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u(x) = ?(?)f (x),

(3.7)
u(x) = ?(?(?) + g(x)),

for (1.1), (1.2), (1.3) respectively. Here ? is an arbitrary differentiable function.
The formulae (2.1)–(2.3) yield
g(x) = ?2 ln ?(x) for (1.1),
f (x) = [?(x)]2/(1?k) (3.8)
for (1.2),
for (1.3).
g(x) = ln ?(x)
Below we present the explicit form of ?(x):
(1) ?(x) = ?? y ? , (2) ?(x) = ?? y ? , (3) ?(x) = ?? y ? ,
(4) ?(x) = ?? y ? , (9) ?(x) = ?? y ? .
In the other cases ?(x) = 1.
4. The exact solutions of the Liouville equation
Substituting (3.5) into (1.1) and using (1)–(10) and (3.8) one obtain the following
PDE:
a2 ?1 ?11 + 4?1 (?2 + a + 1)?12 + 4?2 (?2 ? 1)?22 + a(a ? 1)?1 ?1 +
2
(1)
(4.1)
+2(3?2 ? 1)?2 + 2 + ? exp ? = 0,

where ?1 = ?/b, ?ik = ? 2 /??i ??k , i, k = 1, 2.
b?11 + 4?12 ? 4?2 ?22 ? 2?2 + ? exp ? = 0. (4.2)
(2)

[b2 ? 1/(b2 ?2 ? 1)]?11 ? 4(b2 ?2 ? 1)?12 + 4?2 (b?2 ? 1)?22 ?
(3)
(4.3)
?b2 ?1 + 2(3b2 ?2 ? 1)?2 + 2b2 + ? exp ? = 0.

?11 + 2(2?2 + b1 )?12 + 4?2 ?22 ? ?1 + b?2 ?2 + 2 + (?/b2 ) exp ? = 0. (4.4)
2
(4)

4?1 ?11 ? 4a?12 ? ?22 + 4?1 + ? exp ? = 0. (4.5)
(5)
?1
? 4?1 ?11 + (a2 ?1 + 1)?22 + 4?1 + ? exp ? = 0. (4.6)
(6)

? ?11 + 2(?1 + a2 )?22 + (?/a2 b) exp ? = 0. (4.7)
(7)

(4.8)
(8) b?11 + 4?1 ?12 + 4?2 ?22 + b?2 + ? exp ? = 0.

(a11 ?1 ? 2a12 ?1 + a22 )?11 + 2(a11 ?1 ?2 ? a13 ?1 ? a12 ?2 + a23 )?12 +
2
(9)
+(a11 ?2 ? a13 ?2 + a33 )?22 + 2(a11 ?1 ? a12 )?1 +
2
(4.9)
+2(a11 ?2 ? a13 )?2 + 2a11 + ? exp ? = 0.
288 W.I. Fushchych, N.I. Serov

?11 ? ?22 + ? exp ? = 0. (4.10)
(10)
If one obtains at least one particular solution of any of equations (4.1)–(4.10) then
(3.5) gives a solution of (1.1). Let us consider, (4.1) and (4.10) as an example. If one
supposes that ??/??2 = 0 then (4.1) is reduced to the ordinary differential equation
for the function ?:
a2 ?1 ?11 + a(a ? 1)?1 ?1 + 2 + ?1 exp ? = 0,
2
(4.11)
the general solution of which has the form
?
?1/a
? ?2 ln ??/2bc2 1/2 ?1 1/a
? sinh c1 ?1 + c2 , ?b < 0,
? 1
?
?
?
? ?2 ln ?/2bc2 1/2 ? ?1/a cosh c ? 1/a + c
? , ?b > 0,
11 2
1 1
(4.12)
?(?1 ) =
? ?2 ln ??/2bc2 1/2 ? ?1/a cos c ? 1/a + c
?
? , ?b < 0,
? 11 2
1 1
?
?
?
? ?2 ln ?/2bc2 1/2 ? ?1/a ? 1/a + c , ?b > 0.
2
1 1 1

