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n·m m·m n·n m·n
2 2
y? z + 2 n · k = 0, y? z ? 2 m · k = 0.
? ?
y+
? z+
?
? ? ? ? ? ?
Remark 1. System (18) can be reduced to one ordinary differential equation of second
order. For this aim we introduce new designations:
n·m
y(t) = y · h,
h(t) = exp dt , z (t) = z/h,
?
?
m·m 2 n·n
F 1 (t) = F 2 (t) =
h, .
?h2
?
For functions y and z we obtain the system
? ?
y = F 1 · z, z = ?F 2 y
? ?
? ? ? ?
and hence
·
y?
+ F 2 y = 0. (19)
?
F1
Functions F 1 and F 2 (and vectors m and n respectively) we choose in such manner
that fundamental system of solutions of equation (19) should be known. Then solution
(17) can be written in closed form.
Remark 2. New solutions can be obtained from solutions (16) and (17) by force of (3)
only after transformations generated by operators of type R(m(t)) and Z(?(t)).
Consider system 2 (equations (8)–(11)). Equations (11) immediately gives
? h(?1 ) ? 1
?
v1 = ? (20)
+ ,
2
2? ?2
444 W.I. Fushchych, R.O. Popovych

where h is an arbitrary differentiable function of ?1 . Substituting (20) into the remai-
ning equations (8)–(10), we get
? ?
· 2
? ? ?
? 1 ? ? 1 ? ? ? 2 ? h + (h ? 1) + (v ? s(t)) ,
2 2 2
(21)
q2 = 3 3
2? 4? ?2 ?2 ?2

?
h ?
? ?
2 2 2
(22)
v1 v22 + ?2 v2 = 0,
?2 2?

?
h?2 ?
v1 ? v22 + ?
3 3 3
(23)
?2 v2 = 0.
?2 2?

After change of independent variables

|?(t)|dt, |?(t)|?2 (24)
?= ?=

in equations (22) and (23) we obtain the decomposed system of linear equations
h(t) 2
v? ? v?? +
2 2
(25)
v = 0,
??
h(t) ? 2 3
v? ? v?? +
3 3
(26)
v? = 0.
?
Remark 3. An arbitrary solution of equation (26) can be written down in the form
v 3 = v? /?, where v 3 is a solution of (25).
? ?
From equation (21)
? ?
· 2
? ? (h ? 1)2 (v 2 (?, ?) ? s(t))2
2
?2 ? ? 1? ??
? ? h ln ?2 ?
q= + d?2 .(27)
2 3
4 ? 2? 2?2 ?2

Formulas (20), (24)–(27) and anzats 2 give a solution of the NSEs.
Remark 4. New solutions can be obtained from this solution by force of (4) only
alter transformations generated by operators of type Jab , R(m(t)) and Z(?(t)).
Let us to investigate the symmetry properties of the equation
h(t)
fr ? frr = 0 (28)
ft +
r
and to construct some its exact solutions.
Theorem 2. The maximal, in the sense of Lie, invariance algebra of equation (28)
is the algebra
A1 = f ?f , g(t, r)?f if h(t) = const;
1.
A2 = ?t , D, ?, f ?f , g(t, r)?f if h = const, h ? {0; ?2};
2.
A3 = ?t , D, ?, f ?f , ?r + = 2t?t ? r ?
h ht
3. 2r f ?f , G f ?f , g(t, r)?f
r
if h ? {0; ?2}.
Symmetry reduction of the Navier–Stokes equations 445

