Equivalence Group of
Lie's Second-Order Equation
of Type III
Dai Jiayuan
[email protected]
Thesis for the Degree Master of Science (two years)
in Mathematical Modeling and Simulation
30 credit points (30 ECTS credits)
February 2012
Blekinge Institute of Technology
School of Engineering
Deparment of Mathematics and Science
Supervisor: Prof. Nail H. Ibragimov
Examiner: Dr. Raisa Khamitova
Abstract
Based on symmetry and invariance principles, Lie group analysis is the only
systematic method of analytically solving nonlinear dierential equations.
Nonlinear second-order ordinary dierential equations (ODEs) admitting
two-dimensional Lie algebras can be transformed into one of the four types
of canonical forms via Lie's integration method. In this thesis, Lie's secondorder equation of type III is considered from the point of view of equivalence
transformations. The generators of the equivalence group and the principal
Lie algebra are calculated.
Keywords:
equivalence transformation, second-order ordinary dier-
ential equations, principal Lie algebra.
2
Acknowledgements
First of all, I would like to acknowledge the Master Program in Mathematical Modeling and Simulation at Blekinge Institute of Technology, which
enabled us to have such a wonderful experience in Sweden.
Secondly, I would like to express my gratitude to my supervisor Prof.
Nail H. Ibragimov, who has introduced me into the interesting eld of group
analysis in mathematics and helped me with my thesis. His encouragement
and inspiration were always present during this research.
Furthermore, it is also a pleasure for me to express gratitude to my
previous programme manager, Dr.
Raisa Khamitova, who has given me
valuable advice and shown innite patience. I would also like to thank my
current programme manager Mattias Eriksson, whose support has provided
me a favorable environment for learning.
I would also like to thank all my friends and those who have helped me
during my stay and studies in Sweden.
Last but not least, I want to thank my parents for all of their understanding and support.
Thank you all!
3
Contents
1
2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.1
Second-order ODEs . . . . . . . . . . . . . . . . . . .
5
1.2
Lie's integration method using canonical variables . .
5
1.3
Equivalence transformations . . . . . . . . . . . . . .
6
1.4
Lie's second-order equation of type III
. . . . . . . .
7
Principal Lie algebra
. . . . . . . . . . . . . . . . . . . . . .
7
2.1
Calculation of the principal Lie algebra . . . . . . . .
8
2.2
Checking operators of the equivalence group
. . . . .
11
Equivalence transformations . . . . . . . . . . . . . . . . . .
12
3.1
Calculation of equivalence transformations . . . . . .
12
3.2
Checking operators of the equivalence group
. . . . .
18
4
Projections of the equivalence Lie algebra . . . . . . . . . . .
19
5
Examples
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
6
Summary
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3
4
1
Introduction
1.1
Second-order ODEs
Many mathematical models of real world problems involve second-order
nonlinear ordinary dierential equations. General method for analytically
solving these equations is Lie group analysis, which is particularly simple
and ecient for second-order equations. Lie groups provide mathematical
tools for revealing and using the symmetry of dierential equations. These
methods are based on the theory of continuous groups developed by Sophus
Lie.
Second-order equations have the following form:
u00 = g(t, u, u0 ).
(1.1)
A decisive step in integrating ODEs involves simplifying the frame by changing the variables. Lie group analysis provides a method of determining a
suitable change of variables using symmetries. The process will be further
discussed in the subsequent section.
1.2
Lie's integration method using canonical variables
Denition 1.1.
Let
Lr
be an
r-dimensional
linear space spanned by any
linearly independent operators of a set of operators. The space
Lr
r
is called
a Lie algebra if it is closed under a commutator.
Lie's integration method is based on the canonical coordinates for twodimensional Lie algebras
L2 .
For every
L2 ,
canonical variables provide the
simplest form of its basis and therefore reduce a dierential equation admitting
L2
to an integrable form. Second-order ordinary dierential equations
admitting
L2
can be classied into four types introduced by Lie (see, e.g.
[1] and [2]). The result is formulated as follows.
Theorem 1.1.
Any two-dimensional Lie algebra
L2 can be transformed into
one of the four non-similar standard forms (Table 1.1) through the proper
choice of its basis as well as the suitable variables
variables.
