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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(3): 585-597

Published online September 30, 2020

### Surfaces of Revolution of Type 1 in Galilean 3-Space

Ali Çakmak*, Hasan Es, Murat Kemal Karacan, Sezai KızıltuĞ

Department of Mathematics, Faculty of Sciences and Arts, Bitlis Eren University 13000, Bitlis, Turkey
e-mail : acakmak@beu.edu.tr or alicakmak@yahoo.com
Department of Mathematics Education, Faculty of Education, Gazi University 06500, Ankara, Turkey
e-mail : hasanes@gazi.edu.tr
Department of Mathematics, Faculty of Sciences and Arts, Usak University, 1 Eylul Campus 64200, Usak, Turkey
e-mail : murat.karacan@usak.edu.tr
Department of Mathematics Education, Faculty of Sciences and Arts, Erzincan University 24000, Erzincan, Turkey
e-mail : skiziltug@erzincan.edu.tr

Received: February 22, 2019; Revised: March 23, 2020; Accepted: March 30, 2020

In this study, we classify surfaces of revolution of Type 1 in the three dimensional Galilean space in terms of the position vector field, Gauss map, and Laplacian operator of the first and the second fundamental forms on the surface. Furthermore, we give a classification of surfaces of revolution of Type 1 generated by a non-isotropic curve satisfying the pointwise 1-type Gauss map equation.

Keywords: Laplace operator, Gauss map, Galilean space, surfaces of revolution, pointwise 1-type

Let be an isometric immersion of a connected n-dimensional manifold in the m-dimensional Euclidean space . Denote by H and Δ the mean curvature and the Laplacian of M with respect to the Riemannian metric on M induced from that of , respectively. Takahashi [15] proved that the submanifolds in satisfying Δx = λx, that is, for which all coordinate functions are eigenfunctions of the Laplacian with the same eigenvalue λ ∈ ℝ, are either the minimal submanifolds of or the minimal submanifolds of the hypersphere in [5, 6, 18, 19].

As an extension of Takahashi’s theorem, Garay [11] studied hypersurfaces in whose coordinate functions are eigenfunctions of the Laplacian, but not necessarily associated to the same eigenvalue. He considered hypersurfaces in satisfying the condition

$Δx=Ax,$

where AMat (m,ℝ) is an m×m diagonal matrix, and proved that such hypersurfaces are minimal (H = 0) in and are open pieces of either round hyperspheres or generalized right spherical cylinders.

Related to this, Dillen, Pas and Verstraelen [9] investigated surfaces in whose immersions satisfy the condition

$Δx=Ax+B,$

where AMat (3,ℝ) is a 3 × 3 real matrix and B ∈ ℝ3 [5, 6, 18, 19].

The notion of an isometric immersion x is naturally extended to smooth functions on submanifolds of Euclidean space or pseudo-Euclidean space. The most natural one of them is the Gauss map of the submanifold. In particular, if the submanifold is a hypersurface, the Gauss map can be identified with its unit normal vector field. Dillen, Pas and Verstraelen [10] studied surfaces of revolution in the three dimensional Euclidean space such that its Gauss map G satisfies the condition

$ΔG=AG,$

where AMat (3,ℝ).

In the late 1970’s B.-Y. Chen introduced the notion of Euclidean immersions of finite type. Essentially these are submanifolds whose immersion into the m–dimensional Euclidean space is constructed by making use of a finite number of -valued eigenfunctions of their Laplacian. The first results on this subject are collected in the book [3]. In a framework of the theory of finite type, B.-Y. Chen and P. Piccini [4] made a general study on submanifolds of Euclidean spaces with finite type Gauss maps. Several geometers also studied submanifolds of Euclidean spaces or pseudo-Euclidean spaces with finite type Gauss maps [11].

From the above definition one can see that a submanifold has 1-type Gauss map G if and only if G satisfies the equation

$ΔG=λ (G+C)$

for a constant λ and a constant vector C, where Δ denotes the Laplace operator on a submanifold. A plane, a circular cylinder and a sphere are surfaces with a 1-type Gauss map. Similarly, a submanifold is said to have a pointwise 1-type Gauss map if the Laplacian of its Gauss map takes the form

$ΔG=F (G+C)$

for a non-zero smooth function F and a constant vector C. More precisely, a pointwise 1-type Gauss map is said to be of the first kind if (1.5) is satisfied for C = 0, and of the second kind if C ≠ 0. A helicoid, a catenoid and a right cone are the typical examples of surfaces with pointwise 1-type Gauss maps [7].

