A STUDY ON LIE SYMMETRY ANALYSIS TO DIFFERENTIAL EQUATIONS

by

Maria Akter Student No. 1015093006P

Registration No. 1015093006, Session: October 2015

MASTER OF PHILOSOPHY IN

MATHEMATICS

Department of Mathematics Bangladesh University of Engineering and Technology (BUET), Dhaka-1000

December - 2019

ii

iii

Declaration of Authorship

I declare that A STUDY ON LIE SYMMETRY ANALYSIS TO DIFFERENTIAL

EQUATIONS is my own work which is completed under the supervision of Dr. Khandker Farid

Uddin Ahmed. All the sources that I have used or quoted have been indicated and acknowledged

by means of a complete list of references.

(Maria Akter)

Date: 10th December, 2019

iv

Dedication

I would like to dedicate this dissertation to my parents whom have always been my

pillar of strength and an encouragement for me to work hard in my success of life.

vi

Abstract In this work, travelling wave and a family of invariant solutions of nonlinear filtration partial

differential equation are found. In general, by using Lie symmetry transformations, it is possible

to reduce the partial differential equations into ordinary differential equations, if the symmetries

admitted by target equations allow to determine the Lie point transformations. In the case of

nonlinear filtration equation, we are looking for travelling wave solutions which are invariant

under a particular group of Lie symmetry. By using the invariance surface condition of nonlinear

filtration equation we get a much simpler form of partial differential equation to solve than the

original one. After that applying the method of characteristics to obtain fundamental differential

invariant such that the travelling wave solutions of nonlinear filtration equation can be found. In

case of a solution of any partial differential equation that is invariant under a group of Lie

symmetry it is possible to find new solution from invariant solution by using the other Lie

symmetries of given PDE. Since travelling wave solutions are invariant, therefore with the help

of above strategy new solutions of nonlinear filtration equation are found by utilizing the

remaining symmetries of filtration equation from travelling wave solutions.

v

Acknowledgement

With Allah everything is possible. I would like to affirm the continual mercy, help and blessing

showered by the Almighty Allah without which it would have been impossible to accomplish the

arduous job I was assigned to.

I am indebted to my supervisor, Dr. Khandker Farid Uddin Ahmed for the guidance, patience

and professional assistance he has offered throughout this thesis. It is with gratitude that I have

tapped from his vast wisdom and knowledge.

Special words of thank goes to Dr. Salma Nasrin for having introduced me to Lie group and Lie

algebra. I acknowledge the teaching I received from her in my Master’s Degree program. It has

rounded my understanding of Lie symmetry analysis.

Words of thank go to Dr. Md. Elias and Dr. Md. Abdul Hakim Khan for being on my defense

committee, for their careful review of this thesis, for reading through and correcting this work

meticulously.

I want to show my gratitude to all my colleagues in Department of Mathematics at BUET.

Family members, friends with whom I have a formidable bond, I say thanks for their words of

encouragement, and expressing confidence in my ability.

(Maria Akter)

Date: 10th December, 2019

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Contents

Board of Examiners ii

Declaration of Authorship iii

Dedication iv

Acknowledgement v

Abstract vi

Contents vii

List of figures ix

Chapter 1: Introduction 1

1.1 Historical Background 1

1.2 Literature Review 2

1.3 Framework of the Thesis 4

Chapter 2: Symmetries and Differential Equations 6

2.1 Symmetry 6

2.2 Symmetries of ODE 9

2.3 The Symmetry Condition 12

2.4 Change of Coordinates 15

2.5 Canonical Coordinates 21

2.5.1 Solving ODE with Canonical Coordinates 28

2.5.2 The Linearized Symmetry Condition 31

Chapter 3: Lie Symmetries and Higher Order ODE 37

3.1 Infinitesimal Generator 37

3.2 Symmetry Condition for Higher Order ODE 41

viii

3.2.1 The Linearized Symmetry Condition for Higher Order

ODE 43

3.3 Reduction of Order 47

3.4 Invariant Solutions of Higher Order ODE 49

3.4.1 Differential Invariant and Solutions of Higher Order

ODE 51

3.5 Lie Algebra of Point Symmetry Generators 57

Chapter 4: Lie Symmetries and Partial Differential Equations 65

4.1 Symmetries of PDEs with Two Dependent Variables 65

4.2 Linearized Symmetry Condition of PDEs 67

4.3 Group Invariant Solution of PDEs 71

Chapter 5: Lie Symmetry Analysis of Nonlinear Filtration Equation 77

5.1 Travelling Wave Solution of PDE 77

5.2 Travelling Wave Solution with Lie Symmetry 80

5.3 Travelling Wave Solution of Nonlinear Filtration Equation

with Lie Symmetry 82

5.4 Solution of Nonlinear Filtration Equation with Lie Symmetry

from Travelling Wave Solutions 90

Chapter 6: Conclusion 92

References 94

ix

List of Figures

Figure No. Name of Figure Page No.

Figure 2.1 Some triangles and their symmetries 7

Figure 2.2 Rotation of unit circle 8

Figure 2.3 Solutions of ODE 2.1 11

Figure 2.4 The action of symmetry 18

Figure 2.5 One dimensional orbit 18

Figure 5.1 Travelling Waves: (a) wave front, (b) Pulse wave, (c) spatially

periodic wave

79

Figure 5.2 Solution of equation (5.46) 89

Figure 5.3 Solution of equation (5.50) 91

1

Chapter 1

Introduction

1.1 Historical Background

The Lie group analysis for solving differential equations in the area of mathematics was pioneered

by Sophus Lie in the 19th century (1849-1899). Lie initiated his program on the basis of analogy.

If finite groups were required to decide on the solvability of finite degree polynomial equations,

then infinite groups would probably be involved in the theory of ordinary and partial differential

equations ([19], [20]). Sophus Lie made a profound and deep-rooted discovery that all ad-hoc

techniques designed to solve ordinary differential equations could be explained and deduced

simply by his theory. These techniques were in fact special cases of a general integration procedure

of classifying ordinary differential equations in terms of their symmetry groups. This discovery

led to Lie identifying a full set of equations which could be integrated or reduced to lower order

equations by his method ([30], [31]).

In the 1940s to 1950s, Ellis Kolchin, Armand Borel and Claude Chevalley realized that many

foundational results concerning Lie groups can be developed completely algebraically, giving rise

to the theory of algebraic groups defined over an arbitrary field. This insight opened new

possibilities in pure algebra, by providing a uniform construction for most finite simple groups, as

well as in algebraic geometry. The theory of automorphic forms, an important branch of

modern number theory, deals extensively with analogues of Lie groups over adel rings, p-adic Lie

groups play an important role, via their connections with Galois representations in number theory.

The Lie Symmetry method is analytic and highly algorithmic. The method systematically unifies

and extends exceptional techniques to construct explicit solutions for differential equations. The

emphasis is on explicit computational algorithms to discover symmetries admitted by differential

equations and to construct invariant solutions resulting from the symmetries ([28], [30]). Lie

groups and Lie algebras, because of their manifold, and therefore, differentiability structure, find

very natural applications in areas of physics and mathematics in which symmetry and

2

differentiability play important roles. Lie himself started the subject by analyzing the symmetry of

differential equations in the hope that a systematic method of solving them could be

discovered. Later, Emmy Noether applied the same idea for variational problems involving

symmetries and obtained one of the most beautiful pieces of mathematical physics, which is, the

relation between symmetries and conservation laws [38]. More recently, generalizing the gauge

invariance of electromagnetism, Yang and Mills have considered nonabelian gauge theories in

which gauge invariance is governed by a nonabelian Lie group [39]. Such theories have been

successfully built for three of the four fundamental interactions such as electromagnetism, weak

nuclear and strong nuclear.

Lie group analysis established itself to be an effective method of solving nonlinear differential

equations analytically. In fact, the first general solution of the problem of classification was given

by Sophus Lie for an extensive class of partial differential equations [33]. Since then many

researchers have done work on various families of differential equations. The results of their work

have been captured in several outstanding literary works by amongst others Ibragimov (1999),

Hydon (2000), Bluman and Anco (2002).

1.2 Literature Review

It is generally known fact that many phenomena in fluid dynamics, plasma physics, nonlinear

optics, biology, chemistry, engineering, etc. can be described by nonlinear partial differential

equations. Therefore, finding the exact solutions of the nonlinear partial differential equations is

an important task in the study of many fields of sciences. In order to help engineers and physicist

to better understand the working or to better provide knowledge to the physical problem, many

powerful and direct methods for solving nonlinear PDEs have been derived. Some of the most

efficient method to find exact solutions of nonlinear partial differential equations are Darboux and

Backlund transformation methods [1], Riccati Bernoulli sub-ODE method [2], Hyperbolic tangent

method [3], Homogeneous balance method [4], Lie-symmetry method [5], Inverse scattering

transformation [6], CK method [7] and many more techniques. Among these methods, Lie

symmetry method is most efficient and reliable method to obtain exact solutions of nonlinear

PDEs.

3

In 1997, S. Katz, P. Mayr and C. Vafa, by using classical symmetry in the context of type II strings,

found the exact solution for the module space of the Coulomb branch of all N = 2 gauge theories

in four dimensions involving products of SU gauge groups with arbitrary number of bi-

fundamental matter for chosen pairs, as well as an arbitrary number of fundamental matter for each

factor [8]. Transverse vibrations of a string moving with time-dependent velocity have been

investigated and analytical solutions of this problem are explained by E. Özkaya and M.

Pakdemirli, by using the systematic approach of Lie group theory in 2000 [9]. For a linear ordinary

differential equation, the Lie algebra of its infinitesimal Lie symmetries is compared with its

differential Galois group and for this purpose an algebraic formulation of Lie symmetries is

developed by W. R. Oudshoorn and M. V. D. Put, in 2001. They explained that there is no direct

relation between the two above objects and in connection with this a new algorithm for computing

the Lie symmetries of a linear ordinary differential equation is presented [10].

In the year of 2002, R. Cherniha and J. R. King presented a complete description of the classical Lie

symmetries of a coupled system of partial differential equations comprising a pair of semilinear

reaction diffusion equations with constant diffusivities and arbitrary nonlinearities in the reaction

terms, in any number of spatial dimensions [11]. The classification of all similarity solutions of a

boundary-value problem for a nonlinear diffusion equation arising in the study of a charged power

law Non- Newtonian fluid through a time dependent transverse magnetic field are described by C.

W. Soh, in 2005. The Authors also found the two families of exact invariant solutions for nonlinear

diffusion equation [12]. In 2008, V. D. Djordjevic and T. M. Atanackovic analyzed the self-similar

solutions to a nonlinear fractional diffusion equation and fractional Burgers/Korteweg deVries

equation in one spatial variable by using Lie-group scaling transformation [13].

H. Liu, J. Li, Q. Zhang have performed Lie symmetry analysis for the general Burgers’ equation

and investigated the algebraic structure of the symmetry groups for this equation in 2009. They

also introduce a new method based on the power series method for dealing with partial differential

equations, which is called the generalized power series method [14]. In 2010, Lie group analysis

is employed to derive some exact solutions of a generalized (3+1) dimensional Kadomtsev

Petviashvili equation which describes the dynamics of solitons and nonlinear waves in plasmas

and superfluids by C. M. Khalique and A. R. Adem [15]. A relation among symbolic computation,

Lie symmetry analysis, Painlevé test, conservation laws and similarity solutions are established by

4

H. Zhi, Z. Yang, H. Chang in the year of 2014. By deriving the group classifications and symmetry

reduction via Lie group method some integrable conditions are determined via the Painleve test by

the Authors [16].

Lie symmetry method provides an effective tool for deriving the analytic solutions of the nonlinear

partial differential equations. In recent years, many authors have studied the nonlinear fractional

differential equations because these equations express many nonlinear physical phenomena and

dynamic forms in physics, electrochemistry and viscoelasticity. Time fractional nonlinear partial

differential equations arise from classical nonlinear partial differential equations by replacing its

time derivative with the fractional derivative. Obtaining analytical or numerical solution of

fractional differential equations is one of the troublesome and challenging issue among

mathematicians and engineers, specifically in recent years. In 2017, Lie Symmetry method is

developed by M. Ilie, J. Biazar and Z. Ayati, to solve second-order fractional differential equations,

based on conformable fractional derivative [17].

1.3 Framework of the Thesis

The present work titled “A STUDY ON LIE SYMMETRY ANALYSIS TO DIFFERENTIAL

EQUATIONS” seeks to explore the analysis of nonlinear filtration equation through determining

the symmetries and new invariant solutions of some of these symmetries. The analysis is through

a method developed in [37]. Throughout the thesis we use Lie point symmetries.

The dissertation outline is as follows:

Chapter 1 is all about the introduction of this thesis work. This chapter is organized by explaining

the historical background of Lie group, literature review, initial purpose of Lie theory and recent

state of the Lie group analysis for solving differential equations.

Chapter 2 explores the rigorous explanation of symmetry and its significance for solving

differential equations. We present the concepts of symmetries of ODE, the symmetry condition,

change of coordinates, canonical coordinates in this chapter which serve as tools in the analysis of

the first order ODE. Solving first order ODE with canonical coordinates and the linearized

symmetry condition are discussed at the rest of this chapter ([18], [31], [37]).

5

Chapter 3 is the continuation of previous chapter. In this chapter, we have focused on finding

general solution of higher order ODE by using Lie point symmetry. For this purpose, the chapter

is organized as follows: infinitesimal generator [37], symmetry condition and linearized symmetry

condition for higher order ODE ([18], [22]), reduction of order [25], invariant solutions [30], Lie

algebras [32]. Several examples are given at every section of this chapter to relate the concept of

Lie symmetry with real life phenomena.

Chapter 4 is all about the extension of the chapter 2 and chapter 3, that is, notion of Lie symmetry

is explained at the context of partial differential equations. It introduces the symmetries of PDE

with two dependent variables, linearized symmetry condition of PDEs and group invariant

solutions of PDEs. At the end of this chapter, the detail analyze of the exact solution of one

dimensional heat equation is given, by using Lie point symmetry.

