Translation and homothetical TH-surfaces in the 3-dimensional Euclidean space \(\mathbb{E}^{3}\) and Lorentzian-Minkowski space \(\mathbb{E}_{1}^{3}\)

1. Introduction

The theory of minimal surfaces has found many applications in differential geometry and also in physics. In [1] and [2], H. Liu gave some classification results for translation surfaces. A minimal translation hypersurface in a Euclidean space is either locally a hyperplane or an open part of a cylinder on Scherk's surfaces, as proved in Dillen et al. [3]. In [1] was generalized to translation surfaces with constant mean curvature and constant Gaussian curvature in \(\mathbb{E}^{3}.\) Saglam and Sabuncuoglu proved that every homothetical lightlike hypersurface in a semi-Euclidean \(\mathbb{E}_{q}^{m+2}\) space is minimal [4]. Jiu and Sun studied \(n-\) dimensional minimal homothetical hypersurfces and gave their classification [5]. R. Lopez [6] studied translation surfaces in the 3-dimensional hyperbolic space and classified minimal translation surfaces. Meng and Liu [7] considered factorable surfaces along two lightlike directions and spacelike-lightlike directions in Minkowski 3-space \(\mathbb{E}_{1}^{3}\) and they also gave some classifcation theorems. In [8], Yu and Liu studied the factorable minimal surfaces in \(\mathbb{E}_{1}^{3}\) and \(\mathbb{E}^{3}\), and gave some classification theorems. Guler et al. [9] defined by translation and homothetical TH-surfaces in the three dimensional Euclidean space.

2. Preliminaries

Let \(\mathbb{E}_{1}^{3}\) be a 3-dimensional Minkowski space with the scalar product of index \(1\) given by \begin{equation*} g_{L}=ds^{2}=-dx^{2}+dy^{2}+dz^{2}, \end{equation*} where \((x,y,z)\) is a rectangular coordinate system of \(\mathbb{E}_{1}^{3}.\) A vector \(V\) of \(\mathbb{E}_{1}^{3}\) is said to be timelike if \(g_{L}(V,V)< 0, \) spacelike if \(g_{L}(V,V)>0\) or \(V=0\) and lightlike or null if \(g_{L}(V,V)=0 \) and \(V\neq 0.\) A surface in \(\mathbb{E}_{1}^{3}\) is spacelike, timelike or lightlike if the tangent plane at any point is spacelike, timelike or lightlike, respectively. The Lorentz scalar product of the vectors \(V\) and \(W\) is defined by \(g_{L}(V,W)=-v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3},\) where \(V=(v_{1},v_{2},v_{3}),\) \(W=(w_{1},w_{2},w_{3})\in \) \(\mathbb{E}_{1}^{3}.\) For any \(V,\) \(W\in \) \(\mathbb{E}_{1}^{3}\), the pseudo-vector product of \(V\) and \(W\) is defined as follows: \begin{equation*} V\wedge _{L}W=\big(-v_{2}w_{3}+v_{3}w_{2},\text{ }v_{3}w_{1}-v_{1}w_{3}\text{ },v_{1}w_{2}-v_{2}w_{1}\big). \end{equation*} We denote a surface \(M^{2}\) in \(\mathbb{E}_{1}^{3}\) by \begin{equation*} r(u,v)=\big(r_{1}(u,v),\text{ }r_{2}(u,v),\text{ }r_{3}(u,v)\big). \end{equation*}

Definition 1. [10] A translation surface in Minkowski \(3\)-space is a surface that is parameterized by either \begin{eqnarray*} r(u,v) &=&(u,\text{ }v,\text{ }f(u)+g(v))\;\; if\;\; L\;\; is \;\; timelike, \\ r(u,v) &=&(f(u)+g(v),\text{ }u,\text{ }v)\;\; if \;\; L\;\; is \;\; spacelike, \\ r(u,v) &=&(u+v,\text{ }g(v),\text{ }f(u)+v)\;\; if\;\; L\;\; is \;\; lightlike, \end{eqnarray*} with \(L\) the intersection of the two planes that contain the curves that generate the surface.

Theorem 2. [11]

1. The only translation surfaces with constant Gauss curvature \(K=0\) are cylindrical surfaces.
2. There are no translation surfaces with constant Gauss curvature \(K\neq 0\) if one of the generating curves is planar.

Definition 3. A homothetical (factorable) surface \(M^{2}\) in the 3-dimensional Lorentzian space \(\mathbb{E}_{1}^{3}\) is a surface that is a graph of a function \begin{equation*} z(u,\text{ }v)=f(u)g(v), \end{equation*} where \(f:I\subset \mathbb{R}\rightarrow \mathbb{R}\) and \(g:J\subset \mathbb{R}\rightarrow \mathbb{R}\) are two smooth functions.

Theorem 4. [11] Planes and helicoids are the only minimal homothetical surfaces in Euclidean space.

Accordingly, we define an extended surface in \(\mathbb{E}_{1}^{3}\) using definitions as above and called it TH-type surface as follows [9]:

Definition 5. A surface \(M^{2}\) in the 3-dimensional Lorentzian space \(\mathbb{E}_{1}^{3}\) is a TH-surface if it can be parameterized either by a patch

or where \(A\) and \(B\) are non-zero real numbers.

Remark 1.

1. If \(A\neq 0\) and \(B=0\) in (1), then surface is a translation surface.
2. If \(A=0\) and \(B\neq 0\) in (1), then surface is a homothetical (factorable) surface.

