On the construction of dynamic systems from given integrals

ON THE CONSTRUCTION OF DYNAMIC SYSTEMS

FROM GIVEN INTEGRALS

RAFAEL RAMIREZ Departament d'Enginyeria lnforrnatica, Universitat Rovira i Virgili, Tarragona, Spain

and

NATALIA SADOVSKAIA Departament de Matemgltica Aplicada II, Universitat Politkcnica de Catalunya, Barcelona, Spain

(Received: 29 November 1993; in final form: 2 January 1996; accepted 11 January 1996)

Abstract. In this communication we propose a new approach for studying a particular type of inverse problems in mechanics related to the construction of a force field from given integrals.

An extension of the Danielli problem is obtained. The given results are applied to the Suslov problem, and illustrated in specific examples.

Key words: Inverse problem, mechanical system, Danielli problem, Suslov problem, Minkovski's space, force field

1. Introduction

Book One of Newton's Philosophiae Naturalis Principia Mathematica is totally dominated by the idea of determining the forces capable of generating planetary orbits of the solar system.

The problem of finding the forces which generate a given motion has played a dominant role in the history of dynamics from Newton's time to the present. In fact, this problem received attention by Bertran [1], Suslov [7], Joukovski [5], Darboux [3], Danielli [2], Whittaker [9] and recently by Szebehely [8], Galiullin [4] and their followers (the references can be found in [6]).

Firstly, let us introduce the notations and concepts necessary in the article. Let

X - smooth manifold, N = dim X, (Xl, z2 , . . . , XN) - local coordinates on X, x(X) - Lie algebra of vector fields on X, A(X) - algebra of 1-forms on X, C~(X) - the ring of the smooth functions on X.

By G and V we respectively denote the symmetrical, nondegenerate with an arbitrary signature metric tensor and the application

v : x ( x ) × x ( x ) - . x ( x )

(U, V) ~ VvV,

Celestial Mechanics and Dynamical Astronomy 63" 149-170, 1996. (~) 1996 Kluwer Academic Publishers. Printed in the Netherlands.

150 RAFAEL RAMIREZ AND NATALIA SADOVSKAIA

which satisfies the following conditions:

V is JR linear with respect to V and C °o linear with respect to U, VuG(U, W) = 0, V(U, V, W, ) ~ x (X)

DEFINITION. The vector field V E x (X) is called the integral element of the 1-form w E A(X) , if

N ~(v) _= ~ ~kv k = 0.

k=l

We will denote by ](~(Wl, W2,. . . , 60M) the set of integral elements of the 1-forms wj E A x ( X ) , j = 1 , 2 , . . . , M .

Now we introduce the tensors/3j, j = 1,2 such that

& x(X) × x(X) --, c ~ ( x )

(u, v) ~ &(u, v),

/3~(u, v) = I(Vuoo(V)- Vvw(U)) -= ½dw(U, V)

&(u, v) = ½(vu~(v) + vv~(u)) =/3*(u, v),

and the vector field Tz, z C C:

Tz = det

B{ ~-z B1 ... BIN E1

B 1 u ~ + ~ . . . B~N e ~

• • , , , " •

O1 02 ... ON 0

= det T

or, what is the same,

~N=I zN-kXk Tz = d e t ( z I + / 3 ) - 1 ~ ( ~ ( z f q,- ,~)0), (1.1)

where/3 _= G- l /3 *, X1, X 2 , . . . , X N are the vector fields which can be obtained when we expand det 7" with respect to the components of 0 - (01 ,02 , . . . , ON).

2. The State of the Problem

Let us give a stationary mechanical system with an X configuration space and T(X) phase space, with N degrees of freedom.

CONSTRUCTION OF DYNAMIC SYSTEMS FROM GIVEN INTEGRALS 151

By x = (xl,x2,...,xN),v= (x'l,x'2,.. . ,x)V) and by

T: r ( x ) ~

(x, v) ~ T ( x , v) = ½a(v , v)

co = Z a ( ~ , d x ) E A(X)

we will denote the generalized coordinales, the vector velocity, the kinetic energy and 1-form, respectively.

