Article Isometric Deformation of (m, n)-Type Helicoidal Surface in the Three Dimensional Euclidean Space
Erhan Güler
Faculty of Sciences, Department of Mathematics, Bartın University, 74100 Bartın, Turkey; [email protected]; Tel.: +90-378-5011000-1521
Received: 13 September 2018; Accepted: 26 October 2018; Published: 29 October 2018
Abstract: We consider a new kind of helicoidal surface for natural numbers (m, n) in the three-dimensional Euclidean space. We study a helicoidal surface of value (m, n), which is locally isometric to a rotational surface of value (m, n). In addition, we calculate the Laplace–Beltrami operator of the rotational surface of value (0, 1).
Keywords: euclidean three-space; helicoidal surface of value (m, n); rotational surface of value (m, n); mean curvature; Gaussian curvature; Gauss map
1. Introduction The notion of the finite-type immersion of submanifolds of a Euclidean space has been used in classifying and characterizing well-known Riemannian submanifolds [1]. Chen posed the problem of classifying the finite-type surfaces in the three-dimensional Euclidean space E3. Then, the theory of submanifolds of a finite type was studied by many geometers [1–21]. Lawson [22] gave the general definition of the Laplace–Beltrami operator in his lecture notes. Takahashi [23] stated that minimal surfaces and spheres are the only surfaces in E3 satisfying the condition ∆r = λr, λ ∈ R. Ferrandez, Garay and Lucas [10] proved that the surfaces of E3 satisfying ∆H = AH, A ∈ Mat(3, 3) are either minimal, or an open piece of a sphere, or of a right circular cylinder. Choi and Kim [5] characterized the minimal helicoid in terms of a pointwise one-type Gauss map of the first kind. Dillen, Pas and Verstraelen [7] proved that the only surfaces in E3 satisfying ∆r = Ar + B, A ∈ Mat(3, 3), B ∈ Mat(3, 1) are the minimal surfaces, the spheres and the circular cylinders. Senoussi and Bekkar [24] studied helicoidal surfaces M2 in E3, which are of the finite type in the sense of Chen with respect to the fundamental forms I, II and III. The right helicoid (resp. catenoid) is the only ruled (resp. rotational) surface that is minimal in classical surface geometry in Euclidean space. If we focus on the ruled (helicoid) and rotational characters, we see Bour’s theorem in [25]. The French mathematician Edmond Bour used the semi-geodesic coordinates and found a number of new cases of the deformation of surfaces in 1862. He also gave in [25] a well-known theorem about the helicoidal and rotational surfaces. Kenmotsu [26] focused on the surfaces of revolution with the prescribed mean curvature. Regarding helicoidal surfaces, do Carmo and Dajczer [3] proved that, by using a result of Bour [25], there exists a two-parameter family of helicoidal surfaces isometric to a given helicoidal surface. Hitt and Roussos [27] also studied the helicoidal surfaces with constant mean curvature. Ikawa [14,15] determined pairs of surfaces by Bour’s theorem. Güler [28] also studied the isometric helicoidal and rotational surfaces of value m. Güler and Yaylı [12] focused on the generalized Bour’s theorem in three-space. We consider a new kind of helicoidal surface of value (m, n) in Euclidean three-space E3 in this paper. We give some basic notions of the three-dimensional Euclidean geometry in Section2. In
Mathematics 2018, 6, 226; doi:10.3390/math6110226 www.mdpi.com/journal/mathematics Mathematics 2018, 6, 226 2 of 12
Section3, we give the definition of a helicoidal surface of value (m, n) and obtain isometric helicoidal and rotational surfaces of value (m, n) (resp. of value (0, 1) in Section4) via Bour’s theorem. We also calculate the mean curvature and the Gaussian curvature of the rotational surface of value (0, 1) in Section4. Moreover, in Section5, we calculate the Laplace–Beltrami operator of the rotational surface 3 of value (0, 1). Finally, we give the rotational surface satisfying ∆R0,1 = AR0,1 in E in the last section.
