# Timelike Tangent Developable Surfaces and Bonnet Surfaces

Authors: Soley Ersoy; Kemal Eren
Journal: Abstract and Applied Analysis
 (2016)

---

## Abstract

A criterion was given for a timelike surface to be a Bonnet surface in 3-dimensional Minkowski space by Chen and Li, 1999. In this study, we obtain a necessary and sufficient condition for a timelike tangent developable surface to be a timelike Bonnet surface by the aid of this criterion. This is examined under the condition of the curvature and torsion of the base curve of the timelike developable surface being nonconstant. Moreover, we investigate the nontrivial isometry preserving the mean curvature for a timelike flat helicoidal surface by considering the curvature and torsion of the base curve of the timelike developable surface as being constant.


---

## Body

## 1. Introduction

Surfaces which admit a one-parameter family of isometric deformations preserving the mean curvature are called Bonnet surfaces. In 1867, Bonnet proved that any surface with constant mean curvature inR3 (which is not totally umbilical) is a Bonnet surface [1]. Cartan obtained some detailed results for Bonnet surfaces in [2]. Lawson extended Bonnet’s results to any surface with constant mean curvature in Riemannian 3-manifold of constant curvature. Also, it was proved that any Bonnet surface of nonconstant mean curvature depends on six arbitrary constants [3]. Characterization for isometric deformation preserving the principal curvatures of surfaces was obtained by the aid of differential forms by Chern in [4]. The geometric characterizations of helicoidal surfaces of constant mean curvature, helicoidal surfaces as Bonnet surfaces, and tangent developable surfaces as Bonnet surfaces were studied by Roussos in [5], [6] and [7], respectively. Roussos obtained a characterization for isometric deformation preserving the mean curvature by using the method of Chern. Soyuçok gave the necessary and sufficient condition of a surface to be Bonnet surface, which is to have a special system of isothermal parameters [8]. Moreover, Soyuçok proved that 3-dimensional hyperspace in 4-dimensional space is Bonnet surface if and only if hypersurface has orthogonal net [9]. Bağdatlı and Soyuçok studied hypersurfaces preserving the mean curvature and proved that a hypersurface in Rn+1 is Bonnet surface if and only if hypersurface has orthogonal A-net [10]. On the other hand, Chen and Li studied 3-dimensional Minkowski space and classified timelike Bonnet surfaces [11].


