Construction of 3D potentials from a pre-assigned two-parametric family of orbits

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M.-C. Anisiu, Th. Kotoulas, Construction of 3D potentials from a pre-assigned two-parametric family of orbits, Inverse Problems 22 (2006), 2255-2269, http://10.1088/0266-5611/22/6/021

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Inverse problems

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Construction of 3D potentials from a preassigned two-parametric family of orbits

Mira-Cristiana Anisiu 1 and Thomas A Kotoulas 2
1 T Popoviciu Institute of Numerical Analysis, Romanian Academy, PO Box 68, 400110 Cluj-Napoca, Romania
2 Department of Physics, Aristotle University of Thessaloniki, GR-541 24 Thessaloniki, Greece
E-mail: mira@math.ubbcluj.ro and tkoto@skiathos.physics.auth.gr
Abstract

One of the main problems of astrophysics is to determine the mean field potential of galaxies. The astronomical observations, as well as the numerical simulations, lead to the determination of families of star orbits in a galaxy. On this basis, using the tools of the inverse problem of dynamics, it is possible to find the gravitational potential which gives rise to such motions. The problem can be formulated in various ways; we consider here that the particle trajectories are given by a spatial two-parameter family of curves. From these trajectories we obtain certain functions known as orbital functions, and look for potentials of a special form. If the orbital functions satisfy some differential conditions, a step-by-step procedure offers the expression of the potential. Similar problems arise in thermodynamics and nuclear physics, where axially symmetric potentials are used as models for deformed nuclei.

1. Introduction

The mean field approach is important to the many-body systems, due to the fact that the NN-body problem is practically impossible to handle for large values of NN. The motion of a single particle may be studied in the field generated by many bodies, under the hypothesis that it does not perturb significantly the external field. This can be done, for example, in astrophysics (a star in a galaxy, a galaxy in a large galactic cluster) or in nuclear physics (nucleons in nuclei, atoms in metallic clusters).

In such models, the potential is not known in advance, but it is supposed to have some properties as spherical or axial symmetry. The knowledge of a family of orbits makes possible the determination of the potential, using the tools of the inverse problem of dynamics.

In this paper, we study the possibility of finding 3D potentials which give rise to a family of orbits known in advance. Among others, we construct a homogeneous potential which can produce a family of elliptic orbits (which are frequent in galaxy models), and an axially
symmetric potential compatible with a family of hyperbolae. A related 2D problem was first studied by Szebehely (1974), with the aim to use the results for determining the Earth’s potential on the basis of satellites’ orbits: find all the potentials V(x,y)V(x,y) which can give rise to a given mono-parametric family of curves f(x,y)=cf(x,y)=c traced in the xyxy-Cartesian plane by a material point of unit mass. These potentials satisfy a linear first-order equation in which the total energy appears, known as Szebehely’s equation. A linear second-order partial differential equation in the unknown function VV, involving only the family of orbits and the potential-not the energy dependence, was produced by Bozis (1983). Anisiu (2004a) derived in a unified manner the two basic equations of the inverse problem of dynamics, and the region where real motion of the particle takes place (Bozis and Ichtiaroglou 1994).

Three-dimensional versions of the inverse problem were studied by Érdi (1982) for a monoparametric family of spatial orbits, and, for a two-parametric family, by Bozis (1983) (for general force fields) and by Váradi and Érdi (1983). Puel (1984) presented the intrinsic (independent of the coordinate frame) equations of the 3D inverse problem. Other results have been obtained by Bozis and Nakhla (1986), and Shorokov (1988). These results are summarized in the review paper of Bozis (1995). Recently, Bozis and Kotoulas (2004) have studied the case of two-parametric families of straight lines (FSL) produced by genuine three-dimensional potentials. Moreover, the same authors produced the free-ofenergy equations and derived 3D homogeneous potentials which give rise to two-parametric families of homogeneous orbits in space (Bozis and Kotoulas 2005). At the same time, Anisiu (2004b, 2005) obtained in a direct way the two free-of-energy PDEs of the three-dimensional inverse problem and the region where real motion is allowed, presenting also several families of orbits compatible with 3D potentials.

The free-of-energy equation for the planar inverse problem was used to find the richest set of potentials which can generate a given family of ellipses (Bozis and Caranicolas 1997). The same equation enabled Caranicolas (1998) to produce potentials which allow for figure-eight families of orbits; such families appear in astrophysics, being detected in the NN-body problem simulation in barred galaxies. More recently, Bozis and Anisiu (2005) have dealt with a solvable version of the planar inverse problem and found potentials of special type V=v(γ)V=v(\gamma) where γ=fy/fx\gamma=f_{y}/f_{x}. This approach was motivated by the fact that potentials of the form v(γ)v(\gamma) had already appeared as solutions of the planar inverse problem. Another reason was that, when potentials v(γ)v(\gamma) compatible with the given family γ\gamma do exist, they can be calculated by quadrature.

The present work extends to three dimensions the results concerning the planar potentials of the form V=v(γ)V=v(\gamma) compatible with a pre-assigned family of orbits f(x,y)=cf(x,y)=c. We deal with the following version of the 3D inverse problem of dynamics: given a two-parametric family of regular curves f(x,y,z)=c1,g(x,y,z)=c2f(x,y,z)=c_{1},g(x,y,z)=c_{2} (with orbital functions α,β\alpha,\beta defined in (6)), find the potentials of the form V=v(α)V=v(\alpha) or V=F(α,β)V=F(\alpha,\beta) producing these families of curves as trajectories. For example, it is proved that the family (73) of elliptic orbits can be traced by a material point under the action of the potential V(α)=α2(2+α2)/4V(\alpha)=-\alpha^{2}\left(2+\alpha^{2}\right)/4. Working with the orbital functions α\alpha and β\beta from (62), we obtain the potential V=F(α,β)V=F(\alpha,\beta) from (64), which is an axially symmetric one. As mentioned by Boccaletti and Puccaco (1996, vol 1, p 349), axially symmetric potentials are important for astrophysical applications, since they are generated by mass distributions provided with rotational motions around the symmetry axis, typical of solid celestial bodies, stars, disc galaxies and elliptical galaxies. Besides the applications in galactic dynamics (Contopoulos 1960, Binney and Tremaine 1987, p 121; Caranicolas 2004), they are used in models in thermodynamics (e.g. Olaya-Castro and Quiroga (2000)) and in quantum mechanics (e.g. Heiss et al (1994)).

The problem under consideration does not have solutions for arbitrary families of curves. We determine the necessary and sufficient conditions on α\alpha and β\beta which guarantee the existence of such potentials. A promising feature is that, when the conditions are fulfilled, we can construct the potentials by quadrature. We focus on genuine three-dimensional potentials, i.e. potentials whose analytical expression involves all the variables x,y,zx,y,z. It follows that we take into account only potentials for which no partial derivative is identically zero. All over the paper we shall regard VV as the potential energy function, hence the total energy is given by =(x˙2+y˙2+z˙2)/2+V(x,y,z)\mathcal{E}=\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)/2+V(x,y,z). In section 2 we present the basic equations of the 3D inverse problem of dynamics. In section 3 we study the case V=v(α)V=v(\alpha). The case V=F(α,β)V=F(\alpha,\beta) is examined in section 4. In both sections 3 and 4 we give step-by-step procedures, which lead to the expression of the potential when the orbital functions satisfy certain differential conditions. The mathematics are used as a tool, and the method can be applied whenever a family of orbits will be observed, or generated by a simulation. We refer to the case of straight lines in section 5 and conclude in section 6.

