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if ? = 1 (? = 0) respectively. Here the functions n and ? satisfy the following
equations:
nt + ?(n2 , ?n1 , 0)T = bi mi ,
?t = bi ?i + 1 (nttt · n ? ntt · nt ) + ?(n1 n2 ? n1 n2 ).
tt tt
2

7. A3 (? 1 , ? 2 , ? 3 , ?) = J12 + R(0, 0, ? 3 ), R(? 1 , ? 2 , 0), R(?? 2 , ? 1 , 0) , where
7

? a ? C ? ((t0 , t1 ), R), ?tt ? 2 ? ? 1 ?tt ? 0, ? i ? i ? 0,
1 2
? 3 = 0.
Algebras A3 (? 1 , ? 2 , ? 3 ) and A3 (?1 , ? 2 , ? 3 ) are equivalent if ? ?a ? R, ? ?4 = 0:
7? ? ?
7

? 1 (t) = ?4 (? 1 (t) cos ?3 ? ? 2 (t) sin ?3 ),
??
??
? 2 (t) = ?4 (? 1 (t) sin ?3 + ? 2 (t) cos ?3 ), (A.11)
? 3 (t) = e??1 ? 3 (t),
??
where t = te?2?1 + ?2 .
?
8. A3 (m1 , m2 , m3 ) = R(m1 ), R(m2 ), R(m3 ) , where
8

ma ? C ? ((t0 , t1 ), R3 ), ma · mb ? ma · mb = 0.
rank(m1 , m2 , m3 ) = 3, tt tt
1 2 3
Algebras A3 (m1 , m2 , m3 ) and A3 (m , m , m ) are equivalent if ? ?i ? R3 , ? B ? O(3),
8? ??
8
? {dab } : det{dab } = 0 such that
ma (t) = dab B mb (t),
?? (A.12)
where t = te?2?1 + ?2 .
?
Symmetry reduction and exact solutions of the Navier–Stokes equations 235

B On construction of ansatzes for the Navier–Stokes
field by means of the Lie method
The general method for constructing a complete set of inequivalent Lie ansatzes of
a system of PDEs are well known and described, for example, in [27, 28]. However,
in some cases when the symmetry operators of the system have a special form, this
method can be modified [9]. Thus, in the case of the NSEs, coefficients of an arbitrary
operator
Q = ? 0 ?t + ? a ?a + ? a ?ua + ? 0 ?p
from A(N S) satisfy the following conditions:
? 0 = ? 0 (t, x), ? a = ? a (t, x), ? a = ? ab (t, x)ub + ? a0 (t, x),
(B.1)
? 0 = ? 01 (t, x)p + ? 00 (t, x).

(The coefficients ? a , ? 0 , ? a , and ? 0 also satisfy stronger conditions than (B.1). For
example if Q ? A(N S), then ? 0 = ? 0 (t), ? ab = const, and so on. But conditions (B.1)
are sufficient to simplify the general method.) Therefore, ansatzes for the Navier–
Stokes field can be constructing in the following way:
1. We fix a M -dimensional subalgebra of A(N S) with the basis elements
Qm = ? m0 ?t + ? ma ?a + (? mab ub + ? ma0 )?ua + (? m01 p + ? m00 )?p , (B.2)
where M ? {1; 2; 3}, m = 1, M , and
rank{(? m0 , ? m1 , ? m2 , ? m3 ), m = 1, M } = M. (B.3)
To construct a complete set of inequivalent Lie ansatzes of codimension M for the
Navier–Stokes field, we have to use the set of M -dimensional subalgebras from Sec-
tion A. Condition (B.3) is neeeded for the existance of ansatzes connected with this
subalgebra.
2. We find the invariant independent variables ?n = ?n (t, x), n = 1, N , where
N = 4 ? M, as a set of functionally independent solutions of the following system:
Lm ? = Qm ? = ? m0 ?t ? + ? ma ?a ? = 0, (B.4)
m = 1, M ,
where Lm := ? m0 ?t + ? ma ?a .
3. We present the Navier–Stokes field in the form:
ua = f ab (t, x)v b (? ) + g a (t, x), p = f 0 (t, x)q(? ) + g 0 (t, x), (B.5)
? ?
where v a and q are new unknown functions of ? = {?n , n = 1, N }. Acting on
?
m
representation (B.5) with the operators Q , we obtain the following equations on
functions f ab , g a , f 0 , and g 0 :
Lm f ab = ? mac f cb , Lm g a = ? mab g b + ? ma0 , c = 1, 3,
(B.6)
Lm f 0 = ? m01 f 0 , Lm g 0 = ? m01 g 0 + ? m00 .

If the set of functions f ab , f 0 , g a , and g 0 is a particular solution of (B.6) and satisfies
the conditions rank{(f 1b , f 2,b , f 3b ), b = 1, 3} = 3 and f 0 = 0, formulas (B.5) give an
ansatz for the Navier–Stokes field.
236 W.I. Fushchych, R.O. Popovych

The ansatz connected with the fixed subalgebra is not determined in an unique
manner. Thus, if
? ?l
?
?l = ?l (? ),
? ?? det = 0,
??n l,n=1,N
(B.7)
?ab
f (t, x) = f (t, x)F (? ), g (t, x) = g a (t, x) + f ac (t, x)Gc (? ),
ac cb a
? ? ?
?
f 0 (t, x) = f 0 (t, x)F 0 (? ), g 0 (t, x) = g 0 (t, x) + f 0 (t, x)G0 (? ),
? ? ?

the formulas
? ?
v? ?
ua = f ab (t, x)?b (? ) + g a (t, x), p = f 0 (t, x)q(? ) + g 0 (t, x) (B.8)
? ? ? ?