5
x, y
called canonical
Table 1.1: Four types of nonlinear second-order equations admitting
L2
∂
∂
, X2 =
X1 =
∂x
∂y
∂
∂
X1 =
, X2 = x
∂y
∂y
∂
∂
∂
, X2 = x
+y
X1 =
∂y
∂x
∂y
∂
∂
X1 =
, X2 = y
∂y
∂y
Type
Standard form of
I
II
III
IV
L2
Canonical form of the Equation
y 00 = f (y 0 )
y 00 = f (x)
y 00 =
1
f (y 0 )
x
y 00 = f (x)y 0
Lie proved Theorem 1.1 in order to integrate all second-order equations
admitting a two-dimensional Lie algebra. Lie's method consists of classifying these equations into four types according to Table 1.1. By introducing
the canonical variables
x, y ,
the admitted Lie algebra
L2
is reduced to one
of the standard forms given in Table 1.1. Equation (1.1) is then rewritten
in the canonical variables, and the resulting equation is as follows:
y 00 = f (x, y, y 0 )
(1.2)
where Equation (1.2) has one of the four integrable canonical forms given
in Table 1.1.
The equation is then integrated and rewritten by using the original variables
1.3
t, u
to complete the integration procedure.
Equivalence transformations
Equivalence transformations of dierential equations are useful in investigating various properties of dierential equations.
We take one simple
example to explain this further. An equivalence transformation of homogeneous linear equations
y 00 + a(x)y 0 + b(x)y = 0
(1.3)
or
L2 [y] = 0
involves a change of variables that preserve the linearity and homogeneity
of the equations.
The set of all equivalence transformations comprises an arbitrary change
6
of the independent variable,
x̄ = φ(x),
φ0 6= 0,
(1.4)
and the linear substitution of the dependent variable,
y = σ(x)ȳ,
Denition 1.2.
σ 6= 0.
(1.5)
Two equations of a form similar to Equation (1.3) are said
to be equivalent if they are connected by a combination of the transformations (1.4) and (1.5). Furthermore, two equations are termed equivalent by
function if they can be mapped into each other by using a method of linear
substitution (1.5).
Theorem 1.2.
Any homogeneous linear equation (1.3) is equivalent to the
simplest linear equation,
ȳ 00 = 0,
where
(1.6)
ȳ 00 = d2 ȳ/dx̄2 .
We are going to use this core idea of equivalence transformations to
investigate a second-order nonlinear dierential equation.
1.4
Lie's second-order equation of type III
In this thesis, the equivalence group of Lie's second-order nonlinear equation of type III is considered. The aim of this thesis is mainly to use the
innitesimal method for nding the equivalence Lie algebra. Lie's secondorder equation of type III is written in the form:
y 00 =
where
2
y 0 = dy/dx,
and
f
1
f (y 0 ),
x
(1.7)
is a smooth arbitrary function of
y0.
Principal Lie algebra
For a given family of dierential equations, the Lie group classication begins with determining the principal Lie group, the group admitted by any
equation of the family in question.
The Lie algebra of the principal Lie group is called the principal Lie
algebra of the equations and is denoted by
7
Lp
(see e.g. [3]).
2.1
Calculation of the principal Lie algebra
In order to derive the principal Lie algebra for the Lie's second-order nonlinear equation of type III
y 00 =
1
f (y 0 ),
x
we need to use the invariance condition rstly. The invariance condition for
Equation (1.7) can be written as follows:
1
00
0
= 0.
X(2) y − f (y ) 1
x
y 00 = f (y 0 )
x
Here,
X(2)
is denoted as the second prolongation for the operator
has the form
X = ξ(x, y)
X(2)
(2.8)
∂
∂
+ η(x, y) ·
∂x
∂y
can be considered as the combination of the operator
longed terms. Thus,
X(2)
which
(2.9)
X
and two pro-
can be written in the form
X(2) = X + ζ1
where the coecients
X
∂
∂
+ ζ2 00 ,
0
∂y
∂y
(2.10)
ζ1 and ζ2 are given by the usual prolongation formulae:
ζ1 = Dx (η) − y 0 Dx (ξ),
(2.11)
00
ζ2 = Dx (ζ1 ) − y Dx (ξ).
