Gen. Math. Notes, Vol. 24, No. 1, September 2014, pp.... ISSN 2219-7184; Copyright © ICSRS Publication, 2014

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Gen. Math. Notes, Vol. 24, No. 1, September 2014, pp. 98-108
ISSN 2219-7184; Copyright © ICSRS Publication, 2014
www.i-csrs.org
Available free online at http://www.geman.in
On Characterization of Inextensible Flows of
Dual Curves According to Dual Darboux
Frame
Zehra Ekinci
Celal Bayar University, Faculty of Arts and Sciences
Department of Mathematics, Muradiye Campus
Muradiye, Manisa, Turkey
E-mail: zehra.ari@cbu.edu.tr
(Received: 30-5-14 / Accepted: 12-7-14)
Abstract
In this paper, we study inextensible flows dual curves according to dual
Darboux frame. Necessary and sufficient conditions for an inelastic dual curve
flow are expressed as a partial differential equation involving the dual geodesic
curvature.
Keywords: Dual Darboux Frame, Inextensible flows.
1
Introduction
Recently, the study of the motion of inelastic curves has an important role. The
time evolution of a curve represented by its corresponding flow. The flow of a
curve is said to be inextensible if, firstly its arc length is preserved and secondly
its intrinsic curvature is preserved. Physically, inextensible curve flows give rise
to motions in which no strain energy is induced. The swinging motion of cord of
fixed length, for example, or of a piece of paper carried by the wind, can be
On Characterization of Inextensible Flows of...
99
described by inextensible curve. Some movement in nature is inspired to examine
flow of curves as snake and elephant's trunk movement. For example, both
Chirikjian and Burdick [5] and Mochiyama et al. [9] study the shape control
hyper-redundant, or snake-like robots. Inextensible curve and surface flows
emerge many problems in computer vision [14] and computer animation [10].
Particularly, inextensible time evolution of curves and surfaces is examined
mathematically. Significant methods of this article developed by Gage and
Hamilton [11], and Grayson [13] for studying the shrinking of closed plane curves
to a circle via the heat equation. In [12] Gage also studies area preserving
evolution of inelastic plane curves. Another related study is that [6], which
considers less restrictive mappings that locally preserve volume only. In [2, 3]
Know et al. study evolution of inelastic plane curves, and inextensible flows of
curves and developable surfaces.
In this paper, we study inextensible flows dual curves according to dual Darboux
frame. Necessary and sufficient conditions for an inelastic dual curve flow are
expressed as a partial differential equation involving the dual geodesic curvature.
2
Preliminaries
Let D = IR × IR = {a = ( a, a ∗ ) : a, a ∗ ∈ IR} be the set of the pairs (a, a ∗ ) . For
a = (a, a ∗ ) , b = (b, b∗ ) ∈ D the following operations are defined on D:
Equality:
Addition:
Multiplication:
a = b ⇔ a = b, a ∗ = b ∗
a + b ⇔ ( a + b, a ∗ + b ∗ )
ab ⇔ (ab, ab∗ + a ∗b)
The element ε = (0,1) ∈ D satisfies the relationships
ε ≠ 0 , ε 2 = 0 , ε 1 = 1ε = ε
(2.1)
Let consider the element a ∈ D of the form a = (a, 0) . Then the mapping
f : D → IR, f (a, 0) = a is a isomorphism. So, we can write a = (a, 0) . By the
multiplication rule we have that
a = (a, a ∗ ) = a + ε a∗
Then a = a + ε a∗ is called dual number and ε is called dual unit. Thus the set of
dual numbers is given by
(2.2)
D = {a = a + ε a ∗ : a, a ∗ ∈ IR, ε 2 = 0}
100
Zehra Ekinci
The set D forms a commutative group under addition. The associative laws hold
for multiplication. Dual numbers are distributive and form a ring over the real
number field [4].
Dual function of dual number presents a mapping of a dual numbers space on
itself. Properties of dual functions were thoroughly investigated by Dimentberg
[4]. He derived the general expression for dual analytic (differentiable) function as
follows
f ( x ) = f ( x + ε x∗ ) = f ( x) + ε x∗ f ′( x) ,
(2.3)
where f ′( x ) is derivative of f ( x ) and x, x∗ ∈ IR . This definition allows us to
write the dual forms of some well-known functions as follows

