Nail H. Ibragimov
A Practical Course in
Differential Equations and
Mathematical Modelling
Classical and new methods
Nonlinear mathematical models
Symmetry and invariance principles
ALGA Publications
Blekinge Institute of Technology
Karlskrona, Sweden
Preface
Modern mathematics has over 300 years of history. From the very beginning,
it was focused on differential equations as a major tool for mathematical
modelling. Most of mathematical models in physics, engineering sciences,
biomathematics, etc. lead to nonlinear differential equations.
Todays engineering and science students and researchers routinely confront problems in mathematical modelling involving solution techniques for
differential equations. Sometimes these solutions can be obtained analytically by numerous traditional ad hoc methods appropriate for integrating
particular types of equations. More often, however, the solutions cannot be
obtained by these methods, in spite of the fact that, e.g. over 400 types
of integrable second-order ordinary differential equations were accumulated
due to ad hoc approaches and summarized in voluminous catalogues.
On the other hand, the fundamental natural laws and technological problems formulated in terms of differential equations can be successfully treated
and solved by Lie group methods. For example, Lie group analysis reduces
the classical 400 types of equations to 4 types only! Development of group
analysis furnished ample evidence that the theory provides a universal tool
for tackling considerable numbers of differential equations even when other
means of integration fail. In fact, group analysis is the only universal and effective method for solving nonlinear differential equations analytically. The
old integration methods rely essentially on linearity as well as on constant
coefficients. Group analysis deals equally easily with linear and nonlinear
equations, as well as with constant and variable coefficients. For example,
from the traditional point of view, the linear equation
dn y
dn−1 y
dy
+
a
+ · · · + an−1
+ an y = 0
1
n
n−1
dx
dx
dx
with constant coefficients a1 , . . . , an is different from the equation
x̄n
n−1
dn ȳ
ȳ
dȳ
n−1 d
+
a
x̄
+ · · · + an−1 x̄
+ an ȳ = 0
1
n
n−1
dx̄
dx̄
dx̄
known as Euler’s equation. From the group standpoint, however, these
equations are merely two different representations of one and the same equav
vi
PREFACE
tion with two known commuting symmetries, namely,
X1 =
∂
,
∂x
X2 = y
∂
∂y
and X 1 = x̄
∂
,
∂ x̄
X 2 = ȳ
∂
∂ ȳ
for the first and second equation, respectively. These symmetries span two
similar Lie algebras and readily lead to the transformation x = ln |x̄| converting Euler’s equation to the equation with constant coefficients.
Today group analysis
is becoming part of
curricula in differential
equations and nonlinear mathematical modBTH
elling and attracts more
and more students. For
example, the course
in Partial Differential
Equations at Moscow
Institute of Physics and
Technology attracted
more than 100 students
when I used Lie group
methods, instead of 10
students that we had in the traditional course. The same happened when I
delivered similar lectures for science students in South Africa and Sweden.
The present text is based on these lectures and reflects, to a certain
extent, my own taste and experience. It is designed for the course in differential equations delivered at the Blekinge Institute of Technology for
engineering, mathematics and science students. In my presentation, I have
striven to make the group analysis of differential equations more accessible
for engineering and science students. Therefore, the emphasis in this book
is on applications of known symmetries rather than on their computation.
I gratefully acknowledge the invaluable help from Ms. Elena Ishmakova
during the preparation of this manuscript. She also translated from Russian
into English and typed the notes of my lectures in Moscow. Thanks are also
due to my colleague Claes Jogréus for valuable discussions. Finally, I wish
to thank my wife Raisa for her support and understanding of this venture.
Karlskrona, 31 August 2004
Nail H. Ibragimov
Contents
Preface
v
1 Selected topics from analysis
1.1 Elementary mathematics . . . . . . . . . . . . . . . . .
1.1.1 Numbers, variables and elementary functions . .
1.1.2 Quadratic and cubic equations . . . . . . . . . .
1.1.3 Areas of similar figures. Ellipse as an example .
1.2 Differential and integral calculus . . . . . . . . . . . . .
1.2.1 Rules for differentiation . . . . . . . . . . . . .
1.2.2 The mean value theorem . . . . . . . . . . . . .
1.2.3 Invariance of the differential . . . . . . . . . . .
1.2.4 Rules for integration . . . . . . . . . . . . . . .
1.2.5 The Taylor series . . . . . . . . . . . . . . . . .
1.2.6 Complex functions . . . . . . . . . . . . . . . .
1.2.7 The symbol o. Equivalent functions . . . . . . .
1.2.8 Functional independence. The Jacobian matrix
1.2.9 Linear independence of functions. Wronskian . .
1.2.10 Fundamental differential equation of the integral
culus. Integration by quadrature . . . . . . . .
1.2.11 Differential equations for families of curves . . .
1.3 Vector analysis . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Vector algebra . . . . . . . . . . . . . . . . . . .
1.3.2 Vector functions . . . . . . . . . . . . . . . . . .
1.3.3 Vector fields . . . . . . . . . . . . . . . . . . . .
1.3.4 The Laplace equation as an example . . . . . .
1.3.5 Three classical integral theorems . . . . . . . .
1.4 Notation of differential algebra . . . . . . . . . . . . . .
1.4.1 Differential variables. Total differentiation . . .
1.4.2 Differential functions with several variables . . .
1.4.3 The frame of differential equations . . . . . . .
1.4.4 Transformation of derivatives . . . . . . . . . .
vii
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
cal. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
1
1
1
4
8
9
9
11
11
12
13
15
17
18
19
19
21
23
23
25
26
28
29
29
30
32
33
34
viii
CONTENTS
1.5
Variational calculus . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Principle of least action . . . . . . . . . . . . . . . .
1.5.2 Euler-Lagrange equations with several variables . . .
2 Mathematical models
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . .
2.2 Natural Phenomena . . . . . . . . . . . . . . . . . . .
2.2.1 Population models . . . . . . . . . . . . . . .
2.2.2 Ecology: Radioactive waste products . . . . .
2.2.3 Kepler’s laws. Newton’s gravitation law . . . .
2.2.4 Free fall of a body near the earth . . . . . . .
2.2.5 Meteoroid . . . . . . . . . . . . . . . . . . . .
2.3 Physics and engineering sciences . . . . . . . . . . . .
2.3.1 Newton’s model of cooling . . . . . . . . . . .
2.3.2 Mechanical vibrations. Pendulum . . . . . . .
2.3.3 Collapse of driving shafts . . . . . . . . . . . .
2.3.4 The van der Pol equation . . . . . . . . . . .
2.3.5 Telegraph equation . . . . . . . . . . . . . . .
2.3.6 Electrodynamics . . . . . . . . . . . . . . . .
2.3.7 Fluid dynamics . . . . . . . . . . . . . . . . .
2.3.8 A model of an irrigation system . . . . . . . .
2.3.9 Magnetohydrodynamics . . . . . . . . . . . .
2.4 Diffusion phenomena . . . . . . . . . . . . . . . . . .
2.4.1 Linear heat equation . . . . . . . . . . . . . .
2.4.2 Nonlinear heat equation . . . . . . . . . . . .
2.4.3 The Burgers and Korteweg-de Vries equations
2.4.4 Mathematical modelling in finance . . . . . .
2.5 Biomathematics . . . . . . . . . . . . . . . . . . . . .
2.5.1 Smart mushrooms . . . . . . . . . . . . . . . .
2.5.2 A tumour growth model . . . . . . . . . . . .
2.6 Wave phenomena . . . . . . . . . . . . . . . . . . . .
2.6.1 Small vibrations of a string . . . . . . . . . .
2.6.2 Vibrating membrane . . . . . . . . . . . . . .
2.6.3 Minimal surfaces . . . . . . . . . . . . . . . .
2.6.4 Vibrating slender rods and plates . . . . . . .
2.6.5 Nonlinear waves . . . . . . . . . . . . . . . . .
2.6.6 The Chaplygin and the Tricomi equations . .
36
36
37
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
39
39
40
40
41
42
43
44
46
46
53
57
59
60
61
62
64
64
65
65
67
68
68
69
69
71
72
72
75
77
78
80
82
3 Ordinary differential equations: Traditional approach
3.1 Introduction and elementary methods . . . . . . . . . . . . .
3.1.1 Cauchy’s problem. Existence of solutions . . . . . . .
83
83
83
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
ix
CONTENTS
3.2
3.3
3.4
3.5
3.1.2 Integration of the equation y (n) = f (x) . . . . . . . .
3.1.3 Separable equations . . . . . . . . . . . . . . . . . . .
3.1.4 Exact equations . . . . . . . . . . . . . . . . . . . . .
3.1.5 Integrating factor . . . . . . . . . . . . . . . . . . . .
3.1.6 Homogeneous equations . . . . . . . . . . . . . . . .
3.1.7 Reduction of order . . . . . . . . . . . . . . . . . . .
3.1.8 The Riccati equation . . . . . . . . . . . . . . . . . .
3.1.9 The Bernoulli equation . . . . . . . . . . . . . . . . .
First-order linear equations . . . . . . . . . . . . . . . . . .
3.2.1 Homogeneous equations . . . . . . . . . . . . . . . .
3.2.2 Integration of the non-homogeneous equation by variation of the parameter . . . . . . . . . . . . . . . . .
Second-order linear equations . . . . . . . . . . . . . . . . .
3.3.1 Homogeneous equation: Superposition . . . . . . . .
3.3.2 Homogeneous equation: Equivalence transformation
and the invariant . . . . . . . . . . . . . . . . . . . .
3.3.3 Homogeneous equation: Constant coefficients . . . .
3.3.4 Non-homogeneous equations: Method of variation of
parameters . . . . . . . . . . . . . . . . . . . . . . .
3.3.5 Bessel’s equation and the Bessel functions . . . . . .
3.3.6 Hypergeometric equation . . . . . . . . . . . . . . . .
Higher-order linear equations . . . . . . . . . . . . . . . . .
