Surfaces with Constant Negative Curvature

Abstract

In this paper, we have considered surfaces with constant negative Gaussian curvature in the simply isotropic 3-Space by defined Sauer and Strubeckerr. Firstly, we have studied the isotropic $II$-flat, isotropic minimal and isotropic $II$-minimal, the constant second Gaussian curvature, and the constant mean curvature of surfaces with constant negative curvature (SCNC) in the simply isotropic 3-space. Surfaces with symmetry are obtained when the mean curvatures are equal. Further, we have investigated the constant Casorati, the tangential and the amalgamatic curvatures of SCNC.

1. Introduction

Constant curvature for surfaces was one of the top subjects regarding differential geometry in the 19th century (see [1,2]). Surfaces with curvature $\mathbf{K}=-1$ are denoted as $\mathbf{K}$-surfaces, and this topic is one of the main studies in differential geometry. The hyperbolic plane’s intrinsic geometry is provided on $\mathbf{K}$-surfaces with a model [3,4] and the pseudosphere is the oldest known example of that geometry [5,6].

One of the most substantial problems in differential geometry is to construct a surface with constant negative Gaussian curvature in Euclidean space. From a known surface with $\mathbf{K}=-1$ [7], the Bäcklund’s theorem provides a geometrical method to build a family of surfaces with Gaussian curvature. For the Gaussian $\mathbf{K}$-surfaces, Bäcklund transformation is given by Tian [8]. For pseudospherical surfaces, Bäcklund transformation can be limited to a transformation on space curves [9]. Many studies have been conducted in other spaces, such as Minkowski space [10,11]. $\mathbf{K}$-surfaces in an isotropic 3-space were studied extensively by K.Strubeckerr, as in [12,13,14]. Decu and Verstraelen defined isotropic Casorati curvature [15]. Suceava investigated the tangential and amalgamatic curvatures in Euclidean 3-space [16].

Casorati proposed the Casorati curvature over Gauss and mean curvatures since this correlates better with the general intuition of curvature [16,17]. Human/computer vision and geometry are investigated using Casorati curvature [18,19]. The idea behind the amalgamatic curvature is to expand the ratio $\frac{\tau }{\kappa }$ to the higher dimensions [20]. This idea can be traced back to papers [21,22], with the improvements provided in [23,24]. A recent application can be found in [16]. An important development for the invariant curvature is studied in [25] and named tangential curvature [16].

In this paper, we present the Gaussian, the second Gaussian, the mean, the second mean, the Casorati, the tangential and the amalgamatic curvatures of surfaces with a constant negative curvature defined by K.Strubeckerr and V.R.Sauer. The symmetry on the surfaces can be seen in the figures below.

2. Preliminaries

An absolute figure is an ordered triple $(w,f_1,f_2)$ consisting of an absolute plane w with $f_1$ and $f_2$, which are its two complex–conjugate straight lines from the projective three-space $\mathcal{P}\left({\mathbb{R}}^3\right)$. These are required to define the simply isotropic space ${\mathbb{I}}_3^1$, which is a Cayley–Klein space. $x_0=0$ give the absolute plane w and $x_0=x_1+i{x}_2=0$, $x_0=x_1-i{x}_2=0$ are the absolute lines $f_1$, $f_2$. They are called the homogeneous coordinates or projective coordinates in $\mathcal{P}\left({\mathbb{R}}^3\right)$ [26]. Homogeneous coordinates played an important role in capturing the projection of a 3D view for use in monitors, T.V., etc. [27] Further information about Cayley–Klein spaces can be acquired from [28].

The absolute point $\mathbb{F}(0:0:0:1)$ is defined as the crossing node of these two lines. A motion in ${\mathbb{I}}_3^1$ is a given with corresponding coordinates, as $x=\frac{x_1}{x_0},y=\frac{x_2}{x_0},z=\frac{x_3}{x_0}$ can be found at [26] and given as

$$\begin{array}{cc}\hfill x^{\prime }& =c_1+xcos\alpha -ysin\alpha \hfill \\ \hfill y^{\prime }& =c_2+xsin\alpha +ycos\alpha \hfill \\ \hfill z^{\prime }& =c_3+c_{4}x+c_{5}y+z,\hfill \end{array}$$

(1)

where $c_1,c_2,c_3,c_4,c_5,\alpha \in \mathbb{R}$. These are named isotropic congruence transformations [26]. Isotropic congruence transformations looks like Euclidean motions (combination of a translation and a rotation) in the projection onto the xy-plane. This projection is named as a “top view” [29,30,31]. The combination of a Euclidean motion in the xy-plane and affine transformation with shearing along the z-direction is called an isotropic motion [32].