Chapter 5 is the core of the thesis. This chapter deals with the explanation of traveling wave

solution of PDEs and its relation with Lie symmetries. The travelling wave solution with Lie group

analysis is used for solving nonlinear filtration equation in this chapter. In addition to that, we

derived the new form of solution of nonlinear filtration equation from old one by using remaining

symmetries, at the end of this chapter.

Chapter 6 presents the conclusive remarks of this dissertation. It also describes the impact of the

result of this thesis on various field of science. Finally, where the method of Lie symmetry will be

successfully applicable is explained.

6

Chapter 𝟐

Symmetries and Differential Equations

Symmetry is the key for solving differential equations. There are many well-known techniques for

obtaining exact solutions of differential equations, but most of them are merely special case of a

few powerful symmetry methods. This chapter presents the underlying theory of Lie Symmetry

Analysis for solving first order ordinary differential equations and the tools of this chapter will be

used in subsequent chapters.

2.1 Symmetry

To begin, we consider the following definition:

Definition 2.1.1 A symmetry of a geometrical object is a transformation whose action leaves the

object apparently unchanged. For instance, consider the result of rotating an equilateral triangle

anticlockwise about its center. After a rotation of 2𝜋/3, the triangle looks the same as it did before

the rotation, so this transformation is a symmetry. Rotations of 4𝜋/3 and 2𝜋 are also symmetries

of the equilateral triangle. In fact, rotating by 2𝜋 is equivalent to doing nothing, because each point

is mapped to itself. The transformation mapping each point to itself, is a symmetry of any

geometrical object: it is called the trivial symmetry.

Symmetries are commonly used to classify geometrical objects. Suppose that the three triangles

illustrated in Fig. 2.1 are made from some rigid material, with indistinguishable sides. The

symmetries of these triangles are readily found by experiment. The equilateral triangle has the

trivial symmetry; the rotations described above, and flips about the three axes marked in Fig. 2.1

(a). These flips are equivalent to reflections in the axes. So an equilateral triangle has six distinct

symmetries [21]. The isosceles triangle in Fig. 2.1 (b) has two: a flip (as shown) and the trivial

symmetry. Finally, the triangle with three unequal sides in Fig. 2.1 (c) has only the trivial

symmetry.

7

(a) (b) (c)

Fig. 2.1: Some triangles and their symmetries

There are certain constraints on symmetries of geometrical objects. Each symmetry has a unique

inverse, which is itself a symmetry. The combined action of the symmetry and its inverse upon the

object (in either order) leaves the object unchanged. For example, let Γ denotes a rotation of the

equilateral triangle by 2𝜋/3. Then Γ−1 (the inverse of Γ) is a rotation by 4𝜋/3. Normally,

symmetries are smooth. Suppose that 𝑥 denotes the position of a general point of the object and

Γ: 𝑥 → �̂�(𝑥) is any symmetry, then we assume that 𝑥 is infinitely differentiable with respect

to 𝑥. Moreover, since Γ−1 is also a symmetry, so 𝑥 is infinitely differentiable with respect to 𝑥.

Thus Γ is a 𝐶∞diffeomorphism, that is, a smooth invertible mapping whose inverse is also smooth.

Symmetries preserve structure. It is usual for geometrical objects to have some structure which

describes what the object is made from. Earlier it is considered that symmetries of triangles made

from a rigid material. The only transformations under which a triangle remains rigid are those

which preserve the distance between any two points on the triangle, namely translations, rotations,

and reflections (flips). These transformations are the only possible symmetries, because all other

transformations fail to preserve the rigid structure. However, if the triangles are made from an

elastic material such as rubber, the class of structure-preserving transformations is larger, and new

symmetries may be found. For example, a triangle with three unequal sides can be stretched into

an equilateral triangle, then rotated by 2𝜋/3 about its center, and finally stretched so as to appear

to have its original shape. This transformation is not a symmetry of a rigid triangle. Thus we can

say that a transformation is a symmetry if it satisfies the following three conditions:

(C1) The transformation preserves structures.

(C2) The transformation is a diffeomorphism.

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(C3) The transformation maps the object to itself. That is a planar object in the (𝑥, 𝑦)-plane and

its image in the (�̂�, �̂�)-plane are indistinguishable.

A rigid triangle has a finite set of symmetries. But there are many objects which have an infinite

set of symmetries. For example, the (rigid) unit circle 𝑥2 + 𝑦2 = 1 has symmetry

Γ𝜀: (𝑥, 𝑦) ↦ (�̂�, �̂�) = (𝑥 cos 𝜀 − 𝑦 sin 𝜀 , 𝑥 sin 𝜀 + 𝑦 cos 𝜀)

for each 𝜀 ∈ (−𝜋, 𝜋]. In terms of polar coordinates

Γ𝜀: (cos 𝜃, sin 𝜃) ↦ (cos(𝜃 + 𝜀) , sin(𝜃 + 𝜀)),

as shown in Fig. 2.2. Hence the transformation is a rotation by 𝜀 about the centre of the circle. It

preserves the structure (rotations are rigid), and it is smooth and invertible (the inverse of a rotation

by 𝜀 is a rotation by – 𝜀). To prove that the symmetry condition (C3) is satisfied, note that 𝑥2 +

�̂�2 = 𝑥2 + 𝑦2 and therefore 𝑥2 + �̂�2 = 1 when 𝑥2 + �̂�2 = 1.

Fig. 2.2: Rotation of unit circle

The unit circle has other symmetries such as reflections in each straight line passing through the

center. It is easy to show that every reflection is equivalent to the reflection

Γ𝑟 ∶ (𝑥, 𝑦) ↦ (−𝑥, 𝑦)

0 x

y

0 -1 1

𝜃 x

y

-1 1

9

followed by a rotation Γ𝜀. The infinite set of symmetries Γ𝜀 is an example of a one-parameter Lie

group.

Definition 2.1.2 Suppose that an object occupying a subset of ℝ𝑁 has an infinite set of

symmetries

Γ𝜀 ∶ 𝑥𝑘 ↦ 𝑥𝑘(𝑥1, 𝑥2, … … … … 𝑥𝑁; 𝜀), 𝑘 = 1, 2, … … … 𝑁.

where 𝜀 is a real parameter satisfying the following conditions:

(CI) Γ0 is the trivial symmetry that is when 𝜀 = 0 then 𝑥𝑘 = 𝑥𝑘.

(C2) Γ𝜀 is a symmetry for every 𝜀 in some neighbourhood of zero.

(C3) Γ𝜆Γ𝜀 = Γ𝜆+𝜀 for every 𝜆, 𝜀 sufficiently close to zero.

(C4) Each 𝑥𝑘 may be represented as a Taylor series in 𝜀 and therefore

𝑥𝑘(𝑥1, 𝑥2, … … … … 𝑥𝑁; 𝜀) = 𝑥𝑘 + 𝜀𝜉𝑘(𝑥1, 𝑥2, … … … … 𝑥𝑁) + 𝑂(𝜀2),

𝑘 = 1, 2, … … … 𝑁.

Then the set of symmetries Γ𝜀 is a one-parameter local Lie group.

Symmetries belonging to a one-parameter Lie group depend continuously on the parameter 𝜀. An

object may also have symmetries that belong to a discrete group. These discrete symmetries cannot

be represented by a continuous parameter. For example, the set of symmetries of the equilateral

triangle has the structure of the dihedral group, 𝐷3. Whereas the two symmetries

of the isosceles triangle form the cyclic group ℤ2.

2.2 Symmetries of ODE

In the previous section, satisfying the symmetry condition means that the mapping is a bijection

from an equilateral triangle to itself. Applied to the solution curves of a differential equation,

satisfying the symmetry condition requires that a symmetry maps one solution curve to another.

The resulting solution curve must also satisfy the original differential equation. The symmetry

condition is crucial for solving differential equations with symmetry methods.

10

For example, consider a simple differential equation

0dy

dx (2.1)

The set of all solutions of this ODE is the set of horizontal lines, 𝑦(𝑥) = 𝑐, 𝑐 ∈ ℝ, which fill in the

(𝑥, 𝑦)-plane. Fig 2.3 represents the solution of ODE (2.1).

There are many symmetries for ODE (2.1), such as ( , ) ( , )x y x y and ( , ) ( , )x y x y are

the symmetries of (2.1), where = real parameter. For the parameter , Lie symmetry group for

the ODE (2.1) is:

: ( , ) ( , )P x y x y

The identity 𝐼 in a Lie symmetry group maps a point to itself. The ODE (2.1) is represented

geometrically by the set of all solutions, and so any symmetry of the ODE must necessarily map

the solution set to itself. That is, the symmetry condition (C3) requires that the set of solution

curves in the ( , )x y -plane must be indistinguishable from its image in the ( , )x y -plane. Therefore

0dy

dx when 0dy

dx (2.2)

A smooth transformation of the plane is invertible if its Jacobian is nonzero, so we impose the

further condition

0x y y xx y x y (2.3)

A particular solution curve of ODE (2.1) will be mapped to a solution curve and so

( , ) ( ),y x c c c c (2.4)

11

Fig. 2.3: Solutions of ODE 2.1

Here 𝑥 is regarded as a function of x and 𝑐 that is obtained by inverting ( , )x x x c . Differentiating

(2.4) with respect to 𝑥, we get

( , ) 0xy x c c

Thus, from (2.3) it can be said that, the symmetries of ODE (2.1) are of the form

( , )x y ( ( , ), ( ))f x y g y , 0, 0x yf g (2.5)

Here we use the known general solution of (2.1) to derive (2.2), which led to the result (2.5).

However, this result can be also obtained directly from (2.2), as follows:

On the solution curves, y is a function of 𝑥, and hence ( , )x x y and ( , )y x y may be regarded as

functions of 𝑥. Then, by the chain rule, (2.2) can be rewritten as

0,x

x

D ydy

dx D x when 0dy

dx (2.6)

y(x)

0

n

x

12

where xD denotes the total derivative with respect to 𝑥:

'' '' ...x x y y

D y y (2.7)

Therefore (2.2) becomes

'

' 0,x y

x y

y y y

x y x

when ' 0y

(2.8)

That is,

0x

x

y

x

(2.9)

Hence (2.5) holds. The advantage of using the symmetry condition in the form (2.2) is that one

can obtain information about the symmetries without having to know the solution of the

differential equation in advance.

2.3 The Symmetry Condition

In the previous section we see that symmetry of the ODE ' 0y can easily be visualized but it is

not possible for every ODE. Now consider the general form of first order ODE:

( , ).dyx y

dx (2.10)

Now, the symmetry condition of ODE requires that any symmetry maps that the set of solution

curves in the ( , )x y plane to an identical set of curves in the ( , )x y plane. Thus the symmetry

condition for (2.10) is, as follows:

( , )dyx y

dx when, ( , )dy

x ydx

(2.11)

Now since 𝑦 is a function of 𝑥 and 𝑐 an integration of constant, so (2.11) yields

'

' ( , )x

x

x y

x y

D yx y

D x

y y y

x y x

when, ( , )dy

x ydx

(2.12)

13

which is equivalent to

( , )

( , )( , )

x y

x y

x yy x y y

x x y x

(2.13)

Equation (2.13) is the symmetry condition for equation (2.10).

Example 2.3.1 We will show that the differential equation

21dy y

dx x

(2.14)

has a symmetry

( , ) ( , )x y e x y (2.15)

Here, 21( , ) y

x yx

.

Thus substituting (2.14), (2.15) into the symmetry condition (2.13), we get

Now, right side of (2.16): 21 y

x

21 y

e x

Since, 0,xy 0yx and so left side of (2.16) is,

2 2

2

2

1 11

1

x y

x y

y yy y

yx x

y e x e xx x

x

Therefore, the symmetry condition is satisfied.

2

2

2

11

1

x y

x y

yy y

yx

y xx x

x

(2.16)

14

Example 2.3.2 There are some cases, it is possible to use symmetry condition (2.13) to determine

the symmetries of a differential equation.

For example, consider the ODE

dyy

dx (2.17)

We want to find symmetry of this equation that satisfies equation (2.13). In other words, the

symmetry should satisfy the following:

x y

x y

y yyy

x yx

In order to find the symmetry, one can solve this equation for x and y . To make this easier, one

can make assumptions about x and y . Let’s assume that x is a function of 𝑥 and 𝑦 and y maps to

𝑦, that is

( , ) ( ( , ), )x y x x y y

Then Equation (2.13) becomes:

x y

yy

x yx

(2.18)

Multiplying (2.18) by x yx yx , we get

( )x yy y x yx

Now, if 0y , then

1 x yx yx (2.19)

Now, there are many symmetries of the type (2.19) such as the following Lie symmetries:

( , ) ( , ),x y x y (2.20)

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2.4 Change of Coordinates

Suppose that we are able to find a nontrivial one-parameter Lie group of symmetries of a first order

ODE (2.13). Then the Lie group can be used to determine the general solution of the ODE. Any

ODE that has a symmetry of the form

( , ) ( , )x y x y (2.21)

can be reduced to quadrature. Now, for (2.21), the symmetry condition (2.13) reduces to

Differentiating (2.22) with respect to at 0, on the left side we get

( , ) 0x y

On the right side by using chain rule to differentiate ( , )x y with respect to and therefore,

0 x y

x y

Now, 0, 1x y

and therefore, 0

y

, which imply that is a function of 𝑥 only.

Thus, the original ODE becomes

( )dyx

dx and ( )y x dx c .

While a symmetry of the form in Equation (2.22) does not exist in Cartesian coordinates for all

differential equations, it is possible to find such a symmetry in a different coordinate system.