Sipus and Divjak [17] defined surfaces of revolution in the 3-dimensional pseudo-Galilean space $G31$ and described surfaces of revolution of constant curvature. Yoon [18, 19] characterized surfaces of revolution in $G31$. Dede, Ekici and Goemanse [8] defined and studied three types of surfaces of revolution in Galilean 3-space. They classified the surfaces of revolution with vanishing Gaussian curvature or vanishing mean curvature in Galilean 3-space . Choi, Kim and Yoon [6] gave the classification of surfaces of revolution generated by an isotropic curve satisfying a pointwise 1-type Gauss map equation. Choi [5] completely classified the surfaces of revolution satisfying condition (1.3). Karacan, Yoon and Bukcu [13] classified surfaces of revolution satisfying ΔJxi = λixi, J = 1,2 and ΔIIIxi = λixi.

The main purpose of this paper is a complete classification of surfaces of revolution in the three dimensional Galilean space in terms of the position vector field, Gauss map, pointwise 1-type Gauss map equation and Laplacian operators of the first and the second fundamental forms on the surface.

The Galilean space G3 is a Cayley-Klein space defined from a 3-dimensional projective space ℘(ℝ3) with the absolute figure that consists of an ordered triple {w, f, I}, where w is the ideal (absolute) plane, f is the line (absolute line) in w and I is the fixed elliptic involution of points of f. We introduce homogeneous coordinates in G3 in such a way that the absolute plane w is given by x0 = 0, the absolute line f by x0 = x1 = 0 and the elliptic involution by (0 : 0 : x2 : x3) → (0 : 0 : x3 : −x2). In affine coordinates defined by (0 : x1 : x2 : x3) → (1 : x : y : z), distance between points Pi = (xi, yi, zi), i = 1, 2 is defined by

$d (P1,P2)={∣x2-x1∣,ifx1≠x2(y2-y1)2+(z2-z1)2,ifx1=x2.$

The group of motions of G3 is a six-parameter group given (in affine coordinates) by

$x¯=a+x, y¯=b+cx+y cos θ+z sin θ, z¯=d+ex-y sin θ+z cos θ.$

A Cr-surface S, r ≥ 1, immersed in the Galilean space, x : US, U ⊂ ℝ2, x(u, v) = (x(u, v), y(u, v), z(u, v)), has the following first fundamental form

$I=(g1du+g2dv)2+ɛ (h11du2+2h12dudv+h22dv2),$

where the symbols gi = xi, hij = i. j stand for derivatives of the first coordinate function x(u, v) with respect to u, v and for the Euclidean scalar product of the projections k of vectors xk onto the yz-plane, respectively. Furthermore,

$ɛ={0,if direction du:dv is non-isotropic,1,if direction du:dv is isotropic.$

In every point of a surface there exists a unique isotropic direction defined by g1du + g2dv = 0. In that direction, the arc length is measured by

$ds2=h11du2+2h12dudv+h22dv2=h11g22-2h12g1g2+h22g12g12=W2g12dv2,$

where g1 ≠ 0.

A surface is called admissible if it has no Euclidean tangent planes. Therefore, for an admissible surface either g1 ≠ 0 or g2 ≠ 0 holds. An admissible surface can always locally be expressed as

$z=f(u,v).$

The Gaussian K and mean curvature H are Cr–2 functions, r ≥ 1, defined by

$K=LN-M2W2, H=g22L-2g1g2M+g12N2W2,$

where

$Lij=x1xij-xijx1x1.G, x1=g1≠0.$

We will use Lij, i, j = 1, 2, for L, M, N if more convenient. The vector G defines a normal vector to a surface

$G=1W(0,-x2z1+x1z2,x2y1-x1y2),$

where W2 = (x2x1x1x2)2 [8, 14].

It is well known in terms of local coordinates {u, v} ofMthe Laplacian operators ΔI and ΔII of the first and the second fundamental forms on M are defined by

$ΔIx=-1EG-F2[∂∂u(Gxu-FxvEG-F2)-∂∂v(Fxu-ExvEG-F2)],$$ΔIIx=-1LN-M2[∂∂u(Nxu-MxvLN-M2)-∂∂v(Mxu-LxvLN-M2)]$

[1, 2, 12, 13, 16].

