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?t
W.I. Fushchych, Scientific Works 2002, Vol. 4, 486–491.

Generating solutions for nonlinear equations
by the Legendre transformation
W.I. FUSHCHYCH, V.A. TYCHININ
Предложен новый метод размножения решений нелинейных уравнений, основанный
на использовании их инвариантности относительно преобразований Лежандра.

1. The superposition principle for nonlinear equations is not fulfilled. That is why
it is important to have a method of generating solutions starting from known ones,
i.e. an algorithm of constructing solution when the partial solutions are known.
In the present paper we suggest a new method of generating solutions for nonlinear
equations based on the use of their invariance with respect to Legendre transformati-
on. For this aim, first of all, it is necessary to describe the equations which possess
such property. Here we do not consider solution of this general problem, and list some
first and second order equations invariant with respect to the Legendre transformation
and use this property to construct new solutions from known ones.
In [1] there was accomplished the group-theoretical classification for nonlinear
second order differential equations invariant under the Lorentz group O(1, n) and
the Poincar? group P (1, n) and different their extensions. These equations have the
e
important property due to which there exists simple algorithm of constructing new
solution from known ones, or, in another words, there exists the procedure of gene-
rating solutions. We use the Legendre transformation for further classification of
Lorentz and Poincar? invariant equations. In other words, we distinguish from the
e
set of Lorentz and Poincar? invariant those equations which are additionally invariant
e
with respect to Legendre transformation.
2. The Legendre-invariant equations. The Legendre transformation in the space
of n independent variables we write in the form
?v
u = yµ vµ ? v, (1)
vµ = , xµ = vµ , det(vµ? ) = 0.
?xµ
So, the first and second order derivatives are changing as
uµ? = det ?1 (v?? )aµ? (v?? ) (µ, ?, ?, ? = 0, n ? 1). (2)
u µ = yµ ,
If not declare another, summation is understood over repeated indexes,
? ?
···
u00 u0,n?1
det(u?? ) = det ? · · · ?,
··· ··· (3)
un?1,0 · · · un?1,n?1
aµ? (v?? ) is an algebraic supplement to element vµ? .
The differential equation L(x, u) = 0 is called invariant with respect to Legandre
transformation (1) when the following condition
(4)
L(x, u) = ?L(y, v)
Доклады АН Украины, 1992, № 7, C. 20–25
Generating solutions for nonlinear equations by the Legendre transformation 487

is fulfilled. Here ? is an arbitrary parameter or function of y, v, vµ .
We list several sets of nonlinear equations invariant under the Lorentz group
O(1, n) and the Poincar? group P (1, n) as well as the Legandre transformation.
e
Consider equations

det(uµ? )2u ± tr (uµ? ) = 0, 2u ± det(uµ? ) tr (uµ? ) = 0,
(5)
det m (uµ? ) ± det ?mn (u?? ) det m (aµ? (u?? )) = 0,

uµ uµ + x2 + c = 0, (x2 = xµ xµ ),
(xµ + uµ )(xµ + uµ ) = x2 + 2xµ uµ + uµ uµ = 0,
uµ u? det(u?? ) ± xµ x? aµ? (u?? ) = 0, xµ x? uµ? det(u?? ) ± uµ u? aµ? (u?? ) = 0,
f (x2 ) det m (u?? ) ± f (uµ uµ ) det ?nm (u?? ) det m (aµ? (u?? )) = 0, (6)
f (uµ uµ ) det m (u?? ) ± f (x2 ) det ?nm (u?? ) det m (aµ? (u?? )) = 0,
1 + (x2 ? uµ uµ ) u11 + 2(u0 ? x0 )(u1 ? x1 )u10 ?
? 1 ? (x2 ? uµ uµ ) u00 = 0 (µ = 0, 1),

where 2u = gµ? uµ? is d’Alambert operator,
def
uµ uµ = gµ? uµ u? , tr (uµ? ) = gµ? uµ? ,

(µ, ? = 0, n ? 1), gµ? is metric tensor of the space with signature (+, ?, . . . , ?), f is
an arbitrary smooth function.
Let us list several equations which are invariant nor under the Lorentz group
O(1, n ? 1) not under Poincar? group P (1, n ? 1), but are invariant with respect to
e
Legendre transformation

?µ (xµ + uµ ) = 0, ?µ (u? )xµ + ?µ (x? )uµ = 0,
gµ? [?µ (x? ) + ?µ (u? )][?? (x? ) + ?? (u? )] = 0,
?µ xµ det(u?? )2u ± ?µ uµ tr (u?? ) = 0,
?µ (u? )xµ det(u?? )2u ± ?µ (x? )uµ tr (u?? ) = 0, (7)
?µ xµ 2u ± ?µ uµ det(u?? ) tr (u?? ) = 0,
?µ (u? )xµ det m (u?? ) ± ?µ (x? )uµ det ?nm (aµ? (u?? )) = 0,
u00 u11 ? u2 = u?1 u11 .
10 00