The equation

$$d{s}^2=d{x}^2+d{y}^2$$

is defined as the metric of ${\mathbb{I}}_3^1$. Let U = $(u_1,u_2,u_3)$ and V = $(v_1,v_2,v_3)$ be vectors in ${\mathbb{I}}_3^1$; then, the inner product of U and V is defined as,

$$<U,V{>}_i=\left\{\begin{array}{cc}{u}_3{v}_3\hfill & \mathrm{if}{u}_{1,2}=0\mathrm{and}{v}_{1,2}=0\hfill \\ u_1{v}_1+u_2{v}_2\hfill & \mathrm{if}\mathrm{otherwise}\hfill \end{array}\right.$$

This metric is induced by the absolute figure. If a line is not parallel to the z-direction, it is called non-isotropic; otherwise, it is isotropic. Isotropic planes are the planes that contain an isotropic line. Consider a $C^r$-surface $\mathbf{M}$, $r\ge 1$, in ${\mathbb{I}}_3^1$ parameterized by

$$\mathbf{M}:\mathbf{x}(u,v)=\left(x(u,v),y(u,v),z(u,v)\right).$$

Let an arbitrary surface in ${\mathbb{I}}_3^1$ be called $\mathbf{M}$. If a surface has no isotropic tangent planes, then it is called an admissible surface. The first and second forms $\mathbf{I}$ and $\mathbf{II}$, called fundamental forms, of $\mathbf{M}$, have the coefficients $E,F,G$ and $L,N,M$, respectively, which can be easily stated with the induced matrix, given by [32] as,

$$\begin{array}{cc}\hfill I=& Ed{u}^2+2Fdudv+Gd{v}^2,\hfill \\ \hfill II=& Ld{u}^2+2Mdudv+Nd{v}^2,\hfill \end{array}$$

where (with $\delta =\sqrt{EG-F^2}$)

$$\begin{array}{cc}\hfill E=& <{\mathbf{x}}_u,{\mathbf{x}}_u{>}_i,F=<{\mathbf{x}}_u,{\mathbf{x}}_v{>}_i,G=<{\mathbf{x}}_v,{\mathbf{x}}_v{>}_i,\hfill \\ \hfill L=& \frac{det({\mathbf{x}}_u,{\mathbf{x}}_v,{\mathbf{x}}_{uu})}{\delta },M=\frac{det({\mathbf{x}}_u,{\mathbf{x}}_v,{\mathbf{x}}_{uv})}{\delta },N=\frac{det({\mathbf{x}}_u,{\mathbf{x}}_v,{\mathbf{x}}_{vv})}{\delta }.\hfill \end{array}$$

Then, $\mathbf{K}$ and $\mathbf{H}$, the isotropic Gaussian curvature and mean curvature, can be defined as

$$\mathbf{K}=k_1{k}_2=\frac{LN-M^2}{EG-F^2},2\mathbf{H}=k_1+k_2=\frac{EN-2FM+GL}{EG-F^2},$$

(2)

where $k_1,k_2$ are principal curvatures.Therefore, the extrema of the normal curvatures, $k_1$ and $k_2$, are determined with the non-isotropic section of the surface. Here, when $\mathbf{K}=0,$ and $\mathbf{H}=0,$ the surfaces $\mathbf{M}$ are called, respectively, isotropic flat and isotropic minimal [26]. Isotropic $\mathbf{II}$-flat and isotropic $\mathbf{II}$-minimal are named after the moment when a non-developable surface’s second Gaussian curvature and second mean curvature are zero, respectively. The second Gaussian curvature ${\mathbf{K}}_{\mathbf{II}}$ of $\mathbf{M}$ is given by

$${\mathbf{K}}_{\mathbf{II}}=\frac{1}{{\left(LN-M^2\right)}^2}\left[\begin{array}{c}\left|\begin{array}{ccc}-\frac{L_{uu}}{2}+M_{uv}-\frac{N_{vv}}{2}& \frac{L_u}{2}& M_u-\frac{L_v}{2}\\ M_v-\frac{N_u}{2}& L& M\\ \frac{N_v}{2}& M& N\end{array}\right|\\ -\left|\begin{array}{ccc}0& \frac{L_v}{2}& \frac{N_u}{2}\\ \frac{L_v}{2}& L& M\\ \frac{N_u}{2}& M& N\end{array}\right|\end{array}\right].$$