2. The 3D inverse problem of dynamics: basic facts

The three-dimensional inverse problem of dynamics, as formulated by Bozis and Nakhla (1986), seeks all the potentials V=V(x,y,z)V=V(x,y,z) of C2C^{2}-class (continuous with continuous derivatives up to second order) which are compatible with a pre-assigned family of regular orbits

f(x,y,z)=c1,g(x,y,z)=c2f(x,y,z)=c_{1},\quad g(x,y,z)=c_{2} (1)

By regular orbits we mean that the functions f,gf,g are of C3C^{3}-class on a domain D3D\subset\mathbb{R}^{3} and such that f0\nabla f\neq 0 and g0\nabla g\neq 0, and, in addition f×g0\nabla f\times\nabla g\neq 0 on DD, so that δ00\delta_{0}\neq 0 in (3). The total energy is constant on each curve of the family, namely

=(f,g)\mathcal{E}=\mathcal{E}(f,g) (2)

Differentiating equations (1) with respect to time tt and taking into account the conservation of the energy one can express the velocity vector; differentiating again (1) and using the equations of motion one obtains the two linear PDEs satisfied by the potential VV. These equations, in which the energy dependence function =(f,g)\mathcal{E}=\mathcal{E}(f,g) appears, are

δ2Vxδ1Vy\displaystyle\delta_{2}V_{x}-\delta_{1}V_{y} =2(δ1a2δ2a1)δ0(V),\displaystyle=\frac{2\left(\delta_{1}a_{2}-\delta_{2}a_{1}\right)}{\delta_{0}}(\mathcal{E}-V), (3)
δ3Vxδ1Vz\displaystyle\delta_{3}V_{x}-\delta_{1}V_{z} =2(δ1a3δ3a1)δ0(V),\displaystyle=\frac{2\left(\delta_{1}a_{3}-\delta_{3}a_{1}\right)}{\delta_{0}}(\mathcal{E}-V),

where

δ¯=(δ1,δ2,δ3)=f×g,δ0=|δ¯|2\bar{\delta}=\left(\delta_{1},\delta_{2},\delta_{3}\right)=\nabla f\times\nabla g,\quad\delta_{0}=|\bar{\delta}|^{2} (4)

and

ai=δ¯δi,i=1,2,3.a_{i}=\bar{\delta}\cdot\nabla\delta_{i},\quad i=1,2,3. (5)

The vectorial and scalar products in 3\mathbb{R}^{3} are denoted by ’ ×\times ’, respectively ’ ’, and the subscripts x,y,zx,y,z denote the corresponding partial derivatives. Equations (3) in a vectorial form have been derived by Puel (1984) using Frenet’s reference frame.

The orbital functions

α=δ2δ1,β=δ3δ1\alpha=\frac{\delta_{2}}{\delta_{1}},\quad\beta=\frac{\delta_{3}}{\delta_{1}} (6)

have been defined by Bozis and Kotoulas (2004), in a domain where δ10\delta_{1}\neq 0. With the use of notations

ϵ¯=(1,α,β),Θ=1+α2+β2\displaystyle\bar{\epsilon}=(1,\alpha,\beta),\quad\Theta=1+\alpha^{2}+\beta^{2}
α0=ϵ¯α=αx+ααy+βαz\displaystyle\alpha_{0}=\bar{\epsilon}\cdot\nabla\alpha=\alpha_{x}+\alpha\alpha_{y}+\beta\alpha_{z} (7)
β0=ϵ¯β=βx+αβy+ββz\displaystyle\beta_{0}=\bar{\epsilon}\cdot\nabla\beta=\beta_{x}+\alpha\beta_{y}+\beta\beta_{z}

equations (3) take the simpler form (Bozis and Kotoulas 2005)

αVxVy=2α0Θ(V),βVxVz=2β0Θ(V).\alpha V_{x}-V_{y}=\frac{2\alpha_{0}}{\Theta}(\mathcal{E}-V),\quad\beta V_{x}-V_{z}=\frac{2\beta_{0}}{\Theta}(\mathcal{E}-V). (8)

The region where real motion exists (Shorokhov 1988, Anisiu 2004b) is

αVxVyα00.\frac{\alpha V_{x}-V_{y}}{\alpha_{0}}\geqslant 0. (9)

Furthermore, Bozis and Kotoulas (2005) eliminated the energy dependence function from equations (8) and obtained two energy-free PDEs for the unknown potential function V=V(x,y,z)V=V(x,y,z), one of first order and the other of second order. These equations relate the potential to the given family of curves, supposing that the last one does not consist of straight lines. It was proved by Bozis and Kotoulas (2004) that α0=0\alpha_{0}=0 and β0=0\beta_{0}=0 only when the family (1) consists of straight lines; we shall treat this special case in section 5.

Remark 1. As was mentioned by Bozis and Kotoulas (2005), the transformation xx,yz,zyx\rightarrow x,y\rightarrow z,z\rightarrow y brings αβ\alpha\rightarrow\beta and βα\beta\rightarrow\alpha as well as α0β0\alpha_{0}\rightarrow\beta_{0} and β0α0\beta_{0}\rightarrow\alpha_{0}.

From now on we shall consider α00\alpha_{0}\neq 0. (The case β00\beta_{0}\neq 0 can be treated directly using instead of equation (12) the equation obtained from the second-order equation (25) in the paper of Anisiu (2004b).) The first-order equation is

k1Vx+k2Vy+k3Vz=0k_{1}V_{x}+k_{2}V_{y}+k_{3}V_{z}=0 (10)

where

k1=αβ0βα0,k2=β0,k3=α0k_{1}=\alpha\beta_{0}-\beta\alpha_{0},\quad k_{2}=-\beta_{0},\quad k_{3}=\alpha_{0} (11)

The second-order PDE of the 3D inverse problem reads
k11Vxx+k12Vxy+k13Vxz+k22Vyy+k23Vyz+k01Vx+k02Vy+k03Vz=0k_{11}V_{xx}+k_{12}V_{xy}+k_{13}V_{xz}+k_{22}V_{yy}+k_{23}V_{yz}+k_{01}V_{x}+k_{02}V_{y}+k_{03}V_{z}=0,
where

k11=αΘα0,k12=(α21)Θα0,k13=αβΘα0k22=αΘα0,k23=βΘα0,k01=(Θ+2)α02+αK,k02=2αα02K,k03=2βα02,K=2(αα0+ββ0)α0Θ(α0x+αα0y+βα0z),\begin{array}[]{lll}k_{11}=\alpha\Theta\alpha_{0},&k_{12}=\left(\alpha^{2}-1\right)\Theta\alpha_{0},&k_{13}=\alpha\beta\Theta\alpha_{0}\\ k_{22}=-\alpha\Theta\alpha_{0},&k_{23}=-\beta\Theta\alpha_{0},&\\ k_{01}=(\Theta+2)\alpha_{0}^{2}+\alpha K,&k_{02}=2\alpha\alpha_{0}^{2}-K,&k_{03}=2\beta\alpha_{0}^{2},\\ K=2\left(\alpha\alpha_{0}+\beta\beta_{0}\right)\alpha_{0}-\Theta\left(\alpha_{0x}+\alpha\alpha_{0y}+\beta\alpha_{0z}\right),&\end{array}

We note here that the previous equations (10) and (12) have also been produced on first grounds by Anisiu (2004b). These equations are equivalent to system (3), as was proved by Anisiu (2005), and have the advantage that it is not necessary to give the energy in advance. From (10) and (12) it is easy to check that if VV is a solution, then V~=c1V+c2\tilde{V}=c_{1}V+c_{2} is a solution too ( c1,c2c_{1},c_{2} are constants). So, without loss of generality, we shall omit these constants below.

In the present paper we extend the 2D results of Bozis and Anisiu (2005) and study the following version of the 3D inverse problem of dynamics: given a two-parametric family of regular orbits (1), we try to find the three-dimensional potentials of the form V=v(α)V=v(\alpha) or V=F(α,β)V=F(\alpha,\beta) which are compatible with that family. Having no a priori information on the energy, we deal with the two linear PDEs (10) and (12).

3. Potentials of the form 𝑽=𝒗(𝜶)\boldsymbol{V}=\boldsymbol{v}(\boldsymbol{\alpha})

In this section we shall try to find solutions of the form V=v(α)V=v(\alpha) for the above two equations (10) and (12). In view of remark 1, we shall not consider here the case of V=v(β)V=v(\beta).

