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solution ?(?1 , ?2 ) in explicit form. To demonstrate the solutions of equation (16)
generating process we choose ? in simplest form
?(?) = k?,
where k is an arbitrary constant. Then
1
z = 2k(arctg 2y1 ? arctg 2(y1 ? y2 )) ? c arctg 2y1 .
2
So, corresponding partial solution of equation (21) have the form

k ? 1c 1 k 1
(1)
? ? ?2 ??
4
u = (?1 + ?2 )
?1 + ?2 4 ?2 4
(23)
k? 1
4c
1 1 k 1
? 2 k ? c arctg 2 ? + 2k arctg 2 ?.
4 ?1 + ?2 4 ?2 4
Writing ?(?1 , ?2 ) with ?1 and ?2 determined by (18) in variables ?µ and make use of
Legendre transformation (1), we obtain second solution

k ? 1c
1 1
(2)
u = 4 k ? c ? ?1 ? ?2 ??
4
4 k ? 4 c ? ?1 ? ?2
1
4 4
k 1
? [4k ? ?2 ] ??
4k ? ?2 4
(24)
k? 1
4c
1 1
? 2 k ? c arctg 2 ? +
4 k? ? ?1 ? ?2
1
4 4
4c

k 1
?.
+ 2k arctg 2
4k ? ?2 4
If solution (22) of the equation (21) is ? = 0, we get
1 c 1 c 1
(1)
u = ? ?1 ? + c arctg 2 ?,
2 ?1 4 ?1 4
(25)
?1 ?1
(2)
u = (c ? ?1 ) ? c arctg .
c ? ?1 c ? ?1
Choosing invariants of equation (16) the functions
?1 = ?µ xµ , ?2 = xµ xµ = x2 ,
Generating solutions for nonlinear equations by the Legendre transformation 491

we obtain the solutions
(1) (2)
u = ??1 ?1 ? ?2 , u = ??1 ?1 ? ?2 .
2 2 (26)

i.e. this solution is Legandre-invariant solution of equation (16).
Example 3. The equation

(27)
det(uµ? )2u + tr (uµ? ) = 0

is invariant with respect to Legandre transformation. An ansatz
? = xµ xµ = x2 (28)
u = ?(?),

reduces (27) to ODE

8??2 + ? ?[4(n + 1)?2 ? n + 1] + 2n?3 ? n? = 0.
?? ? ? ? ?

Following the formula (9), from (28)
(1)
(µ = 0, n ? 1),
u = ?(?2 ), ? 2 = ?µ ? µ

we obtain
(2)
u = ?µ ?µ (?2 ) ? ?(?2 ), ?µ (?2 ) = xµ . (29)

Hence we find

x2 = 4?2 (?)2 , (30)
?µ = 2?µ ?,
? ?

and formula (29) take the form
1 2 ?1
(2)
x ? (?(x2 )) ? ?(?(x2 )).
u= (31)
?
2

1. Фущич В.И., Штелень В.М., Серов Н.И., Симметрийный анализ и точные решения нелинейных
уравнений математической физики, Киев, Наук. думка, 1989, 336 с.
W.I. Fushchych, Scientific Works 2002, Vol. 4, 492–499.

Nonlocal symmetry and generating solutions
for Harry–Dym type equations
W.I. FUSHCHYCH, V.A. TYCHININ
Изучена нелиевская симметрия уравнений u0 = f (u)u111 , w0 = g(w1 )w111 , выде-
лены уравнения, допускающие нелокальную линеаризацию; установлены формулы
размножения решений. Для редукции нелинейных уравнений применяется нелиев-
ский анзац u = h(x)?(?) + f (x)?(?) + g(x).
?