3.4.1 Homogeneous equations. Fundamental system . . . .
3.4.2 Non-homogeneous equations. Variation of parameters
3.4.3 Equations with constant coefficients . . . . . . . . . .
3.4.4 Euler’s equation . . . . . . . . . . . . . . . . . . . . .
Systems of first-order equations . . . . . . . . . . . . . . . .
3.5.1 General properties of systems . . . . . . . . . . . . .
3.5.2 First integrals . . . . . . . . . . . . . . . . . . . . . .
3.5.3 Linear systems with constant coefficients . . . . . . .
3.5.4 Variation of parameters for systems . . . . . . . . . .
4 First-order partial differential equations
4.1 Introduction . . . . . . . . . . . . . . . .
4.2 Homogeneous linear equations . . . . . .
4.3 Particular non-homogeneous equations .
4.4 Quasi-linear equations . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
85
85
86
87
88
89
90
92
92
93
93
94
94
95
98
100
102
102
104
104
104
105
107
107
107
108
111
113
115
115
116
117
119
5 Linear partial differential equations of the second order
123
5.1 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.1.1 Characteristics. Three types of equations . . . . . . . 123
x
CONTENTS
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
125
126
127
128
129
130
130
133
138
140
140
143
6 Symmetry of differential equations
6.1 Transformation groups . . . . . . . . . . . . . . . . . . .
6.1.1 One-parameter groups on the plane . . . . . . . .
6.1.2 Group generator and the Lie equations . . . . . .
6.1.3 Invariants . . . . . . . . . . . . . . . . . . . . . .
6.1.4 Canonical variables . . . . . . . . . . . . . . . . .
6.2 Ordinary differential equations . . . . . . . . . . . . . . .
6.2.1 Definition of a symmetry group . . . . . . . . . .
6.2.2 Lie’s integrating factor . . . . . . . . . . . . . . .
6.2.3 Method of canonical variables . . . . . . . . . . .
6.2.4 Invariant solutions . . . . . . . . . . . . . . . . .
6.2.5 General solution provided by invariant solutions .
6.2.6 Two-dimensional Lie algebras . . . . . . . . . . .
6.2.7 Second-order equations: Lie’s integration method
6.2.8 Linearization of second-order equations . . . . . .
6.2.9 Linearization of third-order equations . . . . . . .
6.3 Nonlinear partial differential equations . . . . . . . . . .
6.3.1 Definition of a symmetry group . . . . . . . . . .
6.3.2 Group transformations of solutions . . . . . . . .
6.3.3 Invariant solutions . . . . . . . . . . . . . . . . .
6.3.4 The Burgers equation . . . . . . . . . . . . . . . .
6.3.5 A nonlinear boundary-value problem . . . . . . .
6.3.6 Invariant solutions for an irrigation system . . . .
6.3.7 Invariant solutions for a tumour growth model . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
145
145
145
147
148
149
151
151
153
154
158
159
160
161
167
168
169
169
171
172
174
177
179
182
5.2
5.3
5.1.2 The standard form of the hyperbolic equations
5.1.3 The standard form of the parabolic equations
5.1.4 The standard form of the elliptic equations . .
5.1.5 Equations of a mixed type . . . . . . . . . . .
5.1.6 The type of nonlinear equations . . . . . . . .
Integration of hyperbolic equations . . . . . . . . . .
5.2.1 d’Alambert’s solution . . . . . . . . . . . . . .
5.2.2 Euler’s method . . . . . . . . . . . . . . . . .
5.2.3 Laplace’s cascade method . . . . . . . . . . .
The method of separation of variables . . . . . . . . .
5.3.1 Vibration of a string tied at its ends . . . . .
5.3.2 Mixed problem for the heat equation . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
7 Group analysis of initial value problems
185
7.1 Generalized functions or distributions . . . . . . . . . . . . . 186
7.1.1 Definition and examples . . . . . . . . . . . . . . . . 186
CONTENTS
7.2
7.1.2 Multiplication by a function . . . . . . . . . . . . . .
7.1.3 Differentiation . . . . . . . . . . . . . . . . . . . . . .
7.1.4 Transformations of distributions . . . . . . . . . . . .
7.1.5 Convolution . . . . . . . . . . . . . . . . . . . . . . .
Fundamental solution . . . . . . . . . . . . . . . . . . . . . .
7.2.1 Definition of the fundamental solution for the heat
equation . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.2 An invariance principle . . . . . . . . . . . . . . . . .
7.2.3 Derivation of the fundamental solution using the invariance principle . . . . . . . . . . . . . . . . . . . .
7.2.4 Solution of the initial value problem . . . . . . . . . .
xi
189
190
190
192
193
193
193
194
196
Bibliography
198
Index
199
Chapter 6
Symmetry of differential
equations
The idea of symmetry permeates all mathematical models formulated in
terms of differential equations. Mathematical tools for revealing and using
the symmetry of differential equations are provided by the theory of continuous groups originated and elaborated by an outstanding mathematician of
the nineteenth century, Sophus Lie.
Lie group analysis provides general methods for integration of linear and
nonlinear ordinary differential equations analytically using their symmetries
[9], [7]. Lie group methods are also efficient in finding exact solutions to
nonlinear partial differential equations (see, e.g. [12], [11], [5], [1]).
This chapter focuses on finding solutions of nonlinear ordinary and partial differential equations using their symmetries.
6.1
6.1.1
Transformation groups
One-parameter groups on the plane
Let us consider a change of the variables x, y involving a parameter a :
Ta :
x = ϕ(x, y, a),
y = ψ(x, y, a),
(6.1.1)
ψ(x, y, 0) = y.
(6.1.2)
with functions ϕ and ψ such that
T0 :
ϕ(x, y, 0) = x,
It is assumed that ϕ(x, y, a) and ψ(x, y, a) are functionally independent, i.e.
their Jacobian does not vanish (see Section 1.2.8, Theorem 1.2.4):
¯
¯
¯ ϕx ϕy ¯
¯
¯
¯ ψx ψy ¯ 6= 0.
145
146
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
One can treat the equations Ta (6.1.1) also as a transformation that
carries any point P = (x, y) of the (x, y)-plane into a new position P = (x, y)
and write P = Ta (P ). Accordingly, the inverse transformation Ta−1 given by
Ta−1 :
x = ϕ−1 (x, y, a),
y = ψ −1 (x, y, a)
(6.1.3)
returns P into the original position P, i.e.
Ta−1 (P ) = P.
Furthermore, the equations (6.1.2) means that T0 is the identical transformation I, i.e.
T0 (P ) = P.
Let Ta and Tb be two transformations (6.1.1) with different values a and
b of the parameter. Their composition (or product) Tb Ta is defined as the
consecutive application of these transformations and is given by
³
´
x = ϕ(x, y, b) = ϕ ϕ(x, y, a), ψ(x, y, a), b ,
³
´
y = ψ(x, y, b) = ψ ϕ(x, y, a), ψ(x, y, a), b .
(6.1.4)
The geometric interpretation of the product is as follows. Since Ta carries
the point P to the point P = Ta (P ), which Tb carries to the new position
P = Tb (P ), the product Tb Ta is destined to carry P directly to its final
location P , without a stopover at P . Thus, (6.1.4) means that
def
P = Tb (P ) = Tb Ta (P ).
Definition 6.1.1. The one-parameter family G of transformations (6.1.1)
obeying the initial condition (6.1.2) is called a one-parameter group if G
contains the inverse (6.1.3) and the composition Tb Ta of all its elements:
Tb Ta = Ta+b .
The latter condition, invoking (6.1.4), is written:
³
ϕ ϕ(x, y, a), ψ(x, y, a),
³
ψ ϕ(x, y, a), ψ(x, y, a),
b
´
b
´
= ϕ(x, y, a + b),
(6.1.5)
= ψ(x, y, a + b).
147
6.1. TRANSFORMATION GROUPS
6.1.2
Group generator and the Lie equations
The expansion of the functions ϕ(x, y, a) and ψ(x, y, a) into Taylor’s series
in a near a = 0, taking into account the initial condition (6.1.2), yields the
infinitesimal transformation
x ≈ x + ξ(x, y)a,
y ≈ y + η(x, y)a,
(6.1.6)
where
ξ(x, y) =
∂ϕ(x, y, a) ¯¯
¯ ,
∂a
a=0
η(x, y) =
∂ψ(x, y, a) ¯¯
¯ ·
∂a
a=0
(6.1.7)
The vector (ξ, η) with components (6.1.7) is the tangent vector (at the
point (x, y)) to the curve described by the transformed points (x̄, ȳ), and is
therefore called the tangent vector field of the group G.
The tangent vector field (6.1.7) is associated with the first-order differential operator
∂
∂
+ η(x, y)
(6.1.8)
X = ξ(x, y)
∂x
∂y
called the generator of the group G.
Given an infinitesimal transformation (6.1.6), or the generator (6.1.8),
the transformations (6.1.1) are defined by integrating the following system
of ordinary differential equations called the Lie equations:
dϕ
= ξ(ϕ, ψ),
da
dψ
= η(ϕ, ψ),
da
¯
¯
ϕ¯
a=0
= x,
(6.1.9)
¯
¯
ψ¯
= y.
¯
¯
x¯
= x,
a=0
These equations are written also as follows:
dx
= ξ(x, y),
da
dy
= η(x, y),
da
a=0
¯
¯
y¯
a=0
(6.1.10)
= y.
For example, the group of rotations defined by
x̄ = x cos a + y sin a,
ȳ = y cos a − x sin a ,
has the infinitesimal transformation
x ≈ x + ya,
y ≈ y − xa
148
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
and the generator
X=y
∂
∂
−x ,
∂x
∂y
respectively. You can easily verify the Lie equations:
¯
dx
¯
= y, x¯
= x,
da
a=0
¯
dy
¯
= y.
