? = ?(t) is an arbitrary smooth function of z1 = t. Let us make the transfor-
mation from the symmetry group of the NSEs:

u(t, x) = u(t, x ? l) + lt (t),
?
p(t, x) = p(t, x ? l) ? ltt (t) · x,
?

where ltt · mi ? l · mi = 0 and
tt

k · lt ? ??1 (ni · l)mi + ??1 (k · l)kt + ? = 0.
t

This transformation does not modify ansatz (3.9), but it makes the function ?(t)
vanish, i.e., k · w = 0. Therefore, without loss of generality we may assume, at once,
?
that k · w = 0.
Let f i = f i (z1 , z2 ) = mi · w. Since m1 · m2 ? m1 · m2 = 0, it follows that
tt tt
m1 · m2 ? m1 · m2 = C = const. Let us multiply the scalar equation (3.17) by mi
t t
and k. As a result we obtain the linear system of PDEs with variable coefficients in
the functions f i and s:

f1 ? ?f22 + C??1 (mi · m2 )f 1 ? (mi · m1 )f 2 ? 2C??2 (k ? kt ) · mi z2 = 0,
i i

s2 = 2??2 (ni · kt )f i + ??2 (ktt · k ? 2kt · kt )z2 .

Consider two possible cases.
A. Let C = 0. Then there exist functions g i = g i (?, ?), where ? = ?(t)dt and
? = z2 , such that f i = g? and g? ? g?? = 0. Therefore,
i i i

u = ??1 (g? (?, ?) + mi · x)ni ? ??1 (kt · x)k,
i
t
p = 2??2 (ni · kt )g i (?, ?) + 1 ??2 (ktt · k ? 2kt · kt )? 2 ? (5.1)
2
? 1 ??1 (ni · x)(mi · x) ? 1 ??2 (k · mi )(ni · x)(k · x),
tt tt
2 2

where m1 · m2 ? m1 · m2 = 0, k = m1 ? m2 , n1 = m2 ? k, n2 = k ? m1 , ? = |k|2 ,
t t
? = k · x, ? = ?(t)dt, and g? ? g?? = 0.
i i

For example, if m = (? 1 (t), 0, 0) and n = (0, ? 2 (t), 0) with ? i (t) = 0, it follows that

u1 = (? 1 )?1 (f 1 + ?t x1 ), u2 = (? 2 )?1 (f 2 + ?t x2 ), u3 = ?(? 1 ? 2 )t (? 1 ? 2 )?1 x3 ,
1 2

p = ? 1 ?tt (? 1 )?1 x2 ? 1 ?tt (? 2 )?1 x2 +
1 2
1 2
2 2
2
1 2 ?1
? (? 1 ? 2 )t (? 1 ? 2 )?1
1 12
x2 ,
+ 2 (? ? )tt (? ? ) 3

where f i = f i (?, ?), f? ? f?? = 0, ? = (? 1 ? 2 )2 dt, and ? = ? 1 ? 2 x3 . If m1 =
i i

(? 1 (t), ? 2 (t), 0) and m2 = (0, 0, ? 3 (t)) with ? 3 (t) = 0 and ? i (t)? i (t) = 0, we obtain
Symmetry reduction and exact solutions of the Navier–Stokes equations 203

that
u1 = (? i ? i )?1 ? 1 (g? + ?t xi ) ? ? 2 ?t (? 3 )?2 ? + ?t x1 ? ?t x2
i 3 2 1
,

u2 = (? i ? i )?1 ? 2 (g? + ?t xi ) + ? 1 ?t (? 3 )?2 ? + ?t x1 ? ?t x2
i 3 2 1
,
u3 = (? 3 )?1 (f + ?t x3 ),
3

p = 2(? 3 )?1 (? 1 ?t ? ?t ? 2 )(? i ? i )?2 g + 1 ??1 ?
2 1
2

? ??1 (?tt ? 3 ? 2?t ?t )? i ? i ? 2? 3 ?t ? i ?t ? 2(? 3 )2 ?t ?t ? 2 +
3 33 3 i ii

+ (? 3 )2 (? 2 ?tt ? ? 1 ?tt )(x2 ? x2 ) ? 2(?tt ? 2 + ? 1 ?tt )x1 x2 ? ? i ? i ? 3 ?tt x2 .
2 1 1 2 3
1 2 3

