New solutions in the direct problem of dynamics

Mira Anisiu
(original), paper

Abstract

Given a planar potential \(V\), we look for families of orbits \(f(x,y)=c\) (determined by their slope function \(\U{3b3} =fy/fx\)), traced by a material point of unit mass under the action of that potential. The second-order equation which relates \(\U{3b3}\) and \(V\) is nonlinear in \(\U{3b3}\); to find special solutions, we consider in addition a linear first-order partial differential equation satisfied by \(\U{3b3}\). The problem does not admit always solutions; but when solutions do exist, they can be found by algebraic manipulations. Examples are given for homogeneous families \(\U{3b3}\), and for some special cases which arise in the course of reasoning.

Authors

Cristina Blaga
Faculty of Mathematics and Computer Science, Babes-Bolyai University, Cluj-Napoca, Romania

Mira-Cristiana Anisiu
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy,  Cluj-Napoca, Romania

George Bozis
Department of Physics, Aristotle University of Thessaloniki GR-54006, Greece

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C. Blaga, M.-C. Anisiu, G. Bozis, New solutions in the direct problem of dynamics, PADEU 19 (2007), 17-27 (pdf file here)

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https://ictp.acad.ro/anisiu/papers/2007-Blaga-A-Bozis-NewSolutions.pdf

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2007-Blaga-A-Bozis-NewSolutions

New solutions in the direct problem OF DYNAMICS

C. Blaga 1 1 ^(1){ }^{1}1, M.-C. Anisiu 2 2 ^(2){ }^{2}2,G. Bozis 3 3 ^(3){ }^{3}3 1 1 ^(1){ }^{1}1 Faculty of Mathematics and Computer Science,"Babeş-Bolyai" University, Cluj-Napoca, Romania 2 2 ^(2){ }^{2}2 T. Popoviciu Institute of Numerical Analysis, Romanian Academy P.O. Box 68, 400110 Cluj-Napoca, Romania 3 3 ^(3){ }^{3}3 Department of Physics, Aristotle University of Thessaloniki GR-54006, GreeceE-mail: 1 1 ^(1){ }^{1}1 cpblaga@math.ubbcluj.ro, 2 2 ^(2){ }^{2}2 mira@math.ubbcluj.ro, 3 3 ^(3){ }^{3}3 gbozis@auth.gr

Abstract

Given a planar potential V V VVV, we look for families of orbits f ( x , y ) = c f ( x , y ) = c f(x,y)=cf(x, y)=cf(x,y)=c (determined by their slope function γ = f y / f x γ = f y / f x gamma=f_(y)//f_(x)\gamma=f_{y} / f_{x}γ=fy/fx ), traced by a material point of unit mass under the action of that potential. The second-order equation which relates γ γ gamma\gammaγ and V V VVV is nonlinear in γ γ gamma\gammaγ; to find special solutions, we consider in addition a linear first-order partial differential equation satisfied by γ γ gamma\gammaγ. The problem does not admit always solutions; but when solutions do exist, they can be found by algebraic manipulations. Examples are given for homogeneous families γ γ gamma\gammaγ, and for some special cases which arise in the course of reasoning.

1 Introduction

The planar direct problem of Dynamics consists in finding families of orbits f ( x , y ) = c f ( x , y ) = c f(x,y)=cf(x, y)=cf(x,y)=c traced in the x y x y xyx yxy Cartesian plane by a material point of unit mass, under the action of a given potential V V VVV.
Any family of orbits is determined by its 'slope function' γ = f y / f x γ = f y / f x gamma=f_(y)//f_(x)\gamma=f_{y} / f_{x}γ=fy/fx, the subscripts denoting partial derivatives. There are two equations relating the functions V , γ V , γ V,gammaV, \gammaV,γ (and their derivatives):
(i) the first order equation in V V VVV, given by Szebehely (1974) (equation (8) below), which is associated with the energy dependence on the family f f fff;
(ii) the energy-free second order linear equation in V V VVV, given by Bozis (1984) and written below in the form (6)-(7).
These equations, born in the framework of the inverse problem, are rearranged here in order to face the direct problem, as suggested by Bozis (1995). The difficulty with the second order equation arises from its nonlinearity in the unknown family γ γ gamma\gammaγ. This is why in several papers additional information on the families of orbits (sometimes on the given potentials also) was used in order to obtain solutions of the direct problem. Homogeneous families produced by homogeneous or inhomogeneous potentials were studied by Bozis and Grigoriadou (1993) and by Bozis et al (1997), as well as families of orbits with γ = γ ( x ) γ = γ ( x ) gamma=gamma(x)\gamma=\gamma(x)γ=γ(x), corresponding to families f ( x , y ) = y + h ( x ) = c f ( x , y ) = y + h ( x ) = c f(x,y)=y+h(x)=cf(x, y)=y+h(x)=cf(x,y)=y+h(x)=c (Bozis et al, 2000). Later on (Anisiu et al, 2004), the solutions of equation (6) were looked for in a class of functions verifying a linear PDE
(1) r ( x , y ) γ x + γ y = 0 (1) r ( x , y ) γ x + γ y = 0 {:(1)r(x","y)gamma_(x)+gamma_(y)=0:}\begin{equation*} r(x, y) \gamma_{x}+\gamma_{y}=0 \tag{1} \end{equation*}(1)r(x,y)γx+γy=0
this class contains the homogeneous functions f f fff, for which γ γ gamma\gammaγ is homogeneous of zero degree and r = x / y r = x / y r=x//yr=x / yr=x/y. In all these cases γ γ gamma\gammaγ was found as the common root of certain algebraic equations in γ γ gamma\gammaγ, with coefficients depending on V V VVV and on derivatives of V V VVV.
In what follows we consider a given potential V V VVV and study the existence and the construction of solutions γ γ gamma\gammaγ of the direct problem of dynamics, under the hypothesis that γ γ gamma\gammaγ satisfies an equation of the form
(2) a ( x , y , γ ) γ x + b ( x , y , γ ) γ y = c ( x , y , γ ) (2) a ( x , y , γ ) γ x + b ( x , y , γ ) γ y = c ( x , y , γ ) {:(2)a(x","y","gamma)gamma_(x)+b(x","y","gamma)gamma_(y)=c(x","y","gamma):}\begin{equation*} a(x, y, \gamma) \gamma_{x}+b(x, y, \gamma) \gamma_{y}=c(x, y, \gamma) \tag{2} \end{equation*}(2)a(x,y,γ)γx+b(x,y,γ)γy=c(x,y,γ)
We may suppose b ≠ 0 b ≠ 0 b!=0b \neq 0b≠0 and denote by r = a / b r = a / b r=a//br=a / br=a/b and s = c / b s = c / b s=c//bs=c / bs=c/b.
In the following we replace (2) by the equation
(3) r ( x , y , γ ) γ x + γ y = s ( x , y , γ ) (3) r ( x , y , γ ) γ x + γ y = s ( x , y , γ ) {:(3)r(x","y","gamma)gamma_(x)+gamma_(y)=s(x","y","gamma):}\begin{equation*} r(x, y, \gamma) \gamma_{x}+\gamma_{y}=s(x, y, \gamma) \tag{3} \end{equation*}(3)r(x,y,γ)γx+γy=s(x,y,γ)
with r r rrr and s s sss known functions of x , y , γ x , y , γ x,y,gammax, y, \gammax,y,γ. We then develop the reasoning to check whether the given potential can be compatible with families γ = γ ( x , y ) γ = γ ( x , y ) gamma=gamma(x,y)\gamma=\gamma(x, y)γ=γ(x,y) satisfying the condition (3).
In section 2 we give the basic partial differential equations of the direct problem and add to them two (second order) differential relations derived from (3). Then, in section 3 we obtain the algebraic equations verified by γ x γ x gamma_(x)\gamma_{x}γx. In section 4 we obtain three algebraic equations which the required family must satisfy when γ γ gamma\gammaγ is a homogeneous function of degree m m mmm. The resultants of the two pairs of equations must vanish and this leads to two differential conditions which all adequate potentials must satisfy. In section 5 we present some special cases and examples. A synthesis is presented in section 6.

