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Using the subgroup structure of the group P (1, 2) in papers [1, 2] some classes
of exact solutions of equation (1) in the space R1,2 were built. The analogous results
in the space R1,3 were obtained in [3, 4]. The generalization of results for the n-
dimensional case was considered in [5, 6]. In order to find exact solutions, symmetry
ansatzes reducing equation (1) to ordinary differential equations were applied in above
mentioned papers.
In the present paper in order to build exact solutions of equation (1), symmetry
ansatzes reducing equation (1) to equations of two invariant variables are used. We are
interested in these ansatzes because a reduced equation often has additional symmet-
ries. This fact permits to apply these ansatzes for finding new solutions of the present
equation. Let us cite as an example the ansatz u = u(x0 ? xn , x1 , . . . , xn?1 ) which
was considered in [6]. The corresponding reduced equation has the infinite group of
invariance. Note that this ansatz is built by one-dimensional subalgebra P0 + Pn .
In the present paper the series of ansatzes of such a kind as u = u(?1 , ?2 ), where
?1 = x0 ? xm , ?2 = x2 ? x2 ? · · · ? x2 , 2 ? m ? n, is considered. These ansatzes
m
0 1
are built by the subalgebras AE1 [1, m ? 1] ? AE[m + 1, n], where AE1 [1, m ? 1] =
G1 , . . . , Gm?1 , J12 , . . . , Jm?2,m?1 , AE[m + 1, n] = Pm+1 , . . . , Pn , Jm+1,m+2 , . . .,
Jn?1,n , Ga = J0a ? Jam , a = 1, . . . , m ? 1, and if m = n we think AE[m + 1, n] = 0.
The ansatz u = u(?1 , ?2 ) reduces equation (1) to the equation

4?1 u12 + 4?2 u22 + 2(m + 1)u2 + ?uk = 0. (2)

Let us investigate symmetry of the equation (2).
Äîïîâiäi ÍÀÍ Óêðà¿íè, 1995, ¹ 2, P. 33–37.
366 W.I. Fushchych, A.F. Barannyk, Yu.D. Moskalenko

Theorem 1. The maximal algebra of invariance of equation (2) in the case of k = 0,
m+1
m?1 and m > 1 in the Lie sense is the 4-dimensional Lie algebra A(4) which is
generated by such operators:
? ? 1 ? ? 1 ?
? ?
X 1 = ?1 + ?2 u , X2 = ? 2 u,
k ? 1 ?u k ? 1 ?u
?w1 ??2 ?w2
m?1 ? (m ? 1)(k ? 1)
? ?
? ? 1.
l
M = ?1 ? 1 + ?2 u , l=
?w1 ??2 2 ?u 2
Theorem 2. The maximal algebra of invariance of equation (2) in the case of
m+1
k = m?1 and m > 1 in the sense of Lie is the 4-dimensional Lie algebra B(4) which
is generated by such operators:
m?1
? ? ?
?
S = ?1 ln ?1 + ?2 ln ?1 (ln ?1 + 1)u ,
??1 ??2 2 ?u
m?1 ?
? ?
?
Z1 = ?1 + ?2 u,
??1 ??2 2 ?u
m?1 ?
? ?
?
Z2 = ?2 u , Z3 = ?1 .
??2 2 ?u ??2
Let us consider two cases.
m+1
1. The case k = m?1 . Classify one-dimensional subalgebras of the algebra A(4)
with respect to G-conjugation, where G = exp A(4). Ansatzes, built by these sub-
algebras, reduce the equation (2) to ordinary differential equations. Note that the
operators of the algebra A(4) satisfy the following commutation relations: [X1 , X2 ] =
0, [X1 , X3 ] = 0, [X1 , M ] = lM , [X2 , X3 ] = ?X3 , [X2 , M ] = 0, [X3 , M ] = 0.
Theorem 3. Let K be one-dimensional subalgebra of the algebra A(4). Then K is
conjugated with one of the following algebras: 1) K1 = X1 + ?X2 ; 2) K2 = X2 ;
3) K3 = X1 + ?X3 (? = ±1); 4) K4 = X3 ; 5) K5 = M + ?X2 (? = 0, ±1);
6) K6 = M + ?X3 (? = ±1).
The following ansatzes correspond to the subalgebras K1 –K6 of the theorem 3:
?+1
???1
K1 : u = ? 1?k ?(?), ? = ?2 ? 1 ;
1
1?k
K2 : u = ?2 ?(?), ? = ?1 ;
1
?2
? ? ln ?1 ;
1?k
K3 : u = ?1 ?(?), ? =
?1
K4 : u = ?(?), ? = ?1 ;
? ?l ?2
1
1
K5 : u = (?1 ?2 ) 1?k ?(?), ? = ?1 + ln ;
l ?1
?2 ? ?l
1?m
K6 : u = ?1 2 ?(?), ? = + ?1 .
?1 l
These ansatzes reduce equation (2) to ordinary differential equations with an
unknown function ?(?):
4(l ? ?k)
K1 : ?4?? ? + ? + ??k = 0;
? ?
k?1
4? 4l
K2 : ? ?? ? + ??k = 0;
?
k?1 (k ? 1)2
4l
K3 : ?4?? + ? + ??k = 0;
? ?
k?1
Exact solutions of the multidimensional nonlinear d’Alembert equation 367

K4 : ??k = 0;
4?
K5 : ?4?? + ? + ??k = 0;
? ?
k?1
K6 : ?4?? + ??k = 0.
?

