Equivalent Lagrangians and
Transformation Maps for Differential
Equations
Nicole Wilson
Supervisor: Prof A.H. Kara
A dissertation submitted to the Faculty of Science, University of the
Witwatersrand, in fulfilment of the requirements for the degree of
Master of Science.
Johannesburg, 2012.
DECLARATION
I declare that the contents of this dissertation are original except where due references
have been made. It is being submitted for the degree of Master of Science at the
University of the Witwatersrand in Johannesburg. It has not been submitted before
for any degree to any other institution.
Nicole Wilson
This day of , 2012, at .
1
Acknowledgements
The financial assistance of the National Research Foundation (NRF) towards this
research is hereby acknowledged. Opinions expressed and conclusions arrived at, are
those of the author and are not necessarily to be attributed to the NRF. This work
has also been made possible in part by the University of the Witwatersrand, as well
as the Ernst & Ethel Eriksen Trust.
I gratefully acknowledge the support of my supervisor, Prof A. H. Kara, without
whom this dissertation would not have been possible. I am also grateful to my
father, Anthony Wilson, for his assistance in the proof-reading of this document, as
well as to my mother, Helen Wilson, for her continued support.
2
Abstract
The Method of Equivalent Lagrangians is used to find the solutions of a given
differential equation by exploiting the possible existence of an isomorphic Lie
point symmetry algebra and, more particularly, an isomorphic Noether point
symmetry algebra. Applications include ordinary differential equations such
as the Kummer Equation and the Combined Gravity-Inertial-Rossby Wave
Equation and certain classes of partial differential equations related to the
(1 + 1) linear wave equation. We also make generalisations to the (2 + 1) and
(3 + 1) linear wave equations.
3
Contents
Introduction 7
1 Preliminaries 10
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3 Illustrative Example . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 The Kummer and Combined Gravity-Inertial-Rossby Wave ODEs 20
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2 The Kummer Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3 The Combined Gravity-Inertial-Rossby Wave
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4 Discussion and Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 27
4
3 The (1 + 1) Wave Equation and Related PDEs 29
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.3 Form of Equivalent Lagrangian for the Linear Wave Equation in
(1 + 1) Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.4 Finding Transformations -
Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.5 Main Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.6 Discussion and Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 42
4 Generalisation to PDEs in Three and Four Independent Variables 44
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.2 The Two-Dimensional Linear Wave Equation . . . . . . . . . . . . . . 45
4.2.1 Constructing an Equivalent Lagrangian . . . . . . . . . . . . . 46
4.2.2 Finding Transformations -
The Reverse Procedure . . . . . . . . . . . . . . . . . . . . . . 47
4.3 The Three-Dimensional Linear Wave
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.4 Discussion and Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 54
5
Conclusion 56
6
Introduction
Lie Theory was first formulated by Sophus Lie (1842 - 1899) as a means of unifying
the various integration techniques associated with the solving of differential equations,
[14]. This lead to the study of continuous groups of symmetries (transformations that
leave a given differential equation invariant, see [1]). However, although this field has
its origins in differential equations, Lie Theory’s astounding applicability to fields
such as differential geometry lead to its rapid “abstraction and globalization,”[14],
in particular by Schreier in 1925, who discovered the importance of the topology of
Lie groups, a concern that was taken much further by E´lie Cartan in the 1930’s, [16].
Hence the applicability of Lie Theory to differential equations was largely overlooked
for decades until the subject was resurrected by mathematicians such as Birkhoff
and Ovsiannikov, [14]. Since then, this aspect of Lie Theory has grown considerably
and indeed it is in this arena that this study will make its contribution.
The Method of Equivalent Lagrangians is used to find the solutions of a given
differential equation by exploiting the possible existence of an isomorphic Lie point
symmetry algebra and, more particularly, an isomorphic Noether point symmetry
algebra. The underlying idea of the method is to construct a regular point transformation
which maps the Lagrangian of a ‘simpler’ differential equation (with known solutions)
to the Lagrangian of the differential equation in question. Once determined, this
7
point transformation will then provide a way of mapping the solutions of the simpler
differential equation to the solutions of the equation which we seek to solve. This
transformation can also be used to find conserved quantities for the equation in
question, if the conserved quantities for the simpler differential equation are known.
The advantage of this method is that once a transformation map is found, it can be
used to find all possible solutions and conserved quantities of the differential equation
in question, thereby avoiding the complex integration procedures which are normally
required. In this study, the Method of Equivalent Lagrangians is described for scalar
second-order linear ordinary differential equations (ODEs) and for certain classes of
partial differential equations (PDEs) in two, three and four independent variables.
Previous investigations of this method have been limited. In [6], Kara and Mahomed
apply the method to two cases of the equation of the form
q¨ + p(t)q˙ + r(t)q = µq˙2q−1 + f(t)qn.
This study builds on their work through the generalisation of the method to a
larger class of scalar second-order ODEs, by applying the method to two new ODEs.
Another new contribution of this study is the extension of the method to certain
classes of PDEs in two, three and four independent variables. In [5], Kara uses
the definition of equivalent Lagrangians for PDEs in two independent variables in
order to find an equivalent Lagrangian under a given transformation. In this study
we will describe this procedure in detail as it is the “reverse” of the Method of
Equivalent Lagrangians. Following this, we will apply the method to the same
equivalent Lagrangians in order to recover the known transformations. We shall
also generalise the method to some classes of PDEs in three and four independent
variables.
In all cases, the aim is to demonstrate the method in a comprehensive manner,
which naturally comes at the price of generality. However, the design and layout
8
of the method is such that it may be easily imitated for other classes of differential
equations.
The structure of the dissertation is as follows. In the first chapter we state the
fundamental concepts that are required to describe and apply the Method of Equivalent
Lagrangians. We also provide an illustrative example of the method and its application
in finding solutions and conserved quantities of differential equations. The second
chapter deals with the application of the method to two linear second-order ODEs
in mathematical physics, namely the Kummer Equation and the Combined Gravity-
Inertial-Rossby Wave Equation. In the third chapter we extend the method to
PDEs in two independent variables. Using the standard Lagrangian and previous
knowledge of the (1 + 1) wave equation, we are able to apply the method to certain
PDEs such as the canonical form of the wave equation, the wave equation with
dissipation and a Klein-Gordon equation. The fourth chapter discusses the extension
of the method to certain classes of PDEs in three and four independent variables,
using previous knowledge of the standard Lagrangian for the linear wave equations
in three and four independent variables, respectively.
9
Chapter 1
Preliminaries
1.1 Introduction
In this chapter we look at the theoretical foundations on which the Method of
Equivalent Lagrangians is based. The definitions and concepts stated here are
essential to an understanding of this study and will be used throughout the dissertation.
They are all well known and can be found in [14].
1.2 Fundamentals
Consider an rth-order system of PDEs of n independent variables x = (x1, x2, . . . , xn)
and m dependent variables u = (u1, u2, . . . , um)
Gµ(x, u, u(1), . . . , u(r)) = 0, µ = 1, . . . , m˜, (1.1)
10
where u(1), u(2), . . . , u(r) denote the collections of all first, second, . . . , rth-order
partial derivatives. The total differentiation operator with respect to xi is given by
Di =
∂
∂xi