Let \(\mathbf{N}\) denotes the unit normal vector field of \(M^{2}\) and put \( g_{L}(\mathbf{N},\mathbf{N})=\varepsilon =\pm 1\), so that \(\varepsilon =-1\) or \(\varepsilon =1\) according to \(M^{2}\) is endowed with a Lorentzian or Riemannian metric, respectively.

The mean curvature and the Gauss curvature are \begin{equation*} H=\frac{EN+GL-2FM}{2\left\vert EG-F^{2}\right\vert },\text{ }K=g_{L}(\mathbf{ N},\mathbf{N})\frac{LN-M^{2}}{EG-F^{2}}, \end{equation*} where \(E,\) \(G,\) \(F\) are the coefficients of the first fundamental form, \(L,\) \(M,\) \(N\) \ are the coefficients of the second fundamental form.

In this paper, we define TH-surfaces in the 3-dimensional Euclidean space \( \mathbb{E}^{3}\) and Lorentzian-Minkowski space \(\mathbb{E}_{1}^{3},\) and completely classify minimal or flat TH-surfaces.

3. Minimal TH-surfaces in \(\mathbb{E}_{1}^{3}\)

A surface \(M^{2}\) in the 3-dimensional Lorentzian space \(\mathbb{E}_{1}^{3}\) is called minimal when locally each point on the surface has a neighborhood which is the surface of least area with respect to its boundary [12]. In 1775, J. B. Meusnier showed that the condition of minimality of a surface in \(\mathbb{E}^{3}\) is equivalent with the vanishing of its mean curvature function, \(H=0.\)

Let \(z=f(x,y)\) define a graph \(M^{2}\) in the Euclidean 3-space \(\mathbb{E} ^{3}\). If \(M^{2}\) is minimal, the function \(f\) \ satisfies

which was obtained by J. L. Lagrange in \(1760\).

Let \(M^{2}\) be a TH-surface in \(\mathbb{E}_{1}^{3}\) parameterized by a patch \begin{equation*} r(u,\text{ }v)=(u,\text{ }v,\text{ }A(f(u)+g(v))+Bf(u)g(v)), \end{equation*} where \(A\) and \(B\) are non-zero real numbers.

So \begin{equation*} r_{u}=(1,\text{ }0,\text{ }f^{\prime }\alpha ),\;\; r_{v}=(0,\text{ }1, \text{ }g^{\prime }\gamma ), \end{equation*} where \(\alpha =A+Bg\) and \(\gamma =A+Bf.\)

After eliminating \(f^{\prime }\) and \(g^{\prime }\) we find \begin{equation*} E=\frac{\gamma ^{\prime 2}\alpha ^{2}-B^{2}}{B^{2}},\;F=\frac{\alpha \gamma \alpha ^{\prime }\gamma ^{\prime }}{B^{2}},\text{ }G=\frac{\gamma ^{2}\alpha ^{\prime 2}+B^{2}}{B^{2}}. \end{equation*}

The unit normal vector is given by \begin{equation*} N =\frac{1}{WB}(\alpha \gamma ^{\prime },\text{ }-\gamma \alpha ^{\prime },\text{ }B), \end{equation*} where \(W^{2}=B^{-2}g_{L}(N,N)(\gamma ^{\prime 2}\alpha ^{2}-\alpha ^{\prime 2}\gamma ^{2}-B^{2})\) and \begin{equation*} g_{L}(N,N)=\varepsilon ,\;\varepsilon =\left\{ \begin{array}{l} 1\;\; M^{2}\text{ is spacelike }(\gamma ^{\prime 2}\alpha ^{2}-\alpha ^{\prime 2}\gamma ^{2}-B^{2}>0), \\ -1\;\; M^{2}\text{ is timelike }(\gamma ^{\prime 2}\alpha ^{2}-\alpha ^{\prime 2}\gamma ^{2}-B^{2}< 0)\text{.} \end{array} \right. \end{equation*} The constant \(\varepsilon \) is called the sign of \(M^{2}\). The coefficients of the second fundamental form are given by \begin{equation*} L=\frac{\alpha \gamma ^{\prime \prime }}{BW},\;M=\frac{\alpha ^{\prime }\gamma ^{\prime }}{BW},\;N=\frac{\gamma \alpha ^{\prime \prime }}{BW}. \end{equation*} The expression of \(H\) is Then \(M^{2}\) is a minimal surface if and only if We distinguish the following cases.
Case 1. Let \(\gamma ^{\prime }=0\). In this case (5) gives \(\gamma \alpha ^{\prime \prime }=0.\)

1. If \(\gamma =0\), then \(f(u)=-\frac{A}{B}\), \(M^{2}\) is the horizontal plane of equation \(z=-\frac{A^{2}}{B}\).
2. If \(\alpha ^{\prime \prime }=0\), then \(\alpha (v)=a_{1}v+b_{1},\) \(a_{1},\) \(b_{1}\in\mathbb{R}\), and \(\gamma (u)=c_{1},\) \(c_{1}\in\mathbb{R},\ M^{2}\) is the plane of equation \(z=c_{2}v+c_{3},\) \(c_{2},\) \(c_{3}\in\mathbb{R}\).

Case 2. Let \(\alpha ^{\prime }=0\). In this case (5) gives \(\gamma ^{\prime \prime }\alpha =0.\)

1. ]If \(\alpha =0\), then \(g(v)=-\frac{A}{B}\), \(M^{2}\) is the horizontal plane of equation \(z=-\frac{A^{2}}{B}\).
2. If \(\gamma ^{\prime \prime }=0\), then \(\gamma (u)=a_{2}u+b_{2},\) \( a_{2},\) \(b_{2}\in\mathbb{R}\), and \(\alpha (v)=c_{4},\) \(c_{4}\in\mathbb{R},\ M^{2}\) is the plane of equation \(z=c_{5}u+c_{6},\) \(c_{5},\) \(c_{6}\in\mathbb{R}\).

