<< Ïðåäûäóùàÿ ñòð. 130(èç 135 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>
where D(u) ? C 1 (R1 , R1 ) is an arbitrary function. Further we consider in detail the
case D(u) = ? ? const.
On putting z 2 = R(? ) we get the following equation:
4? 4
R? = ? R + R2 + ?R ? S(R).
2
(75)
3
Integration of (75) yields
z2
dR
= ±(ln v + ln C(u)). (76)
S(R)
0

Here C(u) is an arbitrary smooth function.
On the general solution of the d’Alembert equation 553

Let us represent the polynomial S(R) in the form
4
S(R) = ? ?R(R ? ?1 )(R ? ?2 )(R ? ?3 ),
3
where ?i are the roots of the polynomial S(R) that satisfy equations (the Vieta’s
theorem)
3 3?
?1 ? 2 + ? 2 ? 3 + ? 3 ? 1 = ?
?1 + ?2 + ?3 = 0, , ?1 ? 2 ? 3 = .
4? 4
The explicit form of the integral in the left side of Eq. (76) depends on relations
connecting the roots ?i .
Case 1. ?1 = 0, ?2 = ?3 , ?2 = 0, ?3 = 0. Such a case taken place under ? = 0,
solution of Eq. (72) being given by the formulae
v 1/2
3C(u)
under ? = a2 > 0, (77)
?=
a(1 + C 2 (u)v)2

v 1/2
3C(u)
under ? = ?a2 < 0, (78)
?=
a(1 ? C 2 (u)v)2

Case 2. ?1 = ?2 , ?2 = 0, ?3 = 0, ?3 = ?2 . Such relations are satisfied provided
? = a2 > 0, ? = ±(3a)?1 , solution of Eq. (72) taking the form
1/2
sin(ln(vC(u))) + 1
(79)
?= .
av(2 sin(ln(vC(u))) ? 4)

Case 3. ?1 = ?2 , ?2 = ?3 , ?3 = ?1 . ? = ?a2 < 0. In such a case the polynomial
S(R) has two real and two complex roots. Therefore it is represented in the form

4a2
R(R + ?1 )((R + ?2 )2 + ?3 ),
2
S(R) =
3
solution of Eq. (72) taking the form
? ?1/2
2a v
? ?
p?1 1 ? cn v pq ln(vC(u))
3
(80)
?= .
2a v
? v (p + q)cn +q?p ?
v pq ln(vC(u))
3

Here
2
(p + q)2 + ?1
1
2 2 2
(?1 + ?2 )2 + ?3 ,
p= ?2 + ?3 , q= h= .
2 pq
1
Case 4. ?1 = ?2 , ?2 = ?3 , ?3 = ?1 , 0 < ? < (3?)2 , ? = a2 . The polynomial S(R)
has two real and two complex roots and is given by the formula

4a2
R(?1 ? R)((R + ?2 )2 + ?3 ).
2
S(R) =
3
554 W.I. Fushchych, R.Z. Zhdanov, I.V. Revenko

The solution of Eq. (72) has the form
? ?1/2
2a v
? ?
q?1 1 + cn v3 pq ln(vC(u))
(81)
?= ,
? v (p + q) + (q ? p)cn v vpq ln(vC(u)) ?
2a
3

where
?1 ? (p + q)2
2
1
(?2 ? ?1 2 2 2
)2
p= + ?3 , q= ?2 + ?3 , h= .
2 pq
Case 5. ?1 = ?2 , ?2 = ?3 , ?3 = ?1 , ? = a2 > 0, ?(3?)2 < 1. Is this case the
polynomial S(R) has four real roots ?0 < ?1 < ?2 < ?3 (one of them is equal to zero)
and is represented in the form
4a2
(?0 ? R)(R ? ?1 )(R ? ?2 )(R ? ?3 ).
S(R) =
3
? ?1/2
??0 (?1 ? ?3 ) ? ?3 (?1 ? ?0 ) sn2 v (?0 ? ?2 )(?1 ? ?3 ) ln(vC(u)) ?
a
3
?= .(82)
? v ? ? ? ? (? ? ? ) sn2 v (?0 ? ?2 )(?1 ? ?3 ) ln(vC(u)) ?
a
1 3 1 0 3