= −x, y ¯
da
a=0
Example 6.1.1. Let us find the one-parameter group given by its infinitesimal transformation
x ≈ x + ax2 , y ≈ y + axy,
or, equivalently, by the following generator:
X = x2
∂
∂
+ xy ·
∂x
∂y
(6.1.11)
The Lie equations (6.1.10) have the form
¯
dx
¯
= x,
= x2 , x ¯
da
a=0
¯
dy
¯
= x y, y ¯
= y.
da
a=0
The differential equations of this system are easily solved and yield
x=−
1
,
a + C1
y=
C2
·
a + C1
The initial conditions imply that C1 = −1/x, C2 = −y/x. Consequently we
arrive at the following one-parameter group know as the projective transformation group:
x
y
x=
, y=
·
(6.1.12)
1 − ax
1 − ax
6.1.3
Invariants
A function J(x, y) is called an invariant of the group of transformations
(6.1.1) if the following equation holds:
¡
¢
J(x, y) ≡ J ϕ(x, y, a), ψ(x, y, a) = J(x, y).
6.1. TRANSFORMATION GROUPS
149
Theorem 6.1.1. A necessary and sufficient condition for a function J(x, y)
to be an invariant is that it solves the following partial differential equation:
X(J) ≡ ξ(x, y)
∂J
∂J
+ η(x, y)
= 0.
∂x
∂y
(6.1.13)
Proof. Let us discuss only the necessity. It follows from the Taylor expansion of J(x, y) with respect to a :
³ ∂J
∂J ´
J(x, y) ≈ J(x + aξ, y + aη) ≈ J(x, y) + a ξ
= J(x, y) + aX(J).
+η
∂x
∂y
Indeed, if J(x, y) is an invariant we substitute J(x, y) = J(x, y) in the lefthand side of the above equation and get aX(J) = 0, and hence X(J) = 0.
It follows from Lemma 3.5.1 that every one-parameter group of point
transformations in the plane has one independent invariant, which can be
taken to be the left-hand side of any first integral φ(x, y) = C of the characteristic equation:
dx
dy
=
·
(6.1.14)
ξ(x, y)
η(x, y)
Any other invariant has the form J(x, y) = F (φ(x, y)).
6.1.4
Canonical variables
The following simple result has many applications, e.g. in integration of
differential equations.
Theorem 6.1.2. Every one-parameter group of transformations (6.1.1) with
the generator
∂
∂
X = ξ(x, y)
+ η(x, y)
∂x
∂y
can be reduced to the group of translations t̄ = t + a, ū = u with the
generator
∂
X=
∂t
by introducing new variables t = t(x, y), u = u(x, y) called canonical
variables and defined by the following first-order linear partial differential
equations:
X(t) ≡ ξ(x, y)
∂t(x, y)
∂t(x, y)
+ η(x, y)
= 1,
∂x
∂y
∂u(x, y)
∂u(x, y)
X(u) ≡ ξ(x, y)
+ η(x, y)
= 0.
∂x
∂y
(6.1.15)
150
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
Example 6.1.2. Let us find canonical variables for the dilation group with
the generator
∂
∂
+y ·
(6.1.16)
X=x
∂x
∂y
The first equation (6.1.15) has the form
X(t) = x
∂t
∂t
+y
= 1.
∂x
∂y
Since it is sufficient to find any particular solution to this equation, we can
look, e.g. for a solution t = t(x) depending only on x. Then the above
equation reduces to the ordinary differential equation x t0 (x) = 1, whence
t = ln |x|. The second equation (6.1.15) has the form
X(u) = x
∂u
∂u
+y
= 0.
∂x
∂y
The characteristic equation
dy
dx
=
x
y
has the first integral y/x = C, and hence u = y/x solves the equation
X(u) = 0. Thus, we have the following canonical variables:
t = ln |x|,
u=
y
·
x
In these variables, the dilation group
x̄ = x ea ,
ȳ = y ea
reduces to the translation group
t̄ = t + a,
ū = u.
Indeed,
t̄ = ln |x̄| = ln(|x| ea ) = ln |x| + a = t + a,
ū =
y
y ea
ȳ
=
=
= u.
x̄
x ea
x
6.2. ORDINARY DIFFERENTIAL EQUATIONS
6.2
6.2.1
151
Ordinary differential equations
Definition of a symmetry group
Definition 6.2.1. The group of transformations (6.1.1) is termed a symmetry group of an ordinary differential equation
dy
= f (x, y)
dx
if the form of the differential equation remains the same after the change of
variables (6.1.1). It means that
dy
= f (x, y)
dx
with the same function f as in the original equation. A symmetry group
of a differential equation is also termed a group admitted by this equation.
The definition is the same for higher-order equations.
The generator X of a group admitted by a differential equation is also
termed an admitted operator or an infinitesimal symmetry of the equation
in question.
Example 6.2.1. It is evident that the equation
y 0 = f (y)
does not alter after the transformation x = x + a since the equation does
not explicitly contain the independent variable x. Therefore, the symmetry
of this differential equations is given by the group of translations along the
x-axis,
x=x+a
with the generator
X=
∂
·
∂x
Likewise, the equation
y 0 = f (x)
admits the group of translations along the y-axis,
y =y+a
with the generator
X=
∂
·
∂y
152
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
Example 6.2.2. The equation
y0 = f
³y´
x
(6.2.1)
admits the group of homogeneous dilations (scaling transformations)
x = xea ,
y = yea
with the generator (6.1.16),
X=x
∂
∂
+y ·
∂x
∂y
Example 6.2.3. Consider the following Riccati equation:
y0 + y2 −
2
= 0.
x2
(6.2.2)
Its left-hand side is a rational function in the variables x, y, y 0 . Therefore,
one can try to find an admitted group in the form of a dilation:
x = kx,
y = ly.
Since
2
l 0
1 2
y + l2 y 2 − 2 2 ,
2 =
k
k x
x
the invariance condition requires that
y0 + y2 −
y0 + y2 −
³
2´
2
0
2
=
λ
·
y
+
y
−
,
x2
x2
where
1
l
= l2 = 2 ·
k
k
The latter equations yield l = 1/k, where k > 0 is an arbitrary parameter.
Setting k = ea , we get the non-homogeneous dilation group:
λ=
x = x ea ,
y = y e−a .
Hence, the Riccati equation (6.2.2) has the following infinitesimal symmetry:
X=x
∂
∂
−y ·
∂x
∂y
(6.2.3)
153
6.2. ORDINARY DIFFERENTIAL EQUATIONS
6.2.2
Lie’s integrating factor
Consider a first-order equation written in the symmetric form (3.1.5):
M (x, y)dx + N (x, y)dy = 0.
(6.2.4)
Lie showed that if
∂
∂
+ η(x, y)
∂x
∂y
is a symmetry for Equation (6.2.4) and ξQ − ηP 6= 0, then the function
X = ξ(x, y)
µ=
1
ξM + ηN
(6.2.5)
is an integrating factor for Eq. (6.2.4). It is called Lie’s integrating factor.
Example 6.2.4. Let us rewrite the Riccati equation (6.2.2) in the differential form (6.2.4):
¶
µ
2
2
(6.2.6)
dy + y − 2 dx = 0,
x
find Lie’s integrating factor using the symmetry (6.2.3) and finally integrate
the Riccati equation. Substituting
ξ = x,
η = −y,
M = y2 −
2
,
x2
N =1
in (6.2.5), we obtain the integrating factor
µ=
x2 y 2
x
·
− xy − 2
After multiplication by this factor Equation (6.2.6) becomes:
xdy + (xy 2 − 2/x)dx
= 0.
x2 y 2 − xy − 2
Let us rewrite it in the following form convenient for integration:
³
dx
1 xy − 2 ´
xdy + ydx
= 0.
+
= d ln x + ln
x2 y 2 − xy − 2
x
3 xy + 1
The integration yields:
xy − 2
C
= 3,
xy + 1
x
whence solving for y, we obtain the solution of the Riccati equation:
2x3 + C
y=
,
x(x3 − C)
C = const.
(6.2.7)
154
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
6.2.3
Method of canonical variables
To integrate a linear or nonlinear first-order ordinary differential equation
y 0 = f (x, y)
(6.2.8)
with a known infinitesimal symmetry
X = ξ(x, y)
∂
∂
+ η(x, y)
∂x
∂y
(6.2.9)
by the method of canonical variables, you need to make the following steps.
First step: Find canonical variables t, u by solving the equations (6.1.15)
for the given generator (6.2.9).
Second step: Rewrite Equation (6.2.8) in the canonical variables t and u
by letting u be the new dependent variable of the independent variable t,
i.e. letting u = u(t) and expressing the old derivative y 0 = dy/dx via new
variables t, u and the derivative u0 = du/dt. Then Equation (6.2.8) will have
the integrable form
du
= g(u).
(6.2.10)
dt
Third step: Integrate Equation (6.2.10), substitute in its solution u =
φ(t, C) the expressions t = t(x, y) and u = u(x, y) thus obtaining the solution of Equation (6.2.8).
Example 6.2.5. One can integrate by quadrature any equation of the form
(6.2.1) using its infinitesimal symmetry (6.1.16). To be specific, let us integrate the following particular equation of the form (6.2.1):
y0 =
y y3
+
·
x x3
(6.2.11)
First step: The canonical variables for the infinitesimal symmetry (6.1.16)
are found in Example 6.1.2. They are:
y
t = ln x, u = ·
x
Second step: Let us rewrite the equation (6.2.11) in the canonical variables t and u. Since y = xu and dt/dx = 1/x, we have denoting u0 = du/dt :
y0 ≡
dy
d(xu)
du
du dt
1
=
=u+x
=u+x
= u + xu0 = u + u0 .
dx
dx
dx
dt dx
x
Thus, Equation (6.2.11) becomes a particular equation of the form (6.2.10):
du
= u3 .
dt
6.2. ORDINARY DIFFERENTIAL EQUATIONS
155
Third step: Integration of the above equation yields
u = ±√
1
,
C − 2t
whence, upon substituting t = ln |x| and y = xu :
y = ±√
x
·
C − ln x2
Example 6.2.6. Let us integrate the Riccati equation (6.2.2),
2
dy
+ y 2 − 2 = 0,
dx
x
by the method of canonical variables using its symmetry (6.2.3),
X=x
∂
∂
−y ·
∂x
∂y
First step: Solution of equations (6.1.15) with the above operator X
provides the canonical variables, t = ln x and u = xy.