In the Galilean space there are two types of rotations: Euclidean rotations given by the normal form

$x¯=x, y¯=y cos v+z sin v, z¯=-y sin v+z cos v$

and isotropic rotations with the normal form

$x¯=x+ct, y¯=y+xt+ct22, z¯=z.$

Then the surface of revolution of Type 1 can be written as

$x(u,v)=(f(u),g(u) cos v,-g(u) sin v).$

Suppose that α is parametrized by arc-length. In this case, the parametrization of M is given by

$x(u,v)=(u,g(u) cos v,-g(u) sin v).$

Next, we consider the isotropic rotations. By rotating the isotropic curve α(u) = (0, f(u), g(u)) about the z–axis by isotropic rotation (3.2), we obtain the parametrization of the surface of revolution of Type 2 as

$x(u,v)=(cv,f(u)+cv22,g(u)),$

where f and g are smooth functions and c ≠ 0 ∈ ℝ [12].

Finally, we assume, again without loss of generality, that the profile curve α(u) = (f(u), g(u), 0) lies in the isotropic xy− plane and is parameterized by

$x(u,v)=(f(u)+cv,g(u),vf(u)+cv22),$

where f and g are smooth functions and c ≠ 0 ∈ ℝ. The surface (3.6) is called the surface of revolution of Type 3 [12].

In this section, we classify surface of revolution of Type 1 given in satisfying the equation

$ΔIx=Ax,$

where A = (aij) ∈ Mat(3,R) and

$ΔIxi=(ΔIx1,ΔIx2,ΔIx3),$

where

$x1=u, x2=g(u) cos v, x3=-g(u) sin v.$

For this surface of revolution, the coefficients of the first and second fundamental forms are

$g1=1,g2=0,h11=g′2(u), h12=0, h22=g2(u),L11=L=-g″(u), L22=N=g(u), L12=M=0,E=1, F=0, G=g2(u),$

respectively. The Gaussian curvature K and the mean curvature H are

$K=-g″(u)g(u), H=12g(u).$

### Corollary 4.1

There are no minimal surfaces of revolution (3.4).

### Corollary 4.2

The profile curve of surface of revolution of Type 1 of constant Gaussian curvature inis as follows:

• If$K=1a2$ , then the general solution of the differential equation (4.5) is

$g(u)=c1 cosua+c2 sinua,$

where c1, c2, a ∈ ℝ.

• IfK = 0, then the general solution of the differential equation (4.5) is

$g(u)=c1u+c2,$

where c1, c2 ∈ ℝ.

The Laplacian operator on M with the help of (2.3), (4.3) and (4.4) turns out to be

$ΔIx=(-g′g,cos v (1-g′2-gg″)g,-sin v (1-g′2-gg″)g).$

Suppose that M satisfies (4.1). Then from (4.2) and (4.3), we have

$a11u+a12g(u) cos v-a13g(u) sin v =-g′ga21u+a22g(u) cos v-a23g(u) sin v =cos v (1-g′2-gg″)ga31u+a32g(u) cos v-a33g(u) sin v =-sin v (1-g′2-gg″)g}$

Since the functions cos v, sin v and the constant function are linearly independent, by (4.6) we get a12 = a13 = a21 = a23 = a31 = a32 = 0, a11 = λ, a22 = a33 = μ. Consequently the matrix A satisfies

$A=[λ000μ000μ]$

and (4.6) can be rewritten as

$λu=-g′g,$$μg(u) cos v=cos v (1-g′2-gg″)g,$$μg(u) sin v=sin v (1-g′2-gg″)g.$

From (4.8), (4.9) and (4.10), we obtain

$λu=-g′g or g=-g′λu, λ≠0.μg(u)=(1-g′2-gg″)g.$

Combining the first and the second equation of (4.11), we obtain

$(λu+μλu) g′(u)-λug′(u)-g″(u)=0.$

If we solve ordinary differential equation (4.12) with Mathematica, we get

$g(u)=c1±∫1ueλx22(c2x2μλ+(λx2)μλGamma [1-μλ,λx2]) dx,$

where λ ≠ 0, μ ≠ 0, ci ∈ ℝ. The solution (4.13) does not satisfy (4.8) and (4.9). Let λ ≠ 0, μ = 0, from (4.12), we obtain

$(λu) g′(u)-λug′(u)-g″(u)=0.$

Its general solution is

$g(u)=c1±∫1u1+e2c2+λx2dx,$

where c1, c2 ∈ ℝ. The solution (4.15) does not satisfy (4.8) and (4.9). Since λ ≠ 0, there is no harmonic surface of revolution given by (3.4) in the three dimensional Galilean space .

