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??
3? ?(4) ? ?(3) = e,
64
??
4? ?(4) ? 2 + 4?(3) ? + 2?(2) = e,
16
5? ?(4) = ?e? ,
6? 0 = ?e? .
426 W.I. Fushchych, R.Z. Zhdanov, O.V. Roman

Equation 5? has the partial solution
24
(? + c)?4 ,
? = ln ? > 0,
?
that leads us to the following exact solutions of equation (6):
24
(x0 + c1 )?4 ,
u = ln ? > 0,
?
(29)
24
(x1 + c2 )?4 ,
u = ln ? > 0.
?
Here c, c1 , c2 are arbitrary constants. This solutions are invariant under the operators
P0 and P1 accordingly.
To finish the Section let us note that the solutions (29) can be obtained by making
use of the ansatz in Liouville form [4]:
? ?
2
? ?
? 24 ?1 (?1 )?2 (?2 ) ?
? ?
, ?1 = x0 + x1 , ?2 = x0 ? x1 ,
u = ln
? ? (?1 (?1 ) + ?2 (?2 ))4 ?
? ?

which reduces equation (6) to one of the following systems:
1. ?1 = 0, ?2 = 0;
? ?
2?2 2?2
?1 ?2
2. ?1 =
? , ?2 =
? .
?1 ?2
Here ? and ? mean the first derivative and the second one of the corresponding
? ?
argument.
Finding the general solution of the systems we get the following exact solutions
of equation (6):
(a2 ? b2 )2
24
(30)
u = ln ,
? (ax0 + bx1 + c)4
where a, b, c are arbitrary constants.
Solution (30) can be obtained from (29) by the transformations of the extended
Poincar? group with the generators (24) and (25).
e


4 Reduction and exact solutions
of the equation 22 u = ?uk
It follows from Theorem 2 that when n = 1 the equation (5) is invariant under the
extended Poincar? group P (1, 1) with the generators (24) and
e
? ? 4 ?
(31)
D = x0 + x1 + u.
1 ? k ?u
?x0 ?x1
If an equation admits the symmetry operator
? ?
X = ? µ (x) (32)
+ ?(x)u
?xµ ?u
Symmetry properties, reduction and exact solutions of biwave equations 427

then its solutions can be found in the form [4]:

(33)
u(x) = f (x)?(?)

provided functions ?(x) and f (x) satisfy the following system:
?? ?f (x)
? µ (x) ? µ (x) (34)
= 0, = ?(x)f (x).
?xµ ?xµ
With an allowance for invariance of equation (5) under the changes of variables
(28) we write P (1, 1)-nonequivalent ansatzes of the form (33) in Table 2.
Table 2.
N Algebra Invariant variables ? Ansatz
2
1? D ? J01 x0 + x1 u = (x0 ? x1 ) 1?k ?(?)
4
??1
2? D + ?J01 , ? = ?1 (x0 ? x1 )(x0 + x1 ) ?+1 u = (x0 + x1 ) (1?k)(?+1) ?(?)
(x0 + x1 + 1 )?
2
3? 4
D + J01 + P0 u = exp (x1 ? x0 ) ?(?)
? exp 2(x1 ? x0 ) k?1


4? x2 ? x2 u = ?(?)
J01 0 1
?
5 u = ?(?)
P0 x1
?
6 P0 + P1 x0 + x1 u = ?(?)

Let us note that analogous ansatzes were obtained in [4] for the nonlinear wave
equation

2u = ?uk . (35)

Substituting the ansatzes obtained to equation (5) we get the following equations
for the function ?(?):
1 + k (2) ?k
1? ?= ?,
(1 ? k)2 32
3k + 1
2? (? ? 1)2 ?(4) ? 2 + 2(? ? 1)(? + 1)2 + 2? ??(3) +
1?k
6? ? 10 8 ?
+ 2 ?2 ? 4? + 3 + ?(2) = (? + 1)2 ?k ,
+
1?k (1 ? k)2 16
5k ? 1 (3) 4k 2 ?k
3? ?(4) ? 2 + ?(2) =
? ?+ ?,
k?1 (1 ? k)2 64
?k
4? ?(4) ? 2 + 4?(3) ? + 2?(2) = ?,
16
5? ?(4) = ??k ,
6? ??k = 0.

