ñòð. 131 |

where ?, T are arbitrary functions (not equal simultaneously to zero).

Thus to find exact solutions of (5) in the form (6) it is sufficient to solve the

system (8) and

T (?)? + ?(?)? = F (?, ?? ),

? ?

(9)

T (?)?? + ?(?)?? = F ? (?, ?? ).

? ?

To solve the system (8) we use the results of Collins [2], where similar systems for

the functions of three independent variables were investigated. The partial solutions of

the system (8) when µ = 0, 1, . . . , n, T (?) = 1, ?(?) = N (? ? A)?1 , N = 0, 1, . . . , n,

A = const, are given in Table 1. Evidently when n > 2 they are not general solutions.

Table 1

N Solutions Conditions on parameters

? + ?y + F (?y) (?y = ?0 y0 ? F is an arbitrary function of ?y, y? = x? + a? ,

0

a? = const, ?2 = 1, ? 2 = ?? = 0

??1 y1 ? · · · ? ?n yn )

? ? A = [(?i y)(?i y)]1/2

1, . . . , n ?µ ?µ = ? ij , i, j = 1, . . . , N + 1, y? = x? + a? ,

ij

a? = const

Below we consider systems of the form (5)

2u = ?u(uu? )k ,

(10)

2u? = ?? u? (uu? )k ,

2u = (?1 u + i?2 u? )(u2 + u?2 )k ,

(11)

2u = (?1 u? ? i?2 u? )(u2 + u?2 )k

(?1 , ?2 , k are arbitrary real numbers and ? is an arbitrary complex number), which

are invariant corresponding with respect to the operators Q1 = u?u ? u? ?u? and

Q2 = u? ?u ? u?u? (the operator of charge).

The system (9) with T (?) = 1, ?(?) = N/(? ? A) and ? from Table 1, for (10)

takes the form

N

? + ? = ??(??? )k ,

??

??A

(12)

N

?? + ?? = ?? ?? (??? )k ,

? ?

??A

The symmetry and exact solutions of the non-linear d’Alembert equation 567

where N = 0, 1, . . . , n; for (11) the system (9) takes the form

N

? + ? = (?1 ? + i?2 ?? )(?2 + ??2 )k ,

??

??A

(13)

N

?? + ?? = (?1 ?? ? i?2 ?)(?2 + ??2 )k .

? ?

??A

It is convenient to search for solutions of (12) in the form

?? = re?i?

? = rei? , (14)

for solutions of (13) in the form

1 + i ? 1 ? i ?? 1 ? i ? 1 + i ??

?? = r

ve+ ve ve+ ve (15)

?=r , .

22 22 22 22

For the real functions r = r(?) and ? = ?(?) we obtain the system

N ?

r + r + ? ?2 r = ?1 r2k+1 ,

??

??A

(16)

N?

?? ?

?r + 2r? + r? = ?2 r2k+1 .

??A

Here ? = ?1 + i?2 , ? = ?1 for (12) and ? = 1 for (13).

Let N = 0. With an arbitrary k and ?2 = 0 the system (16) has the general

solution in the parametrical form (? = ?1 ):

?1/2

? 2k+2 ?c2

+ 21 + c2

?= r dr + c3 ,

k+1 r

(17)

?1/2

?c2

? 2k+2

r?2 1

? = c1 r + + c2 dr + c4 ,

2

k+1 r

When k = ?2 we obtain the general solution of (16) in explicit form in elementary

functions

1/2

? ? ?c21

2

r = c2 (? + c3 ) + ,

c2

?

c1 ?1 c2 (? + c3 )

?

? ? ? ?c2 > 0,

? (? ? ?c2 )1/2 tan (? ? ?c2 )1/2 + c4 ,

? 1

?

? 1 1

? (18)

?

? + c3 + c?1 (?c2 ? ?)1/2

c1

, ? ? ?c2 < 0,

1

2

?= ln

? 2(?c ?1

2 ? ?)1/2 1

? ? + c3 ? c2 (?c12 ? ?)1/2

?

? 1

?

?

?? c1

? ? = ?c2 ,

, 1

(c2 (? + c3 )

c1 = 0, c2 , c3 , c4 are arbitrary real numbers. (If c1 = 0, ? = const.)

Note 1. The solvability of systems of ordinary differential equations in quadratures

is connected with their wide symmetry. Systems of the form (12) can be reduced to

systems of four first-order equations and we may suppose that for their solvability in

quadratures it is necessary for the range of basis of their invariance algebra to be

568 W.I. Fushchych, I.A. Yehorchenko

not less than 4 [7]. However, this condition is not sufficient. The system (12) when

k = ?2, N = 0 has the maximal invariance algebra among systems of such form with

the basis operators

1

??? + (??? + ?? ??? ), ? 2 ?? + ?(??? + ?? ??? ), ??? ? ?? ???

