Mathematics

Determining a Metric by Boundary

Single-Time Flow Energy

ARIANA PITEAFaculty of Applied Sciences,Department of Mathematics I,

University Politehnica of Bucharest,Splaiul Independentei 313

RO-060042, Bucharest, [email protected]

Abstract: Our theory of determining a tensor by single-time flow energy is similar to those developed bySharafutdinov. Section 1 refines the theory of potential curves determined by a flow and a Riemannianmetric. Section 2 defines the boundary energy of a potential curve and proves that the problem ofdetermining a metric from a single-time flow boundary energy cannot have a unique solution. Section 3linearizes the above-mentioned problem and defines the notion of single-ray transform.

Key Words: potential curves, least squares Lagrangian, single-ray transform, boundary energy.

1 Potential curves

Let (M, g) be a compact Riemannian mani-fold with the boundary M and of dimension n.We consider x = (xi) local coordinates on themanifold (M, g), (Gjk,`) and (Gijk) its Christof-fel symbols of the first and of the second type,respectively.

Let : [0, 1] M , (t) = x, x(t) =(x1(t), . . . , xn(t)) be a C-curve. We want to ap-

proximate the velocitydx

dtof components

dxi

dtby

a C-distinguished vector field X of componentsXi(x), in the sense of least squares. For that we

build the flowdxi

dt(t) = Xi(x(t)), 1, n, x(0) = p,

x(1) = q, where p and q are two points from theboundary M of the manifold M , and the leastsquares Lagrangian (flow energy density),

L(t, x(t), x(t)) =12gij(x(t))[xi(t)Xi(x(t))]

[xj(t)Xj(x(t)],

where xi(t) =dxi

dt(t).

The geometric dynamics (ODEs or PDEs) isa Lagrangian dynamics (ODEs or PDEs) deter-mined by a least squares Lagrangian attached toa (single-time or multi-time) flow and a pair ofRiemannian metrics, one in the source space andother in the target space.

Let us look for the Euler-Lagrange prolonga-tion of the flow obtained as Euler-Lagrange ODEsproduced by L. The extremals of L are called po-tential curves. These curves are geodesics [4].

Theorem 1.1 The extremals of L are de-scribed by the ODEs:

dtxi = gi`gkj(`Xk)Xj + F ij xj , i = 1, n

x(0) = p, x(1) = q,

where:

dtxi = xi +Gijk x

j xk, i = 1, n,

F ij = jXi gi`gkj`Xk, i, j = 1, n; (1)

jXi = Xi

xj+GijkX

k, i, j = 1, n. (2)

Proof. We computeL

xk=

12gijxk

xi xj gijxk

xiXj +12gijxk

XiXj

gij xj Xi

xk+ gij

Xi

xkXj ;

L

xk= gikxi gikXi;

d

dt

(L

xk

)=

gikx`

x`xi + gikxi gikx`

xiX`

gik(X i

t+

X i

x`x`).

WSEAS TRANSACTIONS on MATHEMATICS Ariana Pitea

ISSN: 1109-276940

Issue 1, Volume 7, January 2008

We take into account formula (1) and

gijxk

= G`kig`j +G`kjg`i, i, j, k = 1, n.

If we replace these relations in Euler-Lagrangeequations,

L

xk d

dt

(L

xk

)= 0, k = 1, n,

we find

gkjxj

dt= gij(kXi)Xj + gkj(`Xj)x`

gij xikXj , k = 1, n.

Transvecting by g`k and using formula (2), weobtain

xi

dt= gikg`j(kX`)Xj + F ij xj , i = 1, n

Definition 1.1 The map C([0, 1],M),(t) = x, which verifies the ODEs from the pre-vious theorem is called potential curve associatedto the d-vector field X.

Definition 1.2 The Riemannian metric g iscalled simple if there is a unique potential curvex: [0, 1]M , x(0) = p, x(1) = q, p, q M .

2 Determining a metric byboundary flow energy

Starting from the boundary flow energy, westudy the recovering of a tensor from the centeredmoments which determine the metric g.

Let g be a simple metric and x: [0, 1]M thecorresponding potential curve, x(0) = p, x(1) = q,p, q M .

Definition 2.1 Let p and q be two points onthe boundary M of the manifoldM .The functionEg: M M R, (p, q) 7 Eg(p, q),

Eg(p, q) =12

10gij(x(t))[xi(t)Xi(x(t))][xj(t)

Xj(x(t))] dt

is called the boundary energy of the flow along thepotential curve x: [0, 1] M ,x(0) = p, x(1) = q,p, q M .

