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. 99
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CFu ? (Cu ? 2B1 )F = C0 + C11 + 2CB1 .
Hereafter subscript, mean differentiation with respect to the corresponding vari-
ables (x0 , x1 , u).
Case 2. A = 0, B = 0
CFu ? Cu F = C0 + C11 + 2CC1u + C 2 Cuu . (40)
Having constructed the general solutions of nonlinear systems (39), (40), we shall
obtain the general operator of conditional symmetry of equation.
Theorem 7 [19]. The equation (36) is Q-conditionally-invariant under the operator
(38) (A = 1, Bu = 0) iff it, is locally equivalent to the equation
u0 + u11 = b3 u3 + b1 u + b0 , (41)
b0 , b1 , b3 = const,
422 W.I. Fushchych

the operator (38) having the form
3 3
b 3 u 3 + b1 u ? b0 ? u . (42)
Q = ?0 + 2b3 u?1 +
2 2
The equation (41) is reduced to one of the following canonical equations:
u0 + u11 = ?u(u2 ? 1), (43)

u0 + u11 = ?(u3 ? 3u + 2), (44)

u0 + u11 = ?u3 , (45)

u0 + u11 = ?u(u2 + 1). (46)

Ans?tzes constructed by means of the operator (42) have the from
a
v
?(?) = 2 tan?1 u + 2?x1 , ? = ? ln(1 ? u?2 ) + 2?x0 , (47)
v
4 u+2 2
? (u ? 1)?1 ? 2?x1 ,
?(?) = ? ln
9 u?1 3
(48)
2 u+2 2
? (u ? 1)?1 ? 3?x0 ,
? = ln
9 u?1 3
v
?(?) = ?2u?1 + 2?x1 , ? = ?u?2 ? 3?x0 , (49)
v
?(?) = 2 tan?1 u ? 2?x1 , ? = ? ln(1 + u?2 ) ? 3?x0 . (50)

The anz?tzes (47)–(50) reduce the equations (43)–(46) to ODE:
a
2? = (?2 ? 1)?, 2? = ?3 ? 3? + 2, (51)
? ? ? ? ? ?

2? = ?3 , 2? = ?(?2 + 1). (52)
? ? ? ??

It is evident from the above equations that ans?tzes generated by the operator
a
of conditional invariance (42) change essentially their nonlinearities in second parts.
This fact allows to integrate the ODE (51), (52) in elementary functions

?(?) = ?2 tan?1 (53)
C1 exp ? + 1 + C2 ,

3 3
ln C1 ? (? + 2?) = ln C2 ? (? ? ?), (54)
2 2

?(?) = 2 C1 ? ? + C2 , (55)

?(?) = 2 tan?1 C1 exp ? ? 1 + C2 , (56)

where C1 , C2 are constants.
Thus, substitute (53)–(56) into (47)–(50), we get families of exact solutions of the
equations (43)–(46). These solutions cannot be obtained within the framework of the
Lie method.
Conditional symmetry of equations of nonlinear mathematical physics 423

Theorem 8 [20]. The equation (37) is Q-conditionally-invariant under the operator
(38) with A = 1 if functions B, C satisfy the following system of equations:
(57)
uCuu = 2(BBu + uBu1 ), Buu = 0,

B0 + uB11 ? CBu?1 ? 2uCu1 + 2BB1 ? 2CBu = 0, (58)

C0 + uC11 ? C 2 u?1 + 2CB1 = 0. (59)

Solving equations (57)–(59), we get an explicit form of the operator (38)
(60)
Q = b1 Q1 + b2 Q2 + b3 D1 + b4 D2 + b5 ?0 + b6 ?1 ,

Q1 = x1 ?0 + u?1 , Q2 = x2 ?0 + 2x1 u?1 + 2u2 ?u ,
0
(61)
D1 = 2x0 ?0 + x1 ?1 , D2 = x1 ?1 + 2u?u , bi = const, i = 1, 6.

Theorem 9 [20]. The equation (37) is Q-conditionally-invariant under the operator
(62)
Q = ?1 + C(x, u)?u ,
if C(x, u) satisfies the condition
C0 + u(C11 + 2CC1u + C 2 Cuu ) + CC1 + C 2 Cu = 0. (63)
Partial solutions or the equation (63) give rise to explicit form of operators of
conditional symmetry. Below we adduce some of them
v
v
(64)
Q3 = x0 ?1 + 2u?u ,
v
(65)
Q4 = 2x0 ?1 + R(u)?u ,

(66)
Q5 = ?1 + ln u?u ,

(67)
Q6 = x0 ?1 + x1 ?u ,

where R(u) a solution of ODE uR + R = R?1 .
? ?
Let us adduce some ans?tzes generated by operators Q1 , Q2 , Q3
a
1
x0 u ? x2 = ?(u), (68)
21
2ux0 u
? x1 = ? (69)
,
x1 x1
2
1 x1
v + ?(x0 ) (70)
u= .
2 x0

Reduced equations have the form
for the ansatz (68),
?(u) = 0
?
u
for the ansatz (69),
?
? =0
x1
2x0 ?(x0 ) + ?(x0 ) = 0 for the ansatz (70).
?
424 W.I. Fushchych

