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Vol. 40, No. 1, 2010, 57-76

ZERMELLO’S CONDITIONS AND ENERGIES OF HIGHER ORDER IN GENERALIZED

LAGRANGE-HAMILTON SPACES

Irena ˇComi´c1

Abstract. Many significant geometers contributed to the generalization of Riemann spaces in different directions. An almost complete list of them can be found in Miron’s books. Here are mention [1], [2], [17] and [18]

in which Hamilton and Finsler spaces are examined, further [3–9], where generalized Hamilton spaces are studied; [10] is most connected with the subject, and in [11–16] this problem also appears. Zermello’s condition in Miron’sOsckM was examined in [6]. Here, the Zermello’s conditions are given in Lagrange-Hamilton spaces, introduced in [9] and presented at the Workshop on Finsler Geometry 2009, Debrecen. It is proved that for a fundamental function for which the Zermello’s conditions are satisfied all energies of higher order are equal to zero.

AMS Mathematics Subject Classification (2000): 53B40, 53C60

Key words and phrases: Lagrange-Hamilton spaces, Zermello’s conditions

1. Some invariants in generalized Lagrange-Hamilton spaces

Group of coordinate transformations. Generalized Lagrange-Hamilton spaces are introduced in [9]. We shall recall only those parts of which are necessary for the understanding of Zermello’s condition in these spaces.

Let us denote by (LH)(nk)the (2k+ 1)ndimensionalCmanifold in which a point (y, p) = (x=y(0), y(1), y(2), . . . , y(k), p(1), p(2), . . . , p(k)) has the coordi- nates

(xa =y0a, y1a, y2a, . . . , yka, p1a, p2a, . . . , pka), a= 1, n.

Some curve c (LH)(nk) is given by c : t [a, b] c(t) (LH)(nk). A point (y, p)∈c(t) has the coordinates (xa(t) =y0a(t), y1a(t), . . . , yka(t), p1a(t), . . . , pka(t)), where

yAa(t) =dAty0a(t) A= 1, k, dAt = dA dtA, (1.1)

pαa(t) =dα−1t p1a(t), α= 1, k, dα−1t = dα−1 dtα−1.

1Faculty of Technical Sciences, 21000 Novi Sad, Serbia, e-mail: [email protected], http://imft.ftn.uns.ac.rs/~irena/

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The allowable coordinate transformations are given by xa0 =xa0(xa)⇔xa=xa(xa0)

(1.2)

y1a0 =Baa0y1a, Baa0 =0axa0=axa0, ∂Aa=

∂yAa A= 0, k,

y2a0 = Ã

1 0

!

(d1tBaa0)y1a+ Ã

1 1

!

Baa0y2a=d1t(Baa0y1a),

y3a0 = Ã

2 0

!

(d2tBaa0)y1a+ Ã

2 1

!

(d1tBaa0)y2a+ Ã

2 2

!

Baa0y3a=d2t(Baa0y1a), . . . ,

yAa0 = Ã

A−1 0

!

(dA−1t Baa0)y1a+ Ã

A−1 1

!

(dA−2t Baa0)y2a+· · ·

· · ·+ Ã

A−1 A−1

!

Baa0yAa=dA−1t (Baa0y1a), . . . ,

yka0 = Ã

k−1 0

!

(dk−1t Baa0)y1a+ Ã

k−1 1

!

(dk−2t Baa0)y2a+· · ·

· · ·+ Ã

k−1 k−1

!

Baa0yka=dk−1t (Baa0y1a),

p1a0 =Baa0p1a Baa0 =0a0xa= ∂xa

∂xa0 =Baa0(t),

p2a0 = Ã

1 0

!

(d1tBaa0)p1a+ Ã

1 1

!

Baa0p2a=d1t(Baa0p1a),

p3a0 = Ã

2 0

!

(d2tBaa0)p1a+ Ã

2 1

!

(d1tBaa0)p2a+ Ã

2 2

!

Baa0p3a=d2t(Baa0p1a),

p4a0 = Ã

3 0

!