## 2. Preliminaries

LetM be a timelike surface in 3-dimensional Minkowski space with nondegenerate metric tensor g=-dx12+dx22+dx32, where x1,x2,x3 is a system of the canonical coordinates in R3. Let Φ:M→R13 be a timelike immersion that admits a nontrivial isometry preserving the mean curvature. Nontriviality means that the immersion in the family is not in the form of r∘Φ, where r:R13→R13 is an immersion of R13. These kinds of surfaces are called timelike Bonnet surfaces by Chen and Li in [11]. Suppose that {e1,e2,e3} is a local orthonormal frame at the point x∈M, where e1 is a timelike tangent vector, e2 is a spacelike unit tangent vector, and e3 is a spacelike unit normal vector field at x∈M. e3 can be regarded as a map e3:M→S12, where S12=x∈R13:x,x=1 is the de Sitter space. Let wi, 1≤i≤3, be dual 1-forms of ei defined by wiej=ei,ej=εiδij, 1≤i,j≤3, and let wji, 1≤i,j≤3, be connection forms; then (1)dx=w1e1+w2e2,de1=w12e2+w13e3,de2=w21e1+w23e3,de3=w31e1+w32e2,such that w13=w31, w23=-w32, and w12=w21.The Weingarten mapA:TxM→TxM is given by (2)Ae1=-ae1-be2,Ae2=be1-ce2,and A has real eigenvector if and only if (a+c)2/4-(ac+b2)≥0; that is, H2-K=(a-c)2/4-b2≥0 [11]. Here, H=a+c/2 and K=ac+b2 are the mean and Gaussian curvatures of M, respectively.Unless otherwise stated throughout this paper, we will assume thatH2>K and e1,e2 are the eigenvectors. Thus, b=0 and (3)w13=-aw1,w23=cw2,where a and c are the principal curvatures throughout e1 and e2. Then, the mean and Gaussian curvatures of the surface M are (4)H=a+c2,K=ac,respectively, and J=a-c/2>0. Suppose that w12=hw1+kw2; then h and k can be determined from the Cartan structure equations given by (5)dw1=w2∧w21,dw2=w1∧w12.The Gaussian and Codazzi equations are (6)dw12=w13∧w32=-Kw1∧w2,(7)dw13=w12∧w23,dw23=w21∧w13,respectively [11]. If we substitute the equations of (3) into the Codazzi equations, respectively, and make them equal to the exterior differentiations of (3), then we give (8)da+a-chw2∧w1=0,dc+c-akw1∧w2=0.After reformulating the equations of (8), we can give (9)da=c-apw1+hw2,dc=a-ckw1+qw2.Since dH=da+dc/2, these last two equations give us (10)2dH=a-ck-pw1+q-hw2.Here, if the functions u and v are defined as u=k-p and v=q-h, then the last equation becomes (11)2dH=a-cuw1+vw2;that is, dH=J(uw1+vw2). Moreover, from (9), we get (12)daa-c=u-kw1-hw2,dca-c=kw1+v+hw2.Thus, by the aid of these last two equations, we see (13)dln⁡a-c=u-2kw1-v+2hw2.If we take into consideration (11), then the gradient of the mean curvature function H is (14)∇H=a-c2-ue1+ve2.Thus, it is easily seen that (15)g∇H,∇H=H2-K-u2+v2.Here, we assume that ∇H is nonnull vector field; that is, ±u≠v. By taking ε=sgn⁡g∇H,∇H=±1, we find (16)ε-u2+v2=g∇H,∇HH2-K.By the fact that Hodge ∗ operator is defined as (17)∗w1=w2,∗w2=w1,∗2=1,the connection form w12 given by (3) becomes (18)∗w12=h∗w1+k∗w2=kw1+hw2.Let us define (19)θ1=uw1+vw2,θ2=vw1+uw2,(20)α1=uw1-vw2,α2=-vw1+uw2.Then, the following equations are obvious: (21)∗θ1=θ2,∗θ2=θ1,(22)∗α1=α2,∗α2=α1.By considering the second equation of (21), if we rearrange (11), we get (23)2dH=a-cθ1.In a similar way, from (18) and (20), (13) becomes (24)dln⁡a-c=uw1-vw2-2kw1+hw2.Thus, we find (25)dln⁡a-c=α1-2∗w12.


