Symmetries of Systems with the Same Jacobi Multiplier

Abstract

The concept of the Jacobi multiplier for ordinary differential equations up to the second order is reviewed and its connection with classical methods of canonical variables and differential invariants is established. We express, for equations of the second order, the Jacobi multiplier in terms of integrating factors for reduced equations of the first order. We also investigate, from a symmetry point of view, how two different systems with the same Jacobi multiplier are interrelated. As a result, we determine the conditions when such systems admit the same two-dimensional Lie algebra of symmetries. Several illustrative examples are given.

1. Introduction

Developing methods for finding solutions of ordinary differential equations (ODEs) has been a problem of interest over many centuries, with significant contributions of great mathematicians such as Euler, Jacobi, Lie, Poincaré and many others. It is very well known that the existence of sufficient number of first integrals (constants of motion) allows one to find the general solution of a given system of ordinary differential equations or reduce their order.

One way to determine a first integral is related with the concept of a group of continuous transformations and their invariants. The existence of such a group, admitted by given ODEs, reveals symmetry properties of differential equations and facilitate the process of integration. The foundations of the symmetry group analysis are contained in the works of Sophus Lie [1,2]. The actual formulation of the methods of modern group analysis of differential equations is described in various classical textbooks such as [3,4,5,6] and many others.

At present, symmetry methods are widely used to construct particular solutions (and sometimes general ones) of both linear and nonlinear equations. Based on the admitted symmetry algebras, one can analyze the integrability, classify the parameters included in the equations, describe conservation laws, search for coordinate systems in which second-order linear equations admit the separation of variables, etc. In the present work, we focus on the well-known but not completely studied concept of the Jacobi multiplier and its application to ODEs.

In Ref. [7], Jacobi introduced the concept of a multiplier M (now called the Jacobi multiplier), which satisfies a first-order partial differential equation, associated with a system of n first-order ODEs for n unknown functions. Moreover, he introduced what is now called the principle of the last multiplier $\mu$, which states the following: if one uses $n-1$ first integrals to express $n-1$ functions, then the last first integral can be expressed in quadratures with the help of an integrating factor (multiplier) $\mu$, that satisfies a similar equation for M.

The generalization and the basic properties of the Jacobi multiplier M and its relation to the concept of symmetry of a system of ODEs are described in [8] (see also [9]). Nucci, Leach and co-authors revived the interest of the Jacobi multiplier in the context of group analysis as a useful tool for calculating symmetries and their corresponding Lagrangians. For example, in Ref. [10], the last multiplier is used to calculate nonlocal symmetries for the Euler–Poinsot system; in Ref. [11], the first integrals of the Kepler problem are obtained by the method of Jacobi’s last multiplier; in a series of papers [12,13,14,15], the connection of the Jacobi multiplier with Lagrangians is analyzed for various mechanical systems. Let us mention also two recent works given in Ref. [16] and Ref. [17]. The first one provides the survey of the geometric theory of Jacobi multipliers in terms of jet manifolds in the spirit of Vinogradov’s school [18,19]. The second paper gives and up-to-date review of the Jacobi multiplier from a geometric perspective and shows some of its physical applications, such as finding constants of motion, which generalizes the so-called Hojman symmetry. Because the literature is vast, we recommend to the interested reader Ref. [20] and references therein for a detailed explanation of the use and applications of the Jacobi last multiplier.

The main objective of this work is to study a set of different Euler–Lagrange ODEs of the second order with different Lagrangians but with the same Jacobi multiplier and analyze the relation of their corresponding symmetries for such ODEs. The article is organized as follows: First, in Section 2, we determine the relationship between the symmetries of two first-order ODEs with the same integrating factor. For the case of a first-order ODE, the Jacobi multiplier is just the integrating factor. This section is more methodological in nature, but it is important for a more immediate understanding. In Section 3, we review the definition, main properties and connection of the Jacobi multiplier with the method of differential invariants for integrating a second-order ODE. Moreover, we establish the relation of the Jacobi multiplier for second-order ordinary differential equations, with integrating factors of two first-order ODEs, expressed in terms of invariants of an admitted symmetry. In this sense, the Jacobi multiplier can be considered as an integrating factor of a second-order ODE. Moreover, we consider the relation between the Lagrangian, obtained by means of the Jacobi multiplier, and the Noether theorem. Finally, for systems with the same Jacobi multiplier but with different Lagrangians, we find the necessary conditions for which two different ODEs share the same symmetries. We summarize our conclusions in Section 4.

2. First-Order Ordinary Differential Equation

2.1. Symmetries and Integrating Factor

First, we review some well-known facts from the theory of point symmetries for ordinary differential equations (ODEs) of the first order given by

$$\beta (x,y)dx-\alpha (x,y)dy=0\sim \frac{dx}{\alpha }=\frac{dy}{\beta }$$

(1)

or in normal form

$$F:=y^{\prime }-f(x,y)=0,f(x,y):=\beta /\alpha .$$

Let us remember that the problem of solving the ODE (1) is equivalent to determining an invariant of the operator $A:=\alpha {\partial }_x+\beta {\partial }_y$, i.e., a function $\omega =\omega (x,y)$ such that

$$A\left(\omega \right)=0,$$

then $\omega =\mathrm{const}.$ is the integral of (1). We use ${\partial }_t$ to denote $\frac{\partial }{\partial t}$ for the rest of the paper.

The standard way to find the point symmetry of (1) [3]

$$S=\xi (x,y){\partial }_x+\eta (x,y){\partial }_y$$

is to apply the first prolongation of S

$$\underset{1}{S}=S+\zeta {\partial }_{y^{\prime }},\zeta =D_x\left(\eta \right)-y^{\prime }{D}_x\left(\xi \right),D_x:={\partial }_x+y^{\prime }{\partial }_y$$

to equation $F=0$

$${\left.\underset{1}{S}\left(F\right)\right|}_{F=0}=0,$$

(2)

and to solve the first-order linear partial equations for the unknown functions $\xi$ and $\eta$.

