3.1. Symmetries and the Jacobi Multiplier
Let us now consider a second-order differential equation given in the following form
for the unknown function
. Introducing a new function
, we obtain the system
of two first-order equations:
or in its symmetric form
We can consider the following operator
, similarly to what we had for the first-order ODEs. The general integral of (
12) is given by a set of two functionally independent first integrals
(constants of motion), such that
From (
13) it follows that
, i.e., functions
are invariants of the operator
A. Since any other solution
of (
12) is an invariant of
A and is a function of
and
, the Jacobian reads
i.e.,
Comparing the above relation with
, and denoting
, we obtain
The function
M is called the Jacobi multiplier [
9]. Due to the relation of
M with Jacobians,
M is a solution of the equation
The form of
M depends on the selection of
. If one has
,
as an independent first integrals, then function
is a Jacobi multiplier, and in general, any function of the form
is a Jacobi multiplier. In particular, any multiplier
M is defined up to an arbitrary nonzero constant factor.
Any point symmetry of system (
12) has the form
and represents the so-called contact symmetry for Equation (
11) if
or
depends on
v (with the nonlinear dependence of
on
v if
). The application of the first prolongation of
S
to system
gives two linear equations for determining
,
and
:
where we have taken into account that
.
Excluding the function
from the commutator relation
leads to (
16). Once again, there is a trivial symmetry
. For simplicity, one can introduce the following operator (evolutionary form of symmetry [
6]):
The function
is called in Ref. [
18] a generating function of symmetry. For the operator
, we have
Thus, to determine the unknown function
, we have to solve one second-order linear partial differential equation (defining equation) given by
As in the case of the first-order ODE, there is an infinite number of point symmetries
for system (
12) and in general, the problem of finding them has the same degree of difficulty as solving the given system. Of course, if we make an additional supposition about the form of the symmetries or about the form of
, then the problem of determining the admitted symmetries is drastically simplified. For example, in order to find all admitted point symmetries of Equation (
11), we can assume that
, which enables us to split Equation (
18) with respect to
v. In the case of the so-called Hojman symmetries [
17,
21], when
, Equation (
18) is simplified to
. The derivation of the above equation with respect to
v leads to the following expression
, which gives the first integral
. The simplest case
leads to the general solution
.
Now, let us consider the general case. Let
be a nontrivial invariant of the symmetry operator
, i.e.,
then, the general invariant of
has the form
. Operator
A is a symmetry of the above equation, because
, therefore
. If
, then
. If
, then taking into account that
, we have the first-order equation
If one is able to determine the general integral of the above equation, then one obtains the first integral
of (
12).
Let us note that
, that is,
obtained from (
20) is an invariant of symmetry
. Moreover,
(if
, then
and from (
19),
, which gives a trivial invariant, or
and so from (
17),
); therefore, we can relate
v with a new variable
. Using
z, we can rewrite system (
13) with only one first-order equation
and the operator
A takes the form
. Thus, the nontrivial symmetry
allows us to reduce the order of the system (
12). Let us note that for Equation (
11) the described procedure is known as a method of differential invariants, because the invariant
(differential invariant of the first order) depends on the derivative
, and
is the differential invariant of the second order.
The integrating factor
of (
21) satisfies the equation
similar to (
14); therefore,
is a Jacobi multiplier, called Jacobi’s
last multiplier, because it permits us to determine the last integral
. Finally, the knowledge of an invariant of symmetry
and Jacobi’s last multiplier
allows us to solve the system (
12) in quadratures.
On the other hand, if it is possible to determine a function
such that
, then the Jacobian of transformation
,
, reads as
where the functions
can be taken as new variables. Moreover,
; therefore,
and
, i.e., in new variables (called canonical coordinates), the system
admits the symmetry
, which represents a translation in
u. If one knows the first integral
obtained from (
20), then
is the second first integral. As an example, let us consider [
22] (it is a particular case of the class of ODEs from [
23]).
, which admits the symmetry
. The canonical variables are
,
, and
; therefore,
It is interesting to note, that in this particular example, there is no symmetry for
A with a linear dependence on
v, that is, Equation (
11) corresponding to
A is completely integrable but does not admit any point symmetry, only the contact one. The importance of contact (dynamical) symmetries for the integration of ODEs of the second order is discussed, for example, in [
23].
3.2. Relations between Jacobi Multiplier, Last Jacobi Multiplier and Symmetries
Incidentally, there is a simple relation between the Jacobi multiplier and the coefficients of two nonproportional symmetries
and
admitted by system (
12).
Let us consider
. Then, [
9] (par. 152)
Any admitted symmetry transforms a solution of the given system into another solution of the same system, that is
, and we obtain
Therefore, from the property that
is a Jacobi multiplier if
, function
is a Jacobi multiplier.
Using the variables
t,
x,
z for Equation (
21), the symmetries
,
take the following form
and the integrating factor (last Jacobi multiplier) reads as follows:
Let
and
be two nonproportional symmetries of system (
12), forming the basis of two-dimensional Lie algebra. If the Lie algebra is not Abelian, then it is possible to determine
,
,
such that
. Here, and after, we assume that the Lie algebra
is Abelian or
.
Let us start to reduce the order of system (
12) using the symmetry
, i.e., selecting its invariant
, then expressing operator
A in terms of the variables
t,
x and
, we obtain
and the given system is reduced to the following one
Note that Equation (
20) is included in the above relations. The second symmetry
transforms to
and obviously commutes with
. Moreover,
(else
). Considering operator
in variables
t and
and applying all the facts from
Section 2 for ODE
, we obtain its integrating factor
.