We will consider the notations

p = O . k T d x k = G(v, dx) p. = G(v* ,dx) , v* E ~(aJ)

H = ½G- l (p ,p ) = ~ G k J p k p j

H, = Hip=p,

We will denote the variational derivative (d/dr) 0~'k T - 0xk T by gk (T). After some calculations it is easy to prove the following

PROPOSITION 2.1. Let

F = Ek(T0)

To = a ( v * , v) - 1 . ~ a ( v ,~*),

so for the vector field F* = Fly=v, E x ( X ) the following relation holds

~t =_ G(F*, dx) = dH* + ~,, dp., (2.!)

where Lv* is the contraction. []

DEFINITION. U E C~(X) function such that f~ = dU.

From (2.1) we evidently get

CONSEQUENCE 2.1. The F* field is gradient iff

cv* dp, = - d h , h C C ~ ( X ) .

In fact in this case we obtain

{ ~2 = dU,

U = H * - h

We will call the vector field a gradient force field if there is a

(2.2)

(2.3)

1 5 2 R A F A E L R A M I R E Z A N D NATALIA S A D O V S K A I A

PROPOSITION 2.2. The differential equations

£k(T) = Fk, k = 1 , 2 , . . . , N (2.4)

are Lagrangian. The proof follows from the fact that (2.4) defined the extremum of the functional

S = 1 G ( v - v * , v - v*)dt.

In the equations (2.4) we have considerable arbitrariness related to the vector field v*. By using this indetermination we set the next problem.

PROBLEM 1 (extended Danielli's problem). Let df~ E A(X) , a = 1, 2 , . . . , M be the independent 1-forms.

v* E /C(dfl, d f 2 , . . . , dfM) must be constructed in such way that the given mechanical system under the action of F* force field has partial integrals

fa(xl,x2,...,X N) ---- C a , O/ = 1 , 2 , . . . , M < N.

PROBLEM 2 (the Suslov problem). To study Problem 1 in the case when

M = N - 1

~2 = dU.

Solving the stated problems depends on the integral element of the 1-forms ~ok, k = 1 , 2 , . . . , M, i.e. depends on the set E(~zl,w2,. . . ,WM)- The purpose of the next chapter is to construct this set.

3. On the Set/C(wl, w2,.. . , WM)

N • V* PROBLEM 3. Let w E A(X) be such that w = ~i=l~i dxi E x ( X ) must be determined from the following condition: w(v*) N . i = ~i=l~,V = 0. Evidently, to find v* we can limit ourselves to determining the fields ( r 0 ) , . . . , T(N_I) ) E x ( X ) :

1, for u = # ,

G(r(~), ~-(~)) = 0, for u ¢ #.

N _ ~ (,) = o,

i = l

CONSTRUCTION OF DYNAMIC SYSTEMS FROM GIVEN INTEGRALS 1 5 3

where u, # = 1 , . . . , N - 1. If this is possible, we then get

V* N - 1 = ~ i = 1 AiT(i) (3.1)

G(v*, v*) N-1 2 = ~ i = l ,X~,

where by A1, . . . , AN-I, we denote an arbitrary smooth function on X. By differentiating (3.1) and considering (2.1), we deduce the equalities

N - 1

F * = V * ( v * ) = ~ v*(Ai)v(i)-p(v*,v*)[, (3.2) i = l

where [ is a vector field with ([1 . . . , iN) E C ~ components such that

N

i=l

p(v*, v*) = v;(~)(v*) _= B*(v*, v*).

By applying the principle of least contractions deduced by Gauss the following problem can be set [6]:

PROBLEM. Find the

extremum p(v, v) T

under the conditions

{ G(~-, ~-) : 1 v',N ¢.Ti /-.,i=1 ~* : O.

The solution to this problem may be used to solve the Danielli and Suslov problem. To resolve this problem we first introduce the L Lagrangian:

L = p(v, v) + u(G(v, v) - 1) + # ~ ¢iv i.

From the equations

o.(L) : o,

O,(L) = O,

we get

P v + #~ = O,

~ N 1 ~i Ti : O, ( 3 . 3 )


where r and P are the matrix:

= co l (T1 , . . . , T N)

P = u I + 1 3 .

CONSEQUENCE 3.1. From (2.4) we get 2p(r, 7-) + u = 0. It is easy to prove that (3.3) can be rewritten as follows

K r / = 0,

where by K and ~1 we denote the matrix:

P~ P) . . . P~v ~l

K = : : " . . :

e~r P 2 p N eN

~1 ~2 ~N 0

r /= col(r l , . . . , r N, #).

