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2?
22. G1 , G2 + P1 + ?P2 , P0 + P3
1 x1
?m??45 ?03 exp ? ?03 (?mx1 ?45 + ?1 ) ?
?(x) = exp
2 2?
?mx2 x2 1
? exp 2
?45 ?03 exp ??45 ?03 ?1 +
2
2(? + ?) (? + ?) 2?
1??
(? + ? ? ???45 )?03 ?1
+ m(?1 + ??2 ) (? + ?) + ?(?).
2?

23. G1 , G2 + P2 , P0 + P3
1 x1
?m?45 ?03 ? exp ? ?03 (?mx1 ?45 + ?1 ) ?
?(x) = exp
2 2?
?mx2
? exp ?45 ?03 ?
2
2(? + 1)
x2 1
? exp ?m(? + 1)?45 ?2 + (??45 ? ? ? 1)?03 ?2 ?(?).
(? + 1)2 2
Non-local ans?tze for the Dirac equation
a 365

24. J03 , G1 , P2
x1 1
?(x) = exp(?m?45 ?2 x2 ) exp ? ?03 ?1 exp ? ?0 ?3 ln ? ? x2 ? x2 ? x2 .
0 1 3
2? 2
25. J03 + ?P1 + ?P2 , G1 , P0 + P3
1 x1
?m?45 ?03 ? exp ? ?03 (?mx1 ?45 + ?1 ) ?
?(x) = exp
2 2?
mx2
? exp [???2 + ?(1 + ?45 )?03 ][??45 (?0 ?3 ? 1) ? ? +
?2

+ ?03 ?45 (??1 + ??2 ) + ?03 ] ?(?).

26. J12 + P0 + P3 , G1 , G2
1
?(x) = exp ? ?03 (?1 x1 + ?2 x2 ) ?
2?
x2 ? x2 1
? exp ?m?45 ?03 + ?1 ?2 (??45 ? 1)
0
?(?).
2? 2
27. J03 + ?J12 , G1 , G2
1
(?1 x1 + ?2 x2 )?03 ?
?(x) = exp
2?
1
? exp ? (?0 ?3 + ??1 ?2 ) ln ? ? x2 ? x2 .
0
2
The following notations were used in the above ans?tze:
a
? = x3 ? x0 ,
Gk = J0k + Jk3 , k = 1, 2, ? = x3 + x0 ,
?03 = ?0 + ?3 , ?45 = ?4 + ?5 ,
? and ? are constants, ?(z) is a new unknown spinor and Q1 , Q2 , Q3 is a subalgebra
of the algebra (3) and (4) having basis elements Q1 , Q2 , Q3 .
Let us adduce an example of reduced ODE. If one substitutes ansatz 8 into (2)
then the equation for ?(z) becomes
d 1
+ z ?1/2 ?2 ? m + 2?m(?4 + ?5 ) ?(z) = 0.
2z 1/2 ?2
dz 2
Note 1. If one puts ? = 0 in (3) then (3) and (4) generate the local Lie group P (1, 3).
That is why, on putting ? = 0 into the ans?tze above, one obtains Poincar?-invariant
a e
ans?tze for the spinor field constructed in [3].
a
Note 2. The above non-local ans?tze can be applied to the construction of exact
a
solutions of non-linear Lorentz-invariant spinor equations admitting the group (5).
One example of such equations is
?
{?µ ? µ + ?[?(?4 + ?5 )?µ ?µ ?](?4 + ?5 )}? = 0,
?
where ? is constant and ? = ?T ?0 ?4 . This problem will be considered in a future
publication.
366 W.I. Fushchych, R.Z. Zhdanov