In this section, we classify surfaces of revolution of Type 1 with non-degenerate second fundamental form in satisfying the equation

$ΔIIx=Ax,$

where A = (aij) ∈ Mat(3,R). By a straightforward computation, the Laplacian ΔII of the second fundamental form II on M is expressible as

$ΔIIxi=((g′(u)g″(u)-g(u)g‴(u)2g(u)g″2(u))cos v (g′2(u)g″(u)+4g(u)g″2(u)-g(u)g′(u)g‴(u)2g(u)g″2(u))-sin v (g′2(u)g″(u)+4g(u)g″2(u)-g(u)g′(u)g‴(u)2g(u)g″2(u))).$

Suppose that M satisfies (5.1). Then from (2.4) and (4.4), we have

$a11u+a12g(u) cos v-a13g(u) sin v=12g″(g′g-g‴g″),a21u+a22g(u) cos v-a23g(u) sin v=cos v (12g″(g′g-g‴g″)+2),a31u+a32g(u) cos v-a33g(u) sin v=-sin v (12g″(g′g-g‴g″)+2).$

Since the functions cos v, sin v and the constant function are linearly independent, by (5.3) we get a12 = a13 = a21 = a23 = a31 = a32 = 0, a11 = λ, a22 = a33 = μ. Consequently matrix A satisfies

$A=[λ000μ000μ].$

Then the system (5.3) reduces now to the equations

$2λug″2 =g′g″-gg‴,-2μg2g″2-4μgg″2 =g′(g′g″-gg‴).$

Combining the first and the second equation of (5.5), we get

$2-μg+λug′=0,$

where g ≠ 0 and g″ ≠ 0. Its general solution is given by

$g(u)=2μ+c1uμλ,$

where c1 ∈ ℝ. The solution (5.7) does not satisfies (5.5). If λ ≠ 0, μ = 0, then we have

$g(u)=c1-2 ln uλ.$

The solution (5.8) does not satisfies (5.5). Let λ = 0, μ = 0, from (5.6), we have a contradiction. Consequently, we have:

### Theorem 5.1

LetMbe a non-isotropic surface of revolution of Type 1 with non-degenerate second fundamental form given by (3.4) in the three dimensional Galilean space. There is no the surfaceMsatisfying the condition ΔIIx = Ax, AMat(3,R).

In this section, we classify surfaces of revolution of Type 1 in satisfying the equation

$ΔIG=AG,$

where A = (aij) ∈ Mat(3,R).

### Theorem 6.1

LetMbe a surface of revolution given by (3.4) in the three dimensional Galilean space. ThenMsatisfies (6.1) if and only if it is an open part of a cylinder.

Proof

Let M be a surface of revolution generated by a unit speed nonisotropic curve in . Then M is parametrized by

$x(u,v)=(u,g(u) cos v,-g(u) sin v).$

where g is a positive function. The Gauss map G of M is obtained by

$G=(0,-cos v,sin v).$

Suppose that M satisfies (6.1). Then from (4.4) and (5.3) we get the system of differential equations

$-a12 cos v+a13 sin v=0,-a22 cos v+a23 sin v=-cos vg2,-a32 cos v+a33 sin v=sin vg2.$

In order to prove the theorem we have to solve (6.4). From (6.4) we easily deduce that

$a12=a13=a21=a23=a32=0, a22=a33, a22=a33=1g2(u)$

and

$ΔIG=1g2(u)G.$

From this g(u) is a constant function. Consequently, M is an open part of a cylinder. It can be easily shown that the converse assertion is also true.

### Theorem 6.2

There is no surfaces of revolution of Type 1 generated by a nonisotropic curve inwith harmonic Gauss map.