Taking into account that ${\left.D_x\right|}_{F=0}=A/\alpha$, criterion (2) takes the form

$$\frac{S\left(\alpha \right)-A\left(\xi \right)}{\alpha }=\frac{S\left(\beta \right)-A\left(\eta \right)}{\beta }.$$

(3)

If one considers the relation for the commutator [5,9]

$$[S,A]=S\circ A-A\circ S=\lambda (x,y)A,$$

(4)

then we obtain Equation (3) exactly, excluding the function $\lambda$. Obviously, there is a trivial symmetry of the form $S^0=\delta (x,y)A$ for any smooth function $\delta$, because the above equality (4) holds for $\lambda =-A\left(\delta \right)$.

If one knows a nontrivial symmetry S, then the application of (4) to integral $\omega$ gives

$$S\left(A\right(\omega \left)\right)-A\left(S\right(\omega \left)\right)=-A\left(S\right(\omega \left)\right)=\lambda A\left(\omega \right)=0;$$

therefore, $S\left(\omega \right)=h\left(\omega \right)$ for some function h, which represents the well-known property of an admitted symmetry to transform a solution of a given equation into another solution of the same equation. Thus, the integral $\omega$ satisfies the following algebraic system of equations for ${\omega }_{,x}:={\partial }_x\left(\omega \right)$, ${\omega }_{,y}:={\partial }_y\left(\omega \right)$, respectively, given by:

$$\begin{array}{ccc}\hfill A\left(\omega \right)& =& \alpha {\omega }_{,x}+\beta {\omega }_{,y}=0,\hfill \\ \hfill S\left(\omega \right)& =& \xi {\omega }_{,x}+\eta {\omega }_{,y}=h\left(\omega \right).\hfill \end{array}$$

The determinant of this system $\Delta :=\beta \xi -\alpha \eta$ is nonzero; therefore, we can write

$$d\omega =h\left(\omega \right)\left(\frac{\beta }{\Delta }dx-\frac{\alpha }{\Delta }dy\right)=0.$$

If we demand $\mu :=1/\Delta$ to be the integrating factor for (1)

$${\left(\frac{\beta }{\Delta }\right)}_{,y}=-{\left(\frac{\alpha }{\Delta }\right)}_{,x}$$

(5)

we arrive at expression (3). Thus, to obtain a symmetry S, one can use the equivalent relations: (2), (4) or (5). This connection reflects the well-known fact that in general, the problem of finding a point symmetry of (1) has the same degree of difficulty as the problem of finding its integrating factor.

If an integrating factor $\mu$ for (1) is known, then expressing $\xi =\Delta /\beta +\alpha \eta /\beta$, we obtain

$$S=\left(\Delta /\beta +\alpha \eta /\beta \right){\partial }_x+\eta {\partial }_y=\frac{\Delta }{\beta }{\partial }_x+\frac{\eta }{\beta }A=:S^x+S^0,$$

and for $\eta =\beta \xi /\alpha -\Delta /\alpha$, by analogy, we have

$$S=\xi {\partial }_x+\left(\beta \xi /\alpha -\Delta /\alpha \right){\partial }_y=-\frac{\Delta }{\alpha }{\partial }_y+\frac{\xi }{\alpha }A=:S^y+S^0,$$

where we have taken into account that $\frac{\eta }{\beta }A$ and $\frac{\xi }{\alpha }A$ are trivial symmetries $S^0$. Thus, any equation of (1) with integrating factor $\mu =\mu (x,y)$ admits the symmetry (which we call the canonical symmetry)

$$S^x={\left(\beta \mu \right)}^{-1}{\partial }_x\mathrm{or}{S}^y={\left(\alpha \mu \right)}^{-1}{\partial }_y.$$

2.2. Symmetries of Two ODEs with the Same Integrating Factor

Let us now consider another ODE different from (1), but with the same integrating factor $\mu (x,y)$

$$\tilde{\beta }(x,y)dx-\tilde{\alpha }(x,y)dy=0,$$

(6)

where $\alpha \tilde{\beta }-\beta \tilde{\alpha }\ne 0$. Then, the relation between the canonical symmetry $S^x$ for (1) and the symmetry ${\tilde{S}}^x={\left(\tilde{\beta }\mu \right)}^{-1}{\partial }_x$ for (6) is very simple

$${\tilde{S}}^x=\frac{\beta }{\tilde{\beta }}{S}^x.$$

Due to the presence of the trivial symmetry $S^0=\delta (x,y)A$, it is easy to show that any two equations with the same integrating factor actually have the same symmetry. Indeed, from the following equality

$$S=S^x+\delta A={\tilde{S}}^x+\tilde{\delta }\tilde{A},$$

(7)

we obtain

$$\tilde{\delta }\tilde{\beta }=\delta \beta ,\delta ={\mu }^{-1}\frac{1-\tilde{\beta }/\beta }{\alpha \tilde{\beta }-\beta \tilde{\alpha }}.$$

(8)

This result, to the best of our knowledge, has not been reported in the literature.