2 Partial differential equations satisfied by γ γ gamma\gammaγ

We consider a planar potential V V VVV under the action of which a monoparametric family of orbits
(4) f ( x , y ) = c (4) f ( x , y ) = c {:(4)f(x","y)=c:}\begin{equation*} f(x, y)=c \tag{4} \end{equation*}(4)f(x,y)=c
can be described by a material point of unit mass. This family can be represented in a unique way by its slope function
(5) γ = f y f x (5) γ = f y f x {:(5)gamma=(f_(y))/(f_(x)):}\begin{equation*} \gamma=\frac{f_{y}}{f_{x}} \tag{5} \end{equation*}(5)γ=fyfx
To each γ γ gamma\gammaγ there corresponds a unique family (4).
The nonlinear second order differential equation relating potentials and orbits in the form suitable for the direct problem (Bozis, 1995) is
(6) γ 2 γ x x − 2 γ γ x y + γ y y = h (6) γ 2 γ x x − 2 γ γ x y + γ y y = h {:(6)gamma^(2)gamma_(xx)-2gammagamma_(xy)+gamma_(yy)=h:}\begin{equation*} \gamma^{2} \gamma_{x x}-2 \gamma \gamma_{x y}+\gamma_{y y}=h \tag{6} \end{equation*}(6)γ2γxx−2γγxy+γyy=h
where
(7) h = γ γ x − γ y V y γ + V x ( − γ x V x + ( 2 γ γ x − 3 γ y ) V y + γ ( V x x − V y y ) + ( γ 2 − 1 ) V x y ) . (7) h = γ γ x − γ y V y γ + V x − γ x V x + 2 γ γ x − 3 γ y V y + γ V x x − V y y + γ 2 − 1 V x y . {:(7)h=(gammagamma_(x)-gamma_(y))/(V_(y)gamma+V_(x))(-gamma_(x)V_(x)+(2gammagamma_(x)-3gamma_(y))V_(y)+gamma(V_(xx)-V_(yy))+(gamma^(2)-1)V_(xy)).:}\begin{equation*} h=\frac{\gamma \gamma_{x}-\gamma_{y}}{V_{y} \gamma+V_{x}}\left(-\gamma_{x} V_{x}+\left(2 \gamma \gamma_{x}-3 \gamma_{y}\right) V_{y}+\gamma\left(V_{x x}-V_{y y}\right)+\left(\gamma^{2}-1\right) V_{x y}\right) . \tag{7} \end{equation*}(7)h=γγx−γyVyγ+Vx(−γxVx+(2γγx−3γy)Vy+γ(Vxx−Vyy)+(γ2−1)Vxy).
Szebehely's equation (1974) involving the total energy E ( f ) E ( f ) E(f)E(f)E(f) is (Bozis, 1983)
(8) V x + γ V y + 2 Γ 1 + γ 2 ( E ( f ) − V ) = 0 (8) V x + γ V y + 2 Γ 1 + γ 2 ( E ( f ) − V ) = 0 {:(8)V_(x)+gammaV_(y)+(2Gamma)/(1+gamma^(2))(E(f)-V)=0:}\begin{equation*} V_{x}+\gamma V_{y}+\frac{2 \Gamma}{1+\gamma^{2}}(E(f)-V)=0 \tag{8} \end{equation*}(8)Vx+γVy+2Γ1+γ2(E(f)−V)=0
where
(9) Γ = γ γ x − γ y (9) Γ = γ γ x − γ y {:(9)Gamma=gammagamma_(x)-gamma_(y):}\begin{equation*} \Gamma=\gamma \gamma_{x}-\gamma_{y} \tag{9} \end{equation*}(9)Γ=γγx−γy
In order to solve (8) for E ( f ) E ( f ) E(f)E(f)E(f), the condition Γ ≠ 0 Γ ≠ 0 Gamma!=0\Gamma \neq 0Γ≠0 must be imposed, hence it follows also that V x + γ V y ≠ 0 V x + γ V y ≠ 0 V_(x)+gammaV_(y)!=0V_{x}+\gamma V_{y} \neq 0Vx+γVy≠0. The case Γ = 0 Γ = 0 Gamma=0\Gamma=0Γ=0 was studied in detail by Bozis and Anisiu (2001) and will be considered in section 5. If for a given V V VVV we can find a solution γ γ gamma\gammaγ of (6), equation (8) will allow us to find the energy along each member of the family, namely
(10) E ( f ) = V − ( V x + γ V y ) ( 1 + γ 2 ) 2 Γ (10) E ( f ) = V − V x + γ V y 1 + γ 2 2 Γ {:(10)E(f)=V-((V_(x)+gammaV_(y))(1+gamma^(2)))/(2Gamma):}\begin{equation*} E(f)=V-\frac{\left(V_{x}+\gamma V_{y}\right)\left(1+\gamma^{2}\right)}{2 \Gamma} \tag{10} \end{equation*}(10)E(f)=V−(Vx+γVy)(1+γ2)2Γ
The real parts of the orbits of the family are lying in the region defined by the inequality (Bozis and Ichtiaroglou, 1994)
(11) V x + γ V y Γ ≤ 0 . (11) V x + γ V y Γ ≤ 0 . {:(11)(V_(x)+gammaV_(y))/(Gamma) <= 0.:}\begin{equation*} \frac{V_{x}+\gamma V_{y}}{\Gamma} \leq 0 . \tag{11} \end{equation*}(11)Vx+γVyΓ≤0.
As we have mentioned in the Introduction, the special families of orbits we are going to consider are those for which equation (3) is also satisfied. We differentiate it with respect to x x xxx and obtain
(12) r γ x x + γ x y = − r 001 γ x 2 − r 100 γ x + s 001 γ x + s 100 . (12) r γ x x + γ x y = − r 001 γ x 2 − r 100 γ x + s 001 γ x + s 100 . {:(12)rgamma_(xx)+gamma_(xy)=-r_(001)gamma_(x)^(2)-r_(100)gamma_(x)+s_(001)gamma_(x)+s_(100).:}\begin{equation*} r \gamma_{x x}+\gamma_{x y}=-r_{001} \gamma_{x}^{2}-r_{100} \gamma_{x}+s_{001} \gamma_{x}+s_{100} . \tag{12} \end{equation*}(12)rγxx+γxy=−r001γx2−r100γx+s001γx+s100.
Then we differentiate (3) with respect to y y yyy
(13) r γ x y + γ y y = − r 001 γ x γ y − r 010 γ x + s 001 γ y + s 010 (13) r γ x y + γ y y = − r 001 γ x γ y − r 010 γ x + s 001 γ y + s 010 {:(13)rgamma_(xy)+gamma_(yy)=-r_(001)gamma_(x)gamma_(y)-r_(010)gamma_(x)+s_(001)gamma_(y)+s_(010):}\begin{equation*} r \gamma_{x y}+\gamma_{y y}=-r_{001} \gamma_{x} \gamma_{y}-r_{010} \gamma_{x}+s_{001} \gamma_{y}+s_{010} \tag{13} \end{equation*}(13)rγxy+γyy=−r001γxγy−r010γx+s001γy+s010
For the functions r r rrr and s s sss, which depend on the three variables x , y , γ x , y , γ x,y,gammax, y, \gammax,y,γ, we adopt the three-subscripts notation, e. g. ∂ i + j + k s / ∂ x i ∂ y j ∂ γ k = s i j k ∂ i + j + k s / ∂ x i ∂ y j ∂ γ k = s i j k del^(i+j+k)s//delx^(i)dely^(j)delgamma^(k)=s_(ijk)\partial^{i+j+k} s / \partial x^{i} \partial y^{j} \partial \gamma^{k}=s_{i j k}∂i+j+ks/∂xi∂yj∂γk=sijk. The system of equations (6), (12) and (13) allows us to obtain the second order derivatives of γ γ gamma\gammaγ in terms of γ γ gamma\gammaγ and its first order derivatives.