Hence from (3.5) and (4.12) one obtains the solution of (1.1)
u = ?2 ln[?P (x) sinh(c1 Q(x) + c2 )], u = ?2 ln[?P (x) cosh(c1 Q(x) + c2 )],
(4.13)
u = ?2 ln[?P (x) cos(c1 Q(x) + c2 )], u = ?2 ln[?P (x)(Q(x) + c2 )],
where P (x) = (?? y ? )?1/a , Q(x) = ?? y ? (?? y ? )1/2 , ? 2 = ?? 2 = ??/2bc2 .
1
Equation (4.10) is the two-dimensional Liouville equation. Its general solution was
found by Liouville [14]:
8 f1 (?1 + ?2 )f2 (?1 ? ?2 )
?(?1 , ?2 ) = ln ? (4.14)
,
? [f1 (?1 + ?2 ) + f2 (?1 ? ?2 )]2
where f1 and f2 are arbitrary differentiable functions.
Note. The two-dimensional Liouville equation can be solved in other ways, e.g. with
the help of the theory of complex variables. But we believe the simplest way is
to linearise the Liouville equation. Fushchych and Tychinin [12], using non-local
substitutions
c1 ? W
u = ln W? W? 1 ? tanh2 v ? = x0 ? x1 ,
, ? = x0 + x1 ,
2
or
u = ln 2W? W? /(W + c2 )2 ,
or
W + c3
v
u = ln W? W? 1 + tan2
2
reduce the Liouville equation to 2W = 0, the general solution of which was obtained
by d’Alembert. Using those formulae we obtain the Liouville solution (4.14).
From (3.5) and (4.14) one obtains a solution of (1.1):
8 f1 (?? x? )f2 (?? x? )
u = ln ? (4.15)
,
? [f1 (?? x? ) + f2 (?? x? )]2
where ?? ? ? = ?? ? ? = 0, ?? ? ? = 2.
The symmetry and some exact solutions 289

The other solutions of (1.1) we have obtained have the form (4.13) with
P (x) = F ?1 (?? y ? ), Q(x) = ?? y ? F (?? y ? ),
(a)
P (x) = F ?1 (?? y ? ), Q(x) = ?? y ? F (?? y ? ) ? ln F (?? y ? ),
(b)
Q(x) = (y? y ? )1/2 (?? y ? )?1 ,
P (x) = ?? y ? ,
(c)
Q(x) = ln ?1 , ?1 = (?? y ? )2 + y? y ? ,
(d) P (x) = ?1 ,
Q(x) = ?? y ? + a ln ?? y ? ,
(e) P (x) = 1,
Q(x) = ?? y ? ,
(f ) P (x) = 1,
Q(x) = ?? y ? + F (?? y ? ),
(g) P (x) = 1,
where F is an arbitrary differentiable function, ?? ?? = ?µ ? ? = 0, ?? ? ? = b = 0.
Besides, from (4.8) we have the particular solution of (1.1) in the form
1
u(x) = ? ln ?x? x? . (4.16)
2
We have obtained the solutions of (1.1) when n = 3 and they are easily generalised
to more general cases n ? 4. For n ? 4 some solutions of (1.1) may be obtained in an
analogous way, integrating (1.5) and determining the invariants ?(x).
5. The exact solutions of the nonlinear d’Alembert equation
Using (3.6) arid the explicit form of invariants ?(x) and function f (x) (1)–(10)
one obtains the tollowing PDE:
a2 ?1 ?11 ? 4a?1 (?2 ? a)?12 + 4?2 (?2 ? b)?22 +
2
(1)
+a[a ? 1 + 4/(1 ? k)]?1 ? + 2(k ? 1)?1 [(3k + 1)?2 ? 2b(k + 1)]?2 + (5.1)
+2(k + 1)(k ? 1)?2 ? + (?/b)?k = 0.

b?11 + 4?12 ? 4?2 ?22 + 2(3 + k)(1 ? k)?1 ?2 + ??k = 0. (5.2)
(2)

[b2 ? 1/(b2 ?2 ? 1)]?11 ? 4(b2 ?2 ? 1)?12 + 4?2 (b2 ?2 ? 1)?22 +
(3)
+b2 (3 + k)(1 ? k)?1 ?1 ? 4(1 ? k)?1 [?2 b(3 ? k) ? (1 + k)]?2 + (5.3)
+2b2 (1 + k)(1 ? k)?2 ? + ??k = 0.