For example, let us consider two equations. The first one is with homogeneous coefficients

$$2dx-\frac{x^2}{y^2}dy=0,$$

(9)

and the second one is the linear equation

$$\left(cosx-2\frac{y}{x}\right)dx+dy=0.$$

(10)

Both equations have the same integrating factor $\mu =x^{-2}$; therefore, their canonical symmetries are given by:

$$S^x=\frac{x^2}{2}{\partial }_x,{\tilde{S}}^x=\frac{x^3}{xcosx-2y}{\partial }_x.$$

It is easy to see, using (7) and (8), that Equations (9) and (10) admit the same following symmetry

$$S=\frac{x^2(x^2+y^2){\partial }_x+x{y}^2(2x+2y-xcosx){\partial }_y}{2y^2-2xy+x^{2}cosx},$$

which gives the same integrating factor $\mu$ for both equations.

3. Second-Order Ordinary Differential Equation

3.1. Symmetries and the Jacobi Multiplier

Let us now consider a second-order differential equation given in the following form

$$\ddot{x}=f(t,x,\dot{x})$$

(11)

for the unknown function $x=x\left(t\right)$. Introducing a new function $v:=\dot{x}=dx/dt$, we obtain the system $F:=(F^1,F^2)=0$ of two first-order equations:

$$F^1:=\dot{x}-v=0,F^2:=\dot{v}-f(t,x,v)=0,$$

(12)

or in its symmetric form

$$dt=\frac{dx}{v}=\frac{dv}{f}.$$

(13)

We can consider the following operator $A:={\partial }_t+v{\partial }_x+f{\partial }_v$, 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 ${\omega }^i={\omega }^i(t,x,v)=c^i$ (constants of motion), such that

$${\left.d{\omega }^i\right|}_{F=0}=0,\left|J^{xv}\right|:=\left|\frac{\partial ({\omega }^1,{\omega }^2)}{\partial (x,v)}\right|\ne 0.$$

From (13) it follows that $A\left({\omega }^i\right)=0$, i.e., functions ${\omega }^i$ are invariants of the operator A. Since any other solution $\Omega$ of (12) is an invariant of A and is a function of ${\omega }^1$ and ${\omega }^2$, the Jacobian reads

$$\left|\frac{\partial (\Omega ,{\omega }^1,{\omega }^2)}{\partial (t,x,v)}\right|=0,$$

i.e.,

$$\left|J^{xv}\right|{\Omega }_{,t}-\left|J^{tv}\right|{\Omega }_{,x}+\left|J^{tx}\right|{\Omega }_{,v}=0,|J^{tv}|:=\left|\frac{\partial ({\omega }^1,{\omega }^2)}{\partial (t,v)}\right|,\left|J^{tx}\right|:=\left|\frac{\partial ({\omega }^1,{\omega }^2)}{\partial (t,x)}\right|.$$

Comparing the above relation with $A(\Omega )=0$, and denoting $M:=\left|J^{xv}\right|$, we obtain

$$Mv=-\left|J^{tv}\right|,Mf=\left|J^{tx}\right|.$$

The function M is called the Jacobi multiplier [9]. Due to the relation of M with Jacobians, M is a solution of the equation

$$M_{,t}+{\left(Mv\right)}_{,x}+{\left(Mf\right)}_{,v}=0.$$

(14)

The form of M depends on the selection of ${\omega }^{1,2}$. If one has ${\tilde{\omega }}^1$, ${\tilde{\omega }}^2$ as an independent first integrals, then function $\tilde{M}=M\left|\frac{\partial ({\tilde{\omega }}^1,{\tilde{\omega }}^2)}{\partial ({\omega }^1,{\omega }^2)}\right|$ is a Jacobi multiplier, and in general, any function of the form $M\Omega$ 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

$$S=\xi (t,x,v){\partial }_t+{\eta }^1(t,x,v){\partial }_x+{\eta }^2(t,x,v){\partial }_v$$

and represents the so-called contact symmetry for Equation (11) if $\xi$ or ${\eta }^1$ depends on v (with the nonlinear dependence of ${\eta }^1$ on v if ${\xi }_{,v}=0$). The application of the first prolongation of S

$$\begin{array}{cc}& \underset{1}{S}=S+{\zeta }^1{\partial }_{\dot{x}}+{\zeta }^2{\partial }_{\dot{v}},\\ & {\zeta }^1=D_t\left({\eta }^1\right)-\dot{x}{D}_t\left(\xi \right),{\zeta }^2=D_t\left({\eta }^2\right)-\dot{v}{D}_t\left(\xi \right),\\ & D_t={\partial }_t+\dot{x}{\partial }_x+\dot{v}{\partial }_v\end{array}$$

(15)

to system $F=0$

$${\left.\underset{1}{S}\left(F^1\right)\right|}_{F=0}=0,{\left.\underset{1}{S}\left(F^2\right)\right|}_{F=0}=0$$

gives two linear equations for determining $\xi$, ${\eta }^1$ and ${\eta }^2$:

$${\eta }^2=A\left({\eta }^1\right)-vA\left(\xi \right),S\left(f\right)=A\left({\eta }^2\right)-fA\left(\xi \right),$$

(16)

where we have taken into account that ${\left.D_t\right|}_{F=0}={\partial }_t+v{\partial }_x+f{\partial }_v=A$.

