<<

. 103
( 135 .)



>>

17. A17 (?, ?) = R(0, 0, ?(t)), R 0, 0, ?(t) + Z(?(t)) , ? = 0, ? = 0,
(?(t))2
where algebras A17 (?1 , ? 1 ) and A17 (?2 , ? 2 ) are equivalent if ? c = 0, ? ?, ? ? R:
2?
(?2 , ? 2 )(t) = ec ?1 , e?? c? 1 (te2? + ?);
18. A18 (?, ?) = Z(?(t)), Z(?(t)) , c1 ? + c2 ? = 0 ? c1 , c2 ? R, where algebras
A18 (?2 , ? 2 ) and A18 (?2 , ? 2 ) are equivalent if ? {akl }k,l=1,2 , det{akl } = 0, ? ?, ? ? R:
(?2 , ? 2 )(t) = (a11 ?1 + a12 ? 1 , a21 ?1 + a22 ? 1 )(te2? + ?).
Theorem 1 is proved with method described in [4, 5].
Lying in a linear span of operators Jab , R(m) and Z(?) are algebras 1, 2 and 3.
Ans?tze constructed with these algebras have the form
a
m·x n·x k·x ?
? ? bij (t)
n?
? ?
1. u = v(?1 , ?2 ) + m+ k, p = q(?1 , ?2 ) + xi xj ,
? ? ? 2
where ?1 = t, ?2 = k · x, k = m ? n, m = n ? k, n = k ? m, ? = |k|2 ,
? ?
k·m i j k·n i j
? ?
1
bij = ? mi mi + ni ni +
?? ?? mk +
? nk ;
?
? ? ?
Symmetry reduction of the Navier–Stokes equations 441

x2 2
u1 = x1 v 1 (?1 , ?2 ) ? 2 (v (?1 , ?2 ) ? s(t)),
2.
?2
x1
u2 = x2 v 1 (?1 , ?2 ) + 2 (v 2 (?1 , ?2 ) ? s(t)),
?2
?
13 ?(t)
u3 = v (?1 , ?2 ) + x3 ,
?(t) ?(t)
?
?(t) x2 x2
p = q(?1 , ?2 ) ? 3
+ ?(t) arctg ,
?(t) 2 x1

x1 + x2 , s(t) =
where ?1 = t, ?2 = ?(t)dt;
2 2

x2 2
u1 = x1 v 1 (?1 , ?2 ) ? 2 (v (?1 , ?2 ) ? s(t)),
3.
?2
x1
u2 = x2 v 1 (?1 , ?2 ) + 2 (v 2 (?1 , ?2 ) ? s(t)),
?2
?
13 ?(t) 1 x2
3
u= v (?1 , ?2 ) + x3 + arctg ,
?(t) ?(t) ?(t) x1
?
?(t) x2 x2
p = q(?1 , ?2 ) ? 3
+ ?(t) arctg ,
?(t) 2 x1

where ?1 = t, ?2 = x1 + x2 , s(t) = ?(t)dt;
2 2
Here numeration of ans?tze correspond to that of algebras in theorem 1. Substitu-
a
ting ans?tze 1–3 into the NSEs, we obtain equations reduced
a
m·v n·v
? ?
1. v 1 ? ?v 22 + q2 k + ·m+ n + e · ? 2 = 0, (6)
? ?
? ?
k · v 2 = 0, (7)

?? ?
2k · k ? k · k
2 ?
(m · n ? n · m)k ? k +
where e = e(t) = ? ? k;
?2 ?2
12 31 1
2. v1 + ((v 1 )2 ? (v ? s)2 + ?2 v 1 v2 ? v22 +
1 1 1
(8)
v + q2 = 0,
?2 2
4
?2 ?2

12
v1 + ?2 v 1 v2 ? v22 +
2 2 2
(9)
v = 0,
?2 2
13
v1 + ?2 v 1 v2 ? v22 +
3 3 3
(10)
v = 0,
?2 2
?
?
1 1
(11)
2v + ? 2 v2 + = 0,
?

12 31 1
3. v1 + (v 1 )2 ? 4 (v ? s) + ?2 v v2 ? v22 + ? v2
1 2 11 1
(12)
+ q2 = 0,
?2 ?2
2

12
v1 + ?2 v 1 v2 ? v22 +
2 2 2
(13)
v = 0,
?2 2
442 W.I. Fushchych, R.O. Popovych

1 3 v2 ? s
v1 + ?2 v 1 v2 ? v22 ?
3 3 3
(14)
v+ = 0,
?2 2 2
?2

?
?
2v 1 + ?2 v2 +
1
(15)
= 0.
?