Equations 1? , 2? , 4? have the partial solutions of the form:
1
64 (k + 1)2 k?1
2
? ? k?1 , k = ?1,
?=
? (k ? 1)4
428 W.I. Fushchych, R.Z. Zhdanov, O.V. Roman

and equation 5? has the partial solution of the form
1
8 (k + 1)(k + 3)(3k + 1) 1
k?1
4
? ? k?1 , k = ?1, ?3, ?
?=
(k ? 1)4
? 3
which lead us to the following solutions of equation (5):
1
2
? k?1
64 (k + 1)2 k?1

(x0 + x1 + c1 )(x0 ? x1 + c2 ) k = ?1,
u= ,
? (k ? 1)4
1
8 (k + 1)(k + 3)(3k + 1) 1
k?1
4
k = ?1, ?3, ? ,
u= (x0 + c3 ) 1?k ,
(k ? 1)4
? 3
1
8 (k + 1)(k + 3)(3k + 1) 1
k?1
4
k = ?1, ?3, ? ,
u= (x1 + c4 ) 1?k ,
(k ? 1)4
? 3
where c1 , c2 , c3 , c4 are arbitrary constants.
Note that equation (35) has analogous solutions (see [4]).


5 Reduction and exact solutions
of the equation 22 u = ?u?3
It follows from Theorems 2 and 3 that when n = 1 the equation
22 u = ?u?3 (36)
is invariant under the conformal group C(1, 1) with the generators (24) and
? ? ?
D(1) = x0 + x1 +u ,
?x0 ?x1 ?u
(37)
?
= 2xµ D(1) ? (x? x? )
(1)
Kµ , µ, ? = 0, 1.
?xµ
By analogy with the previous Section solutions of equation (36) can be found in
the form (33) where functions ?(x) and f (x) are the solutions of the system (34)
provided the operator (32) belongs to the invariance algebra of equation (36).
To obtain all the C(1, 1)–nonequivalent ansatzes we use the one-dimensional
nonequivalent subalgebras of the conformal algebra AC(1, 1) adduced in [9].
Omitting rather cumbersome computations and taking account of equation (36)
being invariant under the changes of variables (28) we write nonequivalent ansatzes
in Table 3.
We omit subalgebras not containing conformal the operator (37) since they were
considered in the previous Section.
Substituting ansatzes obtained in (36) we get the following equations for the
function ?(?):
? ?3
1? ?(4) + 2?(2) + ? = ?;
16
? ?3
2? (?2 ? 1)2 ?(4) + 2(?2 + 1)?(2) + ? = ?;
16
Symmetry properties, reduction and exact solutions of biwave equations 429

Table 3.
N Algebra Invariant variables ? Ansatz
1/2
u = (x0 ?x1 )2 + 1 ?
arctg(x1 ? x0 )+
(1)
1? P 0 + K0 1/2
+arctg(x1 + x0 )
? (x0 +x1 )2 + 1 ?(?)
1/2
u = (x0 ?x1 )2 + 1 ?
(1) (1)
? P1 ) (? ? 1)arctg(x0 ? x1 )+
P0 + +
K0 ?(K1
2? 1/2
+(? + 1)arctg(x1 + x0 )
0<?<1 ? (x0 +x1 )2 + 1 ?(?)
1/2
(x0 ? x1 )2 + 1
u= ?
(1) (1)
?
3 P0 + + ? P1 x0 + x1
K0 K1
??(?)
1/2
x0 + x1 +
(x0 ? x1 )2 + 1
u= ?
(1) (1)
?
1 1 + x0 ? x1
4 2P1 + +
K0 K1
+ ln ??(?)
2 1 ? x0 + x1
1/2
x0 + x1 + (x0 ? x1 )2 + 1
u= ?
(1) (1)
?
5 2P1 ? ?
K0 K1
+arctg(x0 ? x1 ) ??(?)
1/2
u = (x0 ? x1 )2 + 1 ?

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