3 Algebraic equations satisfied by γ x γ x gamma_(x)\gamma_{x}γx

We solve the system of equations (6), (12) and (13) with respect to γ x x , γ x y γ x x , γ x y gamma_(xx),gamma_(xy)\gamma_{x x}, \gamma_{x y}γxx,γxy and γ y y γ y y gamma_(yy)\gamma_{y y}γyy. These second order derivatives depend on γ , γ x , γ y γ , γ x , γ y gamma,gamma_(x),gamma_(y)\gamma, \gamma_{x}, \gamma_{y}γ,γx,γy, on r , s r , s r,sr, sr,s, and their first-order derivatives and, of course, on the first and second order derivatives of V V VVV. In fact, considering (3), we can express γ y γ y gamma_(y)\gamma_{y}γy in terms of γ x γ x gamma_(x)\gamma_{x}γx. We introduce the notations
Π = ( γ + r ) 2 ( V y γ + V x ) K = − 2 ( r 001 − 1 ) V y γ 2 + [ ( 5 r − 2 r r 001 ) V y − ( 2 r 001 + 1 ) V x ] γ + + r [ 3 r V y − ( 1 + 2 r 001 ) V x ] L = V x y γ 3 + [ V x x − V y y + r V x y − 2 ( r 100 − s 001 ) V y ] γ 2 + [ r ( V x x − V y y ) − V x y + 2 ( s 001 − r 100 ) V x + (14) + ( − r r 100 + r 010 + s r 001 − 5 s + 2 r s 001 ) V y ] γ − r V x y + ( − r r 100 + r 010 + s r 001 + s + 2 r s 001 ) V x − 6 r s V y M = ( − s V x y + 2 s 100 V y ) γ 2 + [ s ( V y y − V x x ) + 2 s 100 V x + + ( r s 100 − s 010 − s s 001 ) V y ] γ + s V x y + ( r s 100 − s 010 − s s 001 ) V x + 3 s 2 V y Π = ( γ + r ) 2 V y γ + V x K = − 2 r 001 − 1 V y γ 2 + 5 r − 2 r r 001 V y − 2 r 001 + 1 V x γ + + r 3 r V y − 1 + 2 r 001 V x L = V x y γ 3 + V x x − V y y + r V x y − 2 r 100 − s 001 V y γ 2 + r V x x − V y y − V x y + 2 s 001 − r 100 V x + (14) + − r r 100 + r 010 + s r 001 − 5 s + 2 r s 001 V y γ − r V x y + − r r 100 + r 010 + s r 001 + s + 2 r s 001 V x − 6 r s V y M = − s V x y + 2 s 100 V y γ 2 + s V y y − V x x + 2 s 100 V x + + r s 100 − s 010 − s s 001 V y γ + s V x y + r s 100 − s 010 − s s 001 V x + 3 s 2 V y {:[Pi=(gamma+r)^(2)(V_(y)gamma+V_(x))],[K=-2(r_(001)-1)V_(y)gamma^(2)+[(5r-2rr_(001))V_(y)-(2r_(001)+1)V_(x)]gamma+],[+r[3rV_(y)-(1+2r_(001))V_(x)]],[L=V_(xy)gamma^(3)+[V_(xx)-V_(yy)+rV_(xy)-2(r_(100)-s_(001))V_(y)]gamma^(2)],[+[r(V_(xx)-V_(yy))-V_(xy)+2(s_(001)-r_(100))V_(x)+:}],[(14){:+(-rr_(100)+r_(010)+sr_(001)-5s+2rs_(001))V_(y)]gamma],[-rV_(xy)+(-rr_(100)+r_(010)+sr_(001)+s+2rs_(001))V_(x)-6rsV_(y)],[M=(-sV_(xy)+2s_(100)V_(y))gamma^(2)+[s(V_(yy)-V_(xx))+2s_(100)V_(x)+:}],[{:+(rs_(100)-s_(010)-ss_(001))V_(y)]gamma],[+sV_(xy)+(rs_(100)-s_(010)-ss_(001))V_(x)+3s^(2)V_(y)]:}\begin{align*} \Pi & =(\gamma+r)^{2}\left(V_{y} \gamma+V_{x}\right) \\ K & =-2\left(r_{001}-1\right) V_{y} \gamma^{2}+\left[\left(5 r-2 r r_{001}\right) V_{y}-\left(2 r_{001}+1\right) V_{x}\right] \gamma+ \\ & +r\left[3 r V_{y}-\left(1+2 r_{001}\right) V_{x}\right] \\ L & =V_{x y} \gamma^{3}+\left[V_{x x}-V_{y y}+r V_{x y}-2\left(r_{100}-s_{001}\right) V_{y}\right] \gamma^{2} \\ & +\left[r\left(V_{x x}-V_{y y}\right)-V_{x y}+2\left(s_{001}-r_{100}\right) V_{x}+\right. \\ & \left.+\left(-r r_{100}+r_{010}+s r_{001}-5 s+2 r s_{001}\right) V_{y}\right] \gamma \tag{14}\\ & -r V_{x y}+\left(-r r_{100}+r_{010}+s r_{001}+s+2 r s_{001}\right) V_{x}-6 r s V_{y} \\ M & =\left(-s V_{x y}+2 s_{100} V_{y}\right) \gamma^{2}+\left[s\left(V_{y y}-V_{x x}\right)+2 s_{100} V_{x}+\right. \\ & \left.+\left(r s_{100}-s_{010}-s s_{001}\right) V_{y}\right] \gamma \\ & +s V_{x y}+\left(r s_{100}-s_{010}-s s_{001}\right) V_{x}+3 s^{2} V_{y} \end{align*}Π=(γ+r)2(Vyγ+Vx)K=−2(r001−1)Vyγ2+[(5r−2rr001)Vy−(2r001+1)Vx]γ++r[3rVy−(1+2r001)Vx]L=Vxyγ3+[Vxx−Vyy+rVxy−2(r100−s001)Vy]γ2+[r(Vxx−Vyy)−Vxy+2(s001−r100)Vx+(14)+(−rr100+r010+sr001−5s+2rs001)Vy]γ−rVxy+(−rr100+r010+sr001+s+2rs001)Vx−6rsVyM=(−sVxy+2s100Vy)γ2+[s(Vyy−Vxx)+2s100Vx++(rs100−s010−ss001)Vy]γ+sVxy+(rs100−s010−ss001)Vx+3s2Vy
Then, the second-order derivatives of γ γ gamma\gammaγ can be expressed as
γ x x = 1 Π ( K γ x 2 + L γ x + M ) γ x x = 1 Π K γ x 2 + L γ x + M gamma_(xx)=(1)/(Pi)(Kgamma_(x)^(2)+Lgamma_(x)+M)\gamma_{x x}=\frac{1}{\Pi}\left(K \gamma_{x}^{2}+L \gamma_{x}+M\right)γxx=1Π(Kγx2+Lγx+M)