(3)

The second mean curvature for a surface in simply isotropic 3-space is given by [15]

$${\mathbf{H}}_{\mathbf{II}}=\mathbf{H}-\frac{1}{2\sqrt{LN-M^2}}\sum _{i,j=1}^2\frac{\partial }{\partial u_i}\left(\sqrt{LN-M^2}{L}^{ij}\frac{\partial }{\partial u_i}\left(ln\sqrt{\left|\mathbf{K}\right|}\right)\right),$$

(4)

where $u_1=u$, $u_2=v$ and $\left(L^{ij}\right)$ is the inverse of the matrix $L_{ij}$ of the second fundamental form [33]. The isotropic Casorati curvature is defined by

$$\mathbf{C}=\frac{k_1^2+k_2^2}{2}=2{\mathbf{H}}^2-\mathbf{K}$$

(5)

The tangential curvature and the amalgamatic curvature are given by

$$\tau =\frac{\left|k_1{k}_2\right|-1+\sqrt{\left(1+k_1^2\right)\left(1+k_2^2\right)}}{\left|k_1\right|+\left|k_2\right|}=\frac{\left|\mathbf{K}\right|-1+\sqrt{{\left(\mathbf{K}-1\right)}^2+4{\mathbf{H}}^2}}{\mathbf{H}},$$

(6)

$$A=\frac{2k_1{k}_2}{k_1+k_2}=\frac{\mathbf{K}}{\mathbf{H}},$$

(7)

respectively [16].

3. Curvatures of SCNC in ${\mathbb{I}}_{\mathbf{3}}^{\mathbf{1}}$

In this chapter, we provide the curvatures of SCNC in ${\mathbb{I}}_3^1$. The general form of SCNC can be derived as follows. Let $\mathbf{x}$ be a surface with

$$\mathbf{x}(u,v)=(x_1(u,v),x_2(u,v),x_3(u,v))$$

when we have

$$x_1(u,v)=f_1+g_1$$

$$x_2(u,v)=f_2+g_2$$

where $f_1^{\prime }{g}_2^{\prime }-f_2^{\prime }{g}_1^{\prime }\ne 0$ with $f_1,f_2$ and $g_1,g_2$ are twice continuous differentiable functions of u and v, respectively. The parametric curves are asymptotic if and only if the two conditions

$$\left|\begin{array}{ccc}{f}_1^{\prime }& f_2^{\prime }& {x_3}_u\\ g_1^{\prime }& g_2^{\prime }& {x_3}_v\\ f_1^{\prime \prime }& f_2^{\prime \prime }& {x_3}_{uu}\end{array}\right|=0,\left|\begin{array}{ccc}{f}_1^{\prime }& f_2^{\prime }& {x_3}_u\\ g_1^{\prime }& g_2^{\prime }& {x_3}_v\\ g_1^{\prime \prime }& g_2^{\prime \prime }& {x_3}_{vv}\end{array}\right|=0$$

are fulfilled. From the above equations with $f_1^{\prime }{g}_2^{\prime }-f_2^{\prime }{g}_1^{\prime }\ne 0$ condition, the existence of two functions $a(u,v)$ and $b(u,v)$ can be uniquely determined by $x_3(u,v)$, which satisfy the equations

$${x_3}_u=a{f}_1^{\prime }+b{f}_2^{\prime },{x_3}_v=a{g}_1^{\prime }+b{g}_2^{\prime },$$

$${x_3}_{uu}=a{f}_1^{\prime \prime }+b{f}_2^{\prime \prime },{x_3}_{vv}=a{g}_1^{\prime \prime }+b{g}_2^{\prime \prime }.$$

By solving these equations, we can obtained

$$a_{uv}=b_{uv}=0,a={\mathsf{\Phi }}_1\left(u\right)+{\mathsf{\Psi }}_2\left(v\right),b={\mathsf{\Phi }}_2\left(u\right)+{\mathsf{\Psi }}_2\left(v\right).$$