Example 2.4.1 The following differential equation becomes much simpler to integrate when

written in polar coordinates

3 2

2 3dy y x y y x

dx xy x y x

(2.23)

( , ) ( , )x y x y (2.22)

16

If we convert it into polar coordinates ( , )r , then (2.23) becomes

2(1 )drr r

d (2.24)

The rotational symmetry of (2.23), (�̂�, �̂�) = (𝑥 cos 𝜀 − 𝑦 sin 𝜀 , 𝑥 sin 𝜀 + 𝑦 cos 𝜀) in polar

coordinates ( , ) ( , )r r becomes translation symmetry of (2.24) in .

So the ODE (2.24) is easily solvable.

Definition 2.4.1 Consider the ODE

( , )dyx y

dx (2.25)

Let, ( )y f x be the solution of (2.25). Consider a symmetry which maps ( )y f x to ( )y f x

such that

( , )dyx y

dx (2.26)

Now, the symmetry transform the curve ( )y f x to the set of points ( , )x y such that

( , ( )), ( , ( ))x x x f x y y x f x (2.27)

By solving first equation of (2.27) for the value of 𝑥, we get ( )x x x and substituting it into the

second equation. Then

( ) ( ( ), ( ( )))f x y y x x f x x (2.28)

If this symmetry belongs to a one-parameter Lie group, then f is a function of 𝑥 and parameter

only. Here ( , )x y and ( , )x y plane contain the same set of solution curves. So it is more convenient

to take one plane. Therefore, symmetry is now regarded as a mapping of the ( , )x y plane to itself,

is called the action of the symmetry on the ( , )x y plane. Here, the point ( , )x y mapped to the point

whose coordinates are

( , ) ( ( , ), ( , ))x y x x y y x y .

17

Definition 2.4.2 The solution curve ( )y f x is the set of points with coordinates ( , ( ))x f x . It is

mapped to the set of points with coordinates ( , ( ))x f x , that is, to the solution curve ( )y f x .

Therefore the curve is ( )y f x invariant under the symmetry if f f .

A symmetry is trivial if its action leaves every solution curve invariant.

Example 2.4.1 Consider the ODE

2 ; 0, 0dy yx y

dx x (2.29)

whose solution is 21y c x and symmetry is ( , ) ( , );x y e x e y real parameter. Here the

symmetry ( , )x y map the solution curve 21y c x to the curve 3 2

1y c e x in the ( , )x y plane.

Therefore the action of this symmetry on the quadrant 0, 0x y maps the solution 21y c x to

the solution 3 21y c e x , as shown in Fig. 2.4.

Definition 2.4.3 Orbits are essential tools for solving differential equations using symmetry

methods. Suppose there is a point 𝑃 on a solution curve to a differential equation. Under a given

symmetry, the orbit of 𝑃 is the set of all points that 𝑃 can be mapped to for all possible values of

.

Example 2.4.2 Consider the ODE

0dy

dx

The orbits for the points on the solution curves of this differential equation are vertical lines

under the symmetry

( , ) ( , )x y x y

For instance, under the symmetry, the orbit of the point (1, 0) includes

{(1,1),(1,2),(1,3),...}

18

Fig. 2.4: The action of symmetry

Figure 2.5 shows the orbit of the point (1, 0) under the symmetry ( , ) ( , )x y x y .

( , )x y

Fig. 2.5: One dimensional orbit

Definition 2.4.4 Consider the orbit through a noninvariant point ( , )x y .Then the tangent vector to

the orbit at the point ( , )x y is denoted by ( ( , ), ( , ))x y x y and defined by

x

y

0

0

( , )x y

x

y

0

y=f(x)

( )y f x

19

( , )x y ,dx

d ( , )x y

dy

d (2.30)

Where, real parameter.

In particular, the tangent vector at ( , )x y is

0 0( ( , ), ( , )) ( | , | )dx dy

x y x yd d

(2.31)

The Taylor series for the action of Lie group can be written as

2( , ) ( ),x x x y O 2( , ) ( )y y x y O (2.32)

Now, we know that a point is called invariant if under a Lie symmetry it maps to itself. From

equation (2.32), it is clear that the point (𝑥, 𝑦) is invariant if the tangent vector is zero, that is,

( , ) ( , ) 0x y x y (2.33)

The tangent vectors ( , )x y and ( , )x y can be used to find a new coordinate system. Also, ( , )x y

and ( , )x y can sometimes be used to find solution curves without the use of different coordinates.

The tangent vectors are useful for finding invariant solution curves. An invariant

solution curve is always mapped to itself under a symmetry. In other words, a curve is invariant if

and only if the tangent to the curve at (𝑥, 𝑦) is parallel to the tangent vector ( ( , ), ( , ))x y x y .

This condition can be defined by introducing characteristics

' '( , , ) ( , ) ( , )Q x y y x y y x y

We know that ' ( , ),y x y and ( , ) ( , ) 0x y x y , when the solution curve ( )y f x is invariant.

Thus, for invariant curve, ( , ) 0,Q x y when ( ).y f x

The points on an invariant solution curve are mapped either to themselves or to another point on

the same curve. Therefore, the orbit of a noninvariant point on an invariant solution curve is the

solution curve itself. If a solution curve is invariant that means that the derivative at the point ( , )x y

20

will point in the same direction as the tangent vectors to the orbit. As varies, the point is mapped

to another point on the same solution curve, rather than a different solution curve.

Thus,

( , )( , )( , )

dy x yx y

dx x y

(2.34)

Example 2.4.3 The symmetry

( 1)( , ) ( , )e xx y e x e y

(2.35)

acts trivially on

dyy

dx (2.36)

To show this we use the equation (2.34).

Here,

x e x and ( 1)e xy e y

Thus,

0| ( , )dxx y x

d

0| ( , )dyx y xy

d

Substituting the value of , in the following characteristic equation, we obtain

' '( , , ) ( , ) ( , )

0Q x y y x y y x y

xy xy

21

Therefore, the symmetry (2.35) acts trivially on (2.36) because characteristic equals 0.

Example 2.4.4 Consider the Riccati equation

23

2 1 , 0dy yxy x

dx x x

(2.37)

It has the symmetry

2( , ) ( , )x y e x e y (2.38)

The tangent vectors are

( , ) , ( , ) 2x y x x y y

The characteristic is

23

2 22

' '( ) ( , ) ( , )2 12 ( )

1

, ,Q x y x y

yy xy x

x x

x yx

x y y y

(2.39)

Equation (2.39) is 0 when 2y x . So the symmetry (2.38) acts nontrivially on (2.37) except for

2y x .

2.5 Canonical Coordinates

We know that if the ODE of the form (2.25) has symmetries which includes translation

symmetries

( , ) ( , )x y x y (2.40)

may be integrates directly. If the ODE has a symmetry of type (2.40), then the ODE can be

rewritten in terms of new coordinates. Our target is now to introduce this new coordinates which

is called canonical coordinates. From equation (2.40), it can be observed that all the orbits of this

symmetry has the same tangent vector at random point ( , )x y , as follows:

( ( , ), ( , )) (0,1)x y x y (2.41)

22

Thus our aim is to introduce the coordinates ( , ) ( ( , ), ( , ))r s r x y s x y for given Lie symmetry such

that

( , ) ( ( , ), ( , )) ( , )r s r x y s x y r s (2.42)

Now the tangent vector at any point is (0, 1), that is,

0 0| 0, | 1dr ds

d d (2.43)

Applying the chain rule to equation (2.43), we get

0 0 0| | | ( , ) ( , )dr dr dx dr dy dr dr

x y x yd dx d dy d dx dy

(2.44)

and

0 0 0| | | ( , ) ( , )dr dr dx dr dy dr dr

x y x yd dx d dy d dx dy

(2.45)

Equation (2.44) and (2.45) can be simplified in the form as

( , ) ( , ) 0x yr x y r x y (2.46)

( , ) ( , ) 1x ys x y s x y (2.47)

The equation (2.46) and (2.47) are solvable if the nondegeneracy condition

0x y y xr s r s (2.48)

If a curve of constant 𝑠 and a curve of constant 𝑟 meet at a point, they cross one another

transversely. Any pair of functions ( , ), ( , )r x y s x y satisfying (2.46), (2.47) and (2.48) is called a

pair of canonical coordinates.

By definition, the tangent vector at any noninvariant point is parallel to the curve of constant 𝑟

passing through that point. Therefore, the curve of constant 𝑟 coincides with the orbit through the

point. The orbit is invariant under the Lie group, so 𝑟 is sometimes referred to as an invariant

canonical coordinate. The curves of constant 𝑠 are not invariant, because they cross the

one-dimensional orbits transversely.

23

Canonical coordinates cannot be defined at an invariant point, because the determining equation

for 𝑠 in (2.46), (2.47) has no solution if ( , ) ( , ) 0x y x y .

Now, equation (2.46) and (2.47) can be solved by the method of characteristics [5]. The solution

of these equation can be regarded as a surface ( , )r x y and ( , )s x y . Equation (2.46) satisfy:

, , 1 , ,0 0x yr r (2.49)

We know that for the surface ( , )r x y z , the gradient is , , 1x yr r . So, , , 1x yr r is a normal vector

to the surface, ( , )r x y z . Since, the dot product , , 1 , ,0 0x yr r , thus , ,0 is a vector

which is orthogonal to , , 1x yr r . Therefore, ( , ), ( , ),0x y x y is in the tangent plane to ( , )r x y

.

Consider a parametric curve 𝐶 on the surface ( , )r x y , since ( , ), ( , ),0x y x y is in the tangent

plane to ( , )r x y , so ( ( ), ( )), ( ( ), ( )),0x t y t x t y t is tangent to 𝐶. Thus, we can write symmetric

equation as

, , 0dx dy dz

dt dt dt (2.50)

Equation (2.50) can be rewritten as

dx dy

(2.51)

To find 𝑟, we can use equation (2.51).

Similarly, for equation (2.47), we can write

, , 1dx dy dz

dt dt dt

(2.52)

Therefore

dx dydz

(2.53)

Rename 𝑧 as 𝑠, then (2.53) becomes:

24

( , ) ( , )dx dy

dsx y x y

(2.54)

Let, ( , )g x y be the first integral of the ODE:

( , )dy

f x ydx

(2.55)

Then, ( , )g x y is non-constant but its value along the solution curve of (2.55) is constant. Therefore,

( , )g x y c , where 𝑐 is a constant. Applying the total derivative operator for ( , )g x y , we obtain

( , ) 0, 0x y yg f x y g g (2.56)

Dividing equation (2.46) by ( , )x y , we get

( , ) ( , ) 0( , ) ( , )

dr x y dr x y

dx x y dy x y

and therefore

( , ) 0( , )x y

x yr r

x y

(2.57)

From (2.55) and (2.56), we get

( , ) , ( , ) 0( , )

dy x yx y

dx x y

(2.58)

We use equation (2.58) to find 𝑟, because ( , )r x y is the first integral of (2.56) and equals to the

constant 𝑐. To find 𝑠, we can use the equation (2.54)

( , ) ( , )dx dy

dsx y x y

And therefore,

( , ) ( , )dy dx

dsx y x y

(2.59)

25

( , ) ( , )dy dx

sx y x y

(2.60)

Example 2.5.1 Consider the following one parameter Lie symmetry group

0( , ) ( , );m mx y e x e y (2.61)

We want to find the canonical coordinates for symmetry (2.61). Here, the tangent vectors are

0 0( , ) | ( ) |dx dx y e x x

d d

and 0 0( , ) | ( ) |mdy dx y e y my

d d

.

Now, ( , )r x y is the first integral of

( , )( , )

dy x y my

dx x y x

(2.62)

Equation (2.62) is separable, so by integrating, we get

1ln ln m

m

m

dy dxm y x c

y x

yy cx c

x

Thus, ( , ) .m

yr x y

x

To find 𝑠, we use the equation (1.60) to get

ln( , )dx dx

s xx y x

Therefore, the canonical coordinates are ( , ) ( , ln ).m

yr s x

x

Example 2.5.2 Consider the following one parameter Lie symmetry group

( , ) ( , )

1 1x y

x yx x

(2.63)

26

Here, the tangent vectors are

22

20 0( , ) | |(1 )

xx y x

x

(2.64)

and

20 0( , ) | |

(1 )xy

x y xyx

(2.65)

Now, ( , )r x y is the first integral of

( , )( , )

dy x y y

dx x y x

This equation is separable.

Integrating, we get

1ln ln

( , )

y x c y cx

yc r x y

x

For 𝑠:

2

1dxs

x x

Therefore, canonical coordinates are

1( , ) ( , ), 0.yr s x

x x

From the above two example it can be said that tangent vectors are more important than symmetries

to find canonical coordinates. In this section, we have already seen how to find tangent vectors.

Sometimes it is possible to find tangent vectors without knowing symmetry. Once the tangent

vectors and canonical coordinates are known, then it is possible to find symmetry by using

canonical coordinates. Firstly, we have to solve the canonical coordinates ( , )r x y and ( , )s x y for

𝑥 and 𝑦 to get

27

( , ), ( , )x f r s y g r s

Then, x and y are

( , ) ( ( , ), ( , ) )x f r s f r x y s x y (2.66)

( , ) ( ( , ), ( , ) )y g r s g r x y s x y (2.67)

Example 2.5.3 In this example, we will show how to find symmetries from the following

canonical coordinates:

1( , ) ( , ), 0yr s x

x x

Solving for 𝑟 and 𝑠, we get

1, rx y

s s

Thus, by definition,

1 1( , )

11 1

xs s x y

x

x

x

And

( , )( , )

1 1

r r x yy

s s x y

y

yx

x

x

28

Therefore, the symmetry is

( , ) ( , ).1 1

x yx y

x x

2.5.1 Solving ODE with Canonical Coordinates

Our aim is to write the ODE (2.25) in terms of canonical coordinates, so it is necessary to find

ds

dr. For this reason, apply the total derivative operator to ds

dr, which gives

.

( , )( , )

x y

x y

s x y sds

dr r x y r

(2.68)

This will result an equation ds

dr written in terms of 𝑥 and 𝑦. To write it in terms of 𝑟 and 𝑠, solve

the coordinates ( , )r x y and ( , )s x y for 𝑥 and 𝑦, then simplify. From there, solve the equation and

put the solution back into Cartesian coordinates.