Here D = 2t?t + r?r , ? = 4t2 ?t + 4tr?r ? (r2 + 2(1 ? h)t)f ?f , g(t, r) is an arbitrary
solution of (28).
Theorem 2 is proved by the standard Lie algorithm [4].
Consider case h = const in detail.
Theorem 3. If h = ?2n, n ? N, then any solution of (28) have the form f =
n? ? ? ?
1
f , where f is a solution of the one-dimensional linear heat equation: ft = frr .
r ?r
To prove the theorem 3 one should use the remark 3.
Reducing equation (28) by inequivalent one-dimensional subalgebras of A2 we
construct the following solutions:
by subalgebra ?t + au?u , where a ? {?1; 0; 1}:
f = e?t r? (C1 J? (r) + C2 Y? (r)) if a = ?1,
f = et r? (C1 I? (r) + C2 K? (r)) if a = 1,
f = C1 rh+1 + C2 if h = ?1 and a = 0,
f = C1 ln r + C2 if h = ?1 and a = 0;
here J? , Y? is the Bessel functions of real variable, I? , K? is the Bessel functions of
complex variable, ? = (h + 1)/2;
by subalgebra D + 2au?u , where a ? R:
h?1 h+1 ?
h?1
f = ta e??/8 ? ? a,
W , ,
4
4 4 4
where ? = r2 /t and W (k, m, x) is the general solution of the Whittaker equation
4x2 y = (x2 ? 4kx + 4m2 ? 1)y;
by the subalgebra ?t + ? + au?u , where a ? R:
a
h?1
f = (4t2 + 1) 4 exp ?t? + arctg 2t ?(?),
2
x2
where ? = and ? is a solution of the equation
4t2 +1

4?? + 2(1 ? h)? + (? ? a)? = 0;
if a = 0 then
? ? 1+h
?(?) = ? µ C1 Jµ + C2 Yµ , µ= .
2 2 4
Consider equation (28) when h(t) is an arbitrary function of time.
Theorem 4. Equation (28) is Q-conditionally invariant under the operators
Q = ?t + A(t, r)?r + (B 1 (t, r)u + B 2 (t, r))?u , where
1.
h h h
At ? Ar + 2 A ? Arr + 2Ar A ? 1
+ 2Br = 0,
r r r
hi
Bt + Br ? Brr + 2Ar B i = 0, i = 1, 2;
i i
r
Q = ?r + B(t, r, u)?u , where
2.
h h
Bt ? 2 B + Br ? Brr + 2BBur ? B 2 Buu = 0.
r r
Equation (28) does not have other operators of Q-conditional symmetry.
446 W.I. Fushchych, R.O. Popovych

This theorem is proved by method described in [3].
Therefore, unlike Lie symmetry, Q-conditional symmetry, theorem 4 (28) for arbit-
rary h(t) is rather wide. Thus, in particular, theorem 4 give rise to that equation (28)
is Q-conditional invariant under the operators ?r and G = (2t + C)?r ? rf ?f . By
means of reduction of equation (28) using the operator G we obtain the following
solution
r2
c1 h(t)
f=v exp ? +2 dt ,
4t + 2C 4t + 2C
4t + 2C
and generalizing this we can construct solutions of the form
N
r2
Tk (t)r2k exp ?
f= .
4t + 2C
k=0

Other class of solutions of (28) is given with formula
N
Tk (t)r2k .
f=
k=0

For example, if N = 1 then f = C1 r2 + 2t ? 2 h(t)dt +C2 . We here do not present
results for arbitrary N as they arc very cumbersome.
Consider system 3 (equations (12)–(15)). In this case we get
? h(t) ? 1
?
v1 = ? (29)
+ ,
2
2? ?2

h(t) 2
v? ? v?? +
2 2
(30)
v = 0,
??

h(t) ? 2 3 v 2 ? s(t)
v? ? v?? +
3 3
(31)
v? + = 0,
?2
?
? ?
· 2
? ? (h ? 1)2 v 2 (?, ?) ? s(t))2
2
?2 ? ? 1???
? ? h ln ?2 ? ?
q= d?2 ,(32)
2 3
4 ? 2? 2?2 ?2

where ? = |?(t)|dt, ? = |?(t)|?2 , s(t) = a(t)dt. Formulas (29)–(32) and
ansatz 3 give a solution of the NSEs.
Remark 5. New solutions can be obtained from this solution by force of (5) only
after transformations generated by operators of type Jab , R(m(t)) and Z(a(t)).
Let us to write down system (30)–(31) in the form
?
h(? )
f? ? f?? + (33)
f? = 0,
?
? ??
h(? ) ? 2 f ? s(? )
g? ? g?? + (34)
g? + = 0.
?2
?
Symmetry reduction of the Navier–Stokes equations 447

If (f, g) is a solution of (33)–(34) then (f, g + g 0 ), where function g 0 satisfies the
equation
?
h(? ) ? 2 0
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