(d3tBac0)p1c+ Ã

3 1

!

(d2tBca0)p2c+ Ã

3 2

!

(d1tBac0)p3c+ Ã

3 3

!

Bca0p4c, . . . ,

pαa0 = Ã

α−1 0

!

(dα−1t Baa0)p1a+ Ã

α−1 1

!

(dα−2t Baa0)p2a+· · ·

· · ·+ Ã

α−1 α−1

!

Baa0pαa, . . . ,

pka0 = Ã

k−1 0

!

(dk−1t Baa0)p1a+ Ã

k−1 1

!

(dk−2t Baa0)p2a+· · ·

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· · ·+ Ã

k−1 k−1

! Baa0pka.

Theorem 1.1. The transformations of type (1.3) on the common domain form a group.

Definition 1.1. The generalized Lagrange-Hamilton space of order k, (GLH)(nk), is an(LH)(nk)space, where the group of allowable transformations is given by (1.2) and in which a fundamental function F(x, y(1), y(2), . . . , y(k), p(1), p(2), . . . , p(k))is given, whereF :U →Ris differentiable onU˜ (where rank [y1a] = 1, rank [p1a] = 1) and continuous in those points of U, where y1a and p1a are equal to zero, U is a domain in (GLH)(nk).

The natural and special adapted bases in T(GLH)(nk) and T(GLH)(nk). The natural basis, ¯BLH of T(GLH)(nk) as usual consists of partial derivatives of variables, i.e.

B¯LH ={∂0a, ∂1a, . . . , ∂ka, ∂1a, ∂2a, . . . , ∂ka}, 0a =a=

∂xa =

∂y0a, (1.3)

Aa =

∂yAa A= 1, k, αa=

∂pαa, α= 1, k.

Theorem 1.2. The elements of B¯LH transform in the following way:

(1.4)

0a=(∂0ay0a0)∂0a0+(∂0ay1a0)∂1a0+(∂0ay2a0)∂2a0+(∂0ay3a0)∂3a0+· · ·+(∂0ayka0)∂ka0+ (∂0ap1a0)∂1a0+(∂0ap2a0)∂2a0+(∂0ap3a0)∂3a0+· · ·+(∂0apka0)∂ka0,

1a= (∂1ay1a0)∂1a0+(∂1ay2a0)∂2a0+(∂1ay3a0)∂3a0+· · ·+(∂1ayka0)∂ka0+ (∂1ap2a0)∂2a0+(∂1ap3a0)∂3a0+· · ·+(∂1apka0)∂ka0,

2a= (∂2ay2a0)∂2a0+(∂2ay3a0)∂3a0+· · ·+(∂2ayka0)∂ka0+ (∂2ap3a0)∂3a0+· · ·+(∂2apka0)∂ka0, ...

ka= (∂kayka0)∂ka0

1a= (∂1ap1a0)∂1a0+(∂1ap2a0)∂2a0+(∂1ap3a0)∂3a0+· · ·+(∂1apka0)∂ka0,

2a= (∂2ap2a0)∂2a0+(∂2ap3a0)∂3a0+· · ·+(∂2apka0)∂ka0,

3a= (∂3ap3a0)∂3a0+· · ·+(∂3apka0)∂ka0,

...

ka= (∂kapka0)∂ka0.

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The natural basis ofT(GLH)(nk)is

B¯LH ={dy0a, dy1a, . . . , dyka, dp1a, dp2a, . . . , dpka}.

Theorem 1.3. The elements of B¯LH transform in the following way:

(1.5)

dy0a0 = (∂0ay0a0)dy0a

dy1a0 = (∂0ay1a0)dy0a+ (∂1ay1a0)dy1a, . . . ,

dyka0 = (∂0ayka0)dy0a+ (∂1ayka0)dy1a+· · ·+ (∂kayka0)dyka, dp1a0 = (∂0ap1a0)dy0a+ (∂1ap1a0)dp1a,

dp2a0 = (∂0ap2a0)dy0a+ (∂1ap2a0)dy1a+ (∂1ap2a0)dp1a+ (∂2ap2a0)dp2a, . . . ,

dpka0 = (∂0apka0)dy0a+ (∂1apka0)dy1a+· · ·+ (∂(k−1)apka0)dy(k−1)a+ (∂1apka0)dp1a+· · ·+ (∂kapka0)dpka.