+ uαi
∂
∂uα
+ uαij
∂
∂uαj
+ · · · , i = 1, . . . , n. (1.2)
A current T = (T 1, . . . , T n) is conserved if it satisfies
DiT
i = 0 (1.3)
along the solutions of (1.1). T is also known as a conserved quantity.
The Euler-Lagrange operator is defined by
δ
δuα
=
∂
∂uα
+
∑
s≥1
(−1)sDi1 , . . . , Dis
∂
∂uαi1···is
, α = 1, . . . ,m. (1.4)
Hence, the Euler-Lagrange equations are of the form
δL
δuα
= 0, α = 1, . . . ,m, (1.5)
where L is a Lagrangian of some order. The solutions of equation (1.5) are the
optimizers of the functional
∫
L(x, u, u(1), . . .) dx. (1.6)
A vector field X of the form
X = ξi
∂
∂xi
+ ηα
∂
∂uα
, ξi, ηα ∈ A, (1.7)
which leaves (1.6) invariant (up to, possibly, a gauge vector) is known as a Noether
symmetry, whereA is the space of differential functions. Equivalently, X is a Noether
symmetry of L if there is a vector B = (B1, . . . , Bn) ∈ A such that
X(L) + LDi(ξ
i) = Di(B
i), (1.8)
11
where X is prolonged to the degree of L, [14]. If the vector B is identically zero,
then X is a strict Noether symmetry, [3].
Two linear algebras (e.g. Lie symmetry or Noether symmetry algebras) are said to
be isomorphic if they each possess the same structure constants in their respective
bases, [2].
It is well known that if the Noether symmetry algebras for two Lagrangians, L and
L, are isomorphic, the Lagrangians can be mapped from one to the other. In the
light of this, we define the notion of equivalent Lagrangians.
Definition 1. Let x = (x1, x2, . . . , xn) be a vector of n independent variables and
u = (u1, u2, . . . , um) a vector of m dependent variables, where u(1), u(2), . . . , u(r)
denote the collections of all first, second, . . . , rth-order partial derivatives of u. Let
X and U be similarly defined. Two Lagrangians,
L = L(x, u, u(1), . . . , u(r))
and
L = L(X,U, U(1), . . . , U(r)),
are said to be equivalent up to guage if and only if there exists a transformation,
X = X(x, u) and U = U(x, u), such that
L(x, u, u(1), . . .) = L(X,U, U(1), . . .) J(x, u, u(1)) + div f˜ , (1.9)
where J is the Jacobian, [5], and f˜ = (f1, f2, . . . , fn) is a guage vector such that
fi = fi(x, u) for all i = 1, 2, . . . , n.
For ordinary differential equations, in which u = u(x), i.e. x is scalar, the definition
of equivalence up to gauge becomes
12
Definition 1*. Two Lagrangians, L and L, are said to be equivalent up to guage if
and only if there exists a transformation, X = X(x, u) and U = U(x, u), such that
L(x, u, u′) = L(X,U, U ′)
dU
dx
+ fx + u
′fu, (1.10)
where the guage function, f , is an arbitrary function of x and u, [6].
For functions of a single variable, we shall use the prime notation to denote total
differentiation. For example, if u = u(x), then dudx = u
′. Similarly, d
2u
dx2 = u
′′, and so
on. For functions of more than one variable, partial differentiation will be denoted
by subscript notation. For example, if u = u(x, t), then ∂u∂x = ux and
∂u
∂t = ut.
Similarly, ∂
2u
∂x2 = uxx, and so on.
Remark: The definitions of equivalent Lagrangians imply that given a variational
differential equation with corresponding Lagragian L, we can find a regular point
transformation X = X(x, u) and U = U(x, u) which maps another (equivalent)
Lagrangian L to L. This regular point transformation also maps the solutions of
the differential equation associated with L to the solutions of the original differential
equation.
In addition to this, once we have found the regular point transformationX = X(x, u)
and U = U(x, u) mentioned above, it is possible to use this transformation to map
the (known) conserved quantities of the differential equation associated with L to
the conserved quantities of the equation in question. This is demonstrated in the
following section by means of an illustrative example.
13
1.3 Illustrative Example
Consider the well-known harmonic oscillator ODE
y′′ + y = 0, (1.11)
with Lagrangian
L =
1
2
y′ 2 −
1
2
y2. (1.12)
We wish to solve and find conserved quantities for the ODE (1.11) using the Method
of Equivalent Lagrangians. In order to do so, we must find the regular point
transformation X = X(x, y) and Y = Y (x, y) that maps the Lagrangian
L =
1
2
Y ′ 2, (1.13)
associated with the free particle ODE,
Y ′′ = 0, (1.14)
to the Lagrangian (1.12) associated with the ODE (1.11). This can be done because
equations (1.11) and (1.14) are equivalent to each other, and hence can be transformed
from one to the other by means of invertible maps, [10]. The corollory of this is
that these equations have isomorphic eight-dimensional Lie algebras ([8], [9]) and
isomorphic five-dimensional Noether algebras, [11]. Hence L and L are equivalent
Lagrangians and therefore we can apply Definition 1* in order to find the required
transformation map. Once the transformation map has been found, we will be able
use it to transform the known solutions and conserved quantities of equation (1.14)
to those of equation (1.11).
From Definition 1* we have that
1
2
y′ 2 −
1
2
y2 =
1
2
Y ′ 2
dX
dx
+ fx + y
′fy. (1.15)
14
Now, by the chain rule,
Y ′ =
dY
dx
dX
dx
,
and by the total differentiation operator (see (1.2)),
dY
dx
=
∂Y
∂x
+ y′
∂Y
∂y
.
Similarly, dXdx =
∂X
∂x + y
′ ∂X
∂y . Hence equation (1.15) becomes
1
2
y′ 2 −
1
2
y2 =
1
2
( dY
dx
dX
dx
)2
dX
dx
+
df
dx
=
1
2
(
dY
dx
)2
dX
dx
+
df
dx
=
1
2
(Yx + y′Yy)2
(Xx + y′Xy)
+ fx + y
′fy.
For the sake of simplicity we shall now assume that X is a function of x only (i.e.
X = X(x)). Hence Xy = 0. From this, the above equation simplifies to
1
2
y′ 2 −
1
2
y2 =
(Yx + y′Yy)2
2Xx
+ fx + y
′fy
=
1
2
Y 2x
Xx
+
YxYy
Xx
y′ +
1
2
Y 2y
Xx
y′ 2 + fx + y
′fy.
Since X and Y are dependent on x and y only, and not on their derivatives, it
follows that we can separate the above equation by powers of y′. From this we get
the following system of three PDEs.
y′2 :
1
2
=
1
2
Y 2y
Xx
,
y′ : 0 =
YxYy
Xx
+ fy,
1 : −
1
2
y′ 2 =
1
2
Y 2x
Xx
+ fx.
15
From the first equation we have that
Y 2y = Xx.
Since we have already assumed that X = X(x), we can now make the additional
assumption that X is of the form
X =
∫
a2 dx,
where a is a function of x. Hence
Y 2y = a
2(x),
which implies that
Yy = a(x),
and so
Y =
∫
a(x) dy,
from which it follows that
Y = a(x)y + b(x),
where we can assume, without loss of generality, that b(x) = 0. Differentiating Y
partially with respect to x, we obtain the expression
Yx = a
′y.
Solving for fy in the second equation in the system above, and then substituting
our newly-found expressions for Xx, Yx and Yy, we deduce that
fy = −
a′
a
y.
Integrating the above equation with respect to y gives us the expression
f = −
1
2
a′
a
y2 + c(x),
16
where, without loss of generality, we can take c(x) = 0. Now that we have an
expression for f , we can differentiate this partially with respect to x and substitute
fx into the third equation in our system. From this it follows that
−
1
2
y2 =
1
2
(a′y)2
a
+
1
2
(
a′
a
)′
y2. (1.16)
Equating the coefficients of y2, we arrive at the equation
−
1
2
=
1
2
a′ 2
a
+
1
2
(
a′
a
)′
. (1.17)
Now we can make the substitution A = a
′
a , and multiply through by 2, thus
simplifying the above equation to
−1 = A2 − A′. (1.18)
Rewriting this equation in the form
dA
dx
= 1 + A2, (1.19)
we realise that it is a separable ODE, the solution of which is
arctan(A) = x+ k,
where k is an arbitrary constant which we can assume, for our purposes, to be equal
to 0. Hence,
A =
a′
a
= tanx.
Integrating both sides of this equation yields the result
ln a = ln (sec x) + ln d,
where d is an arbitrary constant. From this we arrive at the solution
a = d secx. (1.20)
17
Thus, the transformation in question is given by the equations
X = d2 tanx, Y = dy secx, (1.21)
and the guage function is given by
f = −
1
2
y2 tanx.
This transformation in turn maps the solutions of (1.14) to the solutions of (1.11).
The solutions of (1.14) are easy to obtain by integrating both sides of the equation
twice. The general solution is therefore of the form
Y = αX + β, (1.22)
where α and β are arbitrary constants. By substituting our transformation X =
X(x, y) and Y = Y (x, y), given by (1.21), into equation (1.22) and solving for y, we
obtain the expression
y = α sinx+ β cosx, (1.23)
which is the general solution for equation (1.11). In order to verify this, we note
that
y′ = α cosx− β sinx,
which implies that
y′′ = −α sinx− β cosx.
Hence
y′′ + y = −α sinx− β cosx+ α sinx+ β cosx = 0, (1.24)
and so equation (1.11) is satisfied.
In addition to finding solutions, we can use our transformation to find the conserved
quantities of (1.11), by applying our transformation map to the conserved quantities
of equation (1.14).
18
Consider, for example, the known conserved quantity
I = XY ′ − Y (1.25)
of equation (1.14). Using transformation X = X(x, y) and Y = Y (x, y), given by
(1.21), it follows that a conserved quantity for equation (1.11) is
I = X
(
dY
dX
)
− Y
= X
( dY
dx
dX
dx
)
− Y
=
tanx secx(y′ + y tanx)
sec2 x
− y secx
= sin x(y′ + y tanx)− y secx
= y′ sinx+ y sinx tanx− y secx.