Example 2.5.4 Again, consider the Riccati equation

23

2 1 , 0dy yxy x

dx x x (2.69)

It has the symmetry

2( , ) ( , ).x y e x e y

Here, the tangent vectors are

( , ) , ( , ) 2 .x y x x y y

To find 𝑟, we use equation

( , ) 2 .( , )

dy x y y

dx x y x

This equation is separable, so by integrating, we get

29

1

2

2 ln 2lndy dxy x c

y x

y cx

Thus, 2( , ) .r x y c x y

Now, to find 𝑠, we use:

.( , )dx dx

dsx y x

Now, integrating above equation:

ln .dxs x

x

Therefore, the canonical coordinates are

2( , ) ( , ln ).r s x y x

Writing 𝑥 and 𝑦 in terms of 𝑟 and 𝑠, we obtain:

2, .s sx e y re

Now, from equation (2.68),

2

4 23 2

1( , )( , ) 2

11 .1 12 2

x y

x y

s x y sds xdydr r x y r

xy xdx

x

x yxy x y xy

x

Putting the values 2,s sx e y re in the above equation, we get

24 2 41 1 .

11s s

ds

dr re r e

30

Now we can integrate the equation

2 2

1 .1 1

ds drds

dr r r

And therefore,

2 .1

drs

r

By using partial fraction to integrate

0

0

1 1 1 12 1 2 1

1 (ln( 1) ln( 1))21 1ln( ) .2 1

s dr drr r

r r p

rp

r

Substituting 2( , ) ( , ln )r s x y x , we get

2

021 1ln ln( ) .2 1

x yx p

x y

Thus, 2

21.1

x yx p

x y

Squaring both sides, we get

22

2

2 2 2

2

4 2

11

( 1) ( 1)

.

x yx p

x y

x x y p x y

p xy

x px

Hence this is the solution of Riccati equation.

31

2.5.2 The Linearized Symmetry Condition

According to the previous discussion of solving differential equations mainly depends on Lie

symmetry. We know the Lie symmetry condition is:

( , )( , )

( , )x y

x y

y x y yx y

x x y x

(2.70)

By solving the above equation we can have the Lie symmetry ( , )x y . But the equation is a

complicated nonlinear partial differential equation of two unknowns; this is why it is not always

possible to find symmetry from (2.70). However, Lie symmetry can be derived from much simpler

tangent vector field, which is called linearized symmetry condition.

If we use the definition of Lie symmetry and use Taylor series expansion to expand ,x y and

( , )x y at 0 , gives:

2( , ) ( )x x x y O (2.71)

2( , ) ( )y y x y O (2.72)

2( , ) ( , ) ( ) ( ).x yx y x y O (2.73)

Substituting (3.71), (3.72), (3.73) in equation (3.70), we get,

( )( ).

1 ( )x y

x y

x y

Then we multiply by the denominator to get:

( ) ( ) ( ).x y x y x y

Neglecting 2 terms and simplifying:

2( ) .x y x y x y (2.74)

The equation (2.74) is the linearized symmetry condition.

32

The next few examples will show how to use linearized symmetry condition to find symmetry,

which is the main tool for solving differential equations.

Example 2.5.5 Consider the ODE

.dy y

xdx x

(2.75)

If we substitute it in linearized symmetry condition (2.74), we get,

2( ) .x y x y x y

22

1( ) ( )( ) ( (1 ) ( )) 0.x y y x

y y yx x

x x x x

It is necessary to find , from the above equation, but the equation is complicated. So, we make

an ansatz about , . Let, 0 and is a function of 𝑥 and 𝑦 only. Then we get

0.xx

This ODE is easy to solve:

0ln ln.

d dx

x

x c

cx

Thus, ( , ) (0, ).cx So, we can find canonical coordinates 𝑟 and 𝑠. When ( , ) 0 .x y r x

Now, .dyds

Integrating,

.dy ys

cx cx

33

If 1,c then we get ( ( , ), ( , )) ( , ).yr x y s x y x

x Thus,

21, .x y

ys s

x x

We know that

21 ( )

1.1

( , )( , )

x y

x y

y yx

x x x

s x y sds

dr r x y r

Therefore, ODE becomes, 1ds

dr , by integrating 1s r c .

Putting 𝑥 and 𝑦 back in, we get

1.y

x cx

Thus the general solution of (2.75) is 21 .y x c x

Example 2.5.6 Consider the following differential equation:

2 .x xdye y y e

dx (2.76)

Here,

2( , ) x xx y e y y e .

So,

2 , 2 1.x x xx ye y e e y

The linearized symmetry condition is

34

2( ) .x y x y x y

Now we will make an ansatz about ( , )x y and ( , )x y . Let , 1 and is a function of 𝑦 only.

Then the linearized condition becomes

0.y x y

2 2( ) ( ) (2 1) 0.x x x x xy e y y e e y e e y (2.77)

Simplifying (2.77) to get,

( 2 ) ( 1) 0.x xy y ye y y y e y

Now,

2 0.y y y (2.78)

0.y y (2.79)

1 0.y (2.80)

From (2.80),

1 .dyy

d

Now, we can find canonical coordinates,

.dyy

dx

This equation is separable, so by integrating

1ln .xdydx y x c y ce

y

35

Therefore, .x xy y

c re e

To find 𝑠: Since 1 , .s dx x

Thus the canonical coordinates are: .( , ) ( , )x

yr s x

e

We know that

2

2 2

( , )( , )

1( )

1 .1

x y

x y

x x x x

x

s x y sds

dr r x y r

ye e e y y e

y e

Therefore,

2 .1

1ds

dr r

(2.81)

Integrating (2.81),

121 tan ( ).

1s dr r

r

Thus,

1tan ( ) tan( ) .x

x

yx y x e

e

The general solution is, .tan( ) xy x e k

Example 2.5.7 Consider an ODE of the form

( ).dy y

Fdx x

(2.82)

ODE of form (2.82) admits the one parameter Lie group of scaling symmetries

36

.( , ) ( , )x y e x e y

For this type of ODE the canonical coordinates are:

., ln , 0yr s x x

x

There are two possibilities:

(1) If ( )F r r , then symmetries are trivial, and the general solution of (2.82) is r c , that is

.y cx

(2) If the symmetries are not trivial then equation (2.82) is equivalent to

1 .( )

ds

dr F r r

Thus, the solution of (2.82) is

/

ln( )

y x drx c

F r r

where, 𝑐 is the integration of constant.

In this chapter, we have explained in detail how to solve a first order ODE by using canonical

coordinates and Lie symmetry. Also, we have seen relevant example of tangent vector field,

linearized symmetry conditions, canonical coordinates, etc. In the next chapter, we will discuss the

basic tool and ingredients for solving higher order ODE, which is the continuation of this chapter.

37

Chapter 𝟑

Lie Symmetries and Higher Order ODE

We have restricted our discussion on first order ODE in chapter two. In this chapter, we need to

extend these ideas to higher order ODEs. This chapter is devoted to the Lie symmetries of higher

order ODE, finite symmetry transformations of determined operators, reduction of order,

differential invariants and Lie algebras.

3.1 Infinitesimal Generator

For extending the ideas of first order ODEs to higher order ODEs, it will no longer possible to use

two-dimensional pictures to represent everything of importance which we did for first order ODEs.

For explaining the Lie symmetries of higher order it is necessary to introduce a compact notation

that is easily extended to deal with differential equations of arbitrary order, with any number of

dependent and independent variables.

To begin we consider the following definition:

Definition 3.1.1 Suppose that a first order ODE has a one parameter Lie group of symmetries

,: ( , ) ( , )P x y x y whose tangent vector at (𝑥, 𝑦) is ( , ) , then the partial differential operator,

( , ) ( , )X x y x y

x y

(3.1)

is called the infinitesimal generator of the Lie group.

The infinitesimal generator can also be written as compact form, as follows:

( , ) ( , )x yX x y x y (3.2)

Example 3.1.1 Consider the following Lie point symmetry:

( , ) ( , ).x y x y

The above symmetry has an infinitesimal generator

38

.Xx y

because ( , ) 1, ( , ) 1x y x y .

Example 3.1.2 Consider the Lie symmetry:

( , ) ( , ).1 1

x yx y

y y

It has the tangent vectors

( , )x y xy and 2( , ) .x y y

The infinitesimal generator for this symmetry is

2X xy yx y

Infinitesimal generator can be used to determine the symmetry, which is explained by the

following example:

Example 3.1.3 Consider the following infinitesimal generator:

.X y

x y

(3.3)

Here, the tangent vectors are

( , ) 1, ( , ) .x y x y y

From the definition of tangent vectors, we know that

0 0( , ) | | ., ( , )dx dy

x yd d

x y

(3.4)

where, ( , )x y represents the Lie symmetry.

Integrating equation (3.4), we get x and y such that

39

, .x x y e y

Therefore, the Lie symmetry for infinitesimal generator (3.3) is:

( , ) ( , ).x y x e y

The next example will explain the relation of canonical coordinates and infinitesimal generator.

Example 3.1.4 Consider the infinitesimal generator:

2(1 ) .X x xyx y

(3.5)

The tangent vectors are:

2( , ) 1x y x and ( , ) .x y xy

To find canonical coordinates:

2

2

1

.1

dy xy

dx x

dy xdx

y x

Integrating,

22

2

2

2

1ln ln(1 )1 2

ln ln( 1 )

1

.1

dy xdxy x

y x

y x

y c x

yc

x

Therefore, 2

.1

yr

x

For 𝑠 we know that 2 .( , ) 1dx dx

dsx y x

40

Integrating,

12 tan .

1dx

s s x cx

By solving 𝑟(𝑥, 𝑦) and 𝑠(𝑥, 𝑦) for 𝑥 and 𝑦:

2tan , 1 tan .x s y r s

We know from previous chapter:

( , ) ( ( , ), ( , ) )( , ) ( ( , ), ( , ) ).

x f r s f r x y s x y

y g r s g r x y s x y

Thus,

1tan tan( ) tan((tan ) ).x s s x

We know that

tan tantan( ) .1 tan tan

a ba b

a b

(3.6)

Therefore,

tan1 tan

sincossin1cos

cos sin .cos sin

xx

x

x

x

x

x

Similarly, to find 𝑠:

2

2

2 1

2

1 tan

1 tan ( )

. 1 tan (tan ).1

y r s

r s

yx

x

41

Now, using equation (3.6) and simplifying gives:

2

2

2 2 2 2

2 22

2

tan. 1 ( )1 tan1

1 tan tan1 2 tan tan1

1 tan1 tan

.cos sin

y xy

xx

y x x

x xx

y

x

y

x

Thus, the symmetry is

cos sin( , ) ( , ).cos sin cos sinx y

x yx x

Now we can establish the relation between canonical coordinates and infinitesimal generator.

Equation (3.2) can be written in terms of canonical coordinate (𝑟, 𝑠) as

0, 1.Xr Xs (3.7)

Now, we will show the effect of change of coordinates for infinitesimal generator. Let (𝑢, 𝑣) be

the new coordinates and 𝐹(𝑢, 𝑣) be a smooth function. Then

( , ) ( ( , ), ( , ){ } { }

( ) ( ) .x u x v y u y v

u v

XF u v XF u x y v x y

u F v F u F v F

Xu F Xv F

If (𝑢, 𝑣) = (𝑟, 𝑠), then

( ) ( ) .r s sX Xr Xs

3.2 Symmetry Condition for Higher Order ODE

Consider the following higher order ODE:

( ) ( 1)'( , , ,........ );m my x y y y where, ( )m

mm

d yy

dx

(3.8)

42

The symmetry condition of higher order works like first order ODE. That is, a symmetry for

equation (3.8) is a diffeomorphism that maps the set of solutions of the ODE to itself. Thus

: ( , ) ( , ).x y x y (3.9)

Now, the action of for the derivative ( )my is the mapping as follows:

( ) ( )' ' .:( , , ,..., ) ( , , ,..., )m mx y y y x y y y (3.10)

where

( )m

mm

d yy

dx , 1,2,...,m n

(3.11)

The mapping (3.10), (3.11) is called the nth prolongation of . The term ( )my is defined by

( 1)( 1)( ) (0) .,

mmm x

x

D ydyy y y

dx D x

(3.12)

Here, xD is the total derivative with respect 𝑥:

'

' '' ...x x y yD y y (3.13)

Therefore, the symmetry condition for ODE (3.8) is;

( ) ( 1)'( , , ,... ).m my x y y y (3.14)

When (3.8) holds and ( )my are given by (3.12).

Most of the time, the equation (3.14) is nonlinear. Thus, it is necessary to linearize the symmetry

condition (3.14) by putting 0 .

Following example will examine whether a given symmetry for a ODE satisfy the symmetry

condition (3.14) or not:

Example 3.2.1 We show that the transformation

1( , ) ( , ).yx y

x x (3.15)

is a symmetry for the following second order ODE,

43

'' 0, 0.y x (3.16)

From (3.12),

'( )( ) '1( ) ( )

xx

xx

yD

D y xy y xyD x

Dx

3'

'' ''( )1( )

x

x

D y xyy x y

Dx

Thus, the symmetry condition is,

'' 0,y when '' 0.y

3.2.1 The Linearized Symmetry Condition for Higher Order ODE

We have already explained the linearized symmetry condition for first order ODE, now we present

it for higher order ODE. Linearized symmetry condition is used for finding the tangent vectors

( , ), ( , )x y x y . As demonstrated in the next example, the linearized symmetry condition is used

to determine infinitesimal generator for a given symmetry. To begin, the Taylor series expansion

for ,x y and ky is given by

2( ),x x O (3.17)

2( ),y y O (3.18)

( ) ( ) ( ) 2( ).k k ky y O (3.19)

Here, ( )k denotes the tangent vector that correspond to the kth derivative of :y

( )( )

0| .kk

y

Substituting equations (3.17), (3.18), (3.19) into (3.14); ( )O terms yield the linearized

symmetry condition:

44

(1) ( 1)

(1) ( 1)( ) ... n

nn

y yx y

(3.20)

The functions ( )k are calculated from equation (3.12) as follows:

2

2

'' ( ) .

