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. 119
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Proof of the Theorem 2. Case 1: ux0 = 0. In this case the hodograph transformation
(15) reducing the system (2) with n = 5 to the form
4 4
wz0 ? 0
2
(21)
wza za = 0, wza = 1,
a=1 a=1

is defined.
The general solution of nonlinear complex Eqs. (21) was constructed in [15]. It is
given by one of the following formulas:
4
(22)
(1) w(z) = ?a (?, z0 )za + ?1 (?, z0 ),
a=1

where ? = ? (z0 , . . . , z4 ) is a function determined in implicit way
4
(23)
?a (?, z0 )za + ?2 (?, z0 ) = 0
a=1

and ?a , ?a , ?1 , ?2 ? C 2 (C2 , C1 ) are arbitrary smooth functions satisfying the relations
4 4 4 4
??a
2 2
(24)
?a = 1, ?a ?a = ?a = 0, ?a = 0;
??
a=1 a=1 a=1 a=1
502 R.Z. Zhdanov, I.V. Revenko, W.I. Fushchych

4
(25)
(2) w(z) = ?a (z0 )za + ?1 (z0 ),
a=1

where ?a , ?1 ? C 2 (C1 , C1 ) are arbitrary functions satisfying the relation
4
2
(26)
?a = 1.
a=1

Rewriting the formulas (23), (24) in the initial variables x, u(x), we have
4
(27)
x0 = ?a (?, u)xa + ?1 (?, u),
a=1

where ? = ? (u, x) is a function determined in implicit way
4
(28)
?a (?, u)xa + ?2 (?, u) = 0,
a=1

and the relations (24) hold.
Evidently, the formulas (27), (28) are obtained from (6)–(8) with a particular
choice of functions Aµ , Bµ , C1 , C2
C1 = ??1 ,
A0 = 1, Aa = ?a ,
(29)
C2 = ??2 ,
B0 = 0, Ba = ?a ,
where a = 1, 4.
Next, by force of inequality wz0 ? 0 we get from (22)
4
(30)
(?az0 + ?a? ?z0 )xa + ?1z0 + ?1? ?z0 = 0.
a=1

Differentiation of (23) with respect to z0 yields the following expression for ?z0 :
?1
4 4
?z0 = ? ?az0 xa + ?2z0 ?a? xa + ?2?
a=1 a=1

Substitution of the above result into (30) yields the relation
4 4
?az0 xa + ?1z0 ?a? xa + ?1?
?1
4
a=1 a=1
?a? xa + ?2? = 0.
4 4
a=1
?az0 xa + ?2z0 ?a? xa + ?2?
a=1 a=1

As the direct check shows, the above inequality is equivalent to (9) provided the
conditions (29) hold.
Now we turn to solutions of the system (21) of the form (25). Rewriting the
formulas (25), (26) in the initial variables x, u(x) we get
4 4
2
x0 = ?a (u)xa + ?1 (u), ?a (u) = 1.
a=1 a=1
General solution of the d’Alembert equation with a nonlinear eikonal constraint 503

Making in the equalities obtained the change ?a = Aa A?1 , a = 1, 4, ?1 =
0
?C1 A?1 , we arrive at the formulas (10), (11).
0
Thus, under ux0 ? 0 the general solution of the system (2) is contained in the
class of functions u(x) given by the formulas (6)–(9) or (10), (11).
Case 2: ux0 ? 0, u = const. It is a common knowledge that the system of PDE
(2) is invariant under the generalized Poincar? group P (1, n ? 1) (see, e.g. [16])
e
xµ = ?µ? x? + ?µ , u (x ) = u(x),
where ?µ? , ?µ are arbitrary complex parameters satisfying the relations ??µ ?? ? =
gµ? , µ, ? = 0, n ? 1. Hence, it follows that the transformation
u(x) > u(x ) = u(?µ? x? ) (31)
leaves the set of solutions of the system (2) invariant. Consequently, provided u(x) =
const we can always transform u to such a form that ux0 = 0. Thus, in the case 2 the
general solution is also given by the formulas (6)–(11) within the transformation (31).
Case 3: u = const. Choosing in (10), (11) Aµ = 0, µ = 0, 4, C1 = u = const we
come to the conclusion that this solution is described by the formulas (6)–(11).
Thus, we have proved that, within a transformation from the group P (1, 4) (31),
the general solution of the system of PDE (2) with n = 5 is given by the formulas
(6)–(11). But these formulas are represented in a manifestly covariant form and are
not altered with the transformation (31). Consequently, to complete the proof of the
theorem it is enough to demonstrate that each function u = u(x) determined by the
equalities (6)–(11) is a solution of the system of equations (2).
Differentiating the relations (6), (7) 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 =
? (32)
1
Aµ (B?u x? + C1u ) ? Bµ (A?u x? + C2u ) ,
?xµ =
?
where ? = 0 by force of (9). Consequently,
uxµ uxµ = ??2 Bµ B µ (A?? x? + C1? )2 ? 2Aµ B µ (A?? x? + C1? )(B?? x? + C2? ) +
+ Aµ Aµ (B?? x? + C2? )2 = 0.
Analogously, differentiating (32) with respect to x? and convoluting the expression
obtained with the metric tensor gµ? , we get g µ? uxµ x? ? 25 u = 0.
Next, differentiating (10) with respect to xµ we have

uxµ = ?Aµ (A? x? + C1 )?1 ,
? ? µ = 0, 4,
whence
uxµ x? = ?(Aµ A? + A? Aµ )(A? x? + C1 )?2 + Aµ A? (A? x? + C1 )(A? x? + C1 )?2 .
? ? ? ? ? ? ? ?
504 R.Z. Zhdanov, I.V. Revenko, W.I. Fushchych