For the arbitrary functions ${\mathsf{\Phi }}_1,{\mathsf{\Phi }}_2,{\mathsf{\Psi }}_1,{\mathsf{\Psi }}_2$, from the solution of the previous equation, we obtain,

$${\mathsf{\Phi }}_1^{\prime }{f}_1^{\prime }+{\mathsf{\Phi }}_2^{\prime }{f}_2^{\prime }=0,{\mathsf{\Psi }}_1^{\prime }{g}_1^{\prime }+{\mathsf{\Psi }}_1^{\prime }{g}_2^{\prime }=0,$$

results in

$${\mathsf{\Phi }}_1^{\prime }=\mathsf{\Lambda }\left(u\right)f_2^{\prime },{\mathsf{\Phi }}_2^{\prime }=-\mathsf{\Lambda }\left(u\right)f_1^{\prime },{\mathsf{\Psi }}_1^{\prime }=M\left(v\right)g_2^{\prime },{\mathsf{\Psi }}_2^{\prime }=-M\left(v\right)g_1^{\prime }.$$

Then,

$${x_3}_{uv}=M\left(v\right)(g_2^{\prime }{f}_1^{\prime }-g_1^{\prime }{f}_2^{\prime }),$$

$${x_3}_{vu}=\mathsf{\Lambda }\left(u\right)(g_1^{\prime }{f}_2^{\prime }-g_2^{\prime }{f}_1^{\prime })$$

by choosing

$$\mathsf{\Lambda }\left(u\right)=-M\left(v\right)=const.=k_1$$

we obtain

$$a=k_1(f_2-g_2)+k_2,b=-k_1(f_1-g_1)+k_3$$

by using these in the equations,

$${x_3}_u=k_1(f_1^{\prime }{f}_2-f_1{f}_2^{\prime })+k_1(f_2^{\prime }{g}_1-f_1^{\prime }{g}_2)+k_2{f}_1^{\prime }+k_3{f}_2^{\prime },$$

$${x_3}_v=k_1(g_2^{\prime }{g}_1-g_1^{\prime }{g}_2)+k_1(g_1^{\prime }{f}_2-g_2^{\prime }{f}_1)+k_2{g}_1^{\prime }+k_3{g}_2^{\prime }.$$

The integrability condition ${x_3}_{uv}={x_3}_{vu}$ is fulfilled and, from integration, results in $x_3=k_1\{(f_2^{\prime }{g}_1-f_1^{\prime }{g}_2)+\int (f_1^{\prime }{f}_2-f_1{f}_2^{\prime })du+\int (g_2^{\prime }{g}_1-g_1^{\prime }{g}_2)dv\}+k_2(f_1+g_1)+k_3(f_2+g_2)+k_4$, from this, $k_1=1,k_2=k_3=k_4=0$, we can obtained $\mathbf{x}(u,v)=(x_1,x_2,x_3)$ as

$$\mathbf{x}(u,v)=\left(\begin{array}{c}{f}_1+g_1,\\ f_2+g_2,\\ \left(f_2{g}_1-f_1{g}_2\right)+\int \left(f_1^{\prime }{f}_2-f_2^{\prime }{f}_1\right)du\\ +\int \left(g_1{g}_2^{\prime }-g_2{g}_1^{\prime }\right)dv\end{array}\right)$$

(8)

where $f_1,f_2,f$ are functions of u and $g_1,g_2,g$ are functions of v [12,34]. The isotropic curvature $\mathbf{K}$ and the mean curvature $\mathbf{H}$ of the surface (8) is given by

$$\mathbf{K}=-1,\mathbf{H}=\frac{f_1^{\prime }{g}_1^{\prime }+f_2^{\prime }{g}_2^{\prime }}{f_2^{\prime }{g}_1^{\prime }-f_1^{\prime }{g}_2^{\prime }}.$$

(9)

Assume the mean curvature of (8) is constant. Then,

$${\mathbf{H}}_0=\frac{f_1^{\prime }{g}_1^{\prime }+f_2^{\prime }{g}_2^{\prime }}{f_2^{\prime }{g}_1^{\prime }-f_1^{\prime }{g}_2^{\prime }},$$

