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where ?µ? , ?µ are arbitrary complex parameters satisfying the relations ?µ? ??? =
gµ? , µ, ? = 0, n ? 1. Hence, it follows that, the transformation

(32)
u(x) + u(x ) = u(?µ? x? )

leaves the set of solutions of the system (1) invariant. So when u(x) ? const we can
obtain ux0 ? 0 by using the transformation (32). Consequently, in the case 2 the
general solution is also given by the formulae (7)–(12) up to the transformation (32).
Case 3. u = const. Choosing in (11), (12) Aµ = 0, µ = 0, 4, C1 = u + const we
come to the condition that this solution is described by the formulae (7)–(12).
Thus, we have proved that, up to transformations from the group P (1, 4) (32), the
general solution of the system of PDE (1) with n = 5 is given by the formulae (7)–
(12). But these formulae are not changed with the transformation (32). So to complete
the proof of the theorem it is enough to demonstrate that each function u = u(x),
determined by the equalities (7)–(12), is a solution of the system of equations (1).
Differentiating the relations (7), (8) with respect to xµ , we have

Aµ + ?xµ (A?? x? + C1? ) + uxµ (A?u x? + C1u ) = 0,
B µ + ?xµ (B?? x? + C2? ) + uxµ (B?u x? + C2u ) = 0.

Resolving the above system of linear algebraic equations with respect to uxµ , ?xµ ,
we get
1
(Bµ (A?? x? + C1? ) ? Aµ (B?? x? + C2? )),
uxµ =
? (33)
1
= (Aµ (B?u x? + C1u ) ? Bµ (A?u x? + C2u )),
?xµ
?
where ? = 0 by force of (10). Consequently,

uxµ uxµ = ??2 Bµ Bµ (A?? x? + C1? )2 ?

? 2Aµ Bµ (A?? x? + C1? )(B?? x? + C2? )+ Aµ Aµ (B?? x? + C2? )2 = 0.

Analogously, differentiating (33) with respect to x? and convoluting the obtained
expression with the metric tensor gµ? , we get

g µ? uxµ x? = 25 u = 0.

Further, differentiating (11) with respect to xµ , we have

uxµ = ?Aµ (A? x? + C1 )?1 ,
? ? µ = 0, 4,
On the general solution of the d’Alembert equation 545

whence
uxµ x? = ?(Aµ A? + A? Aµ )(A? x? + C1 )?2 + Aµ A? (A? x? + C1 )(A? x? + C1 )?2 .
? ? ? ? ? ? ? ?

Consequently,

uxµ uxµ = Aµ Aµ (A? x? + C1 )?2 = 0,
? ?
25 u ? ux x = ?(Aµ Aµ )(A? x? + C1 )?2 +
?? ? ?
µ µ

+ Aµ Aµ (A? x? + C1 )(A? x? + C1 )?2 = 0.
? ? ? ?

Theorem 3 is proved.
4. Applications: reduction of the nonlinear wave equation (2). Following [7,
8] we look for a solution of the nonlinear wave equation
24 w = F (w), F ? C 1 (R1 , R1 ) (34)
in the form
(35)
w = ?(w1 , w2 ),
where wi = wi (x) ? C 2 (R4 , R1 ) are functionally-independent. The functions w1 (x),
w2 (x) are determined by the demand that the substitution of (35) into (34) yields two-
dimensional PDE for a function ?(w1 , w2 ). As a result we obtain an over-determined
system of PDE [8]
24 w1 = f1 (w1 , w2 ), 24 w2 = f2 (w1 , w2 ),
(36)
w1xµ w1xµ = g1 (w1 , w2 ), w2xµ w2xµ = g2 (w1 , w2 ),
w1xµ w2xµ = g3 (w1 , w2 ), rank ?wi /?xµ 2 3 = 2
i=1 µ=0

and besides the function ?(w1 , w2 ) satisfies the two-dimensional PDE
(37)
g1 ?w1 w1 + g2 ?w2 w2 + 2g3 ?w1 w2 + f1 ?w1 + f2 ?w2 = F (?).
Let us consider the following problem: to describe all smooth real functions w1 (x),
w2 (x) such that the ansatz (35) reduces Eq. (34) to ordinary differential equation
(ODE) with respect to the variable w1 . It means that one has to put coefficients g2 ,
g3 , f2 in (37) equal to zero. In other words, it is necessary to construct the general
solution of the system of nonlinear PDE
24 w1 = f1 (w1 , w2 ), w1xµ w1xµ = g1 (w1 , w2 ),
(38)
w1xµ w2xµ = 0, w2xµ w2xµ = 0, 24 w2 = 0.
The above system contains Eqs. (1) as a subsystem. So, the d’Alembert–eikonal
system (1) arises in a natural way when solving the problem of reduction of Eq. (34)
be PDE having the smaller dimension (see, also [7, 9]).
Under the appropriate choice of the function G(w1 , w2 ) the change of variables
v = G(w1 , w2 ), u = w2
reduces the system (38) to the form
24 v = f (v, u), (39a)
vxµ vxµ = ?,
546 W.I. Fushchych, R.Z. Zhdanov, I.V. Revenko

24 u = 0, (39b)
vxµ uxµ = 0, uxµ uxµ = 0,

vx0 vx1 vx2 vx3
(39c)
rank = 2,
ux0 ux1 ux2 ux3

where ? is a real parameter taking the values ?1, 0, 1.
Before formulating the principal assertion, we shall prove an auxiliary lemma.
Lemma. Let a = (a0 , a1 , a2 , a3 ), b = (b0 , b1 , b2 , b3 ) be four-vectors defined in the real
Minkowski space M (1, 3). Suppose they satisfy the relations
3
b2 = 0. (40)
aµ bµ = bµ bµ = 0, µ
µ=0