1. Let us consider two classes of one-dimensional third order nonlinear equations
u0 ? f (u)u111 = 0, (1)

w0 ? g(w11 )w111 = 0, (2)
?nu
?u ?w
n
u µ = ?µ u = , u1 . . . 1 = ?1 u = , wµ = ?µ w = ,
?xn
?xµ ?xµ
1
n

?nw
(µ = 0, 1, n ? N),
n
w1 . . . 1 = ?1 w =
?xn
1
n

where f (u), g(w11 ) are arbitrary smooth functions.
In the present paper linearizable equations are picked out from the sets of equa-
tions (1) and (2) by means of nonlocal transformations. Non-Lie symmetry of equa-
tions (1), (2) is investigated. The formulas of generating solutions for nonlinear equati-
ons belonging to classes (1), (2) are obtained. Non-Lie ansatz
d?
(3)
u = h(x)?(?) + f (x)?(?) + g(x),
? x = (x0 , x1 ), ?(?) =
? ,
d?
which should be consider as the generalization of ansatz [1]
u = f (x)?(?) + g(x)
is used for reducing equations (1), (2) to ODE. Some sets of exact partial solutions
for nonlinear equations are constructed.
Note 1. The Eq. (1) is equivalent to equation
z0 ? ?1 c(z) = 0
3
(4)
The connection between these equations is given by transformation
(4a)
c(z) = u.
Thereby, the equality
f (u) = c(c?1 [u])
?
Доклады АН Украины, 1992, № 9, С. 24–30.
Nonlocal symmetry and generating solutions for Harry–Dym type equations 493

holds. Here c?1 [u] is the function inverse to c(u). In that case, when f (u) = u3 ,
1
c(z) = z ? 2 , the Eq. (4) coincides with the known Harry–Dym equation [2].
2. Nonlocal symmetry. Consider the Eq. (4)
3 2
z0 = ?1 c(z) = ?1 (c(z)z1 ).
?
The substitution
(5a)
z = w11
reduces (4) to equation
(5)
w0 = c(w11 )w111 .
?
Making use the Euler–Ampere transformation
w = y1 v1 ? v, (6)
x1 = v1 , x0 = y0 , v = v(y0 , y1 ), v11 = 0,
under Eq. (5), we get
? ?1 ?3 (7)
v0 = c(v11 )v11 v111 .
Using the substitution
(7a)
v11 = z(y0 , y1 )
in equation (7), twice differentiated on y1 , we get
z0 = ?1 c(z ?1 )z ?3 z1 .
2
(8)
?
It follows from (8), that transformations (5a), (6), (7a) do not take out any Eq. (4)
beyond the this class of equations, none the less the set of Eq. (4) is not invariant
under these transformations. If function c(z ?1 )z ?3 in (8) satisfies the condition
?
c(z ?1 )z ?3 = ?, (9a)
? ? = const.
then Eq. (4) is linearisable. When the condition
c(z) = c(z ?1 )z ?3 (9b)
? ?
holds, then the Eq. (8) coincides with initial equation (4), i.e. these equations are
invariant with respect to nonlocal transformations (5a), (6), (7a).
The condition (9b) allows to describe all the equations of the class (4) which are
invariant with respect to transformations (5a), (6), (7a).
Theorem 1. The Eq. (4) is invariant with respect to transformations. (5a), (6), (7a),
if it has the form
1
z0 = ?1 z ? 2 ?(ln z)z1 .
2
(10)

Here ?(?) is an arbitrary smooth even function.
Corollary 1. The Eq. (1) is invariant with respect to transformations (4a), (5a), (6),
(7a), (4a) if it has the form
?3
u0 = c?1 [u] ? ln c?1 [u] u111 ,
2
(11)
494 W.I. Fushchych, V.A. Tychinin

where c?1 [u] is the function inverce to c(u) and it is determined implicitly from the
formula
3
z ? 2 ?(ln z)dz. (12)
u=

Example 1. From the theorem 1 and the corollary 1 under ?(?) = 1 we get the
following invariant equations
13 1
z0 = ?1 ? z ? 2 z 1 = ?1 z ? 2 ,
2 3
(13)
2

u0 = u3 u111 . (14)

So, Eq. (13) is known as Harry–Dym equation. Letting ?(?) = cos ?, we obtain the
equation
3
z0 = ?1 z ? 2 cos ln zz1
2
(15)

and corresponding to it equation of the class (1)
?3
?1
cos ln c?1 [u] u111 .
2
(16)
u0 = c [u]
Here c?1 [u] is determined implicitly by formula
4 1 1
sin ln z ? cos ln z z ? 2 . (17)
u=
5 2
So, we establish that the equations
3
?3
z0 = ?1 z ?2 ,
3
(18a,b,c)
u0 = u 2 u111 , w0 = w11 w111
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