Excluding the function $\lambda$ from the commutator relation

$$[S,A]=\lambda (t,x,v)A$$

leads to (16). Once again, there is a trivial symmetry $S^0=\delta (t,x,v)A$. For simplicity, one can introduce the following operator (evolutionary form of symmetry [6]):

$$\widehat{S}:=S-\xi A=:{\widehat{\eta }}^1{\partial }_x+{\widehat{\eta }}^2{\partial }_v.$$

The function ${\widehat{\eta }}^1$ is called in Ref. [18] a generating function of symmetry. For the operator $\widehat{S}$, we have

$$[\widehat{S},A]=0\sim {\widehat{\eta }}^2=A\left({\widehat{\eta }}^1\right),\widehat{S}\left(f\right)=A\left({\widehat{\eta }}^2\right),{\widehat{\eta }}^1\not\equiv 0.$$

(17)

Thus, to determine the unknown function ${\widehat{\eta }}^1$, we have to solve one second-order linear partial differential equation (defining equation) given by

$$\widehat{S}\left(f\right)=A\left(A\left({\widehat{\eta }}^1\right)\right).$$

(18)

As in the case of the first-order ODE, there is an infinite number of point symmetries $\widehat{S}$ 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 $f(t,x,v)$, 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 ${\widehat{\eta }}^1=a(t,x)v+b(t,x)$, which enables us to split Equation (18) with respect to v. In the case of the so-called Hojman symmetries [17,21], when $f_{,v}=0$, Equation (18) is simplified to $A\left(A\left({\widehat{\eta }}^1\right)\right)-{\widehat{\eta }}^1{f}_{,x}=0$. The derivation of the above equation with respect to v leads to the following expression $A({\widehat{\eta }}_{,x}^1+{\widehat{\eta }}_{,v}^2)=0$, which gives the first integral $\mathrm{div}\widehat{S}:={\widehat{\eta }}_{,x}^1+{\widehat{\eta }}_{,v}^2=\mathrm{const}$. The simplest case $f=0$ leads to the general solution ${\widehat{\eta }}^1=a(v,x-tv)t+b(v,x-tv)$.

Now, let us consider the general case. Let $\omega =\omega (t,x,v)$ be a nontrivial invariant of the symmetry operator $\widehat{S}$, i.e.,

$$\widehat{S}\left(\omega \right)={\widehat{\eta }}^1{\omega }_{,x}+{\widehat{\eta }}^2{\omega }_{,v}=0,\omega \ne t,\omega \not\equiv \mathrm{const}.;$$

(19)

then, the general invariant of $\widehat{S}$ has the form $I=I(t,\omega )$. Operator A is a symmetry of the above equation, because $[A,\widehat{S}]=0$, therefore $A\left(\omega \right)=:h(t,\omega )$. If $h\equiv 0$, then ${\omega }^1=\omega$. If $h\not\equiv 0$, then taking into account that $A\left(\omega \right)=d\omega /dt$, we have the first-order equation

$$\frac{d\omega }{dt}=h(t,\omega ).$$

(20)

If one is able to determine the general integral of the above equation, then one obtains the first integral ${\omega }^1={\omega }^1(t,\omega )=c^1$ of (12).

Let us note that $\widehat{S}\left({\omega }^1\right)={\omega }_{,\omega }^1\widehat{S}\left(\omega \right)=0$, that is, ${\omega }^1$ obtained from (20) is an invariant of symmetry $\widehat{S}$. Moreover, ${\omega }_{,v}^1\ne 0$ (if ${\omega }_{,v}^1=0$, then ${\omega }_{,v}=0$ and from (19), ${\omega }_{,x}=0$, which gives a trivial invariant, or ${\widehat{\eta }}^1=0$ and so from (17), ${\widehat{\eta }}^2=0$); therefore, we can relate v with a new variable $z:={\omega }^1(t,x,v)=c^1$. Using z, we can rewrite system (13) with only one first-order equation

$$v(t,x,z)dt-dx=0,$$

(21)

and the operator A takes the form $A^z={\partial }_t+v(t,x,z){\partial }_x$. Thus, the nontrivial symmetry $\widehat{S}$ 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 $\omega$ (differential invariant of the first order) depends on the derivative $v=\dot{x}$, and $A\left(\omega \right)$ is the differential invariant of the second order.

The integrating factor $\mu =\mu (t,x,z)$ of (21) satisfies the equation ${\mu }_{,t}+{\left(\mu v(t,x,z)\right)}_{,x}=0$ similar to (14); therefore, $\mu$ is a Jacobi multiplier, called Jacobi’s last multiplier, because it permits us to determine the last integral ${\omega }^2=c^2$. Finally, the knowledge of an invariant of symmetry $\widehat{S}$ and Jacobi’s last multiplier $\mu$ allows us to solve the system (12) in quadratures.

On the other hand, if it is possible to determine a function $u(t,x,v)$ such that $\widehat{S}\left(u\right)=1$, then the Jacobian of transformation $u=u(t,x,v)$, $\omega =\omega (t,x,v)$, reads as

$$|J^{u\omega }|:=\left|\frac{\partial (u,\omega )}{\partial (x,v)}\right|=u_{,x}{\omega }_{,v}-u_{,v}{\omega }_{,x}=\frac{{\omega }_{,v}}{{\widehat{\eta }}^1}\ne 0,$$

where the functions $(u,\omega )$ can be taken as new variables. Moreover, $\widehat{S}\left(A\left(u\right)\right)=A\left(\widehat{S}\left(u\right)\right)=0$; therefore, $A\left(u\right)=:a(t,\omega )$ and $A^{u\omega }={\partial }_t+a(t,\omega ){\partial }_u+h(t,\omega ){\partial }_{\omega }$, i.e., in new variables (called canonical coordinates), the system $F=0$ admits the symmetry ${\widehat{S}}^{u\omega }={\partial }_u$, which represents a translation in u. If one knows the first integral ${\omega }^1(t,\omega )=c^1$ obtained from (20), then $u-\int a(t,\omega (t,c^1))dt={\omega }^2=c^2$ is the second first integral. As an example, let us consider [22] (it is a particular case of the class of ODEs from [23]). $A={\partial }_t+v{\partial }_x+\left(v\frac{t^2+t+v}{x}-2t-1\right){\partial }_v$, which admits the symmetry $\widehat{S}=exp\left\{t(t^2+t+v)/x\right\}\left({\partial }_x+\frac{t^2+t+v}{x}{\partial }_v\right)$. The canonical variables are $u=x{e}^{-t\omega }$, $\omega =(t^2+t+v)/x$, and $A^{u\omega }={\partial }_t-t(t+1)e^{-t\omega }{\partial }_u$; therefore,

$${\omega }^1=\omega =c^1,{\omega }^2=x{e}^{-c_{1}t}+\int t(t+1)e^{-c_{1}t}dt=c^2.$$

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 ${\widehat{S}}^1:={\widehat{\eta }}^{11}{\partial }_x+{\widehat{\eta }}^{12}{\partial }_v$ and ${\widehat{S}}^2:={\widehat{\eta }}^{21}{\partial }_x+{\widehat{\eta }}^{22}{\partial }_v$ admitted by system (12).