Second step: We have:
u
1 du
1 du dt
u0
d ³u´
u
u
dy
=− 2 +
=
=− 2 +
=− 2 + 2·
dx
dx x
x
x dx
x
x dt dx
x
x
Therefore,
¢
dy
2
u0
u
u2
2
1 ¡
+ y 2 − 2 = 2 − 2 + 2 − 2 = 2 u0 + u2 − u − 2 = 0.
dx
x
x
x
x
x
x
Thus, the Riccati equation is rewritten in the canonical variables in the
following integrable form (6.2.10):
du
= −(u2 − u − 2).
dt
Third step: Let us integrate the above equation. Separating the variables:
du
= −dt,
u2 − u − 2
decomposing here the rational fraction into elementary fractions:
·
¸
1
1
1
1
,
=
−
u2 − u − 2
3 u−2 u+1
156
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
and integrating we have:
ln
µ
u−2
u+1
¶
= −3t + ln C ·
Now solve this equation with respect to u,
u=
C + 2e3t
,
e3t − C
substitute t = ln x, u = xy and arrive again to the solution (6.2.7) of the
Riicati equation:
2x3 + C
·
y=
x(x3 − C)
Example 6.2.7. The equation
y0 =
y 1 ³y´
+ F
x x
x
(6.2.12)
with an arbitrary function F admits the generator (6.1.11)
X = x2
∂
∂
+ xy
∂x
∂y
of the projective transformation group (6.1.12),
x=
x
,
1 − ax
y=
y
·
1 − ax
Indeed, the equations
Dx (x) =
1
,
(1 − ax)2
Dx (y) =
ay + (1 − ax)y 0
(1 − ax)2
yield:
y0 =
dȳ
Dx ȳ
=
,
x̄
Dx x̄
whence, substituting
Dx (x) =
1
,
(1 − ax)2
Dx (y) =
(1 − ax)y 0 + ay
(1 − ax)2
we obtain
y 0 = (1 − ax)y 0 + ay.
(6.2.13)
157
6.2. ORDINARY DIFFERENTIAL EQUATIONS
Therefore,
y
1
y − − F
x x
0
or
µ ¶
y 1 − ax ³ y ´
y
= (1 − ax)y 0 + ay − −
F
x
x
x
x
y
1
y − − F
x x
0
µ ¶
·
¸
y
1 ³y´
y
0
= (1 − ax) y − − F
.
x
x x
x
Hence, Equation (6.2.12) yields:
1
y
y − − F
x x
0
µ ¶
y
= 0.
x
Let us integrate a particular equation of the form (6.2.12), namely that
with the function F (σ) = σ 2 , i.e. the following equation:
y y2
y = + 3·
x x
0
(6.2.14)
First step: The equations (6.1.15) with the operator (6.2.13) are written:
X(t) = x2
∂t
∂t
+ xy
= 1,
∂x
∂y
X(u) = x2
∂u
∂u
+ xy
= 0.
∂x
∂y
Taking a particular solution of these equations with t = t(x), one obtains
the canonical variables
1
y
t=− , u= ·
(6.2.15)
x
x
Second step: We have:
y0 =
1
d(xu)
du
du dt
1
=u+x
=u+x
= u + xu0 2 = u + u0 ,
dx
dx
dt dx
x
x
and the equation (6.2.14) takes the following integrable form (6.2.10):
du
= u2 .
dt
Third step: Integration of the above equation yields
u=−
1
,
C +t
whence, substituting t = 1/x and u = y/x :
y=
x2
·
1 − Cx
(6.2.16)
158
6.2.4
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
Invariant solutions
An essential feature of a symmetry group G of an ordinary differential equation is that it converts any solution (integral curve) of the equation in question into a solution. In other words, the symmetry transformations merely
permute the integral curves among themselves. It may happen that some of
the integral curves are individually unaltered under G. Such integral curves
are termed invariant solutions.
Example 6.2.8. Consider again the Riccati equation (6.2.2)
2
dy
+ y2 − 2 = 0
dx
x
and find its invariant solutions with respect to the infinitesimal symmetry
X=x
∂
∂
−y ·
∂x
∂y
The invariant test (6.1.13), X(J) = 0, provides a single independent invariant xy. Therefore xy = const. is the only relation written in terms of
invariants. Hence, the general form of invariant solutions is xy = λ, or
y = λ/x, with an arbitrary constant λ. The substitution into the Riccati
equation reduces the latter to an algebraic, namely the quadratic equation,
λ2 − λ − 2 = 0, whence λ1 = 2 and λ2 = −1. Thus, one arrives at the
following two invariant solutions:
y1 =
2
x
1
and y2 = − ·
x
Example 6.2.9. The general homogeneous linear ordinary differential equation with constant coefficients
Ln [y] ≡ y (n) + A1 y (n−1) + · · · + An−1 y 0 + An y = 0,
admits the translation group with the generator X1 = ∂/∂x (since the
coefficients are constant) and multiplication of y by any parameter, i.e. the
dilation group with the generator X2 = y∂/∂y (because of the homogeneity
of the equation). Therefore, the equation admits the linear combination of
these generators, X = X1 + λX2 :
X=
∂
∂
+ λy ,
∂x
∂y
λ = const.
The equation dy/y = λdx yields the only independent invariant u = ye−λx .
Invariant solutions are obtained by setting u = C. Hence, Euler’s ansatz:
y = Ceλx .
6.2. ORDINARY DIFFERENTIAL EQUATIONS
6.2.5
159
General solution provided by invariant solutions
The technique of invariant solutions furnishes a simple way for obtaining
the general integral of first-order ordinary differential equations with two
known infinitesimal symmetries. Consider the following example.
Example 6.2-8. Consider the equation (6.2.12) with F (σ) = σ n , i.e.
the equation
yn
y
y 0 = + n+1 ·
x x
It admits, along with the projective group with the generator (6.2.13), the
dilation group with the generator
X = (n − 1)x
∂
∂
+ ny ·
∂x
∂y
Let us take, e.g. n = 2. Thus, we consider the equation (6.2.14),
y0 =
y y2
,
+
x x3
with two known infinitesimal symmetries:
X1 = x 2
∂
∂
+ xy ,
∂x
∂y
X2 = x
∂
∂
+ 2y ·
∂x
∂y
The equation X2 (J) = 0 provides one independent invariant y/x2 . Consequently, the invariant solution is obtained by letting y/x2 = λ, or y = λx2
with an arbitrary constant λ 6= 0. Then our differential equation reduces to
y0 −
y2
y
− 3 = 2λx − λx − λ2 x = λ(1 − λ)x = 0.
x x
Hence λ = 1, and the invariant solution simply is
y = x2 .
(6.2.17)
Let us take now the projective transformation (6.1.12)
x=
x
,
1 − ax
y=
y
1 − ax
generated by X1 , rewrite the invariant solution (6.2.17) in the new variables
in the form
y = x2
160
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
and substitute here the expressions for x and y :
y
x2
=
·
1 − ax
(1 − ax)2
Denoting the parameter a by C, one obtains the general solution (6.2.16):
x2
·
y=
1 − Cx
6.2.6
Two-dimensional Lie algebras
Given two linearly independent differential operators of the form (6.1.8):
X1 = ξ 1
∂
∂
+ η1 ,
∂x
∂y
X2 = ξ 2
∂
∂
+ η2 ,
∂x
∂y
(6.2.18)
their commutator [X1 , X2 ] is defined by the formula
³
´∂
³
´∂
[X1 , X2 ] = X1 (ξ2 ) − X2 (ξ1 )
+ X1 (η2 ) − X2 (η1 )
·
∂x
∂y
(6.2.19)
Denote by L2 the two-dimensional linear space spanned by the operators
(6.2.18), i.e. the set of the operators of the form
X = C 1 X1 + C 2 X2
with arbitrary constant coefficients C1 , C2 .
Definition 6.2.2. The space L2 is called a two-dimensional Lie algebra if
[X1 , X2 ] ∈ L2 . The operators X1 and X2 provide a basis of L2 .
Lie’s method of integration of second-order ordinary differential equations is based on so-called canonical coordinates in two-dimensional Lie algebras. These variables provide, for every L2 , the simplest form of its basis
and therefore reduce a differential equation admitting the L2 , to an integrable form. The basic statements are as follows.
Theorem 6.2.1. Any two dimensional Lie algebra can be transformed, by
a proper choice of its basis and suitable variables t, u, called canonical variables, to one of the four non-similar standard forms presented in the following table.