?1 ?11 + 2?1 ?2 [1 ? (b1 /2b2 )?2 ]?12 + ?2 ?22 + 4(1 ? k)?1 (?1 ?1 +
2 2 2
(4)
(5.4)
+?2 ?2 ) + 2(1 + k)(1 ? k)?2 ? + (?/b2 )?k = 0.

4?1 ?11 ? 4a?12 ? ?22 + 4?1 + ??k = 0. (5.5)
(5)
?1
? 4?1 ?11 + (1 + a2 ?1 )?22 + 4?1 + ??k = 0. (5.6)
(6)

? ?11 + (2?1 + a2 )?22 + (?1 /a2 )?k = 0. (5.7)
(7)

b?11 + 4?1 ?12 + 4?2 ?22 + b?2 + ??k = 0. (5.8)
(8)

(a11 ?1 ? 2a12 ?1 + a22 )?11 + 2(a11 ?1 ?2 ? a13 ?1 ? a12 ?2 + a23 )?12 +
2
(9)
+(a11 ?2 ? 2a13 ?2 + a33 )?22 + 2(k + 1)(k ? 1)?1 [(a11 ?1 ? a12 )?1 + (5.9)
2

+(a11 ?2 ? a13 )?2 ] + 2a11 (k + 1)(k ? 1)?2 ? + ??k = 0.
290 W.I. Fushchych, N.I. Serov

?11 ? ?22 + ??k = 0. (5.10)
(10)

Equation (5.1) when ??/??2 = 0 becomes the Emden–Fowler one
? 2 V?? + 2?V? + (?/b)? k+1 V k = 0 (5.11)
via the substitution
1/a
? = ? (k+1)/(k?1) V (?), (5.12)
? = ?1 .
We have found some particular solutions of (5.2)–(5.10) and then we have the
following solutions of (1.2):
2/(1?k)
u = [?? y ? + ?? y ? (c2 + ln a? y ? )] (5.13)
,
where ?? ?? = ?? ? ? = 0, ?? ? ? = b = ? 2 ?(1 ? k)2 /(1 + k).
1

1/(1?k)
1
(k ? 1)2 /(k ? 3) y? y ? (5.14)
u= ,
2
1/(1?k)
1
? ? 1 ? k2 ?2 ?
(5.15)
u= (?? y ) + y? y ,
2
where ?? ? ? = ?1.
2/(1?k)
1/2
1
c2 ± (1 ? k( ?(1 + k)?1 (?? y ? + a ln ?? y ? ) (5.16)
u= ,
2

where ?? ?? = ?? ? ? = 0, ?? ? ? = ?1.
2/(1?k)
u = c2 + (2a)?1 (?? y ? )2 + ?? y ? (5.17)
,
where ?? ?? = ?? ? ? = 0, ?? ? ? = ? 2 ?(1 ? k)2 (1 + k).
1

From (5.13)–(5.17) one can see that all the solutions of (1.2) obtained have the
form
u = [F (y) + G(z)]? , (5.18)
where ? takes the values 1/(1 ? k) and 2/(1 ? k), and
y = y 1 , . . . , y n?1 , z = z 1 , . . . , z n?1 ,
a = 1, n ? 1, x ? Rn .
y a = y a (x), z a = z a (x),
If one searches for the solutions of (1.2) in the form (5.18), then the substitution
of (5.18) in (1.2) leads to the equation for the functions F , G, y a , z a :
(? ? 1)Aµ Aµ + (F + G) Fab yµ y bµ + Gab zµ z bµ + Fa 2y a + Ga 2z a +
a a
(5.19)
+(?/?)(F + G)?(k?1)+2 = 0,
where
Fa ? ?F/?y a ,
a a
Fab = ? 2 F/?y a ?y b ,
Aµ = Fa yµ + Ga zµ ,
Ga ? ?G/?z a , yµ ? ?y a /?xµ ,
Gab = ? 2 G/?z a ?z b , a
(5.20)
zµ ? ?z a /?xµ , µ = 0, n ? 1, a, b = 1, n ? 1.
a
The symmetry and some exact solutions 291

Below we list some particular solutions of (5.19):

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