This is verified by the fact that
dI
dx
= y′′ sinx+ y sinx
= sin x(y′′ + y)
= 0.
Hence we have demonstrated how to find transformation maps using the definition of
equivalent Lagrangians. Furthermore, once the transformation has been found, we
have shown how it can be used to find solutions and conserved quantities of a given
differential equation. In the next chapter, the Method of Equivalent Lagrangians will
be applied to two scalar second-order linear ODEs, namely the Kummer Equation
and the Combined Gravity-Inertial-Rossby Wave Equation.
19
Chapter 2
The Kummer and Combined
Gravity-Inertial-Rossby Wave ODEs
2.1 Introduction
Second-order ODEs can be divided into equivalence classes based on their Lie
symmetries, [11]. Two equations belong to the same equivalence class if there exists
a diffeomorphism that transforms one of the equations to the other, [11]. If a second-
order ODE admits eight Lie symmetries (the maximum number of Lie symmetries
of a scalar second-order ODE, by Lie’s ‘Counting Theorem,’ [7]), it belongs to the
equivalence class of the equation Y ′′ = 0, [11]. Hence it can be mapped to this
equation by means of a regular point transformation. All linear second-order ODEs
belong to the equivalence class of the equation Y ′′ = 0 (see [10] and [15]). It is
an important result from the work of Lie ([8], [9]) that such equations possess the
maximum eight-dimensional Lie symmetry algebra, [10].
20
In [11], Mahomed et al. prove that the maximum dimension of the Noether symmetry
algebra for a scalar second-order ODE is five, and that the equation (1.14) with
standard Lagrangian (1.13) attains this maximum. This five-dimensional Noether
algebra is therefore isomorphic to any other five-dimensional Noether algebra for
which the associated Euler-Lagrange equation is of the same equivalence class as
(1.14), [11]. Therefore it follows that for any scalar second-order linear ODE with
Lagrangian, L, generating a five-dimensional Noether algebra, L can be mapped
to L by means of a regular point transformation X = X(x, y) and Y = Y (x, y)
(this transformation evidently also transforms the corresponding Euler-Lagrange
equations, for L and L respectively, from one to the other, [11]).
We use the Method of Equivalent Lagrangians detailed above to find the transformation
maps that can be used to find solutions and conserved quantities for two scalar
second-order linear ODEs, namely the Kummer Equation and the Combined Gravity-
Inertial-Rossby Wave Equation.
2.2 The Kummer Equation
The Kummer Equation has several applications in theoretical physics. Inter alia,
it models the velocity distribution of electrons in a high frequency gas discharge.
Using the solutions of this equation, which are called the Confluent Hypergeometric
Functions, [12], together with kinetic theory, it is thus possible to predict the high
frequency breakdown electric field for gases, [12]. The differential equation is given
by
xy′′ + (2k − x)y′ − ky = 0, (2.1)
21
where k is an arbitrary constant. By rearranging this equation and multiplying by
an integrating factor, we discover that a Lagrangian for this equation is
L =
1
2
x2ke−x(y′ +
k
x
y2). (2.2)
Equation (2.1) has eight Lie symmetries (this is because it is a second-order linear
ODE, see [10]). Therefore there is a point transformation, X = X(x, y) and Y =
Y (x, y), which maps equation (1.14), with Lagrangian (1.13), to equation (2.1). The
Lagrangian for equation (1.14) is known to have five Noether symmetries, [11]. From
the existence of the point transformation above, it follows that the Lagrangian for
the Kummer Equation (2.1), given by (2.2), also has five Noether symmetries, [10].
Therefore Lagrangians (2.2) and (1.13) are equivalent. Invoking Definition 1* and
substituting L and L into equation (1.10), we can find the point transformation
X = X(x, y) and Y = Y (x, y) that maps (1.13) to (2.2), and hence equation (1.14)
to equation (2.1).
Equation (1.10) gives us
1
2
x2ke−x(y′ 2 +
k
x
y2) =
1
2
Y ′ 2
dX
dx
+ fx + y
′fy
=
1
2
(
dY
dX
)2
dX
dx
+ fx + y
′fy
=
1
2
( dY
dx
dX
dx
)2
dX
dx
+ fx + y
′fy
=
1
2
(
Yx + 2YxYyy′ + y′ 2Y 2y
Xx + y′Xy
)
+ fx + y
′fy.
In order to simplify the above equation, we assume that X is a function of x only
and is of the form
X =
∫
a2 dx,
22
where a is a function of x. That is, Xy = 0, and the above equation becomes
1
2
x2ke−x(y′ 2 +
k
x
y2) =
1
2
(
Y 2x + 2YxYyy
′ + y′ 2Y 2y
a2
)
+ fx + y
′fy. (2.3)
Now, since X and Y are dependent on x and y only, and not on their derivatives,
we can separate equation (2.3) by powers of y′, after which we obtain a system of
three equations,
y′ 2 :
1
2
x2ke−x =
1
2
(
Y 2y
a2
)
,
y′ : 0 =
1
2
(2YxYy
a2
)
+ fy,
1 :
k
2x
x2ke−xy2 =
1
2
(
Y 2x
a2
)
+ fx.
From the first equation we determine that
Yy = ax
ke−
1
2x.
Integrating with respect to y results in the expression
Y = axke−
1
2xy + b(x), (2.4)
for which we can assume, without loss of generality, that b(x)=0. We can differentiate
equation (2.4) partially with respect to x, and substitute expressions for Yx and
Yy (given above) into the second equation in the system, in order to obtain the
expression
fy = −x
2k−1e−xy
(
a′
a
x+ k −
1
2
x
)
,
from which we conclude that
f = −
1
2
x2k−1e−xy2
(
a′
a
x+ k −
1
2
x
)
+ c(x), (2.5)
where we again assume, without loss of generality, that c(x)=0.
23
For the third equation, we can substitute our expression for Yx to obtain
xk =
(
a′
a
)
x2 + 2
(
a′
a
)
x(k −
1
2
x) + (k −
1
2
x)2 + fx.
Making the substitution
A =
a′
a
simplifies the above equation to
y2e−xx2k−1k = (A2x2 + 2Ax(k −
1
2
x) + (k −
1
2
x)2)x2ke−xy2 + fx.
We then differentiate (2.5) partially with respect to x, and obtain an expression for
fx, which we can substitute into the above equation. This simplifies to
0 = A2 − A′ −
k
x
(
k
x
− 1
)
−
1
4
. (2.6)
Integrating the equation A = a
′
a gives us the expression
a = Be
∫
A dx,
where A satisfies equation (2.6) and B is an arbitrary constant not equal to zero.
Hence we have that
X =
∫
B2e2
∫
A dx dx, (2.7)
and
Y = Be
∫
A dxxke−
1
2xy. (2.8)
Equations (2.7) and (2.8), whereA satisfies (2.6), define our regular point transformation
X = X(x, y) and Y = Y (x, y), which transforms equation (1.14) to equation (2.1).
We know that the general solution of equation (1.14) is given by
Y = αX + β,
24
where α and β are arbitrary constants. Therefore, we can substitute expressions (2.7)
and (2.8), for X and Y respectively, to obtain an expression for y which is the general
solution of the Kummer Equation (2.1). The regular point transformation found
above can also be used to find the conserved quantities of the Kummer Equation in
the same way in which we found the conserved quantities for the harmonic oscillator
equation in the previous section.
2.3 The Combined Gravity-Inertial-Rossby Wave
Equation
The Combined Gravity-Inertial-Rossby Wave Equation is given by
y′′ + g(x)y = 0, (2.9)
where g(x) is an arbitrary function of x. The derivation of this equation is outlined
in [13]. Very briefly, the governing equations for the Combined Gravity-Inertial-
Rossby Waves on a β-plane reduce to a partial differential equation, which, with
Fourier plane wave analysis, becomes a second-order ODE describing the latitudinal
structure of the perturbations. In equation (2.9), x and y are local Cartesian
coordinates and g(x) is the wave number, see [13]. By inspection, we find that
L =
1
2
y′ 2 −
1
2
g(x)y2 (2.10)
is a Lagrangian for equation (2.9). As for the Kummer Equation and its corresponding
Lagrangian, it can be shown that equation (2.9) with Lagrangian (2.10) has an eight-
dimensional Lie symmetry algebra (this is because it is a linear second-order ODE
for both arbitrary and specific g(x), see [10]) and a five-dimensional Noether algebra
(which follows from the equivalence of (2.9) and (1.14), see [10]). Therefore equation
25
(1.14) can be mapped to this equation using the Method of Equivalent Lagrangians.
We follow the same procedure as for the Kummer Equation in the previous section,
with our aim being to find the regular point transformation X = X(x, y) and
Y = Y (x, y) that maps equation (1.14) to equation (2.9).
As before, we begin by substituting expressions for L and L (given by (2.10) and
(1.13) respectively) into equation (1.10). This gives the equation
1
2
y′ 2 −
1
2
g(x)y2 =
1
2
(
Yx + 2YxYyy′ + y′ 2Y 2y
Xx + y′Xy
)
+ fx + y
′fy. (2.11)
Again we assume that X is of the form
X =
∫
a2 dx,
where a = a(x), which simplifies the above equation to
1
2
y′ 2 −
1
2
g(x)y2 =
1
2
(
Y 2x + 2YxYyy
′ + y′ 2Y 2y
a2
)
+ fx + y
′fy. (2.12)
Separating by powers of y′, we obtain the following system of three equations,
y′ 2 :
1
2
=
1
2
Y 2y
a2
,
y′ : 0 =
YxYy
a2
+ fy,
1 : −
1
2
g(x)y2 =
1
2
Y 2x
a2
+ fx.
From the first equation we deduce that
Y = ay + b(x), (2.13)
where we can assume, without loss of generality, that b(x) = 0. Substituting
expressions for Yy and Yx into the second equation in the system, and then integrating
