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js? (Zv ?) = 0, (7)
with Zv an associated vertical Lie symmetry generator for the equation. This ansatz
will reduce the dimension of (2) by one. The solution of the reduced equation is known
as a similarity solution of (2). Thus, the existence of a symmetry provides us with
a similarity ansatz and a possible exact solution can be calculated. The converse is
however not and true, i.e., any exact solution of a partial differential equation is not
associated with a symmetry of the equation. For such solutions one can introduce
conditional symmetries, i.e. symmetries that leave the equation invariant under some
additional condition.
2. Q-symmetry generators
Following [9–13] we give the definition for conditional invariance of (2).
Q-symmetry generators and exact solutions 153

Definition. Equation (3) is called Q-conditionally invariant if

(8)
LQv F = 0
?
?

under the condition

(9)
Qv ? = 0.
?
Q is called the Q-symmetry generator and Qv the prolonged vertical Q-symmetry
generator.
Here = indicates the restriction to solution of F = 0 and (9) together with their
?
prolongations. Q is considered in the form of a Lie symmetry generator.
Let us now study (1) by the used of the above definition. We are interested only
in nonlinear functions f . From the definition it follows that the Lie derivative (8), for
the equation

F ? u0 ? ?u11 ? f (u) = 0 (10)

under the condition

Qv ? ? ? ? ?0 u0 ? ?1 u1 = 0, (11)

has to be studied. Let us consider the Q-symmetry generator in the form
? ? ?
(12)
Q=c + ?1 (u) + ?(u) ,
?x0 ?x1 ?u
where c is an arbitrary real constant. We can state the following
Theorem 1. The generator
? ?
(13)
Q = k1 + ?(u)
?x1 ?u
is a Q-symmetry generator for (1) if an only if
? d?
f (u) = ?(u) ? (14)
+ c1 ,
2
k1 du
where ? is an arbitrary differentiate function of u and k1 , c1 are arbitrary real
constants.
Proof. By applying the Lie derivative (8) and condition (9), with generator (13), we
obtain the following determining equations using computer algebra [15, 16]:

d2 ?1 d2 ?1 2 d2 ?
d?1 2
c ? 2 = 0, ?3c? 2 ? + (3cf + 2?)
3
? + 2c? 2 ?1 ? = 0,
du 1
du du du
2 2
d ?1 d?1 2 d?
c 3c? 2 ? ? 2 ?1 ? c? 2 ?1 = 0
du du du
and
d2 ?1 3 d2 ?
d?1 2 d? 3 df 3
? ? 3f ?1 ? + f ?1 ? ? 2 ?1 ? 2 ?
? ? ? = 0.
du 1
du2 du du du
154 N. Euler, A. K?hler, W.I. Fushchych
o

For c = 0 the general solution of the above four equations gives only a linear function
f (u) = a1 u + a2 , where a1 and a2 are arbitrary real constants. For c = 0 the general
solution
? d?
?1 (u) = k1 , f (u) = ?(u) ? 2 + c1 ,
k1 du
follows.
Consider the Q-symmetry generator in the form
? ? ?
(15)
Q = ?0 (u) +c + ?(u) ,
?x0 ?x1 ?u
where c is an arbitrary constant. We can state the following
Theorem 2. The generator
2 c2
k2 ? ? 12 k3 ?
+? u + k1 u ? (16)
Q= +c + k4
u + k1 ?x0 ?x1 3 ?k2 2 u + k1 ?u
is a Q-symmetry generator for (1) if an only if
2?
(17)
f (u) = ,
3 ?0
where
k2
?0 =
u + k1
and
2 c2 12 k3
?=? u + k1 u ? + k4 .
3 ?k2 2 u + k1
Here k1 , . . . , k4 are arbitrary real constants.
Proof. Applying the Lie derivative (8) and condition (9), with generator (15), the
determining equations are given by
c(3f ?0 ? 2?) = 0,
2
d2 ?0 d2 ? 2
d?0 d?0 d? d?0
? 2 ?0 ? ? 2? ? ? 2 ?0 = 0,
?0 + 2c2
? + 2?
du du du du du du
2
d2 ?0 d?0
c ? 2 ?0 + 2 =0
du du
and
d?0 d? df
?f ? + f ?0 ? ?0 ? = 0.
du du du
For c = 0 only linear functions f are obtained for the general solution of the above
system. If c = 0, f follows from the first determining equation and the condition on ?
is in the form of a linear second order equation, namely
d2 ? 2c2
2 d?
+ + 2 = 0.
du2 u + k1 du ?k
The general solution, given in theorem 2, follows.
Q-symmetry generators and exact solutions 155