?? ,

2

but when ?2 = 0 it reduces to a Riccati equation not solvable in quadratures.

The system (16) when N = 1, N = 2, k = (N ? 2)(N ? 1)?1 by the substitution

t = (? ? A)N ?1 , r = (? ? A)1?N ? can be reduced to the form

? ?? ?

? + ? ?2 ? = ?1 ?2k+1 , 2?? + ?? = ?2 ?2k+1 .

?

We obtain its solutions in parametrical form (?2 = 0) and from them we obtain

the solutions of (16)

?1/2

? 2k+2 ?c2

r = ? c3 + (N ? 1) + 21 + c2

? d? ,

k+1 ?

?1/2

? 2k+2 ?c2

?2

? = c1 (N ? 1) + 21 + c2

2

(19)

? ? d? + c4 ,

k+1 ?

1/(N ?1)

?1/2

? 2k+2 ?c2

? = c3 + (N ? 1) + 21 + c2

? d? + A,

k+1 ?

c1 = 0, c2 , c3 , c4 are arbitrary constants chosen for r, ? to be real.

From solutions (17)–(19) and substitutions (14) and (15) we obtain the solutions of

(12) and (13) respectively. With ? from Table 1 we get solutions of the initial systems

(10) and (11).

As (10) and (11) are invariant with respect to the scale transformations it is

possible to find ansatze reducing them to the first-order differential equations which

have more chances to be solved in quadratures. We search for such ansatze in the

form

u? = f (x)?? (?). (20)

u = f (x)?(?),

The corresponding conditions on f and ? are as follows:

2f (x) = F (?)f 2k+1 (x),

f 2? + 2fµ ?µ = G(?)f 2k+1 (x), (21)

?µ ?µ = 0,

where F , G are arbitrary functions.

It is interesting enough to investigate the system (21) itself but here we do not go

into this matter and adduce only some solutions:

?x

f (x) = [(? i x)(? i x)]a , ? = ,

[(? i x)(? i x)]b

(22)

1

a = ? , ? = ?? = 0, ?µ ?µ = ? , b = 0, 1,

2 i ii ij

2k

the sum by i is implied, i = 1, . . . , m, m ? n, 1 ? 2a = m, when b = 1.

The symmetry and exact solutions of the non-linear d’Alembert equation 569

For the ansatz (20) with f , ? from (22) we obtain the reduced equations and their

solutions.

For equations (10)

(i) b = 0, m + 2a ? 1 = 0

m + 2a ? 1 ?

?(??? )k ,

??+? =

2 4a

?1/2k

? = c ? k(m+2a?1) ? c1 ?

(23)

m + 2a ? 1

i?2 1

? exp ? ln ? k(m+2a?1) ? c1

(ln ?) ,

?1 2 2k

1/k

1

?

2a(m + 2a ? 1)

cc = ;

?1

(ii) b = 0, m + 2a ? 1 = 0

? = c(?1 k 2 ln ? + c1 )?(?1 +i?2 )/2k?1 , cc? = 1; (24)

(iii) b = 1

?

?(??? )k ,

? ? ? a? =

2(m + 2a ? 1)

(25)

1/k

2a(m + 2a ? 1)

? = c? ?i?2 /2k?1 |1 ? c1 ?|??2 /2k?1 , cc? = ? .

?1

In a similar way solutions of (11) can be obtained; if ? has the form (15) then

(i) b = 0, m + 2a ? 1 = 0

?1/2k

[(m + 2a ? 1)4a]1/2k (?1 + ?)?1/2k ,

r = 1 ? c1 ? k(m+2a?1)

(26)

?1 ? ?2 m + 2a ? 1 1

ln ? ? ln 1 ? c1 ? k(m+2a?1)

?= ;

?1 + ?2 2 2k

(ii) b = 0, m + 2a ? 1 = 0

r = (2k 2 (?1 + ?2 ) ln ? + c1 )?1/2k ,

(27)

1 ?1 ? ?2

?=? ln |k 2 (?1 + ?2 ) ln ? + c1 |;

2k ?1 + ?2

(iii) b = 1, m + 2a ? 1 = 0

1/2k

a(m + 2a ? 1)

(1 ? c1 ?)?1/2k ,

?

r=

?1 + ?2

(28)

1 ?1 ? ?2

?=? (ln ? + ln |1 ? c1 ?|).

2k ?1 + ?2

Substituting the obtained solutions (23)–(28) of the reduced equations into the

ansatz (20) and (22) we get the multiparametrical families of exact solutions of (10)

and (11) correspondingly.

570 W.I. Fushchych, I.A. Yehorchenko

The ansatz (20) and (22) when b = 0, b = 1 allows us to obtain the reduced

ñòð. 131 |