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PROBLEM OF METRIZABILITY FOR THE DYNAMICAL SYSTEMS

arXiv:solv-int/9404003v1 18 Apr 1993PROBLEM OF METRIZABILITY FOR THE DYNAMICAL SYSTEMSACCEPTING THE NORMAL SHIFT.Sharipov R.A.June, 1993.Abstract. The problem of metrizability for the dynamical systems accepting the normal shift is formulated and solved.

Theexplicit formula for the force field of metrizable Newtonian dynamical system ¨r = F(r, ˙r) is found.1. Introduction.The class of Newtonian dynamical systems accepting the normal shift was first defined in [1] (see also [2] and [3]).It’s the class of dynamical systems in Rn given by the differential equations of the form(1.1)¨r = F(r, ˙r)and possessing some additional geometrical property: the property of conserving the orthogonality of trajectoriesand the hypersurfaces shifted along these trajectories.

The idea for considering such systems was found as a resultof generalizing the classical construction of normal shift which is known also as a Bonnet transformation. Let S bethe hypersurface in Rn.

From each point M on S we draw the segment MM ′ with the fixed length l along thenormal vector to S. The points M ′ then form another hypersurface S′, the segment MM ′ being perpendicular to S′.The transformation f : S −→S′ just described is the classical Bonnet transformation. It has the generalization fornon-Euclidean situation: for the Riemannian metric gij one should replace the segment MM ′ of a straight line bythe segment of geodesic line with the length l. The transformation f : S −→S′ in this case is the metrical Bonnettransformation or the normal shift with respect to the metric gij.

Trajectories of this shift are defined by the equationof geodesic lines(1.2)¨rk = −Γkij ˙ri ˙rjThis equation can be treated as a particular case of the Newtonian dynamical system (same indices on different levelsin (1.2) and everywhere below imply summation).In [1] (see also [2] and [3]) the classical Bonnet transformation was generalized for the case of dynamical systems(1.1) in Rn. Other generalization is a metrical Bonnet transformation.

So quite natural question is: how do these twogeneralizations relate each other? In order to treat (1.2) as a dynamical system accepting the normal shift in standardEuclidean metric in Rn we should have the coincidence of the concept of orthogonality with respect to both Euclideanand non-Euclidean metrics δij and gij in Rn.

That is gij = e−2fδij should be conformally Euclidean metric. Problemof metrizability then can be stated as follows.Problem of metrizability.

Under which circumstances the dynamical system (1.1) accepting the normal shift inthe sense of [1] and [4] is equivalent to the metrical normal shift for some conformally Euclidean metric in Rn.Studying this problem and solving it is the goal of present paper. As we shall see below its solution is explicit andconstructive.Typeset by AMS-TEX1

2SHARIPOV R.A.2. Geodesic lines of conformally Euclidean metric in RnLet δij be the metric tensor for the standard Euclidean metric in Rn.

Let’s consider conformally Euclidean metric(2.1)gij = e−2fδijwhere f = f(r) = f(r1, . .

. , rn) is some scalar function in Rn.

Metrical connection for the metric (2.1) is given by(2.2)Γkij = δijδks∂sf −(δki ∂jf + δkj ∂if)Here by ∂s we denote the partial derivative with respect to rs. The equation of geodesic lines for (2.2) has the followingform(2.3)¨rk = −δks∂sf ˙riδij ˙rj + 2∂if ˙ri ˙rkIn a vectorial form like (1.1) the force field for the dynamical system corresponding to (2.3) is written as follows(2.4)F(r, v) = −∇f|v|2 + 2 ⟨∇f, v⟩vHere ∇f is a gradient of the function f considered as a vector in Rn and ⟨∇f, v⟩is a standard Euclidean scalar productof ∇f with the vector of velocity v. The modulus of the velocity vector in (2.4) is also calculated in a standard metricδij.Because of its geometrical origin the dynamical system (1.1) with force field (2.4) accepts the normal shift.

Howeverit is curious to check this fact directly. Following the receipt of [4] and [5] we choose the coordinates u1, .

. .

, un−1 ona unit sphere |v| = 1 in the velocity space. Let’s denote v = |v| and let N be the unit vector directed along the vectorv.