Then the inequality aµ aµ ? 0 holds.
Proof. It is known that any isotropic vector b in the space M (1, 3) can be reduced to
the form b = (?, ?, 0, 0), ? = 0 by means of transformations from the group P (1, 3).
Substituting b = (?, ?, 0, 0) into the first equality from (40), we get
?(a0 ? a3 ) = 0 <> a0 = a3 .
Consequently, the vector a has the following component: a0 , a1 , a2 , a0 . That is
why aµ aµ = a2 ? a2 ? a2 ? a ? 02 = ?(a2 + a ? 22 ) ? 0.
0 1 2 1
Let us note that aµ aµ = 0 iff a2 = a3 , i.e. aµ aµ = 0 iff the vectors a and b are
parallel.
Theorem 4. Eqs. (39a-c) are compatible iff
f = ?N (v + h(u))?1 ,
? = ?1, (41)
where h ? C 1 (R1 , R1 ) is an arbitrary function, N = 0, 1, 2, 3.
Theorem 5. The general solution of the system of Eqs. (39a-c) being determined up
to the transformation from the group P (1, 3) is given by the following formulae:
a) under f = ?3(v + h(u))?1 , ? = ?1
(v + h(u))2 = ?(A? A? )?1 (Aµ xµ + B)2 +
?? ? ?
+ (A? A? )?3 (Eµ??? Aµ A? A? x? + C)2 ,
?? ?? (42)
Aµ (u)xµ + B(u) = 0;
b) under f = ?2(v + h(u))?1 , ? = ?1
(v + h(u))2 = ?(A? A? )?1 (Aµ xµ + B),
?? ? ? (43a)
Aµ xµ + B = 0,
where Aµ (u), B(u), C(u) are arbitrary smooth functions satisfying the relations
?? (43b)
Aµ Aµ = 0, Aµ Aµ = 0;
c) under under f = ?(v + h(u))?1 , ? = ?1
u = C0 (x0 ? x3 ),
(44)
(v + h(x0 ? x3 ))2 = (x1 + C1 (x0 ? x3 ))2 + (x2 + C2 (x0 ? x3 ))2 ,
where C0 , C1 , C2 are arbitrary smooth functions;
On the general solution of the d’Alembert equation 547

d) under f = 0, ? = ?1

v = (?A? A? )?3/2 Eµ??? Aµ A? A? x? + C,
?? ?? (45)
1) Aµ xµ + B = 0,

where Aµ (u), B(u), C(u) are arbitrary smooth functions satisfying the relations
(43b);

u = C0 (x0 ? x3 ), (46)
2)

v = x1 cos C1 (x0 ? x3 ) + x2 sin C1 (x0 ? x3 ) + C2 (x0 ? x3 ), (47)

where C0 , C1 , C2 are arbitrary smooth functions.
In the above formulae (42), (43a), (45) we denote by the symbol Eµ??? the
components of antisymmetrical fourth-order tensor, i.e.
?
? 1, (µ, ?, ?, ?) = cycle (0, 1, 2, 3),
?
?1, (µ, ?, ?, ?) = cycle (1, 0, 2, 3), (48)
Eµ??? =
?
?
0, in the remaining cases.

Proof of Theorems 4, 5. By force of (39c) u ? const. Consequently, up to
transformations from the group P (1, 3) ux0 ? 0. That is why one can apply to
Eqs. (39) the hodograph transformation

z0 = u(x), za = xa , a = 1, 3,
(49)
w(z) = x0 , v = v(z0 , za ).

As a result the system (39a,b) reads
3 3
2
(50a)
wza = 1, wza za = 0,
a=1 a=1

3
(50b)
vza wza = 0,
a=1

3 3
?1
= ??, (vza za + 2wz0 vza wza z0 ) = ?f (v, z0 ).
2
(50c)
vza
a=1 a=1

Since v(z) is a real-valued function, ? = ?1 or ? = 0.
Case 1. ? = ?1. As it is shown in the Section 1, the general solution of the system
(50a) in the class of real-valued functions w(z) is given by the formulae (19), (20)
with n = 4. On substituting (19) into (50b), we obtain the linear first-order PDE
3
(51)
?a (z0 )vza = 0,
a=1

the general solution of which is represented in the form

(52)
v = v(z0 , ?1 , ?2 ).
548 W.I. Fushchych, R.Z. Zhdanov, I.V. Revenko

In (52)
?1/2
3 3
2
z0 , ?1 = ?a ?a za + ? ,
?
a=1 a=1
?1/2
3 3
?2
?2 = ?a Eabc za ?b ?c
?
a=1 a,b,c=1

3
2
are first integrals of Eq. (51) and what is more ?a = 0 (the case ?a = const,
a=1
a = 1, 3 will be considered separately).
Substitution of the expression (52) into (50c) yields the system of two PDE for
a function v = v(z0 , ?1 , ?2 )

v?1 ?2 + v?2 ?2 + 2??1 v?1 = ?f (v, z0 ), (53a)
1

2 2
(53b)
v?1 + v?2 = 1.

Let us exclude function f (v, z0 ) from (53) by considering of the third-order diffe-
rential consequence of (53)

v?2 (v?1 ?1 + v?2 ?2 + 2??1 v?1 )?1 ? v?1 (v?1 ?1 + v?2 ?2 + 2??1 v?1 )?2 = 0, (54a)
1 1

2 2
(54b)
v?1 + v?2 = 1.

Further we shall consider the cases v?2 ?2 = 0 and v?2 ?2 = 0 separately.
A. v?2 ?2 = 0. Then

(55)
v = g1 (z0 , ?1 )?2 + g2 (z0 , ?1 ),

where g1 , g2 ? C 2 (R2 , R1 ) are arbitrary functions.
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