Definition 1.2. The special adapted basis BLH of T(GLH)(nk)

(1.6) BLH =0a, δ1a, . . . , δka, δ1a, δ2a, . . . , δka}

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is defined by (1.7)

δ0a0

0

¢0a ¡1

0

¢N0a1b1b ¡2

0

¢N0a2b2b ¡3

0

¢N0a3b3b − · · · − ¡k

0

¢N0akbkb

¡0

0

¢N0a1b1b¡1

0

¢N0a2b2b¡2

0

¢N0a3b3b− · · · − ¡k−1

0

¢N0akbkb

δ1a= ¡1

1

¢1a ¡2

1

¢N0a1b2b ¡3

1

¢N0a2b3b − · · · − ¡k

1

¢N0a(k−1)bkb

¡1

1

¢N0a1b2b¡2

1

¢N0a2b3b− · · · − ¡k−1

1

¢N0a(k−1)bkb

δ2a= ¡2

2

¢2a ¡3

2

¢N0a1b3b − · · · − ¡k

2

¢N0a(k−2)bkb

¡2

2

¢N0a1b3b− · · · − ¡k−1

2

¢N0a(k−2)bkb

δ3a= ¡3

3

¢3a − · · · − ¡k

3

¢N0a(k−3)bkb

− · · · −¡k−1

3

¢N0a(k−3)bkb, . . . ,

δka= ¡k

k

¢ka

δ1a= ¡0

0

¢1a ¡1

0

¢N2b0a2b ¡2

0

¢N3b0a3b − · · · − ¡k

0

¢Nkb0akb

δ2a= ¡1

1

¢2a ¡2

1

¢N2b0a3b − · · · − ¡k−1

1

¢N(k−1)b0a kb

δ3b= ¡2

2

¢3b − · · · − ¡k−2

2

¢N(k−2)b0a kb, . . . ,

δkb= ¡k−1

k−1

¢kb.

Definition 1.3. The special adapted basisBLH of T(GLH)(nk) is (1.8) BLH ={δy0a, δy1a, . . . , δyka, δp1a, δp2a, . . . , δpka}, where

δy0a=dy0a=dxa (1.9)

δy1a= µ1

1

dy1a+

µ1 0

M0b1ady0b

δy2a= µ2

2

dy2a+

µ2 1

M0b1ady1b+ µ2

0

M0b2ady0b,

δy3a= µ3

3

dy3a+

µ3 2

M0b1ady2b+ µ3

1

M0b2ady1b+ µ3

0

M0b3ady0b, . . . ,

δyka= µk

k

dyka+

µ k k−1

M0b1ady(k−1)b+ µ k

k−2

M0b2ady(k−2)b+· · ·

· · ·+ µk

0

M0bkady0b,

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δp1a= µ0

0

M0a1bdy0b+ µ0

0

dp1a,

δp2a= µ1

0

M0a2bdy0b+ µ1

1

M0a1bdy1b+ µ1

0

M0a1bdp1b+ µ1

1

dp2a,

δp3a= µ2

0

M0a3bdy0b+ µ2

1

M0a2bdy1b+ µ2

2

M0a1bdy2b+ µ2

0

M0a2bdp1b+ µ2

1

M0a1bdp2b+ µ2

2

dp3a, . . . ,

δpka= µk−1

0

M0akbdy0b+ µk−1

1

M0a(k−1)bdy1b+· · ·

· · ·+ µk−1

k−1

M0a1bdy(k−1)b+ µk−1

0

M0a(k−1)bdp1b

+ µk−1

1

M0a(k−2)bdp2b+· · ·+ µk−1

k−1

dpka.

