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are reduced to the linear equation
(19)
v0 = v111 (? = 1)
and that, in particular, the Harry–Dym equation and connected with it equations
?3
1
z0 = ? 1 z ? 2 ,
u0 = u3 u111 , 3
w0 = w112 w111 (20a,b,c)

are invariant with respect to corresponding nonlocal transformations.
3. The nonlocal superposition and the generating solutions.
Theorem 2. The solutions superposition formula for Eq. (18a)
3
(18a)
u0 = u 2 u111
has the form

(3) (1) (2) (1) (2)
u (x0 , x1 ) = u (x0 , ? 1 ) + u (x0 , ? 2 ) + 2 u (x0 , ? 1 ) u (x0 , ? 2 ), (21a)

d? 1 d? 2
(21b)
= ,
(1) (2)
u (x0 , ? 1 ) u (x0 , ? 2 )
Nonlocal symmetry and generating solutions for Harry–Dym type equations 495

? 1 + ? 2 = x1 , (21c)

(1) (2)
u (x0 , ? 1 ) u (x0 , ? 2 )
1 (1) (2)
1
u 11 (x0 , ? 1 ) + u 11 (x0 , ? 2 ) . (21d)
?0 =
2 (1) (2)
u (x0 , ? 1 ) + u (x0 , ? 2 )

Let us illustrate the efficiency of the formula (21).
Example 2. Let us take, as initial, the simplest stationary solutions of Eq. (18a)
(1) 1 (2) 2
1 2
= (x1 )2 , = 4(x1 )2 .
u (x1 ) u (x1 )
1 2
Replace x1 and x1 in this solutions for parameters ? 1 , ? 2
(1) (2)
u = (? 1 )2 , u = 4(? 2 )2 .
The differential Eq. (21b) takes the form
d? 2 ?2
(22)
=2 1
d? 1 ?
and has the general solution
(? 1 )2
?2 = ? (23)
.
2?(x0 )
Here ?(x0 ) is an arbitrary smooth function. The equation for ? 1
(? 1 )2 ? 2?? 1 + 2?x1 = 0 (24)
we obtain making use of (21c) and replacing in (23) ? 2 for the expression x1 ? ? 1 .
From (24) we find
2
(3)
u (x0 , x1 ) = [? 1 + 2? 2 ]2 = [2x1 ? ? 1 ]2 = 2x1 ? ? ± ? 2?x1
?2 (25)
.

The function ?(x0 ) can be defined more precisely from the condition that ? 1 is the
solution of Eq. (21d). As a result, we get the equation for ?(x0 )
?
? = ?6?.
Therefore
? = c exp(?6x0 ),
(3)
where c is an arbitrary constant. So, the new solution u , which is constructed from
(1) (2)
u and u , is of the form
2
(3)
u (x0 , x1 ) = 2x1 ? ce?6x0 ± c2 e?12x0 ? 2cx1 e?6x0 (26)
.

Example 3. Let us choose, as initial, the following two solutions of Eq. (18a):
(1) (2)
u = x2 , u = 9x2
1 1
496 W.I. Fushchych, V.A. Tychinin

and rewrite them in variables ? 1 and ? 2
(1) (2)
u = (? 1 )2 , u = 9(? 2 )2 .
Unlike the previous example when solving ODE (21b), one obtains the cubic equation
for ? 1
(? 1 )3 ? ?? 1 + ?x1 = 0, (27)
? = ?(x0 ).
The real solution of the Eq. (27) can be written in the form
1 v ?1
1
? 1 = ?3??1 cos (27a)
arccos ?x1 , ?= 3 3? 2 (x0 ).
3 2
(3)
The solution u
2 2
2 1
(3)
u (x0 , x1 ) = [3x1 ? 2? ] = 9 x1 ? ? 1 = 9 x1 + 2??1 cos arccos ?x1 (28)
12
3 3
we find from the formula (21a). The condition on ?(x0 ) is of the form
?
? = 12?.
Hence
? = c exp(12x0 ).
(3)
c is an arbitrary constant. Finally, one can write solution u in the form
2
1
(3) ?12x0
cos arccos cx1 e12x0
u (x0 , x1 ) = 9 x1 + 2ce (29)
.
3
4. The non-group generating of solutions. For equations of the class (11) we
(1)
can write formula of generating solutions. Let u (x0 , x1 ) be a known partial solution
(2)
of nonlinear Eq. (11) and u (x0 , x1 ) is its new solution, then the following assertion
holds true.
Theorem 3. The formula of generating solutions for Eq. (11), giving by nonlocal
symmetry (4a), (5a), (6), (7a), (4a) has the form
1
2
(2) (1)
[ u (x0 , ? )]?2 d?
u (x0 , x1 ) = x1 ? ? (30a)
,