3.1. Conditions on the orbital functions

We suppose that V=v(α)V=v(\alpha) with vv of C2C^{2}-class and prepare the derivatives of first order of the potential function VV with respect to x,y,zx,y,z respectively Vx=vαx,Vy=vαy,Vz=vαzV_{x}=v^{\prime}\alpha_{x},V_{y}=v^{\prime}\alpha_{y},V_{z}=v^{\prime}\alpha_{z}, where "’ denotes the derivative of vv with respect to its unique variable α\alpha. We insert Vx,VyV_{x},V_{y} and VzV_{z} into (10) and get

v(α0(βαxαz)β0(ααxαy))=0.v^{\prime}\left(\alpha_{0}\left(\beta\alpha_{x}-\alpha_{z}\right)-\beta_{0}\left(\alpha\alpha_{x}-\alpha_{y}\right)\right)=0. (14)

Since we do not admit constant potentials, v(α)v^{\prime}(\alpha) will not be zero on nonvoid open sets. It follows a first necessary condition on the orbital function α\alpha, namely

α0(βαxαz)β0(ααxαy)=0\alpha_{0}\left(\beta\alpha_{x}-\alpha_{z}\right)-\beta_{0}\left(\alpha\alpha_{x}-\alpha_{y}\right)=0 (15)

with α0\alpha_{0} and β0\beta_{0} given in (7). Now, we shall work with the second-order PDE (12). Firstly, we prepare the second-order derivatives of the potential function VV with respect to x,y,zx,y,z,

Vxx=v′′αx2+vαxx,Vyy=v′′αy2+vαyyVxz=v′′αxαz+vαxz,Vyz=v′′αyαz+vαyzVxy=v′′αxαy+vαxy\begin{array}[]{ll}V_{xx}=v^{\prime\prime}\alpha_{x}^{2}+v^{\prime}\alpha_{xx},&V_{yy}=v^{\prime\prime}\alpha_{y}^{2}+v^{\prime}\alpha_{yy}\\ V_{xz}=v^{\prime\prime}\alpha_{x}\alpha_{z}+v^{\prime}\alpha_{xz},&V_{yz}=v^{\prime\prime}\alpha_{y}\alpha_{z}+v^{\prime}\alpha_{yz}\\ V_{xy}=v^{\prime\prime}\alpha_{x}\alpha_{y}+v^{\prime}\alpha_{xy}&\end{array}

and insert them into equation (12). Thus, we end up with the following relation,

𝒜v′′+v=0,\mathcal{A}v^{\prime\prime}+\mathcal{B}v^{\prime}=0, (17)

where

𝒜=Θα02(ααxαy)\displaystyle\mathcal{A}=\Theta\alpha_{0}^{2}\left(\alpha\alpha_{x}-\alpha_{y}\right)
=Θα0σ+2α03+(ααxαy)K\displaystyle\mathcal{B}=\Theta\alpha_{0}\sigma+2\alpha_{0}^{3}+\left(\alpha\alpha_{x}-\alpha_{y}\right)K (18)

and

σ=ϵ¯(ααxαy)\sigma=\bar{\epsilon}\cdot\nabla\left(\alpha\alpha_{x}-\alpha_{y}\right) (19)

Remark 2. Keeping in mind our assumption that α00\alpha_{0}\neq 0, it follows that if 𝒜=0\mathcal{A}=0, then ααxαy=0\alpha\alpha_{x}-\alpha_{y}=0. Consequently, =2α030\mathcal{B}=2\alpha_{0}^{3}\neq 0 and equation (17) has no nonconstant solution.

From (17) we obtain for 𝒜0\mathcal{A}\neq 0

v′′v=𝒞,\frac{v^{\prime\prime}}{v^{\prime}}=\mathcal{C}, (20)

where

𝒞=𝒜=[σ(ααxαy)α0+2α0Θ(ααxαy)+KΘα02],\mathcal{C}=-\frac{\mathcal{B}}{\mathcal{A}}=-\left[\frac{\sigma}{\left(\alpha\alpha_{x}-\alpha_{y}\right)\alpha_{0}}+\frac{2\alpha_{0}}{\Theta\left(\alpha\alpha_{x}-\alpha_{y}\right)}+\frac{K}{\Theta\alpha_{0}^{2}}\right], (21)

with α0\alpha_{0} and Θ\Theta given in (7), KK in (13) and σ\sigma in (19). Since the ratio v′′/vv^{\prime\prime}/v^{\prime} depends only on the orbital function α\alpha, the expression of 𝒞\mathcal{C} must have the same property. Thus, it must be

𝒞=𝒞(α)\mathcal{C}=\mathcal{C}(\alpha) (22)

or, equivalently,

𝒞xαx=𝒞yαy=𝒞zαz\frac{\mathcal{C}_{x}}{\alpha_{x}}=\frac{\mathcal{C}_{y}}{\alpha_{y}}=\frac{\mathcal{C}_{z}}{\alpha_{z}} (23)

This happens if and only if α\alpha and β\beta satisfy

 (i) 𝒞xαy𝒞yαx=0; (ii) 𝒞yαz𝒞zαy=0\text{ (i) }\mathcal{C}_{x}\alpha_{y}-\mathcal{C}_{y}\alpha_{x}=0;\quad\text{ (ii) }\mathcal{C}_{y}\alpha_{z}-\mathcal{C}_{z}\alpha_{y}=0 (24)

Conversely, let us suppose that conditions (15) and (24i, ii) are satisfied by the functions α\alpha and β\beta, and that 𝒜0\mathcal{A}\neq 0. Then 𝒞=𝒞(α)\mathcal{C}=\mathcal{C}(\alpha) and we integrate (20) two times with respect to α\alpha and obtain

V=v(α)=d1e𝒞(α)dαdα+d2,d1,d2 const. V=v(\alpha)=d_{1}\int\mathrm{e}^{\int\mathcal{C}(\alpha)\mathrm{d}\alpha}\mathrm{~d}\alpha+d_{2},\quad d_{1},d_{2}\quad\text{ const. } (25)

We shall omit the constants d1,d2d_{1},d_{2} in the following as we explained above. The meaning of the solution (25) is that we can find the potential function VV by quadratures.

Now, we can formulate the following
Proposition 1. If 𝒜0\mathcal{A}\neq 0 and the three conditions (15) and (24i, ii) are satisfied for the given functions α\alpha and β\beta, then a potential of the form V=v(α)V=v(\alpha) does exist and it is determined uniquely from (25) up to two arbitrary constants.

Synthesis of the problem

(1) We start with the functions α\alpha and β\beta (or with ff and gg, and then obtain α,β\alpha,\beta from (6)).
(2) We calculate the expression in (15); if it is different from zero, the problem has no solution.
(3) If (15) is satisfied, we estimate 𝒜\mathcal{A} from (18); if 𝒜=0\mathcal{A}=0, we have no solution.
(4) If 𝒜0\mathcal{A}\neq 0, with \mathcal{B} from (18) we calculate 𝒞\mathcal{C} from (21) and verify conditions (24i, ii). If at least one of them is not satisfied, there is no solution.
(5) If (24i, ii) are verified, we obtain the potential V=v(α)V=v(\alpha) from (25).

3.2. Suitable pairs of orbital functions and examples

We look for suitable pairs ( α,β\alpha,\beta ), with α\alpha a function of three variables satisfying α00\alpha_{0}\neq 0, for which conditions (15) and (24i, ii) are fulfilled.

A natural trial is for α=β\alpha=\beta; condition (15) becomes αyαz=0\alpha_{y}-\alpha_{z}=0, with the solution

α=A(x,y+z),\alpha=A(x,y+z), (26)

where AA is a two-variable function. The function α\alpha from (26) satisfies condition (24ii) too. Condition (24i) is fulfilled for several special forms of the function AA in (26) mentioned below ( MM is an arbitrary one-variable function):

  • α=β=M(x+y+z)\alpha=\beta=M(x+y+z);

  • α=β=M(xyz)\alpha=\beta=M(x-y-z);

  • α=β=M((y+z)/x)\alpha=\beta=M((y+z)/x).

We shall present in detail an example from the first class, for which the potential function V=v(α)V=v(\alpha) is given explicitly.

