A Note on Inextensible Flows of Curves on Oriented Surface

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2 Preliminaries

Let S be an oriented surface in three-dimensional Euclidean space E3 and α (s) be a curve lying on the surface S. Suppose that the curve α (s) is spatial then −→ −→ −→ −→ there exists the Frenet frame T, N, B at each points of the curve where T −→ −→ is unit tangent vector, N is principaln normalo vector and B is binormal vector, respectively. The Frenet equation of the curve α (s) is given by −→ −→ T ′ = κN −→ −→ −→ N ′ = −κ T + τ B −→ −→ B′ = −τ N where κ and τ are curvature and torsion of the curve α (s), respectively. Since the curve α (s) lies on the surface S there exists another frame of the curve −→ −→ −→ α (s) which is called Darboux frame and denoted by T, g , n . In this frame −→ −→ T is the unit tangent of the curve, n is the unit normaln of theo surface S and −→ −→ −→ −→ −→ g is a unit vector given by g = n × T . Since the unit tangent T is common −→ −→ −→ −→ element of both Frenet frame and Darboux frame, the vectors N, B, g and n lie on the same plane. So that the relations between these frames can be given as follows −→ −→ T 1 0 0 T −→ −→ g = 0 cosϕ sin ϕ N  −→     −→  n 0 − sin ϕ cos ϕ B       −→ −→   where ϕ is the angle between the vectors g and N . The derivative formulae of

2 the Darboux frame is . −→ −→ T 0 kg kn T −→. −→  g  = − kg 0 τg g .    −→  −→ kn − τg 0 n  n          where kg, kn and τg are called the geodesic curvature, the normal curvature and the geodesic torsions, respectively. Here and in the following, we use ”dot” to denote the derivative with respect to the arc length parameter of a curve. The relations between the geodesic curvature, normal curvature, geodesic torsion and κ, τ are given as follows, [11]

dϕ kg = κ cos ϕ , kn = κ sin ϕ , τg = τ + ds .

Furthermore, the geodesic curvature kg and geodesic torsion τg of curve α (s) can be calculated as follows, [11]

−→ 2−→ d α d α × −→ kg = ds , ds2 n −→ −→ Dd α −→ × d n E τg = ds , n ds . In the diﬀerential geometry of surfaces,D for a curveE α (s) lying on a surface S the following relationships are well-known, [11] i- α (s) is a geodesic curve if and only if kg = 0, ii- α (s) is a asymptotic line if and only if kn = 0, iii- α (s) is a principal line if and only if τg = 0. Through the every point of the surface a geodesic passes in every direction. A geodesic is uniquely determined by an initial point and tangent at that point. All straight lines on a surface are geodesics. Along all curved geodesics the principal normal coincides with the surface normal. Along asymptotic lines osculating planes and tangent planes coincide, along geodesics they are normal. Through a point of a non-developable surface pass two asymptotic lines which can be real or imaginary.

3 Inextensible Flows of Curve Lying on Oriented Surface

Throughout this paper, we suppose that

3 α : [0,l] × [0, w) → M ⊂ E is a one parameter family of diﬀerentiable curves on orientable surface M in E3, where l is the arclength of the initial curve. Let u be the curve parameterization −→ ≤ ≤ ∂ α variable, 0 u l. If the speed of curve α is denoted by v = ∂u then the arclength of α is

3 u −→ u ∂ α S (u)= du = v du. (3.1) ∂u Z0 Z0

∂ The operator ∂s is given in terms of u by

∂ 1 ∂ ∂s = v ∂u . (3.2) Thus, the arclength is ds = v du.

Definition 3.1 Let M be an orientable surface and α be a differentiable curve 3 −→ −→ −→ on M in E . Any flow of the curve α with respect to Darboux frame T, g , n can be expressed following form: n o

−→ ∂ α −→ −→ −→ ∂t = f1 T + f2 g + f3 n . (3.3) Here, f1, f2 and f3 are scalar speed of the curve α. Let the arclength variation be

u S (u,t)= v du. (3.4) 0 R In the Euclidean space the requirement that the curve not be subject to any elongation or compression can be expressed by the condition

u ∂ ∂v ∈ ∂t S (u,t)= ∂t du =0 , u [0, 1]. (3.5) 0

R −→ ∂ α Deﬁnition 3.2 A curve evolution α (u,t) and its ﬂow ∂t on the oriented surface M in E3 are said to be inextensible if −→ ∂ ∂ α =0. ∂t ∂u

Now, we research the necessary and suﬃcient condition for inelastic curve ﬂow. For this reason, we need to the following Lemma.