161
6.2. ORDINARY DIFFERENTIAL EQUATIONS
Table 6.2.1 Structure and non-similar realizations of L2
Type
Structure of L2
Standard form of L2
I
[X1 , X2 ] = 0, ξ1 η2 − η1 ξ2 6= 0
II
[X1 , X2 ] = 0, ξ1 η2 − η1 ξ2 = 0
III
[X1 , X2 ] = X1 , ξ1 η2 − η1 ξ2 6= 0
IV
[X1 , X2 ] = X1 , ξ1 η2 − η1 ξ2 = 0
6.2.7
∂
∂
, X2 =
∂t
∂u
∂
∂
, X2 = t
X1 =
∂u
∂u
∂
∂
∂
X1 =
, X2 = t + u
∂u
∂t
∂u
∂
∂
X1 =
, X2 = u
∂u
∂u
X1 =
Second-order equations: Lie’s integration method
Lie proved Theorem 6.2.1 in order to integrate all second-order equations
y 00 = f (x, y, y 0 )
(6.2.20)
admitting a two-dimensional Lie algebra. Lie’s method consists in classifying these equations into four types in accordance with Table 6.2.1. Namely,
we introduce canonical variables, t, u reducing the admitted Lie algebra L2
to its standard form from Table 6.2.1. Then Equation (6.2.20) written in
the canonical variables,
u00 = g(t, u, u0 )
(6.2.21)
takes one of the integrable canonical forms given in the following table.
Table 6.2.2 Four types of second-order equations admitting L2
Type
I
II
III
IV
Standard form of L2
∂
∂
, X2 =
∂t
∂u
∂
∂
, X2 = t
X1 =
∂u
∂u
∂
∂
∂
X1 =
, X2 = t + u
∂u
∂t
∂u
∂
∂
X1 =
, X2 = u
∂u
∂u
X1 =
Canonical form of the Equation
u00 = f (u0 )
u00 = f (u)
1
u00 = f (u0 )
t
u00 = f (t)u0
162
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
Thus, the method is as follows. Provided that we know an admitted
algebra L2 with a basis (6.2.18), the integration requires the following steps.
Step 1. Determine the type of L2 according to the Structure column of
Table 6.2.1. A change of the basis of L2 may be required to accord the
table.
Type I : X1 (t) = 1, X2 (t) = 0;
X1 (u) = 0, X2 (u) = 1.
Type II : X1 (t) = 0, X2 (t) = 0;
X1 (u) = 1, X2 (u) = t.
Type III : X1 (t) = 0, X2 (t) = t;
X1 (u) = 1, X2 (u) = u.
Type IV : X1 (t) = 0, X2 (t) = 0;
X1 (u) = 1, X2 (u) = u.
(6.2.22)
Now rewrite the differential equation in the canonical variables choosing t
as a new independent variable and u as a dependent one. It will have one
of the integrable forms given in Table 6.2.2. Integrate the equation.
Step 3. Rewrite the resulting solution in the original variables x, y, thus
completing the integration procedure.
Example 6.2.10. The equation
y 00 = yy 02 − xy 03
(6.2.23)
admits the two-dimensional Lie algebra with the basis
X1 = y
∂
,
∂x
X2 = x
∂
·
∂x
(6.2.24)
Step 1. The operators (6.2.24) satisfy the equations
[X1 , X2 ] = X1 ,
ξ1 η2 − η1 ξ2 = 0,
and hence, L2 has type IV in Table 6.2.1.
Step 2. The equations (6.2.22) for type IV:
X1 (t) = 0, X2 (t) = 0; X1 (u) = 1, X2 (u) = u,
yield the canonical variables
t = y,
u=
x
·
y
From the definition of t, we have Dx = y 0 Dt and its action on the variable
u yields
y − xy 0 = y 2 y 0 u0 .
163
6.2. ORDINARY DIFFERENTIAL EQUATIONS
Solving the latter equation for y 0 and differentiating it again, we obtain:
y0 =
1
y
=
,
2
0
x+y u
u + tu0
y 00 = −y 0
2u0 + tu00
2u0 + tu00
−
·
(u + tu0 )2 (u + tu0 )3
Consequently equation (6.2.23) assumes the linear form
³
2´ 0
u.
u00 = − t +
t
Its solution is given by quadrature and has the form:
Z
1 −t2 /2
e
dt.
u = C2 − C1
t2
Step 3. Returning to the original variables, we have the solution of our
equation:
µ
¶
Z
1 −y2 /2
x = y C2 − C 1
e
dy .
y2
Example 6.2.11. The nonlinear second-order equation
y 00 =
y0
1
−
2
y
xy
(6.2.25)
admits L2 with the following basis:
X1 = x 2
∂
∂
+ xy ,
∂x
∂y
X2 = −x
∂
y ∂
−
·
∂x 2 ∂y
(6.2.26)
Step 1. One can readily verify that
[X1 , X2 ] = X1 ,
ξ 1 η2 − η 1 ξ 2 = −
x2 y
6= 0.
2
Hence, the algebra L2 has the structure of type III in Table 6.2.1.
Step 2. Let us find the canonical variables and integrate our equation
these variables. The system of equations X1 (t) = 0, X2 (t) = t for the new
variable t yields
³ y ´2
t=
,
(6.2.27)
x
and the system X1 (u) = 1, X2 (u) = u for u yields:
1
u=− ·
x
(6.2.28)
164
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
In the canonical variables t, u the operators X1 , X2 become
X1 =
∂
,
∂u
X2 = t
∂
∂
+u ·
∂t
∂u
Under the change of the independent variable (6.2.27), the total differentiation Dx transforms into the total differentiation Dt defined by the
following equation :
µ 2¶
y(xy 0 − y)
y
Dx = D x
Dt ,
· Dt = 2
x2
x3
or
Dx = 2u(t −
√
t y 0 ) Dt .
(6.2.29)
Now we differentiate both sides of equation (6.2.28) by using the equation
(6.2.29) and invoking that the left-hand side of (6.2.28) depends on t and
its right-hand side depends on x, to obtain:
¶
µ
√ 0
1
1
= 2 = u2 ,
2u(t − t y ) Dt (u) = Dx −
x
x
or
√
u
·
2u0
For calculation of the transformation of the second derivative, it is convenient to write the transformation of the total differentiation and of the first
derivative as
u2
Dx = 0 Dt
u
and
√
u
y0 = t − √ 0 ,
2 tu
respectively. Then one readily obtains the following transformation of the
second derivative:
u3
u3 u00
y 00 = √ 02 + √ 03 ·
4t t u
2 tu
Equations (6.2.27)-(6.2.28) yield:
t−
t y0 =
1
u2
=√ ·
xy
t
Furthermore, the expression for y 0 and equations (6.2.27)-(6.2.28) yield:
u2
y0
u3
√
√
=
−
·
y2
t 2t t u0
6.2. ORDINARY DIFFERENTIAL EQUATIONS
165
After substituting the above expressions, equation (6.2.25) assumes the integrable form:
µ
¶
1 0
1
00
0
u =− u u +
.
(6.2.30)
t
2
Is equation (6.2.30) equivalent to the original equation (6.2.25)? More
specifically, the question is whether or not all solutions of equation (6.2.25)
are obtained from the solutions of (6.2.30) by the change of variables (6.2.27)(6.2.28), and visa versa. The answer is not self evident since the variable t
defined by (6.2.27) involves the dependent variable y of the original equation
(6.2.25) and therefore t can be regarded as a new independent variable only
if (6.2.25) does not have solutions along which t is not identically constant.
The direct inspection shows, however, that (6.2.25) has indeed such singular
solutions where t = const., namely the solutions given by the straight lines:
y = Kx,
K = const.
All other solutions of equation (6.2.25) are obtained from solutions of (6.2.30)
by the change of variables (6.2.27)-(6.2.28).
Furthermore, one should also inspect whether or not all solutions of
equation (6.2.30) are related with solutions of (6.2.25). We notice that
equation (6.2.30) is obviously satisfied by u0 = 0 as well as by u0 = −1/2,
the corresponding solutions being u = A, u = C − 2t , where A and C are
arbitrary constants. According to (6.2.28), the first of the above solutions,
u = A, means x = const. Hence it is not related to any solution of equation
(6.2.25) and should be ignored. But the second solution, u = C − t/2,
provides a solution of equation (6.2.25). Namely, substituting there the
expressions (6.2.27) and (6.2.28) for t and u, respectively, one obtains the
following solution of equation (6.2.25):
√
y = ± 2x + Cx2 .
Now we integrate equation (6.2.30) excluding the above singular solutions, i.e. assuming u0 6= 0 and u0 + 1/2 6= 0. Then (6.2.30) yields:
¯
¯ 0
Z
Z
Z
¯ √¯
¯ 2u + 1 ¯
du0
2du0
du0
¯
¯
¯,
¯
ln ¯K1 t¯ = −
= ln ¯
=
−
u0 (2u0 + 1)
2u0 + 1
u0
u0 ¯
or
√
2u0 + 1
,
K1 t =
u0
whence
u0 =
1
√
,
2(C1 t − 1)
166
where
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
C1 = K1 /2 6= 0. Finally, the elementary integration yields:
´
¯ √
¯
1 ³ √
¯
¯
u = 2 C1 t + ln C1 t − 1 + C2 .
C1
Step 3. Let us rewrite the solution in the original variables. Replacing
in the last equation t and u by the expressions (6.2.27) and (6.2.28), respectively, one arrives at the following implicit representation of the solution
y(x) of equation (6.2.25) involving two constants, C1 6= 0 and C2 :
¯ y
¯
¯
¯
C1 y + C2 x + x ln ¯C1 − 1¯ + C12 = 0.
x
Adding to the latter two singular solutions discussed above one ultimately
arrives at the general solution of equation (6.2.25) represented by the following three distinctly different formulae with arbitrary constants K, C, C 1 , C2
with C1 6= 0 :
y = Kx,
(6.2.31)
√
y = ± 2x + Cx2 ,
(6.2.32)
¯ y
¯
¯
¯
C1 y + C2 x + x ln ¯C1 − 1¯ + C12 = 0.
(6.2.33)
x
Exercise 6.2.1. Solve the following Cauchy problems for equation (6.2.25):
(i) y 00 =
1
y0
−
,
y 2 xy
(ii) y 00 =
1
y0
−
,
2
y
xy
1
y0
,
(iii) y = 2 −
y
xy
00
¯
¯
y ¯x=1 = 1, y 0 ¯x=1 = 1;
¯
¯
y ¯x=1 = 1, y 0 ¯x=1 = 0;
¯
¯
y ¯x=1 = 1, y 0 ¯x=1 = 2.