(10)

where ${\mathbf{H}}_0\in \mathbb{R}.$ If we use the separation of variables method, the mean curvature and the mean curvature ${\mathbf{H}}_0\ne 0\in \mathbb{R}$ if and only if

$$\frac{f_2^{\prime }}{f_1^{\prime }}=\frac{g_1^{\prime }+{\mathbf{H}}_0{g}_2^{\prime }}{{\mathbf{H}}_0{g}_1^{\prime }-g_2^{\prime }},$$

where $u,v$ are independent variables and both sides of the Equation (10) are constant. If we show that this constant is equal to p, we can obtain

$$\frac{f_2^{\prime }}{f_1^{\prime }}=p=\frac{g_1^{\prime }+{\mathbf{H}}_0{g}_2^{\prime }}{{\mathbf{H}}_0{g}_1^{\prime }-g_2^{\prime }}.$$

(11)

Hence, we can write

$$\left\{\begin{array}{c}{f}_1=c_1+\frac{f_2}{p}\\ f_2=c_2+p{f}_1\\ g_1=c_3+\frac{\left({\mathbf{H}}_0+p\right)g_2}{{\mathbf{H}}_{0}p-1}\\ g_2=c_4+\frac{\left({\mathbf{H}}_{0}p-1\right)g_1}{{\mathbf{H}}_0+p}\end{array}\right.,$$

(12)

where $c_i\in \mathbb{R}.$

If the surface (8) is isotropic minimal, then, from (10), we have

$$f_1^{\prime }{g}_1^{\prime }+f_2^{\prime }{g}_2^{\prime }=0$$

and as a result of the solution of this equation,

$$\left\{\begin{array}{c}{f}_1=c_1+p{f}_2\\ f_2=c_2+\frac{f_1}{p}\\ g_1=c_3-\frac{g_2}{p}\\ g_2=c_4-p{g}_2\end{array}\right..$$

Figure 1 and Figure 2 are drawn for Equation (12); their functions are shown in their respective figures.

Figure 1. $f_1=sinu$ and $g_1=cosv$ with $c_2=c_4=1$, $H_0=5/2$, $p=2$.

Figure 2. $f_1=e^u$ and $g_1=e^v$ with $c_2=c_4=1$, $H_0=5/2$, $p=2$.

If we choose

$$\left\{\begin{array}{c}{f}_1=u\\ f_2=f^{\prime }\\ g_1=v\\ g_2=g^{\prime }\end{array}\right.,$$

(13)

the surface (8) turns into the following form

$$\mathbf{x}(u,v)=\left(u+v,f^{\prime }+g^{\prime },2\left(f-g\right)+\left(v-u\right)\left(f^{\prime }+g^{\prime }\right)\right).$$

(14)

The Figure 3 with symmetry is drawn for (14) as an example where the constants $c_2,c_4,H_0$ and p are from Equation (12) and $f_1,g_1$ are from (13) [12,13,14].

Figure 3. $f_1=u$ and $g_1=v$ with $c_2=c_4=1$, $H_0=5/2$, $p=2$.

Let us consider that the surface (14). Then, using the surface (14), the coefficients of the first and the second fundamental forms are given by

$$E=1+f^{\prime {\prime }^2},G=1+g^{\prime {\prime }^2},F=1+f^{\prime \prime }{g}^{\prime \prime },$$

(15)

and

$$L=0,N=0,M=-\left(f^{\prime \prime }-g^{\prime \prime }\right),$$

(16)

respectively. The isotropic Gaussian curvature $\mathbf{K},{\mathbf{K}}_{\mathbf{II}}$ and the mean curvatures $\mathbf{H},{\mathbf{H}}_{\mathbf{II}}$, the isotropic Casorati curvature, the tangential curvature and the amalgamatic curvature of the surface (14) are given by

$$\mathbf{K}=-1,{\mathbf{K}}_{\mathbf{II}}=\frac{f^{\prime \prime \prime }{g}^{\prime \prime \prime }}{{\left(f^{\prime \prime }-g^{\prime \prime }\right)}^3},$$

(17)

$$\mathbf{H}={\mathbf{H}}_{\mathbf{II}}=\frac{1+f^{\prime \prime }{g}^{\prime \prime }}{f^{\prime \prime }-g^{\prime \prime }},$$