1 ( )x

x

x

x

D y y D Oy

D x D O

(3.21)

Now multiplying equation (3.21) by

11

x

x

D

D

and ignore the higher order term

2(1) '( ) ( ).'x xy y D y D O (3.22)

If we compare equation (3.20) and (3.22), then

(1) ' .x xD y D (3.23)

Similarly

( ) ( 1) 2( )

2 .( )1 ( )

k kk x

x

y D Oy

D O

and hence

( ) ( 1) ( ) .k k kx xD y D (3.24)

Now for 𝑘 = 1,

(1) 2 ,' ' ' ' ' '( ) ( ) ( ) ( )x x x y x y x y x yD y D y y y y y

(2) 2 3 ''' ' ' '(2 ) ( ) ( 2 ) ( ) ( 2 3 ) .xx xy xx yy xy yy y x yy y y y y

Definition 3.2.1 We know that the infinitesimal generator is defined by:

( , ) ( , )x yX x y x y

45

Then for the action of Lie symmetries on derivatives of order 𝑛 or smaller, the prolonged

infinitesimal generator is defined by

( ) (1) ( )( )' ... .n n

y nxXyy

(3.25)

where ( )n is defined by equation (3.24). We can use the prolonged infinitesimal generator to

write the linearized symmetry condition (3.20) in compact form as follows:

( ) (1) ( 1)( ) ( ( , , ,..., ) 0.n nnX y x y y y (3.26)

Example 3.2.2 Consider the second order ODE which arises in the study of swimming of micro-organism:

22'' ( ) .

'yy y

y

(3.27)

The linearized symmetry condition is:

If we compare the power of 'y , then the determining equations:

1 0,yy yy

(3.29)

2

1 12 0,yy xy yy y

(3.30)

2 22 3 0,xy xx y xyy

(3.31)

2 .( 2 ) 2 0xx y xy y (3.32)

Integrating (3.29), we get

( )ln ( ).a x y b x (3.33)

22 3 2

2

2'2

''

' ''

' ( )' '(2 ) ( ) ( 2 ) ( ) ( 2 3 )( )

2( 2 ) { ( ) }( )

xx xy xx yy xy yy y x y

x y x y

yy y y y y

y

y yy y y

y y

(3.28)

46

From (3.30), we get

2' .( )(ln ) ( ) ln ( )a x y c x y y d x y (3.34)

where, 𝑎, 𝑏, 𝑐, 𝑑 are unknown functions which we have to determine from (3.31), (3.32).

Substituting (3.33), (3.34) into (3.31) gives

'' ' '' .3 ( )ln 3 ( ) 2 ( ) ( ) 0a x y a x y c x b x

Therefore

'' ' .( ) 0, ( ) 2 ( )a x b x c x (3.35)

If we substitute (3.33), (3.34) into (3.32), we get

2 2'' ' ''( ) ln ( ) ln (2 ( ) ( ) ( )) ( ) 0.c x y y c x y y b x c x d x y d x y

Equating the coefficients, we get

( ) 0,c x '( ) 2 ( ),d x b x '' .( ) 0d x

From (3.34),

1 2( ) ,b x c c x 2( ) .d x c

where, 1 2,c c are arbitrary constants. Hence

1 2( ) ,x c c x 2 .2c y (3.36)

Thus infinitesimal generator is:

1 2 2( ) 2x y x yX c c x c y

1 2 .( 2 )x x yc c x y

(3.37)

Every infinitesimal generator is of the form:

1 21 2 .X X Xc c

47

From (3.37),

1 ,xX 2 2 .x yX x y

3.3 Reduction of Order

To solve a first order ODE, we use canonical coordinates which are also useful for higher order

ODE. It can be used to reduce the order of higher order ODE, as follows:

Let 𝑋 be an infinitesimal generator of one parameter Lie group of symmetries of the ODE

( ) ( 1)'( , , ,..., );m my x y y y where, 2.m (3.38)

Suppose that (𝑟, 𝑠) be the canonical coordinates such that

.sX (3.39)

If we write (3.38) in terms of canonical coordinates, then it becomes

( 1)( ) ( , , ,..., );mms r s ss where, ( ) .k

k

k

d ss

dr

(3.40)

The ODE (3.40) is invariant under the translation of 𝑠, so 0.s

Thus equation (3.40) becomes:

( ) ( 1)( );, , ...,,m ms r ss s where, ( ) .

kk

k

d ss

dr

(3.41)

If we set, u s , then from (3.41)

( 1) ( 2)( );, , ,...,m mu r u u u where, 1

( )1 .

kk

k

d su

dr

(3.42)

Suppose that the ODE (3.42) has the general solution

1 2 1,( ).; , ..., mu g r c c c (3.43)

Thus the general solution of the original ODE (3.38) is:

48

( , )1 2 1),( , ) ( .; , ...,r x y

mm dr cs x y g r c c c (3.44)

Normally, if 𝑢 is a function of 𝑟 and s such that 0,( , )su r s then the ODE (3.43) becomes:

( 1) ( 2)( , , ,..., );m mu r u u u when ( ) .

kk

k

d uu

dr

(3.45)

If it is possible to find the solution of (3.45), then the relationship

( , ).s s r u

gives the general solution of (3.38):

1 2 1

( , ), )( , ) ( ; , ..., m m

r x ydr cs x y g r c c c (3.46)

The next example will explain the process of reduction of order:

Example 3.3.1 Consider the following second order nonlinear ODE:

2''' '.( ) 1( )y

y y yy y

(3.47)

This ODE has a one-parameter Lie group of symmetries whose infinitesimal generator is:

.X x

The canonical coordinates are:

( , ) ( , ).r s y x

Now,

1'( ) .dsy

drs

If we choose

' 1( ) .u s y

49

Then the ODE (3.47) becomes:

23

''

'1( ) .

( )du y u

r udr r ry

This is a Bernoulli equation, whose general solution is:

212 1.u r c r

Therefore,

21

1 .( ) 2 1

drdr

u r r c rs

After simplifying, we get

2 21 1 1 2

11 2

2 21 1 1 2

1 tanh( 1( )),( ) ,

1 tanh( 1 ( )),

c c c x c

y c x c

c c c x c

212

12

1

111

c

c

c

which is the solution of (3.47).

3.4 Invariant Solutions of Higher Order ODE

There are many ODEs which cannot be solved by Lie symmetry. In that case, it is possible to find

invariant solution. From chapter one we know that any curve 𝐶 on the (𝑥, 𝑦) plane is invariant

under the group generated by 𝑋 satisfies

' '( , , ) .Q x y y y (3.48)

Example 3.4.1 Consider the Blasius equation

''''' .y yy (3.49)

which has the following symmetry generated by

1 ,xX 2 .x yX x y (3.50)

50

The symmetries (3.50) reduced the Blasius equation to a first order ODE whose solution is not

known. Therefore, if we want to find exact solution it is better to find invariant solution.

For 1 xX X , equation (3.48) reduces to:

' ' '( , , ) 0.Q x y y y y (3.51)

Therefore every curve that is invariant under 1X is:

,y c where, c constant. (3.52)

Similarly, characteristic equation for 2 x yX X x y is:

' ' '( , , ) 0.Q x y y y y xy (3.53)

Thus,

' 0y xy

dyy x

dx

dy dx

y x

Integrating,

ln ln ln

, .

dy dx

y x

y x c

cy c

x

The general solution Blasius equation is:

,y c .c

yx

(3.54)

Example 3.4.2 Consider the ODE:

3

''' ,1y

y 0.x (3.55)

51

which has the following symmetry generated by

1 ,xX 23 .4x yX x y

Now, there are no solution that are invariant under 1X . So, the invariant curve under 2X is:

'3 0.4

Q y xy

Therefore, on every invariant curve

' 3 .4

yy

x (3.56)

Differentiating (3.56) to get

2 2 2 2

'3 3 9 3 34 4 16 4 16

'' y y y y yy

x x x x x

Similarly,

2 3 3

'''' 3 3 15

16 8 64y y y

yx x x

(3.57)

on every invariant curve. Comparing (3.57) and (3.55), we get

1 34 464( )

15y x

(3.58)

which is the invariant solution.

3.4.1 Differential Invariant and Solutions of Higher Order ODE

In section 3.3 we have seen that a single generator is needed to reduce the order of an ODE once.

In this section, we will discuss the reduction of nth order ODE with R n Lie point symmetries

to an ODE of order (𝑛 − 𝑅).

52

Definition 3.4.1 Let 𝑋 be the generator of Lie symmetry for the ODE:

( ) ' ( 1)( ), 2., , ,...,m m my x y y y (3.59)

In terms of canonical coordinates, equation (3.59) becomes:

( 1) ( 2)'( ),, , ,...,m mu r u u u (3.60)

where, ( , )u u r s such that .0su

Now the ODE (3.60) consists of functions that are invariants under the action generated by .sX

These type of functions are called differential invariants. A nonconstant function ( )'( , , ,..., )k

I x y y y is a kth order differential invariant of the group generated by 𝑋 if

( ) 0.kX I (3.61)

For canonical coordinates ( ) ,ksX therefore every kth order differential invariants carry the

following form:

( )( ), ,..., kI F r s s or ( 1)( ), ,..., kI F r u u (3.62)

for some function 𝐹.

Definition 3.4.2 From equation (3.62) it can be said that 𝑟(𝑥, 𝑦) is the only zero order differential

invariant and the differential invariant of order one is '( , , ).u x y y All others differential invariants

of order two or greater are functions of 𝑟, 𝑢 and derivative of 𝑢 with respect to 𝑟, that’s why 𝑟 and

𝑢 are called the fundamental differential invariant.

From equation (3.61), it can be written that:

(1) ( )' ( )... 0.k

kx y y yI I I I

That is 𝐼 is the first integral of

( )

( )...k

k

dx dy dy

(3.63)

From (3.63), it can be said that 𝑟 is the first integral of

53

.dx dy

Also, 𝑢 is the first integral of

'

(1) .dx dy dy

Example 3.4.3 Consider the infinitesimal generator

.x yX y x (3.64)

We will find the fundamental differential invariant. Here, ( , ) , ( , ) .x y y x y x Thus, 𝑟(𝑥, 𝑦)

is the first integral of

.dx dy dx dy

y x

Integrating,

2 2 .xdx ydy x y c

Thus,

2 2( , ) .r x y x y

Similarly, '( ), ,u x y y is the first integral of

2

'

' .1 ( )

dx dy d

y x

y

y

(3.65)

Consider the equation,

2 2.dy dy

x r y

54

Now, taking last two terms of (3.65), we get

2 2

'

2

'

' ' 21 ( ) 1 ( )dy dy

x r y

dy

y

dy

y

1 1

1 1

1 1

'

'

'

sin ( ) tan .

tan sin ( ).

tan tan ( ).

yc

r

yc y

r

yc y

x

y

Thus, the first integral of (3.65) is of the form:

1 1'( , tan tan ( )).yI F r

xy (3.66)

Let

1 1'

''tan(tan tan ) .y xy y

u yx x yy

Suppose that ℒ is the set of all infinitesimal generators of one parameter Lie groups of point

symmetries of an ODE of order 2.n Then, if x yX , which can also be written as

1.

R

i ii

X c X

Therefore, if

1 2 1 1 2 2, .X X c X c X

Thus, ℒ is a vector space. Suppose that 1 2{ , ,..., }RX X X is a basis of ℒ, then an ODE which has

more than one Lie point symmetry generator can be written in terms of differential invariant as

follows:

55

( )

( )1 1 1

( )2 2 2

( )

. . .0

. . .0

. . . .. .

. . . .. .

. . . .. .

. . .0

x

R

R

R

R

R R R

y

y

I

I

I

(3.67)

The above system has two independent solutions which can be found by Gaussian elimination

and by the method of characteristic. One solution is independent of ( )Ry , which we denote by Rr

and Ru denote the other solution which depends nontrivially on ( )Ry . Thus the ODE (3.59)

reduces to

( 1)( )., ,...,m R m Rr R R Ru r u u

Thus an 𝑅 parameter symmetry group enables us to reduce the order of the ODE by 𝑅.

Example 3.4.4 Consider the ODE

( ) ''''2 (1 ) .iv y yy

y (3.68)

which has three-parameter Lie group of point symmetries generated by

1 ,xX 2 ,x yX x y 3

2 2 .x yX x xy (3.69)

We want to find the fundamental differential invariants. Using (3.67),

[1 0 0𝑥 𝑦 0

𝑥2 2𝑥𝑦 2𝑦

0 0−𝑦′′ −2𝑦′′′

2(𝑦′ − 𝑥𝑦′′) −4𝑥𝑦′′′]

[

𝐼𝑥𝐼𝑦𝐼𝑦′

𝐼𝑦′′

𝐼𝑦′′′]

= [000]

(3.70)

By using Gaussian elimination, we get

56

[1 0 00 𝑦 00 0 𝑦

0 0−𝑦′′ −2𝑦′′′

𝑦′ 0]

[

𝐼𝑥𝐼𝑦𝐼𝑦′

𝐼𝑦′′

𝐼𝑦′′′]

= [000]

(3.71)

Now, the third equation of (3.71) gives

' ''' .0

y yyI y I (3.72)

The characteristics equation of (3.72) is

' ''

' .dx dy

x y

dy dy

y y

This equation gives

'' '2 '''( , ,2 , ).I I x y yy y y

Substituting this into the second equation of (3.71), we get

2'' '2 '''( ).,2 ,I I x yy y y y

Finally, the first equation of (3.71) gives

2'2'' '''(2 , ).I I y y yyy

Thus the fundamental differential invariants are

3'2'' ,2r yy y 2

3'''.u y y (3.73)

We can now calculate higher order differential invariants

3 3

3 3

( )'

''' .2

x

x

ivdu D u yyy

dr D r y

Now, the ODE (3.68) is equivalent to

3

3

1.du

dr

57

whose general solution is

3 3 1.u r c

After simplifying this equation, we get

''' 12

'' '2,2

yyy y c

y

which is invariant under the three-parameter Lie group generated by ℒ.