Let us consider $\Delta ={\widehat{\eta }}^{11}{\widehat{\eta }}^{22}-{\widehat{\eta }}^{12}{\widehat{\eta }}^{21}\ne 0$. Then, [9] (par. 152)

$$\begin{array}{ll}\Delta & M=\left|\begin{array}{cc}{\widehat{\eta }}^{11}& {\widehat{\eta }}^{12}\\ {\widehat{\eta }}^{21}& {\widehat{\eta }}^{22}\end{array}\right|\left|J^{xv}\right|=\left|\left(\begin{array}{cc}{\widehat{\eta }}^{11}& {\widehat{\eta }}^{12}\\ {\widehat{\eta }}^{21}& {\widehat{\eta }}^{22}\end{array}\right)\left(\begin{array}{cc}{\omega }_{,x}^1& {\omega }_{,x}^2\\ {\omega }_{,v}^1& {\omega }_{,v}^2\end{array}\right)\right|\hfill \\ & \hfill =\left|\begin{array}{cc}{\widehat{S}}^1\left({\omega }^1\right)& {\widehat{S}}^1\left({\omega }^2\right)\\ {\widehat{S}}^2\left({\omega }^1\right)& {\widehat{S}}^2\left({\omega }^2\right)\end{array}\right|={\widehat{S}}^1\left({\omega }^1\right){\widehat{S}}^2\left({\omega }^2\right)-{\widehat{S}}^1\left({\omega }^2\right){\widehat{S}}^2\left({\omega }^1\right).\end{array}$$

Any admitted symmetry transforms a solution of the given system into another solution of the same system, that is ${\widehat{S}}^i\left({\omega }^j\right)=h^{ij}({\omega }^1,{\omega }^2)$, and we obtain

$$\Delta M=H({\omega }^1,{\omega }^2),A\left(H\right)=0.$$

Therefore, from the property that $M\Omega$ is a Jacobi multiplier if $A(\Omega )=0$, function ${\Delta }^{-1}$ is a Jacobi multiplier.

Using the variables t, x, z for Equation (21), the symmetries ${\widehat{S}}^1$, ${\widehat{S}}^2$ take the following form

$${\widehat{Z}}^1:={\widehat{S}}^1\left(x\right){\partial }_x,{\widehat{Z}}^2:={\widehat{S}}^2\left(x\right){\partial }_x+{\widehat{S}}^2\left(z\right){\partial }_z,$$

and the integrating factor (last Jacobi multiplier) reads as follows:

$${\mu }^{-1}(t,x,z)=\left|\begin{array}{cc}{\widehat{S}}^1\left(x\right)& 0\\ {\widehat{S}}^2\left(x\right)& {\widehat{S}}^2\left(z\right)\end{array}\right|={\widehat{S}}^1\left(x\right){\widehat{S}}^2\left(z\right)=:{\Delta }^z,{\Delta }^z={\omega }_{,v}^1\Delta .$$

Let ${\widehat{S}}^1$ and ${\widehat{S}}^2$ 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 ${\overline{S}}^i=a^i{\widehat{S}}^1+b^i{\widehat{S}}^2$, $a^i$, $b^i\in \mathbb{R}$ such that $[{\overline{S}}^1,{\overline{S}}^2]={\overline{S}}^1$. Here, and after, we assume that the Lie algebra $\langle {\widehat{S}}^1,{\widehat{S}}^2\rangle$ is Abelian or $[{\widehat{S}}^1,{\widehat{S}}^2]={\widehat{S}}^1$.

Let us start to reduce the order of system (12) using the symmetry ${\widehat{S}}^1$, i.e., selecting its invariant ${\widehat{S}}^1\left(\omega \right)=0$, then expressing operator A in terms of the variables t, x and $\omega$, we obtain

$$A^{\omega }:={\partial }_t+v(t,x,\omega ){\partial }_x+h(t,\omega ){\partial }_{\omega },$$

and the given system is reduced to the following one

$$dt=\frac{dx}{v(t,x,\omega )}=\frac{d\omega }{h(t,\omega )}.$$

Note that Equation (20) is included in the above relations. The second symmetry ${\widehat{S}}^2$ transforms to ${\widehat{S}}^{2\omega }:={\widehat{S}}^2\left(x\right){\partial }_x+{\widehat{S}}^2\left(\omega \right){\partial }_{\omega }$ and obviously commutes with $A^{\omega }$. Moreover, ${\widehat{S}}^2\left(\omega \right)\ne 0$ (else $\Delta =0$). Considering operator ${\widehat{S}}^{2\omega }$ in variables t and $\omega$ and applying all the facts from Section 2 for ODE $h(t,\omega )dt-d\omega =0$, we obtain its integrating factor ${\mu }^{\omega }:=-1/{\widehat{S}}^2\left(\omega \right)$.