1. Fushchych W.I., Nikitin A.G., Symmetries of Maxwell’s equations, Dordrecht, Reidel, 1987.
2. Fushchych W.I., Zhdanov R.Z. J. Phys. A: Math. Gen., 1987, 20, 4173.
3. Fushchych W.I., Zhdanov R.Z., Fiz. Elem. Cast. Atom. Jadra (USSR), 1988, 19, 1154–1196.
4. Patera J., Winternitz P., Zassenhaus H., J. Math. Phys., 1975, 16, 1597.
5. Kamran N., Legare M., McLenaghan R.G., Winternitz P., J. Math. Phys., 1988, 29, 403.
6. Fushchych W.I., Teor. Mat. Fiz., 1971, 7, 3.
7. Fushchych W.I., Lett. Nuovo Cimento, 1974, 11, 508.
8. Fushchych W.I., Zhdanov R.Z., in Symmetry and Solutions of Nonlinear Equations of Mathematical
Physics, Kiev, Mathematical Institute, 1987, 17.
W.I. Fushchych, Scientific Works 2001, Vol. 3, 367–371.

On some new exact solutions of nonlinear
d’Allembert and Hamilton equations
W.I. FUSHCHYCH, R.Z. ZHDANOV
Some new exact solutions of d’Allembert–Hamilton and d’Allembert equations are obtai-
ned. The necessary conditions of integrability of over-determined d’Allembert–Hamilton
system of nonlinear differential equations are established.

1. It was the Euler’s idea (1734–1740 y.) that problem of integrating partial di-
fferential equations (PDE) could be solved by reducing them to ordinary equations
(ODE). But one can not apply this idea to arbitrary PDE. Therefore it was suggested
by Fushchych [4, 5] to restrict oneself by PDE possessing wide symmetry groups.
This program was realized for some nonlinear wave equations by Fushchych and
Serov [7], Fushchych and Shtelen [8] and Fushchych and Zhdanov [9] (see also [1,
11, 12]). The vast list of references on this point can be found in Fushchych and
Nikitin [6].
When reducing PDE to ODE one has always to deal with the problem of investi-
gating compatibility of some systems of PDE. For example, nonlinear d’Allembert
equation
2u = F1 (u), 2 = ?x0 ? ?x1 ? ?x2 ? ?x3
2 2 2 2
(1)
with the aid of ansatz [4]
(2)
u = ?(?), ? = ?(x0 , x1 , x2 , x3 ),
is reduced to the ODE having variable coefficients [7]
?µ ? µ ? + 2? ? = F1 (?), (3)
? ?

where ?µ ? ?xµ , µ = 0, 3, ? ? d? . Hereafter the summation over repeated indices
d?
??
?
in Minkowsky space having the metric gµ? = diag (1, ?1, ?1, ?1) is supposed, i.e.
?µ ? µ ? g µ? ?µ ?? = ?0 ? ?1 ? ?2 ? ?3 .
2 2 2 2

We demand new variable ? to satisfy d’Allembert and Hamilton equations simul-
taneously
2? = F2 (?), (4)

?µ ? µ = F3 (?). (5)

As a result equation (3) takes the form
(6)
F3 (?)? + F2 (?)? = F1 (?).
? ?
Winternitz and collaborators (see [1, 11]) construct new variables ? by using
subgroup structure of the Poincar? group P (1, 3). One can be easily convinced that
e
invariants obtained in this way satisfy system (4), (5).
Preprint 468, Institute for Mathematics and its Applications, University of Minnesota, 1988, 5 p.
368 W.I. Fushchych, R.Z. Zhdanov

So to obtain set of variables ? making possible to reduce multi-dimensional PDE
(1) to ODE one has to consider the problem of compatibility of system (4), (5) and
then to integrate it.
In the present paper compatibility of equations (4), (5) is investigated, i.e. all
smooth functions ensuring the compatibility of d’Allembert–Hamilton system are
described.
The direct application of Cartan’s method of investigation of compatibility of over-
determined PDE [2] is rather difficult. To avoid arising difficulties we essentially use
symmetry properties of system (4), (5) [7, 8].
System (4), (5) via the change of dependent variable z = z(?) can be reduced to
the following system
2? = F (?), (7)

