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Substituting (55) into (54b) and splitting the obtained quality by the powers of ?2 ,
we have

g1 + (g2?2 )2 = 1,
2
g1?1 = 0,

whence

v = ??1 ± 1 ? ?2 ?2 ? h(z0 ). (56)

Here ? ? R1 is an arbitrary smooth function. v
Substituting (56) into (53a), we get an algebraic equation ? 1 ? ?2 = 0, whence
? = 0, ±1.
Finally, substitution of (56) into (53a) yields an equation for f (v, z0 )

2???1 = ?f ??1 ± 1 ? ?2 ?2 ? h(z0 ), z0 . (57)
1

From Eq. (57) it follows that under ? = 0

v = ±?2 ? h(z0 ) (58)
f = 0,
On the general solution of the d’Alembert equation 549

and under ? = ±1
f = ?2(v + h(z0 ))?1 , v = ±?1 ? h(z0 ). (59)
B. v?2 ?2 = 0. In such a case one can apply the Euler transformation to Eqs. (54)
v?1 = ?Gy1 ,
z0 = y0 , ?1 = y1 , ?2 = Gy2 , v + G = ?2 y2 , v?2 = y2 ,
v?2 ?2 = (Gy2 y2 )?1 , v?1 ?2 = ?Gy1 y2 (Gy2 y2 )?1 , (60)
v?1 ?1 = (G21 y2 ? Gy1 y1 Gy2 y2 )(Gy2 y2 )?1 .
y

Here y0 , y1 , y2 are new independent variables, G = G(y0 , y1 , y2 ) is a new function.
In the new variables y, G(y) the equation (54b) is linearized

Gy1 = ± 1 ? y2 ,
2

whence

G = ±y1 1 ? y2 + H(y0 , y2 ), H ? C 2 (R2 , R1 ).
2 (61)

The equation (54a) after the change of variables (60) and substitution of the
formula (61) takes the form
?2
y1 ? (1 ? y2 )3/2 Hy2 y2 3y2 Hy2 y2 + (y2 ? 1)Hy2 y2 y2 + 2y1 Hy2 y2 = 0.
2 2 2
(62)

Splitting (62) by the powers of y1 and integrating the obtained equations, we get
H = h1 (y0 )y2 + h2 (y0 ).
Substituting the above result into (61) and returning to the initial variables z0 , ?1 ,
?2 , v(z0 , ?1 , ?2 ), we have the general solution of the system of PDE (54)
1/2
v + h2 (z0 ) = ± (?2 ? h1 (z0 ))2 + ?2 (63)
.
1

At last, substituting (63) into the equation (53a), we come to conclusion that the
function f is determined by the formula
f (v, z0 ) = ?3(v + h2 (z0 ))?1 .
2 2 2
Let, us consider now the case ?a = const, a = 1, 3. Then the equality ?1 +?2 +?3 =
1 holds. That is why, using transformations from the group P (1, 3), one can obtain
?1 = ?2 = 0, ?3 = 1, i.e. u = C0 (x0 ? x3 ), C0 ? C 2 (R1 , R1 ). Then, from Eqs. (39b) it
follows that v = v(?, x1 , x2 ), ? = x0 ? x3 and what is more Eqs. (39a) take the form
vx1 x1 + vx2 x2 = ?f (v, C0 (?)).
2 2
(64)
vx1 + vx2 = 1,
It, is known [7, 10] that Eqs. (64) are compatible iff f = 0 or f = ?(v + h(?))?1 ,
h ? C 1 (R1 , R1 ). And besides the general solution of (64) is given by the formulae
(46) and (44) respectively.
Thus we have completely investigated the case ? = ?1.
Case 2. ? = 0. By force of the fact that the function v is a real one, from (50b)
it follows that v = v(z0 ). Consequently, the equality v = v(u) holds that breaks the
condition (39c). So under ? = 0 the system (39a-c) is incompatible.
550 W.I. Fushchych, R.Z. Zhdanov, I.V. Revenko

So, we have proved that the system of nonlinear PDE (39a-c) is compatible iff the
relations (41) hold and its general solution is given by one of the formulae (44), (46),
(58), (59), (63). To complete the proof, one has to rewrite the expressions (58), (59),
(63) in the manifestly covariant form (42), (43a), (45).
Let us consider as an example the formula (59)

?1/2
3 3
v = ±?1 ? h(z0 ) = ± ? h(u),
?2 (65)
?a xa ?a (u) + ?(u)
? ?
a=1 a=1

the function u(x) being determined by the formula (12)

3 3
2
(66)
?a (u)xa + ?(u) = 0, ?a (u) = 1.
a=1 a=1

Let us make in (65), (66) the change ?a = Aa A?1 , ? = ?BA?1 , whence
0 0

Aµ (u)xµ + B(u) = 0, Aµ Aµ = 0,
?1/2
3
Aa A?1 ? Aa A0 A?2
? ?
h(u) + v = ± ?
0 0
a=1
3
xa (Aa A?1 ? Aa A0 A?2 ) + B A0 A?2 ? BA?1 =
? ? ? ?
? 0 0 0 0
a=1
3
(A2 A?2 + A2 A2 A?1 ? 2Aa Aa A0 A?3 )?1/2 ?
?a ? ? ?
=± a00
0 0
a=1
3
xa (Aa A?1 ? Aa A0 A?2 ) + B A0 A?2 ? BA0 =
? ? ? ?
? 0 0 0
a=1
?1/2
= ± ?Aµ Aµ A?2 ? Aµ Aµ A2 A?4 + 2Aµ Aµ A0 A?3
?? ?0 ? ? ?
0 0 0

? ?A?1 (xµ Aµ + B) + A?2 A0 (xµ Aµ + B) =
? ? ?
0 0

= ?(?Aµ Aµ )?1/2 (xµ Aµ + B).
?? ? ?