## 3. Timelike Bonnet Surfaces in Minkowski Space

Since we will make use of it in the following section, let us briefly recall construction of the criterion given by [11] for a timelike surface to be a Bonnet surface in 3-dimensional Minkowski space.LetM- be another timelike surface in R13 with the principal direction vectors such that it is an isometric deformation of M preserving the first fundamental form and principal curvatures. Suppose that w-1,w-2 is a principal coframe corresponding to orthonormal e-1,e-2 of M-; then the first fundamental form of M- is (26)-w-12+w-22=-w12+w22,and the principal curvatures throughout e-1 and e-2 are (27)a-=a,c-=c,respectively. It is seen from (26) that there is a function φ on M- such that (28)w-1=cosh⁡φw1+sinh⁡φw2,w-2=sinh⁡φw1+cosh⁡φw2.By direct calculations and the exterior derivative of (28), we get dw-1=w-2∧(-dφ+w12) and dw-2=w-1∧(-dφ+w12). From the first Cartan structure equations, we obtain (29)w-12=w-21=w12-dφ.On the other hand, from (25), we write dln⁡a--c-=α-1-2∗w-12. Equations (27) and (29) and the last equation give us α1-2∗w12=α-1-2∗w-12. If we apply ∗ operator to this equality, we find ∗α1-2w12=∗α-1-2w-12. By the fact that ∗α1=α2, ∗α2=α1, ∗2=1 we see α2-2w12=α-2-2w-12. After the necessary arrangements, we find w-12-w12=1/2α-2-α2. If we consider (29), we get (30)dφ=12α2-α-2.2dH=(α--c-)θ-1 is satisfied for M- with respect to (23). Considering (27) and comparing the last equation with (23), we see θ-1=θ1. Thus, u-w-1+v-w-2=uw1+vw2 and if we consider (28) we write (31)u-=ucosh⁡φ-vsinh⁡φ,v-=-usinh⁡φ+vcosh⁡φ.By taking (20), if we substitute (28) and (31) into α-2=-v-w-1+u-w-2, we obtain (32)α-2=sinh⁡2φα1+cosh⁡2φα2.Let us define T=coth⁡φ. The differentiation of T is (33)dT=Tα1+α2.The total differential equation (33) is satisfied by rotations of the principal directions with hyperbolic angles φ during isometric deformations. The deformation is nontrivial if and only if (33) is completely integrable [11].Theorem 1.
Every timelike constant mean curvature surfaceM with H2>K in R13 has one-parameter family of nontrivial isometric deformations preserving the mean curvature; that is, M is timelike Bonnet surface [11].In order to investigate the situations ofH being constant or nonconstant, let (34)dα1=Pα1∧α2,dα2=Qα1∧α2,thus defining P and Q. By substituting (34) into the exterior derivative of (33), we get (35)TP+Q-1α1∧α2=0.Thus, the following classification is satisfied:C1:H= constant.C2:H≠ constant, P=0, and Q=1.C3:H≠ constant, P≠0, and Q≠1.If these categories are investigated separately, thenC1,C2, and C3 can be interpreted as follows.Since the mean curvature is constant in the case ofC1 from (34) it is obvious that u=v=0. Thus, by (20), we see α1=α2=0. Consequently, from (33), T is constant.Since the mean curvature is nonconstant andP=0 and Q=1 in C2, (35) is satisfied for all T.The mean curvature is nonconstant andP≠0 and Q≠1 in C3. In this case, by considering (35), we get(36)T=1-QP.T can be computed for any timelike surface with nonconstant mean curvature, nonnull ∇H and H2>K but in order to find the nontrivial isometry Φ preserving the mean curvature T given in (36) must satisfy (33). Thus, (37)d1-QP=1-QPα1+α2constitutes a criterion for being timelike Bonnet surface.In the following section, we will check this criterion for timelike tangent developable surfaces and obtain the necessary and sufficient condition for timelike tangent developable surfaces to belong to caseC3.