2. Preliminaries We shall identify a vector (a,b,c) with its transpose. In this section, we will obtain the rotational and helicoidal surfaces in E3. The reader can find basic elements of differential geometry in [29,30]. We define the rotational surface and helicoidal surface in E3. For an open interval I ⊂ R, let γ : I −→ Π be a curve in a plane Π, and let ` be a straight line in Π. A rotational surface in E3 is defined as a surface rotating the curve γ around the line ` (these are called the profile curve and the axis, respectively). Suppose that when a profile curve γ rotates around the axis `, it simultaneously displaces parallel lines orthogonal to the axis `, so that the speed of displacement is proportional to the speed of rotation. Then, the resulting surface is called the helicoidal surface with axis ` and pitch + a ∈ R . Let ` be the line spanned by the vector (0, 0, 1). The orthogonal matrix that fixes the above vector is given by cos θ − sin θ 0 M(θ) = sin θ cos θ 0 , θ ∈ R. 0 0 1
t t The matrix M is found by solving the following equations: M` = `, M M = MM = I3, det M = 1, simultaneously. When the axis of rotation is `, there is a Euclidean transformation by which the axis is ` transformed to the z-axis of E3. The profile curve is given by γ(r) = (r, 0, ϕ (r)), where ϕ (r) : I ⊂ R −→ R is a differentiable function for all r ∈ I. A helicoidal surface in three-dimensional Euclidean space which is spanned by the vector (0, 0, 1) with pitch a is as follows
H(r, θ) = M(θ)γ(r) + aθ`.
When a = 0, the helicoidal surface is just a rotational surface.
3. Helicoidal Surfaces of Value (m, n) We define a new type of helicoidal surface. Using Bour’s theorem on the helicoidal surface, we obtain an isometric rotational surface in this section.
Definition 1. A helicoidal surface of value (m, n) is given by
1 2 Hm,n (r, θ) = Hm (r, θ) + Hm,n (r, θ) , (1)
1 1 1 1 2 2 2 1 1 where m, n ∈ N, Hm (r, θ) = ` = (0, 0, 1) is the rotating axis, and the profile curves are rm+1 1 rm+2n+1 1 γ1 (r) = , 0, ϕ (r) , γ2 (r) = − , 0, ϕ (r) , m m + 1 2 m,n m + 2n + 1 2 1 2 + + m ∈ R − {−1} in γm, m ∈ R − {−1 − 2n} in γm,n, r ∈ R , 0 ≤ θ ≤ 2π and the pitch a ∈ R . Since the helicoidal surface is given by rotating the profile curves γ around the axis ` and simultaneously displacing parallel lines orthogonal to the axis `, the speed of displacement is proportional to the speed of rotation. Next, we give a theorem about locally isometric helicoidal-rotational surfaces of value (m, n). Theorem 1. A helicoidal surface of value (m, n) rm+1 rm+2n+1 m+1 cos [(m + 1) θ] − m+2n+1 cos [(m + 2n + 1) θ] H (r, θ) = rm+1 rm+2n+1 (2) m,n − m+1 sin [(m + 1) θ] − m+2n+1 sin [(m + 2n + 1) θ] ϕ (r) + aθ is isometric to the rotational surface of value (m, n) m+1 m+2n+1 rR rR m+1 cos [(m + 1) θR] − m+2n+1 cos [(m + 2n + 1) θR] rm+1 rm+2n+1 Rm,n(rR, θR) = R R (3) − m+1 sin [(m + 1) θR] − m+2n+1 sin [(m + 2n + 1) θR] ϕR (rR) by Bour’s theorem, where " #2 Gm+n−1{(m + 3n) G2n + 2 (m + 2n) Gn cos[2(m + n + 1)θ] det I −2Gn sin[2(m + n + 1)θ] + m + n} ϕ02 = R " #2 r2m+2n{(2m + 6n) r4n−1 + (4m + 8n) r2n cos[2(m + n + 1)θ] G + (2m + 2) r−1} 4r4m+4n+2 sin2[2(m + n + 1)θ ] + R R 2m+2 4n 2n rR rR + 2rR cos[2(m + n + 1)θR] + 1 2m h 4n 2n i −rR rR − 2rR cos (2 (m + n + 1) θR) + 1 , √ rR = G, Z F θ = θ + dr, R G E = r2m r4n − 2r2n cos[2(m + n + 1)θ] + 1 + ϕ02, F = 2r2m+2n+1 sin[2(m + n + 1)θ] + aϕ0, G = r2m+2 r4n + 2r2n cos[2(m + n + 1)θ] + 1 + a2, det I = EG − F2, + + m ∈ R − {−1, −1 − 2n} , n ∈ R, r ∈ R , θ ∈ I ⊂ R and the pitch a ∈ R . Proof. The line element of the the helicoidal surface Hm,n(r, θ) is ds2 = {r2m(r4n − 2r2n cos[2(m + n + 1)θ] + 1) + ϕ02}dr2 +2{2r2m+2n+1 sin[2(m + n + 1)θ] + aϕ0}drdθ (4) + {r2m+2 r4n + 2r2n cos[2(m + n + 1)θ] + 1 + a2}dθ2. Mathematics 2018, 6, 226 4 of 12 Helices in Hm,n are curves defined by r = const. Therefore curves in Hm,n that are orthogonal to helices R F supply the orthogonality condition F dr + G dθ = 0. Thus, we obtain θ = − G dr + c, where c is R F constant. Hence, if we put θ = θ + G dr, then curves orthogonal to helices are given by θ = const. F Substituting the equation dθ = dθ − G dr into the line element (4), we have Q 2 ds2 = dr2 + G dθ , (5) G √ q Q where Q := det I. Setting r := G dr, Ω(r) := G,(5) becomes 2 ds2 = dr2 + Ω2 (r) dθ . (6) The rotational surface (3) has the line element 2 QR 2 2 dsR = drR + GR dθR, (7) GR where 2m 4n 2n 02 ER = rR (rR + 1 − 2rR cos[2(m + n + 1)θR]) + ϕR , 2m+2n+1 FR = 2rR sin[2(m + n + 1)θR], 2m+2 4n 2n GR = rR rR + 1 + 2rR cos[2(m + n + 1)θR] . q √ Again, setting r := QR dr , Ω (r ) := G , then (7) becomes R GR R R R R 2 2 2 2 dsR = drR + ΩR (rR) dθR. (8) Comparing (6) with (8), if we take r = rR, θ = θR, Ω (r) = ΩR (rR) , then we have an isometry between Hm,n(r, θ) and Rm,n(rR, θR). Therefore, it follows that r s Q QR dr = drR. (9) G GR Substituting the equation v u 2 u G [(G) ] = R r drR t 2 dr [(GR)r] G into (9), we get the function ϕR. 4. Helicoidal Surface of Value (0, 1) We give the helicoidal surface of value (0, 1) using Bour’s theorem in this section. Proposition 1. A