Example 1. We consider the family

f(x,y,z)=(x+y+z+32)e2x,g(x,y,z)=14(2y2z3)f(x,y,z)=\left(x+y+z+\frac{3}{2}\right)\mathrm{e}^{-2x},\quad g(x,y,z)=\frac{1}{4}(2y-2z-3)

which gives

α=x+y+z+1,β=x+y+z+1.\alpha=x+y+z+1,\quad\beta=x+y+z+1. (27)

For this family we have α0=β0=2(x+y+z)+3=2α+1\alpha_{0}=\beta_{0}=2(x+y+z)+3=2\alpha+1 and the three conditions (15) and (24i, ii) are satisfied. From (25) we calculate the potential function v(α)=(14α)/(2(α1)2)v(\alpha)=(1-4\alpha)/\left(2(\alpha-1)^{2}\right); hence

V(x,y,z)=4(x+y+z)+32(x+y+z)2.V(x,y,z)=-\frac{4(x+y+z)+3}{2(x+y+z)^{2}}. (28)

From (8) we calculate the energy =1\mathcal{E}=1 and from (9) the allowed region of the motion of the test particle 1/(x+y+z)201/(x+y+z)^{2}\geqslant 0, which means that the motion of the test particle can take place everywhere in the space, except for the plane x+y+z=0x+y+z=0.

Counter-examples. We consider the two-parametric family of curves (1) with

f(x,y,z)=eyz,g(x,y,z)=12(x2eyz+e2z),f(x,y,z)=\mathrm{e}^{y-z},\quad g(x,y,z)=-\frac{1}{2}\left(x^{2}\mathrm{e}^{y-z}+\mathrm{e}^{-2z}\right),

for which

α=xey+z,β=α\alpha=x\mathrm{e}^{y+z},\quad\beta=\alpha (29)

and α0=β0=ey+z(1+2x2ey+z)\alpha_{0}=\beta_{0}=\mathrm{e}^{y+z}\left(1+2x^{2}\mathrm{e}^{y+z}\right). We checked conditions (15) and (24i, ii) for this pair of orbits and ascertained that conditions (15) and (24ii) are verified but condition (24i) is not satisfied for this family. So, this family is not compatible with any potential V=v(α)V=v(\alpha). The same conclusion holds for the orbital functions α=zex+y,β=α\alpha=z\mathrm{e}^{x+y},\beta=\alpha, for which (24i) is satisfied, but (15) and (24ii) are not.

Another choice for the pair (α,β)(\alpha,\beta) is to make β0=0\beta_{0}=0. This happens, for example, for β=\beta= constant, let it be β=k\beta=k. Condition (15) reduces to kαxαz=0k\alpha_{x}-\alpha_{z}=0, with the solution

α=A(y,x+kz)\alpha=A(y,x+kz) (30)

where AA is again a two-variable function. Generally, the functions α\alpha given by (30) and β=k\beta=k do not satisfy conditions (24i, ii). Nevertheless, we can find classes of functions which satisfy these conditions too, like:

  • α=M(x+e1y+z),β=k(e1\alpha=M\left(x+e_{1}y+z\right),\beta=k\left(e_{1}\right. free constant ););

  • α=M((f1(x+kz)+f2y+f3)/y),β=k(f1,f2,f3\alpha=M\left(\left(f_{1}(x+kz)+f_{2}y+f_{3}\right)/y\right),\beta=k\left(f_{1},f_{2},f_{3}\right. free constants).

Example 2. For the two-parametric family of curves (1) with

f(x,y,z)=ex(xy+kzk21),g(x,y,z)=kxz,f(x,y,z)=\mathrm{e}^{x}\left(x-y+kz-k^{2}-1\right),\quad g(x,y,z)=kx-z,

we have

α=xy+kz,β=k,\alpha=x-y+kz,\quad\beta=k, (31)

and α0=1x+ykz+k2,β0=0\alpha_{0}=1-x+y-kz+k^{2},\beta_{0}=0. From (25) we get v(α)=(2αk2)/(α+1)2v(\alpha)=\left(2\alpha-k^{2}\right)/(\alpha+1)^{2}, and from (8) and (9) the energy =1\mathcal{E}=1 and the allowed region 2/(xy+kz+1)202/(x-y+kz+1)^{2}\geqslant 0; hence the motion can take place everywhere in space except for the plane xy+kz+1=0x-y+kz+1=0.

It is not necessary that α\alpha and β\beta are equal, or β\beta is constant. For example, from the class α=x+e1y+e2z,β=x+e1y+e2z+e3\alpha=x+e_{1}y+e_{2}z,\beta=x+e_{1}y+e_{2}z+e_{3}, the functions which satisfy condition (15) are those with e3=e2e1e_{3}=e_{2}-e_{1}; these functions satisfy also conditions (24i, ii) and the potential can be determined by quadratures.

4. Potentials of the form 𝑽=𝑭(𝜶,𝜷)\boldsymbol{V}=\boldsymbol{F}(\boldsymbol{\alpha},\boldsymbol{\beta})

In this section we shall find potentials of the form V=F(α,β)V=F(\alpha,\beta), with α,β\alpha,\beta functionally independent and FF of C2C^{2}-class. The condition α00\alpha_{0}\neq 0 is maintained. We impose necessary conditions on the orbital functions α,β\alpha,\beta and then we shall determine, when possible, the potential function V=F(α,β)V=F(\alpha,\beta) so that it satisfies the two linear PDEs (10) and (12).

4.1. Conditions on the orbital functions α,β\alpha,\beta

We look for potentials which are functions of α\alpha and β\beta. To this end, the necessary and sufficient condition for V,α,βV,\alpha,\beta is

|VxαxβxVyαyβyVzαzβz|=0\left|\begin{array}[]{lll}V_{x}&\alpha_{x}&\beta_{x}\\ V_{y}&\alpha_{y}&\beta_{y}\\ V_{z}&\alpha_{z}&\beta_{z}\end{array}\right|=0

or, equivalently,

l1Vx+l2Vy+l3Vz=0l_{1}V_{x}+l_{2}V_{y}+l_{3}V_{z}=0 (32)

where

l1=αyβzαzβy,l2=αzβxαxβz,l3=αxβyαyβxl_{1}=\alpha_{y}\beta_{z}-\alpha_{z}\beta_{y},\quad l_{2}=\alpha_{z}\beta_{x}-\alpha_{x}\beta_{z},\quad l_{3}=\alpha_{x}\beta_{y}-\alpha_{y}\beta_{x} (33)

Equation (32) must be compatible with (10). Thus, we have another non-trivial PDE to ensure the compatibility of these two PDEs (Favard 1963, Smirnov 1964). This new PDE reads

m1Vx+m2Vy+m3Vz=0m_{1}V_{x}+m_{2}V_{y}+m_{3}V_{z}=0 (34)

where

mi=k1lix+k2liy+k3lizl1kixl2kiyl3kiz,i=1,2,3.m_{i}=k_{1}\frac{\partial l_{i}}{\partial x}+k_{2}\frac{\partial l_{i}}{\partial y}+k_{3}\frac{\partial l_{i}}{\partial z}-l_{1}\frac{\partial k_{i}}{\partial x}-l_{2}\frac{\partial k_{i}}{\partial y}-l_{3}\frac{\partial k_{i}}{\partial z},\quad i=1,2,3. (35)

Thus, the necessary and sufficient condition for finding the nonzero derivatives Vx,Vy,VzV_{x},V_{y},V_{z} from (10), (32) and (34) is

|k1l1m1k2l2m2k3l3m3|=0\left|\begin{array}[]{lll}k_{1}&l_{1}&m_{1}\\ k_{2}&l_{2}&m_{2}\\ k_{3}&l_{3}&m_{3}\end{array}\right|=0

or, equivalently,

k1(l2m3l3m2)k2(l1m3l3m1)+k3(l1m2l2m1)=0k_{1}\left(l_{2}m_{3}-l_{3}m_{2}\right)-k_{2}\left(l_{1}m_{3}-l_{3}m_{1}\right)+k_{3}\left(l_{1}m_{2}-l_{2}m_{1}\right)=0 (36)