## 4. Timelike Tangent Developable Surfaces

Timelike tangent developable surfaces can be investigated in two subcases when the curvature and torsion of base curve of these surfaces are nonconstant and constant.


### 4.1. Timelike Tangent Developable Surfaces with Base Curve with Nonconstant Curvature and Torsion

LetM be a timelike tangent developable surface given by (38)X:I×R⟶R13s,t⟶Xs,t=ηs+te1s,where ηs is timelike curve parametrized by its arc-length s and e1s=η˙(s) is timelike unit tangent vector field. Also assume that t>0. (In the same way, t<0 gives the second sheet of this surface.) Let e2(s) be principal normal vector field of ηs and of course it is spacelike. Then, e˙1s=κ(s)e2(s) such that κ(s)≥0 is the curvature of the base curve ηs. From (38), it is seen that Xss,t=e1s+tκ(s)e2(s) and Xts,t=e1s. The first fundamental form of this surface is (39)I=-1+t2κ2sds2-2dsdt-dt2.Then, dXs,t=(ds+dt)e1s+(tκsds)e2s can be given from dXs,t=Xsds+Xtdt. Also, keeping in mind dXs,t=w1e1s+w2e2s, we see (40)w1=ds+dt,w2=tκsds.By the fact that w12=gde1s,e2s, we get w12=κ(s)ds. On the other hand, since w2=tκsds, that is, κsds=w2/t, we find w12=1/tw2. By defining e3s=e1s∧e2s, e3s is binormal vector field of base curve ηs and spacelike. Let us denote the torsion of ηs by τs. Considering the Serret-Frenet formulae, we obtain(41)w13=gde1s,e3s=0=0w1,w23=gde2s,e3s=-τsds.Moreover, we get w23=-τsw2/tκ(s) since ds=w2/tκ(s). From the fact that w1,w2 is the dual coframe corresponding to principal frame field e1,e2, the principal curvatures throughout the principal direction vector fields a and c satisfy a=0>c=-τ(s)/tκ(s). That is, since t>0, it is assumed that τs>0. By taking the principal curvatures in terms of curvature and torsion of base curve, H and J can be given as (42)H=12-τstκs,J=12τstκs,respectively. On the other hand, from (11), we get (43)d-τs2tκs=12τstκsuw1+vw2.If we substitute (40) into this last equation, we find (44)d-τstκs=τstκsuds+dt+vtκsds.If we call (45)Fs=ln⁡τsκs′by considering (42) from (44) we obtain (46)u=1t,v=-Fstκs-1t2κs.If we substitute (40) and (46) into (20), we get (47)α1=Fs+2tds+1tdt,α2=κs+Fstκs+1t2κsds+Fstκs+1t2κsdt.The exterior differentiations of the equalities given in (47) are(48)dα1=2t2ds∧dt,dα2=t2F′s-Fsln⁡κs′+tFs-ln⁡κs′+2t3κsds∧dt,respectively. The exterior product of the equalities given in (47) is (49)α1∧α2=t2F2s-κ2s+2tFs+1t3κsds∧dtsuch that it never vanishes. By putting (48) and (49) into the equalities in (34), we find (50)P=2tκst2F2s-κ2s+2tFs+1,Q=t2F′s-Fsln⁡κs′+tFs-ln⁡κs′+2t2F2s-κ2s+2tFs+1.In order to check whether the criterion of being a timelike Bonnet surface given in the case of C3 is satisfied or not, let us substitute the equations given in (50) into (36) and find (51)T=t2F2s-κ2s-F′s+Fsln⁡κs′+tFs+ln⁡κs′-12tκs.By arranging (45), we get (52)Fs=ln⁡τs′-ln⁡κs′.Thus, (51) becomes(53)T=t2-κ2s-F′s+Fsln⁡τs′+tln⁡τs′-12tκs.From exterior derivative of (53), we obtain (54)dT=t2κs-F′s+Fsln⁡τs′-κ2s′-t2κ2s-F′s+Fsln⁡τs′-κ2s+12κsln⁡τs′′-t2tκ2sln⁡τs′+12tκ2sds+t2κs-F′s+Fsln⁡τs′-κ2s+12t2κsdt.In order to find the right side of (33), we consider (47) and (53) and find (55)Tα1+α2=t2Fs-F′s+Fsln⁡τs′-κ2s+t3Fsln⁡τs′-2F′s+Fs+2ln⁡τs′2tκsds+t2-F′s+Fsln⁡τs′-κ2s+tln⁡τs′+2Fs+12t2κsdt.If (54) and (55) are compared, (56)ln⁡τs′=-2Fsis obtained since dT=Tα1+α2. If we write this equation into (52), we get (57)ln⁡τs′=-3Fs.If we substitute (56) into (52) and solve it, we find (58)Aτ3s=κ2s,where A is a constant value such that Aτ>0. If (56) and (57) are written in (54), (59)dT=t2-F′s-2F2s-κ2s′+3Fs-F′s-2F2s-κ2s-t2F′s+6F2s-3Fs2tκsds+t2Fsln⁡τs′-κ2s-F′s+12t2κsdtis obtained. By comparing (55) with (59) and substituting into (56), we get -F′s-2F2s-κ2s′+3Fs-F′s-2F2s-κ2s=F(s)-F′s-2F2s-κ2s. It is easily seen that(60)-F′s-2F2s-κ2sτs′=0;that is,(61)-F′s-2F2s-κ2sτs=B,where B is constant. From this last equation, we get (62)-F′s-2F2s=κ2s+Bτs.If we put (56) and (58) into (62), then we find (63)τ′′sτs-2τ′s2=2Aτ5s+2Bτ3s.By multiplying each side of the last equation by 2/τ(s), we get (64)2τ′′s-4τsτ′s2=4Aτ4s+4Bτ2s.Here, assume that dτ/ds=τ′(s)=σ(τ). Thus, (68) becomes (65)σ2′-4τsσ2=4Aτ4s+4Bτ2s.The solution of this differential equation is (66)σ2=τ′2=4Aτ5+Cτ4-4Bτ3,C=constant.Here, τ′2≥0 and Aτ3s=κ2s>0. Thus, there is the inequality (67)4τ2+CAτ-4BA≥0.If Δ=C2-44A(-4B)>0, that is, C2>-64AB, then τ<r1<r2 or r1<r2<τ and τ≠0, where r1 and r2 are the roots of the quadratic trinomial 4τ2+C/Aτ-4B/A=0.Thus, the following theorem can be given.Theorem 2.
Timelike tangent developable surfaceX(s,t) is a timelike Bonnet surface if and only if (68)iAτ3s=k2s,Aτs>0,A=constant,iiτ′′sτs-2τ′s2=2Aτ5s+2Bτ3s,B=constantorτ′2=Aτ34τ2+CAτ-4BA,iiiC2>-64AB,C=constant.