It is well known that (3.5) has a nonvanishing solution if and only if

det(K) = p N - 1 + HpN-2 q_ . . . q_ R = O,

where by H and R we denote

N - 1

H = - E / , , i , R = ( -1 )N-1 1-- [ "i. i=1

For H and R there is the following expression

E}V=l Oj(~/Idet(a)l)~ cj H = sp(K)~,:0 =

R det(K)~=0.

(3.4)

(3.5)

(3.6)

From the above results we obtain the important

PROPOSITION 3.1. Let (uj, un) be the roots of(3.6) and ( ~(J), rl (i)) the solutions of the equations

K(j)rl (j) = O,

where K(j) = Ku=ua. Hence we get:

CONSTRUCTION OF DYNAMIC SYSTEMS FROM GIVEN INTEGRALS 15 5

(i) G(~(j),~(.))=o, if ( . j - . k ) # o . p - 1 (ii) I f P(j) = PI~=,j is a nondegeneratematrix, then v(j) = -IZ(j) (j) ~, where

#(j) is chosen such that G(T(j), r(j)) = 1.

C O N S E Q U E N C E 3.2. Let g be a function of the complex variable z such that

where Tz is the vector field introduced in the first chapter. If the zeroes of g are simple, then the vj, j = 1, 2 , . . . , M vector fields can be determined by the formula

~-j = Tzlz=~.

C O N S E Q U E N C E 3.3. If the N - 2 zeroes of the function g are differents then any v* C/C(aJ) vector field can be represented as follows

N - 2 N-1

v* = ~ ~j(j) + ,~N-lV± = ~ ~r(j), j = l j = l

(3.7)

where v± = T(N_I), is a vector field which is orthogonal to (T(1), . . . , T(N_2) , ~).

Of course if all zeroes of g are different, then any v* E/C (a~) vector field permits the representation

N-1 N

V* : E /~jTj = E OlkXk j = l k=l

(3.8)

where Xk, [ ( = 1, 2 , . . . , N are the vector fields given by the formula (1.1). From (3.2) and (3.1) we get:

N N

j = l n= l

(3.9)

The mechanical mean of (3.1) and (3.9) is easy to deduce.

4. The Danielli and Suslov Problem

The publication of Suslov's partial differential equations for the potential of a given family of orbits was a considerable development in the inverse problem in dynamics. This family is obtained as an intersection of the N - 1 hypersurface

f j ( x l , x 2 , . . . , X N) = Cj, j = 1 , 2 , . . . , N - 1

for a mechanical system with N degrees of freedom ( N > 1).


Danielli set, in essence, the more general problem of finding the force field which is capable of generating the orbits in the plane f ( x , y) = c. It seems important to be able to develop the Danielli problem. This allows to start from a family orbits and to build a dynamical system around them. This essentially inverse problem is important in the study of fundamental questions such as control theory.

We now propose the solution to the extension of the Danielli and Suslov problems.

PROPOSITION 4.1 (particular solution to the extended Danielli problem)• Let the 9 zeroes be different• So the solution to the extended Danielliproblem is the force field F* with the v* E / C ( d f ) integral element such that

N v* = E A k ( d f ( X k ) X 1 - d f ( X 1 ) X k ) .

k=2

Likewise we can study the case when function g has N - 2 different zeros.

PROPOSITION 4.2 (solution to the Suslov problem). Let the 1-forms df l , d r2 , . . . d f N - 1 be independent. So the solution of the Suslov problem is F = grad(U) vector

field with v* E K;(dfl, d f2 , . . . , dfN-1) such that:

v* = A det { Olf l Olf2 . . . O l f N - 1 O1 ) • : . . . : :

\ ONfl ON& .. . ONfN-1 ON

f~o -= ~." dp, = - d h ( f l , f 2 , . . . , fm-1)

Function U in this case is the following

v =

h = - fro.

(4.1)

(4.2)

The aim o f the next assertions is to apply these results, in particular when X is a region o f the N-dimensional Euclidean space E N and w = d f .

The matrix K given by formula (3.5) in this case takes the form

K =

I + u fx2xl

fXNX

L,

f xll;2 . . • f xlxN f x 1

: "•. :

f ~2 . . . f xN 0

(4.3)

where by v we denote vCEN=l(Oz~(f)) 2 and fx~ - OxJ(f).