The operator
Dx
in the above equation denotes the operator of total dier-
entiation with respect to
x:
Dx =
∂
∂
+ y0 ·
∂x
∂y
Thus, the prolongation coecients
ζ
can be both obtained by formulae
(2.11) and (2.12). They have the following form:
ζ1 =ηx + y 0 ηy − y 0 (ξx + y 0 ξy ),
ζ2 =ηxx + y 0 ηxy + y 00 ηy + y 0 (ηxy + y 0 ηyy )
− y 00 (ξx + y 0 ξy ) − y 0 [ξxx + y 0 ξxy + y 00 ξy
+ y 0 (ξxy + y 0 ξyy )] − y 00 (ξx + y 0 ξy ),
8
(2.12)
or
ζ1 =ηx + (ηy − ξx )y 0 − ξy y 02 ,
(2.13)
ζ2 =ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02
(2.14)
− ξyy y 03 + (ηy − 2ξx − 3ξy y 0 )y 00 .
Using (2.10) we derive the following determing equation
f0
f
ζ2 − ζ1 + 2 ξ 1
= 0.
x
x
y 00 = f (y 0 )
x
Substituting
ζ1
and
ζ2
(2.15)
into Equation (2.15) we obtain
ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 − ξyy y 03
+(ηy − 2ξx − 3ξy y 0 )y 00 −
−ξy y
f0
(ηx + (ηy − ξx )y 0
x
02
(2.16)
f = 0.
+ 2ξ 1
x
y 00 = f (y 0 )
x
Substituting
y 00 =
1
f (y 0 ),
x
we derive the nal form of the determing equation:
f0
3f
ηxx + 2ηxy − ξxx −
ξy − (ηy − ξx ) y 0
x
x
(2.17)
+(ηyy − 2ξxy +
The functions
ξ
and
f0
f
f0
f
ξy )y 02 − ξyy y 03 + (ηy − 2ξx ) − ηx + 2 ξ = 0.
x
x
x
x
η
do not depend on
be satised identically for any
ξ
and
η,
9
y0.
Since Equation (2.17) should
we can split the equation into the
following system:
y 03 : ξyy = 0,
f0
ξy = 0,
x
f
f0
y 0 : 2ηxy − ξxx − 3 ξy − (ηy − ξx ) = 0,
x
x
f
ξ
f0
: ηxx + (−ηy + 2ξx + ) − ηx = 0.
x
x
x
y 02 : ηyy − 2ξxy +
Since
f
is an arbitrary smooth function, the coecient in front of
(2.18)
f
and
f0
should be equal to zero. Therefore, the equations can be splitted further.
Hence, from (2.18) we obtain the following system:
y 03 : ξyy = 0,
y 02 : ηyy − 2ξxy = 0, ξy = 0,
y 0 : 2ηxy − ξxx = 0, ξy = 0, ξx − ηy = 0,
ξ
: ηxx = 0, −ηy + 2ξx + = 0, ηx = 0.
x
From the above equations,
ξy = 0
and
ηx = 0,
(2.19)
it follows that
ξ = ξ(x),
η = η(y).
In addition, the functions
ξ
and
η
satisfy the equation
ηy − ξx = 0.
(2.20)
It is possible only if
ηy = ξx = C1 ,
Thus the funtions
ξ
and
η
C1 = const.
have the following form:
ξ = C1 x + C2,
η = C1 y + C3 .
(2.21)
The following step is to solve the remaining equation in (2.19), namely,
−ηy + 2ξx +
ξ
= 0.
x
Using (2.21) we derive
C2
= 0,
x
10
(2.22)
which gives
C2 = 0.
Thus,
ξ = C1 x,
η = C1 y + C3 .
Hence, Equation (1.7) with an arbitary function
dimensional Lie algebra
L2
f (y 0 )
admits the two-
with the basis
∂
∂
+y ,
∂x
∂y
∂
X2 =
·
∂y
X1 = x
Finally, we can say that
L2
(2.23)
is the two-dimensional Lie algebra known as the
principal Lie algebra for Equation (1.7).
2.2
Checking operators of the equivalence group
We have obtained the two-dimensional Lie algebra, we need to check if it is
correct.
