Differential Equations for a Space Curve According to the Unit Darboux Vector

Turk. J. Math. Comput. Sci. 9(2018) 91–97 c MatDer http://dergipark.gov.tr/tjmcs http://tjmcs.matder.org.tr MATDER Differential Equations for a Space Curve According to The Unit Darboux Vector Osman Çakıra , Süleyman Şenyurta,∗ a Department of Mathematics, Faculty of Arts and Sciences, Ordu University, 52200, Ordu, Turkey. Received: 04-09-2018 • Accepted: 09-10-2018 Abstract. In this work, the differential equation of a differentiable curve is expressed, by making use of Laplace and normal Laplace operators, as a linear combination of the unit Darboux vector defined as C = sinϕT + cosϕB of that curve. Later, the necessary and sufficient conditions are given for the space curves to be a 1-type Darboux vector. 2010 AMS Classification: 14H45, 14H50, 53A04. Keywords: Darboux vector, Laplacian operator, helix, space curve, differential equation. 1. Introduction It is really important to find a relation between a special curve and its curvatures in differential geometry. One of such special curves of this kinds is an helix. It is well-known that the necessary and sufficient condition for a curve to be an helix, in the Euclidean 3-space E 3 , is that the ratio of the curvature to the torsion of the given curve must be constant [9]. So many researchers have studied on helices and there are lots of papers focusing exclusively on helices. There have been so many studies in literature, to cite some examples, Chen and Ishikawa classified the biharmonic curves [3, 5]. Later Kocayigit and Hacisalihoglu have studied the space curves and biharmonic curves in the Euclidean 3-space E 3 and Minkowski 3-space E13 [7, 8]. Also Arslan and et al. [2] have given some characterizations of 1-type Darboux vector by using Laplacian and normal Laplacian operators. In this paper, by taking Fenchel’s work [4] into account, the differential equation of a space curve, in the Euclidean 3-space, is given first according to the unit Darboux vector and then according to the normal connexion. In the case of helix of the curve, the differential equation obtained from Laplace and normal Laplace operators, is also given. *Corresponding Author Email addresses: (O. Çakır), (S. Şenyurt) Differential Equations for a Space Curve According to The Unit Darboux Vector 92 Figure 1. Darboux Vector 2. Preliminaries Let α : I → E , α(s) = α1 (s), α2 (s), α3 (s) be a differentiable curve with a unit speed. The Frenet frame of this curve is given as 00 α (s) 0 T (s) = α (s), N(s) = , B(s) = T (s) ∧ N(s). 00 k α (s) k If we denote the curvature of the curve α by κ(s) and the torsion by τ(s) then we have 3
0 00 000 hα (s) ∧ α (s), α (s)i . k α0 ∧ α00 k2 Frenet vectors T, N, B and their derivative vectors satisfy the following Frenet-Serret formulae along the curve α, 00 κ(s) =k α (s) k , ∇α0 T (s) = τ(s) = κ(s)N(s) , ∇α0 N(s) = −κ(s)T (s) + τ(s)B(s) , ∇α0 B(s) = −τ(s)N(s) (2.1) d and s is the arc length parameter of the curve α [6]. The where ∇ is the Levi-Civita connection given by ∇α0 = ds vector fields T , N , B are called unit tangent vector field, principle normal vector field and binormal vector field of α respectively. The Frenet formulae given in (2.1) may be interpreted as follows: If a moving point traverses the curve in such a way that s is the time parameter, then the moving frame {T , N , B} moves according to equations (2.1). This motion contains, apart from an instantaneous translation, instantaneous