Case 3. Let \(\gamma ^{\prime \prime }=0\) and \(\gamma ^{\prime }\neq 0\). Then \(\gamma (u)=\lambda u+\delta ,\) \((\lambda ,\) \(\delta )\in\mathbb{R}\setminus \left\{ 0\right\} \times\mathbb{R}\) and \(\alpha \) is a solution of the following ODE Then the general solution of (6) is given by \begin{equation*} \alpha (v)=-\frac{B}{\lambda }\coth (\lambda _{1}v+\lambda _{2}),\text{ } \lambda _{1},\text{ }\lambda _{2}\in\mathbb{R}. \end{equation*} Hence \begin{equation*} g(v)=-\frac{1}{\lambda }\coth (\lambda _{1}v+\lambda _{2})-\frac{A}{B},\text{ }\lambda _{1},\text{ }\lambda _{2}\in\mathbb{R}. \end{equation*} Case 4. Let \(\alpha ^{\prime \prime }=0\) and \(\alpha ^{\prime }\neq 0\). Then \(\alpha (v)=\lambda v+\delta ,\) \((\lambda ,\) \(\delta )\in\mathbb{R}\setminus \left\{ 0\right\} \times\mathbb{R}\) and \(\gamma \) is a solution of the following ODE Then the general solution of (7) is given by \begin{equation*} \gamma (u)=\frac{B}{\lambda }\tan (\lambda _{1}u+\lambda _{2}),\text{ } \lambda _{1},\text{ }\lambda _{2}\in\mathbb{R}. \end{equation*} Hence \begin{equation*} f(u)=\frac{1}{\lambda }\tan (\lambda _{1}u+\lambda _{2})-\frac{A}{B},\text{ } \lambda _{1},\text{ }\lambda _{2}\in\mathbb{R}. \end{equation*} Case 5. Let \(\gamma ^{\prime \prime }\neq 0.\) By symmetry in the discussion of the case, we also suppose \(\alpha ^{\prime \prime }\neq 0\). If we divide (5) by \(\alpha \gamma \alpha ^{\prime 2}\gamma ^{\prime 2},\) we obtain \begin{equation*} \frac{B^{2}\gamma ^{\prime \prime }}{\gamma \alpha ^{\prime 2}\gamma ^{\prime 2}}+\frac{\gamma \gamma ^{\prime \prime }}{\gamma ^{\prime 2}}- \frac{B^{2}\alpha ^{\prime \prime }}{\alpha \alpha ^{\prime 2}\gamma ^{\prime 2}}+\frac{\alpha \alpha ^{\prime \prime }}{\alpha ^{\prime 2}}-2=0. \end{equation*} Thus, after a derivation with respect to \(u\), followed by a derivation with respect to \(v,\) we obtain \begin{equation*} \left( \frac{\gamma ^{\prime \prime }}{\gamma \gamma ^{\prime 2}}\right) _{,u}\left( \frac{1}{\alpha ^{\prime 2}}\right) _{,v}-\left( \frac{\alpha ^{\prime \prime }}{\alpha \alpha ^{\prime 2}}\right) _{,v}\left( \frac{1}{ \gamma ^{\prime 2}}\right) _{,u}=0. \end{equation*} Hence we deduce the existence of a real number \(k\in\mathbb{R}\) such that The first equation of (8) can integrate obtaining From the second equation in (8), we obtain Substituting the above in (5), we get \begin{equation*} \alpha \gamma ((k+c\gamma ^{\prime 2})(B^{2}+\alpha ^{\prime 2}\gamma ^{2})-2\alpha ^{\prime 2}\gamma ^{\prime 2}+(k+b\alpha ^{\prime 2})(\gamma ^{\prime 2}\alpha ^{2}-B^{2}))=0. \end{equation*} If we simplify by \(\alpha \gamma \) and then we divide by \(\alpha ^{\prime 2}\gamma ^{\prime 2}\), we get \begin{equation*} \frac{bB^{2}-k\gamma ^{2}}{\gamma ^{\prime 2}}-c\gamma ^{2}+2=\frac{ cB^{2}+k\alpha ^{2}}{\alpha ^{\prime 2}}+b\alpha ^{2}. \end{equation*} Hence, we deduce the existence of a real number \(\lambda \in\mathbb{R}\) such that Differentiating with respect to \(u\) and \(v\), respectively, we have Let us compare these expressions of \(\alpha ^{\prime \prime }\) and \(\gamma ^{\prime \prime }\) with those ones that appeared in (9) and (10) and replace the values of \(\gamma ^{\prime 2}\) and \(\alpha ^{\prime 2}\) obtained in (11).
We get \begin{equation*} \left\{ \begin{array}{l} (\lambda k+bcB^{2})(\lambda -1-b\alpha ^{2})=0 \\ ((\lambda -2)k+bcB^{2})(\lambda -1+c\gamma ^{2})=0. \end{array} \right. \end{equation*} We discuss all possibilities.