4.1. VECTOR FIELD V* AND F * IN THE PLANE

For N = 2 the problem of constructing fields v* and F* is trivial. In fact, the vector fields v* and F* in this case are

v* = A(fyOx - fxOy) -- A{f},

t;' = {f, ½~2}V* _ ~2X2 '

where ~ E C~°(X) is an arbitrary function, X2 is a vector field obtained from the formula (1.1) which in this case takes the form

X 2 = X 2 ( x ) O x q- X2(y)Oqy

Xa(x) = (fxfyy - f y f~y )= A ( f ) f ~ - ( f 2 + .1,2) 2 - z

X2(y) = fyfi:~ - fxfi:y - A ( f ) f y - -~ , y

where A ( f ) = f~x + fyy. As a consequence we obtain that the most general force field capable of gen-

erating the orbits (Danielli's problem [2]) { f ( x , y) = c} is F* vector field such that

{ ~ = {f , 1 2 r ( x ) ~:~ } h - X2(x) :~2 - (4.4)

~) = - { f , ½A2}f~ - Xz(y)A 2 =_ F(y) .

By considering Proposition 4.2 we deduce, that the vector field F* is gradient if and only if A is such that

q = fx(,~2)x -~- L(~2)y -~ 2A(f))~ 2 = 2 0 ( f )

where 0 is a function of f . In fact, in this case the formula (2.2) takes the form:

~* dp, = ½qdf.

Hence, it is easy to deduce the following expression for U(x, y)

U(x ,y ) = # 2f2 + ]2 h ( f ) 2 (4.5)

a( f ) = j o ( f ) d f,

Szebehely [8] has derived a linear, first-order partial differential equation for the potential field generating a family of period orbits in the plane. This equation can

158 RAFAEL RAMIREZ AND NATALIA SADOVSKA1A

easily be obtained from (4.4) by considering that A 2 = 2(U + h). In fact, if (4.4) is multiplied by fx and fu and added we get

f~:Ux + fyUy + 2 ( fxxf2 + f y y f ~ - 2 f x y f x f y ) ( U ( x , y ) + h( f ) ) = 0 . (4.6) y +g

Evidently, (4.5) is a general solution to (4.6).

EXAMPLE 1. To illustrate these ideas, we consider the field v* in the plane with components:

v * ( x ) = A ( a y - 2 b x y ) v*(y) +

This vector field is generated by f ( x , y) = 21-(x 2 -4- ay 2) 4- bl(x 3 - 3y2x) = c. From here it can easily be deduced that in this case force field F* is such that:

ii { f , ½A2}fy A 2 ( ( l + a ) f - 1 2 F(x) , = _ ~(f{ 4- f2))~) =_

~) {f , ½A2}fx A2((1 + a ) f - 1 2 2 = _ _ ~(fz + f~))y) _ F(y)

Evidently, F* is potential iff A 2 : const. (we set/~2 : 1). Hence

h ( f ) = (1 4- a) f U(x ,y) = 1 2 g(fz 4- fu2) _ (1 4- a) f = U1 4- (b(x 2 4- y2))2

where by U1 we denote Henon-Heil 's potential function:

Ul(x,y) = - ( a ( x 2 4- y2) 4- 2b(3axy2 _ x3).

4.2. VECTOR FIELDS v* AND F * IN ]E 2

In this section we consider the case in which N -- 3. At first we observe that equation (3.6) takes the form

u 2 + Hu + R = 0, (4.7)

where H and R are the mean and the Gaussian curvature respectively of surface f ( x , y, z) = el, i.e.,

H = div grffd f q j 2 + f 2 y + f 2 '

Evidently, the roots of (4.7) are

.l = -½(H + VH2-4R),

R = d e t K f2 , -0"

_

u 2 = - I ( H - V H 2 - 4 R ) .

By applying the results given in the first chapter, we can prove the following


PROPOSITION 4.3. Let function g have two different zeroes. So the vector fields X1, X2, X3 are such that

Xl = fxox + fyoy + f~o~,

X2 = ( A f f x - lx)O~ + ( A f f u - Iv)Oy + ( A f f ~ - l~)O.,,

X3 = det K]z=0,

where A f = fx~ + fry + f ~ and l 1 2 2 = ~(f~ + f~ + f2 ) . Function g in this case is the fo l lowing

412Z 2 + H Zo + I ( g( Zo) d e t ( B + Z l ) ' z = v / ~ Z o

From (3.4) we deduce that if the zeroes of g are different then the F* force field with

v* = ) ~ l ( d f ( X 1 ) X 2 - d f ( X 2 ) X 1 ) + )~2(df (X1)X3 - d f ( X 3 ) X 1 )

is the solution to the extended Danielli problem in this case. If function g does not have two different zeroes, then vector fields X1, X2, X3

are dependent. Should this be the case, as we mention above, the solution to the Danielli problem is field/;* with

v* = ) ~ 2 ( d f ( X l ) X 2 - d f ( X 2 ) X l ) + •oXo

where Xo is a vector field which is orthogonal to fields X1, X2.

