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Curves of Constant Breadth According to Darboux Frame

2015

In this paper, we investigate constant breadth curves on a surface according to Darboux frame and give some characterizations of these curves.

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 7 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) References [1] Akdoğan Z., Maden A., Some characterization of curves of constant breadth in E n space, Turk. J. Math., 25, 433-444, 2001. 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