For the nonlinear equation

?2u
?u
= auk , (18)
+
?x2
?x0 1

with a and k arbitrary real constants, and k = 1, we can state the following
Theorem 3. The generator
? ? ?
(19)
Q= + ?1 (x0 , x1 ) + ?(x0 , x1 )u ,
?x0 ?x1 ?u
is a Q-symmetry generator for (18) if and only if the following conditions on ?1 and
? are satisfied:

?2?
??
2 = (k ? 1)? ,
2
(20)
+
?x0 ?x1

??1 k + 3 ??
? (k ? 1)??1 = (21)
?x0 2 ?x1

and
1?k
??1
(22)
= ?.
?x1 2
The proof follows directly from the invariance condition (8) together with (9).
Note that the above condition on ?1 reduces to the following third order ordinary
differential equation:

2 d3 ?1 d2 ?1
+ ?1 2 = 0,
k ? 1 dx3 dx1
1

which can be transformed to the Abel equation of the second kind
k?1
dy
y + 6x2 + (k ? 1)x = 0
+ y 2 + 7x +
xy
dx 2
where
?1 dx
2
= P (?1 ), P (?1 ) = ?1 x(z), z = ln(?1 ), = y(x).
x1 dz
With other special ans?tze for ?0 , ?1 and ? we obtain the following results for (1)
a
with ? = ?1.
Theorem 4. 1. The generator
v ? ?
(23)
Q = 2 x0 + f (u)
?x1 ?u
is a Q-symmetry generator for (1) (? = ?1), if and only if f satisfies the equation

d2 f
(24)
f = 2.
du2
156 N. Euler, A. K?hler, W.I. Fushchych
o

The general solution of (24) is given by
df ?
± = u + k2 ,
?
4 ln f + k1

? ?
where k1 and k2 are integrating constants.
2. The generator
? ?
(25)
Q = x1 + f (u)
?x1 ?u
is a Q-symmetry generator for (1) (? = ?1), if and only if f satisfies the equation

d2 f df
?1 . (26)
f =2
du2 du
The general solution of (26) is given by

?
f (u) = ± (w ? 1) exp(w)/k1 ,

where w is obtained from

??1 ?
k1 (w ? 1)?1 exp(w)dw = u + k2 .
±

? ?
Here k1 and k2 are integrating constants.
The proof follows by applying the invariance condition (8) together with (9).
Let us make some remarks on Q-symmetries. The determining equations for Q-
generators are nonlinear over-determined systems of differential equations. This is
in contrast to Lie symmetry generators where the determining equations are linear
differential equations. It is obvious that every Lie symmetry of an equation is also a
Q-symmetry but that the converse is not true, so that the above Q-symmetries do not
generate Lie transformation groups that leave the equation invariant. If we multiply a
Q-symmetry (or Lie symmetry) of a particular equation by an arbitrary function, we
again find a Q-symmetry for that equation.
3. Q-similarity solutions
Let us now make use of theorems 1 to 4 to construct exact solutions of (1). The
similarity ansatz is obtained by solving the linear partial differential equation
?u ?u
js? (Q ?) ? ?0 ? ? = 0. (27)
+ ?1
?x0 ?x1
We seek the general solution of (27) in the form

?(x0 , x1 , u) = ?{?[x0 , x1 , u(x0 , x1 )]},

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