In [9], there are given the conditions for M’s and N’s such that the elements ofBLH andBLH are tensors and when these bases are dual to each other.

The J¯structure in (GLH)(nk)

Definition 1.4. The k-tangent structure J¯is aF linear mapping J¯:T(GLH)(nk)→T(GLH)(nk)

defined by

Jdy¯ 0a= 0,Jdy¯ 1a=dy0a,Jdy¯ 2a= 2dy1a, . . . ,Jdy¯ ka=kdy(k−1)a (1.10)

Jdp¯ 1a = 0,Jdp¯ 2a =dp1a,Jdp¯ 3a= 2dp2a, . . . ,Jdp¯ ka= (k1)dp(k−1)a, from which it follows

J¯ = dy0a⊗∂1a+ 2dy1a⊗∂2a+· · ·+kdy(k−1)a⊗∂ka+ (1.11)

dp1a⊗∂2a+ 2dp2a⊗∂3a+· · ·+ (k1)dp(k−1)a⊗∂ka. In [9], it is proved that the ¯J structure in the special adapted bases BLH

andBLH is given by

J¯ = δy0a⊗δ1a+2δy1a⊗δ2a+3δy2a⊗δ3a+· · ·+kδy(k−1)a⊗δka+ (1.12)

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δp1a⊗δ2a+2δp2a⊗δ3a+3δp3a⊗δ4a+· · ·+(k1)δp(k−1)a⊗δka. Remark 1.1. From (1.11) and (1.12) it follows that thek-tangent structure J¯in the natural and special adapted bases has the same coordinates.

It is also proved that the following relations are valid

Jδy¯ 0a = 0,Jδy¯ 1a =δy0a,Jδy¯ 2a= 2δy1a, . . . ,Jδy¯ ka=kδy(k−1)a (1.13)

Jδp¯ 1a= 0,Jδp¯ 2a=δp1a,Jδp¯ 3a= 2δp2a, . . . ,Jδp¯ ka= (k1)δp(k−1)a, The Liouville vector field. If M(y0a, y1a, . . . , yka, p1a, p2a, . . . , pka) and M0(y0a+dy0a, y1a+dy1a, . . . , yka+dyka, p1a+dp1a, p2a+dp2a, . . . , pka+dpka) are two points in (GLH)(nk), then the vector M M0 expressed in the natural basisT(GLH)(nk)has the form [9]

M M0=dr = dy0a0a+dy1a1a+· · ·+dykaka+ (1.14)

dp1a1a+dp2a2a+· · ·+dpkaka. It is proved ([9]) thatdris coordinate invariant, i.e.

dr = δy0aδ0a+δy1aδ1a+· · ·+δykaδka+ (1.15)

δp1aδ1a+δp2aδ2a+· · ·+δpkaδka.

Definition 1.5. The Liouville vector fieldsΓ¯0,Γ¯1,Γ¯2, . . . ,Γ¯k are defined by (1.16)

Γ¯k=dr, J¯Γ¯A= ¯ΓAJ¯= (k(A1))¯ΓA−1, A¯= 1, k, J¯¯Γ0= ¯Γ0J¯= 0.

Remark 1.2. From (1.14)-(1.15) it is obvious thatdr has the same com- ponents in the natural and special adapted bases. The same property has the structure ¯J (see Remark 1.1). This fact allows that the action ¯J on drcan be written by the equations of the same form in both coordinate systems.

In (GLH)(nk) it is difficult to construct vector fields, but using dr, the structure ¯J, one family of the Liouville vector field can, be constructed.

From (1.16) it follows

J¯Γ¯0= ¯Γ0J¯= 0, J¯¯Γ1= ¯Γ1J¯=kΓ¯0, J¯Γ¯2= ¯Γ2J¯= (k1)¯Γ1, (1.17)

. . . ,J¯Γ¯k−1= ¯Γk−1J¯= 2¯Γk−2, J¯Γ¯k = ¯ΓkJ¯= ¯Γk−1.

