стр. 28 |

1?k

S(?) being of the form

1 ? + B1 + B2

S(?) = ? 2,

1 ? k (? + B1 )(? + B2 ) ? B3

Equation (13) for arbitrary constants B1 , B2 , B3 , k = 1 can be solved in quadratures:

1

v ?(1 ? k)2 1?k

k?1

(14)

?= ? ?(?) d? ,

2

2

where ? = (? + B1 )(? + B2 ) ? B3 .

2

Substituting (14) into (3) with f of the form (6), (7), we can obtain a class of

solutions for the non-linear d’Alembert equation (1).

5. Compatibility and solutions of the system (5) with ?µ ?µ = 0. In this case

2

R(?) = 0, so T (?) must not vanish. We can take T (?) = 1?k and obtain the system

1 1

fµ ?µ + f 2? = 2f = f k S(?).

f k, (15)

1?k

2

If 2? = 0, then from the first equation of (15)

1

f = h(?, ?1 , ?2 ) + ?3 ) 1?k

(16)

,

where ?1 , ?2 , ?3 are functions on x,

?µ ?µ = ?µ ?µ = ?1,

11 22

1 2 33 12

(17)

?µ ?µ = ?µ ?µ = ?µ ?µ = ?µ ?µ = 0,

13

?µ ?µ = 1.

With the substitution (16) the second equation (15) reduces to the form

? 2?? ? ?21 ? ?22 = 0. (18)

??1 ?1 + ??2 ?2 = S(?), ? ?

The compatibility and solutions of the system of Laplace and Hamilton–Jacobi

equations were considered in detail in [8]. The system (18) is compatible iff

?

? where ?

S(?) = , = 0.

?

If we take for the solutions of the system (17)

?1 = ?x, ?2 = ?x, ?3 = ?x,

where ?µ , ?µ , ?µ are parameters satisfying (6), we shall get the solutions (6), (7)–(9)

of the system (15).

New conditionally invariant solutions for non-linear d’Alembert equation 115

Note 2. A system similar to (15) arose in [8] while searching for ansatzes of the form

u = exp(if (x))?(?) for a nonlinear Schr?dinger equation 2iut + uaa ? uF (|u|) = 0.

o

It is known [9] that complex n-dimensional non-linear d’Alembert equation can be

reduced by similarity methods to (n ? 1)-dimensional Schr?dinger equation.

o

Note 3. The ansatz (3), (6), (7) can be used to get solutions also for complex non-

linear d’Alembert equation, the function ? being complex-valued.

For the equation

2u = ?u(uu? )

k?1

,

2

we get the reduced equation

?

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

k?1

2? ?

?

where ? = (? + B1 )(? + B2 ) ? B3 .

2

From the reduced equation we can find ?:

1

?(1 ? k)2

v 1?k

k?1

?= ? ? d? exp i?,

2

2

where ? is an arbitrary constant.

1. Fushchych W.I., Serov N.I., J. Phys. A, 1983, 16, 3645–3656.

2. Fushchych W.I., Shtelen W.M. and Serov N.I., Symmetry analysis and exact solutions of nonlinear

equations of mathematical physics, Kyiv, Naukova Dumka, 1989 (in Russian); Dordrecht, Kluwer

Publishers, 1993 (in English).

3. Fushchych W.I., Yegorchenko I.A., J. Phys. A, 1989, 22, 2643–2652.

4. Fushchych W.I., In the book Symmetry and Solutions of Nonlinear Equations of Mathematical

Physics, Kiev, Institute of Mathematics, 1987, 4–16.

5. Fushchych W.I., Serov N.I., In the book Symmetry and Solutions of Equations of Mathematical

Physics, Kiev, Institute of Mathematics, 1988, 96–103.

6. Fushchych W.I., Ukr. Math. Zhurn., 1991, 43, 1456–1470.

7. Ovsyannikov L.V., Group analysis of differential equations, New York, Academic Press, 1982.

8. Fushchych W.I., Yegorchenko I.A., Ukr. Math. Zhurn., 1991, 43, 1620–1628.

9. Fushchych W.I., Barannik L.F. and Barannik A.F., Subgroup analysis of the Galilei and Poincar?

e

group and reduction of nonlinear equations, Kiev, Naukova Dumka, 1991.

W.I. Fushchych, Scientific Works 2003, Vol. 5, 116–119.

Anti-reduction of the nonlinear wave

equation

W.I. FUSHCHYCH, R.Z. ZHDANOV

Ми запропонували конструктивний метод зведення рiвняння з частинними похiдни-

ми до декiлькох рiвнянь з меншим числом незалежних змiнних. Застосувавши цей

пiдхiд до багатовимiрного нелiнiйного хвильового рiвняння, ми побудували низку

принципово нових анзацiв, якi редукують його до двох звичайних диференцiальних

рiвнянь.

The wide class of solutions of the multi-dimensional wave equation

2u ? ux0 x0 ? ?3 u = F (u) (1)

can be obtained by means of the following ansatz [1–3]:

(2)

u = ?(?),

where ? is an arbitrary smooth function and ? = ?(x) is the absolute invariant of

some three-dimensional subgroup of the Poincar? group P (1, 3). As a result, one

e

gets ordinary differential equation (ODE) for a function ?(?). That is why, the term

“reduction” is used: a number of dependent and independent variables is decreased.