3 −→ −→ −→ Lemma 3.1 In E , let M be an orientable surface and T, g , n be a Dar- boux frame of α on M. There exists following relation betweenn theo scalar speed functions f1, f2, f3 and the normal curvature kn, geodesic curvature kg of α the curve ∂v ∂f1 − − ∂t = ∂u f2vkg f3vkn. (3.6)

4 −→ −→ ∂ ∂ 2 ∂ α ∂ α Proof. Since ∂u and ∂t commute and v = ∂u , ∂u , we have D E −→ −→ ∂v ∂ ∂ α ∂ α 2v ∂t = ∂t ∂u , ∂u −→ −→ D∂ α ∂ E −→ −→ =2 ∂u , ∂u f1 T + f2 g + f3 n D ∂f1 − − E =2v ∂u f2vkg f3vkn . Thus, we reach

∂v ∂f1 = − f2vk − f3vk . ∂t ∂u g n If we take in the conditions of being geodesic and asymptotic of a curve and Lemma 3.1, we give the following

Corollary 3.1 If the curve is a geodesic curve or asymptotic curve, then there is following equations

∂v ∂f1 − ∂t = ∂u f3vkn or ∂v ∂f1 − ∂t = ∂u f2vkg, respectively. −→ −→ −→ Theorem 3.1 Let T, g , n be Darboux frame of the curve α on M and −→ −→ ∂ α −→ n −→ o R3 ∂t = f1 T + f2 g + f3 n be a differentiable flow of α in . Then the flow is inextensible if and only if

∂f1 ∂s = f2kg + f3kn. (3.7) Proof. Suppose that the curve ﬂow is inextensible. From equations (3.4) and (3.6) for u ∈ [0,l] , we see that

u u ∂ ∂v ∂f1 S (u,t)= du = − f2vk − f3vk du =0. (3.8) ∂t ∂t ∂u g n Z0 Z0

Thus, it can be seen that

∂f1 = f2vk + f3vk . (3.9) ∂u g n Considering the last equation and (3.2), we reach

∂f1 ∂s = f2kg + f3kn. Conversely, following similar way as above, the proof is completed. From Theorem 3.1, we have following corollary.

5 Corollary 3.2 i- Let the curve α is a geodesic curve on M. Then the curve ∂f1 ﬂow is inextensible if and only if ∂s = f3kn. ii- Let the curve α is a asymptotic line on M. Then the curve ﬂow is inex- ∂f1 tensible if and only if ∂s = f2kg. Now, we restrict ourselves to the arclength parameterized curves. That is, v =1 and the local coordinate u corresponds to the curve arclength s. We require the following Lemma.

3 −→ −→ −→ Lemma 3.2 Let M be an orientable surface in E and T, g , n be a Dar- −→ −→ −→ boux frame of the curve α on M. Then, the diﬀerentiationsn of T, og , n with respect to t is n o −→ ∂ T ∂f2 − −→ ∂f3 −→ ∂t = f1kg + ∂s f3τg g + f1kn + ∂s + f2τg n −→ −→ ∂ g − ∂f2 − −→ ∂t = f1kg + ∂s f3τg T + ψ n −→ −→ ∂ n − ∂f3 − −→ ∂t = f1kn + ∂s + f2τg T ψ g

−→ ∂ g −→ where ψ = ∂t , n . D E ∂ ∂ Proof. Since ∂t and ∂s are commutative, it seen that

−→ −→ −→ ∂ T ∂ ∂ α ∂ ∂ α ∂ −→ −→ −→ ∂t = ∂t ∂s = ∂s ∂t = ∂s f1 T + f2 g + f3 n −→ −→ −→ −→ ∂f1 ∂ T ∂f2 −→ ∂ g ∂f3 −→ ∂ n = ∂s T + f1 ∂s + ∂s g + f2 ∂s + ∂s n + f3 ∂s . Substituting the equation (3.7) into the last equation and using Theorem 3.1, we have −→ ∂ T ∂f2 −→ ∂f3 −→ = f1k + − f3τ g + f1k + + f2τ n . ∂t g ∂s g n ∂s g Now, let us diﬀerentiate the Darboux frame with respect to t as follows; −→ −→ ∂ −→ −→ ∂ T −→ −→ ∂ g 0= T, g = , g + T, ∂t * ∂t + ∂t D E −→ ∂f2 −→ ∂ g = f1k + − f3τ + T, (3.10) g ∂s g ∂t −→ −→ ∂ −→ −→ ∂ T −→ −→ ∂ n 0 = T, n = , n + T, ∂t * ∂t + ∂t D E −→ ∂f3 −→ ∂ n = f1k + + f2τ + T, (3.11) n ∂s g ∂t

