ñòð. 120 |

where Aµ = Aµ (u), B = B(u), C = C(u) are arbitrary smooth functions satisfying

the relations

??

Aµ Aµ = 0, Aµ Aµ = 0, (45)

?1

c) under f = ? v + h(u) , ? = ?1

2 2 2

v + h(x0 ? x3 ) = x1 + C1 (x0 ? x3 ) + x2 + C2 (x0 ? x3 ) ,

(46)

u = C0 (x0 ? x3 ),

where C0 , C1 , C2 are arbitrary smooth functions;

d) under f = 0, ? = ?1

v = (?A? A? )?3/2 ?µ??? Aµ A? A? x? + C,

?? ?? Aµ xµ + B = 0, (47)

(1)

where Aµ = Aµ (u), B = B(u), C = C(u) are arbitrary smooth functions satisfying

the relations (45);

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

(2)

(48)

u = C0 (x0 ? x3 ),

where C0 , C1 , C2 are arbitrary smooth functions.

In the above formulas (43), (47) we denote by ?µ??? the completely anti-symmet-

ric fourth-order tensor (the Levi-Civita tensor), i.e.

?

? 1, (µ, ?, ?, ?) = cycle (0, 1, 2, 3),

?

?µ??? = ?1, (µ, ?, ?, ?) = cycle (1, 0, 2, 3),

?

?

0, in the remaining cases.

Proof of the Theorems 4, 5. By force of (40) u = const. Consequently, within

a transformation from the group P (1, 3) ux0 = 0. That is why, one can apply to

Eqs. (38)–(40) the hodograph transformation

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

As a result, the system (38), (39) reads

3 3

2

(49)

wza = 1, wza za = 0,

a=1 a=1

3

(50)

vza wza = 0,

a=1

3 3

?1

= ??, (vza za + 2wz0 vza wza z0 ) = ?f (v, z0 ).

2

(51)

vza

a=1 a=1

General solution of the d’Alembert equation with a nonlinear eikonal constraint 507

As v(z) is a real-valued function, ? ? 0. Scaling, if necessary, the function v we

can put ? = ?1 or ? = 0.

Case 1: ? = ?1. As it is shown in the Section 2, the general solution of the system

(49) in the class of real-valued functions w(z) is given by the formulas (18), (19) with

n = 4. Substituting (18) into (50), we obtain a first-order linear PDE

3

(52)

?a (z0 )vza = 0,

a=1

whose general solution is represented in the form

(53)

v = v(z0 , ?1 , ?2 ).

In (53),

?1/2

3 3

?2

z0 , ?1 = ?a ?a za + ? ,

? ?

a=1 a=1

?3/2

3 3

?2

?2 = ?a ?abc za ?b ?c

?

a=1 a,b,c=1

3

?2

are the first integrals of Eq. (52) and what is more, ?a = 0 (the case ?a = const,

a=1

a = 1, 3 will be treated separately), ?abc is the third-order anti-symmetric tensor with

?123 = 1.

Substitution the expression (53) into (51) yields the system of two PDE for

a function v = v(z0 , ?1 , ?2 )

v?1 ?1 + v?2 ?2 + 2??1 v?1 = ?f (v, z0 ), (54)

1

2 2

(55)

v?1 + v?2 = 1.

To get rid of an arbitrary element (function) f (v, z0 ) from (54) we consider instead

of system (54), (55) its differential consequence

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 )?1 = 0, (56)

1 1

2 2

(57)

v?1 + v?2 = 1,

that is obtained by differentiating the first equation with respect to ?1 , ?2 , multiplying

the expressions obtained by v?2 and ?v?1 , respectively, and summing.

Further, we will consider the subcases v?2 ?2 = 0 and v?2 ?2 = 0 separately.

Subcase 1.A: v?2 ?2 = 0. Then,

(58)

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

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

Substituting (58) into (57) and splitting an equality obtained by the powers of ?2 ,

we have

g1 + (g2?2 )2 = 1,

2

g1?1 = 0,

508 R.Z. Zhdanov, I.V. Revenko, W.I. Fushchych

whence

v = ??1 ± 1 ? ?2 ?2 ? h(z0 ). (59)

Here ? ? R1 , h is an arbitrary smooth function. v

Inserting (59) into (56) we get an algebraic equation ? 1 ? ?2 = 0, whence

? = 0, ±1.