Proof

Let M be a surface of revolution of Type 1 defined by (3.4) in . If M has harmonic Gauss map, that is, M satisfies ΔIG =0, then g−2(u)G = 0. It is impossible because g(u) is a positive function and G is the unit normal vector field of M.

### Theorem 6.3

LetMbe a surface of revolution of Type 1 generated by a nonisotropic curve in the three dimensional Galilean space. ThenMhas point wise 1-type Gauss map of the first kind.

Proof

Let M be a surface of revolution of Type 1 generated by a non-isotropic curve in . Suppose that M has pointwise 1-type Gauss map. Combining (1.5) and (6.6), one gets F(u) = g−2(u) and C = 0. Thus the Gauss map G of M is of pointwise 1-type of the first kind.

### Theorem 6.4

There is no surface of revolution of Type 1 generated by a nonisotropic curve inwith pointwise 1-type Gauss map of the second kind.

Proof

Let M be a surface of revolution of Type 1 defined by (3.4) in . By Theorem 6.3, M has only pointwise 1-type Gauss map of the first kind. Thus, the theorem is proved.

### Remark 6.5

We consider a surface defined by

$x(u,v)=(u,(a2u+b2) cos v,-(a2u+b2) sin v),$

where a, b ∈ ℝ and $u>-b2a2$. The surface is a cone satisfying the (a2x + b2)2 = y2 + z2. From (6.6) the Laplacian ΔIG of the Gauss map G of the surface is obtained by

$ΔIG=1(a2u+b2)2G.$

Thus, a cone in has pointwise 1-type Gauss map of the first kind.

In this section, we classify surfaces of revolution of Type 1 in satisfying the equation

$ΔIIG=AG,$

where A = (aij) ∈ Mat(3,R).

### Theorem 7.1

There is no non-isotropic surfaces of revolution of Type 1 given by (4.2) satisfying (7.1) in the three dimensional Galilean space.

Proof

Let M be a surface of revolution generated by a unit speed nonisotropic curve in . Suppose that M satisfies (7.1). Then from (4.4) and (5.3) we get the system of equations

$-a12 cos v+a13 sin v=0,-a22 cos v+a23 sin v=-cos vg,-a32 cos v+a33 sin v=sin vg.$

In order to prove the theorem we have to solve (7.2). From (7.2) we easily deduce that

$a12=a13=a21=a23=a32=0, a22=a33, a22=1g(u)$

and

$ΔIIG=1g(u)G.$

From this g(u) is a constant function. For the nondegeneracy of the second fundamental form of M, we assume that g″ is nonvanishing everywhere. If a non-isotropic surface of revolution of Type 1 satisfies (7.1), then the function g is constant. It is a contradiction.

### Theorem 7.2

There is no surfaces of revolution of Type 1 generated by a nonisotropic curve inwith harmonic Gauss map.

Proof

Let M be a surface of revolution of Type 1 defined by (4.1) in . If M has harmonic Gauss map, that is, M satisfies ΔIIG =0, then g−1(u) G = 0. It is impossible because g(u) is a positive function and G is the unit normal vector field of M.

### Theorem 7.3

LetMbe a surface of revolution of Type 1 generated by a nonisotropic curve in the three dimensional Galilean space. ThenMhas point wise 1-type Gauss map of the first kind.

Proof

Let M be a surface of revolution of Type 1 generated by a non-isotropic curve in . Suppose that M has pointwise 1-type Gauss map. Combining (1.5) and (7.3), one gets F(u) = g−1(u) and C = 0. Thus the Gauss map G of M is of pointwise 1-type of the first kind.

### Theorem 7.4

There is no surface of revolution of Type 1 generated by a nonisotropic curve inwith pointwise 1-type Gauss map of the second kind.

Proof

Let M be a surface of revolution of Type 1 defined by (4.1) in . By Theorem 7.3, M has only pointwise 1-type Gauss map of the first kind. Thus, the theorem is proved.

### Remark 7.5

We consider a surface defined by

$x(u,v)=(u, (a2u+b2) cos v,-(a2u+b2) sin v),$

where a, b ∈ ℝ and $u>-b2a2$. The surface is a cone satisfying the (a2x + b2)2 = y2 + z2. From (7.3) the Laplacian ΔIIG of the Gauss map G of the surface is obtained by

$ΔIIG=1(a2u+b2)G.$

Thus, a cone in has pointwise 1-type Gauss map of the first kind.

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