If the symmetries ${\widehat{S}}^i$ commute, $[{\widehat{S}}^1,{\widehat{S}}^2]=0$, then ${\widehat{S}}^2$ is the symmetry of equation ${\widehat{S}}^1\left(\omega \right)=0$, so ${\widehat{S}}^2\left(\omega \right)=h^2(t,\omega )$. In the case when $[{\widehat{S}}^1,{\widehat{S}}^2]={\widehat{S}}^1$, we have $[{\widehat{S}}^1,{\widehat{S}}^2]\left(\omega \right)={\widehat{S}}^1\left({\widehat{S}}^2\left(\omega \right)\right)=0$, that is, function ${\widehat{S}}^2\left(\omega \right)$ is an invariant of operator ${\widehat{S}}^1$; therefore, once again, ${\widehat{S}}^2\left(\omega \right)=h^2(t,\omega )$. In any case, ${\mu }^{\omega }$ does not depend on x explicitly. Therefore, the first integral does not either: ${\omega }^1={\omega }^1(t,\omega )=z$. Using the variables t, x and $v(t,x,z)$ the symmetry ${\widehat{S}}^1$ takes the form ${\widehat{S}}^1\left(x\right){\partial }_x$ and produces the integrating factor ${\mu }^x:=1/{\widehat{S}}^1\left(x\right)=1/{\widehat{\eta }}^{11}$ for Equation $v(t,x,z)dt-dx=0$. Collecting all integrating factors, we have the following relations between them:

$${\Delta }^{-1}=-{\omega }_{,v}{\mu }^x{\mu }^{\omega },\mu =-{\mu }^x{\mu }^{\omega }/{\omega }_{,\omega }^1.$$

(22)

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]

$$d{\omega }^1={\Delta }^{-1}\left|\begin{array}{ccc}dt& dx& dv\\ 1& v& f\\ 0& {\eta }^1& {\eta }^2\end{array}\right|={\Delta }^{-1}{\omega }_{,\omega }^1\left(-A\left(\omega \right)dt+d\omega \right)=0.$$

Taking into account (22), the above equation is equivalent to ${\mu }^{\omega }(h(t,\omega )dt-d\omega )=0$ and is therefore integrable.

Let us note that in canonical coordinates $(u,\omega )$, such that ${\widehat{S}}^1\left(u\right)=1$, ${\widehat{S}}^1\left(\omega \right)=0$, we have

$${\Delta }^{u\omega }=\left|\begin{array}{cc}1& 0\\ {\widehat{S}}^2\left(u\right)& {\widehat{S}}^2\left(\omega \right)\end{array}\right|={\widehat{S}}^2\left(\omega \right)=-1/{\mu }^{\omega }=\left|J^{u\omega }\right|\Delta .$$

The quantity $S^1\vee S^2:={\xi }^1{\eta }^2-{\eta }^1{\xi }^2$ for two point symmetries $S^1={\xi }^1(t,x){\partial }_t+{\eta }^1(t,x){\partial }_x$ and $S^2={\xi }^2(t,x){\partial }_t+{\eta }^2(t,x){\partial }_x$ of Equation (11) is used in the general scheme for the integration in quadratures of such ODEs [1,9] with Lie algebra $\langle S^1,S^2\rangle$. In [24], expression $S^1\vee S^2$ is called the pseudo scalar product. It is easy to see that $\Delta =(S^1\vee S^2)f+{\widehat{\eta }}^{11}{\zeta }^{12}-{\widehat{\eta }}^{21}{\zeta }^{11}$, with $\Delta$ formed by two nonproportional canonical symmetries ${\widehat{S}}^{1,2}$ corresponding to $S^{1,2}$, and ${\zeta }^{1i}$ are coefficients of the first prolongation of $S^i$.

Another interesting relation between the Jacobi multiplier M and symmetry $\widehat{S}$ was pointed out by Bianchi [9] (par. 152) and is given by

$$A(\widehat{S}(lnM)+\mathrm{div}\widehat{S})=0.$$

In the case of a divergence-free symmetry $\widehat{S}$, one obtains $\widehat{S}(lnM)=\Omega ({\omega }^1,{\omega }^2)$. 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., $\widehat{S}\left(H({\omega }^1,{\omega }^2)\right)=\Omega ({\omega }^1,{\omega }^2)$ we conclude that $A\left(M\right)=0$.

As an example, let us consider $A={\partial }_t+v{\partial }_x+2v(v+1)/(x-t){\partial }_v$ with two commuting symmetries ${\widehat{S}}^1=\frac{v+1}{t-x}\left({\partial }_x+\frac{v+1}{x-t}{\partial }_v\right)$, ${\widehat{S}}^2=(1-v){\partial }_x+2\frac{v+v^2}{t-x}{\partial }_v$ and $\Delta ={(v+1)}^3/{(t-x)}^2$. It is easy to see that $\omega =(v+1)/(x-t)$ is an invariant of ${\widehat{S}}^1$ and ${\widehat{S}}^2\left(\omega \right)=-{\omega }^2$. Therefore, ${\mu }^{\omega }={\omega }^{-2}$ is the integrating factor of Equation (20) for $h(t,\omega )=A\left(\omega \right)={\omega }^2$. The integration gives the first integral ${\omega }^1=t+{\omega }^{-1}$. For Equation (21) with $v(t,x,z)=\frac{x-z}{z-t}$, the integrating factor is ${\mu }^x=1/{\widehat{S}}^1\left(x\right)=t-z$. Let us note that the last Jacobi multiplier $\mu =1/{\Delta }^z$ in this case coincides with ${\mu }^x$, because ${\widehat{S}}^2\left(z\right)=1$.