??
The only thing left is to prove that Aµ Aµ < 0. Since Aµ Aµ = 0, the equality
? ?
Aµ Aµ = 0 holds. Consequently, by force of the lemma ?Aµ Aµ ? 0 and what is
?? ?
more the equality Aµ Aµ = 0 holds iff Aµ = k(u)Aµ . The general solution of the
? ?
above system of ordinary differential equations reads Aµ = k(u)?µ , where k(u) is an
?1 ?1
arbitrary function, ?µ ? R , ?µ ?µ = 0. Whence it follows that ?a = Aa A0 = ?a ?0 =
1
3
?2
const and the condition ?a = 0 does not hold. We come to the contradiction
a=1
??
whence it follows that Aµ Aµ < 0.
Thus we have obtained the formula (43a). Derivation of the remaining formulae
from (42), (45) is carried out in the same way. The theorems are proved.
On the general solution of the d’Alembert equation 551

Substitution of the above obtained results into the formula w = ?(v, u) yields the
following collection of ans?tzes for the nonlinear wave equation (34)
a

w = ? ? h(u) ± (?A? A? )?1 (Aµ xµ + B)2 ?
?? ? ?
1.
1/2
? (A? A? )?3 (Eµ??? Aµ A? A? x? + C(u))2
?? ?? ,u ;
?? ? ?
w = ? ? h(u) ± (?A? A? )1/2 (Aµ xµ + B), u ;
2.

w = ? h(x0 ? x3 ) ± ([x1 + C1 (x0 ? x3 )]2 + (67)
3.

+ [x2 + C2 (x0 ? x3 )]2 )1/2 , x0 ? x3 ;

w = ? (?A? A? )?3/2 (Eµ??? Aµ A? A? x? + C(u)), u ;
?? ??
4.

w = ? x1 cos C1 (x0 ? x3 ) + x2 sin C1 (x0 ? x3 ) + C2 (x0 ? x3 ), x0 ? x3 .
5.

Here u = u(x) is determined by JSSF (8) with n = 4.
Substitution of the expressions (67) into (34) gives the following equations for
? = ?(u, v):
1. ?vv + 3(v + h(u))?1 ?v = ?F (?), (68)
2. ?vv + 2(v + h(u))?1 ?v = ?F (?), (69)
3. ?vv + (v + h(u))?1 ?v = ?F (?),
4. ?vv = ?F (?), (70)
5. ?vv = F (?).
Eqs. 4, 5 from (68)–(70) are known to be integrable in quadratures. Therefore,
any solution of the d’Alembert–eikonal system (1) corresponds to some class of exact
solutions of the nonlinear wave equation (34) that contains arbitrary functions. Saying
it in another way, the formulae (67) make it possible to construct wide families
of exact solutions of the nonlinear PDE (34) using exact solutions of the linear
d’Alembert equation 24 u = 0 satisfying the additional constraint uxµ uxµ = 0.
It is interesting to compare our approach to the problem of reduction of Eq. (34)
with classical Lie approach. In the framework of the Lie approach the functions w1 (x),
w2 (x) from (35) are looked for as invariants of the symmetry group of the equation
under study (in the case involved it is the Poincar? group P (1, 3)). Since the group
e
P (1, 3) is a finite-parameter group, its invariants cannot contain an arbitrary function
(complete description of invariants of the group P (1, 3) had been carried out in [11]).
So the ans?tzes (67) cannot be obtained by means of Lie symmetry of the PDE (34).
a
The ans?tzes (67) correspond to conditional invariance of the nonlinear wave
a
equation (34). It means that there exist two differential operators Qa = ?aµ (x)?xµ ,
a = 1, 2 such that
Qa w ? Qa ?(w1 , w2 ) = 0, a = 1, 2
and besides the system of PDE
24 w ? F (w) = 0
Qa w = 0, a = 1, 2,
552 W.I. Fushchych, R.Z. Zhdanov, I.V. Revenko

is invariant in Lie’s sense under the one-parameter groups having generators Q1 , Q2
(on the conditional invariance of mathematical and theoretical physics equations see
[8, 12, 13]).
It is worth noting that the ans?tzes 2, 5 from (67) were obtained in [14] without
a
using the concept of conditional invariance.
5. On the new exact, solutions of the nonlinear wave equation. The general
solution of Eqs. (70) is given by the following quadrature [15]:
?1/2
?(u,v) ?
? (71)
v + D(u) = F (z)dz + C(u) d?,
0 0

where D(u), C(u) ? C 2 (R1 , R1 ) are arbitrary functions.
Substituting into (71) expressions for u(x), v(x) given by the formulae 4, 5 from
(67), we get two classes of exact solutions of the nonlinear wave equation (34) that
contain several arbitrary functions of one variable.
Eqs. (68), (69) are Emden–Fauler type equations. They were investigated by many
authors see, e.g. [15]). In particular, it is known that the equations
?vv + 2v ?1 ?v = ???5 , (72)

?vv + 3v ?1 ?v = ???3 (73)

are integrated in quadratures. In the paper [11] it had been established that Eqs. (72),
(73) possess the Painleve property. This fact made it possible to integrate them
by applying rather complicated technique. We shall demonstrate how to integrate
Eqs. (72), (73) by using their symmetry properties.
It occurs that Eq. (72) admits the symmetry operator Q = 2v?v ? ??? . Follo-
wing [15] we find the change of the variables
? = z(? )v ?1/2 , ? = ln v
that reduce the operator Q to the form Q = ?? . Eq. (72) in the new variables reads
1
z ? ?z 5 ,
z? ? =
4
whence
12 ?6 1
z ? z + D(u),
2
(74)
z? =
4 3 4
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