cos( x ) = cos( x + ε x∗ ) = cos( x) − ε x∗ sin( x),

∗
∗
sin( x ) = sin( x + ε x ) = sin( x) + ε x cos( x),

∗
 x = x + ε x∗ = x + ε x , ( x > 0).
2 x

(2.4)
Let D 3 = D × D × D be the set of all triples of dual numbers, i.e.,
D3 = {aɶ = (a1 , a2 , a3 ) : ai ∈ D, i = 1, 2,3}
(2.5)
Then the set D 3 is called dual space. The elements of D 3 are called dual vectors.
Similar to the dual numbers, a dual vector aɶ may be expressed in the form
aɶ = a + ε a ∗ = (a , a ∗ ) , where a and a ∗ are the vectors of IR 3 . Then for any
vectors aɶ = a + ε a ∗ and bɶ = b + ε b ∗ of D 3 , the scalar product and the vector
product are defined by
(2.6)
aɶ , bɶ = a , b + ε a , b ∗ + a ∗ , b ,
(
and
)
(
)
aɶ × bɶ = a × b + ε a × b * + a ∗ × b ,
(2.7)
Respectively, where a , b and a × b are the inner product and the vector product
of the vectors a and a ∗ in IR 3 , respectively.
The norm of a dual vector aɶ is given by
aɶ = a + ε
a, a∗
a
, (a ≠ 0) .
(2.8)
On Characterization of Inextensible Flows of...
101
A dual vector aɶ with norm 1 + ε 0 is called dual unit vector. The set of dual unit
vectors is given by
Sɶ 2 = {aɶ = ( a1 , a2 , a3 ) ∈ D 3 : aɶ , aɶ = 1 + ε 0} ,
(2.9)
and called dual unit sphere[8,15].
In the Euclidean 3-space IR3 , an oriented line L is determined by a point p ∈ L
and a unit vector a . Then, one can define a ∗ = p × a which is called moment
vector. The value of a ∗ does not depend on the point p , because any other point
q in L can be given by q = p + λ a and then a ∗ = p × a = q × a . Reciprocally,
when such a pair (a , a ∗ ) is given, one recovers the line L as
L = {( a × a ∗ ) + λ a : a , a ∗ ∈ E 3 , λ ∈ IR } , written in parametric equations. The
vectors a and a ∗ are not independent of one another and they satisfy the
following relationships
a , a = 1, a , a ∗ = 0
(2.10)
The components ai , ai∗ (1 ≤ i ≤ 3) of the vectors a and a ∗ are called the
normalized Plucker coordinates of the line L . We see that the dual unit vector
aɶ = a + ε a ∗ corresponds to the line L . This correspondence is known as E. Study
Mapping: There exists a one-to-one correspondence between the vectors of dual
unit sphere Sɶ 2 and the directed lines of the space IR 3 . By the aid of this
correspondence, the properties of the spatial motion of a line can be derived.
Hence, the geometry of ruled surface is represented by the geometry of dual
curves lying on the dual unit sphere Sɶ 2 .
The angle θ = θ + εθ * between two dual unit vectors aɶ , bɶ is called dual angle
and defined by
(2.11)
aɶ , bɶ = cos θ = cos θ − εθ * sin θ .
By considering the E. Study Mapping, the geometric interpretation of dual angle
is that, θ is the real angle between the lines L1 , L2 corresponding to the dual unit
vectors aɶ , bɶ , respectively, and θ * is the shortest distance between those lines [9].
Let now ( xɶ ) be a dual curve represented by the dual vector eɶ (u ) = e (u ) + ε e ∗ (u ) .
The unit vector e draws a curve on the real unit sphere S 2 and is called the (real)
indicatrix of ( xɶ ) . We suppose throughout that it is not a single point. We take the
parameter u as the arc-length parameter s of the real indicatrix and denote the
differentiation with respect to s by primes. Then we have e′, e′ = 1 . The vector
102
Zehra Ekinci
e′ = t is the unit vector parallel to the tangent of the indicatrix. The equation
e ∗ ( s ) = p ( s ) × e ( s ) has infinity of solutions for the function p ( s ) . If we take
po ( s ) as a solution, the set of all solutions is given by p ( s ) = po ( s ) + λ ( s ) e ( s ) ,
where λ is a real scalar function of s . Therefore we have p′, e′ = po′ , e′ + λ .