Solution. Substituting x = 1, y = 1, y 0 = 1 in all three solution formulae
(6.2.31)-(6.2.33) one can readily verify that the initial conditions (i) can
be satisfied only by (6.2.31) with K = 1. Similar reasoning singles out the
second solution formula (6.2.32) with the plus sign and with C = −1 for
x = 1, y = 1, y 0 = 0. Likewise, the substitution x = 1, y = 1, y 0 = 2 singles
out the solution formula (6.2.33) with C1 = 2, C2 = −6. Hence, the solution
to the above Cauchy problems is given by:
¯
¯ y
√
¯
¯
(i) y = x, (ii) y = 2x − x2 , (iii) 2y − 6x + x ln ¯2 − 1¯ + 4 = 0.
x
6.2. ORDINARY DIFFERENTIAL EQUATIONS
6.2.8
167
Linearization of second-order equations
Lie1 found the general form of all second-order equations (6.2.20) that can
be reduced to the linear equation
du
d2 u
=
A(t)
+ B(t)u + C(t)
dt2
dt
by a change of variables
t = ϕ(x, y),
u = ψ(x, y).
(6.2.34)
(6.2.35)
He showed that the linearizable second-order equations should be at most
cubic in the first-order derivative, i.e. belong to the family of equations of
the form
y 00 + F3 (x, y)y 0 3 + F2 (x, y)y 0 2 + F1 (x, y)y 0 + F (x, y) = 0,
(6.2.36)
The form (6.2.36) can be obtained by using the transformation of derivatives
given by Theorem 1.4.1 in Section 1.4.4.
Lie’s linearization test can be formulated as follows (see, e.g. [7]).
Theorem 6.2.2. Equation (6.2.36) is linearizable if and only if its coefficients satisfy the following equations:
3(F3 )xx − 2(F2 )xy + (F1 )yy = (3F1 F3 − F22 )x − 3(F F3 )y − 3F3 Fy + F2 (F1 )y ,
3Fyy − 2(F1 )xy + (F2 )xx = 3(F F3 )x + (F12 − 3F F2 )y + 3F (F3 )x − F1 (F2 )x .
Lie’s linearization test is simple and convenient in practice. Consider
examples.
Example 6.2.12. The equation
y 00 + F (x, y) = 0
has the form (6.2.36) with F3 = F2 = F1 = 0. The linearization test yields
Fyy = 0. Hence, the equation y 00 = F (x, y) cannot be linearized unless it is
already linear.
Example 6.2.13. The equations
1
y 00 − (y 0 + y 0 3 ) = 0
x
and
1
y 00 + (y 0 + y 0 3 ) = 0
x
also have the form (6.2.36). Their coefficients are F3 = F1 = −1/x, F2 =
F = 0 and F3 = F1 = 1/x, F2 = F = 0, respectively. The linearization test
shows that the first equation is linearizable, whereas the second one is not.
1
S. Lie, Archiv for Matematik og Naturvidenskab, vol. 8, Heft 4, pp. 371-458, 1883.
168
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
6.2.9
Linearization of third-order equations
Recently we found the general form of all third-order equations2
y 000 = f (x, y, y 0 , y 00 )
(6.2.37)
that can be linearized by a change of variables (6.2.35),
t = ϕ(x, y),
u = ψ(x, y).
We showed that every linearizable third-order equation belongs to one of
the families of equations given in the following theorem and obtained the
necessary and sufficients conditions for linearization for each family.
Theorem 6.2.3. Every equation (6.2.37) obtained from a linear equation
by a change of variables (6.2.35) belongs to the family of equations
1 h
000
− 3(y 00 )2 + (C2 y 02 + C1 y 0 + C0 )y 00
y + 0
y +r
(6.2.38)
i
05
04
03
02
0
+D5 y + D4 y + D3 y + D2 y + D1 y + D0 = 0
if ϕy 6= 0 or to the family of equations
y 000 + (A1 y 0 + A0 )y 00 + B3 y 03 + B2 y 02 + B1 y 0 + B0 = 0,
(6.2.39)
if ϕy = 0. The coefficients
r = r(x, y),
Ci = Ci (x, y),
Di = Di (x, y)
and
Ai = Ai (x, y),
Bi = Bi (x, y)
are determined by the conditions for linearization of Equation (6.2.38) and
Equation (6.2.39), respectively.
Example 6.2.14. The non-linear equation
i
1h
002
05
000
y + 0 − 3y − xy = 0
y
has the form (6.2.38) with the following coefficients:
r = 0,
C0 = C1 = C2 = 0,
D0 = D1 = D2 = D3 = D4 = 0,
2
D5 = −x.
N.H. Ibragimov and S.V. Meleshko, Archives of ALGA, vol. 1, pp. 71-92, 2004.
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
169
Applying the linearization test, we find that the equation in quiestion can
be reduced to the linear equation
u000 + u = 0
by the change of variables
t = y,
6.3
6.3.1
u = x.
Nonlinear partial differential equations
Definition of a symmetry group
Definition of symmetry group for partial differential equations is the same as
that for ordinary differential equations. Let us consider evolutionary partial
differential equations of the second order:
ut = F (t, x, u, ux , uxx ),
∂F/∂uxx 6= 0.
(6.3.1)
Definition 3. A set G of invertible transformations of the variables t, x, u :
t̄ = f (t, x, u, a),
x̄ = g(t, x, u, a),
ū = h(t, x, u, a),
(6.3.2)
is called a one-parameter group admitted by the equation (6.3.1), or a symmetry group of the equation (6.3.1), if G contains the inverse to its transformations, the identity t = t, x = x, u = u, as well as the composition:
t ≡ f (t, x, u, b) = f (t, x, u, a + b),
x ≡ g(t, x, u, b) = g(t, x, u, a + b),
u ≡ h(t, x, u, b) = h(t, x, u, a + b),
and if the equation (6.3.1) has the same form in the new variables t, x, u :
ut̄ = F (t, x, u, ux̄ , ux̄x̄ ).
(6.3.3)
The function F has the same form in both equations (6.3.1) and (6.3.3).
Again, the construction of the symmetry group G is equivalent to determination of its infinitesimal transformations
t ≈ t + aτ (t, x, u),
x ≈ x + aξ(t, x, u),
u ≈ u + aη(t, x, u)
(6.3.4)
170
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
obtained from (6.3.2) by expanding into Taylor series with respect to the
group parameter a and keeping only the terms linear in a. The infinitesimal transformation (6.3.4) provides the generator of the group G, i.e. the
differential operator
X = τ (t, x, u)
∂
∂
∂
+ ξ(t, x, u)
+ η(t, x, u)
∂t
∂x
∂u
(6.3.5)
acting on any differentiable function J(t, x, u) as follows:
X(J) = τ (t, x, u)
∂J
∂J
∂J
+ ξ(t, x, u)
+ η(t, x, u)
·
∂t
∂x
∂u
The generator (6.3.5) is called an operator admitted by equation (6.3.1) or
an infinitesimal symmetry for equation (6.3.1).
The group transformations (6.3.2) corresponding to the generator (6.3.5)
are found by solving the Lie equations
dt
= τ (t, x, u),
da
dx
= ξ(t, x, u),
da
du
= η(t, x, u),
da
(6.3.6)
with the initial conditions:
¯
t ¯a=0 = t,
¯
x ¯a=0 = x,
¯
u ¯a=0 = u.
I will begin by illustrating the basic methods by the linear heat equation
ut − uxx = 0.
Then we will consider the following nonlinear equation, known as the Burgers equation:
ut = uxx + uux .
Any symmetry transformation of a differential equation carries over any
solution of the differential equation into its solution. It means that, just
like in the case of ordinary differential equations, the solutions of a partial
differential equation are permuted among themselves under the action of a
symmetry group. The solutions may also be individually unaltered, then
they are called invariant solutions. Accordingly, group analysis provides
two basic ways for construction of exact solutions: group transformations
of known solutions and construction of invariant solutions.
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
6.3.2
171
Group transformations of solutions
The method is based on the fact that a symmetry group transforms any
solutions of the equation in question into solution of the same equation.
Namely, let (6.3.2) be a symmetry transformation group of the equation
(6.3.1), and let a function
u = Φ(t, x)
solve the equation (6.3.1). Since (6.3.2) is a symmetry transformation, the
above solution can be also written in the new variables:
ū = Φ(t̄, x̄).
Replacing here ū, t̄, x̄ from (6.3.2), we get
¡
¢
h(t, x, u, a) = Φ f (t, x, u, a), g(t, x, u, a) .
(6.3.7)
Having solved equation (6.3.7) with respect to u, one obtains a one-parameter
family (with the parameter a) of new solutions to the equation (6.3.1).
In this and the next sections, the illustrations are given for the linear
heat equation. A nonlinear partial differential equation, namely the Burgers
equations, will be discussed in Section 6.3.4.
The heat equation
ut − uxx = 0
(6.3.8)
is invariant, e.g. under the heat representation of the Galilean transformation:
2
t = t, x = x + 2at, u = u e−(ax+a t) .
(6.3.9)
The generator of this one-parameter group has the form
X = 2t
∂
∂
− xu ·
∂x
∂u
(6.3.10)
Any solution
u = Φ(t, x)
of the heat equation can be converted into a new solution by the transformation (6.3.9). Since the heat equation is invariant under this transformation,
we write it in the form ut̄ − ux̄x̄ = 0 using the variables t, x, u, and take the
solution in the same variables:
u = Φ(t, x)
Then we substitute here the expressions (6.3.9) for t, x, u, and upon solving
the resulting equation
ue−(ax+a
2 t)
= Φ(t, x + 2at)
172
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
for u, we obtains the new solution
2
u = eax+a t Φ(t, x + 2at)
(6.3.11)
involving the parameter a.