3.5 Lie Algebra of Point Symmetry Generators

This section will explain structure of Lie algebra and how to use it for solving higher order

differential equations.

Definition 3.5.1 Let, 1 2,X X , then the commutator of 1X with 2X is defined and denoted by

1 2 1 2 2 1[ , ] .X X X X X X (3.74)

where

( , ) ( , ) ;i i x i yX x y x y 1,2.i

and

1 22 2

1 2 1 2 2 1 1 2 1 2 1 2( ) ( ) ( ) .x x y y x yX X X X

Therefore,

1 2 1 2 2 1 1 2 2 1[ , ] ( ) ( ) .x yX X X X X X (3.75)

The commutator has following properties:

(1) It is antisymmetric, that is,

1 2 2 1[ , ] [ ].X X X X (3.76)

(2) The commutator is bilinear such that

58

1 1 2 2 3 1 1 3 2 2 3

1 2 2 3 3 2 1 2 3 1 3

[ , ] [ , ] [ , ],[ , ] [ , ] [ , ].c X c X X c X X c X X

X c X c X c X X c X X

(3.77)

(3) It satisfies the following Jacobi identity:

1 2 3 2 3 1 3 1 2[ ,[ , ]] [ ,[ , ]] [ ,[ , ]] 0.X X X X X X X X X (3.78)

(4) The commutator is independent of change in coordinates. That is, if 1 2 3[ , ]X X X be any

commutator in (𝑥, 𝑦) coordinates and 1 2 3[ , ]X X X in (𝑢, 𝑣) coordinates, then 3 3.X X

(5) The commutator of the prolonged generators

( )(1) ( )

'( ) ... k

k

x y yy

kiX

is defined by

( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 2 1[ , ] .k k k k k kX X X X X X

Now if 1 2 3[ , ]X X X , then ( ) ( ) ( )1 2 3[ , ] .k k kX X X

(6) Each generator in ℒ satisfies the linearized symmetry condition

( ) ( ) ( )( )( ) 0.n n ni i

niX y X , when ( ) .ny (3.79)

Since, ( )ni is linear in highest derivative ( )ny , is independent of highest derivative, thus the

condition (3.79) is satisfied if and only if

( ) ( ) ( )( ) ( ) 0,n n ni iX y y (3.80)

where

( )( 1)'

( )( , , ,........., ) .n

n ii n

x y y yy

(7) The vector space ℒ is closed under commutator, that is, if

1 2 1 2[, , ] .X X X X

59

Therefore the commutator of any two generators in the basis is a linear combination of the basis

generators such that

[ , ] .jk

i ij kX X Xc (3.81)

The constants kijc are called structure constant. If [ , ] 0,i jX X then iX and

jX are said to be

commute. Every generator commute with itself, that is, 1 1 2 2 0[ [, ] , ]X X X X .

Example 3.5.1 Consider the ODE:

3''' .y y (3.82)

whose Lie point symmetries are generated by

1 ,xX 234

.x yX x y (3.83)

Now the commutator of 1X with 2X is

1 2 1 2 1 2 13[ , ] ( ( ) (1)) ( ( ) (0)) .4 xx yX X X x X X y X X

Since, every generator commute with itself, so

1 1[ , ] 0,X X 2 2[ , ] 0.X X

Again, commutator is antisymmetric, thus

2 1 1 2 1[ , ] [ , ] .X X X X X

Thus the only nonzero structure constants are

112 1,c 1

21 1.c

Now we define Lie algebra:

60

Definition 3.5.2 A Lie algebra is a vector space that is closed under [ , ] which is bilinear, anti-

symmetric, and satisfies the Jacobi identity.

Lie algebras occur in many branches of applied mathematics and physics. Commonly, one is

interested in the action of a multiparameter Lie group, whose linearization about the identity yields

the Lie algebra that generates the group. Lie algebras may also exist without reference to an

underlying Lie group. A Lie algebra is defined abstractly by its structure constants, but it may

appear in many different forms (or realizations), as the following example shows.

Example 3.5.2 The set of all vectors in space 3x under cross product is a Lie algebra if the

commutator is defined by:

1 2 1 2[ , ] .x x x x

We know the standard basis of 3 are 1 2 3(1,0,0) , (0,1,0) , (0,0,1) .t t tx x x

Then the nontrivial commutators are:

1 2 3 2 3 1 3 1 2, , .x x x x x x x x x

The only nonzero structure constants are

3 1 2 3 1 212 23 31 21 32 131, 1.c c c c c c

Let, [ℳ, ℵ] denote the set of all commutators of generators in ℳ ⊂ ℒ with generators in ℵ ⊂ ℒ,

that is,

, {[ , ]: , }.i j i jX X X X

Now a subspace ℳ ⊂ ℒ is called a subalgebra if it is closed under the commutator:

, .

A subalgebra ℳ ⊂ ℒ is called an ideal if , .

61

For any Lie algebra ℒ, one ideal can be constructed is called the derived algebra (1) ,which

consists of all commutators of elements of ℒ such that

(1) [ , ] (3.84)

Clearly, [ℒ (1), ℒ] is a subset of ℒ (1), therefore ℒ (1) is an ideal. If ℒ (1) ≠ ℒ, then the derived

subalgebra is,

ℒ (2) = [ℒ (1), ℒ (1)] (3.85)

Proceeding in this way, we get

ℒ (𝑘) = [ℒ (𝑘−1), ℒ (𝑘−1)] (3.86)

If the series of derived subalgebra terminates with ℒ (𝑘) = 0, then ℒ is said to be solvable.

Therefore, a 𝑅-dimensional Lie algebra is solvable if there is a chain of subalgebras:

{0} = ℒ (0) ⊂ ℒ (1) ⊂ ℒ (2) ⊂∙∙∙⊂ ℒ (𝑅) = ℒ (3.87)

Where, dim(ℒ (𝑘)) = 𝑘, such that ℒ (𝑘−1) is an ideal of ℒ (𝑘).

Let, ℒ be a solvable Lie algebra, then if we choose a basis such that

1 ., , 1,2,...,k k

k kX X k R

(3.88)

Hence

1 2( , ,..., ).k

kSpan X X X (3.89)

Any basis defined by (3.88) is called canonical basis. That is, a basis is equivalent if the structure

constants satisfy

.0,kijc i j k (3.90)

Example 3.5.3 Suppose that the fourth order ODE

4( ) ''' 3( ) .ivy y

(3.91)

has a five dimensional Lie algebra ℒ, which has a basis

62

21 2 3

54

, ,, .

y y y

x x

X X x X x

X X x

(3.92)

The nonzero commutators [ , ]i jX X , with i j are,

2 4 1 2 5 2 3 4 2

3 5 3 4 5 4

[ , ] , [ , ] , [ , ] 2 ,[ , ] 2 , [ , ] .

X X X X X X X X X

X X X X X X

(3.93)

Thus, the derived algebra is

1 2 3 4

1 ( , , , ).Span X X X X (3.94)

Since ℒ (1) ≠ ℒ, therefore for finding ℒ (2), the nonzero commutators of elements are

2 4 1 3 4 2[ , ] , [ , ] 2 .X X X X X X (3.95)

Therefore,

21 2( , ).Span X X (3.96)

Finally, 𝑋1 and 𝑋2 commute, thus 3 .0 Hence, ℒ is solvable.

The basis (3.92) is canonical.

We now solve an ODE systematically by the following method:

(1) Using symmetry condition to get Lie point symmetries.

(2) Calculating the commutators of the basis generators to find the series of derived subalgebras.

(3) Determining the sufficiently large solvabe subalgebra, finding canonical coordinates and

fundamental differential invariants.

(4) Rewriting the ODE in terms of differential invariants to use each generator in turn to carry

out one integration.

63

Example 3.5.4 Consider the ODE

''2'' '' ., 0y

y y yy

(3.97)

It has the following Lie point symmetry generated by

1 2, .2x x yX X x y (3.98)

The group generated by 1X has the following fundamental differential invariants

'1 1, ,r y u y

and the fundamental differential invariants of the group generated by 1X and 2X are

' ''

2 2 23/2 , .y yr u

yy

Hence the equation (3.97) reduces to

22 2 1.u r (3.99)

The Lie algebra is solvable and (3.98) is a canonical basis. Thus,

(1) '2 1 1 2 1 12 2 , 3 3 .X r y r X u y u

The restriction of 2X to 1 1( , )r u is

1 12 1 12 3 .r uX r u

Now, 12 3/2

1,u

rr

and for this, let 2 1 .1 ln( )2

s r Then,

2

2 22

22 2 2.

3 2

dsds rdx

drdr r u

dx

From (3.99),

64

2 22

2 2.

2ds r

dr r

Integrating,

22 22

2

22 2 .

21 ln( 2)2

rds dr

r

s r c

After rewriting it in terms of 1 1( , )r u , we get

12 2

1 1 1 1 1(4 2 ) ,u r c r c constant. (3.100)

If we complete one step of iteration and repeat the same process, then we can use the generator

1X to solve (3.100). Using canonical coordinates 1 1( , ) ( , )r s y x to obtain

12 1/2

1 1 1 1'

1 1 .(4 2 )

ds

dr r c ry

Integrating,

11 2 1/2

1 1 1

1 11 2 1 1

1.

(4 2 )2cosh ( )

drds

r c r

s c c cr

Rewriting the ODE in original variables, we obtain the general solution

2 21 1 22 sech ( ( )).y c c x c

In this chapter, we have seen the detail explanation of infinitesimal generator, symmetry condition

of higher order ODE, reduction of order, Lie algebras, differential invariant, which are the tools of

solving higher order differential equations. We have also arranged this chapter with several

examples of relevant topics. In the next chapter we will explore everything which is discussed in

previous chapter but for the purpose of solving partial differential equation.

65

Chapter 𝟒

Lie Symmetries and Partial Differential Equations

The purpose of this chapter is to obtain some exact solution of partial differential equations (PDE)

by Lie symmetry analysis. In this chapter, we have focused on finding symmetries of PDE, which

is the main tool for solving any differential equations. The rest of the chapter is organized in the

following manner: defining linearized symmetry condition of PDE, illustration of group invariant

solution of PDE. A brief explanation of every topic is given in each section by illustrating it with

several examples.

4.1 Symmetries of PDEs with Two Dependent Variables

Symmetries of PDE are similar to symmetries of ODE. Firstly, we consider the simple PDE with

two independent variables 𝑥, 𝑡 and one dependent variable 𝑢. The following diffeomorphism,

: ( , , ) ( ( , , ), ( , , ), ( , , )).x t u x x t u t x t u u x t u (4.1)

is called a point transformation if it maps the surface ( , )u u x y to the following surface:

( , , ( , )),( , , ( , )),( , , ( , )).

x x x t u x t

t t x t u x t

u u x t u x t

(4.2)

Consider the following total derivatives:

......

x t

x t

x x x u xx u xt u

t t t u xt u tt u

D

D

u u u

u u u

(4.3)

The first two equation of (4.2) may be used to find 𝑥 and 𝑡 in terms of x and t if the following

Jacobian is nonzero,

0,x x

t t

D x D t

D x D t when ( , ).u u x t

(4.4)

66

If equation (4.4) is true, then the last equation of (4.2) can be written as,

( , ).u u x t (4.5)

Applying chain rule to equation (4.5), we get

.xx x x

t t t t

uD u D x D t

uD u D x D t

(4.6)

Then by Cramer’s rule,

1 1, .x t

x xx x

t tt t

D x D uD u D tu u

D x D uD u D t

(4.7)

Now if ju is the derivative of u with respect to x and t respectively, then

1 ,

1 .

jj

j

jj

j

jx

jt

x x

t t

x x

t t

D u D tuu

D u D tx

D x D uuu

D x D ut

(4.8)

Consider a nth order PDE of the following form:

( , , , , ,...) 0x tF x t u u u (4.9)

To define the point symmetry of PDE we write the equation (4.9) in following form:

( , , , , ,...) 0.x tF u x t u u u (4.10)

Where, u is the nth order derivative of 𝑢 and is independent of .u Now the point

transformation is a point symmetry of (4.9) if

( , , , , ,...) 0,x tF x t u u u when (4.9) holds. (4.11)

Example 4.1.1 We will show that the following symmetry condition

1( , , ) ( , ,2( ))2 4x

x t u ut xt t

(4.12)

is a point symmetry of Burgers’ equation

.t x xxu uu u (4.13)

67

The Jacobian of (4.12) is

3

2 2

1 012 .

1 82 4

t

x t

t t

Thus,

32

2( 1) 01 4 ( 1),11/ 8 2( )

4

x

x x

t

tu

u t tut tu u

t

23

2

1 2( 1)1 2 8 ( ),

1/ 8 2( )2

x

t xt

t

tut

u t t u xtu tu xxt

tu ut

2

33 2

2

4 01 8 .11/ 8 4( 2 1)4

xx

x x xx

xt x

t u

u t ut t u tu

t

Now,

38 ( ) 0,x t xtu uu t u uu when, 0.t xu uu

Therefore, point symmetry (4.12) satisfies the symmetry condition.

4.2 Linearized Symmetry Condition of PDEs

In this section the systematic procedure of finding symmetries of PDE will explore and the

procedure is same as ODE. Consider the point symmetries of the form

2

2

2

( , , ) ( ),( , , ) ( ),( , , ) ( ).

x x x t u O

t t x t u O

u u x t u O

(4.14)

68

Now, the infinitesimal generator is

.x t uX (4.15)

If we solve

( , , ), ( , , ), ( , , )dx dt dux t u x t u x t u

d d d

subject to initial condition 0( , , ) | ( , , )x t u x t u , then we get ( , , ).x t u

Now a surface ( , )u u x t is mapped to itself by the group of transformations generated by 𝑋 if

( ( , )) 0,X u u x t when ( , ).u u x t (4.16)

The characteristic of group is defined by

,x tQ u u when ( , ).u u x t (4.17)

The surface (4.16) is invariant if

0,Q when ( , ).u u x t (4.18)

Equation (4.18) is called invariant condition.