By taking λ = λo = − po′ , e′ we see that po ( s ) + λo ( s ) e ( s ) = c ( s ) is the unique
solution for p ( s ) with c′, e′ = 0 . Then, the given dual curve ( xɶ ) corresponding
to the ruled surface
ϕe = c ( s ) + ve ( s )
(2.12)
may be represented by
eɶ ( s ) = e + ε c × e
(2.13)
where
e , e = 1 , e′, e′ = 1 , c′, e′ = 0 .
(2.14)
Then we have
eɶ′ = t + ε det(c′, e , t ) = 1 + ε∆
(2.15)
where ∆ = det(c′, e , t ) . The dual arc-length s of the dual curve ( xɶ ) is given by
s
s
s
0
0
0
s = ∫ eɶ′(u ) du = ∫ (1 + ε∆)du = s + ε ∫ ∆du
(2.16)
From (16) we have s′ = 1 + ε∆ . Therefore, the dual unit tangent to the curve eɶ ( s )
is given by
deɶ eɶ′
eɶ′
= =
= tɶ = t + ε (c × t )
(2.17)
ds s′ 1 + ε∆
Introducing the dual unit vector gɶ = eɶ × tɶ = g + ε c × g we have the dual frame
{eɶ, tɶ, gɶ} which is known as dual geodesic trihedron or dual Darboux frame of
ϕe (or (eɶ ) ). Also, it is well known that the real orthonormal frame {e , t , g} is
called the geodesic trihedron of the indicatrix e ( s ) with the derivations
e′ = t , t ′ = γ g − e , g ′ = −γ t
(2.18)
where γ is called the conical curvature [1]. Similar to (2.18), the derivatives of
the vectors of the dual frame {eɶ, tɶ, gɶ } are given by
deɶ ɶ dtɶ
dgɶ
=t,
= γ gɶ − eɶ,
= −γ tɶ
ds
ds
ds
(2.19)
γ = γ + ε (δ − γ∆) , δ = c′, e
(2.20)
where
On Characterization of Inextensible Flows of...
103
and the dual darboux vector of the frame is dɶ = γ eɶ + gɶ . From the definition of ∆
and (2.20) we also have
c′ = δ e + ∆g
(2.21)
The dual curvature of dual curve (ruled surface) eɶ ( s ) is
R=
1
(2.22)
(1 + γ 2 )
The unit vector dɶo with the same sense as the Darboux vector dɶ = γ eɶ + gɶ is
given by
γ
1
dɶo =
eɶ +
gɶ
(2.23)
(1 + γ 2 )
(1 + γ 2 )
Then, the dual angle between dɶo and eɶ satisfies the followings
cos ρ =
γ
(1 + γ 2 )
, sin ρ =
1
(1 + γ 2 )
(2.24)
where ρ is the dual spherical radius of curvature. Hence R = sin ρ , γ = cot ρ [7].
3
Inextensible Flows of Dual Curves according to Dual
Darboux Frame
Throughout this study, we assume that xɶ :[0,1] × [0, ω ] → D3 is a one parameter
family of smooth dual curves in dual space D3 . Let u
parametrization variable, 0 ≤ u ≤ 1 .
be the curve
The arclenght of xɶ is given by
u
s =∫
0
∂xɶ
du = ∫ v du
∂u
0
u
(3.1)
where
∂xɶ
∂xɶ ∂xɶ
=
,
∂u
∂u ∂u
12
.
and we assume that s ∗ is a parameter in terms of s . The operator
terms of u by
∂ 1 ∂
=
∂s v ∂u
(3.2)
∂
is given in
∂s
104
Zehra Ekinci
where
v=
∂xɶ
.
∂u
The arclenght parameter is ds = v du .
Any flow of xɶ can be represented as
∂xɶ
= f1eɶ + f 2tɶ + f 3 gɶ .
∂t
(3.3)
The arclength is given up to a constant by
u
s (u , t ) = ∫ v du .
0
In the dual space the requirement that the curve not be exposed to any elongation
or compression can be expressed by the condition
∂
∂v
s (u , t ) = ∫ du = 0
∂t
∂t
0
u
(3.4)
for u ∈ [0,1] .
Definition 3.1: A dual curve evolution xɶ (u , t ) and
∂xɶ
flow of this curve in D3
∂t
are said to be inextensible if
∂ ∂xɶ
=0.
∂ t ∂u
∂xɶ
be a smooth flow of the dual curve xɶ . The flow is
∂u
inextensible if and only if
Lemma 3.2: Let
∂v ∂f 2
=
+ f1v − γ f3v
∂t ∂u
∂f 2∗
= γ f 3∗v + γ ∗ f3v + γ f3v∗ − f1∗v − f1 v∗ .
∂u
Proof: Suppose that
(3.5)
∂xɶ
be a smooth flow of the curve xɶ . Considering definition
∂u
of xɶ , we have
v2 =
∂xɶ ∂xɶ
,
∂u ∂u
(3.6)
On Characterization of Inextensible Flows of...
105
∂
∂
and
commute since and are independent coordinates. Then, by
∂u
∂t
differentiating of formula (3.6) we get
2v
On the other hand, changing
∂v
∂
=
∂t ∂t
∂xɶ ∂xɶ
.
,
∂u ∂u
∂
∂
and
we
∂u
∂t
v
∂v
∂xɶ ∂  ∂xɶ 
=
,  
∂t
∂u ∂u  ∂ t 
From equation (3.3), we obtain
v
∂v
∂xɶ ∂
=
, ( f1eɶ + f 2tɶ + f3 gɶ ) .
∂t
∂u ∂u
By the formula of dual darboux, we have