Open problem 1. Given an energy E, isthere a metric g that realizes this energy? Howcan these metrics be found?

Let us show that the existence problem of themetrics with the property that E: MMRrepresents the boundary energy cannot have aunique solution.

Let :M M be a diffeomorphism with theproperty |M = id. This diffeomorphism trans-forms the simple metric g0 into a simple metricg1 = g0, because we have

g1(x)(, ) = g0((dx), (dx))(x),

where dx:TxM T(x)M is the differential of.

It can be noticed that

X i=

xi

xiXi,

whereXirepresents the distinguished vector field

X with respect to x = (x).The metrics g0 and g1 give different families of

potential curves with the same boundary energyE.

Open problem 2. The problem of recover-ing a metric by boundary energy can be changedin the following way. Let g0 and g1 be simplemetrics on M . Does the equality Eg0 = Eg1 im-pliy the existence of a diffeomorphism :M M ,|M = id and g1 = g0?

3 Linearization of the problemof finding a metric from theboundary energy

Let us linearize the open problem 1. Let(g ) be a family of simple metrics which dependssmoothly on parameter (, ), > 0. Letp and q be two points of the border M of themanifold M and a = E(p, q), where E: M M R is the given boundary energy. Considerx : [0, a] M the potential curve correspondingto the pair (p, q), x = (x,i), x,i(t) = xi(t, ),i = 1, n. We denote g = (gij).

The energy of deformation x is

Eg (p, q) =12

a0gij(x

(t))[xi(t, )Xi(x(t, ), )]

[xj(t, )Xj(x(t, ), )] dt.


ISSN: 1109-276941


Differentiating with respect to we obtain

Eg (p, q)=

a0

{gij

(x (t))[xi(t, )Xi(x (t), )]

[xj(t, )Xj(x (t), ) + gij

xk(x (t))

[1

2xi(t, )xj(t, )]

xi(t, )Xj(x (t), ) + 12Xi(x (t), )

Xj(x (t), )

]xk

(t, ) + gij(x

(t))

[xi

(t, )xj(t, )

xi

(t, )Xj(x (t), ) xi(t, )X

j

(x (t), )

xi(t, )Xj

xk(x (t), )

xk

(t, ) +Xi(x (t), )

Xj

(x (t), ) +Xi(x (t), )

Xj

xk(x (t), )

xk

(t, )

]}dt =

a0

gij

(x(t, ))

[xi(t, )xj(t, ))

2xi(t, )Xj(x (t), +Xi(x (t), )Xj(x (t), )]dt

+

a0

{gijxk

(x (t))

[1

2xi(t, )xj(t, )

xi(t, )Xj(x (t), ) + 12Xi(x (t), )Xj(x (t), )

]gij(x (t))[xi(t, )Xi(x (t), )]

Xj

xk(x (t), )

}xk

(t, ) dt+

a0

gij(x (t))[xj(t, )

Xj(x (t), )]xi

(t, ) dt

a0

gij(x (t))[xi(t, )

Xi(x (t), )]Xj

xk(x (t), )

}xk

(t, ) dt+

a0

gij(x (t))

[xi(t, )Xi(x (t), )]Xj

(x (t), ) dt. (3)

Integrating by parts, then considering = 0and using the fact that

xi

a0

= 0,

the third integral from relation (3) becomes

I3= a0

d

dt

=0

gij(x (t))[xj(t, )Xj(x (t),)]x

i

(t, 0) dt

= a0

{g0ijx`

(x0(t))x`(t, 0)[xj(t,0)Xj(x0(t), 0)]

+g0ij(x0(t))

[xj(t, 0)X

j

x`(x0(t), 0)

]x`(t,0)

}xi

(t,0) dt.