1. If \begin{equation*} \left\{ \begin{array}{l} \lambda k+bcB^{2}=0 \\ (\lambda -2)k+bcB^{2}=0, \end{array} \right. \end{equation*} then \(k=0\) and \(bc=0\). Then (12) gives \(\gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
2. If \begin{equation*} \left\{ \begin{array}{l} \lambda k+bcB^{2}=0 \\ c=0 \\ \lambda =1, \end{array} \right. \end{equation*} we obtain \(k=0\). Then (12) gives \(\gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
3. If \begin{equation*} \left\{ \begin{array}{l} (\lambda -2)k+bcB^{2}=0 \\ b=0 \\ \lambda =1, \end{array} \right. \end{equation*} we obtain \(k=0\). Then (12) gives \(\gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
4. If \begin{equation*} \left\{ \begin{array}{c} \lambda -1-b\alpha ^{2}=0 \\ \lambda -1+c\gamma ^{2}=0 \end{array} \right. \end{equation*} we deduce that \(\alpha \), \(\gamma \) are both constant functions, and so, \( \gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
5. If \(b=0,\) \(c=0\) and \(\lambda =1,\) Equation (11) writes as
   \begin{equation} \left\{ \begin{array}{c} \gamma ^{\prime 2}=k\gamma ^{2} \\ \alpha ^{\prime 2}=k\alpha ^{2}. \end{array} \right. \label{surface minimal6} \end{equation}

   (13)
   The equations (13) have the following solutions \begin{equation*} \alpha (v)=k_{1}e^{\sqrt{k}v},\gamma (u)=k_{2}e^{\sqrt{k}u},\text{ }k>0, \end{equation*} where \(k_{1}\), \(k_{2}\in\mathbb{R}\) are integration constants.

Hence \begin{equation*} g(v)=\lambda _{1}e^{\sqrt{k}v}-\frac{A}{B}, f(u)=\lambda _{2}e^{ \sqrt{k}u}-\frac{A}{B},\text{ }k>0. \end{equation*} Therefore, we have the following:

Theorem 6. Let \(M^{2}\) be a TH-surface in \(\mathbb{E}_{1}^{3}.\) If \(M^{2}\) is minimal surface, then \(M^{2}\) can be parameterized as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }A(f(u)+g(v))+Bf(u)g(v)), \end{equation*} where
\(1)\) either \(f(u)=-\frac{A}{B}\) and \(g(v)\) is a smooth function in \(v.\) \(2)\) \(g(v)=-\frac{A}{B}\) and \(f(u)\) is a smooth function in \(u.\)
\(3)\) \(f(u)=\lambda _{1}u+\lambda _{2}\) and \(g(v)=\lambda _{3}\coth (\lambda _{4}v+\lambda _{5})-\lambda _{6}\), \(\lambda _{i}\in\mathbb{R}.\)
\(4)\) \(f(u)=\frac{1}{\lambda }\tan (\lambda _{1}u+\lambda _{2})-\frac{A}{B},\) \(\lambda _{1},\) \(\lambda _{2}\in\mathbb{R}\) and \(g(v)=\delta _{5}v+\delta _{6}\), \(\delta _{i}\in\mathbb{R}.\)
\(5)\) \(f(u)=\lambda _{2}e^{\sqrt{k}u}-\frac{A}{B}\) and \(g(v)=\lambda _{1}e^{ \sqrt{k}v}-\frac{A}{B}.\)

Let \(M^{2}\) be a TH-surface in \(\mathbb{E}_{1}^{3}\) parameterized by a patch \begin{equation*} r(u,\text{ }v)=(A(f(u)+g(v))+Bf(u)g(v),\text{ }u,\text{ }v), \end{equation*} where \(A\) and \(B\) are non-zero real numbers.
So \begin{equation*} r_{u}=(f^{\prime }\alpha ,\text{ }1,\text{ }0),\;r_{v}=(g^{\prime }\gamma ,\text{ }0,\text{ }1), \end{equation*} where \(\alpha =A+Bg\) and \(\gamma =A+Bf.\)
We have \begin{equation*} E=\frac{-\gamma ^{\prime 2}\alpha ^{2}+B^{2}}{B^{2}},\; F=-\frac{ \alpha \gamma \alpha ^{\prime }\gamma ^{\prime }}{B^{2}},\text{ }G=\frac{ -\gamma ^{2}\alpha ^{\prime 2}+B^{2}}{B^{2}}. \end{equation*} The coefficients of the second fundamental form on \(M^{2}\) are obtained by \begin{equation*} L=\frac{\alpha \gamma ^{\prime \prime }}{BW}, M=\frac{\alpha ^{\prime }\gamma ^{\prime }}{BW},N=\frac{\gamma \alpha ^{\prime \prime }}{BW}. \end{equation*} Then \(M^{2}\) is a minimal surface if and only if where \(\alpha =A+Bg\) and \(\gamma =A+Bf.\)
Using the same algebraic techniques as in the case of surfaces (1 ), we get:

Theorem 7. Let \(M^{2}\) be a TH-surface in \(\mathbb{E}_{1}^{3}.\) If \(M^{2}\) is minimal surface, then \(M^{2}\) can be parameterized as \begin{equation*} r(u,\text{ }v)=(A(f(u)+g(v))+Bf(u)g(v),\text{ }u,\text{ }v), \end{equation*} where
\(1)\) either \(f(u)=\frac{\zeta }{B}u+\alpha \) and \(g(v)=-\frac{1}{\zeta } \coth (\lambda _{3}v+\lambda _{4})-\frac{A}{B}.\)
\(2)\) \(f(u)=-\frac{A}{B}\) and \(g(v)\) is a smooth function in \(v.\)
\(3)\) \(g(v)=-\frac{A}{B}\) and \(f(u)\) is a smooth function in \(u.\)
\(4)\) or \(g(v)=\frac{\delta }{B}v+\mu \) and \(f(u)=-\frac{1}{\delta }\coth (\lambda _{1}u+\lambda _{2})-\frac{A}{B}.\)

4. TH-surfaces with zero Gaussian curvature in \(\mathbb{E}_{1}^{3}\)

A non-degenerate surface in \(\mathbb{E}_{1}^{3}\) is called flat, if its Gaussian curvature vanishes identically.