Then(2.5)N = N(u1, . .

. , un−1)Mi = ∂N∂uiFor the derivatives of the vectors Mi defined by (2.5) one has a Weingarten derivation formula(2.6)∂Mi∂uj = ϑkijMk −GijNwhere Gij = ⟨Mi, Mj⟩is an induced metric on the unit sphere.

Let’s consider the expansion of the force field (2.4) inthe basis formed by N and the vectors Mi(2.7)F = AN + BiMiLet’s find the spatial gradients of the coefficients of the expansion (2.7) and then let’s expand them in the same basis.As a result we get(2.8)∂A∂rk = aNk + αpMpk∂Bi∂rk = biNk + βipMpkAccording to the results of [4] and [5] the weak normality condition for the dynamical system with the force field (2.7)is given by the following equationsBi = −Gik ∂A∂uk(2.9)αi + BqBkv2ϑiqk −BiAv2+ bi++ Av∂Bi∂v + ∂Bi∂ukBkv2 −biv∂A∂v = 0(2.10)

PROBLEM OF METRIZABILITY . .

.3In order to check the equations (2.9) and (2.10) in the case of force field (2.4) we calculate the coefficients of expansion(2.7) explicitly. Because of the orthogonality ⟨N, Mi⟩= 0 we have(2.11)A = ⟨F, N⟩= ⟨∇f, N⟩v2Bi = Gik ⟨F, Mk⟩= −Gik ⟨∇f, Mk⟩v2From (2.11) we can see that the relationship (2.9) becomes the identity due to (2.5).

For the coefficients αi and bi in(2.8) we findαi = Gik ∂A∂rq δqsMks = Gik∂2f∂rq∂rm δmpNpδqsMksv2bi = ∂Bi∂rq δqsNs = −Gik∂2f∂rq∂rm δmpMkpδqsNsv2From these two equalities we see that αi and bi differ only by sign. When substituting them into (2.10) they vanish.Let’s calculate the sixth term in (2.10) using the formula (2.6)∂Bi∂ukBkv2 = BiAv2+∂Giq∂uk + GisϑqskGqpBpBkv2Taking into account the concordance of the metric Gij and the metrical connection ϑkij we can bring this equationinto the following form∂Bi∂ukBkv2 = BiAv2−GsqϑiskGqpBpBkv2= BiAv2−BsBkv2ϑiskWhen substituting this form of sixth term into (2.10) it cancels the second and the third terms in (2.10).

Because ofquadratic dependence of A and Bi in (2.11) upon v fifth and seventh terms in (2.10) cancel each other. Resumingall above we conclude that the equations (2.9) and (2.10) hold identically for the components of the expansion (2.7).Weak normality condition for (2.4) is fulfilled.Next step consists in substituting the geodesic flows of the form (2.4) by some dynamical systems similar to them.Let’s start with the Euclidean metric gij = δij.

Force field (2.4) then is zero F(r, v) = 0, trajectories are straightlines. For the dynamical system with the force field F(r, v) = v they are also straight lines.

From geometrical pointof view these two dynamical systems realize the same normal shift. This example shows that for to solve the problemof metrizability one should find all dynamical systems accepting the normal shift in Rn for which the trajectories arethe geodesic lines of conformally Euclidean metrics.

The problem of geometrical coincidence of trajectories for twodifferent dynamical systems was first stated in [6]. There the following pairs of dynamical systems were considered(2.12)¨rk + Γkij ˙ri ˙rj = F k(r, ˙r)¨rk + ˜Γkij ˙ri ˙rj = ˜F k(r, ˙r)In the case of F k = ˜F k = 0 the condition of coincidence of trajectories for the dynamical systems (2.12) is known asa condition of geodesical equivalence for the affine connections Γkij and ˜Γkij.

The detailed discussion of the questionsconnected with geodesical equivalence and geodesical maps can be found in the monograph [7] (see also [8] and [9]).For nonzero F k and ˜F k if the trajectories of the dynamical systems (2.12) coincide then one say that one of thesesystems is a modeling system for another. This case was considered in [10-13].