with respect to y, we have that
f = −
1
2
a′
a
y2 + c(x), (2.14)
26
for which we again assume, without loss of generality, that c(x) = 0. Finally,
after substituting expressions for Yx and fx into the third equation and making the
substitution A = a
′
a , we obtain the equation
−
1
2
g(x)y2 =
1
2
A2y2 −
1
2
A′y2, (2.15)
which simplifies to
0 = A′ − A2 − g(x). (2.16)
Thus, as before, a = e
∫
A dx, where A satisfies equation (2.16). Hence we have that
our regular point transformation, X = X(x, y) and Y = Y (x, y), which transforms
equation (1.14) to equation (2.9), is given by
X =
∫
e2
∫
A dx dx, Y = e
∫
A dxy,
where A satisfies (2.16). This enables us to find the solutions of the Combined
Gravity-Inertial-Rossby Wave Equation in the same way that we found the solutions
of the harmonic oscillator equation in the illustrative example. This regular point
transformation can in turn be used to find the conserved quantities of the Combined
Gravity-Inertial-Rossby Wave Equation.
2.4 Discussion and Conclusion
In this chapter, we have expanded on the work of Kara and Mahomed in [6] by
applying the Method of Equivalent Lagrangians to the second-order Kummer Equation
and the Combined Gravity-Inertial-Rossby Wave Equation. Because both of these
equations possess Lagrangians whose Noether algebras are isomorphic to that of
the Lagrangian (1.13), we were able to find transformations that map (1.13) to the
Lagrangians of these two equations respectively. Once a transformation is known,
27
it can be used to retrieve all solutions and conserved quantities of the equation in
question, since all solutions and conserved quantities for equation (1.14), associated
with Lagrangian (1.13), are known. This is extremely useful as it means that the
lengthy calculations usually associated with the integration techniques used to solve
these equations can be avoided. A limitation of this method is that the equation in
question must possess a Lagrangian, which is not always the case. In addition, this
Lagrangian must be equivalent to another Lagrangian whose associated ODE has
known solutions (and conserved quantities), which in our case was (1.13). If it is
not known whether or not two Lagrangians are equivalent, their Noether symmetries
must be computed. If their Noether symmetries are of the same dimension, then
it follows that their Noether symmetry algebras are isomorphic, [2], and hence
the Lagrangians are equivalent. Under these conditions, the Method of Equivalent
Lagrangians can be applied to other classes of ODEs using the method demonstrated
above. In the next chapter we shall turn our attention to equivalent Lagrangians of
PDEs in two independent variables.
28
Chapter 3
The (1 + 1) Wave Equation and
Related PDEs
3.1 Introduction
We now study the application of the method to some classes of PDEs in two
independent variables. We first demonstrate, through the use of two illustrative
examples, that given a Lagrangian, L, and a known transformation, one can construct
an equivalent Lagrangian, L, using Definition 1 given in the first chapter. Following
this, we turn our attention to the construction of a standard form for the Lagrangian
equivalent to the usual Lagrangian of the one-dimensional linear wave equation.
This will enable us to apply the Method of Equivalent Lagrangians to PDEs whose
Lagrangians are known to be equivalent to that of the one-dimensional linear wave
equation. In this latter situation, the aim of the method is to construct a transformation
that maps one Lagrangian, L, to its equivalent, L. As before, this will enable us
29
to find solutions and conserved quantities of the PDE associated with L, as long as
those of the PDE associated with L are known. In the final section of this chapter,
we apply the Method of Equivalent Lagrangians to map the one-dimensional linear
wave equation with dissipation to a particular Klein-Gordon equation which we seek
to solve. This is the main application of this chapter.
3.2 Illustrative Examples
Example 1. In the first example, we use a given Lagrangian, L, and a given
transformation, X = X(x, t, u), T = T (x, t, u) and U = U(x, t, u), in order to
construct an equivalent Lagrangian, L.
Consider the (1 + 1) linear wave equation with unit wave speed,
UTT − UXX = 0. (3.1)
In this equation, X and T are independent variables, and U is the dependent
variable. In other words, U = U(X,T ). Equation (3.1) is known to have the
Lagrangian
L =
1
2
(U2T − U
2
X), (3.2)
see [1]. Suppose we are given the transformation
X = t+ x, T = t− x, U = u,
which is the standard transformation to canonical form, [5]. By making the correct
substitutions into equation (1.9), we can calculate L.
30
Firstly, the Jacobian, J , is given by
J =
∣
∣
∣
∣
∣
∣
∣
dX
dx
dT
dx
dX
dt
dT
dt
∣
∣
∣
∣
∣
∣
∣
, (3.3)
for two independent variables x and t, [5].
The Lagrangian L is a function of the variables x, t and u, where u = u(x, t). Hence,
using our canonical transformation above, we have that
J =
∣
∣
∣
∣
∣
∣
∣
1 −1
1 1
∣
∣
∣
∣
∣
∣
∣
= 2.
From equation (1.9), assuming zero guage, it follows that
L =
1
2
(U2T − U
2
X)2 (3.4)
In order to find UT and UX , we use the two equations
dU
dt
=
∂T
∂t
∂U
∂T
+
∂X
∂t
∂U
∂X
(3.5)
and
dU
dx
=
∂T
∂x
∂U
∂T
+
∂X
∂x
∂U
∂X
, (3.6)
which are derived from the multivariable case of the chain rule, and solve for ∂U∂T and
∂U
∂X .
Using our standard transformation to canonical form, (3.5) and (3.6) become
ut = UT + UX , ux = −UT + UX .
Solving these simultaneously, we obtain
UT =
1
2
(ut − ux), UX =
1
2
(ut + ux).
31
We now substitute these expressions into equation (3.4), from which we have
L =
1
4
(u2t − 2uxut + u
2
x)−
1
4
(u2t + 2uxut + u
2
x)
= −uxut.
Hence L is equivalent to L in the sense of Definition 1. The Euler-Lagrange Equation
associated with L is
uxt = 0, (3.7)
which is the canonical form of the wave equation given in (3.1), see [1].
Example 2. In the previous example, we made use of a canonical transformation
in order to find a Lagrangian equivalent to L. In this example, we follow the
procedure detailed in [5], in which an equivalent Lagrangian is constructed for the
wave equation (3.1), but where the transformed variables are a consequence of the
underlying symmetry structure concluded from a generator in the Noether algebra.
It can be verified that
G(X,T, U) =
∂
∂T
(3.8)
is a Noether point symmetry generator for the Lagrangian L given by (3.2), [1].
Suppose we wish to map G to the dilation symmetry generator
G(x, t, u) = x
∂
∂x
+
1
2
u
∂
∂u
. (3.9)
Once this mapping is found, it can be used in formula (1.9) to determine L. The
formula for change of variables is given by
G(X,T, U) = G(X)
∂
∂X
+G(T )
∂
∂T
+G(U)
∂
∂U
. (3.10)
Substituting (3.8) into equation (3.10), and equating coefficients, we obtain the three
equations
G(X) = 0, G(T ) = 1, G(U) = 0.
32
We solve these equations using the method of invariants, from which we conclude
that
X = F(t ,
u2
x
), T = ln x + G(t ,
u2
x
), U = H(t ,
u2
x
),
where F , G and H are arbitrary functions. For particular choices of F , G and H,
we arrive at
X = t, T = ln x, U =
u2
x
.
This gives us our transformation. Now, using the expression given in (3.3) for J , we
have that
J =
∣
∣
∣
∣
∣
∣
∣
0 1x
1 0
∣
∣
∣
∣
∣
∣
∣
= −
1
x
.
As in the first example, we find UT and UX by solving equations (3.5) and (3.6)
simultaneously. From this we obtain
UX =
2uut
x
and
UT = 2uux −
u2
x
.
We now substitute the expressions for J , UX and UT into equation (1.9), again
assuming zero guage, which gives us
L =
1
2
(U2T − U
2
X).
(
−
1
x
)
=
−1
2x
(
4u2u2x −
4u3ux
x
+
u4
x2
−
4u2u2t
x2
)
=
−u2
2x3
(4u2xx
2 − 4uuxx+ u
2 − 4u2t ).
Hence we have shown how, for a given PDE in two independent variables, given a
Lagrangian, L, and a transformation (which may be constructed using a symmetry
generator in the Noether algebra, as is demonstrated in this example), one can find
a Lagrangian L, which is equivalent to L.
33
In the applications that follow, we show that if we have two equivalent Lagrangians of
PDEs in two independent variables, then we can find the mapping which transforms
one Lagrangian to the other. This mapping can be used to find solutions of the
PDEs associated with the Lagrangians, as well as to find conserved quantities.
3.3 Form of Equivalent Lagrangian for the Linear
Wave Equation in (1 + 1) Dimensions
We now find an expression for the form of a Lagrangian, L, which is equivalent to
the usual Lagrangian, L, of the one-dimensional linear wave equation. Once we have
this form, given any L equivalent to L, we can find the transformation that maps
L to L, and hence the solutions and conserved quantities of the (1 + 1) linear wave
equation to those of the differential equation associated with L.
Solving equations (3.5) and (3.6) simultaneously for UT and UX , we obtain
UT =
dU
dt Xx −Xt
dU
dx
TtXx −XtTx
(3.11)
and
UX =
Tt dUdx − Tx
dU
dt
TtXx −XtTx
. (3.12)
Therefore,
1
2
(U2T − U
2
X) =
1
2