helicoidal surface of value (0, 1): (see Figure1) r3 r cos (θ) − 3 cos (3θ) H (r, θ) = r3 , (10) 0,1 −r sin (θ) − 3 sin (3θ) ϕ (r) + aθ Mathematics 2018, 6, 226 5 of 12 is isometric to the rotational surface of value (0, 1) √ h i √ h i R F 1 3 R F G cos θ + G dr − 3 G cos 3 θ + G dr √ h R i √ h R i R0,1(rR, θR) = − G sin θ + F dr − 1 G3 sin 3 θ + F dr , (11) G 3 G ϕR (rR) where 2 2 02 {3G + 4G cos (4θ) − 2G sin (4θ) + 1} det I ϕR = [r2{6r3 + 8r2 cos (4θ) + 2r−1}]2 G 3 2 h R F i 4G sin 4 θ + G dr + h h i i 2 R F G G + 2G cos 4 θ + G dr + 1 Z F − G2 − 2G cos 4 θ + dr + 1 , G E = r4 − 2r2 cos (4θ) + 1 + ϕ02, F = 2r3 sin (4θ) + aϕ0, h i G = r2 r4 + 2r2 cos (4θ) + 1 + a2, + det I = EG − F2, r, a ∈ R , 0 ≤ θ ≤ 2π. Proof. Taking m = 0, n = 1 in the previous theorem, we easily get the results. Figure 1. Two views of the helicoidal surface of value (0, 1), ϕ (r) = r2. Corollary 1. When a = 0, ϕ (r) = r in (10), we obtain a rotational surface of value (0, 1) (see Figure2). Mathematics 2018, 6, 226 6 of 12 Figure 2. Two views of the rotational surface of value (0, 1), ϕ (r) = r. Corollary 2. When a = 0 and ϕ (r) = r2 cos (2θ) in (10) , we have Enneper’s minimal surface (see Figure3). Figure 3. Two views of the Enneper minimal surface of value (0, 1), a = 0, ϕ (r) = r2 cos (2θ). Proposition 2. The mean curvature and the Gaussian curvature of (10) are as follows 1 H = {r(r4 − 1)(r6 + 2r4 cos(4θ) + r2 + a2)ϕ00 2 (det I)3/2 +r2(3r4 + 2r2 cos (4θ) − 1)ϕ03 − 6ar3 sin(4θ)ϕ02 +[2r2(−5r6 + r2 − 3a2) cos (4θ) + r10 − 8r6 + 4a2r4 −r2 − 2a2]ϕ0 + 2ar 6r6 − 2r2 + a2 sin(4θ)} and 1 K = ({[3r11 + 2r5(r4 − 1) cos(4θ) − 4r7 + r3]ϕ0 (det I)2 −2ar5(r4 − 1) sin(4θ)}ϕ00 + [−2r6(3r4 + 1) −2r4(r6 − 1) cos(4θ)]ϕ02 − 2ar3(3 − 11r4) sin(4θ)]ϕ0 +4a2r2(3r4 − 1) cos(4θ) − a2 9r8 − 2r4 + 1 . 2 0 dϕ + respectively, where det I = EG − F , ϕ = dr , r, a ∈ R , 0 ≤ θ ≤ 2π. Mathematics 2018, 6, 226 7 of 12 Proof. Taking the differential with respect to r, θ to H0,1, we have cos (θ) − r2 cos (3θ) ( ) = − ( ) − 2 ( ) H0,1 r sin θ r sin 3θ . ϕ0 and −r sin(θ) + r3 sin(3θ) ( ) = − ( ) − 3 ( ) H0,1 θ r cos θ r cos 3θ , a The coefficients of the first fundamental form of the surface are h i E = r4 − 2r2 cos (4θ) + 1 + ϕ02, F = 2r3 sin (4θ) + aϕ0, h i G = r2 r4 + 2r2 cos (4θ) + 1 + a2, Then, we get det I = [r6 + 2r4 cos (4θ) + r2]ϕ02 − 4ar3 sin (4θ) ϕ0 +r10 − 2r6 + a2r4 − 2a2r2 cos (4θ) + r2 + a2. Using the second differentials −2r cos (3θ) − sin(θ) + 3r2 sin(3θ) ( ) = − ( ) ( ) = − ( ) − 2 ( ) H0,1 rr 2r sin 3θ , H0,1 rθ cos θ 3r cos 3θ , ϕ00 0 −r cos(θ) + 3r3 