(15) γ x y = − 1 Π { ( r K + r 001 Π ) γ x 2 + [ r L + ( r 100 − s 001 ) Π ] γ x + r M − s 100 Π } γ y y = 1 Π { ( r 2 K + 2 r r 001 Π ) γ x 2 + [ r 2 L + ( r r 100 − r 010 − s r 001 − 2 r s 001 ) Π ] γ x + r 2 M − ( r s 100 + s s 001 + s 010 ) Π } (15) γ x y = − 1 Π r K + r 001 Π γ x 2 + r L + r 100 − s 001 Π γ x + r M − s 100 Π γ y y = 1 Π r 2 K + 2 r r 001 Π γ x 2 + r 2 L + r r 100 − r 010 − s r 001 − 2 r s 001 Π γ x + r 2 M − r s 100 + s s 001 + s 010 Π {:[(15)gamma_(xy)=-(1)/(Pi){(rK+r_(001)Pi)gamma_(x)^(2)+[rL+(r_(100)-s_(001))Pi]gamma_(x)+rM-s_(100)Pi}],[gamma_(yy)=(1)/(Pi){(r^(2)K+2rr_(001)Pi)gamma_(x)^(2)+[r^(2)L+(rr_(100)-r_(010)-sr_(001)-2rs_(001))Pi]gamma_(x):}],[{:+r^(2)M-(rs_(100)+ss_(001)+s_(010))Pi}]:}\begin{align*} \gamma_{x y} & =-\frac{1}{\Pi}\left\{\left(r K+r_{001} \Pi\right) \gamma_{x}^{2}+\left[r L+\left(r_{100}-s_{001}\right) \Pi\right] \gamma_{x}+r M-s_{100} \Pi\right\} \tag{15}\\ \gamma_{y y} & =\frac{1}{\Pi}\left\{\left(r^{2} K+2 r r_{001} \Pi\right) \gamma_{x}^{2}+\left[r^{2} L+\left(r r_{100}-r_{010}-s r_{001}-2 r s_{001}\right) \Pi\right] \gamma_{x}\right. \\ & \left.+r^{2} M-\left(r s_{100}+s s_{001}+s_{010}\right) \Pi\right\} \end{align*}(15)γxy=−1Π{(rK+r001Π)γx2+[rL+(r100−s001)Π]γx+rM−s100Π}γyy=1Π{(r2K+2rr001Π)γx2+[r2L+(rr100−r010−sr001−2rs001)Π]γx+r2M−(rs100+ss001+s010)Π}
Remark 1 As we have already mentioned, we have V y γ + V x ≠ 0 V y γ + V x ≠ 0 V_(y)gamma+V_(x)!=0V_{y} \gamma+V_{x} \neq 0Vyγ+Vx≠0. The case γ + r = 0 γ + r = 0 gamma+r=0\gamma+r=0γ+r=0 will be studied later. For the moment we suppose that the denominator Π Π Pi\PiΠ in (15) is different from zero.
Working with (15) we find that the two compatibility conditions ( γ x x ) y = ( γ x y ) x γ x x y = γ x y x (gamma_(xx))_(y)=(gamma_(xy))_(x)\left(\gamma_{x x}\right)_{y}= \left(\gamma_{x y}\right)_{x}(γxx)y=(γxy)x and ( γ x y ) y = ( γ y y ) x γ x y y = γ y y x (gamma_(xy))_(y)=(gamma_(yy))_(x)\left(\gamma_{x y}\right)_{y}=\left(\gamma_{y y}\right)_{x}(γxy)y=(γyy)x produce one single relation which, after substituting γ x x , γ x y γ x x , γ x y gamma_(xx),gamma_(xy)\gamma_{x x}, \gamma_{x y}γxx,γxy and γ y y γ y y gamma_(yy)\gamma_{y y}γyy given by (15) and γ y γ y gamma_(y)\gamma_{y}γy from (3), reduces to a third-degree algebraic equation in γ x γ x gamma_(x)\gamma_{x}γx
(16) Γ 3 γ x 3 + Γ 2 γ x 2 + Γ 1 γ x + Γ 0 = 0 (16) Γ 3 γ x 3 + Γ 2 γ x 2 + Γ 1 γ x + Γ 0 = 0 {:(16)Gamma_(3)gamma_(x)^(3)+Gamma_(2)gamma_(x)^(2)+Gamma_(1)gamma_(x)+Gamma_(0)=0:}\begin{equation*} \Gamma_{3} \gamma_{x}^{3}+\Gamma_{2} \gamma_{x}^{2}+\Gamma_{1} \gamma_{x}+\Gamma_{0}=0 \tag{16} \end{equation*}(16)Γ3γx3+Γ2γx2+Γ1γx+Γ0=0
The coefficient Γ 3 Γ 3 Gamma_(3)\Gamma_{3}Γ3 of γ x 3 γ x 3 gamma_(x)^(3)\gamma_{x}^{3}γx3 in (16) is given by
(17) Γ 3 = ( γ + r ) 2 ( V y γ + V x ) [ ( V y γ + V x ) ( γ + r ) r 002 − 2 ( V y γ + V x ) r 001 2 + ( 2 V y γ + 3 r V y − V x ) r 001 ] . (17) Γ 3 = ( γ + r ) 2 V y γ + V x V y γ + V x ( γ + r ) r 002 − 2 V y γ + V x r 001 2 + 2 V y γ + 3 r V y − V x r 001 . {:[(17)Gamma_(3)=(gamma+r)^(2)(V_(y)gamma+V_(x))[(V_(y)gamma+V_(x))(gamma+r)r_(002)-2(V_(y)gamma+V_(x))r_(001)^(2):}],[{:+(2V_(y)gamma+3rV_(y)-V_(x))r_(001)].]:}\begin{align*} \Gamma_{3} & =(\gamma+r)^{2}\left(V_{y} \gamma+V_{x}\right)\left[\left(V_{y} \gamma+V_{x}\right)(\gamma+r) r_{002}-2\left(V_{y} \gamma+V_{x}\right) r_{001}^{2}\right. \tag{17}\\ & \left.+\left(2 V_{y} \gamma+3 r V_{y}-V_{x}\right) r_{001}\right] . \end{align*}(17)Γ3=(γ+r)2(Vyγ+Vx)[(Vyγ+Vx)(γ+r)r002−2(Vyγ+Vx)r0012+(2Vyγ+3rVy−Vx)r001].
In the last factor of Γ 3 Γ 3 Gamma_(3)\Gamma_{3}Γ3, all the terms contain the derivatives of r r rrr with respect to its third variable γ γ gamma\gammaγ. It follows that, if r r rrr depends merely on x x xxx and y y yyy, equation (16) is in fact at most of second degree in γ x γ x gamma_(x)\gamma_{x}γx. There are significant situations when this condition is fulfilled, as in the case of functions γ γ gamma\gammaγ homogeneous of order m ≠ 0 m ≠ 0 m!=0m \neq 0m≠0, which verify
(18) x γ x + y γ y = m γ (18) x γ x + y γ y = m γ {:(18)xgamma_(x)+ygamma_(y)=m gamma:}\begin{equation*} x \gamma_{x}+y \gamma_{y}=m \gamma \tag{18} \end{equation*}(18)xγx+yγy=mγ
The coefficient Γ 0 Γ 0 Gamma_(0)\Gamma_{0}Γ0 in (16) reads
(19) Γ 0 = a 1 s 3 + a 2 s 2 + a 3 s − ( V y γ + V x ) ( a 4 s 200 + a 5 s 110 + a 6 s 020 + a 7 s 100 + a 8 s 010 ) . (19) Γ 0 = a 1 s 3 + a 2 s 2 + a 3 s − V y γ + V x a 4 s 200 + a 5 s 110 + a 6 s 020 + a 7 s 100 + a 8 s 010 . {:(19)Gamma_(0)=a_(1)s^(3)+a_(2)s^(2)+a_(3)s-(V_(y)gamma+V_(x))(a_(4)s_(200)+a_(5)s_(110)+a_(6)s_(020)+a_(7)s_(100)+a_(8)s_(010)).:}\begin{equation*} \Gamma_{0}=a_{1} s^{3}+a_{2} s^{2}+a_{3} s-\left(V_{y} \gamma+V_{x}\right)\left(a_{4} s_{200}+a_{5} s_{110}+a_{6} s_{020}+a_{7} s_{100}+a_{8} s_{010}\right) . \tag{19} \end{equation*}(19)Γ0=a1s3+a2s2+a3s−(Vyγ+Vx)(a4s200+a5s110+a6s020+a7s100+a8s010).
It follows that the coefficient Γ 0 = 0 Γ 0 = 0 Gamma_(0)=0\Gamma_{0}=0Γ0=0 if s ( x , y , γ ) = 0 s ( x , y , γ ) = 0 s(x,y,gamma)=0s(x, y, \gamma)=0s(x,y,γ)=0. After a factorization by γ x γ x gamma_(x)\gamma_{x}γx, equation (16) is again of second degree.