(1)
= [ u (x0 , ? )]?1 , (30b)
(1)
[ u (x0 , ? )]?2 d?, (30c)
x1 =

?3
?0 = ?1 ?1 2 ?11 . (30d)

Let us demonstrate the efficiency of the formula (30) for Eq. (20a) on several
simple examples.
Nonlocal symmetry and generating solutions for Harry–Dym type equations 497

(1)
Example 4. Let u . Then
1
2
(2)
u (x0 , ? ) = x1 ? ? d? d? , x1 = d? = ? + ?1 (x0 ).

?1 (x0 ) is an arbitrary function. Calculating the integral in the first equality and
resolving the second one with respect to ? , we get
1
2
1
(2)
u (x0 , ? ) = x1 ? ? ? 2 ? ?1 ? + ?2 (x0 ) (31)
,
2
? = x1 ? ?1 (x0 ). (32)
Having excluded parameter ? from equalities of the system (31), (32) we get the
(2)
solution u (x0 , x1 ) in explicit form
1
2
1
(2)
?2 ? ?2 ? ?3 = const.
u (x0 , x1 ) = (33)
21
Example 5. The function
1
(1)
(?1 ? x1 )2
u (x0 , x1 ) = (34)
4
is the solution of the Eq. (20a). ?1 is an arbitrary constant. It follows from relations
(30b,c), that
(2)
u (x0 , ? ) = 4(?1 ? ? )?2 , (35)
16
(?1 ? ? )?3 + ?2 (x0 ). (36)
x1 =
3
Resolving (36) with respect to ? , one obtain
?1
3
3
?1
? = h(x1 ? ?2 (x0 )) h=? (37)
+ ?1 , .
3
16
Substituting ? from the formula (37) into condition (30d), one gets
?
?2 = ?1.
(2)
Let us substitute specified value of ? into the formula (35) and find the solution u
2
3
3
(2) 2
u (x0 , x1 ) = k(x0 + x1 ) , (38)
k= .
3
2
5. The non-Lie ans?tze. Let us consider the ansatz of the form
a
d?
(39)
w = h(x)?(?) + f (x)?(?) + g(x), x = (x0 , x1 ), ?(?) =
? ?
d?
for constructing of solutions for Eq. (20c):
?3
w0 ? w112 w111 = 0. (20c)
Let us summarize the results obtained for Eq. (20c) in table.
498 W.I. Fushchych, V.A. Tychinin

? h(x) f (x) g(x)
x?2 ?3
1
1 x0 0 ? (?) ?1 (x0 )x1 + ?2 (x0 ),
6
x?1 ?2x1
2 x0 + 1 ?1 (x0 )x1 + ?2 (x0 ),
1
?1 ?1
ln x0 + x?1 ?2x1 x0
3 3
3 x0 ?1 (x0 )x1 + ?2 (x0 ),
1
?1 ?1 ?1
?x0 ?2
3 3 3
4 ln x0 + arth x1 2x1 x0 x0 ?(?)dx1 + 2 ?(?)?
x2 +1
? x2 ?1 dx1 + 8 ?(?) (x2x1 2 dx1 ?
1
?1)
1 1

?dx1 + ?1 (x0 )x1 + ?2 (x0 ) ,
?1 ?1 ?1
ln x0 ? arctg x1 ?2x1 x0 ?(?)dx1 ? 2
3 3 3
5 x0 x0 2 ?(?)?
1?x2
? 1+x2 dx1 ? 8 ?(?) (1+x2 )2 dx1 ?
x1
1
1 1

?dx1 + ?1 (x0 )x1 + ?2 (x0 ) ,
1+x2
?2x1 ?2 ?(?) 1?x1 dx1 ?
6 x0 + arth x1 1 ?(?)dx1 + 2 2
1
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