EXAMPLE 2. Let the 1-form d f be such that d f = x dx + y dy - dz. Vector field X1, )(2, X3 in this case are

X1 = xOx "Jc yOy - Oz,

X 2 • xOx + yOy - 20z,

X3 ": Oz.

These field are clearly dependent. Vector field Xo is

Xo = xOu - yaK,

and as a consequence we get that the most general integral element of the given 1-form is

v* = (Alx - A2y)o~ + (Aly + A2x)O~ + (x 2 + y2)AlOz


where A1, A2 C Cc°(X) and force field F* in this case is such that

{ x = (J~l + A:~ - A2)x - (A2 + 2A1A2)Y ~ F(x),

~) (AI + AI 2 A2)y q- (A2 + 2A1A2)x ~ F(y) ,

(/~-1 + 2A12)( x2 + y2) = F(z),

These equations describe the behaviour of a point with unit mass moving on the 1 2 surface z = ~(x + y2).

EXAMPLE 3. Let us suppose that surface is such that f ( x , y, z) = r + blx + b2y + b3z = cl, where r = x/x 2 + y2 + z 2, and bl, b2, b3 are some parameters.

For this case the mean and Gaussian curvatures are:

H = (2(r lgrad(f ) l )2 + f2 ) R = 2f2 (r lgrad(f) l) 3 (r lgrad(f)l) 4'

where [grad(f)t 2 b~ + b 2 + b~ - 1 + 2y The Vector fields X1, X2, X3 in this case are:

Xz - f(x'-rY' W) (xO~ + yOy + wO~),

X3 • 1X1 ~-rX2, r

from here we easy deduce that dim(X1, X2, X3) -~ 2. Vector field X0 in this case is

Xo = (b3y - b2w)O~ + (blw - b3x)Oy + (b2x - bly)O.o

It is easy to obtain the expression for the force field F*. In particular the vector field F* = aX2, a C C ° ( X ) is the force field capable of generating the orbits

r + blx + b2y + b3w = cl,

Ax + By + Cw = O, Abl + Bb2 + Cb3 = O.

4.3. VECTOR FIELDS V* AND F* IN E N

We now apply the results of the first section for N > 3. In order to illustrate proposition I in E N we study the following example:


EXAMPLE 4. Let f be:

N

f ( x i , x2 , . . . ,XN) = ~ a j x f = 1. j = l

We still assume that the parameters a l , a2, . . . , aN are such that a l < a2 < • • •

aN. Vector field Tz and function g in this case are the following:

- E 0k, (4.8)

g(z) = df(~Cz) = - ~ (akxk)2 a k + z '

from here we obtain

g'(z) = E \ z + ak,] '

so we deduce that all zeroes of g are simple. It is easy to prove that the vector fields X1, X2, . . . , XN form an Abelian Lie Algebra. In particularly for N = 3 we get that X1, X2, X3 are the following

X1 = alxOx + a2yOy + a3wOw, X 2 = a l ( a 2 + a3)xOx + a 2 ( a l + a3)yOy 4- a 3 ( a l + a2)Ow,

X3 = ala2a3(xOz + yOy -I- WOw)

Let us introduce complex function g*:

g*(z) = g(z) - g(z) 2 (4.9)

Moreover, due to (4.8) and equality g(uj) = 0 , j = 1 , 2 , . . . , N - 1 we get

g*(O) = g*(/21) . . . . . g*(YN-1) = O,

so g* permits the representation

I I y ~ ' z(z - - j ) g * (z) = ~-~--__o-~+a- ~ . (4.10)

If we now consider (4.8), (4.10), we deduce that

N - 1 "N-I I ' I j = l ak(ak + vj) (4.11) ( akxk)2= res g , (z) = = (--1) ~----V---;,------Z'.--

z=--ak lly=l(ak aj)ljck

Therefore, we get the following


PROPOSITION 4.4. By taking (4.8), then:

(i) All zeroes o f 9 are different. (ii) The roots V l , . . . , UN-1 are the Jacobian coordinates on the surface

P, aj(xJ) 2 = 1.