Let us suppose that condition (36) is fulfilled and V=F(α,β)V=F(\alpha,\beta). We prepare the first-order derivatives of VV with respect to x,y,zx,y,z,
Vx=Fααx+Fββx,Vy=Fααy+Fββy,Vz=Fααz+FββzV_{x}=F_{\alpha}\alpha_{x}+F_{\beta}\beta_{x},\quad V_{y}=F_{\alpha}\alpha_{y}+F_{\beta}\beta_{y},\quad V_{z}=F_{\alpha}\alpha_{z}+F_{\beta}\beta_{z},
and replace them into (10). We obtain

𝒫Fα+𝒬Fβ=0\mathcal{P}F_{\alpha}+\mathcal{Q}F_{\beta}=0 (38)

where

𝒫=α0(βαxαz)β0(ααxαy)\displaystyle\mathcal{P}=\alpha_{0}\left(\beta\alpha_{x}-\alpha_{z}\right)-\beta_{0}\left(\alpha\alpha_{x}-\alpha_{y}\right) (39)
𝒬=α0(ββxβz)β0(αβxβy)\displaystyle\mathcal{Q}=\alpha_{0}\left(\beta\beta_{x}-\beta_{z}\right)-\beta_{0}\left(\alpha\beta_{x}-\beta_{y}\right)

Remark 3. If 𝒫=0\mathcal{P}=0 and 𝒬0\mathcal{Q}\neq 0, we have Fβ=0F_{\beta}=0 and the potential VV will be a function V=v(α)V=v(\alpha), case studied in section 3 (condition (15) is precisely 𝒫=0\mathcal{P}=0 ). For 𝒬=0\mathcal{Q}=0 and 𝒫0\mathcal{P}\neq 0, it follows that Fα=0F_{\alpha}=0 and VV will depend merely on β\beta. This case is similar to that studied in section 3 (see remark 1). We shall examine separately the cases 𝒫=0\mathcal{P}=0 and 𝒬=0\mathcal{Q}=0.

From (38) it follows for 𝒫0\mathcal{P}\neq 0 and 𝒬0\mathcal{Q}\neq 0 that

FαFβ=𝒬𝒫=𝒟\frac{F_{\alpha}}{F_{\beta}}=-\frac{\mathcal{Q}}{\mathcal{P}}=\mathcal{D} (40)

Since V=F(α,β)V=F(\alpha,\beta), the ratio Fα/FβF_{\alpha}/F_{\beta} in (40) must depend only on α,β\alpha,\beta, i.e. 𝒟=𝒟(α,β)\mathcal{D}=\mathcal{D}(\alpha,\beta). To this end, the necessary and sufficient condition is

l1𝒟x+l2𝒟y+l3𝒟z=0,l_{1}\mathcal{D}_{x}+l_{2}\mathcal{D}_{y}+l_{3}\mathcal{D}_{z}=0, (41)

where l1,l2,l3l_{1},l_{2},l_{3} were defined in (33). So, condition (41) is the second necessary one for α,β\alpha,\beta in order that the problem has a solution V=F(α,β)V=F(\alpha,\beta).

Now, we proceed further and prepare the second-order derivatives of VV with respect to x,y,zx,y,z. We replace them into (12) and obtain

Γ11Fαα+Γ12Fαβ+Γ22Fββ+Γ10Fα+Γ01Fβ=0,\Gamma_{11}F_{\alpha\alpha}+\Gamma_{12}F_{\alpha\beta}+\Gamma_{22}F_{\beta\beta}+\Gamma_{10}F_{\alpha}+\Gamma_{01}F_{\beta}=0, (42)

where

Γ11=α0(ααxαy),Γ12=β0(ααxαy)+α0(αβxβy),\displaystyle\Gamma_{11}=\alpha_{0}\left(\alpha\alpha_{x}-\alpha_{y}\right),\quad\Gamma_{12}=\beta_{0}\left(\alpha\alpha_{x}-\alpha_{y}\right)+\alpha_{0}\left(\alpha\beta_{x}-\beta_{y}\right),
Γ22=β0(αβxβy),\displaystyle\Gamma_{22}=\beta_{0}\left(\alpha\beta_{x}-\beta_{y}\right),
Γ10=σ+2α02/Θ+(ααxαy)K/(Θα0),\displaystyle\Gamma_{10}=\sigma+2\alpha_{0}^{2}/\Theta+\left(\alpha\alpha_{x}-\alpha_{y}\right)K/\left(\Theta\alpha_{0}\right), (43)
Γ01=ρ+2α0β0/Θ+(αβxβy)K/(Θα0),\displaystyle\Gamma_{01}=\rho+2\alpha_{0}\beta_{0}/\Theta+\left(\alpha\beta_{x}-\beta_{y}\right)K/\left(\Theta\alpha_{0}\right),

with KK given in (13), σ\sigma in (19) and

ρ=ϵ¯(αβxβy).\rho=\bar{\epsilon}\cdot\nabla\left(\alpha\beta_{x}-\beta_{y}\right). (44)

If for the given α\alpha and β\beta we have Γ11=Γ12=Γ22=0\Gamma_{11}=\Gamma_{12}=\Gamma_{22}=0, equation (42) is a first-order one. We shall consider this case in subsection 4.1.2.

We suppose now that equation (42) contains indeed some of the second-order derivatives of FF. From (40) we have

Fα=𝒟FβF_{\alpha}=\mathcal{D}F_{\beta} (45)

and estimate the derivatives FααF_{\alpha\alpha} and FαβF_{\alpha\beta},

Fαα=𝒟αFβ+𝒟Fαβ,Fαβ=𝒟βFβ+𝒟Fββ.F_{\alpha\alpha}=\mathcal{D}_{\alpha}F_{\beta}+\mathcal{D}F_{\alpha\beta},\quad F_{\alpha\beta}=\mathcal{D}_{\beta}F_{\beta}+\mathcal{D}F_{\beta\beta}. (46)

We replace (45) and (46) into (42) and get

Fββ+𝒮Fβ=0,\mathcal{R}F_{\beta\beta}+\mathcal{S}F_{\beta}=0, (47)

where

=\displaystyle\mathcal{R}= (α0𝒟+β0)((ααxαy)𝒟+αβxβy)\displaystyle\left(\alpha_{0}\mathcal{D}+\beta_{0}\right)\left(\left(\alpha\alpha_{x}-\alpha_{y}\right)\mathcal{D}+\alpha\beta_{x}-\beta_{y}\right)
𝒮=\displaystyle\mathcal{S}= ((ααxαy)𝒟+αβxβy)(α0𝒟β+K/(Θα0))\displaystyle\left(\left(\alpha\alpha_{x}-\alpha_{y}\right)\mathcal{D}+\alpha\beta_{x}-\beta_{y}\right)\left(\alpha_{0}\mathcal{D}_{\beta}+K/\left(\Theta\alpha_{0}\right)\right)
+(ααxαy)(α0𝒟α+β0𝒟β)+(σ+2α02/Θ)𝒟+ρ+2α0β0/Θ\displaystyle+\left(\alpha\alpha_{x}-\alpha_{y}\right)\left(\alpha_{0}\mathcal{D}_{\alpha}+\beta_{0}\mathcal{D}_{\beta}\right)+\left(\sigma+2\alpha_{0}^{2}/\Theta\right)\mathcal{D}+\rho+2\alpha_{0}\beta_{0}/\Theta (48)

If =0\mathcal{R}=0 and 𝒮0\mathcal{S}\neq 0, the potential VV will depend only on α\alpha, which is excluded. The case =𝒮=0\mathcal{R}=\mathcal{S}=0 will be considered in subsection 4.1.3.

From (47) we obtain for 0\mathcal{R}\neq 0

FββFβ=𝒮=H.\frac{F_{\beta\beta}}{F_{\beta}}=-\frac{\mathcal{S}}{\mathcal{R}}=H. (49)

Since V=F(α,β)V=F(\alpha,\beta), the ratio Fββ/FβF_{\beta\beta}/F_{\beta} in (49) must depend only on α,β\alpha,\beta. Thus, we have H=H(α,β)H=H(\alpha,\beta). To this end, the necessary and sufficient condition is

l1Hx+l2Hy+l3Hz=0,l_{1}H_{x}+l_{2}H_{y}+l_{3}H_{z}=0, (50)

where l1,l2,l3l_{1},l_{2},l_{3} were defined in (33). Condition (50) is the third necessary one for α,β\alpha,\beta so that the problem has such a solution.