### 4.2. Timelike Tangent Developable Surfaces with Base Curve with Constant Curvature and Torsion

If the curvature and torsion of timelike base curve of timelike tangent developable surfaceXs,t=ηs+te1(s) are constants, then the base curve ηs is timelike circular helix. All t-constant curves are helices and Xs,t is a timelike flat helicoidal surface. Since w1,w2,w12,w13,w23, a, and c are determined as in Section 4.1, we have (69)Fs=0,T=coth⁡φ=-t2κ2-12tκ,sinh⁡φ=2tκt2κ2-1,cosh⁡φ=-t2κ2+1t2κ2-1,φ∈0,∞.By considering the possibility of reflection, the forms of the image surface are (70)I¯=-1+t2κ2sds2-2dsdt-dt2,w-1=±cosh⁡φw1+sinh⁡φw2=±ds+t2κ2+1t2κ2-1dt,w-2=±sinh⁡φw1+cosh⁡φw2=±-tκds+2tκt2κ2-1dt,w-12=±-dφ+w12=±-2κt2κ2-1dt+κds=±1tw-2,w-13=0w-1=0,w-23=-τstκsw-2=±τds-2tκt2κ2-1dt.Moreover, the image surface ΦXs,t is flat but it is not a cylinder or a cone because of dα1=0; that is, P=0 is satisfied for cylinders and cones. Thus, a timelike developable surface cannot be a cylinder or a cone. Therefore, image surface is timelike developable surface given by Ys~,t~=η~s~+t~e~1s~, t~>0. Here, s~=-s-2/κarctanh⁡κt is the arc-length of the timelike base curve η~s~ and η~˙s~=e~1s~ is timelike unit tangent vector. The forms of timelike surfaces Ys~,t~ are (71)I~=-1+t~2κ~2s~ds~2-2ds~dt~-dt~,w~1=ds~+dt~,w~2=t~κ~s~ds~,w~12=κ~s~ds~=w~2t~,w~13=0w~1,w~23=-τ~s~t~κ~s~w~2.By comparing the forms given in (70) with (71), one can see that w-1 should be taken with the sign + whereas w-2 should be taken with the sign −; that is, the reflection is necessary and t~=t, κ~=κ, τ~=τ, s~=-s-2/κarctanh⁡κt. Thus, the image surface is the original surface and the nontrivial isometry preserving the mean curvature is (72)Φ:Xs,t⟶Xs,ts,t⟶Φs,t=-s-2κarctanh⁡κt,tor (73)Φηs+te1s=η-s-2κarctanh⁡κt+te1-s-2κarctanh⁡κt.Thus, the following corollaries can be given.Corollary 3.
The mappingΦ:Xs,t→Xs,t is a timelike isometry preserving the mean curvature. The second fundamental forms of corresponding surfaces are different; that is, Φ is nontrivial.Corollary 4.
s
~
=
-
s
-
2
/
κ
arctanh
⁡
κ
t
,
t
~
=
-
t
<
0, is also a timelike nontrivial isometry from X to its other sheet preserving mean curvature.