The terms inheriting the trajectoriesand trajectory equivalence below seem to be more preferable than the term modeling from purely linguistical point ofview.3. Inheriting the trajectories and trajectory equivalence of dynamical systems.Let’s consider the pair of Newtonian dynamical systems (1.1) of the second order∂ttr = F1(r, ∂tr)(3.1)∂ττr = F2(r, ∂τr)(3.2)

4SHARIPOV R.A.Trajectories of these systems are defined by the initial position and initial velocityr|t=0 = r0∂tr|t=0 = v0(3.3)r|τ=0 = r0∂τr|τ=0 = w0(3.4)Let r = R1(t, r0, v0) and r = R2(τ, r0, w0) are the solutions of the equations (3.1) and (3.2) with the initial conditions(3.3) and (3.4).Definition 1. Say that the dynamical system (3.2) inherits the trajectories of the system (3.1) if for any pair ofvectors r0 and w0 ̸= 0 one can find the vector v0 and the function T (τ) such that T (0) = 0 and the following equality(3.5)R1(T (τ), r0, v0) = R2(τ, r0, w0)holds identically by τ in some neighborhood of zero τ = 0.Definition 2.

Two dynamical systems are called trajectory equivalent if each of them inherits the trajectories of theother.Differentiating (3.5) by τ for τ = 0 we get the relation between the vectors w0 and v0 in the following form(3.6)v0 ∂τT (0) = w0From (3.6) we see that v0 ̸= 0 and ∂τT (0) ̸= 0. For nonzero τ we have(3.7)∂tR1(T (τ), r0, v0) ∂τT = ∂τR2(τ, r0, w0)Let’s differentiate the relationship (3.7) by τ.

Then for τ = 0 we get(3.8)F1(r0, v0) ∂τT (0)2 + v0 ∂ττT (0) = F2(r0, w0)The relationship (3.8) bind the force field of the dynamical system (3.2) with the force field of the system (3.1)trajectories of which are inherited according to the definition 1. Because of v0 ̸= w0 this relationship is nonlocal.However, if F1(r, v) is homogeneous function by v this relationship becomes local.

It’s the very case we shall considerbelow.Let γ be the degree of homogeneity for the function F1(r, v) with respect to its vectorial argument v.Therelationship (3.8) then has the following form(3.9)F1(r, w) ∂τT (0)2−γ + w ∂ττT (0) ∂τT (0)−1 = F2(r, w)Since the initial data r0 and w0 in (3.4) are arbitrary we omit the index 0 everywhere in (3.9). Because of (3.9) thevectors w, F1(r, w) and F2(r, w) are linearly dependent.

Two cases are possible:(1) special case when the vectors F1(r, w) and w are linearly dependent,(2) generic case when vectors F1(r, w) and w are linearly independent.Let’s start with the first case. Here for the force field of the dynamical system (3.1) we have(3.10)F1(r, w) = wH1(r, w)|w|where H1(r, w) is a scalar function homogeneous by w with order of homogeneity γ.

Because of (3.9) the force fieldF2(r, w) has the similar form(3.11)F2(r, w) = wH2(r, w)|w|

PROBLEM OF METRIZABILITY . .

.5However the function H2(r, w) in (3.11) shouldn’t be homogeneous by w. Trajectories of the dynamical systems (3.1)and (3.2) with the force fields (3.10) and (3.11) are straight lines, therefore any two such systems are inheriting thetrajectories of each other even when the function H1(r, w) is not homogeneous by w.The second case is more complicated. Here vector F2(r, w) can be decomposed by the vectors F1(r, w) and w(3.12)F2(r, w) = C(r, w)F1(r, w) + wH(r, w)|w|Differentiating (3.7) by τ for τ ̸= 0 we get the equation for the function T (τ)(3.13)F1(r, w) ∂τT 2−γ + w ∂ττT ∂τT −1 = F2(r, w)where r and w are the functions of τ defined by (3.2) and (3.4)r = r(τ) = R2(τ, r0, w0)w = w(τ) = ∂τrSince the expansion of F2(r, w) by two linearly independent vectors F1(r, w) and w is unique from (3.12) and (3.13)we get(3.14)(∂τT )2−γ = C(r, w)∂ττT = H(r, w)|w|∂τTBecause of (3.14) we should consider two different subcases depending on the value of γ: γ ̸= 2 and γ = 2.