cos(3θ) ( ) = ( ) + 3 ( ) H0,1 θθ r sin θ 3r sin 3θ , 0 and the Gauss map (the unit normal) (r3 cos (3θ) + r cos (θ))ϕ0 − a(r2 sin (3θ) + sin (θ)) 1 e = √ (r3 sin (3θ) − r sin (θ))ϕ0 + a(r2 cos (3θ) − cos (θ)) det I r5 − r of the surface H0,1, we have the coefficients of the second fundamental form of the surface as follows 1 L = √ [(r5 − r)ϕ00 − r2(r2 + cos (4θ))ϕ0 − 2ar sin (4θ)], det I 1 M = √ [2r3 sin (3θ) ϕ0 − 3ar4 + 2ar2 cos (4θ) + a], det I 1 N = √ [(3r6 − r2 + 2r4 cos (4θ))ϕ0 − 2ar3 sin (4θ)]. det I Therefore, we can see the results easily. Mathematics 2018, 6, 226 8 of 12 Corollary 3. If the helicoidal surface of value (0, 1) is minimal, then we get the differential equation as follows r(r4 − 1)(r6 + 2r4 cos(4θ) + r2 + a2)ϕ00 +r2(3r4 + 2r2 cos (4θ) − 1)ϕ03 − 6ar3 sin(4θ)ϕ02 +[2r2(−5r6 + r2 − 3a2) cos (4θ) + r10 − 8r6 + 4a2r4 −r2 − 2a2]ϕ0 + 2ar 6r6 − 2r2 + a2 sin(4θ) = 0. The helicoidal surface of value (1, 1) is the same surface for value three in [28]. 5. Laplace–Beltrami Operator 3 2 The Laplace–Beltrami operator of a smooth function φ = φ(r, θ) |D (D ⊂ R ) of class C with respect to the first fundamental form of surface M is the operator ∆, which is defined as follows 1 2 ∂ √ ∂φ = ij ∆φ √ ∑ i gg j , (12) g i,j=1 ∂x ∂x ij −1 where g = (gkl) and g = det gij . Clearly, we write ∆φ as follows √ √ ∂ gg11 ∂φ − ∂ gg12 ∂φ 1 ∂x1 ∂x1 ∂x1 ∂x2 ∆φ = √ √ √ . (13) g − ∂ gg21 ∂φ + ∂ gg22 ∂φ ∂x2 ∂x1 ∂x2 ∂x2 Using a more transparent notation, we get 1 ∂ Gφ − Fφ ∂ Fφ − Eφ ∆φ = √ √u v − √u v , (14) det I ∂u det I ∂v det I where g = det gij = det I. Now, we consider the rotational surface r3 r cos (θ) − 3 cos (3θ) R (r, θ) = r3 . (15) 0,1 −r sin (θ) − 3 sin (3θ) ϕ (r) The first fundamental matrix of the surface is as follows ! r4 − 2r2 cos (4θ) + 1 + ϕ02 2r3 sin (4θ) I = . 2r3 sin (4θ) r2(r4 + 2r2 cos (4θ) + 1) The inverse matrix of I is as follows ! 1 r2 r4 + 2r2 cos 4θ + 1 −2r3 sin 4θ I−1 = , det I −2r3 sin 4θ r4 − 2r2 cos 4θ + 1 + ϕ02 where det I = r2( r4 − 2 r4 + r4 + 2r2 cos 4θ + 1 ϕ02 + 1). The Laplace–Beltrami operator ∆R0,1 of a rotational surface R0,1 is given by 1 ∂ ∂ ∆R = √ U − V , det I ∂r ∂θ Mathematics 2018, 6, 226 9 of 12 where GR − FR FR − ER U = √r θ , V = √r θ . det I det I Then, we obtain the following results −r2 r4 − 1 cos θ + r2 cos 3θ 1 U = √ r2 r4 − 1 sin θ − r2 sin 3θ , det I r2 ϕ0(r4 + 2r2 cos (4θ) + 1) r 1 + ϕ02 − r4 sin θ + r2 1 − r4 − ϕ20 sin 3θ 1 V = √ r 1 + ϕ20 − r4 cos θ − r2 1 − r4 − ϕ02 cos 3θ . det I 2r3 ϕ0 sin (4θ) Using differentials of r, θ on U, V, respectively, we get 1 ∂ ∂ ∆I R = √ (U) − (V) det I ∂r ∂θ J ϕ0 ϕ00 + J ϕ04 + J ϕ02 + J r3 1 2 3 4 = 0 00 + 04 + 02 + 2 T1 ϕ ϕ T2 ϕ T3 ϕ T4 , (det I) 03 00 02 00 0 00 03 Q1 ϕ ϕ + Q2 ϕ ϕ + Q3 ϕ ϕ + Q4 ϕ + Q5 where 4 8 3 4 8 3 4 5 4 J1 = r 1 + r − 2r cos θ + r 2 − r − r cos 3θ + r 1 − r cos 5θ + r 1 − r cos 7θ, 4 6 2 4 J2 = − cos θ + 4r cos θ + 3r cos 3θ + r cos 5θ + r cos 7θ, 4 8 2 4 8 2 4 J3 = −3 + r + 14r cos θ + r −6 + 7r + 11r cos 3θ + r 1 + 3r cos 5θ + r4 −3 + 7r4 cos 7θ, 4 8 12 2 4 8 12 J4 = 2 −1 + 3r − 3r + r cos θ + 6r −1 + 3r − 3r + r cos 3θ, 8 3 8 3 4 5 4 T1 = r 1 − r sin θ + r −1 + r sin 3θ + 2r 1 − r cos 4θ sin θ + 2r −1 + r sin 3θ cos 4θ, 4 6 2 4 T2 = −1 + 4r cos θ + 3r cos 3θ + r cos 5θ + r cos 7θ, 4 8 4 8 2 4 8 T3 = −2 − 3r + 5r cos θ + −1 + 6r + 3r sin θ + r 3 − 6r − 5r sin 3θ + 3r2 −1 − r4 + 2r8 cos 3θ + r2 1 − r4 cos 5θ + 8r6 sin θ cos 4θ + 4r4 1 − 3r4 sin 3θ cos 4θ + r4 −1 + r4 cos 7θ, 4 8 12 T4 = −1 + 3r − 3r + r (cos θ + sin θ + cos 3θ − sin 3θ) , 3 5 9 5 2 2 Q1 = r + 4r + 2r + r + 4r r + cos 4θ cos 4θ, 5 9 3 4 2 Q2 = −r − 2r − r − 4r 1 + r + r cos 4θ cos 4θ, 5 9 13 3 4 8 Q3 = r − r − r + r + 2r 1 − 2r + r cos 4θ, 4 8 2 4 Q4 = 1 − 4r + 3r − 2r 1 − r cos 4θ, 4 8 12 2 4 8 Q5 = 1 + 7r − 9r + r + r −2 + 12r − 10r cos 4θ. Remark 1. When the rotational surface R0,1 has the equation ∆R0,1= 0, we have to solve the system of equations as follows 0 00 04 02 J1 ϕ ϕ + J2 ϕ + J3 ϕ + J4 = 0, 0 00 04 02 T1 ϕ ϕ + T2 ϕ + T3 ϕ + T4 = 0, 03 00 02 00 0 00 03 Q1 ϕ ϕ + Q2 ϕ ϕ + Q3 ϕ ϕ + Q4 ϕ + Q5 = 0. here, finding the function ϕ is a problem. Corollary 4. When ϕ = c = const., then we get 2 4 8 12 2 cos θ + 6r cos 3θ −1 + 3r − 3r + r 2 ∆R0 1 = cos θ + sin θ + 3r (cos 3θ − sin 3θ) . , r2 [(r4 − 2) r4 + 1] 0 Mathematics 2018, 6, 226 10 of 12 3 6. Rotational Surface Satisfying ∆R0,1 = AR0,1 in E 2 3 Theorem 2. Let R0,1 : M −→ E be an isometric immersion given by (15). Then, ∆R0,1 = AR0,1 if and only if M2 has zero mean curvature. Proof. The Gauss map of the rotational surface R0,1 is (cos θ + r2 cos 3θ)ϕ0 1 e = (− sin θ + r2 sin 3θ)ϕ0 , W −1 + r4 where q W = (r4 − 2) r4 + (r4 + 2r2 cos 4θ + 1) ϕ02 + 1. We use − 2He = AR0,1, (16) where A = aij is a 3 × 3 matrix. The equation ∆R0,1 = AR0,1 by means of ! r4 − 2r2 cos (4θ) + 1 + ϕ02 2r3 sin (4θ) I = . 