The prolongation of (4.14) is defined by,

2

2

( , , , , ) ( ),( , , , , ) ( ).

xx x tx

tt x tt

u u x t u u u O

u u x t u u u O

(4.19)

From (4.6),

( , , , , ) ,( , , , , ) .

xx t x x x t x

tx t t x t t t

x t u u u D u D u D

x t u u u D u D u D

(4.20)

Suppose that

2( ),jj j Ou u (4.21)

where

1 2 1 2

1 2 1 2.,j j j j

j jj j j j

u uu u

x t x t

69

Then, (4.21) becomes

2

2

( ),

( ).

jxjxjx

jtjtjt

u u O

u u O

(4.22)

where

,

.

jx jx jx x jt x

jt jt jx t jt t

D u D u D

D u D u D

(4.23)

Now, the prolonged formula

(1) ,x t x t

x t x tx t u u u u uX X (4.24)

(2) (1) .xx xt tt

xx xt ttu u uX X (4.25)

From (4.23) we get the Lie point symmetries

2( ) ,xx u x x x t u x u x tu u u u u (4.26)

2( ) ,tt u t t x x u x t u tu u u u u (4.27)

and

2

3 2

(2 ) ( 2 )2 ( 2 )2 3 2 .

xxxx xu xx x xx t uu xu x

xu x t uu x uu x t u x xx

x xt u x xx u t xx u x xt

u u u

u u u u u u

u u u u u u u

(4.28)

Similarly, we can define ,tt xt . Now, the Lie point symmetries is obtained by differentiating

(4.11) with respect to at 0. Thus we obtain the linearized symmetry condition

0,XF when 0.F (4.29)

The equation (4.10) enables us to eliminate u from (4.29), then we split the remaining terms to

obtain a linear system of determining equations for , and . The vector space ℒ of all Lie point

symmetry generated by a PDE is a Lie algebra but it is not finite dimensional.

70

Example 4.2.1 Consider the PDE

2.t xu u (4.30)

The symmetry condition is

2 .t xxu (4.31)

Simplifying

2 3 4

2 3 .

( )2 ( ( ) ( ) )

t t x u t x u x u x

x x u x x u x x u x

u u u u

u u u u

Equating the power of xu , we get

0,u (4.32)

2 0,u x (4.33)

2 0,u t x (4.34)

2 0,t x (4.35)

.0t (4.36)

Integrating (4.32),

( , ),a x t a constant.

Therefore, the general solution of (4.33) is

.2 ( , )xa u b x t

From (4.34),

2 .2 (2 ) ( , )xx x ta u b a u c x t

Substituting these results in (4.35) and (4.36), we obtain

71

24 4( ) 2 0,xxx xx xt t xa u b a u b c (4.37)

22 (2 ) 0xxt xt tt ta u b a u c (4.38)

From (4.37), (4.38),

0,tc (4.39)

2 0,t xb c (4.40)

2 0,xt ttb a (4.41)

0,xx xtb a

(4.42)

0,xxta

(4.43)

.0xxxa

(4.44)

Solving equation (4.39) through (4.41), we get

'

'' 2

( ), 2 ( ) ( ),2 ( ) ( ) ( ).

c x b x t x

a x t x t x

(4.45)

To find the value of unknown , , , ; we substitute equation (4.45) into equation (4.42),

(4.43) and (4.44),

271 2 4 6 8 9

2 251 4 8 9 10

2 251 2 3 4 6 8

14 2 ( 2 ) 4 2 ,2

4 ,2 4 .

c tx c t c x tu c x c c xu c u

c t c xt c t c x c x c

c x c x c c xu c u c u c u

(4.46 )

From equation (4.46) it can be said that the Lie algebra is ten dimensional.

4.3 Group Invariant Solution of PDEs

We have seen in previous section, the detail procedure of finding Lie symmetry of PDEs. Now

we will explore the idea of determining exact solution by using Lie symmetry. To find invariant

solution consider the following PDE with two independent variables,

0( , , , , ,...)x tF x t u u u (4.47)

A solution ( , )u u x t is invariant under the group generated by

72

.x t uX (4.48)

Every invariant solution satisfies the invariant surface condition

0.x uQ (4.49)

The equation (4.49) is easier than (4.47) to solve. The equation (4.49) can be solved by the

method of characteristic. The characteristics equation is:

.dx dt du

(4.50)

Now, if ( , , ), ( , , )r x t u v x t u are two functionally independent first integrals of (4.50), then every

invariant of the group is a function of 𝑟 and 𝑣. Let, 0,uv then the general solution of the

invariant surface condition is

( ).v G r (4.51)

If we substitute this solution into equation (4.47), we can determine the function 𝐺. If 𝑟 and 𝑣

both depend on 𝑢, then we have to find the solution

.r c (4.52)

These are the only solutions of the invariant surface condition that are not of the form (4.51). If 𝑟

is a function of independent variables 𝑥 and 𝑡 only, then there is no solution of the form (4.52).

Now we will solve the one dimensional heat equation by Lie symmetry analysis, as follows:

Example 4.3.1 Consider the one dimensional heat equation

.t xxu u (4.53)

Here, 𝑥 and 𝑡 are independent variables and 𝑢 is dependent variable. Now, we know that the total

derivative is defined by

......

x t

x t

x x x u xx u xt u

t t t u xt u tt u

D

D

u u u

u u u

(4.54)

73

The infinitesimal generator is

( , , ) ( , , ) ( , , ) .X x t u x t u x t u

x t u

(4.55)

The second prolongation of 𝑋 is

(2) (1) .xx xt tt

xx xt ttu u uX X

(4.56)

When

(1) .x t x t

x t x tx t u u u u uX X

Let

.( , , , , , ) 0x t xx t xxF x t u u u u u u

Now the determining equation is obtained by invariance condition:

(2) ( ) | 0.xxtu uX F

That is,

.( ) | 0x t xx xt tt xxt

x t xx xt ttx t u u u u u u t xx u uu u

By simplifying it, we get

.0t xx

(4.57)

After substituting the value of t and xx from equation (4.27) and (4.28), we get,

2 2

3 2 .

{ ( ) } (2 ) ( 2 )2 ( 2 ) 2 3 2 0t u t t x x u x t u t xx xu xx x xx t uu xu x

xu x t uu x uu x t u x xx x xt u x xx u t xx u x xt

u u u u u u u u

u u u u u u u u u u u u u

Equating the coefficient of 𝑢 and its derivatives, we get

74

: 0,t xxC (4.58)

: ( ) ( ) 0,u t u xxu (4.59)

0: 2 ,t t xx xu (4.60)

0: 2( ) ,x t xx u xu (4.61)

.: 0xt xu (4.62)

If ,fu g then by solving the equations (4.58) to (4.62), we get

51 2 4c tx c x c t c (4.63)

21 2 32c t c t c (4.64)

21 4 1 6

1 14 2

f c x c x c t c (4.65)

Therefore,

( , , ) ( , , ) ( ) .X x t u x t u fu gx t u

(4.66)

Thus the generators are,

1 2 3

2 24

5 6

, , ,1 1( ) ,4 2

1 , , .2

x t x t

x t u

x u u u

X X X x t

X xt t x u

X t xu X u X g

(4.67)

For finding invariant solution we consider two-parameter Lie group of symmetry which is

generated by

3 62 , .x t uX x t X u (4.68)

Every generator of a one-parameter Lie group of scaling is of the form:

3 6 ,X mX nX where 𝑚, 𝑛 are constants.

75

If 1,m then

2 .x t uX x t nu

The invariant condition is

,2x tQ nu xu tu (4.69)

which is solved by the method of characteristics

.2

dx dt du

x t nu

(4.70)

Fundamental differential invariants are

1/2 /2, .nr xt v ut

Since, 𝑟 is independent of 𝑢, so every invariant solution is of the form:

/2 1/2 .( ) ( )nv G r u t G xt (4.71)

Differentiating (4.71), we get

( 2)/2 '

( 2)/2 ''

1 1( ( ) ( )),2 2( ).

nt

nxx

nu t rG r G r

u t G r

Substituting it into heat equation (4.53), we get

'' '1 1 0,2 2

G rG nG (4.72)

which is the second order ODE. The general solution of (4.72) is,

1/2 1/21 2

1 1( ) ( ,2 ) ( ,2 ),2 2

G r cU n r c V n r

where ,U V are parabolic cylinder functions. If 𝑛 = −1, then we get

2 2 2/4 /4 /41 2 0

. r

r r zG e e e dzc c

76

Substituting this into equation (4.71), we obtain a large family of solution of heat equation, as

follows,

21/2 /4 .x tu t e

In this chapter, we have performed Lie symmetry analysis for PDE. We also investigated the

algebraic structure of symmetry group of PDE, and invariant solution of any PDE. After applying

invariant condition on PDE, we got reduced ODE which are solved by the method of

characteristics. Moreover, several examples are explored for which we solved the well-known heat

equation by Lie invariant method in the last section of this chapter. From all the discussion of this

chapter it can be said that Lie symmetry analysis is a very powerful, effective method to find exact

solution of an PDE.

77

Chapter 5

Lie Symmetry Analysis of Nonlinear Filtration Equation

The Lie symmetry method plays a very important role for solving higher order differential

equations and physical models. To construct the exact solution of nonlinear partial differential

equation, various methods like Backlund transform method, inverse scattering method, the

extended tanh-function method, Hirota method and Lie transformation method have been

introduced. Among the above-mentioned methods, the Lie group analysis method plays an

effective role to get exact solution of nonlinear PDE. In this chapter, we use Lie group analysis for

finding the travelling wave solution of nonlinear filtration equation. This chapter deals with the

rigorous explanation of travelling wave solution for PDE with Lie transformation. In addition to

that this chapter discussed the main result of this dissertation, which is, how to get a family of new

solutions for nonlinear filtration equation from some invariant solutions.

5.1 Travelling Wave Solution of PDE

In this section, the theory behind travelling waves is introduced, where two PDEs are solved

analytically, in favor of illustrating the different forms of travelling waves that exist. The difficulty

behind solving non-linear PDEs is addressed, resulting in the exploration of many powerful

methods for solving non-linear PDEs numerically and analytically.

A travelling wave is a wave that advances in a particular direction, with the addition of retaining

a fixed shape. Moreover, a travelling wave is associated with having a constant velocity throughout

its course of propagation. Such waves are observed in many areas of science, like in combustion,

which may occur as a result of a chemical reaction. In mathematical biology, the impulses that are

apparent in nerve fibers are represented as travelling waves. Also, in conservation laws associated

problems in fluid dynamics, shock profiles are characterized as travelling waves. Furthermore, the

structures present in solid mechanics are typically modeled as standing waves. Hence, it is

important to determine the dynamics of such solutions.

A travelling wave solution is obtained upon solving a model that corresponds to a system.

Generally, these models take the forms of partial differential equations (PDEs), where the

78

dynamics of the systems are comprehended upon solving for solutions. These travelling wave

solutions are expressed as

( , ) ( ),u x t U z where, .z x ct (5.1)

Here, the spatial and time domains are represented as x and t, with the velocity of the wave given

as c.

If c = 0, the resulting wave is named a stationary wave. Such waves do not propagate, and are

typically observed when inducing a fixed boundary. In fact, there are three types of travelling

waves that are attributed to having certain properties, which is showed in Fig. 5.1.

(1) For a travelling wave that approaches constant states given by

( ) , ( )l rU u U u with ,l ru u (5.2)

is called a wave front.

(2) However, if the constant states are equal with rlu u , then the corresponding wave is known

as a pulse wave.

(3) If a wave exhibits periodicity with

( ) ( ),U z F U z when 0,F

then, the wave is called a spatially periodic wave.

One of the fundamental questions in the theory of nonlinear PDEs is whether a given PDE admit

such a travelling wave as a solution. This question is generally asked without regard to initial

conditions so that the wave is assumed to have existed for all times. However, boundary

conditions of the form

( , )u t constant, ( , )u t constant, (5.3)

are usually imposed. We will explain the detail situation by following example:

79

(a) (b)

(c)

Fig. 5.1: Travelling waves: (a) wave front (b) pulse wave (c) spatially periodic wave

Examples 5.1.1 We want to find the travelling wave solution of the following PDE:

2 0,tt xxu a u a constant. (5.4)

According to the definition any solution of the form

( , ) ( ),u x t f x ct (5.5)

is called the travelling wave solution.

Thus, differentiating (5.4) partially, we get,

' '' 2

' ''

( )( ), ( )( ) ,

( ), ( ).t tt

x xx

u f x ct c u f x ct c

u f x ct u f x ct

Substituting these into equation (5.4), we get,

2 2 '' 2 ''

'' 2 2

0

( ) 0.tt xxu a u c f a f

f c a

By integrating, ( ) ( ), ,f s A B s A B constant or .c a In first case, the general solution is

( , ) ( ),u x t A B x ct but the boundary condition (5.3) is not satisfied unless 0.B So, the only

travelling wave solution is constant. In the second case, for any twice differentiable function,

80

lim .s

d

Thus ( , ) ( )u x t f x ct is a travelling wave solution.

Examples 5.1.2 Consider the diffusion equation:

.t xxu mu (5.6)

Let, the travelling wave solution be

( , ) ( );u x t f x ct c constant. (5.7)

Differentiating (5.7) and then substituting it into (5.6) gives,

' '' 0.cf mf

The general solution of above differential equation is

( ) exp( ).cs

f s a bm

(5.8)

From (5.8), it is clear that for 𝑓 to be constant at both plus and minus infinity it is necessary that

0.b Thus there is no nonconstant travelling wave solution of diffusion equation.