A surface in \(\mathbb{E}_{1}^{3}\) parameterized by (1), after eliminating \(f,\) \(g\) and their derivatives, has Gaussian curvature \begin{equation*} K=g_{L}(N,N)\frac{\alpha \gamma \alpha ^{\prime \prime }\gamma ^{\prime \prime }-\gamma ^{\prime 2}\alpha ^{\prime 2}}{ B^{2}W^{4}}. \end{equation*}

Suppose that \(M^{2}\) has zero Gaussian curvature. Then we have Case 1. Let \(\gamma ^{\prime }=0\). In this case \(\gamma \) is a constant function \(\gamma (u)=u_{0}\) and the parametrization of (1) writes as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }\delta _{1}g(v)+\delta _{2});\text{ }\delta _{1},\delta _{2}\in\mathbb{R}. \end{equation*} This means that \(M^{2}\) is a cylindrical surface with base curve a plane curve in the \(vz-\) plane.
Case 2. Let \(\alpha ^{\prime }=0\). In this case \(\alpha \) is a constant function \(\alpha (v)=v_{0}\) and the parametrization of (1) writes as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }\delta _{3}f(u)+\delta _{4});\text{ }\delta _{3},\delta _{4}\in\mathbb{R}. \end{equation*} This means that \(M^{2}\) is a cylindrical surface with base curve a plane curve in the \(uz-\) plane.
Case 3. Let \(\gamma ^{\prime \prime }=0\) and \(\gamma ^{\prime }\neq 0\). Then \(\gamma (u)=\lambda _{1}u+\lambda _{2},\) \((\lambda _{1},\) \(\lambda _{2})\in\mathbb{R}\backslash \left\{ 0\right\} \times\mathbb{R}.\) Moreover, (15) gives \(\alpha ^{\prime }=0\) and \(\alpha (v)=\) \(v_{0}\) is a constant function. Now \(M^{2}\) is the plane of equation \( z(u,v)=\lambda _{3}u+\lambda _{4};\) \(\lambda _{3},\) \(\lambda _{4}\in\mathbb{R}\)
Case 4. Let \(\alpha ^{\prime \prime }=0\) and \(\alpha ^{\prime }\neq 0\). Then \(\alpha (v)=\lambda v+\delta _{1},\) \((\lambda ,\) \(\delta _{1})\in\mathbb{R}\setminus \left\{ 0\right\} \times\mathbb{R}.\) Moreover, (15) gives \(\gamma ^{\prime }=0\) and \(\gamma (u)=\) \(u_{0}\) is a constant function. Now \(M^{2}\) is the plane of equation \( z(u,v)=\lambda _{5}u+\lambda _{6};\) \(\lambda _{5},\) \(\lambda _{6}\in\mathbb{R}.\)
Case 5. Let \(\gamma ^{\prime \prime }\neq 0\) and \(\alpha ^{\prime \prime }\neq 0\).
Equation (15) writes as \begin{equation*} \frac{\gamma \gamma ^{\prime \prime }}{\gamma ^{\prime 2}}=\frac{\alpha ^{\prime 2}}{\alpha \alpha ^{\prime \prime }}. \end{equation*} Therefore, there exists a real number \(\lambda \in\mathbb{R}\setminus \left\{ 0\right\} \) uch that \begin{equation*} \frac{\gamma \gamma ^{\prime \prime }}{\gamma ^{\prime 2}}=\lambda =\frac{ \alpha ^{\prime 2}}{\alpha \alpha ^{\prime \prime }}. \end{equation*} Integrate these equations where \(k_{1}\) and \(k_{2}\) are constants of integration.

1. If \(\lambda =1\), the general solution of (16) is given by \begin{equation*} \left\{ \begin{array}{c} \gamma (u)=\lambda _{1}e^{k_{1}u} \\ \alpha (v)=\lambda _{2}e^{k_{2}v}, \end{array} \right. \end{equation*} where \(\lambda _{1}\) and \(\lambda _{2}\) are constants of integration.
   Hence \begin{equation*} \left\{ \begin{array}{l} f(u)=\lambda _{3}e^{k_{1}u}+\lambda _{4} \\ g(v)=\lambda _{5}e^{k_{2}v}+\lambda _{6}, \end{array} \right. \end{equation*} where \(\lambda _{3}\), \(\lambda _{4},\) \(\lambda _{5}\), \(\lambda _{6}\in\mathbb{R}\).
2. If \(\lambda \neq 1\), the general solution of (16) is given by \begin{equation*} \left\{ \begin{array}{c} \gamma (u)=((1-\lambda )k_{1}u+c_{1})^{\frac{1}{1-\lambda }} \\ \alpha (v)=((\frac{\lambda -1}{\lambda })k_{2}v+c_{2})^{\frac{\lambda }{ \lambda -1}}, \end{array} \right. \end{equation*} where \(c_{1}\) and \(c_{2}\) are constants of integration.
   Hence \begin{equation*} \left\{ \begin{array}{l} f(u)=c_{3}((1-\lambda )k_{1}u+c_{1})^{\frac{1}{1-\lambda }}+c_{4} \\ g(v)=c_{5}((\frac{\lambda -1}{\lambda })k_{2}v+c_{2})^{\frac{\lambda }{ \lambda -1}}+c_{6}, \end{array} \right. \end{equation*} where \(c_{3}\), \(c_{4},\) \(c_{5}\), \(c_{6}\in\mathbb{R}\).