2r3 sin (4θ) r2(r4 + 2r2 cos (4θ) + 1) and ∆R0,1 = −2He give rise to the following system of ODEs r3 r3 Ω(cos θ + r2 cos 3θ)ϕ0 − a r cos (θ) − cos (3θ) − a −r sin (θ) − sin (3θ) = ϕa , 11 3 12 3 13 r3 r3 Ω(− sin θ + r2 sin 3θ)ϕ0 − a r cos (θ) − cos (3θ) − a −r sin (θ) − sin (3θ) = ϕa , 21 3 22 3 23 r3 r3 Ω −1 + r4 = a r cos (θ) − cos (3θ) + a −r sin (θ) − sin (3θ) + ϕa , 31 3 32 3 33 2H where Ω (r) = W . Differentiating the ODEs with respect to θ, we have a13 = a23 = a33 = 0, Ω (r) = 0. (17) From (17), we get r3 r3 −a r cos (θ) − cos (3θ) − a −r sin (θ) − sin (3θ) = 0, 11 3 12 3 r3 r3 −a r cos (θ) − cos (3θ) − a −r sin (θ) − sin (3θ) = 0, 21 3 22 3 r3 r3 a r cos (θ) − cos (3θ) + a −r sin (θ) − sin (3θ) = 0. 31 3 32 3 2H here, cos and sin are linearly independent functions of θ, then we have that aij = 0. From Ω (r) = W , we obtain H = 0. Finally, R0,1 is a minimal rotational hypersurface. Author Contributions: The author gave the idea for the isometric deformation problem of an (m, n)-type helicoidal surface in three-space. Then, he checked and polished the draft. Funding: This research received no external funding. Conflicts of Interest: The author declares that there is no conflict of interest regarding the publication of this paper. Mathematics 2018, 6, 226 11 of 12 References 1. Chen, B.Y. Total Mean Curvature and Submanifolds of Finite Type; World Scientific: Singapore, 1984. 2. Arslan, K.; Kılıç Bayram, B.; Bulca, B.; Öztürk, G. Generalized Rotation Surfaces in E4. Results Math. 2012, 61, 315–327. [CrossRef] 3. Do Carmo, M.; Dajczer, M. Helicoidal surfaces with constant mean curvature. Tohoku Math. J. 1982, 34, 351–367. [CrossRef] 4. Chen, B.Y.; Choi, M.; Kim, Y.H. Surfaces of revolution with pointwise 1-type Gauss map. Korean Math. Soc. 2005, 42, 447–455. [CrossRef] 5. Choi, M.; Kim, Y.H. Characterization of the helicoid as ruled surfaces with pointwise 1-type Gauss map. Bull. Korean Math. Soc. 2001, 38, 753–761. 6. Choi, M.; Yoon, D.W. Helicoidal surfaces of the third fundamental form in Minkowski 3-space. Bull. Korean Math. Soc. 2015, 52, 1569–1578. [CrossRef] 7. Dillen, F.; Pas, J.; Verstraelen, L. On surfaces of finite type in Euclidean 3-space. Kodai Math. J. 1990, 13, 10–21. [CrossRef] 8. Dursun, U.; Turgay, N.C. Minimal and pseudo-umbilical rotational surfaces in Euclidean space E4. Mediterr. J. Math. 2013, 10, 497–506. [CrossRef] 3 9. Kim, Y.H.; Turgay, N.C. Surfaces in E with L1-pointwise 1-type Gauss map. Bull. Korean Math. Soc. 2013, 50, 935–949. [CrossRef] 10. Ferrandez, A.; Garay, O.J.; Lucas, P. On a Certain Class of Conformally at Euclidean Hypersurfaces. In Global Analysis and Global Differential Geometry; Springer: Berlin, Germany, 1990; pp. 48–54. 11. Güler, E.; Turgut Vanlı, A. Bour’s theorem in Minkowski 3-space. J. Math. Kyoto Univ. 2006, 46, 47–63. [CrossRef] 12. Güler, E.; Yaylı, Y. Generalized Bour theorem. Kuwait J. Sci. 2015, 42, 79–90. 