As a consequence of these results, we obtain the most general vector field v* E /C(df): which generates the following differential equations

N-1 Aj ack = akxk E

j=l vj + ak

These equations can be written in Jacobian coordinates, in particular for N = 3, we get

{ £'1 • P(Vl,/Q)~I + P(Ul, V2))~2 Pl - - P2

/)2 = P(/-/2,/"2)A1 + P(P2,/Jl)A2 //1 - 122

where P denote the polynomial such that

P ( u , v ) = ala2 + (al + a2)u + uv.

From (4.11), for N = 3 we get that

Pl,/22 = ala2a3((xl) 2 -}- (x2) 2 + (x3) 2)

= a2[x3~2~ Vl + l/2 al + a2 + a3 -- (al2(xl) 2 + a2(x2) 2 + 3\ ) ]'

so, if we suppose that (x l )2 + (x2)2 + (X3) 2 = c o n s t . , t h e n )l 1 q- A 2 = 0. T h e r e f o r e ,

the above equations can be rewritten respectively in the form:

{ xl = ~(a3 _ a2)x2x3

x2 ~ ( a l a3)xlx 3

x3 ~(a2 al)x2x 1

where # = A2(v 2 - Vl ) / z l x2x 3 and

01 = ~2Pl

fi2 = --A2V2


Remark. For N = 3 there exists the following expression for Jacobian coordinates:

{ l/1 = ~ / ~ j = l ( a j x J ) 2 ( - U + ~/H 2 - 4R),

1 3 Vl = -~ /2 j=l (a jxJ)2( -H - ~/H 2 - 4R),

where H and R are the mean and Gaussian curvature of the given ellipsoid. Of course, in this case force field F* can be represented by the formula F* =

~ozjXj where aj C C°°(X). So the extended Danielli problem can be solved in this case.

5. The Vector Field v and F* with M Invariants

In what follows we will assume that vector field v* also has M invariants such that

f j (x 1,x2, . . . ,x N ) = c j , where j = 1 , . . . , M . M < N - 1 . (5.1)

We consider that f l , f2,. • •, fM are independent functions. By orthogonalizing vectors grad(fl); grad(f2);...;grad(gM)we can obtain

the system of vectors e~; e~; . . . ; e~ :

1, f o r a = fl (e~,e-'/3)= 0, f o r a # f l .

By denoting

rad(f) -Igrad(f)l

we get the following representation for e]

N

= Z rj/3e/3 (5.2) /3=1

where rj;~ is a matrix:

rj/3 = ( ~ , ~/3) - (ej, ~/3) (5,3)

By setting(5.2) in the equalities (el, v*-(x)) = 0 and considering (3.1) we get the following conditions for the arbitrary functions Aj, j = 1, 2 , . . . , N - 1

N - I

£j#a# = 0, (5.4) ¢~=1


where j = 1 ; . . . ; M < N - 1. Solving respect to A~ denoting by F = det(Fj~) the Gramma determinant and by substituting A1; A2; . . . ; AM into (3.1) we have,

N-1

E -/=M+I

(e l , T1) . . . . . . (e l , 7"M) (E1,T'/)

(~2, ~ ) . . . . . . (~2,,-v) (~2, ~-~) A-/

det : : ".. •

( e v , r,) . . . . . . ( e v , r M ) (eM, r-r)

(5.5)

This is the most general integral element of the df~ , ~ = 1 , 2 , . . . , M + 1.1-forms. From here we can easily construct force field F which gu.arantees that the

movements of the material point with kinetic energy T = 21-E(xk) 2 take place in the submanifold

{ fv (x 1, X2 , . . . , X N) = ev, l; = 1 , . . . , M ,

fM+l(X 1,x2,. . . ,x N) = CM+l,

Based on the above results we prove the following

PROPOSITION 5.1. Let the hypersurface be

fM+I(xl,x2,.. . ,X N) = CM+I,

such that function g has at least N - 2 different zeroes.

So the vector field tangent to the submanifold

fv(x 1, X2,..., x N) = Cv t.' = 1 , . . . , M,

is defined by formula (5.5).