(18)

$$\mathbf{C}=1+\frac{2{\left(1+f^{\prime \prime }{g}^{\prime \prime }\right)}^2}{{\left(f^{\prime \prime }-g^{\prime \prime }\right)}^2},$$

(19)

$$\tau =\frac{\sqrt{{\left(1+f^{\prime \prime }\right)}^2{\left(1+g^{\prime \prime }\right)}^2}}{1+f^{\prime \prime }{g}^{\prime \prime }},$$

(20)

$$\mathbf{A}=\frac{\sqrt{{\left(1+f^{\prime \prime }\right)}^2{\left(1+g^{\prime \prime }\right)}^2}}{1+f^{\prime \prime }{g}^{\prime \prime }},$$

(21)

respectively. Let us assume that the surface (14) has a constant mean curvature. Then,

$$\frac{1+f^{\prime \prime }{g}^{\prime \prime }}{f^{\prime \prime }-g^{\prime \prime }}={\mathbf{H}}_0,$$

(22)

where ${\mathbf{H}}_0\in \mathbb{R}.$ If we use the separation of variables method, the mean curvature ${\mathbf{H}}_0$ is a constant but 0, if and only if

$$\frac{1-{\mathbf{H}}_0{f}^{\prime \prime }}{-\left({\mathbf{H}}_0+f^{\prime \prime }\right)}=p_1=g^{\prime \prime },$$

(23)

where $p_1\ne 0\in \mathbb{R}.$ We can easily obtain

$$\left\{\begin{array}{c}f=c_1+u{c}_2+\frac{\frac{u^2}{2}\left(1+{\mathbf{H}}_0{p}_1\right)}{{\mathbf{H}}_0-p_1}\\ g=c_3+v{c}_4+\frac{p_1{v}^2}{2}\end{array}\right.,$$

(24)

where $c_i\in \mathbb{R}.$

We can obtain the next Figure 4 by choosing $c_1=c_2=c_3=c_4=1$, $H_0=5/2$ and $p_1=2$ at (24) and by using (14).

Figure 4. $f=1+u+6u^2$ and $g=1+v+v^2$.

Suppose that the mean curvature $\mathbf{H}$ of the constant negative curvature surface vanishes identically; then, from (22), we have

$$1+f^{\prime \prime }{g}^{\prime \prime }=0.$$

(25)

Here, u and v are independent variables, so each side of (25) is equal to a constant, called $p_2$. Hence, the two equations are

$$-\frac{1}{f^{\prime \prime }}=p_2=g^{\prime \prime }.$$

(26)

With appropriate solutions to these differential equations, we have

$$\left\{\begin{array}{c}f=c_1+u{c}_2-\frac{u^2}{2p_2}\\ g=c_3+v{c}_4+\frac{p_2{v}^2}{2}\end{array}\right.,$$

(27)

where $c_i\in \mathbb{R}.$

Thus, we have the following results:

Theorem 1.

Let $\mathbf{M}$ be the isotropic surface (14) with the constant isotropic mean curvature $\mathbf{H}\ne 0$ in ${\mathbb{I}}_3^1$. Then, the functions f and g are given by (24).

Theorem 2.

Let $\mathbf{M}$ be the isotropic surface (14) with zero isotropic mean curvature (isotropic minimal $\mathbf{H}\ne 0$) in ${\mathbb{I}}_3^1.$ Then, the functions f and g are given by (27).

Theorem 3.

If the isotropic surface parametrized by (14) in ${\mathbb{I}}_3^1$ has ${\mathbf{K}}_{\mathbf{II}}=0,$ then

$$\left\{\begin{array}{c}f=c_1+c_{2}u+c_3{u}^2\\ g=c_4+c_{5}v+c_6{v}^2\end{array}\right.,$$

(28)

and if ${\mathbf{K}}_{\mathbf{II}}$ is a non-zero constant,

$$\left\{\begin{array}{c}f=\frac{4\sqrt{2}\left(u+c_7\right)\sqrt{-c_2\left(u+c_7\right)}-3pu{c}_8\left(u+2c_7\right)}{6c_2}+c_9+u{c}_{10}\\ g=c_{11}+v{c}_{12}-\frac{p{v}^2}{2}\end{array}\right.,$$

where $c_i,p\in \mathbb{R}$ where $c_2\ne 0$ and $c_2(u+c_7)\in {\mathbb{R}}^{+}$.