• X1 = x
∂
∂
+y
∂x
∂y
Firstly, we prolong the operator until the second order according to the
prolongation formulae (2.11). Here
η=y
and
ξ = x.
ζ1 = Dx (y) − y 0 Dx (x) = y 0 − y 0 = 0,
ζ2 = −y 00 Dx (x) = −y 00 ,
therefore
X1(2) = x
∂
∂
∂
+y
− y 00 00 ·
∂x
∂y
∂y
Then, the invariance condition for Equation (1.7) can be written as
1
00
0
X1(2) y − f (y ) = 0,
1
x
y 00 = f (y 0 )
x
which yields
f −y +
= 0.
x y00 = 1 f (y0 )
x
00
11
Hence, the operator
• X2 =
X1
is admitted by Equation (1.7).
∂
∂y
In this case we can not prolong
X2
further. Thus, the invariance condition
(2.8) is apparently satised.
3
Equivalence transformations
Equivalence transformations play a core role in the theory of invariants.
The set of all equivalence transformations of a given family of dierential
equations forms a group, the so called equivalence group, and is denoted by
E.
The continuous group of equivalence transformations is a subgroup of
the group
E
and therefore denoted by
E c.
There are two main methods for calculating the equivalence transformations for a family of equations. The rst one is direct search for equivalence
transformations which allows one to calculate the most general equivalence
group
E.
The second method, suggested by L. V. Ovsyannikov, is to de-
termine generators of a continuous equivalence group
E c.
In general, the
innitesimal method is simpler for computing than the direct method.
3.1
Calculation of equivalence transformations
An equivalence transformation is a non-degenerate change of the variables
x, y, f
taking any equation of the form (1.7) into an equation of the same
form. Thus, the following denition is used:
Denition 3.1.
An equivalence transformation of the family of Equations
(1.7) is a change of variables
x̄ = ϕ(x, y),
ȳ = ψ(x, y),
f¯ = φ(x, y, f ),
such that it transfers Equation (1.7) with an arbitrary function
(3.24)
f (y 0 )
into
an equation of the same form
y 00 =
where the function
f,
1¯ 0
f (y ),
x̄
(3.25)
f¯ may, in general, be dierent from the original function
while the equations (1.7) and (3.25) are said to be equivalent.
12
In order to calculate the equivalence transformation (3.24) for Equation
(1.7), we write this equation in the following
1
y 00 − f = 0,
x
Now we look for the generator
Y
extended form
fx = 0,
of the group
fy = 0.
Ec
(3.26)
of the equivalence trans-
formations (3.24). The generator can be written in the space of variables
x, y
and
f
as follows:
Y
= ξ(x, y)
∂
∂
∂
+ η(x, y)
+ µ(x, y, f ) ·
∂x
∂y
∂f
(3.27)
The prolongation of the operator (3.27) to all variables involved in Equations (3.26) has the form:
∂
∂
∂
∂
∂
∂
∂
Ye = ξ
+ ω2
·
+η
+µ
+ ζ1 0 + ζ2 00 + ω1
∂x
∂y
∂f
∂y
∂y
∂fx
∂fy
The condition that
Y
(3.28)
is a generator of an equivalence group is equivalent
to the statement that
Ye
satises the innitesimal invariance test for the
extended system (3.26).
The invariance conditions for the system (3.26) are
Ye
f 00
y −
= 0,
x y00 = f
x
(3.29)
and
Ye (fx ) = 0, Ye (fy ) = 0.
According to the formulae (2.13)-(2.14), calculated in Chapter 2,
have the form
ζ1 =ηx + (ηy − ξx )y 0 − ξy y 02 ,
ζ2 =ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02
− ξyy y 03 + (ηy − 2ξx − 3ξy y 0 )y 00 .
13
(3.30)
ζ1
and
ζ2
We use the notation
Dx =
∂
∂
∂
+ y0
+ y 00 0
∂x
∂y
∂y
for the usual operator of total dierentiation with respect to
x,
while the
two forms
∂
∂
+ fx
∂x
∂f
D̃x =
(3.31)
and
D̃y =
∂
∂
+ fy
∂y
∂f
(3.32)
denote the new operators of total dierentiations necessary for the extended System (3.26).