dU
dt
2
(X2x − T
2
x ) +
dU
dx
2
(X2t − T
2
t ) + 2
dU
dx
dU
dt (TtTx −XtXx)
T 2t X2x − 2TtTxXtXx +X
2
t T 2x

 ,
from equation (1.9), assuming that guage is zero. Then
J =
∣
∣
∣
∣
∣
∣
∣
dX
dx
dT
dx
dX
dt
dT
dt
∣
∣
∣
∣
∣
∣
∣
34
=
dX
dx
dT
dt
−
dX
dt
dT
dx
.
By the chain rule, dXdx = Xx + uxXu, and similarly for
dT
dt ,
dX
dt and
dT
dx . Hence
J = (Xx + uxXu)(Tt + utTu)− (Xt + utXu)(Tx + uxTu)
= Xx(Tt + utTu)−Xt(Tx + uxTu) +Xu(uxTt − utTx),
and so
L =
1
2
(U2T − U
2
X)J
=
1
2


dU
dt
2
(X2x − T
2
x ) +
dU
dx
2
(X2t − T
2
t ) + 2
dU
dx
dU
dt (TtTx −XtXx)
T 2t X2x − 2TtTxXtXx +X
2
t T 2x

 [Xx(Tt
+ utTu)−Xt(Tx + uxTu) +Xu(uxTt − utTx)].
If we assume that Xu = 0 and Tu = 0 (i.e. we assume that X = X(x, t) and
T = T (x, t)), then the above expression for the Lagrangian reduces to
L =
1
2


dU
dt
2
(X2x − T
2
x ) +
dU
dx
2
(X2t − T
2
t ) + 2
dU
dx
dU
dt (TtTx −XtXx)
(XxTt −XtTx)2

 (XxTt −XtTx)
=
1
2


dU
dt
2
(X2x − T
2
x ) +
dU
dx
2
(X2t − T
2
t ) + 2
dU
dx
dU
dt (TtTx −XtXx)
XxTt −XtTx

