ñòð. 48 |

By assumption, one of the coe?cients ? 2 , ? 1 , ? 2 is not equal to zero. Without loss

of generality, we can suppose that ? 2 = 0 (if this is not the case, we make a change

x > u, u > x or x > v, v > x). Performing the transformation

t = t + F 1 (x, u, v), x = F 2 (x, u, v), u = G1 (x, u, v), v = G2 (x, u, v),

where the functions F 1 , F 2 are solutions of PDEs

P1 F 2 + ? 1 = 0, P1 F 2 = 1

and G1 , G2 are functionally independent ?rst integrals of the PDE P1 F = 0, we

reduce the operators P0 , P1 to become P0 = ?t , P1 = ?x .

Next, as the operator M commutes with P0 , P1 , its coe?cients do not depend on

t, x. Consequently, it has the form

M = ? 1 (u, v)?t + ? 2 (u, v)?x + ? 1 (u, v)?u + ? 2 (u, v)?v .

Suppose ?rst that (? 1 )2 + (? 2 )2 = 0. Then, the change of variables

t = t + F 1 (u, v), x = x + F 2 (u, v), u = G1 (u, v), v = G2 (u, v),

where F 1 , F 2 , G1 are solutions of PDEs

M F 1 + ? 1 = 0, M F 2 + ? 2 = 0, M G1 = 1

and G2 is a ?rst integral of the PDE M F = 0, reduces the operators P0 , P1 , M to

the form (8).

If ? 1 = 0, ? 2 = 0, then formulae (9) are obtained.

Case 2. rank R = 1. If we make transformation (3) reducing the operator P0 to

the form P0 = ?t , then the operator P1 becomes P1 = ?(x, u, v)?t (the function ? does

not depend on t because P0 and P1 commute). As ? = const (otherwise the operators

P0 and P1 are linearly dependent), making the change of variables

t = t, x = ?(x, u, v), u = u, v =v

transforms the operator P1 to be P1 = x?t .

It follows from the commutation relations [P0 , M ] = 0, [P1 , M ] = 0 that

?

M = ?(x, u, v)?t + ? 1 (x, u, v)?u + ? 2 (x, u, v)?v .

? ?

? ?

Subcase 2.1. ? 1 = ? 2 = 0. Provided the equalities ?u = ?v = 0 hold, formulae

? ?

? ?

(11) are obtained. If (?u )2 + (?v )2 = 0, then making the transformation

?

t = t, x = x, u = ?(x, u, v), v =v

we arrive at formulae (12).

Subcase 2.2. (?1 )2 + (?2 )2 = 0. Performing the change of variables

? ?

u = G1 (x, u, v), v = G2 (x, u, v),

t = t + F (x, u, v), x = x,

214 R.Z. Zhdanov, W.I. Fushchych

where F , G1 , G2 satisfy PDEs

? M G1 = 2, M G2 = 0,

M F + ? = 0,

we rewrite the operators P0 , P1 , M in the form (10). The lemma is proved.

Theorem 1. Inequivalent representations of the Galilei algebra by LVFs (1) are ex-

hausted by those given in (6) and by the following ones:

1. P0 = ?t , P1 = ?x , M = 2u?x ,

(13)

G = (t + xu)?x + u2 ?u ;

v

P1 = ?x , M = 2 u?t ± x ?u2 ? 2u ?x ,

2. P0 = ?t ,

v v (14)

G = xu?t + t ± x ?u2 ? 2u ?x ± u ?u2 ? 2u ?u ;

3. P0 = ?t , P1 = ?x , M = 2(u?t + v?x ),

(15)

G = xu?t + (t + xv)?x + uv?u + (u + v 2 )?v ;

M = ?? 1 ±

2

1 + ?x2 ?t ,

4. P0 = ?t , P1 = x?t ,

(16)

1±

2 1

?x2

G = tx?t + x + 1+ ?x + ??u ;

?

5. P0 = ?t , P1 = x?t , M = 2u?t ,

(17)

G = tx?t + (x2 ? u)?x + xu?u ,

where ? is an arbitrary real parameter, ? = 0, 1.

