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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 , ?)
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. Гуpсa E., Iнтегрування рiвнянь з частинними похiдними першого порядку, Київ, Рад. шк., 1941,
415 c.
2. Фущич В.И., Тычинин В.А., Жданов Р.З., Нелокальная линеаризация и точные решения неко-
торых уравнений Монжа–Ампера, Дирака, Препринт № 85.88, Киев, Институт математики АН
УССР, 1985, 28 с.
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
Перетворення годографа однiєї скалярної функцiї в R(1, 1) та R(1, 3), а також двох
скалярних функцiй в R(1, 1) використанi для розмноження розв’язкiв нелiнiйних
piвнянь; побудованi класи годограф-iнварiантних piвнянь другого порядку.

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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