Now, we integrate (49) with respect to β\beta and get

Fβ=w(α)exp[βH(α,b)db]F_{\beta}=w(\alpha)\exp\left[\int^{\beta}H(\alpha,b)\mathrm{d}b\right] (51)

where ww is an arbitrary function of one argument. We replace (51) into (45) and obtain

Fα=𝒟(α,β)w(α)exp[βH(α,b)db]F_{\alpha}=\mathcal{D}(\alpha,\beta)w(\alpha)\exp\left[\int^{\beta}H(\alpha,b)\mathrm{d}b\right] (52)

The compatibility condition Fαβ=FβαF_{\alpha\beta}=F_{\beta\alpha} becomes

w(α)w(α)=𝒟H+𝒟ββHα(α,b)db=u\frac{w^{\prime}(\alpha)}{w(\alpha)}=\mathcal{D}H+\mathcal{D}_{\beta}-\int^{\beta}H_{\alpha}(\alpha,b)\mathrm{d}b=u (53)

Since the lhs of (53) depends only on α\alpha, the expression of uu must depend only on α\alpha too. Thus, we have another condition on the orbital functions α,β\alpha,\beta, namely u/β=0\partial u/\partial\beta=0, or

𝒟ββ+𝒟βH+𝒟HβHα=0\mathcal{D}_{\beta\beta}+\mathcal{D}_{\beta}H+\mathcal{D}H_{\beta}-H_{\alpha}=0 (54)

The differential condition (54) is the fourth one which must be satisfied by the pre-assigned set of functions α,β\alpha,\beta. Then the arbitrary function w(α)w(\alpha) is determined as

w(α)=d1eu(α)dα,d1= const. w(\alpha)=d_{1}\mathrm{e}^{\int u(\alpha)\mathrm{d}\alpha},\quad d_{1}=\text{ const. } (55)

At last, the potential function will be calculated from (51) and (52).
We shall examine now some special cases which can arise for the orbital functions α\alpha and β\beta.
4.1.1. 𝒫=0\mathcal{P}=0 and 𝒬=0\mathcal{Q}=0. We have to deal with equation (42) alone and try to find its suitable solutions.

Keeping in mind that α00\alpha_{0}\neq 0, we have 𝒫=0\mathcal{P}=0 and 𝒬=0\mathcal{Q}=0 if and only if

βαxαz=β0α0(ααxαy),ββxβz=β0α0(αβxβy).\beta\alpha_{x}-\alpha_{z}=\frac{\beta_{0}}{\alpha_{0}}\left(\alpha\alpha_{x}-\alpha_{y}\right),\quad\quad\beta\beta_{x}-\beta_{z}=\frac{\beta_{0}}{\alpha_{0}}\left(\alpha\beta_{x}-\beta_{y}\right). (56)

Not all the pairs α\alpha and β\beta which satisfy (56) are suitable for our problem. For example, let α\alpha and β\beta satisfy βαxαz=0,ααxαy=0,ββxβz=0\beta\alpha_{x}-\alpha_{z}=0,\alpha\alpha_{x}-\alpha_{y}=0,\beta\beta_{x}-\beta_{z}=0 and αβxβy=0\alpha\beta_{x}-\beta_{y}=0. The matrix

(αxαyαzβxβyβz)=(αxααxβαxβxαβxββx)\left(\begin{array}[]{lll}\alpha_{x}&\alpha_{y}&\alpha_{z}\\ \beta_{x}&\beta_{y}&\beta_{z}\end{array}\right)=\left(\begin{array}[]{lll}\alpha_{x}&\alpha\alpha_{x}&\beta\alpha_{x}\\ \beta_{x}&\alpha\beta_{x}&\beta\beta_{x}\end{array}\right)

does not have the rank equal to two, so α\alpha and β\beta are not functionally independent.
Nevertheless, there are functionally independent functions with α00\alpha_{0}\neq 0 which satisfy (56), as for example

α=y((yz1)2+x2y2)x(yz1),β=xy1yz.\alpha=\frac{y\left((yz-1)^{2}+x^{2}y^{2}\right)}{x(yz-1)},\quad\beta=\frac{xy}{1-yz}.

4.1.2. Γ11=0,Γ12=0\Gamma_{11}=0,\Gamma_{12}=0 and Γ22=0\Gamma_{22}=0. We have Γ11=0,Γ12=0\Gamma_{11}=0,\Gamma_{12}=0 and Γ22=0\Gamma_{22}=0 if and only if

ααxαy=0 and αβxβy=0.\alpha\alpha_{x}-\alpha_{y}=0\quad\text{ and }\quad\alpha\beta_{x}-\beta_{y}=0. (57)

It follows that σ\sigma and ρ\rho, defined in (19), respectively in (44), are both zero. Equation (42) is now of first order, namely

α0Fα+β0Fβ=0\alpha_{0}F_{\alpha}+\beta_{0}F_{\beta}=0 (58)

Equation (45) reads

(βαxαz)Fα+(ββxβz)Fβ=0\left(\beta\alpha_{x}-\alpha_{z}\right)F_{\alpha}+\left(\beta\beta_{x}-\beta_{z}\right)F_{\beta}=0 (59)

In view of (57), we have α0=(1+α2)αx+βαz\alpha_{0}=\left(1+\alpha^{2}\right)\alpha_{x}+\beta\alpha_{z} and β0=(1+α2)βx+ββz\beta_{0}=\left(1+\alpha^{2}\right)\beta_{x}+\beta\beta_{z}. We substitute these values into (60) and obtain that Θ(αzβxαxβz)=0\Theta\left(\alpha_{z}\beta_{x}-\alpha_{x}\beta_{z}\right)=0; hence

αzβxαxβz=0.\alpha_{z}\beta_{x}-\alpha_{x}\beta_{z}=0. (61)

From (57) we obtain that αxβy=ααxβx=αyβx\alpha_{x}\beta_{y}=\alpha\alpha_{x}\beta_{x}=\alpha_{y}\beta_{x}. This together with (61) implies that

rank(αxαyαzβxβyβz)=1\operatorname{rank}\left(\begin{array}[]{ccc}\alpha_{x}&\alpha_{y}&\alpha_{z}\\ \beta_{x}&\beta_{y}&\beta_{z}\end{array}\right)=1

and α\alpha and β\beta are functionally dependent. Therefore in the case Γ11=0,Γ12=0\Gamma_{11}=0,\Gamma_{12}=0 and Γ22=0\Gamma_{22}=0 we do not obtain acceptable solutions for our problem.
4.1.3. =0\mathcal{R}=0 and 𝒮=0\mathcal{S}=0. For α\alpha and β\beta with =0\mathcal{R}=0 and 𝒮=0\mathcal{S}=0, the function FF has to satisfy only the first-order equation (45). On the other hand, α\alpha and β\beta are bound to fulfil several differential conditions.

From (48) we have =0\mathcal{R}=0 if and only if either ααxαy=αβxβy=0\alpha\alpha_{x}-\alpha_{y}=\alpha\beta_{x}-\beta_{y}=0, or 𝒟=(αβxβy)/(ααxαy)\mathcal{D}=-\left(\alpha\beta_{x}-\beta_{y}\right)/\left(\alpha\alpha_{x}-\alpha_{y}\right), or 𝒟=β0/α0\mathcal{D}=-\beta_{0}/\alpha_{0}. For the first situation we obtain Γ11=Γ12=Γ22=0\Gamma_{11}=\Gamma_{12}=\Gamma_{22}=0 (see section 4.1.2), which contradicts the actual supposition that equation (42) is of second order.