Here f = f (?, ?), f? ? f?? = 0, g = g(?, ?), g? ? g?? = 0, ? = (? 3 )2 ? i ? i dt,
? = ? 3 (? 2 x1 ? ? 1 x2 ), and ? = (? 3 )2 ? i ? i .
Note 5.1 The equation
m 1 · m 2 ? m1 · m 2 = 0 (5.2)
t t

can easily be solved in the following way: Let us fix arbitrary smooth vector-functions
m1 , l ? C ? ((t0 , t1 ), R3 ) such that m1 (t) = 0, l(t) = 0, and m1 (t) · l(t) = 0 for all
t ? (t0 , t1 ). Then the vector-function m2 = m2 (t) is taken in the form
m2 (t) = ?(t)m1 + l(t). (5.3)
Equation (5.2) implies
?(t) = (m1 · m1 )?1 (m1 · l ? m1 · lt )dt. (5.4)
t

B. Let C = 0. By means of the transformation mi > aij mj , where aij = const and
det{aij } = C, we make C = 1. Then we obtain the following solution of the NSEs
(1.1)
u = ??1 ?ij (t)g? (?, ?) + ?i0 (t)? + mi · x? ??1 ((k ? mi ) · x) ni ? ??1 (kt · x)k,
j
t

p = 2??2 (ni · kt )(?ij (t)g i (?, ?) + 1 ?i0 (t)? 2 ) + 1 ??2 (ktt · k ? 2kt · kt )? 2 ? (5.5)
2 2
? 1 ??1 (ni · x)(mi · x) ? 1 ??2 (k · mi )(ni · x)(k · x).
tt tt
2 2

Here m1 · m2 ? m1 · m2 = 1, k = m1 ? m2 , n1 = m2 ? k, n2 = k ? m1 , ? = |k|2 ,
t t
? = k · x, ? = ?(t)dt, and g? ? g?? = 0. (?1i (t), ?2i (t)) (i = 1, 2) are linearly
i i

independent solutions of the system
?t + ??1 (mi · m2 )?1 ? ??1 (mi · m1 )?2 = 0,
i
(5.6)
and (?10 (t), ?20 (t)) is a particular solution of the nonhomogeneous system
?t + ??1 (mi · m2 )?1 ? ??1 (mi · m1 )?2 = 2??2 (k ? kt ) · mi .
i
(5.7)
For example, if m1 = (? cos ?, ? sin ?, 0) and m2 = (?? sin ?, ? cos ?, 0), where
? = ?(t) = 0 and ? = ? 1 (?)?2 dt (therefore, m1 · m2 ? m1 · m2 = 1), we obtain
t t
2

u1 = ? ?1 f 1 cos ? ? f 2 sin ? + ?t x1 ? 1 ? ?1 x2 ,
2
?1
f sin ? + f cos ? + ?t x2 + 1 ? ?1 x1 ,
2 1 2
u =? 2
?1
u = ?2?t ? x3 ,
3

p = (?tt ? ? 3?t ?t )? ?2 x2 ? 1 (?tt ? ?1 ? 1 ? ?4 )xi xi .
3 2 4
204 W.I. Fushchych, R.O. Popovych

Here f i = f i (?, ?), f? ? f?? = 0, ? = (?)4 dt, and ? = (?)2 x3 .
i i

Note 5.2 As in the case C = 0, the solutions of the equation
m1 · m2 ? m1 · m2 = 1 (5.8)
t t

can be sought in form (5.3). As a result we obtain that
|m1 |?2 (m1 · l ? m1 · lt ? 1)dt. (5.9)
?(t) = t

Note 5.3 System (5.6) can be reduced to a second-order homogeneous differential
equation either in ?1 , i.e.,

?|m1 |?2 ?t (m1 · m2 )|m1 |?2 + |m1 |?2 ?1 = 0
1
(5.10)
+ t
t

(then ?2 = |m1 |?2 (??t + (m1 · m2 )?1 )), or in ?2 , i.e.,
1

?|m2 |?2 ?t + ? (m1 · m2 )|m2 |?2 + |m2 |?2 ?2 = 0
2
(5.11)
t
t

(then ?1 = |m2 |?2 (???t + (m1 · m2 )?2 )). Under the notation of Note 5.1 equation
2

(5.10) has the form:
+ |m1 |?2 (m1 · l ? m1 · lt )?1 = 0.
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