give an ansatz which is equivalent to ansatz (B.5). The reduced system of PDEs
on the functions v a and q is obtained from the system on v a and q by means of
? ?
a local transformation. Our problem is to find or “to guess”, at once, such an ansatz
that the corresponding reduced system has a simple and convenient form for our
investigation. Otherwise, we can obtain a very complicated reduced system which
will be not convenient for investigation and we can not simplify it.
Consider a simple example.
Let M = 1 and let us give the algebra ?t + ?J12 , where ? ? {0; 1}. For this
algebra, the invariant independent variables ya = ya (t, x) are functionally independent
solutions of the equation Ly = 0 (see (B.4)), where

L := ?t + ?(x1 ?x2 ? x2 ?x1 ). (B.9)

There exists an infinite set of choices for the variables ya . For example, we can give
the following expressions for ya :
x1
? ?t, y2 = (x2 + x2 )1/2 ,
y1 = arctan y3 = x3 .
1 2
x2
However choosing ya in such a way, for ? = 0 we obtain a reduced system which
strongly differs from the “natural” reduced system for ? = 0 (the NSEs for steady
flows of a viscous fluid in Cartesian coordinates). It is better to choose the following
variables ya :

y1 = x1 cos ?t + x2 sin ?t, y2 = ?x1 sin ?t + x2 cos ?t, y3 = x3 .

The vector-functions f b = (f 1b , f 2b , f 3b ), b = 1, 3, should be linearly independent
solutions of the system

Lf 1 = ??f 2 , Lf 2 = ?f 1 , Lf 3 = 0

and the function f 0 should satisfy the equation Lf 0 = 0 and the condition f 0 =
0. Here the operator L is defined by (B.9). We give the following values of these
functions:

f 1 = (cos ?t, sin ?t, 0), f 2 = (? sin ?t, cos ?t, 0), f 3 = (0, 0, 1), f 0 = 1.

The functions g a and g 0 are solutions of the equations

Lg 1 = ??g 2 , Lg 2 = ?g 1 , Lg 3 = 0, Lg 0 = 0.
Symmetry reduction and exact solutions of the Navier–Stokes equations 237

We can make, for example, g a and g 0 vanish. Then the corresponding ansatz has the
form:
u1 = v 1 cos ?t ? v 2 sin ?t, u2 = v 1 sin ?t + v 2 cos ?t, u3 = v 3 , p = q , (B.10)
? ? ? ? ? ?
where v a = v a (y1 , y2 , y3 ) and q = q (y1 , y2 , y3 ) are the new unknown functions. Substi-
? ? ??
tuting ansatz (B.10) into the NSEs, we obtain the following reduced system:
v a va ? vaa + q1 + ?y2 v1 ? ?y1 v2 ? ??2 = 0,
? ?1 ?1 ?1 ?1
? v
v a va ? vaa + q2 + ?y2 v1 ? ?y1 v2 + ??1 = 0,
? ?2 ?2 ?2 ?2
? v
(B.11)
v a va ? vaa + q3 + ?y2 v1 ? ?y1 v2 = 0,
? ?3 ?3 ?3 ?3
?
?a
va = 0.
Here subscripts 1,2, and 3 of functions in (B.11) denote differentiation with respect
to y1 , y2 , and y3 accordingly. System (B.11), having variable coefficients, can be
simplified by means of the local transformation
v 1 = v 1 ? ?y2 , v 2 = v 2 + ?y1 , q = q + 1 (y1 + y2 ).
v3 = v3 , 2 2
(B.12)
? ? ? ? 2

Ansatz (B.10) and system (B.11) are transformed under (B.12) into ansatz (2.2) and
system (2.7), where
g 1 = ??x2 , g 2 = ?x1 , g 0 = 1 ? 2 (x2 + x2 ), (B.13)
g3 = 0, 1 2
2

?1 = ?2?, and ?2 = 0. Therefore, we can give the values of g a and g 0 from (B.13)
and obtain ansatz (2.2) and system (2.7) at once.
The above is a good example how a reduced system can be simplified by means of
modifying (complicating) an ansatz corresponding to it. Thus, system (2.7) is simpler
than system (B.11) and ansatz (2.2) is more complicated than ansatz (B.10).
Finally, let us make several short notes about constructing other ansatzes for the
Navier–Stokes field.
Ansatz corresponding to the algebra A1 (m, ?) (see Subsection A.2) can be cons-
4
tructed only for such t that m(t) = 0. For these values of t, the parameter-function ?
can be made to vanish by means of equivalence transformations (A.7).
Ansatz corresponding to the algebra A2 (?, ? 1 , ?, ? 2 ) (see Subsection A.3) can be
8
constructed only for such t that ?(t) = 0. For these values of t, the parameter-function
? 2 can be made to vanish by means of equivalence transformations (A.8). Moreover,
it can be considered that ?t ? ? ??t ? {0; 1}. The algebra obtained finally is denoted
by A2 (?, ?, ?, 0).
8
Ansatz corresponding to the algebra A2 (m1 , ?1 , m2 , ?2 ) (see Subsection A.3) can
9
be constructed only for such t that rank(m1 , m2 ) = 2. For these values of t, the
parameter-functions ?i can be made to vanish by means of equivalence transforma-
tions (A.9).
The algebras A2 (?, ?), A2 (?), and A2 (?) can not be used to construct ansatzes
10 11 12
by means of the Lie algorithm.
In view of equivalence transformation (A.11), the functions ? i in the algebra
A3 (? 1 , ? 2 , ? 3 ) (see Subsection A.4) can be considered to satisfy the following conditi-
7
on:
?t ? 2 ? ? 1 ?t ? {0; 1 }.
1 2
2
238 W.I. Fushchych, R.O. Popovych

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