Curves of Constant Breadth According to Darboux
Frame
arXiv:1510.08712v1 [math.GM] 28 Oct 2015
Bülent Altunkaya 1 , Ferdag KAHRAMAN AKSOYAK 2
1
2
Ahi Evran University, Division of Elementary Mathematics Education Kırşehir, TURKEY
Ahi Evran University, Division of Elementary Mathematics Education Kırşehir, TURKEY
Abstract
In this paper, we investigate constant breadth curves on a surface according to
Darboux frame and give some characterizations of these curves.
Key words and Phrases: Darboux frame, constant breadth curve, Euclidean space.
2010 Mathematics Subject Classification: 53B25 ; 53C40 .
1 Introduction
Since the first introduction of constant breadth curves in the plane by L. Euler in 1778
[4], many researchers focused on this subject and found out a lot of properties about
constant breadth curves in the plane [11], [3], [15]. Fujiwara has introduced constant
breadth curves, by taking a closed curve whose normal plane at a point P has only one
more point Q in common with the curve and for which the distance d (P, Q) is constant.
After the development of cam design, researchers have shown strong interest to this
subject again and many interesting properties have been discovered. For example Köse
has defined a new concept called space is a pair of curve of constant breadth in [9], a
pair of unit speed space curves of class C3 with non-vanishing curvature and torsion
in E 3 , which have parallel tangents in opposite directions at corresponding points, and
the distance between these points is always constant by using the Frenet frame.
The characterizations of Köse’s paper on constant breadth curves in the space has
led us to investigate this topic according to Darboux frame on a surface.
2 Basic Concepts
Now, we introduce some basic concepts about our study. Let M be an oriented surface
and β be a unit speed curve of class C3 on M. As we know, β has a natural frame called
Frenet frame {T, N, B} with properties below:
T′ =
N′ =
B′ =
κN,
−κT + τ B
−τ N
(1)
where κ is the curvature, τ is the torsion, T is the unit tangent vector field, N is the
principal normal vector field and B is the binormal vector field of the curve β .
0 E-mail: 1 bulent
[email protected] (B. Altunkaya ); 2
[email protected] (F. Kahraman
Aksoyak)
1
Let us take the unit tangent vector field of the curve β and the unit normal vector
field n of the surface M. If we define unit vector field g as g = (n ◦ β )× T where × cross
product. We will have a new frame called Darboux frame {T, g, n ◦ β }. The relations
between these two frames can be given as follows:
T
T
1
0
0
g = 0 cos α
sin α N
(2)
0 − sin α cos α
n◦β
B
where α (s) is the angle between the vector fields n ◦ β and B. If we take the derivatives
of T, g, n with respect to s, we will have
T′
0
kg kn
T
= −kg
g′
0
tg g
(3)
−kn −tg 0
n◦β
(n ◦ β )′
where kg , kn and tg are called the geodesic curvature, the normal curvature and the
geodesic torsion respectively. Then, we will have following relations
kg
= κ cos α
kn
tg
= κ sin α
= τ − α′
(4)
In the differential geometry of surfaces, for a curve β (s) lying on a surface, there are
following cases:
i) β is a geodesic curve if and only if kg = 0.
ii) β is an asymptotic line if and only if kn = 0.
iii) β is a principal line if and only if tg = 0.
3 Curves of Constant Breadth According to Darboux
Frame
Let β (s) and β ∗ (s∗ ) be a pair of unit speed curves of class C3 with non-vanishing
curvature and torsion in E 3 which have parallel tangents in opposite directions at corresponding points and the distance between these points is always constant. We will
call (β ∗ , β ) as curve pair of constant breath.
If β lies on a surface, it has Darboux frame in addition to Frenet frame with properties (1), (2), (3) and (4). So we may write for β ∗
β ∗ (s∗ ) = β (s) + m1 (s)T (s) + m2 (s)g(s) + m3 (s) (n ◦ β )(s)
If we differentiate this equation with respect to s and use (3), we will have
(β ∗ )′
=
d β ∗ ds∗
= 1 + m′1 − m3 kn − m2 kg T + m1 kg + m′2 − m3tg g
∗
ds ds
+ m1 kn + m2tg + m′3 (n ◦ β )
2
(5)
and
dβ ∗
d β ∗ ds∗
ds∗
= ∗
= T∗
ds
ds ds
ds
As we know hT, T ∗ i = −1. Then
−
ds∗
= 1 + m′1 − m3 kn − m2 kg .
ds
m′1
= m2 k g + m3 k n − 1 −
So we find from (5),
m′2
m′3
ds∗
ds
(6)
= m3tg − m1 kg
= −m1 kn − m2tg .