Theorem 1.4. The Liouville vector fields Γ¯0,¯Γ1, . . . ,Γ¯k from (GLH)(nk) ex- pressed in the special adapted basis B of T(GLH)(nk), have the form [9]

(1.18) Γ¯0 =

µk 0

δy0aδka,

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Γ¯1 = µk

1

δy1aδka+ µk−1

0

δy0aδ(k−1)a+ µk−1

0

δp1aδka,

Γ¯2 = µk

2

δy2aδka+ µk−1

1

δy1aδ(k−1)a+ µk−2

0

δy0aδ(k−2)a+ µk−1

1

δp2aδka+ µk−2

0

δp1aδ(k−1)a,

Γ¯3 = µk

3

δy3aδka+ µk−1

2

δy2aδ(k−1)a+ µk−2

1

δy1aδ(k−2)a+ µk−3

0

δy0aδ(k−3)a+ µk−1

2

δp3aδka+ µk−2

1

δp2aδ(k−1)a+ µk−3

0

δp1aδ(k−2)a, . . . ,

Γ¯k−1 = µ k

k−1

δy(k−1)aδka+ µk−1

k−2

δy(k−2)aδ(k−1)a+· · ·+ µ2

1

δy1aδ2a+ µ1

0

δy0aδ1a+ µk−1

k−2

δp(k−1)aδka+ µk−2

k−3

δp(k−2)aδ(k−1)a+· · ·+ µ1

0

δp1aδ2a,

Γ¯k = µk

k

δykaδka+ µk−1

k−1

δy(k−1)aδ(k−1)A+· · ·+ µ1

1

δy1aδ1a+ µ0

0

δy0aδ0a+ µk−1

k−1

δpkaδka+ µk−2

k−2

δp(k−1)aδ(k−1)a+· · ·+ µ0

0

δp1aδ1a.

Theorem 1.5. The Liouville vector fields Γ¯0,Γ¯1, . . . ,Γ¯k in (GLH)(nk) in the natural basis B¯ of T(GLH)(nk) have the form obtained from (1.18) if δyAa, δpαa, δAa, δαa, are substituted by dyAa, dpαa, ∂Aa, ∂αa respectively for ev- ery A= 0, k,α= 1, k.

Proof. The proof follows from Definition 1.5 and Remark 1.2.

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2. The Zermello’s conditions

Letc be a curve in (GLH)(nk)such that

c:t∈[0,1]→xa(t)∂a+d1txa(t)∂1a+· · ·+dktxa(t)∂ka+ (2.1)

p1a(t)∂1a+· · ·+dk−1t p1aka=

=y0a(t)∂0a+y1a1a+· · ·+ykaka+ p1a1a+· · ·+pkaka andImc⊂U.

The integral of actionIc is (2.2) Ic=

Z1

0

F(x, y(1), y(2), . . . , y(k), p(1), p(2), . . . , p(k))dt.

Ic does not depend on the parametrization of the curvec: xa=y0a(t) =y0a, yAa =dAtxa =dAxa

dtA , A= 1, k, p1a =p1a(t), pαa =dα−1t p1a =dα−1p1a

dtα−1 α= 1, k if

Z1

0

F(x, y1, y2, . . . , yk, p1, p2, . . . , pk)dt= (2.3)

= Z1

0

F(x, y10, y20, . . ., yk0, p10, p20, . . . , pk0)ds,

wheres=s(t) is at leastCk function,s0(t)>0 fort∈[0,1],s(0) = 0,s(1) = 1 and

yAa0 =dsAxa =dAxa

dsA , A= 1, k, (2.4)

pαa0 =dsαp1a =dα−1p1a

dsα−1 , α= 1, k.

The equations, which give the conditions when (2.3) is satisfied are called Zer- mello’s conditions. The equality (2.3) will be satisfied if along the curvec we have

(2.5)

F(x, y1, y2, . . . , yk, p1, p2, . . . , pk) =F(x, y10, y20, . . . , yk0, p10, p20, . . . , pk0)s0,

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where s0 = dsdt. To express (2.5) in invariant form we need some relations. We shall use the notation

s(α)= dαs

dtα, α= 1, k.

