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where ? is an arbitrary function of its arguments and ? is an arbitrary function of the
similarity variable ?. We call solutions, obtained by Q-symmetries, the Q-similarity
solutions.
Let us consider the following cases:
Q-symmetry generators and exact solutions 157

Case l(a): Consider the equation
?2u
?u ?
? ? 2 = ? 2 exp(2u) + k2 exp(u). (28)
?x0 ?x1 k1
This corresponds to ? = exp(u) for the Q-symmetry given in theorem 1. By solving
(27) for the Q-symmetry
? ?
Q = k1 + exp(u)
?x1 ?u
we obtain the similarity ansatz
x1
u = ? ln ?(?) ? and ? = x0 .
k1
On insertion into (28) we obtain the reduced equation
d?
? k2 = 0. (29)
d?
An exact solution of (28) is thus given by
x1
u(x0 , x1 ) = ? ln ? + k2 x0 + k3 ,
k1
where k1 , k2 , k3 are arbitrary real constants.
Case l(b): Consider the equation
?2u
?u
? ? 2 = ?(b5 u5 + b3 u3 + b1 u). (30)
?x0 ?x1
This corresponds to ? = a3 u3 + a1 u for the Q-symmetry given in theorem 1 with
k1 = 1 and
? ?
2
1? 3? 3?b3 b5
? 12b1 ?? , a3 =
a1 = b3 + ,
6? b5 b5 3?
? ?
1 3?b2 3b2 ?
2
+? ? 12b1 ?? .
c1 = ? 3 3
3 b5 3 b5

By solving (27) for the Q-symmetry
? ?
+ (a3 u3 + a1 u) ,
Q = k1
?x1 ?u
we obtain the similarity ansatz
v
a1
exp(a1 x1 /c) 1 ? a3 ?2 (?) exp(2a1 x1 /c) and ? = x0 .
u=
?(?)
The reduced equation is given by
d?
? a1 c1 ? = 0, (31)
d?
158 N. Euler, A. K?hler, W.I. Fushchych
o

with the general solution

?(?) = c exp(a1 c1 ?).
?

Here c is an arbitrary real constant. An exact solution of (30) is thus given by
?
v
c a1 exp[a1 (x1 + c1 x0 )]
?
u= .
1 ? a3 c2 exp[2a1 (x1 + c1 x0 )]
?

Case 2: Consider the equation

?2u
?u
? ? 2 = ?b3 u3 ? b2 u2 + b1 u + b0 . (32)
?x0 ?x1

This corresponds to ?(u) = q3 u2 + q2 u + q1 for the Q-symmetry in theorem 2. Here
v2
b3 c 2 c b1 2b2
q1 = v v+
q3 = ?c q2 = ?
, b2 , .
2? 3 ?b3 9b3
?b3 2

The real constants b1 , b2 , b3 are related to the constants k1 , k2 , k3 and k4 in the
Q-symmetry given in theorem 2 by the relations

2cb2
b2 c 2 b1 c
k4 = v +v 2 ,
k1 = , k2 = , k3 = 0,
3b3 3 ?b3 2?b3 9 2?b3 b3
where
2b3
b1 b2 2
b0 = + .
27b2
3b3 3

In terms of b2 and b3 , ?0 is given by
v
c 2b3
?0 (u) = v .
?(3b3 u + b2 )
In order to solve (27) for the above given ?0 and ? we must solve the equation

d2 y dy
? q2 + q3 q1 y = 0,
d?2 d?
where
1d
u=? (ln y).
q3 d?
? is the group parameter. Thus, there are three cases to be studied:

q2 ? 4q1 q3 = 0, q2 ? 4q1 q3 < 0, q2 ? 4q1 q3 > 0.
2 2 2


Case 2(a): Consider q2 ? 4q1 q3 = 0, i.e.,
2


1 b2
b1 = ? 2 .
3 b3
Q-symmetry generators and exact solutions 159

The similarity ansatz is given by
v
b2 [x1 ? ?(?)] ? 3 2b3 ? 3b3
and ? = x0 ?
u= .
3b3 [?(?) ? x1 ] (3ub3 + b2 )2
The reduced equation then takes the form
3
d2 ? 1 d?
? (33)
= 0,
2
d? 3? d?
which has the general solution
2
?(?) = ?3? ? ? + c1 + c2 .
? ?
3?
Here c1 , c2 are arbitrary real constants. Solving for u, an exact solution of (32) takes
??
the form
v
6 2b3 ?(x1 ? c2 ) ? b2 (x2 ? 2?2 x1 ? 9?2 c1 + 6?x0 + c2 )
? c ? ?2
1
u= .
2 ? 2? x ? 9?2 c + 6?x + c2 )
3b3 (x1 c2 1 ?1 ?2
0

Case 2(b): Consider q2 ? 4q1 q3 < 0, i.e.
2


b2
b1 + 2 < 0.
3b3
The similarity ansatz is then given by
? 1 ?
u=? ?,
q3 tan{?[x1 ? ?(?)]/c} q3
?1/2
c2 ?2
? = x0 + ln 1+ ,
3?? 2 (q3 u + ?)2

where
q2 1
4q1 q3 ? q2 .
2
?= , ?=
2 2
The reduced equation takes the form
3
d2 ? d? d?
(34)
A 2 +B +C = 0,
d? d? d?
where
A = 6b3 (81b4 b4 + 108b3 b2 b3 + 54b2 b4 b2 + 12b1 b6 b3 + b8 ),
13 123 123 2 2
B = 9b1 b3 (81b1 b3 + 135b1 b2 b3 + 90b1 b2 b3 + 30b1 b2 b3 + 5b8 ) + 3b10 ,
44 323 242 6
2 1

and
1
C = ? A.
3
The general solution of (34) is

A B
exp(2B?/A) ? C?1 /A + c2 ,
?(?) = arctan c ?
B C
160 N. Euler, A. K?hler, W.I. Fushchych
o

where c1 , c2 are arbitrary real constants. An implicit solution of (32) can then be
??
given in the form
?c2 B/(2A? 2 ?)
?2
A B 2Bx0 C?1
c
?
arctan exp 1+ +
(q3 u + ?)2
B C A A
c ?
arctan ? = x1 ? c2 .
+ ?
? q3 u
Case 2(c): Consider q2 ? 4q1 q1 > 0, i.e.
2

b2
2
b1 + > 0.
3b3
The similarity ansatz is then given by
v v
?(?) exp(A1 + A2 )(q2 + ?) ? 4q3 ?b2
v
u=
12q3 ?b3 ? 2q3 ?(?) exp(A1 + A2 )
and
v
v
2uq3 + q2 ? ?
v exp(?x1 ?/c),
?=?
2uq3 + q2 + ?
where
v
b2 q 2
x0 ?
v ?b2 q2 q3 + 3b3 q1 q3 + 2 3
A1 = ,
3b3
cq3 2b3
v
x1 q2 ? b2 q 3
?
A2 = +
c 2 2 3b3
and
2c2 b2
2

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