Proof. To prove the theorem it su?ces to solve the commutation relations for the

basis operators P0 , P1 , M , G of the Galilei algebra in the class of LVF (1) within

di?eomorphisms (3). All inequivalent realizations of the three-dimensional commuta-

tive algebra having the basis operators P0 , P1 , M are given by formulae (8)–(12). What

is left is to solve the commutation relations for the generator of Galilei transformations

G = ? 1 (t, x, u, v)?t + ? 2 (t, x, u, v)?x + ? 1 (t, x, u, v)?u + ? 2 (t, x, u, v)?v

[P1 , G] = 1 M, (18)

[P0 , G] = P1 , [M, G] = 0

2

for each set of operators P0 , P1 , M listed in (8)–(12). Since case (8) has been studied

in detail in [8] and shown to yield representations (5), we will restrict ourselves to

considering cases (9)–(12).

Case 1. Operators P0 , P1 , M have the form (9). It is easy to establish that, using

transformations (3), it is possible to reduce the operator M from (9) to one of the

forms

M = 2(??t + u?x ), M = 2(u?t + ?(u)?x ), M = 2(u?t + v?x ),

where ? is an arbitrary smooth function and ? is an arbitrary real constant.

Subcase 1.1. M = 2(??t + u?x ). Inserting the formulae for P0 , P1 , M into (18)

and equating the coe?cients of linearly independent operators ?t , ?x , ?u , ?v yield the

following over-determined system of PDEs for coe?cients of the operator G:

1 2 1 2 1 2

?t = 0, ?t = 1, ?t = 0, ?t = 0, ?x = ?, ?x = u,

??t + u?x = 0, ??t + u?x ? ? 1 = 0

1 2 1 1 2 2

?x = 0, ?x = 0,

On new representations of Galilei groups 215

As a compatibility condition of the above system, we get ? = 0 and what is more

? 1 = F 1 (u, v), ? 2 = t + xu + F 2 (u, v), ? 1 = u2 , ? 2 = F 3 (u, v),

where F 1 , F 2 , F 3 are arbitrary smooth functions.

Making the change of variables

(19)

t = t + T (u, v), x = x + X(u, v), u = u, v = V (u, v),

where t, X, V are solutions of the system of PDEs

u2 Tu + F 3 Tv + F 1 = 0, u2 Tu + F 3 Tv + F 1 = 0, u2 Vu + F 3 Vv = 0.

we transform the operator G to become

G = (t + xu)?x + u2 ?u ,

thus getting formulae (13).

Subcase 1.2. M = 2(u?t + ?(u)?x ). Substituting the expressions for P0 , P1 ,

M into (18) and equating the coe?cients of the linearly-independent operators ?t ,

?x , ?u , ?v give the following over-determined system of PDEs for coe?cients of the

operator G:

1 2 1 2 1 2

?t = 0, ?t = 1, ?t = 0, ?t = 0, ?x = u, ?x = ?(u),

?

u?t + ?(u)?x ? ? 1 = 0, u?t + ?(u)?x ? ?(u)? 1 = 0.

1 2 1 1 2 2

?x = 0, ?x = 0,

The general solution of the above system reads

? 1 = xu + F 1 (u, v), ? 2 = t + x?(u) + F 2 (u, v),

? 1 = u?(u), ? 2 = F 3 (u, v),

where

?(u) = ± ?u2 ? 2u,

F 1 , F 2 , F 3 are arbitrary smooth functions and ? is an arbitrary real parameter.

Performing, if necessary, the change of variables (19), we can put the functions

F , F 2 , F 3 equal to zero. Thus, the operator G is of the form

1

G = xu?t + t ± ?u2 ? 2u ?x ± u ?u2 ? 2u ?u

and we arrive at representation (14).

Subcase 1.3 M = 2(u?t +v?x ). With this choice of M , the commutation relations

(18) give the following system of PDEs for coe?cients of the operator G:

1 2 1 2 1 2 1 2

?t = 0, ?t = 1, ?t = 0, ?t = 0, ?x = u, ?x = v, ?x = 0, ?x = 0,

u?t + v?x ? ? 1 = 0, u?t + v?x ? ? 2 = 0,

1 1 2 2

which general solution reads

? 1 = xu + F 1 (u, v), ? 2 = t + xv + F 2 (u, v), ? 1 = uv, ?2 = u + v2 .

Here F 1 , F 2 are arbitrary smooth functions.

216 R.Z. Zhdanov, W.I. Fushchych

Making the transformation (19) with V ? v, we reduce the operator G to the form

G = xu?t + (t + xv)?x + uv?u + (u + v 2 )?v ,

thus getting representation (15).

Case 2. Operators P0 , P1 , M have the form (10). An easy check shows that

the system of PDEs obtained by substitution of P0 , P1 , M from (10) into (18) is

incompatible.