Exercise 6.3.1. Obtain the solutions of the heat equation by the transformation (6.3.9) and by the formula (6.3.11) applied to the following two
simple solutions:
(i) u = 1,
(ii) u = x.
Solution. (i) Inserting u = 1 into (6.3.9), i.e. letting
u e−(ax+a
2 t)
= 1,
one obtains the following new solution (6.3.11):
2
u = eax+a t .
The same result is obtained from (6.3.11) by letting Φ(t, x + 2at) = 1.
(ii) Inserting u = x into (6.3.9), i.e. letting
u e−(ax+a
2 t)
= x + 2at,
one obtains the following new solution (6.3.11):
2
u = (x + 2at) eax+a t .
The same result is obtained from (6.3.11) by letting Φ(t, x + 2at) = x + 2at.
6.3.3
Invariant solutions
If a group transformation maps a solution into itself, we arrive at what is
called a self-similar or group invariant solution. Consider, e.g. evolutionary equations (6.3.1). Given an infinitesimal symmetry (6.3.5) of equation
(6.3.1), the invariant solutions under the one-parameter group generated
by X are obtained as follows. One calculates two independent invariants
J1 = λ(t, x) and J2 = µ(t, x, u) by solving the equation
X(J) ≡ τ (t, x, u)
∂J
∂J
∂J
+ ξ(t, x, u)
+ η(t, x, u)
= 0,
∂t
∂x
∂u
173
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
or its characteristic system:
dx
du
dt
=
=
·
τ (t, x, u)
ξ(t, x, u)
η(t, x, u)
(6.3.12)
Then one designates one of the invariants as a function of the other, e.g.
µ = φ(λ),
(6.3.13)
and solves equation (6.3.13) with respect to u. Finally, one substitutes the
expression for u in equation (6.3.1) and obtains an ordinary differential
equation for the unknown function φ(λ) of one variable. This procedure
reduces the number of independent variables by one.
Exercise 6.3.2. Find the invariant solution of the heat equation (6.3.8)
under the group (6.3.9) with the generator (6.3.10):
X = 2t
∂
∂
− xu ·
∂x
∂u
Solution. There are two independent invariants for X. One of them is t,
while the other is obtained from the characteristic equation
dx du
+
= 0,
2t
xu
or
xdx du
+
= 0.
2t
u
Integration of the latter equation yields the invariant
x2
J = u e 4t .
Consequently, one seeks the invariant solution in the form J = φ(t), or
x2
u = φ(t) e− 4t .
Now substitute this expression into the heat equation. We have:
µ
µ 2
¶
¶
2
2
2
x
x
x2
1
0
− x4t
− x4t
− x4t
ut = φ + 2 φ e , ux = − φ e , uxx =
,
−
φ
e
4t
2t
4t2 2t
and the second-order partial differential equation ut − uxx = 0 reduces to
the first-order ordinary differential equation:
dφ
φ
+
= 0.
dt
2t
√
It follows that φ(t) = C/ t, C = const. Hence the invariant solution is
x2
C
u = √ e− 4t .
t
174
6.3.4
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
The Burgers equation
The Burgers equation
ut = uxx + uux
(6.3.14)
has five linearly independent infinitesimal symmetries:
X1 =
X4 = 2t
∂
,
∂t
X2 =
∂
∂
∂
+x
−u ,
∂t
∂x
∂u
∂
,
∂x
X5 = t 2
X3 = t
∂
∂
−
,
∂x ∂u
∂
∂
∂
+ tx
− (x + tu) ·
∂t
∂x
∂u
(6.3.15)
Consider, e.g. the generator X5 from (6.3.15). The Lie equations have
the form
dx̄
dū
dt̄
= t̄ 2 ,
= t̄ x̄,
= −(x̄ + t̄ ū).
da
da
da
Integration of these equations yields:
t=
t
,
1 − at
x=
x
1 − at
(6.3.16)
and
u = u(1 − at) − ax.
(6.3.17)
Substituting the transformations (6.3.16)-(6.3.17) in equation (6.3.7),
one maps any known solution u = Φ(t, x) of the Burgers equation to the
following one-parameter set of new solutions:
³ t
ax
1
x ´
u=
.
(6.3.18)
+
Φ
,
1 − at 1 − at
1 − at 1 − at
Example 6.3.1. One can obtain many examples, by choosing as an initial
solution u = Φ(t, x), any invariant solution. Let us take, e.g. the invariant
solution under the space translation generated by X2 from (6.3.15). In this
case the invariants are λ = t and µ = t, and equation (6.3.13) is written
u = φ(t). Substitution in the Burgers equation yields the obvious constant
solution u = k. It is mapped by (6.3.18) into the following one-parameter
set of solutions:
k + ax
u=
·
1 − at
Example 6.3.2. One of the physically significant types of solutions is obtained by assuming the invariance under the time translation group generated by X1 . This assumption provides the stationary solution
u = Φ(x)
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
175
for which the Burgers equation yields
Φ00 + ΦΦ0 = 0.
(6.3.19)
Integrate it once:
Φ2
= C1 ,
2
and integrate again be setting C1 = 0, C1 = ν 2 > 0, C1 = −ω 2 < 0 to
obtain:
2
Φ(x) =
,
x+C
³
ν ´
Φ(x) = ν th C + x ,
(6.3.20)
2
³
ω ´
Φ(x) = ω tg C − x .
2
The Galilean transformation t = t, x = x + at, u = u − a generated by X3
maps the stationary solutions (6.3.20) to travelling waves u = u(x − ct).
Φ0 +
Example 6.3.3. If one applies the transformation (6.3.18) to the stationary solutions (6.3.20), one obtains the following new non-stationary solutions:
2
ax
+
,
u=
1 − at x + C(1 − at)
·
µ
¶¸
νx
1
ax + ν th C +
,
(6.3.21)
u=
1 − at
2(1 − at)
·
µ
¶¸
ωx
1
u=
ax + ω tg C −
.
1 − at
2(1 − at)
Example 6.3.4. Let us find the invariant solutions under the projective
group generated by X5 . The characteristic system
dx
dt
du
=
=−
2
t
tx
x + tu
provides the invariants λ = x/t and µ = x+tu. Hence, the general expression
(6.3.13) for invariant solution takes the form
x 1
u = − + Φ(λ),
t
t
λ=
x
·
t
(6.3.22)
Substituting this expression in the Burgers equation (6.3.14), one obtains for
Φ(λ) precisely the equation (6.3.19). Hence, its general solution is obtained
from (6.3.20) where x is replaced by λ. The corresponding invariant solutions
176
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
are obtained by substituting in (6.3.22) the resulting expressions for Φ(λ).
For example, using for Φ(λ) the second formula (6.3.20) by letting there
ν = π, one obtains the solution
x π ³
πx ´
u = − + th C +
.
(6.3.23)
t
t
2t
It is important in non-linear acoustics and was derived by R.I. Khokhlov in
1961 by physical reasoning.
Example 6.3.5. The invariant solutions under the group of dilations generated by X4 lead to what is called in physics similarity solutions because of
their connection with the dimensional analysis. The characteristic system
dt
dx
du
=
=−
2t
x
u
√
√
provides the following invariants: λ = x/ t, µ = t u. Consequently, one
seeks the invariant solutions in the form
1
u = √ Φ(λ),
t
x
λ= √ ,
t
and arrives at the following equation for the similarity solutions of the Burgers equation:
1
(6.3.24)
Φ00 + ΦΦ0 + (λ Φ0 + Φ) = 0.
2
Integrating once, one has:
1
Φ0 + (Φ2 + λ Φ) = C,
2
Letting C = 0, one obtains the solution (found in physics by O.V. Rudenko)
2
2
e−x /(4t)
√ ,
u= √
πt B + erf(x/(2 t)
where B is and arbitrary constant and
2
erf(z) = √
π
is the error function.
Z
z
2
e−s ds
0
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
6.3.5
177
A nonlinear boundary-value problem
Consider the following nonlinear equation
∆u = eu ,
(6.3.25)
where u = u(x, y) and ∆u = uxx +uyy is the Laplacian with two independent
variables. Equation (6.3.25) admits the operator
X=ξ
∂
∂
∂
+η
− 2ξx
,
∂x
∂y
∂u
(6.3.26)
where ξ(x, y) and η(x, y) are arbitrary solutions of the Cauchy-Riemann
system
ξx − ηy = 0, ξy + ηx = 0.
(6.3.27)
Consequently, one can express the general solution of the nonlinear equation
(6.3.25) via the solution of the Laplace equation
∆v = 0
in the form
vx2 + vy2
u = ln 2
v2
µ
(6.3.28)
¶
.
(6.3.29)
In other words, the nonlinear equation (6.3.25) is mapped to the linear
equation (6.3.28) by the transformation (6.3.29).
We will consider here the following boundary-value problem in the circle
of radius r = 1 :
¯
¯
∆u = eu , u¯
= 0,
(6.3.30)
r=1
p
where r = x2 + y 2 . The general solution (6.3.29) is not suitable for solution of our problem since it leads to the nonlinear boundary-value problem
1 ¯¯
= 0.
∆v = 0, (vx2 + vy2 − v 2 )¯
2
r=1
In order to solve the problem (6.3.30) it is convenient to use the polar
coordinates
x = r cos ϕ, y = r sin ϕ.
(6.3.31)
In these coordinates equation (6.3.25) is written
1
1
urr + ur + 2 uϕϕ = eu .
r
r
(6.3.32)
178
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
Since the differential equation and the
boundary condition in the problem (6.3.30)
are invariant with respect to the rotation
group, we will seek the solution depending only on the variable r. Then equation
(6.3.31) is written
1
urr + ur = eu .
r
(6.3.33)
Figure 6.1: Solution
(6.3.40)
√
2
We will assume that the function u(r) is with c = 5 − 2 6 is bounded.
bounded at the “singular” point r = 0 and
formulate the boundary conditions of the problem (6.3.30) in the following
form:
u(1) = 0, u(0) < ∞.