Let us denote the angle between the tangents at the points β (s) and β (s + △s) with
△θ . If we denote the vector T (s + △s) − T (s) with △T, we know lim△s→0 △T
△s =
θ
dθ
lim△s→0 △
△s = ds = κ. We called the angle of contingency to the angle △θ [15]. Let
us denote the differentiation with respect to θ with ” · ”. By using the equation ddsθ = κ,
we can writre (6) as follows:
ṁ1
ṁ2
ṁ3
ρ (m2 kg + m3 kn ) − f (θ )
= ρ (m3tg − m1 kg )
= ρ (−m1 kn − m2tg ) .
=
(7)
where ρ = κ1 , ρ ∗ = κ1∗ and ρ + ρ ∗ = f (θ ).
Now we investigate curves of constant breadth according to Darboux frame for
some special cases:
3.1 Case (For geodesic curves)
Let β be non straight line geodesic curve on a surface. Then kg = κ cos α = 0 and
κ 6= 0, we get cos α = 0. So it implies that kn = κ, tg = τ . By using (7), we have
following differential equation system
ṁ1
ṁ2
= m3 − f (θ )
= m3 ϕ
ṁ3
= −m1 − m2 ϕ
(8)
where ϕ = κτ . By using (8), we obtain a differential equation as follows:
dϕ 1
...
m1 + f¨ −
m̈1 + m1 + f˙ + 1 + ϕ 2 ṁ1 + ϕ 2 f = 0
dθ ϕ
We assume that (β ∗ , β )is a curve pair of constant breadth , then
kβ ∗ − β k2 = m21 + m22 + m23 = constant
3
(9)
which imlplies that
m1 ṁ1 + m2ṁ2 + m3 ṁ3 = 0
(10)
By combining (8) and (10) then we get
m1 f (θ ) = 0
3.1.1 Case f (θ ) = 0.
We assume that f (θ ) = 0. By using (9), we get
dϕ 1
...
m1 −
(m̈1 + m1) + 1 + ϕ 2 ṁ1 = 0
dθ ϕ
(11)
If β is a helix curve then ϕ = ϕ0 =constant. From (11), we have
...
m1 + 1 + ϕ02 ṁ1 = 0
whose solution is
1
m1 = q
c1 sin
1 + ϕ02
1 + ϕ02 θ − c2 cos
1 + ϕ02 θ
So we can find as m2 = − ϕ10 (m1 + m̈1 ) and m3 = ṁ1 .
3.1.2 Case m1 = 0.
We assume that m1 = 0, then by using (9), we get
dϕ 1 ˙
f¨ −
f + ϕ2 f = 0
dθ ϕ
(12)
If β is a helix curve, then ϕ = ϕ0 =constant. From (12) we obtain
f¨ + ϕ02 f = 0
whose solution is
f (θ ) = c1 cos (ϕ0 θ ) + c2 sin (ϕ0 θ )
Since m1 = 0 it implies that
m3 = f (θ )
m2 = −
ṁ3
ϕ0
Theorem 1. Let β be a geodesic curve and a helix curve. Let (β , β ∗ ) be a pair of
constant breadth curve. In that case β ∗ can be expressed as one of the following cases:
i)
β ∗ (s∗ ) = β (s) + m1 (s)T (s) −
1
(m1 (s) + m̈1 (s)) g(s) + ṁ1 (s) (n ◦ β )(s)
ϕ0
4
where m1 = √
1
1+ϕ02
c1 sin
ii)
1 + ϕ02 θ − c2 cos
β ∗ (s∗ ) = β (s) −
1 + ϕ02 θ .
f˙(θ )
g(s) + f (θ ) (n ◦ β )(s)
ϕ0
where f (θ ) = c1 cos(ϕ0 θ ) + c2 sin (ϕ0 θ ) .
3.2 Case (For asymptotic lines)
Let β be non straight line asymptotic line on a surface. Then kn = κ sin α = 0 and
κ 6= 0, we have sin α = 0. So we get kg = ε κ, tg = τ , where ε = ±1. By using (7), we
have following differential equation system
ṁ1
ṁ2
ṁ3
= ε m2 − f (θ )
= m3 ϕ − ε m1
(13)
= −m2 ϕ
where ϕ = κτ . By using (13), we obtain a differential equation as follows:
dϕ 1
...
m1 + f¨ −
m̈1 + m1 + f˙ + 1 + ϕ 2 ṁ1 + ϕ 2 f = 0
dθ ϕ
(14)
We assume that (β ∗ , β ) is a curve pair of constant breadth then
kβ ∗ − β k2 = m21 + m22 + m23 = constant
which imlplies that
m1 ṁ1 + m2ṁ2 + m3 ṁ3 = 0
(15)
By combining (13) and (15) then we get
m1 f (θ ) = 0
3.2.1 Case f (θ ) = 0
We assume that f (θ ) = 0. By using (14), we get
dϕ 1
...
m1 −
(m̈1 + m1) + 1 + ϕ 2 ṁ1 = 0
dθ ϕ
If β is a helix curve then ϕ = ϕ0 =constant. From (16), we have
...
m1 + 1 + ϕ02 ṁ1 = 0
whose solution is
1
m1 = q
c1 sin
1 + ϕ02
1 + ϕ02 θ − c2 cos
So we can find as m2 = ε ṁ1 and m3 = ε ϕ10 (m1 + m̈1 ).