 ∂f
 
  ∂f
∂f
∂v
= tɶ,  1 − f 2 v  eɶ +  f1v + 2 − γ f 3v  tɶ +  3 + f 2γ v  gɶ .
∂t
∂u
 ∂u
 
  ∂u

If above equation is separated real and dual part then we have

∂v 
∂f
=  f1v + 2 − γ f 3v 
∂t 
∂u

and
∂v∗ ∂f 2∗
=
− γ f 3∗v − γ ∗ f 3v − γ f 3v∗ + f1∗v + f1 v∗ .
∂t
∂u
respectively. Desired expression is obtained with definition 3.1.
∂xɶ
= f1eɶ + f 2tɶ + f 3 gɶ be a smooth flow of the curve xɶ . The flow
∂u
is inextensible if and only if
∂f 2
= γ f 3 + f1
∂s
(3.7)
∂f 2∗
∗
∗
∗
= γ f 3 + γ f 3 + f1
∂s
∂xɶ
Proof: Let
be extensible. From Eq. (3.4), we have
∂u
Theorem 3.3: Let
106
Zehra Ekinci
∂
∂v
∂f
s (u , t ) = ∫ du = ∫ ( 2 + f1v − γ f 3v )du = 0
∂t
∂t
∂u
0
0
u
u
(3.8)
∀u ∈ [0,1] . Substituting (3.5) in (3.8) and this expression separating dual and real
part complete the proof of the theorem. We suppose that v = 1 and the local
coordinate u corresponds to the curve arc length s . Now we give following
lemma that necessary.
Lemma 3.4:
 