Remark 2 When γ γ gamma\gammaγ satisfies the condition (1) (case studied by Anisiu et al, 2004), in equation (16) Γ 3 = Γ 0 = 0 Γ 3 = Γ 0 = 0 Gamma_(3)=Gamma_(0)=0\Gamma_{3}=\Gamma_{0}=0Γ3=Γ0=0. Therefore γ x γ x gamma_(x)\gamma_{x}γx is the solution of an equation of first degree. This happens, for example, for γ γ gamma\gammaγ homogeneous of order 0.
In what follows, to ease the algebra, we shall assume that the functions r r rrr and/or s s sss are of a form that makes equation (16) of second degree, i. e.
(20) G 2 γ x 2 + G 1 γ x + G 0 = 0 . (20) G 2 γ x 2 + G 1 γ x + G 0 = 0 . {:(20)G_(2)gamma_(x)^(2)+G_(1)gamma_(x)+G_(0)=0.:}\begin{equation*} G_{2} \gamma_{x}^{2}+G_{1} \gamma_{x}+G_{0}=0 . \tag{20} \end{equation*}(20)G2γx2+G1γx+G0=0.
We differentiate (20) with respect to x x xxx and substitute the second-order derivatives of γ γ gamma\gammaγ from (15) and γ y γ y gamma_(y)\gamma_{y}γy from (3); the result will be an equation of third order in γ x γ x gamma_(x)\gamma_{x}γx
(21) H 3 γ x 3 + H 2 γ x 2 + H 1 γ x + H 0 = 0 (21) H 3 γ x 3 + H 2 γ x 2 + H 1 γ x + H 0 = 0 {:(21)H_(3)gamma_(x)^(3)+H_(2)gamma_(x)^(2)+H_(1)gamma_(x)+H_(0)=0:}\begin{equation*} H_{3} \gamma_{x}^{3}+H_{2} \gamma_{x}^{2}+H_{1} \gamma_{x}+H_{0}=0 \tag{21} \end{equation*}(21)H3γx3+H2γx2+H1γx+H0=0
Our calculations have shown that equation (21) is of second degree if s = 0 s = 0 s=0s=0s=0; but it will be of third degree for homogeneous functions of order m m mmm. In order that (20) and (21) have a common solution, the necessary and sufficient condition is that their resultant is null. This is a first condition that γ γ gamma\gammaγ has to fulfil.
Let us suppose that the resultant of (20) and (21) is null. We express γ x 2 γ x 2 gamma_(x)^(2)\gamma_{x}^{2}γx2 from (20) and substitute it in the first two terms of (21), then again in the result. It follows that γ x γ x gamma_(x)\gamma_{x}γx is given by
(22) ( H 3 G 1 2 − H 3 G 2 G 0 − H 2 G 2 G 1 + H 1 G 2 2 ) γ x + H 3 G 1 G 0 − H 2 G 2 G 0 + H 0 G 2 2 = 0 . (22) H 3 G 1 2 − H 3 G 2 G 0 − H 2 G 2 G 1 + H 1 G 2 2 γ x + H 3 G 1 G 0 − H 2 G 2 G 0 + H 0 G 2 2 = 0 . {:(22)(H_(3)G_(1)^(2)-H_(3)G_(2)G_(0)-H_(2)G_(2)G_(1)+H_(1)G_(2)^(2))gamma_(x)+H_(3)G_(1)G_(0)-H_(2)G_(2)G_(0)+H_(0)G_(2)^(2)=0.:}\begin{equation*} \left(H_{3} G_{1}^{2}-H_{3} G_{2} G_{0}-H_{2} G_{2} G_{1}+H_{1} G_{2}^{2}\right) \gamma_{x}+H_{3} G_{1} G_{0}-H_{2} G_{2} G_{0}+H_{0} G_{2}^{2}=0 . \tag{22} \end{equation*}(22)(H3G12−H3G2G0−H2G2G1+H1G22)γx+H3G1G0−H2G2G0+H0G22=0.
If the coefficient of γ x γ x gamma_(x)\gamma_{x}γx is different from zero, we can express γ x γ x gamma_(x)\gamma_{x}γx as a function of γ γ gamma\gammaγ
(23) γ x = − H 3 G 1 G 0 − H 2 G 2 G 0 + H 0 G 2 2 H 3 G 1 2 − H 3 G 2 G 0 − H 2 G 2 G 1 + H 1 G 2 2 , (23) γ x = − H 3 G 1 G 0 − H 2 G 2 G 0 + H 0 G 2 2 H 3 G 1 2 − H 3 G 2 G 0 − H 2 G 2 G 1 + H 1 G 2 2 , {:(23)gamma_(x)=-(H_(3)G_(1)G_(0)-H_(2)G_(2)G_(0)+H_(0)G_(2)^(2))/(H_(3)G_(1)^(2)-H_(3)G_(2)G_(0)-H_(2)G_(2)G_(1)+H_(1)G_(2)^(2))",":}\begin{equation*} \gamma_{x}=-\frac{H_{3} G_{1} G_{0}-H_{2} G_{2} G_{0}+H_{0} G_{2}^{2}}{H_{3} G_{1}^{2}-H_{3} G_{2} G_{0}-H_{2} G_{2} G_{1}+H_{1} G_{2}^{2}}, \tag{23} \end{equation*}(23)γx=−H3G1G0−H2G2G0+H0G22H3G12−H3G2G0−H2G2G1+H1G22,
and γ y γ y gamma_(y)\gamma_{y}γy from (3) as
(24) γ y = s − r γ x . (24) γ y = s − r γ x . {:(24)gamma_(y)=s-rgamma_(x).:}\begin{equation*} \gamma_{y}=s-r \gamma_{x} . \tag{24} \end{equation*}(24)γy=s−rγx.
We write the compatibility condition ( γ x ) y = ( γ y ) x γ x y = γ y x (gamma_(x))_(y)=(gamma_(y))_(x)\left(\gamma_{x}\right)_{y}=\left(\gamma_{y}\right)_{x}(γx)y=(γy)x, in which we replace γ x γ x gamma_(x)\gamma_{x}γx by (23) and γ y γ y gamma_(y)\gamma_{y}γy by (24); we obtain a second condition on γ γ gamma\gammaγ.
From (23) and (24) we can express, after differentiation, γ x x , γ x y , γ y y γ x x , γ x y , γ y y gamma_(xx),gamma_(xy),gamma_(yy)\gamma_{x x}, \gamma_{x y}, \gamma_{y y}γxx,γxy,γyy in terms of γ γ gamma\gammaγ and derivatives of V V VVV up to the fifth order. We insert these values in the basic equation (6), and then the values of γ x γ x gamma_(x)\gamma_{x}γx and γ y γ y gamma_(y)\gamma_{y}γy from (23) and (24). We obtain a third condition on γ γ gamma\gammaγ. In order to obtain solutions of the problem under consideration, these three necessary conditions must be satisfied.
If the coefficient of γ x γ x gamma_(x)\gamma_{x}γx in (22) is zero and the other term is not zero, we have no solution for our problem. If both coefficients in (22) are null, we are left with equation (20).
As an application to the reasoning developed in this section, we shall study first the case of functions γ γ gamma\gammaγ which are homogeneous of order m m mmm.