Thus the ans?tzes (68)–(70) reduce nonlinear heat equations to linear ODE.
a
6. An equation of the Korteweg-de Vries type. Let us consider a non-linear
equation [23]
u0 + F (u)uk + u111 = 0, (71)
1
3
u111 = ? u , k is an arbitrary real parameter. With F (u) = u, k = 1 (71) coincides
?x3
with the classical KdV equation.
Theorem 10 [23]. The equation (71) is Q-conditionally invariant under the following
Galilei-type operator:
Q = xr ?1 + H(x, u)?u , (72)
0

r is an arbitrary real parameter, if
2?k 1?k
1) F (u) = ?1 u + ?2 u ,
u 2

?1/k (73)
k?1 1/2
H(x, u) = u ,
2
F (u) = (?1 ln u)1?k ,
2)
(74)
H(x, u) = (k?1 )?1/k u,
1?k
F (u) = (?1 arcsin u + ?2 )(1 ? u2 )
3) ,
2
(75)
H(x, u) = (k?1 )?1/k (1 ? u2 )1/2 ,
1?k
F (u) = (?1 Arsh u + ?2 )(1 ? u2 )
4) ,
2
(76)
H(x, u) = (k?1 )?1/k (1 + u2 )1/2 ,
5) F (u) = ?1 u,
(77)
H(x, u) = (k?1 )?1/k ,
where r = k ?1 , k = 0, ?1 , ?2 are arbitrary constants.
By means of operators of conditional invariance (72) we reduce the equation (71)
to ODE and construct the following exact solutions:
2
?1/k
x1 k?1 x0 ?2
?1/k
?
u= + ?x0 ,
2 2 ?
when F (u) is of the form (73);
k(k?1 )?3/k ? k +1 ?2
3
?1/k
+ (k?1 x0 )?1/k x1 ?
u = exp ? x0 + ?x0 ,
k?2 ?
when k = 2, F (u) is of the form (74); when k = 2
?2
?1/2 ?1/2
u = exp ?(2?1 )?3/2 x0 + (2?1 x0 )?1/2 x1 ?
ln x0 + ?x0 ;
?
k(k?1 )?3/k ? k +1 ?2
3
?1/k
+ (k?1 x0 )?1/k x1 ?
u = sin x0 + ?x0 , k = 2,
k?2 ?
ln x0 ?2
?1/2
u = sin (2?1 )?3/2 v + ?x0 + (2?1 x0 )?1/2 x1 ? , k = 2,
x0 ?
Conditional symmetry of equations of nonlinear mathematical physics 425

when F (u) is of the form (75);
k(k?1 )?3/k ? k +1
3
?1/k
+ (k?1 x0 )?1/2 x1 ,
u = sh ? x0 + ?x0 k = 2,
k?2
?1/2 ?1/2
u = sh ?(2?1 )?3/2 x0 + (2?1 x0 )?1/2 x1 ,
ln x0 + ?x0 k = 2,
when F (u) is of the form (76). In all formulae ? is an arbitrary parameter.
Thus, having investigated the conditional symmetry of the equation (71), we
construct nontrivial classes of exact solutions.
7. Nonlinear wave equation. An equation of the form
u00 ? (F (u)u1 )1 = 0 (78)
is widely used for description of nonlinear wave processes. The group properties of the
equation (78) were investigated in detail by means of Lie method in [24]. Depending
on explicit form of the function F (u) the equation (78) has wide conditional symmetry.
Theorem 11 [25]. The equation (78) is Q-conditionally invariant under the operator
Q = A(x, u)?0 + B(x, u)?1 + H(x, u)?u ,
if functions A(x, u), B(x, u), H(x, u), F (u) satisfy the following systems of equati-
ons:
Case 1: A = 1, D = F ? B 2
(Bu D?1 )u = 0, F (H1 D?1 )1 ? (H0 D?1 )0 ? H 2 = 0,
(Hu D?1 )u ? H(H0 D?1 )u ? H(Hu D?1 )0 +
+ D2 {2F (B0 D1 ? B1 H0 + H[Bu H1 ? B1 Hu ]) ? BHH1 F } = 0,
?
D2 Huu + D{(H F )u + 2B(Bu Hu ? Buu H) ? 2F B1u ? 2B0u } ?
?
? HD2 + 2BB0 Du + 2BB1 (BF ? 2Bu F ) = 0,
D{B00 + 2(B0 H)u ? 2(BH0u ? Bu H0 ) + 2(H1 F )u ? B11 F + Buu H 2 +
+ 2BHHuu } ? Du {B0 H + Bu H 2 + 2BHHu } +
?
+ B{B1 H F + 2B 2 + 2B0 Bu H + 4BB0 Hu + 4B1 Hu F ? 2B 2 F } = 0.
0 1

Case 2: A = 1, B = F 1/2
?
1) BH + 2BHu = 0, H0 + HHu ? BH1 = 0;
?
2) BH + 2BHu = 0, H0 + HHu ? BH1 = 0;
? ?
[BH 2 + 2B(BH1 + HHu ) + 2B(H0u + HHuu + BH1u )] = (H0 + HHu ?
? ?
? HH1 ) ? [H00 + H 2 Huu ? B 2 H11 + 2HH0u ? 2BHH1 ](BH + 2BHu ) = 0.
Case 3: A = 0, B = 1
? ?

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. 99
( 135 .)



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