The equation corresponding to the subalgebra K1 , in case ? = 0 has the solution
?(k ? 1)2
1?k
? = (? + C).
4l
In consequence we obtain the following solution of equation (1)
?(k ? 1)2
1?k
(3)
u = (?2 + C?1 ).
4l
2l
If ? = k+1 , then the equation corresponding to the subalgebra K1 assumes
8l? 4l
? ?? ? + ??k = 0.
? ?
k+1 k+1
The particular solution of this equation is
?(k ? 1)2 1
?1?k = (? 2 + C)2 .
4l
Therefore, equation (1) has the following solution:
2
?(k ? 1)2 k?1
1
2(k+1)
1?k 2
(4)
u = ?2 + C?1 .
4l
l(k+1)
If ? = , then the equation corresponding to the subalgebra K1 assumes
2

?2l(k + 1)?? ? 2(k + 2)? + ??k = 0.
? ?
This equation has the solution
?(k ? 1)2 2
1 k?1
1?k
? 2 + Cw 2(k+1)
? = .
4l
Therefore, equation (1) has the following solution
2
?(k ? 1)2 l(k+1)+2 k?1
1
2(k+1) 2(k+1)
1?k 2
(5)
u = ?2 + C?1 ?2 .
4l
The equations corresponding to the subalgebras K5 and K6 have such solutions:
?(k ? 1)2 ?(k ? 1)2
1?k
(1 + C? l ), 1?k
(? + C)2 .
? = ? =
4l 8?(k + 1)
Therefore, equation (1) has the following solutions:
?(k ? 1)2
u1?k = l
(6)
?2 (1 + C?1 ),
4l
?(k ? 1)2 l?1 2
? 1?l
u1?k = (7)
?1 ?2 + ?1 + C?1 .
8?(k + 1) l
368 W.I. Fushchych, A.F. Barannyk, Yu.D. Moskalenko

m+1
2. The case k = m?1 . The basis elements of the algebra B(4) satisfy the following
commutation relations: [S, Z1 ] = ?Z1 , [S, Z2 ] = 0, [S, Z3 ] = 0, [Z1 , Z2 ] = 0, [Z1 , Z3 ] =
0, [Z2 , Z3 ] = ?Z3 .
Theorem 4. Let L be one-dimensional subalgebra of the algebra B(4). Then L is
conjugated with one of the following algebras: 1) L1 = Z1 + ?Z2 (? = 0, ±1);
2) L2 = Z2 ; 3) L3 = Z1 + ?Z3 (? = ±1); 4) L4 = Z3 ); 5) L5 = S + ?Z2 ;
6) L6 = S + ?Z3 (? = ±1).
The following ansatzes correspond to the subalgebras L1 –L6 of theorem 4:
1?m
?d?1
2
L1 : u = ?2 ?(?), ? = ?2 ? 1 ;
1?m
2
L2 : u = ?2 ?(?), ? = ?1 ;
?2
1?m
? ? ln ?1 ;
u = ?1 2 ?(?), ? =
L3 :
?1
L4 : u = ?(?), ? = ?1 ;
?1 ln? ?1
1?m
?+1
L5 : u = (?1 ln ?1 ) 2 ?(?), ?= ;
?2
?2
1?m
? ? ln(ln ?1 ).
L6 : u = (?1 ln ?1 ) 2 ?(?), ? =
?1
These ansatzes reduce equation (2) to ordinary differential equations with an
unknown function ?(?):
m+1
L1 : ?4?? 2 ? + 2?(m ? 3)? ? + ?? m?1 = 0;
? ?
m+1
L2 : ?2(m ? 1)? ? + ?? m?1 = 0;
?
m+1
L3 : ?4?? + ?? m?1 = 0;
?
m+1
L4 : ?? m?1 = 0;
m+1
L5 : ?4?? 3 ? + 2((m ? 3)? + m ? 1)? 2 ? + ?? m?1 = 0;
? ?
m+1
L6 : ?4?? ? 2(m ? 1)? + ?? m?1 = 0.
? ?
The equation corresponding to the subalgebra L2 has the solution
?
2
? 1?m = ? (ln ? + C).
(m ? 1)2
Therefore, the equation (2) has the following solution
?
2
u 1?m = ? (8)
(?2 ln ?1 + C?2 ).
(m ? 1)2
The equation corresponding to the subalgebra L3 has the particular solution
?
2
? 1?m = ? (? + C)2 .
4?m(m ? 1)
Therefore, equation (2) has the following solution
?
2
u 1?m = ? (?2 ? ??1 ln ?1 + C?1 )2 . (9)
4?m(m ? 1)?1
Exact solutions of the multidimensional nonlinear d’Alembert equation 369

1?m
In the case of an equation corresponding to the subalgebra L5 ? = 0 or ? = m?3
(m = 3) we obtain the equations:
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