 .
Now, by total differentiation (see (1.2)), we have that
dU
dt
= Ut + utUu
and
dU
dx
= Ux + uxUu.
Therefore we conclude that L takes the following form.
35
Form of the Equivalent Lagrangian. The general form for a Lagrangian equivalent
to the usual Lagrangian of the one-dimensional linear wave equation is
L = 12(XxTt−XtTx) [(Ut + utUu)
2(X2x − T
2
x )
+ (Ux + uxUu)2(X2t − T
2
t ) + 2(Ux + uxUu)(Ut + utUu)(TtTx −XtXx)]
=
[
1
2(XxTt−XtTx)
]
{uxut[2U2u(TtTx −XtXx)]
+ u2x[U
2
u(X
2
t − T
2
t )] + u
2
t [U
2
u(X
2
x − T
2
x )]
+ ux[2UuUx(X2t − T
2
t ) + 2UtUu(TtTx −XtXx)]
+ ut[2UtUu(X2x − T
2
x ) + 2UuUx(TtTx −XtXx)]
+ [U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)]},
(3.13)
where U = U(x, t, u), X = X(x, t) and T = T (x, t).
Once we have a Lagrangian which we know to be equivalent to the Lagrangian
L given by (3.2), we can reverse the process of the two examples above, and
use the form of the Lagrangian given in the above equation in order to find the
transformation that maps the solutions of the one-dimensional wave linear equation
(3.1) to the solutions of the differential equation associated with the equivalent
Lagrangian. As a means of demonstrating and validating this method, we use the
Lagrangians found in the previous two examples (as we know these to be equivalent
to (3.2) by construction) and attempt to recover the required transformations.
3.4 Finding Transformations -
Illustrative Examples
Example 1. Consider equation (3.7) with its Lagrangian
L = −uxut, (3.14)
36
which we found to be equivalent to (3.2) in Example 1 of the previous section. We
use the Form of the Equivalent Lagrangian found in the previous section in order to
find the transformation that maps the Lagrangian (3.2) to this Lagrangian.
Substituting (3.14) for L in equation (3.13), we obtain the equation
−uxut =
1
2(XxTt −XtTx)
[(Ut + utUu)
2(X2x − T
2
x )
+ (Ux + uxUu)
2(X2t − T
2
t ) + 2(Ux + uxUu)(Ut + utUu)(TtTx −XtXx)]
=
[
1
2(XxTt −XtTx)
]
{uxut[2U
2
u(TtTx −XtXx)]
+ u2x[U
2
u(X
2
t − T
2
t )] + u
2
t [U
2
u(X
2
x − T
2
x )]
+ ux[2UuUx(X
2
t − T
2
t ) + 2UtUu(TtTx −XtXx)]
+ ut[2UtUu(X
2
x − T
2
x ) + 2UuUx(TtTx −XtXx)]
+ [U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)]}.
Since the transformation X, T and U is dependent on x, t and u alone, and not on
their derivatives, it follows that we can separate the above equation into six different
equations by the derivative terms of u. We arrive at the equations
utux : −1 =
2U2u(TtTx −XtXx)
2(XxTt −XtTx)
,
u2x : 0 =
U2u(X
2
t − T
2
t )
2(XxTt −XtTx)
,
u2t : 0 =
U2u(X
2
x − T
2
x )
2(XxTt −XtTx)
,
ux : 0 =
2UuUx(X2t − T
2
t ) + 2UtUu(TtTx −XtXx)
2(XxTt −XtTx)
,
ut : 0 =
2UtUu(X2x − T
2
x ) + 2UuUx(TtTx −XtXx)
2(XxTt −XtTx)
,
1 : 0 =
U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)
2(XxTt −XtTx)
.
Using the mathematical software Mathematica (see [17]) to solve this over-determined
37
system of equations, we recover the transformation
X = f(x) + g(t), T = f(x)− g(t), U = u,
for f and g arbitrary functions of x and t respectively.
A particular case of the transformation is
X = t+ x, T = t− x, U = u.
This is the standard transformation to canonical form which we know, from Section
3.2, maps L, given by (3.2), to L, given by (3.14).
Example 2. In order to further test this method, we use the Lagrangian
L =
−u2
2x3
(4u2xx
2 − 4uuxx+ u
2 − 4u2t ), (3.15)
which was constructed to be equivalent to (3.2) in Example 2 of the previous section.
We wish to use the Form of the Equivalent Lagrangian, given by equation (3.13),
in order to recover the transformation used in the construction of (3.15). This
transformation maps (3.2) to (3.15) and hence maps the associated differential
equations from one to the other. Thus we can use this transformation to find
solutions of the differential equation associated with (3.15), if the solutions of (3.1)
are known.
Using the Form of the Equivalent Lagrangian, we substitute (3.15) for L in equation
(3.13), which gives us the equation
−u2
2x3
(4u2xx
2 − 4uuxx+ u
2 − 4u2t ) =
[
1
2(XxTt −XtTx)
]
{uxut[2U
2
u(TtTx −XtXx)]
+ u2x[U
2
u(X
2
t − T
2
t )] + u
2
t [U
2
u(X
2
x − T
2
x )]
+ ux[2UuUx(X
2
t − T
2
t ) + 2UtUu(TtTx −XtXx)]
+ ut[2UtUu(X
2
x − T
2
x ) + 2UuUx(TtTx −XtXx)]
+ [U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)]}.
38
As before, we can separate the above equation into six different equations by separating
by the derivative terms of u. We can do this because X, T and U are dependent on
x, t and u only, and not on their derivatives (this will henceforth be referred to as
point dependency). We arrive at the six equations
utux : 0 =
2U2u(TtTx −XtXx)
2(XxTt −XtTx)
,
u2x :
−2u2
x
=
U2u(X
2
t − T
2
t )
2(XxTt −XtTx)
,
u2t :
2u2
x3
=
U2u(X
2
x − T
2
x )
2(XxTt −XtTx)
,
ux :
2u3
x2
=
2UuUx(X2t − T
2
t ) + 2UtUu(TtTx −XtXx)
2(XxTt −XtTx)
,
ut : 0 =
2UtUu(X2x − T
2
x ) + 2UuUx(TtTx −XtXx)
2(XxTt −XtTx)
,
1 :
−u4
2x3
=
U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)
2(XxTt −XtTx)
.
Using software (see [17]) to solve these equations simultaneously, we discover that
one solution is the transformation
X = f(t+ ln x)− g(t− ln x),
T = f(t+ ln x) + g(t− ln x),
U = ±
u2
x
,
where f and g are arbitrary functions of x and t. If we choose f such that
f(t+ ln x) = 12(t+ ln x) and g such that g(t− ln x) = −
1
2(t− ln x), we have the
same transformation as the one used in Example 2 of the previous section that
resulted in the construction of the Lagrangian (3.15). Thus the Method of Equivalent
Lagrangians is validated for this example. In the next section, we apply the method
to a particular Klein-Gordon equation, to which we recover a transformation map
from the one-dimensional linear wave equation with dissipation.
39
3.5 Main Application
The equation
UTT + UT − UXX = 0 (3.16)
is the one-dimensional case of the linear wave equation with dissipation, [1]. It has
the well-known Lagrangian, after multiplication by a variational factor,
L =
1
2
eT (U2T − U
2
X). (3.17)
We map L to the Lagrangian
L =
1
2
u2t −
1
2
u2x −
1
2
uut +
1
8
u2, (3.18)
giving rise to the Euler equation
utt − uxx −
1
4
u = 0, (3.19)
which we note to be a Klein-Gordon equation, see [2]. The Noether symmetries of
L are of the form X = ξ∂x + τ∂t + φ∂u, with gauge (f, g), and satisfy the equation
0 = uφ4 −
φut
2 +
1
2uuxξt − utuxξt +
1
2uutuxξu − u
2
tuxξu + u
3
xξu
+u2xξx +
1
2uutτt − u
2
t τt +
1
2uu
2
t τu − u
3
t τu
+utu2xτu + utuxτx −
uφt
2 + utφt −
1
2uutφu + u
2
tφu − u
2
xφu − uxφx
+
(
1
2u
2
t −
1
2u
2
x −
1
2uut +
1
8u
2
)
(τt + utτu + ξx + uxξu)− (ft + utfu + gx + uxgu) ,
[17]. This separates into an over-determined system of partial differential equations
whose solution is
X1 = ∂t, f1 = 0, g1 = 0,
X2 = ∂x, f2 = 0, g2 = 0,
X3 = t∂x + x∂t, f3 = 0, g3 = 14u
2,
X∞ = F (x, t)∂u, f∞ = −12uF + uFt, g∞ = −uFx,
40
where F satisfies 14F + Fxx − Ftt = 0, [17]. This Lie algebra is isomorphic to
the Noether algebra corresponding to the Lagrangian L, [4]. Hence L and L are
equivalent Lagrangians.
We can therefore use the equation (1.9) given in Definition 1 (again assuming zero
guage) in order to find the transformation X = X(x, t, u), T = T (x, t, u) and U =
U(x, t, u) that maps L to L.
Assuming that Xu = 0 and Tu = 0 as before, we have that J = XxTt−XtTx. Again,
we use the equations (3.5) and (3.6) and solve simultaneously for UX and UT . Using
the same procedure as before, we eventually deduce that
L =
[
eT
2(XxTt −XtTx)
]
{uxut[2U
2
u(TtTx −XtXx)]
+ u2x[U
2
u(X
2
t − T
2
t )] + u
2
t [U
2
u(X
2
x − T
2
x )]
+ ux[2UuUx(X
2
t − T
2
t ) + 2UtUu(TtTx −XtXx)]
+ ut[2UtUu(X
2
x − T
2
x ) + 2UuUx(TtTx −XtXx)]
+ [U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)]}.
Substituting (3.18) for L and then separating by derivative terms of u (which is
possible by point dependency), we arrive at the six equations
utux : 0 =
2U2ue
T (TtTx −XtXx)
2(XxTt −XtTx)
,
u2x : −
1
2
=
U2ue
T (X2t − T
2
t )
2(XxTt −XtTx)
,
u2t :
1
2
=
U2ue
T (X2x − T
2
x )
2(XxTt −XtTx)
,
ux : 0 =
eT [2UuUx(X2t − T
2
t ) + 2UtUu(TtTx −XtXx)]
2(XxTt −XtTx)
,
ut : −
1
2
u =
eT [2UtUu(X2x − T
2
x ) + 2UuUx(TtTx −XtXx)]
2(XxTt −XtTx)
,
41
1 :
1
8
u2 =
eT [U2t (X
2
x − T
2
x ) + U
2
x(X
2
t − T
2
t ) + 2UtUx(TtTx −XtXx)]
2(XxTt −XtTx)
.
For this overdetermined system of equations, the software (see [17]) yields the result
X = f1(t+ x)− f2(t− x),
T = f1(t+ x) + f2(t− x),
U = −e−
1
2 tu.
A particular choice of the functions f1 and f2 is
f1(t+ x) =
1
2
(t+ x), f2(t− x) =
1
2
(t− x),
from which we recover the transformation
X = x, T = t, U = −e−
1
2 tu.
This is a known transformation mapping (3.16) to (3.19), as is noted in [1]. We note
that this transformation also transforms the Noether symmetries of L to those of L.
3.6 Discussion and Conclusion
In this chapter, we set about extending the Method of Equivalent Lagrangians to
some classes of PDEs in two independent variables. We began by elaborating on
the work done in [5], in which a Lagrangian equivalent to a given Lagrangian was
constructed using a known transformation and Definition 1. This aspect of the
work took the form of two Illustrative Examples. Secondly, we found the Form of
the Equivalent Lagrangian, which is the general form for a Lagrangian equivalent
42
to the usual Lagrangian of the one-dimensional linear wave equation. In the next
two Illustrative Examples, we made use of this form in order to apply the Method
of Equivalent Lagrangians. Using the equivalent Lagrangians constructed in Section
3.2, we sought to recover the transformations, X, T and U , that map the usual
Lagrangian of the (1 + 1) linear wave equation to those equivalent Lagrangians,
namely the transformations used to construct the equivalent Lagrangians in Section
3.2. The successful recovery of these two transformations through the application of
our method serves to confirm the validity of the Method of Equivalent Lagrangians
for these two examples. In the final part of this chapter, we applied the Method of
Equivalent Lagrangians to a Klein-Gordon equation, whose Lagrangian is equivalent
to that of the one-dimensional linear wave equation with dissipation. The method
yielded a transformation which is noted in [1] as mapping these two equations from
one to the other. Thus the Method of Equivalent Lagrangians has been successful
for the main application of this chapter. The transformation found can be used to
find the solutions and conserved quantities of our particular Klein-Gordon equation,
since those of the linear wave equation with dissipation are well known. This section
has extended the Method of Equivalent Lagrangians to some classes of PDEs in two
independent variables, and validated the method for specific examples. In the same
way, the method can be easily extended to other classes of PDEs in two independent
variables, as long as Lagrangians for these equations exist. In the next chapter, we
extend the Method of Equivalent Lagrangians to some classes of PDEs in three and
four independent variables.
43
Chapter 4
Generalisation to PDEs in Three
and Four Independent Variables
4.1 Introduction
We now generalise the Method of Equivalent Lagrangians to some classes of PDEs
in three and four independent variables. As the number of independent variables
increases, the work becomes, computationally, increasingly complex. Therefore we
shall restrict our investigation to the standard linear wave equations with the usual
Lagrangians in three and four independent variables respectively. We begin by
looking at the case of three independent variables. It is well known that the
two-dimensional linear wave equation can be written in polar co-ordinates using
the regular transformation from Cartesian to polar co-ordinates. Applying this
transformation to Definition 1, we are able to construct a Lagrangian which is
equivalent to the usual Lagrangian of the two-dimensional linear wave equation.
44
This is the Lagrangian associated with the two-dimensional wave equation in polar
co-ordinates. Following this, we reverse the procedure, applying the Method of
Equivalent Lagrangians in an attempt to recover the transformation that maps
one Lagrangian to the other, namely the transformation from Cartesian to polar
co-ordinates. For the three-dimensional linear wave equation, we show that the
well-known transformation from Cartesian to spherical co-ordinates satisfies the
equations derived from Definition 1 when two known equivalent Lagrangians for
this wave equation are used.
4.2 The Two-Dimensional Linear Wave Equation
The two-dimensional linear wave equation is well known and is given by
UTT − UXX − UY Y = 0, (4.1)
where T , X and Y are the independent variables, and U = U(T,X, Y ) is the
dependent variable, [1]. The usual Lagrangian for this equation is
L =
1
2
U2T −
1
2
U2X −
1
2
U2Y , (4.2)
[1]. We begin by constructing a Lagrangian L = L(t, x, y) equivalent to L, using the
well-known transformation from Cartesian to polar co-ordinates, which is given by
T = t, X = x cos y, Y = x sin y.
After this we work backwards, starting with equation (1.9) that relates L and L as
equivalent Lagrangians, and using it to derive the transformation from Cartesian to
polar co-ordinates given above.
45
4.2.1 Constructing an Equivalent Lagrangian
In order to find a Lagrangian L which is equivalent to L, we use Definition 1. To
this we apply the transformation from Cartesian to polar co-ordinates, in order to
find J , and substitute (4.2) for L.
Hence we have that
J =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
Tt Xt Yt
Tx Xx Yx
Ty Xy Yy
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0
0 cos y sin y
0 −x sin y x cos y
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= x cos2 y + x sin2 y
= x.
By the multivariate case of the chain rule, we have the matrix equation