Proof. 

From (14), we have

$$\frac{f^{\prime \prime \prime }{g}^{\prime \prime \prime }}{{\left(f^{\prime \prime }-g^{\prime \prime }\right)}^3}=c,$$

(29)

where $c\in \mathbb{R}$.

If ${\mathbf{K}}_{\mathbf{II}}=0,$ from the solution of the Equation (29), we have

$$f^{\prime \prime \prime }{g}^{\prime \prime \prime }=0.$$

(30)

By solving (30), the functions f and g are obtained as follows

$$\left\{\begin{array}{c}f=c_1+c_{2}u+c_3{u}^2\\ g=c_4+c_{5}v+c_6{v}^2\end{array}\right.,$$

where $c_i\in \mathbb{R}.$

When we choose the constants at (28) as $c_1=c_2=c_3=c_4=c_5=c_6=1$, we can obtain the figure with two different intervals as Figure 5.

Figure 5. $f=1+u+u^2$ and $g=1+v+v^2$.

Now, suppose that the second Gaussian curvature ${\mathbf{K}}_{\mathbf{II}}=c$ is a non-zero constant. The partial derivative of (29) with respect to u gives

$$g^{\prime \prime \prime }\left(-\frac{3f^{\prime \prime {\prime }^2}}{f^{\prime \prime }-g^{\prime \prime }}+f^{\left(IV\right)}\right)=0.$$

(31)

Therefore, either, i.e., $g^{\prime \prime \prime }=0$ or

$$\left(-\frac{3f^{\prime \prime {\prime }^2}}{f^{\prime \prime }-g^{\prime \prime }}+f^{\left(IV\right)}\right)=0.$$

(32)

If $g^{\prime \prime \prime }=0$, then we have

$$g=c_1+c_{2}v+c_3{v}^2,$$

where $c_i\in \mathbb{R}.$ If $\left(-\frac{3f^{\prime \prime {\prime }^2}}{f^{\prime \prime }-g^{\prime \prime }}+f^{\left(IV\right)}\right)=0$, then we have

$$-g^{\prime \prime }=\frac{3f^{\prime \prime {\prime }^2}-f^{\prime \prime }{f}^{\left(IV\right)}}{f^{\left(IV\right)}}.$$

(33)

By solving (33), we find

$$\left\{\begin{array}{c}f=\frac{4\sqrt{2}\left(u+c_1\right)\sqrt{-c_2\left(u+c_1\right)}-3p_{3}u{c}_2\left(u+2c_1\right)}{6c_2}+c_3+u{c}_4\\ g=c_5+v{c}_6-\frac{p_3{v}^2}{2}\end{array}\right.,$$

(34)

where $c_i,p_3\in \mathbb{R}$, where $c_2\ne 0$ and $c_2(u+c_7)\in {\mathbb{R}}^{-}$.

As we choose $c_1=-c_2=c_3=c_4=c_5=c_6=p_3=1$ and $H_0=5/2$, we can obtain the Figure 6, shown from two different angles. □

Figure 6. $f=1+u-\frac{1}{6}(4\sqrt{2}{(1+u)}^{\frac{3}{2}}+3u(2+u))$$g=1+v-\frac{v^2}{2}$.

Now, we consider the isotropic surface (14) in ${\mathbb{I}}_3^1$ as satisfying $\mathbf{C}=2{\mathbf{H}}^2-\mathbf{K}$ as constant. From (21), we have

$$1+\frac{2{\left(1+f^{\prime \prime }{g}^{\prime \prime }\right)}^2}{{\left(f^{\prime \prime }-g^{\prime \prime }\right)}^2}={\mathbf{C}}_0,$$

(35)

where ${\mathbf{C}}_0\in \mathbb{R}.$ Taking the partial derivative of (35) with respect to u gives

$$\left((1+{\mathbf{C}}_0)g^{\prime \prime }+f^{\prime \prime }(1-{\mathbf{C}}_0+2g^{\prime {\prime }^2})\right)f^{\prime \prime \prime }=0.$$

(36)