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

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

The Theorem 2 is proved.
The Theorem 3 is a direct consequence of the Theorem 2. Really, solutions of the
system of PDE (2) with n = 4 are obtained from solutions of the system of PDE (2)
with n = 5 provided ux4 ? 0. Imposing on functions u(x) determined by the formulas
(6)–(11) a condition ux4 ? 0 we arrive at the following restrictions on the functions
Aµ , Bµ , C1 , C2 :
A4 = 0, B4 = 0
the same as what was to be proved.


4 Applications: reduction of the nonlinear
d’Alembert equation
Following [8, 15, 16], we look for a solution of the nonlinear d’Alembert equation
24 w = F (w), F ? C 1 (R1 , R1 ) (33)
in the form
(34)
w = ?(?1 , ?2 ),
where ?i = ?i (x) ? C 2 (R4 , R1 ) are supposed to be functionally-independent. The
functions ?1 (x), ?2 (x) are determined by the requirement that the substitution of
(34) into (33) yields two-dimensional PDE for a function ? = ?(?1 , ?2 ). As a result,
we obtain an over-determined system of PDE [16]
24 ?1 = f1 (?1 , ?2 ), 24 ?2 = f2 (?1 , ?2 ),
?1xµ ?1xµ = g1 (?1 , ?2 ), ?2xµ ?2xµ = g2 (?1 , ?2 ),
(35)
2 3
??i
?1xµ ?2xµ = g3 (?1 , ?2 ), rank = 2,
?xµ i=1µ=0

and besides, the function ?(?1 , ?2 ) satisfies a two-dimensional PDE,
(36)
g1 ??1 ?1 + g2 ??2 ?2 + 2g3 ??1 ?2 + f1 ??1 + f2 ??2 = F (?).
Consider the following problem: to describe all smooth real functions ?1 (x), ?2 (x)
such that the Ansatz (34) reduces Eq. (33) to an ordinary differential equation (ODE)
with respect to the variable ?1 . It means that one has to put coefficients g2 , g3 , f2
in (36) equal to zero. In other words, it is necessary to construct a general solution
of the system of nonlinear PDE
24 ?1 = f1 (?1 , ?2 ), ?1xµ ?1xµ = g1 (?1 , ?2 ),
(37)
?1xµ ?2xµ = 0, ?2xµ ?2xµ = 0, 24 ?2 = 0.
General solution of the d’Alembert equation with a nonlinear eikonal constraint 505

The above system includes Eqs. (2) as a subsystem. So, the d’Alembert-eikonal
system (2) arises in a natural way when solving the problem of reduction of Eq. (33)
to PDE having a smaller dimension (see, also [15, 17]).
With an appropriate choice of a function G(?1 , ?2 ) the change of variables

v = G(?1 , ?2 ), u = ?2

reduces the system (37) to the form

24 v = f (u, v), (38)
vxµ vxµ = ?,

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

vx0 vx1 vx2 vx3
(40)
rank = 2,
ux0 ux1 ux2 ux3

where ? is a real parameter taking the values ?1, 0, 1.
Before formulating the principal assertion, we will prove an auxiliary lemma.
Lemma 1. 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
µ µ
b2 = 0. (41)
aµ b = bµ b = 0, µ
µ=0

Then, an inequality aµ aµ ? 0 holds.
Proof. It is known that any isotropic non-null vector b in the space M (1, 3) can be
reduced to the form b = (?, ?, 0, 0), ? = 0 by means of a transformation from the
group P (1, 3). Substituting b = (?, ?, 0, 0) into the first equality from (41), we get

?(a0 ? a2 ) = 0 ? a0 = a3 .

Consequently, the vector a has the following components: a0 , a1 , a2 , a0 . That is
why, aµ a µ = a02 ? a12 ? a22 ? a02 = ?(a12 + a22 ) ? 0. As the quadratic form aµ aµ is
invariant with respect to the group P (1, 3), hence it follows that aµ aµ ? 0.
Let us note that aµ aµ = 0 if and only if a2 = a3 , i.e. aµ aµ = 0 if and only if the
vectors a and b are parallel.
Theorem 4. Eqs. (38)–(40) are compatible if and only if
?1
? = ?1, f = ?N v + h(u) (42)
,

where h ? C 1 (R1 , R1 ) is an arbitrary function, N = 0, 1, 2, 3.
Theorem 4. The general solution of the system of Eqs. (38)–(40) being determined
within a transformation from the group P (1, 3) is given by the following formulas:
?1
a) under f = ?3 v + h(u) , ? = ?1
2
= (?A? A? )?1 (Aµ xµ + B)2 +
?? ? ?
v + h(u)
+ (?A? A? )?3 (?µ??? Aµ A? A? x? + C)2 ,
?? ?? (43)
Aµ xµ + B = 0;
506 R.Z. Zhdanov, I.V. Revenko, W.I. Fushchych

?1
b) under f = ?2 v + h(u) , ? = ?1
2
= (?A? A? )?1 (Aµ xµ + B)2 ,
?? ? ? Aµ xµ + B = 0, (44)

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( 122 .)



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