Theorem 8. Let \(M^{2}\) be a TH-surface in \(\mathbb{E}_{1}^{3}\) with constant Gauss curvature \(K\). If \(M^{2}\)\ has zero Gaussian curvature, then \(M^{2}\) can be parameterized as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }z(u,v)=A(f(u)+g(v))+Bf(u)g(v)), \end{equation*} where
\(1)\) either \(f(u)=\lambda _{1}e^{k_{1}u}+\lambda _{2}\) and \(g(v)=\lambda _{3}e^{k_{2}v}+\lambda _{4},\)
\(2)\) or \(f(u)=\mu _{1}u+\mu _{2}\) and \(g(v)=\mu _{3},\)
\(3)\) or \(g(v)=\nu _{1}v+\nu _{2}\) and \(f(u)=\nu _{3},\)
\(4)\) or \(f(u)=\zeta _{1}((1-\lambda )k_{1}u+\zeta _{2})^{\frac{1}{1-\lambda } }+\zeta _{3}\) and \(g(v)=\zeta _{4}((\frac{\lambda -1}{\lambda })k_{2}v+\zeta _{5})^{\frac{\lambda }{\lambda -1}}+\zeta _{6}\).

5. Minimal TH-surfaces in \(\mathbb{E}^{3}\)

Let \(M^{2}\) be a TH-surface in the Euclidean 3-space \(\mathbb{E}^{3}.\) Then, \(M^{2}\) is parameterized by \begin{equation*} r(u,\text{ }v)=(u,\text{ }v,\text{ }A(f(u)+g(v))+Bf(u)g(v)), \end{equation*} where \(A\) and \(B\) are non-zero real numbers.
We have the natural frame \(\left\{ r_{u},\text{ }r_{v}\right\} \) given by \begin{equation*} r_{u}=(1,\text{ }0,\text{ }f^{\prime }\alpha ),\text{ \ }r_{v}=(0,\text{ }1, \text{ }g^{\prime }\gamma ), \end{equation*} where \(\alpha =A+Bg\) and \(\gamma =A+Bf.\)
From this, the unit normal vector field \(N\) of \(M^{2}\) is given by \begin{equation*} N=\frac{1}{W}(-\alpha f^{\prime },\text{ }-\gamma g^{\prime }, \text{ }1), \end{equation*} where \(W=\sqrt{1+f^{\prime 2}\alpha ^{2}+g^{\prime 2}\gamma ^{2}}.\)
The coefficients of the first fundamental form of \(M^{2}\) are given by \begin{equation*} E=1+f^{\prime 2}\alpha ^{2},\text{ }G=1+g^{\prime 2}\gamma ^{2},\text{ } F=f^{\prime }g^{\prime }\alpha \gamma . \end{equation*} The coefficients of the second fundamental form of the surface are \begin{equation*} L=\frac{\alpha f^{\prime \prime }}{W},\text{ \ }M=\frac{Bf^{\prime }g^{\prime }}{W},\text{ }N=\frac{\gamma g^{\prime \prime }}{W}. \end{equation*} Hence, the mean curvature \(H\) and the Gaussian curvature \(K\) are given by, respectively If the surface is minimal, that is, \(H=0\) on \(M^{2}\), we have from (17) \begin{equation*} \alpha f^{\prime \prime }(1+g^{\prime 2}\gamma ^{2})-2B\alpha \gamma f^{\prime 2}g^{\prime 2}+\gamma g^{\prime \prime }(1+f^{\prime 2}\alpha ^{2})=0. \end{equation*} The previous equation may be rewritten as Since the roles of \(\alpha \) and \(\gamma \) in (19) are symmetric, we only discuss the cases according to the function \(\gamma \). We distinguish cases.
Case 1. Let \(\gamma ^{\prime }=0\). In this case (19) gives \(B^{2}\gamma \alpha ^{\prime \prime }=0.\)

1. If \(\gamma =0\), then \(f(u)=-\frac{A}{B}\), \(M^{2}\) is the horizontal plane of equation \(z=-\frac{A^{2}}{B}\).
2. If \(\alpha ^{\prime \prime }=0\), then \(g(v)=av+b,\) \(a,\) \(b\in\mathbb{R}\), and \(f(u)=c,\) \(c\in\mathbb{R}, M^{2}\) is the plane of equation \(z=c_{1}v+c_{2},\) \(c_{1},\) \(c_{2}\in\mathbb{R}\).