CONSEQUENCE 5.1. I f in (5.5) M = N - 2 then we get

V* -- AN-1 F

(e l , 7-1) . . . . . . (e l , TN-2) (e l , TN-1)

(.2, ~,) . . . . . . (~2, ~N.~) (*~, ~N-,)

- - det " " ".. •

( e N - 2 , T1) . . . . . . (eN-2, TN-2) ( eN-2 , TN-I)

. . . . . . TN~_2 TN~_I

(5.6)

In this case it is possible to choose the coordinates in such a form that v* can be represented by the formula (4.1).


These results can be applied to deduce a new approach to solving Suslov's problem [7].

In particular, if N -- 3, we get from (5.6) that:

v*-(x) = )~[graJ(f2), grad(fm)],

where A is an arbitrary smooth function. The force field F* in this case is such that

P = grad(1A2(l[graJ(j~), grad(j~)]/) 2) + A2((gr~d f l , rot(~))grgd f2

- (grgd f2, rot(~))grgd f l ) )

CONSEQUENCE 5.2. Force field F* is gradient iff the 1-form dp,:

tv* dp, = A2trotv* dfl A d f2,

is such that

~2o = -dh(f l , f2) (5.7)

Hence, we get the following

CONSEQUENCE 5.3. The potential function U(x 1 , x 2, x 3) can be calculated by the formula

{U(xl ,x2, x3)il~2'[grad(fm),gra'(f2)]'2-h(fl,f2),

h(fl, f2) = ft0. (5.8)

Joukovski's result given in [5] is a specific case of (5.9). It should be pointed out that in general the dependence of h on the f l , f2 is not arbitrary, as we can see from formula (5.7). In the following example we illustrate this fact.

EXAMPLE 5. Vector field v*

w x v* = #yOx - #xOy + # 0~,, (5.9) Y

where # x = )~, is generated by two invariants

f l (x ,y ,z) = (x2+ y2) = Cl ' (5.10)

f 2 ( . , v , z ) - = c2 .

By setting in (5.9) # = 1 we can easily obtain that

~ . . d p . = 2x 75 ] d x + 2 y + x 2 ] ~ )


1 2 = d(f l + 2f l f~) = - d ( h ) .

As a consequence, we get a two-parametric family of elliptic orbits given by (5.10) which belongs to the potential

1( y2 z2y2~ U = ~ x 2 + + 7 - - ] - ( f l + l f ~ f ~ ) = - ½ ( x 2 + y 2 + 2 2 ) •

Studying (5.6) once more, we first observe that in the suitable coordinates y l , . • . , yN such that

yJ = f j ( x l , . . . , X N) j = 1 , . . . , N - 1, yN = xN

we can get

v- 1 times ¢%

= (b ,o , . . . , d, 1, o, . . •,o),

TN~_~ = (0,0,•• , 0 , 1)•

So the differential equations which are generated by (5.7) in these coordinates can be rewritten as follows

ff'l = 0 ,

y2V = # ( y l y2 . . . , y N ) .

From here we get EN=I(kJ) 2 = EGjkyJy k.

PROPOSITION 5.2. Let us give a conservative mechanical system with kinetical energy T:

• 1 N . .

T ( y l , . . . , y U , y l , y 2 , . . . , y N ) = 2 E GJ kyjyk' j

with particular integrals yl = e l , . . . , yN-1 = CN-1, then the force field F* is the gradient iff the 1-form p, = G(v*, dy) is such that

1 N

fro = tv* dp. = ~ E ( G k N O N ( # 2) -- GNNOk(#2)) dx k k=l

= _dh (y l , y2 , . . . , yN-1). (5.11)


The proof can be obtained from the above results.

CONSEQUENCE 5.4. Let f~0 be a closed 1-form. So the potential function U is such that

U(yl y2 ...,yN) _.~ 1GNN(yl ,y2 ...,yN)#2-- h(yl ,y2 . . . , y N - 1 ) ,

h ( y l , y2, . . . , y N - 1 ) = j [ frO.

An example will help to illustrate these results.

EXAMPLE 6. Let N = 3 and f l , f2:

f l (X, y, Z) = x z = Cl,

r e ( x , y, z) y z cZ, (5.12)

where (x, y, z) are the Cartesian coordinates.