Let’sconsider the first subcase. Here the function T (τ) is defined by the equation of the first order ∂τT = C(r, w)γ−2.Therefore the second equation (3.14) should be the consequence of the first one.

Differentiating the first equation(3.14) with respect to τ we get∂τ ln(∂τT ) = (γ −2)wi ∂ln(C)∂ri+ F i2∂ln(C)∂wiComparing this equation with the second equation (3.14) and substituting F i2 in it by (3.13) we get the relationshipH|w| =wi ∂ln(C)∂ri+ F i1∂C∂wi 1γ −2 −wi ∂ln(C)∂wi−1which express H(r, w) through C(r, w). Force field F2(r, w) of the dynamical system (3.2) inheriting the trajectoriesof the system (3.1) in this case is defined by one arbitrary scalar function C(r, w).Now let’s consider the second subcase γ = 2 Here the first equation (3.14) is trivial and C(r, w) = 1.

The functionT (τ) is defined only by the second equation (3.14). The force field of the system (3.2) defined by (3.12) contains onearbitrary function H(r, w).

This case γ = 2 is the most interesting since the force fields of the form (2.4) are coveredby this case. Dynamical systems inheriting their trajectories are defined by the following force fields(3.15)F(r, v) = −∇f|v|2 + 2 ⟨∇f, v⟩v + v|v|H(r, v)where f = f(r).

However not all the dynamical systems with the force field (3.15) are accepting the normal shift.4. Problem of metrizability.Let the dynamical system (1.1) with the force field (3.15) be accepting the normal shift.

The force field (3.15) differsfrom (2.4) by the term NH(r, v). Therefore the components of the expansion (2.7) for (3.15) are slightly differentfrom that of (2.4)(4.1)˜A = A + H˜Bi = Bi

6SHARIPOV R.A.But both they are satisfying the equations (2.9) and (2.10). Substituting (4.1) into (2.9) we get the vanishing of thederivatives∂H∂uk = 0H = H(r, v) = H(r, |v|)Therefore the function H in (3.15) doesn’t depend on the direction of the vector of velocity.

It depends only on themodulus of velocity and on the position of the mass point in the space. For the components of the expansion (2.8)which are used in (2.10) we have(4.2)˜αi = αi + hi˜bi = biThe parameters hi are defined by the function H(r, v) according to the formula(4.3)hi = Gik ⟨∇H, Mk⟩= Gik ∂H∂rq δqsMksLet’s substitute (4.1) and (4.2) into the equation (2.10).

As a result we have(4.4)hi −BiHv2+ Hv∂Bi∂v −Biv∂H∂v = 0The relationships (4.4) form the system of n −1 equations for the function H(r, v). Taking into account (2.11) and(4.3) we can transform them to the following ones(4.5)∂H∂rk +v ∂f∂rk ∂H∂v = ∂f∂rk HLet’s consider the vector fields which are defined by the differential equations (4.5)(4.6)Xk =∂∂rk +v ∂f∂rk ∂∂vIt’s easy to check that the vector fields (4.6) are commuting, therefore the equations (4.5) are compatible.

It’s wonderfulthat the common solution for the system of equations (4.5) can be written explicitly(4.7)H(r, v) = H(ve−f)efUsing (4.7) the final result can be formulated as a following theorem giving the solution for the problem of metrizability.Theorem 1. The dynamical system (1.1) in Rn is accepting the normal shift if its force field has the following formF(r, v) = −∇f|v|2 + 2 ⟨∇f, v⟩v + v|v|H(|v|e−f)efwhere f = f(r) = f(r1, .

. .

, rn) and H = H(v) are two arbitrary functions. When it is metrizable the dynamicalsystem (1.1) realizes the metrical normal shift for some conformally Euclidean metric in Rn.The comparison of the force field from the theorem 1 with the examples of [1] shows that even in R2 one can findthe non-metrizable dynamical systems accepting the normal shift.

So the concept of the dynamical systems acceptingthe normal shift is more wide, it cannot be reduced to the metrical normal shift.

PROBLEM OF METRIZABILITY . .

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