4 Functions γ γ gamma\gammaγ homogeneous of order m m mmm

Let us suppose that γ γ gamma\gammaγ satisfies (18), hence we have r = x / y r = x / y r=x//yr=x / yr=x/y and s = m γ / y s = m γ / y s=m gamma//ys=m \gamma / ys=mγ/y. As stated above, the first equation in γ x ( 20 ) γ x ( 20 ) gamma_(x)(20)\gamma_{x}(20)γx(20) is of second degree; its coefficients are in this case polynomials in γ γ gamma\gammaγ. This will happen for the coefficients of the third-degree equation (21) too.
The three conditions on γ γ gamma\gammaγ are in this case polynomials in γ γ gamma\gammaγ. For a common solution to exist, a necessary condition is that the resultants of the two pairs of polynomials vanish. The resultants are equal to their Sylvester determinants (Mishina and Proskuryakov, 1965, p. 164). Thus we obtain two necessary conditions to be satisfied by the potential V V VVV and the function γ γ gamma\gammaγ.
When we start working with a given potential V V VVV and a fixed degree of homogeneity for γ γ gamma\gammaγ, we do not expect the problem to have always a solution. It is advisable to try to factor the first polynomial in γ γ gamma\gammaγ (the resultant of (20) and (21)) and to check directly if the homogeneous functions γ γ gamma\gammaγ are compatible with the potential. Proceeding this way we avoid lengthy calculations.
Example 1 Let us consider V ( x , y ) = − x 4 − y 2 V ( x , y ) = − x 4 − y 2 V(x,y)=-x^(4)-y^(2)V(x, y)=-x^{4}-y^{2}V(x,y)=−x4−y2 and look for functions γ γ gamma\gammaγ homogeneous of first order. The polynomials (20) and (21) are of second, respectively third, degree and their resultant is
(25) R 1 = γ 5 ( y γ − x 2 ) ( y γ + x 2 ) ( y γ + x ) 3 ( y γ + 2 x 3 ) 4 P 10 . (25) R 1 = γ 5 y γ − x 2 y γ + x 2 ( y γ + x ) 3 y γ + 2 x 3 4 P 10 . {:(25)R_(1)=gamma^(5)(y gamma-x^(2))(y gamma+x^(2))(y gamma+x)^(3)(y gamma+2x^(3))^(4)P_(10).:}\begin{equation*} R_{1}=\gamma^{5}\left(y \gamma-x^{2}\right)\left(y \gamma+x^{2}\right)(y \gamma+x)^{3}\left(y \gamma+2 x^{3}\right)^{4} P_{10} . \tag{25} \end{equation*}(25)R1=γ5(yγ−x2)(yγ+x2)(yγ+x)3(yγ+2x3)4P10.
The index of P P PPP denotes in these examples the degree of the respective polynomial in γ γ gamma\gammaγ. The second condition, which follows from the compatibility ( γ x ) y = ( γ y ) x γ x y = γ y x (gamma_(x))_(y)=(gamma_(y))_(x)\left(\gamma_{x}\right)_{y}=\left(\gamma_{y}\right)_{x}(γx)y=(γy)x, reads
(26) R 2 = ( y γ − x 2 ) ( y γ + x 2 ) ( y γ + x ) ( y γ + 2 x 3 ) P 12 . (26) R 2 = y γ − x 2 y γ + x 2 ( y γ + x ) y γ + 2 x 3 P 12 . {:(26)R_(2)=(y gamma-x^(2))(y gamma+x^(2))(y gamma+x)(y gamma+2x^(3))P_(12).:}\begin{equation*} R_{2}=\left(y \gamma-x^{2}\right)\left(y \gamma+x^{2}\right)(y \gamma+x)\left(y \gamma+2 x^{3}\right) P_{12} . \tag{26} \end{equation*}(26)R2=(yγ−x2)(yγ+x2)(yγ+x)(yγ+2x3)P12.
Finally, the condition obtained from the basic equation (6) is
(27) R 3 = ( y γ − x 2 ) ( y γ + x 2 ) ( y γ + 2 x 3 ) P 22 . (27) R 3 = y γ − x 2 y γ + x 2 y γ + 2 x 3 P 22 . {:(27)R_(3)=(y gamma-x^(2))(y gamma+x^(2))(y gamma+2x^(3))P_(22).:}\begin{equation*} R_{3}=\left(y \gamma-x^{2}\right)\left(y \gamma+x^{2}\right)\left(y \gamma+2 x^{3}\right) P_{22} . \tag{27} \end{equation*}(27)R3=(yγ−x2)(yγ+x2)(yγ+2x3)P22.
The three polynomials in γ γ gamma\gammaγ have in common two homogeneous solutions of first order, namely γ 1 = x 2 / y γ 1 = x 2 / y gamma_(1)=x^(2)//y\gamma_{1}=x^{2} / yγ1=x2/y and γ 2 = − x 2 / y γ 2 = − x 2 / y gamma_(2)=-x^(2)//y\gamma_{2}=-x^{2} / yγ2=−x2/y, which correspond to the families f 1 = y e − 1 / x f 1 = y e − 1 / x f_(1)=ye^(-1//x)f_{1}=y e^{-1 / x}f1=ye−1/x, and f 2 = y e 1 / x f 2 = y e 1 / x f_(2)=ye^(1//x)f_{2}=y e^{1 / x}f2=ye1/x and are compatible with the given potential.