?µ ? µ = ?, ? = const, (8)

ODE (6) taking the form
(9)
?? + F (?)? = F1 (?).
? ?
Before formulating the principal result of the paper we adduce without proof some
auxiliary statements.
Lemma 1. Solutions of system (7), (8) satisfy the identities
?
?µ?1 ? µ?1 = ??F (?),
1 2?
?µ?1 ? ?1 ?2 ??2 =
µ
? F (?), (10)
2!
dn F (?)
1
n ? 0,
?µ?1 ? ?1 ?2 ? ?n µ = (??)n
? ,
d? n
n!
?2?
where ??? ? ?x? ?x? , ?, ? = 0, 3.
Lemma 2. Solutions of the system (7), (8) satisfy the following equality:
(10 )
det(?µ? ) = 0.
Let us now formulate the principal statement.
Theorem 1. The necessary condition of compatibility of overdetermined system (7),
(8) is as follows
?
? 0,
?
?
? ?(? + C )?1 ,
1
(11)
F (?) =
? 2?(? + C1 )[(? + C1 )2 + C2 ]?1 ,
?
?
?
3?((? + C1 )2 + C2 )[(? + C1 )3 + 3C2 (? + C1 ) + C3 ]?1 ,
where C1 , C2 , C3 are arbitrary constants.
Proof. By direct (and rather tiresome) verification one can be convinced that the
following identity holds
6(?µ?1 ? ?1 ?2 ??2 ?3 ? ?3 µ ) ? 8(?µ )(?µ?1 ? ?1 ?2 ??2 ) ?
µ µ
(12)
? 3(?µ?1 ? µ?1 )2 + 6(?µ )2 (?µ?1 ? µ?1 ) ? (?µ ? µ )4 = 24 det(?µ? ).
µ
New exact solutions of nonlinear d’Allembert and Hamilton equations 369

Substituting (10), (10 ) into (12) one obtains nonlinear ODE for F (?)
... ...
? ?
?3 F + 4?2 F F + 3?2 F 2 + 6?F F 2 + F 4 = 0, (13)
?
where F ? dF .
d?
General solution of equation (13) is given by formulae (11). Theorem is proved.
Note 1. Compatibility of three-dimensional d’Allembert–Hamilton system has been
investigated in detail by Collins [3]. Collins essentially used geometrical methods
which could not be generalized to higher dimensions.
Using Lie’s method (see e.g. [10]) one can prove the following statement.
Theorem 2. The sytem of PDE (7), (8) is invariant under the 15-parameter confor-
mal group C(1, 3) iff

F (?) = 3?(? + C)?1 , C = const. (14)
? > 0,

Note 2. Formula (14) can be obtained from (11) by putting C2 = C3 = 0. So Theo-
rem 2 demonstrates close connection between compatibility of a system of PDE and
its symmetry.
Note 3. It is common knowledge that PDE (7) is invariant under the group C(1, 3)
iff F (?) = ?? 3 [7]. Consequently, an additional constraint (8) changes essentially
symmetry properties of d’Allembert equation (choosing F3 (?) in a proper way one
can obtain conformally-invariant system of the form (4), (5) under arbitrary F2 (?)).
2. Let us list explicit form of some exact solutions of d’Allembert–Hamilton system
and reduced ODE for function ?(?).

ODE for ?(?)
? F (?) ? = ?(x)
aµ xµ
1. 1 0 ? = F1 (?)
?
1/2
? ?1 ? + ? ?1 ? = F1 (?)
(aµ xµ )2 ? (bµ xµ )2
2. 1 ? ?
1/2
2? ?1 ? + 2? ?1 ? = F1 (?)
(aµ xµ )2 ? (bµ xµ )2 ? (cµ xµ )2
3. 1 ? ?
3? ?1 ? + 3? ?1 ? = F1 (?)
(xµ xµ )1/2
4. 1 ? ?
?1 ? = ?F1 (?)

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