(6.3.34)
One can integrate equation (6.3.33) by means of Lie’s method. Namely,
it has two infinitesimal symmetries
X1 = r
∂
∂
−2 ,
∂r
∂u
X2 = r ln r
∂
− 2(1 + ln r).
∂r
(6.3.35)
We reduce the first operator to the form
X1 = ∂/∂t by the change of variables
t = ln r,
z = u + 2 ln r.
Equation (6.3.33) is written in these
variables as follows
d2 z
= ez .
dt2
(6.3.36)
Integration by means of the standard
substitution dz/dt = p(v) yields
Z
dz
√
= t + C2 .
(6.3.37)
C1 + 2ez
Evaluating the integral in (6.3.36) one
can verify that the condition u(0) < ∞
is not satisfied if C1 ≤ 0. Therefore we
√
calculate the integral for C1 > 0. For the
Figure 6.2:
c2 = 5 + 2 6,
2
sake of convenience we set C1 = λ and
u(0) = u(r0 )√ = u0 ≈ 4.37;
C2 = ln C. Then we evaluate the inter∗ = 1/c, r0 = 2/c.
gral, rewrite the result in the old variables and obtain:
179
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
u = ln
It follows that
2λ2 (cr)λ
·
r2 [1 − (cr)λ ]2
u ≈ (λ − 2) ln r
(6.3.38)
(r → 0).
Therefore the condition u(0) < ∞ entails that λ = 2. Furthermore, the
boundary condition u(1) = 0 takes the form
8c2 = (1 − c2 )2 ,
whence
√
c2 = 5 ± 2 6.
(6.3.39)
Hence, the problem (6.3.30) has two solutions
u = ln(8c2 ) − ln(1 − c2 r2 )2
with
√
c2 = 5 − 2 6
and
√
c2 = 5 + 2 6,
(6.3.40)
√
respectively. The first solution corresponding to the case c2 = 5 − 2 6 is
bounded everywhere in the circle
x2 + y 2 ≤ 1
√
whereas the second solution corresponding to c2 = 5 + 2 6 is unbounded
on the circle
x2 + y 2 = r∗2
with the radius r∗ = 1/c ≈ 0.33 (see Fig.6.1 and Fig.6.2).
6.3.6
Invariant solutions for an irrigation system
Consider the nonlinear partial differential equation (2.3.36) modelling soil
water motion in an irrigation system.
The infinitesimal symmetries of equation (2.3.36) with arbitrary coefficients form a Lie algebra called the principal Lie algebra LP for equation
(2.3.36). It is a three-dimensional algebra spanned by
X1 =
∂
,
∂t
X2 =
∂
,
∂x
X3 =
∂
·
∂z
180
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
There are 29 particular types of the coefficients C(ψ), K(ψ), S(ψ) when an
extension of the algebra LP occurs. Let us consider here one of cases when
LP extends by three operators. Namely, consider the equation (M = const.)
4
M e4ψ
−1
¡
¢
¡
¢
ψt = e−4ψ ψx x + e−4ψ ψz z + 4e−4ψ ψz + M − e−4ψ .
(6.3.41)
Equation (6.3.41) admits a six-dimensional Lie algebra L6 obtained by
adding to the basis X1 , X2 , X3 of LP the following three operators:
∂
1
∂
− (M e4ψ − 1)
,
∂t 4
∂ψ
∂
∂
1
∂
X5 = sin x e−z
− cos x e−z
+ cos x e−z (M e4ψ − 1)
,
∂x
∂z 2
∂ψ
∂
∂
1
∂
X6 = cos x e−z
+ sin x e−z
− sin x e−z (M e4ψ − 1)
·
∂x
∂z 2
∂ψ
X4 = t
Let us find the invariant solutions based on the two-dimensional subalgebra
L2 ⊂ L6 spanned by X4 , X5 . Invariants J(t, x, z, ψ) of L2 are defined by the
system of linear partial differential equations
X4 (J) = 0,
X5 (J) = 0.
(6.3.42)
Solution of this system yields the basis of invariants:
¡
¢
v = te2z e−4ψ − M , λ = ez sin x.
The invariant solutions have the form
¡
¢
te2z e−4ψ − M = Φ(λ),
whence, upon solving for ψ:
¯
e−2z
1 ¯¯
¯
ψ = − ln ¯M +
Φ(λ)¯.
4
t
Substitution into equation (6.3.41) yields
Φ00 (λ) = 4,
whence
Φ(λ) = 2λ2 + l1 λ + l2 .
Thus, the invariant solution is given by
´¯
1 ¯¯
e−2z ³ 2z 2
¯
z
ψ = − ln ¯M +
2e sin x+l1 e sin x+l2 ¯,
4
t
where M, l1 , l2 = const.
l1 , l2 = const. (6.3.43)
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
181
Figure 6.3: Plot of the solution (6.3.43), M = 4, l1 = −2, l2 = −4, t = 0, 01.
Compare the plot of the solution
(6.3.43) given in Figure 6.3 with a
real irrigation system in the 1950s.
Line source drip systems produce
a continuous wetted band along
the y-axis, and the phenomenon
actually involves all three space
coordinates, x, y, and z.
Figure 6.4: An irrigation system.
182
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
6.3.7
Invariant solutions for a tumour growth model
Consider the tumour growth model (2.5.3):
ut = f (u) − (ucx )x ,
ct = −g(c, u),
(6.3.44)
where f (u) and g(c, u) satisfy the conditions
f (u) > 0,
gc (c, u) > 0,
gu (c, u) > 0.
(6.3.45)
If f (u) and g(c, u) are arbitrary functions, the system (6.3.44) is invariant
only under the translations in t and x. In other words, it admits only the
two-dimensional Lie algebra spanned by
X1 =
∂
,
∂t
X2 =
∂
·
∂x
(6.3.46)
There are, however, many particular functions f (u) and g(c, u) when the
system (6.3.44) has more symmetries3 . For example, if
f = αu,
g = G(ue−c ),
where α is an arbitrary constant and G an arbitrary function, the corresponding system
ut = αu − (ucx )x ,
ct = −G(ue−c )
has, along with X1 , X2 , the additional symmetry
X3 =
∂
∂
+u ·
∂c
∂u
Let us take, e.g. G(ue−c ) = ue−c and consider the system
ut = αu − (ucx )x ,
ct = −ue−c ,
α = const.
(6.3.47)
Let us find the invariant solutions under the one-parameter group generated
by the operator
∂
∂
∂
X1 + X 3 =
+
+u ·
∂t ∂c
∂u
The equiation (X1 + X3 )J = 0 gives three independent invariants:
x,
ψ1 = c − t,
ψ2 = ue−t .
3
N.H. Ibragimov and N. Säfström, Communications in Nonlinear Science and Numerical Simulation, vol. 9(1), 61-68, 2004.
183
6.3. NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
The corresponding invariant solutions are defined by
c = t + ψ1 (x),
u = et ψ2 (x).
ut = et ψ2 (x),
ux = et ψ20 (x),
(6.3.48)
We have:
ct = 1,
cx = ψ10 (x),
cxx = ψ100 (x).
(6.3.49)
Substitution (6.3.48) and (6.3.49) in the first equation (6.3.47) yields:
et ψ2 (x) = αet ψ2 (x) − et ψ10 (x)ψ20 (x) − et ψ2 (x)ψ100 (x),
or
(1 − α)ψ2 (x) + ψ10 (x)ψ20 (x) + ψ2 (x)ψ100 (x) = 0.
(6.3.50)
The second equation (6.3.47) yields
1 = −ψ2 (x)e−ψ1 (x)
whence
ψ2 (x) = −eψ1 (x) .
(6.3.51)
Now Equation (6.3.50) becomes
ψ100 (x) + ψ102 + (1 − α) = 0.
It follows that
ψ1 (x) = ln | A2 (x + A1 )|
for α = 1,
¯ ³
´¯
√
√
¯
¯
ψ1 (x) = x α − 1 + ln ¯A2 1 ± e2 α−1 (A1 −x) ¯
¯
¡√
¢¯
ψ1 (x) = ln ¯A2 cos 1 − α (A1 − x) ¯
for α > 1,
for α < 1,
(6.3.52)
(6.3.53)
(6.3.54)
where A1 and A2 are arbitrary constants.
Substituting (6.3.51)–(6.3.54) in (6.3.48), we obtains the following three
different invariant solutions:
c(t, x) = t + ln | A2 (x + A1 )| ,
u(t, x) = −et | A2 (x + A1 )| ,
¯ ³
´¯
√
√
¯
¯
c(t, x) = t + x α − 1 + ln ¯A2 1 ± e2 α−1 (A1 −x) ¯,
¯ ³
´¯
√
√
¯
¯
u(t, x) = −et+x α−1 ¯A2 1 ± e2 α−1 (A1 −x) ¯,
(6.3.55)
(6.3.56)
184
6. SYMMETRY OF DIFFERENTIAL EQUATIONS
and
¯
¡√
¢¯
c(t, x) = t + ln ¯A2 cos 1 − α (A1 − x) ¯ ,
¯
¡√
¢¯
u(t, x) = −et ¯A2 cos 1 − α (A1 − x) ¯ ,
(6.3.57)
for the system (6.3.47) with α = 1, α > 1 and α < 1, respectively.
The conditions (6.3.45) select, however, the solution (6.3.57) with α < 0.
Indeed, only in this case the functions f (u) = αu and g(c, u) = ue−c satisfy
the conditions (6.3.45):
f (u) = αu > 0,
gc (c, u) = −ue−c > 0,
gu (c, u) = e−c > 0
and hence the solution (6.3.57) with α < 0 is relevant to the model (6.3.47).