Case 2. Let \(\gamma ^{\prime \prime }=0\) and \(\gamma ^{\prime }\neq 0\), and by symmetry, \(\alpha ^{\prime }\neq 0\). Then \(\gamma (u)=\lambda u+\delta _{1},\) \((\lambda ,\) \(\delta )\in\mathbb{R}^{\ast }\times\mathbb{R}\) and \(\alpha \) is a solution of the following ODE Then the general solution of (20) is given by \begin{equation*} \alpha (v)=\frac{B}{\lambda }\tan (\lambda _{1}v+\lambda _{2}),\text{ } \lambda _{1},\text{ }\lambda _{2}\in\mathbb{R}. \end{equation*} Hence \begin{equation*} g(v)=\frac{1}{\lambda }\tan (\lambda _{1}v+\lambda _{2})-\frac{A}{B},\text{ } \lambda _{1},\text{ }\lambda _{2}\in\mathbb{R}. \end{equation*} So, the parametrization of \(M^{2}\) can be written in the form \begin{equation*} r(u,\text{ }v)=(u,\text{ }v,\text{ }\lambda _{3}u+\delta _{2}+\frac{A}{ \lambda }\tan (\lambda _{1}v+\lambda _{2})-\frac{A^{2}}{B}+B(\lambda _{3}u+\delta _{2})(\frac{1}{\lambda }\tan (\lambda _{1}v+\lambda _{2})-\frac{ A}{B})), \end{equation*} where \((\lambda _{3},\) \(\delta _{2})\in\mathbb{R}^{\ast }\times\mathbb{R}.\)
Case 3. Let \(\gamma ^{\prime \prime }\neq 0.\) By symmetry in the discussion of the case, we also suppose \(\alpha ^{\prime \prime }\neq 0\). If we divide (19) by \(\alpha \gamma \alpha ^{\prime 2}\gamma ^{\prime 2},\) we obtain \begin{equation*} \frac{B^{2}\gamma ^{\prime \prime }}{\gamma \alpha ^{\prime 2}\gamma ^{\prime 2}}+\frac{\gamma \gamma ^{\prime \prime }}{\gamma ^{\prime 2}}+ \frac{B^{2}\alpha ^{\prime \prime }}{\alpha \alpha ^{\prime 2}\gamma ^{\prime 2}}+\frac{\alpha \alpha ^{\prime \prime }}{\alpha ^{\prime 2}}-2=0. \end{equation*} Thus, after a derivation with respect to \(u\), followed by a derivation with respect to \(v,\) we obtain \begin{equation*} \left( \frac{\gamma ^{\prime \prime }}{\gamma \gamma ^{\prime 2}}\right) _{,u}\left( \frac{1}{\alpha ^{\prime 2}}\right) _{,v}+\left( \frac{\alpha ^{\prime \prime }}{\alpha \alpha ^{\prime 2}}\right) _{,v}\left( \frac{1}{ \gamma ^{\prime 2}}\right) _{,u}=0. \end{equation*} Hence we deduce the existence of a real number \(k\in\mathbb{R}\) such that The first equation of (21) can integrate obtaining From the second equation in (21), we obtain Substituting the above in (19), we get \begin{equation*} \alpha \gamma ((k+b_{1}\gamma ^{\prime 2})(B^{2}+\alpha ^{\prime 2}\gamma ^{2})-2\alpha ^{\prime 2}\gamma ^{\prime 2}-(k+b_{2}\alpha ^{\prime 2})(B^{2}+\gamma ^{\prime 2}\alpha ^{2}))=0. \end{equation*} If we simplify by \(\alpha \gamma \) and then we divide by \(\alpha ^{\prime 2}\gamma ^{\prime 2}\), we get \begin{equation*} \frac{k\gamma ^{2}-b_{2}B^{2}}{\gamma ^{\prime 2}}-2+b_{1}\gamma ^{2}=\frac{ k\alpha ^{2}-b_{1}B^{2}}{\alpha ^{\prime 2}}+b_{2}\alpha ^{2}. \end{equation*} Hence, we deduce the existence of a real number \(\lambda \in\mathbb{R}\) such that Differentiating with respect to \(u\) and \(v\), respectively, we have Let us compare these expressions of \(\alpha ^{\prime \prime }\) and \(\gamma ^{\prime \prime }\) with those ones that appeared in (22) and ( 23) and replace the value of \(\gamma ^{\prime 2}\) and \(\alpha ^{\prime 2}\) obtained in (24). We get \begin{equation*} (\lambda k+2k-b_{1}b_{2}B^{2})(1+\lambda -b_{1}\gamma ^{2})=0, \end{equation*} \begin{equation*} (\lambda k-b_{1}b_{2}B^{2})(\lambda -1-b_{2}\alpha ^{2})=0. \end{equation*} We discuss all possibilities.

1. If \(\lambda k+2k-b_{1}b_{2}B^{2}=0\) and \(\lambda k-b_{1}b_{2}B^{2}=0\), then \(k=0\) and \(b_{1}b_{2}=0\). Then (25) gives \(\gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
2. If \(\lambda k+2k-b_{1}b_{2}B^{2}=0\), \(\lambda =1\) and \(b_{2}=0,\) we obtain \(k=0\). Then (25) gives \(\gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
3. If \(\lambda k-b_{1}b_{2}B^{2}=0\), \(\lambda =-1\) and \(b_{1}=0,\) we obtain \(k=0\). Then (25) gives \(\gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.
4. If \(1+\lambda -b_{1}\gamma ^{2}=0\) and \(\lambda -1-b_{2}\alpha ^{2}=0,\) we deduce that \(\alpha \), \(\gamma \) are both constant functions, and so, \( \gamma ^{\prime \prime }=0\) and \(\alpha ^{\prime \prime }=0\), a contradiction.

Therefore, we have the following:

Theorem 9. Let \(M^{2}\) be a TH-surface in \(\mathbb{E}^{3}.\) If \(M^{2}\) is minimal surface, then \(M^{2}\) is plane or parameterized as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }A(f(u)+g(v))+Bf(u)g(v)), \end{equation*} where

1. either \(f(u)=\frac{\lambda _{1}}{B}u+\frac{\lambda _{2}-A}{B}\) and \( g(v)=\frac{1}{\lambda _{1}}\tan (\lambda _{3}v+\lambda _{4})-\frac{A}{B}\) or
2. [ii)] \(f(u)=\frac{1}{\lambda _{1}}\tan (\lambda _{2}u+\lambda _{3})-\frac{A}{ B}\) and \(g(v)=\frac{\lambda _{1}}{B}v+\frac{\lambda _{4}-A}{B}\).

