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All Lie symmetry operators of (11) are induced by operators from A(NS). Namely,
the operators Jab , D1 are induced by Jab , D and the operators ca ?a (ca = const), ?q
1

is done by

Z(|t|?1 )
R(|t|1/2 (c1 cos ? ? c2 sin ?, c1 sin ? + c2 cos ?, c3 )),

for the anzatz (5) and by

R(c1 cos ?t ? c2 sin ?t, c1 sin ?t + c2 cos ?t, c3 ), Z(1)

for the anzatz (6) respectively. Therefore, Lie reduction of the system (11) gives only
solutions that can be obtained by reducing the NSEs with two- and three-dimensional
subalgebras of A(NS). Let us proceed to the system (12). Let Amax be the maximal,
in the sense of Lie, invariance algebra of (12). Studying symmetry properties of (12),
one has to consider the following cases.
A. ?, ? ? 0. Then

Amax = ?1 , D2 , R1 (?(y1 )), Z 1 (?(y1 )) ,
1

where D2 = 2y1 ?1 + y2 ?2 + y3 ?3 ? v a ?va ? 2q?q , Z 1 (?(y1 )) = ?(y1 )?q , R1 (?(y1 )) =
1

??3 + ?1 ?v2 ? ?11 y3 ?q ; here and from now on ? = ?(y1 ), ? = ?(y1 ) are arbitrary
smooth functions of y1 = t.
B. ? ? 0, ? = 0. In this case expansion of Amax is for ? = (C1 y1 + C2 )?1 , where
C1 , C2 = const. Let C1 = 0. It can be done with the equivalence transformation (3)
so that the constant C2 will vanish, i.e. ? = Cy ?1 where C = const. Then

Amax = D2 , R1 (?(y1 )), Z 1 (?(y1 )) .
1

If C1 = 0, ? = C = const and

Amax = ?1 , R1 (?(y1 )), Z 1 (?(y1 )) .

For other values of ?, i.e. when ?11 ? = ?1 ?1 ,

Amax = R1 (?(y1 )), Z 1 (?(y1 )) .

C. ? = 0. With the equivalence transformation (3), we do ? = 0. In this case
expansion of Amax is for ? = ±|C1 y1 + C2 |1/2 , where C1 , C2 = const. Let C1 = 0.
244 W.I. Fushchych, R.O. Popovych, G.V. Popovych

It can be done with the equivalence transformation (3) so that the constant C2 will
vanish, i.e. ? = C|y1 |1/2 , where C = const. Then
Amax = D2 , Z 1 (?(y1 )), R2 (|y1 |1/2 ), R2 (|y1 |1/2 ln |y1 |) ,
1

where R2 (?(y1 )) = ??3 + ?1 ?v3 . If C1 = 0, i.e. ? = C = const;
Amax = ?1 , Z 1 (?(y1 )), ?3 , y1 ?3 + ?v3 .
For other values of ?, i.e. when (? 2 )11 = 0,

(?(y1 ))?2 dy1
Amax = Z 1 (?(y1 )), R2 (?(y1 )), R2 ?(y1 ) .

In all cases considered above, Lie symmetry operators of (12) are induced by
operators from A(NS). Namely, the operators ?1 , D2 , Z 1 (?(y1 )) are induced by ?t ,
1

D, Z(?(t)) respectively. In case ? ? 0 the operator R1 (?(y1 )) and in case ? = 0
the operator R1 (?(y1 )) where ? ? ? ? ? = 0 are done by R(0, 0, ?(t)). Therefore, Lie
? ?
reduction of the system (12) gives only solutions that can be obtained by reducing the
NSEs with two- and three-dimensional subalgebras of A(NS).
When ? = ? = 0 the system (12) describes axisymmetric motion of a fluid and can
be transformed into a system of two equations for a stream function ?1 and a function
?2 that are determined by
?1 = ?y2 v 3 ,
?1 = y2 v 1 , ?2 = y2 v 2 .
3 2

The transformed system has been studied by L.V. Kapitanskiy [8].
Consider the system (13). Let us introduce the notations

R3 (? 1 (t), ? 2 (t)) = ? i ?i + ?t ?vi ? ?tt yi ?q ,
?3 (t)dt, i i
t = y3 , ?=
?

S = ?v3 ? ?i (t)yi ?q ,
Z 1 (?(t)) = ?(t)?q ,
1
E(?(t)) = 2? ?t + ?t yi ?yi + (?tt yi ? ?t v i )?vi ? 2?t q + ?ttt yj yj ?q ,
2
J12 = y1 ?2 ? y2 ?1 + v 1 ?v2 ? v 2 ?v1 .
1

Theorem 3. The maximal, in the sense of Lie, invariance algebra of (13) is the
algebra
R3 (? 1 (t), ? 2 (t)), Z 1 (?(t)), S, E(?1 (t)), E(?2 (t)), v 3 ?v3 , J12 ,
1
1)

where ?1 = e??(t) e?(t) dt, ?2 = e??(t) , if ?1 = ?2 = 0,
? ? ?

R3 (? 1 (t), ? 2 (t)), Z 1 (?(t)), S, E(?(t)) + 2a1 v 3 ?v3 + 2a2 J12 ,
1
2)

where a1 , a2 , a3 are fixed constants, ? = e??(t)
?
e?(t) dt + a3 if
?