5 Special cases and other examples

The case r = − γ , s = 0 ( Γ = 0 ) r = − γ , s = 0 ( Γ = 0 ) r=-gamma,s=0(Gamma=0)r=-\gamma, s=0(\Gamma=0)r=−γ,s=0(Γ=0)
From the equation (8) it follows that
(28) γ = − V x V y (28) γ = − V x V y {:(28)gamma=-(V_(x))/(V_(y)):}\begin{equation*} \gamma=-\frac{V_{x}}{V_{y}} \tag{28} \end{equation*}(28)γ=−VxVy
and only potentials V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) satisfying the differential condition
(29) V x V y ( V x x − V y y ) = ( V x 2 − V y 2 ) V x y (29) V x V y V x x − V y y = V x 2 − V y 2 V x y {:(29)V_(x)V_(y)(V_(xx)-V_(yy))=(V_(x)^(2)-V_(y)^(2))V_(xy):}\begin{equation*} V_{x} V_{y}\left(V_{x x}-V_{y y}\right)=\left(V_{x}^{2}-V_{y}^{2}\right) V_{x y} \tag{29} \end{equation*}(29)VxVy(Vxx−Vyy)=(Vx2−Vy2)Vxy
are generating families having Γ = 0 Γ = 0 Gamma=0\Gamma=0Γ=0 (Bozis and Anisiu, 2001). So then, for our problem, we see immediately if the given potential satisfies or not the condition (29) and, if the potential is admissible, we readily check whether or not the pertinent γ γ gamma\gammaγ, given by (28), satisfies the pre-assigned condition (3).
As another viewpoint, let us discuss briefly the following two alternatives, possibly leading to an affirmative answer of our problem:
(i) Let us fix the condition (3) but allow the potential V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) to be free. In this case we must inquire whether there exist common solutions for the PDE (29) and the PDE
(30) r ∗ V y V x x + ( V y − r ∗ V x ) V x y − V x V y y + s ∗ V y 2 = 0 (30) r ∗ V y V x x + V y − r ∗ V x V x y − V x V y y + s ∗ V y 2 = 0 {:(30)r^(**)V_(y)V_(xx)+(V_(y)-r^(**)V_(x))V_(xy)-V_(x)V_(yy)+s^(**)V_(y)^(2)=0:}\begin{equation*} r^{*} V_{y} V_{x x}+\left(V_{y}-r^{*} V_{x}\right) V_{x y}-V_{x} V_{y y}+s^{*} V_{y}^{2}=0 \tag{30} \end{equation*}(30)r∗VyVxx+(Vy−r∗Vx)Vxy−VxVyy+s∗Vy2=0
where
(31) r ∗ ( x , y ) = r ( x , y , γ = − V x V y ) and s ∗ ( x , y ) = s ( x , y , γ = − V x V y ) . (31) r ∗ ( x , y ) = r x , y , γ = − V x V y  and  s ∗ ( x , y ) = s x , y , γ = − V x V y . {:(31)r^(**)(x","y)=r(x,y,gamma=-(V_(x))/(V_(y)))" and "s^(**)(x","y)=s(x,y,gamma=-(V_(x))/(V_(y))).:}\begin{equation*} r^{*}(x, y)=r\left(x, y, \gamma=-\frac{V_{x}}{V_{y}}\right) \text { and } s^{*}(x, y)=s\left(x, y, \gamma=-\frac{V_{x}}{V_{y}}\right) . \tag{31} \end{equation*}(31)r∗(x,y)=r(x,y,γ=−VxVy) and s∗(x,y)=s(x,y,γ=−VxVy).
The compatibility of these two equations may be checked in a straightforward way.
(ii) Let us consider a potential V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) satisfying the condition (29), i.e. a potential which produces the family (28) of straight lines and let the functions r r rrr and s s sss in (3) be at our disposal. In this case we are led to infinitely many choices for r r rrr and s s sss for which the condition (30) is satisfied. Indeed, we can take
(32) r ( x , y , γ ) = − γ + ( γ + V x V y ) A ( x , y , γ ) and s ( x , y , γ ) = ( γ + V x V y ) B ( x , y , γ ) , (32) r ( x , y , γ ) = − γ + γ + V x V y A ( x , y , γ )  and  s ( x , y , γ ) = γ + V x V y B ( x , y , γ ) , {:(32)r(x","y","gamma)=-gamma+(gamma+(V_(x))/(V_(y)))A(x","y","gamma)" and "s(x","y","gamma)=(gamma+(V_(x))/(V_(y)))B(x","y","gamma)",":}\begin{equation*} r(x, y, \gamma)=-\gamma+\left(\gamma+\frac{V_{x}}{V_{y}}\right) A(x, y, \gamma) \text { and } s(x, y, \gamma)=\left(\gamma+\frac{V_{x}}{V_{y}}\right) B(x, y, \gamma), \tag{32} \end{equation*}(32)r(x,y,γ)=−γ+(γ+VxVy)A(x,y,γ) and s(x,y,γ)=(γ+VxVy)B(x,y,γ),
where A A AAA and B B BBB are arbitrary functions with the unique provision that the pertinent functions A ∗ ( x , y ) A ∗ ( x , y ) A^(**)(x,y)A^{*}(x, y)A∗(x,y) and B ∗ ( x , y ) B ∗ ( x , y ) B^(**)(x,y)B^{*}(x, y)B∗(x,y) (defined as indicated in (31)) do not become infinite. By choosing the functions r r rrr and s s sss as in (32), we have r ∗ ( x , y ) = V x / V y r ∗ ( x , y ) = V x / V y r^(**)(x,y)=V_(x)//V_(y)r^{*}(x, y)=V_{x} / V_{y}r∗(x,y)=Vx/Vy and s ∗ ( x , y ) = 0 s ∗ ( x , y ) = 0 s^(**)(x,y)=0s^{*}(x, y)=0s∗(x,y)=0, hence condition (30) is identical to (29).
The case r = − γ , s ≠ 0 r = − γ , s ≠ 0 r=-gamma,s!=0r=-\gamma, s \neq 0r=−γ,s≠0
In this case Π = 0 Π = 0 Pi=0\Pi=0Π=0 in the first of the equations (14) and the formulae (15) are meaningless. Let us suppose that r ( x , y , γ ) = γ r ( x , y , γ ) = γ r(x,y,gamma)=gammar(x, y, \gamma)=\gammar(x,y,γ)=γ identically. The condition (3) becomes
(33) γ γ x − γ y = − s (33) γ γ x − γ y = − s {:(33)gammagamma_(x)-gamma_(y)=-s:}\begin{equation*} \gamma \gamma_{x}-\gamma_{y}=-s \tag{33} \end{equation*}(33)γγx−γy=−s
where s s sss may depend on all three variables x , y x , y x,yx, yx,y and γ γ gamma\gammaγ. We suppose here that s s sss is not identically null, to avoid that (33) coincides with Γ = 0 Γ = 0 Gamma=0\Gamma=0Γ=0 (treated above).
From the derivatives of (33) with respect to x x xxx and y y yyy, we find
(34) γ 2 γ x x − 2 γ γ x y + γ y y = s 010 − γ s 100 + s γ x + s s 001 (34) γ 2 γ x x − 2 γ γ x y + γ y y = s 010 − γ s 100 + s γ x + s s 001 {:(34)gamma^(2)gamma_(xx)-2gammagamma_(xy)+gamma_(yy)=s_(010)-gammas_(100)+sgamma_(x)+ss_(001):}\begin{equation*} \gamma^{2} \gamma_{x x}-2 \gamma \gamma_{x y}+\gamma_{y y}=s_{010}-\gamma s_{100}+s \gamma_{x}+s s_{001} \tag{34} \end{equation*}(34)γ2γxx−2γγxy+γyy=s010−γs100+sγx+ss001
So, in view of (7) and (34), equation (6) may be written as
s ( V x + γ V y ) s 001 = ( V x + γ V y ) ( γ s 100 − s 010 ) (35) + s [ γ ( V y y − V x x ) + ( 1 − γ 2 ) V x y + 3 s V y ] s V x + γ V y s 001 = V x + γ V y γ s 100 − s 010 (35) + s γ V y y − V x x + 1 − γ 2 V x y + 3 s V y {:[s(V_(x)+gammaV_(y))s_(001)=(V_(x)+gammaV_(y))(gammas_(100)-s_(010))],[(35)+s[gamma(V_(yy)-V_(xx))+(1-gamma^(2))V_(xy)+3sV_(y)]]:}\begin{align*} & s\left(V_{x}+\gamma V_{y}\right) s_{001}=\left(V_{x}+\gamma V_{y}\right)\left(\gamma s_{100}-s_{010}\right) \\ & +s\left[\gamma\left(V_{y y}-V_{x x}\right)+\left(1-\gamma^{2}\right) V_{x y}+3 s V_{y}\right] \tag{35} \end{align*}s(Vx+γVy)s001=(Vx+γVy)(γs100−s010)(35)+s[γ(Vyy−Vxx)+(1−γ2)Vxy+3sVy]
The above equation (35) replaces the PDE (6) and its meaning is the following: In order that the given potential V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) supports a family γ γ gamma\gammaγ, the "given" function s ( x , y , γ ) s ( x , y , γ ) s(x,y,gamma)s(x, y, \gamma)s(x,y,γ) in (33) must satisfy the PDE (35). In other words, for our problem to admit of an affirmative answer, the required function γ ( x , y ) γ ( x , y ) gamma(x,y)\gamma(x, y)γ(x,y) and the "given" function s ( x , y , γ ) s ( x , y , γ ) s(x,y,gamma)s(x, y, \gamma)s(x,y,γ) must satisfy both equations (33) and (35). To check if these equations have common solutions γ γ gamma\gammaγ we proceed as follows: From (35) we can express (by differentiation) γ x γ x gamma_(x)\gamma_{x}γx and γ y γ y gamma_(y)\gamma_{y}γy in terms of γ γ gamma\gammaγ and insert them into (33), which then will become an equation of the form
(36) F ( x , y , γ ) = 0 (36) F ( x , y , γ ) = 0 {:(36)F(x","y","gamma)=0:}\begin{equation*} F(x, y, \gamma)=0 \tag{36} \end{equation*}(36)F(x,y,γ)=0
Finally we check whether equations (36) and (35) have or do not have common solutions γ ( x , y ) γ ( x , y ) gamma(x,y)\gamma(x, y)γ(x,y).
Example 2 Let us find solutions of (33) with s ( x , y , γ ) = − 6 x / y 2 s ( x , y , γ ) = − 6 x / y 2 s(x,y,gamma)=-6x//y^(2)s(x, y, \gamma)=-6 x / y^{2}s(x,y,γ)=−6x/y2 which represent families compatible with the potential
(37) V ( x , y ) = 4 x 2 + y 2 + 8 x 4 − 2 x 2 y 2 − y 4 + x 3 (37) V ( x , y ) = 4 x 2 + y 2 + 8 x 4 − 2 x 2 y 2 − y 4 + x 3 {:(37)V(x","y)=4x^(2)+y^(2)+8x^(4)-2x^(2)y^(2)-y^(4)+x^(3):}\begin{equation*} V(x, y)=4 x^{2}+y^{2}+8 x^{4}-2 x^{2} y^{2}-y^{4}+x^{3} \tag{37} \end{equation*}(37)V(x,y)=4x2+y2+8x4−2x2y2−y4+x3
Condition (35) is in this case a second-degree polynomial equation in γ γ gamma\gammaγ, which has the solutions
(38) γ = 2 x y and γ = 2 x ( 17 x 2 + 7 y 2 ) y ( 2 x 2 − 2 y 2 + 1 ) (38) γ = 2 x y  and  γ = 2 x 17 x 2 + 7 y 2 y 2 x 2 − 2 y 2 + 1 {:(38)gamma=(2x)/(y)" and "gamma=(2x(17x^(2)+7y^(2)))/(y(2x^(2)-2y^(2)+1)):}\begin{equation*} \gamma=\frac{2 x}{y} \text { and } \gamma=\frac{2 x\left(17 x^{2}+7 y^{2}\right)}{y\left(2 x^{2}-2 y^{2}+1\right)} \tag{38} \end{equation*}(38)γ=2xy and γ=2x(17x2+7y2)y(2x2−2y2+1)
The first one is a solution of our problem.
It may happen that Π = 0 Π = 0 Pi=0\Pi=0Π=0 for some particular functions γ γ gamma\gammaγ. In such a case, we have to check if this particular γ γ gamma\gammaγ satisfies equation (3). In the affirmative case, we put the values of V V VVV and γ γ gamma\gammaγ in (6)-(7) and, if we obtain an identity, we have a solution of our problem.
Example 3 Let us look for families γ γ gamma\gammaγ which are compatible with the HénonHeiles potential
(39) V ( x , y ) = 1 2 ( x 2 + 16 y 2 ) + x 2 y + 16 3 y 3 (39) V ( x , y ) = 1 2 x 2 + 16 y 2 + x 2 y + 16 3 y 3 {:(39)V(x","y)=(1)/(2)(x^(2)+16y^(2))+x^(2)y+(16)/(3)y^(3):}\begin{equation*} V(x, y)=\frac{1}{2}\left(x^{2}+16 y^{2}\right)+x^{2} y+\frac{16}{3} y^{3} \tag{39} \end{equation*}(39)V(x,y)=12(x2+16y2)+x2y+163y3
and which satisfy the equation (3) with r ( x , y , γ ) = x / y + 3 γ r ( x , y , γ ) = x / y + 3 γ r(x,y,gamma)=x//y+3gammar(x, y, \gamma)=x / y+3 \gammar(x,y,γ)=x/y+3γ and s ( x , y , γ ) = − 3 γ / ( 4 y ) s ( x , y , γ ) = − 3 γ / ( 4 y ) s(x,y,gamma)=-3gamma//(4y)s(x, y, \gamma)= -3 \gamma /(4 y)s(x,y,γ)=−3γ/(4y). The equality γ + r = 0 γ + r = 0 gamma+r=0\gamma+r=0γ+r=0 holds if and only if γ = − x / ( 4 y ) γ = − x / ( 4 y ) gamma=-x//(4y)\gamma=-x /(4 y)γ=−x/(4y). This function verifies the equation (3) for the specified values of r r rrr and s s sss, and, together with the potential (39), equation (6)-(7), hence it is a solution of our problem. The same family has been found by Bozis et al (1997) as a homogeneous family generated by the inhomogeneous potential (39).
Remark 3 If V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) and s ( x , y , γ ) s ( x , y , γ ) s(x,y,gamma)s(x, y, \gamma)s(x,y,γ) are left free (to be adequately determined) the possibly existing common solutions γ ( x , y ) γ ( x , y ) gamma(x,y)\gamma(x, y)γ(x,y) of (35) and (36) will be expressed in terms of partial derivatives of the second order in s ( x , y , γ ) s ( x , y , γ ) s(x,y,gamma)s(x, y, \gamma)s(x,y,γ) and of the third order in V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y).