6. TH-surfaces with zero Gaussian curvature in \(\mathbb{E}^{3}\)

A surface in Euclidean 3-space parameterized by (1) has Gaussian curvature \begin{equation*} K=\frac{\alpha \gamma f^{\prime \prime }g^{\prime \prime }-B^{2}f^{\prime 2}g^{\prime 2}}{EG-F^{2}}. \end{equation*} Hence that if \(K=0\), then Since the roles of the function \(\gamma \) and \(\alpha \) are symmetric in ( 26), we discuss the different cases according the function \(\gamma .\)
Case 1. Let \(\gamma ^{\prime }=0\). In this case \(\gamma \) is a constant function \(\gamma (u)=u_{0}\) and the parametrization of (1) writes as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }\delta _{1}g(v)+\delta _{2}). \end{equation*} This means that \(M^{2}\) is a cylindrical surface with base curve a plane curve in the \(vz-\) plane.
Case 2. Let \(\gamma ^{\prime \prime }=0\) and \(\gamma ^{\prime }\neq 0\). Then \(\gamma (u)=\lambda u+\delta _{1},\) \((\lambda ,\) \(\delta )\in\mathbb{R}^{\ast }\times\mathbb{R}.\) Moreover, (26) gives \(\alpha ^{\prime }=0\) and \( \alpha (v)=\) \(v_{0}\) is a constant function. Now \(M^{2}\) is the plane of equation \(z(u,v)=\lambda u+\delta _{1},\) \(\lambda ,\) \(\delta _{1}\in\mathbb{R}.\)
Case 3. Let \(\gamma ^{\prime \prime }\neq 0.\) By the symmetry on the arguments, we also suppose \(\alpha ^{\prime \prime }\neq 0\).
Equation (26) writes as \begin{equation*} \frac{\gamma \gamma ^{\prime \prime }}{\gamma ^{\prime 2}}=\frac{\alpha ^{\prime 2}}{\alpha \alpha ^{\prime \prime }}. \end{equation*} Therefore, there exists a real number \(\lambda \in\mathbb{R}^{\ast }\) such that \begin{equation*} \frac{\gamma \gamma ^{\prime \prime }}{\gamma ^{\prime 2}}=\lambda =\frac{ \alpha ^{\prime 2}}{\alpha \alpha ^{\prime \prime }}. \end{equation*} Integrate these equations where \(k_{1}\) and \(k_{2}\) are constants of integration.

1. If \(\lambda =1\), the general solution of (27) is given by \begin{equation*} \left\{ \begin{array}{c} \gamma (u)=\lambda _{1}e^{k_{1}u} \\ \alpha (v)=\lambda _{2}e^{k_{2}v}, \end{array} \right. \end{equation*} where \(\lambda _{1}\) and \(\lambda _{2}\) are constants of integration.
   Hence \begin{equation*} \left\{ \begin{array}{l} f(u)=\lambda _{3}e^{k_{1}u}+\lambda _{4} \\ g(v)=\lambda _{5}e^{k_{2}v}+\lambda _{6}, \end{array} \right. \end{equation*} where \(\lambda _{3}\), \(\lambda _{4},\) \(\lambda _{5}\), \(\lambda _{6}\in\mathbb{R}\).
2. If \(\lambda \neq 1\), the general solution of (27) is given by \begin{equation*} \left\{ \begin{array}{c} \gamma (u)=((1-\lambda )k_{1}u+c_{1})^{\frac{1}{1-\lambda }} \\ \alpha (v)=((\frac{\lambda -1}{\lambda })k_{2}v+c_{2})^{\frac{\lambda }{ \lambda -1}}, \end{array} \right. \end{equation*} where \(c_{1}\) and \(c_{2}\) are constants of integration.
   Hence \begin{equation*} \left\{ \begin{array}{l} f(u)=c_{3}((1-\lambda )k_{1}u+c_{1})^{\frac{1}{1-\lambda }}+c_{4} \\ g(v)=c_{5}((\frac{\lambda -1}{\lambda })k_{2}v+c_{2})^{\frac{\lambda }{ \lambda -1}}+c_{6}, \end{array} \right. \end{equation*} where \(c_{3}\), \(c_{4},\) \(c_{5}\), \(c_{6}\in\mathbb{R}\).

Theorem 10. Let \(M^{2}\) be a TH-surface in Euclidean \(3-\) space \(\mathbb{E}^{3}\) with constant Gauss curvature \(K\). Then \(K=0\). Furthermore, the surface is plane or is a cylindrical surface over a plane curve or parameterized as \begin{equation*} r(u,v)=(u,\text{ }v,\text{ }A(f(u)+g(v))+Bf(u)g(v)), \end{equation*} where

1. either \(f(u)=\lambda _{3}e^{k_{1}u}+\lambda _{4}\) and \(g(v)=\lambda _{5}e^{k_{2}v}+\lambda _{6}\) or
2. \(f(u)=c_{3}((1-\lambda )k_{1}u+c_{1})^{\frac{1}{1-\lambda }}+c_{4}\) and \(g(v)=c_{5}((\frac{\lambda -1}{\lambda })k_{2}v+c_{2})^{\frac{\lambda }{ \lambda -1}}+c_{6}\).

Acknowledgments

The authors would like to express their thanks to the referee for his useful remarks.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing Interests

The author(s) do not have any competing interests in the manuscript.