Finally, substitution of (59) into (54) yields the equation for f (v, z0 )

2???1 = ?f ??1 ± 1 ? ?2 ?2 ? h(z0 ), z0 . (60)

1

From Eq. (60) it follows that, under ? = 0,

v = ±?2 ? h(z0 ) (61)

f = 0,

and under ? = ±1,

?1

f = ?2 v + h(z0 ) v = ±?1 ? h(z0 ). (62)

,

Subcase 1.B: v?2 ?2 = 0. In this case one can apply to Eqs. (56), (57) the Euler–

Amp?re transformation

e

v?1 = ?Gy1 ,

z0 = y 0 , ?1 = y1 , ?2 = Gy2 , v + G = ?2 y2 , v?2 = y2 ,

?1 ?1

v?2 ?2 = (Gy2 y2 ) , v?1 ?2 = ?Gy1 y2 (Gy2 y2 ) (63)

,

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.

Being rewritten in the new variables y, G(y) the Eq. (57), becomes linear

Gy1 = ± 1 ? y2 ,

2

whence

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

2 (64)

Making in the Eq. (56) the change of variables (63) and inserting the expression

(64), we transform it as follows

?2 ?2

y2 ? (1 ? y2 )3/2 Hy2 y2 3y2 Hy2 y2 + (y2 ? 1)Hy2 y2 y2 + 2y1 y2 Hy2 y2 = 0.(65)

2 2

Splitting (65) by the powers of y1 and integrating the equations obtained, we get

H = h1 (y0 )y2 + h2 (y0 ).

Substituting the above result into (64) and returning to the initial variables z0 , ?1 ,

?2 , v(z0 , ?1 , ?2 ) we obtain the general solution of the system of PDE (56), (57)

1/2

v + h2 (z0 ) = ± [?2 ? h1 (z0 )]2 + ?2 (66)

.

1

At last, inserting (66) into the equation (54), we arrive at the conclusion that the

function f is determined by the formula (42) with N = 3.

General solution of the d’Alembert equation with a nonlinear eikonal constraint 509

2 2 2

If ?a = const, a = 1, 3, then the equality ?1 + ?2 + ?3 = 1 holds. Applying, if

necessary, a transformation from the group P (1, 3) one can put ?1 = ?2 = 0, ?3 = 1,

i.e. u = C0 (x0 ? x3 ), C0 ? C 2 (R1 , R2 ).

As a consequence of Eqs. (39) we get v = v(?, x1 , x2 ), where ? = x0 ? x3 , and

what is more, Eqs. (38) take the form

vx1 x1 + vx2 x2 = ?f v, C0 (?) .

2 2

(67)

vx1 + vx2 = 1,

It is known [15, 18] that Eqs. (67) are compatible if and only if f = 0 or f =

?(v + h(u))?1 , h ? C 1 (R1 , R1 ). And besides, the general solution of (67) is given by

the formulas (48) and (46), 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, it follows

from (51) that v = v(z0 ). Consequently, an equality v = v(u) holds that breaks the

condition (40) which means that under ? = 0 the system (38)–(40) is incompatible.

Thus, we have proved that the system of nonlinear PDE (38)–(40) is compatible if

and only if the relations (42) hold and that its general solution is given by one of the

formulas (46), (48), (61), (62), and (66). To complete the proof, one has to rewrite

the expressions (61), (62), (66) in the manifestly covariant from (43), (44), (47).

Consider, as an example, the formula (62)

?1/2

3 3

v = ±?1 ? h(z0 ) ? ± ? h(u),

?2 (68)

?a (u) xa ?a (u) + ?(u)

? ?

a=1 a=1

the function u(x) being determined by the formula (20),

3 3

2

(69)

?a (u)xa + ?(u) = x0 , ?a (u) = 1.

a=1 a=1

Let us make in (68), (69) a substitution ?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 )2

? ?

v=± ?

0 0

a=1

3

xa (Aa A?1 ? Aa A0 A?2 ) + B A0 A?2 ? BA?1

? ? ? ?

? ? h(u) =

0 0 0 0

a=1

3

(A2 A?2 + A2 A2 A?4 ? 2Aa Aa A0 A?3 )?1/2

? ? ? ?

=± ?

a0 a00 0

a=1

3

xa (Aa A?1 ? Aa A0 A?2 ) + B A0 A?2 ? BA0

? ? ? ?

? ? h(u) =

0 0 0

a=1

?1/2

= ± ?Aµ Aµ A?2 ? Aµ Aµ A2 A?4 + 2Aµ Aµ A0 A?3

?? ? ? ? ?

00

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