The other two coecients
ω1
and
ω2
in (3.28) are obtained by applying
the secondary prolongation procedure to the dierential variables
0
the independent variables y , namely,
fy 0
with
ω1 = D̃x (µ) − fy0 D̃x (ζ1 ),
ω2 = D̃y (µ) − fy0 D̃y (ζ1 ).
From System (3.26) it follows, in particular, that we can set fx
0
Therefore, f = f (y ) and the operators (3.31), (3.32) as well as
= f y = 0.
ω1 and ω2
can be simplied:
D̃x =
∂
,
∂x
ω1 = µx − f 0 ζ1x ,
(3.33)
D̃y =
∂
,
∂y
ω2 = µy − f 0 ζ1y .
Now we notice that the invariance condition (3.30) can be written as
ω1 = 0,
ω2 = 0,
and by using (3.33) in Equation (3.34), we obtain
µx − f 0 ζ1x = 0,
µy − f 0 ζ1y = 0.
14
(3.34)
These equations are held for an arbitrary function f , which depends only
0
on y . Hence, we can split these equations as follows:
This means that
µ
µx = 0,
µy = 0,
ζ1x = 0,
ζ1y = 0.
is a function depending only on
f,
i.e,
µ = µ(f ).
(3.35)
Recalling the formula (3.28) for the operator
Ye ,
we rewrite the invariance
condition (3.29):
1
1
ζ2 − µ + 2 ξf 1
= 0.
x
x
y 00 = f (y 0 )
x
(3.36)
Thus, we have
ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 − ξyy y 03
µ ξf
+(ηy − 2ξx − 3ξy y )y − + 2
x x
0
Substituting
form
y 00 =
1
f
x
00
y 00 =
1 0 = 0.
f (y )
x
(3.37)
in (3.37) and simplifying it, we obtain the reduced
ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02
−ξyy y 03 + (ηy − 2ξx − 3ξy y 0 )
As
ζ1
µ ξf
f
− + 2 = 0.
x x x
is obtained from the prolongation formula (2.13)
ζ1x = ηxx + y 0 ηxy − y 0 (ξxx + y 0 ξxy ),
ζ1y = ηxy + y 0 ηyy − y 0 (ξxy + y 0 ξyy ),
15
(3.38)
ζ1x = 0 and ζ1y = 0 obtained above,
we have the equations
yield the follow-
ing equations:
ηxx + y 0 (ηxy − ξxx ) + y 00 ξxy = 0,
(3.39)
ηxy + y 0 (ηyy − ξxy ) + y 00 ξyy = 0.
(3.40)
From Equations (3.39) and (3.40), it is easy to obtain
ξxy = 0,
ηxy = ξxx ,
ηxx = 0,
(3.41)
ξyy = 0,
The equations
ξxy = 0
and
ηyy = ξxy ,
ηxy = 0
give
ξy = C1 ,
whence, we can determine
ξ
and
η
ηxy = 0.
ηx = C2 ,
as
ξ = C1 y + A(x),
η = C2 x + B(y).
Thus,
ξxx = ηxy = 0, ηyy = ξxy = 0.
(3.42)
Using (3.42) we obtain from these
equations
A00 = 0,
B 00 = 0.
Based on the above equations, the functions
(3.43)
A
and
B
can be dened as
linear functions:
A = C3 x + C5 ,
B = C4 y + C6 .
Hence, we have obtained
ξ = C1 y + C3 x + C5 ,
(3.44)
η = C2 x + C4 y + C6 .
(3.45)
16
From (3.44) and (3.45) it follows that
ξx = C3 ,
ξxx = 0,
ξy = C1 ,
ξyy = 0,
ξxy = 0,
(3.46)
ηx = C2 ,
ηxx = 0,
ηy = C4 ,
ηyy = 0,
ηxy = 0.
According to (3.46), the determing equation (3.38) can be reduced to the
following form
f
µ ξf
(ηy − 2ξx − 3ξy y 0 ) − + 2 = 0,
x
x x
or
µ f (C1 y + C3 x + C5 )
f
(C4 − 2C3 − 3C1 y 0 ) − +
= 0.
x
x
x2
(3.47)
We rewrite it into a new form
µ = (C4 − 2C3 − 3C1 y 0 )f +
= C4 − C3 +
y
x
− 3y
0
(C1 y + C3 x + C5 )f
x
C5
C1 +
f.
x
(3.48)
µ = µ(f ), it does not depend on x and
y . We can conclude that the constants C1 and C5 in Equation (3.48) should
However, from (3.35) we know that
be equal to zero.