3 3
?1 = e 2 ?(t) (?(t))? 2 ?a1 (C1 cos(a2 ln ?(t)) ? C2 sin(a2 ln ?(t))),
?
? ?
(15)
3 3
?2 = e 2 ?(t) (?(t))? 2 ?a1 (C1 sin(a2 ln ?(t)) + C2 cos(a2 ln ?(t))),
?
? ?

e?(t) dt + a3 , C1 , C2 = const, (C1 , C2 ) = (0, 0);
?
where ?(t) =
?

R3 (? 1 (t), ? 2 (t)), Z 1 (?(t)), S, E(?(t)) + 2a1 v 3 ?v3 + 2a2 J12 ,
1
3)
Ans?tze of codimension one for the Navier–Stokes field
a 245

where a1 , a2 are fixed constants, ? = e??(t) if
?

3
?1 = e 2 ?(t)?a1 ?(t) (C1 cos(a2 ?(t)) ? C2 sin(a2 ?(t))),
? ?
? ?
3
?2 = e 2 ?(t)?a1 ?(t) (C1 sin(a2 ?(t)) + C2 cos(a2 ?(t))),
? ?
? ?
e?(t) dt, C1 , C2 = const, (C1 , C2 ) = (0, 0);
?
where ?(t) =
?
R3 (? 1 (t), ? 2 (t)), Z 1 (?(t)), S in all other cases.
4)
Here ? i = ? i (t), ? = ?(t) are arbitrary smooth functions of t = y3 .
Remark 3. If functions ?b = ?b (t) are determined by (14), e?(t) = C|m(t)|, where
?
?
C = const and it follows from the condition ? = ? = 0 that m = |m(t)|e, where
1 2

|e| = 1, e = const.
Remark 4. Vector-functions ni from remark 2 are determined up to the transforma-
tion
n1 = n1 cos ? ? n2 sin ?, n2 = n1 sin ? + n2 cos ?,
where ? = const. Therefore, choosing ?, we can do so that C2 = 0 (then C1 = 0).
The operators R3 (? 1 , ? 2 ) + ?S, Z 1 (?) are induced by R(l) + Z(?), Z(?) respec-
tively, where l = ? i ni + ? 3 m, ?t (m · m) + 2? i (ni · m) = ?, ? ? 3 (m · m)?1 (? i (mt ·
3
t 2
n )) ? 2 ? ? (mtt · n ) + 2 ? (ltt · n ) = 0,
13i 1i
i2 i i

If m = |m(t)|e, where e = const, |e| = 1, the operator J12 is induced by e1 J23 +
1

eJ31 + e3 J12 . For
? = ?t + ?,
m = ?3 e?t (?2 cos ?, ?2 sin ?, ?1 )T , 2 2
?1 + ?2 = 1,
the operator ?t + ?J12 induces the operator ?y3 ? ?1 ?J12 + ?v 3 ?v2 if such vector-
functions ni are chosen:
n2 = ?k 1 sin ?1 ? + k 2 cos ?1 ?,
n1 = k 1 cos ?1 ? + k 2 sin ?1 ?, (16)
where k 1 = (? sin ?, cos ?, 0)T , k 2 = (?1 cos ?, ?1 sin ?, ??2 )T . For
m = ?3 |t + ?4 |?+1/2 (?2 cos ?, ?2 sin ?, ?1 )T , 2 2
?1 + ?2 = 1,
? = ? ln |t + ?4 | + ?,
the operator D + 2?4 ?t + 2?J12 induces the operator
D3 + 2?4 ?y3 ? 2??J12 + 2?v 3 ?v3
1

if vector-functions ni are chosen in the form (15). In all other cases the basis elements
of the maximal, in the sense of Lie, invariance algebra of (13) are not induced by
operators from A(NS).
Remark 5. The invariance algebra of a system of the form (13) with a parameter-
function ?3 = ?3 (t) is like one with a different parameter-function ?3 = ?3 (t). It
?
suggest an idea that there is a local transformation of variables with which one can
make ?3 to vanish. Indeed, let us transform variables in the way
1
1 1
v i + yi ?3 (t) e? 2 ?(t) ,
yi = yi e 2 ?(t) ,
?
e?(t) dt,
?
vi = ?
v3 = v3 ,
? y3 =
? ? ?
2
1
q = qe??(t) + yi yi [(?3 (t))2 ? 2?3 (t)]e??(t) .
? ?
? ?
8
246 W.I. Fushchych, R.O. Popovych, G.V. Popovych

As a result, we obtain the system
v3 + v j vjj + qi + ?i (?3 )?3 = 0,
?i ? ?i ? ?yv
v3 + v j vj ? vjj = 0,
?3 ? ?3 ?3
?i
vj = 0,
3
for functions v a = v a (?1 , y2 , y3 ), q = q (?1 , y2 , y3 ), where ?i (?3 ) = ?i (t)e? 2 ?(t) , sub-
?
? ?y?? ? ?y ? ? ?y
scripts 1, 2, 3 mean differentiation with respect to y1 , y2 , y3 accordingly.
???

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W.I. Fushchych, Scientific Works 2003, Vol. 5, 247–250.

Antireduction and exact solutions
of nonlinear heat equations
W.I. FUSHCHYCH, R.Z. ZHDANOV

We construct a number of ansatzes that reduce one-dimensional nonlinear heat equations
to systems of ordinary differential equations. Integrating these, we obtain new exact
solution of nonlinear heat equations with various nonlinearities.

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