In the above formulae (80)–(82) cn, sn are elliptic functions of the order k.
Substituting the formulae (77)–(82) into the ansatz 2 from (67) with h ? 0, where
u = u(x) is determined by JSSF (43a) we obtain wide families of new exact solutions
of the nonlinear PDE (34) under F (w) = ?w5 .
Eq. (73) is integrated in analogous way. As a result we have
? = ?a2 < 0,
1.
v
(83)
1 2
tg ± 2 ln(vC(u)) ;
?=
av a

? = a2 > 0,
2.
v
(84)
2 2C(u)
?= ;
a(1 + v 2 C 2 (u))
? = ?a2 < 0,
3.
v
(85)
2 2C(u)
?= ;
a(1 ? v 2 C 2 (u))
? = 2a?2 > 0, a > 0,
4.
v
(86)
b2 + d 2
b
? = cn ln(vC(u)) ,
v a

where
1/2 1/2
d = ?a + a
2 2
a2 a2
b= a +a + 4? , + 4? ,
k = b?1
? ? R1 , b2 ? d 2 ;
On the general solution of the d’Alembert equation 555

? = 2a?2 > 0, a > 0,
5.
(87)
b b
? = dn ln(vC(u)) ,
v a
where
1/2 1/2
k = b?1
d = a2 ? a a2 + 4? b2 ? d 2 ;
b = a2 + a a2 + 4? , ,

? = ?2a?2 < 0, a > 0,
6.
v
b2 ? d 2 (88)
b
?= cn ln(vC(u)) ,
v a

where
1/2 1/2
d = a a2 + 4? ? a2
b = a2 + a a2 + 4? , ,
k = d(b2 + d2 )?1/2 ;
? > 0,
a2
? = ?2a?2 < 0, ?
7. < ? < 0, a > 0,
4
(89)
b b
? = tn ln(vC(u)) ,
v a
where
1/2 1/2
k = b?1
d = a2 ? a a2 + 4? b2 ? d 2 ;
b = a2 + a a2 + 4? , ,

? = ?2a?2 < 0,
8.
1/2
2b (90)
1 + cn ln(vC(u))
b a
?= ,
1 ? cn 2b
v ln(vC(u))
a

where
v
b2 ? d 2
a2
4
k= v
?4?a2 , ?<? ,
b= .
4 2b
In the above formulae (83)–(90) cn, dn, tn are elliptic functions of the order k.
Substituting the formulae (83)–(90) into the ansatz 1 from (67) with h = 0, where
u = u(x) is determined by JSSP (43a) we get ad families of exact solutions of the
nonlinear Eq. (34) under F (w) = ?w3 .
Let us emphasize once more that solutions of nonlinear PDE (34) obtained in the
above described manner contain several arbitrary functions and cannot in principle be
constructed by means of symmetry reduction procedure.
In conclusion, we adduce two examples of exact solutions of Eq. (34) with F (w) =
3
?w that can be written down in the explicit form
v
1/2 1/2
u(x) = x2 + x2 + x2 ? x2 2 ln x2 + x2 + x2 ? x2
tg +
1 2 3 0 1 2 3 0

x0 x1 ± x2 x2 + x2 ? x2
1 2 0
+ ln C ,
2 + x2
x1 2
556 W.I. Fushchych, R.Z. Zhdanov, I.V. Revenko

v
1/2 1/2
u(x) = x2 + x2 + x2 ? x2 2 ln x2 + x2 + x2 ? x2
tg +
1 2 3 0 1 2 3 0
(91)
x1 x2 ± x0 x2 + x2 ? x2
1 2 0
+ ln C .
2 + x2
x1 2
 << Ïðåäûäóùàÿ ñòð. 130(èç 135 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>