∂f 
∂eɶ  ∂f1
=
− f 2  eɶ +  f 2γ + 3  gɶ ,
∂ t  ∂s
∂s 
 
where ψ =
(3.9)
∂tɶ
= −eɶ + ψ gɶ ,
∂t
(3.10)

∂f 
∂gɶ
= −  f 2γ + 3  eɶ −ψ tɶ ,
∂t
∂s 

(3.11)
∂tɶ
, gɶ .
∂t
Proof: Considering definition of xɶ , we have
∂eɶ
∂ ∂xɶ ∂
=
=
( f1eɶ + f 2tɶ + f 3 gɶ ).
∂ t ∂ t ∂s ∂s
Using the Darboux equations, we get
 
 
∂f 
∂eɶ  ∂f1
∂f
=
− f 2  eɶ +  f1 + 2 − γ f 3  tɶ +  f 2γ + 3  gɶ .
∂ t  ∂s
∂s
∂s 
 
 
Substituting (3.7) in (3.12), we have
 
∂f 
∂eɶ  ∂f1
=
− f 2  eɶ +  f 2γ + 3  gɶ .
∂ t  ∂s
∂s 
 
Now differentiate the dual Darboux frame by t:
(3.12)
On Characterization of Inextensible Flows of...
0=
107
∂
∂eɶ ɶ
∂tɶ
∂tɶ
eɶ, tɶ =
, t + eɶ,
= 1 + eɶ,
,
∂t
∂t
∂t
∂t
∂f
∂
∂eɶ
∂gɶ
∂gɶ
eɶ, gɶ =
, gɶ + eɶ,
= f 2γ + 3 + eɶ,
,
∂t
∂t
∂t
∂s
∂t
∂ ɶ
∂tɶ
∂gɶ
∂gɶ
0=
t , gɶ =
, gɶ + tɶ,
= ψ + tɶ,
.
∂t
∂t
∂t
∂t
0=
Considering
∂tɶ ɶ
∂gɶ
,t =
, gɶ = 0 and from above statement, we obtain
∂t
∂t
∂tɶ
= −eɶ +ψ gɶ ,
∂t

∂f 
∂gɶ
= −  f 2γ + 3  eɶ −ψ tɶ,
∂t
∂s 

where ψ =
∂tɶ
, gɶ .
∂t
The following theorem states the conditions on the dual geodesic curvature for the
dual curve flow xɶ ( s , t ) to be inextensible.
∂xɶ
= f1eɶ + f 2tɶ + f 3 gɶ is inextensible. Then, the following
∂u
system of partial differential equations holds:
Theorem 3.5: Suppose
∂f ∂ψ
∂γ
= f 2γ + 3 +
.
∂t
∂s ∂s
Proof: Using (3.11), we have

∂f3 
∂ ∂gɶ ∂  
=
 −  f 2γ +
 eɶ −ψ tɶ 
∂s ∂ t ∂s  
∂s 

 ∂ ( − f 2γ ) ∂ 2 f 3  
∂f 
=
+ 2  eɶ −  f 2γ + 3  tɶ
∂s  
∂s 
 ∂s
∂ψ ɶ
−
t −ψ (eɶ + γ gɶ ).
∂s
Therefore from dual Darboux frame
∂ ∂gɶ ∂
∂γ ɶ
=
(−γ tɶ ) = −
t − γ (−eɶ + ψ gɶ ).
∂s ∂ t ∂s
∂s
(3.13)
108
Zehra Ekinci
Thus from two equations we have (3.13).
4
Conclusion
Inextensible time evolutions of curves and surfaces have an important role in
computer vision, robotics and physical science. In this paper inextensible flows of
dual curves according to dual darboux frame have given by considering important
role of dual geometry. Since dual geometry have a significant role robotics,
mechanism and dynamics, this study can be shed light on these areas.
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