Defining (yl ,y2, y3) as follows yl = xz , y2 expression for the metric tensor G

= yz , y3 = z, we get the next

1 2(ff3)~ 0

1 G = 0 2(y3) 2

- y l _y2

(ff3)3 (y3)3

/ (y3)3

__y2 (y3)3

(yl)2 + (y2)2+ (y3)4

2(y3) 4

Hence, considering (5.11) we obtain that cv* dp, 1-form:

~0 (ffl)2 .3 c (ff2)2 .71_ (ff3)4 (01(# 2) dY 1 q- 02(# 2) dff 2) ~" (y3)4

03(#2)/ 1 ( - ~ l y dy 1 --1- y2dy2)

is closed if

(i) .2__ ( 3i4 ' ) (ii) #2 fit (yl)2 -t- (y2)2 _ (y3)4.

(y3)2


Functions h and U are respectively

. e 1 2x ( y l ) 2 q_ (y2)2 (i) rtt, y ,y ) = - - T - - - - '

U(y l , ye , y 3) = 1 (y3 )4 ,

(ii) h(y 1, y 2, y3) = h0,

U yl,y2, y3 ~ ( ( y l ) 2 k ( y 2 ) 2 - ( y 3 ) 4 ) ( ) = (y3)2 . ( y 3 ) 2 ( ( y l ) 2 + (y2)2 q_ 1).

The last potential was deduced by Joukovski in paper [5]. The possibility of carrying out motion (5.12) with the force field F = 2(y3)30y 3

is easy to prove. In fact, in the coordinates (x, y, z) vector field v* generates the equations

2 = )~zx, i1 = Azy,

= - A z 2,

and by deriving we get

- ' (~)z 2 + 2~2z 3,

By choosing ), = 1, we deduce the veracity of the affirmation. Finally, it should be pointed out that the above results are independent of the signature of the metric tensor G, so we can extend all affirmations to the pseudo-Euclidean space. In order to illustrate these ideas we give the following.

EXAMPLE 7. space:

x - - _ _ - - C l ~

W

By transformation

y l x

w

y2 = Y, w

y3 z w

y 4 = w.

Let us give a three-parametric family of orbits in Minkovski 's

y z - - ~ C 2 ~ - - ~ C 3. W W

CONSTRUCTION OF DYNAMIC SYSTEMS FROM GIVEN INTEGRALS 1 69

we easily obtain that G is

G

(y4)2 0 0 2y4y 1

0 (y4)2 0 2ff4ff 2

0 0 (ff4)2 2ff4y3

2ff4yl 2y4y2 294p3 (fit)2 q_ (ff2)2 q_ (ff3)2 _ 1

After some calculations we can deduce that the 1-form ~2o is closed in particular if

(i) #2 = 1I/2(G44),

(ii) # 2 = _y4G44,

so we get

(i) h = / G 4 4 d ( G 4 4 ) ,

= - ~ ( G 2 4 ) ( 1 G 4 4 ) ~ - ] G 4 4 d(G44). U

(ii) h = ho,

U = _ y 4 (G44) 2 _+_ ho 2

6. Conclus ions

In this paper we propose a new approach for studying the inverse problem in dynamics related with to construction of force fields by given partial integrals. The Danielli problem is generalized and an intrinsic solution to the well-known Suslov problem is given.

References

1. Bertrand, M. J.: Sur la possibilitd de deduire d'une seule des lois de Kepler le principe de l'attraction, C.R. Acad. Sci. Paris 9 (1877).

2. Danielli, V.: Giornale di Mat., 18 (1880), 271. 3. Darboux, M. G.: Recherche de la loi que dolt suivre une force centrale pour que la trajectorie

quelle determine soit toujours une conique, C.R. Acad. Sci. Paris 16 (1877). 4. Galiullin, A. S.: Inverse Problems in Dynamics, Nauka, Moscow, 1981 (in Russian). 5. Joukovski, N. E.: The determination of the potential function by given trajectories, Sobranie

Sochinieni, Vol. 1, M.-L., Gostexizdat, (1948), pp. 227-242 [in Russian]. 6. Ramirez, Rafael and Sadovskaia, Natalia: Construction of vector fields by given first integrals,

Collectanea Mathematica 40(3) (1989) [in Spanish]. 7. Suslov, G. K.: The potential function with given partial integrals, Kiev (1890) [in Russian].


8. Szebely, V.: On the determination of the potential, E. Proverbio, Proc. Int. Mtg. Rotation of the Earth, Bologna (1974).

9. Whittaker, E. T.: Analytical Dynamics of Particles and Rigid Bodies, Cambridge University, Cambridge, MA (1959).

Documents

On the construction of dynamic systems from given integrals