On the other hand, there are examples of ansatzes reducing one nonlinear partial

differential equation (PDE) to two or even to three equations [4]. Such procedure

leads to an increase of the number of dependent variables and is called an “anti-

reduction” [4].

In the present paper we suggest a regular approach to the anti-reduction of the

nonlinear differential equation (1).

Consider the ansatz

(3)

u(x) = f (x, ?1 (?1 ), ?2 (?2 ), . . . , ?N (?N ))

and the following ordinary differential equations:

(4)

?i = Ri (?i , ?i , ?i ),

? ? i = 1, N ,

where f , Ri are smooth enough functions, ?i = ?i (x) ? C 2 (Rn , R1 ), i = 1, N . If

substitution of (3) into Eq. (1) with subsequent exclusion of the second derivatives ?i ,

?

i

i = 1, N according to (4) yields an identity with respect to variables ? , ?i , i = 1, N

?

then we say that the anti-reduction of nonlinear PDE (1) to N ODE takes place.

In fact, the above definition contains an algorithm of the anti-reduction. We are

going to realize it, provided N = 2.

Theorem. The equation (1) with a logarithmic nonlinearity

2u = ?u ln u, ? ? R1 (5)

Доповiдi АН України, 1993, № 11, С. 37–41.

Anti-reduction of the nonlinear wave equation 117

is the only nonlinear wave equation belonging to the class of PDE (1) that admits

anti-reduction to two second-order ODE and that is more the ansatz (2) has the

form

(6)

u(x) = a(x)?1 (?1 )?2 (?2 ),

where a(x), ?1 (x), ?2 (x) are smooth functions satisfying the system of PDE

1) ?1xµ ?2xµ = 0,

2) a2?i + 2axµ ?ixµ = 0, i = 1, 2,

(7)

3) ?ixµ ?ixµ = Qi (?i ), i = 1, 2,

2a = ? ln a.

4)

3

Here Qi are arbitrary smooth functions, hxµ gxµ = hx0 gx0 ? hxa gxa .

a=1

Omitting intermidiate computations, we adduce main steps of the proof.

Substituting (3) with N = 2 into Eq. (1), we get

2

{f?i (?i ?ixµ ?ixµ + ?i 2?i ) + f?i ?i ?2 ?ixµ ?ixµ + 2f?i xµ ?ixµ ?i } +

fxµ xµ + ? ? ?i

i=1

+ 2f?1 ?2 ?1 ?2 ?1xµ ?2xµ = F (f (x, ?1 , ?2 )).

??

Replacing ?i by Ri (?i , ?i , ?i ) and splitting the obtained equality with respect to

? ?

?1 , ?2 , we have

??

Ri = Ai (?i , ?i )?2 + Bi (?i , ?i )?i + Ci (?i , ?i ),

?i ? i = 1, 2,

?1xµ ?2xµ f?1 ?2 = 0.

Since the equality f?1 ?2 = 0 leads to the case Fuu = 0, we can put f?1 ?2 = 0

whence ?1xµ ?2xµ = 0.

By force of the above facts we get

1) f?i ?i + Aif?i = 0, i = 1, 2,

f?i (Bi ?ixµ ?ixµ + 2?i ) + 2f?i xµ ?ixµ = 0,

2)

2

(8)

3) fxµ xµ + Ci f?i ?ixµ ?ixµ = F (f ),

i=1

4) ?1xµ ?2xµ = 0.

From the first two equations of the system (8) it follows that

f = H1 (?1 , ?1 )H2 (?2 , ?2 )a(x) + b(x),

where Hi , a(x), b(x) are arbitrary smooth functions.

By redefining functions ?i : ?i > ?i Hi (?i , ?i ), i = 1, 2, we may choose

?

(9)

f = a(x)?1 (?1 )?2 (?2 ) + b(x),

whence A1 = A2 = 0.

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

From the Eq. 2 of the system (8) by force of (9) it follows that Bi = Bi (?i ),

i = 1, 2. Consequently, by redefining functions ?i

?i > ?i = Wi (?i ),

? i = 1, 2,

we may choose B1 = B2 = 0. As a result, the system (8) is read

1) ?1xµ ?2xµ = 0,

2) a2?i + 2axµ ?ixµ = 0, i = 1, 2,

(10)

(2a)?1 ?2 + 2b + a[C1 (?1 , ?1 )?2 ?1xµ ?1xµ +

3)

+ C2 (?2 , ?2 )?1 ?2xµ ?2xµ ] = F (a?1 ?2 + b).

The only thing left is to split Eq. 3 from (10) with respect to variables ?1 , ?2 .

Dividing Eq. 3 into ?1 ?2 and differentiating it with respect to variables ?1 ?2 we get

{(?1 ?2 )?1 [F (a?1 ?2 + b) ? 2b]}?1 ?2 = 0, whence

2

2 2d F dF

? ax + F = 2b, (11)

ax x = ?1 ?2 , ? = ax + b.

2

d? d?

Differentiation of (11) with respect to x yields

d3 F d2 F

ax 3 + = 0.

d? 2

d?

?

Since we are interested in a nonlinear case, the inequality F = 0 holds. Hence, it

follows that

...

? ?1 ?1

F (F ) = ?(ax)

or

?1

...

?

F (F ) = ?? + b.

Differentiating the above equality with respect to ? we obtain nonlinear ODE for

... ...

?2

?

F (?): F F ?2(F )2 = 0, which general solution reads F (?) = ?1 (?1 ? + ?2 ) ln(?1 ? +

?1

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