If one knows one symmetry ${\widehat{S}}^1$ and $\Delta$, then, to obtain the second admitted symmetry ${\widehat{S}}^2$, we have the following three equations:

$$\begin{array}{cc}& {\widehat{\eta }}^{22}=A\left({\widehat{\eta }}^{21}\right),\Delta ={\widehat{\eta }}^{11}{\widehat{\eta }}^{22}-{\widehat{\eta }}^{12}{\widehat{\eta }}^{21},\end{array}$$

(23)

$$\begin{array}{cc}& {\widehat{S}}^2\left(f\right)=A\left({\widehat{\eta }}^{22}\right),\end{array}$$

(24)

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 ${\widehat{\eta }}^{21}$

$$A\left(\frac{{\widehat{\eta }}^{21}}{{\widehat{\eta }}^{11}}\right)=\frac{\Delta }{{\left({\widehat{\eta }}^{11}\right)}^2}$$

(25)

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 $L=L(t,x,v)$; then, Equation (11) is obtained by means of the Euler–Lagrange equation

$$L_{,x}-D_t\left(L_{,v}\right)=0,$$

(26)

where operator $D_t$ is the operator given in (15). For the system (13), we have

$$E\left(L\right)=0,E:={\partial }_x-A\circ {\partial }_v;$$

in the case when $L_{,vv}\ne 0$, we can obtain an expression for $f(t,x,v)$ in terms of the Lagrangian

$$f(t,x,v)=\frac{L_{,x}-L_{,tv}-v{L}_{,xv}}{L_{,vv}}.$$

The derivation of the above expression with respect to v leads to Equation (14) [25] (par. 123) with $M=L_{,vv}$. That is, in the case of a nonzero Jacobi multiplier, one can obtain a Lagrangian by quadratures

$$L^M:=\int \int Mdvdv+Q(t,x)v+R(t,x),$$

with

$$Q_{,t}-R_{,x}=\int \int M_{,x}dvdv-A\left(\int Mdv\right),$$

where the right side of the above relation does not depend on v due to (14).

For two nonproportional symmetries ${\widehat{S}}^{1,2}$, the function ${\Delta }^{-1}$ is a Jacobi multiplier, so one can take $M={\Delta }^{-1}\ne 0$ and obtain Lagrangian $L^M$ at least in quadratures. The relaxed condition for a point symmetry ${\widehat{S}}^1$ to be a variational symmetry, i.e., ${\widehat{S}}^1\left(L^M\right)=A\left(B\right)$ for some function $B(t,x,v)$, gives the relation

$${\widehat{\eta }}^{11}{L}_{,x}^M+{\widehat{\eta }}^{12}{L}_{,v}^M=A\left(B\right)\sim A\left({\widehat{\eta }}^{11}{L}_{,v}^M-B\right)=0.$$

That is, we get the first integral

$$\Omega ({\omega }^1,{\omega }^2)={\widehat{\eta }}^{11}\left(\int Mdv+Q(t,x)\right)-B(t,x,v)=\mathrm{const}.$$

(27)

This is the main idea of Noether’s theorem, which relates a variational symmetry with a constant of motion. Let us note that function $B(t,x,v)$ is more general than gauge function $G(t,x)$, 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 $L=L^M+A\left(G(t,x)\right)$, then

$${\widehat{S}}^1\left(L\right)={\widehat{S}}^1\left(L^M\right)+{\widehat{S}}^1\left(A\left(G\right)\right)={\widehat{S}}^1\left(L^M\right)+{\widehat{\eta }}^{11}A\left(G_{,x}\right)+A\left({\widehat{\eta }}^{11}\right)G_{,x}={\widehat{S}}^1\left(L^M\right)+A\left({\widehat{\eta }}^{11}{G}_{,x}\right).$$

Comparing the above relation with the relaxed condition, we obtain

$$B(t,x,v)=-{\widehat{\eta }}^{11}{G}_{,x}(t,x),$$

which is a very particular form, and actually, $G_{,x}$ enters into the function Q in (27).

Let us consider as an example the classical harmonic oscillator $\ddot{x}=-k^{2}x$, which admits the scale symmetry ${\widehat{S}}^1=x{\partial }_x+v{\partial }_v$ and the time-translation symmetry ${\widehat{S}}^2={\partial }_t-A=-v{\partial }_x+k^{2}x{\partial }_v$ with $\Delta =k^2{x}^2+v^2=M^{-1}$. Introducing “polar” coordinates $v=rsin\varphi$, $kx=rcos\varphi$, we obtain $A^{r\varphi }={\partial }_t-k{\partial }_{\varphi }$, ${\widehat{S}}^{1r\varphi }=r{\partial }_r$, ${\widehat{S}}^{2r\varphi }=k{\partial }_{\varphi }$, $M=r^{-2}$. Therefore,

$$L^M=\frac{v}{kx}arctan\frac{v}{kx}-\frac{1}{2}ln(v^2+k^2{x}^2)=\varphi tan\varphi -lnr.$$

This Lagrangian was mentioned in [26,27]. Condition ${\widehat{S}}^{1r\varphi }\left(L\right)=-1=A^{r\varphi }\left(B\right)$ is fulfilled for $B=-t$. From (27), we obtain the first integral

$${\Omega }^1=x\frac{\varphi }{rcos\varphi }+t=\mathrm{const}.\sim kt+\varphi =c^1.$$

Taking into account that in “polar” coordinates, $\dot{r}=0\sim r=c^2=\mathrm{const}$, we have the well-known solution $v=c^{2}sin(c^1-kt)$ and $kx=c^{2}cos(c^1-kt)$.