Relation (3) becomes, at = 0:

=0

Eg (p, q) =

a0

fij(x0(t))[xi(t, 0)xj(t, 0)

2xi(t, 0)Xj(x0(t), 0) +Xi(x0(t), 0)Xj(x0(t), 0)] dt

+

a0

{g0ijxk

(x0(t))

[1

2xi(t, 0)xj(t, 0)

xi(t, 0)Xj(x0(t), 0) + 12Xi(x0(t), 0)Xj(t, x0(t), 0)

]

g0ij(x0(t))[xi(t, 0)Xi(x0(t), 0)]Xj

xk(x0(t), 0)

}xk

(t, 0) dt+

{ a0

g0ijx`

(x0(t))x`(t, 0)[xj(t, 0)

Xj(x0(t), 0)]+g0ij(x0(t))[xj(t, 0) X

j

x`(x0(t), 0)

x`(t, 0)

]xi

(t, 0) dt

} a0

g0ij(x0(t))[xi(t, 0)

Xi(x0(t), 0)]Xj

(x0(t), 0) dt, (4)

where fij =12

=0

gij .

Making the sum of the second and third inte-grals of (4), we obtain

I2 + I3 = 2g0kj

xi(x0(t))xi(t, 0)xj(t, 0) +

g0ijxk

(t, 0)

[2xi(t, 0)Xj(x0(t), 0) +Xi(x0(t), 0)Xj(x0(t), 0)]

2g0ij(x0(t))[xi(t, 0)

Xj

xk(x0(t), 0) X

i

xk(x0(t), 0)

Xj(x0(t), 0)

]+ 2

g0kjx`

(x0(t))x`(t, 0)Xj(x0(t), 0)

2g0kj(x0(t))[xj(t, 0) X

j

x`(x0(t), 0)x`(t, 0)

]}xk

(t, 0)dt

=1

2

a0

{2g0`k(x0(t))G`ij(x0(t), 0)xi(t, 0)xj(t, 0)

+[Gki,j(x0(t), 0)+Gkj,i(x

0(t), 0)][2xi(t, 0)Xj(x0(t), 0)

+Xi(x0(t), 0)Xj(t, x0(t), 0)]2g0ij(x0(t))[xi(t, 0)

Xj

xk(x0(t), 0) X

i

xk(x0(t), 0)Xj(x0(t), 0)

]+2[Gk`,j(x

0(t), 0) +G`j,k(x0(t), 0)]x`(t, 0)Xj(x0(t), 0)

2g0kj(x0(t))[xj(t, 0) X

j

t(x0(t), 0) X

j

x`(x0(t), 0)


ISSN: 1109-276942


x`(t, 0)

]}xk

(t, 0)dt=

1

2

a0

{2g0`k(x0(t))G`ij(x0(t), 0)

xi(t, 0)xj(t, 0) 2g0iq(x0(t))xi(t, 0)kXq(x0(t), 0)

+2g0jq(x0(t))

[Gqk`(x

0(t), 0)X`(x0(t), 0)Xj(x0(t), 0)

+Xq

xk(x0(t), 0)Xj(x0(t), 0)

]+ 2g0qk(x

0(t))

Gqj`(x0(t), 0)x`Xj(x0(t), 0) + 2g0kj(x

0(t))

Xj

x`(x0(t), 0)x`(t, 0)

}xk

(t, 0) dt

=

a0

{ g0jk(x0(t))[xj(t, 0) g0iq(x0(t))

xi(t, 0)kXq(x0(t), 0) + g0qk(x0(t))Gqj`(x0(t), 0)

x`(t, 0)Xj(x0(t), 0) + g0kj(x0(t))

Xj

x`(x0(t), 0)

x`(t, 0)

}xk

(t, 0) dt =

a0

[ g0`j(x0(t))

(kX`(x0(t), 0)Xj(x0(t), 0)g0ki(x0(t))F ij (x0(t), 0)

xj(t, 0) g0iq(x0(t))xi(t, 0)kXq(x0(t), 0)

+g0jq(t)kXq(x0(t), 0)Xj(x0(t), 0)

+g0qk(x0(t))Gqj`(x

0(t), 0)x`(t, 0)

Xj(x0(t), 0)

]xk

(t, 0) dt =

a0

[ g0ki(x0(t))

F ij (x0(t), 0)xj(t, 0) g0iq(x0(t))xi(t, 0)

(kXq)(x0(t), 0) + g0qk(x0(t))Gqj`(x0(t), 0)x`(t, 0)

Xj(x0(t), 0) + g0kj(x0(t))

Xj

x`(x0(t), 0)

x`(t, 0)

]xk

(t, 0) dt =

a0

xj(t, 0)

{ g0ki(x0(t))

[(jXi)(x0(t), 0) giq0 (x0(t))g0`j(x0(t))

(qX`(x0(t), 0))] g0jq(x0(t))(kXq)(x0(t), 0)

+g0qk(x0(t))Gq`j(x

0(t), 0)X`(x0(t), 0)

+g0k`(x0(t))

X`

xj(x0(t), 0)

}xk

(t, 0) dt

=

a0

xj(t, 0)

[ g0ki(x0(t))(jXi)(x0(t), 0)

+g0`j(x0(t))(kX`)(x0(t), 0) g0jq(x0(t))

(kXq)(x0(t), 0)+g0qk(x0(t))Gq`j(x0(t))X`(x0(t), 0)

+g0k`(x0(t))

X`

xj(x0(t), 0)

]xk

(t, 0) dt = 0.