Asxa=xa(s),s=s(t), we have y1a = dxa

ds s0=y1a(s, s0), (2.6)

y2a = ∂y1a

∂s s0+∂y1a

∂s0 s00=y2a(s, s0s00), y3a = ∂y2a

∂s s0+∂y2a

∂s0 s00+∂y2a

∂s00s000 =y3a(s, s0, s00, s000), . . . , yka = ∂y(k−1)a

∂s s0+∂y(k−1)a

∂s0 s00+· · ·+∂y(k−1)a

∂s(k−1) s(k), p1a = p1a(s), s=s(t),

p2a = ∂p1a

∂s s0=p2a(s, s0), p3a = ∂p2a

∂s s0+∂p2a

∂s0 s00=p3a(s, s0, s00), . . . , pka = ∂p(k−1)a

∂s s0+∂p(k−1)a

∂s0 s00+· · ·+∂p(k−1)a

∂s(k−2) s(k−1)

= pka(s, s0. . . , s(k−1)).

Using the notations (2.7) Aa0 =dy0a

ds =dxa

ds =y1a0, AaA= dA

dtAAa0=dAtAa0 A= 1, k1,

(2.8) Ba1=dp1a

ds =p2a0, Baα= dα−1Ba1

dtα−1 =dα−1t Ba1, α= 1, k1 and the Leibniz rule for differentiation we can prove the following theorem.

Theorem 2.1. yAa,pαa ands(α),A,α= 1, k are connected by formulae:

y1a =Aa0s0, (2.9)

y2a = µ1

0

Aa1s0+

µ1 1

Aa0s00,

y3a = µ2

0

Aa2s0+

µ2 1

Aa1s00+

µ2 2

Aa0s000, . . . ,

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yAa=

µA−1 0

AaA−1s0+

µA−1 1

AaA−2s00+· · ·+

µA−1 A−1

Aa0s(A), . . . ,

yka= µk−1

0

Aak−1s0+ µk−1

1

Aak−2s00+· · ·+ µk−1

k−1

Aa0s(k), p1a =p1a(s), s=s(t)

p2a =B1as0 p3a =

µ1 0

Ba2s0+

µ1 1

Ba1s00, . . . ,

pαa= µα−2

0

Baα−1s0+ µα−2

1

Bα−2a s00+· · ·+ µα−2

α−2

Ba1s(α−1), . . . ,

pka= µk−2

0

Bak−1s0+ µk−2

1

Bk−2a s00+· · ·+ µk−2

k−2

Ba1s(k−1). Theorem 2.2. The following relations are valid:

(2.10) AaA=

µA−1 0

∂AaA−1

∂s s0+ µA−1

1

∂AA−2

∂s s00+· · ·+

µA−1 A−1

∂Aa0

∂s s(A),

(2.11) AaA= dyAa

ds , A= 1, k1,

(2.12) Baα= µα−2

0

∂Bα−1a

∂s s0+ µα−2

1

∂Baα−2

∂s s00+· · ·+ µα−2

α−2

B1as(α−1),

(2.13) Baα=dpαa

ds , α= 2, k1.

Proof.

AaA = dAtAa0=dA−1t (d1tAa0) =dA−1t µ∂Aa0

∂s s0

=

dA−1t

∂s(Aa0s0) =

∂sdA−1t (Aa0s0) =

∂s

·µA−1 0

AaA−1s0+ µA−1

1

AaA−2s00+· · ·+

µA−1 A−1

Aa0s(A)

¸ . If we take∂/∂sfrom the sum in the middle bracket we obtain (2.10), and if we substituteyAa from (2.9) we obtain (2.11). In the similar way we have

Baα = dα−1t B1a=dα−2t (d1tBa1) =dα−2t µ

d1tdp1a

ds

=

(12)

dα−2t

·

∂s µ∂p1a

∂s s0

¶¸

=

∂sdα−2t (Ba1s0) =

∂s

·µα−2 0

Bα−1a s0+ µα−2

1

Baα−2s00+· · ·+ µα−2

α−2

Ba1s(α−1)

¸ . If we take ∂s from the sum in the middle bracket, we obtain (2.12) and if we substitutepαa from (2.9) we obtain (2.13).