rotation with angular velocity vector given by the Darboux vector W = τT + κB, [1]. So the unit Darboux vector is defined as : τ κ W= √ T+ √ B. 2 2 2 κ +τ κ + τ2 If the angle between the Darboux vector W, whose direction is that of instantaneous axis of the rotation and the binormal vector B, is ϕ , then the unit Darboux vector can be given as C = sinϕT + cosϕB , sinϕ = τ κ , cosϕ = kWk kWk (2.2) O. Çakır, S. Şenyurt, Turk. J. Math. Comput. Sci., 9(2018), 91–97 93 citefenchel-1951. Let α : I → E 3 be a differentiable curve then the Laplacian operator of α ,the normal connection of α and the normal Laplacian operator of α are defined as [3, 5] ∆ = −∇2α0 = −∇α0 ∇α0 , → − → − → − → − → − → − ∇⊥α0 ξ = ∇α0 ξ − h∇α0 ξ , T i T , where ∀ ξ ∈ χ(α(I))⊥ (2.3) ∆⊥ = −∇⊥(2) = −∇⊥α0 ∇⊥α0 , α0 (2.4) respectively. Let C be the unit Darboux vector and ∆ be the Laplacian operator of the curve α. Then α is said to be an harmonic Darboux vector if and only if ∆C = 0 and if ∆C = λC holds this time we call C as an harmonic 1-type provided that λ is constant, [2]. 3. Differential Equations for A Space Curve According to The Unit Darboux Vector In this section, we give the differential equations which characterize a curve α in E 3 , as a linear combination of both the unit Darboux vector C and the normal unit Darboux vector C ⊥ . Theorem 3.1. Let α : I → E 3 be a Frenet curve with curvature κ , torsion τ and unit Darboux vector C then the differential equation characterizing the curve α is given by ∇3α0 C − µ3 ∇2α0 C − µ2 ∇α0 C − µ1C = 0 0 µ3 = µ2 = µ1 = ϕ00 (ϕ0 k W k) + , ϕ0 kWk ϕ00 0 ϕ00 (ϕ0 k W k)0 ( 0 ) − ((ϕ0 )2 + k W k2 ) − 0 , ϕ ϕ kWk ϕ0 (ϕ0 k W k)0 − ((ϕ0 )2 )0 . kWk Proof. When we take the derivative of C = sinϕT + cosϕB with respect to s, we get ∇α0 C = ϕ0 (cosϕT − sinϕB) By using the equations (2.2) and (3.1) we can evaluate the values of T and B as cosϕ sinϕ T = sinϕC + 0 ∇α0 C, B = cosϕC − 0 ∇α0 C ϕ ϕ If we take the derivative of equation (3.1) with respect to s, this time we have (3.1) (3.2) ∇2α0 C = ϕ00 (cosϕT − sinϕB) + ϕ0 (cosϕT − sinϕB)0 it follows ∇2α0 C =
ϕ00 (cosϕT − sinϕB) − (ϕ0 )2 sinϕT + cosϕB
+ϕ0 κcosϕ + τsinϕ N and taking the equation (2.2) into account with the second derivative of C we obtain ∇2α0 C = ϕ00 (cosϕT − sinϕB) − (ϕ0 )2C + ϕ0 k W k N (3.3) If we put the values of T and B from (3.2) into the equation (3.3) we get     ∇α0 C ∇α0 C cosϕ − ϕ00 sinϕ cosϕC − sinϕ ∇2α0 C = ϕ00 cosϕ sinϕC + 0 0 ϕ ϕ 0 2 0 −(ϕ ) C + ϕ k W k N So the second derivative of C is given as ∇2α0 C = ϕ00 ∇α0 C − (ϕ0 )2C + ϕ0 k W k N ϕ0 (3.4) Differential Equations for a Space Curve According to The Unit Darboux Vector 94 and from here we can deduce the normal vector N as N= 1 (ϕ0 )2 k W k   ϕ0 ∇2α0 C − ϕ00 ∇α0 C + (ϕ0 )3C (3.5) Finally we take the derivative of (3.4) in this case we get ∇3α0 C = ϕ00 0 ϕ00 ) ∇α0 C + 0 ∇2α0 C − ((ϕ0 )2 )0C − (ϕ0 )2 ∇α0 C + (ϕ0 k W k)0 N 0 ϕ ϕ +ϕ0 k W k (−κT + τB) ( (3.6) Now let’s put the values of T and B, taken from (3.2), into (3.6) we obtain ∇3α0 C  ϕ00  ϕ00 ( 0 )0 − ((ϕ0 )2 + k W k2 ) ∇α0 C + 0 ∇2α0 C ϕ ϕ 0 2 0 0 0 −(ϕ ) ) C + (ϕ k W k) N = (3.7) It is about to finish to obtain the desired differential equation that we substitute (3.5) into the equation (3.7) it becomes  ϕ00 ϕ00 ∇3α0 C = ( 0 )0 − ((ϕ0 )2 + k W k2 )∇α0 C + 0 ∇2α0 C − ((ϕ0 )2 )0C ϕ ϕ  1 0 2 0 0 (ϕ ∇α0 C − ϕ00 ∇α0 C + (ϕ0 )3C)) +(ϕ k W k) ( 0 2 (ϕ ) k W k If we rearrange the above expression we have  ϕ00 (ϕ0 k W k)0  + 0 ∇2α0 C ∇3α0 C (...truncated)