We have obtained that

=0

Eg (p, q)=

a0

fij(x0(t))[xi(t, 0)Xi(x0(t), 0)]

[xj(t, 0)(Xj(x0(t), 0)] dt a0

g0ij(x0(t))[xi(t, 0)

Xi(x0(t), 0)]Xj

(x0(t), 0) dt,

where fij(x0(t)) =12

=0

gij .

Considering thatXj

(x0(t), 0) = 0, and us-

ing the functional

If (x0) =

a0

fij(x0(t))[xi(t, 0)Xi(x0(t), 0)]

[xj(t, 0)Xj(x0(t), 0)]dt

the previous relation becomes

=0

Eg (p, q) = If (x0), (5)

where x0 is the potential curve corresponding tothe points p, q from the border M of the mani-fold M .

The function If is called the single-ray trans-form of the tensor field (fij).

The existence of solutions of the open problem1 for the family (g ) implies the existence of a oneparameter group of diffeomorphisms (x) suchthat g = ( )g0. Explicitly

gij = (g0k` )

xk

xix`

xj, (6)

where (x)=(1(x, ), . . . , n(x, )), x= (x).

Theorem 3.1 Let vk(x)=

=0

(xk)(x, ),

k = 1, n, vi = g0ijvj and vi;j be the covariant

derivative of (vi). Then the following relationholds

fij =12(vi;j + vj;i), i, j = 1, n. (7)

Proof. Differentiating the relation (6) withrespect to and then considering = 0, we find

2fij=

=0

gij =g0k`xm

vmxk

xix`

xj

+g0k`

xi

(

=0

xk)x`xj


ISSN: 1109-276943


+g0k`xk

xi

xj

(

=0

x`)

=g0k`xm

vmki `j + g

0k`

vk

xi`j + g

0k`

ki

v`

xj

=g0ijxq

vq + g0jqvq

xi+ g0iq

vq

xj.

On the other hand,

vi;j + vj;i =vi

xjGmij vm +

vjxi

Gmjivm =g0imxj

vm

+g0imvm

xj+

g0jmxi

vm + g0jmvm

xi

gmq0(g0jqxi

+g0iqxj

g0ij

xq

)g0msv

s

=g0ijxq

vq + g0jqvq

xi+ g0iq

vq

xj,

and relation (7) is provedTherefore, the following generalization of the

open problem 1 appears. To what extent do theintegrals

If (x) = a0fij(x(t))[xi(t)Xi(x(t))]

[xj(t)Xj(x(t))]dt

determine the tensor (fij)?

References:

[1] M. Neagu: Riemann-Lagrange Geometry on 1-JetSpaces, Matrix Rom, Bucharest, 2005.

[2] A. Sharafutdinov: Integral Geometry of TensorFields, VSPBV Utrecht, 1994.

[3] C. Udriste: Convex Functions and OptimizationMethods on Riemannian Manifolds, Kluwer Aca-demic Publishers, 1994.

[4] C. Udriste, M. Ferrara and D. Opris: EconomicGeometry Dynamics, Geometry Balkan Press,2004.

[5] C. Udriste, Ariana Pitea and J. Mihaila: Deter-mination of Metrics by Boundary Energy, BalkanJ. Geom. Appl., vol. 11, no. 1 (2006), pp. 131-143.

[6] C. Udriste, Ariana Pitea and J. Mihaila: KineticPDEs System on the First Order Jet Bundle,Proc. 4-th Int. Coll. Math. Engng&Num. Phys.

[7] C. Udriste and M. Postolache: Atlas of MagneticGeometric Dynamics, Geometry Balkan Press,2001.

[8] H. Urakawa: Calculus of Variations and Har-monic Maps, Shokabo Publishing Co. Ltd.,Tokyo, 1990.


ISSN: 1109-276944


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