By substituting
C1 = 0 and C5 = 0 into System (3.44)-(3.45),
we obtain
that
ξ = C3 x,
η = C2 x + C4 y + C6 .
Introducing
K1 , K2 , K3
and
K4
instead of
C3 , C2 , C4
(3.49)
and
C6 ,
respectively,
Equations (3.49) can be rewritten into a new form
ξ = K1 x,
η = K2 x + K3 y + K4 ,
(3.50)
µ = (K3 − K1 )f
(3.51)
and
where
K1 , K2 , K3
and
K4
are arbitrary constants.
Thus, we have arrived at the following result:
17
Theorem 3.1.
The equivalence algebra
LE
for the equation (1.7) is a
four-dimensional Lie algebra spanned by
∂
∂
−f ,
∂x
∂f
∂
= x ,
∂y
∂
∂
= y
+f ,
∂y
∂f
∂
=
·
∂y
Y1 = x
Y2
Y3
Y4
3.2
Checking operators of the equivalence group
We have got four operators for the equivalence Lie algebra, now we check
them one by one.
• Y1 = x
∂
∂
−f
∂x
∂f
We prolong it to the second order
Y1(2) = x
∂
∂
∂
∂
−f
− y 0 0 − 2y 00 00 ·
∂x
∂f
∂y
∂y
According to the invariant condition
Y1(2)
we have
on any solution
1 y − f 1 = 0,
x
y 00 = f
x
00
1
f
1
−2y 00 + f + f x 2 = −2y 00 + 2 = 0
x
x
x
1
00
of the equation y =
f (y 0 ). Hence, Y1
x
Equations (3.26).
• Y2 = x
∂
∂y
We can prolong it to the rst order only:
Y2(1) = x
∂
∂
+ 0·
∂y ∂y
18
is admitted by
It satises the invariant condition
1 00
Y2(1) y − f 1 = 0,
x
y 00 = f
x
Thus,
Y2
is admitted by System (3.26).
• Y3 = y
Y3
∂
∂
+f
∂y
∂f
prolonged to the second order has the form
Y3(2) = y
∂
∂
∂
∂
+f
+ y 0 0 + y 00 00 ·
∂y
∂f
∂y
∂y
The invariant condition is satised, indeed
Y3(2)
Hence,
Y3
• Y4 =
f
f 00
= y −
= 0.
y −
x y00 = 1 f
x y00 = 1 f
x
x
00
is admitted by Equations (3.26).
∂
∂y
This operator is also admitted by System (3.26).
It is the operator
X2
obtained in the previous chapter.
4
Projections of the equivalence Lie algebra
The Lie algebra of the continuous equivalence group
equivalence algebra and is denoted by
Ec
is the so called
LE .
Most of the extensions of a principal Lie algebra
LF
(an algebra admitted
by every equation of the family of equations under consideration) are taken
from an equivalence algebra
LE .
These extensions are so called
E
extensions.
Now we use the equivalence Lie algebra to simplify the calculation of
symmetry groups without solving the determining equation. If we calculate
the equivalence algebra,
E
extensions for equations of a given family are
obtained by solving the simple algebraic equations only.
19
Consider a continuous group of equivalence transformations
Y =ξ
where the coecients
and
f.
ξ
The coordinate
∂
∂
∂
∂
+η
+µ
+ ζ1 0 ,
∂x
∂y
∂f
∂y
and
ζ1
η
depend on
x
and
y,
while
(4.52)
µ
depends on
x, y
is given by the prolongation formula:
ζ1 = Dx (η) − y 0 Dx (ξ).
We denote by Z and X the projections of the equivalence operator (4.52)
0
to the (f, y )-space and to the (x, y)-space, respectively (see [4]). Namely,
Z ≡ pr(f,y0 ) (Y ) = µ
X ≡ pr(x,y) (Y ) = ξ
∂
∂
+ ζ1 0 ,
∂f
∂y
∂
∂
+η ·
∂x
∂y
(4.53)
The signicance of these projections is determined by the following statement [4], pp. 69-70,
Theorem 4.1.