It follows that if =0\mathcal{R}=0 and 𝒮=0\mathcal{S}=0, the functions α\alpha and β\beta must satisfy in addition to (36) and (41), two relations obtained from (40) and 𝒮=0\mathcal{S}=0 ( 𝒮\mathcal{S} is given in (48)):

𝒫(αβxβy)𝒬(ααxαy)=0 and\displaystyle\mathcal{P}\left(\alpha\beta_{x}-\beta_{y}\right)-\mathcal{Q}\left(\alpha\alpha_{x}-\alpha_{y}\right)=0\quad\text{ and }
(ααxαy)(α0𝒟α+β0𝒟β)+(σ+2α02/Θ)𝒟+ρ+2α0β0/Θ=0;\displaystyle\left(\alpha\alpha_{x}-\alpha_{y}\right)\left(\alpha_{0}\mathcal{D}_{\alpha}+\beta_{0}\mathcal{D}_{\beta}\right)+\left(\sigma+2\alpha_{0}^{2}/\Theta\right)\mathcal{D}+\rho+2\alpha_{0}\beta_{0}/\Theta=0;

or
𝒫β0𝒬α0=0\mathcal{P}\beta_{0}-\mathcal{Q}\alpha_{0}=0\quad and
((ααxαy)𝒟+αβxβy)(α0𝒟β+K/(Θα0))+(ααxαy)(α0𝒟α+β0𝒟β)+σ𝒟+ρ=0\left(\left(\alpha\alpha_{x}-\alpha_{y}\right)\mathcal{D}+\alpha\beta_{x}-\beta_{y}\right)\left(\alpha_{0}\mathcal{D}_{\beta}+K/\left(\Theta\alpha_{0}\right)\right)+\left(\alpha\alpha_{x}-\alpha_{y}\right)\left(\alpha_{0}\mathcal{D}_{\alpha}+\beta_{0}\mathcal{D}_{\beta}\right)+\sigma\mathcal{D}+\rho=0.
Now, we can formulate the following
Proposition 2. If 𝒫,𝒬\mathcal{P},\mathcal{Q} and \mathcal{R} are different from zero and the four conditions (36), (41), (50) and (54) are satisfied for the given orbital functions α\alpha and β\beta, then a potential of the form V=F(α,β)V=F(\alpha,\beta) always exists and it is determined uniquely from (51) and (52) up to two arbitrary constants.

Synthesis of the problem
(1) We start with the functions α\alpha and β\beta (functionally independent), or with ff and gg, and then obtain α,β\alpha,\beta from (6).
(2) We calculate the expression in (36); if it is different from zero, the problem has no solution.
(3) If (36) is satisfied, we calculate 𝒫\mathcal{P} and 𝒬\mathcal{Q} from (39). If 𝒫=0\mathcal{P}=0 and 𝒬0\mathcal{Q}\neq 0, or 𝒫0\mathcal{P}\neq 0 and 𝒬=0\mathcal{Q}=0, there is no solution VV depending on both α\alpha and β\beta. If 𝒫=0\mathcal{P}=0 and 𝒬=0\mathcal{Q}=0, we have at our disposal only equation (42) and we must consider it directly (see subsection 4.1.1).
(4) If 𝒫0\mathcal{P}\neq 0 and 𝒬0\mathcal{Q}\neq 0, we calculate 𝒟\mathcal{D} and verify (41). If it is not satisfied, we have no solution.
(5) If (41) is satisfied, we consider equation (42). If Γ11=0,Γ12=0\Gamma_{11}=0,\Gamma_{12}=0 and Γ22=0\Gamma_{22}=0, this is a first-order equation. The system of homogeneous equations (42) and (38) has no acceptable solution (see subsection 4.1.2).
(6) If (42) is of second order, we calculate \mathcal{R} and 𝒮\mathcal{S}. If =0\mathcal{R}=0 and 𝒮0\mathcal{S}\neq 0, we have no acceptable solution.
(7) If =0\mathcal{R}=0 and 𝒮=0\mathcal{S}=0, we have to deal only with equation (45) and find its solutions directly (see subsection 4.1.3).
(8) If 0\mathcal{R}\neq 0, we determine HH and verify condition (50). If it is not satisfied, we have no solution.
(9) If (50) is true, we verify (54). If it is not satisfied, we have no solution.
(10) If (54) is verified, we obtain ww from (55), then from (51) and (52) we determine F(α,β)F(\alpha,\beta).

4.2. Examples

In this section we shall present pertinent examples which cover the general and the special cases too. Let us begin with

Example 1. We consider the two-parametric family of regular orbits

f(x,y,z)=xy=c1,g(x,y,z)=z/x=c2;f(x,y,z)=xy=c_{1},\quad g(x,y,z)=z/x=c_{2};

the orbital functions are

α=yx,β=zx,\alpha=-\frac{y}{x},\quad\beta=\frac{z}{x}, (62)

and α0=2y/x2,β0=0\alpha_{0}=2y/x^{2},\beta_{0}=0. We checked that the four differential conditions are satisfied for this set of regular orbits. From (51) and (52), we have

Fβ=α2β(1α2+β2)2,Fα=α(1+β2)(1α2+β2)2F_{\beta}=\frac{\alpha^{2}\beta}{\left(1-\alpha^{2}+\beta^{2}\right)^{2}},\quad F_{\alpha}=-\frac{\alpha\left(1+\beta^{2}\right)}{\left(1-\alpha^{2}+\beta^{2}\right)^{2}} (63)

The potential is determined uniquely from (63)

V=F(α,β)=α22(1α2+β2)V=F(\alpha,\beta)=-\frac{\alpha^{2}}{2\left(1-\alpha^{2}+\beta^{2}\right)}

or, equivalently,

V(x,y,z)=y22(x2y2+z2)V(x,y,z)=-\frac{y^{2}}{2\left(x^{2}-y^{2}+z^{2}\right)} (64)

We mention that V=w(r,y)=y2/2(y2r2)V=w(r,y)=y^{2}/2\left(y^{2}-r^{2}\right), with r=x2+z2r=\sqrt{x^{2}+z^{2}}, is an axially symmetric potential.

The family (62) is traced isoenergetically by a test particle of unit mass under the influence of the potential (64) (with energy =1/4\mathcal{E}=-1/4 ) in the domain of the xyzxyz space defined by x2+z2yx2+z2-\sqrt{x^{2}+z^{2}}\leqslant y\leqslant\sqrt{x^{2}+z^{2}}.

Example 2. For the family

f(x,y,z)=y2/x=c1,g(x,y,z)=z/x2=c2,f(x,y,z)=y^{2}/x=c_{1},\quad g(x,y,z)=z/x^{2}=c_{2},

the orbital functions are

α=y2x,β=2zx,\alpha=\frac{y}{2x},\quad\beta=\frac{2z}{x}, (65)

and α0=y/4x2,β0=2z/x2\alpha_{0}=-y/4x^{2},\beta_{0}=2z/x^{2}. The four differential conditions are satisfied and

Fβ=2β(3α2+1)(4α2+β2+2)2,Fα=2α(3β2+2)(4α2+β2+2)2F_{\beta}=-\frac{2\beta\left(3\alpha^{2}+1\right)}{\left(4\alpha^{2}+\beta^{2}+2\right)^{2}},\quad F_{\alpha}=\frac{2\alpha\left(3\beta^{2}+2\right)}{\left(4\alpha^{2}+\beta^{2}+2\right)^{2}} (66)

The potential is determined uniquely from (66) V=F(α,β)=(3α2+1)/(4α2+β2+2)V=F(\alpha,\beta)=\left(3\alpha^{2}+1\right)/\left(4\alpha^{2}+\beta^{2}+2\right) or, equivalently,

V(x,y,z)=4x2+3y24(2x2+y2+4z2).V(x,y,z)=\frac{4x^{2}+3y^{2}}{4\left(2x^{2}+y^{2}+4z^{2}\right)}. (67)

The family (65) is traced isoenergetically by a test particle of unit mass under the influence of the potential (67) (with energy =1\mathcal{E}=1 ). Inequality (9) is 2x2/(2x2+y2+4z2)02x^{2}/\left(2x^{2}+y^{2}+4z^{2}\right)\geqslant 0.

The above potentials (64) and (67) are homogeneous of zero-degree, since they depend on the orbital functions α,β\alpha,\beta which are in this case homogeneous of zero-degree.