Case 3. Operators P0 , P1 , M have the form (11). In this case, the commutation

relations (18) give rise to the following system of PDEs for the coe?cients of the

operator G:

x?t ? ? 2 = ?(x),

1 2 1 2 1

?(x)?t + ?(x)? 2 = 0.

1

?t = x, ?t = 0, ?t = 0, ?t = 0, ?

Solving it, we have

? 2 = x2 ? ?(x),

? 1 = xu + F 1 (x, u, v), ? 1 = F 2 (x, u, v), ? 2 = F 3 (x, u, v),

where

?(x) = ? ? 1 ±

1

1 + ?x2 ,

F 1 , F 2 , F 3 are arbitrary smooth functions and ? is an arbitrary real constant.

Making the change of variables

(20)

t = t + T (x, u, v), x = x, u = U (x, u, v), v = V (x, u, v)

transforms the operator G as follows

1±

G = tx?t + x2 + 1

1 + ?x2 + ??u , ? = 0, 1.

?

Consequently, representation (16) is obtained.

Case 4. Operators P0 , P1 , M have the form (12). Inserting these into commutation

relations (18) we get the system of PDEs for coe?cients of the operator G

x?t ? ? 2 = u,

1 2 1 2 1

u?t + ? 1 = 0

1

?t = x, ?t = 0, ?t = 0, ?t = 0,

having the following general solution:

? 2 = x2 ? u,

? 1 = tx + F 1 (x, u, v), ? 1 = xu, ? 2 = F 2 (x, u, v),

where F 1 , F 2 are arbitrary smooth functions.

The change of variables (20) with U ? u reduces the operator G to the form

G = tx?t + (x2 ? u)?x + xu?u , which yields representation (17). The theorem has

been proved.

Below we give without proof the assertions describing extensions of the Galilei

algebra in the class of LVFs (1).

On new representations of Galilei groups 217

Theorem 2. Inequivalent representations of the extended Galilei algebra AG1 (1, 1)

by LVFs (1) are exhausted by those given in (7) and by the following ones:

1. P0 = ?t , P1 = ?x , M = 2u?x ,

G = (t + xu)?x + u2 ?u , D = 2t?t + x?x + u?u + ??v ;

v

P0 = ?t , P1 = ?x , M = 2 ?u?t ± 2u ?x ,

2.

v v

G = ?xu?t + t ± x 2u ?x ± 2 u3/2 ?u ,

D = 2t?t + x?x + 2u?u + ??v ;

3. P0 = ?t , P1 = ?x , M = 2(u?t + v?x ),

G = xu?t + (t + xv)?x + uv?u + (u + v 2 )?v ,

D = 2t?t + x?x + 2u?u + v?v ;

P0 = ?t , P1 = x?t , M = x2 ?t ,

4.

G = tx?t + 1 x2 ?x , D = 2t?t + x?x + ??u ;

2

P0 = ?t , P1 = x?t , M = x2 ?t ,

5.

G = tx?t + 1 x2 ?x + ?u , D = 2t?t + x?x ? u?u ;

2

6. P0 = ?t , P1 = x?t , M = 2u?t ,

G = tx?t + (x2 ? u)?x + xu?u , D = 2t?t + x?x + 2u?u ,

where ? = 0, 1.

Theorem 3. Inequivalent representations of the generalized Galilei algebra AG2 (1, 1)

by LVFs (1) are exhausted by those given in (8) and by the following one:

v

M = 2 ?u?t ± 2u ?x ,

P 0 = ?t , P1 = ? x ,

v v

G = ?xu?t + t ± x 2u ?x ± 2 u3/2 ?u , D = 2t?t + x?x + 2u?u ,

v v

A = t2 ? 1 ux2 ?t + tx ± 1 x2 2u ?x + 2tu ± x 2 u3/2 ?u .

2 2

Proof of Theorems 2, 3 is analogous to that of Theorem 1 but computations are

much more involved.

Let us note that the list of inequivalent representations of the Lie algebra of

the Poincar? group P (1, 1) and its natural extensions in the class of LVF with two

e

independent and one dependent variables given in [9] is also not complete. The reason

is that these representations are constructed under assumption that the generators of

time and space translations can be reduced to the form P0 = ?x0 , P1 = ?x1 , which is

not always possible. If we skip the above constraint, one more representation of the

Lie algebra of the Poincar? group is obtained

e

ñòð. 48 |