An operator
X
belongs to the principal Lie algebra
LF
for
a system if and only if
X = pr(x,y) (Y )
where
Y
(4.54)
is an equivalence generator such that
pr(x,y) (Y ) = 0.
(4.55)
Let us determine the principal Lie algebra by using Theorem 4.1 for the
general operator of the equivalence algebra of Equation (1.7),
Y =K1 x
∂
∂
∂
+ (K2 x + K3 y + K4 )
+ f (K3 − K1 )
∂x
∂y
∂f
+ [K2 + y 0 (K3 − K1 )]
Here,
Z
and
X
∂
·
∂y 0
(4.56)
are dened according to (4.53). They can be presented as
follows:
Z = pr(f,y0 ) (Y ) = f (K3 − K1 )
∂
∂
+ [K2 + y 0 (K3 − K1 )] 0
∂f
∂y
20
and
X = pr(x,y) (Y ) = K1 x
Let the operator
Z
∂
∂
+ (K2 x + K3 y + K4 ) ·
∂x
∂y
be equal to zero. Hence, we obtain the relation between
the constants:
K3 − K1 = 0,
whence
K1 = K3 .
Substituting
K1 , K2
K2 = 0,
and
K3
in Equation (4.56) we obtain
the reduced form for the equivalence generator
Y = K1 x
(4.57)
Y:
∂
∂
+ (K1 y + K4 ) ·
∂x
∂y
(4.58)
We rearrange the terms, collecting together the terms with the same constants:
Y = K1 (x
∂
∂
∂
+ y ) + K4 ·
∂x
∂y
∂y
Thus, we have obtained, that the principal Lie algebra
LF
is represented
by a two-dimensional Lie algebra which is spanned by
∂
∂
+y ,
∂x
∂y
∂
X2 =
·
∂y
X1 = x
This result is exactly the same as the standard form of
L2
of type III pre-
sented in Table 1.1.
5
Examples
The use of the projection
about the combination of
of
Y1
and
Z is illustrated here by examples. The rst case is
Y1 and Y4 , and the second case is the combination
Y2 .
Example 1.
In the rst case we consider the following equivalence gener-
ator:
Y = Y1 + Y4 ≡ x
The rst prolongation of
Y
∂
∂
∂
∂
+
−f
+ ζ1 0 ·
∂x ∂y
∂f
∂y
can be written as:
∂
∂
∂
∂
Ye = x
+
−f
− y0 0 ·
∂x ∂y
∂f
∂y
21
In the rst prolongation equation, the operator
tion
Y
coincides with its projec-
pr(y0 ,f ) (Y )
Z1 = pr(y0 ,f ) (Y ) = −f
The invariance condition for
f
∂
∂
− y0 0 ·
∂f
∂y
yields
Z1 (f − F (y 0 ))|f =F (y0 ) = 0,
which gives the relation between
f
and
F (y 0 )
(5.59)
as
0 ∂F
−f + y 0
= 0.
∂y f =F (y0 )
Using
f = F (y 0 ),
we rewrite the above relation into the new form:
−F + y 0
∂F
= 0.
∂y 0
Therefore, we obtain the relation between variables
y 0 and f
in the following
form
f = Cy 0 ,
where
C
(5.60)
is an arbitrary constant.
Using the additional operator
X3 = pr(x,y) (Y ) = x
∂
∂
+
,
∂x ∂y
we can easily obtain the form of the relation between
x
and
y
dx
= dy.
x
Thus, a non-linear relation between these two variables
x
and
dened as:
y = Klnx.
Therefore, the rst and second derivatives can be obtained as:
y0 =
K
,
x
y 00 = −
22
K
·
x2
y
can be
By substituting
f = xy 00
from Equation (1.7) into Equation (5.60), it is easy
to notice that the relationship between
x
and
K
is simple:
−K = CK,
whence
C = −1.
Thus, we have
f = −y 0 .
Now we substitute it into Equation (1.7) and obtain
y0
y 00 = − ·
x
(5.61)
Hence, Equation (5.61) admits along with the operators (2.23) the additional operator
X3 = x
Example 2.