Example 3. For the family

f=(e1x+y+e1)ex=c1,g=(e2x+y+e2)ex=c2f=\left(e_{1}x+y+e_{1}\right)\mathrm{e}^{-x}=c_{1},\quad g=\left(e_{2}x+y+e_{2}\right)\mathrm{e}^{-x}=c_{2}

the orbital functions are

α=e1x+y,β=e2x+z,e1,e2= const ,\alpha=e_{1}x+y,\quad\beta=e_{2}x+z,\quad e_{1},e_{2}=\text{ const }, (68)

with α0=α+e1,β0=β+e2\alpha_{0}=\alpha+e_{1},\beta_{0}=\beta+e_{2}. The three conditions (36), (41), (50) are satisfied for any values of e1,e2e_{1},e_{2} but the fourth one, namely condition (54), is satisfied only in the case 1+e12+e22=01+e_{1}^{2}+e_{2}^{2}=0. If we admit complex families, we shall obtain a complex potential (see Contopoulos and Bozis (2000)).

Counter-examples

  • For α=(x+z2)/(1y),β=2z\alpha=\left(x+z^{2}\right)/(1-y),\beta=-2z functionally independent we have 𝒫=0\mathcal{P}=0 and 𝒬=2α0\mathcal{Q}=-2\alpha_{0}. For α=2xy/(y2z),β=x/(y2z)\alpha=-2xy/\left(y^{2}-z\right),\beta=x/\left(y^{2}-z\right) we obtain 𝒫=2xβ0/(y2z)\mathcal{P}=-2x\beta_{0}/\left(y^{2}-z\right) and 𝒬=0\mathcal{Q}=0 (case 3 in synthesis of the problem, subsection 4.1).

  • For the family of regular orbits f(x,y,z)=z/x=c1,g(x,y,z)=x2+y2=c2f(x,y,z)=z/x=c_{1},g(x,y,z)=x^{2}+y^{2}=c_{2}, the orbital functions are

α=xy,β=zx\alpha=-\frac{x}{y},\quad\beta=\frac{z}{x} (69)

and α0=(x2+y2)/y3,β0=0\alpha_{0}=-\left(x^{2}+y^{2}\right)/y^{3},\beta_{0}=0. Conditions (36), (41), (50) are fulfilled, but not the last condition (54).

5. The case of straight lines

As was shown by Bozis and Kotoulas (2004), the family (1) consists of straight lines if and only if the orbital functions satisfy

αx+ααy+βαz=0,βx+αβy+ββz=0\alpha_{x}+\alpha\alpha_{y}+\beta\alpha_{z}=0,\quad\beta_{x}+\alpha\beta_{y}+\beta\beta_{z}=0 (70)

The equations for the potential VV are (Bozis and Kotoulas 2004, Anisiu 2004b)

αVxVy=0,βVxVz=0\alpha V_{x}-V_{y}=0,\quad\beta V_{x}-V_{z}=0 (71)

as it follows easily from (8) in view of (70). We substitute V=v(α)V=v(\alpha) into (71), and recall that v(α)v^{\prime}(\alpha) is not zero on open sets. It follows that ααxαy=0\alpha\alpha_{x}-\alpha_{y}=0 and βαxαz=0\beta\alpha_{x}-\alpha_{z}=0; we insert αy=ααx\alpha_{y}=\alpha\alpha_{x} and αz=βαx\alpha_{z}=\beta\alpha_{x} in the first equation from (70) and get αx=0\alpha_{x}=0, hence also αy=αz=0\alpha_{y}=\alpha_{z}=0. This is not possible for a nonconstant function α\alpha, hence no potential V=v(α)V=v(\alpha) can be compatible with a family of straight lines (70).

Now we look for a solution V=F(α,β)V=F(\alpha,\beta) of (71), with α,β\alpha,\beta functionally independent. We get the system

(ααxαy)Fα+(αβxβy)Fβ=0\displaystyle\left(\alpha\alpha_{x}-\alpha_{y}\right)F_{\alpha}+\left(\alpha\beta_{x}-\beta_{y}\right)F_{\beta}=0 (72)
(βαxαz)Fα+(ββxβz)Fβ=0\displaystyle\left(\beta\alpha_{x}-\alpha_{z}\right)F_{\alpha}+\left(\beta\beta_{x}-\beta_{z}\right)F_{\beta}=0

whose discriminant, in view of (70), is Δ=(1+α2+β2)(αyβzαzβy)\Delta=\left(1+\alpha^{2}+\beta^{2}\right)\left(\alpha_{y}\beta_{z}-\alpha_{z}\beta_{y}\right). If Δ0\Delta\neq 0, it follows that V=V= const, which is excluded. If Δ=0\Delta=0, when we eliminate the terms containing α\alpha from equations (70), we obtain αxβyαyβx=0\alpha_{x}\beta_{y}-\alpha_{y}\beta_{x}=0. This condition, together with αyβzαzβy=0\alpha_{y}\beta_{z}-\alpha_{z}\beta_{y}=0, shows that α\alpha and β\beta must be functionally dependent, which contradicts our hypothesis. It follows that the families of straight lines are not compatible with potentials V=F(α,β)V=F(\alpha,\beta).

6. Conclusions

In the present study we have examined a solvable version of the three-dimensional inverse problem of dynamics. Given a two-parametric family of orbits (1), the potential function VV, which is compatible with it, must satisfy two free-of-energy PDEs (10) and (12). So, our aim is to find special common solutions of these two PDEs for the potential function V=V(x,y,z)V=V(x,y,z).

In fact, instead of the two-parametric family (1), we deal with the pair ( α,β\alpha,\beta ) and try to find solutions of the form V=v(α)V=v(\alpha) or V=F(α,β)V=F(\alpha,\beta). We have to impose differential conditions on α\alpha and β\beta in order that the problem has solutions of one of these forms. The conditions are not difficult to check once the analytical form of the orbital functions is known. When the conditions are fulfilled, we obtain by quadrature the potential as a function of α\alpha, or of α\alpha and β\beta correspondingly. Afterwards, we can determine the function of energy dependence from (8) and the allowed area of motion of a test particle of unit mass from (9) (examples 1 and 2 in sections 3.2 and 4.2).

The theoretical results can be applied to a variety of real-world problems. For example, elliptic orbits have been observed in galaxy models (Miller and Smith (1979), figure 5, p 789). The two-parametric family of elliptic orbits

f(x,y,z)=x2+2y2+z2=c1,g(x,y,z)=zx=c2f(x,y,z)=x^{2}+2y^{2}+z^{2}=c_{1},\quad g(x,y,z)=z-x=c_{2} (73)

leads to the pair

α=x+z2y,β=1.\alpha=-\frac{x+z}{2y},\quad\beta=1. (74)

Adopting the methodology for potentials of the form V=v(α)V=v(\alpha) which was presented in section 3 , we find that the potential (homogeneous of zero-degree) V(α)=α2(2+α2)/4V(\alpha)=-\alpha^{2}\left(2+\alpha^{2}\right)/4, or equivalently,

V(x,y,z)=(x+z)2(x2+8y2+z2+2xz)64y4V(x,y,z)=-\frac{(x+z)^{2}\left(x^{2}+8y^{2}+z^{2}+2xz\right)}{64y^{4}} (75)

produces this family of orbits.
Our hypothesis on the expression of the potential has allowed us to obtain pairs of families of curves and potentials under whose action they are traced in specified regions of the space. Thus we offer a method which can generate various examples, which were not numerous until now due to the complexity of calculations, substantiating the 3D inverse problem of dynamics. It can be apllied to various fields of physics where one disposes of information on particle motions, and the form of the potential is needed.

Acknowledgments

Thomas Kotoulas would like to thank the ’Tiberiu Popoviciu’ Institute of Numerical Analysis, Cluj, Romania for the hospitality from 7 to 12 November 2005. His work was financially supported by the scientific program ’EPEAEK II, PYTHAGORAS’, no 21878, of the Greek Ministry of Education and EU, and the work of Mira-Cristiana Anisiu by 2CEEX0611-96 of
the Romanian Ministry of Education and Research. The authors express their gratitude to the Board Member and to the referees for helping to improve the quality of the paper.

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