6 General comments

In the framework of the inverse problem of Dynamics, a monoparametric family of orbits is uniquely represented by its slope function γ γ gamma\gammaγ defined in (5). For a given potential V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y), the finding of some or all families generated by V V VVV amounts to the solution of the nonlinear in γ γ gamma\gammaγ second order PDE (6). This is a task more or less impossible.
In this paper, in order to ease and make possible the solution of the problem (even by finding a subset of solutions), we add the restriction on γ γ gamma\gammaγ expressed by the differential condition (3). In so doing, we come to have to deal with two PDEs, one of the first and one of the second order in the unknown function γ ( x , y ) γ ( x , y ) gamma(x,y)\gamma(x, y)γ(x,y). Therefore the very existence of a solution is not guaranteed. Yet, we showed that, if such a solution does exist, its finding may be accomplished by algebraic manipulation.
We deal basically with the direct problem, i.e. the potential is given and the orbits are to be found. The functions r ( x , y , γ ) r ( x , y , γ ) r(x,y,gamma)r(x, y, \gamma)r(x,y,γ) and s ( x , y , γ ) s ( x , y , γ ) s(x,y,gamma)s(x, y, \gamma)s(x,y,γ) are also generally given. One then might suggest to face the problem by solving for γ γ gamma\gammaγ the first order PDE (3) and then proceed to find, among its solutions, those which are
compatible with the given potential. However, this last task (possible in some of the examples presented in this paper) does not seem to be easier or performable by a straightforward way. Besides that, the finding of the general solution of (3) is not always possible.
The above strictly direct problem does not generally have a solution. For this reason, we may profitably deal with the two equations (3) and (6) in various ways. We can e.g. allow tentatively the potential V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) to be free and find compatibility conditions on it so that a solution γ ( x , y ) γ ( x , y ) gamma(x,y)\gamma(x, y)γ(x,y) can be found. Or, keeping V ( x , y ) V ( x , y ) V(x,y)V(x, y)V(x,y) fixed, we may allow the functions r r rrr and s s sss in (3) to be free and then adjust them properly so that we obtain a solution.

References

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2007

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