6 From (3.10) and (3.11), we have obtain −→ ∂ g ∂f2 −→ −→ = − f1k + − f3τ T + ψ n ∂t g ∂s g and −→ ∂ n ∂f3 −→ −→ = − f1k + + f2τ T − ψ g ∂t n ∂s g −→ ∂ g −→ respectively, where ψ = ∂t , n . If we take into considerationD lastE Lemma, we have following corollary. Corollary 3.3 Let M be an orientable surface in E3. i- If the curve α is a geodesic curve, then −→ ∂ T ∂f2 − −→ ∂f3 −→ ∂t = ∂s f3τg g + f1kn + ∂s + f2τg n , −→ −→ ∂ g − ∂f2 − −→ ∂t = ∂s f3τg T + ψ n , −→ −→ ∂ n − ∂f3 − −→ ∂t = f1kn + ∂s + f2τg T ψ g , −→ ∂ g −→ where ψ = ∂t , n . ii- If theD curve αE is a asymptotic line, then −→ ∂ T ∂f2 − −→ ∂f3 −→ ∂t = f1kg + ∂s f3τg g + ∂s + f2τg n , −→ −→ ∂ g − ∂f2 − −→ ∂t = f1kg + ∂s f3τg T + ψ n , −→ −→ ∂ n − ∂f3 − −→ ∂t = ∂s + f2τg T ψ g , −→ ∂ g −→ where ψ = ∂t , n . iii- If theD curveE is a curvature line, then −→ ∂ T ∂f2 −→ ∂f3 −→ ∂t = f1kg + ∂s g + f1kn + ∂s n , −→ −→ ∂ g − ∂f2 −→ ∂t = f1kg + ∂s T + ψ n , −→ −→ ∂ n − ∂f3 − −→ ∂t = f1kn + ∂s T ψ g , −→ ∂ g −→ where ψ = ∂t , n . D E −→ ∂ α −→ −→ −→ Theorem 3.2 Suppose that the curve ﬂow ∂t = f1 T + f2 g + f3 n is inextensible on the orientable surface on M. In this case, the following partial diﬀerential equation are held: 2 ∂kg ∂f1 ∂kg ∂ f2 − ∂f3 − ∂τg − − ∂f3 − 2 ∂t = ∂s kg + f1 ∂s + ∂s2 ∂s τg f3 ∂s f1knτg ∂s τg f2τg , 2 ∂kn ∂f1 ∂kn ∂ f3 ∂f2 ∂τg ∂f2 − 2 ∂t = ∂s kn + f1 ∂s + ∂s2 + ∂s τg + f2 ∂s + f1kgτg + ∂s τg f3τg , ∂τg − − ∂f2 ∂ψ ∂t = f1kgkn ∂s kn + f3knτg + ∂s , − − ∂f3 − ψkn = f1kn ∂s f2τg τg, − − ∂f3 ψkg = f1kg ∂s + f3τg τg . 7 −→ −→ ∂ ∂ T ∂ ∂ T Proof. Since ∂s ∂t = ∂t ∂s we get −→ ∂ ∂ T ∂ ∂f2 − −→ ∂f3 −→ ∂s ∂t = ∂s f1kg + ∂s f3τg g + f1kn + ∂s + f2τg n 2 −→ ∂fh1 ∂kg ∂ f2 − ∂f3 − ∂τg −→ i ∂f2 − ∂ g = ∂s kg + f1 ∂s + ∂s2 ∂s τg f3 ∂s g + f1kg + ∂s f3τg ∂s 2 −→ ∂f1 ∂kn ∂ f3 ∂f2 ∂τg −→ ∂f3 ∂ n + ∂s kn + f1 ∂s + ∂s2 + ∂s τg + f2 ∂s n + f1kn + ∂s + f2τg ∂s i.e.,