13. Güler, E.; Yaylı, Y.; Hacı saliho˘glu,H.H. Bour’s theorem on the Gauss map in 3-Euclidean space. Hacettepe J. Math. Stat. 2010, 39, 515–525. 14. Ikawa, T. Bour’s theorem and Gauss map. Yokohama Math. J. 2000, 48, 173–180. 15. Ikawa, T. Bour’s theorem in Minkowski geometry. Tokyo J. Math. 2001, 24, 377–394. [CrossRef] 16. Ji, F.; Kim, Y.H. Mean curvatures and Gauss maps of a pair of isometric helicoidal and rotation surfaces in Minkowski 3-space. J. Math. Anal. Appl. 2010, 368, 623–635. [CrossRef] 17. Ji, F.; Kim, Y.H. Isometries between minimal helicoidal surfaces and rotation surfaces in Minkowski space. Appl. Math. Comput. 2013, 220, 1–11. [CrossRef] 18. The Hieu, D.; Ngoc Thang, N. Bour’s theorem in 4-dimensional Euclidean space. Bull. Korean Math. Soc. 2017, 54, 2081–2089. 19. Ganchev, G.; Milousheva, V. General rotational surfaces in the 4-dimensional Minkowski space. Turk. J. Math. 2014, 38, 883–895. [CrossRef] 20. Güler, E.; Magid, M.; Yaylı, Y. Laplace Beltrami operator of a helicoidal hypersurface in four space. J. Geom. Sym. Phys. 2016, 41, 77–95. [CrossRef] 21. Güler, E.; Hacısaliho˘glu,H.H.; Kim, Y.H. The Gauss Map and the third Laplace–Beltrami operator of the rotational hypersurface in 4-Space. Symmetry 2018, 10, 398. [CrossRef] 22. Lawson, H.B. Lectures on Minimal Submanifolds, 2nd ed.; Mathematics Lecture Series, 9; Publish or Perish, Inc.: Wilmington, NC, USA, 1980; Volume I. 23. Takahashi, T. Minimal immersions of Riemannian manifolds. J. Math. Soc. Jpn. 1966, 18, 380–385. [CrossRef] 24. Senoussi, B.; Bekkar, M. Helicoidal surfaces with ∆Jr = Ar in 3-dimensional Euclidean space. Stud. Univ. Babe¸s-BolyaiMath. 2015, 60, 437–448. 25. Bour, E. Théorie de la déformation des surfaces. J. Êcole Imperiale Polytech. 1862, 22–39, 1–148. 26. Kenmotsu, K. Surfaces of revolution with prescribed mean curvature. Tohôku Math. J. 1980, 32, 147–153. [CrossRef] 27. Hitt, L.; Roussos, I. Computer graphics of helicoidal surfaces with constant mean curvature. An. Acad. Brasil. Ciênc. 1991, 63, 211–228. 28. Güler, E. A new kind helicoidal surface of value m. Int. Elec. J. Geom. 2014, 7, 154–162. Mathematics 2018, 6, 226 12 of 12 29. Eisenhart, L. A Treatise on the Differential Geometry of Curves and Surfaces; Palermo 41 Ginn and Company: Orland, MA, USA, 1909. 30. Spivac, M. A Comprehensive Introduction to Differential Geometry III; Interscience: New York, NY, USA, 1969. c 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).