5.2 Travelling Wave Solution with Lie Symmetry

Any method for finding exact solution of a PDE relate to one or more ODE. For example, the

general solution of a first-order quasilinear PDE is constructed by integrating the characteristic

equations. The general solution of most PDEs depend on various assumption and the solution is

always not explicit. We may want to find similarity solutions, travelling waves, separable

solutions, and so on. Many of these methods involve nothing more than looking for solutions that

are invariant under a particular group of symmetries. Therefore, whenever it’s not possible to find

the explicit solution by Lie symmetry analysis it is more convenient to find the invariant solution

of a given PDE. For example, we can use travelling wave solution for a given PDE with Lie

symmetry.

Consider a PDE whose solution is of the form 𝑢(𝑥, 𝑡). Now if the Lie symmetry generators of

this PDE include

81

1 2,x tX X (5.9)

then generally it has the travelling wave solution of the form

( , ) ( ),u x t f x ct where, c constant. (5.10)

Also, these solutions are invariant under the group generated by

1 2 2 ,xX cX X c (5.11)

because both 𝑢 and 𝑥 − 𝑐𝑡 are invariants. This definition of travelling wave solution is true for

any Lie symmetry of a given PDE of the form:

( , , , , ,...) 0.x tF x t u u u (5.12)

Here, 𝑥, 𝑡 are two independent variables and u is a dependent variable. We know that if 𝑢(𝑥, 𝑡) is

the solution of PDE (5.12), then it is invariant under the group generated by

,x t uX (5.13)

if and only if the characteristics vanishes on the solution. That is, every invariant solution

satisfies the invariant condition

0.x tQ u u (5.14)

Usually, equation (5.14) is much easier to solve that (5.12). By solving (5.14), we will find

which solution of them also satisfy (5.12).

For example, for equation (5.11) characteristic is

.x tQ cu u (5.15)

Now the characteristic equations are

.1 0

dx dt du

dx dt du

c

(5.16)

By integrating first two terms of (5.16)

,

1dx dt

x ct k kc

constant

82

.x ct k r x ct (5.17)

Again, if we integrate the last two terms of (5.17),

,

1 0dt du

u m m constant

.v u

(5.18)

From (5.17) and (5.18) it can be writing,

( ) ( )( ).

v G r v G x ct

u G x ct

(5.19)

Thus, we get the travelling wave solution ( )u G x ct for the Lie group generator (5.11).

5.3 Travelling Wave Solution of Nonlinear Filtration Equation with

Lie Symmetry

In this section, we will explore the travelling wave solution of nonlinear filtration equation. The

partial differential equation of the form,

2 .

1xx

t

x

uu

u

(5.20)

is called the nonlinear filtration equation.

Equation (5.20) can be written as

2(1 ).xx xtu uu (5.21)

The linearized symmetry condition is,

2

(2)2

2

2

2(1 )

(1 )[ 0

[ 0

2 .

(1 )] | .

(1 )] | .x t xx xt tt xx t x

xx xt

x t xx xt tt

x t u u u u u u

xx t x t

x t x

xx x

xx x u u u

t u u u

t

X

u u u

u u

u u

u

u

Substituting the values of , ,xx t x into the above equation, we get

83

2

2 3 2

2

2

2 [ ]

[ ].

(2 ) ( 2 ) 2 ( 2 )2 3 2 ( )

( )x t

x

xx xu xx x xx t uu xu x xu x t uu x uu x t u x xx

x xt u x xx u t xx u x xt x u x x x t u x u x t

t u t t x x u x t u t

u u

u

u u u u u u u u u

u u u u u u u u u u u u

u u u u u

Equating the term which involve 3xu gives

3 00.

uu u t x

x u

u

(5.22)

Again, equating the term which involve 2xu gives

2 2

2 2

2 2 ( ) 2 { ( ) } 0

2 2 { ( ) } 0uu xu uu t t u x t u t u t t u t

uu uu t t u t u t u t t u t

u u u u u

u u u u u

(5.23)

Now the term which involve 2tu are

23 00.

t u

u

u

(5.24)

By proceeding this way, we get many determining equation and by solving these we get the

value of , , , as follows

1 4 5 .c c x c u (5.25)

2 42 .c c t (5.26)

3 4 5 .c c u c x (5.27)

Thus, we know that the infinitesimal generator can be written as

1 4 5 2 4 3 4 5

1 2 3 4 5

1 1 2 2 3 3 4 4 5 55

1

( ) ( 2 ) ( )( 2 ) ( )

.

x t u

x t u

x t u x t u x u

i i

i

X

c c x c u c c t c c u c x

c c c c x t u c u x

c X c X c X c X c X

c X

(5.28)

84

Therefore, from equation (5.28), we obtain the following five-parameter Lie group generators of

nonlinear filtration equation (5.21),

1 2 3

4 5

, , ,2 , .

x t u

x t u x u

X X X

X x t u X u x

(5.29)

Now, we will find the travelling wave solution of nonlinear filtration equation. By definition of

travelling wave solution with Lie symmetries, we know that if a solution 𝑢(𝑥, 𝑡) of PDE has the

symmetry generators

1 2, .x tX X (5.30)

generally have the travelling wave solution of the form

( ),u f x ct where, 𝑐 =constant (5.31)

Also, these solutions will be invariant under the group generated by

1 2 .x tX cX X c (5.32)

We apply this theory for equation (5.21) because it has such symmetry generators.

We know that if 𝑢(𝑥, 𝑡) is the solution of PDE (5.21), then it is invariant under the group

generated by

,x t uX (5.33)

if and only if the characteristic vanishes on the solution. That is, every invariant solution satisfy

the invariant condition

0.x tQ u u (5.34)

By solving (5.34), we will find which solution of them also satisfy (5.21).

For example, equation (5.32) characteristic is

.x tQ cu u (5.35)

Now the characteristic equations are

85

.1 0

dx dt du

dx dt du

c

(5.36)

By integrating first two terms of (5.35), we obtain

,

1dx dt

x ct k kc

constant

.x ct k r x ct

(5.37)

Again, if we integrate the last two terms of (5.35),

,

1 0dt du

u m m constant

.v u

(5.38)

From (5.37) and (5.38) it can be writing,

( ) ( )( ).

v G r v G x ct

u G x ct

(5.39)

Now, differentiate partially (5.39) to obtain,

' '

'' ''

' '

( ) ( ),

( ) ( ),

( ) ( ) ( ) ( ).

x

xx

t

u G x ct G r

u G x ct G r

u c G x ct c G r

(5.40)

Substitute (5.40) into (5.21), we get,

'' ' ' 2

'' ' ' 3

2

( ) ( )[1 ( ( )) ]( ) ( ) [ ( )] 0.

(1 )xx xt

r cG r G r

r c r c r

u u

G

G G G

u

(5.41)

Thus we get a nonlinear ODE. For avoiding complication, we just use ' '', ,G G G in replacement

of ' ''( ), ( ), ( )G r G r G r . So, (5.41) becomes

'' ' ' 3( ) 0.c cG G G (5.42)

Here, G dependent variable, and r independent variable of ODE (5.41).

86

Let,

' .dGG p

dr

Therefore, (5.42) becomes

3 0,dpcp cp

dr where, '' .dp

Gdr

(5.43)

Now, from (5.43)

2

2

(1 )

(1 )

dpcp p

dr

dpc dr

p p

(5.44)

By integrating

2

2

21

12

21

(1 )1( )

11ln ln 12

ln1

1 .cr

dpc dr

p p

pdp c dr

p p

p p cr c

pcr c

p

p c e p

Squaring both sides of above equation gives

2 2 212

212

22

12 21

(1 )

11 1

cr

cr

cr

cr

p c e p

pc e

p

pc e

p c e

87

212

212

22

1

21

1 1

1 1

11

11

cr

cr

cr

cr

c ep

c ep

pc e

pc e

But, we know that

21

21

11

1

cr

cr

dGp

dr c e

drdG

c e

Integrating, both sides give

21

21

21

1

1

cr

cr

cr

cr

drdG

c e

drG

c e

e dr

c e

Now, let

21

,cr

crI

e dr

c e

(consider (+ve) sign)

(5.45)

Let,

1 .

cr cr

cr

w e dw ce dr

e dr dwc

88

Substituting it into equation (5.45), we get,

21

21

1

21

1

1

1

1

1 .1

dwI

c c w

dw

c c w

c

dw

c c w

c

Again, let

1 1

1 .wm dm dw

c c

Thus 𝐼 becomes

1

21

11

c dmI

c c m

2

1

11

1 sin ( ).

dm

c m

mc

Putting the value of 𝑚 and 𝑤 in the above equation, we obtain,

1

1

1

1

11 2

1 sin ( )

1 sin ( )

1 sin ( ) .

cr

cr

wI

c c

e

c c

c e cc

Where, 1 21

1 ,c cc

constant of integration. Thus 𝐺 takes the form

89

11 2

1 sin ( ) .crG c e cc

Thus, we get the solution of the form

12 1

1( ) sin ( exp{ ( )}).u G r c c c x ctc

(5.46)

The above equation is the travelling wave solution of nonlinear filtration equation. Figure (5.1)

shows the solution of equation (5.46).

Fig. 5.2: Solution of equation (5.46)

90

5.4 Solution of Nonlinear Filtration Equation with Lie Symmetry

from Travelling Wave Solution

In this section we will calculate some new solution of nonlinear filtration equation. If we have

found a solution by using Lie symmetry analysis which is invariant under a group of Lie point

symmetries, if there are other symmetries under which this solution is not invariant, we can use

them to map it to a family of new solutions. The total procedure is explained, as follows:

From the previous section we find the travelling wave solution of nonlinear equation of the form

12 1

1( , ) sin ( exp{ ( )}).u x t c c c x ctc

(5.47)

This solution is invariant under the group generated by 1 2 x tcX X c . We can construct

new solution from (5.47) by using other symmetries (5.29) of nonlinear filtration equation. For

example, the symmetry generated by 1 xX are

( , , ) ( , , ).x t u x t u (5.48)

Now equation (5.47) is equivalent to

12 1

1 sin ( exp{ [( ) ]}).u c c c x ctc

(5.49)

So the group generated by 1X maps (5.47) to a four-parameter family of solutions of the form:

12 1

1 sin ( exp{ [( ) ]}).u c c c x ctc

(5.50)

Figure (5.3) shows the solution of equation (5.50).

Now we consider one parameter group generated by 3 uX , which is,

( , , ) ( , , ).x t u x t u (5.51)

This generator (5.51) map (5.50) to the following five-parameter family,

91

12 1

1 sin ( exp{ [( ) ]}) .u c c c x ctc

(5.52)

Fig. 5.3: Solution of equation (5.50).

The remaining two generators 2 4, 2t x t uX X x t u , do not produce any further solution

from (5.52) because their action only changes the values of 1 2, , , ,c c c . Therefore, the above

method cannot be applied for these two generators.

In this chapter, travelling wave solutions of nonlinear filtration equation is obtained by Lie

symmetry analysis and by using this solution two different solutions are determined. Using the

criterion of travelling wave solutions with Lie symmetry we find the invariant condition, then by

the method of characteristics we solve PDE and by simplifying, it converted to an ODE, which is

solved by suitable method. Finally, new solutions are obtained from travelling wave solution of

nonlinear filtration equation by using remaining infinitesimal generator.

92

Chapter 6

Conclusion

In this study, the travelling wave solutions of nonlinear filtration equation are analyzed. After

applying linearized symmetry condition and prolongation formula, symmetry of nonlinear

filtration equation is obtained, then by using the criteria of traveling wave solution with Lie

symmetry we get fundamental differential invariant. After that we determine the solution by the

method of characteristics from fundamental differential invariant [39]. Then we use these solutions

to get new solutions from different symmetry of nonlinear filtration equation. We explained that

for any solution of given PDE that is invariant under a one parameter symmetry group, it is possible

to derive a family of solutions by applying the remaining symmetries of PDE. From the result of

this desertion, it can be seen that the Lie symmetry approach as a very powerful, effective and

direct method to obtain the exact analytic solutions of nonlinear partial differential equation [40].

So far we restricted our attention to local Lie point transformations that leave the system of

differential equations invariant. They are usually called classical symmetries of the system. There

are several uses of them in the context of differential equations like group invariant solutions for

PDEs and integration for ODEs among others. One the important generalizations of classical

symmetries is the notion of generalized symmetries. They were originally introduced by E.

Noether in her celebrated paper [42] which related variational symmetries and conservation laws.

They also play a central role in soliton theory of completely integrable nonlinear partial differential

equations [43].

Evolution equations like KdV (Korteweg–de Vries) and KP (Kadomtsev–Petviashvili) are known

to admit infinite hierarchies of generalized symmetries and conservation laws ([23], [24]). An

approach to classification of integrable systems has been to identify systems admitting generalized

symmetries up to differential substitutions preserving the differential structure of the system. The

infinitesimal generators of generalized symmetries depend on derivatives of dependent variables.

In the special case of one dependent variable, every first order generalized symmetry determines

a contact symmetry and vice versa ([26], [27]). Over the last few decades, numerous useful

93

extensions of the classical Lie approach to group-invariant solutions have been developed. Among

others, they include the method of partially invariant solutions ([44], [29]) non-classical method,

methods of conditional ([43], [34]) and of symmetries of potential systems [41].

Future Works

The symmetries, linearizing transformations and invariant solutions of many well-known PDEs

have been classified during the past three decades. But there is a less work where initial and

boundary condition are considering for solving PDE. Actually, an invariant solution to a boundary

value problem exists if the PDE, the domain, and the boundary conditions are all invariant under

the symmetry group. For example, one can apply Lie symmetry analysis for the following model:

The Lie symmetry analysis of Poiseuille flow in a cylindrical pipe is possible because the equation

of motion, the domain (a cylinder), and the boundary conditions (zero flow at the pipe wall) are

invariant under the rotations about the pipe’s axis.

Noncalsssical symmetries have attracted much research have attracted much research effort

recently because reliable computer algebra package is available. So there is scope to use the result

of this thesis for nonclassical symmetries.

94

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