## 4.1. Timelike Tangent Developable Surfaces with Base Curve with Nonconstant Curvature and Torsion

LetM be a timelike tangent developable surface given by (38)X:I×R⟶R13s,t⟶Xs,t=ηs+te1s,where ηs is timelike curve parametrized by its arc-length s and e1s=η˙(s) is timelike unit tangent vector field. Also assume that t>0. (In the same way, t<0 gives the second sheet of this surface.) Let e2(s) be principal normal vector field of ηs and of course it is spacelike. Then, e˙1s=κ(s)e2(s) such that κ(s)≥0 is the curvature of the base curve ηs. From (38), it is seen that Xss,t=e1s+tκ(s)e2(s) and Xts,t=e1s. The first fundamental form of this surface is (39)I=-1+t2κ2sds2-2dsdt-dt2.Then, dXs,t=(ds+dt)e1s+(tκsds)e2s can be given from dXs,t=Xsds+Xtdt. Also, keeping in mind dXs,t=w1e1s+w2e2s, we see (40)w1=ds+dt,w2=tκsds.By the fact that w12=gde1s,e2s, we get w12=κ(s)ds. On the other hand, since w2=tκsds, that is, κsds=w2/t, we find w12=1/tw2. By defining e3s=e1s∧e2s, e3s is binormal vector field of base curve ηs and spacelike. Let us denote the torsion of ηs by τs. Considering the Serret-Frenet formulae, we obtain(41)w13=gde1s,e3s=0=0w1,w23=gde2s,e3s=-τsds.Moreover, we get w23=-τsw2/tκ(s) since ds=w2/tκ(s). From the fact that w1,w2 is the dual coframe corresponding to principal frame field e1,e2, the principal curvatures throughout the principal direction vector fields a and c satisfy a=0>c=-τ(s)/tκ(s). That is, since t>0, it is assumed that τs>0. By taking the principal curvatures in terms of curvature and torsion of base curve, H and J can be given as (42)H=12-τstκs,J=12τstκs,respectively. On the other hand, from (11), we get (43)d-τs2tκs=12τstκsuw1+vw2.If we substitute (40) into this last equation, we find (44)d-τstκs=τstκsuds+dt+vtκsds.If we call (45)Fs=ln⁡τsκs′by considering (42) from (44) we obtain (46)u=1t,v=-Fstκs-1t2κs.If we substitute (40) and (46) into (20), we get (47)α1=Fs+2tds+1tdt,α2=κs+Fstκs+1t2κsds+Fstκs+1t2κsdt.The exterior differentiations of the equalities given in (47) are(48)dα1=2t2ds∧dt,dα2=t2F′s-Fsln⁡κs′+tFs-ln⁡κs′+2t3κsds∧dt,respectively. The exterior product of the equalities given in (47) is (49)α1∧α2=t2F2s-κ2s+2tFs+1t3κsds∧dtsuch that it never vanishes. By putting (48) and (49) into the equalities in (34), we find (50)P=2tκst2F2s-κ2s+2tFs+1,Q=t2F′s-Fsln⁡κs′+tFs-ln⁡κs′+2t2F2s-κ2s+2tFs+1.In order to check whether the criterion of being a timelike Bonnet surface given in the case of C3 is satisfied or not, let us substitute the equations given in (50) into (36) and find (51)T=t2F2s-κ2s-F′s+Fsln⁡κs′+tFs+ln⁡κs′-12tκs.By arranging (45), we get (52)Fs=ln⁡τs′-ln⁡κs′.Thus, (51) becomes(53)T=t2-κ2s-F′s+Fsln⁡τs′+tln⁡τs′-12tκs.From exterior derivative of (53), we obtain (54)dT=t2κs-F′s+Fsln⁡τs′-κ2s′-t2κ2s-F′s+Fsln⁡τs′-κ2s+12κsln⁡τs′′-t2tκ2sln⁡τs′+12tκ2sds+t2κs-F′s+Fsln⁡τs′-κ2s+12t2κsdt.In order to find the right side of (33), we consider (47) and (53) and find (55)Tα1+α2=t2Fs-F′s+Fsln⁡τs′-κ2s+t3Fsln⁡τs′-2F′s+Fs+2ln⁡τs′2tκsds+t2-F′s+Fsln⁡τs′-κ2s+tln⁡τs′+2Fs+12t2κsdt.If (54) and (55) are compared, (56)ln⁡τs′=-2Fsis obtained since dT=Tα1+α2. If we write this equation into (52), we get (57)ln⁡τs′=-3Fs.If we substitute (56) into (52) and solve it, we find (58)Aτ3s=κ2s,where A is a constant value such that Aτ>0. If (56) and (57) are written in (54), (59)dT=t2-F′s-2F2s-κ2s′+3Fs-F′s-2F2s-κ2s-t2F′s+6F2s-3Fs2tκsds+t2Fsln⁡τs′-κ2s-F′s+12t2κsdtis obtained. By comparing (55) with (59) and substituting into (56), we get -F′s-2F2s-κ2s′+3Fs-F′s-2F2s-κ2s=F(s)-F′s-2F2s-κ2s. It is easily seen that(60)-F′s-2F2s-κ2sτs′=0;that is,(61)-F′s-2F2s-κ2sτs=B,where B is constant. From this last equation, we get (62)-F′s-2F2s=κ2s+Bτs.If we put (56) and (58) into (62), then we find (63)τ′′sτs-2τ′s2=2Aτ5s+2Bτ3s.By multiplying each side of the last equation by 2/τ(s), we get (64)2τ′′s-4τsτ′s2=4Aτ4s+4Bτ2s.Here, assume that dτ/ds=τ′(s)=σ(τ). Thus, (68) becomes (65)σ2′-4τsσ2=4Aτ4s+4Bτ2s.The solution of this differential equation is (66)σ2=τ′2=4Aτ5+Cτ4-4Bτ3,C=constant.Here, τ′2≥0 and Aτ3s=κ2s>0. Thus, there is the inequality (67)4τ2+CAτ-4BA≥0.If Δ=C2-44A(-4B)>0, that is, C2>-64AB, then τ<r1<r2 or r1<r2<τ and τ≠0, where r1 and r2 are the roots of the quadratic trinomial 4τ2+C/Aτ-4B/A=0.Thus, the following theorem can be given.Theorem 2.
Timelike tangent developable surfaceX(s,t) is a timelike Bonnet surface if and only if (68)iAτ3s=k2s,Aτs>0,A=constant,iiτ′′sτs-2τ′s2=2Aτ5s+2Bτ3s,B=constantorτ′2=Aτ34τ2+CAτ-4BA,iiiC2>-64AB,C=constant.