Here subscripts 1 and 2 mean differentiation by variables ?1 and ?2 respectively. Let
us show that nonlinear system 1–3 can be transformed to linear PDEs.
Consider system 1 (equations (6)–(7)). After integration of equation (7) by ?2 :
k · v = h(t). Further we make the transformation from the symmetry group of the
NSEs

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

where ? · m ? l · m = ? · n ? l · n = 0,
? ?
l l

? m · l m ? n · l n + k · l k + h = 0.
? ? ?
k· l? ? ?
? ? ?

?
This transformation does not change ansatz 1 and besides k · v = 0, that is h(t) ? 0.
?
Therefore, without loss of generality we can assume that h(t) ? 0.
Let f = f (?1 , ?2 ) = m · v, g = g(?1 , ?2 ) = n · v. Since m · n ? n · m = 0 then
? ?
m · n ? n · m = C = const. Case C = 0 is reduced by means of change of the basis
? ?
of the algebra A1 (m, n) to case C = 1. Let us multiply the scalar equation (6) by m,
n and k. As result we obtain the linear system of PDEs with variable coefficient for
functions f , g and q:

m·n m·m 2
f1 ? ?f22 + C f? g + 2 C?2 (n · k) = 0,
?
? ? ?
n·n m·n 2
g1 ? ?g22 + C f? g ? 2 C?2 (m · k) = 0,
?
? ? ?
2 ?2 ?
?? ?? ??
q2 = 2 ((m · k)f + (n · k)g) + 2 (k · k ? 2k · k).
? ?
Consider two possible cases.
a) Let C = 0. Then there are functions ?i = ?i (?, ?), i = 1, 2, where ? = ?(t)dt,
? = ?2 , that f = ?1 , g = ?2 and ?i ? ?i = 0, i = 1, 2. Therefore
? ? ? ??

?
m·x n·x k·x
? ?
n?
?1 (?, ?) + m + ?2 (?, ?) +
? ?
u= k,
? ?
? ? ?
? ??
1 k · k ? 2k · k 2
2 2 (16)
?? ??
+ (m · k)?1 (?, ?) + 2 (n · k)?2 (?, ?) + ??
p= 2
? ? 2? ?
k·m k·n
? ?
?(m · x)(m · x) ? (n · x)(n · x) ? (m · x)(k · x) ? (n · x)(k · x) ,
? ? ? ? ? ?
? ?

where m · n ? n · m = 0, k = m ? n, m = n ? k, n = k ? m, ? = |k|2 , ? =
? ? ? ? ?(t)dt,
? = k · x, ?? ? ??? = 0, i = 1, 2.
i
Symmetry reduction of the Navier–Stokes equations 443

b) Let C = 1. Then we obtain the following solutions of the NSEs:
m·x n·x
? ?
?2
y i (t)?i (?, ?) + y 0 (t)? + ?
u= m+
? ?
?
n·x m·x k·x
? ?
n?
+ z i (t)?i (?, ?) + z 0 (t)? + ?
+ k,
?2
? ?
?2
2
(m · k) y i (t)?i (?, ?) + y 0 (t)
?
p= + (17)
?2 2
? ??
k · k ? 2k · k 2
2
2 ? z i (t)?i (?, ?) + z 0 (t) ? 1
+ 2 (n · k) ??
? +
? 2 2? ?
k·m k·n
? ?
?(m · x)(m · x) ? (n · x)(n · x) ? (m · x)(k · x) ? (n · x)(k · x) ,
? ? ? ? ? ?
? ?
where m · n ? n · m = 1, k = m ? n, m = n ? k, n = k ? m, ? = |k|2 , ? = ?(t)dt,
? ? ? ?
? = k · x, ?? ? ??? = 0, i = 1, 2, (y (t), z (t)), i = 1, 2 is a fundamental system of
i i i

solutions for equations
m·n m·m n·n m·n
y? y? (18)
y+
? z = 0, z +
? z=0
? ? ? ?
and (y 0 (t), z 0 (t)) is a particular solutions of the system

<<

. 103
( 135 .)



>>