If $f^{\prime \prime \prime }=0,$ then we have

$$f=c_1+c_{2}u+c_3{u}^2,$$

(37)

where $c_i\in \mathbb{R}.$ If $(1+{\mathbf{C}}_0)g^{\prime \prime }+f^{\prime \prime }(1-{\mathbf{C}}_0+2g^{\prime {\prime }^2})=0,$ then we have

$$\left\{\begin{array}{c}f=c_4+c_{5}u-p_4\frac{u^2}{2}\\ g=c_6+c_{7}v+\frac{\left(1+{\mathbf{C}}_0-\sqrt{{\left(1+{\mathbf{C}}_0\right)}^2+8p_4^2\left({\mathbf{C}}_0-1\right)}\right)v^2}{8p_4}\end{array}\right.,$$

(38)

where $c_i,p_4\ne 0,{\mathbf{C}}_0\in \mathbb{R}$, where ${\left(1+{\mathbf{C}}_0\right)}^2+8p_4^2\left({\mathbf{C}}_0-1\right)$ is positive.

$c_4=c_5=c_6=c_7=p_4=1$ and ${\mathbf{C}}_0=5/2$ are chosen for a Figure 7 at Equation (38).

Figure 7. $f=1+u-u^2/2$ and $g=1+v+1/8(7/2-\sqrt{97}/2)v^2$.

If the surface (14) has zero Casorati curvature ($\mathbf{C}=0$), then, by using a similar technique for the solution, we have

$$\left\{\begin{array}{c}f=c_4+c_{5}u+p_5\frac{u^2}{2}\\ g=c_6+c_{7}v+\frac{\left(-1+\sqrt{1-8p_5^2}\right)v^2}{8p_5}\end{array}\right.,$$

(39)

where $c_i,p_5\in \mathbb{R}.$

$c_4=c_5=c_6=c_7=1$ and $p_5=0.35$ values are provided for the Figure 8 at Equation (39).

Figure 8. $f=1+u-0.175u^2$ and $g=1+v+0.306635v^2$.

Let us assume that the isotropic surface (14) has a constant tangential curvature. Then from (22), we have

$${\tau }_0=\frac{\sqrt{{\left(1+f^{\prime \prime }\right)}^2{\left(1+g^{\prime \prime }\right)}^2}}{1+f^{\prime \prime }{g}^{\prime \prime }},$$

(40)

where ${\tau }_0\in \mathbb{R}.$ By solving (22) we can find

$$\left\{\begin{array}{c}f=c_1+c_{2}u+c_3{u}^2\\ f=c_4+c_{5}u+p_6\frac{u^2}{2}\\ g=c_7\pm v\left(c_8+\frac{v{p}_6}{{\tau }_0^2+\sqrt{-4p_6^2+4{\tau }_0^2{p}_6^2+{\tau }_0^4}}\right)\end{array}\right.,$$

(41)

where $c_i,p_6\in \mathbb{R}$ where $-4p_6^2+4{\tau }_0^2{p}_6^2+{\tau }_0^4\ge 0$

Figure 9 for (41) is constructed with the constants $c_4=c_5=c_6=c_7=c_8=1,p_6=0.8$ and ${\tau }_0=1$.

Figure 9. $f=1+u+0.4u^2$ and $g=1+v+0.4v^2$.

If the (14) has the constant amalgamatic curvature. Then, from (40), we obtain

$${\mathbf{A}}_0=\frac{\sqrt{{\left(1+f^{\prime \prime }\right)}^2{\left(1+g^{\prime \prime }\right)}^2}}{1+f^{\prime \prime }{g}^{\prime \prime }},$$

(42)

where $A_0\in \mathbb{R}.$ By solving (42), we find

$$\left\{\begin{array}{c}f=c_1+c_{2}u+p_7\frac{u^2}{2}\\ g=c_3+c_{4}v-\frac{v^2\left(2p_7+{\mathbf{A}}_0\right)}{2\left(p_7{\mathbf{A}}_0-2\right)}\end{array}\right.,$$

(43)

where $c_i,p_7\in \mathbb{R}$ with $p_7{\mathbf{A}}_0-2\ne 0$.

The Figure 10 is for Equation (43), where the constants are $c_1=c_2=c_3=c_4={\mathbf{A}}_0=1$ and $p_7=0.8$.

Figure 10. $f=1+u+0.4u^2$ and $g=1+v+1.08333v^2$.