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u0 u11 ? u2 + 1 = 0, (7)
11

which is invariant under the transformation (1). The function
(1)
u (x0 , x1 ) = ?(?), ? = ?0 x0 + ?1 x1
102 W.I. Fushchych, V.A. Tychynin

is a solution of Eq. (7), when ? satisfies the first order ODE

q + 2(k ?)?1 (k ?)2 + 1 + 4k? + c1 = 0,
? ?
(8)
q ? ln k ? ? (k ?)2 + 1 ? ln k ? + (k ?)2 + 1 .
? ? ? ?

Here k = 1 ?0 , ?1 = 1, ?1 = 1, ?0 , c1 are arbitrary constants. In this case the
2
generating of new solutions is realized according to the formulae
(2)
u (x0 , x1 ) = ? · ?(?) ? ?(?), (9)
? x1 = ?(?),
? ? = 2kx0 + ?.

From the second equation of the system (9) we get

? = [?]?1 (x1 ) ? 2kx0 .
?

Here [?]?1 (x) is the inverse function to ?(x). Note, that
? ?

[?]?1 (x1 ) = ?, ?(?) = ?(?) = ?([?]?1 (x1 )) = x1 .
? ? ? ??

Then from the first equation of the system (9) we obtain
(2)
u (x0 , x1 ) = ? · x1 + ?(2kx0 + ? ). (10)

Here ?(2kx0 + ? ) is a solution of ODE (8) of argument 2kx0 + ? . Due to equality
(1)
x1 = u (x0 , ? ) = ?(2kx0 + ? ) we get ? from the correlation (8)
?

? = ? g ? + 2(kx1 )?1 (kx1 )2 + 1 + 2kx0 + (4k)?1 · c1 ,
(11)
g ? ? ln kx1 ? (kx1 )2 + 1 ? ln kx1 + (kx1 )2 + 1 .

Thus, the solution u(x0 , x1 ) is determined by the parametric system of equations
(2)
u (x0 , x1 ) = ? ?q ? ? 2(kx1 )?1 (kx1 )2 + 1 ? (4k)?1 · c1 ?
(12)
? ?1 ?1
? x1 g + 2(kx1 ) · c1 ,
)2
(kx1 + 1 + 2kx0 + (4k)

q + 2(k ?)?1 (k ?)2 + 1 + 4k? + c1 = 0, (13)
? ? ? = ?(?).

Example 2. The equation

(u0 ? ?(2) u)(u11 u22 ? u2 ) ? ?(2) u = 0 (14)
12

is (1)-invariant, when the condition u11 u22 ? u2 = 0 is satisfied. The partial solution
12
of Eq. (13) is
(1)
u = ln r2 , r2 = x2 + x2 .
1 2

(1)
Let us replace xa , a = 1, 2 in u for parameters ? a
(1)
u = ln ?2 , ?2 = (? 1 )2 + (? 2 )2 .
Solutions for nonlinear equations via the Euler–Amper? transformation
e 103

and substitute this result into to the formula (6). We obtain
(2)
xa = 2? a ??2 ,
u (x0 , x1 , x2 ) = 2 ? ln ?2 , (15)
a = 1, 2.

Let us express ? 1 , ? 2 through x1 , x2 from the last two conditions of the system (15)

? a = 2xa r?2 , (16)
a = 1, 2.

Substituting ? from (16) into the first equation of the system (15), we get the soluti-
(2)
on u :
(2)
u = 2(1 ? ln 2) + ln r2 .

2. Nonlocal linearization and nonlinear superposition formula. Let us apply the
transformation (1) to a general second order linear PDE

bµ? (y0 , y)vµ? + bµ (y0 , y)vµ + b(y0 , y)v + c(y0 , y) = 0. (17)

y = (y1 , y2 , . . . , yn?1 ), bµ? = b?µ , bµ , b, c are arbitrary smooth functions of y0 , y. As
a result we get the nonlinear equation

{b00 (x0 , u) det(uµ? ) ? 2b0a (x0 , u)u0b aba (ucd ) +
1 1
?1
(ucd ) + b0 (x0 , u) · u0 +
ab
+ b (x0 , u)aab (ucd )} det (18)
1 1
+ ba (x0 , u)xa ? b(x0 , u)[xa ua ? u] ? c(x0 , u) = 0.
1 1 1

Here u = (u1 , u2 , . . . , un?1 ), a, b = 1, n ? 1. Eq. (18) possesses the solutions super-
1
position property, which arises from the superposition of solutions of the linear equa-
tion (17)
(3) (1) (2)
v (y0 , y) = v (y0 , y) + v (y0 , y).
(k) (3)
Let u , k = 1, 2 be known solutions of Eq. (18) and u (x0 , x) be a new solution of the
(3) (1) (2)
same equation. Let us express u through u and u . Making use of Euler–Amper?
e
transformation (1), we get
(3) (3) (1) (2)
(3) (1) (2)
u (x0 , x) = ya va ? v = ya (va + va ) ? v ? v ,
(19)
(3) (1) (2)
xa = va = va + va , x0 = y0 .