−→ 2 −→ ∂ ∂ T ∂f1 ∂kg ∂ f2 − ∂f3 − ∂τg −→ ∂f2 − − −→ ∂s ∂t = ∂s kg + f1 ∂s + ∂s2 ∂s τg f3 ∂s g + f1kg + ∂s f3τg kg T + τg n 2 1 3 2 ∂τg −→ 3 −→ −→ ∂f ∂kn ∂ f ∂f ∂f − − + ∂s kn + f1 ∂s + ∂s2 + ∂s τg + f2 ∂s n + f1kn + ∂s + f2τg kg T τg g while −→ −→ −→ ∂ ∂ T ∂ −→ −→ ∂kg −→ ∂ g ∂kn −→ ∂ n = (k g + k n )= g + k + n + k . ∂t ∂s ∂t g n ∂t g ∂t ∂t n ∂t Thus, from the both of above two equations, we reach 2 ∂kg ∂f1 ∂kg ∂ f2 ∂f3 ∂τg ∂f3 2 = k +f1 + − τ −f3 −f1k τ − τ −f2τ (3.12) ∂t ∂s g ∂s ∂s2 ∂s g ∂s n g ∂s g g and 2 ∂kn ∂f1 ∂kn ∂ f3 ∂f2 ∂τg ∂f2 2 = k +f1 + + τ +f2 +f1k τ + τ −f3τ . (3.13) ∂t ∂s n ∂s ∂s2 ∂s g ∂s g g ∂s g g −→ −→ ∂ ∂ g ∂ ∂ g Noting that ∂s ∂t = ∂t ∂s , it is seen that −→ −→ ∂ ∂ g ∂ − ∂f2 − −→ ∂s ∂t = ∂s f1kg + ∂s f3τg T + ψ n 2 −→ − h∂f1 ∂kg ∂ f2 − ∂f3 −i ∂τg − ∂f2 − −→ −→ = ∂s kg + f1 ∂s + ∂s2 ∂s τg f3 ∂s T f1kg + ∂s f3τg (kg g + kn n ) −→ ∂ψ − − −→ + ∂s n + ψ kn T τg g while −→ −→ −→ ∂ ∂ g ∂ −→ −→ ∂kg −→ ∂ T ∂τg −→ ∂ n = −k T + τ n = − T − k + n + τ . ∂t ∂s ∂t g g ∂t g ∂t ∂t g ∂t Thus, we obtain 2 ∂kg ∂f1 ∂kg ∂ f2 ∂f3 ∂τg = k + f1 + − τ − f3 + ψk (3.14) ∂t ∂s g ∂s ∂s2 ∂s g ∂s n and ∂τg ∂f2 ∂ψ = −f1k k − k + f3k τ + . (3.15) ∂t g n ∂s n n g ∂s From the equations (3.12) and (3.14), it is seen that

− − ∂f3 − 2 ψkn = f1knτg ∂s τg f2τg − − ∂f3 − = f1kn ∂s f2τg τg. 8 −→ −→ ∂ ∂ n ∂ ∂ n By same way as above and considering ∂s ∂t = ∂t ∂s , we reach

2 ∂kn ∂f1 ∂kn ∂ f3 ∂f2 ∂τg = k + f1 + + τ + f2 − ψk . (3.16) ∂t ∂s n ∂s ∂s2 ∂s g ∂s g Hence, from the equations (3.13) and (3.16), we get

∂f2 ψk = −f1k − + f3τ τ . g g ∂s g g

Thus, we give the following corollary from last theorem.

Corollary 3.4 Let M be an orientable surface in E3. i- If the curve α is a geodesic curve on M, then we have

2 ∂kn ∂f1 ∂kn ∂ f3 ∂f2 ∂τg ∂f2 2 = k + f1 + + τ + f2 + τ − f3τ ∂t ∂s n ∂s ∂s2 ∂s g ∂s ∂s g g

∂τg ∂f2 ∂ψ = − k + f3k τ + ∂t ∂s n n g ∂s and ∂f3 ψk = −f1k − − f2τ τ . n n ∂s g g ii- If the curve α is a asymptotic line, we have

2 ∂kg ∂f1 ∂kg ∂ f2 ∂f3 ∂τg ∂f3 2 = k + f1 + − τ − f3 − τ − f2τ ∂t ∂s g ∂s ∂s2 ∂s g ∂s ∂s g g ∂τ ∂ψ g = ∂t ∂s and ∂f2 ψk = −f1k − + f3τ τ . g g ∂s g g iii- If the curve α is a curvature line, then we have

2 ∂kg ∂f1 ∂kg ∂ f2 = kg + f1 + 2 ∂t ∂s ∂s ∂s2 ∂kn ∂f1 ∂kn ∂ f3 ∂t = ∂s kn + f1 ∂s + ∂s2 . References

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