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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
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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.
WSEAS TRANSACTIONS on MATHEMATICS Ariana Pitea
ISSN: 1109-276941
Issue 1, Volume 7, January 2008
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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)
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ISSN: 1109-276942
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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
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ISSN: 1109-276943
Issue 1, Volume 7, January 2008
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+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.
WSEAS TRANSACTIONS on MATHEMATICS Ariana Pitea
ISSN: 1109-276944
Issue 1, Volume 7, January 2008