?µ are vector components. The method of generating solutions for equations (7) is
unknown.
To prove invariance of equations (5), (6), (7) under Legendre transformation (1) it
is necessary to verify fulfilment of relations (4). For example, equation

uµ uµ + x2 + c = 0

is transformed under Legendre transformation into y 2 + vµ v µ + c = 0.
3. The generating solutions. Suppose we have a solution of nonlinear equation
(1) (1)
u = u (x), det( u µ? ) = 0. (8)
x = (x0 , x1 , . . . , xn?1 ),
488 W.I. Fushchych, V.A. Tychinin

Let us rewrite this solution, replacing xµ for parameters ?µ (µ = 0, n ? 1). The
formula of generating solutions resulting from Legendre transformation takes the
form
(2) (1) (1) (1)
u = ?µ u µ (?) ? u (?), u µ (?) (9)
= xµ (? = (?0 , ?1 , . . . , ?n?1 )).
(2)
To find exact solution u (x) it is necessary to eliminate parameters ? in (9). So, the
formula (9) gives the method for generating solutions.
Example 1. Consider an equation from the list given above
u00 u11 ? u2 = u?1 u11 . (10)
10 00

This equation admits Lie algebra P0 , P1 , P2 , J0 , J1 , D1 , D2 :
P0 = ?0 , P1 = ?1 , P2 = ?u , J0 = x0 ?u , J1 = x1 ?u ,
(11)
D1 = x1 ?1 , D2 = x0 ?0 + 2u?u
and consequently Lie symmetry of equation (10) gives the following formula of gene-
rating solutions:
(2) (1)
u = e?2b ( u (eb x0 + ?0 , ea x1 + ?1 ) + ?0 x0 + ?1 x1 + ?2 ), (12)
where a, b, ?0 , ?1 , ?0 , ?1 , ?2 are group parameters.
One of partial solutions of equation (10) have the form
x2
(1)
u = 0 cth x1 . (13)
2
(1)
It is easily convinced that det( u µ? ) = 0. Let us construct a new solution of equa-
tion (10) by using the Legandre transformation. Rewrite (13), replacing xµ , (µ = 0, 1)
for parameters ?µ , i.e.
2
?0
(1)
u= cth ?1 .
2
and make use of formulas (9) to obtain
12 1
(2)
?0 cth ?1 ? ?1 ?0 sh?2 ?1 ,
2
u=
2 2
(14)
?0 cth ?1 = x0 ,
12
? ?0 sh?2 ?1 = x1 .
2
Exclude parameters by expressing them through xµ from two lest equations of sys-
tem (14). We obtain
x2
? 0 = ch2 ?1 ,
2x1
or
1
x2 + 2x1 , x2 + 2x1 .
th ?1 = ?0 =
0 0
x0
Generating solutions for nonlinear equations by the Legendre transformation 489

Substituting ?0 and ?1 into first equation of system (14), we get

x2 + 2x1 x2 + 2x1
1
(2) 0 0
u = x2 (15)
+ x1 arth .
20 x0 x0
Designating

x1
?= 1+2 ,
x0

we write the solution in another form
12
(2)
u= (15 )
x ? + x1 arth ?.
20
It is not difficult to verify that function (15) satisfies the equation (10). Note that this
solution cannot be constructed from (13) by generating it according to formula (12).
Example 2. Consider the equation

(µ = 0, n ? 1).
uµ uµ + x2 + c = 0 (16)

c is an arbitrary constant. The maximal invariance algebra of equation (16) is P0 , J0a ,
Jab :

Jab = xa ?b ? xb ?a (a, b = 1, n ? 1). (17)
P 0 = ?0 , J0a = xa ?0 + x0 ?a ,

An ansatz of the form [1]

?1 = xµ xµ = x2 , ?2 = (?µ xµ )2 + (?µ xµ )2 ,
u = ?(?1 , ?2 ),
(18)
? 2 = ? 2 = ?1, ?? = 0,

reduces n-dimensional PDE (16) to 2-dimensional PDE
1
?1 ?2 ? ?2 ?2 + 2?2 ?1 ?2 + (?1 + c) = 0. (19)
1 2
4
Construct the solution of equation (19) by using Legendre transformation

z = ?1 ?1 + ?2 ?2 ? ?, (20)
y1 = ?1 , y2 = ?2 .

Then the equation (19) is transformed to the linear

1 1
z1 ? (y2 ? 2y1 y2 )z2 + c = 0.
2
(21)
y1 +
4 4

The general solution of (21) is:
1
z = ?(2 arctg 2y1 ? 2 arctg 2(y1 ? y2 )) ? c arctg 2y1 . (22)
2
One gets the general solution of equation (19) by inverting (20)

? = y1 z1 + y2 z2 ? z, ? 1 = z1 , ?2 = z2 ,
490 W.I. Fushchych, V.A. Tychinin

or, substituting z from (22),
1 1 1c
? ? ?
?1 = ? ,
(y1 ? y2 )2 +
1 1 4 y1 + 1
2 2
y1 + 4 4 4
1
?
?2 = ? ,
(y1 ? y2 )2 + 1
4
1
? = y1 ?1 + y2 ?2 ? ? + c arctg 2y1 ,
2
where y1 , y2 play the role of parameters and must be eliminate for obtaining the

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