∂
∂
+
·
∂x ∂y
(5.62)
In the second case we consider the equivalence generator as
Y = Y1 + Y2 ≡ x
The rst prolongation of
Y
∂
∂
∂
∂
+x
−f
+ ζ1 0 ·
∂x
∂y
∂f
∂y
can be rewritten as
∂
∂
∂
∂
Ye = x
+x
−f
+ (1 − y 0 ) 0 ·
∂x
∂y
∂f
∂y
The operator
as
Y
coincides with its projection
pr(y0 ,f ) (Y ), which we denote it
Z2 :
Z2 = pr(y0 ,f ) (Y ) = −f
The invariance condition for
f
∂
∂
+ (1 − y 0 ) 0 ·
∂f
∂y
has the form:
Z2 (f − F (y 0 ))|f =F (y0 ) = 0,
which gives the relation between
f
and
−f − (1 − y 0 )
F (y 0 ):
∂F
∂y 0
23
= 0.
f =F (y 0 )
(5.63)
Based on the condition
f = F (y 0 ),
the relation we obtained above could be rewritten as
F + (1 − y 0 )
∂F
= 0.
∂y 0
y0
Therefore, we derive the relation between
and
f
as
f = C(y 0 − 1),
where
C
(5.64)
is an arbitrary constant.
In order to check this result, let us determine the additional operator
X3 = pr(x,y) (Y ) = x
The characteristic system for
X3
∂
∂
+x ·
∂x
∂y
has the form
dx
dy
= ·
x
x
Therefore, we can get a linear relation between
y = x + K,
The rst and second derivatives of
y
y 00 =
by substituting
f
and
y 00
y
are
y 00 = 0.
f = C(y 0 − 1)
second-order equation
and
K = const.
y 0 = 1,
Since we have we have
x
and
y 00 = 0,
in addition we have Lie's
1
f (y 0 ),
x
into the equation, then we can obtain
1 0
(y − 1)C = 0.
x
Finally, we have
f = C(y 0 − 1).
Substituting this result into Equation (1.7) we have
y 00 =
C 0
(y − 1)·
x
24
(5.65)
Hence, the Equation (5.65) admits along with the operators (2.23) the additional operator
X3 = x
∂
∂
+x ·
∂x
∂y
(5.66)
Since both Equations (5.61) and (5.65) are linear, we could make a conclusion that they admit, in fact, an eight-dimensional Lie algebra.
6
Summary
The following results have been obtained in this thesis. The innitesimal
method was used for nding the equivalence Lie algebra for the second-order
equation of type III:
y 00 =
1
f (y 0 ).
x
Its principal Lie algebra is a two-dimensional Lie algebra that spanned by
∂
∂
+y ,
∂x
∂y
∂
X2 =
·
∂y
X1 = x
The equivalence algebra for Lie's second-order equation of type III is a
four-dimensional Lie algebra spanned by
∂
∂
−f ,
∂x
∂f
∂
Y2 = x ,
∂y
∂
∂
+f ,
Y3 = y
∂y
∂f
∂
Y4 =
·
∂y
Y1 = x
Two examples demonstrating the use of the equivalence algebra are presented in Chapter 5.
25
Bibliography
A Practical Course in Dierential Equations and
Mathematical Modelling: Classical and new methods, nonlinear mathematical models, symmetry and invariance principles, ALGA Publica-
[1] Nail H. Ibragimov,
tions, Karlskrona, Sweden, 2006 or Higher Education Press, China and
World Scientic, Singapore, 2009.
[2] Nail H. Ibragimov, Equivalence groups and invariants of linear and
non-linear equations,
Archives of ALGA,
vol. 1, ALGA Publications,
Karlskrona, Sweden, pp. 1638, 2004.
CRC Handbook of Lie group analysis of differential equations. Vol. 2: Applications in engineering and physical sciences. CRC Press Inc., Boca Raton, 1995.
[3] Nail H. Ibragimov, editor.
[4] Nail H. Ibragimov, Theorem on projections of equivalence Lie algebras,
Selected Works, vol. 2, pp. 6779, ALGA Publications, Karlskrona, Swe-
den, 2006.
26