If the symmetries
commute,
, then
is the symmetry of equation
, so
. In the case when
, we have
, that is, function
is an invariant of operator
; therefore, once again,
. In any case,
does not depend on
x explicitly. Therefore, the first integral does not either:
. Using the variables
t,
x and
the symmetry
takes the form
and produces the integrating factor
for Equation
. Collecting all integrating factors, we have the following relations between them:
To the best of our knowledge, the above relation has not been reported in the literature. Relations (
22) allow us to write down Equation (
20) as a differential one-form [
23]
Taking into account (
22), the above equation is equivalent to
and is therefore integrable.
Let us note that in canonical coordinates
, such that
,
, we have
The quantity
for two point symmetries
and
of Equation (
11) is used in the general scheme for the integration in quadratures of such ODEs [
1,
9] with Lie algebra
. In [
24], expression
is called the pseudo scalar product. It is easy to see that
, with
formed by two nonproportional canonical symmetries
corresponding to
, and
are coefficients of the first prolongation of
.
Another interesting relation between the Jacobi multiplier
M and symmetry
was pointed out by Bianchi [
9] (par. 152) and is given by
In the case of a divergence-free symmetry
, one obtains
. Using the well-known property of a symmetry to transform any solution of a given system to a solution of the same system, i.e.,
we conclude that
.
As an example, let us consider
with two commuting symmetries
,
and
. It is easy to see that
is an invariant of
and
. Therefore,
is the integrating factor of Equation (
20) for
. The integration gives the first integral
. For Equation (
21) with
, the integrating factor is
. Let us note that the last Jacobi multiplier
in this case coincides with
, because
.
If one knows one symmetry
and
, then, to obtain the second admitted symmetry
, we have the following three equations:
but it is easy to verify that Equation (
24) holds if (
23) is valid. Therefore, from (
23), we obtain one equation of the first order to determine
instead of an equation of the second order (
18).
3.3. Jacobi Multiplier and Noether Theorem
It is known [
18] that the Euler–Lagrange equation gives a system of differential equations for some Lagrangian if and only if the so-called operator of universal linearization of the system is autoconjugated (see [
19] Chapter 5, par. 4). Noether’s first theorem establishes a connection between the conservation laws of equations arising from the variational principle with Lagrangian symmetries. Until now, the use of this theorem has provided a practical tool for calculating the conservation laws of the Euler–Lagrange equations. In the case when a given equation is not related with a Lagrangian, one can use the concept of the generating function of conservation laws [
18]. Since any second-order ODE can be seen as the Euler–Lagrange equation for some Lagrangian, we now study the relation between the Jacobi multiplier and Noether’s theorem.
Suppose we have the following Lagrangian
; then, Equation (
11) is obtained by means of the Euler–Lagrange equation
where operator
is the operator given in (
15). For the system (
13), we have
in the case when
, we can obtain an expression for
in terms of the Lagrangian
The derivation of the above expression with respect to
v leads to Equation (
14) [
25] (par. 123) with
. That is, in the case of a nonzero Jacobi multiplier, one can obtain a Lagrangian by quadratures
with
where the right side of the above relation does not depend on
v due to (
14).
For two nonproportional symmetries
, the function
is a Jacobi multiplier, so one can take
and obtain Lagrangian
at least in quadratures. The relaxed condition for a point symmetry
to be a variational symmetry, i.e.,
for some function
, gives the relation
That is, we get the first integral
This is the main idea of Noether’s theorem, which relates a variational symmetry with a constant of motion. Let us note that function
is more general than gauge function
, which can be added to any Lagrangian in the form of a total derivative with respect to
t. If we consider a Lagrangian of the form
, then
Comparing the above relation with the relaxed condition, we obtain
which is a very particular form, and actually,
enters into the function
Q in (
27).
Let us consider as an example the classical harmonic oscillator
, which admits the scale symmetry
and the time-translation symmetry
with
. Introducing “polar” coordinates
,
, we obtain
,
,
,
. Therefore,
This Lagrangian was mentioned in [
26,
27]. Condition
is fulfilled for
. From (
27), we obtain the first integral
Taking into account that in “polar” coordinates,
, we have the well-known solution
and
.
3.4. Systems with the Same Jacobi Multiplier
Let us now consider a modified Lagrangian given by
where
is an arbitrary function. Note that
and
have the same Jacobi multiplier
M by construction. Using Euler–Lagrange Equation (
26), we obtain
Thus, we have two operators
A and
, which represent two different systems with different Lagrangians but with the same Jacobi multiplier. A natural question arises: if
is formed by two nonproportional symmetries
of
A, is it possible to get two symmetries
of
as functions of
?
Let us determine at least when symmetry
of
A is a symmetry for
, i.e., when
. From the commutator
, we obtain
and
. Condition
leads to the following relation
where
is a general invariant of operator
, i.e.,
(see also (
19)). The specific form of function
I should be selected to avoid the dependence of
g on
v. If there exists a symmetry
which commutes with
, then we obtain
. Using the relation
, we have
, i.e.,
Thus, there is a symmetry
if function
in (
29) satisfies the following equation (similar to (
23))
Let us consider the same example with a harmonic oscillator (see previous subsection). The general invariant of
is
. To determine the modified operator
which commutes with
and has the same
, we start from (
28) and obtain the necessary condition for
g
Taking into account that
, we obtain
. Thus, any equation
admits the scale symmetry
.
To satisfy Equation (
30), which takes the form
we assume
, so
and
. Thus,
, which defines the second symmetry of the modified equation, satisfies the relation (
29), i.e.,
. Finally, we obtain the equation of the “hidden” oscillator
, which admits the symmetry
which is nothing more than
.