ut
ux
uy






=






Tt Xt Yt
Tx Xx Yx
Ty Xy Yy












UT
UX
UY






. (4.3)
Hence,






UT
UX
UY






=






Tt Xt Yt
Tx Xx Yx
Ty Xy Yy






−1





ut
ux
uy






=






1 0 0
0 cos y sin y
0 −x sin y x cos y






−1





ut
ux
uy






46
=





1 0 0
0 cos y − 1x sin y
0 sin y 1x cos y












ut
ux
uy






,
which gives us the following expressions for UT , UX and UY .
UT = ut, UX = ux cos y − uy
1
x
sin y, UY = ux sin y + uy
1
x
cos y.
Substituting these into equation (1.9), along with our expression for J , we have that
a Lagrangian equivalent to (4.2) is
L =
1
2
xu2t −
1
2
xu2x −
1
2x
u2y. (4.4)
It is well known that this Lagrangian is isomorphic to the usual Lagrangian of the
two-dimensional linear wave equation, namely (4.2).
Now that we have used our definition of equivalent Lagrangians in order to find L,
we reverse the process, applying the Method of Equivalent Lagrangians in order to
recover the required transformation.
4.2.2 Finding Transformations -
The Reverse Procedure
In the previous section, we found that the Lagrangian L, given by (4.4), is equivalent
to L, the usual Lagrangian of the two-dimensional linear wave equation, given
by (4.2). Now, using the Method of Equivalent Lagrangians, we wish to find the
transformation, T = T (t, x, y), X = X(t, x, y) and Y = Y (t, x, y), that maps L to
L. The transformation is known to us by construction. Thus in order to verify the
method for this particular example, we wish to recover the usual transformation of
Cartesian to polar co-ordinates.
47
Note: We are using two Lagrangians which we know to be equivalent by construction.
If this were not the case, however, we could check that L is equivalent to L by
verifying that they have isomorphic 10-dimensional Noether symmetry algebras.
We begin by computing J . For the sake of simplification, we assume that Tx = Ty =
Xt = Yt = 0. In other words, T = T (t), X = X(x, y) and Y = Y (x, y). From this
assumption it follows that
J =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
Tt Xt Yt
Tx Xx Yx
Ty Xy Yy
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
Tt 0 0
0 Xx Yx
0 Xy Yy
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= Tt(XxYy −XyYx).
The multivariate case of the chain rule, given by (4.3), relates ut, ux and uy to UT ,
UX and UY . Using this equation we have that






UT
UX
UY






=






Tt 0 0
0 Xx Yx
0 Xy Yy






−1





ut
ux
uy






=






1
Tt
0 0
0 Yy(XxYy−XyYx)
−Yx
(XxYy−XyYx)
0 −Xy(XxYy−XyYx)
Xx
(XxYy−XyYx)












ut
ux
uy






=
1
(XxYy −XyYx)






ut(XxYy−XyYx)
Tt
−(Yxuy − Yyux)
(Xxuy −Xyux)






,
which gives us expressions for UT , UX and UY . By Definition 1, making the
48
substitutions for L, L and J , we have the equation
1
2
xu2t −
1
2
xu2x −
1
2x
u2y =
1
2
(U2T − U
2
X − U
2
Y )Tt(XxYy −XyYx). (4.5)
Writing UT , UX and UY in terms of ut, ux and uy, using the expressions in the
matrix above, gives us
1
2
xu2t −
1
2
xu2x −
1
2x
u2y =
1
2Tt(XxYy −XyYx)
(u2t (XxYy −XyYx)
2
− u2x(Y
2
y +X
2
y )− u
2
y(Y
2
x +X
2
x) + 2uxuy(YxYy +XxXy)).
Since T , X and Y are point dependent, we can separate by derivatives of u to obtain
the following over-determined system of four equations.
u2t : Ttx = (XxYy −XyYx),
u2x : Ttx(XxYy −XyYx) = Y
2
y +X
2
y ,
u2y : Tt(XxYy −XyYx) = x(Y
2
x +X
2
x),
uxuy : 0 = 2(YxYy +XxXy).
Solving this system using mathematical software (see [17]) yields the result
T = t+ a, X = F (t) + x cos y, Y = G(t) + x sin y,
where a is a constant and F and G are arbitrary functions of t. If we let F = G = 0,
we recover the transformation
T = t, X = x cos y, Y = x sin y,
which is the standard transformation from Cartesian to polar co-ordinates, as anticipated.
This transformation can in turn be used to find solutions and conserved quantities
of the PDE associated with (4.4), since those associated with the two-dimensional
linear wave equation are known. In the next section we consider the three-dimensional
linear wave equation.
49
4.3 The Three-Dimensional Linear Wave
Equation
We now turn our attention to some classes of PDEs in four independent variables.
The three-dimensional linear wave equation is the well-known second-order PDE
UTT − UXX − UY Y − UZZ = 0, (4.6)
see [1]. Here T , X, Y and Z are the independent variables, and U = U(T,X, Y, Z)
is the dependent variable. The usual Lagrangian for this equation is
L =
1
2
U2T −
1
2
U2X −
1
2
U2Y −
1
2
U2Z , (4.7)
[1]. It is well known that the Lagrangian for the three-dimensional wave equation
in spherical co-ordinates is given by
L =
1
2
x2 sin y u2t −
1
2
sin y x2 u2x −
1
2
sin y u2y −
1
2 sin y
u2z. (4.8)
L and L are naturally equivalent because they are Lagrangians of the same equation.
However, it can also be verified that they generate isomorphic Noether algebras of
point symmetries.
In this section, we shall apply the Method of Equivalent Lagrangians to L and L, in
an attempt to recover the transformation that maps L to L. We shall demonstrate
that equation (1.9) of Definition 1 is satisfied by the transformation from Cartesian
to polar co-ordinates, which is to be expected from our choice of Lagrangians.
As before, we make use of Definition 1, which relates two equivalent Lagrangians to
each other by means of the transformation that maps one to the other. Substituting
50
L and L into equation (1.9) gives us
1
2x
2 sin y u2t −
1
2 sin y x
2 u2x −
1
2 sin y u
2
y −
1
2 sin y u
2
z
= (12U
2
T −
1
2U
2
X −
1
2U
2
Y −
1
2U
2
Z)J. (4.9)
We first calculate the Jacobian, J . We assume that Tx = Ty = Tz = Xt = Yt = Zt =
0. In other words, T = T (t), X = X(x, y, z), Y = Y (x, y, z) and Z = Z(x, y, z).
Furthermore, we assume that T = t. Hence Tt = 1. From these assumptions it
follows that
J =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
Tt Xt Yt Zt
Tx Xx Yx Zx
Ty Xy Yy Zy
Tz Xz Yz Zz
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 0 0
0 Xx Yx Zx
0 Xy Yy Zy
0 Xz Yz Zz
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
= XxYyZz −XxYzZy −XyYxZz +XyYzZx +XzYxZy −XzYyZx.
Now, by the multivariate case of the chain rule (in four dimensions this time), we
have the matrix equation









ut
ux
uy
uz









=









Tt Xt Yt Zt
Tx Xx Yx Zx
Ty Xy Yy Zy
Tz Xz Yz Zz


















UT
UX
UY
UZ









. (4.10)
51
Hence,









UT
UX
UY
UZ









=









Tt Xt Yt Zt
Tx Xx Yx Zx
Ty Xy Yy Zy
Tz Xz Yz Zz









−1








ut
ux
uy
uz









=









1 0 0 0
0 Xx Yx Zx
0 Xy Yy Zy
0 Xz Yz Zz









−1








ut
ux
uy
uz









=
1
J









ut(XxYyZz −XxYzZy −XyYxZz +XyYzZx +XzYxZy −XzYyZx)
uz(YxZy − YyZx)− uy(YxZz − YzZx) + ux(YyZz − YzZy)
uy(XxZz −XzZx)− uz(XxZy −XyZx)− ux(XyZz −XzZy)
uz(XxYy −XyYx)− uy(XxYz −XzYx) + ux(XyYz −XzYy)