5
1 + ϕ02 θ
(16)
3.2.2 Case m1 = 0
We assume that m1 = 0.Then by using (14), we get
dϕ 1 ˙
f¨ −
f + ϕ2 f = 0
dθ ϕ
(17)
If β is a helix curve, then ϕ = ϕ0 =constant. From (17) we obtain
f¨ + ϕ02 f = 0
whose solution is
f (θ ) = c1 cos (ϕ0 θ ) + c2 sin (ϕ0 θ )
Since m1 = 0 it implies that
m2 = ε f (θ )
m3 =
ṁ2
ϕ0
where θ = κds.
R
Theorem 2. Let β be an asymptotic line and a helix curve. Let (β , β ∗ ) be a pair of
constant breadth curve. In that case β ∗ can be expressed as one of the following cases:
i)
ε
(m1 (s) + m̈1 (s)) (n ◦ β )(s)
ϕ0
1 + ϕ02 θ − c2 cos 1 + ϕ02 θ .
β ∗ (s∗ ) = β (s) + m1 (s)T (s) + ε ṁ1 (s) g(s) −
where m1 = √
1
1+ϕ02
c1 sin
ii)
β ∗ (s∗ ) = β (s) + ε f (θ )g(s) + ε
f˙(θ )
(n ◦ β )(s)
ϕ0
where f (θ ) = c1 cos(ϕ0 θ ) + c2 sin (ϕ0 θ ) .
3.3 Case (For principal line)
We assume that β is a principal line. Then we have tg = 0 and it implies that τ = α ′ .
By using (7), we get
ṁ1
ṁ2
ṁ3
= m2 cos α + m3 sin α − f (θ )
(18)
= −m1 cos α
= −m1 sin α
By using (18), we obtain following differential equation
Z
Z
...
(m1 + ṁ1 ) + (ṁ1 + f ) α̇ 2 − sin α m1 cos α d θ − cos α m1 sin α d θ α̈ + f¨ = 0
(19)
6
since tg = 0 we obtain α̇ =
breadth. In that case
τ
κ.
We assume that (β ∗ , β ) is a pair of curve is constant
kβ ∗ − β k2 = m21 + m22 + m23 = constant
which imlplies that
m1 ṁ1 + m2ṁ2 + m3 ṁ3 = 0
(20)
By combining (18) and (20) then we get
m1 f (θ ) = 0
3.3.1 Case f (θ ) = 0
We assume that f (θ ) = 0. By using (19), we get
Z
Z
...
2
(m1 + ṁ1 ) + ṁ1 α̇ − sin α m1 cos α d θ − cos α m1 sin α d θ α̈ = 0
If β is a helix curve then α̇ =
Then we get
m1 (s) = √
1
1 + α̇ 2
τ
κ
(21)
=constant. From (21), we have
...
m1 + 1 + α̇ 2 ṁ1 = 0
c1 sin
1 + α̇ 2 θ − c2 cos
1 + α̇ 2 θ
By using (18) we obtain
ṁ2
ṁ3
=
=
−m1 cos α
−m1 sin α
where α = τ ds.
R
Theorem 3. Let β be a principal line and a helix curve. Let (β , β ∗ ) be a pair of
constant breadth curve such that hβ ∗ , T i = m1 6= 0. In that case β ∗ can be expressed as:
β ∗ = β + m1 (s)T (s) − m1 (s) cos α g(s) − m1 sin α (n ◦ β )(s)
where m1 (s) = √ 1 2 c1 sin 1 + α̇ 2 θ − c2 cos 1 + α̇ 2 θ .
1+α̇
3.3.2 Case m1 = 0
If m1 = 0, then from (19)
f¨ + α̇ 2 f = 0
(22)
where α̇ = κτ .On the other hand since m1 = 0 from (18) we have m2 = c2 =constant,
m3 = c3 =constant and
(23)
f = c2 cos α + c3 sin α
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By combining (22) and (23)
α̈ (−c2 sin α + c3 cos α ) = 0
In that case, if α̈ = 0 then we obtain that α̇ = κτ =constant. β becomes a helix curve.
If −c2 sin α + c3 cos α = 0 then we have α =constant. This means that β is a planar
curve.
Theorem 4. Let β be a principal line. Let (β , β ∗ ) be a pair of constant breadth
curve such that hβ ∗ , T i = m1 = 0. In that case β is a helix curve or a planar curve and
β ∗ can be expressed as:
β ∗ = β + c2 g(s) + c3 (n ◦ β )(s)
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