The explicit form of (2.6) as follows

y1a =y1a0s0, yAa0 =dAsy0a(s), A= 1, k, (2.14)

y2a =y2a0(s0)2+y1a0s00,

y3a =y3a0(s0)3+y2a03s0s00+y1a0s000,

y4a =y4a0(s0)4+y3a06(s0)2s00+y2a0(3(s00)2+ 4s0s000) +y1a0s0v, . . . , p1a =p1a(s), pαa0 =dα−1s p1a, α= 2, k,

p2a =p2a0s0,

p3a =p3a0s02+p2a0s00,

p4a =p4a0(s0)3+ 3p3a0s0s00+p2a0s000,

p5a =p5a0(s0)4+p4a06(s02)s00+p3a0(3(s00)2+ 4s0s000) +p2a0s0v, . . . . From (2.14) it follows

Theorem 2.3. The following relations are valid:

(2.15) ∂y1a

∂s0 =∂y2a

∂s00 =· · ·= ∂yka

∂s(k) =y1a0 = dy0a ds

(2.16) ∂yAa

∂s(B) = A B

∂y(A−1)a

∂s(B−1) =· · ·= µA

B

∂y(A−B)a

∂s

(2.17) ∂p2a

∂s0 = ∂p3a

∂s00 =· · ·= ∂pka

∂s(k−1) =dp1a

ds =p2a0

∂pαa

∂s(β) = (α1) β

∂p(α−1)a

∂s(β−1) =· · ·= µα−1

β

∂p(α−β)a

∂s , (2.18)

k≥A≥B≥0, k≥α≥β+ 11.

(13)

Now we return to the purpose of this examination.

If we take the partial derivatives of F given by (2.5) with respect to s0, s00, . . . , s(k), taking into account (2.6), we get

(2.19) (∂1aF)∂y1a

∂s0 + (∂2aF)∂y2a

∂s0 + (∂3aF)∂y3a

∂s0 +· · ·+ (∂kaF)∂yka

∂s0 + (∂2aF)∂p2a

∂s0 + (∂3aF)∂p3a

∂s0 +· · ·+ (∂kaF)∂pka

∂s0 =F(x, y10, . . . , yk0, p1, . . . , pk0) (∂2aF)∂y2a

∂s00 + (∂3aF)∂y3a

∂s00 +· · ·+ (∂kaF)∂yka

∂s00 + (∂3aF)∂p3a

∂s00 + (∂4aF)∂p4a

∂s00 +· · ·+ (∂kaF)∂pka

∂s00 = 0, . . . , (∂(k−1)aF)∂y(k−1)a

∂s(k−1) + (∂kaF) ∂yka

∂s(k−1)+ (∂kaF) ∂pka

∂s(k−1) = 0, (∂kaF)∂yka

∂s(k) = 0.

On the left-hand side of (2.19) in all equationsF =F(x, y1, . . . , yk, p1, . . . , pk).

If in (2.19) we substitute (2.16) and (2.18) in the form

∂yAa

∂s(B) = µA

B

∂y(A−B)a

∂s =

µA B

y(A−B+1)adt ds

∂pαa

∂s(β) = µα−1

β

∂p(α−β)a

∂s =

µα−1 β

p(α−β+1)adt ds we obtain

[ µ1

1

y1a1a+ µ2

1

y2a2a+· · ·+ µk

1

ykaka](F) + (2.20)

[ µ1

1

p2a2a+ µ2

1

p3a3a+· · ·+ µk−1

1

pkaka](F) =

F(x, y10, . . . , yk0, p01, . . . , p0k)ds

dt =F(x, y1, . . . , yk, p1, . . . , pk) =F {[

µ2 2

y1a2a+ µ3

2

y2a3a+· · ·+ µk

2

y(k−1)aka] +

参照

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