,
from which we can read the expressions for UT , UX , UY and UZ in terms of ut, ux,
uy and uz. Substituting these into equation (4.9), along with our expression for J ,
we have the equation
1
2
x2 sin y u2t −
1
2
sin y x2 u2x −
1
2
sin y u2y −
1
2 sin y
u2z
=
1
2
(−XzYyZx +XyYzZx +XzYxZy −XxYzZy −XyYxZz +XxYyZz)[u
2
t
−
(uzXyYx − uyXzYx − uzXxYy + uxXzYy + uyXxYz − uxXyYz)2
(XzYyZx −XyYzZx −XzYxZy +XxYzZy +XyYxZz −XxYyZz)2
−
(uzYyZx − uyYzZx − uzYxZy + uxYzZy + uyYxZz − uxYyZz)2
(XzYyZx −XyYzZx −XzYxZy +XxYzZy +XyYxZz −XxYyZz)2
−
(uzXyZx − uyXzZx − uzXxZy + uxXzZy + uyXxZz − uxXyZz)2
(−XzYyZx +XyYzZx +XzYxZy −XxYzZy −XyYxZz +XxYyZz)2
].
Since X, Y and Z are point dependent, we can separate by derivatives of u to obtain
the following over-determined system of equations.
52
u2t :
1
2
x2 sin y =
1
2
(−XzYyZx +XyYzZx +XzYxZy −XxYzZy −XyYxZz +XxYyZz),
u2x :
−
1
2
x2 sin y
=
X2z
(
Y 2y + Z
2
y
)
+ (YzZy − YyZz)2 − 2XyXz(YyYz + ZyZz) +X2y (Y
2
z + Z
2
z )
2(XzYyZx −XyYzZx −XzYxZy +XxYzZy +XyYxZz −XxYyZz)
,
u2y :
−
1
2
sin y
=
X2z (Y
2
x + Z
2
x) + (YzZx − YxZz)
2 − 2XxXz(YxYz + ZxZz) +X2x (Y
2
z + Z
2
z )
2(XzYyZx −XyYzZx −XzYxZy +XxYzZy +XyYxZz −XxYyZz)
,
u2z :
−
1
2 sin y
=
X2y (Y
2
x + Z
2
x) + (YyZx − YxZy)
2 − 2XxXy(YxYy + ZxZy) +X2x
(
Y 2y + Z
2
y
)
2(XzYyZx −XyYzZx −XzYxZy +XxYzZy +XyYxZz −XxYyZz)
,
uxuy :
0 =
[
1
−XzYyZx +XyYzZx +XzYxZy −XxYzZy −XyYxZz +XxYyZz
]
×
[
X2z (YxYy + ZxZy) + (YzZx − YxZz)(YzZy − YyZz)
− Xz(XyYxYz +XxYyYz +XyZxZz +XxZyZz) +XxXy
(
Y 2z + Z
2
z
) ]
,
uxuz :
0 =
[
1
−XzYyZx +XyYzZx +XzYxZy −XxYzZy −XyYxZz +XxYyZz
]
×
[
XxXz
(
Y 2y + Z
2
y
)
+ (YyZx − YxZy)(−YzZy + YyZz)
+ X2y (YxYz + ZxZz)−Xy(XzYxYy +XxYyYz +XzZxZy +XxZyZz)
]
,
53
uzuy :
0 =
[
1
−XzYyZx +XyYzZx +XzYxZy −XxYzZy −XyYxZz +XxYyZz
]
×
[
−XxXz(YxYy + ZxZy) + (YyZx − YxZy)(YzZx − YxZz)
+ X2x(YyYz + ZyZz) +Xy
(
Xz
(
Y 2x + Z
2
x
)
−Xx(YxYz + ZxZz)
) ]
.
It is easy to check that these equations are satisfied by the transformation map from
Cartesian to spherical co-ordinates given by
T = t,
X = x sin y cos z,
Y = x sin y sin z,
Z = x cos y.
4.4 Discussion and Conclusion
In this chapter we have generalised the Method of Equivalent Lagrangians to PDEs in
three and four independent variables, using the standard two- and three-dimensional
linear wave equations respectively. We began by constructing a Lagrangian equivalent
to the usual Lagrangian of the two-dimensional linear wave equation, using the
known transformation from Cartesian to polar co-ordinates. This gave us the
Lagrangian associated with the two-dimensional wave equation in polar co-ordinates.
Following this, we applied our method as before, and successfully recovered the
transformation from Cartesian to polar co-ordinates. For the case of four independent
variables, we started with two Lagrangians which we know to be equivalent, and
attempted to use Definition 1 to find the known transformation from Cartesian to
54
spherical co-ordinates. Although we did not go as far as to solve the system of
equations resulting from the application of our method, we demonstrated that they
would be satisfied by the transformation from Cartesian to spherical co-ordinates,
as anticipated. Hence we have verified the Method of Equivalent Lagrangians for
two PDEs in three and four independent variables. In the same way, the method
can be applied to other PDEs in three and four independent variables, provided
that Lagrangians exist which are equivalent to other Lagrangians associated with
PDEs possessing known solutions and conserved quantities. If these conditions
are satisfied, then the transformations mapping one Lagrangian to the other can
be found, and these can be used to transform the known solutions and conserved
quantities to those of the PDE in question.
55
Conclusion
In this study, we have applied the notion of Equivalent Lagrangians to determine
transformations that map differential equations one to another in order generate
solutions, conservation laws, inter alia. In the first chapter, we presented the
fundamental results and definitions required for the application of the method,
along with an illustrative example in which we used the method to find solutions
and conserved quantities of the harmonic oscillator ODE. In the second chapter, we
built on the work done in [6] by applying the method to two second-order linear
ODEs: the Kummer Equation and the Combined Gravity-Inertial-Rossby Wave
Equation. For these two ODEs, the Method of Equivalent Lagrangians successfully
yielded transformation maps which can be used to find solutions and conserved
quantities. The procedure also holds for PDEs of any number of independent
variables. In the following chapters, we demonstrated this for PDEs in two, three
and four independent variables. In the third chapter, we began by following the
procedure detailed in [5], in which an equivalent Lagrangian is constructed, given
an original Lagrangian and a transformation map. We then found the form of a
Lagrangian equivalent to the usual Lagrangian of the linear wave equation in (1+1)
dimensions. This form was then used in two illustrative examples demonstrating the
Method of Equivalent Lagrangians, as well as in the main application of this section,
in which a transformation was recovered that can be used to find the solutions
56
and conserved quantities of a particular Klein-Gordon equation. In the fourth
chapter, we generalised the method to certain classes of PDEs in three and four
independent variables, using the two- and three- dimensional linear wave equations
respectively. As a consequence of the procedure, we recovered two well-known
transformations, namely the mapping from Cartesian to polar co-ordinates, and the
mapping from Cartesian to spherical co-ordinates. The advantages of the Method of
Equivalent Lagrangians are that it enables one to avoid the integration techniques
normally required to solve differential equations and that once a transformation map
is recovered, it can be used to find all solutions and conserved quantities for a given
differential equation. This powerful application makes up for the cumbersome nature
of the procudure. A possible drawback, however, is that a necessary requirement of
the method is the existence of a Lagrangian for the equation under investigation.
57
Bibliography
[1] N. H. Ibragimov, ed. 1994.“CRC Handbook of Lie Group Analysis of Differential
Equations,” Vol. 1 Symmetries and Exact Solutions and Conservation Laws,
CRC Press, Inc., Boca Raton.
[2] N. H. Ibragimov, ed. 1995.“CRC Handbook of Lie Group Analysis of Differential
Equations,” Vol. 2 Applications in Engineering and Physical Sciences, CRC
Press, Inc., Boca Raton.
[3] N. H. Ibragimov, A. H. Kara, F. M. Mahomed, 1998. Lie-Ba¨cklund and Noether
Symmetries with Applications, Nonlinear Dynamics, 15, 115-136.
[4] S. Jamal, A. H. Kara, 2010. New Higher-Order Conservation Laws of Some
Classes of Wave and Gordon-type Equations, Nonlinear Dynamics, 67, 97-102.
[5] A. H. Kara, 2004. Equivalent Lagrangians and the Inverse Variational Problem
with Applications, Quaestiones Mathematicae, 27, 207-216.
[6] A. H. Kara and F. M. Mahomed, 1992. Equivalent Lagrangians and the Solution
of Some Classes of Non-Linear Equations q¨ + p(t)q˙ + r(t)q = µq˙2q−1 + f(t)qn,
J. Nonlinear Mechanics, 27 (6), 919-927.
[7] S. Lie, 1967. “Differentialgleichungen,” Chelsea, New York.
58
[8] S. Lie, 1883. Klassifikation und Integration von gewo¨nlichen
Differentialgleichungen zwischen x, y, die eine Gruppe von Transformationen
gestaten, Archiv der Mathematik, VIII, IX, 187.
[9] S. Lie, 1891. Lectures on Differential Equations with Known Infinitesimal
Transformations, Teubner, Leipzig (in German, Lie’s lectures by G. Sheffers).
[10] F. M. Mahomed, 2007. Symmetry Group Classification of Ordinary Differential
Equations: Survey of Some Results, Mathematical Methods in the Applied
Sciences, 30, 1995-2012.
[11] F. M. Mahomed, A. H. Kara and P. G. L. Leach, 1993. Lie and Noether
Counting Theorems for One-Dimensional Systems, J. Mathematical Analysis
and Applications, 178, 116-129.
[12] A.D. McDonald, 1948. Properties of the Confluent Hypergeometric Function,
Technical Report no. 84, Research Laboratory of Electronics, M.I.T..
[13] J. McKenzie, 2010. Instability of Combined Gravity-Inertial-Rossby Waves in
Atmospheres and Oceans, submitted to Ann. Geophysics.
[14] P. Olver, 1993. Application of Lie Groups to Differential Equations, Springer,
New York.
[15] W. Sarlet, F. M. Mahomed and P. G. L. Leach, 1987. Symmetries of Non-
Linear Differential Equations and Linearisation, J. Physics A: Mathematical
and General, 20, 277-292.
[16] J. Stillwell, 2008. Naive Lie Theory, Springer, New York.
[17] Wolfram Research, Inc, 2008. Mathematica, Version 7.0, Wolfram Research,
Inc, Champaign, Illinois.
59