3.4. Systems with the Same Jacobi Multiplier

Let us now consider a modified Lagrangian given by

$$\tilde{L}:=L^M-\int g(t,x)dx,$$

where $g(t,x)$ is an arbitrary function. Note that $\tilde{L}$ and $L^M$ have the same Jacobi multiplier M by construction. Using Euler–Lagrange Equation (26), we obtain

$$\tilde{f}=f-\frac{g(t,x)}{M}.$$

Thus, we have two operators A and $\tilde{A}:=A-\Delta g(t,x){\partial }_v$, which represent two different systems with different Lagrangians but with the same Jacobi multiplier. A natural question arises: if $\Delta$ is formed by two nonproportional symmetries ${\widehat{S}}^{1,2}$ of A, is it possible to get two symmetries ${\tilde{S}}^{1,2}$ of $\tilde{A}$ as functions of ${\widehat{\eta }}^{ij}$?

Let us determine at least when symmetry ${\widehat{S}}^1$ of A is a symmetry for $\tilde{A}$, i.e., when ${\tilde{S}}^1={\widehat{S}}^1$. From the commutator $[{\widehat{S}}^1,\tilde{A}]=0$, we obtain ${\widehat{\eta }}_{,v}^{11}=0$ and ${\widehat{\eta }}_{,x}^{11}={\widehat{S}}^1\left(lng\Delta \right)$. Condition $g_{,v}=0$ leads to the following relation

$$g(t,x)=\frac{I(t,\omega )}{{\Delta }_{,v}},$$

(28)

where $I(t,\omega )$ is a general invariant of operator ${\widehat{S}}^1$, i.e., ${\widehat{S}}^1\left(I\right)=0$ (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 ${\tilde{S}}^2={\tilde{\eta }}^{21}{\partial }_x+{\tilde{\eta }}^{22}{\partial }_v$ which commutes with ${\widehat{S}}^1$, then we obtain ${\widehat{S}}^1\left({\tilde{\eta }}^{21}\right)={\tilde{S}}^2\left({\eta }^{11}\right)$. Using the relation $\Delta ={\widehat{\eta }}^{11}{\tilde{\eta }}^{22}-{\tilde{\eta }}^{21}{\widehat{\eta }}^{12}$, we have ${\widehat{S}}^1\left({\tilde{\eta }}^{21}/{\widehat{\eta }}^{11}\right)=0$, i.e.,

$${\tilde{\eta }}^{21}={\widehat{\eta }}^{11}I(t,\omega ).$$

(29)

Thus, there is a symmetry ${\tilde{S}}^2$ if function $I(t,\omega )$ in (29) satisfies the following equation (similar to (23))

$$\tilde{A}\left(I\right)=I_{,t}+I_{,\omega }(v{\omega }_{,x}+\tilde{f}{\omega }_{,v})=\frac{\Delta }{{\left({\widehat{\eta }}^{11}\right)}^2}.$$

(30)

Let us consider the same example with a harmonic oscillator (see previous subsection). The general invariant of ${\widehat{S}}^1$ is $I(t,\omega )=I(t,v/x)$. To determine the modified operator $\tilde{A}$ which commutes with ${\widehat{S}}^1$ and has the same $\Delta$, we start from (28) and obtain the necessary condition for g

$$g(t,x)=-\frac{I(t,v/x)}{2v}.$$

Taking into account that $g_{,v}=0$, we obtain $g=\tau \left(t\right)/x$. Thus, any equation $\ddot{x}=\tilde{f}=-k^{2}x(1+\tau \left(t\right))-\tau \left(t\right)v^2/x$ admits the scale symmetry ${\widehat{S}}^1$.

To satisfy Equation (30), which takes the form

$$I_{,t}-(1+\tau )I_{,\omega }(k^2+{\omega }^2)=k^2+{\omega }^2,$$

we assume $I_{,t}=0$, so $\tau =c=\mathrm{const}.$ and $I=-\omega /(c+1)$. Thus, ${\tilde{\eta }}^{21}$, which defines the second symmetry of the modified equation, satisfies the relation (29), i.e., ${\tilde{\eta }}^{21}=-v/(c+1)$. Finally, we obtain the equation of the “hidden” oscillator $\ddot{x}=-k^{2}x(1+c)-c{\dot{x}}^2/x$, which admits the symmetry

$${\tilde{S}}^2=-\frac{v}{c+1}{\partial }_x+\left(k^{2}x+\frac{c}{c+1}\frac{v^2}{x}\right){\partial }_v,$$

which is nothing more than ${\partial }_t=(c+1){\tilde{S}}^2+\tilde{A}$.

4. Conclusions and Discussion

Firstly, we analyzed how the symmetries of two first-order ODEs with the same integrating factor were related. We found out, as a general result, that two such equations admitted the same point symmetry.

Secondly, we established the relation between the Jacobi multiplier for second-order ordinary differential equations and the integrating factors of two first-order ODEs, expressed in terms of the invariant of an admitted symmetry. We pointed out that the knowledge of one admitted symmetry and the Jacobi multiplier allowed us to reduce the order of the equation for the second admitted symmetry.

Moreover, we found out that there existed a set of essentially different Euler–Lagrange ODEs of the second order with different Lagrangians, but with the same Jacobi multiplier. For such systems, we obtained the conditions for which the admitted symmetries were the same. The general problem of relating the symmetries of such systems is still open. Of course, for every second-order ODE, there are infinitely many contact transformations transforming it into any other given second-order equation. However, how to determine such transformations? Probably, the indicated connection between the pseudo scalar product and a nonzero Jacobi multiplier and the use of canonical variables may shed some light on this issue, and it will be the subject of our further investigation.

The obtained results are important for exploring the quantization of Hamiltonian systems in which one can use the Lie symmetries to obtain the quantum-correct Schrödinger equation by preserving the Noether symmetries [28,29,30,31]. Moreover, one can use these results to introduce new nonlinear systems to explore their classical and quantum dynamics, as was recently done in Ref. [32].