## 4.2. Timelike Tangent Developable Surfaces with Base Curve with Constant Curvature and Torsion

If the curvature and torsion of timelike base curve of timelike tangent developable surfaceXs,t=ηs+te1(s) are constants, then the base curve ηs is timelike circular helix. All t-constant curves are helices and Xs,t is a timelike flat helicoidal surface. Since w1,w2,w12,w13,w23, a, and c are determined as in Section 4.1, we have (69)Fs=0,T=coth⁡φ=-t2κ2-12tκ,sinh⁡φ=2tκt2κ2-1,cosh⁡φ=-t2κ2+1t2κ2-1,φ∈0,∞.By considering the possibility of reflection, the forms of the image surface are (70)I¯=-1+t2κ2sds2-2dsdt-dt2,w-1=±cosh⁡φw1+sinh⁡φw2=±ds+t2κ2+1t2κ2-1dt,w-2=±sinh⁡φw1+cosh⁡φw2=±-tκds+2tκt2κ2-1dt,w-12=±-dφ+w12=±-2κt2κ2-1dt+κds=±1tw-2,w-13=0w-1=0,w-23=-τstκsw-2=±τds-2tκt2κ2-1dt.Moreover, the image surface ΦXs,t is flat but it is not a cylinder or a cone because of dα1=0; that is, P=0 is satisfied for cylinders and cones. Thus, a timelike developable surface cannot be a cylinder or a cone. Therefore, image surface is timelike developable surface given by Ys~,t~=η~s~+t~e~1s~, t~>0. Here, s~=-s-2/κarctanh⁡κt is the arc-length of the timelike base curve η~s~ and η~˙s~=e~1s~ is timelike unit tangent vector. The forms of timelike surfaces Ys~,t~ are (71)I~=-1+t~2κ~2s~ds~2-2ds~dt~-dt~,w~1=ds~+dt~,w~2=t~κ~s~ds~,w~12=κ~s~ds~=w~2t~,w~13=0w~1,w~23=-τ~s~t~κ~s~w~2.By comparing the forms given in (70) with (71), one can see that w-1 should be taken with the sign + whereas w-2 should be taken with the sign −; that is, the reflection is necessary and t~=t, κ~=κ, τ~=τ, s~=-s-2/κarctanh⁡κt. Thus, the image surface is the original surface and the nontrivial isometry preserving the mean curvature is (72)Φ:Xs,t⟶Xs,ts,t⟶Φs,t=-s-2κarctanh⁡κt,tor (73)Φηs+te1s=η-s-2κarctanh⁡κt+te1-s-2κarctanh⁡κt.Thus, the following corollaries can be given.Corollary 3.
The mappingΦ:Xs,t→Xs,t is a timelike isometry preserving the mean curvature. The second fundamental forms of corresponding surfaces are different; that is, Φ is nontrivial.Corollary 4.
s
~
=
-
s
-
2
/
κ
arctanh
⁡
κ
t
,
t
~
=
-
t
<
0, is also a timelike nontrivial isometry from X to its other sheet preserving mean curvature.



---

*Source: 6837543-2016-02-01.xml*