(1) (2) (1) (2)
One can express v and v via u and u accordingly, where x are replaced for
parameters ? = (? 1 , ? 2 , . . . , ? n?1 ) in the first and ? = (?1 , ?2 , . . . , ?n?1 ) in the second
ones:
(k) (k) (k) (k) (k) (1)
? u, ? ? ?,
au
y0 = x0 = ? 0 ,
v=? k = 1, 2,
a
(20)
(k) (k) (k) (2) (k) (k)
? ? ?,
u a ( ? 0 , ? ), ?a
va
ya = = .
104 W.I. Fushchych, V.A. Tychynin

Substituting the relations (20) into (19) we obtain the solutions superposition formula
for Eq. (18)
(3) (1) (2) (1) (2)
u (x0 , x) = u (x0 , ? ) + u (x0 , x ? ? ), ? = x ? ?.(21)
u a (x0 , ? ) = u a (x0 , ?),
Here the second equation of the system (19) is used essentially xa = ? a + ?a for elimi-
(2) (2)
nating parameters ?a in the formula (21) and the designation u a (x0 , ?) ? u ?a (x0 , ?)
is adopted as well.
Example 3. Let us use as initial partial solutions
2v
1
(1) (2) 3
u (x0 , x1 ) = x0 ? x2 , u (x0 , x1 ) = k[1 + x1 ? x0 ] 2 , k = ? 2
21 3
of the Euler–Amper?-linearizable equation
e
(22)
u0 u11 + 1 = 0, u11 = 0.
(1) (2)
Replacing the argument x1 for parameter ? in u and for parameter ? = x1 ? ? in u
and making use of the formula (21), we obtain a new solution of Eq. (22)
2v 3
(3)
u (x0 , x1 ) = x0 ? h ? 2h 2 , h ? 2 + x1 ? x0 ± 2(x1 ? x0 ) + 3. (23)
3
Example 4. The nonlinear heat conduction equation
u0 det(uab ) + ?(2) u = 0,
(24)
?(2) ? ?1 + ?2 , det(uab ) = 0
2 2

admits the linearization under the transformation (1) to the equation
v0 ? ?(2) v = 0.
This Eq. (24) possesses the partial solution in parametric form
(1)
u (x0 , x1 , x2 ) = ?x?1 r2 + 2x0 x1 ??1 , r2 = x2 + x2 ,
1 2 (25)
1
x1 (8?x2 ) = ±? exp{??2 r2 (8x0 x2 )?1 }.
0 1
Let the second solution of Eq. (24) take the form
(2)
u (x0 , x1 , x2 ) = x2 ? x2 . (26)
1 2
(3)
Making use of the formula (21) we obtain the new solution u :
1
(3)
u (x0 , x1 , x2 ) = ?(x1 ? ?)?2 x1 ? ? (r2 + ?2 ? 2x1 ?) +
2
2
1 1 1
?1
x1 ? ? + ?2 ? x2 (x1 ? ?)?2 x1 ? ? , (27)
+ 2x0 ? 2
2 4 2
(r + ? ? 2x1 ?)
2 2
1
x1 ? ? = ±? exp ??2
8?x2 ,
4x0 (x1 ? ?)2
0
2
? is the parameter to be eliminated.
1. pa E., I i i , , . ., 1941,
415 c.
2. .., .., .., -
–, , 85.88, ,
, 1985, 28 .
W.I. Fushchych, Scientific Works 2003, Vol. 5, 105–111.

Hodograph transformations and generating of
solutions for nonlinear differential equations
W.I. FUSHCHYCH, V.A. TYCHYNIN
i i R(1, 1) R(1, 3),
i R(1, 1) i ’i ii
pi; i -ii pi .

The results of using the hodograph transformations for solution of applied problems
are well-known. One can find them for example in [1, 2, 3]. We note also the paper [4],
in which a number of invariants for hodograph transformation as well as hodograph-
invariant equations were constructed.
1. Hodograph-invariant and -linearizable equations in R(1, 1). Let us consi-
der the hodograph transformation for one scalar function (M = 1) of two independent
variables x = (x0 , x1 ), n = 2:
u(x) = y1 , x0 = y0 , x1 = v(y),
(1)
?v
? = v1 = ?1 v = = 0, y = (y0 , y1 ).
?y1
Differential prolongations of the transformation (1) generate such expressions for the
first and second order derivatives:
?1 ?1
u0 = ?v0 v1 , (2)
u 1 = v1 ,
?3 ?3
u11 = ?v1 v11 , u10 = ?v1 (v1 v10 ? v0 v11 ),
(3)
?3 2
u00 = ?v1 [v0 v11 ? 2v0 v1 v10 + v1 v00 ].
2


It is clear that (1) is an involutory transformation. This allows to write a set of
differential expressions of order ? 2, which are absolutely invariant under the trans-
formation (1):
f 2 (u1 , u?1 ), f 3 (u0 , ?u0 u?1 ), f 4 (u11 , ?u?3 u11 ),
f 0 (x0 ), f 1 (x1 , u), 1 1 1
(4)
f 5 (u10 , ?u?3 (u1 u10 ? u0 u11 )), f 6 (u00 , ?u?3 [u2 u11 ? 2u0 u1 u10 + u2 u00 ]).
0 1
1 1

Here f 0 is an arbitrary smooth function, f i , i = 1, 6 are arbitrary functions symmetric
on arguments, i.e. f i (x, z) = f i (z, x). So, the second order PDE invariant under the
transformation (1) has the form
F ({f ? }) = 0, {f ? } = {f 0 , f 1 , . . . , f 6 }, (5)

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